Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Proof of $\sin^2 x+\cos^2 x=1$ using Euler's Formula How would you prove $\sin^2x + \cos^2x = 1$ using Euler's formula?
$$e^{ix} = \cos(x) + i\sin(x)$$
This is what I have so far:
$$\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$$
$$\cos(x) = \frac{1}{2} (e^{ix}+e^{-ix})$$
| Multiply $\mathrm e^{\mathrm ix}=\cos(x)+\mathrm i\sin(x)$ by the conjugate identity $\overline{\mathrm e^{\mathrm ix}}=\cos(x)-\mathrm i\sin(x)$ and use that $\overline{\mathrm e^{\mathrm ix}}=\mathrm e^{-\mathrm ix}$ hence $\mathrm e^{\mathrm ix}\cdot\overline{\mathrm e^{\mathrm ix}}=\mathrm e^{\mathrm ix-\mathrm ix}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove: $a\equiv b\pmod{n} \implies \gcd(a,n)=\gcd(b,n)$ Proof: If $a\equiv b\pmod{n}$, then $n$ divides $a-b$. So $a-b=ni$ for some integer $i$. Then, $b=ni-a$. Since $\gcd(a,n)$ divides both $a$ and $n$, it also divides $b$. Similarly, $a=ni+b$, and since $\gcd(b,n)$ divides both $b$ and $n$, it also divides $a$.
Since $\gcd(a,n)$ is a divisor of both $b$ and $n$, we know $\gcd(a,n)\leq\gcd(b,n)$. Similarly, since $\gcd(b,n)$ is a divisor of both $a$ and $n$, we know $\gcd(b,n)\leq\gcd(a,n)$. Therefore, $\gcd(a,n)=\gcd(b,n)$.
Thanks for helping me through this one!
| HINT : Let $d = \gcd(b,n)$, then $b = dx$ and $n = dy$ for some $x$ and $y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
If $g$ is in $N_G(P)$ then $g\in P$ where $P$ is a $p$-Sylow subgroup Please help me to solve this problem. Let $P$ is a $p$-Sylow subgroup of the finite group $G$ and $g$ is an element such that $\lvert g \rvert=p^k$ then if $g$ is in $N_G(P)$ then $g\in P$. Where to start?
| If $H \leq G$ is a $p$-group, then $H \leq N_G(P)$ if and only if $H \leq P$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Graph with the smallest diameter. Consider a graph with $N$ vertexes where each vertex has at most $k$ edges.
I assume that $k < N$. What is the graph which have above property and has the smallest diameter?
Also, could you suggest good books in graph theory. Thanks.
| This question is quite difficult. The upper bound on the number of vertices given above by @Boris Novikov coincides with Moore's bound in the case of the Petersen graph. Moore's bound is not only achieved by the Petersen graph, but also by the Hoffman-Singleton graph, and in general, by the so-called Moore graphs. Unfortunately, there are very few Moore graphs. In most other cases, i.e. when Moore's bound cannot be reached, the optimal graph is not known. For more details see "Moore Graphs and Beyond: A survey of the Degree/Diameter Problem", by Mirka Miller and Josef Siran (Elec. J. Combinatorics, Dynamic Survey 14).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Does a 3-Dimensional coordinate transformation exist such that its scale factors are equal? Let $\vec r=(x,y,z) $ be the position vector expressed in Cartesian coordinates. Let us define the coordinate transformation as
$\vec r(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w)) $
The scale factors are defined by
$h_u=\vert \partial \vec r/\partial u \vert, h_v=\vert \partial \vec r/\partial v \vert, h_w=\vert \partial \vec r/\partial w \vert$
I wonder if a transformation can be defined such that
$h_u=h_v=h_w$
Now a pair of examples in the two dimentional case.
The transformation between elliptic and cartesian coordinates:
$\vec r(u,v)=(cosh(u)cos(v)/2,sinh(u)sin(v)/2) $
$h_u=h_v=\sqrt{cosh^2(u)-cos^2(v)}/2$
The transformation between parabolic and cartesian coordinates.
$\vec r(u,v)=((u^2-v^2)/2,u v) $
$h_u=h_v=\sqrt{u^2+v^2}$
| Suppose we add another condition: not only $$\left\lVert\frac{\partial\mathbf r}{\partial u}\right\rVert=\left\lVert\frac{\partial\mathbf r}{\partial v}\right\rVert=\left\lVert\frac{\partial\mathbf r}{\partial w}\right\rVert,$$ but also $$\frac{\partial\mathbf r}{\partial u}\cdot\frac{\partial\mathbf r}{\partial v}=\frac{\partial\mathbf r}{\partial v}\cdot\frac{\partial\mathbf r}{\partial w}=\frac{\partial\mathbf r}{\partial w}\cdot\frac{\partial\mathbf r}{\partial u}=0.$$
That is, the coordinate system is orthogonal, which is usually desirable. (In particular, both of your two-dimensional examples have this property.) This means that the Jacobian matrix of the transformation is a multiple of the identity, and the transformation $(u,v,w)\mapsto(x,y,z)$ is a conformal map.
