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Finding angles from some other angles related to incircle Let $ABC$ be a triangle and $O$ the center of its enscribed circle.
Let $M = BO \cap AC$ and $N=CO \cap AB$ such that $\measuredangle NMB = 30°, \measuredangle MNC = 50°$.
Find $\angle ABC, \angle BCA$ and $\angle CAB$.
I also posted this here at the Art of Problem Solving.
| At first, I took $O$ to denote the center of the incircle inscribed into the triangle. Only a comment below clarified that you were actually taling about the center of the circumcircle circumscribed around the triangle. Therefore, I have two solutions, one for each interpretation.
Circumcircle
Angles without a proof
I constructed your situation using Cinderella. At first I had one point chosen freely on a line, which I later moved into a position where everything fit the way it should. From the resulting figure, I obtained the following measurements:
\begin{align*}
\measuredangle ABC &= 60° \\
\measuredangle BCA &= 70° \\
\measuredangle CAB &= 50°
\end{align*}
Warning: I accidentially swapped $M$ and $N$ as well as $B$ and $C$ in the following image. So take care to look more at the actual angles than the denoted point names.
I guess that once you know these angles at the corners, you can choose suitable coordinates and using these proove that the angles specified in your question are indeed as required.
Oriented angles
In the above, I interpreted the angles in your question as unoriented, and measured one of the clockwise and the other counter-clockwise. I furthermore assumed $M$ and $N$ to lie between the corresponding corners of the triangle. Strictly speaking, your angles are given in an oriented fashion, and as both of them are positive, $B$ and $C$ must lie on different sides of $MN$. this leads to rather ugly triangles. Of the two possible solutions I found, the better one is the following:
The angles for this triangle are rather ugly, compared to the nice numbers resulting from the interpretation above.
Incircle
Angles without a proof
This result I obtained in a similar way to the one outlined above, using Cinderella with one free point on a line adjusted till things line up as intended.
\begin{align*}
\measuredangle ABC &= 40° \\
\measuredangle BCA &= 120° \\
\measuredangle CAB &= 20°
\end{align*}
Construction
Let us call the half corner angles like this:
\begin{align*}
\alpha &= \angle BAO = \angle OAC \\
\beta &= \angle CBO = \angle OBA \\
\gamma &= \angle ACO = \angle OCB
\end{align*}
From $\triangle BCO$ you see that $\angle BCO = 180° - \beta - \gamma$. That is the same as $\angle MON$, so from $\triangle MNO$ you can conclude that $\beta+\gamma = 30° + 50°$. From $\triangle ABC$ you know that $2\alpha + 2\beta + 2\gamma = 180°$, so $\alpha = 90° - (\beta + \gamma) = 10°$.
Based on this angle, you can construct the triangle even without knowing the corner angles up front:
*
*Choose $M$ and $N$ arbitrarily
*Draw the rays $MO$ and $NO$ using the given angles to obtain $O$
*Draw lines at $90° - \alpha = 80°$ from $MO$ through $M$ and $O$ as indicated in the figure below (green). Around their intersection, draw a circle containing all points that see $M$ and $O$ under an angle of $10°$.
*Repeat the previous step for $NO$. The intersection of these two circles is $A$.
*Now $C = AM \cap NO$ and $B = AN \cap MO$.
Measuring the angles in this figure will give you the desired result up to the precision with which you performed your drawings and measurements.
| {
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Stone Weierstrass on noncompact subset I would like to ask whether we can create a function $f:\mathbb{R}\rightarrow\mathbb{R}$ which is continuous on $\mathbb{R}$ but $f$ is not the pointwise limit of any sequence of polynomial function $\{p_n(x)\}_{n\in\mathbb{N}}$.
Thank you for all helping.
| By Stone-Weierstass applied to $f$ restricted to interval $[-n,n]$ there is a polynomial $p_n$ such that $\sup_{x\in [-n,n]} |f(x)-p_n(x)|< \frac{1}{n}$. The sequence $(p_n)$ converges pointwise to $f$.
On the other hand, there is no sequence of polynomials converging uniformly to $f(x)=\sin x$, since polynomials are unbounded or constant.
| {
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Closed subsets $A,B\subset\mathbb{R}^2$ so that $A+B$ is not closed I am looking for closed subsets $A,B\subset\mathbb{R}^2$ so that $A+B$ is not closed.
I define
$A+B=\{a+b:a\in A,b\in B\}$
I thought of this example, but it is only in $\mathbb{R}$. Take:
$A=\{\frac{1}{n}:n\in\mathbb{Z^+}\}\cup\{0\}$ and $B=\mathbb{Z}$
both of these are closed (is this correct?). But their sum $A+B=\mathbb{Q}$ which is not closed.
| Let $A=\mathbb{Z}$ and $B=p\mathbb{Z}:=\{pn: n\in\mathbb{Z}\}$ where $p$ is any irrational number.
So $A$ and $B$ are two closed subsets of $\mathbb{R}$,
but, $A+B :=\{m+pn: m,n\in\mathbb{Z}\}$ is not closed in $\mathbb{R}$.
| {
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Centroids of triangle On the outside of triangle ABC construct equilateral triangles $ABC_1,BCA_1, CAB_1$, and inside of ABC, construct equilateral triangles $ABC_2,BCA_2, CAB_2$. Let $G_1,G_2,G_3$ $G_3,G_4,G_6$be respectively the centroids of triangles $ABC_1,BCA_1, CAB_1$, $ABC_2,BCA_2, CAB_2$.
Prove that the centroids of triangle $G_1G_2G_3$ and of triangle $G_4G_5G_6$ coincide
| The easiest way to get at the result is to apply an affine transformation to the original triangle so as to make it equilateral. The constructed "interior" triangles will all then coincide with the original triangle, and the conclusion can be easily drawn using basic geometry.
| {
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A question about the definition of fibre bundle The canonical definition of fibre bundle is the following:
Let $B,X,F$ be three topological spaces and $\pi:X\rightarrow B$ a continuous surjective map; then $(X,F,B,\pi)$ is a fibre bundle on $B$ if for all $b\in B$ exist an open neighbourhood of $U$ of $b$ and a homeomorphism $\phi_U:\pi^{-1}(U)\rightarrow U\times F$
such that $proj_U\circ \phi_U=\pi_{|U}$ (where $proj_U$ is the canonical projection on $U$).
A consequence of the previuous definition is that the set $\pi^{-1}(p)$ is homeomorphic to $F$ for all $p\in U$, but I don't understant why this is true. Lets try to resctrict the function $\phi_U$ to the set $\pi^{-1}(p)$ (under the hypothesis that $p\in U$), so we have:
$$proj_U\bigg(\phi_U(\pi^{-1}(p))\bigg)=\pi_{|U}(\pi^{-1}(p)) = p $$
but now we can't multiply both sides for $proj_U^{-1}$ because the function is not injective, and we can't conclude that $\phi_U(\pi^{-1}(p))=proj^{-1}(p)=\{p\}\times F$.
| If $\phi_U$ is a homeomorphism, then $\phi_U$ restricted to $\pi^{-1}(p)$ will also be a homeomorphism. But this is just a map from $\pi^{-1}(p)$ to $p\times F$ and $p\times F$ is homeomorphic to $F$.
| {
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How do I prove this using the definition of a derivative? If $f$ is a differentiable function and $g(x)=xf(x)$, use the definition of a derivative to show that $g'(x)=xf'(x)+f(x)$.
| Just do it: set up the difference quotient and take the limit as $h\to 0$. I’ll get you started:
$$\begin{align*}
g'(x)&=\lim_{h\to 0}\frac{g(x+h)-g(x)}h\\
&=\lim_{h\to 0}\frac{(x+h)f(x+h)-xf(x)}h\\
&=\lim_{h\to 0}\frac{x\big(f(x+h)-f(x)\big)+hf(x+h)}h\\
&=\lim_{h\to 0}\frac{x\big(f(x+h)-f(x)\big)}h+\lim_{h\to 0}\frac{hf(x+h)}h\;.
\end{align*}$$
Now just finish working out what those last two limits are; it shouldn’t be hard, especially when you already know what they must be.
| {
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Prove $ax - x\log(x)$ is convex? How do you prove a function like $ax - x\log(x)$ is convex? The definition doesn't seem to work easily due to the non-linearity of the log function.
Any ideas?
| A function $f(x) \in C^2(\Omega)$ is convex if its second derivative is non-negative.
$$f(x) = x \log(x) \implies f'(x) = x \cdot \dfrac1x + \log(x) \implies f''(x) = \dfrac1x > 0$$
EDIT
If $f(x) = ax - x\log(x)$, then
$f''(x) = - \dfrac1x$ and hence the function is concave.
| {
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A continuous function with measurable domain.
Let $D$ and $E$ be measurable sets and $f$ a function with domain $D\cup E$. We proved that $f$ is measurable on $D \cup E$ if and only if its restrictions to $D$ and $E$ are measurable. Is the same true if measurable is replaced by continuous.
I wrote the question straight out of the book this time to make sure I did this correctly. I mechanically replaced measurable with continuous starting with the measurable after $f$.
Below here is what I wrote before I changed the question
I've proven the case where continuous is switched for measurable. I'm just not sure of a meaningful relationship between measurable sets for domains and continuity.
| That's wrong. Let $f\colon[0,1] \to \mathbb R$ be given by
\[
x\mapsto \begin{cases} 1 & x = 0\\ 0 & x > 0
\end{cases} \]
Then $f$ is not continuous, but it is, restricted to $D := \{0\}$ and $E := (0,1]$, both of which are measurable.
Let me add something concerning your edit: As the example above shows, for just measurable parts $D$ and $E$ continuity of $f|_D$ and $f|_E$ doesn't imply that of $f$ on $D \cup E$. But if both $D$ and $E$ are open in $D \cup E$ or both closed, then your result holds:
For open $D$ and $E$ we argue as follows: Let $U \subseteq \mathbb R$ be open, then we have $f^{-1}[U] = f|_D^{-1}[U] \cup f|_E^{-1}[U]$. Now, $f|_D^{-1}[U]$ is open in $D$ by continuity of $f|_D$, as $D$ is open in $D\cup E$, $f|_D^{-1}[U]$ is also. By the same argument, replacing $D$ by $E$, $f|_E^{-1}[U]$ is open in $D\cup E$, hence $f^{-1}[U]$ is, proving the continuity of $f$.
The case of closed sets can be proved by exactly the same argument, replacing every "open" above by "closed".
| {
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Number of songs sung. There were 750 people when the first song was sung. After each song, 50 people are leaving the hall. How many songs are sung to make them zero?
The answer is 16, I am unable to understand it. I am getting 15 as the answer. Please explain.
| Listen to the question carefully: "There were $750$ people when the first song was sung"
First song was already sung $+1$;
remaining people is $750$.
$750/50=15$;
Answer is $15+1=16$
| {
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Write down a proof for $\bot\Rightarrow q$ in proposition calculus I am given the hint in the question that I will need to use the axiom $(((s\Rightarrow \bot)\Rightarrow \bot)\Rightarrow s)$.
The axioms I am using are $$(s\Rightarrow (t \Rightarrow s)) \\((s\Rightarrow(t\Rightarrow u))\Rightarrow((s\Rightarrow t)\Rightarrow(s\Rightarrow u)) \\ (((s\Rightarrow \bot)\Rightarrow \bot)\Rightarrow s)$$
In a proof every step is either an axiom or deduced by modus ponens.
| Using your first two axioms you can prove the deduction theorem. So to prove $\vdash \bot \Rightarrow q$, it's enough to prove $\bot \vdash q$. The hint suggests using the third axiom. With that, you can show that $((q \Rightarrow \bot)\Rightarrow \bot)\vdash q$. So you're done if you can prove $\bot \vdash ((q \Rightarrow \bot)\Rightarrow \bot)$. Can you see how to do this?
| {
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Find the value of f(343, 56)? I have got a problem and I am unable to think how to proceed.
$a$ and $b$ are natural numbers. Let $f(a, b)$ be the number of cells that the line joining $(a, b)$ to $(0, 0)$ cuts in the region $0 ≤ x ≤ a$ and $0 ≤ y ≤ b$. For example $f(1, 1)$ is $1$ because the line joining $(1, 1)$ and $(0, 0)$ cuts just one cell. Similarly $f(2, 1)$ is $2$ and $f(3, 2) = 4$.
Find $f(343, 56)$.
I have tried by making the equation of line joining $(0,0)$ and $(343, 56)$.
I got the equation as $8x = 49y$.
Now I tried it by randomly putting the values of $x$ and $y$ which are both less that $343$ and $56$ respectively.
But I am unable to get it.
Is there is any better approach?
Thanks in advance.
| A start: Note that $343=7\cdot 49$, and $56=7\cdot 8$. First find $f(49,8)$.
More: If $a$ and $b$ are relatively prime, draw an $a\times b$ chessboard, and think of the chessboard squares you travel through as you go from the beginning to the end.
| {
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Show that $f$ has at most one fixed point Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function. $x\in\mathbb{R}$ is a fixed point of $f$ if $f(x)=x$. Show that if $f'(t)\neq 1\;\forall\;t\in\mathbb{R}$, then $f$ has at most one fixed point.
My biggest problem with this is that it doesn't seem to be true. For example, consider $f(x)=x^2$. Then certainly $f(0)=0$ and $f(1)=1 \Rightarrow 0$ and $1$ are fixed points. But $f'(x)=2x\neq 1 \;\forall\;x\in\mathbb{R}$. Is there some sort of formulation that makes this statement correct? Am I missing something obvious? This is a problem from an old exam, so I'm assuming that maybe there's some sort of typo or missing condition.
| This problem is straight out of baby Rudin.
Assume by contradiction that $f$ has more than one fixed point. Select any two distinct fixed points, say, $x$ and $y$.
Then, $f(x) = x$ and $f(y) = y$. By the Mean Value Theorem, there exists some $\alpha \in (x,y)$ such that $f'(\alpha) = \frac{f(x)-f(y)}{x-y} = \frac{x-y}{x-y} = 1$, contradicting the hypothesis.
| {
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Reference for topology and fiber bundle I am looking for an introductory book that explains the relations of topology and bundles.
I know a basic topology and algebraic topology. But I don't know much about bundles. I want a book that
*
*explains the definition of bundles carefully and give some intuition on bundles
*explains relations between topology and bundles
*has physics motivation (if possible)
If you know a good books, please let me know. Thank you very much in advance.
| Steenrod's "The Topology Of Fibre Bundles" is a classic. It isn't particularly modern but it does the basics very well.
Husemoller's "Fibre Bundles" is a bit more modern and has a bit more of a physics-y outlook but still very much a book for mathematicians. I find it not as pleasant to read as Steenrod's book but it's fine.
Peter May has some nice notes on bundles and fibrations, available on his webpage. They're quite modern but not written with a physics outlook, very much the outlook of a topologist.
| {
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If $\#A$ and $m$ are relatively prime, then $a\mapsto ma$ is automorphism? Is it true if $A$ is a finite, abelian group and $m$ is some integer relatively prime to the order of $A$, then the map $a\mapsto ma$ is an automorphism?
It's left as an exercise in some course notes, but I cannot verify it.
In particular it is used to prove that Hall complements exist for abelian Hall subgroups of finite groups.
| Hint: since $\gcd(m,\#A)=1$, you can find an integer $d$ such that $dm\equiv1\pmod{\#A}$. Show that the map $a \mapsto da$ is the inverse of $a \mapsto ma$.
| {
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How do we draw the number hierarchy from natural to complex in a Venn diagram? I want to make a Venn diagram that shows the complete number hierarchy from the smallest (natural number) to the largest (complex number). It must include natural, integer, rational, irrational, real and complex numbers.
How do we draw the number hierarchy from natural to complex in a Venn diagram?
