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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
The second virial coefficient describes the contribution of the pair-wise potential to the pressure of the gas. The third virial coefficient depends on interactions between three molecules, and so on and so forth.
Introduction
As the density is increased the interactions between gas molecules become non-negligible. Deviations from the ideal gas law have been described in a large number of equations of state. The
virial equation of state expresses the deviation from ideality in terms of a power series in the density.
\[ \dfrac{P}{kT} = \rho + B_2(T)\rho^2 + B_3(T)\rho^3 + ...\]
\(B_2(T)\) is the second virial coefficient, \(B_3(T)\) is called the third virial coefficient, etc.
The j
th virial coefficient can be calculated in terms of the interaction of j molecules in a volume \(V\). The second and third virial coefficients give most of the deviation from ideal (P/rkT) up to 100 atm.
The
second virial coefficient is usually written as B or as \(B_2\). The second virial coefficient represents the initial departure from ideal-gas behavior. The second virial coefficient, in three dimensions, is given by
\[B_{2}(T)= - \dfrac{1}{2} \int \left( \exp\left(-\dfrac{\Phi_{12}({\mathbf r})}{k_BT}\right) -1 \right) 4 \pi r^2 dr \]
where \(\Phi_{12}({\mathbf r})\) is the intermolecular pair potential, T is the temperature and \(k_B\) is the Boltzmann constant. Notice that the expression within the parenthesis of the integral is the
Mayer f-function. In practice, the integral is often very hard to integrate analytically for anything other than, say, the hard sphere model, thus one numerically evaluates
\[B_{2}(T)= - \dfrac{1}{2} \int \left( \left\langle \exp\left(-\dfrac{\Phi_{12}({\mathbf r})}{k_BT}\right)\right\rangle -1 \right) 4 \pi r^2 dr \]
calculating
\[ \left\langle \exp\left(-\dfrac{\Phi_{12}({\mathbf r})}{k_BT}\right)\right\rangle\]
Calculation of virial coefficients
The configuration integrals for \(Z_1\), \(Z_2\), and \(Z_3\) are
\( Z_1 = \int dr_1 = V\) \( Z_2 = \int e^{-U_2/kT}dr_1\;dr_2\) \( Z_3 = \int e^{-U_3/kT}dr_1\;dr_2\;dr_3\)
The series method allows the calculation of a number of virial coefficients. Recall that the second and third virial coefficient can account for the properties of gases up the hundreds of atmospheres. We will discuss the calculation of the second virial coefficient for a monatomic gas to illustrate the procedure. To calculation \(B_2(T)\) we need \(U_2\). For monatomic particles it is reasonable to assume that the potential depends only on the separation of the two particles so \(U_2 = u(r_{12})\), where \(r_{12} = |r_2 - r_1|\). We have
Using a change of variables we can write this integral \(r_{12} = r_2 - r_1\) and after integration over \(r_1\) we can transform variables from \(dr_{12}\) to \(4\pi r^2dr\). The result is
\[ B_2(T) = -2 \pi \int [ e^{-\beta u(r)} - 1 ] r^2 \; dr\]
This expression can be used to obtain parameters from experiment. The second virial coefficient is tabulated for a number of gases.
For a hard-sphere potential
there is an infinite repulsive wall at a particle radius \(\sigma\). There is no attractive part.
\[ B_2(T) = -2 \pi \int_0^{\sigma} [ - 1 ] r^2 \; dr = \dfrac{2 \pi \sigma^2}{3}\]
The Lennard-Jones potential cannot be calculated analytically, but the integral can be computed numerically. The second virial coefficient was a useful starting point for obtaining Lennard-Jones parameters that were used in simulations.
Isihara-Hadwiger Formula
The Isihara-Hadwiger formula was discovered simultaneously and independently by Isihara and the Swiss mathematician Hadwiger in 1950. The second virial coefficient for any hard convex body is given by the exact relation
\[B_2=RS+V\]
or
\[\dfrac{B_2}{V}=1+3 \alpha\]
where
\[\alpha = \dfrac{RS}{3V}\]
where \(V\) is the volume, \(S\), the surface area, and \(R\) the mean radius of curvature.
Hard spheres
For the hard sphere model one has
[9]
\[B_{2}(T)= - \dfrac{1}{2} \int_0^\sigma \left(\langle 0\rangle -1 \right) 4 \pi r^2 dr \] leading to \[B_{2}= \dfrac{2\pi\sigma^3}{3}\]
Note that \(B_{2}\) for the hard sphere is independent of temperature.
Van der Waals equation of state
For the Van der Waals equation of state one has:
\[B_{2}(T)= b -\dfrac{a}{RT} \]
Excluded volume
The second virial coefficient can be computed from the expression
\[B_{2}= \dfrac{1}{2} \iint v_{\mathrm {excluded}} (\Omega,\Omega') f(\Omega) f(\Omega')~ {\mathrm d}\Omega {\mathrm d}\Omega'\]
where \(v_{\mathrm {excluded}}\) is the excluded volume.
References Reid, B. P. J. Chem. Educ. 1996, 73, 612-615. Harold Conroy "Molecular Schrödinger Equation. VIII. A New Method for the Evaluation of Multidimensional Integrals", Journal of Chemical Physics 47pp. 5307 (1967) I. Nezbeda, J. Kolafa and S. Labík "The spherical harmonic expansion coefficients and multidimensional integrals in theories of liquids", Czechoslovak Journal of Physics 39pp. 65-79 (1989) Akira Isihara "Determination of Molecular Shape by Osmotic Measurement", Journal of Chemical Physics 18pp. 1446-1449 (1950) Akira Isihara and Tsuyoshi Hayashida "Theory of High Polymer Solutions. I. Second Virial Coefficient for Rigid Ovaloids Model", Journal of the Physical Society of Japan 6pp. 40-45 (1951) Akira Isihara and Tsuyoshi Hayashida "Theory of High Polymer Solutions. II. Special Forms of Second Osmotic Coefficient", Journal of the Physical Society of Japan 6pp. 46-50 (1951) H. Hadwiger "Einige Anwendungen eines Funkticnalsatzes fur konvexe Körper in der räumichen Integralgeometrie" Mh. Math. 54pp. 345- (1950) H. Hadwiger "Der kinetische Radius nichtkugelförmiger Moleküle" Experientia 7pp. 395-398 (1951) H. Hadwiger "Altes und Neues über Konvexe Körper" Birkäuser Verlag (1955) Donald A. McQuarrie "Statistical Mechanics", University Science Books (2000) ISBN 978-1-891389-15-3 Eq. 12-40 I. Amdur and E. A. Mason "Properties of Gases at Very High Temperatures", Physics of Fluids 1pp. 370-383 (1958) Related reading W. H. Stockmayer "Second Virial Coefficients of Polar Gases", Journal of Chemical Physics 9pp. 398- (1941) G. A. Vliegenthart and H. N. W. Lekkerkerker "Predicting the gas–liquid critical point from the second virial coefficient", Journal of Chemical Physics 112pp. 5364-5369 (2000) |
Geometric series is very important in analysis. It denition is
\( \sum _{n=0}^{\infty}(z)^n \) Calculate this infinite sum is easier as seems if we note following \( S_{n} = \sum _{n=0}^{\infty}(z)^n = 1 + z + z^2 + ... + z^n \Rightarrow zS_{n}= z + z^2 + ... + z^{n+1} \) this implies \( S_{n}(1-z) = 1 + z + z^2 + ... + z^n - (z + z^2 + ... + z^{n+1}) = 1 + z^{n+1}\) thus \(S_{n} = \frac{1 + z^{n+1}}{1-z}\) Now is easy taking limit when \( n\mapsto \infty \) \( \sum _{n=0}^{\infty}(z)^n = \frac{1}{1-z}\) Geometric Series Sum of geometric series Convergence radius of geometric series
We apply the ratio test
\( \lim_{n\to \infty} |\frac{S_{n+1}}{Z_{n}}| = \lim_{n\to \infty} |\frac{x^{n+1}}{z{n}}| = |z| \)
Thus, geometric series converges if |z| < 1 and diverges if |z| > 1
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I have to solve a set of nonlinear optimization problems in the subspace defined as the orthogonal space to a given vector.
More precisely, $$ \arg\min f(\vec x) \qquad \text{with} \qquad \vec x \cdot \vec n =0 $$
I am thinking of applying the nonlinear conjugate gradient method projecting the direction of descent but I am wondering in which step I should do this projection operation.
One idea is that the projection should take place only in the update of the position like $$ \vec d_k = - \vec \nabla f(\vec x_{k-1}) + \beta \vec d_{k-1} \\ \alpha_k = \arg \min f(\vec x_{k-1} + \alpha \vec d_k) \\ \vec x_{k} = \vec x_{k-1} + (\alpha_k \vec d_k) - [(\alpha_k \vec d_k)\cdot \vec n ]\vec n $$
The other idea is to apply projection directly to the gradient itself such that even the past directions are subjected to the constraint of living in the orthogonal subspace to $\vec n$
$$ \vec g_k = \vec \nabla f(\vec x_{k-1}) - [\vec \nabla f(\vec x_{k-1})\cdot \vec n]\vec n \\ \vec d_k = - \vec g_k + \beta \vec d_{k-1} \\ \alpha_k = \arg \min f(\vec x_{k-1} + \alpha \vec d_k) \\ \vec x_{k} = \vec x_{k-1} + (\alpha_k \vec d_k) $$
What is the best way to solve this problem? |
Oct 12th 2018, 01:48 AM
# 1
Junior Member
Join Date: Sep 2018
Posts: 2
Problem with different reference frames - special relativity
I have the following conditions: A plane with length L in its own rest frame moves with constant velocity with respect
to the inertial system S. The inertial system of the plane is called S'. Two flight attendants A and B start simultaneously from the middle of the plane and move with constant velocity
toward the front end and back end of the plane, respectively, such that the magnitude of their velocities, u, is the same, as seen in S'. In reference frame S' it takes a time τ before they reach the ends of the plane.
The questions goes:
Seen from S, A reaches the back end of the train a time T before B reaches the front end.
a) Find the velocity v of the plane with respect to S in terms of L,T and c.
b) Seen from S, find how far the plane moves in the time it takes for A to reach the back end of the plane.
In A, I used that the spacetime interval in the two reference frames are invariant. in S' $\Delta x=L$ and $\Delta ct=0$ since the attendants reach the ends simultaneously. for S I am given that $\Delta t=T$ and the length of the plane can be found by a Lorentz transform. By setting the two spacetime intervals equal I get:
$(cT)^2-\frac{L^2}{(1-\frac{v^2}{c^2})}=-L^2\Longrightarrow v=\pm\sqrt{c^2+\frac{c^2L^2}{-L^2-(cT)^2}}$
In question B, my idea was that since I have found (I hope) the velocity of the plane relative to S, I could make a Lorentz transform of $\tau$ to get the time it takes the attendant to reach the back seen from S and then multiply by the velocity. I tried this out, but the calculations get very nasty and I think the result is supposed to be a bit prettier than what I got. Is it possible to calculate the distance the plane travels in the time A reach the back end simply by making a Lorentz-transform of $\tau$ and multiplying by the velocity of reference frame S' relative to S? |
Question
How does BayesiaLab calculate the Means and Values in the
Monitors? What is the difference? Answer
For each node that has values associated with its states, an Expected Value $v$ is computed by using the associated values and the marginal probability distribution of the node $$v = \sum_{s \in S} p_s \times V_s$$ where $p_s$ is the marginal probability of state $s$ and $V_s$ is its associated value.
This
Expected Value is displayed in the monitor.
Let's take a discrete node
Age with three categorical states:
The
Node Editor allows associating numerical values with these states.
v = 0.23 \times 25 + 0.415 \times 45 + 0.355 \times 80 = 52.825
Let's suppose now that the variable
Age has three numerical states.
As it's a numerical node, its monitor will have a
Mean value, a Standard Deviation and an Expected Value.
When the states do not have any associated values, $V_s$ is automatically set to the numerical value of the state.
Otherwise, the state values defined by the user are used:
The Mean value $m$ is computed with the following equation: $$m = \sum_{s \in S} p_s \times c_s$$ where $c_s$ is the numerical value of the state.
Let's consider now a continuous variable
Age defined on the domain [15 ; 99], and discretized into three states: Young Adult: [15 ; 30] Adult: ]30 ; 60] Senior: ]60 ; 99]
Again, as it's a numerical node, its monitor will have a
Mean value, a Standard Deviation and an Expected Value.
The Mean value $m$ is computed with the following equation: $$m = \sum_{s \in S} p_s \times c_s$$ where $c_s$ is the central tendency of the state defined as:the mid-range of the state when no data is associated,the arithmetic mean of the data points that are associated with the state.When the states do not have any associated values, $V_s$ is automatically set to the central tendency of the state.
If a dataset is associated to a continuous variable, clicking on the Generate Values buttons sets the values $V_s$ to the current arithmetic means.
When new pieces of evidence are set, a the delta value is displayed in the monitor:
This delta is the difference between the current
Expected Value v
and:
the previous one, the one corresponding to the reference probability distribution set with in the toolbar.
When only some states have an associated value, the Expected Value is computed on the states $S^*$ that have associated values $$v = \sum_{s \in S^*} \frac{p_s}{P^*} \times V_s$$ where $S^*$ is only made of one state, the node is considered as not having any associated values. |
For exercises 1-10, consider points \(P(−1,3), Q(1,5),\) and \(R(−3,7)\). Determine the requested vectors and express each of them \(a.\) in component form and \(b.\) by using standard unit vectors.
1) \( {\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{PQ}}} \)
Answer: \(a. \vec{PQ}=⟨2,2⟩ \quad b. \vec{PQ}=2\hat{\mathbf i}+2\hat{\mathbf j}\)
2) \(\vec{PR}\)
3) \(\vec{QP}\)
Answer: \(a. \vec{QP}=⟨−2,−2⟩ \quad b. \vec{QP}=−2\hat{\mathbf i}−2\hat{\mathbf j}\)
4) \(\vec{RP}\)
5) \(\vec{PQ}+\vec{PR}\)
Answer: \(a. \vec{PQ}+\vec{PR}=⟨0,6⟩ \quad b. \vec{PQ}+\vec{PR}=6\hat{\mathbf j}\)
6) \(\vec{PQ}−\vec{PR}\)
7) \(2\vec{PQ}−2\vec{PR}\)
Answer: \(a. 2\vec{PQ}→−2\vec{PR}=⟨8,−4⟩ \quad b. 2\vec{PQ}−2\vec{PR}=8\hat{\mathbf i}−4\hat{\mathbf j}\)
8) \(2\vec{PQ}+\frac{1}{2}\vec{PR}\)
9) The unit vector in the direction of \(\vec{PQ}\)
Answer: \(a. ⟨\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}⟩ \quad b. \frac{1}{\sqrt{2}}\hat{\mathbf i}+\frac{1}{\sqrt{2}}\hat{\mathbf j}\)
10) The unit vector in the direction of \(\vec{PR}\)
11) A vector \({\overset{\scriptstyle\rightharpoonup}{\mathbf v}}\) has initial point \((−1,−3)\) and terminal point \((2,1)\). Find the unit vector in the direction of \({\overset{\scriptstyle\rightharpoonup}{\mathbf v}}\). Express the answer in component form.
Answer: \(⟨\frac{3}{5},\frac{4}{5}⟩\)
12) A vector \(\vec{\mathbf v}\) has initial point \((−2,5)\) and terminal point \((3,−1)\). Find the unit vector in the direction of \(\vec{\mathbf v}\). Express the answer in component form.
13) The vector \(\vec{\mathbf v}\) has initial point \(P(1,0)\) and terminal point \(Q\) that is on the
y-axis and above the initial point. Find the coordinates of terminal point \(Q\) such that the magnitude of the vector \(\vec{\mathbf v}\) is \(\sqrt{5}\). Answer: \(Q(0,2)\)
14) The vector \(\vec{\mathbf v}\) has initial point \(P(1,1)\) and terminal point \(Q\) that is on the
x-axis and left of the initial point. Find the coordinates of terminal point \(Q\) such that the magnitude of the vector \(\vec{\mathbf v}\) is \(\sqrt{10}\).
For exercises 15 and 16, use the given vectors \(\vec{\mathbf a}\) and \(\vec{\mathbf b}\).
a. Determine the vector sum \(\vec{\mathbf a}+\vec{\mathbf b}\) and express it in both the component form and by using the standard unit vectors.
b. Find the vector difference \(\vec{\mathbf a}−\vec{\mathbf b}\) and express it in both the component form and by using the standard unit vectors.
c. Verify that the vectors \(\vec{\mathbf a}, \, \vec{\mathbf b},\) and \(\vec{\mathbf a}+\vec{\mathbf b}\), and, respectively, \(\vec{\mathbf a}, \, \vec{\mathbf b}\), and \(\vec{\mathbf a}−\vec{\mathbf b}\) satisfy the triangle inequality.
d. Determine the vectors \(2\vec{\mathbf a}, −\vec{\mathbf b},\) and \(2\vec{\mathbf a}−\vec{\mathbf b}.\) Express the vectors in both the component form and by using standard unit vectors.
15) \(\vec{\mathbf a}=2\hat{\mathbf i}+\hat{\mathbf j}, \vec{\mathbf b}=\hat{\mathbf i}+3\hat{\mathbf j}\)
Answer: \(a.\, \vec{\mathbf a}+\vec{\mathbf b}=⟨3,4⟩, \quad \vec{\mathbf a}+\vec{\mathbf b}=3\hat{\mathbf i}+4\hat{\mathbf j}\) \(b.\, \vec{\mathbf a}−\vec{\mathbf b}=⟨1,−2⟩, \quad \vec{\mathbf a}−\vec{\mathbf b}=\hat{\mathbf i}−2\hat{\mathbf j}\) \(c.\) Answers will vary \(d.\, 2\vec{\mathbf a}=⟨4,2⟩, \quad 2\vec{\mathbf a}=4\hat{\mathbf i}+2\hat{\mathbf j}, \quad −\vec{\mathbf b}=⟨−1,−3⟩, \quad −\vec{\mathbf b}=−\hat{\mathbf i}−3\hat{\mathbf j}, \quad 2\vec{\mathbf a}−\vec{\mathbf b}=⟨3,−1⟩, \quad 2\vec{\mathbf a}−\vec{\mathbf b}=3\hat{\mathbf i}−\hat{\mathbf j}\)
16) \(\vec{\mathbf a}=2\hat{\mathbf i}, \vec{\mathbf b}=−2\hat{\mathbf i}+2\hat{\mathbf j}\)
17) Let \(\vec{\mathbf a}\) be a standard-position vector with terminal point \((−2,−4)\). Let \(\vec{\mathbf b}\) be a vector with initial point \((1,2)\) and terminal point \((−1,4)\). Find the magnitude of vector \(−3\vec{\mathbf a}+\vec{\mathbf b}−4\hat{\mathbf i}+\hat{\mathbf j}.\)
Answer: \(15\)
18) Let \(\vec{\mathbf a}\) be a standard-position vector with terminal point at \((2,5)\). Let \(\vec{\mathbf b}\) be a vector with initial point \((−1,3)\) and terminal point \((1,0)\). Find the magnitude of vector \(\vec{\mathbf a}−3\vec{\mathbf b}+14\hat{\mathbf i}−14\hat{\mathbf j}.\)
19) Let \(\vec{\mathbf u}\) and \(\vec{\mathbf v}\) be two nonzero vectors that are nonequivalent. Consider the vectors \(\vec{\mathbf a}=4\vec{\mathbf u}+5\vec{\mathbf v}\) and \(\vec{\mathbf b}=\vec{\mathbf u}+2\vec{\mathbf v}\) defined in terms of \(\vec{\mathbf u}\) and \(\vec{\mathbf v}\). Find the scalar \(λ\) such that vectors \(\vec{\mathbf a}+λ\vec{\mathbf b}\) and \(\vec{\mathbf u}−\vec{\mathbf v}\) are equivalent.
Answer: \(λ=−3\)
20) Let \(\vec{\mathbf u}\) and \(\vec{\mathbf v}\) be two nonzero vectors that are nonequivalent. Consider the vectors \(\vec{\mathbf a}=2\vec{\mathbf u}−4\vec{\mathbf v}\) and \(\vec{\mathbf b}=3\vec{\mathbf u}−7\vec{\mathbf v}\) defined in terms of \(\vec{\mathbf u}\) and \(\vec{\mathbf v}\). Find the scalars \(α\) and \(β\) such that vectors \(α\vec{\mathbf a}+β\vec{\mathbf b}\) and \(\vec{\mathbf u}−\vec{\mathbf v}\) are equivalent.
21) Consider the vector \(\vec{\mathbf a}(t)=⟨\cos t, \sin t⟩\) with components that depend on a real number \(t\). As the number \(t\) varies, the components of \(\vec{\mathbf a}(t)\) change as well, depending on the functions that define them.
a. Write the vectors \(\vec{\mathbf a}(0)\) and \(\vec{\mathbf a}(π)\) in component form.
b. Show that the magnitude \(∥\vec{\mathbf a}(t)∥\) of vector \(\vec{\mathbf a}(t)\) remains constant for any real number \(t\).
c. As \(t\) varies, show that the terminal point of vector \(\vec{\mathbf a}(t)\) describes a circle centered at the origin of radius \(1\).
Answer: \(a.\, \vec{\mathbf a}(0)=⟨1,0⟩, \quad \vec{\mathbf a}(π)=⟨−1,0⟩\) \(b.\) Answers may vary \(c.\) Answers may vary
22) Consider vector \(\vec{\mathbf a}(x)=⟨x,\sqrt{1−x^2}⟩\) with components that depend on a real number \(x∈[−1,1]\). As the number \(x\) varies, the components of \(\vec{\mathbf a}(x)\) change as well, depending on the functions that define them.
a. Write the vectors \(\vec{\mathbf a}(0)\) and \(\vec{\mathbf a}(1)\) in component form.
b. Show that the magnitude \(∥\vec{\mathbf a}(x)∥\) of vector \(\vec{\mathbf a}(x)\) remains constant for any real number \(x\)
c. As \(x\) varies, show that the terminal point of vector \(\vec{\mathbf a}(x)\) describes a circle centered at the origin of radius \(1\).
23) Show that vectors \(\vec{\mathbf a}(t)=⟨\cos t, \sin t⟩\) and \(\vec{\mathbf a}(x)=⟨x,\sqrt{1−x^2}⟩\) are equivalent for \(x=r\) and \(t=2kπ\), where \(k\) is an integer.
Answer: Answers may vary
24) Show that vectors \(\vec{\mathbf a}(t)=⟨\cos t, \sin t⟩\) and \(\vec{\mathbf a}(x)=⟨x,\sqrt{1−x^2}⟩\) are opposite for \(x=r\) and \(t=π+2kπ\), where \(k\) is an integer.
For exercises 25-28, find a vector \(\vec{\mathbf v}\) with the given magnitude and in the same direction as the vector \(\vec{\mathbf u}\).
25) \(\|\vec{\mathbf v}\|=7, \quad \vec{\mathbf u}=⟨3,4⟩\)
Answer: \(\vec{\mathbf v}=⟨\frac{21}{5},\frac{28}{5}⟩\)
26) \(‖\vec{\mathbf v}‖=3,\quad \vec{\mathbf u}=⟨−2,5⟩\)
27) \(‖\vec{\mathbf v}‖=7,\quad \vec{\mathbf u}=⟨3,−5⟩\)
Answer: \(\vec{\mathbf v}=⟨\frac{21\sqrt{34}}{34},−\frac{35\sqrt{34}}{34}⟩\)
28) \(‖\vec{\mathbf v}‖=10,\quad \vec{\mathbf u}=⟨2,−1⟩\)
For exercises 29-34, find the component form of vector \(\vec{\mathbf u}\), given its magnitude and the angle the vector makes with the positive x-axis. Give exact answers when possible.
29) \(‖\vec{\mathbf u}‖=2, θ=30°\)
Answer: \(\vec{\mathbf u}=⟨\sqrt{3},1⟩\)
30) \(‖\vec{\mathbf u}‖=6, θ=60°\)
31) \(‖\vec{\mathbf u}‖=5, θ=\frac{π}{2}\)
Answer: \(\vec{\mathbf u}=⟨0,5⟩\)
32) \(‖\vec{\mathbf u}‖=8, θ=π\)
33) \(‖\vec{\mathbf u}‖=10, θ=\frac{5π}{6}\)
Answer: \(\vec{\mathbf u}=⟨−5\sqrt{3},5⟩\)
34) \(‖\vec{\mathbf u}‖=50, θ=\frac{3π}{4}\)
For exercises 35 and 36, vector \(\vec{\mathbf u}\) is given. Find the angle \(θ∈[0,2π)\) that vector \(\vec{\mathbf u}\) makes with the positive direction of the x-axis, in a counter-clockwise direction.
35) \(\vec{\mathbf u}=5\sqrt{2}\hat{\mathbf i}−5\sqrt{2}\hat{\mathbf j}\)
Answer: \(θ=\frac{7π}{4}\)
36) \(\vec{\mathbf u}=−\sqrt{3}\hat{\mathbf i}−\hat{\mathbf j}\)
37) Let \(\vec{\mathbf a}=⟨a_1,a_2⟩, \vec{\mathbf b}=⟨b_1,b_2⟩\), and \(\vec{\mathbf c}=⟨c_1,c_2⟩\) be three nonzero vectors. If \(a_1b_2−a_2b_1≠0\), then show there are two scalars, \(α\) and \(β\), such that \(\vec{\mathbf c}=α\vec{\mathbf a}+β\vec{\mathbf b}.\)
Answer: Answers may vary
38) Consider vectors \(\vec{\mathbf a}=⟨2,−4⟩, \vec{\mathbf b}=⟨−1,2⟩,\) and \(\vec{\mathbf c}=\vec{\mathbf 0}\) Determine the scalars \(α\) and \(β\) such that \(\vec{\mathbf c}=α\vec{\mathbf a}+β\vec{\mathbf b}\).
39) Let \(P(x_0,f(x_0))\) be a fixed point on the graph of the differential function \(f\) with a domain that is the set of real numbers.
a. Determine the real number \(z_0\) such that point \(Q(x_0+1,z_0)\) is situated on the line tangent to the graph of \(f\) at point \(P\).
b. Determine the unit vector \(\vec{\mathbf u}\) with initial point \(P\) and terminal point \(Q\).
Answer: \(a. \quad z_0=f(x_0)+f′(x_0); \quad b. \quad \vec{\mathbf u}=\frac{1}{\sqrt{1+[f′(x_0)]^2}}⟨1,f′(x_0)⟩\)
40) Consider the function \(f(x)=x^4,\) where \(x∈R\).
a. Determine the real number \(z_0\) such that point \(Q(2,z_0)\) s situated on the line tangent to the graph of \(f\) at point \(P(1,1)\).
b. Determine the unit vector \(\vec{\mathbf u}\) with initial point \(P\) and terminal point \(Q\).
41) Consider \(f\) and \(g\) two functions defined on the same set of real numbers \(D\). Let \(\vec{\mathbf a}=⟨x,f(x)⟩\) and \(\vec{\mathbf b}=⟨x,g(x)⟩\) be two vectors that describe the graphs of the functions, where \(x∈D\). Show that if the graphs of the functions \(f\) and \(g\) do not intersect, then the vectors \(\vec{\mathbf a}\) and \(\vec{\mathbf b}\) are not equivalent.
42) Find \(x∈R\) such that vectors \(\vec{\mathbf a}=⟨x, \sin x⟩\) and \(\vec{\mathbf b}=⟨x, \cos x⟩\) are equivalent.
43) Calculate the coordinates of point \(D\) such that \(ABCD\) is a parallelogram, with \(A(1,1), B(2,4)\), and \(C(7,4)\).
Answer: \(D(6,1)\)
44) Consider the points \(A(2,1), B(10,6), C(13,4)\), and \(D(16,−2)\). Determine the component form of vector \(\vec{AD}\).
45) The
speed of an object is the magnitude of its related velocity vector. A football thrown by a quarterback has an initial speed of \(70\) mph and an angle of elevation of \(30°\). Determine the velocity vector in mph and express it in component form. (Round to two decimal places.)
Answer: \(⟨60.62,35⟩\)
46) A baseball player throws a baseball at an angle of \(30°\) with the horizontal. If the initial speed of the ball is \(100\) mph, find the horizontal and vertical components of the initial velocity vector of the baseball. (Round to two decimal places.)
47) A bullet is fired with an initial velocity of \(1500\) ft/sec at an angle of \(60°\) with the horizontal. Find the horizontal and vertical components of the velocity vector of the bullet. (Round to two decimal places.)
Answer: The horizontal and vertical components are \(750\) ft/sec and \(1299.04\) ft/sec, respectively.
48) [T] A 65-kg sprinter exerts a force of \(798\) N at a \(19°\) angle with respect to the ground on the starting block at the instant a race begins. Find the horizontal component of the force. (Round to two decimal places.)
49) [T] Two forces, a horizontal force of \(45\) lb and another of \(52\) lb, act on the same object. The angle between these forces is \(25°\). Find the magnitude and direction angle from the positive
x-axis of the resultant force that acts on the object. (Round to two decimal places.)
Answer: The magnitude of resultant force is \(94.71\) lb; the direction angle is \(13.42°\).
50) [T] Two forces, a vertical force of \(26\) lb and another of \(45\) lb, act on the same object. The angle between these forces is \(55°\). Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to two decimal places.)
51) [T] Three forces act on object. Two of the forces have the magnitudes \(58\) N and \(27\) N, and make angles \(53°\) and \(152°\), respectively, with the positive
x-axis. Find the magnitude and the direction angle from the positive x-axis of the third force such that the resultant force acting on the object is zero. (Round to two decimal places.) Answer: The magnitude of the third vector is \(60.03\)N; the direction angle is \(259.38°\).
52) Three forces with magnitudes
80 lb, 120 lb, and 60 lb act on an object at angles of \(45°, 60°\) and \(30°\), respectively, with the positive x-axis. Find the magnitude and direction angle from the positive x-axis of the resultant force. (Round to two decimal places.)
53) [T] An airplane is flying in the direction of \(43°\) east of north (also abbreviated as \(N43E\) at a speed of \(550\) mph. A wind with speed \(25\) mph comes from the southwest at a bearing of \(N15E\). What are the ground speed and new direction of the airplane?
Answer: The new ground speed of the airplane is \(572.19\) mph; the new direction is \(N41.82E.\)
54) [T] A boat is traveling in the water at \(30\) mph in a direction of \(N20E\) (that is, \(20°\) east of north). A strong current is moving at \(15\) mph in a direction of \(N45E\). What are the new speed and direction of the boat?
55) [T] A 50-lb weight is hung by a cable so that the two portions of the cable make angles of \(40°\) and \(53°\), respectively, with the horizontal. Find the magnitudes of the forces of tension \(\vec{\mathbf T_1}\) and \(\vec{\mathbf T_2}\) in the cables if the resultant force acting on the object is zero. (Round to two decimal places.)
Answer: \(\|\vec{\mathbf T_1}\|=30.13 \, lb, \quad \|\vec{\mathbf T_2}\|=38.35 \, lb\)
56) [T] A 62-lb weight hangs from a rope that makes the angles of \(29°\) and \(61°\), respectively, with the horizontal. Find the magnitudes of the forces of tension \(\vec{\mathbf T_1}\) and \(\vec{\mathbf T_2}\) in the cables if the resultant force acting on the object is zero. (Round to two decimal places.)
57) [T] A 1500-lb boat is parked on a ramp that makes an angle of \(30°\) with the horizontal. The boat’s weight vector points downward and is a sum of two vectors: a horizontal vector \(\vec{\mathbf v_1}\) that is parallel to the ramp and a vertical vector \(\vec{\mathbf v_2}\) that is perpendicular to the inclined surface. The magnitudes of vectors \(\vec{\mathbf v_1}\) and \(\vec{\mathbf v_2}\) are the horizontal and vertical component, respectively, of the boat’s weight vector. Find the magnitudes of \(\vec{\mathbf v_1}\) and \(\vec{\mathbf v_2}\). (Round to the nearest integer.)
Answer: \(\|\vec{\mathbf v_1}\|=750 \, lb, \quad \|\vec{\mathbf v_2}\|=1299 \, lb\)
58) [T] An 85-lb box is at rest on a \(26°\) incline. Determine the magnitude of the force parallel to the incline necessary to keep the box from sliding. (Round to the nearest integer.)
59) A guy-wire supports a pole that is \(75\) ft high. One end of the wire is attached to the top of the pole and the other end is anchored to the ground \(50\) ft from the base of the pole. Determine the horizontal and vertical components of the force of tension in the wire if its magnitude is \(50\) lb. (Round to the nearest integer.)
Answer: The two horizontal and vertical components of the force of tension are \(28\) lb and \(42\) lb, respectively.
60) A telephone pole guy-wire has an angle of elevation of \(35°\) with respect to the ground. The force of tension in the guy-wire is \(120\) lb. Find the horizontal and vertical components of the force of tension. (Round to the nearest integer.)
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.
Exercises and LaTeX edited by Paul Seeburger |
One of the earliest models I looked at with the information equilibrium (IE) framework was the relationship $P : N \rightleftarrows L$ where $P$ is the price level, $N$ is nominal output, and $L$ is the level of employment. The relationship effectively captures Okun's law (i.e. changes in real GDP are related to changes in employment), and is a component of what I called the "quantity theory of labor" (and capital). The IE notation is shorthand for the equation:
\text{(1) } \; P \equiv \frac{dN}{dL} = k \; \frac{N}{L}
$$
with information transfer (IT) index (a free parameter) $k$. The price level is shown as the "detector" of information flow, and represents an abstract price in the information equilibrium framework. The solution to this differential equation is
\begin{align}
\frac{N}{N_{0}} = & \left( \frac{L}{L_{0}}\right)^{k}\\
P = & \; k \frac{N_{0}}{L_{0}} \left( \frac{L}{L_{0}}\right)^{k-1}
\end{align}
$$
Note that if $k = 2$, we have $P \sim L$ (hence the "quantity theory of labor" moniker), and we can obtain (one form of) Okun's law
\frac{d}{dt} \log L = \frac{d}{dt} \log R
$$
by simply differentiating the equation (1) above. (We take $R \equiv N/P$ = RGDP.)
Dynamic equilibrium
Now dynamic equilibrium (in the presence of shocks) is another way of looking at the same equation where we look at the growth of the (abstract) price on the left hand side and the growth of the ratio on the right hand side. Symbolically:
\begin{align}
\frac{d}{dt} \log P = & \; (k - 1) \frac{d}{dt} \log L\\
\frac{d}{dt} \log \frac{N}{L} = & \; (k - 1) \frac{d}{dt} \log L
\end{align}
$$
If we take $L \sim e^{\lambda t}$, $P \sim e^{\pi t}$, and $N/L \sim e^{\gamma t}$ then
\text{(2) } \; \pi = \gamma = (k - 1) \lambda
$$
I have previously looked at the core PCE price level, and noticed that it is well-described by a single shock centered at 1978.7 (I hypothesized this was related to a demographic shock of women entering the workforce). My question to myself was: what about $N/L$? The answer turned out to tell us that the abstract price $P$ isn't the measured by PCE (i.e. equation (2) is wrong), but rather something in information equilibrium with PCE. We also gain some new perspective on shocks.
The same procedure comes up with a similar result for $N/L$:
The biggest shock is centered at 1977.5, which is equal to the location of the shock to PCE inflation within the error. However, the first issue is that there is an additional shock associated with the aftermath of the great recession (centered at 2014.7). This shock is not obvious in the PCE data. This can be cleared up a bit by forcing the model to have an additional shock:
The model places a tiny shock (also centered at 2014.7 [2]) that is basically buried in the noise. Note that this shock may explain part of the low post-recession inflation and the more recent rise (the equilibrium inflation rate is 1.7%). More on this below in the forecasting section. In any case, this shows the small post-recession shock is not a serious issue.
What is a serious issue is that $\pi =$ 1.7% equilibrium inflation rate compared with the ratio $N/L$ growth rate of $\gamma =$ 3.8%. This tells us that the IE relationship $PCE : NGDP \rightleftarrows L$ is
an approximate effective theory. However, it's not a serious issue that can't be dealt with via a simple fix; much like the interest rate model [1] instead of PCE at best beingthe abstract price $P$, we just assume PCE is in information equilibrium withthe abstract price $P$:
\begin{align}
PCE & \rightleftarrows P\\
P : NGDP & \rightleftarrows L
\end{align}
$$
This introduces a second IT index we'll call $c$ and take $P \sim e^{\gamma t}$ (required by the theory above) and $PCE \sim e^{\pi t}$ so that
\text{(2') } \; \frac{\pi}{c} = \gamma = (k - 1) \lambda
$$
Now $\pi = $ 1.7% per year and $\gamma = $ 3.8% so $c = $ 0.45. Note that $\lambda \sim $ 2% meaning that $k \sim 2$ (i.e. the "quantity theory of labor").
What does this mean? Well, if we thought this model was accessing some fundamental "truth", we could say that the growth rate of $P$ (i.e. $\gamma$) is the "true" inflation rate that we only measure crudely with measures like PCE (actually CPI would be a slightly "better" measure with $c$ closer to 1).
But such a simple model is unlikely to be the best possible description of a macroeconomy, so it's probably best to take an agnostic "effective theory" approach. In that view, we just take the model above to be a decent approximation for some scope (e.g. average values over time scales of 10s of years, with errors for e.g. oil shocks).
Addendum: forecasting
The dynamic equilibrium results do lend themselves to (conditional [3]) forecasting, so here is the recent data for $N/L$ and core PCE (and their derivatives) along with a forecast through 2020:
This forecast predicts growth will rise back towards the equilibrium growth rate of 3.8%. Note that this is the growth rate of $N/L$,
NGDP growth. not
Here is PCE including the tiny post-recession shock:
The horizontal line is at 1.7% inflation; we can just make out the tiny dip centered at 2014.7 accounting for a 0.25 percentage point drop below 1.7% at the peak of the shock (i.e. about 1.4% inflation during 2014, which is in fact the average during that time ... 1.36%).
For completeness, here is the forecast without this shock:
Footnotes
[1] Instead of being the price of money, the interest rate is assumed to be in information equilibrium with the price of money.
[2] I previously associated the positive inflation shock with a demographic shift (women entering the workforce). The second shock occurs nearly 40 years later (potentially associated with the cohort that entered the workforce in the 1970s leaving the workforce). It could also be that we are seeing the demographic effect of people affected by the great recession leaving the workforce (the drop in the employment-population ratio), or of the baby boomer cohort retiring (the peak of which was centered in in the mid-1950s, meaning the mid-2010s shock is 60 years later).