Liouville's theorem states that in three or more dimensions, all such maps are compositions of translations, similarities, orthogonal transformations and inversions. So the space of such coordinate systems is much more restricted than in two dimensions. Nevertheless, we do have a non-Cartesian example, namely inversion:
$$(x,y,z)=\frac{(u,v,w)}{u^2+v^2+w^2}.$$
This has $\lVert\partial\mathbf r/\partial u\rVert=\lVert\partial\mathbf r/\partial v\rVert=\lVert\partial\mathbf r/\partial w\rVert=1/(u^2+v^2+w^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Linear algebra - Dimension theorem. Suppose we have a vector space $V$, and $U$, $W$ subspaces of $V$.
Dimension theorem states:
$$ \dim(U+W)=\dim U+ \dim W - \dim (U\cap W).$$
My question is:
Why is $U \cap W$ necessary in this theorem?
| $U \cap W$ is the intersection of the vector spaces $U$ and $W$, that is, the set of all vectors of the space $V$ which are in both subspaces $U$ and $W$.
As $U$ and $W$ are both subspaces of $V$, their intersection $U \cap W$ is also a subspace of $V$ (this assertion can be easily proved). Because $U \cap W$ is a subspace, it is also a vector space itself, and as such it has a basis. The number of elements in this basis will be the space's dimension, $\dim (U \cap W)$.
Loosely speaking, one could think that summing $\dim(U)$ and $\dim(W)$ would yield $\dim(U+W)$. But as $(U \cap W) \subset U$ and $(U \cap W) \subset W$, the sum $\dim(U) + \dim(W)$ "counts" two times the dimension of $U \cap W$ - once in $\dim(U)$ and once more in $\dim(W)$. To make it sum up to $\dim(U+W)$ accurately, we must then subtract the dimension of $U \cap W$, so that it is "counted" only once. This way, we obtain:
$$ \dim(U+W) = \dim(U) + \dim(W) - \dim(U \cap W).$$
Note that this is not, by any means, a formal proof. It is only an informal explanation of why $U \cap W$ is needed in this formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
There exists only two groups of order $p^2$ up to isomorphism. I just proved that any finite group of order $p^2$ for $p$ a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is $G=\Bbb Z/p\Bbb Z\oplus \Bbb Z/p\Bbb Z$. This is abelian and has order $p^2$. I think the other is $\Bbb Z/p^2 \Bbb Z$.
Now, it should follow from the fact that there is only one cyclic group of order $n$ up to isomorphism that these two are unique up to isomorphism. All I need to show is these two are in fact not isomorphic. It suffices to show that $G$ as before is not cyclic. But this is easy to see, since we cannot generate any $(x,y)$ with $x\neq y$ by repeated addition of some $(z,z)$.
Now, it suffices to show that any other group of order $p^2$ is isomorphic to either one of these two groups. If the group is cyclic, we're done, so assume it is not cyclic. One can see that $G=\langle (1,0) ,(0,1)\rangle$. How can I move on?
| A proof of that could be:
The center of a group is a subgroup, so it's order must divide $p^2$, but it's a known fact that if a group has order $p^m$, with $p$ prime, then the center of the group is different from $p^{m-1}$ and different from $1$, so in our case, the center has order $p^2$, so it's abelian.
That's what you already have, know, by the theorem of structure of finite abelian groups, the group can be either $\mathbb{Z}_p\times\mathbb{Z}_p$ or $\mathbb{Z}_{p^2}$. But from that same theorem, one can deduce that a group $\mathbb{Z}_m\times\mathbb{Z}_n$ is isomorphic to a group $\mathbb{Z}_{nm}$ iff $\gcd(m,n)=1$, so in our case, those two groups are not isomorphic, and there are only two groups of order $p^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 0
} |
Is there a simple algorithm for factoring polynomials over the reals? Any real polynomial can be expressed as a product of quadratic and binomial factors like $(x+a)$ and $(x^2 + bx + c)$. Given a polynomial, is there an algorithm which will find such factors?
For example, how can I express $x^4 +1$ in the form $(x^2 + bx + c)(x^2 +dx + f)?$
| If you can find roots, you can find factors. As others have pointed out, this needs to be done using numerical methods for polynomials with degree greater than 4. In fact it's often a good idea to use numerical methods (rather than closed-form formulae) even in the degree 3 and degree 4 cases, too. There's lots of good software for numerically finding roots of polynomials. The best-known is the Jenkins Traub method
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How many ways are there to consider $\Bbb Q$ as an $\Bbb R$-module? How many ways are there to consider $\Bbb Q$ as an $\Bbb R$-module?