Edit 1:
I found a diagram as follows, but it does not include the complex number.
My doubt is that shoul I add one more rectangle, that is a litte bit larger, to enclose the real rectangle? But I think the gap is too large enough only for i, right?
Edit 2:
Is it correct if I draw as follows?
| Emmad's second link is just perfect, IMHO. For something right in front of you, here's this:
| {
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Jordan Form of a matrix I'm trying to find a matrix $P$ such that $J=P^{-1}AP$, where $J$ is the Jordan Form of the matrix:
$$A=\begin{pmatrix}
-1&2&2\\
-3&4&3\\
1&-1&0
\end{pmatrix}
$$
The characteristic polynomial is: $p(\lambda)=(\lambda-1)^3$, and a eigenvector for $A-I$ is $\begin{pmatrix}
0 \\ 1 \\-1 \end{pmatrix}$. Now, how can I find other $2$ vectors?
Thanks for your help.
| Maybe your fault is, that there is a second eigenvector, the Jordan normal form is $$\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}.$$
You can find a second eigenvector $w$ and a vector $z$ such that $(A-I)z=w$.
| {
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Continuity of function which is Lipschitz with respect to each variables separately Let a function $f: I\times J \rightarrow \mathbb R$, where $I,J$ are intervals in $\mathbb R$, be Lipschitz with respect of each variable separately. Is it then $f$ continuous with respect of both variables?
Thanks
| It depends on what you mean with Lipschitz with respect to each variable separately.
Consider the function $f\colon\mathbb R^2\to \mathbb R$, $(x,y)\mapsto xy$.
Then for fixed $y$, the function $x\mapsto f(x,y)$ is Lipschitz (with Lipschitz constant $y$ depending on $y$) and similarly $y\mapsto f(x,y)$ for fixed $x$.
However $f$ itself is not Lipschitz as $f(t,t)=t^2$ shows.
On the other hand, if there is a single Lipschitz constant $L$ working for all functions of the form $x\mapsto f(x,y)$ and $y\mapsto f(y,x)$, then $f$ is Lipschitz with constant $L\sqrt 2$.
| {
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Let $G=(V,E)$ be a connected graph with $|E|=17$ and for all vertices $\deg(v)>3$. What is the maximum value of $|V|$? Let $G=(V,E)$ be a connected graph with $|E|=17$ and for all vertices $\deg(v)>3$. What is the maximum value of $|V|$? (What is the maximum possible number of vertices?)
| HINT: Suppose that $V=\{v_1,\dots,v_n\}$. Then $$\sum_{k=1}^n\deg(v_k)=34\;;\tag{1}$$ why?
If $\deg(v_k)\ge 4$ for $k=1,\dots,n$, then $$\sum_{k=1}^n\deg(v_k)\ge\sum_{k=1}^n4\;.\tag{2}$$ Now combine $(1)$ and $(2)$.
| {
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showing $\nexists\;\beta\in\mathbb N:\alpha<\beta<\alpha+1$ I want to prove that $\nexists\; \beta\in\mathbb N$ such that $\alpha<\beta<\alpha+1$ for all $\alpha\in\mathbb N$. I just want to use the Peano axioms and $+$ and $\cdot$
If $\alpha<\beta$ then there is a $\gamma\in\mathbb N$ such that $\beta=\alpha+\gamma$.
If $\beta<\alpha+1$ then there is a $\delta\in\mathbb N$ such that $\alpha+1=\beta+\delta$.
Now I tried to equalize the two equations and I got $\gamma\le0$ which is contradictory to $\gamma\in\mathbb N$. But I used $\alpha+1-\delta=\beta$ in which the $-$ is problematic because I am not allowed to use it.
Anbody knows a better solution? Thanks a lot!
| I'm not sure if the following is allowed but:
Inserting the first equation in the second we get:
$\alpha+1=\alpha+\gamma+\delta$.
Now we can substract $\alpha$ at both sides to get $1=\gamma+\delta$, which is a contradiction.
| {
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How do you find the vertex of a (Bézier) quadratic curve? Before I elaborate, I do not mean a quadratic function! I mean a quadratic curve as seen here.
With these curves, you are given 3 points: the starting point, the control point, and the ending point. I need to know how to find the vertex of the curve. Also, I am talking about Bézier curves, but just quadratic Bézier curves-the ones with only 3 points.
| A nice question! And it has a nice answer. I arrived at it by a series of hand-drawn sketches and scribbled calculations, so I don't have time right now to present the derivation. But here is the answer:
We are given three point $P_0$, $P_1$, and $P_2$ (the start-, control-, and end-points). The Bézier curve for these points is the parabola through $P_0$ and $P_2$ whose tangents at $P_0$ and $P_2$ coincide with the lines $P_0P_1$ and $P_1P_2$ respectively. It has the parametric form
$$B(t) = (1-t)Q_0(t) + tQ_1(t)$$
where
$$Q_0(t) = (1-t)P_0+tP_1$$
and
$$Q_1(t)=(1-t)P_1+tP_2$$
Here is what you have to do to find the vertex of the parabola:
Complete the parallelogram $P_0P_1P_2P_3$ by setting $P_3 = P_0 + P_2 - P_1$. Find the parameter $t$ such that $P_0X(t)P_1$ is a right angle, where $X(t)$ is the point on $P_1P_3$ equal to $(1-t)P_1+tP_3$. Then the vertex of the parabola is the point $B(t)$.
Note that $t$ is not necessarily in $[0,1]$.
Briefly, the idea is that we follow the tangent $Q_0(t)Q_1(t)$ around the curve until the length $|Q_0(t')B(t')|$ is equal to the length $|P_0Q_0(t')|$. Then the symmetry dictates that the vertex of the parabola is reached when $t = t'/2$. You can obtain this value of $t$ by the above procedure.
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Proof needed for a function $x^2$ for irrationals and $x$ for rationals How can I prove that the function $f:[0,1] \rightarrow \mathbb{R}$, defined as
$$ f(x) = \left\{\begin{array}{l l} x &\text{if }x \in \mathbb{Q} \\ x^2 & \text{if } x \notin \mathbb{Q}
\end{array} \right. $$
is continuous on $0$ and $1$, but nowhere else?
I really don't know where to start.
I know the (equivalent) definitions of continuous functions (epsilon delta, $\lim f(x) = f(c)$, topological definition with epsilon and delta neighbourhoods, and the definition where as $x_n$ goes to $c$, it implies that $f(x_n)$ goes to $f(c)$).
| Hint: Choose a sequence of rationals converging to an irrational and vice-versa and recall
that continuity also implies sequential continuity, to conclude what you want.
Move your mouse over the gray area for a complete solution.
Consider $a \in [0,1] \backslash \mathbb{Q}$. For this $a$, choose a sequence of rationals converging to $a$ i.e. $\{a_n\}_{n=1}^{\infty}$, where $a_n \in [0,1] \cap\mathbb{Q}$. One such choice for this sequence is $a_n = \dfrac{\lfloor 10^n a\rfloor}{10^n}$. If $f$ were to be continuous (recall that continuity also implies sequential continuity), then $$\lim_{n \to \infty} f(a_n) = f\left(\lim_{n \to \infty} a_n \right) = f(a)$$ But this gives us that $a = a^2$, which is not true for any $a \in [0,1] \backslash \mathbb{Q}$. Similarly, argue when $a \in [0,1] \cap\mathbb{Q}$, by picking a sequence of irrationals converging to $a$ i.e. $\{a_n\}_{n=1}^{\infty}$, where $a_n \in [0,1] \backslash \mathbb{Q}$. One such choice is $a_n = \left(1 - \dfrac{\sqrt{2}}{2n} \right)a$. If $f$ were to be continuous (recall that continuity also implies sequential continuity), then $$\lim_{n \to \infty} f(a_n) = f \left(\lim_{n \to \infty} a_n \right) = f(a)$$ But this gives us that $a^2 = a$, which is true only for any $a =0$ or $a=1$. Hence, the function is continuous only at $0$ and $1$.
| {
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"url": "https://math.stackexchange.com/questions/217595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A question about infinite utility streams At the end of Diamond's Evaluation of Infinite Utility Streams he proves a theorem (which he doesn't give a name to, but it's at the very end of the article). There is a step in which he jumps from $(u,0)_{rep}\succ (0,u)_{rep}$ to $(u,0)\succ_t (0,u)$, and I don't understand where that comes from. It seems like it's the opposite direction of axiom A2, and I don't see how that's derived.
| Step: $(u,0)_{rep}\succ (0,u)_{rep}\implies (u,0)\succ_2(0,u)$
Proof: Suppose not. Since $\succeq_2$ is complete, we would have $(0,u)\succeq_2 (u,0)$ otherwise, and hence by A2 $(0,u)_{rep}\succeq (u,0)_{rep}$. This cannot be.
I don't see how the rest of Diamond's proof works out though. $(u,0,U)\succeq(0,u,U)$ follows now from A1, but I don't see why this should be strict.
| {
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All possible combinations of x letters (what is this called in mathematics) Firstly, thank you for looking at my question.
I would like to know what this kind of problem is called in mathematics:
Given a set of letters, find all possible 'words' you can make with those letters. For example for 'abc' the solution would be:
a, b, c, ab, ac, abc, ba, bac, bca, ca, cab, cba
Some background, I am writing a computer program to play Scrabble and need to generate all possible words given from a set of letters. I'm researching algorithms for this problem but couldn't quite figure out what the general name is for this type of problem. I'm curious to find out so I thought I would ask.
I thought this was a type of permutation problem but reading up on Permutations I see that the length of the result is set, not variable. And it's not a Combination since the order matters.
| All possible outcomes of a probability problem are called the "sample space". Is that what you were looking for?
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is intersection of two independent set probability a multiplication process? Why is the probability of intersection of two independent sets $A$ and $B$, a multiplication of their respective probabilities i.e. Why is
$$\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)?$$
this question is about the intuition behind the definition of independence of sets in a probability space
| Perhaps, one way of looking at it is the fact that intersection means to add conditions. In that sense, for every individual satisfying the condition A you have to compute how many satisfy condition B, thus the multiplication issue. Note that satisfying two conditions is scarcer than satisfying just one, so the "weight" of individuals satisfying both conditions respect the whole population must be less than the weight of the ones satisfying just one condition. Then, given the fact that probabilities are less than or equal one, multiplication seems a reasonable way of describing the process.
I know it is not the same in a formal sense, but is like saying for every row, compute how many columns, then you get the total number of pieces.
| {
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"timestamp": "2023-03-29T00:00:00",
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Benford's law with random integers I tried testing random integers for compliance with Benford's law, which they are apparently supposed to do. However, when I try doing this with Python,
map(lambda x:str(x)[0], [random.randint(0, 10000) for a in range(100000)]).count('1')
I get approximately equal frequencies for all leading digits. Why is this the case? Might it have something to do with how the pseudorandom number generator, the Mersenne twister, works?
| Benford's law applies only to distributions that are scale-invariant and thus applies approximately to many real-life data sources, especially when we measure with arbitrary units: If the leading-digit distribution of a sample is essentially the same whether we measure in inches or centimeters, this is only possible if the logarithm is equidistributed (or approximately so over a range wide enough to cover several orders of magnitudes).
| {
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Triple integration in cylindrical coordinates Determine the value of $ \int_{0}^{2} \int_{0}^{\sqrt{2x - x^2}} \int_{0}^{1} z \sqrt{x^2 +y^2} dz\,dy\,dx $
My attempt: So in cylindrical coordinates, the integrand is simply $ \rho$.
$\sqrt{2x-x^2} $ is a circle of centre (1,0) in the xy plane. So $ x^2 + y^2 = 2x => \rho^2 = 2\rho\cos\theta => \rho = 2\cos\theta $
Therfore, I arrived at the limit transformations, $ 0 < \rho < 2\cos\theta,\,\, 0 < z < 1, \text{and}\,\,0 < \theta < \frac{\pi}{2} $
Bringing this together gives $ \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\cos\theta} \int_{0}^{1} z\,\,\rho^3\,dz\,d\rho\,d\theta $ in cylindrical coordinates. Is this correct?
| As joriki noted completely; your integral would be $$ \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\cos\theta} \int_{0}^{1} z\,\,\rho^2\,dz\,d\rho\,d\theta=\bigg(\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\cos\theta}\rho^2\,d\rho\,d\theta \bigg)\times \int_{0}^{1} z\ dz\\\ =\bigg(\int_{0}^{\frac{\pi}{2}} \frac{\rho^3}{3}\bigg|_0^{2\cos(\theta)}d\theta \bigg)\times \frac{1}{2}=\frac{8}{9}$$
| {
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Constructing Riemann surfaces using the covering spaces In the paper "On the dynamics of polynomial-like mappings" of Adrien Douady and John Hamal Hubbard, there is a way of constructing Riemann surfaces. I recite it as follow:
A polynomail-like map of degree d is a triple $(U,U',f)$ where $U$ and $U'$ are open subsets of $\mathbb{C}$ isomorphic to discs, with $U'$ relatively compact in $U$, and $f: U'\rightarrow U $ a $\mathbb{C}$-analytic mapping, proper of degree $d$. Let $L \subset U' $be a compact connect subset containing $f^{-1}\left(\overline{U'}\right)$ and the critical points of $f$, and such that $X_0=U-L$ is connected. Let $X_n$ be a covering space of $X_0$ of degree $d^n$, $\rho_n:X_{n+1}\rightarrow X_n$ and $\pi_n:X_n\rightarrow X_0$ be the projections and let $X$ be the disjoint union of the $X_n$. For each $n$ choose a lifting $$\widetilde{f}_n\colon \pi_n^{-1}(U'-L)\rightarrow X_{n+1},$$ of $f$. Then $T$ is the quotient of $X$ by the equivalence relation identifying $x$ to $\widetilde{f}_n(x)$ for all $x\in \pi_n^{-1}(U'-L)$ and all $n=0,1,2,\ldots$. The open set $T'$ is the union of the images of the $X_n, n=1,2,\ldots$, and $F:T'\rightarrow T$ is induced by the $\rho_n$.
Why $T$ is a Riemann surface and isomorphic to an annulus of finit modulus? Is there anything special about the $\pi_n,\rho_n$? What kind of background do I need?
| I will give an informal answer. Your covering space is just a collection of holed spaces right? The equivalence relation just projects the holed space down onto a space isomorphic to $U'-L$. It does this by pasting the image and the preimage of $f$ together. (In an informal sense with each iteration the covering space gets bigger. To see this consider the map $z \mapsto z^2$ on the punctured disk (disk without the origin) with radius 1 from the origin. With 1 iteration you cover the disk twice. Applying it another time on the double covering of the disk you cover it four times etc. Quotienting by all these iterations you get the punctured disk.) Hence you get something isomorphic to a space with an annulus of finite modulus. The latter space is just a classical (Parabolic) Riemann surface (Google it!).
| {
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Relations between p norms The $p$-norm on $\mathbb R^n$ is given by $\|x\|_{p}=\big(\sum_{k=1}^n
|x_{k}|^p\big)^{1/p}$. For $0 < p < q$ it can be shown that $\|x\|_p\geq\|x\|_q$ (1, 2). It appears that in $\mathbb{R}^n$ a number of opposite inequalities can also be obtained. In fact, since all norms in a finite-dimensional vector space are equivalent, this must be the case. So far, I only found the following: $\|x\|_{1} \leq\sqrt n\,\|x\|_{2}$(3), $\|x\|_{2} \leq \sqrt n\,\|x\|_\infty$ (4). Geometrically, it is easy to see that opposite inequalities must hold in $\mathbb R^n$. For instance, for $n=2$ and $n=3$ one can see that for $0 < p < q$, the spheres with radius $\sqrt n$ with $\|\cdot\|_p$ inscribe spheres with radius $1$ with $\|\cdot\|_q$.