[3] The condition is that no additional shock occurs during the forecast period. |
Definition:Total Ordering Definition
$\mathcal R$ is a
total ordering on $S$ if and only if: $\forall x, y \in S: x \mathop {\mathcal R} y \lor y \mathop {\mathcal R} x$
$\mathcal R$ is a
total ordering on $S$ if and only if: $(1): \quad \mathcal R \circ \mathcal R = \mathcal R$ $(2): \quad \mathcal R \cap \mathcal R^{-1} = \Delta_S$ $(3): \quad \mathcal R \cup \mathcal R^{-1} = S \times S$ Also known as
Some sources call this a
linear ordering, or a simple ordering. If it is necessary to emphasise that a total ordering $\preceq$ is not strict, then the term weak total ordering may be used. Also see Results about total orderingscan be found here. |
Fermat spiral
A planar transcendental curve the equation of which in polar coordinates has the form
$$\rho=a\sqrt\phi.$$
To each value of $\phi$ correspond two values of $\rho$ — a positive and a negative one. The Fermat spiral is centrally symmetric relative to the pole, which is a point of inflection. It belongs to the class of so-called algebraic spirals.
They were first studied by P. Fermat (1636).
Figure: f038420a
References
[1] A.A. Savelov, "Planar curves" , Moscow (1960) (In Russian) Comments References
[a1] J.D. Lawrence, "A catalog of special plane curves" , Dover, reprint (1972) How to Cite This Entry:
Fermat spiral.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Fermat_spiral&oldid=32537 |
"Off-Policy Evaluation for Slate Recommendation" by Swaminathan et al. 2017 Off-policy evaluation for slate recommendation. Swaminathan, Adith, Akshay Krishnamurthy, Alekh Agarwal, Miro Dudik, John Langford, Damien Jose, and Imed Zitouni. In Advances in Neural Information Processing Systems, pp. 3635-3645. 2017. Summary
Slate recommendation concerns a set of recommendations that are exposed to a user at once. For example, when you do a search, you get a page of ranked results. When you visit the home page of a music, movie, or book app, you usually see a whole page of recommendations. This paper is about how to do off-policy evaluation and optimization with data collected with a known logging policy.
They set up the problem as a contextual bandit where each action is a slate of sub-actions:$$s = (s_1, \dots, s_l)$$
where \(s\) refers to the slate and \(s_j\) refers to the \(j^\mathrm{th}\) item in the slate.
They key idea is that they estimate the reward for particular context \(x\) and slate \(s\) using a linear regression:$$\hat{V}(x, s) = \mathbb{1}_s^\top \phi_{x}$$ where \(\mathbb{1}_s\) is an indicator vector of size \(lm\) telling us which items (out of a possible \(m\) items) appears in the \(l\) slots. \(\phi_x\) represents the contribution towards to the overall reward for each item at each slot. The assumption is therefore that, in any given context, the contribution of an item to the overall reward must be linear.
The attraction of this approach is that it does not try to model the rewards and suffer the model mismatch bias. Instead, they impose a linearity assumption on how the individual item rewards relate to the overall rewards for a slate (and make the typical importance sample reweighting assumption that the target policy \(\pi\) is non-zero wherever the logging policy \(\mu\) is non-zero for all contexts). After that, they replace the predicted reward with a single-sample Monte Carlo estimate from the logged data to arrive at the
where the weights with the smallest norm can be estimated using the closed form solution for linear regression. One thing that was not clear from the paper was how to estimate the uncentered covariance matrix but I guess that can be done in the same way using a single-sample empirical estimate.
This is a pretty neat idea and they go on to show that the PI estimator is unbiased and has low variance in relation to the number of items and slots in the slate.
Discussion
I like the approach of avoiding bias in the estimator by making assumptions about how the reward decomposes and how it relates to the context. The single sample estimate, on the face of it, looks pretty crude but due to the unbiasedness, the errors cancel out over a large number of logged impressions. The same cannot be said for the model-based approach (what they call the
direct method or DM). However, as they show in their experiments, DM does well for smaller sets of impressions (< 10^5) which makes sense because that's where the bias introduced by model assumptions can compensate for the small data issues.
While they show that the linear reward assumption has common metrics like NDCG as special cases, I wonder how much interaction exists between items within a slate. An item that is a bad match for a context could tank the reward for the whole page but it might also be bad in how it interacts with other items. Interactions cannot be captured by the linearity, only indirectly through the context.
And I also wonder what the payoff is for introducing a better models in DM. Specifically, they seem to have considered only regression trees and lasso regression as their baselines for DM. Model mismatch bias can be addressed by better models, reducing the error in the DM estimate. |
Birthday Paradox Contents Paradox
Let there be $23$ or more people in a room.
The probability that at least $2$ of them have the same birthday is greater than $50 \%$.
Proof
Let there be $n$ people in the room.
Let $\map p n$ be the probability that no two people in the room have the same birthday.
Let the birthday of person $1$ be established.
The probability that person $2$ shares person $1$'s birthday is $\dfrac 1 {365}$.
Thus, the probability that person $2$ does not share person $1$'s birthday is $\dfrac {364} {365}$.
Similarly, the probability that person $3$ does not share the birthday of either person $1$ or person $2$ is $\dfrac {363} {365}$.
And further, the probability that person $n$ does not share the birthday of any of the people indexed $1$ to $n - 1$ is $\dfrac {365 - \paren {n - 1} } {365}$.
Hence the total probability that none of the $n$ people share a birthday is given by:
$\map p n = \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}$
\(\displaystyle \map p n\) \(=\) \(\displaystyle \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}\) \(\displaystyle \) \(=\) \(\displaystyle \dfrac {365!} {365^n} \binom {365} n\)
Setting $n = 23$ and evaluating the above gives:
$\map p {23} \approx 0.493$
Hence the probability that at least $2$ people share a birthday is $1 = 0.492 = 0.507 = 50.7 \%$
$\blacksquare$
Conclusion
This is a veridical paradox.
Counter-intuitively, the probability of a shared birthday amongst such a small group of people is surprisingly high.
Let $n$ be a set of people.
Let the probability that at least $3$ of them have the same birthday be greater than $50 \%$.
Then $n \ge 88$. |
This was really driving me nuts. The so-called "barometric formula". Here's what it is all about: an ideal gas in a container of a given volume would fill that volume at constant pressure $p$ and temperature $T$ so that the
ideal gas law applies:
\[pV=RnT,\]
where $T$ is the temperature, $n$ is the number of gas molecules (or number of moles, it's just a matter of picking a consistent set of units) and $R$ is the universal gas constant.
So what happens if you put this container in a homogeneous gravitational field characterized by the acceleration constant $g$? Something has to change.
Assuming that the temperature of the gas remains constant, the difference in pressure is simply the difference in the weight of a column of gas, which is expressed by the so-called aerostatic equation:
\[dp=-g\rho~dz,\]
where $\rho=\rho(z)$ is the density of the gas at height $z$. The infinitesimal version of the ideal gas law is $p~dV=RT~dn$ but $dn=(dn/dV)dV=(\mu dn/dV)/\mu~dV=\rho/\mu~dV$ where $\mu$ is the particle (or molar) mass, so
\begin{align}p~dV&=\frac{RT}{\mu}\rho~dV,\\\rho&=\frac{\mu p}{RT}.\end{align}
Putting this expression into the aerostatic equation gives
\begin{align}dp&=-\frac{g\mu p}{RT}dz,~~~~{\rm or}\\\frac{1}{p}dp&=-\frac{g\mu}{RT}dz.\end{align}
This last equation can be trivially integrated, producing the barometric formula:
\[p=p_0e^{-g\mu/RT}.\]
This looks nice, but why should we believe that $T$ remains constant with altitude, in this case? Indeed, Feynman himself says, when he introduces the barometric formula, that it really doesn't apply to our atmosphere since the temperature of air varies with altitude. There are at least two intuitive reasons to think that temperature would vary with altitude even in an equilibrium. First, if one imagines a gas as a bunch of little rubber balls that are being shaken, it stands to reason that only the faster balls reach beyond a certain altitude; so while high up, there will be fewer balls, they'd also tend to be "hotter" (keeping in mind that temperature is just the average kinetic energy of a particle.) Another way of looking at it is to think about a house in the summer: it's hot in the attic, cold in the basement, and there really is no reason to believe that, if the house were wrapped with a perfect insulating material, the temperatures would necessarily equalize. Or would they?
So let me rephrase the question. Instead of asking about the change in pressure in an isothermal column of gas, let me ask this: What is the equilibrium state of a column of gas in a homogeneous gravitational field? Equilibrium in this case need not mean isothermal. It may mean isothermal, but that has to be proven. What equilibrium means is that the system is in a state of maximum entropy.
How can we find the equilibrium state? We have two differential equations. One is the ideal gas law, the other is the aerostatic equation, but now both need to be modified to account for the variability of $T$. Two equations are not sufficient to solve for three unknown functions ($T$, $p$ and $\rho$), but we also have a third condition: in the equilibrium state, entropy is maximal.
Using a fundamental equation from thermodynamics,
\[T~dS=dU+p~dV,\]
we can express the infinitesimal change in entropy, $dS$ (here, $dU$ is the infinitesimal potential energy, $dV$ is the volume element, both of which are calculable from $\rho$ and $z$). The condition for equilibrium is that the integral $\int~dS$ be maximal, which is a problem from the calculus of variations. Unfortunately, I was not able to obtain a solution this way. To get to this point, I'd have to be able to express $p$, $\rho$ and $T$ using a function that has appropriate boundary conditions (e.g., $m(z)$ representing the mass of gas in the column up to a height $z$, so that $m(0)=0$ and $m(H)=M$ is the total mass of gas) so that I could have an equation like
\[\int~dS=\int L(m,dm/dz,z)~dz,\]
from which an Euler-Lagrange equation can be obtained and solved for $m$. Well, I couldn't do it, and it is possible that no such nice, clean solution exists.
Fortunately, shortly before I was ready to go bonkers, I came across a paper
1 that addressed precisely this problem, i.e., why is $T={\rm const.}$?
Here's the essence of the argument: The particles of the gas would indeed be bouncing about, and only those with more kinetic energy will bounce up high enough to get to the upper reaches of the column. But,
by the time they get up there, they'll have lost much of their kinetic energy! So whatever particles get all the way up there will have the same velocity distribution (hence, the same temperature) as the particles at ground level. So, the barometric formula is, in fact, the correct formula, because an equilibrium system will be isothermal.
Unlike my failed attempt, which was to use the axioms of thermodynamics, this reasoning starts with statistical physics. The starting point is the idea that the number of particles located at altitude between $z_1$ and $z_1+dz_1$, and moving at velocities $v_1$ and $v_1+dv_1$ can be expressed using a distribution function $f(z_1,v_1,t_1)$:
\[N=f(z_1,v_1,t_1)~dz_1dv_1.\]
Let us denote the area element $dz_1dv_1$ as $d\tau_1$.
As the system evolves from $t_1$ to $t_2$, the particles move from $z_1$ to $z_2$ and their velocities change from $v_1$ to $v_2$, in accordance with the dynamical equations of the system. However, the particle number remains unchanged, therefore
\[f(z_1,v_1,t_1)d\tau_1=f(z_2,v_2,t_2)d\tau_2.\]
If we could somehow show that $d\tau_1=d\tau_2$, that would mean that the distribution function is independent of time.
Let us take the special case of the homogeneous gravitational field. Initially, the area element is located in the $[z,v]$ phase space at coordinates $[z_1,v_1]$, $[z_1+dz_1,v_1]$, $[z_1,v_1+dv_1]$, $[z_1+dz_1,v_1+dv_1]$. As the system evolves and $t$ time elapses ($t=t_2-t_1$), the four corners of this rectangle move in accordance with the equations
\begin{align}z_2&=z_1+v_1t+\frac{1}{2}gt^2,\\v_2&=v_1+gt.\end{align}
Therefore, the four corners of the area element are mapped as follows (see also Fig. 1):
\begin{align}[z_1,v_1]&\rightarrow[z_2,v_2],\\ [z_1+dz_1,v_1]&\rightarrow[z_2+dz_1,v_2],\\ [z_1,v_1+dv_1]&\rightarrow[z_2+dv_1t,v_2+dv_1],\\ [z_1+dz_1,v_1+dv_1]&\rightarrow[z_2+dz_1+dv_1t,v_2+dv_1].\end{align}
These corners define a parallelogram with area $dz_1dv_1$, hence
\[d\tau_2=dz_1dv_1=d\tau_1,\]
which means that
\[f(z_1,v_1,t_1)=f(z_2,v_2,t_2).\]
Now let us make an assumption, a very reasonable one on thermodynamical grounds: namely that at some point $t_1$ in time, at some height $z_1$, the distribution function is the Maxwell-distribution:
\[f(z_1,v_1,t_1)=Ce^{-mv_1^2/2RT}.\]
Here, $C$ is a constant, $T$ is the temperature, and $m$ is the mass of a particle.
If the particles move in a collisionless manner, energy for each particle is conserved. The particle energy is a sum of its potential and kinetic energies:
\[E=mgz+\frac{1}{2}mv^2.\]
Energy conservation means that
\[mgz_1+\frac{1}{2}mv_1^2=mgz_2+\frac{1}{2}mgv_2^2.\]
From this,
\[\frac{1}{2}mv_1^2=mgz_2-mgz_1+\frac{1}{2}mgv_2^2.\]
Putting this expression into the Maxwell distribution equation, we get
\[f(z_1,v_1,t_1)=Ce^{mg(z_2-z_1)/RT}e^{-mv_2^2/2RT}.\]
As per the equality established earlier, this also means that
\[f(z_2,v_2,t_2)=Ce^{mg(z_2-z_1)/RT}e^{-mv_2^2/2RT}.\]
The first part of the right-hand side is a function of positions only. The velocity-dependent part of the distribution (hence, the relative probabilities of different velocities) is the same as before. In other words, the temperature is the same, regardless of the altitude; only the density of the medium varies with height.
One obvious question is whether this can be generalized to an arbitrary potential function, not just a homogeneous potential. A key step in the reasoning was the proof that the area element
dτ remains unchanged over time. We proved this in the special case of an homogeneous field, but that method doesn't work in the general case.
There is, however, a very powerful theorem in mechanics that is exactly about this issue: Liouville's theorem, that states that a volume element in phase space is a constant of the motion, hence it remains unchanged as we advance time and move along a particle's trajectory. This is exactly what we are saying: $d\tau_1=d\tau_2$.
In the general case, our distribution function will therefore take the form,
\[f(z_2,v_2,t_2)=Ce^{[V(z_2)-V(z_1)]/RT}e^{-mv_2^2/2RT},\]
where $V$ is a potential function. So long as $V$ is a function of only the coordinates, not velocities or time, the result stands: a column of ideal gas in equilibrium will be isothermal.
And, although we only considered the one-dimensional case here, the reasoning can be easily extended to more dimensions, and fields such as the central gravitational field of a planet like the Earth.
So why is the Earth's atmosphere not isothermal, then? Mainly because it is not an equilibrium system! It is constantly heated by the Earth itself from below, it exchanges heat with the oceans, it is heated by the Sun during the day, cooled radiatively during the night... a very complex system indeed, which is why global warming, for instance, remains such a contentious issue.
1Charles A. Coombes and Hans Laue: "A paradox concerning the temperature distribution of a gas in a gravitational field", Am. J. Phys, 53 (3) March 1985 |
Fitting models to large datasets and/or models involving a large number of random effects (for the
rma.mv() function) can be time consuming. Admittedly, some routines in the metafor package are not optimized for speed and efficient memory usage by default. However, there are various ways for speeding up the model fitting, which are discussed below.
Most meta-analytic datasets are relatively small, with less than a few dozen studies (and often even much smaller than that). However, occasionally, one may deal with a much larger dataset, at which point the
rma() (i.e.,
rma.uni()) function may start to behave sluggishly. To illustrate this, let's simulate a large dataset with $k = 4000$ studies based on a random-effects model (with $\mu = 0.50$ and $\tau^2 = 0.25$).
We then measure how long it takes to fit a random-effects model to these data.
library(metafor) system.time(res <- rma(yi, vi)) user system elapsed 565.14 1.87 567.03
The values are given in seconds and the one we are interested in is the last one (the elapsed time). So, it took almost 10 minutes to fit this model, although this was on a pretty outdated workstation with an Intel Xeon E5430 CPU running at 2.67 Ghz. A more modern/faster CPU will crunch through this much quicker, but even larger values of $k$ will eventually lead to the same problem. In fact, the model fitting time will tend to increase at a roughly quadratic rate as a function of $k$, as can be seen in the figure below.
The reason for this is that
rma() carries out some computations that involve $k \times k$ matrices. As these matrices grow in size, the time it takes to carry out the matrix algebra increases at a roughly quadratic rate. Note that this generally won't be an issue unless the number of studies is in the thousands, but at that point, model fitting can become really slow.
There are two ways of speeding up the model fitting in such a scenario. The first involves switching to the
rma.mv() function, which we can also use for fitting the random-effects model (see also this comparison of the rma.uni() and rma.mv() functions for more details). Here, we can make use of the fact that the marginal variance-covariance matrix of the observed outcomes is actually very sparse (to be precise, it is diagonal). By setting the argument
sparse=TRUE, sparse matrix objects are used to the extent possible, which can speed up model fitting substantially for certain models. The syntax for fitting the random-effects model would be:
id <- 1:k system.time(res <- rma.mv(yi, vi, random = ~ 1 | id, sparse=TRUE)) user system elapsed 40.09 1.05 41.14
So, instead of almost ten minutes, it now only took about 40 seconds, which is quite an improvement.
An alternative approach for speeding up the model fitting is to make use of optimized routines for the matrix algebra. The BLAS (Basic Linear Algebra Subprograms) library supplied with "vanilla R" is quite good, but enhanced linear algebra routines can be quite a bit faster. It is beyond the scope of this note to discuss how we can get R to make use of such enhanced routines, but the interested reader should take a look at the relevant section in the R Installation and Administration manual.
However, one of the easiest ways for getting the benefits of enhanced math routines is to install Microsoft R Open (MRO), which is a 100% compatible distribution of R (with a few added components), but most importantly, it provides the option to automatically install Intel's Math Kernel Library (MKL) alongside MRO. As the name implies, MKL will be most beneficial on Intel CPUs, so those with AMD processors may not see as much of a performance boost (or none at all? haven't tested this). At any rate, here are the results when fitting the model above with MRO+MKL using the
rma() function.
system.time(res <- rma(yi, vi)) user system elapsed 42.83 1.93 44.77
So, about 45 seconds, as opposed to almost 10 minutes with vanilla R. Interestingly, MRO+MKL is about as quick for this example as using vanilla R with
rma.mv(..., sparse=TRUE). However, we can get even better performance with MKL if we allow for multicore processing (one of the benefits of MKL is that it can run multithreaded and hence make use of multiple cores). The workstation I am using for these analyses actually has two quad-core CPUs, so there are 8 cores available we can make use of. Using the
setMKLthreads() command, we can set the number of cores that MKL is allowed to make use of. Let's see what happens if we give it all 8 cores.
setMKLthreads(8) system.time(res <- rma(yi, vi)) user system elapsed 68.72 10.77 11.63
So now we are down to less than 12 seconds! That's almost 50 times faster than what we started out with (567 seconds). I also examined the model fitting time as a function of the number of cores we make available to MKL. The results are shown in the graph below.
With more cores, things do speed up, but the gains diminish as we add more cores. Also, with just a single core, MKL took about 45 seconds as opposed to vanilla R, which took almost 10 minutes. So the largest gains come from switching to MKL in the first place, not the use of multiple cores.
One can even try combining the use of MRO+MKL with
rma.mv(..., sparse=TRUE). In this particular example, this didn't yield any additional performance benefits (and actually slowed things a little bit). I suspect that the performance benefits of MKL in this example are actually related to using numerical routines that can take advantage of the sparseness of the matrices automatically. Hence, there is no benefit in trying to exploit this characteristic of the data twice. However, with even larger values of $k$, use of
rma.mv(..., sparse=TRUE) together with MKL can yield some additional benefits due to more efficient storage of the underlying matrices.
Fitting complex models with the
rma.mv() function can also be time consuming. The most crucial aspect here is the number (and types) of random effects that are added to the model (the number of fixed effects is pretty much irrelevant for the underlying optimization routines). For illustration purposes, I simulated a large dataset based on a rather complex random effects structure. This dataset can be loaded directly into R with:
dat <- read.table("http://www.metafor-project.org/lib/exe/fetch.php/tips:data_complex_model.dat", header=TRUE, sep="\t")
For reasons to be explained shortly, we will also load two correlation matrices:
R.plant <- read.table("http://www.metafor-project.org/lib/exe/fetch.php/tips:data_complex_model_R_plant.dat", header=TRUE, sep="\t") R.fungus <- read.table("http://www.metafor-project.org/lib/exe/fetch.php/tips:data_complex_model_R_fungus.dat", header=TRUE, sep="\t")
First, let's look at the first 10 rows of the dataset.
head(dat, 10) study experiment plant fungus plant.phyl fungus.phyl yi vi 1 1 1 P07 F09 P07 F09 0.737 0.4985 2 1 2 P34 F22 P34 F22 1.947 0.5874 3 1 3 P14 F06 P14 F06 -0.573 0.4418 4 2 1 P27 F02 P27 F02 -0.266 0.1107 5 2 2 P27 F06 P27 F06 -0.317 0.0847 6 3 1 P19 F25 P19 F25 -0.051 0.2088 7 3 2 P14 F12 P14 F12 0.685 0.2195 8 3 3 P14 F06 P14 F06 1.374 0.0640 9 4 1 P25 F24 P25 F24 0.979 0.2103 10 4 2 P14 F24 P14 F24 1.288 0.3204
So, we have experiments nested within studies that were conducted with various combinations of plant species and fungi. We can familiarize ourselves a bit more with this dataset using the following commands.
nrow(dat) # number of rows (i.e., observed outcomes) length(unique(dat$study)) # number of studies table(table(dat$study)) # table of the number of experiments within studies length(unique(dat$plant)) # number of plants table(dat$plant) # table of the number of outcomes per plant length(unique(dat$fungus)) # number of fungi table(dat$fungus) # table of the number of outcomes per fungus table(dat$plant, dat$fungus) # table of plant-fungus frequencies
The output (not shown) indicates that there are 2000 observed outcomes, 868 studies, and most studies included between 1 and 4 experiments. Furthermore, 35 different plant species were studied and 25 different fungi, each with different frequencies.
We will fit a three-level meta-analytic model to these data (with experiments nested within studies) with crossed random effects for the plant and fungus factors. In addition, the
R.plant and
R.fungus correlation matrices loaded earlier reflect phylogenetic correlations for the various plant species and fungi studied. Hence, we will include phylogenetic random effects in the model based on these correlation matrices. For more details on models of this type, see Konstantopoulos (2011) and Nakagawa and Santos (2012). The code for fitting such a model is shown below.
system.time( res <- rma.mv(yi, vi, random = list(~ 1 | study/experiment, ~ 1 | plant, ~ 1 | fungus, ~ 1 | plant.phyl, ~ 1 | fungus.phyl), R = list(plant.phyl = R.plant, fungus.phyl = R.fungus), data = dat) ) user system elapsed 5696.19 59.11 5755.53
So, using vanilla R, fitting this model on my workstation took about 95 minutes. Ouch!
Let's try speeding things up. Given the discussion above, one may be inclined to try the
sparse=TRUE option. However, this yields:
user system elapsed 23755.76 362.74 24120.41
So now it took almost 7 hours to fit the same model, so this attempt really backfired. The reason for this is that the underlying matrices that the optimization routine has to deal with are not sparse at all (due to the crossed and correlated random effects). Forcing the use of sparse matrices then creates additional (and unnecessary) overhead, leading to a substantial slowdown.
Instead, we can give MRO+MKL a try. As above, I examined the time it took to fit the model as a function of cores made available. These are the results:
So, even with just one core, MRO+MKL took about 20 minutes. Again, we see diminishing returns as we make more cores available to MKL, but with all 8 cores, the model fitting was down to less than 8 minutes. This is more than 12 times faster than vanilla R. Quite a difference.
A few other adjustments can be tried to speed up the model fitting. First of all, the
rma.mv() function provides a lot of control over the optimization routine (see
help(rma.mv) and especially the information about the
control argument). For example, switching to a different optimizer may speed up the model fitting – but could also slow things down. So, whether this is useful is a matter of trial and error. Furthermore, the starting values for the optimization are not chosen in a terribly clever way at the moment and could be far off, in which case convergence may be slow. One can set the starting values manually via the
control argument. This could be useful, for example, when fitting several different but similar models to the same dataset.
One thing I would like to clarify. The comparison between "vanilla R" and "MRO+MKL" is really a comparison between the reference BLAS library of R versus MKL (so vanilla R versus MRO is not the issue here). Note that there are other linear algebra libraries (especially ATLAS and OpenBLAS) that can also be used with R and that may provide similar speedups. If anybody has made comparisons between these different options in conjunction with metafor, I would be curious to hear about it.
Konstantopoulos, S. (2011). Fixed effects and variance components estimation in three-level meta-analysis.
Research Synthesis Methods, 2(1), 61–76.
Nakagawa, S., & Santos, E. S. A. (2012). Methodological issues and advances in biological meta-analysis.
Evolutionary Ecology, 26(5), 1253–1274. |
A blog of Python-related topics and code.
To produce a children's sticker chart from a provided image, the following code divides it into squares (which can be cut out) and produces further images with matching labels for the reverse side of the printed image and a piece of card onto which the squares can be stuck (you need to provide your own glue).
Loosely speaking, in the Gregorian calendar, Easter falls on the first Sunday following the first full Moon on or after 21 March. There are various algorithms (
Computus) which can be used to calculate its date, and the cycle of dates repeats every 5.7 million years.The following code produces a bar chart of the distribution of Easter dates using one of these algorithms.
A linear transformation in two dimensions can be visualized through its effect on the two orthonormal basis vectors $\hat{\imath}$ and $\hat{\jmath}$. In general, it can be represented by a $2 \times 2$ matrix, $\boldsymbol{T}$, which acts on a vector $v$ to map it from a vector space spanned by one basis onto a different vector space spanned by another basis: $\boldsymbol{v'} = \boldsymbol{T}\boldsymbol{v}$. This change of basis can be visualized by drawing the basis vectors in the two-dimensional plane, along with equally-spaced "grid lines" parallel to each of them. A linear transformation keeps the grid lines evenly spaced, and the origin fixed.
The mildly controversial geocoding system What3words encodes the geographic coordinates of a location (to a resolution of 3 m) on the Earth's surface into three dictionary words, through some proprietary algorithm. The idea is that human's find it easier to remember and communicate these words than the sequence of digits that makes up the corresponding latitude and longitude. For example, the Victoria Memorial in front of Buckingham Palace in London is located at (51.50187, -0.14063) in decimal latitude, longitude coordinates, but simply
using.woods.laws in the language of What3words.
The magnetic field due to a magnetic dipole moment, $\boldsymbol{m}$ at a point $\boldsymbol{r}$ relative to it may be written $$ \boldsymbol{B}(\boldsymbol{r}) = \frac{\mu_0}{4\pi r^3}[3\boldsymbol{\hat{r}(\boldsymbol{\hat{r}} \cdot \boldsymbol{m}) - \boldsymbol{m}}], $$ where $\mu_0$ is the vacuum permeability. In geomagnetism, it is usual to write the radial and angular components of $\boldsymbol{B}$ as: $$ \begin{align*} B_r & = -2B_0\left(\frac{R_\mathrm{E}}{r}\right)^3\cos\theta, \\ B_\theta & = -B_0\left(\frac{R_\mathrm{E}}{r}\right)^3\sin\theta, \\ B_\phi &= 0, \end{align*} $$ where $\theta$ is polar (colatitude) angle (relative to the magnetic North pole), $\phi$ is the azimuthal angle (longitude), and $R_\mathrm{E}$ is the Earth's radius, about 6370 km. See below for a derivation of these formulae. |
A note on cohomological dimension over Cohen-Macaulay rings
Bull. Korean Math. Soc. Published online August 20, 2019
Iraj Bagheriyeh, Kamal Bahmanpour, and Ghader GhasemiUniversity of Mohaghegh Ardabili
Abstract : Let $(R,\m)$ be a Noetherian local Cohen-Macaulay ring and $I$ be a proper ideal of $R$.Assume that $\beta_R(I,R)$ denotes the constant value of ${\rm depth}_R(R/I^{n}R )$ for $n\gg0$.In this paper we introduce the new notion $\gamma_R(I,R)$ and the we prove the following inequalities:$$\beta_R(I,R)\leq\gamma_R(I,R)\leq \dim R - \cd(I,R)\leq \dim R/I.$$ Also, some applications of these inequalities will be included.
Keywords : canonical module, Cohen-Macaulay ring, cohomological dimension, local cohomology, Noetherian ring |
In a paper by Joos and Zeh, Z Phys B 59 (1985) 223, they say:This 'coming into being of classical properties' appears related to what Heisenberg may have meant by his famous remark [7]: 'Die "Bahn" entsteht erst dadurch, dass wir sie beobachten.'Google Translate says this means something ...
@EmilioPisanty Tough call. It's technical language, so you wouldn't expect every German speaker to be able to provide a correct interpretation—it calls for someone who know how German is used in talking about quantum mechanics.
Litmus are a London-based space rock band formed in 2000 by Martin (bass guitar/vocals), Simon (guitar/vocals) and Ben (drums), joined the following year by Andy Thompson (keyboards, 2001–2007) and Anton (synths). Matt Thompson joined on synth (2002–2004), while Marek replaced Ben in 2003. Oli Mayne (keyboards) joined in 2008, then left in 2010, along with Anton. As of November 2012 the line-up is Martin Litmus (bass/vocals), Simon Fiddler (guitar/vocals), Marek Bublik (drums) and James Hodkinson (keyboards/effects). They are influenced by mid-1970s Hawkwind and Black Sabbath, amongst others.They...
@JohnRennie Well, they repeatedly stressed their model is "trust work time" where there are no fixed hours you have to be there, but unless the rest of my team are night owls like I am I will have to adapt ;)
I think u can get a rough estimate, COVFEFE is 7 characters, probability of a 7-character length string being exactly that is $(1/26)^7\approx 1.2\times 10^{-10}$ so I guess you would have to type approx a billion characters to start getting a good chance that COVFEFE appears.
@ooolb Consider the hyperbolic space $H^n$ with the standard metric. Compute $$\inf\left\{\left(\int u^{2n/(n-2)}\right)^{-(n-2)/n}\left(4\frac{n-1}{n-2}\int|\nabla u|^2+\int Ru^2\right): u\in C^\infty_c\setminus\{0\}, u\ge0\right\}$$
@BalarkaSen sorry if you were in our discord you would know
@ooolb It's unlikely to be $-\infty$ since $H^n$ has bounded geometry so Sobolev embedding works as expected. Construct a metric that blows up near infinity (incomplete is probably necessary) so that the inf is in fact $-\infty$.
@Sid Eating glamorous and expensive food on a regular basis and not as a necessity would mean you're embracing consumer fetish and capitalism, yes. That doesn't inherently prevent you from being a communism, but it does have an ironic implication.
@Sid Eh. I think there's plenty of room between "I think capitalism is a detrimental regime and think we could be better" and "I hate capitalism and will never go near anything associated with it", yet the former is still conceivably communist.
Then we can end up with people arguing is favor "Communism" who distance themselves from, say the USSR and red China, and people who arguing in favor of "Capitalism" who distance themselves from, say the US and the Europe Union.
since I come from a rock n' roll background, the first thing is that I prefer a tonal continuity. I don't like beats as much as I like a riff or something atmospheric (that's mostly why I don't like a lot of rap)
I think I liked Madvillany because it had nonstandard rhyming styles and Madlib's composition
Why is the graviton spin 2, beyond hand-waiving, sense is, you do the gravitational waves thing of reducing $R_{00} = 0$ to $g^{\mu \nu} g_{\rho \sigma,\mu \nu} = 0$ for a weak gravitational field in harmonic coordinates, with solution $g_{\mu \nu} = \varepsilon_{\mu \nu} e^{ikx} + \varepsilon_{\mu \nu}^* e^{-ikx}$, then magic? |
This question already has an answer here:
I know that diminishing marginal returns even to all factors of production doesn't imply decreasing returns to scale. But could you please give me just an example of such production function?
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This question already has an answer here:
I know that diminishing marginal returns even to all factors of production doesn't imply decreasing returns to scale. But could you please give me just an example of such production function?
Decreasing marginal returns to a factor means that keeping the other factors fixed, the marginal output generated by this factor is decreasing. When looking at returns to scale, we change all outputs. Increasing a factor with decreasing marginal returns can have an indirect effect in increasing the marginal productivity of other factors. If we increase all factors at the same time, the indirect effects may outweigh the direct effect. The production function $F:\mathbb{R}_+^2$ given by $$F(x,y)=(x+1)^{2/3}(y+1)^{2/3}$$ has decreasing marginal factor productivities everywhere but not decreasing returns to scale (it doesn't have increasing returns to scale either).
For decreasing marginal returns we require second partial derivatives to be negative, since we examine what happens if we vary only one input
So any function $$y = \prod_{i=1}^{m}x_i^{a_i},\;\;\; 0<a_i<1\;\; \forall \,i$$
with $\sum a_i >1$ for increasing returns to scale, and with $\sum a_i <1$ for decreasing returns to scale, since here we examine what happens of we increase by the same proportion all inputs.
Responding to comments
We check returns to scale for this function by examining, for $k>1$, the expression
$$ \prod_{i=1}^{m}(kx_i)^{a_i} = k^{\sum a_i}\cdot y$$
If the sum of the alphas is higher than unity, output increases more than $k$ so we have increasing returns to scale, and correspondingly for deceasing returns to scale when the sum of the alphas is smaller than unity.
Regarding decreasing marginal returns, the rate of change of marginal output generated by a factor keeping the others fixed is given by its own second partial derivative, which here is
$$\frac {\partial^2 y}{\partial x_i^2}= a_i(a_i-1)\cdot \frac {y}{x_i^2}$$
When $0<a_i<1$, these second partials are all negative. |
Consider $A =\left( \begin{array}{ccc} -1 & 2 & 2\\ 2 & 2 & -1\\ 2 & -1 & 2\\ \end{array} \right)$. Find the eigenvalues of $A$.
So I know the characteristic polynomial is:
$$f_A(\lambda) = (-\lambda)^n+(trA)(-\lambda)^{n-1}+...+\det A$$
I found the $\det A = 27$, so the characteristic polynomial of $A$ is:
$$-\lambda^3+3\lambda^2-c\lambda+27$$
However, the textbook I'm using doesn't give any method for finding the value of $c$. I have the solution to the problem, and the value of $c$ is in fact $9$, but is there any method to numerically solve for it?
If we have a $4\times 4$ matrix, we'll end up with a characteristic polynomial of the form $$\lambda^4-tr A(\lambda)^3+c_1\lambda^2-c_2\lambda+\det A$$
Similarly, is there a method to solve for $c_1,c_2$ in this case? |
Let $x_1,...,x_n $ are distinct real numbers.
Is it a formula for the Vandermonde type determinant $V(x_1, \cdots,x_n)$ whose last column is $x_1^k,\ \cdots,\ x_n^k$, where $k \geq n$, instead of $x_1^{n-1},\ \cdots,\ x_n^{n-1}$?
Thanks
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Sure, at least you can find such a formula for any fixed $k \geqslant n$. Not sure about a general formula for an unknown $k$ though.
Here is a hint: if all the $x_i$ are distinct, vector $(x_1^k,\ldots,x_n^k)$ is a linear combination of vectors $(x_1^i,\ldots,x_n^i)$, $i=0,\,1,\ldots,n-1$. If $$ (x_1^k,\ldots,x_n^k) = \sum_{i=0}^{n-1} \lambda_i (x_1^i,\ldots,x_n^i), $$ then your determinant is simply equal to $\lambda_{n-1}$ times the usual Vandermonde determinant. So all you need to do is find $\lambda_{n-1}$.
To continue Dan's answer, we want $$ x_i^k = \sum_{j=0}^{n-1} \lambda_j x_i^j, \qquad 1 \le i \le n $$ That is the polynomial $p(x) := \sum_{j=0}^{n-1} \lambda_j x^j$ interpolates $x^k$ at $x_0, \ldots, x_{n-1}$. Lagrange interpolation gives $$ p(x) = \sum_{j=0}^{n-1} x_j^k \cdot \prod_{\ell \ne j} \frac{x-x_\ell}{x_j - x_\ell} $$ $\lambda_{n-1}$ is the coefficient of $x^{n-1}$, so it is $$ \lambda_{n-1} = \sum_{j=0}^{n-1} x_j^k \prod_{\ell\ne j} \frac 1{x_j - x_\ell}. $$
Perform Laplace expansion along the last column. As the deletion of the $\ell$-th row and the last column gives a $(n-1)\times(n-1)$ Vandermonde matrix (in the original flavour), we get $$ \sum_{\ell=1}^n (-1)^{\ell+n} x_\ell^k\prod_{i<j\,\textrm{ and }\,i,j\not=\ell}(x_j-x_i). $$
Actually there is a general formula but not an easy way to describe it.
Let me first "define" the polynomials $f_m(x_1,x_2,\ldots,x_k)$ through some examples.
Let $V(x_1,x_2,\ldots,x_n,k)$ denote the Vandermonde type determinant whose last column is $x_1^k,x_2^k,\ldots,x_n^k$ where $k\geq n-1$. So $V(x_1,x_2,\ldots,x_n,n-1)$ is the usual Vandermonde. Then $$V(x_1,x_2,\ldots,x_n,k)=V(x_1,x_2,\ldots,x_n,n-1)\cdot f_{k-n+1}(x_1,x_2,\ldots,x_n).$$ |
I’ll post what I have so far, as requested and because someone might pick up a way forward from the end result. It is also too long for a comment but should not be considered a solution to the problem. I feel that there might actually be a closed from worth chasing.