I guess there is only one way, and that is the trivial case. i.e.
$$\forall r\in \Bbb R,\, \forall a,b\in \Bbb Q \qquad r\cdot \frac{a}{b}=0$$
With an idea I can proof this:
$$\exists r\not=s\in \Bbb R \quad\text{s.t.}\quad r\cdot \frac{a}{b}=s\cdot \frac {a}{b}$$
but I think this idea is wrong.
| There are no ways to consider $\mathbb{Q}$ as an $\mathbb{R}$-module (i.e. $\mathbb{R}$-vector space), if you follow the usual convention of requiring that modules be unital, i.e. $1\cdot q=q$ for any $q\in\mathbb{Q}$. This is because a vector space over $\mathbb{R}$ (or indeed, any field) is free; because $\mathbb{R}$ is uncountable, any $\mathbb{R}$-vector space must consist of either one element or uncountably many elements, neither of which is the case for $\mathbb{Q}$.
If you do not require your modules to be unital, then the trivial module structure (multiplication by any real number gives 0) is the only possible one. This is because, if there is some $r\in\mathbb{R}$ and some $q\in\mathbb{Q}$ such that $r\cdot q\neq 0$, then we reach the same problem as before, because for any non-zero real number $s$, we have that
$$0\neq rq=\left(\frac{r}{s}\right)\cdot sq$$
so that $sq\neq 0$ for any non-zero $s\in\mathbb{R}$, and we have $rq\neq sq$ if $r\neq s$ (because $(r-s)q\neq 0$), thereby making $\mathbb{Q}$ uncountable again (contradiction).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
If the sum of the digits of $n$ are divisible by 9, then $n$ is divisible by 9; help understanding part of a proof
Let $n$ be a positive integer such that $n<1000$. If the sum of the digits of $n$ is divisible by 9, then $n$ is divisible by 9.
I got up to here:
$$100a + 10b + c = n$$
$$a+b+c = 9k,\quad k \in\mathbb{Z}$$
I didn't know what to do after this, so I consulted the solution
The next step is:
$$100a+10b+c = n = 9k +99a+9b = 9(k +11a+b)$$
I don't get how you can add $99a + 9b$ randomly, can someone please explain this for me?
| Simple way to see it: Take the number $N = \sum_{0 \le k \le n} d_k 10^k$ where the $d_k$ are the digits, modulo 9 you have:
$$
N = \sum_{0 \le k \le n} d_k 10^k
\equiv \sum_{0 \le k \le n} d_k \pmod{9}
$$
since $10 \equiv 1 \pmod{9}$, and so $10^k \equiv 1 \pmod{9}$ for all $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is the set of integers with respect to the p-adic metric compact? Given the integers and a prime $p$. I thought I had successfully shown that $\mathbb{Z}$ was compact with respect to the metric $|\cdot |_p$, by showing that the open ball centered at zero contained all integers with more than a certain number of factors of $p$, and then showing that the remaining integers took on a finite number of possible p-adic absolute values and thus fell into a finite number of balls.
Now if the integers are compact with respect to $|\cdot |_p$, then that means they are complete with respect to $|\cdot |_p$.
But then I read that the p-adic integers $\mathbb{Z}_p$ are defined to be the completion of the integers with respect to $|\cdot |_p$, and include in their completion all the rational numbers with p-adic absolute value less than or equal to one. So this means that the integers with respect to the p-adic metric are not complete, and thus not compact, and hence there must be something wrong with my proof, correct?
Edit: Ok upon typing this up I realized that my proof is most likely wrong as there's no reason to conclude that two elements with the same absolute value are necessarily in the same ball.
| You don't prove compactness "by showing that the open ball centered at zero contained all integers with ..., and then showing that the remaining integers ... fell into a finite number of balls", i.e. by showing that there is a finite number of open balls covering the space.
Actually you can show more : $\Bbb Z$ with the $p$-adic metric is a bounded metric space : every integer is at a distance less than $1$ from $0$.
Instead, to prove compactness you have to show that for any covering of $\Bbb Z$ by open balls, you can select a finite number of those open balls and still cover $\Bbb Z$. For example, pick the covering of $\Bbb Z$ by placing an open ball on $n$ with radius $p^{-|n|}$. For most $p$ ($p \ge 5$) , you can't extract a finite cover of $\Bbb Z$ from this cover.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Book Recommendations and Proofs for a First Course in Real Analysis I am taking real analysis in university. I find that it is difficult to prove some certain questions. What I want to ask is:
*
*How do we come out with a proof? Do we use some intuitive idea first and then write it down formally?