It is not hard to prove the inequality (4). According to Wikipedia, inequality (3) follows directly from Cauchy-Schwarz, but I don't see how. For $n=2$ it is easily proven (see below), but not for $n>2$. So my questions are:
*
*How can relation (3) be proven for arbitrary $n\,$?
*Can this be generalized into something of the form $\|x\|_{p} \leq C \|x\|_{q}$ for arbitrary $0<p < q\,$?
*Do any of the relations also hold for infinite-dimensional spaces, i.e. in $l^p$ spaces?
Notes:
$\|x\|_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + \big(|x_{1}|^2 + |x_{2}|^2\big) = 2|x_{1}|^2 + 2|x_{2}|^2$, hence $=2\|x\|_{2}^{2}$
$\|x\|_{1} \leq \sqrt 2 \|x\|_{2}$.
This works because $|x_{1}|^2 + |x_{2}|^2 \geq 2|x_{1}\|x_{2}|$, but only because $(|x_{1}| - |x_{2}|)^2 \geq 0$, while for more than two terms $\big(|x_{1}| \pm |x_{2}| \pm \dotsb \pm |x_{n}|\big)^2 \geq 0$ gives an inequality that never gives the right signs for the cross terms.
| *
*Using Cauchy–Schwarz inequality we get for all $x\in\mathbb{R}^n$
$$
\Vert x\Vert_1=
\sum\limits_{i=1}^n|x_i|=
\sum\limits_{i=1}^n|x_i|\cdot 1\leq
\left(\sum\limits_{i=1}^n|x_i|^2\right)^{1/2}\left(\sum\limits_{i=1}^n 1^2\right)^{1/2}=
\sqrt{n}\Vert x\Vert_2
$$
*Such a bound does exist. Recall Hölder's inequality
$$
\sum\limits_{i=1}^n |a_i||b_i|\leq
\left(\sum\limits_{i=1}^n|a_i|^r\right)^{\frac{1}{r}}\left(\sum\limits_{i=1}^n|b_i|^{\frac{r}{r-1}}\right)^{1-\frac{1}{r}}
$$
Apply it to the case $|a_i|=|x_i|^p$, $|b_i|=1$ and $r=q/p>1$
$$
\sum\limits_{i=1}^n |x_i|^p=
\sum\limits_{i=1}^n |x_i|^p\cdot 1\leq
\left(\sum\limits_{i=1}^n (|x_i|^p)^{\frac{q}{p}}\right)^{\frac{p}{q}}
\left(\sum\limits_{i=1}^n 1^{\frac{q}{q-p}}\right)^{1-\frac{p}{q}}=
\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}
$$
Then
$$
\Vert x\Vert_p=
\left(\sum\limits_{i=1}^n |x_i|^p\right)^{1/p}\leq
\left(\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}\right)^{1/p}=
\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{1}{q}} n^{\frac{1}{p}-\frac{1}{q}}=\\=
n^{1/p-1/q}\Vert x\Vert_q
$$
In fact $C=n^{1/p-1/q}$ is the best possible constant.
*For infinite dimensional case such inequality doesn't hold. For explanation see this answer.
| {
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$X_n\overset{\mathcal{D}}{\rightarrow}X$, $Y_n\overset{\mathbb{P}}{\rightarrow}Y \implies X_n\cdot Y_n\overset{\mathcal{D}}{\rightarrow}X\cdot Y\ ?$ The title says it. I know that if limiting variable $Y$ is constant a.s. (so that $\mathbb{P}(Y=c)=1)$ then the convergence in probability is equivalent to the convergence in law, i.e. $$Y_n\overset{\mathbb{P}}{\longrightarrow}c \iff Y_n\overset{\mathcal{D}}{\longrightarrow}c,$$ and then Slutsky's theorem asserts that $X_n\cdot Y_n\overset{\mathcal{D}}{\longrightarrow}X\cdot c$. But what about the case when $Y$ is not constant? Does $X_n\overset{\mathcal{D}}{\longrightarrow}X$, $Y_n\overset{\mathbb{P}}{\longrightarrow}Y$ imply $X_n\cdot Y_n\overset{\mathcal{D}}{\longrightarrow}X\cdot Y$ ?
I would appreciate any hints.
| Let $Y$ represent a fair coin with sides valued $0$ (zero) and $1$ (one). Set $Y_n = Y$, $X = Y$, $X_n = 1-Y$. The premise is fulfilled, but $X_n\cdot Y_n = 0\overset{\mathcal{D}}{\nrightarrow}Y = X\cdot Y$.
| {
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Evaluate congruences with non-prime divisor with Fermat's Little Theorem I can evaluate $ 17^{2012}\bmod13$ with Fermat's little theorem because $13$ is a prime number. (Fermat's Little theorem says $a^{p-1}\bmod p\equiv1$.)
But what if when I need to evaluate for example $12^{1729}\bmod 36$? in this case, $36$ is not a prime.
| Your example is slightly trivial, because already $12^2\equiv0\bmod36$. If the base and modulus were coprime, you could use Euler's theorem. In cases in between, where the base contains some but not all factors of the modulus, you can reduce by the common factors and then apply Euler's theorem.
| {
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Showing $\mathbb{Z}_6$ is an injective module over itself I want to show that $\mathbb{Z_{6}}$ is an injective module over itself. I was thinking in using Baer's criterion but not sure how to apply it. So it suffices to look at non-trivial ideals, the non-trivial ideals of $\mathbb{Z_{6}}$ are:
(1) $I=\{0,3\}$
(2) $J=\{0,2,4\}$
So take a $\mathbb{Z_{6}}$-map $f: I \rightarrow \mathbb{Z_{6}}$. Since $f$ is a group homomorphism it must map generators to generators right? so $3 \mapsto 1$ and $0 \rightarrow 0$. Now can we say suppose $f(1)=k$ then define $g: \mathbb{Z_{6}} \rightarrow \mathbb{Z_{6}}$ by sending the remaining elements, (those distinct from 0 and 3), say n, to $nk$?
| I found a solution for the 'general' case:
Let I be a ideal of $\mathbb{Z}/n\mathbb{Z}$ then we know that $I=\langle \overline{k} \rangle$ for some $k$ such that $k\mid n$.
If $f:I\rightarrow \mathbb{Z}/n \mathbb{Z}$ is a $\mathbb{Z}/n \mathbb{Z}$-morphism then $im f\subset I$. To show this we note that if $\overline{x}\in im f$ then there exist $\overline{c}=\overline{lk}$ with $\overline{l}\in\mathbb{Z}/n \mathbb{Z}$ s.t. $f(\overline{c})=\overline{x}$, but $n=ks$ for some $s$, so $\overline{0}=f(\overline{ln})=\overline{s}\cdot f(\overline{lk})=\overline{sx}$, then $sx=nt$ for some $t$, but again since $n=ks$, we get $x=tk$.
In particular for $\overline{k}\in I$ we have $f(\overline{k})=b\overline{k}$ for some $b$. So for any $x\in I$ we have $f(x)=bx$ then we just take the extension of $f$ to be the map $\mathbb{Z}/n\mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$ where $x\mapsto bx$.
| {
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About the homework of differentiation This problem is not solved.
$$
\begin{align}
f(x) &=\log\ \sqrt{\frac{1+\sqrt{2}x +x^2}{1-\sqrt{2}x +x^2}}+\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right) \cr
\frac{df}{dx}&=\mathord?
\end{align}
$$
| The answer is $$\frac{2\sqrt{2}}{1+x^4}$$
Hints: $$\frac{d\left(\log(1+x^2\pm\sqrt{2}x\right)}{dx}=\frac{2x\pm\sqrt{2}}{1+x^2\pm\sqrt{2}x}$$ and $$\left(1+x^2+\sqrt{2}x\right)\left(1+x^2-\sqrt{2}x\right)=\left(1+x^2\right)^2-\left(\sqrt{2}x\right)^2$$
Similarly,
$$\frac{d\left(\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right)\right)}{dx}=\frac{1}{1+\left(\frac{\sqrt{2}x}{1-x^2}\right)^2}\frac{d\left(\frac{\sqrt{2}x}{1-x^2}\right)}{dx}$$
$$=\frac{1}{1+\left(\frac{\sqrt{2}x}{1-x^2}\right)^2}\frac{\sqrt{2}}{2}\left(\frac{1}{\left(1-x\right)^2}+\frac{1}{\left(1+x\right)^2}\right)$$
The rest is by calculations.
| {
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How does $\sqrt{|e^{-y}\cos x + ie^{-y}\sin x|}= e^{-y}$ How does $\sqrt{|e^{-y}\cos x + ie^{-y}\sin x|} = e^{-y}$ which is less than $1$?
This is a step from a question I am doing but I am not sure how the square root equaled & $e^{-y}$
| I'd start by using properties of the modulus:
$$|e^{-y} \cos x + i e^{-y} \sin x|=|e^{-y}||\cos x + i \sin x|=e^{-y}|e^{ix}|=e^{-y}$$
| {
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Linear algebra: finding a Tikhonov regularizer matrix A more general soft constraint is the Tikhonov regularization constraint
$$
\mathbf{w}^\text{T}\Gamma^\text{T}\Gamma\mathbf{w} \leq C
$$
which can capture relationships among the $w_i$ (the matrix $\Gamma$ is the Tikhonov regularizer).
(a) What should $\Gamma$ be to obtain a constraint of the form $\sum_{q=0}^Q w_q^2 \leq C$?
I think this is just the identity matrix since $\sum_{q=0}^Q w_q^2 = \mathbf{w}^\text{T}\mathbf{w}$
(b) What should $\Gamma$ be to obtain a constraint of the form $\left(\sum_{q=0}^Q w_q\right)^2 \leq C$?
To me, this is saying $\mathbf{ww} \leq C$. How is it possible to get $\mathbf{ww} = \mathbf{w}^\text{T}\mathbf{w}$ just by multiplying by some $\Gamma^\text{T}\Gamma$? Where am I going wrong here?
| (a) You are right, in order to obtain $\mathbf{w}^T\Gamma^T \Gamma \mathbf{w}=\sum_{q=0}^Q w_q^2$, you should use $\Gamma=I$, where $I$ is the identity matrix.
(b) Be careful with your dimensions, $\mathbf{w}\mathbf{w}$ is not defined. You are trying to multiply a $Q\times 1$ matrix by a $Q\times 1$ matrix, which is not possible. To obtain $\mathbf{w}^T\Gamma^T \Gamma \mathbf{w}=\left(\sum_{q=0}^Q w_q\right)^2$, you should use $\Gamma=(1~ 1~\ldots~1)$, i.e. a row of ones. This implies that $\mathbf{w}^T\Gamma^T=\sum_{q=0}^Q w_q$, and thus that $\mathbf{w}^T\Gamma^T \Gamma \mathbf{w}=\left(\sum_{q=0}^Q w_q\right)^2$.
| {
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Why the spectral theorem is named "spectral theorem"? "If $V$ is a complex inner product space and $T\in \mathcal{L}(V)$. Then $V$ has an orthonormal basis Consisting of eigenvectors of T if and only if $T$ is normal".
I know that the set of orthonormal vectors is called the "spectrum" and I guess that's where the name of the theorem. But what is the reason for naming it?
| I think the top voted answer is very helpful:
Since the theory is about eigenvalues of linear operators, and Heisenberg and other physicists related the spectral lines seen with prisms
It might be worth explaining a bit more precisely how the theory is related to optics, especially for people not familiar with how a prism works (like me)
As shown in the figure, a prism automatically decomposes a light into a series of monochromatic lights and refracts each of them by a different angle.(cf Cauchy's transmission equation)
We could consider the following analogies with the Spectrum Theorem:
*
*the Matrix Operator $A$ as a prism
*the input vector $v$ as a polychromatic light
*each eigenvector of $A$ as monochromatic light, and the associated eigenvalue as the refractive index of the monochromatic light.
With spectral theorem, we know that we can decompose symmetric matrices(I stay in $\mathbb{R}$ for simplicity) into $A = V^{-1}\Lambda V = \sum{\lambda_i \vec{v_i} \cdot \vec{v_i}^t }$.
And $A \vec{x}=\sum{\lambda_i \vec{v_i} \cdot \vec{v_i}^t } \vec{x}= \sum{\lambda_i \vec{v_i} x_i }$
This is very similar to what a prism does with light: decompose, refract, and output. So, the spectrum theorem tells us we could find a spectrum of eigenvectors of a given symmetric matrix S.
References:
Spectrum of a matrix
| {
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harmonic conjugates and cauchy riemann eqns I'm trying to find function $v(x,y)$ such that the pair $(u,v)$ satisfies the Cauchy-Riemann equations for the following functions $u(x,y)$:
a) $u = \log(x^2+y^2)$
$$
u_x = v_y \Rightarrow \frac{2x}{x^2+y^2} = v_y \Rightarrow v = \frac{2xy}{x^2+y^2}?
$$
b) $u = \sin x \cosh y$
$$
u_x = \cos x \cosh y = v_y \Rightarrow v = \sinh y \cos x + C
$$
c) $u = \frac{x}{x^2+y^2}$
$u_x = v_y$, but I am getting a mess with integration. The reason is that is there a way to do this by integration, or is the way I have started this seem correct? Thanks!
| Here's how to find the corresponding imaginary parts by educated guessing. Maybe not the most systematic method, but maybe it improves your intuition about how these things behave.
For the first, remember that $e^{a + ib} = e^a(\cos b + i\sin b)$ (assuming $a,b \in \mathbb{R}$). In other words, $e^z$ sort of maps from polar coordinates to cartesian coordinates. $\log(z)$ thus does the reverse mapping, i.e. $\log(a+ib) = \log(|z|) + i\arg(a+ib) + i2\pi n$. This motivates the assumption that your first $u$ is the real part of $$
f(z) = \log(z^2)
$$
Which it actually is, since $\Re(f(x+iy)) = \log(|(x+iy)^2|) = \log(|x+iy|^2) = log(x^2+y^2)$. The corresponding $v$ is thus $v(x,y) = \Im(f(x+iy))$, i.e. $$
v(x,y) = \arg(z^2) + 2\pi n= 2\arg(z) + 2\pi n= 2\left(arctan\left(\frac{y}{x}\right) + \pi \tilde{n}\right)
$$
Your integration yields the wrong result because you forgot to take the denominator (which depends on $y$!) into account when finding the antiderivative.
For the second, use that $\sin(x+iy)$ = $\sin(x)\cos(iy) + \cos(x)\sin(iy)$ = $\sin(x)\cosh(y) +i\cos(x)\sinh(y)$ . From that, you get that $$
f(z) = \sin(z)
$$
and $$
v(x,y) = \cos(x)\sinh(y)
$$
which is the same as your integration yields.
For the third, observe that $x^2+y^2 = z\bar{z}$ if $z = x+iy$. Thus, one guess for $f(z)$ could be $\frac{z}{z\bar{z}}$. That's no good, however, because $z \to \bar{z}$ is not holomorphic. But since the real parts of $z$ and $\bar{z}$ are identical, you can also try $$
f(z) = \frac{\bar{z}}{z\bar{z}} = \frac{1}{z}
$$ And voilá, since $$
\frac{1}{z} = \frac{1}{x+iy} = \frac{x-iy}{(x+iy)(i-iy)} = \frac{x - iy}{x^2 + y^2}
$$
you indeed have $\Re(f(x+iy)) = \frac{x}{x^2+y^2}$, and thus $$
v(x,y) = \frac{-y}{x^2+y^2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Ratio of Boys and Girls In a country where everyone wants a boy, each family continues having babies till they have a boy. After some time, what is the proportion of boys to girls in the country? (Assuming probability of having a boy or a girl is the same)
| Obviously the ratio of boys to girls could be any rational number, or infinite. If you either fix the number of families or, more generally, specify a probability distribution over the number of families, then B/G is a random variable with infinite expected value (because there's always some non-zero chance that G=0).
| {
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Smooth function on $\mathbb R$ whose small increments are not controlled by the first derivative at infinity I need some help in finding a (as simple as possible) smooth function $f:\mathbb R \rightarrow \mathbb R$ which does NOT satisfy the following:
There exist a constant $C>0$, a compact $K\subset\mathbb R$ and $h_0>0$ such that for every $|h| \leq h_0$ and every $x\in\mathbb R\setminus K$
$|h|^{-1}|f(x+h) - f (x)| \leq C |f'(x)|$
EDIT: and there exists a $\tilde C>0$ such that $|f'(x)|>\tilde C$ for $x\in\mathbb R\setminus K$.