Firstly the function of interest has the integral representation of$$u\left( x,y,t \right)=\int\limits_{0}^{\infty }{{{e}^{-ts}}\sinh \left( \pi s \right){{K}_{is}}\left( x \right){{K}_{is}}\left( y \right)ds}.$$It should be noted that this function appears to satisfy a diffusion like differential equation, so it might help to post some background information regarding where all this has come from. The reason I say that is that these sort of integrals – in terms of Lebedev’s transform / index transforms – have a very natural setting in eigenfunction expansions (see Titchmarsh - eigenfunction expansions vol 1). So there might be a natural interpretation of it, and it’s properties, within that framework. Note the identity $${{e}^{-ts}}=\frac{2}{\pi }\int\limits_{0}^{\infty }{\frac{w}{{{w}^{2}}+{{s}^{2}}}\sin \left( wt \right)du}\ \ \ \ \operatorname{Re}\left( t,s \right)>0$$And so we have $$\frac{\partial }{\partial t}u\left( x,t \right)=-\int\limits_{0}^{\infty }{\frac{s}{{{w}^{2}}+{{s}^{2}}}\sinh \left( \pi s \right){{K}_{is}}\left( x \right){{K}_{is}}\left( y \right)ds}\frac{2}{\pi }\int\limits_{0}^{\infty }{w\sin \left( wt \right)dw}$$Now from Bateman’s Integral Transforms vol. 2 pg 176 12.1.8 we have$$\int\limits_{0}^{\infty }{\frac{s}{{{n}^{2}}+{{s}^{2}}}\sinh \left( \pi s \right){{K}_{is}}\left( x \right){{K}_{is}}\left( y \right)ds}=\left\{ \begin{matrix} \frac{{{\pi }^{2}}}{2}{{I}_{n}}\left( y \right){{K}_{n}}\left( x \right) & 0<y<x \\ \frac{{{\pi }^{2}}}{2}{{I}_{n}}\left( x \right){{K}_{n}}\left( y \right) & x<y<\infty \\\end{matrix} \right.$$This is for n, a natural number. Suppose we can analytically continue it then we would have for example
$$\frac{\partial }{\partial t}u\left( x,y,t \right)=-\pi \int\limits_{0}^{\infty }{w{{I}_{w}}\left( y \right){{K}_{w}}\left( x \right)\sin \left( wt \right)dw}$$
for $0<y<x$, and a similar representation for $x<y$ etc. Integrating we find $$u\left( x,y,t \right)=\pi \int\limits_{0}^{\infty }{{{I}_{w}}\left( y \right){{K}_{w}}\left( x \right)\cos \left( wt \right)dw}+{{c}_{1}}\left( x,y \right)\ \ \ \ 0<y<x$$ $$u\left( x,y,t \right)=\pi \int\limits_{0}^{\infty }{{{I}_{w}}\left( x \right){{K}_{w}}\left( y \right)\cos \left( wt \right)dw}+{{c}_{2}}\left( x,y \right)\ \ \ \ 0<x<y$$
Note:${{K}_{w}}\left( x \right)\sim \sqrt{\frac{\pi }{2w}}{{\left( \frac{ex}{2w} \right)}^{-w}}$ and ${{I}_{w}}\left( x \right)\sim \sqrt{\frac{\pi }{2w}}{{\left( \frac{ex}{2w} \right)}^{w}}$ for large w. Now observe that from the original representation that $\underset{t\to \infty }{\mathop{\lim }}\,u\left( x,y,t \right)=0$, by the Riemann-Lebesgue lemma. Therefore also by RL the first term in the above representations becomes zero in the limit, and hence we must have therefore ${{c}_{1}}={{c}_{2}}=0$. We have then$$u\left( x,y,t \right)=\pi \int\limits_{0}^{\infty }{{{\kappa }_{w}}\left( x,y \right)\cos \left( wt \right)dw}$$
where $\kappa$ takes on the different forms depending on $x<y$ or $y<x$. At the moment the reason why I think all this works and is worth pursuing is that a numerical integration comparison of this representation to that of the original representation is spot-on to many decimal places for a wide range of x,y that I have tested. Obviously not a proof, but I think one can ‘feel' that the justification to move n away from the naturals and to a real number wouldn’t be a stretch. Another reason is that if you now use Lebedev’s formula (pg 140 from special functions), you will arrive at the formula you have already derived (second last formula). Since Lebedev's formula is valid for arbitrary indices, then this might end up constituting a 'proof' of continuation.
Note from the form above we have$${{\hat{u}}_{C}}\left( x,y,w \right)=\pi {{K}_{w}}\left( x \right){{I}_{w}}\left( y \right)$$where ${{\hat{u}}_{C}}$ is the cosine transform of u, which may be useful (?). I have various other representations but nothing that is heading to anything I’d call ‘closed’. |
I have read the following theorem:
If $p_1,p_2,\dots,p_n$ are distinct prime numbers, then$$\left(\mathbb Q\left[\sqrt p_1,\dots,\sqrt p_n\right]:\mathbb Q\right)=2^n.$$
I have tried to prove a more general statement but I have a problem at one point. (I still don't know how to prove the theorem above, too, because I don't know how not to use linear independence, which I do in the more general statement below.) Could you please help me overcome the obstacle I've encountered? I will post the intended proof and make it clear where I'm having trouble.
I want to prove the following statement:
Let $n\geq 1$. The set $B_n:=\left\{\sqrt {p_1^{\epsilon_1}}\sqrt {p_2^{\epsilon_2}}\cdots\sqrt {p_n^{\epsilon_n}}\,|\,(\epsilon_1,\epsilon_2,\cdots,\epsilon_n)\in\{0,1\}^n\right\}$ has $2^n$ elements and is a $\mathbb Q-$basis of $\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_n\right].$
The proof will be by induction.
For $n=1,$ we have $B_n=\left\{1,\sqrt {p_1}\right\}.$ It is clear that $\sqrt{p_1}\neq 1,$ so the set has $2=2^1$ elements. It is the basis of $\mathbb Q[\sqrt{p_1}]$ because the minimal polynomial of $\sqrt {p_1}$ over $\mathbb Q$ has degree $2,$ and there is a theorem that $K[a]$ has $a^0,\cdots,a^{d-1}$ as a basis, where $d$ is the degree of the minimal polynomial of $a$ over $K$.
Suppose the statement is true for $n-1$, where $n\geq 2.$ We have
$$ \left(B_n=B_{n-1}\cup\sqrt{p_n}B_{n-1}\right)\text { and } \left(B_{n-1}\cap\sqrt{p_n}B_{n-1}=\emptyset\right), $$
which is easy to see. It is also easy to see that $\operatorname{card}(B_{n-1})=\operatorname{card}(\sqrt{p_n}B_{n-1}),$ and therefore
$$ \operatorname{card}B_{n}=2^n. $$
Let
$$ \sum_{x\in B_{n}}q_xx=0 $$
for some $\{q_x\}_{x\in B_n}\subset\mathbb Q.$ Let $p(x):=\sqrt{p_n}x$ for all $x\in B_{n-1}.$ We have
$$ \sum_{x\in B_{n}}q_xx=\sum_{x\in B_{n-1}} q_xx+\sum_{x\in \sqrt{p_n}B_{n-1}} q_xx=\sum_{x\in B_{n-1}} q_xx+\sum_{x\in B_{n-1}} q_{p(x)}\sqrt{p_n}x. $$
Therefore
$$ \sum_{x\in B_{n-1}} q_xx=-\sqrt{p_n}\sum_{x\in B_{n-1}} q_{p(x)}x,\tag1 $$
and we can make the following division iff $q_{p(x)}\neq 0$ for all $x\in B_{n-1}$ (because $B_{n-1}$ is linearly indepentent over $\mathbb Q$):
$$ \sqrt{p_n}=-\frac{\sum_{x\in B_{n-1}} q_xx}{\sum_{x\in B_{n-1}} q_{p(x)}x}, $$
The right-hand side belongs to $\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_{n-1}\right],$ so we have
$$ \sqrt{p_n}\in \mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_{n-1}\right]. $$
Therefore we can write $\sqrt{p_n}$ uniquely in the basis $B_{n-1}$.
$$ \sqrt{p_n}=\sum_{y\in B_{n-1}}c_yy $$
for some $\{c_y\}_{y\in B_{n-1}}\subset \mathbb Q.$
After squaring this equation we will obtain
$$ p_n=\sum_{y\in B_{n-1}}c_y^2y^2+2\sum_{y,z\in B_{n-1}}c_yc_zyz. $$
The last sum must be zero because it is not in $\mathbb Q$ and because after reducing it, we obtain a representation of $p_n$ in the basis $B_{n-1},$ which is unique. Thus
$$p_n=\sum_{y\in B_{n-1}}c_y^2y^2.$$
Unfortunately, I can't prove that $c_yc_z$ is always zero. This was my first thought, but clearly there's trouble with the possibility of reductions in $$ \sum_{y,z\in B_{n-1}}c_yc_zyz. $$
Different pairs $y,z$ may yield the same element of $B_{n-1}$ in the product $yz.$ This happens for example when $y=\sqrt 5\sqrt 3,$ $z=\sqrt 5\sqrt 2,$ and $y'= \sqrt 11\sqrt 2,$ $z'=\sqrt 11\sqrt 3$.
If it were true that $c_yc_z$ is always zero, I would be able to continue my proof as follows. We would have only one $y_0$ such that $c_{y_0}\neq 0$ and we'd get
$$p_n=c_{y_0}^2y_0^2.$$
Let $c_{y_0}=\frac kl$. We can write $$l^2p_n=k^2y_0^2.$$
But $y_0^2$ is the product of some primes different from $p_n$. Therefore the greatest power of $p_n$ that divides the right-hand side is even. However, the greatest power of $p_n$ that divides the left-hand side is odd. A contradiction.
The contradiction proves that $q_{p(x)}=0$ for all $x\in B_{n-1}.$ Hence $(1)$ gives us that
$$ \sum_{x\in B_{n-1}} q_xx=0 $$
and linear independence of $B_{n-1}$ gives us that $q_x=0$ for all $x\in B_{n-1}.$
This gives us that $B_n$ is linearly independent. It generates the whole $\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_n\right]$ because
$$ \mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_n\right]=\left(\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_{n-1}\right]\right)\left[\sqrt{p_n}\right]. $$
This
would end the proof. |
2018-09-11 04:29
Proprieties of FBK UFSDs after neutron and proton irradiation up to $6*10^{15}$ neq/cm$^2$ / Mazza, S.M. (UC, Santa Cruz, Inst. Part. Phys.) ; Estrada, E. (UC, Santa Cruz, Inst. Part. Phys.) ; Galloway, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; Gee, C. (UC, Santa Cruz, Inst. Part. Phys.) ; Goto, A. (UC, Santa Cruz, Inst. Part. Phys.) ; Luce, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; McKinney-Martinez, F. (UC, Santa Cruz, Inst. Part. Phys.) ; Rodriguez, R. (UC, Santa Cruz, Inst. Part. Phys.) ; Sadrozinski, H.F.-W. (UC, Santa Cruz, Inst. Part. Phys.) ; Seiden, A. (UC, Santa Cruz, Inst. Part. Phys.) et al. The properties of 60-{\mu}m thick Ultra-Fast Silicon Detectors (UFSD) detectors manufactured by Fondazione Bruno Kessler (FBK), Trento (Italy) were tested before and after irradiation with minimum ionizing particles (MIPs) from a 90Sr \b{eta}-source . [...] arXiv:1804.05449. - 13 p. Preprint - Full text Details des Eintrags - Ähnliche Datensätze 2018-08-25 06:58
Charge-collection efficiency of heavily irradiated silicon diodes operated with an increased free-carrier concentration and under forward bias / Mandić, I (Ljubljana U. ; Stefan Inst., Ljubljana) ; Cindro, V (Ljubljana U. ; Stefan Inst., Ljubljana) ; Kramberger, G (Ljubljana U. ; Stefan Inst., Ljubljana) ; Mikuž, M (Ljubljana U. ; Stefan Inst., Ljubljana) ; Zavrtanik, M (Ljubljana U. ; Stefan Inst., Ljubljana) The charge-collection efficiency of Si pad diodes irradiated with neutrons up to $8 \times 10^{15} \ \rm{n} \ cm^{-2}$ was measured using a $^{90}$Sr source at temperatures from -180 to -30°C. The measurements were made with diodes under forward and reverse bias. [...] 2004 - 12 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 533 (2004) 442-453 Details des Eintrags - Ähnliche Datensätze 2018-08-23 11:31 Details des Eintrags - Ähnliche Datensätze 2018-08-23 11:31
Effect of electron injection on defect reactions in irradiated silicon containing boron, carbon, and oxygen / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Yakushevich, H S (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) Comparative studies employing Deep Level Transient Spectroscopy and C-V measurements have been performed on recombination-enhanced reactions between defects of interstitial type in boron doped silicon diodes irradiated with alpha-particles. It has been shown that self-interstitial related defects which are immobile even at room temperatures can be activated by very low forward currents at liquid nitrogen temperatures. [...] 2018 - 7 p. - Published in : J. Appl. Phys. 123 (2018) 161576 Details des Eintrags - Ähnliche Datensätze 2018-08-23 11:31 Details des Eintrags - Ähnliche Datensätze 2018-08-23 11:31
Characterization of magnetic Czochralski silicon radiation detectors / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) Silicon wafers grown by the Magnetic Czochralski (MCZ) method have been processed in form of pad diodes at Instituto de Microelectrònica de Barcelona (IMB-CNM) facilities. The n-type MCZ wafers were manufactured by Okmetic OYJ and they have a nominal resistivity of $1 \rm{k} \Omega cm$. [...] 2005 - 9 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 548 (2005) 355-363 Details des Eintrags - Ähnliche Datensätze 2018-08-23 11:31
Silicon detectors: From radiation hard devices operating beyond LHC conditions to characterization of primary fourfold coordinated vacancy defects / Lazanu, I (Bucharest U.) ; Lazanu, S (Bucharest, Nat. Inst. Mat. Sci.) The physics potential at future hadron colliders as LHC and its upgrades in energy and luminosity Super-LHC and Very-LHC respectively, as well as the requirements for detectors in the conditions of possible scenarios for radiation environments are discussed in this contribution.Silicon detectors will be used extensively in experiments at these new facilities where they will be exposed to high fluences of fast hadrons. The principal obstacle to long-time operation arises from bulk displacement damage in silicon, which acts as an irreversible process in the in the material and conduces to the increase of the leakage current of the detector, decreases the satisfactory Signal/Noise ratio, and increases the effective carrier concentration. [...] 2005 - 9 p. - Published in : Rom. Rep. Phys.: 57 (2005) , no. 3, pp. 342-348 External link: RORPE Details des Eintrags - Ähnliche Datensätze 2018-08-22 06:27
Numerical simulation of radiation damage effects in p-type and n-type FZ silicon detectors / Petasecca, M (Perugia U. ; INFN, Perugia) ; Moscatelli, F (Perugia U. ; INFN, Perugia ; IMM, Bologna) ; Passeri, D (Perugia U. ; INFN, Perugia) ; Pignatel, G U (Perugia U. ; INFN, Perugia) In the framework of the CERN-RD50 Collaboration, the adoption of p-type substrates has been proposed as a suitable mean to improve the radiation hardness of silicon detectors up to fluencies of $1 \times 10^{16} \rm{n}/cm^2$. In this work two numerical simulation models will be presented for p-type and n-type silicon detectors, respectively. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 2971-2976 Details des Eintrags - Ähnliche Datensätze 2018-08-22 06:27
Technology development of p-type microstrip detectors with radiation hard p-spray isolation / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) ; Díez, S (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) A technology for the fabrication of p-type microstrip silicon radiation detectors using p-spray implant isolation has been developed at CNM-IMB. The p-spray isolation has been optimized in order to withstand a gamma irradiation dose up to 50 Mrad (Si), which represents the ionization radiation dose expected in the middle region of the SCT-Atlas detector of the future Super-LHC during 10 years of operation. [...] 2006 - 6 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 566 (2006) 360-365 Details des Eintrags - Ähnliche Datensätze 2018-08-22 06:27
Defect characterization in silicon particle detectors irradiated with Li ions / Scaringella, M (INFN, Florence ; U. Florence (main)) ; Menichelli, D (INFN, Florence ; U. Florence (main)) ; Candelori, A (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Bruzzi, M (INFN, Florence ; U. Florence (main)) High Energy Physics experiments at future very high luminosity colliders will require ultra radiation-hard silicon detectors that can withstand fast hadron fluences up to $10^{16}$ cm$^{-2}$. In order to test the detectors radiation hardness in this fluence range, long irradiation times are required at the currently available proton irradiation facilities. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 589-594 Details des Eintrags - Ähnliche Datensätze |
Electronic Communications in Probability Electron. Commun. Probab. Volume 18 (2013), paper no. 72, 10 pp. The impact of selection in the $\Lambda$-Wright-Fisher model Abstract
The purpose of this article is to study some asymptotic properties of the $\Lambda$-Wright-Fisher process with selection. This process represents the frequency of a disadvantaged allele. The resampling mechanism is governed by a finite measure $\Lambda$ on $[0,1]$ and selection by a parameter $\alpha$. When the measure $\Lambda$ obeys $\int_{0}^{1}-\log(1-x)x^{-2}\Lambda(dx)<\infty$, some particular behaviour in the frequency of the allele can occur. The selection coefficient $\alpha$ may be large enough to override the random genetic drift. In other words, for certain selection pressure, the disadvantaged allele will vanish asymptotically with probability one. This phenomenon cannot occur in the classical Wright-Fisher diffusion. We study the dual process of the $\Lambda$-Wright-Fisher process with selection and prove this result through martingale arguments.
There is an Erratum in ECP volume 19 paper 15 (2014). Article information Source Electron. Commun. Probab., Volume 18 (2013), paper no. 72, 10 pp. Dates Accepted: 24 August 2013 First available in Project Euclid: 7 June 2016 Permanent link to this document https://projecteuclid.org/euclid.ecp/1465315611 Digital Object Identifier doi:10.1214/ECP.v18-2838 Mathematical Reviews number (MathSciNet) MR3101637 Zentralblatt MATH identifier 1337.60179 Rights This work is licensed under a Creative Commons Attribution 3.0 License. Citation
Foucart, Clément. The impact of selection in the $\Lambda$-Wright-Fisher model. Electron. Commun. Probab. 18 (2013), paper no. 72, 10 pp. doi:10.1214/ECP.v18-2838. https://projecteuclid.org/euclid.ecp/1465315611 |
In the last set of notes we talked about how to
differentiate \(k\)-forms using the exterior derivative \(d\). We’d also like some way to integrate forms. Actually, there’s surprisingly little to say about integration given the setup we already have. Suppose we want to compute the total area \(A_\Omega\) of a region \(\Omega\) in the plane:
If you remember back to calculus class, the basic idea was to break up the domain into a bunch of little pieces that are easy to measure (like squares) and add up their areas:
\[ A_\Omega \approx \sum_i A_i. \]
As these squares get smaller and smaller we get a better and better approximation, ultimately achieving the true area
\[ A_\Omega = \int_\Omega dA. \]
Alternatively, we could write the individual areas using differential forms — in particular, \(A_i = dx^1 \wedge dx^2(u,v)\). Therefore, the area element \(dA\) is really nothing more than the standard volume form \(dx^1 \wedge dx^2\) on \(\mathbb{R}^2\). (Not too surprising, since the whole point of \(k\)-forms was to measure volume!)
To make things more interesting, let’s say that the contribution of each little square is weighted by some scalar function \(\phi\). In this case we get the quantity
\[ \int_\Omega \phi\ dA = \int_\Omega \phi\ dx^1 \wedge dx^2. \]
Again the integrand \(\phi\ dx^1 \wedge dx^2\) can be thought of as a 2-form. In other words, you’ve been working with differential forms your whole life, even if you didn’t realize it! More generally, integrands on an \(n\)-dimensional space are always \(n\)-forms, since we need to “plug in” \(n\) orthogonal vectors representing the local volume. For now, however, looking at surfaces (i.e., 2-manifolds) will give us all the intuition we need.
Integration on Surfaces
If you think back to our discussion of the Hodge star, you’ll remember the
volume form
\[ \omega = \sqrt{\mathrm{det}(g)} dx^1 \wedge dx^2, \]
which measures the area of little parallelograms on our surface. The factor \(\sqrt{\mathrm{det}(g)}\) reminds us that we can’t simply measure the volume in the domain \(M\) — we also have to take into account any “stretching” induced by the map \(f: M \rightarrow \mathbb{R}^2\). Of course, when we integrate a function on a surface, we should also take this stretching into account. For instance, to integrate a function \(\phi: M \rightarrow \mathbb{R}\), we would write
\[ \int_\Omega \phi \omega = \int_\Omega \phi \sqrt{\mathrm{det}(g)}\ dx^1 \wedge dx^2. \]
In the case of a
conformal parameterization things become even simpler — since \(\sqrt{\mathrm{det}(g)} = a\) we have just
\[ \int_\Omega \phi a\ dx^1 \wedge dx^2, \]
where \(a: M \rightarrow \mathbb{R}\) is the scaling factor. In other words, we scale the value of \(\phi\) up or down depending on the amount by which the surface locally “inflates” or “deflates.” In fact, this whole story gives a nice geometric interpretation to good old-fashioned integrals: you can imagine that \(\int_\Omega \phi\ dA\) represents the area of some suitably deformed version of the initially planar region \(\Omega\).
Stokes’ Theorem
The main reason for studying integration on manifolds is to take advantage of the world’s most powerful tool:
Stokes’ theorem. Without further ado, Stokes’ theorem says that
\[ \int_\Omega d\alpha = \int_{\partial\Omega} \alpha, \]
where \(\alpha\) is any \(n-1\)-form on an \(n\)-dimensional domain \(\Omega\). In other words, integrating a differential form over the boundary of a manifold is the same as integrating its derivative over the entire domain.
If this trick sounds familiar to you, it’s probably because you’ve seen it time and again in different contexts and under different names: the
divergence theorem, Green’s theorem, the fundamental theorem of calculus, Cauchy’s integral formula, etc. Picking apart these special cases will really help us understand the more general meaning of Stokes’ theorem. Divergence Theorem
Let’s start with the divergence theorem from vector calculus, which says that
\[ \int_\Omega \nabla \cdot X dA = \int_{\partial\Omega} N \cdot X d\ell, \]
where \(X\) is a vector field on \(\Omega\) and \(N\) represents the (outward-pointing) unit normals on the boundary of \(\Omega\). A better name for this theorem might have been the “what goes in must come out theorem”, because if you think about \(X\) as the flow of water throughout the domain \(\Omega\) then it’s clear that the amount of water being pumped into \(\Omega\) (via pipes in the ground) must be the same as the amount flowing out of its boundary at any moment in time:
Let’s try writing this theorem using exterior calculus. First, remember that we can write the divergence of \(X\) as \(\nabla \cdot X = \star d \star X^\flat\). It’s a bit harder to see how to write the right-hand side of the divergence theorem, but think about what integration does here: it takes tangents to the boundary and sticks them into a 1-form. For instance, \(\int_\Omega X^\flat\) adds up the
tangential components of \(X\). To get the normal component we could rotate \(X^\flat\) by a quarter turn, which conveniently enough is achieved by hitting it with the Hodge star. Overall we get
\[ \int_\Omega d \star X^\flat = \int_{\partial\Omega} \star X^\flat, \]
which, as promised, is just a special case of Stokes’ theorem. Alternatively, we can use Stokes’ theorem to provide a more geometric interpretation of the divergence operator itself: when integrated over
any region \(\Omega\) — no matter how small — the divergence operator gives the total flux through the region boundary. In the discrete case we’ll see that this boundary flux interpretation is the only notion of divergence — in other words, there’s no concept of divergence at a single point.
By the way, why does \(d \star X^\flat\) appear on the left-hand side instead of \(\star d \star X^\flat\)? The reason is that \(\star d \star X^\flat\) is a 0-form, so we have to hit it with another Hodge star to turn it into an object that measures areas (i.e., a 2-form). Applying this transformation is no different from appending \(dA\) to \(\nabla \cdot X\) — we’re specifying how volume should be measured on our domain.
Fundamental Theorem of Calculus
The fundamental theorem of calculus is in fact
so fundamental that you may not even remember what it is. It basically says that for a real-valued function \(\phi: \mathbb{R} \rightarrow \mathbb{R}\) on the real line
\[ \int_a^b \frac{\partial \phi}{\partial x} dx = \phi(b) - \phi(a). \]
In other words, the total change over an interval \([a,b]\) is (as you might expect) how much you end up with minus how much you started with. But soft, behold! All we’ve done is written Stokes’ theorem once again:
\[ \int_{[a,b]} d\phi = \int_{\partial[a,b]} \phi, \]
since the boundary of the interval \([a,b]\) consists only of the two endpoints \(a\) and \(b\).
Hopefully these two examples give you a good feel for what Stokes’ theorem says. In the end, it reads almost like a Zen kōan: what happens on the outside is purely a function of the change within. (Perhaps it is Stokes’ that deserves the name, “fundamental theorem of calculus!”) |
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@Mathphile I found no prime of the form $$n^{n+1}+(n+1)^{n+2}$$ for $n>392$ yet and neither a reason why the expression cannot be prime for odd n, although there are far more even cases without a known factor than odd cases.
@TheSimpliFire That´s what I´m thinking about, I had some "vague feeling" that there must be some elementary proof, so I decided to find it, and then I found it, it is really "too elementary", but I like surprises, if they´re good.
It is in fact difficult, I did not understand all the details either. But the ECM-method is analogue to the p-1-method which works well, then there is a factor p such that p-1 is smooth (has only small prime factors)
Brocard's problem is a problem in mathematics that asks to find integer values of n and m for whichn!+1=m2,{\displaystyle n!+1=m^{2},}where n! is the factorial. It was posed by Henri Brocard in a pair of articles in 1876 and 1885, and independently in 1913 by Srinivasa Ramanujan.== Brown numbers ==Pairs of the numbers (n, m) that solve Brocard's problem are called Brown numbers. There are only three known pairs of Brown numbers:(4,5), (5,11...
$\textbf{Corollary.}$ No solutions to Brocard's problem (with $n>10$) occur when $n$ that satisfies either \begin{equation}n!=[2\cdot 5^{2^k}-1\pmod{10^k}]^2-1\end{equation} or \begin{equation}n!=[2\cdot 16^{5^k}-1\pmod{10^k}]^2-1\end{equation} for a positive integer $k$. These are the OEIS sequences A224473 and A224474.
Proof: First, note that since $(10^k\pm1)^2-1\equiv((-1)^k\pm1)^2-1\equiv1\pm2(-1)^k\not\equiv0\pmod{11}$, $m\ne 10^k\pm1$ for $n>10$. If $k$ denotes the number of trailing zeros of $n!$, Legendre's formula implies that \begin{equation}k=\min\left\{\sum_{i=1}^\infty\left\lfloor\frac n{2^i}\right\rfloor,\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\right\}=\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\end{equation} where $\lfloor\cdot\rfloor$ denotes the floor function.
The upper limit can be replaced by $\lfloor\log_5n\rfloor$ since for $i>\lfloor\log_5n\rfloor$, $\left\lfloor\frac n{5^i}\right\rfloor=0$. An upper bound can be found using geometric series and the fact that $\lfloor x\rfloor\le x$: \begin{equation}k=\sum_{i=1}^{\lfloor\log_5n\rfloor}\left\lfloor\frac n{5^i}\right\rfloor\le\sum_{i=1}^{\lfloor\log_5n\rfloor}\frac n{5^i}=\frac n4\left(1-\frac1{5^{\lfloor\log_5n\rfloor}}\right)<\frac n4.\end{equation}
Thus $n!$ has $k$ zeroes for some $n\in(4k,\infty)$. Since $m=2\cdot5^{2^k}-1\pmod{10^k}$ and $2\cdot16^{5^k}-1\pmod{10^k}$ has at most $k$ digits, $m^2-1$ has only at most $2k$ digits by the conditions in the Corollary. The Corollary if $n!$ has more than $2k$ digits for $n>10$. From equation $(4)$, $n!$ has at least the same number of digits as $(4k)!$. Stirling's formula implies that \begin{equation}(4k)!>\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\end{equation}
Since the number of digits of an integer $t$ is $1+\lfloor\log t\rfloor$ where $\log$ denotes the logarithm in base $10$, the number of digits of $n!$ is at least \begin{equation}1+\left\lfloor\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)\right\rfloor\ge\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right).\end{equation}
Therefore it suffices to show that for $k\ge2$ (since $n>10$ and $k<n/4$), \begin{equation}\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)>2k\iff8\pi k\left(\frac{4k}e\right)^{8k}>10^{4k}\end{equation} which holds if and only if \begin{equation}\left(\frac{10}{\left(\frac{4k}e\right)}\right)^{4k}<8\pi k\iff k^2(8\pi k)^{\frac1{4k}}>\frac58e^2.\end{equation}
Now consider the function $f(x)=x^2(8\pi x)^{\frac1{4x}}$ over the domain $\Bbb R^+$, which is clearly positive there. Then after considerable algebra it is found that \begin{align*}f'(x)&=2x(8\pi x)^{\frac1{4x}}+\frac14(8\pi x)^{\frac1{4x}}(1-\ln(8\pi x))\\\implies f'(x)&=\frac{2f(x)}{x^2}\left(x-\frac18\ln(8\pi x)\right)>0\end{align*} for $x>0$ as $\min\{x-\frac18\ln(8\pi x)\}>0$ in the domain.
Thus $f$ is monotonically increasing in $(0,\infty)$, and since $2^2(8\pi\cdot2)^{\frac18}>\frac58e^2$, the inequality in equation $(8)$ holds. This means that the number of digits of $n!$ exceeds $2k$, proving the Corollary. $\square$
We get $n^n+3\equiv 0\pmod 4$ for odd $n$, so we can see from here that it is even (or, we could have used @TheSimpliFire's one-or-two-step method to derive this without any contradiction - which is better)
@TheSimpliFire Hey! with $4\pmod {10}$ and $0\pmod 4$ then this is the same as $10m_1+4$ and $4m_2$. If we set them equal to each other, we have that $5m_1=2(m_2-m_1)$ which means $m_1$ is even. We get $4\pmod {20}$ now :P
Yet again a conjecture!Motivated by Catalan's conjecture and a recent question of mine, I conjecture thatFor distinct, positive integers $a,b$, the only solution to this equation $$a^b-b^a=a+b\tag1$$ is $(a,b)=(2,5).$It is of anticipation that there will be much fewer solutions for incr... |
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$
Is $~\pi^2\approx g~$ a coincidence ?
Some have answered
, others said yes , and no yet others considered $(!)$ as perfectly viable options. Personally, I cannot help but chuckle, as this question reminds me of both Newton’s famous disc, which can be said to be both white and colored at the same time, depending on whether it is either rotating, or at rest. To add even more to the already mystifying fog of confusion, I shall hereby venture yet a opinion fourth :
$\qquad\qquad\qquad\qquad\qquad\quad$
We don’t know, and we never shall !
Granted, such a statement, when taken at face value, would undoubtedly appear as an impious affront to
Hilbert’s celebrated adage, wir mussen wissen, wir werden wissen, but before anyone accuses me of embracing either philosophical pessimism or epistemological agnosticism, let me assure you, dear reader, that such is simply the case; rather, I am basing this short assertion purely on mathematical foundations. Basically, there are four main ways in which a measuring unit can be created, that is both practical or anthropocentric, as well as universally meaningful, at the same time $($not to mention reproducible$)$ not :
the length of the pendulum with a half-period of exactly
, since the length of a pendulum with a half-period of one second will be exceedingly long; one minute
the
, the ten-millionth , or even the hundred-millionth part of either a terrestrial meridian, or the Earth’s equator, since the other two adjacent options, i.e., the billionth and the millionth part, would be either way too big, or way too small; ten-billionth
the distance traveled by light in the
, the hundred-millionth , or even the billionth part of a second; again, the other two adjacent options i.e., the ten- billionth and the ten-millionth part, would be either way too long, or way too short; hundred-billionth
the length of a so-called
$($ i.e., the sixtieth part of a third $)$ of the Earth’s meridian or equator. second
Of course, someone might, at this point, easily be tempted to say that I have committed a hideous and unpardonable abuse by painstakingly enumerating all those powers of ten listed above, since the metric system, as we have it today, is
decimal, but such would not necessarily have been the case, given an alternate course of human history $($thus, for instance, if one were to take the distance traveled by light in $10^{-9}$ seconds, such a length could easily have been interpreted as representing a “new foot”, to be further subdivided into $12$ “new inches”, ultimately yielding a “new yard” of $0.9$ metres$)$. coincidentally
Now, the shocking surprise, which astounded many at the time of its first discovery, and still does so even today, is as follows
: the ratio of the first three units is $1:4:3$, , the sheer “niceness” of the numbers involved being utterly uncanny, to say the almost exactly very least. $($Spooky, thought-provoking, challenging, bewildering, and mesmerizing also come to mind$)$. Adding insult to injury, as the proverb goes, we also notice that twice the value of the latter unit, representing the $3~600^\text{th}$ part of a nautical mile, equals $103$ centimetres, with an error of ; speak- ing of which, the thousandth part of a nautical mile is also conspicuously close to the length of a fathom, measuring the distance between the fingertips of a man’s outstretched arms. less than $\pm1$ millimetre
Furthermore, even if one were quite purposefully to go out of one’s way, and intentionally try to avoid the two coincidences above, by $($repeatedly$)$ dividing, based purely on number-theoretical principles, the aforementioned non-metric unit into, say, sevenths, $($since the powers of all other previous primes already appear abundantly in its sexagesimal creation$)$, one would arrive at the eerie conclusion that it adds up to $5.4$ metres, with an error of
. less than half a millimetre
As an aside, as $($even further$)$ coincidence would have it, my own personal fathom is $1.8$ metres almost exactly, with an error of no more than a few millimetres, making the above length my own personal rod; indeed, I am a rather metric person, since even my own height towers at just slightly over $1.7$ metres, and does
exceed $171\rm~cm$ — but I digress $\ldots$ not
Some of the above relations are $($easily$)$ explained $($away$)$ by means of basic arithmetic, such as, for instance, the fact that $3\cdot7^3\simeq2^{10}\simeq10^3$, or $2^7\simeq5^3\simeq11^2$, and $2^8\simeq3^5$, the latter two “culprits” being responsible for the beautiful approximation $3000_{12}\simeq5000_{10}$, or, equivalently, $12^4\simeq2\cdot10^4$, which relate duodecimal thousands and myriads to their decimal counterparts; others, however, are $($much$)$ harder to dispel. Nevertheless, this is
what we shall endeavour to achieve ! precisely
Let us therefore fearlessly approach the most awe-inspiring of all the above-listed coincidences, and merrily $($and mercilessly$)$
the life out of it $-$ in the name of science ! :-$)$ debunk
Now, the way I see it, if the ratio in question were
$3:4$, then dividing the distance traveled by light in a day’s time $($since this is the smallest naturally occurring time unit which is also easily observable by man$)$ to the length of an Earth’s meridian should yield a result of truly $648~000.~$ However, by employing the most accurate measurements known to date, namely that of $$c=299~792~458~\rm\dfrac ms~,$$ and a quarter of a terrestrial meridian being $\ell\simeq10~001~965~.~7293\rm~m$, we ultimately arrive at the exactly and dull figure of $~\dfrac{24\cdot60^2\cdot c}\ell~\simeq~647~424~\dfrac49,~$ which is roughly $~575~\dfrac59~$ times less than expected. uninspiring
In other words, by
the enhancing of our lengths and ratios, the ghosts of modern superstitions are forever shattered in the cold light of day by the power of reason, and our minds can resolution rest assured that the whole thing was nothing more than a finally , tempest in a teapot or much ado about nothing, as Shakespeare so wonderfully put it all those centuries ago ! Now all that is left to do is no one notices that the previous ratio can also be expressed as $27~27\rm~BB$ in base $12$, with an error of less than one and a half units. :-$)$ praying
On a more serious note, it all boils down to divisors $($usually powers of $2,~3,$ and $5)$ and numeration systems. $-~$
it ?$\ldots$ In the words of Thomas More, Doesn’t I trust I make myself obscure. :-$)$ |
Nov 10th 2008, 04:58 AM
# 2
Member
Join Date: Jul 2008
Posts: 60
Ok, I know what the problem is, I made a mistake in the moment of inertia...
I do have another question due.
First of all, when they say this is the angular acceleration of the poll, does it mean the angular acceleration of the center of mass?
In the next part of the question, they ask for the angular velocity as a function of $\displaystyle \varphi$.
I tried to use energy conservation and came up with this equation:
$\displaystyle mgL=mgL(1-1/2\cos(\varphi))+1/2 mv^2+1/2 I\omega^2$
When
$\displaystyle v = \omega*(L/2)$
Is this right?
10x. |
Textbook solution
The first thing to do is to define the center of mass and relative coordinates:
$$ R(t) = {m_1 r_1 + m_2 r_2 \over m_1 + m_2} $$
$$ r(t) = r_2 - r_1 $$
You invert this to find
$$ r_1= R - {m_2\over M} r$$$$r_2=R+ {m_1\over M} r$$
The equation of motion for R is trivial, since center of mass is a conservation law:
$$ {d^2 R \over dt^2 } = 0 $$
and it is solved by $$ R(t) = V_0 t + R_0 $$ where $V_0$ and $R_0$ are the initial center of mass velocity and position respectively (which are calculated from the given initial conditions).
The nontrivial equation is for the relative coordinate:$$ m{d^2 r \over dt^2} = - {m_1 m_2 r\over |r|^3}$$where m is the reduced mass:${1\over m} ={1\over m_1} + {1\over m_2}$.
Or:$$ {d^2 r \over dt^2} = - {M r\over |r|^3}$$
Where $M=m_1 + m_2$ is the total mass.
The problem is reduced to solving the Kepler motion in a 1/r potential. From now on, I will rescale time to make the mass parameter in the r equation 1.
You can choose the x axis to lie along the initial r, and the y-axis to lie along the component of the initial $\dot{r}$ perpendicular to the initial r. Another way of saying this is that you rotate the coordinates to make the angular momentum vector $r\times p$ where $p=m\dot{r}$ to lie along the z-axis. This rotation reduces the problem to a plane, and the rotation matrix columns are given by the normalized initial r (now along the x-axis), the component of the initial velocity perpendicular to r, normalized (along the y-axis), and normalized L along the z-axis.