*What books do you recommended for an undergraduate who is studying real analysis? Are there any books which explain the motivation of theorems?
| While this doesn't speak, directly, to Real Analysis, it is a recommendation that will help you there, and in other courses you're encounter, or will encounter soon:
In terms of both reading and writing proofs, in general, an excellent book to work through and/or have as a reference is Velleman's great text How to Prove It: A Structured Approach. The best way to overcome doubt and apprehension about proofs, whether trying to understand them or to write them, is to be patient, persist, and dig in and do it! (often, write and rewrite, read and reread, until you're convinced and you're convinced you can convince others!)
One helpful (and free-to-use) online resource is the website maintained by MathCS.org: Interactive Real Analysis.
"Interactive Real Analysis is an online, interactive textbook for Real Analysis or Advanced Calculus in one real variable. It deals with sets, sequences, series, continuity, differentiability, integrability (Riemann and Lebesgue), topology, power series, and more."
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
Norm of the operator $Tf=\int_{-1}^0f(t)\ dt-\int_{0}^1f(t)\ dt$ Consider the operator $T:(C[-1, 1], \|\cdot\|_\infty)\rightarrow \mathbb R$ given by, $$Tf=\int_{-1}^0f(t)\ dt-\int_{0}^1f(t)\ dt,$$ is $\|T\|=2$. How to show $\|T\|=2$? On the one hand it is easy,
$$\begin{align}
|Tf|&=\left|\int_{-1}^0f(t)\ dt-\int_{0}^1f(t)\ dt\right|\\&\leq \int_{-1}^0|f(t)|\ dt+\int_{0}^1|f(t)|\ dt\\ &\leq \|f\|_\infty \left(\int_{-1}^0\ dt+\int_{0}^1\ dt\right)\\ &=2.
\end{align}$$ How to show $\|T\|\geq 2$?
| Try $f$ piecewise linear with $f(x)=-1$ if $x\leqslant -a$, $f(x)=x/a$ if $-a\leqslant x\leqslant a$ and $f(x)=+1$ if $x\geqslant a$, when $a\to0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Total Orders and Minimum/Maximum elements How can I prove that for any given Poset $(A,\preceq)$, $\preceq$ is a total order implies that $\forall a\in\preceq$, if a is a maximal, then a is maximum? Same goes for minimal/minimum.
| I assume that you are asking the following:
In a totally ordered set, why is every maximal element a greatest element?
The answer is simple: Assume $a$ is maximal. Let $b$ be an arbitrary element of our set. Since the set is totally ordered set we have either $a\leq b$ or $b\leq a$. Since $a$ is maximal, we must have $b\leq a$. Since $b$ was arbitrary, it follows that $a$ is a greatest element.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Range and Kernel of a $3\times 3$ identity matrix? geometrically describe both.
the kernell would be just the zero vector, correct? and would also merely live in the 1st dimension, but geometrically speaking be non existent?
and for the range of the matrix, it would just be $\{(1,0,0),(0,1,0),(0,0,1)\}$. geometrically this would span the whole 3 dimensional space, correct?
| The kernel is a subspace of the domain, regarding the matrix as a transformation. So the kernel is written as $\{ (0, 0, 0) \}$. Now, you can certainly regard this kernel as a space as well, but it is properly called 0-dimension, not non-existant.
Your range is correct. All of $\mathbb{R}^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/317985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How is GLB and LUB different than the maximum and minimum of a poset? As the subject asks, how is the greatest lower bound different than the minimum element in a poset, and subsequently, how is the least upper bound different than the minimum? How does a set having no maximum but multiple maximal elements affect the existence of a LUB?
Edit:
Here's an assignment question I have on the topic...
Consider the subset $A=\{4,5,7\}$ of $W$.
Does $\sup(A)$ exist? What about $\inf(A)$?
I'm guessing that Sup is Suprema (or LUB), and inf is Infimum (GLB)... but I have no idea because I missed that day in class.
All I can guess is that the subset $A$ is ordered such that $7\preceq 5$, but $4$ and $5$ are incomparable. I don't know what to do with it.
| The supremum coincides with maximum in the finite case. That said, a great deal of basic example are not finite, so supremum is a different operation in general. The most notorious thing here is that supremum is an element not necesarilly in the original set.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Surface area of a cylinders and prisms A cylinder has a diameter of 9cm and a height of 25cm. What is the surface area of the cylinder if it has a top and a base?
| The top and the bottom each have area $\pi(9/2)^2$.