EDIT 2: my intuition is that such an $f$ may look like this: the first derivative stays always positive and oscillates (around g(x)=|x| for example), the oscillations becoming both faster and larger in amplitude when $x$C goes to infinity.
Many thanks.
| Try e.g. $f(x) = \cos(x)$. All you need is that $f$ is not constant on any interval and $f'$ has arbitrarily large zeros.
EDIT: With the new condition, take $$f(x) = \int_0^x (1 + t^2 \cos^2(t^2))\ dt = x+\frac{x \sin \left( 2\,{x}^{2} \right)}{8}-\frac{\sqrt {\pi }}{16}{\rm
FresnelS} \left( 2\,{\frac {x}{\sqrt {\pi }}} \right) +\frac{{x}^{3}}{6}
$$
Note that for $x > 0$
$$\eqalign{\frac{f(x+h) -f(x)}{h} &= \frac{1}{h} \int_x^{x+h} (1 + t^2 \cos^2(t^2))\ dt\cr
&> \frac{x^2}{h} \int_x^{x+h} \cos^2(t^2)\ dt \cr
&= \frac{x^2}{h} \left(\frac{h}{2} + \frac{\sqrt{\pi}}{4} \left({\rm FresnelC}\left(
\frac{2(x+h)}{\sqrt{\pi}}\right) - {\rm FresnelC}\left(
\frac{2x}{\sqrt{\pi}}\right) \right)\right)
\cr
&= \frac{x^2}{2} + o(x^2)}$$
as $x \to \infty$ for any fixed $h$ (since ${\rm FresnelC}(t) \to 1/2$ as $t \to \infty$), while $f'(\sqrt{(n+1/2)\pi})=1$ for positive integers $n$.
| {
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Question about exponential functions (easy one) Can someone please explain me, why when we are looking at the function $f(x)=a^x $ , we should remember that $1 \neq a >0 $ ? (And not saying that we can't put an x that satisfies:
$ 0 < x < 1 $ ?
Any understandable explanation will be great!
Thanks !
| Besides to @André's theoretical answer(+1); imagine what would be happen if $a<0$? For example if $a=-4$, what would you do with $(-4)^{\frac{m}{n}}$ wherein $n$ is even and $m$ is odd? Would we have a real number? How many fractions of this type are there in $\mathbb R$? I see the close story can be when $a=1$.
| {
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Continued Fractions Approximation I have come across continued fractions approximation but I am unsure what the steps are.
For example How would you express the following rational function in continued-fraction form:
$${x^2+3x+2 \over x^2-x+1}$$
| This might be what you are looking for:
$$
\begin{align}
\frac{x^2+3x+2}{x^2-x+1}
&=1+\cfrac{4x+1}{x^2-x+1}\\[4pt]
&=1+\cfrac1{\frac14x-\frac5{16}+\cfrac{\frac{21}{16}}{4x+1}}\\[4pt]
&=1+\cfrac1{\frac14x-\frac5{16}+\cfrac1{\frac{64}{21}x+\frac{16}{21}}}
\end{align}
$$
At each stage, we are doing a polynomial division instead of an integer division, but otherwise, the process is the same as with continued fractions with integers.
We can get the Bezout polynomials by truncating the continued fraction:
$$
1+\cfrac1{\frac14x-\frac5{16}}=\frac{4x+11}{4x-5}
$$
That is, we can write the polynomial GCD (a constant since they are relatively prime) as
$$
(4x+11)(x^2-x+1)-(4x-5)(x^2+3x+2)=21
$$
| {
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Showing that $f$ is differentiable at $x=0$, using the mean value theorem The following exercise is again about the Mean Value Theorem :)
Let $f : [0,1] \rightarrow \mathbb{R}$ be continuous and differentiable on $(0,1)$.
Assume that $$ \lim_{x\rightarrow 0^+} f'(x)= \lambda.$$
Show that $f$ is differentiable (from the right) at $0$ and that $f'(0)=\lambda$.
Hint: Mean Value Theorem.
What exactly is 'differentiable from the right'? Do I have to show that the limit as $x$ approaches $0$ from the right side exists? How can I do that?
How can I use the Mean Value Theorem to show that the derivative on $0$ equals $\lambda$?
Mean Value Theorem: If $f:[a,b] \rightarrow \mathbb{R}$ is continuous on $ [a,b] $ and differentiable on $ (a,b)$, then there exists a point $ c \in (a,b)$ where
$$ f'(c)= \frac{f(b)-f(a)}{b-a}.$$
| Being differentiable at $0$ from the right means $\displaystyle{\lim_{x \to 0^+} \frac{f(x)-f(0)}{x-\alpha}}$ exist. Its value is $f'_+(0)$.
From the Mean Value Theorem, for each $x>0$ there exists a $y$ with $0<y<x$, such that $$f'(y)=\lambda=\frac{f(x)-f(0)}{x} = \frac{f(x)-f(0)}{x-0}$$
This implies that:$$\displaystyle{\lim_{x \to 0^+}} f'(y) = \displaystyle{\lim_{x \to 0^+}} \lambda= \displaystyle{\lim_{x \to 0^+}}\frac{f(x)-f(0)}{x-0} = f'_+(0) = \lambda$$
Is this a good proof? Thanks for your help all :-)
| {
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functional analysis complementary subspace Let $Y$ and $Z$ be closed subspaces in a Banach space $X$. Show that each $x \in X$ has a unique decomposition $x = y + z$, $y\in Y$, $z\in Z$ iff $Y + Z = X$ and $Y\cap Z = \{0\}$. Show in this case that there is a constant $\alpha>0$ such that $ǁyǁ + ǁzǁ \leq\alphaǁxǁ$ for every $x \in X$
| Hint: Assume that $Y\cap Z=\{0\}$. If $z+y=x=z'+y'$ s.t. $z,z'\in Z$, $y,y'\in Y$ then $z-z'=y'-y$.
For the other direction, if $y\in Y\cap Z$ then $0+0=y+(-y)$.
| {
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Calculus 1- Find directly the derivative of a function f. The following limit represents the derivative of a function $f$ at a point $a$. Evaluate the limit.
$$\lim\limits_{h \to 0 } \frac{\sin^2\left(\frac\pi 4+h \right)-\frac 1 2} h$$
| Let $f(x)=\sin^2x$. We have, $f^{\prime}(x)=2\sin x\cos x$. In the other hand
\begin{equation}
\begin{array}{lll}
\lim_{h\rightarrow 0}\frac{\sin^2\left(\frac{\pi}{4}+h\right)-\frac{1}{2}}{h}&=&\lim_{h\rightarrow 0}\frac{f\left(\frac{\pi}{4}+h\right)-f\left(\frac{\pi}{4}\right)}{h}\\
&=&f^{\prime}(\pi/2)\\
&=&2\sin(\pi/4)\cos(\pi/4)\\
&=&2(1/\sqrt{2})(1/\sqrt{2})=1.
\end{array}
\end{equation}
| {
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Discrete Subgroups of $\mbox{Isom}(X)$ and orbits Let $X$ be a metric space, and let $G$ be a discrete subgroup of $\mbox{Isom}(X)$ in the compact-open topology. Fix $x \in X$. If $X$ is a proper metric space, it's not hard to show using Arzela-Ascoli that $Gx$ is discrete. However, is there an easy example of a metric space that is not proper so that $Gx$ is not discrete for a discrete $G \subset \mbox{Isom}(X)$?
I thought about permutations of an orthonormal basis in $l^2(\mathbb{N})$, but no luck there.
| Let $X = \mathbb{R}^2$ with the following metric: $$d((x_1,y_1),(x_2,y_2)) = \begin{cases} |y_1-y_2| & \text{if $x_1=x_2$,} \\ |y_1|+|y_2|+|x_1-x_2| & \text{if $x_1\ne x_2$}. \end{cases}$$ I don't know if there is a name for this metric, it is the length of the shortest path if we only allow arbitrary vertical segments and horizontal segments along the $x$-axis. Now let $\mathbb{R}$ act on $X$ by horizontal translation, i.e., $t \circ (x,y) = (x+t,y)$. This is obviously an isometry for every $t$, by definition of the metric. Since the $x$-axis is an orbit, and the metric restricted to the $x$-axis coincides with the usual metric, the orbit of any point $(x,0)$ under this action is non-discrete. In order to see that the group action is discrete, let $s,t \in \mathbb{R}$ with $s\ne t$, and observe that $d(t\circ(0,1),s\circ(0,1)) = d((t,1),(s,1)) = 2+|t-s| >2$, so the orbit of $(0,1)$ has no accumulation point in $X$.
| {
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Simple Tensor Product Question about Well-definedness If I want to define a homomorphism, $f$, from $A\otimes_R B$ into some $R$ module $M$. If I defined it on simple tensors $a\otimes b$ what are the conditions I need to check to make this is well defined.
Does it suffice to check that $f(r(a\otimes b))=f((ra)\otimes b)=f(a\otimes (rb))$ or is it more complicated than that.
Thank you all.
| You simply need to define an $R$-bilinear map $\tilde{f} : A \times B \rightarrow M$. The universal property of the tensor product then induces an $R$-module homomorphism $f : A \otimes_R B \rightarrow M$. To check that $\tilde{f}$ is $R$-bilinear, you must show:
(1) $\tilde{f}(ra,b) = \tilde{f}(a,rb) = r\tilde{f}(a,b)$
(2) $\tilde{f}(a_1 + a_2,b) = \tilde{f}(a_1,b) + \tilde{f}(a_2,b)$
(3) $\tilde{f}(a,b_1 + b_2) = \tilde{f}(a,b_1) + \tilde{f}(a,b_2)$
| {
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If a function is bounded almost everywhere, then globally bounded? Let $f: \mathbb{R} \to \mathbb{R}$ be a function that is bounded almost everywhere. Then is it bounded?
If so, what is the main idea or method in tis proof, and can I generalize this for upto what?
| Let $f$ map the irrationals to $0$ and the rationals to themselves. Since the rational numbers form a countable set, it has measure $0$. Then f is bounded almost everywhere but is not bounded.
More generally, on any infinite set, one can define a function that is bounded almost everywhere but is not bounded. Simply take a countable subset $x_1,x_2,x_3,\ldots$ and send $x_i$ to $i$ and other elements to $0$.
| {
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Show that $(f_n)$ is equicontinuous, given uniform convergence
Let $f_n: [a,b] \rightarrow, n \in \mathbb{N}$, be a sequence of functions converging uniformly to $f: [a,b] \rightarrow \mathbb{R}$ on $[a,b]$. Suppose that each $f_n$ is continuous on [a,b] and differentiable on (a,b), and that the sequence of derivatives $(f'_n)$ is uniformly bounded on (a,b). This means that there exists an $M>0$ such that $|f'_n(x)| \le M$ for all $x \in (a,b)$ and all $n \in \mathbb{N}$
Question: Show that $(f_n)$ is equicontinuous.
Known definitions:
*
*A sequence of functions $(f_n)$ converges uniformly to a limit function $f$ on a set $A$, if, for every $\epsilon >0$ , there exists an $N\in \mathbb{N}$. such that$|f_n(x) - f(x)| < \epsilon$ whenever $n \ge N$ and $x \in A$
*Cauchy Criterion for Uniform Convergence: A sequence of functions $(f_n)$ converges uniformly on a set $A$, if and only if, for every $\epsilon >0$ , there exists an $N\in \mathbb{N}$. such that$|f_n(x) - f_m(x)| < \epsilon$ for all $n,m \ge N$ and all $x \in A$
*A sequence of functions $(f_n)$ defined on a set $E$, is called equicontinuous if for every $\epsilon >0$ , there exists a $\delta>0$ such that $N\in \mathbb{N}$. such that$|f_n(x) - f_n(y)| < \epsilon$ for all $n \in N$ and $|x-y| \lt \delta$ in $E$
*A sequence of derivatives $(f_n')$ is uniformly bounded on (a,b) if there exists an $M>0$ such that $|f_n'(x)|≤M$ for all $x∈(a,b)$ and all $n∈N$
| Hint
$$ |f_n(x)-f_n(y)|=|(f_n(x)-f(x))+(f(x)-f(y))+(f(y)-f_n(y))| $$
$$ \leq |f_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-f_n(y)| \,.$$
Now, use the assumptions you have been given.
| {
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Show that f is uniformly continuous and that $f_n$ is equicontinuous
$f_n: A \rightarrow \mathbb{R}$,$n \in \mathbb{N}$ is a sequence of functions defined on $A \subseteq \mathbb{R} $. Suppose that $(f_n)$ converges uniformly to $f: A \rightarrow \mathbb{R}$, and that each $f_n$ is uniformly continuous on $A$.
1.) Can you show that $f$ is uniformly continuous on A?
2.) Can you show that $(f_n)$ is equicontinuous?
*
*We are given that$(f_n)$ converges uniformly to $f$. This means that for every $\epsilon>0$ there exists an $N\in \mathbb{N}$, such that $|f_n(x)-f(x)| <\epsilon$ whenever $n \ge N$ and $x \in A$. We have to show that $f$ is uniformly continuous on $A$, which means that for every $\epsilon >0$ there exists a $\delta>0$ such that $|x-y|<\delta$ implies |$f(x)-f(y)|<\epsilon$
*We need to show that for every $\epsilon>0$ there exists a $\delta>0$ such that $|f_n(x)-f_n(y)|< \epsilon$ for all $n\in \mathbb{N}$ and $\|x-y|< \delta$ in $A$
| For 1, let $\epsilon >0$. Then pick $n$ such that $|f_n(x) - f(x)| < \epsilon/3$ on $A$. By uniform continuity of $f_n$, there exists a $\delta$ such that $|x-y| < \delta \Longrightarrow |f_n(x)-f_n(y)| < \epsilon/3$. Now if $|x-y| < \delta$,
$$ |f(x)-f(y)| = |f(x)-f_n(x)+f_n(x)-f_n(y)+f_n(y) -f(y)| \leq $$
$$|f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y) -f(y)| < \epsilon$$
| {
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Continuous extension of a real function Related;
Open set in $\mathbb{R}$ is a union of at most countable collection of disjoint segments
This is the theorem i need to prove;
"Let $E(\subset \mathbb{R})$ be closed subset and $f:E\rightarrow \mathbb{R}$ be a contiuous function. Then there exists a continuous function $g:\mathbb{R} \rightarrow \mathbb{R}$ such that $g(x)=f(x), \forall x\in E$."
I have tried hours to prove this, but couldn't. I found some solutions, but ridiculously all are wrong. Every solution states that "If $x\in E$ and $x$ is not an interior point of $E$, then $x$ is an endpoint of a segment of at most countable collection of disjoint segments.". However, this is indeed false! (Check Arthur's argument in the link above)
Wrong solution Q4.5;
http://www.math.ust.hk/~majhu/Math203/Rudin/Homework15.pdf
Just like the argument in this solution, i can see that $g$ is continuous on $E^c$ and $Int(E)$. But how do i show that $g$ is continuous on $E$?
| This is a special case of the Tietze extension theorem. This is a standard result whose proof can be found in any decent topology text. A rather different proof can be found here.
| {
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prove an analytic function has at least n zeros I`m confused about this problem:
Let G be a bounded region in C whose boundary consists of n circles.