You then use units to set the total over reduced mass to 1, and use polar coordinates in the x-y plane of the motion, and note that the angular momentum is constant:
$$ r^2 {d\theta\over dt} = L $$
This tells you that equal areas are swept out by the r-vector in equal times. The equation of motion for r(t) (no longer a vector, now a scalar radial coordinate) is:
$$ {d^2 r \over dt^2} = {L^2 \over r^3} - {1 \over r^2} $$
Then you change time out for $\theta$, expressing everything in terms of $r(\theta)$, which you can do using the equal area law, whenever the anguar momentum is nonzero (if the initial angular momentum is zero, or very close to zero, this is a one-dimensional two-body problem which can be solved directly by more elementary means). The equation of motion for $r(\theta)$ simplifies when you make a coordinate transformation to $u={1\over r}$:
$$ {d^2 u \over d\theta^2} = C - u $$
Where C is some unimportant constant, and this is solved by
$$u(\theta) = {1\over A} (1 + a \cos(\theta -\theta_0)) $$
Where A is the semi-major axis of the ellipse (if the orbit is an ellipse), $\theta_0$ determines the orientation in the x-y plane, and a is the eccentricity of the ellipse (if a<1), or determines the angle of the hyperbola (if a>1) or tells you the orbit is a parabola (a=1).
the only result you need is that
$$ r(\theta) = {A\over 1+ a\cos(\theta-\theta_0)} $$
This gives you the solution of r as a function of $\theta$, which gives the shape of the orbit. This is where textbooks stop.
Finding $\theta$ as a function of $t$
But you then want the solution for $\theta$ as a function of time, to get the r and $\theta$ as functions of time. This is determined from the area law, conceptually:
$$ r^2 {d\theta\over dt} = L$$
$$ {d\theta \over (1+ a \cos(\theta-\theta_0))^2 }= {L \over A^2} dt $$
and integrating this from time 0 to time t, tells you in principle what $\theta(t)$ is. The result can be written as:
$$ F(a,\theta) = F(a,\theta_0) + {L\over A^2} t $$
Where $F(a,\theta)$ is the special function that gives you the area of a conic section of parameter a, in a wedge from the focus where one half-line is along the major axis, and the other half-line makes an angle $\theta$ with the first. This special function is not expressible in terms of elementary functions.
This function is defined by the integral above, and you can calculate it numerically using any numerical integration method. Finding this function, and inverting it, is the only difficult part of this problem. There are three limits which are necessary for perturbations:
for a=0, $ F(0,\theta) = \theta$ for a=1, $ F(1,\theta) = {y\over 4} + {y^3\over 12} $ where $y=r\sin{\theta} = {\sin(\theta)\over 1+\cos(\theta)}$ for $a=\infty$, $ F(a,\theta) \approx {1\over a} \tan(\theta) $
Each of these are elementary degenerations: the first is the circle, the second is the parabola, and the third is a straight line. The important thing is that each of these degenerations gives you x(t) and y(t) which are simple, and further, you can perturb around each of these three limits in a nice way. In the following, the parameter t is rescaled to absorb ${L\over A^2}$
circle: $x(t) = \cos(t)$ $y(t) = \sin(t)$ parabola: $y(t)= (\sqrt{1+36t^2} + 6t)^{1\over 3} - (\sqrt{1+36t^2} - 6t)^{1\over 3} $ $x(t) = {1\over 2} - {y^2\over 2} $ line: $x(t) = {1\over a}$, $y(t) = t$
The line and circle are obvious, the parabola is found by inverting the cubic for y as a function of t using the cubic equation.
Near the circle, time is periodic with the orbital period, which is the area inside the ellipse $aA^2$ divided by the area sweep-rate $L/2$. So you have once-winding function from a circle to a circle, which can always be written as a Fourier series with a linear term, which is found from the power series of the integrand in a, integrated term by term. Near the straight-line hyperbola, you can similarly perturb in a series, and the only interesting degeneration is the parabola. Near the parabola, the perturbation theory is a little more complicated. |
I was doing a couple of problems for homework:
Calculate $K_\mathrm{sp}$ of $\ce{AgI}$ at $55.0\ \mathrm{^\circ C}$
Calculate $K_\mathrm{b}$ of $\ce{NH3}$ at $36.0\ \mathrm{^\circ C}$
I have to use $\Delta G^\circ= -RT\ln K$ and $\Delta G= \Delta H-T\,\Delta S$
When I did this $\Delta G^\circ$ is positive ($89.59\ \mathrm{kJ/mol}$ and $28.037831\ \mathrm{kJ/mol}$ respectively), yet $K_\mathrm{sp}$ for $\ce{AgI}$ is $5.5\times10^{-15}$ and $K_\mathrm{b}$ for $\ce{NH3}$ is $1.8\times10^{-5}$, indicating that there are some products and the reactions do occur. Plus, $1.0\ \mathrm M$ $\ce{NH3}$ in solution has a $\mathrm{pH}$ of $11.6$ so it must react a little.
According to the second law of thermodynamics, if $\Delta G$ is positive, the reaction is not spontaneous, right? But clearly, they, in fact, are to a certain extent. What is going on? |
On phaseless compressed sensing with partially known support
School of Mathematics, Tianjin University, Tianjin 300072, China
We establish a theoretical framework for the problem of phaseless compressed sensing with partially known signal support, which aims at generalizing the Null Space Property and the Strong Restricted Isometry Property from phase retrieval to partially sparse phase retrieval. We first introduce the concepts of the Partial Null Space Property (P-NSP) and the Partial Strong Restricted Isometry Property (P-SRIP); and then show that both the P-NSP and the P-SRIP are exact recovery conditions for the problem of partially sparse phase retrieval. We also prove that a random Gaussian matrix $ A\in \mathbb{R}^{m\times n} $ satisfies the P-SRIP with high probability when $ m = O(t(k-r)\log(\frac{n-r}{t(k-r)})). $
Keywords:Phase retrieval, compressed sensing, phaseless compressed sensing, partial null space property, partial strong restricted isometry property. Mathematics Subject Classification:Primary: 90C90; Secondary: 94A12. Citation:Ying Zhang, Ling Ma, Zheng-Hai Huang. On phaseless compressed sensing with partially known support. Journal of Industrial & Management Optimization, doi: 10.3934/jimo.2019014
References:
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R. Balan, B. Bodmann, P. G. Casazza and D. Edidin,
Saving phase: injectivity and stability for phase retrieval,
[3]
A.S. Bandeira, J. Cahill, D. Mixon and A. Nelson,
Painless reconstruction from magnitudes of frame coefficients,
[4] [5] [6]
O. Bunk, A. Diza, F. Pfeiffer, C. David, B. Schmitt, D. K. Satapathy and J. F. van der Veen,
Diffractive imaging for periodic samples: Retrieving one-dimensional concentration profiles across microfluidic channels,
[7]
T. Cai and A. Zhang,
Sparse representation of a polytope and recovery of sparse signals and low-rank matrices,
[8] [9] [10]
E. J. Candès, T. Strohmer and V. Voroninski,
Exact and stable signal recovery from magnitude measurements via convex programming,
[11]
A. Conca, D. Edidin, M. Hering and C. Vinzant,
An algebraic characterization of injectivity in phase retrieval,
[12] [13] [14]
M. P. Friedlander, H. Mansour, R. Saab and O. Yilmaz,
Recovering compressively sampled signals using partial support information,
[15] [16] [17] [18] [19]
J. Miao, T. Ishikawa, Q. Shen and T. Earnest,
Extending X-ray crystallography to allow the imagine of non-crystalline materials, cells and single protein complexes,
[20] [21]
D. T. Peng, N. H. Xiu and J. Yu,
$S_{1/2}$ regularization methods and fixed point algorithms for affine rank minimization problems,
[22]
H. Qiu, X. Chen, W. Liu, G. Zhou, Y. J. Wang and J. Lai,
A fast $l_1$-solver and its applications to robust face recognition,
[23]
N. Vaswani and W. Lu,
Modified-CS: Modifying compressive sensing for problems with partially known support,
[24]
V. Voroninski and Z. Q. Xu,
A strong restricted isometry property, with an application to phaseless compressed sensing,
[25] [26] [27]
Y. Wang, W. Liu, L. Caccetta and G. Zhou,
Parameter selection for nonnegative $l_1$ matrix/tensor sparse decomposition,
[28]
Y. Wang, G. Zhou, L. Caccetta and W. Liu,
An alternative Lagrange-dual based algorithm for sparse signal reconstruction,
[29] [30]
G. W. You, Z. H. Huang and Y. Wang,
A theoretical perspective of solving phaseless compressive sensing via its nonconvex relaxation,
[31]
L. J. Zhang, L. C. Kong, Y. Li and S. L. Zhou,
A smoothing iterative method for quantile regression with nonconvex $l_p$ penalty,
show all references
References:
[1] [2]
R. Balan, B. Bodmann, P. G. Casazza and D. Edidin,
Saving phase: injectivity and stability for phase retrieval,
[3]
A.S. Bandeira, J. Cahill, D. Mixon and A. Nelson,
Painless reconstruction from magnitudes of frame coefficients,
[4] [5] [6]
O. Bunk, A. Diza, F. Pfeiffer, C. David, B. Schmitt, D. K. Satapathy and J. F. van der Veen,
Diffractive imaging for periodic samples: Retrieving one-dimensional concentration profiles across microfluidic channels,
[7]
T. Cai and A. Zhang,
Sparse representation of a polytope and recovery of sparse signals and low-rank matrices,
[8] [9] [10]
E. J. Candès, T. Strohmer and V. Voroninski,
Exact and stable signal recovery from magnitude measurements via convex programming,
[11]
A. Conca, D. Edidin, M. Hering and C. Vinzant,
An algebraic characterization of injectivity in phase retrieval,
[12] [13] [14]
M. P. Friedlander, H. Mansour, R. Saab and O. Yilmaz,
Recovering compressively sampled signals using partial support information,
[15] [16] [17] [18] [19]
J. Miao, T. Ishikawa, Q. Shen and T. Earnest,
Extending X-ray crystallography to allow the imagine of non-crystalline materials, cells and single protein complexes,
[20] [21]
D. T. Peng, N. H. Xiu and J. Yu,
$S_{1/2}$ regularization methods and fixed point algorithms for affine rank minimization problems,
[22]
H. Qiu, X. Chen, W. Liu, G. Zhou, Y. J. Wang and J. Lai,
A fast $l_1$-solver and its applications to robust face recognition,
[23]
N. Vaswani and W. Lu,
Modified-CS: Modifying compressive sensing for problems with partially known support,
[24]
V. Voroninski and Z. Q. Xu,
A strong restricted isometry property, with an application to phaseless compressed sensing,
[25] [26] [27]
Y. Wang, W. Liu, L. Caccetta and G. Zhou,
Parameter selection for nonnegative $l_1$ matrix/tensor sparse decomposition,
[28]
Y. Wang, G. Zhou, L. Caccetta and W. Liu,
An alternative Lagrange-dual based algorithm for sparse signal reconstruction,
[29] [30]
G. W. You, Z. H. Huang and Y. Wang,
A theoretical perspective of solving phaseless compressive sensing via its nonconvex relaxation,
[31]
L. J. Zhang, L. C. Kong, Y. Li and S. L. Zhou,
A smoothing iterative method for quantile regression with nonconvex $l_p$ penalty,
[1]
Steven L. Brunton, Joshua L. Proctor, Jonathan H. Tu, J. Nathan Kutz.
Compressed sensing and dynamic mode decomposition.
[2]
Yingying Li, Stanley Osher.
Coordinate descent optimization for
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Mikhail Krastanov, Michael Malisoff, Peter Wolenski.
On the strong invariance property for non-Lipschitz dynamics.
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Zdzisław Brzeźniak, Paul André Razafimandimby.
Irreducibility and strong Feller property for stochastic evolution equations in Banach spaces.
[9] [10]
Michael Röckner, Jiyong Shin, Gerald Trutnau.
Non-symmetric distorted Brownian motion: Strong solutions, strong Feller property and
non-explosion results.
[11]
Masahito Ohta.
Strong instability of standing waves for nonlinear Schrödinger equations with a partial confinement.
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Vianney Perchet, Marc Quincampoix.
A differential game on Wasserstein space. Application to weak approachability with partial monitoring.
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Kazumine Moriyasu, Kazuhiro Sakai, Kenichiro Yamamoto.
Regular maps with the specification property.
[18] [19] [20]
2018 Impact Factor: 1.025
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Difference between revisions of "Inertia"
(→Normalised Inertia Constants)
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:: <math>S_{b}</math> is the rated power of the machine (VA)
:: <math>S_{b}</math> is the rated power of the machine (VA)
−
Based on actual
+ + + + + + + + + + + + + + + +
Based on actual data, the normalised inertia constants for different types and sizes of generators are summarised in the table below:
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| Combustion engine || 26 || 0.3 || 1.25 || 2.5 || 0.6 || 0.95 || 1.6
| Combustion engine || 26 || 0.3 || 1.25 || 2.5 || 0.6 || 0.95 || 1.6
|-
|-
− − − − − − − − − − − − − − −
|}
|}
[[Category:Fundamentals]]
[[Category:Fundamentals]]
Revision as of 05:51, 27 August 2018
In power systems engineering, "inertia" is a concept that typically refers to rotational inertia or rotational kinetic energy. For synchronous systems that run at some nominal frequency (i.e. 50Hz or 60Hz), inertia is the energy that is stored in the rotating masses of equipment electro-mechanically coupled to the system, e.g. generator rotors, fly wheels, turbine shafts.
Derivation
Below is a basic derivation of power system rotational inertia from first principles, starting from the basics of circle geometry and ending at the definition of moment of inertia (and it's relationship to kinetic energy).
The length of a circle arc is given by:
[math] L = \theta r [/math]
where [math]L[/math] is the length of the arc (m)
[math]\theta[/math] is the angle of the arc (radians) [math]r[/math] is the radius of the circle (m)
A cylindrical body rotating about the axis of its centre of mass therefore has a rotational velocity of:
[math] v = \frac{\theta r}{t} [/math]
where [math]v[/math] is the rotational velocity (m/s)
[math]t[/math] is the time it takes for the mass to rotate L metres (s)
Alternatively, rotational velocity can be expressed as:
[math] v = \omega r [/math]
where [math]\omega = \frac{\theta}{t} = \frac{2 \pi \times n}{60}[/math] is the angular velocity (rad/s)
[math]n[/math] is the speed in revolutions per minute (rpm)
The kinetic energy of a circular rotating mass can be derived from the classical Newtonian expression for the kinetic energy of rigid bodies:
[math] KE = \frac{1}{2} mv^{2} = \frac{1}{2} m(\omega r)^{2}[/math]
where [math]KE[/math] is the rotational kinetic energy (Joules or kg.m
2/s 2 or MW.s, all of which are equivalent) [math]m[/math] is the mass of the rotating body (kg)
Alternatively, rotational kinetic energy can be expressed as:
[math] KE = \frac{1}{2} J\omega^{2} [/math]
where [math]J = mr^{2}[/math] is called the
moment of inertia (kg.m 2).
Note that in physics, the moment of inertia [math]J[/math] is normally denoted as [math]I[/math]. In electrical engineering, the convention is for the letter "i" to always be reserved for current, and is therefore often replaced by the letter "j", e.g. the complex number operator i in mathematics is j in electrical engineering.
Normalised Inertia Constants
The moment of inertia can be expressed as a normalised quantity called the
inertia constant H, calculated as the ratio of the rotational kinetic energy of the machine at nominal speed to its rated power (VA): [math]H = \frac{1}{2} \frac{J \omega_0^{2}}{S_{b}}[/math]
where [math]\omega = 2 \pi \frac{n}{60}[/math] is the nominal mechanical angular frequency (rad/s)
[math]n[/math] is the nominal speed of the machine (revolutions per minute) [math]S_{b}[/math] is the rated power of the machine (VA) Generator Inertia
The moment of inertia for a generator is dependent on its mass and apparent radius, which in turn is largely driven by its prime mover type.
Based on actual generator data, the normalised inertia constants for different types and sizes of generators are summarised in the table below:
Machine type Number of samples MVA Rating Inertia constant H Min Median Max Min Median Max Steam turbine 45 28.6 389 904 2.1 3.2 5.7 Gas turbine 47 22.5 99.5 588 1.9 5.0 8.9 Hydro turbine 22 13.3 46.8 312.5 2.4 3.7 6.8 Combustion engine 26 0.3 1.25 2.5 0.6 0.95 1.6 |
ISSN:
1534-0392
eISSN:
1553-5258
All Issues
Communications on Pure & Applied Analysis
March 2009 , Volume 8 , Issue 2
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Abstract:
We study the long-time behavior of non-negative solutions to the Cauchy problem
(
P)
$\qquad \rho(x) \partial_t u= \Delta u^m\qquad$
in $Q$:$=\mathbb R^n\times\mathbb R_+$
$u(x, 0)=u_0$
in dimensions $n\ge 3$. We assume that $m> 1$ (slow
diffusion) and $\rho(x)$ is positive, bounded and behaves
like $\rho(x)$~$|x|^{-\gamma}$ as $|x|\to\infty$, with
$0\le \gamma<2$. The data $u_0$ are assumed to be nonnegative
and such that $\int \rho(x)u_0 dx< \infty$.
Our asymptotic analysis leads to the associated singular equation $|x|^{-\gamma}u_t= \Delta u^m,$ which admits a one-parameter family of selfsimilar solutions $ U_E(x,t)=t^{-\alpha}F_E(xt^{-\beta})$, $E>0$, which are source-type in the sense that $|x|^{-\gamma}u(x,0)=E\delta(x)$. We show that these solutions provide the first term in the asymptotic expansion of generic solutions to problem ( P)
for large
times, both in the weighted $L^1$ sense
$u(t)=U_E(t)+o(1)\qquad$ in $L^1_\rho$
and in the uniform sense $u(t)=U_E(t)+o(t^{-\alpha})$ in $L^\infty $ as $t\to \infty$ for the explicit rate $\alpha=\alpha(m,n,\gamma)>0$ which is precisely the time-decay rate of $U_E$. For a given solution, the proper choice of the parameter is $E=\int \rho(x)u_0 dx$.
Abstract:
A nonlinear, density-dependent system of diffusion-reaction equations describing development of bacterial biofilms is analyzed. It comprises two non-standard diffusion effects, degeneracy as in the porous medium equation and fast diffusion. The existence of a unique bounded solution and a global attractor is proved in dependence of the boundary conditions. This is achieved by studying a system of non-degenerate auxiliary approximation equations and the construction of a Lipschitz continuous semigroup by passing to the limit in the approximation parameter. Numerical examples are included in order to illustrate the main result.
Abstract:
We study elliptic problems at critical growth under Steklov boundary conditions in bounded domains. For a second order problem we prove existence of nontrivial nodal solutions. These are obtained by combining a suitable linking argument with fine estimates on the concentration of Sobolev minimizers on the boundary. When the domain is the unit ball, we obtain a multiplicity result by taking advantage of the explicit form of the Steklov eigenfunctions. We also partially extend the results in the ball to the case of fourth order Steklov boundary value problems.
Abstract:
In this survey, our aim is to emphasize the main known limitations to the use of Wigner measures for Schrödinger equations. After a short review of successful applications of Wigner measures to study the semi-classical limit of solutions to Schrödinger equations, we list some examples where Wigner measures cannot be a good tool to describe high frequency limits. Typically, the Wigner measures may not capture effects which are not negligible at the pointwise level, or the propagation of Wigner measures may be an ill-posed problem. In the latter situation, two families of functions may have the same Wigner measures at some initial time, but different Wigner measures for a larger time. In the case of systems, this difficulty can partially be avoided by considering more refined Wigner measures such as two-scale Wigner measures; however, we give examples of situations where this quadratic approach fails.
Abstract:
This paper deals with the bounded and blowup solutions of the quasilinear parabolic system $u_t = u^p ( \Delta u + a v) + f(u, v, Du, x)$ and $v_t = v^q ( \Delta v + b u) + g(u, v, Dv, x)$ with homogeneous Dirichlet boundary condition. Under suitable conditions on the lower order terms $f$ and $g$, it is shown that all solutions are bounded if $(1+c_1) \sqrt{ab} < \l_1$ and blow up in a finite time if $(1+c_1) \sqrt{ab} > \lambda_1$, where $\lambda_1$ is the first eigenvalue of $-\Delta $ in $\Omega$ with Dirichlet data and $c_1 > -1$ related to $f$ and $g$.
Abstract:
We study an abstract inverse problem of reconstruction of the solution of a semilinear mixed integrodifferential parabolic problem, together with a convolution kernel. The supplementary information required to solve the problem also involves a convolution term with the same unknown kernel. The abstract results are applicable to the identification of a memory kernel in a
strongly dampedwave equation using a flux condition. Abstract:
In this paper the mountain--pass theorem and the Ekeland variational principle in a suitable Orlicz space are employed to establish the existence of positive standing wave solutions for a quasilinear Schrödinger equation involving a combination of concave and convex terms. The second order nonlinearity considered in this paper corresponds to the superfluid equation in plasma physics.
Abstract:
This paper deals with the rigorous study of the diffusive stress relaxation in the multidimensional system arising in the mathematical modeling of viscoelastic materials. The control of an appropriate high order energy shall lead to the proof of that limit in Sobolev space. It is shown also as the same result can be obtained in terms of relative modulate energies.
Abstract:
We study the stability of the dynamics of a network of $n$ formal neurons interacting through an asymmetric matrix with independent random Gaussian elements of the type introduced by Rajan and Abbott ([1]). The neurons are represented by the values of their electric potentials $x_i, i=1,\cdots,n$. Using the approach developed in a previous paper by us ([6]) we obtain sufficient conditions for diverging synchronized behavior and for stability.
Abstract:
Let $B$ denote the unit ball in $\mathbb R^N$, $N\geq 3$. We consider the classical Brezis-Nirenberg problem
$ \Delta u+\lambda u+u^{\frac{N+2}{N-2}} =0 \quad$ in$\quad B$
$ u>0 \quad$ in $\quad B $
$ u=0 \quad$ on $\quad \partial B$
where $\lambda$ is a constant. It is proven in [3] that this problem has a classical solution if and only if $\underline \lambda < \lambda < \lambda _1$ where $\underline \lambda = 0$ if $N\ge 4$, $\underline \lambda = \frac{\lambda _1}4$ if $N=3$. This solution is found to be unique in [17]. We prove that there is a number $\lambda_*$ and a continuous function $a(\lambda)\ge 0$ decreasing in $(\underline \lambda, \lambda_*]$, increasing in $[\lambda_*, \lambda_1)$ such that for each $\lambda$ in this range and each $\mu\in (a(\lambda),\infty)$ there exist a $\mu$-periodic function $w_\mu(t)$ and two distinct radial solutions $u_{\mu j}$, $j=1,2$, singular at the origin, with $u_{\mu j}(x)$~$|x|^{-\frac{N-2}2}w_\mu$( log $|x|$) as $x\to 0$. They approach respectively zero and the classical solution as $\mu\to +\infty$. At $\lambda =\lambda_*$ there is in addition to those above a solution ~$ c_N|x|^{-\frac{N-2}2}$. This clarifies a previous result by Benguria, Dolbeault and Esteban in [2], where a existence of a continuum of singular solutions for each $\lambda\in (\underline\lambda, \lambda_1)$ was found.
Abstract:
We provide a general framework of inequalities induced by the Aubry-Mather theory of Hamilton-Jacobi equations. This framework deals with a sufficient condition on functions $f\in C^1(\mathbb R^n)$ and $g\in C(\mathbb R^n)$ in order that $f-g$ takes its minimum over $\mathbb R^n$ on the set {$x\in \mathbb R^n |Df(x)=0$}. As an application of this framework, we provide proofs of the arithmetic mean-geometric mean inequality, Hölder's inequality and Hilbert's inequality in a unified way.
Abstract:
We consider a phase-field system of Caginalp type on a three-dimensional bounded domain. The order parameter $\psi $ fulfills a dynamic boundary condition, while the (relative) temperature $\theta $ is subject to a boundary condition of Dirichlet, Neumann, Robin or Wentzell type. The corresponding class of initial and boundary value problems has already been studied by the authors, proving well-posedness results and the existence of global as well as exponential attractors. Here we intend to show first that the previous analysis can be redone for larger phase-spaces, provided that the bulk potential has a fourth-order growth at most whereas the boundary potential has an arbitrary polynomial growth. Moreover, assuming the potentials to be real analytic, we demonstrate that each trajectory converges to a single equilibrium by means of a Łojasiewicz-Simon type inequality. We also obtain a convergence rate estimate.
Abstract:
In this paper, we study the blowup rate estimate for a system of semilinear parabolic equations. The blowup rate depends on whether the two components of the solution of this system blow up simultaneously or not.
Abstract:
This work is devoted to the existence of positive entire solutions for semilinear elliptic systems. With the aid of a degree theory argument, we use the shooting method and Pohozaev-type identity to show the existence of positive radial solutions.
Abstract:
We present two new families of polynomial differential systems of arbitrary degree with centers, a two--parameter family and a four--parameter family.
Abstract:
We study the Schrödinger operator $ H = - \Delta + V $ on the product of two copies of an infinite blowup of the Sierpinski gasket, where $ V$ is the analog of a Coulomb potential ($\Delta V$ is a multiple of a delta function). So $H$ is the analog of the standard Hydrogen atom model in nonrelativistic quantum mechanics. Like the classical model, we show that the essential spectrum of $H$ is the same as for $ - \Delta $, and there is a countable discrete spectrum of negative eigenvalues.
Abstract:
For inhomogeneous diagonal system with distinct characteristics or with characteristics with constant multiplicity, under the assumption that the system is linearly degenerate and the $C^1$ norm of the initial data is bounded, we show that the mechanism of the formation of singularities of classical solution to its Cauchy problem must be of ODE type. Similar results are also obtained for corresponding mixed initial-boundary value problems on a semi-unbounded domain.
Abstract:
In this article we prove some sharp regularity results for the stationary and the evolution Navier-Stokes equations with shear dependent viscosity, under the no-slip boundary condition. This is a classical turbulence model, considered by von Neumann and Richtmeyer in the 50's, and by Smagorinski in the beginning of the 60's (for $p=3$). The model was extended to other physical situations, and deeply studied from a mathematical point of view, by Ladyzhenskaya in the second half of the 60's. In the sequel we consider the case $ p> 2 $. We are interested in regularity results in Sobolev spaces,
up to the boundary, in dimension $n= 3$, for the second order derivatives of the velocity and the first order derivatives of the pressure. In spite of the very rich literature on this subject, sharp regularity results up to the boundary are quite new. In the sequel we improve in a very substantial way all the known results in the literature. In order to emphasize the very new ideas, we consider a flat boundary (the so called "cubic-domain" case). However, all the regularity results stated here hold in the presence of smooth boundaries, by following [3]. Readers Authors Editors Referees Librarians More Email Alert
Add your name and e-mail address to receive news of forthcoming issues of this journal:
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(Sorry was asleep at that time but forgot to log out, hence the apparent lack of response)
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details
http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference |
(Sorry was asleep at that time but forgot to log out, hence the apparent lack of response)
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details
http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference |
I came across John Duffield Quantum Computing SE via this hot question. I was curious to see an account with 1 reputation and a question with hundreds of upvotes.It turned out that the reason why he has so little reputation despite a massively popular question is that he was suspended.May I ...
@Nelimee Do we need to merge? Currently, there's just one question with "phase-estimation" and another question with "quantum-phase-estimation". Might we as well use just one tag? (say just "phase-estimation")
@Blue 'merging', if I'm getting the terms right, is a specific single action that does exactly that and is generally preferable to editing tags on questions. Having said that, if it's just one question, it doesn't really matter although performing a proper merge is still probably preferable
Merging is taking all the questions with a specific tag and replacing that tag with a different one, on all those questions, on a tag level, without permanently changing anything about the underlying tags
@Blue yeah, you could do that. It generally requires votes, so it's probably not worth bothering when only one question has that tag
@glS "Every hermitian matrix satisfy this property: more specifically, all and only Hermitian matrices have this property" ha? I though it was only a subset of the set of valid matrices ^^ Thanks for the precision :)
@Nelimee if you think about it it's quite easy to see. Unitary matrices are the ones with phases as eigenvalues, while Hermitians have real eigenvalues. Therefore, if a matrix is not Hermitian (does not have real eigenvalues), then its exponential will not have eigenvalues of the form $e^{i\phi}$ with $\phi\in\mathbb R$. Although I'm not sure whether there could be exceptions for non diagonalizable matrices (if $A$ is not diagonalizable, then the above argument doesn't work)
This is an elementary question, but a little subtle so I hope it is suitable for MO.Let $T$ be an $n \times n$ square matrix over $\mathbb{C}$.The characteristic polynomial $T - \lambda I$ splits into linear factors like $T - \lambda_iI$, and we have the Jordan canonical form:$$ J = \begin...
@Nelimee no! unitarily diagonalizable matrices are all and only the normal ones (satisfying $AA^\dagger =A^\dagger A$). For general diagonalizability if I'm not mistaken onecharacterization is that the sum of the dimensions of the eigenspaces has to match the total dimension
@Blue I actually agree with Nelimee here that it's not that easy. You get $UU^\dagger = e^{iA} e^{-iA^\dagger}$, but if $A$ and $A^\dagger$ do not commute it's not straightforward that this doesn't give you an identity
I'm getting confused. I remember there being some theorem about one-to-one mappings between unitaries and hermitians provided by the exponential, but it was some time ago and may be confusing things in my head
@Nelimee if there is a $0$ there then it becomes the normality condition. Otherwise it means that the matrix is not normal, therefore not unitarily diagonalizable, but still the product of exponentials is relatively easy to write
@Blue you are right indeed. If $U$ is unitary then for sure you can write it as exponential of an Hermitian (time $i$). This is easily proven because $U$ is ensured to be unitarily diagonalizable, so you can simply compute it's logarithm through the eigenvalues. However, logarithms are tricky and multivalued, and there may be logarithms which are not diagonalizable at all.
I've actually recently asked some questions on math.SE on related topics
@Mithrandir24601 indeed, that was also what @Nelimee showed with an example above. I believe my argument holds for unitarily diagonalizable matrices. If a matrix is only generally diagonalizable (so it's not normal) then it's not true
also probably even more generally without $i$ factors
so, in conclusion, it does indeed seem that $e^{iA}$ unitary implies $A$ Hermitian. It therefore also seems that $e^{iA}$ unitary implies $A$ normal, so that also my argument passing through the spectra works (though one has to show that $A$ is ensured to be normal)
Now what we need to look for is 1) The exact set of conditions for which the matrix exponential $e^A$ of a complex matrix $A$, is unitary 2) The exact set of conditions for which the matrix exponential $e^{iA}$ of a real matrix $A$ is unitary
@Blue fair enough - as with @Semiclassical I was thinking about it with the t parameter, as that's what we care about in physics :P I can possibly come up with a number of non-Hermitian matrices that gives unitary evolution for a specific t
Or rather, the exponential of which is unitary for $t+n\tau$, although I'd need to check
If you're afraid of the density of diagonalizable matrices, simply triangularize $A$. You get $$A=P^{-1}UP,$$ with $U$ upper triangular and the eigenvalues $\{\lambda_j\}$ of $A$ on the diagonal.Then$$\mbox{det}\;e^A=\mbox{det}(P^{-1}e^UP)=\mbox{det}\;e^U.$$Now observe that $e^U$ is upper ...
There's 15 hours left on a bountied question, but the person who offered the bounty is suspended and his suspension doesn't expire until about 2 days, meaning he may not be able to award the bounty himself?That's not fair: It's a 300 point bounty. The largest bounty ever offered on QCSE. Let h... |
The aim of the following is to address the intuition side of the question - I doubt that I have more experience than anybody else, and the following is certainly not rigorous - still... One way to the answer - at least in this case! - is to use calculus, i.e., assume everything in sight is differentiable, and perhaps try "to sweep up the loose ends" afterwards.
As pointed out by Holonomia, the function $g(x,y) = f(x)f(y)$ is constant on circles. This means that the gradient of the differentiable $g$ is parallel to the vector $(2x,2y)$, as the latter is normal to $x^2 +y^2 = c$. So $$ {\rm grad}\ g = \lambda\cdot (2x, 2y), $$ where where $\lambda = \lambda(x,y)$ is a scalar function. Comparing the components, one gets$$f'(x) f(y) = \lambda 2 x,$$and $$f'(y) f(x) = \lambda 2 y.$$
Doing the algebra (formally), one obtains $$ {f'(x) \over 2x f(x)} = {f'(y) \over 2y f(y)}.$$Therefore, both sides of the equality are constant, i.e.,$$ {f'(x) \over 2x f(x)} = \beta,$$with $\beta$ some constant. Cross-multiplying by $2x$ and integrating, oneends up with $$ f(x) = \alpha e^{\beta x^2},$$for some constant $\alpha$ - i.e., Holonomia's answer.
Edit * Some "sweeping up," by request, to show that any $f$ satisfying the conditions of the problem (continuity, functional equation) is differentiable at $x=c$, for every $c$. Fix $c$, set $a = |c|$, and consider $$ f(x) \int_{a+10}^{a+20} f(y)\, dy = \int_{a+10}^{a+20} f(x) f(y) \,dy = \int_{a+10}^{a+20} \psi( x^2 +y^2) \, dy,$$where $\psi$ is Holonomia's $\psi$.The integral multiplying $f(x)$ is not zero if $f$ is not identically zero (using a Holonomia-style argument and $f(\sqrt 2 x)f(0) = f(x)^2 $, for instance, to conclude that the continuous $f$ is nowhere $0$ if not identically $0$).With the change of variables $y= \sqrt{r^2-x^2}$, the integral on the right becomes$$ \int_{\sqrt {(a+10)^2 +x^2}}^{\sqrt {(a+20)^2 +x^2}} \psi ( r^2) {r\over \sqrt{r^2-x^2} }\, dr,$$which is differentiable at (in a neighborhood of) $x=c$, because $r^2-x^2 \ge (a+10)^2 >0$, $\psi$ is continuous, and the limits of the integral are differentiable. Thus $f(x)$ is differentiable. |
Why set of real numbers not a set of ordered pairs ?
We write $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$, then we define addition and multiplication on this new set. Together with those definitions we call $\mathbb{R}^2$ as the set of complex numbers $\mathbb{C}$.
This is the gist of what I know from my book.
(I don't know much about complex numbers)
We can write any real number as $r = x + y$ ,where $x$ is a rational number and $y$ is a purely irrational number ($2\pi$ is correct value of y but not $2 + \pi$) $\qquad{(1})$.
A complex number is written as $z = x+ iy$, where $x$ and $y$ is real numbers $\qquad{(2})$.
Statement (1) and (2) are similar. We can say, by drawing a little bit of inspiration from the the definition of complex numbers, that $\mathbb{R} = \{(x,y): \ \ x\in \mathbb{Q}, y \in \mathbb{I}\} = \mathbb{Q}\times \mathbb{I}$.
Hence the set of complex numbers is a set of ordered pairs.
But this something that I have never seen anywhere.
I want know what is the fault in defining set of real numbers like this ? |
This is something I've been trying to figure out for a long time, and all I have is vague numerical results. I'm trying to answer the following question analytically:
Suppose I have a time dependent system:
$u_t = L(x, u, u_x, u_{xx}) , u(t=0,x) = u_0 $
I approximate differences with some finite difference scheme, for instance $u_{xx}^j = \frac{1}{dx^2} (u^{j-1} +u^{j+1} -2u^{j} )$
And I approximate each time step with an implicit FD method, for instance Runge-Kutta.
The Question: Given an orthonormal polynomial base $\{ p_j (x) \}_{j=0}^{\infty } $, and given that $ u_0 $ is as smooth as necessary, what can we say about the spectral coefficients $u(x,t_f) = \sum\limits_j a_j p_j (x)$ for large $t_f$ ?
I think this is closely tied, if not determined by, how differentiable is $u(x,t)$ , but I'm not sure about that either. |
The idea behind the information transfer model is that what is called "demand" in economics is essentially a source of information that is being transmitted to the "supply", a receiver, and the thing measuring the information transfer is what we call the "price".
Choosing constant information sources (i.e. keeping demand constant) or constant information destinations (i.e. a fixed quantity supplied) allows you to trace out supply and demand curves and the movement of those curves (allowing the information source and destination to vary) recovers the Marshall model. We can see diminishing marginal utility in the downward sloping demand curves; this comes from the definition of the "detector" measuring the price having supply in the denominator which follows from the identification of the demand as the information
source.
Note that since Fielitz and Borchardt originally described physical processes with this information transfer model, we can make an analogy between economics and thermodynamics, specifically ideal gasses:
Price is analogous to the pressure of an ideal gas Demand is analogous to the work done by an ideal gas (and is related to temperature and energy content) Supply is analogous to the volume of an ideal gas
The equation relating the price, supply and demand is analogous to the ideal gas law. One point to make here is that we haven't made any description of how the demand behaves over time, just how it behaves under small perturbations from "equilibrium" -- by which we mean a constant price defined by the intersection of a given supply and demand curve (with ideal information transfer). Humans will decide they don't want e.g. desktop PC's anymore (because of tablets or laptops or whatever reason) and the demand will drop. The economy (and population) grow. Economists frequently use supply and demand diagrams to describe models or specific shocks and this information transfer framework recovers that logic. In the future, I would like to see where else we can take model. In the next post, I will show how you could get (downward) sticky prices from this model by looking at non-ideal information transfer.
PS If we use the linearized version of the supply and demand relationship near the equilibrium price, we can find the (short run) price elasticities from$$
Q^d =Q_{\text{ref}}^d +\frac{Q_0^d}{\kappa }-Q_{\text{ref}}^s P
$$
$$
Q^s = Q_{\text{ref}}^s-\kappa Q_0^s+\frac{Q_0^s{}^2\kappa ^2}{Q_{\text{ref}}^d}P
$$
Such that$$
e^d = \frac{dQ^d/Q^d}{dP/P} =\frac{\kappa Q^d- Q^d_0 - \kappa Q_\text{ref}^d}{\kappa Q^d}
$$
Expanding around
$$
\Delta
Q^d=Q^d-Q_{\text{ref}}^d
$$
$$
e^d \simeq - \frac{Q^d_0}{\kappa Q^d_\text{ref}} + O(\Delta Q^d)
$$
And analogously
$$
e^s \simeq \frac{\kappa Q^s_0}{Q^s_\text{ref}} + O(\Delta Q^s)
$$
From which we can measure $ \kappa $.
(Note, I said
(Note, I said
fewerequations.) |
According to Kolmogorov, the energy spectrum function of a turbulent fluid is given as,
$E(k)=C\epsilon^{\frac{2}{3}}k^{\frac{-5}{3}}$
where $\epsilon$ is the energy flux and $k=\frac{2\pi}{r}$ where $r$ is the length scale.
The normal explanation I see in most physics texts and articles is that the -5/3 exponent is found purely through dimensional analysis. Obviously dimensional analysis is very useful, but often times there is a more physical explanation as well. Does anyone have any insights? |
No, the damping coefficient will not vary with mass.