For the rest, use a can opener to remove the top and bottom of the can. Then use metal shears to cut straight down, and flatten out the metal. We get a rectangle of height $25$, and width the circumference of the top. So the width is $9\pi$, and therefore the area of the rectangle is $(9\pi)(25)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Change of Variables in a 3 dimensional integral Let $\int_0^{\infty}\int_0^{\infty}\int_{-\infty}^{\infty}f(x_1,x_2,x_3)dx_1dx_2dx_3$ be a 3 dimensional variable ( i.e. $0\leq x_1,x_2\leq \infty,-\infty\leq x_3\leq \infty.)$ I am defining the following change of variables : $$x_1'=x_1,x_2'=x_2, x_3'=c_1x_1+c_2x_2+c_3x_3$$
Question : What will be my new domain expressed in $x_1',x_2',x_3'.$ I understand the Jacobian concept but it is not clear how to define my new boundaries.
| New limits coincide with the old ones.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Derivatives of univalent functions must converge to derivative of univalent function? This is probably something basic that I am missing. I am reading the article Normal Families: New Perspectives by Lawrence Zalcman, and in one of his examples he makes the following assertion (I am paraphrasing, not quoting, for brevity - I hope I didn't ruin the correctness of the statement):
Let $f_n : D \to \mathbb C$ be a sequence of analytic univalent functions on a domain $D$. Suppose the sequence of first derivatives $g_n := f_n\prime$ converges locally uniformly to an analytic function $g$. Then $g$ is also the first derivative of a univalent function on $D$, or zero.
Why is this true? It looks like a twist on Hurwitz's Theorem, but I don't get it.
| Yes, this is a twist on Hurwitz's theorem: the limit of non-vanishing functions $f_n'$ is either identically zero, or nowhere zero. The first case is clear. In the second case, fix a point $z_0\in D$ and consider the functions $\tilde f_n=f_n-f_n(z_0)$. Since $\tilde f_n$ is pinned down at $z_0$, and the derivatives $\tilde f_n'=f_n'$ converge locally uniformly, the functions $\tilde f_n$ converge locally uniformly. Let $f$ be their limit. Since $f'=\lim \tilde f_n' = g$ is nowhere zero, $f$ is locally univalent.
To prove that $f$ is univalent in $D$, argue by contradiction: suppose there is $w\in \mathbb C$ and points $z_1\ne z_2$ such that $f-w$ vanishes at $z_1,z_2$. Pick $r>0$ such that the $r$-neighborhoods of $z_1,z_2$ are disjoint. By Hurwitz's theorem, for all sufficiently large $n$ the function $\tilde f_n-w$ vanishes in the aforementioned neighborhoods. This is a contradiction, because $\tilde f_n$ is univalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Any open subset of $\Bbb R$ is a countable union of disjoint open intervals
Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals.
This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.
| In a locally connected space $X$, all connected components of open sets are open. This is in fact equivalent to being locally connected.
Proof: (one direction) let $O$ be an open subset of a locally connected space $X$. Let $C$ be a component of $O$ (as a (sub)space in its own right). Let $x \in C$. Then let $U_x$ be a connected neighbourhood of $x$ in $X$ such that $U_x \subset O$, which can be done as $O$ is open and the connected neighbourhoods form a local base. Then $U_x,C \subset O$ are both connected and intersect (in $x$) so their union $U_x \cup C \subset O$ is a connected subset of $O$ containing $x$, so by maximality of components $U_x \cup C \subset C$. But then $U_x$ witnesses that $x$ is an interior point of $C$, and this shows all points of $C$ are interior points, hence $C$ is open (in either $X$ or $O$, that's equivalent).
Now $\mathbb{R}$ is locally connected (open intervals form a local base of connected sets) and so every open set if a disjoint union of its components, which are open connected subsets of $\mathbb{R}$, hence are open intervals (potentially of infinite "length", i.e. segments). That there are countably many of them at most, follows from the already given "rational in every interval" argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "285",
"answer_count": 17,
"answer_id": 11
} |
$ \int_{0}^{1}x^m.(1-x)^{15-m}dx$ where $m\in \mathbb{N}$ $=\displaystyle \int_{0}^{1}x^m.(1-x)^{15-m}dx$ where $m\in \mathbb{N}$
My Try:: Put $x=\sin^2 \theta$ and $dx = 2\sin \theta.\cos \theta.d\theta$ and changing limit, We Get
$ = \displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2m}\theta.\cos^{30-2m}\theta.2\sin \theta.\cos \theta d\theta$
$ = \displaystyle 2\int_{0}^{\frac{\pi}{2}} \sin^{2m+1}\theta.\cos^{31-2m}\theta d\theta$
Now How can i proceed after that.
Is there is any method to Calculate the Given Integral Then plz explain here.
Thanks
| $$f(m,n)=\int_0^1 x^m(1-x)^ndx$$
Repeated partial integration on the right hand side reveals that:
$$(m+1)f(m,n)=n\,f(m+1,n-1)$$
$$(m+2)(m+1)f(m,n)=n(n-1)\,f(m+2,n-2)$$
$$\cdots$$
$$(m+n)\cdots (m+1)\,f(m,n)=n!\,f(m+n,0)$$
$$\text{i.e}$$
$$(m+n)!f(m,n)=n!\,m!\,f(m+n,0)$$
Since
$$f(m+n,0)=\int_0^1 t^{m+n}dt=\frac{1}{m+n+1}$$
We have:
$$f(m,n)=\frac{n!\,m!}{(m+n+1)!}$$
Now simply let $n=15-m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve $ 13x \equiv 1 ~ (\text{mod} ~ 17) $?