Suppose that f is a non-constant function analytic on G: Show that
if absolute value of f(z) = 1 for all z in the boundary of G then f has at least n zeros (counting multiplicities) in G.
What does it mean that boundary consists of n circles? How can I start solving the problem?
Any help please...
| I don't know what tools you have at your disposal, but this follows from some basic topology. The assumptions imply that $f$ is a proper map from the region $G$ to the unit disk $\mathbb{D}$. As such the map has a topological degree, and since the preimage of the boundary $|z|=1$ contains $n$ components, this degree has to be at least $n$, meaning that every $z\in\mathbb{D}$ has to have at least $n$ preimages, counted with multiplicity.
| {
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How many ways can the letters of the word TOMORROW be arranged if the Os can't be together? How many ways can the letters of the word TOMORROW be arranged if the Os cant be together?
I know TOMORROW can be arranged in $\frac{8!}{3!2!} = 3360$ ways. But how many ways can it be arranged if the Os can't be together? And what is the intuition behind this process?
| First, you have to remove the permutations like this TOMOORRW and TMOOORRW, so see OO as an element, then we have $3360-\frac{7!}{2!}$.
EDITED
Now, Notice you remove words like this TOMOORRW and TMOOORRW with OO and OOO, but you remove a little bit more you want, why? can you see this? how to fix this problem?
| {
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Determinant of rank-one perturbations of (invertible) matrices I read something that suggests that if $I$ is the $n$-by-$n$ identity matrix, $v$ is an $n$-dimensional real column vector with $\|v\| = 1$ (standard Euclidean norm), and $t > 0$, then
$$\det(I + t v v^T) = 1 + t$$
Can anyone prove this or provide a reference?
More generally, is there also an (easy) formula for calculating $\det(A + wv^T)$ for $v,w \in \mathbb{K}^{d \times 1}$ and some (invertible) matrix $A \in \Bbb{K}^{d \times d}$?
| I solved it. The determinant of $I+tvv^T$ is the product of its eigenvalues. $v$ is an eigenvector with eigenvalue $1+t$. $I+tvv^T$ is real and symmetric, so it has a basis of real mutually orthogonal eigenvectors, one of which is $v$. If $w$ is another one, then $(I+tvv^T)w=w$, so all the other eigenvalues are $1$.
I feel like I should have known this already. Can anyone provide a reference for this and similar facts?
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Prove that if X and Y are Normal and independent random variables, X+Y and X−Y are independent Prove that if X and Y are Normal and independent random variables, X+Y and X−Y are independent. Note that X and Y also have the same mean and standard deviation.
Note that this is a duplicate of Prove that if $X$ and $Y$ are Normal and independent random variables, $X+Y$ and $X-Y$ are independent, however, there isn't a complete solution to the answer given and I do not understand exactly what the hints are suggesting.
My attempt was to check if $f_{x+y,x-y}(u,v) = f_{x+y}(u)f_{x-y}(v)$, however, this does not seem to be working out too nicely.
| Define $U = X + Y, V = X - Y$. Then, $X = (U + V)/2, Y = (U - V)/2$. Find the Jacobian $J$ for the transformation.
Then, $f_{U,V}(u,v)=f_{X}(x=(u+v)/2)f_{Y}(y=(u-v)/2)|J|$.
You will find that $f_{U,V}(u,v)$ factors into a function of $u$ alone and a function of $v$ alone. Thus, by the Factorization thm, $U$ and $V$ are independent.
| {
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A collection of Isomorphic Groups
So the answer is
Questions
1) How exactly is $<\pi>$ isomorphic to the other integer groups? I mean $\pi$ itself isn't even an integer.
2)What is exactly is the key saying for the single element sets? Are they trying to say they are isomorphic to themselves?
3) How exactly is $\{\mathbb{Z_6}, G \}$ and $\{ \mathbb{Z_2}, S_2\}$ isomorphic?
| 1) Notice that $\langle \pi \rangle=\{\pi^n \mid n \in \mathbb Z\}$ and we also have $\pi^n\pi^m=\pi^{n+m}$ what group does this remind you of?
2) The single element sets are not isomorphic to any of the other groups. As an aside it's not particularly meaningful to say a group is isomorphic to itself. You generally speak of two groups that are distinct in some meaningful way as being isomorphic.
| {
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How to Decompose $\mathbb{N}$ like this?
Possible Duplicate:
Partitioning an infinite set
Partition of N into infinite number of infinite disjoint sets?
Is it possible to find a family of sets $X_{i}$, $i\in\mathbb{N}$, such that:
*
*$\forall i$, $X_i$ is infinite,
*$X_i\cap X_j=\emptyset$ for $i\neq j$,
*$\mathbb{N}=\bigcup_{i=1}^{\infty}X_i$
Maybe it is an easy question, but I'm curious about the answer and I couldnt figure out any solution.
Thanks
| Let $p_n$ be the $n$-th prime number. That is $p_1=2; p_2=3; p_3=5; p_4=7$ and so on.
For $n>0$ let $X_n=\{(p_n)^k\mid k\in\mathbb N\setminus\{0\}\}$.
For $X_0$ take all the rest of the numbers available, namely $k\in X_0$ if and only if $k$ can be divided by two distinct prime numbers, or if $k=1$.
| {
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Weakly convex functions are convex Let us consider a continuous function $f \colon \mathbb{R} \to \mathbb{R}$. Let us call $f$ weakly convex if
$$
\int_{-\infty}^{+\infty}f(x)[\varphi(x+h)+\varphi(x-h)-2\varphi(x)]dx\geq 0 \tag{1}
$$
for all $h \in \mathbb{R}$ and all $\varphi \in C_0^\infty(\mathbb{R})$ with $\varphi \geq 0$.
I was told that $f$ is weakly convex if, and only if, $f$ is convex; although I can imagine that (1) is essentially the statement $f'' \geq 0$ in a weak sense, I cannot find a complete proof.
Is this well-known? Is there any reference?
| By a change of variables (translation-invariance of Lebesgue measure) the given inequality can be equivalently rewritten as
$$
\int [f(x+h)+f(x-h)-2f(x)]\varphi(x)\,dx \geq 0 \qquad \text{for all }0 \leq \varphi \in C^{\infty}_0(\mathbb{R})\text{ and all }h \gt 0.
$$
If $f$ were not midpoint convex then there would be $x \in \mathbb{R}$ and $h \gt 0$ such that $f(x+h) + f(x-h) - 2f(x) \lt 0$. By continuity of $f$ this must hold in some small neighborhood $U$ of $x$, so taking any nonzero $\varphi \geq 0$ supported in $U$ would yield a contradiction to the assumed inequality.
Thus, $f$ is midpoint convex and hence convex because $f$ is continuous.
Edit: The converse direction should be clear: it follows from $f(x+h) + f(x-h) - 2f(x) \geq 0$ for convex $f$ and all $h \gt 0$ so that the integrand in the first paragraph is non-negative.
Finally, the argument above works without essential change for continuous $f \colon \mathbb{R}^n \to \mathbb{R}$.
| {
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Question about how to implicitly differentiate composite functions I have a question. How in general would one differentiate a composite function like $F(x,y,z)=2x^2-yz+xz^2$ where $x=2\sin t$ , $y=t^2-t+1$ , and $z = 3e^{-1}$ ? I want to find the value of $\frac{dF}{dt}$ evaluated at $t=0$ and I don't know how. Can someone please walk me through this?
I tried a couple of things, including chain rules and jacobians. I know that $\frac{dF}{dt}$ should equal $\frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y} \frac{dy}{dt} + \frac{\partial F}{\partial z} \frac{dz}{dt}$ but for some reason this doesn't work, or I am doing something wrong. I start out by differentiating to get $\frac{\partial F}{\partial x}=4x+z^2$, $\frac{\partial F}{\partial y}= -z$, $\frac{\partial F}{\partial z} = 2xz-y$, $\frac{dz}{dt}=0$, $\frac{dx}{dt}=2\cos t$, $\frac{dy}{dt}=2t-1$ but this doesn't match the answer, which my book says is $24$.
How do they get this, and where is my error? Thanks.
Update:
What I get is as follows:
$F(x,y,z)=2x^2-yz+xz^2$,
$\frac{\partial F}{\partial t}=\frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y} \frac{dy}{dt} + \frac{\partial F}{\partial z} \frac{dz}{dt}$,$\frac{\partial F}{\partial t}=(4x+z^2)(2cos(t))-z(2t-1)$ Which for $t=0$ gives $x=0$ and
$\left. \frac{\partial F}{\partial t} \right|_{t=0} = 2z^2+z=9e^{-2}+3e^{-1}$ which clearly isn't $24$ so I must be doing something completely wrong.
Edit: I want to rephrase the question. Since everyone else I have talked to thinks there was an error in the book, does everyone here agree?
| $$\frac{\partial F}{\partial t}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}=$$
$$=(4x+z^2)\cdot 2\cos t-z(2t-1)+(2xz-y)\cdot 0$$
You'll now to substitute:
$$t=0\Longrightarrow\,x=0\,,\,y=1\,,\,z=3\,e^{-1}$$
The final result is, if I'm not wrong,
$$9\,e^{-2}+3\,e^{-1}$$
which has nothing to do with $\,24\,$, so either the book (which one, btw?) has a mistake or you miscopied the exercise.
Ps. Please do you check that according to what you wrote $\,z\,$ is independent of $\,t\,$...
| {
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An inequality for all natural numbers Prove, using the principle of induction, that for all $n \in \mathbb{N}$, we have have the following inequality:
$$1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n} \leq 2\sqrt n$$
| Suppose
$1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n} \leq 2\sqrt n$
and
$1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt {n+1}} > 2\sqrt {n+1}$
(i.e., that the induction hypothesis is false).
Subtracting these,
$\frac{1}{\sqrt {n+1}} > 2\sqrt {n+1} - 2\sqrt n
= 2(\sqrt {n+1} - \sqrt n)\frac{\sqrt {n+1} +\sqrt n}{\sqrt {n+1} +\sqrt n}
= \frac{2}{\sqrt {n+1} +\sqrt n}
$
or
$\sqrt {n+1} +\sqrt n > 2 \sqrt {n+1}$, which is false.
So the induction hypothesis is true (once a base case is established).
| {
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Minimum Number of Nodes for Full Binary Tree with Level $\lambda$ If the level ($\lambda$) of a full binary tree at zero is just a root node, than I know that I can get the maximum possible number of nodes (N) for a full binary tree using the following:
N = $2^{\lambda+1}$- 1
Is the minimum possible number of nodes the following?
N = 2*$\lambda$ + 1
| Sorry to say but what all of you are discussing. AFAIK for full binary tree nodes = [(2^(h+1)) - 1] (fixed).
*
*For strict binary tree, max node = [2^h + 1] and min node = [2h + 1].
| {
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Show $x^2 +xy-y^2 = 0$ is only true when $x$ & $y$ are zero. Show that it is impossible to find non-zero integers $x$ and $y$ satisfying $x^2 +xy-y^2 = 0$.
| The quadratic form factors into a product of lines $$0 = x^2 + xy - y^2 = -(y-\tfrac{1-\sqrt{5}}{2}x)(y-\tfrac{1+\sqrt{5}}{2}x),$$ equality holds if either
*
*$y=\tfrac{1-\sqrt{5}}{2}x$
*$y=\tfrac{1+\sqrt{5}}{2}x$
but this can't happen for $x,y$ integers unless they're both zero.
| {
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Proving the 3-dimensional representation of S3 is reducible The 3-dimensional representation of the group S3 can be constructed by introducing a vector $(a,b,c)$ and permute its component by matrix multiplication.
For example, the representation for the operation $(23):(a,b,c)\rightarrow(a,c,b)$ is
$
D(23)=\left(\begin{matrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \end{matrix}\right)
$
and so forth.
The exercise is to prove this representation is reducible. The hint tells me to find a common eigenvector for all 6 matrices which is just $(1,1,1)$. How do I proceed from here? Any help is appreciated.
| Here's another way to prove it's reducible, although it may depend on stuff you haven't learned yet. The order of the group is the sum of the squares of the degrees of the irreducible representations. So a group of order 6 can't have an irreducible representation of degree 3; $3^2\gt6$.
| {
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Orders of the Normal Subgroups of $A_4$
Prove that $A_4$ has no normal subgroup of order $3.$
This is how I started:
Assume that $A_4$ has a normal subgroup of order $3$, for example $K$.
I take the Quotient Group $A_4/K$ with $4$ distinct cosets, each of order $3$.
But I want to prove that these distinct cosets will not contain $(12)(34),(13)(24)$ and $(14)(23)$> Therefore a contradiction.
Please help, I'm really stuck!!
| In fact, one can show that all normal subgroups of $A_4$ are $1$, $K_4$ (Klein four group) and $A_4$. Note that two permutations of $S_n$ are conjugate iff they have the same type. So we can write down (by some easy calculations) all conjugate classes of $A_4$ are the following $4 $ classes:
*
*type $1^4$: {(1)}
*type $2^2$: {(12)(34),(13)(24),(14)(23)}
*type $3^1$:{(123),(142),(134),(243)} and {(132),(124),(143),(234)}
So all normal subgroups of $A_4$ are $1$, $K_4$ and $A_4$.
| {
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Finding the limit of this function as n tends to infinity... $$\lim_{n\rightarrow\infty}\frac{n}{3}\left[\ln\left(e-\frac{3}{n}\right)t-1\right]$$
I'm having little trouble figuring this out.
I did try to differentiate it about 3 times and ended up with something like this
$$f'''(n) = \frac{1}{3} \left(\frac{1}{e - \frac{3}{n}}\right) + \left(\frac{1}{3e - \frac{9}{n}}\right) - \left(\frac{9}{3e - \frac{9}{n}}\right)$$
So I wonder if the limit of this would be calculated as
$$lim_{n\rightarrow\infty}\frac{1}{3} \left(\frac{1}{e - \frac{3}{n}}\right) + \left(\frac{1}{3e - \frac{9}{n}}\right) - \left(\frac{9}{3e - \frac{9}{n}}\right) = \frac{-7}{3e}$$
Which feels terribly wrong.
I suspect that I did the differentiation wrong.
Any pointers would be cool and the provision of a simpler method would be dynamite.
| Your limit is equal to
$$\lim_{n\to +\infty}\frac{n}{3}\,\log\left(1-\frac{3}{en}\right),$$
but since:
$$\lim_{x\to 0}\frac{\log(1-x)}{x}=-1,$$
(by squeezing, by convexity or by De l'Hopital rule) you have:
$$\lim_{n\to +\infty}\frac{n}{3}\,\log\left(1-\frac{3}{en}\right)=-\frac{1}{e}.$$
| {
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If $ A$ is open and $ B$ is closed, is $B\setminus A$ open or closed or neither? If $ A$ is open and $ B$ is closed, is $B\setminus A$ open or closed or neither?
I think it is closed, is that right? How can I prove it?
| Yes, if $A$ is open and $B$ is closed, then $B\setminus A$ is closed. To prove it, just note that $X\setminus A$ is closed (where $X$ is the whole space), and $B\setminus A=B\cap(X\setminus A)$, so $B\setminus A$ is the intersection of two closed sets and is therefore closed.