Based on the back in forth in the comments, you are confusing a few concepts here.
The damping coefficient (subscript $c$) is a measure of applied force compared to velocity. In terms of the equations of simple harmonic motion, this is a constant which has no terms dictated by mass.
Your mention of "under-damping" in the comments leads me to believe you are confusing damping coefficient $c$ with damping ratio $\zeta$.
$\zeta$ will determine the characteristics of the damped harmonic motion (i.e. under-damped, over-damped, critically damped). $\zeta$ is given by the equation $$\zeta = \frac{c}{2 \sqrt {mk}}$$ where $c$, $m$ and $k$ are all constants.
You can see that mass will absolutely affect the damping ratio, but not the damping coefficient (since that is assumed constant).
In a real life scenario the damper may not perform the same if different masses are used; but traditionally with the idealized equations it just has a constant value.
I find the Wikipedia page has some good information on harmonic oscillation. Also, if I misunderstood what your question was really about please let me know. |
In a paper by Joos and Zeh, Z Phys B 59 (1985) 223, they say:This 'coming into being of classical properties' appears related to what Heisenberg may have meant by his famous remark [7]: 'Die "Bahn" entsteht erst dadurch, dass wir sie beobachten.'Google Translate says this means something ...
@EmilioPisanty Tough call. It's technical language, so you wouldn't expect every German speaker to be able to provide a correct interpretation—it calls for someone who know how German is used in talking about quantum mechanics.
Litmus are a London-based space rock band formed in 2000 by Martin (bass guitar/vocals), Simon (guitar/vocals) and Ben (drums), joined the following year by Andy Thompson (keyboards, 2001–2007) and Anton (synths). Matt Thompson joined on synth (2002–2004), while Marek replaced Ben in 2003. Oli Mayne (keyboards) joined in 2008, then left in 2010, along with Anton. As of November 2012 the line-up is Martin Litmus (bass/vocals), Simon Fiddler (guitar/vocals), Marek Bublik (drums) and James Hodkinson (keyboards/effects). They are influenced by mid-1970s Hawkwind and Black Sabbath, amongst others.They...
@JohnRennie Well, they repeatedly stressed their model is "trust work time" where there are no fixed hours you have to be there, but unless the rest of my team are night owls like I am I will have to adapt ;)
I think u can get a rough estimate, COVFEFE is 7 characters, probability of a 7-character length string being exactly that is $(1/26)^7\approx 1.2\times 10^{-10}$ so I guess you would have to type approx a billion characters to start getting a good chance that COVFEFE appears.
@ooolb Consider the hyperbolic space $H^n$ with the standard metric. Compute $$\inf\left\{\left(\int u^{2n/(n-2)}\right)^{-(n-2)/n}\left(4\frac{n-1}{n-2}\int|\nabla u|^2+\int Ru^2\right): u\in C^\infty_c\setminus\{0\}, u\ge0\right\}$$
@BalarkaSen sorry if you were in our discord you would know
@ooolb It's unlikely to be $-\infty$ since $H^n$ has bounded geometry so Sobolev embedding works as expected. Construct a metric that blows up near infinity (incomplete is probably necessary) so that the inf is in fact $-\infty$.
@Sid Eating glamorous and expensive food on a regular basis and not as a necessity would mean you're embracing consumer fetish and capitalism, yes. That doesn't inherently prevent you from being a communism, but it does have an ironic implication.
@Sid Eh. I think there's plenty of room between "I think capitalism is a detrimental regime and think we could be better" and "I hate capitalism and will never go near anything associated with it", yet the former is still conceivably communist.
Then we can end up with people arguing is favor "Communism" who distance themselves from, say the USSR and red China, and people who arguing in favor of "Capitalism" who distance themselves from, say the US and the Europe Union.
since I come from a rock n' roll background, the first thing is that I prefer a tonal continuity. I don't like beats as much as I like a riff or something atmospheric (that's mostly why I don't like a lot of rap)
I think I liked Madvillany because it had nonstandard rhyming styles and Madlib's composition
Why is the graviton spin 2, beyond hand-waiving, sense is, you do the gravitational waves thing of reducing $R_{00} = 0$ to $g^{\mu \nu} g_{\rho \sigma,\mu \nu} = 0$ for a weak gravitational field in harmonic coordinates, with solution $g_{\mu \nu} = \varepsilon_{\mu \nu} e^{ikx} + \varepsilon_{\mu \nu}^* e^{-ikx}$, then magic? |
This is a continuation of my previous question.
I have two classes, $C_1$ and $C_2$.
$C_1$ is a bivariate Gaussian with mean $\mu = (0,0)$ and covariance $\Sigma = I$
$C_2$ is a bivariate Gaussian with mean $\mu = (1,3)$ and covariance $\Sigma = 2I$, where $I$ is the identity matrix.
I am trying to calculate $P(x|C_1)$ and $P(x|C_2)$ so I can eventually calculate $P(C|x)$
To calculate $P(x|C_1)$ and $P(x|C_2)$ this I'm using the formula for a bivariate normal distribution found here.
My covariance is zero, which makes this a little bit easier.
When I use this calculate $P(x|C_1)$ I have...
$z=x_1^2 - x_2^2$
$p=0$
$p(x_1,x_2) = (\frac{1}{2\pi})e^{-z/2}$
When I use this to calculate $P(x|C_2)$ I have...
$z=\frac{1}{4}( (x_1-1)^2 + (x_2-3)^2 )$
$p=0$
$p(x_1,x_2) = (\frac{1}{8\pi})e^{z/8}$
Did I do this correctly? Also, I'm a bit confused as to whether what I'm doing even gives me $P(x|C_1)$ and $P(x|C_2)$. I'm a bit over my head in the class I'm in, so if I'm totally wrong there please correct me.
Anyways, with these two values I'm supposed to calculate $P(C|x)$ using Bayes rule (I think). I have the priors of $C_1$ and $C_2$ (they are 0.4 and 0.6 respectively), but I'm lost exactly on how to calculate $P(C|x)$ with this.
Could somebody basically just check over some of my work and help me out with the process using Bayes rule to calculate $P(C|x)$?
EDIT: My end goal here is to calculate an optimal decision boundary between $C_1$ and $C_2$, if I'm going the complete wrong way here let me know, but from what I've gathered through time spent on this site, I think I'm headed the right way. |
Although we will have practically no occasion to use the quantum microcanonical ensemble (we relied on it more heavily in classical statistical mechanics), for completeness, we define it here. The function \(f\), for this ensemble, is
\[f(E_i)\delta E = \theta(E_i-(E+\delta E))-\theta(E_i-E)\]
where \(\theta (x) \) is the Heaviside step function. This says that \(f(E_i)\delta E\) is 1 if \(E<E_i<(E+\delta E)\) and 0 otherwise. The partition function for the ensemble is \({\rm Tr}(\rho) \), since the trace of \(\underline {\rho } \) is the number of members in the ensemble:
\[\Omega(N,V,E) = {\rm Tr}(\rho) = \sum_i\left[\theta(E_i-(E+\delta E))-\theta(E_i-E)\right]\]
The thermodynamics that are derived from this partition function are exactly the same as they are in the classical case:
\[S (N, V, E ) \] \(-k\ln \Omega(N,V,E)\) \[{1 \over T}\] \(-k\left({\partial \ln \Omega \over \partial E } \right)_{N,V} \) etc. |
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
Castellón, 12/12/2017
Take an isolated surface singularity.
By Milnor fibration theorem, the germ is the topological cone over the link.
Let \(\pi: M \rightarrow \Sigma\) be a \(\mathbb{S}^1\) bundle
Given two \(\mathbb{S}^1\) bundles \(\pi_1: M_1 \rightarrow \Sigma_1\), \(\pi_2: M_2 \rightarrow \Sigma_2\)
Are obtained by several plumbings
Are the result of adding handles to closed balls (equivalently, identifying pairs of disks in the boundary)
There are special curves that identify the handles (
cutting curves).
Two handle bodies of the same genus can be glued along the boundary, obtaining a closed \(3\)-manifold.
Every \(3\)-manifold can be obtained in this way (although not in a unique way).
The gluing surface, together with the two systems of cutting curves is called the
Heegaard diagram of the splitting.
Heegaard-Floer homology is defined after such diagrams.
Let \(\mathbb{T}\) be a solid torus, \(c\) a closed curve in \(\partial \mathbb{T}\) isotopic to its core. \(M\) a \(3\)-manifold with boundary, and \(s\) a closed curve in \(\partial M\). Consider \(A_c\), \(A_s\) small neighborhoods in the boundaries of \(c\) and \(s\).
Lemma: The result of gluing \(\mathbb{T}\) with \(M\) through an orientation reversing homeomorphism \(A_c\to A_s\) is homeomorphic to \(M\).
This operation is called a
float gluing Lemma: Let \(\Sigma'\) be an oriented surface of genus \(g\) with one boundary component. Then \(\Sigma'\times I\) is a handle body of genus \(2g\). Lemma: Let \(\Sigma'\) be an oriented surface of genus \(g\) with \(n\) boundary components. Then \(\Sigma'\times I\) is a handle body of genus \(2g+n-1\). Lemma: Let \(\Sigma'\) be an oriented surface of genus \(g\) with \(n\) boundary components. Then \(\Sigma'\times I\) is a handle body of genus \(2g+n-1\).
Let's start with a \((g,e)\) \(\mathbb{S}^1\) bundle \(\pi\).
Over the \(D_i\), choose trivializations such that local Euler numbers are the \(\pm 1\) chosen before.
Our Heegaard splitting is \(M_s\), \(M_b\cup \pi^{-1}(D_1\cup\cdots\cup D_n)\)
Our Heegaard surface is formed by two copies of a genus \(g\) surface, joined by \(n\) tubes.
One system of cutting curves:
The other system goes parallel except in the boundary of the floats:
The other system goes parallel except in the boundary of the floats:
Lemma: Let \(M_1,M_2\) be handle bodies of geni \(g_1,g_2\). \(c_i\) is a curve in \(\partial M_i\) that meets only une cutting curve, transversally, and only at one point. \(A_i\) is an annulus in \(\partial M_i\) that is a regular neighborhood of \(c_i\). The gluing of \(M_1\) and \(M_2\) along an orientation-reversing homeomorphism \(A_1\to M_2\) is a handle body of genus \(g_1+g_2-1\).
If we have chosen the same sign in the two tubes, the pieces \(M_s,M_b\) match, so we just have to move the cutting curves in the common part until they match.
So we have take away the cylinders of the tori being glued, and glue the cutting curves accordingly.
One of them are just glued, the other are glued in one end, and make a turn in the other one.
Theorem: Let \(M\) be a handle body of genus \(g\). \(c_1,c_2\) two disjoint curves in \(\partial M\) that intersect transversally two different cutting curves in only one point each, and touch no other one. Consider \(A_1,A_2\) annuli in \(\partial M\) that are regular neighborhoods of them. The result of gluing along an orientation reversing homeomorphism \(A_1\to A_2\) is again a handle body of genus \(g\).
The resulting cutting curves are:
The resulting cutting curves are:
The Poincare sphere |
Let $\pi:A\longrightarrow M$ be a vector bundle and $E\subseteq A$ a vector sub-bundle. Recall, a $k$-form on $A$ is a section of $\Lambda^k A^*$. Let us write $\Omega^k(A):=\Gamma(\Lambda^k A^*)$.
The inclusion $\jmath: E\longrightarrow A$ is a vector bundle map and therefore induces $\jmath^*:\Omega^k(A)\longrightarrow \Omega^k(E)$ so that we can ''restrict" forms on $A$ to forms on $E$.
Conversely, is it possible to extend forms on $E$ to forms on $A$?
Thanks. |
In most type systems, the type rules work together to define judgements of the form:
$$\Gamma\vdash e:\tau$$
This states that in context $\Gamma$ the expression $e$ has type $\tau$.
$\Gamma$ is a mapping of the free variables of $e$ to their types.
A type system will consist of a set of axioms and rules (a formal system of rules of inference, as Raphael points out).
An axiom is of the form
$$\dfrac{}{\Gamma \vdash e:\tau}$$
This states that the judgement $\Gamma \vdash e:\tau$ holds (always).
An example is
$$\dfrac{}{x:\tau\vdash x:\tau}$$
which states that under the assumption that the type of variable $x$ is $\tau$, then theexpression $x$ has type $\tau$.
Inference rules take facts that have already been determined and build larger facts from them. For instance the inference rule
$$\dfrac{\Gamma\vdash e_1:\tau\to\tau' \quad \Gamma\vdash e_2:\tau}{\Gamma\vdash e_1\ e_2:\tau'}$$
says that if I have a derivation of the fact $\Gamma\vdash e_1:\tau\to\tau'$ and a derivation of the fact $\Gamma\vdash e_2:\tau$, then I can obtain a derivation of the fact $\Gamma\vdash e_1\ e_2:\tau'$. In this case, this is the rule for typing function application.
There are two ways of reading this rule:
top-down - given two expressions (a function and another expression) and some constraints on their type, we can construct another expression (the application of the function to the expression) with the given type. bottom-up - given an expression that is, in this case, the application of a function to some expression, the way this is typed is by first typing the two expressions, ensuring that their types satisfy some constraints, namely that the first is a function type and that the second has the argument type of the function.
Some inference rules also manipulate $\Gamma$ by adding new ingredients into it (view-ed bottom up). Here is the rule for $\lambda$-abstraction:
$$\dfrac{\Gamma x:\tau\vdash e:\tau'}{\Gamma\vdash \lambda x.e:\tau\to \tau'}$$
The inference rules are applied inductively based on the syntax of the expression being considered to form a derivation tree. At the leaves of the tree (at the top) will be axioms, and branches will be formed by applying inference rules. At the very bottom of the tree is the expression you are interested in typing.
For example, a derivation of the typing of expression $\lambda f.\lambda x.f\ x$ is
$$\dfrac{\dfrac{}{f:\tau\to\tau',x:\tau\vdash f:\tau\to\tau'} \qquad \dfrac{}{f:\tau\to\tau',x:\tau\vdash x:\tau}}{\dfrac{f:\tau\to\tau',x:\tau\vdash f\ x:\tau'}{\dfrac{f:\tau\to\tau'\vdash \lambda x.f\ x:\tau'}{\vdash \lambda f.\lambda x.f\ x:\tau'}}}$$Two very good books for learning about type systems are:
Both books are very comprehensive, yet they start slowly, building a solid foundation. |
I am trying to calculate an integral involving multiple indicator functions, such as:
$$ h(u,v,w) = -\int_0^1 J^{\prime\prime}(s) (I_{(0,s]}(u) - s)(I_{(0,s]}(v) - s)(I_{(0,s]}(w) - s)\, dF^{-1}(s)$$
where $0 < u, v, w < 1$ and
$$\begin{align} J(s) &= 6s(1-s)\\ F(s) &= 1/[1 + \exp[-s]]\\ F^{-1}(s) &= -\log[-1 + 1/s]\\ \end{align} $$
Here is my code:
Js = 6 s (1 - s) (*weight function*)FDJs = D[Js, s] (*first derivative of Js*)SDJs = D[FDJs, s] (*second derivative of Js*)Fs = 1/(1 + Exp[-(s)]) (*Fs*)Finv = -Log[-1 + 1/s] (*F inverse*)DFinv = D[Finv, s] (*derivative of F inverse*)h = -Integrate[ SDJs*(Boole[u <= s] - s)*(Boole[v <= s] - s)*(Boole[w <= s] - s) DFinv, {s, 0, 1}, Assumptions -> {0 < u < 1 , 0 < v < 1, 0 < w < 1}]
I get some output, but I am not sure if the syntax is correct especially for the indicator function. |
We will call a complementary set of $$A$$, and denote it as $$A^c$$, the set difference $$(U - A)$$, $$U$$ being the universal set. This is: $$$A^c=\{x: \ x\in U \ and \ x\notin A\}$$$
The complementary set of $$A$$ is the set of the elements $$x$$ that satisfy $$x$$ belongs to $$U$$, and $$x$$ does not belong to $$A$$.
Some basic properties of the complement are:
$$U^c=\emptyset$$ and $$\emptyset^c=U$$ $$A-B=A\cap B^c$$ $$(A^c)^c=A$$ $$A\cup A^c=U$$ and $$A\cap A^c=\emptyset$$ $$(A\cup B)^c=A^c\cap B^c$$ and $$(A\cap B)^c=A^c\cup B^c$$
Property 5 is known by the name of De Morgan's Laws. |
Let $\{p_n\}$ denote a sequence such that: $$ S_n = {1\over p_1} + {1\over p_2} + \cdots + {1\over p_n} $$ converges. Prove that: $$ \sigma_n = \left(1+{1\over p_1}\right)\left(1+{1\over p_2}\right)\cdots\left(1+{1\over p_n}\right) $$ converges, where $n, p_n \in \Bbb N$.
Consider each bracket from $\sigma_n$. By ${1\over p_n} > 0$: $$ \forall k \in \Bbb N: \left(1+{1\over p_k}\right) > 1 $$ So $\sigma_n$ must be monotonically increasing by: $$ {\sigma_{n+1} \over \sigma_n} = \left(1+{1\over p_{n+1}}\right) > 1 $$
To show a monotonic sequence is convergent it's sufficient to show that it's bounded above. Lets try to find the bound. Recall: $$ \ln(1+x) \le x \iff (1+x) \le e^x \tag1 $$ So by $(1)$ we have: $$ \sigma_n \le e^{S_n} $$ But $S_n$ is convergent! And thus: $$ \sigma_n \le e^L $$ where: $$ L = \lim_{n\to\infty}S_n $$ Which by monotone convergence theorem proves $\sigma_n$ is convergent.
I'm kindly asking to verify my proof and point to the mistakes in case of any. Thank you! |
The presentation of the homology version of Cauchy's theorem in Ahlfors is slick, but sweeps a lot of the topology under the rug using clever arguments. This question is an attempt to reconcile Ahlfors' analytic notion of a curve being homologous to zero (presented in his book
Complex Analysis and originally due to E. Artin, I believe) with the standard definition in homology as found in Hatcher.
We work in the complex plane and fix $a\in \mathbb C$. Let $\gamma$ be a continuous map $[0,1]\rightarrow C\backslash \{ a\}$. Following Munkres in his book
Topology, we define the winding number of $\gamma$ with respect to the point $a$ by considering
$$g(t)=\frac{\gamma(t)-a}{|\gamma(t)-a|}.$$
This is clearly a loop in $S^1$ and corresponds to some multiple of of the generator of the fundamental group of $S^1$. If the generator is $\tau$ and $g(t)$ corresponds to $m\tau$, $m\in\mathbb Z$, we define the winding number $n(\gamma, a)$ to be $m$. This is the definition of winding number I will use in this question, but in case $\gamma$ is piecewise differentiable, it corresponds to the analytic definition by integration given in Ahlfors.
Ahlfors calls a curve contained in an open region $\Omega$ homologous to zero if $n(\gamma,a)=0$ for all $a\in \Omega^c$.
In homology theory, as I understand it, we would call a curve homologous to zero if it represents the zero element in $H_1(\Omega, \mathbb Z)$. That is, $\gamma$ is the boundary of some singular $2$-chain.
My question: Why do all of these notions agree? Why is Ahlfors' definition of being homologous to zero (using Munkres' definition of winding number) agree with the usual homological one?
I would like to use Munkres' definition because it works for continuous curves, not just piecewise differentiable ones, and it seems to me the equivalence should hold in this generality.
Edit: This result appears as proposition 1.9.13 in Berenstein and Gay's book on complex analysis. |
As others have pointed out, it is important to justify that $6 \mid q^2 \Rightarrow 6\mid q$. And it isn't totally clear from your proof.
I suggest the following.
First show that $\sqrt{6}$ is not an integer. It's not difficult to do that. Since $4<6<9$, it follows that $2<\sqrt{6}<3$ and that means that $\sqrt{6}$ is not an integer.
Now assume that $\sqrt{6}$ is a rational number, $\frac{p}{q}$ where $p$ and $q$ are co-prime positive integers and $q>1$.
Now you can write
$$6=\frac{p^2}{q^2}$$
$$\Rightarrow 6q=\frac{p^2}{q}$$
It is clear that the left hand side is an integer. But the right hand side isn't since $p^2$ and $q$ share no common factors.
So this equality can not hold. And $\sqrt{6}$ can not equal $\frac{p}{q}$.
So it has to be an irrational number.
There's an incredibly short proof of this if you know the rational root theorem. Just notice that $\sqrt{6}$ is a root of the monic polynomial $x^2-6$. The proof is almost immediate.
EDIT: Here's a messy justification of why $q$ does not divide $p^2$. Let $p=\prod {p_i}^{x_i}$ and $q=\prod {p_j}^{y_j}$ such that $x_i$ and $y_j$ are positive integers. This notation is incredibly informal but it gets the message across.
Now since $p$ and $q$ are co-prime, $p_i\neq p_j$ for any $i$ & $j$. Now $p^2=\prod {p_i}^{2{x_i}}$. Notice that $p^2$ has the same prime divisors as $p$. Since $p$ and $q$ share no common prime factors, it follows that $p^2$ and $q$ share no common prime factors.
That means $q\nmid p^2$. |
$\def\d{\mathrm{d}}$There was a hint in the book, use intregation by parts in this way: $$\lim_{x\to 0^+} \frac{1}{x} \int_0^{x} \sin\frac{1}{t} \,\d t = \lim_{x\to 0^+} \frac{1}{x} \int_0^{x} t^2 \left(\frac{1}{t^2} \sin\frac{1}{t}\right)\,\d t.$$
When we integrate by parts we find this integral:
$$\int_0^{x} t\cos\frac{1}{t}\,\d t .$$
In every symbolic calculator it says is a special function and gives the value of the integral as $\mathrm{Ci}(x)$, but I'm using calculus 1 knowledge, any hint or help? Thanks in advance. |
The dodecahedron is a regular polyhedron of $$12$$ faces. If the mentioned faces are regular pentagons we call it a regular dodecahedron:
To find the area of edge a of a dodecahedron $$a=10 \ m$$.
To calculate the area of edge $$a$$ of the regular dodecahedron it will be necessary to first find the area of side $$a$$ of a regular pentagon.
We need to use trigonometry to find the area of the pentagon from $$a$$. This can be very useful, since $$a$$ will not be a regular occurrence in problems like finding the area of a pentagon, or a dodecahedron.
The five triangles with height $$ap$$ that compose the pentagon are equal, $$$A_{pentagon}=5 \cdot (\dfrac{a \cdot ap}{2}) \\ b^2=ap^2+\Big(\dfrac{a}{2}\Big)^2$$$ We have to, first, find $$b$$. As, $$\beta=\dfrac{360^\circ}{5}=72^\circ$$ and using the definition of the sinus of the triangle, $$$\sin \dfrac{\beta}{2}=\sin 36^\circ= \dfrac{opposite}{hypotenuse}=\dfrac{\dfrac{a}{2}}{b} \\ b=\dfrac{5}{\sin 36^\circ}= 8,5 \ m \\ 8,5^2=ap^2+5^2 \\ ap= 6,9 \ m$$$
Then, $$$A_{pentagon}=5 \cdot \dfrac{10 \cdot 6,9}{2}= 172 \ m^2 \\ A_{dodecahedron}= 2063,5 \ m^2$$$
Finally, the following expressions allow us to find the area and volume of the dodecahedron of edge $$a$$: $$$A=30 \cdot a \cdot ap \\ V=\dfrac{1}{4}(15+7\sqrt{5})a^3$$$ |
Latex distinguishes between three different enumeration/itemization environments. Each of them provide four levels, which means you can have nested lists of up to four levels.
Enumerate: \begin{enumerate} \item ... \end{enumerate}
The enumerate-environment is used to create numbered lists.
If you like to change the appearance of the enumerator, the simplest way to change is to use the enumerate-package, giving you the possibility to optionally choose an enumerator. \usepackage{enumerate} ... \begin{enumerate}[I]%for capital roman numbers. \item \end{enumerate}
\begin{enumerate}[(a)]%for small alpha-characters within brackets. \item \end{enumerate} Itemize: \begin{itemize} \item ... \end{itemize}
Itemization is probably the mostly used list in Latex. It also provides four levels. The bullets can be changed for each level using the following command:
\renewcommand{\labelitemi}{$\bullet$} \renewcommand{\labelitemii}{$\cdot$} \renewcommand{\labelitemiii}{$\diamond$} \renewcommand{\labelitemiv}{$\ast$}
Amongst the more commonly used ones are
$\bullet$ (),
$\cdot$ (),
$\diamond$ (),
$-$ (),
$\ast$ () and
$\circ$ ().
Description: \begin{description} \item[] ... \end{description}
The description list might be the least known. It comes in very handy if you need to explain notations or terms. Its neither numbered nor bulleted.
Example: \begin{description} \item[Biology] Study of life. \item[Physics] Science of matter and its motion. \item[Psychology] Scientific study of mental processes and behaviour. \end{description}
And in a PDF it would look like this:
Nested lists:
Lists can be nested. In other words, it is possible to have a sub-list for an item of a list. Usage is straight forward, different environments can be mixed (see example) and the maximum depth (number of levels) is 4. Here is an example:
\begin{itemize} \item First level, itemize, first item \begin{itemize} \item Second level, itemize, first item \item Second level, itemize, second item \begin{enumerate} \item Third level, enumerate, first item \item Third level, enumerate, second item \end{enumerate} \end{itemize} \item First level, itemize, second item \end{itemize}
The space between different items can be controlled with the \itemsep command (can only be added just after “begin”):
\begin{itemize}\itemsep2pt \item \end{itemize} |
You are asking all the right questions!
Let me give you an extreme example. Then I'll answer you question.
Let $f(x)$ then the function that.
If $x$ is irrational then $f(x) = 2+x$.
If $x = \frac ab$ where $\frac ab$ is a rational number in "lowest terms", then $f(x) = 2 + \frac 1{b}$. (We'll assume the denominator is not negative although the numerator might be.)
Then $f(0)$ is undefined (as $2 + \frac 1n)$ is undefined) and this weird function is certainly not continuous. I won't (at this time) go into the exact meaning of contininuous but you can see it jumps from $2$ to $2 \frac 1{b}$ as $x$ goes from rational to irational values and as rationals with low denominators or infinitely close to rationals with high denominators, it jumps about like a flea.
Claim 1: $\lim_{x\to 0} f(x) = 2$.
Claim 2: $\lim_{x \to 1}f(x) \ne 3$
Claim 3: $\lim_{x \to 1}f(x) = k$ will be false no matter what $k$ we pick.
To prove that $\lim_{x\to 0} f(x) = 2$ we want to show that we can "force" the $f(x)$s to get "very close to $2$" if we force the $x$s to get "very close" to $0$.
If we want the $f(x)$s to be within $\frac 14$ of $2$ we can force the $x$s to be within a $\frac 14$ of $0$. If $|x| < \frac 14$ then either $x$ is irrational or $x =\frac ab$ for some $a,b$.
If $x$ is irrational, then $f(x) = 2+x$. $x$ is within $\frac 14$ of zero, so $f(x)$ is within $\frac 14$ of $2$.
If $x= 0$ we have a problem in that $f(x)$ undefined. Well, that's not
really a problem because we are interested in the behavior of $f(x)$ near $0$. Not at $0$.
If $x = \frac ab$ is rational and $x \ne 0$ and $|x| < \frac 14$ then $b \ge 4$ and $\frac 1b < \frac 14$. That means $f(x) = 2 + \frac 1b$ is such that $2 < f(x) < 2 + \frac 14$ so $f(x)$ is with $\frac 14$ of $2$.
So we forced $f(x)$ to be within $\frac 14$ of $2$ by forcing $x$ to be within $\frac 14$ of $0$.
That was one example, can we come up with something for
all choices of how close we want to be.
If we want $f(x)$ to be within $\epsilon$ of $2$ for any very very small value of $\epsilon$, we can do this by forcing $x$ to be within $\delta = \epsilon$ of $0$.
If $x$ is irrational, $f(x) = 2+x$. But $x$ is within $\epsilon$ of $0$ so $f(x)$ is within epsilon of $2$.
If $x$ is rational and $x = \frac ab; a\ne 0$ and $|\frac ab | < \delta = \epsilon$. Then $b > \frac 1{\epsilon}$. SO $f(x) = 2 + \frac 1b < 2 + \epsilon$. So $f(x)$ is within $\epsilon$ of $2$.
So that does it. We can force $f(x)$ to be as close to $2$ as we like by forcing $x$ to be within a certain distance of $0$.
Claim 2: Trying to force $f(x)$ to be close to $3$ by forcing $x$ to be close to $1$.
Let's try to force $f(x)$ to be within $\frac 14$ of $3$.
Well matter how close we chose $x$ to be $1$ we will find a rational, $r$ with a denominator more than $4$ and $f(x) = 2 +\frac 14 < 2+\frac 14$ and that will not be within $\frac 14$ of $3$.
Indeed for any $x < 1$ or $x > 1$ that we pick, we will find we can always find that there is an integer $N > \frac 1{|x-1|}$ and either $x < 1 - \frac 1N$ or $1+ \frac 1N<x$ and so $f(1 \pm \frac 1N ) = 2 + \frac 1N$ but there will also be an irration $y$ so that $x < y < 1$ or $1 < y < x$. and $f(x)$ is within $|x-1|$ of $3$.
So we can't force $f(x)$ to get close to
anything at $x=1$.
....
Hopefully that makes the definition clearer.
$\lim_{x\to a} f(x) = L$ means
For any distance $\epsilon > 0$, no matter how small, we can force $f(x)$ to be within $\epsilon$ of $L$ (i.e. force $|f(x) - L | <\epsilon$), by finding a $\delta$ so that whenever we have $x$ within $\delta$ of $a$ (i.e. $|x - a| < \delta$) we will
have to have $|f(x) - L | < \epsilon$.
So on to your questions:
1) " I assume open interval just means something of the form (a,b) that doesn't include endpoints, but don't we also need that interval to be continuous?"
The open interval is around the
inputs near $a$. It has nothing to do with the outputs $f(x)$, which can hop about like fleas (provided the hop in increasingly smaller hop for $x$s near $a$.
So all this is saying is there's a small area around $a$ where all the $x$ around $a$ will have $f(x)$ well-defined.
2) "I am confused about the usage of "for all". Like do they mean literally for all x?
No. They mean all $x$ so that $|x - a| < \delta$. In other words they mean all the $x \in (a-\delta,a+\delta)$. |
Assume, we have an $m\times n$ block matrix $M=\left[\begin{array}{c c}A&C\\B&D \end{array}\right]$, where
$A$ is an $m_1 \times n_1$ matrix of rank $k_A$. $B$ is an $m_2 \times n_1$ matrix of rank $k_B$. $C$ is an $m_1 \times n_2$ matrix of rank $k_C$. $D$ is an $m_2 \times n_2$ matrix of rank $k_D$.
Obviously, $m_1+m_2=m$ and $n_1+n_2=n$.
In the paper by Carl D. Meyer "Generalized inverses and ranks of block matrices",
SIAM J. Appl. Math, vol. 25, no. 4, pp. 597-602, Dec. 1973, I found a way to bound the rank of a block matrix $M$ by the rank of its subblocks, as follows (Corollary 4.1):
$$ \begin{equation} \newcommand{\rank}[1]{\text{rank}(#1)} \rank{M}\leq\rank{A}+\rank{B}+\rank{C}+\rank{Z}, \tag{1}\label{1} \end{equation}$$
where $Z=D-BA^-C$, thus leading to
$$ \begin{equation} \rank{M}\leq k_A+k_B+k_C+\underbrace{\big(k_D+\min(k_A,k_B,k_C)\big)}_{\rank{Z}}, \tag{2}\label{2} \end{equation}$$ as $\rank{Z}\leq \rank{D}+\rank{BA^-C}=k_D+\min(k_A,k_B,k_C)$ by the properties of matrix ranks: rank of a sum and rank of a product. So, we can convert an original result $\eqref{1}$ to $$ \begin{equation} \rank{M}\leq k_A+k_B+k_C+k_D+\min(k_A,k_B,k_C)\leq 5\underbrace{\max{(k_A,k_B,k_C,k_D)}}_{k_\max}. \tag{3}\label{3} \end{equation}$$
I have the following questions:
Is anybody aware of any way to come up with a tighter (than $\eqref{1}$ or $\eqref{3}$) bound to a rank of a $2\times 2$ block matrix in terms of the ranks matrices of its subblocks? The reason is that in my numerical experiments (matrices arise from boundary value problem) I cannot come up even close to this bound (usually staying safely within $2k_\max$) - which proves nothing, but certainly makes me think. The original paper talks about exact ranks. Would this or any other bound change significantly if we start talking about the numerical rank $\tilde{k}_M(\epsilon)$ in terms of numerical ranks of its subblocks $\tilde{k}_{A,B,C,D}(\epsilon)$? Here $\epsilon$ is the relative tolerance for a truncated SVD. In other words, if $\tilde{k}_{M}(\epsilon)=5$, then $\sigma_6/\sigma_1<\epsilon$. ($\sigma_{1},\ldots,\sigma_{\min(m,n)}$ being the singular values of $M$) Is there an efficient way to construct a truncated SVD of $M$ from already computed $\epsilon$-truncated SVD's of subblocks $A,B,C,D$? and how to estimate the error in the resultant truncated SVD of the original matrix $M$? |
Let's start considering the following example:
A screw factory has two machines, the $$M1$$, which is old, and does $$75\%$$ of all the screws, and the $$M2$$, newer but small, that does $$25\%$$ of the screws. The $$M1$$ does $$4\%$$ of defective screws, while the $$M2$$ just does $$2\%$$ of defective screws. If we choose a screw at random: what is the probability that ir turns out to be defectively?
As we have seen, we could solve the problem using the theorem of the total probability. Remember that it was the same as to use a tree to solve it.
Let's look now at the problem from another point of view. If we know that a screw is defective: what is the probability that it has been made by machine $$M1$$?
In other words, we are wondering for the conditional probability $$P(M1/D)$$. On the one hand, for the definition of conditional probability, we have that:
$$$P(M1/D)=\dfrac{P(M1\cap D)}{P(D)}$$$
On the other hand, if we represent our problem in a tree, we see that we can compute $$P(M1\cap D)$$, since it is the probability of the colored branch: to be made by $$M1$$ and to be defective.
Remember that the theorem of total probability is: $$$P(D) = P(M1)\cdot P(D/M1) + P(M2)\cdot P(D/M2)$$$ So we obtain: $$$P(M1/D)=\dfrac{P(M1)\cdot P(D/M1)}{P(M1)\cdot P(D/M1) + P(M2)\cdot P(D/M2)}$$$
In our example, $$$P(M1/D)=\dfrac{0,75\cdot 0,04}{0,75\cdot 0,04 + 0,25\cdot 0,02}=0,857$$$
What we have done to solve this problem can be generalized through the Bayes' theorem.
Bayes' Theorem:
Let $$A_1,A_2,\ldots, A_n$$ be a partition of the sample space and $$B$$ uan event associated with the same experiment. Then, we have that: $$$ P(A_i/B)=\dfrac{P(A_i)\cdot P(B/A_i)}{P(A_1)\cdot P(B/A_1)+\ldots+P(A_n)\cdot P(B/A_n)}$$$
Let's see how to apply it.
We have three boxes with light bulbs. The first one contains $$10$$ bulbs, with $$4$$ of them broken; in the second one there are $$6$$ light bulbs, and only one broken, and in the third one there are three broken light bulbs out of eight. Now we want to ask ours: If we take a broken bulb what is the probability that comes from box $$1$$?
Let's remember that $$C1$$, $$C2$$, $$C3$$ represent boxes $$1$$, $$2$$ and $$3$$. Also $$F = $$ "broken bulb", for what $$\overline{F}=$$ "not broken bulb".
Now, we are only interested in the top branch of our tree.
We are interested in $$P(C1/F)$$. From Bayes' theorem, $$$ P(C1/F)=\dfrac{P(C1)\cdot P(F/C1)}{P(C1)\cdot P(F/C1)+P(C2)\cdot P(F/C2)+P(C3)\cdot P(F/C3)}$$$
In our case, $$$P(C1/F)=\dfrac{\dfrac{1}{3}\cdot \dfrac{4}{10}}{\dfrac{1}{3}\cdot \dfrac{4}{10} + \dfrac{1}{3}\cdot \dfrac{1}{6}+\dfrac{1}{3}\cdot \dfrac{3}{8}}=\dfrac{\dfrac{4}{30}}{\dfrac{113}{360}}=\dfrac{48}{113}=0,425$$$
that is to say, $$42,5\%$$.
Suppose we have $$250$$ doctors from Europe meeting in a conference. Among these $$115$$ are Germans; $$65$$, French, and $$70$$ Englishmen. We also know that, $$75\%$$ of the Germans, $$60\%$$ of the French and $$65\%$$ of the Englishmen are in favour of using a new vaccine for the flu. In order to decide whether the vaccine is finally used they agree on the following: among all the doctors they select at random three doctors, who answer if they are in favour or not (with replacement). Remember that with replacement means that a doctor can be selected all three times (or two times). The vaccine is approved out of these three picks, at least two agree on using the vaccine. We are now interested in: If we pick a doctor randomly and he is in favor of using the vaccine: what is the probability that he is French?
Let's consider the following events: $$A =$$ "German doctor", $$F =$$ "French doctor", $$I =$$ "English doctor", and $$V =$$ "to be in favor of the vaccine" (and therefore, $$ \overline{V}=$$ "o be against the vaccine").
We represent our problem in a tree.
We are looking for $$P(F/V)$$. From Bayes' theorem, $$$ P(F/V)=\dfrac{P(F)\cdot P(V/F)}{P(A)\cdot P(V/A)+P(F)\cdot P(V/F)+P(I)\cdot P(V/I)}$$$
In our case, $$$P(F/V)=\dfrac{\dfrac{65}{250}\cdot 0,6}{\dfrac{115}{250}\cdot 0,75 + \dfrac{65}{250}\cdot 0,6+\dfrac{70}{250}\cdot 0,65} =0,228$$$ |
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Azimuthally differential pion femtoscopy relative to the second and thrid harmonic in Pb-Pb 2.76 TeV collisions from ALICE
(Elsevier, 2017-11)
Azimuthally differential femtoscopic measurements, being sensitive to spatio-temporal characteristics of the source as well as to the collective velocity fields at freeze-out, provide very important information on the ...