How to solve $ 13x \equiv 1 ~ (\text{mod} ~ 17) $?
Please give me some ideas. Thank you.
| $$\frac{17}{13}=1+\frac4{13}=1+\frac1{\frac{13}4}=1+\frac1{3+\frac14}$$
The last but one convergent of $\frac{17}{13}$ is $1+\frac13=\frac43$
Using the relationship of the successive convergents of a continued fraction, $17\cdot3-13\cdot4=-1\implies 13\cdot4\equiv1\pmod{17}\implies x\equiv4\pmod{17}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
Is it necessary to know a lot of advance math to become a good junior high/high school teacher? By "advance math" I refer to Real Analysis, Abstract Algebra and Linear Algebra (to the level of Axler). I received mainly Bs in these courses with the exception of the intro-level Linear Algebra. Since I intend to be a teacher, is it necessary to have mastered these subjects at the elementary level?
|
"Is it necessary to know a lot of advanced math to become a good junior high/high school teacher?"
It is neither necessary nor sufficient.
One of the best math teachers I've ever had probably doesn't remember much (if any) advanced math beyond linear algebra. I don't know how he performed as a math student himself, but I'm not sure it really matters. He is almost universally loved by his students, is very involved in the math team, and consistently challenges students to go beyond the routine algorithms and formulas.
On the flip side, I've also had teachers who certainly knew advanced math very well, but were unable to communicate effectively even the basics to students.
Of course, as Qiaochu points out, having an understanding of advanced math certainly couldn't hurt. For example, it might help provide a deeper context for some of the math you're teaching.
Now I'm not a teacher myself yet, so maybe take what I'm saying with a grain of salt. But as far as I can tell, good teaching is ultimately about dedication, effective communication, a responsiveness to student needs, and an interest in mathematics that goes beyond the textbook. I think these are the important things.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
composition of continuous functions I was wondering if a function $f:[a,b]\rightarrow[c,d]$ is continuous, $g:[c,d]\rightarrow\mathbb{R}$ is continuous, does it necessarily imply that $g\circ f$ is continuous? Are there counterexamples? What is the necessary and sufficient condition for $g\circ f$ to be continuous?
This is not HWQ. I am just wondering if that is possible.
| With the sequence definition of continuity it is obvious that $g\circ f$ is continous, because
$$\lim_{n\rightarrow \infty} g(f(x_n))=g(\lim_{n\rightarrow \infty} f(x_n)) = g(f(\lim_{n\rightarrow \infty} x_n))$$
because $f$ and $g$ are continuous.
It is hard to say what is necessary that the composition of function is continuous, taking
$$D(x)=\left\{
\begin{array}{rl}
0 & x\in \mathbb{R}\setminus \mathbb{Q}\\
1 & x \in \mathbb{Q}\\
\end{array}
\right.$$
is discontinuous in every $x\in \mathbb{R}$ but $D(D(x))=1$ is $C^\infty$.
$C^\infty$ means the function is arbitrary often continuous differentiable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Lambda Calculus Equivalence I'm a bit new to lambda calculus and was wondering about the equivalence of two expressions
$$(\lambda x.\lambda y.xy)\lambda z.z\overset{?}=(\lambda x.\lambda y.xy)(\lambda z.z)$$
Can anyone help out?
| By convention the outer most parenthesis are dropped for minimal clutter.
$$\color{red}{(\lambda x.\lambda y.xy)}\color{blue}{\lambda z.z}\iff\color{red}{(\lambda x.\lambda y.xy)}\color{blue}{(\lambda z.z)}$$
The same thing is done in algebra:
$$\color{red}{(z)}\color{blue}{(x+y)}\iff \color{red}z\color{blue}{(x+y)}$$
In lambda calculus there is a similar "order of operations" as in conventional mathematics. Things to note are parenthesis are evaluated first.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Ordinal exponentiation - $2^{\omega}=\omega$ This is my understanding of ordinal arithmetic - two ordinals are the same as one another if there is an order-preserving bijection between them. So for instance
$$1+\omega = \omega$$
because if
$$f(\langle x,y\rangle)=\begin{cases}y+1 & x=1\\ 1 &\text{otherwise}\end{cases}$$
Then $f$ is an order-preserving bijection between $\{ 0 \} \times 1 \cup \{ 1 \} \times \omega$ and $\omega$, where $\{ 0 \} \times 1 \cup \{ 1 \} \times \omega$ is endowed with the addition order.