Alternatively, you can observe that $X\setminus(B\setminus A)=(X\setminus B)\cup A$ is the union of two open sets and therefore open, so its complement, $B\setminus A$, is closed.
| {
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Is every subset of a metric space a metric subspace? Is every subset of a metric space a metric subspace? A simple proof does justify that all are subspaces, still, wanted to know if I missed something.
| Let $(x,p)$ be metric space and letting $Y$ be a non empty subset of $X$. Define the function $§$ on $Y$. $Y$ by $§(x,y)=p(x,y)$ for all $x,y$ in $Y$.Then $(Y,§)$ is also a metric space called a subspace of the metric space $(x,p)$.
| {
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Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$ Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$. Please show me the way you find it.
The answer in my textbook is $f(x)=\frac{1+x^2+x^4}{x\cdot \sqrt{1-x^2}}$
| $$f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2-2$$
Let $x+\frac{1}{x}=z$.
Then we get, $$f(z)=z^2-2.$$
Hence we put x on the place of z. And we get
$f(x)=x^2-2$.
| {
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How do i solve this double integral $$\int_0^1\int_{-\pi}^\pi x\sqrt{1-x^2\sin^2(y)}\mathrm{d}y\mathrm{d}x$$
How do I solve this question here?
| Switch the order of integration.
Integrating first over $x$, we obtain ($u=1-x^2 \sin^2y$)
$$\int_0^1\!dx\, x \sqrt{1-x^2 \sin^2 y}
= \frac{1}{2 \sin^2 y} \int_{\cos^2y}^1\!du\,\sqrt{u}
= \frac{1}{3\sin^2 y} ( 1 - |\cos^3 y|).$$
What is missing is the integral over $y$. Using one of the standard method to integral rational functions of trigonometric function over a full period, we obtain finally
$$\int_0^1\!dx\int_{-\pi}^\pi \!dy\,x\sqrt{1-x^2\sin^2(y)} = \frac{8}{3}.$$
| {
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Lie Groups question from Brian Hall's Lie Groups, Lie Algebras and their representations. In page 60 of Hall's textbook, ex. 8 assignment (c), he asks me to prove that if $A$ is a unipotent matrix then $\exp(\log A))=A$.
In the hint he gives to show that for $A(t)=I+t(A-I)$ we get
$$\exp(\log A(t)) = A(t) , \ t<<1$$
I don't see how I can plug here $t=1$, this equality is true only for $t \rightarrow 0$, right?
| If $A$ is unipotent, then $A - I$ is nilpotent, meaning that $(A-I)^n = 0$ for all sufficiently large $n$. This will turn your power series into a polynomial.
| {
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A tough differential calculus problem This is a question I've had a lot of trouble with. I HAVE solved it, however, with a lot of trouble and with an extremely ugly calculation. So I want to ask you guys (who are probably more 'mathematically-minded' so to say) how you would solve this. Keep in mind that you shouldn't use too advanced stuff, no differential equations or similair things learned in college:
Given are the functions $f_p(x) = \dfrac{9\sqrt{x^2+p}}{x^2+2}$. The line $k$ with a slope of 2,5 touches $f_p$ in $A$ with $x_A = -1$. Get the function of k algebraically.
*
*I might have used wrong terminology, because English is not my native language, I will hopefully clear up doubts on what this problem is by showing what I did.
First off, I got $[f_p(x)]'$. This was EXTREMELY troublesome, and is the main reason why I found this problem challenging, because of all the steps. Can you guys show me the easiest and especially quickest way to get this derivative?
After that, I filled in $-1$ in the derivative and made the derivative equal to $2\dfrac{1}{2}$, this was also troublesome for me, I kept getting wrong answers for a while, again: Can you guys show me the easiest and especially quickest way to solve this?
After you get p it is pretty straightforward. I know this might sound like a weird question, but it basically boils down to: I need quicker and easier ways to do this. I don't want to make careless mistakes, but because the length of these types of question, it ALWAYS happens. Any tips or tricks regarding this topic in general would be much appreciated too.
Update: A bounty will go to the person with the most clear and concise way of solving this question!
| By "touches" I assume you mean that the line is tangent to the graph of $f_p$. You can try implicit differentiation. Start with
$$ y = \frac{9\sqrt{x^2 + p}}{x^2 + 2}. $$
Multiply by $x^2 + 2$ to get
$$ y(x^2 + 2) = 9\sqrt{x^2 + p}. $$
Squaring, you get
$$ y^2 (x^2 + 2)^2 = 81 (x^2 + p). $$
Differentiate both sides implicitly by $x$:
$$ 2yy'(x^2 + 2)^2 + y^2 2 (x^2 + 2) 2x = 162x. $$
Now, plug in all the data ($x = -1$, $y'(-1) = 2.5$) to get a quadratic equation for $y(-1) = y$:
$$ -12y^2 + 45y = -162. $$
Solutions are $y = 6$ and $y = \frac{-9}{4}$, but notice your function is always positive, so $y(-1) = 6$ and the line is $2.5x + 8.5$.
| {
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Do these $\delta-\epsilon$ proofs work? I'm new to $\delta, \epsilon$ proofs and not sure if I've got the hand of them quite yet.
$$ \lim_{x\to -2} (2x^2+5x+3)= 1 $$
$|2x^2 + 5x + 3 - 1| < \epsilon$
$|(2x + 1)(x + 2)| < \epsilon$
$|(2x + 4 - 3)(x + 2)| < \epsilon$
$|(2(x+2)^2 -3(x + 2)| \leq 2|x+2|^2 +3|x + 2| < \epsilon$ (via the triangle inequality)
Let $|x+2| < 1$
Then $\delta=\min\left(\dfrac{\epsilon}{5}, 1\right)$
and
$$\lim_{x\to -2} (3x^2+4x-2)= 2$$
$|3x^2 + 4x - 2 - 2| < \epsilon$
$|3x^2 - 12 + 4x + 8| < \epsilon$
$|3(x+2)(x-2) + 4(x + 2)| < \epsilon$
$|3(x+2)(x + 2 -4) + 4(x + 2)| < \epsilon$
$|3[(x+2)^2 -4(x+2)] + 4(x + 2)| < \epsilon$
$|3(x+2)^2 - 8(x+2)| \leq 3|x+2|^2 + 8|x+2| < \epsilon$
Let $|x + 2| < 1$
Then $\delta =\min\left(\dfrac{\epsilon}{11}, 1\right)$
and to make sure I'm understanding this properly, when we assert that $|x+2| < 1$, this means that $\delta \leq 1$ as well, because if $\delta \geq 1$, this would allow for $|x+2| \geq 1$, which violates that condition we just imposed?
Edit: Apologies for the bad tex
| $$ \lim_{x\to -2} (2x^2+5x+3)= 1 $$
Finding $\delta$: $|2x^2 + 5x + 3 - 1| < \epsilon$
$|(2x + 1)(x + 2)| < \epsilon$
$|x - (-2)| < \delta $, pick $\delta = 3$
$|x+2| < 3 \Rightarrow -5 < x < 1 \Rightarrow -9 < 2x + 1 < 3$
This implies $|2x + 1||x + 2| < 3 \cdot |x + 2| < \epsilon \Rightarrow |x+2| < \frac{\epsilon}{3} $
$\\[22pt]$
Actual Proof: Let $\epsilon > 0 $. Choose $\delta = min\{3,\frac{\epsilon}{3}\}$
and assume that $0 < |x + 2| < \delta \Rightarrow |2x + 1||x+2| < 3 \cdot |x + 2| < 3 \cdot \frac{\epsilon}{3} = \epsilon$
| {
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Prove that $1-x/3\le\frac{\sin x}x\le1.1-x/4, \forall x\in(0,\pi]$ Prove that
$$1- \frac{x}{3} \le \frac{\sin x}x \le 1.1 - \frac{x}{4}, \quad \forall x\in(0,\pi].$$
| From the concavity of $f(x)=\cos x$ over $[0,\pi/2]$, we have:
$$ \forall x\in[0,\pi/2],\quad \cos x\geq 1-\frac{2}{\pi}x, $$
from which
$$ \forall x\in[0,\pi/2],\quad \sin x\geq \frac{1}{\pi}x(\pi-x) = x-\frac{1}{\pi}x^2$$
follows, by integration. Now, both the RHS and the LHS are symmetric wrt $x=\frac{\pi}{2}$, so we can extend the inequality over the whole $[0,\pi]$ interval:
$$ \forall x\in[0,\pi],\quad \frac{\sin x}{x}\geq 1-\frac{x}{\pi}\geq 1-\frac{x}{3}, $$
proving the "easier" inequality. Using an analogue tecnique, it's possible to establish something a little weaker than the other desired inequality. Since:
$$\operatorname{argmax}_{[0,\pi]}\left(\cos x+\frac{x}{2}\right)=\frac{\pi}{6},$$
we have:
$$ \forall x\in[0,\pi],\quad \cos x+\frac{1}{2}-\left(\frac{\pi}{12}+\cos\frac{\pi}{6}\right)\leq 0,$$
so, by integration,
$$\forall x\in[0,\pi],\quad \sin x +\frac{x^2}{4} - \left(\frac{\pi}{12}+\cos\frac{\pi}{6}\right)x \leq 0,$$
holds, that is:
$$\forall x\in[0,\pi],\quad \frac{\sin x}{x}\leq 1.1278247\ldots-\frac{x}{4}.$$
| {
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Question about complex numbers (what's wrong with my reasoning)? Can someone point out the flaw here?
$$e^{-3\pi i/4} = e^{5\pi i/4}$$
So raising to $\frac{1}{2}$, we should get
$$e^{-3\pi i/8} = e^{5\pi i/8}$$
but this is false.
| Paraphrase using $e^0=1$ and $e^{\pi i}=-1$. We can write
$$
e^{-3\pi i/4}\;1^2=e^{-3\pi i/4}\;(-1)^2
$$
Raising to the $\frac12$ power yields
$$
e^{-3\pi i/8}\;1=e^{-3\pi i/8}\;(-1)
$$
The problem is that without proper restrictions (e.g. branch cuts), the square root is not well-defined on $\mathbb{C}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does the category framework permit new logics? It appears to me that a topos permits a broader concept of subsets than the yes/no decission of a characteristic function in a set theory setting. Probably because the subobject classifier doesn't have to be {0,1}.
But I wonder, aren't all the multivalued logics also part of/can be modeled in set theory? Is there some new logic coming in with topoi which weren't there before? Did it just help discovering new ideas? Fuzzy stuff etc. are all existent in "conventional set theory mathematics" already, right?
| No, you can't model all multivalued logics in set theory. Set theory models classical propositional logic, but it does not model a logic where say the principle of contradiction fails and its negation fail also. All formal theorems of any multivalued logic exist within classical logic in the sense that if A comes as a formula in multivalued logic, it will also happen in classical logic and thus can get modeled by set theory (the converse does seem to hold for some multivalued logics, but hardly all that many of them). But, the domain of truth values differs for a formula in a multivalued logic than in classical logic.
Fuzzy stuff does NOT exist withing conventional set theory mathematics. The axiom of extensionality does not hold for fuzzy sets.
| {
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"source": "stackexchange",
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Why is every representation of $\textrm{GL}_n(\Bbb{C})$ completely determined by its character? I know that every (Lie group) representation of $\textrm{GL}_n(\Bbb{C})$ is completely reducible; this I believe comes from the fact that every representation of the maximal compact subgroup $\textrm{U}(n)$ is completely reducible. More explicitly, suppose $V$ is a representation of $\textrm{GL}_n(\Bbb{C})$. Then $V$ is also a representation of $\textrm{U}(n)$, by complete reducibility of the unitary group we know that there is a $\textrm{U}(n)$ invariant inner product such that if $U$ is any $\textrm{GL}_n$ - invariant subspace of $V$ (and hence $\textrm{U}(n)$ invariant), there is an orthogonal complement $W$ such that
$$V = U \oplus W$$
with $W$ invariant under $\textrm{U}(n)$. Now $W$ as a representation of the real Lie algebra $\mathfrak{u}(n)$ is invariant and hence under the complexified Lie algebra
$$\mathfrak{gl}_n = \mathfrak{u}_n \oplus i \hspace{1mm} \mathfrak{u}(n).$$
Since $\textrm{GL}_n(\Bbb{C})$ is connected $W$ is also invariant under $\textrm{GL}_n$ showing that every representation of it is completely reducible.
Now I have read several textbooks on representation theory (e.g. Bump's Lie Groups, Procesi's book of the same name) and they all seem to tacitly assume that every representation of $\textrm{GL}_n$ is completely determined by its character; i.e. if two representations have the same character then they are isomorphic.
Now in the finite groups case, we concluded this fact based on 1) Maschke's Theorem and 2) Linear independence of characters.
We do not necessarily have 2) so how can we conclude the fact I said about about $\textrm{GL}_n$?
Thanks.
| This boils down to facts about the representation theory of compact groups: there every complex representation is determined by its character. Now a representation of $GL(n,\mathbb C)$ is determines by a representation of its Lie algebra. But this is the complexification of $u(n)$ and complex representations of a Lie algebra are in one to one correspondence with complex representations of its complexification.
| {
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If every proper quotient is finite, then $G\cong\mathbb Z$ Here is my problem:
Let $G$ is an infinite abelian group. Prove that if every proper quotient is finite, then $G\cong\mathbb Z$.
And here is my incompleted approach:
I know that the quotient subgroup $\frac{G}{tG}$ wherein $tG$ is torsion subgroup of $G$ is always torsion-free. So, if $tG\neq\{0\}$ then here we have $\frac{G}{tG}$ torsion-free and finite simultonously which is a contradiction. Then $G$ is itself a torsion-free group.
Moreover, I assume $G$ be a divisible group, so: $$G\cong\sum\mathbb Q\oplus\sum_{p\in P}\mathbb Z(p^{\infty})$$ As any proper quotient of $G$ is infinite, so I concluded it is not divisible. I confess that I am missing the final part. If my way to this problem untill my last conclusion is valid logically, please help me about the last part of the proof. Thanks
| Let $a_0\in G$ be a nonzero element.
Then $\langle a\rangle\cong \mathbb Z$ as $G$ is torsion-free (which you have shown).
Now $Q_0=G/\langle a_0\rangle$ is a finite abelian group.
If $Q_0\cong 1$, we are done.
Otherwise, select $a_1\in G\setminus\langle a_0\rangle$. Then let $Q_1=G/\langle a_0, a_1\rangle$, etc.
The orders of the finite groups $Q_0, Q_1, \ldots$ are strictly decreasing as long as they are $>1$, hence we ultimately find an $a_n$ with $G=\langle a_1, \ldots,a_n\rangle$.
Thus $G$ is a finitely generated abelian group. A quick check with the classification theorem shows that $G\cong \mathbb Z$.
| {
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Comparing Two Sums with Binomial Coefficients How do I use pascals identity:
$${2n\choose 2k}={2n-1\choose 2k}+{2n-1\choose 2k-1}$$ to prove that
$$\displaystyle\sum_{k=0}^{n}{2n\choose 2k}=\displaystyle\sum_{k=0}^{2n-1}{2n-1\choose k}$$
for every positive integer $n$ ?
| Other than Pascal's identity, we just notice that the sums on the right are the same because they cover the same binomial coefficients (red=even, green=odd, and blue=both).
$$
\begin{align}
\sum_{k=0}^n\binom{2n}{2k}
&=\sum_{k=0}^n\color{#C00000}{\binom{2n-1}{2k}}+\color{#00A000}{\binom{2n-1}{2k-1}}\\
&=\sum_{k=0}^{2n-1}\color{#0000FF}{\binom{2n-1}{k}}
\end{align}
$$
Note that $\binom{2n-1}{-1}=\binom{2n-1}{2n}=0$.
| {
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Laplace integral and leading order behavior Consider the integral:
$$
\int_0^{\pi/2}\sqrt{\sin t}e^{-x\sin^4 t} \, dt
$$
I'm trying to use Laplace's method to find its leading asymptotic behavior as $x\rightarrow\infty$, but I'm running into problems because the maximum of $\phi(t)$ (i.e. $-\sin^{4}t$) is $0$ and occurs at $0$ (call this $c$). In my notes on the Laplace Method, it specifically demands that $f(c)$ (in this case $f(t)=\sqrt{\sin t}$) cannot equal zero--but it does. How do I get around this?
| Plot integrand for few values of $x$:
It is apparent that the maximum shifts closer to the origin as $x$ grows.