Investigations of anisotropic collectivity using multi-particle correlations in pp, p-Pb and Pb-Pb collisions
(Elsevier, 2017-11)
Two- and multi-particle azimuthal correlations have proven to be an excellent tool to probe the properties of the strongly interacting matter created in heavy-ion collisions. Recently, the results obtained for multi-particle ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
Yesterday I asked about $\lim_\limits{\sigma_A, \sigma_B \to \infty }e^{-\sigma_A-\sigma_B} \sum_{k=0}^{\infty} \frac{{\sigma_A}^k}{k!}\cdot\frac{{\sigma_B}^k}{k!}$, and some fellow users helped me understand that this expression goes to zero. I hoped that I could use the logic of this proof to also understand this, slightly different, limit and ideally prove that it goes to zero as well. It didn't work out.
I see two possibilities to prove that this second limit goes to zero. Firstly, one could show that the ratio of the two probabilities is bounded. Secondly, one could autonomously prove that the second expression goes to zero. Unfortunately, it seems there is no clear result as to that one of the two expression is bounded by the other, or at least I can't find one, as the exponent for $\sigma_B$ is higher for one, but it also has $(k+1)!$ instead of $k!$.
The interpretation of the expressions are that we have two independent poisson distributed random variables, one with mean $\sigma_A$ (say X) and one with mean $\sigma_B$ (say Y). One of the expressions is the probability that X=Y, as the means go to infinity, the other one that X=Y+1 for the means going to infinity.
TL;DR: Ideally, I need to show that both expressions are zero. Here are two proofs that the one not in the heading is zero.
Thanks so much in advance, rm |
I understand that using a unit vector (of a vector say $\vec{a}$ ) and computing the directional derivative gives the slope (or rate of change of the function) in the direction of the vector.
I have three questions :
If I use the vector itself rather than it's unit vector what will I getwhen I compute it's dot product with the gradient of the function? It wouldn't give the slope of the curve(formed by the of the function with the slicing containing the vector $\vec{a}$ ), would it? plane Note: The function is scalar.
Also going by it's formal definition:
$\displaystyle \nabla _{\mathbf {v}}{f}({\mathbf {x}})=\lim _{h\rightarrow 0}{\frac {f({\mathbf {x}}+h{\mathbf {v}})-f({\mathbf {x}})}{h}}$
$\mathbf {v}$ is a vector
Quoting from Wikipedia
This definition is valid in a broad range of contexts, for example where the norm of a vector (and hence a unit vector) is undefined.
What does that mean?
Also quoting from Wikipedia:
If the function f is differentiable at x, then the directional derivative exists along any vector v, and one has
$\displaystyle \nabla _{\mathbf {v} }{f}({\mathbf {x} })=\nabla f({\mathbf {x} })\cdot {\mathbf {v} }$
Intuitively, the directional derivative of f at a point x represents the rate of change of f, in the direction of v with respect to time, when moving past x.
Why is it mentioned with respect to time isn't it with respect to the change in x (or/and y ) in the direction of the vector ? |
Let $S$ be a vector space of functions from $\mathbb{R}^n$ to $\mathbb{R}$, say $S := \{ f:\mathbb{R}^n \rightarrow \mathbb{R} \}$.
I am looking for some examples in which the dimension of $S$ is known.
For instance, trivial examples are the following.
Linear functions $f(x) := a^\top x$ implies that $\text{dim}(S) = n$.
Quadratic functions $f(x) := x^\top A x$ implies that $\text{dim}(S) = n^2$, or probably just $n(n+1)/2$ because we can take $A$ symmetric.
What is the dimension of the space of:
Sinusoidal functions $f(x) := a \sin( b^\top x + c) $? Is it just $n+2$?
Other known less-trivial examples?
Then, if $S_1$ has dimension $d_1$ and $S_2$ has dimension $d_2$, what is the dimension of $S:= \{ f := f_1 \circ f_2 : \ f_1 \in S_1, f_2 \in S_2 \}$?
Also, given a $d$-dimensional vector space $S_0$ of functions $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^m$, and a function $g: \mathbb{R}^m \rightarrow \mathbb{R}$, what is the dimension of the space $S := \{ g \circ f : \ f \in S_0 \}$? |
The other answers are nice, but none address the question: what numeric base(s) might quantum computers use? I will answer in two parts: first, the question is a little subtle, and second, you may use any numeric base, and then you work with qutrits or in general with qudits, which lead to qualitatively new intuitions! Or at any rate, I will try to make the case that they do.
A quantum bit isn't just a $0$ or a $1$, it's a bit more complex than that. For example, a quantum bit may be in the state $\sqrt{\frac{1}{4}}|0\rangle+\sqrt{\frac{3}{4}}|1\rangle$. When measured, you will measure the outcome $0$ with probability $\frac{1}{4}$ and the outcome $1$ with probability $\frac{3}{4}$. The 'superposition' you talked about is $\sqrt{\frac{1}{2}}|0\rangle+\sqrt{\frac{1}{2}}|1\rangle$, but in general any pair of complex numbers $a$ and $b$ will do, as long as $a^2+b^2=1$. If you have three qubits, then you can entangle them, and the state will be
$$a_0|000\rangle + a_1|001\rangle+a_2|010\rangle+a_3|011\rangle+a_4|100\rangle +a_5|101\rangle+a_6|110\rangle +a_7|111\rangle$$
But when you measure this three-qubit system, your measurement outcome is one out of these 8 states, that is, three bits. This is this really weird dichotomy where on the one hand quantum systems seem to have this exponential state space, but on the other hand we only seem to be able to 'get at' a logarithmic part of the state space. In 'Quantum Computing Since Democritus', Scott Aaronson probes this question by matching off several complexity classes to try and understand how much of this exponential state space we can exploit for computation.
Having said that, there is an obvious complaint to the answer above: all the notation is in binary. Qubits are in a superposition of
two base states, and entangling them doesn't change that much, because three qubits are in a superposition of $2^3$ base states. It's a legitimate complaint, because one usually thinks of $\texttt{unsigned int}$ as a number, and only remembers that it is implemented as a 32-bit string as an afterthought.
Enter the qutrit. It is a vector in $\mathbb{C}^3$, in other words, it consists of
three base states rather than two. You operate on this vector with a $3\times 3$ matrix, and all the usual things done in quantum computing don't change much, because any operation expressed in terms of qutrits can be expressed in terms of qudits, so it's really just syntactic sugar. But some problems are much easier to write down and/or think about when expressed as qudits instead of entangled qubits. For example, a variation of the Deutsch-Josza problem might ask, given an oracle for a function $f\colon \{0,\ldots, kn-1\}\rightarrow \{0,\ldots, k-1\}$, is this function constant or balanced, given that one is promised to be the case? This function naturally takes one $k$-qudit register as input. To solve it, you must apply a Fourier transform to this $k$-qudit, like so: (if this goes over your head, don't worry, it's just for illustration)
$$|a\rangle \mapsto \sum_{u=0}^{k-1}e^{i2\pi\frac{au}{k}}|u\rangle$$
If you want to express this in binary, you end up with a gate that does this on numbers $0\ldots k-1$ and acts trivially (does nothing) on all numbers $\geq k$, which is slightly less contrived than doing it this way. Similarly, consider a Bernstein-Vazirani variation where the oracle computes an in-product in some radix $r$. If $r=2$, then we know how to do it. But if $r=5$, the problem is much easier to solve by hand using several $5$-qudit registers. Some problems are easier if you have several different qudit registers, e.g. one $5$-qudit register and one $2$-qudit register.
In summary, yes, you are free to consider other numeric bases, and in the right setting that will make your life easier, for the same reason that thinking about numbers in terms other than their binary expansion helps you with normal computers. I felt compelled to answer because while most answers explained that a qubit has something to do with two base states when measuring but infinite in principle, no answer mentioned that the OPs suggestion of using other bases is legitimate and in fact really happens (for example, in Quantum walks on graphs, Aharonov et al. use a subroutine that takes a qubit and an $n$-qudit as input) |
Let $X$ be a topological space, and let $\{K_\alpha\}_{\alpha\in A}$ be a family of closed compact subsets of $X$. Show that $\bigcap_{\alpha\in A} K_\alpha$ is compact.
Proof:
Let $\mathcal{T}$ be the given topology on $X$. And let $\mathcal{T}_\alpha$ be the corresponding subspace topology on $K_\alpha$.
Let $K=\bigcap_{\alpha\in A} K_\alpha$ and let $\mathcal{T}_K$ be its subspace topology.
First of all we note that $K$ is closed since it is an intersection of closed sets.
Now, pick any family of closed sets $F:=\{F_\beta\}_{\beta\in B}$ of $K$ such that it has the finite intersection property. We want to show that $\bigcap F\neq\emptyset$.
For each $\beta\in B$, since $F_\beta$ is closed in $K$, $K\setminus F_\beta$ is open in $K$, and hence by the definition of the subspace topology on $K$, $F_\beta=K\setminus(U_i\cap K)=(K\setminus U_i)\cup(K\setminus K)=K\setminus U_i$ for some $U_i\in\mathcal{T}$.
Then, for each $\alpha\in A$, \begin{equation*} K_\alpha\setminus F_\beta=K_\alpha\setminus(K\setminus U_i)=K_\alpha\cap(K\cap U_i')'=K_\alpha\cap(K'\cup U_i)=(K_\alpha\cap K')\cup(K_\alpha\cap U_i) \end{equation*} is open in $K_\alpha$, since $K'$ and $U_i$ are open in $X$, and hence $K_\alpha\cap K'$ and $K_\alpha\cap U_i$ are open in $K_\alpha$ by the definition of subspace topology. (note: we denoted $A'$ be the complement of $A$ in $X$ where $A$ is a subset of $X$.)
Therefore the family $F$ is a family of closed sets in $K_\alpha$ such that it has the finite intersection property for each $\alpha\in A$.
\Then Since each $K_\alpha$ is compact, we conclude that $\bigcap F\neq\emptyset$.
Would this proof be false? |
This post will be more speculative than the derivation of supply and demand -- it will give one possible take of how sticky prices appear in the model (which are key to at least some schools of modern macroeconomic theory). If we return to non-ideal information transfer $I_{Q^s} \leq I_{Q^d}$ such that Equations (4) and (5) become$$
(10) \space P\leq \frac{1}{\kappa }\frac{Q^d}{Q^s}
$$
$$
(11) \space \frac{dQ^d}{dQ^s}\leq\frac{1}{\kappa }\frac{Q^d}{Q^s}
$$
and we can use Gronwall's inequality [2] (for ODEs/stochastic differential equations) to use our supply and demand systems of equations (8a,b) and (9a,b) as upper bounds on the
perceivedsupply and demand curves. These upper bounds are the ideal supply and demand curves that intersect at the ideal price $P^*$. By perceived supply and demand curve, we mean the supply and demand curves that have the observedprice $P$ as the equilibrium price. In the case of ideal information transfer, the ideal price is the observed price. However, in the case of non-ideal information transfer the observed price can occur anywhere in the area below the the ideal supply or demand curves.
We first look at the case where we have a constant supply source analgous to Eq. (9a,b). In the case of imperfect information transfer, the equations become (via Gronwall's inequality):$$
(12a) \space P\leq\frac{1}{\kappa }\frac{\left\langle Q^d\right\rangle }{Q_0^s}
$$
$$
(12b)\space \Delta Q^s\geq \kappa Q_0^s\log \left(\frac{\left\langle Q^d\right\rangle }{Q_{\text{ref}}^d}\right)
$$
This creates a situation where all allowed prices for a given demand curve fall in the red shaded area below the ideal supply curve defined by the equality in Eq. (12b) in the figure below and the observed price $P$ will fall below the "ideal price" $P^*$.
If we have imperfect information transfer, then all prices along the demand curve (gray) beneath the supply curve are valid. Specifically, the prices marked in dark blue are valid and they remain valid price solutions under downward shifts in the demand curve until it reaches the bounding supply curve and then will be forced to follow the supply curve to price $P'$ and beyond. This looks different from the standard economics perspective in the next figure below. The price $P$ appears to be sticky downward under small shifts of the demand curve, but could revert to ordinary (non-sticky) behavior for large shifts (compared to how far we are from ideal information transfer). The upper portion of the supply curve (here shown in dashed gray) is
-- but is technically unknown in this picture. assumed
The size of downward shifts in the demand curve for which prices remain sticky could
ceterus paribus(including keeping the ideal supply and demand curves constant) give an indication of the magnitude of the "information gap" between the observed imperfect information transfer and perfect information transfer.
Note: The dashed gray line is assumed in the economic narrative. The counterfactual of a rightward shift in the demand curve is never observed when prices are observed to be sticky downward with a leftward shift in the demand curve.
If we bound the information transfer from below, we see we can actually get prices that are sticky upward. See next figure below. In general, a point appearing inside the red shaded area will experience sticky prices for shifts upward and downward over some region of changes in demand, only to revert to ordinary (non-sticky) supply and demand behavior as shifts become large and we approach the boundaries of the shaded region.
One interesting point is that given the preponderance of sticky downward prices vs sticky upward prices would imply that our observed price (dark red point) tends to be nearer the lower bound. This makes intuitive sense as we could imagine what we know at the time of observing the price $P$ as being a lower bound on the information we have about the supply curve.
Again, we show how this figure looks in the standard economics narrative in the figure below. We have a price that is sticky upward and downward for small shifts in the demand curve, but becomes "unstuck" for larger shifts.
These results imply a large deviation from ideal information transfer for a given good will result in stickier prices. What causes the non-ideal information transfer? One could imagine cases where the good is difficult to assess (from lots of variables) or is rarely assessed (an inefficient market) -- both situations that come into play in the employment market (people are complex and they may have annual salary reviews but this doesn't mean they are being fully assessed in an open market).
The problem is that while the sticky price trajectory remains
with Equations (12a,b) -- it is a solution to the differential inequality -- it may not be consistent by Equations (12a,b). It is consistent for a pencil standing on its end to fall with the point facing west, but it is not required to fall with the point facing west. A potential direction would be to look at all possible trajectories using some kind of stochastic process and see if they converge to some value. required
There is the additional issue that the sticky price trajectory implies that the size of the "information gap" is decreasing assuming the ideal supply curve remains constant in order for the trajectory to reach the edge. But why should it reach the edge? Maybe the information gap is preserved in the short run so that the ideal supply curve moves to the left in tandem. Maybe in the long run, more efficient markets could cause the gap to shrink? I don't know, but again I think a stochastic model might shed some light here.
In particular, Markov information sources are frequently used in communication theory and Markov models are of interest in economics; the information transfer model may just create constraints ... which is my goal. The actual dynamics of how a price sticks and how the information gap will be potential subjects for future posts. As you can see, this is an incomplete picture and hence why I decided to go the blog/working paper route. I don't have all the answers!
In particular, Markov information sources are frequently used in communication theory and Markov models are of interest in economics; the information transfer model may just create constraints ... which is my goal. The actual dynamics of how a price sticks and how the information gap will be potential subjects for future posts. As you can see, this is an incomplete picture and hence why I decided to go the blog/working paper route. I don't have all the answers!
References
[1] Information transfer model of natural processes: from the ideal gas law to the distance dependent redshift P. Fielitz, G. Borchardt http://arxiv.org/abs/0905.0610v2
[2] http://en.wikipedia.org/wiki/Gronwall's_inequality
[3] http://en.wikipedia.org/wiki/Noisy_channel_coding_theorem#Mathematical_statement
[4] http://en.wikipedia.org/wiki/Entropic_force
[5] http://en.wikipedia.org/wiki/Sticky_(economics) |
Q. Let $X$ be a random variable with the probability density function
$ f_{X}(x) = \begin{cases} 1 &\text{ if} \quad 0 < x < 1 \\ 0 &\text{ otherwise} \end{cases} $
Let $Y$ be a random variable with the conditional probability density function
$ f_{Y|X}(y|x) = \begin{cases} 1/x &\text{ if} \quad 0 < y < x \\ 0 &\text{ otherwise} \end{cases} $
What is the marginal probability density function for $Y$?
I see the answer must be $f_{Y}(y) = 2(1 - y)$ for $0 < y < 1$. (Edit: see below.) However, I am having trouble deriving this result. My approach has been to apply the law of total probability in the form
\begin{align*} f_{Y}(y) &= \int\limits_{-\infty}^{+\infty} f_{Y|X}(y|x) f_{X}(x) \text{ d}x \\ &= \int\limits_{0}^{1} f_{Y|X}(y|x) \text{ d} x \\ &= \int\limits_{0}^{x} \frac{\text{d} x}{x} \end{align*}
But here I run into the difficulty that $x$ appears in the limit. Is there another way of going about this problem? There is likely something obvious I am missing. Thanks in advance for your help.
UPDATE. Starting with J.G.'s advice on the limits, I found the easier way to approach the problem is first to write the joint density
$ f_{X,Y}(x,y) = \begin{cases} 1/x &\quad \text{if } 0 < y < x < 1 \\ 0 &\quad \text{otherwise} \end{cases} $
Then the marginal density for $Y$ is \begin{align*} f_{Y}(y) &= \int\limits_{-\infty}^{+\infty} f_{X,Y}(x, y) \text{ d} x \\ &= \int\limits_{y}^{1} \frac{\text{ d} x}{x} \\ &= - \ln (y) \end{align*}
Pretty weird result. Turns out my earlier intuition that $f_{Y}(y) = 2 (1 - y)$ was wrong. |
I've got cumulative distribution function given: $F_X(t) = 0 $ for $t<0$, $\frac{1}{3} $ for $t=0$ , $\frac{1}{3} + \frac{t}{90} $ for $ t\in (0,60)$ and $1$ for $t \ge 60$. I am to find expexted value ($EX$). So, what to do? I wanted to find a density function, but I can't, because cumulative distribution function is not continuous. So what am I supposed to do here?
You do
not need to find a density function. If a random variable takes onpositive values only (or is positive with probability $1$), then itsexpected value is $$E[X] = \int_0^\infty [1-F_X(t)]\,\mathrm dt.$$
Since you are given $F_X(t)$, just compute the value of the above integral. In this instance, since $F(t)$ increases linearly from $\frac{1}{3}$ at $t=0$ to $1$ at $t=60$, $[1-F_X(t)]$ is decreasing linearly from $\frac{2}{3}$ at $t=0$ to $0$ at $t=60$, that is, you are finding the area of a (right) triangle which you can do without necessarily formally integrating.
Read more about this useful method in the answers to this question.
You can certainly write the density function for each interval of interest, and make sure you account for the normalization constant (such that your pdf integrates to 1). Then, use the definition of $E(X)$.
You can find a pdf. It's given by $$p_X(t) = \frac{1}{3}\delta(t) + \frac{1}{90},\quad 0<t<60$$ (zero otherwise). So $E\{X\}$ is given by
$$E\{X\} = \int_{0}^{60}t\left ( \frac{1}{3}\delta(t) + \frac{1}{90}\right )\;dt$$
So we get
$$E\{X\} = \frac{1}{3}\int_{0}^{60}t\delta(t)\;dt + \frac{1}{90}\int_{0}^{60}t\;dt = \frac{1}{90}\frac{60^2}{2} = 20$$
Another way of looking at it is using the Riemann-Stieltjes integral. In this case, we have
$$E[X] = \int_{-\infty}^\infty t\ dF_X(t).$$
The advantage here is that $F_X(t)$ satisfies all the conditions necessary for the Riemann-Stieltjes integral to exist: it is monotonic, and even though it has discontinuities, the function $f(t) = t$ is everywhere continuous.
By using properties of the Riemann-Stieljes integral, we can easily compute your solution simply by separating the integral into piecewise components corresponding to the breakpoints of your piecewise linear CDF. Since $F_X(t)$ is constant above $t=60$ and below $t=0$, it is continuous and we can simply note that $dF_X(t) = 0$ in these regions.
This gets you exactly to Matt L.'s solution, but maybe it seems a little less like magic. |
I have a list of data existing as {{{x, y, z}, f},...} from which I construct a 3D interpolation function using Interpolation[ ]. I am able to construct a vector field from the gradient of this scalar field, I have verified this looks correct by plotting it.
I wish to find the 'attractors' of this vector field.
By 'attractor' I mean the points at which a particle would end up at $t \rightarrow \infty$ by following the field. I.e. they flow down the gradient field to these points. I think attractor is the right name for this, please let me know if I am using the wrong word. I have also seen them referred to as $\omega$ limits.
I would like to use mathematica to find all of these points and here I begin to struggle. My initial attempt was to use:
interp = Table[ {{dat[[i, 1]], dat[[i, 2]], dat[[i, 3]]}, 1/(1 + dat[[i, 7]])}, {i, 1, Length[dat[[All, 1]]]}]intf = Interpolation[interp]intfd[x_,y_,z_] :=Evaluate[D[intf[x,y,z],{{x,y,z}}]]Table[ NMinimize[ {Norm[intfd[x,y,z]],-4<=x<=4,-4<=y<=4,-4<=z<=4}, {{x,-5,5},{y,-5,5},{z,-5,5}}, Method->{"NelderMead","RandomSeed"->i} ], {i,1,20}]
To try to find the points of 0 gradient. I'm aware this will also find the $\alpha$ limits (or sources for the field, where the particle is at $t \rightarrow -\infty$) and I'm not even 100% sure that Norm[V[x,y,z]] == 0 is a definition for one of these limits.
Also this is a crappy stochastic method for finding as many as I can. Though I can intuit where the points should be ahead of time to verify I have found them all I would like a general solution.
Can anyone suggest a better method for finding the attractors of the field? |
T Singh
Articles written in Pramana – Journal of Physics
Volume 63 Issue 5 November 2004 pp 937-945
We have studied five-dimensional homogeneous cosmological models with variable
Volume 68 Issue 5 May 2007 pp 721-734 Research Articles
We have studied the evolution of a homogeneous, anisotropic universe given by a Bianchi Type-V cosmological model filled with viscous fluid, in the presence of cosmological constant 𝛬. The role of viscous fluid and 𝛬-term in the Bianchi Type-V universe has been studied.
Volume 69 Issue 2 August 2007 pp 159-166 Research Articles
In modified generalized scalar–tensor (GST) theory, the cosmological term 𝛬 is a function of the scalar field 𝜙 and its derivatives $\dot{\phi}^{2}$. We obtain exact solutions of the field equations in Bianchi Type-I, V and VIo space–times. The evolution of the scale factor, the scalar field and the cosmological term has been discussed. The Bianchi Type-I model has been discussed in detail. Further, Bianchi Type-V and VIo models can be studied on the lines similar to Bianchi Type-I model.
Volume 71 Issue 3 September 2008 pp 447-458 Research Articles
The Bianchi Type-I Universe filled with dark energy from a wet dark fluid has been considered. A new equation of state for the dark energy component of the Universe has been used. It is modeled on the equation of state $p = \gamma(\rho − \rho_{*})$ which can describe a liquid, for example water. The exact solutions to the corresponding field equations are obtained in quadrature form. The solution for constant deceleration parameter have been studied in detail for both power-law and exponential forms. The cases $\gamma = 1$ and $\gamma = 0$ have also been analysed.
Current Issue
Volume 93 | Issue 5 November 2019
Click here for Editorial Note on CAP Mode |
I came across a problem thatwhile doing some review that states:
Consider the transformation $\textit{T}$:$\mathbb{R}^2\rightarrow\mathbb{R}^2$ defined by the matrix $$ \begin{pmatrix} 2 & 0 \\ 1 & 3 \\ \end{pmatrix} $$ Prove that the transformation is linear, state the kernel of T and find the image of the vector (1, 2) under this transformation. For showing the linear transformation, I wrote in a more familiar fashion: T($x_1$, $x_2$) = (2$x_1$, $x_1$+$3x_2$). However, I am not stuck as to how to define the vectors/matrices for showing closure under addition and scalar multiplication. Any advice/solutions?
I came across a problem thatwhile doing some review that states:
Let $u,v \in \mathbb R^2$ and $\alpha \in \mathbb R$. If we prove that $T(\alpha u + v) = \alpha T(u)+T(v)$ then $T$ its a linear transformation.
Let $u = (u_1, u_2)$ and $v = (v_1,v_2)$ for any $u_1,u_2,v_1, v_2, \alpha \in \mathbb R$ then: $$\alpha u + v = \alpha(u_1,u_2)+(v_1,v_2)=(\alpha u_1+v_1,\alpha u_2 + v_2)$$
We calculate $T(\alpha u + v)$: $$T(\alpha u + v)=T(\alpha u_1+v_1,\alpha u_2 + v_2)$$
$$=(2(\alpha u_1+v_1),(\alpha u_1+v_1)+3(\alpha u_2 + v_2))$$
$$=(2 \alpha u_1 + 2v_1,(\alpha u_1+3 \alpha u_2)+(v_1+3v_2))$$
$$=(2\alpha u_1,\alpha u_1+3 \alpha u_2)+(2v_1,v_1+3v_2)$$
$$=\alpha (2u_1, u_1+3u_2)+(2v_1,v_1+3v_2)=\alpha T(u)+T(v)$$
So $T$ is a linear transformation.
Forthe kernel of $T$, you only need to do Gauss elimination on the matrix and then find the free variables. In this case after doing Gauss elimination we find that
$$\begin{pmatrix} 1 & 3 \\ 0 & -6 \\ \end{pmatrix}$$
This means that $ker(T) = \{0 \}$
You can calculate the image of $(1,2)$ directly simply put $T(1,2)$ or make the matrix multiplication: $$\begin{pmatrix} 1 & 3 \\ 0 & -6 \\ \end{pmatrix} \begin{pmatrix} 1\\ 2 \end{pmatrix}$$
I hope it helps. |
I don't know if this is research level or math se level so I try posting here first.
The problem is to prove that the left hand side equals the right hand side for $t>0$ when leaving out singularities such as the points $t$ equal to the imaginary part of a Riemann zeta zero:
$$\left\lfloor \frac{t \log \left(\frac{t}{2 e \pi }\right)}{2 \pi }+\frac{7}{8}\right\rfloor +\frac{1}{2} \left(-1+\text{sgn}\left(\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)\right)=\frac{\vartheta (t)+\Im\left(\log \left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)}{\pi }$$
Can you prove that the left hand side essentially equals the right hand side for $t>0$ when not considering the singularities?
$\text{sgn}$ is the sign function and the floor function is the down square brackets. $\zeta(s)$ is the Riemann zeta function and $\vartheta(t)$ is the Riemann Siegel theta function.
Two plots to verify the similarity between the left hand side and the right hand side:
(*Mathematica 8*)Plot[(Sign[Im[Zeta[1/2 + I*t]]] - 1)/2 + Floor[t/(2*Pi)*Log[t/(2*Pi*Exp[1])] + 7/8], {t, 0, 60}, Filling -> 0]
Left hand side:
(*Mathematica 8*) Plot[(RiemannSiegelTheta[t] + Im[Log[Zeta[1/2 + I*t]]])/Pi, {t, 0, 60}, Filling -> 0]
Right hand side:
Related: https://oeis.org/A135297 |
Let $\pi:X\to X/\!\!\!\sim\;$ denote the projection map associated with $\sim$. (That is, for any $x\in X$, $\pi(x)$ is the $\sim$-equivalence class that $x$ belongs to.) Let $\nsim\; \subseteq X \times X$ be shorthand for the complement of $\;\;\sim\;\;$ in $X \times X\;$, i.e. $\nsim\;\;=\;(X \times X\;) \;\;-\; \sim\;$.
Suppose that $\pi(x) \neq \pi(y)\;$. (Here I'm relying on the fact that, since $\pi$ is surjective, any element $\widetilde{z}\in X/\!\!\!\sim\;$ may be written in the form $\pi(z)$, for some $z \in X$.) We must show that there exist open sets $U_{\pi(x)}, U_{\pi(y)} \subseteq X/\!\!\!\sim\;$ such that ${\pi(x)} \in U_{\pi(x)}$, ${\pi(y)} \in U_{\pi(y)}$, and $U_{\pi(x)} \cap U_{\pi(y)} = \varnothing\;$.
By assumption, $\;\sim\; \subseteq X \times X$ is closed, so $\nsim\; \subseteq X \times X$ is open. Therefore there exist open neighborhoods $N_x$ and $N_y$ of $x$ and $y$, respectively, such that $(x,\;y)\in N_x \times N_y \subseteq$$\;\;\nsim\;$. (This is because the family of all pairwise products of open subsets of $X$ is a basis for the product topology on $X \times X$.)
For any $v, w \in X$,
$$
(v,\;w) \;\in \;\nsim \;\;\;\;\;\Leftrightarrow\;\;\;\;\;
\pi(v) \neq \pi(w)\;\;.
$$
Therefore,
$$
N_x \times N_y \subseteq \;\;\nsim\;\;\;\;\Leftrightarrow\;\;\;\;
\forall (v, w) \in N_x \times N_y \;[\pi(v) \neq \pi(w)] \;\;\;\;\Leftrightarrow\;\;\;\;
\pi[N_x] \cap \pi[N_y] = \varnothing
$$
Furthermore, since $\pi$ is open (by assumption), the image sets $\pi[N_x], \pi[N_y] \subseteq X/\!\!\!\sim\;$ are open neighborhoods of ${\pi(x)}$ and ${\pi(y)}$, respectively. Therefore, $\pi[N_x]$ and $\pi[N_y]$ are the desired open neighborhoods $U_{\pi(x)} \ni {\pi(x)}, U_{\pi(y)} \ni {\pi(y)}$. |
Here, I mostly referred to the Cobb-Douglas production function piece, not the piece of the Solow model responsible for creating the equilibrium level of capital. That part is relatively straight-forward. Here we go ...
Let's assume two additional information equilibrium relationships with capital $K$ being the information source and investment $I$ and depreciation $D$ (include population growth in here if you'd like) being information destinations. In the notation I've been using: $K \rightarrow I$ and $K \rightarrow D$.
This immediately leads to the solutions of the differential equations:
\frac{K}{K_{0}} = \left( \frac{D}{D_{0}}\right)^{\delta}
$$
$$
\frac{K}{K_{0}} = \left( \frac{I}{I_{0}}\right)^{\sigma}
$$
Therefore we have (the first relationship coming from the Cobb-Douglas production function)
Y \sim K^{\alpha} \text{ , }\;\;\;\; I \sim K^{1/\sigma} \text{ and }\;\;\;\; D \sim K^{1/\delta}
$$
If $\sigma = 1/\alpha$ and $\delta = 1$ we recover the original Solow model, but in general $\sigma > \delta$ allows there to be an equilibrium. Here is a generic plot:
Assuming the relationships $K \rightarrow I$ and $K \rightarrow D$ hold simultaneously gives us the equilibrium value of $K = K^{*}$:
K^{*} = K_{0} \exp \left( \frac{\sigma \delta \log I_{0}/D_{0}}{\sigma - \delta} \right)
$$
As a side note, I left the small $K$ region off on purpose. The information equilibrium model is not valid for small values of $K$ (or any variable). That allows one to choose parameters for investment and depreciation that could be e.g. greater than output for small $K$ -- a nonsense result in the Solow model, but just an invalid region of the model in the information equilibrium framework.
An interesting add-on is that $Y$ and $I$ have a supply and demand relationship in partial equilibrium with capital being demand and investment being supply (since $Y \rightarrow K$, by transitivity they are in information equilibrium). If $s$ is the savings rate (the price in the market $Y \rightarrow I = Y \rightarrow K \rightarrow I$), we should be able to work out how it changes depending on shocks to demand. There should be a direct connection to the IS-LM model as well. |
Simulating a Classical Computer Using a Quantum Computer An easier way to think of quantum computers
I was very fortunate my last quarter as a graduate student at UCSD to have helped organize a quantum computing seminar. I was strongly influenced by Aaronson’s excellent book Quantum Computing Since Democritus where Aaronson makes a strong point that Quantum Computing is easy if you abstract away the physics.
That may sound controversial but it really isn’t.
Do algorithmists think of the electromagnetic waves storing bits on disk when they write pseudocode?
Theorists are generally unaware of how computers work and more applied computer scientists are similarly unaware of how electronics work and justly so! Abstraction is powerful and allows one to focus on problems that exist at different conceptual levels. Why should quantum computing be any different? The approach that makes the most sense is to take it as a given that physicists will provide us with a basic set of quantum logic gates whether they are spinning or freezing or backflipping elementary particles in various gases is irrelevant to us. We can make more progress if we work with quantum logic gates as opposed to quantum physics.
What is a Quantum Computer?
I’ll assume that my reader has no knowledge of quantum computing and attempt to draw an analogy with a more well known device: the classical computer.
Boolean logic fugue
At the most abstract level, a computer is a function \(f \) of \(n \) bits that returns either 0 or 1.
\(f(b_1, \dots , b_n) = {0,1 } \)
Any binary function on \(n \) bits can be computed using combinations of the well known “AND”, “OR” and “NOT” gates or just a “NAND”. The fantastic book The Elements of Computing Systems by Nisan and Schocken shows how it is possible to start with a “NAND” gate and move on to assemblers, compilers, virtual machines and finally tetris.
I leave it as an exercise to the reader to see how a “NAND” gate can be implemented using the basic “AND”, “OR” and “NOT” gates. The hardware implementation of “AND”, “OR” gates are easy if one remembers the distinction between in-series and in-parallel circuits.
If you’ve ever wondered why binary is the base of choice for computers the above is a strong argument for that choice, the other being that it is simple to represent one of two states 0 or 1 using a ground or 5V.
Back to Quantum Computers
A quantum computer is then a function \(f \) of \(n \) quantum bits that when measured (think of this as returning for now) returns either 0 or 1.
Quantum Computers are at least as powerful as Classical Computers
This is a neat result that I proved in the introductory lecture of the seminar, of course this result is not novel in any way: I’m not sure if it was even formally stated in a paper or just taken as a known trivial result but regardless it’s a good exercise to prove your first result about quantum computers.
Proving that quantum computers are at least as powerful as classical computers using truth tables
Let’s take a look at the truth table of a NAND gate where the first two columns represent the two inputs and the last colum presents the output.
IN1 IN2 OUT 0 0 1 0 1 1 1 0 1 1 1 0
Now let’s take a look at the truth table for a basic quantum gate the Toffoli gate. A Toffoli gate takes in three inputs and produces three outputs.
IN1 IN2 IN3 OUT1 OUT2 OUT3 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 1 1 1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 0
Now inspect the third output in the rows for which the third input is 1. Do those rows look familiar? They should! Because they correspond exactly to the NAND gate! So we can fix the third input bit of a Toffoli gate to 1 to simulate a NAND gate. Therefore we can simulate any NAND circuit or any boolean circuit using Toffoli gates.
Are computers more powerful if they are allowed to make coin flips?
The short answer is nobody knows for sure but it is hypothesized that it indeed is. This seems counter-intuitive, nobody would expect that two-face could make superior decisions by flipping a coin so why would the same not be true for computers.
Polynomial identity testing: Case study for randomized algorithms
Suppose you were given two large polynomials \(P_1 \) and \(P_2 \) with many variables each. A natural question is: is \(P_1 \equiv P_2\)? There is no known deterministic algorithm for it but there is an incredibly simple randomized one!
While you are not sure if Pick a point uniformily at random \(x \in \) Domain Evaluate \(P_1(x) - P_2(x) \) If \(P_1(x) - P_2(x) \) == 0 continue Else Return: \(P_1 ! \equiv P_2 \) If \(P_1(x) - P_2(x) \) == 0
Why does this work? Well, if for any \(x\), \(P_1(x) != P_2(x)\) then they are not equivalent. In addition, the more times we iterate over the loop the more confident we are that the two polynomials are equivalent!
Simulating a coin flip using a quantum computer
First we’re going to have to define what a quantum bit is. We represent an arbitrary quantum bit \(a\) as \( | a > = \alpha | 0 > + \beta | 1 > \)
The above definition is read as a quantum bit \(a\) is in state 0 with probability \(\alpha^2\) and in state 1 with probability \(\beta^2\). When people talk about how a quantum bit can be both 1 and 0 at the same time they are refering to the above expression yet they are incorrectly interpreting it. There is nothing weird, magical or massively parallel about a distribution over two variables.
To continue, we note that that any logical operation can be represented algebraically, for instance: \(A \wedge B = A . B \).
We will use, the basic quantum Hadamard Gate. \(H = \frac{1}{\sqrt{2}} \begin{bmatrix}1 & 1 ; 1 & -1 \end{bmatrix} \). We then apply the Hadamard gate on a \(|0>\) bit:
\(H |0> = \frac{1}{\sqrt{2}}|0> + \frac{1}{\sqrt{2}}|1> \) which when measured gives us either 0 or 1 with probability 1/2.
In other words using a Hadamard gate allows us to simulate a fair coin
This doesn’t seem like a lot but it is quite significant! Think of how you’d simulate a uniform distribution over \(n \) variables using a uniform distribution over 2 variables. Also related, I hear simulating a fair coin using an unbiased coin is a popular interview question. Another natural question to think about is then how would you simulate a biased coin using an unbiased one?
Epilogue
Good news is that quantum computers could probably run your favorite videogame. This is however not the most interesting property of quantum computers, if there is any interest then I can probably have some followup posts on Grover’s search algorithm and Shor’s factoring algorithm. Although those are arguably the most well known quantum algorithms there are many results that are incredibly funky and I’d recommend you check out the topics we had in the theory reading group at UCSD.
The two big open questions are perhaps: can quantum computers actually be built? If they can, can they solve more problems than classical computers? The complexity theory community seems to think so. As Aaronson would say: all possibilities are equally intriguing since either we learn some pretty fundamental limitations of phyics or we get swanky new quantum computers. |
An RLC circuit is a simple electric circuit with a resistor, inductor and capacitor in it -- with resistance
R, inductance Land capacitance C, respectively. It's one of the simplest circuits that displays non-trivial behavior.
You can derive an equation for the behavior by using Kirchhoff's laws (conservation of the stocks and flows of electrons) and the properties of the circuit elements. Wikipedia does a fine job.
You arrive at a solution for the current as a function of time that looks generically like this (not the most general solution, but a solution):
i(t) = A e^{\left( -\alpha + \sqrt{\alpha^{2} - \omega^{2}} \right) t}
$$
with $\alpha = R/2L$ and $\omega = 1/\sqrt{L C}$. If you fill in some numbers for these parameters, you can get all kinds of behavior:
As you can tell from that diagram, the Kirchhoff conservation laws don't in any way nail down the behavior of the circuit. The values you choose for
R, Land Cdo. You could have a slowly decaying current or a quickly oscillating one. It depends on R, Land C.
Now you may wonder why I am talking about this on an economics blog. Well, Cullen Roche implicitly asked a question:
Although [stock flow consistent models are] widely used in the Fed and on Wall Street it hasn’t made much impact on more mainstream academic economic modeling techniques for reasons I don’t fully know.