Likwise if
$$g(\langle x,y \rangle)=2 \times x+y$$
Then $g$ is an order-preserving bijection between $2 \times \omega$ and $\omega$, where $2 \times \omega$ is endowed with the multiplication order, and so $2 \cdot \omega =\omega$ , whereas $\lnot 2 \cdot \omega =\omega \cdot 2$ because $< 0,1 >$ is a limit of $\omega \times 2$ under the multiplication order whereas $2 \cdot \omega$ has no limit ordinals.
On Wikipedia's page, Exponentiation is described for ordinals, where in particular, it says that $2^{\omega} = \omega$. How can this be when $\omega$ does not even have the same cardinality as $2^{\omega}$ - to wit, isn't $2^{\omega}$ uncountable, with the same cardinality as the reals?
| Ordinal exponentiation is not cardinal exponentiation.
The cardinal exponentiation $2^\omega$ is indeed uncountable and has the cardinality of the continuum.
The ordinal exponentiation $2^\omega$ is the supremum of $\{2^n\mid n\in\omega\}$ which in turn is exactly $\omega$ again.
Also related:
*
*How is $\epsilon_0$ countable?
*Do $\omega^\omega=2^{\aleph_0}=\aleph_1$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Derivative of $e^{\ln(1/x)}$ This question looks so simple, yet it confused me.
If $f(x) = e^{\ln(1/x)}$, then $f'(x) =$ ?
I got $e^{\ln(1/x)} \cdot \ln(1/x) \cdot (-1/x^2)$.
And the correct answer is just the plain $-1/x^2$. But I don't know how I can cancel out the other two function.
| Hint: the exponential and logarithm are inverse functions:
$$e^{\ln u}=u$$
for any $u>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Are all integers fractions? In a college class I was asked this question on a quiz in regards to sets:
All integers are fractions. T/F.
I answered False because if an integer is written in fraction notation it is then classified as a rational number. The teacher said the answer was True and gave me the link http://www.purplemath.com/modules/numtypes.htm. As a teacher of mathematics in the K-12 system I have always taught that integers were all whole numbers above and below zero, and including zero. All of the resources I have used agree to my definition. Please clarify this for me.
What is the truth, or are she and I just mincing words?
| Every integer $x \in \mathbb Z$ can be expressed as the fraction $x \over 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/318948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 0
} |
Is $f(x)= \cos(e^x)$ uniformly continuous? As in the topic, my quest is to check (and prove) whether the given function $$f(x)= \cos(e^x)$$is uniformly continuous on $\left\{\begin{matrix}x \in (-\infty;0]
\\
x \in [0; +\infty)
\end{matrix}\right.$ .
My problem is that I have absolutely no idea how to do it. Any hints will be appreciated and do not feel offended if I ask you a question which you consider stupid, but such are the beginnings. Thank you in advance.
| Note that
$f$ is differentiable on $\mathbb{R}$ and $f'(x)=-e^x\sin{(e^x)}.$
Let $x_n=\ln{\left(\dfrac{\pi}{6}+2\pi n\right)}, \;\; y_n= \ln{\left(\dfrac{\pi}{3}+2\pi n\right)}.$
Then $$|x_n-y_n| = {\ln{\left(\dfrac{\pi}{3}+2\pi n\right)} -\ln{\left(\dfrac{\pi}{6}+2\pi n\right)}}=\ln{\dfrac{\dfrac{\pi}{3}+2\pi n}{{\dfrac{\pi}{6}+2\pi n}}}=\\ = \ln{\left(1+\dfrac{1}{1+12 n }\right)}\sim \dfrac{1}{1+12 n }, \;\; n\to \infty .$$
By the mean value ( Lagrange's) theorem $$|f(x_n)-f(y_n|=|f'(\xi_n)|\cdot|x_n-y_n|=e^{\xi_n}|\sin(e^{\xi_n})|\cdot|x_n-y_n|\geqslant \\ \geqslant e^{x_n}\cdot\dfrac{1}{2}\cdot\dfrac{1}{1+12 n }=\dfrac{\dfrac{\pi}{6}+2\pi n}{2+24n} \underset{n\to\infty}{\rightarrow} \dfrac {\pi}{12},$$
which proves that $f$ is not uniformly continuous on $[0,\;+\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Value of limsup i? This is a part of my question.
$\lim \sup \cos(n\pi/12)$ as n goes to infinity
What is the value of this limit?
| When a sequence is bounded, the limsup is the largest limit of all convergent subsequences.
For all $n$, $\cos(n\pi/2)\leq 1$ so $\limsup\cos(n\pi/2) \leq 1$.
And for the extraction $n=4k$, $\cos(n\pi/2)=\cos(2k\pi)=1$. So $\limsup\cos(n\pi/2)\geq 1$.