Let's rewrite the integrand as follows:
$$
\int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t = \int_0^{\pi/2} \exp\left(\frac{1}{2} \log(\sin(t))-x \sin^4(t)\right) \mathrm{d}t
$$
The maximum of the integrand is determined by
$$
0 = \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{1}{2} \log(\sin(t))-x \sin^4(t)\right) = \cot(t) \left( \frac{1}{2} - 4 x \sin^4(t)\right)
$$
that is at $t_\ast = \arcsin\left((8 x)^{-1/4}\right)$. Then using Laplace's method:
$$
\int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t \approx \int_{0}^{\pi/2} \exp\left(\phi(t_\ast) + \frac{1}{2} \phi^{\prime\prime}(t_\ast) (t-t_\ast)^2 \right) \mathrm{d}t = \exp\left(\phi(t_\ast)\right) \sqrt{\frac{2\pi}{-\phi^{\prime\prime}(t_\ast)}}
$$
Easy algebra gives $\exp\left(\phi(t_\ast)\right) = (8 \mathrm{e} x)^{-1/8}$, $-\phi^{\prime\prime}(t_\ast) = 4 \sqrt{2 x} - 2$, giving
$$
\int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t \approx (8 \mathrm{e} x)^{-1/8} \sqrt{ \frac{\pi}{2 \sqrt{2 x} -1}}
$$
| {
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Check my workings: Show that $\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$ Let $f''$ be continuous on $\mathbb{R}$. Show that
$$\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$$
My workings
$$\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=\lim_{h\to0}\frac{f(x+h)-f(x)-[f(x)-f(x-h)]}{h^2}=\frac{\lim_{h\to0}\frac{f(x+h)-f(x)}{h}-\lim_{h\to0}\frac{f(x)-f(x-h)}{h}}{\lim_{h\to0}h}$$
By the definition of derivative, I move on to the next step. Also, I observe that everything in this question as continuous and differentiable up to $f''(x)$.
$$=\frac{f'(x)-f'(x-h)}{\lim_{h\to0}h}$$
I do not know how to justify the next move but,
$$=\lim_{h\to0}\frac{f'(x)-f'(x-h)}{h}$$
Then by the definition of derivative again,
$$=f''(x-h)$$
Which is so close to the answer. So I shall assume that since $h\to0$ for $x-h$, therefore $x-h=x$? And so,
$$=f''(x)$$
I think i made a crapload of generalization and fallactic errors... I also have another way, which was to work from $f''(x)$ to the LHS. But I realised I assume that the h were the same for $f'(x)$ and $f''(x)$.
Is it normal to be unable to solve this question at the first try? Or am I just too weak in mathematics?
| Try applying L'Hospital's Rule to $h$, that is, differentiate with respect to $h$.
| {
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Evaluate $\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$.
Evaluate $\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$
I sense the answer has some connection with $e$, but I don't know how it is. Please help. Thank you.
| For $\frac{P(n)}{(n-r)!},$ where $P(n)$ is a polynomial.
If the degree of $P(n)$ is $m>0,$ we can write $P(n)=A_0+A_1(n-r)+A_2(n-r)(n-r-1)+\cdots+A_m(n-r)(n-r-1)\cdots\{(n-r)-(m-1)\}$
Here $k^2=C+B(k-1)+A(k-1)(k-2)$
Putting $k=1$ in the above identity, $C=1$
$k=2,B+C=4\implies B=3$
$k=0\implies 2A-B+C=0,2A=B-C=3-1\implies A=1$
or comparing the coefficients of $k^2,A=1$
So, $$\frac{k^2}{(k-1)!}=\frac{(k-1)(k-2)+3(k-1)+1}{(k-1)!}=\frac1{(k-3)!}+\frac3{(k-2)!}+\frac1{(k-1)!}$$
$$\sum_{k=1}^\infty \frac{k^2}{(k-1)!}=\sum_{k=1}^\infty \left(\frac{(k-1)(k-2)+3(k-1)+1}{(k-1)!}\right)$$
$$=\sum_{k=3}^\infty \frac1{(k-3)!}+\sum_{k=2}^\infty \frac3{(k-2)!}+\sum_{k=1}^\infty \frac1{(k-1)!}$$ as $\frac1 {(-r)!}=0$ for $r>0$
$=e+3e+e=5e$
| {
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Galois theory (Showing $G$ is not abelain) Suppose $G$ is the Galois group of an irreducible degree $5$ polynomial $f \in \mathbb{Q}[x]$ such that $|G| = 10$. Then $G$ is non-abelian.
Proof: Suppose $G$ is abelian. Let $M$ be the splitting field of $f$. Let $\theta$ be a root of $f$. Consider $\mathbb{Q}(\theta) \subseteq M$. Since $G$ is abelian every subgroup is normal. This means $\mathbb{Q}(\theta) \subseteq M$ is a normal extension. So $f$ splits completely in $\mathbb{Q}(\theta)$. Then what how to complete the proof. How would I get a contradiction?
| The only abelian group of order $10$ is cyclic. Since $G$ is a subgroup of $S_5$, it's enough to show that there's no element of order $10$ in $S_5$.
If you decompose a permutation in $S_5$ as a product of disjoint cycles, then the order is the LCM of the cycle lengths - and these can be any partition of $5$.
Since $5 = 1 + 4 = 1 + 1 + 3 = 2 + 3 = 1 + 1 + 1 + 2 = 1 + 2 + 2 = 1
+ 1 + 1 + 1 + 1$ are the only partitions, the only orders that appear are $1,2,3,4,5,6$ and in particular not $10$.
| {
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Is this differe equation (and its solution) already known? While studying for an exam, I met the following nonlinear differential equation
$a\ddot{x}+b\dot{x}+c\sin x +d\cos x=k$
where $a,b,c,d,k$ are all real constants. My teacher says that this differential equation does not admit closed form solution, but on this I would like to compare myself with you. Is this equation (and its solution) already known?
Thank you very much
| $a\ddot{x}+b\dot{x}+c\sin x+d\cos x=k$
$a\dfrac{d^2x}{dt^2}+b\dfrac{dx}{dt}+c\sin x+d\cos x-k=0$
This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0317.pdf
Let $\dfrac{dx}{dt}=u$ ,
Then $\dfrac{d^2x}{dt^2}=\dfrac{du}{dt}=\dfrac{du}{dx}\dfrac{dx}{dt}=u\dfrac{du}{dx}$
$\therefore au\dfrac{du}{dx}+bu+c\sin x+d\cos x-k=0$
$au\dfrac{du}{dx}=-bu-c\sin x-d\cos x+k$
$u\dfrac{du}{dx}=-\dfrac{bu}{a}-\dfrac{c\sin x+d\cos x-k}{a}$
This belongs to an Abel equation of the second kind.
In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $u=\dfrac{1}{v}$,
Then $\dfrac{du}{dx}=-\dfrac{1}{v^2}\dfrac{dv}{dx}$
$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dx}=-\dfrac{b}{av}-\dfrac{c\sin x+d\cos x-k}{a}$
$\dfrac{dv}{dx}=\dfrac{(c\sin x+d\cos x-k)v^3}{a}+\dfrac{bv^2}{a}$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
| {
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Find the volume of the solid under the plane $x + 2y - z = 0$ and above the region bounded by $y=x$ and $y = x^4$. Find the volume of the solid under the plane $x + 2y - z = 0$ and above the region bounded by $y = x$ and $y = x^4$.
$$
\int_0^1\int_{x^4}^x{x+2ydydx}\\
\int_0^1{x^2-x^8dx}\\
\frac{1}{3}-\frac{1}{9} = \frac{2}{9}
$$
Did I make a misstep? The answer book says I am incorrect.
| I think it should be calculated as
\begin{eqnarray*}
V&=&\int_0^1\int_{x^4}^x\int_0^{x+2y}dzdydx\\
&=&\int_0^1\int_{x^4}^x(x+2y)dydx\\
&=&\int_0^1\left.\left(xy+y^2\right)\right|_{x^4}^xdx\\
&=&\int_0^1(2x^2-x^5-x^8)dx\\
&=&\left.(\frac{2}{3}x^3-\frac{1}{6}x^6-\frac{1}{9}x^9)\right|_0^1\\
&=&\frac{7}{18}
\end{eqnarray*}
| {
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problem on entire function Let $f$ be an entire function. For which of the following cases $f$ is not necessarily a constant
*
*$\operatorname{im}(f'(z))>0$ for all $z$
*$f'(0)=0$ and $|f'(z)|\leq3$ for all $z$
*$f(n)=3$ for all integer $n$
*$f(z) =i$ when $z=(1+\frac{k}{n})$ for every positive integer $k$
I think 1 is true since $f'(z)=$constant so $f(z)=cz$ for some $c$
For 2, $f=0$
For 3, $f$ is not constant since $f(z)=3 \cos2\pi z$
I have no idea for 4
am i right for other three options
| You're right for $1$, where the $c$ you mention should have positive imaginary part. For example $f(z)=iz$ does the job.
For $2$, $f$ need not be identically $0$. $f(z)=2$ satisfies the requirements, for instance. $f$ does have to be constant, though, because if $f$ is entire, $f'$ is too, and so by Liouville's theorem since $f'$ is bounded it's constant, and the other condition forces it to be $0$ everywhere.
You're right on $3$.
For $4$, I'm still not sure what $n$ is. My guess is that this will be an identity theorem application, and that $f$ will be $i$ on a set with an accumulation point, which will force it to be $i$ everywhere.
But as it's written now, for $n$ is a fixed integer you could get $f$ non-constant in the same way as you did in number $3$: set
$$f(z)=i\cos(2n\pi(z-1))$$
| {
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convergent series, sequences? I want to construct a sequence of rational numbers whose sum converges to an irrational number and whose sum of absolute values converges to 1.
I can find/construct plenty of examples that has one or the other property, but I am having trouble find/construct one that has both these properties.
Any hints(not solutions)?
| Find two irrational numbers $a > 0$ and $b < 0$ such that $a-b = 1$ but $a+b$ is irrational. Create a series with positive terms that sum to $a$ and another series with negative terms that sum to $b$. Combine the two series.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is a rational $0$-cycle on an algebraic variety $X$ over $\mathbb{Q}$? I found the following assertion in a paper about Hilbert modular forms that I'm trying to read.
Let $X$ be an algebraic variety over $\mathbb{Q}$, and let $\Psi$ be a rational function on $X$ and $C = \sum n_P P$ be a rational $0$-cycle on $X$. Then $\Psi(C) = \prod \Psi(P)^{n_P}$ is a rational number.
By searching online I found some definitions of an algebraic cycle, but I haven't found what a rational $0$-cycle is. So the questions I have are:
*
*What does it mean that $C = \sum n_P P$ is a rational $0$-cycle? Does it mean that the coefficients $n_P \in \mathbb{Q}$ and that $\sum n_P = 0$? And what would be a good reference for these basic definitions?
*How do we prove that $\Psi(C) = \prod \Psi(P)^{n_P}$ is a rational number?
Thank you very much for any help.
| To say that $C = \sum n_P P$ is a rational $0$-cycle means that $n_P\in \mathbb Z$ and that $P\in X$ is a rational point i.e. a closed point with residue field $\kappa (P)=\mathbb Q$.
If the rational function $\Psi$ is defined at $P$ its value at $P$ is a rational number $\Psi(P) \in \mathbb Q$ and if $\Psi$ is defined at all $P$'s with $n_P\neq 0$ we have $\Psi(C) = \prod \Psi(P)^{n_P}\in \mathbb Q$.
As an illustration of what it means for $P$ to be rational, take the simplest example $X =\mathbb A^1_\mathbb Q=Spec ( \mathbb Q[T])$.
Then the point $P$ corresponding to the prime ideal $J_P= \langle T-1/2\rangle\subset \mathbb Q[T]$ is rational since $\mathbb Q[T]/ \langle T-1/2\rangle=\mathbb Q$.
However the closed point $Q$ corresponding to the prime ideal $J_Q= \langle T^3-2\rangle\subset \mathbb Q[T]$ is not rational since the canonical morphism $\mathbb Q \to \mathbb Q[T]/ \langle T^3-2 \rangle $ is not an isomorphism.
| {
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Relationship between Legendre polynomials and Legendre functions of the second kind I'm taking an ODE course at the moment, and my instructor gave us the following problem:
Derive the following formula for Legendre functions $Q_n(x)$ of the second kind:
$$Q_n(x) = P_n(x) \int \frac{1}{[P_n(x)]^2 (1-x^2)}dx$$
where $P_n(x)$ is the $n$-th Legendre polynomial.
He introduced Legendre functions in the context of second order ODEs, but we haven't really used them for anything - moreover, this is the only problem we were assigned that has anything to do with them. As a result, I'm sort of at a loss of where to start.
I've tried a couple of things (like using the actual Legendre ODE
$$(1-x^2)y^{\prime \prime} - 2xy^{\prime} + n(n+1)y = 0$$
and plugging in the solution $y(x)=a_1P_n(x)+a_2Q_n(x)$ and proceeding from there) but so far, haven't been able to go anywhere.
Any help (preferably as elementary as possible) would be much appreciated. Thanks!
| The most general solution of Legendre equation is
$$y = A{P_n} + B{Q_n}.$$
Let $y(x) = A(x){P_n}(x)$. Then $y' = AP' + A'P$ and $y'' = AP'' + 2A'P' + A''P$. So
$$(1 - {x^2})(AP'' + 2A'P' + A''P) - 2x(AP' + A'P) + n(n + 1)AP = 0.$$
Note that
$$(1 - {x^2})(AP'') - 2x(AP') + n(n + 1)AP = 0$$
which means some terms in the above equation vanish.
Now let $A' = u$ and reduce the order so that
$$2\frac{{dP}}{P} + \frac{{du}}{u} - \frac{{2xdx}}{{1 - {x^2}}} = 0$$
and
$$u = \frac{{{\text{const}}}}{{(1 - {x^2}){P^2}}}$$
so
$$A = {C_n}\int {\frac{1}{{(1 - {x^2}){P^2}}}dx}.$$
See if you can do the rest.
| {
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Compute $\lim_{n\to\infty}\int_0^n \left(1+\frac{x}{2n}\right)^ne^{-x}\,dx$. I'm trying to teach myself some analysis (I'm currently studying algebra), and I'm a bit stuck on this question. It's strange because of the $n$ appearing as a limit of integration; I want to apply something like LDCT (I guess), but it doesn't seem that can be done directly.
I have noticed that the change of variables $u=1+\frac{x}{2n}$ helps. With this, the problem becomes
$$
\lim_{n\to\infty}\int_1^{3/2}2nu^ne^{-2n(u-1)}\,du.
$$
This at least solves the issue of the integration limits. Let's let $f_n(u):=2nu^ne^{-2n(u-1)}$ for brevity. I believe it can be shown that
$$
\lim_{n\to\infty}f_n(u)=\cases{\infty,\,u=1\\0,\,1<u\leq 3/2}
$$
using L'Hopital's rule and the fact that $u^n$ intersects $e^{2n(u-1)}$ where $u=1$, and so the exponential function is larger than $u^n$ for $n>1$.