The reason is that the content of stock flow consistent modeling is identical to Kirchhoff's laws. Currents are flows of electrons (flows of money); voltages are stocks of electrons (stocks of money).
Krichhoff's laws do not in any way nail down the behavior of an RLC circuit.
SFC models do not nail down the behavior of the economy.
If you asked what the impact of some policy was and I gave you the graph above, you'd probably never ask again.
What SFC models do in order to hide the fact that anything could result from an SFC model is effectively assume
R = L = C = 1, which gives you this:
I'm sure to get objections to this. There might even be legitimate objections. But I ask of any would-be objector:
How is accounting for money different from accounting for electrons?
Before saying this circuit model is in continuous time, note that there are circuits with clock cycles -- in particular the device you are currently reading this post with.
I can't for the life of me think of any objection, and I showed exactly this problem with a SFC model from Godley and Lavoie:
But to answer Cullen's implicit question -- as the two
Nick Rowe is generally better than me at these things. Mathematicanotebooks above show, SFC models don't specify the behavior of an economy without assuming R = L = C = 1 ... that is to say Γ = 1. Update:
Nick Rowe is generally better than me at these things. |
I'm solving the 3D Diffusion equation $$u_t=k(u_{xx}+u_{yy}+u_{zz})$$
in MATLAB using Fourier techniques. I assume a 3D Fourier expansion $(e^{-ipx},e^{-imy},e^{-imz})$of the solution.
Physical space: $u(x,y,z,t)$. Fourier Space: $c(m,n,p,t)$.
Substitution and differentiation result in: $$c(m,n,p)^{N+1}-c(m,n,p)^{N} = -\frac{k\Delta T}{2} (p^2+m^2+n^2)(c(m,n,p)^{N+1}+c(m,n,p)^{N})$$ after applying a Crank-Nicolson scheme.
I'm using
fftn() and
ifftn() to forward my coefficients in time and bring them back to physical space. However I achieve universal decay from the initial condition to zero with no heat flux in any direction, for all time. Typical time step:
0.00001. Typical
k=0.005.
Is the problem with my application of
fftn() or the stability of my finite-difference?
Edit: I've taken an initial condition of $50 \sin(2x)$. I merely took the boundaries of that initial condition and imposed them as the boundary conditions for all time. No good I'm guessing? Edit 2: I forgot the wavenumber squared on the rhs. Thanks James! |
doi: 10.1685/journal.caim.531
Stochastic processes related to time-fractional diffusion-wave equation Abstract
It is known that the solution to the Cauchy problem:
$$ D^\beta_* u(x,t)= R^\alpha u(x,t) \,, \quad u(x,0)=\delta(x) \,, \quad \frac{\partial}{\partial x}u(x,t=0) \equiv 0 \,, \quad -\infty < x < \infty \,, \quad t > 0 \,, $$ is a probability density if $$ 1 < \beta \le \alpha \le 2 $$ where $$ D^\beta_*$$ is the time fractional Caputo derivative of order \beta whereas $$R^\alpha$$ denotes the spatial Riesz fractional pseudo-differential operator. In the present paper it is considered the question if u(x,t) can be interpreted in a natural way as the sojourn probability density (in point x, evolving in time t) of a randomly wandering particle starting in the origin x=0 at instant t=0. We show that this indeed can be done in the extreme case \alpha=2, that is $$R^\alpha=\displaystyle{\frac{\partial^2}{\partial x^2}}$$ Moreover, if \alpha=2 we can replace $$D^\beta_*$$ by an operator of distributed orders with a non-negative (generalized) weight function b(\beta): $$ \displaystyle{\int_{(1,2]} \!\!\! b(\beta) \, D^\beta_* \dots d\beta} $$ For this case u(x,t) is a probability density.
$$
D^\beta_* u(x,t)= R^\alpha u(x,t) \,, \quad u(x,0)=\delta(x) \,,
\quad \frac{\partial}{\partial x}u(x,t=0) \equiv 0 \,,
\quad -\infty < x < \infty \,, \quad t > 0 \,,
$$
is a probability density if
$$ 1 < \beta \le \alpha \le 2 $$
where
$$ D^\beta_*$$
is the time fractional Caputo derivative of order \beta whereas $$R^\alpha$$ denotes the spatial Riesz fractional pseudo-differential operator.
In the present paper it is considered the question if u(x,t) can be interpreted in a natural way as the sojourn probability density (in point x, evolving in time t) of a randomly wandering particle starting in the origin x=0 at instant t=0. We show that this indeed can be done in the extreme case \alpha=2, that is $$R^\alpha=\displaystyle{\frac{\partial^2}{\partial x^2}}$$ Moreover, if \alpha=2 we can replace $$D^\beta_*$$ by an operator of distributed orders with a non-negative (generalized) weight function b(\beta):
$$
\displaystyle{\int_{(1,2]} \!\!\! b(\beta) \, D^\beta_* \dots d\beta}
$$
For this case u(x,t) is a probability density.
Refbacks There are currently no refbacks. |
Given that $n>2$. Prove that if $2^n-1$ is prime then $2^n+1$ is composite or vice versa. I looked on wikipedia on Fermat number and Mersenne prime, but I still don't know how they work.
Hint $\ a\mid (a\!-\!1)^n\!\pm 1\ $ since $ $ mod $\ a\!:\ (a\!-\!1)^n \equiv (-1)^n\equiv \pm 1$
e.g. $\,\ 10\mid 9^n\pm1\, $ is a well-known case (power's of $9$ end with digit $1$ or $9)$
Your case can be viewed as the analogy in radix $\,3.$
Clearly you can't prove that $2^n-1$ composite implies $2^n+1$ is prime. Take $n=6$, for example. Going the other way, think about divisibility by $3$.
At least one of $2^n-1$ and $2^n+1$ must be divisible by $3$, and neither of them is $3$. Thus, at least one of them is composite. |
Difference between revisions of "Droop Control"
Line 3: Line 3:
== Background ==
== Background ==
−
Recall that the [[AC_Power_Transmission#Lossless_Line_.28Classical_Approach.29|active and reactive power transmitted across a lossless line]] are:
+ + + + + + +
Recall that the [[AC_Power_Transmission#Lossless_Line_.28Classical_Approach.29|active and reactive power transmitted across a lossless line]] are:
: <math>P = \frac{V_{1} V_{2}}{X} \sin\delta </math>
: <math>P = \frac{V_{1} V_{2}}{X} \sin\delta </math>
Line 17: Line 23:
From the above, we can see that active power has a large influence on the power angle and reactive power has a large influence on the voltage difference. Restated, by controlling active and reactive power, we can also control the power angle and voltage. We also know from the [[Swing Equation|swing equation]] that frequency in synchronous power systems is related to the power angle, so by controlling active power, we can therefore control frequency.
From the above, we can see that active power has a large influence on the power angle and reactive power has a large influence on the voltage difference. Restated, by controlling active and reactive power, we can also control the power angle and voltage. We also know from the [[Swing Equation|swing equation]] that frequency in synchronous power systems is related to the power angle, so by controlling active power, we can therefore control frequency.
− + + + + +
forms the basis of frequency and voltage droop control where active and reactive power are adjusted according to linear characteristics, based on the following control equations:
: <math>f = f_{0} - r_{p} (P - P_{0}) \, </math> ... Eq. 1
: <math>f = f_{0} - r_{p} (P - P_{0}) \, </math> ... Eq. 1
Line 45: Line 55:
The same logic above can be applied to the voltage droop characteristic.
The same logic above can be applied to the voltage droop characteristic.
−
== Alternative Droop Equations ==
+
== Alternative Droop Equations ==
The basic per-unit droop equations in Eq. 1 and Eq. 2 above can be expressed in natural quantities and in terms of deviations as follows:
The basic per-unit droop equations in Eq. 1 and Eq. 2 above can be expressed in natural quantities and in terms of deviations as follows:
Revision as of 05:15, 12 August 2018
Droop control is a control strategy commonly applied to generators for primary frequency control (and occasionally voltaqe control) to allow parallel generator operation (e.g. load sharing).
Contents Background Physical Intuition
TBA
Generic Formulation
A more generic formulation of the droop control concept stems from the coupling between active power and frequency, and similarly reactive power and voltage. Recall that the active and reactive power transmitted across a lossless line are:
[math]P = \frac{V_{1} V_{2}}{X} \sin\delta [/math] [math]Q = \frac{V_{2}}{X} (V_{2} - V_{1} \cos\delta) [/math]
Since the power angle [math]\delta \,[/math] is typically small, we can simplify this further by using the approximations [math]\sin\delta \approx \delta \,[/math] and [math]\cos\delta \approx 1 \,[/math]:
[math]\delta \approx \frac{PX}{V_{1} V_{2}} [/math] [math](V_{2} - V_{1}) \approx \frac{QX}{V_{2}} [/math]
From the above, we can see that active power has a large influence on the power angle and reactive power has a large influence on the voltage difference. Restated, by controlling active and reactive power, we can also control the power angle and voltage. We also know from the swing equation that frequency in synchronous power systems is related to the power angle, so by controlling active power, we can therefore control frequency.
Droop Equations Per-Unit Droop Equations
The coupling of active power to frequency and reactive power to voltage forms the basis of frequency and voltage droop control where active and reactive power are adjusted according to linear characteristics, based on the following control equations:
[math]f = f_{0} - r_{p} (P - P_{0}) \, [/math] ... Eq. 1 [math]V = V_{0} - r_{q} (Q - Q_{0}) \, [/math] ... Eq. 2
where [math]f \, [/math] is the system frequency (in per unit)
[math]f_{0} \, [/math] is the base frequency (in per unit) [math]r_{p} \, [/math] is the frequency droop control setting (in per unit) [math]P \, [/math] is the active power of the unit (in per unit) [math]P_{0} \, [/math] is the base active power of the unit (in per unit) [math]V \, [/math] is the voltage at the measurement location (in per unit) [math]V_{0} \, [/math] is the base voltage (in per unit) [math]Q \, [/math] is the reactive power of the unit (in per unit) [math]Q_{0} \, [/math] is the base reactive power of the unit (in per unit) [math]r_{q} \, [/math] is the voltage droop control setting (in per unit)
These two equations are plotted in the characteristics below:
The frequency droop characteristic above can be interpreted as follows: when frequency falls from [math]f_{0}[/math] to [math]f[/math], the power output of the generating unit is allowed to increase from [math]P_{0}[/math] to [math]P[/math]. A falling frequency indicates an increase in loading and a requirement for more active power. Multiple parallel units with the same droop characteristic can respond to the fall in frequency by increasing their active power outputs simultaneously. The increase in active power output will counteract the reduction in frequency and the units will settle at active power outputs and frequency at a steady-state point on the droop characteristic. The droop characteristic therefore allows multiple units to share load without the units fighting each other to control the load (called "hunting").
The same logic above can be applied to the voltage droop characteristic.
Alternative Droop Equations
The basic per-unit droop equations in Eq. 1 and Eq. 2 above can be expressed in natural quantities and in terms of deviations as follows:
[math]r_{p} = \frac{\Delta f}{\Delta P} \times \frac{P_{n}}{f_{n}} [/math] [math]r_{q} = \frac{\Delta V}{\Delta Q} \times \frac{Q_{n}}{V_{n}} [/math]
where [math]\Delta f \, [/math] is the frequency deviation (in Hz)
[math]f_{n} \, [/math] is the nominal frequency (in Hz), e.g. 50 or 60 Hz [math]\Delta P \, [/math] is the active power deviation (in kW or MW) [math]P_{n} \, [/math] is the rated active power of the unit (in kW or MW) [math]r_{p} \, [/math] is the frequency droop control setting (in per unit) [math]\Delta V \, [/math] is the voltage deviation at the measurement location (in V) [math]V_{n} \, [/math] is the nominal voltage (in V) [math]\Delta Q \, [/math] is the reactive power deviation (in kVAr or MVAr) [math]Q_{n} \, [/math] is the rated reactive power of the unit (in kVAr or MVAr) [math]r_{q} \, [/math] is the voltage droop control setting (in per unit) Droop Control Setpoints
Droop settings are normally quoted in % droop. The setting indicates the percentage amount the measured quantity must change to cause a 100% change in the controlled quantity. For example, a 5% frequency droop setting means that for a 5% change in frequency, the unit's power output changes by 100%. This means that if the frequency falls by 1%, the unit with a 5% droop setting will increase its power output by 20%.
The short video below shows some examples of frequency (speed) droop:
Limitations of Droop Control
Frequency droop control is useful for allowing multiple generating units to automatically change their power outputs based on dynamically changing loads. However, consider what happens when there is a significant contingency such as the loss of a large generating unit. If the system remains stable, all the other units would pick up the slack, but the droop characteristic allows the frequency to settle at a steady-state value
below its nominal value (for example, 49.7Hz or 59.7Hz). Conversely, if a large load is tripped, then the frequency will settle at a steady-state value above its nominal value (for example, 50.5Hz or 60.5Hz).
Other controllers are therefore necessary to bring the frequency back to its nominal value (i.e. 50Hz or 60hz), which are called secondary and tertiary frequency controllers. |
2018-09-11 04:29
Proprieties of FBK UFSDs after neutron and proton irradiation up to $6*10^{15}$ neq/cm$^2$ / Mazza, S.M. (UC, Santa Cruz, Inst. Part. Phys.) ; Estrada, E. (UC, Santa Cruz, Inst. Part. Phys.) ; Galloway, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; Gee, C. (UC, Santa Cruz, Inst. Part. Phys.) ; Goto, A. (UC, Santa Cruz, Inst. Part. Phys.) ; Luce, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; McKinney-Martinez, F. (UC, Santa Cruz, Inst. Part. Phys.) ; Rodriguez, R. (UC, Santa Cruz, Inst. Part. Phys.) ; Sadrozinski, H.F.-W. (UC, Santa Cruz, Inst. Part. Phys.) ; Seiden, A. (UC, Santa Cruz, Inst. Part. Phys.) et al. The properties of 60-{\mu}m thick Ultra-Fast Silicon Detectors (UFSD) detectors manufactured by Fondazione Bruno Kessler (FBK), Trento (Italy) were tested before and after irradiation with minimum ionizing particles (MIPs) from a 90Sr \b{eta}-source . [...] arXiv:1804.05449. - 13 p. Preprint - Full text Notice détaillée - Notices similaires 2018-08-25 06:58
Charge-collection efficiency of heavily irradiated silicon diodes operated with an increased free-carrier concentration and under forward bias / Mandić, I (Ljubljana U. ; Stefan Inst., Ljubljana) ; Cindro, V (Ljubljana U. ; Stefan Inst., Ljubljana) ; Kramberger, G (Ljubljana U. ; Stefan Inst., Ljubljana) ; Mikuž, M (Ljubljana U. ; Stefan Inst., Ljubljana) ; Zavrtanik, M (Ljubljana U. ; Stefan Inst., Ljubljana) The charge-collection efficiency of Si pad diodes irradiated with neutrons up to $8 \times 10^{15} \ \rm{n} \ cm^{-2}$ was measured using a $^{90}$Sr source at temperatures from -180 to -30°C. The measurements were made with diodes under forward and reverse bias. [...] 2004 - 12 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 533 (2004) 442-453 Notice détaillée - Notices similaires 2018-08-23 11:31 Notice détaillée - Notices similaires 2018-08-23 11:31
Effect of electron injection on defect reactions in irradiated silicon containing boron, carbon, and oxygen / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Yakushevich, H S (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) Comparative studies employing Deep Level Transient Spectroscopy and C-V measurements have been performed on recombination-enhanced reactions between defects of interstitial type in boron doped silicon diodes irradiated with alpha-particles. It has been shown that self-interstitial related defects which are immobile even at room temperatures can be activated by very low forward currents at liquid nitrogen temperatures. [...] 2018 - 7 p. - Published in : J. Appl. Phys. 123 (2018) 161576 Notice détaillée - Notices similaires 2018-08-23 11:31 Notice détaillée - Notices similaires 2018-08-23 11:31
Characterization of magnetic Czochralski silicon radiation detectors / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) Silicon wafers grown by the Magnetic Czochralski (MCZ) method have been processed in form of pad diodes at Instituto de Microelectrònica de Barcelona (IMB-CNM) facilities. The n-type MCZ wafers were manufactured by Okmetic OYJ and they have a nominal resistivity of $1 \rm{k} \Omega cm$. [...] 2005 - 9 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 548 (2005) 355-363 Notice détaillée - Notices similaires 2018-08-23 11:31
Silicon detectors: From radiation hard devices operating beyond LHC conditions to characterization of primary fourfold coordinated vacancy defects / Lazanu, I (Bucharest U.) ; Lazanu, S (Bucharest, Nat. Inst. Mat. Sci.) The physics potential at future hadron colliders as LHC and its upgrades in energy and luminosity Super-LHC and Very-LHC respectively, as well as the requirements for detectors in the conditions of possible scenarios for radiation environments are discussed in this contribution.Silicon detectors will be used extensively in experiments at these new facilities where they will be exposed to high fluences of fast hadrons. The principal obstacle to long-time operation arises from bulk displacement damage in silicon, which acts as an irreversible process in the in the material and conduces to the increase of the leakage current of the detector, decreases the satisfactory Signal/Noise ratio, and increases the effective carrier concentration. [...] 2005 - 9 p. - Published in : Rom. Rep. Phys.: 57 (2005) , no. 3, pp. 342-348 External link: RORPE Notice détaillée - Notices similaires 2018-08-22 06:27
Numerical simulation of radiation damage effects in p-type and n-type FZ silicon detectors / Petasecca, M (Perugia U. ; INFN, Perugia) ; Moscatelli, F (Perugia U. ; INFN, Perugia ; IMM, Bologna) ; Passeri, D (Perugia U. ; INFN, Perugia) ; Pignatel, G U (Perugia U. ; INFN, Perugia) In the framework of the CERN-RD50 Collaboration, the adoption of p-type substrates has been proposed as a suitable mean to improve the radiation hardness of silicon detectors up to fluencies of $1 \times 10^{16} \rm{n}/cm^2$. In this work two numerical simulation models will be presented for p-type and n-type silicon detectors, respectively. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 2971-2976 Notice détaillée - Notices similaires 2018-08-22 06:27
Technology development of p-type microstrip detectors with radiation hard p-spray isolation / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) ; Díez, S (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) A technology for the fabrication of p-type microstrip silicon radiation detectors using p-spray implant isolation has been developed at CNM-IMB. The p-spray isolation has been optimized in order to withstand a gamma irradiation dose up to 50 Mrad (Si), which represents the ionization radiation dose expected in the middle region of the SCT-Atlas detector of the future Super-LHC during 10 years of operation. [...] 2006 - 6 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 566 (2006) 360-365 Notice détaillée - Notices similaires 2018-08-22 06:27
Defect characterization in silicon particle detectors irradiated with Li ions / Scaringella, M (INFN, Florence ; U. Florence (main)) ; Menichelli, D (INFN, Florence ; U. Florence (main)) ; Candelori, A (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Bruzzi, M (INFN, Florence ; U. Florence (main)) High Energy Physics experiments at future very high luminosity colliders will require ultra radiation-hard silicon detectors that can withstand fast hadron fluences up to $10^{16}$ cm$^{-2}$. In order to test the detectors radiation hardness in this fluence range, long irradiation times are required at the currently available proton irradiation facilities. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 589-594 Notice détaillée - Notices similaires |
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June 2009 , Volume 2 , Issue 2
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Abstract:
In October, 2007 the AMS Central Sectional Meeting was held at DePaul University in Chicago, IL. At that time, the guest editors of this issue of DCDS-S organized two special sessions dedicated to dynamical systems: “Smooth dynamical systems” and “Ergodic theory and symbolic dynamical systems”. The meeting was preceded by a workshop titled “Applications of measurable and smooth dynamical systems to number theory”, hosted jointly by DePaul and Northeastern Illinois universities, where M. Einsiedler, A. Katok, and A. Venkatesh gave expository lectures. The goal of the workshop was to disseminate the tools and ideas of their work on the Littlewood conjecture and related topics to the larger dynamical systems community. This confluence of events encouraged us to put together this volume which gives a small snapshot of research conducted in dynamical systems around the time of the workshop and conference.
The volume does not represent the entire scope of what is a very large and active field but does span a variety of distinct areas in dynamical systems. It is worth noting, however, that there are natural groupings around some central ideas which cut across sub-disciplines.
For more information please click the “Full Text” above.
Abstract:
Suppose $X$ and $Y$ are Polish spaces each endowed with Borel probability measures $\mu$ and $\nu$. We call these Polish probability spaces. We say a map $\phi$ is a
nearly continuousif there are measurable subsets $X_0\subseteq X$ and $Y_0\subseteq Y$, each of full measure, and $\phi:X_0\to Y_0$ is measure-preserving and continuous in the relative topologies on these subsets. We show that this is a natural context to study morphisms between ergodic homeomorphisms of Polish probability spaces. In previous work such maps have been called almost continuousor finitary. We propose the name measured topological dynamicsfor this area of study. Suppose one has measure-preserving and ergodic maps $T$ and $S$ acting on $X$ and $Y$ respectively. Suppose $\phi$ is a measure-preserving bijection defined between subsets of full measure on these two spaces. Our main result is that such a $\phi$ can always be regularizedin the following sense. Both $T$ and $S$ have full groups ($FG(T)$ and $FG(S)$) consisting of those measurable bijections that carry a point to a point on the same orbit. We will show that there exists $f\in FG(T)$ and $h\in FG(S)$ so that $h\phi f$ is nearly continuous. This comes close to giving an alternate proof of the result of del Junco and Şahin, that any two measure-preserving ergodic homeomorphisms of nonatomic Polish probability spaces are continuously orbit equivalent on invariant $G_\delta$ subsets of full measure. One says $T$ and $S$ are evenly Kakutani equivalent if one has an orbit equivalence $\phi$ which restricted to some subset is a conjugacy of the induced maps. Our main result implies that any such measurable Kakutani equivalence can be regularized to a Kakutani equivalence that is nearly continuous. We describe a natural nearly continuous analogue of Kakutani equivalence and prove it strictly stronger than Kakutani equivalence. To do this we introduce a concept of nearly unique ergodicity. Abstract:
We present an example of Kakutani equivalent and strong orbit equivalent substitution systems that are not conjugate.
Abstract:
Let $X$ be a Polish space and $T_t$ a jointly Borel measurable action of $\mathbb{R}^+ = [0, \infty)$ on $X$ by surjective maps preserving some Borel probability measure $\mu$ on $X$. We show that if each $T_t$ is countable-to-1 and if $T_t$ has the "discrete orbit branching property'' (described in the introduction), then $(X, T_t)$ is isomorphic to a "semiflow under a function''.
Abstract:
We introduce a family of adic transformations on diagrams that are nonstationary and nonsimple. This family includes some previously studied adic transformations such as the Pascal and Euler adic transformations. We give examples of particular adic transformations with roots of unity and we show as well that the Euler adic is totally ergodic. We show that the Euler adic and a disjoint subfamily of adic transformations are loosely Bernoulli.
Abstract: Heavinessrefers to a sequence of partial sums maintaining a certain lower bound and was recently introduced and studied in [11]. After a review of basic properties to familiarize the reader with the ideas of heaviness, general principles of heaviness in symbolic dynamics are introduced. The classical Morse sequence is used to study a specific example of heaviness in a system with nontrivial rational eigenvalues. To contrast, Sturmian sequences are examined, including a new condition for a sequence to be Sturmian. Abstract:
We show that in the category of effective $\mathbb{Z}$-dynamical systems there is a universal system, i.e. one that factors onto every other effective system. In particular, for $d\geq3$ there exist $d$-dimensional shifts of finite type which are universal for $1$-dimensional subactions of SFTs. On the other hand, we show that there is no universal effective $\mathbb{Z}^{d}$-system for $d\geq2$, and in particular SFTs cannot be universal for subactions of rank $\geq2$. As a consequence, a decrease in entropy and Medvedev degree and periodic data are not sufficient for a factor map to exists between SFTs.
We also discuss dynamics of cellular automata on their limit sets and show that (except for the unavoidable presence of a periodic point) they can model a large class of physical systems.
Abstract:
We show that given any tiling of Euclidean space, any geometric pattern of points, we can find a patch of tiles (of arbitrarily large size) so that copies of this patch appear in the tiling nearly centered on a scaled and translated version of the pattern. The rather simple proof uses Furstenberg's topological multiple recurrence theorem.
Abstract:
In this article we study the ergodic Hilbert transform modulated by bounded sequences. We prove that sequences satisfying some variation conditions and are universally good for ordinary ergodic averages, such as the sequences defined by the Fourier coefficients of $L_p$ functions, are universally good modulating sequences for the ergodic Hilbert transform. We also prove that sequences belonging to the subfamily $B_1^{\alpha} $ of the two-sided bounded Besicovitch class $B_1$ are good modulating sequences for the ergodic Hilbert transform.
Abstract:
We announce ultrametric analogues of the results of Kleinbock-Margulis for shrinking target properties of semisimple group actions on symmetric spaces. The main applications are $S$-arithmetic Diophantine approximation results and logarithm laws for buildings, generalizing the work of Hersonsky-Paulin on trees.
Abstract:
This note contains some remarks about the homologies that can be associated to a foliation which is invariant and uniformly expanded by a diffeomorphism. We construct a family of 'dynamical' closed currents supported on the foliation which help us relate the geometric volume growth of the leaves under the diffeomorphism with the map induced on homology in the case when these currents have nonzero homology.
Abstract:
We develop a general theory of implicit generating forms for volume-preserving diffeomorphisms on a manifold. Our results generalize the classical formulas for generating functions of symplectic twist maps and examples of Carroll for volume-preserving maps on $\R^n$.
Abstract:
We study the isosceles three body problem with fixed symmetry line for arbitrary masses, as a subsystem of the N-body problem. Our goal is to construct minimizing noncollision periodic orbits using a symmetric variational method with a finite order symmetry group. The solution of this variational problem gives existence of noncollision periodic orbits which realize certain symbolic sequences of rotations and oscillations in the isosceles three body problem for any choice of the mass ratio. The Maslov index for these periodic orbits is used to prove the main result, Theorem 4.1, which states that the minimizing curves in the three dimensional reduced energy momentum surface can be extended to periodic curves which are generically hyperbolic. This reminds one of a theorem of Poincaré [8], concerning minimizing periodic geodesics on orientable 2D surfaces. The results in this paper are novel in two directions: in addition to the higher dimensional setting, the minimization in the current problem is over a symmetry class, rather than a loop space.
Abstract:
We consider a dynamical system whose phase space contains a two-dimensional normally hyperbolic invariant manifold diffeomorphic to an annulus. We assume that the dynamics restricted to the annulus is given by an area preserving monotone twist map. We assume that in the annulus there exist finite sequences of primary invariant Lipschitz tori of dimension $1$, with the property that the unstable manifold of each torus has a topologically crossing intersection with the stable manifold of the next torus in the sequence. We assume that the dynamics along these tori is topologically transitive. We assume that the tori in these sequences, possibly with the exception of the tori at the ends of the sequences, can be $C^0$-approximated from both sides by other primary invariant tori in the annulus. We assume that the region in the annulus between two successive sequences of tori is a Birkhoff zone of instability. We prove the existence of orbits that follow the sequences of invariant tori and cross the Birkhoff zones of instability.
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Given two linear functions $f(x)$ and $g(x)$ defined on real values, let's say that I want to show that $f(x) > g(x)$ for all real $x > t > 0$. According to the order-1 Taylor expansion at the origin, these two functions can be written as \begin{equation*} f(x) = f(0) + (x-0)f'(0)\quad \text{and}\quad g(x) = g(0) + (x-0)g'(0) \end{equation*} because the higher-order derivatives will be zero for linear polynomials.
If I can assume that $f(0) = g(0)$, then $f'(0) > g'(0)$ will guarantee that $f(x) > g(x)$ for all $x>t>0$. (Well, it may guarantee $f(x)>g(x)$ for all $x > 0$, but let's just say that the lower bound of interest is some positive number $t$.)
I know multivariable Taylor expansion, but don't have much background on it. So I'd like to ask whether the above approach still works for real-coefficients multilinear polynomials $g,f:\mathbb{R}^n\to\mathbb{R}$.As you know,
multilinear means that no variable $x_i$ appears squared, cubed, etc. And my goal is to show $f(x_1,\cdots,x_n)> g(x_1,\cdots,x_n)$ for $x_1 > t, x_2, > t, \cdots, x_n> t$ where $t$ is some positive constant assuming $f(0,\cdots,0) = g(0,\cdots, 0)$.
Extending from the simple case above, I may say the order-$(1,\cdots,1)$ Taylor expansion at the origin of these two polynomials are \begin{align*} f(x_1,\cdots,x_n) &= \left(\prod_{i=1}^n\left(1 + x_i\frac{\partial}{\partial z_i}\right)\right) f(z_1,\cdots,z_n)\big\vert_{z_1 = \cdots = z_n = 0} \\ g(x_1,\cdots,x_n) &= \left(\prod_{i=1}^n\left(1 + x_i\frac{\partial}{\partial z_i}\right)\right) g(z_1,\cdots,z_n)\big\vert_{z_1 = \cdots = z_n = 0} \\ \end{align*} because the simple case corresponds to $f(x) = (1+x \frac{\partial}{\partial z})f(z)\big\vert_{z=0}$.
The simple case of linear polynomials is all about comparing the slopes$f'(0)$ and $g'(0)$. What about the multilinear cases? Similar to the linear case with one variable, is it OK just to compare the all the partial derivatives in each variable at the origin\begin{equation*}\frac{\partial}{\partial z_i}f(z_1,\cdots,z_n)\vert_{z_1=\cdots=z_n=0} > \frac{\partial}{\partial z_i}g(z_1,\cdots,z_n)\vert_{z_1=\cdots=z_n=0}\end{equation*}to conclude that $f(x) > g(x)$ in the domain of interest? If the above doesn't make sense at all, then what is the generalway to compare two multilinear polynomials over some domain $x_1 > t, x_2 > t, \cdots, x_n >t$ for some positive constant $t$? Is there any textbook explaining this problem? |
Consider for large integers $n$ the expression $\sin \left(\pi \sqrt{4 n^2+n}\right)$.
Since
$\sqrt{4 n^2+n}=2 n \sqrt{1 + \frac{1}{4 n}}$
we can use the standard series for the square root and next the standard series for $sin$ to find a series expansion of this expression around $\infty$. That is a simple first year exercise.
I was unable to do this computation in an elegant way with
Mathematica. I expected the following to work:
Series[Sin[Sqrt[n+4 n^2] \[Pi]], {n,\[Infinity],3}, Assumptions->{n\[Element]Integers}]
$\sin \left(2 \pi n+\frac{\pi }{4}-\frac{\pi }{64 n}+\frac{\pi }{512 n^2}-\frac{5 \pi }{16384 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)\right)$
To my surprise, the expansion is only inside
Sin and the fact that
n is an integer is not used.
I tried to use
Normal and
TrigExpand without success. Finally I found a rather dirty trick to obtain the result I was looking for:
Series[Sin[Sqrt[n + 4 n^2] \[Pi]], {n, \[Infinity], 3}] /. Sin[x_] :> Sin[x - 2 \[Pi] n]
$\frac{1}{\sqrt{2}}-\frac{\pi }{64 \sqrt{2} n}+\frac{\frac{\pi }{512 \sqrt{2}}-\frac{\pi ^2}{8192 \sqrt{2}}}{n^2}+\frac{-\frac{5 \pi }{16384 \sqrt{2}}+\frac{\pi ^2}{32768 \sqrt{2}}+\frac{\pi ^3}{1572864 \sqrt{2}}}{n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$
Can this result be found in a more straightforward way?
Edit
Let us reconsider
Series[Sin[Sqrt[n + 4 n^2] \[Pi]], {n, \[Infinity], 3}, Assumptions -> {n \[Element] Integers}]
$\sin \left(2 \pi n+\frac{\pi }{4}-\frac{\pi }{64 n}+\frac{\pi }{512 n^2}-\frac{5 \pi }{16384 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)\right)$
Henric Schumacher is quite right in his assumption that we do not find a series expansion, but only a series expansion wrapped in
Sin, because
Sin has an essential singularity at
Infinity. The documentation of
Series, under
Possible Issues, shows something similar. So this should not have surprised me.
The assumption that
nis an integer has not prevented the appearance of the term
2\[Pi]n in the argument of
Sin. Without that term, the result would immediately further evaluate to the result we are looking for. So we have to get rid of this term.
The argument of
Sin is a
SeriesData expression, which is atomic. Therefore, the substitution
2\[Pi]n->0 does not work. But assignment is possible:
sd=Series[Sin[Sqrt[n+4 n^2] \[Pi]],{n,\[Infinity],3}];sd[[1,3,1]]=0;sd
$\frac{1}{\sqrt{2}}-\frac{\pi }{64 \sqrt{2} n}+\frac{\frac{\pi }{512 \sqrt{2}}-\frac{\pi ^2}{8192 \sqrt{2}}}{n^2}+\frac{-\frac{5 \pi }{16384 \sqrt{2}}+\frac{\pi ^2}{32768 \sqrt{2}}+\frac{\pi ^3}{1572864 \sqrt{2}}}{n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$
Another way of getting rid of the term
2\[Pi]n is to use
Normal and
FullSimplify with the assumption that
n is an integer, as done in the answer of Mikado. After that, we need another call of
Series to arrive at the desired result.
To summarize: at the moment, it seems that there is no straightforward way for finding this series expansion by
Mathematica, while it is easily found with pen and paper. |
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December 2009 , Volume 2 , Issue 4
A special issue on Bifurcation Delay
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The theory of slow-fast systems is a challenging field both from the viewpoint of theory and applications. Advances made over the last decade led to remarkable new insights and we therefore decided that it is worthwhile to gather snapshots of results and achievements in this field through invited experts. We believe that this volume of DCDS-S contains a varied and interesting overview of different aspects of slow-fast systems with emphasis on 'bifurcation delay' phenomena. Unfortunately, as could be expected, not all invitees were able to sent a contribution due to their loaded agenda, or the strict deadlines we had to impose.
Slow-fast systems deal with problems and models in which different (time- or space-) scales play an important role. From a dynamical systems point of view we can think of studying dynamics expressed by differential equations in the presence of curves, surfaces or more general varieties of singularities. Such sets of singularities are said to be critical. Perturbing such equations by adding an $\varepsilon$-small movement that destroys most of the singularities can create complex dynamics. These perturbation problems are also called singular perturbations and can often be presented as differential equations in which the highest order derivatives are multiplied by a parameter $\varepsilon$, reducing the order of the equation when $\varepsilon\to 0$.
For more information please click the “Full Text” above.
Abstract:
In this paper we consider singular perturbation problems occuring in planar slow-fast systems $(\dot x=y-F(x,\lambda),\dot y=-\varepsilon G(x,\lambda))$ where $F$ and $G$ are smooth or even real analytic for some results, $\lambda$ is a multiparameter and $\varepsilon$ is a small parameter. We deal with turning points that are limiting situations of (generalized) Hopf bifurcations and that we call slow-fast Hopf points. We investigate the number of limit cycles that can appear near a slow-fast Hopf point and this under very general conditions. One of the results states that for any analytic family of planar systems, depending on a finite number of parameters, there is a finite upperbound for the number of limit cycles that can bifurcate from a slow-fast Hopf point.
The most difficult problem to deal with concerns the uniform treatment of the evolution that a limit cycle undergoes when it grows from a small limit cycle near the singular point to a canard cycle of detectable size. This explains the title of the paper. The treatment is based on blow-up, good normal forms and appropriate Chebyshev systems. In the paper we also relate the slow-divergence integral as it is used in singular perturbation theory to Abelian integrals that have to be used in studying limit cycles close to the singular point.
Abstract:
A singularly perturbed planar system of differential equations modeling an autocatalytic chemical reaction is studied. For certain parameter values a limit cycle exists. Geometric singular perturbation theory is used to prove the existence of this limit cycle. A central tool in the analysis is the blow-up method which allows the identification of a complicated singular cycle which is shown to persist.
Abstract:
We study the organization of mixed-mode oscillations (MMOs) in the Olsen model for the peroxidase-oxidase reaction, which is a four-dimensional system with multiple time scales. A numerical continuation study shows that the MMOs appear as families in a complicated bifurcation structure that involves many regions of multistability. We show that the small-amplitude oscillations of the MMOs arise from the slow passage through a (delayed) Hopf bifurcation of a three-dimensional fast subsystem, while large-amplitude excursions are associated with a global reinjection mechanism. To characterize these two key components of MMO dynamics geometrically we consider attracting and repelling slow manifolds in phase space. More specifically, these objects are surfaces that are defined and computed as one-parameter families of stable and unstable manifolds of saddle equilibria of the fast subsystem. The attracting and repelling slow manifolds interact near the Hopf bifurcation, but also explain the geometry of the global reinjection mechanism. Their intersection gives rise to canard-like orbits that organize the spiralling nature of the MMOs.
Abstract:
Acker
et al( J. Comp. Neurosci., 15, pp.71-90, 2003) developed a model of stellate cells which reproduces qualitative oscillatory patterns known as mixed mode oscillations observed in experiments. This model includes different time scales and can therefore be viewed as a singularly perturbed system of differential equations. The bifurcation structure of this model is very rich, and includes a novel class of homoclinic bifurcation points. The key to the bifurcation analysis is a folded node singularity that allows trajectories known as canards to cross from a stable slow manifold to an unstable slow manifold as well as a node equilibrium of the slow flow on the unstable slow manifold. In this work we focus on the novel homoclinic orbits within the bifurcation diagram and show that the return of canards from the unstable slow manifold to the funnel of the folded node on the stable slow manifold results in a horseshoe map, and therefore gives rise to chaotic invariant sets. We also use a one-dimensional map to explain why many homoclinic orbits occur in "clusters'' at exponentially close parameter values. Abstract:
The FitzHugh-Nagumo equation has been investigated with a wide array of different methods in the last three decades. Recently a version of the equations with an applied current was analyzed by Champneys, Kirk, Knobloch, Oldeman and Sneyd [5] using numerical continuation methods. They obtained a complicated bifurcation diagram in parameter space featuring a C-shaped curve of homoclinic bifurcations and a U-shaped curve of Hopf bifurcations. We use techniques from multiple time-scale dynamics to understand the structures of this bifurcation diagram based on geometric singular perturbation analysis of the FitzHugh-Nagumo equation. Numerical and analytical techniques show that if the ratio of the time-scales in the FitzHugh-Nagumo equation tends to zero, then our singular limit analysis correctly represents the observed CU-structure. Geometric insight from the analysis can even be used to compute bifurcation curves which are inaccessible via continuation methods. The results of our analysis are summarized in a singular bifurcation diagram.
Abstract:
Mixed mode oscillations (MMO's) composed of subthreshold oscillations (STO's) and spikes appear via a variety of mechanisms in models of neural dynamics. Two key elements that can influence the prominence of the STO's are multiple time scales and time varying parameters near critical points. These features can lead to dynamics associated with bifurcation delay, and we consider three systems with this behavior. While it is well known that bifurcation delay related to a slow time scale is sensitive to noise, we compare other aspects of the noise-sensitivity in the context of MMO's, where not only bifurcation delay, but also coherence resonance and dynamics in the interspike interval play a role. Noise can play a role in amplifying the STO's but it can also drive the system into repetitive spiking without STO's. In particular we compare integrate and fire models with models that capture both spike and STO dynamics. The interplay of the underlying bifurcation structure and the modeling of the return mechanism following the spike are major factors in the robustness and noise sensitivity of the STO's in the context of multiple time scales.
Abstract:
In the classical bifurcation theory, behavior of systems depending on a parameter is considered for values of this parameter close to some critical, bifurcational one. In the theory of dynamical bifurcations a parameter is changing slowly in time and passes through a value that would be bifurcational in the classical static theory. Some arising here phenomena are drastically different from predictions derived by the static approach. Let at a bifurcational value of a parameter an equilibrium or a limit cycle loses its asymptotic linear stability but remains non-degenerate. It turns out that in analytic systems the stability loss delays inevitably: phase points remain near the unstable equilibrium (cycle) for a long time after the bifurcation; during this time the parameter changes by a quantity of order 1. Such delay is not in general found in non-analytic (even infinitely smooth) systems. A survey of some background on stability loss delay phenomenon is presented in this paper.
Abstract:
We consider a two dimensional family of real vector fields. We suppose that there exists a stationary point where the linearized vector field has successively a stable focus, an unstable focus and an unstable node. It is known that when the parameter moves slowly, a bifurcation delay appears due to the Hopf bifurcation. The main question investigated in this article is the continuation of the delay in the region of the unstable node. Another problem is to determine the input-output relation which characterizes all the possible delays.
Abstract:
We give a non-exhaustive overview of the problem of bifurcation delay from its appearance in France at the end of the eighties to the most recent contributions. We present the bifurcation delay for differential equations as well as for discrete dynamical systems.
Abstract:
The construction of orbits with specific asymptotic properties, such as orbits that are heteroclinic or homoclinic to certain invariant sets, involves tracking stable and unstable manifolds around the system's phase space. This work addresses how, in some generality, the tracking can be achieved during the passage near a distinguished invariant manifold in the phase space. This leads to a very general form of the Exchange Lemma and it is further shown how the lemma can be used in the construction of distinguished homoclinic and heteroclinic orbits.
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Suppose that there is an algorithm which sorts a sequence of $n$ elements
$$a_1, a_2, ..., a_n$$
Each of the $a_i$ is chosen with probability $1/k$ from a set of $k$ distinct integer numbers.
Is it true, given that $k \to \infty$, that:
The probability that any two of incoming sequence elements are equal, tends to $0$? The probability that the incoming sequence is already sorted, tends to $\frac{1}{n!}$?
Why / why not? |
Condition on Connectedness by Clopen Sets Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.
Then: $T$ admits no separation the only clopen sets of $T$ are $S$ and $\varnothing$. Thus both conditions can be used to define a connected topological space. Proof Necessary Condition
Then by definition $T$ admits a separation, $A \mid B$ say.
Then both $A$ and $B$ are clopen sets of $T$, neither of which is either $S$ or $\varnothing$.
$\Box$
Sufficient Condition
Suppose $\exists H \subseteq S$ which is clopen.
Hence by definition, $T$ is not connected.
$\blacksquare$
Also see Definition:Connected Topological Space: Definition from Separation Definition:Connected Topological Space: Definition from Clopen Sets |
I saw the following claim in some book without a proof and couldn't prove it myself.
$\dfrac{d}{dp}\mathbb{P}\left(\text{Bin}\left(n,\,p\right)\leq d\right)=-n\cdot\mathbb{P}\left(\text{Bin}\left(n-1,\,p\right)=d\right)$
So far I got:
$\begin{array}{l} \dfrac{d}{dp}\mathbb{P}\left(\text{Bin}\left(n,\,p\right)\leq d\right)=\\ \dfrac{d}{dp}\sum\limits _{i=0}^{d}\left(\begin{array}{c} n\\ i \end{array}\right)p^{i}\left(1-p\right)^{n-i}=\\ -n\cdot\left(1-p\right)^{n-1}+\sum\limits _{i=1}^{d}\left(\begin{array}{c} n\\ i \end{array}\right)\left[ip^{i-1}\left(1-p\right)^{n-i}-p^{i}\left(n-i\right)\left(1-p\right)^{n-i-1}\right] \end{array}$
But I am not very good playing with binomial coefficients and don't know how to proceed. |
I'm reading quantum chemistry. The book says that the orbital angular momentum of a $\pi$ electron along the symmetry axis of a molecule made up of two atoms is $\pm 1$. I think this is a primary question, but I do not konw why.
I currently have a preliminary understanding of this:
In a molecule inculding two atoms, potential energy axisymmetric about z' axis(the line connecting the two atoms). So, the angular momentum along z axis is quantized. That is to say $m_z$ is good quantum number. Let's consider $\pi$ orbit made up of two $p_z$. The orbital angular momentum along z axis(the symmetry axis of $p_z$) of an electron in $p_z$ is 0. So considering orbital angular momentum along z' axis of this electron, the electron is in $\frac 1 {\sqrt 2}(|+\rangle-|-\rangle)$. So the orbital angular momentum along z' axis is either 1 or -1. So the orbital angular momentum along z' axis of a $\pi$ electron is either 1 or -1.
Is my understanding right? |
I have a system of ODEs which is (at least moderately) stiff.
Consider the class of spectral collocation methods https://en.wikipedia.org/wiki/Spectral_method or the related class of weighted residual methods.
Basically, take a system of ODEs like $$ y'(t) = f(t, y(t)) $$ where $f(\cdot,\cdot)$ is nonlinear in my case and leads to a stiff ODE, and I want to solve it on $t \in [0,M]$. Then approximate the solution $$y(t) \approx \sum_{i=0}^N a_i T_i(t)$$ where $T_i(\cdot)$ are a basis (e.g. Chebyshev polynomials), $N$ is the maximum order, and the basis is over $[0,M]$ if appropriate. Note that given $\{a_i\}$, $y'(t)$ can be calculated by differentiating the polynomial basis.
Now, evaluate the ODE given a $\{a_i\}$ at the roots of the polynomial basis (which can be shown to be optimal) and find the residual. Choose $\{a_i\}$ to minimize this residual, or just solve a system of equations with the residual = 0.
Question: how well does spectral collocation with a Chebyshev basis handle systems that seem stiff when solved with finite differences? The worry is that the same issues that cause finite differences to have issues (e.g. convergence to a very flat function with no slope) could cause major issues with a finite dimensional approximation of the function.
If Chebyshev has problems, then is there a better basis which might have less issues.
A little more context: I have implemented spectral collocation with Chebyshev polynomials to solve a stationary Kolmogorov Forward equation in CDFs. These, of course, need to be strictly positive and weakly monotonically increasing and asymptotically go towards $1$ if it is a proper probability distribution. I am seeing weird behavior where the pdf drops below $0$ (i.e., it becomes decreasing) or strange jumps at the corners. It is very hard for me to tell if these are artifacts in the solution scheme or if I have just chosen parameters for the model where no solution exists. Increasing the number of Chebyshev basis functions doesn't eliminate them, but that could be because they can't, or only slowly, converge. Even more context: For what it is worth, in the simplest form of my problem there are two sets of ODEs (which are ultimately coupled): Parameters: $\lambda_{\ell}, \lambda_h, r,$ and $\chi$. In some ways $F_{\ell}'(0), F_h'(0),$ and $g$ are also parameters for the purposes here. There are two discrete states which leads to the system: $i \in \{\ell,h\}$
The variable $z\in[0,\infty)$ in reality, but these converge pretty quickly so choosing some $z \in [0,\bar{z}]$ for collocation methods is reasonable.
System (1) The following comes from a stationary Kolmogorov Forward Equation (given a $\gamma(z)$ function such that $\gamma(0) = 0$ and $\lim\limits_{z\to\infty} \gamma(z) = g$. The latter condition leads to a vanishing derivative term in the ODE (i.e., a singular mass matrix if put in that canonical form.)$$\begin{align}0 &= g F_{\ell}'(z) + \lambda_h F_h(z) - \lambda_{\ell}F_{\ell}(z) + g (F_{\ell}'(0) + F_h'(0))(F_{\ell}(z) + F_h(z)) - g F_{\ell}'(0)\\0 &= (g - \gamma(z))F_h'(z) + \lambda_{\ell}F_{\ell}(z) - \lambda_h F_h(z) - g F_h'(0)\end{align}$$ Forget that this is a linear ODE, as the actual one is nonlinear and much trickier. However, it can still be written in a $M(z, F(z))\cdot F'(z) = \Phi(z, F(z))$ term for some operators and a mass matrix. The trouble seems to come out of the term on the derivative going to $0$ (i.e., a singular mass matrix). It might seem odd that there is a $F_{\ell}'(0)$ and a $F_h'(0)$ in the ODE as parameters but this is not a mistake. You can easily show that any solution to the ODE will have the $F_i'(0)$ matching these constants by construction. The initial condition is $F_i(0) = 0$. Furthermore, from any $F_i'(0)$, one can analytically find $\lim\limits_{z\to\infty}F_h(z)$. With $F_{\ell}(\infty) + F_h(\infty) = 1$, this means we can write it as a BVP if we wish. System (2) The following comes from a Hamilton-Jacobi-Bellman equation
\begin{align} 0 &= 1 - (r + \lambda_{\ell})w_{\ell}(z) - g w'_{\ell}(z) + \lambda_h w_h(z)\\ 0 &= 1 - (r + \lambda_h)w_h(z) - (g - \frac{\chi}{2} w_h(z))w_h'(z) + \lambda_h w_{\ell}(z) + \frac{\chi}{4} w_h(z)^2 \end{align} subject to, $w_{\ell}(0) = w_h(0) = 0$ and defined on $z \in [0,\infty)$. You can analytically find $w_{\ell}(\infty)$ and $w_h(\infty)$ which can help with various methods to set it as a BVP.
One can show that $\lim\limits_{z\to\infty}(g - \frac{\chi}{2} w_h(z)) = 0$ which leads to a vanishing derivative term in the ODE.
Coupled: In reality, the $\gamma(z)$ from the first solution is $$\gamma(z) \equiv \frac{\chi}{2}w_h(z)$$ Leaving Out: A whole bunch of equilibrium conditions in terms of integrals of $w_i(z)$ and $F_i(z)$ which pin down $g, F'_{\ell}(0)$ and $F'_h(0)$. Evaluating these concurrently with the solution to the ODEs is the reason that spectral methods are preferable since I can add in more equations and solve everything at the same time.
... is what I have written useful? Maybe not, but you can see the asymptotic singularity in the mass matrix. Is this stiff? I think so. When you solve the second system with finite differences, for example, it has all the hallmarks of a stiff system. |
Most people start out by calculating the volume of liquid. As with many mathematical tasks some thought in advance may save a lot of work . Failing that, if you review your method you may find a neater and more efficient way to do it. Try to evaluate your own work, think about it and ask yourself questions like: "what is the key issue here?", "does my answer suggest a connection in the problem I did not use?","have I done it the best way?" . Very often the best approach leads to a really pretty bit of maths.
The quick method is to look at the scale factor.
Consider the enlargement of the conical space above the liquid to the whole cone. The scale factor is $2$ (linearly) so the volume scale factor is $2^3=8$
The space above the liquid has an eighth of the volume of the whole cone and the liquid takes up seven eighths of the volume.
When it is inverted the volume of water is still seven eighths of the volume of the cone so we use the fact again that the cube of the linear scale factor gives the volume scale factor to get: $$ (\frac{h}{x})^3 =\frac{7}{8}$$
and so $$h = { \frac{\sqrt[3] 7x}{2}}$$
That is, the height of the liquid in the upturned cone is $\frac {\sqrt[3] 7} {2}$ or 0.9565 of the original height
Other method is to equate the 2 results for the volume of the liquid.
No-one had any trouble in showing that the volume of liquid was $7 \frac{\pi r^2x}{24}$ (*) where $r$ was the base radius and x the height of the cone.
All used the properties of similar triangles to find the radius of the base of liquid in the upturned cone, which is (rh/x). Hence the volume of liquid in the upturned cone is $$ \pi/3 \times (\frac {rh}{x})^2 \times h (**) $$
(*) and (**) were thus equated and the height of the liquid in the upturned cone was found, by cancelling, to be:$$h = { \frac{\sqrt[3] 7x}{2}}$$
or approximately $0.9565$ of the original height.
Easy to follow solutions to this problem were received from: Sam, Jonathan and Kevin, Tom, Euan, Michael and James of Madras College and Moray and Richard of Wellingborough School |
Minkowski space is a real
affine space of dimension $4$ whose space of translations is equipped with a metric of Lorentzian type.
A (real)
affine space is a triple $(\mathbb A, V, \vec{})$, where $\mathbb A$ is a set whose elements are said points, $V$ is a (real) vector space and $\vec{}$ is a map $\vec{} : \mathbb A \times \mathbb A \to V$ with the following properties,
$$\forall q \in \mathbb A\:, \forall v \in V\:, \exists \mbox{ and is unique } p \in \mathbb A\quad \mbox{such that}\quad \vec{qp} = v\:,\tag{1}$$
$$\vec{pq} + \vec{qr} = \vec{pr}\quad \forall p, q,r \in \mathbb A\:.\tag{2}$$
by definition, the
dimension of the affine space is that of $V$, whose elements are said translations.
From now on, if $p,q\in \mathbb A$ a $v \in V$, $$p= q+ v$$
means
$$\vec{qp}=v\:.$$
Form (1) this notation is well posed. $q+v$ is the action of the translation $v$ on the point $q$. This action is transitive and free, its existence physically corresponds to homogeneity of both space and time in special relativity.
Assuming that $V$ is finite dimensional, if one fixes $o \in \mathbb A$ and a basis $e_1,\ldots, e_n \in V$, a
Cartesian coordinate system on the affine space $\mathbb A$ with origin $o$ and axes $e_1,\ldots, e_n$ is the bijective map$$\mathbb R^n \ni (x^1,\ldots, x^n) \mapsto o + \sum_{j=1}^n x^je_j \in \mathbb A$$ Once again, using (1) one sees that, in fact, the map above is bijectve and thus identifies $\mathbb A$ with $\mathbb R^n$.
Changing $o$ to $o'$ and the basis $e_1,\ldots, e_n$ to the basis $e'_1,\ldots, e_n'$, one obtains a different Cartesian coordinate system $x'^1,\ldots, x'^n$. It is simply proved that the rule to pass form the latter coordinate system to the former has the form$$x'^a = c^a+ \sum_{j=1}^n {A^a}_j x^j \tag{3}$$for $n$ constant coefficients $c^j$ and a nonsingular $n\times n$ matrix of coefficients ${A^a}_j$.
The said matrix verifies
$$e_k = \sum_{i=1}^n {A^i}_k e'_i\tag{3'}$$
whereas the coefficients $c^k$ are the components of the vector $\vec{oo'}$.
(As a matter of fact the affine structure gives rise to a natural differentiable real analytic structure on $\mathbb A$ of dimension $n$.)
A real affine space equipped with a (pseudo)scalar product in $V$ is called
(pseudo)Euclidean space.
Minkowski spacetime $\mathbb M^4$ is a (real) four dimensional affine space equipped with a pseudo scalar product $g: V\times V \to \mathbb R$ of
Lorentzian type.
"Of Lorentzian type" means that there exist bases, $e_0,e_1,e_2,e_3$, in $V$ such that (I adopt here the convention $-+++$)
$$g(e_0,e_0)=-1 \:,\quad g(e_i,e_i) = 1 \mbox{ if $i=1,2,3$}\:,\quad g(e_i,e_j) =0 \mbox{ if $i\neq j$.}\tag{4}$$
These bases are called
Minkowskian bases. Lorentz group $O(1,3)$ is nothing but the group of matrices $\Lambda$ connecting pairs of Minkowskian bases.It is therefore defined by
$$O(1,3) := \left\{ \lambda \in M(4,\mathbb R) \:|\: \Lambda \eta \Lambda^t = \eta \right\}$$
where $\eta = diag(-1,1,1,1)$ is the matrix representing the metric $g$ in (4) in every Minkowskian basis.
A
Minkowskian coordinate system on $\mathbb M^4$ is a Cartesian coordinate system whose axes are a Minkowskian basis.
Lorentz transformations are transformations of coordinates between pairs of Minkowskian coordinate systems with the same origin (so that $c^k=0$ in (3)). Thus they have the form
$$x'^a = \sum_{j=1}^n {\Lambda^a}_j x^j $$
for some $\Lambda \in O(1,3)$.If we admit different origins we obtain the so-called
Poincaré transformations
$$x'^a = c^a+ \sum_{j=1}^n {\Lambda^a}_j x^j \:.$$
When viewing Lorentz transformations as transformation of coordinates, their formal linearity does not play a relevant physical role, since it only reflects the arbitrary initial choice of the same origin for both reference frames. However, these transformations are also transformations of bases (3') in the space of translations (the tangent space), in this case linearity is natural because it reflects the natural linear space structure of the translations. |
The definition you are given is equivalent to
$$\lim_{x \to a} \frac{f(x)-g(x)}{(x-a)^n}=0 $$
We are given $g$ defined as the sum
$$g(x) = \sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i$$
so we want to check wether or not
$$\lim_{x \to a} \frac{f(x)-\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i}{(x-a)^n}=0 $$
Since the last term has $(x-a)^n$ in it, we can take it out and get
$$\lim_{x \to a} \frac{f(x)-\sum_{i=0}^{n-1} \frac{f^{(i)}(a)}{i!}(x-a)^i}{(x-a)^n}-\frac{f^{(n)}(a)}{n!}= $$
The sum is a polynomial in $(x-a)$ of degree $n-1$, which, as well as $f$, has its $k^{th}$ derivative defined for $k=0,1,\dots,n$, so we can use L'Hôpital Bernoulli to find the indeterminate limit. Applying it $n-1$ times gives
$$\frac 1 {n!}\mathop {\lim }\limits_{x \to a} \frac{{{f^{\left( {n - 1} \right)}}(x) - {f^{\left( {n - 1} \right)}}\left( a \right)}}{{\left( {x - a} \right)}} - \frac{{{f^{(n)}}(a)}}{{n!}} = $$
And by definition of the derivative you get
$$\frac{{{f^{(n)}}(a)}}{{n!}} - \frac{{{f^{(n)}}(a)}}{{n!}} = 0$$
Note we can apply L'Hôpital in every $n-1$ step because it is the case in every step we face a $\frac 0 0 $ indeterminate form.
What you are asked to prove is usually stated as
$$f(x) = T_n^a(x)+o((x-a)^n)\text{ ; when } x \to a$$
and means the error is of greater order (it goes to zero
faster) than $(x-a)^n$ when $x \to a$. Though confusing, the meaning of "greater order" might be cleared up with this example: $(x-a)^{n+1}$ is of greater order than $(x-a)^n$, and thus goes to $0$ "faster" than it for $x \to a$, since
$$\mathop {\lim }\limits_{x \to a} \frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{{{\left( {x - a} \right)}^n}}} = \mathop {\lim }\limits_{x \to a} \left( {x - a} \right) = 0$$ |
I'm answering my own question. It is related to the answers given by both AJMansfield and Tony J., but slightly different, needing the inclusion of a vectorization transpose reordering operator, which is more involved than a simple transpose.
Long story short, the answer is that the desired matrix is given by,$$(B^T \otimes A)P,$$
where $\otimes$ is the Kronecker (outer) product, and $P$ is the permutation matrix that converts $\mathrm{vec}(X^T)$ into $\mathrm{vec}(X)$.
Reframe problem in terms of vectorizing matrix equation
The matrix needed, $(A \times B)$, is the matrix satisfying the following vectorized matrix equation,$$(A \times B)\mathrm{vec}(X) = \mathrm{vec}(AX^TB).$$
To see this, let the columns of $A,X,B$ be denoted $a_i,x_i,b_i$ respectively and compute,$$\mathrm{vec}(AX^TB) = \begin{bmatrix}AX^T b_1 \\AX^T b_2 \\\vdots \\AX^T b_n\end{bmatrix} = \begin{bmatrix}A \begin{bmatrix}x_1^Tb_1 \\x_2^T b_1 \\\vdots \\x_n^T b_1\end{bmatrix} \\A \begin{bmatrix}x_1^Tb_2 \\x_2^T b_2 \\\vdots \\x_n^T b_2\end{bmatrix} \\\vdots\end{bmatrix} = \begin{bmatrix}\sum_i a_i x_i^T b_1 \\\sum_i a_i x_i^T b_2 \\\vdots \\\sum_i a_i x_i^T b_n\end{bmatrix}= \\ \begin{bmatrix}\sum_i a_i b_1^T x_i \\\sum_i a_i b_2^T x_i \\\vdots \\\sum_i a_i b_n^T x_i\end{bmatrix} =\begin{bmatrix}a_1 b_1^T & a_2 b_1^T & \dots & a_n b_1^T \\a_1 b_2^T & a_2 b_2^T & \dots & a_n b_2^T \\\vdots & \vdots & \ddots & \vdots \\a_1 b_n^T & a_2 b_n^T & \dots & a_n b_n^T\end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\\vdots \\x_n\end{bmatrix} = (A \times B)\mathrm{vec}(X).$$
The matrix expression $\mathrm{vec}(AX^TB)$ is actually where the original question came from before I asked it.
Expressing matrix product in terms of Kronecker product and permutation matrix
Using the vectorized matrix expression interpretation of the matrix, and recalling the relationship between the outer (kronecker) product and vectorization yields,$$\mathrm{vec}(AX^TB) = (B^T \otimes A)\mathrm{vec}(X^T).$$This is almost what we want, except we have $X^T$ within the vectorization instead of $X$. Introducing the "transpose reordering operator" $P$ defined such that$$\boxed{P\mathrm{vec}(X) := \mathrm{vec}(X^T)},$$we have\begin{align}(A \times B)\mathrm{vec}(X) =& \mathrm{vec}(AX^TB) \\=& (B^T \otimes A)\mathrm{vec}(X^T) \\=& (B^T \otimes A)P\mathrm{vec}(X),\end{align}and so$$\boxed{(A \times B) = (B^T \otimes A)P}.$$
Vectorization transpose operator (more details)
The transpose reordering permutation matrix $P$ is uniquely defined by the above relation, but it also has a simple construction as follows. Let $Y$ be the matrix of the same size as $X$, but with value $k$ at the $k$'th linear index - $\mathrm{vec}(Y) = (1,2,3,4,5,6,\dots,nm)^T$. For example in the 4-by-5 case we have,$$Y = \begin{bmatrix}1 & 5 & 9 & 13 & 17 \\2 & 6 & 10 & 14 & 18\\3 & 7 & 11 & 15 & 19\\4 & 8 & 12 & 16 & 20\end{bmatrix}.$$Then $P$ is the permutation matrix where the $1$ in the $i$'th row is in the column given by the $i'th$ entry of $\mathrm{vec}(Y^T)$. Ie,$$P_{ij} = \begin{cases}1, \quad j \mathrm{~is~the~}i'th\mathrm{~entry~of~}\mathrm{vec}(Y^T), \\0, \quad \mathrm{else}\end{cases}$$In picture form, it looks like this:
Matlab code
The following code generates the vectorization permutation matrix. I release it to the public domain.
function P = vecpermute(n,m)
%Gives the permutation matrix to take the vectorization of a matrix to the
% vectorization of it's transpose. Ie, P*vec(X) = vec(X^T).
Y = zeros(n,m);
for k=1:n*m
Y(k) = k;
end
YT = Y';
P = sparse(Y(:),YT(:),ones(length(Y(:)),1), n*m, n*m);
end |
I want to find an example of Polish space which is not locally compact. I am thinking about the space of all continuous function from $[0,1]$ to $R$, endowed with the metric $d(f,g) = \sup_{x\in [0,1]}|f(x)-g(x)|$.
I know this space is complete. And by Weierstrass Approximation Theorem, all the polynomials with rationals coefficients are a countable dense subset of it, so it is Polish.
Then suppose the function $f=0$ has a compact neighbourhood, then there exists $r >0$ such that all the continuous functions bounded by $r$ are in the neighbourhood. But then we can define a sequence of functions such as $g_n(x) = \begin{cases} 0, x<a_n\\ r,x>a_{n+1}\\r \frac{x-a_n}{a_{n+1} - a_n}, a_n \leq x\leq a_{n+1}\end{cases}$, where $(a_n)_n$ increases to(but never reaches) 1. Then for $m,n$ different, we have $d(g_n, g_m) = r$, so the function $f=0$ has no compact neighbourhood. Therefore this space is not locally compact.
Did I miss something? |
@egreg It does this "I just need to make use of the standard hyphenation function of LaTeX, except "behind the scenes", without actually typesetting anything." (if not typesetting includes typesetting in a hidden box) it doesn't address the use case that he said he wanted that for
@JosephWright ah yes, unlike the hyphenation near box question, I guess that makes sense, basically can't just rely on lccode anymore. I suppose you don't want the hyphenation code in my last answer by default?
@JosephWright anway if we rip out all the auto-testing (since mac/windows/linux come out the same anyway) but leave in the .cfg possibility, there is no actual loss of functionality if someone is still using a vms tex or whatever
I want to change the tracking (space between the characters) for a sans serif font. I found that I can use the microtype package to change the tracking of the smallcaps font (\textsc{foo}), but I can't figure out how to make \textsc{} a sans serif font.
@DavidCarlisle -- if you write it as "4 May 2016" you don't need a comma (or, in the u.s., want a comma).
@egreg (even if you're not here at the moment) -- tomorrow is international archaeology day: twitter.com/ArchaeologyDay , so there must be someplace near you that you could visit to demonstrate your firsthand knowledge.
@barbarabeeton I prefer May 4, 2016, for some reason (don't know why actually)
@barbarabeeton but I have another question maybe better suited for you please: If a member of a conference scientific committee writes a preface for the special issue, can the signature say John Doe \\ for the scientific committee or is there a better wording?
@barbarabeeton overrightarrow answer will have to wait, need time to debug \ialign :-) (it's not the \smash wat did it) on the other hand if we mention \ialign enough it may interest @egreg enough to debug it for us.
@DavidCarlisle -- okay. are you sure the \smash isn't involved? i thought it might also be the reason that the arrow is too close to the "M". (\smash[t] might have been more appropriate.) i haven't yet had a chance to try it out at "normal" size; after all, \Huge is magnified from a larger base for the alphabet, but always from 10pt for symbols, and that's bound to have an effect, not necessarily positive. (and yes, that is the sort of thing that seems to fascinate @egreg.)
@barbarabeeton yes I edited the arrow macros not to have relbar (ie just omit the extender entirely and just have a single arrowhead but it still overprinted when in the \ialign construct but I'd already spent too long on it at work so stopped, may try to look this weekend (but it's uktug tomorrow)
if the expression is put into an \fbox, it is clear all around. even with the \smash. so something else is going on. put it into a text block, with \newline after the preceding text, and directly following before another text line. i think the intention is to treat the "M" as a large operator (like \sum or \prod, but the submitter wasn't very specific about the intent.)
@egreg -- okay. i'll double check that with plain tex. but that doesn't explain why there's also an overlap of the arrow with the "M", at least in the output i got. personally, i think that that arrow is horrendously too large in that context, which is why i'd like to know what is intended.
@barbarabeeton the overlap below is much smaller, see the righthand box with the arrow in egreg's image, it just extends below and catches the serifs on the M, but th eoverlap above is pretty bad really
@DavidCarlisle -- i think other possible/probable contexts for the \over*arrows have to be looked at also. this example is way outside the contexts i would expect. and any change should work without adverse effect in the "normal" contexts.
@DavidCarlisle -- maybe better take a look at the latin modern math arrowheads ...
@DavidCarlisle I see no real way out. The CM arrows extend above the x-height, but the advertised height is 1ex (actually a bit less). If you add the strut, you end up with too big a space when using other fonts.
MagSafe is a series of proprietary magnetically attached power connectors, originally introduced by Apple Inc. on January 10, 2006, in conjunction with the MacBook Pro at the Macworld Expo in San Francisco, California. The connector is held in place magnetically so that if it is tugged — for example, by someone tripping over the cord — it will pull out of the socket without damaging the connector or the computer power socket, and without pulling the computer off the surface on which it is located.The concept of MagSafe is copied from the magnetic power connectors that are part of many deep fryers...
has anyone converted from LaTeX -> Word before? I have seen questions on the site but I'm wondering what the result is like... and whether the document is still completely editable etc after the conversion? I mean, if the doc is written in LaTeX, then converted to Word, is the word editable?
I'm not familiar with word, so I'm not sure if there are things there that would just get goofed up or something.
@baxx never use word (have a copy just because but I don't use it;-) but have helped enough people with things over the years, these days I'd probably convert to html latexml or tex4ht then import the html into word and see what come out
You should be able to cut and paste mathematics from your web browser to Word (or any of the Micorsoft Office suite). Unfortunately at present you have to make a small edit but any text editor will do for that.Givenx=\frac{-b\pm\sqrt{b^2-4ac}}{2a}Make a small html file that looks like<!...
@baxx all the convertors that I mention can deal with document \newcommand to a certain extent. if it is just \newcommand\z{\mathbb{Z}} that is no problem in any of them, if it's half a million lines of tex commands implementing tikz then it gets trickier.
@baxx yes but they are extremes but the thing is you just never know, you may see a simple article class document that uses no hard looking packages then get half way through and find \makeatletter several hundred lines of trick tex macros copied from this site that are over-writing latex format internals. |
Nehari Manifold and Multiplicity Results for a Class of Fractional Boundary Value Problems with $p$-Laplacian
Bull. Korean Math. Soc. Published online August 6, 2019
Abdeljabbar Ghanmi and Ziheng ZhangUniversity of Jeddah, Tianjin Polytechnic University
Abstract : In this work, we investigate the following fractional boundary value problems\begin{eqnarray*}\left\{\begin{array}{ll}_{t}D_{T}^{\alpha}\left(|_{0}D_{t}^{\alpha}(u(t))|^{p-2} {_0 D}_{t}^{\alpha}u(t)\right)=\nabla W(t,u(t))+\lambda g(t) |u(t)|^{q-2}u(t),\;t\in (0,T),\\[0.25cm]u(0)=u(T)=0,\end{array}\right.\end{eqnarray*}where $\nabla W(t,u)$ is the gradient of $W(t,u)$ at $u$ and $W\in C([0,T]\times \mathbb{R}^{n},\mathbb{R})$ is homogeneous of degree $r$, $\lambda$ is a positive parameter, $g\in C([0,T])$, $1<r<p<q$ and $\frac{1}{p}<\alpha<1$. Using the Fibering map and Nehari manifold, for some positive constant $\lambda_0$ such that$0<\lambda<\lambda_0$, we prove the existence of at least two non-trivial solutions.
Keywords : Nonlinear fractional differential equations; Boundary value problem; Existence of solutions; Nehari Manifold |
Could someone explain the correspondence between lines in twistor space and minkowski space-time points? a basic derivation would suffice
The ordinary twistor space is parameterized by $(\lambda^\alpha,\mu_{\dot\alpha})$. Here, the $\alpha$ is a 2-valued $SL(2,C)$ spinor index of one chirality and the dotted index is its complex conjugate, the index of the opposite chirality.
At the level of spinors, vectors are equivalent to "spintensors" with one undotted and one dotted index. $$ V_\mu = \sigma_\mu^{\alpha \dot\alpha} V_{\alpha \dot \alpha}.$$ This is a basic fact about the Lie algebras. $SO(3,1)$ is locally isomorphic to $SL(2,C)$ - they're the same 6-dimensional Lie groups - and the 4-vector is the tensor product of ${\bf 2}$ and ${\bf \overline{2}}$.
If you're unfamiliar with this equivalence of vectors of "spintensors" with two indices, notice that the components of a 4-vector may be organized as $$v^{\alpha\dot\alpha} = \left( \begin{array}{cc} v^0+v^3 & v^1-i v^2\\ v^1+i v^2 & v^0-v^3 \end{array} \right) $$ Note that the determinant of this matrix - a natural function of the matrix elements - is simply $v^\mu v_\mu$. The whole matrix may be understood as the appropriate combination of the three Pauli matrices - plus the "time-like Pauli" (identity) matrix multiplied by the time component of the vector. Sorry if my signs deviate from the prevailing convention.
So far, it has only been a story about the vectors or spinors. What about the twistors? Well, it's a simple one-line formula. If I have a point $x^\mu$ which is equivalent to the $x^{\alpha \dot\alpha}$ matrix - as any 4-vector - I may simply write an equation $$ \lambda^\alpha = x^{\alpha \dot\alpha} \mu_{\dot\alpha} $$ Note that it is a set of two complex linear equations - for $\alpha=0,1$ - so it defines a linear object. For a different value of $x^\mu$, I get different equations. Moreover, the equations link the $\lambda$ and $\mu$ objects which are coordinates on the twistor space.
Now, I have to explain why the equations above define a line. First, the $\lambda$ and $\mu$ objects are pairs of complex numbers, so in total, we have four complex coordinates. However, the twistor space is a projective space. Note that if the equations above are satisfied for some $\lambda$ and $\mu$ - four complex numbers - they will also be satisfied if you multiply both $\lambda$ and $\mu$ by an arbitrary complex number (the same one for both). This projectivity holds universally: $\mu$ is naturally scaled in the same way as $\lambda$ because it may be understood as a dimensionally inverse object to $\lambda^{\dot\alpha}$ (an object that appeared for the first time in this answer) that scales in the inverse way relatively to $\lambda^\alpha$ in order to keep e.g. the vector $\lambda^\alpha \lambda^{\dot\alpha}$ constant.
So the twistor space is really a complex projective space, $CP^3$, and the equations above are two complex conditions, so we're left with a one(-complex)-dimensional object, a complex line. Spacetime points are in one-to-one correspondence to complex lines in the twistor space.
Many more things may be translated. For example, if two spacetime points are separated by a null interval, the corresponding two lines in the twistor space intersect. The intersection - a point in the twistor space - may be identified with a null line in the Minkowski space, and I could derive many other things of this kind.
I would like to add some further points to the answers above on the Twistor Space <--> Spacetime correspondence.
The Twistor space
T is a four complex dimensional space with elements described by $(Z^0,Z^1,Z^2,Z^3)$ or $Z^{\alpha}=(\omega^A,\pi_{A'})$ in spinor terms. The incidence relation between Minkowski points and Twistors is given (in spinor form) by:
$\omega^A=ix^{AA'}\pi_{A'}$
(In addition to a different representation of $x^{AA'}$ to that of Lubos it should also be noted that there is a 4-fold ambiguity in the representation of a spacetime point by a Twistor $(Z^{\alpha},-Z^{\alpha},iZ^{\alpha},-iZ^{\alpha})$ are all the same point as there are two sets of double cover involved.)
This Twistor Space contains a norm
$||Z||=Z^AZ^c_{A}=\omega^A\pi^c_{A} + \omega^{cA'}\pi_{A'}$ where $^c$ denotes spinor/twistor complex conjugation.
There are two sets of reductions to Twistor space to obtain the Minkowski space correspondence:
||Z||=0 these are the
nullTwistors - the space sometimes denoted N. Tmodulo $\lambda Z^{\alpha}$ - the space denoted PT-projective Twistor Space.
Putting these two together we get
PN - projective Null Twistor Space. It is PN which is mapped to Minkowski space via the incidence relation ie projective null twistors.
This raises the question as to what to more general cases of Twistor mappings correspond to.
In general
PT maps to Complexified Minkowski space. This space has the $(x^0,x^1,x^2,x^3)$ become complex numbers and the metric becomes a complex analytic extension of the Minkowski metric. Non-null Twistors become complex planes in that space.
The mapping to Complexified Minkowski space had several attractions in the original Twistor theory including the idea that in a curved space context maybe the disappearance of a point as a Singularity could be seen as just its disappearance from real Minkowski space, but not the more basic Twistor space.
Finally in the Incidence relation we are assuming that $\pi_{A'}$ is non-zero. In that case the Incidence relation could be geometrically considered to be a mapping to a Compactified completion of Complex Minkowski Space which includes the null cone at Infinity and thereby the other constructs of Compactified Minkowski space used in Penrose diagrams.
Given Luboš answer, a good place to learn about this stuff is straight from the horse's mouth:Spinors and Space-Time: Volume 1, Two-Spinor Calculus and Relativistic Fields and Spinors and space-time: Spinor and twistor methods in space-time geometry. |
Suppose I have a 2D mesh consisting of nonoverlapping triangles $\{T_k\}_{k=1}^N$, and a set of points $\{p_i\}_{i=1}^M \subset \cup_{k=1}^N T_K$. What is the best way to determine which triangle each of the points lies in?
For example, in the following image we have $p_1 \in T_2$, $p_2 \in T_4$, $p_3 \in T_2$, so I would like a function $f$ that returns the list $f(p_1,p_2,p_3)=[2,4,2]$.
Matlab has the function pointlocation which does what I want for Delaunay meshes, but it fails for general meshes.
My first (dumb) thought is, for all nodes $p_i$, loop through all the triangles to find out which triangle $p_i$ is in. However, this is is extremely inefficient - you might have to loop through every triangle for every point, so it could take $O(N \cdot M)$ work.
My next thought is, for all points $p_i$, find the nearest mesh node via nearest-neighbor search, then look through triangles attached to that nearest node. In this case, the work would be $O(a\cdot M\cdot log(N))$, where $a$ is the maximum number of triangles attached to any node in the mesh. There are a couple solvable but annoying issues with this approach,
It requires implementing an efficient nearest-neighbor search (or finding a library that has it), which could be a nontrivial task. It requires storing a list of which triangles are attached to each node, which my code is currently not set up for - right now there is just a list of node coordinates and a list of elements.
Altogether it seems inelegant, and I think there should be a better way. This must be a problem that arises a lot, so I was wondering if anyone could recommend the best way to approach finding what triangles the nodes are in, either theoretically or in terms of available libraries.
Thanks! |
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