I think I was also able to show that $\{f_n\}$ is eventually decreasing on $(1,3/2]$, and so Dini's Theorem says that the sequence is uniformly convergent to $0$ on $[u_0,3/2]$ for any $u_0\in (1,3/2]$. Since each $f_n$ is continuous on the closed and bounded interval $[u_0,3/2]$, each is bounded; as the convergence is uniform, the sequence is uniformly bounded.
Thus, the Lebesgue Dominated Convergence Theorem says
$$
\lim_{n\to\infty}\int_{u_0}^{3/2}2nu^ne^{-2n(u-1)}\,du=\int_{u_0}^{3/2}0\,du=0.
$$
So it looks like I'm almost there, I just need to extend the lower limit all the way to $1$. I think this amounts to asking whether we can switch the order of the limits in
$$\lim_{n\to\infty}\lim_{u_0\to 1^+}\int_{u_0}^{3/2}2nu^ne^{-2n(u-1)}\,du,
$$
and (finally!) this is where I'm stuck. I feel like this step should be easy, and it's quite possible I'm missing something obvious. That happens a lot when I try to do analysis because of my practically nonexistent background.
| HINT Note that $$\left(1 + \dfrac{x}{2n} \right)^n < e^{x/2}$$ for all $n$. Hence, $$ \left(1 + \dfrac{x}{2n} \right)^n e^{-x} < e^{-x/2}$$
Your sequence $$f_n(x) = \begin{cases} \left(1 + \dfrac{x}{2n} \right)^n e^{-x} & x \in [0,n]\\ 0 & x > n\end{cases}$$ is dominated by $g(x) = e^{-x/2}$. Now apply LDCT.
| {
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"url": "https://math.stackexchange.com/questions/222485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Studying $ u_{n}=\frac{1}{n!}\int_0^1 (\arcsin x)^n \mathrm dx $ I would like to find a simple equivalent of:
$$ u_{n}=\frac{1}{n!}\int_0^1 (\arcsin x)^n \mathrm dx $$
We have:
$$ 0\leq u_{n}\leq \frac{1}{n!}\left(\frac{\pi}{2}\right)^n \rightarrow0$$
So $$ u_{n} \rightarrow 0$$
Clearly:
$$ u_{n} \sim \frac{1}{n!} \int_{\sin(1)}^1 (\arcsin x)^n \mathrm dx $$
But is there a simpler equivalent for $u_{n}$?
Using integration by part:
$$ \int_0^1 (\arcsin x)^n \mathrm dx = \left(\frac{\pi}{2}\right)^n - n\int_0^1 \frac{x(\arcsin x)^{n-1}}{\sqrt{1-x^2}} \mathrm dx$$
But the relation
$$ u_{n} \sim \frac{1}{n!} \left(\frac{\pi}{2}\right)^n$$
seems to be wrong...
| The change of variable $x=\cos\left(\frac{\pi s}{2n}\right)$ yields
$$
u_n=\frac1{n!}\left(\frac\pi2\right)^{n+2}\frac1{n^2}v_n,
$$
with
$$
v_n=\int_0^n\left(1-\frac{s}n\right)^n\,\frac{2n}\pi \sin\left(\frac{\pi s}{2n}\right)\,\mathrm ds.
$$
When $n\to\infty$, $\left(1-\frac{s}n\right)^n\mathbf 1_{0\leqslant s\leqslant n}\to\mathrm e^{-s}$ and $\frac{2n}\pi \sin\left(\frac{\pi s}{2n}\right)\mathbf 1_{0\leqslant s\leqslant n}\to s$. Both convergences are monotonic hence $v_n\to\int\limits_0^\infty\mathrm e^{-s}\,s\,\mathrm ds=1$. Finally,
$$
u_n\sim\frac1{(n+2)!}\left(\frac\pi2\right)^{n+2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/222555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to show that this function is bijective
Possible Duplicate:
Proving the Cantor Pairing Function Bijective
Assume I define
$$ f: \mathbb N \times \mathbb N \to \mathbb N, (a,b) \mapsto a + \frac{(a + b ) ( a + b + 1)}{2} $$
How to show that this function is bijective? For injectivity I tried to show that if $f(a,b) = f(n,m) $ then $(a,b) = (n,m)$ but I end up getting something like $3(n-a) + (n+m)^2 -(a+b)^2 + m - b = 0$ and don't see how to proceed from there. There has to be something cleverer than treating all possible cases of $a \leq n, b \leq m$ etc.
For surjectivity I'm just stuck. If I do $f(0,n)$ and $f(n,0)$ it doesn't seem to lead anywhere.
Thanks for your help.
| The term $(a+b)(a+b+1)/2$ is the sum of the numbers from $1$ to $a+b$. For a fixed value of $s=a+b$, $a$ ranges from $0$ to $s$, so we need $s+1$ different results for these arguments. Now you can prove by induction that the range from $s(s+1)/2$ to $(s+1)(s+2)/2-1$ contains precisely the images of the pairs with sum $s$.
And I truly empathize with the loss of your teddy bear. My teddy bear would be the first thing I'd save in case of fire.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Number of integral solutions for $|x | + | y | + | z | = 10$ How can I find the number of integral solution to the
equation
$|x | + | y | + | z | = 10.$
I am using the formula,
Number of integral solutions for $|x| +|y| +|z| = p$ is $(4P^2) +2 $, So the answer is 402.
But, I want to know, How we can find it without using formula.
any suggestion !!!
Please help
| (1) z=0, $40$ patterns
z=1, $36$
z=2, $32$
・・・
z=9,$4$
but z=10,$2$ the total is
$S=2(4+8+\dots+36)+40+2=402$
(2) Another counting
There are 8 areas by plus and minus of x,y,z, 40 patterns at x=0, 2 patterns at y=z=0, therefore
$40+8*10C2+2=402$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/222690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 8,
"answer_id": 6
} |
Absolute value and sign of an elasticity In my microeconomics book, I read that when we have $1+\dfrac{1}{\eta}$ where $\eta$ is an elasticity coefficient, we can write $1-\dfrac{
1}{|\eta|}$ "to avoid ambiguities stemming from the negative sign of the elasticity".
What does this mean? Is it always legitimate to perform such a transformation?
| If the elasticity coefficient $\eta$ is negative, then $|\eta|=-\eta$. The ambiguity arises because some people may suppress the negative sign and write it as a positive number instead. In this case, using $1+\frac{1}{\eta}$ becomes ambiguous.
For example, if $\eta=-2$ but people write it as $\eta=2$, then $1+\frac{1}{\eta}$ can mean $1-\frac{1}{2}$ or $1+\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/222765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
existence of hyperbolic groups I perfectly understand that the Milnor-Schwarz lemma tells me that cocompact lattice in semisimple Lie groups of higher rank are not hyperbolic (in the sense of Gromov).
But do there exist noncocompact lattices in higher rank semisimple Lie groups which are hyperbolic? (wikipedia is saying "no" without any reference, but I do not trust that article since it contains mistakes...)
| You asked a related question in another post, and you erased the question while I was posting an answer. So I post it here. The question was complementary to the one above so I think it's relevant to include the answer: why are non-uniform lattices in rank 1 symmetric spaces of noncompact type not hyperbolic except in the case of the hyperbolic plane? Here is my answer.
I think it's a result of Garland and Raghunathan (Annals 1970 "Fundamental domains..."). They show that given such a lattice, for some point $\omega$ at infinity, the stabilizer of $\omega$ in the lattice acts cocompactly on each horosphere based at $\omega$. This horosphere is modeled on a $(n-1)$-dimensional simply connected nilpotent Lie group, where $n$ is the real dimension of the rank 1 symmetric space of non-compact type. Thus the nonuniform lattice contains a f.g. nilpotent group of Hirsch length $n-1$. This is possible in a hyperbolic group only if $n\le 2$. (Note: if $n\ge 3$, it follows that the lattice contains a free abelian group of rank 2.) More precise results about the structure of these lattices were formalized into the concept of relatively hyperbolic groups, see Gromov, Farb, etc. They indicate that intuitively, these "peripheral" subgroups are the only obstruction to hyperbolicity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/222813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Simple matrix equation I believe I'm missing an important concept and I need your help.
I have the following question:
"If $A^2 - A = 0$ then $A = 0$ or $A = I$"
I know that the answer is FALSE (only because someone told me) but when I try to find out a concrete matrix which satisfies this equation (which isn't $0$ or $I$) I fail.
Can you please give me a direction to find a concrete matrix? What is the idea behind this question?
Guy
| If for a polynomial $p$ and a matrix $A$ you have $p(A)=0$ then for every invertible matrix $W$ you have $$p(W^{-1}AW)=W^{-1}p(A)W=0 . $$
Here $p=x^2-x$, you can take $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $W$ any invertible matrix to make a lot of examples.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/222886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Understanding what this probability represents Say I want the probability that a five card poker hand contains exactly two kings. This would be
$$\frac{{4\choose 2}{48 \choose 3}}{52\choose 5}$$
Now if I drop the $48 \choose 3$, which represents the 3 non king cards, what can
the probability $\frac{4\choose 2}{52\choose 5}$ be taken to represent?
Is it the number of hands containing at least 2 kings?
| It would be the probability of a hand containing exactly two kings and three specified non-kings, e.g. the ace, 2 and 3 of spades.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/222928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
modular multiplicative inverse I have a homework problem that I've attempted for days in vain... It's asking me to find an $n$ so that there is exactly one element of the complete residue system $\pmod n$ that is its own inverse apart from $1$ and $n-1$. It also asks me to construct an infinite sequence of $n's$ so that the complete residue system $\pmod n$ has elements that are their own inverses apart from $1$ and $n-1$.
For the first part, I tried all $n$ from $3$ up to $40$, but none worked...
For the second part, I'm really confused...
Could someone please help me with this? Thanks!
| For the second: What about $3$ and $5 \pmod8$?
For the first: if $x^2\equiv 1$ then $(-x)^2\equiv 1$, too, so if there is only one (besides $\pm 1$), then $x\equiv -x \pmod{n}$, that is $2x=n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/222973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find all Laurent series of the form... Find all Laurent series of the form $\sum_{-\infty} ^{\infty} a_n $ for the function
$f(z)= \frac{z^2}{(1-z)^2(1+z)}$
There are a lot of problems similar to this. What are all the forms? I need to see this example to understand the idea.
| Here is related problem. First, convert the $f(z)$ to the form
$$f(z) = \frac{1}{4}\, \frac{1}{\left( 1+z \right)}+\frac{3}{4}\, \frac{1}{\left( -1+z \right) }+\frac{1}{2}\,\frac{1}{\left( -1+z \right)}$$
using partial fraction. Factoring out $z$ gives
$$ f(z)= \frac{1}{4z}\frac{1}{(1+\frac{1}{z})}- \frac{3}{4z}\frac{1}{(1-\frac{1}{z})}-\frac{1}{2z^2}\frac{1}{(1-\frac{1}{z})^2}\,. $$
Now, using the series expansion of each term yields the Laurent series for $|z|>1$
$$ = \frac{1}{4z}(1-\frac{1}{z}+ \frac{1}{z^2}+\dots) -\frac{3}{4z}(1+\frac{1}{z}+ \frac{1}{z^2}+\dots)-\frac{1}{2z}\sum_{k=0}^{\infty}{-2\choose k}\frac{(-1)^k}{z^k} \,, $$
$$ \sum_{k=0}^{\infty}\left( \frac{(-1)^k}{4}-\frac{3}{4}-\frac{(-1)^k{-2\choose k}}{2} \right)\frac{1}{z^{k+1}}\,. $$
Note that,
$$(1+x)^{-1}=1-x+x^2-\dots \,,$$
$$ (1-x)^{-1}=1+x+x^2+\dots \,,$$
$$ (1-x)^{m}=\sum_{k=0}^{\infty}{m\choose k}(-1)^kx^k $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/223037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show there is no measure on $\mathbb{N}$ such that $\mu(\{0,k,2k,\ldots\})=\frac{1}{k}$ for all $k\ge 1$
For $k\ge1$, let $A_k=\{0,k,2k,\ldots\}.$ Show that there is no measure $\mu$ on $\mathbb{N}$ satisfying $\mu(A_k)=\frac{1}{k}$ for all $k\ge1$.
What I have done so far:
I am trying to apply Borel-Cantelli lemma ($\mu(\mathbb{N})=1)$. Let $(p_n)_{n\in \mathbb{N}}=(2,3,5,\ldots)$ be the increasing sequence of all prime numbers. It is the case that for all $k\in\mathbb{N}$ and $i_1\lt i_2 \lt \ldots \lt i_k$ we have $\mu\left(A_{p_{i_1}}\cap\ldots\cap A_{p_{i_k}}\right)=\mu\left(A_{p_{i_1}}\right)\ldots\mu\left(A_{p_{i_k}}\right)$
so all $A_n$ are independent. We know that the sum of the reciprocals of the primes $\sum\limits_{n=1}^\infty \frac{1}{p_n}$diverges and hence, by B-C second lemma, $$\mu\left(\bigcap\limits_{n=1}^\infty\bigcup\limits_{m=n}^\infty A_{p_m}\right)=1$$ holds. However, I cannot figure out how to conclude the reasoning. I would appreciate any help.
| You are basically there. You proved in your last line that $\mu$-a.e. number is divisible by infinitely many primes, which is an obvious contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/223129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How Many Ways to Make a Pair Given Five Poker Cards I'm confused at the general method of solving this type of problem. The wikipedia page says that there are:
${13 \choose 1} {4 \choose 2} {12 \choose 3} {4 \choose 1}^{3}$ ways to select a pair when 5 cards are dealt. Can someone outline what each calculation means? For example, is $13 \choose 1$ the process of selecting 1 card in the beginning?
| $13\choose 1$ is the number of ways of choosing the denomination of the pair, whether it is a pair of kings, or a pair of threes, or whatever. Then there are $4\choose 2$ ways to choose suits for the two cards of the pair.
Then there are $12\choose 3 $ ways to choose the three different denominations of the remaining three cards from the twelve that are different from the denomination of the pair. Each of those other three cards can be of ${4\choose 1}=4$ different suits, for ${4\choose 1}^3$ choices of three suits in all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/223203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How many distinct functions can be defined from set A to B? In my discrete mathematics class our notes say that between set $A$ (having $6$ elements) and set $B$ (having $8$ elements), there are $8^6$ distinct functions that can be formed, in other words: $|B|^{|A|}$ distinct functions. But no explanation is offered and I can't seem to figure out why this is true. Can anyone elaborate?
| Let's say for concreteness that $A$ is the set $\{p,q,r,s,t,u\}$, and $B$ is a set with $8$ elements distinct from those of $A$. Let's try to define a function $f:A\to B$.
What is $f(p)$? It could be any element of $B$, so we have 8 choices.
What is $f(q)$? It could be any element of $B$, so we have 8 choices.
...
What is $f(u)$? It could be any element of $B$, so we have 8 choices.
So there are $8\cdot8\cdot8\cdot8\cdot8\cdot8 = 8^6$ ways to choose values for $f$, and each possible set of choices defines a different function $f$. So that's how many functions there are.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/223240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 6,
"answer_id": 3
} |
Linear Algebra Fields question I have the following statement (which is false) and I'm trying to understand why, I can't find a concrete example.
If $a$ belongs to $Z_n$ and $a^2 = 1$ then $a=1$ or $a=-1$
Can someone give me a direction?
Guy
| The statement does hold when the structure is a field, i.e. when $n$ is prime. The statement is false for some composite $n$. For example, consider any $n$ of the form $x(x+2)$ for some $x\in\mathbb{N}^+$ then we will have
$$(x+1)^2 \equiv x^2 + 2x + 1 \equiv n + 1 \equiv 1 \pmod n$$
so that $\pm(x+1)$ is a solution in addition to $\pm1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/223359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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