Hence
$$
\limsup_{n\rightarrow +\infty}\cos(n\pi/2)=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A ‘strong’ form of the Fundamental Theorem of Algebra Let $ n \in \mathbb{N} $ and $ a_{0},\ldots,a_{n-1} \in \mathbb{C} $ be constants. By the Fundamental Theorem of Algebra, the polynomial
$$
p(z) := z^{n} + \sum_{k=0}^{n-1} a_{k} z^{k} \in \mathbb{C}[z]
$$
has $ n $ roots, including multiplicity. If we vary the values of $ a_{0},\ldots,a_{n-1} $, the roots will obviously change, so it seems natural to ask the following question.
Do the $ n $ roots of $ p(z) $ depend on the coefficients in an analytic sort of way? More precisely, can we find holomorphic functions $ r_{1},\ldots,r_{n}: \mathbb{C}^{n} \to \mathbb{C} $ such that
$$
z^{n} + \sum_{k=0}^{n-1} a_{k} z^{k} = \prod_{j=1}^{n} [z - {r_{j}}(a_{0},\ldots,a_{n-1})]?
$$
The definition of a holomorphic function of several complex variables is given as follows:
Definition Let $ n \in \mathbb{N} $ and $ \Omega \subseteq \mathbb{C}^{n} $ be a domain (i.e., a connected open subset). A function $ f: \Omega \to \mathbb{C} $ is said to be holomorphic if and only if it is holomorphic in the usual sense in each of its $ n $ variables.
The existence of $ r_{1},\ldots,r_{n}: \mathbb{C}^{n} \to \mathbb{C} $ that are continuous seems to be a well-known result (due to Ostrowski, perhaps?), but I am unable to find anything in the literature that is concerned with the holomorphicity of these functions.
Any help would be greatly appreciated. Thank you very much!
| To give a little expansion to @Andreas’s answer, let’s examine a little more closely the way the coefficients depend on the roots. Let’s take an $n$-tuple of roots, say $\rho=(\rho_1,\cdots,\rho_n)$ and form the corresponding $n$-tuple whose entries are the coefficients $a=(a_0,\cdots,a_{n-1})$ of the monic polynomial whose roots are the $\rho_i$’s. You have the map $C\colon\rho\mapsto a$, and you can ask what the Jacobian determinant is of this map, call it $J$. Then the fact is that $J^2$ is the discriminant of the polynomial $F(x)=x^n+\sum_0^{n-1}a_{n-i}x^i$, which as you probably know is a polynomial in the $a_i$’s. This fact makes it very clear, via the Inverse Function Theorem, how and when and why the roots depend on the coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Number of ways to seat students at a round table subject to certain conditions. In an Olympic contest, there are $n$ teams. Each team is composed of $k$ students attending different subjects. How many ways are there to seat all the students at a round table such that $k$ students in a team sit together and there are no two students who attend the same subject seat next to one another?
My attempt:
Let $S_n$ denote the total way to seat all the student in $n$ teams with $k$ students on each team in a way that satisfies the problem.
Then I find that $\forall n \geq 2$ $S_{n+1}=\alpha.nS_n$ with $\alpha = 2(k-1)!-(k-2)!$
But there is problem for me to find $S_2$ because it may be non-relative to $S_1$. Help me!
| I think that your recurrence isn’t quite right. If you start with an acceptable arrangement of $n$ teams, you can insert an $(n+1)$-st team in any of the $n$ slots between adjacent teams. The members of the new team can be permuted in $k!$ ways; $(k-1)!$ of these have an unacceptable person at one end, $(k-1)!$ have an unacceptable person at the other end, and $(k-2)!$ have an unacceptable person at both ends, so
$$S_{n+1}=n\Big(k!-2(k-1)!+(k-2)!\Big)S_n\;,$$
i.e., we should have $\alpha=k!-2(k-1)!+(k-2)!$.
I’m assuming now that arrangements that differ only by a rotation of the table are considered the same. Then $S_1=(k-1)!$. There are at least two ways to see that $S_2=k!\alpha$.
*
*Start with any of the $(k-1)!$ arrangements of one team. There are $k$ slots into which we can insert the second team, and the argument given above shows that within its slot it can be arranged in $\alpha$ ways, for a total of $(k-1)!k\alpha=k!\alpha$ arrangements.
*There are $k!$ ways to seat the first team around half of the table, and by the argument given above there are $\alpha$ acceptable ways to seat the second team around the other half of the table.
Combine this with the recurrence $S_n=(n-1)\alpha S_{n-1}$ for $n\ge 3$, and you can easily get a closed expression for $S_n$ in terms of $n$ and $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Elementary topology problem. Let $ ((Y_{\alpha},\tau_{\alpha}) \mid \alpha \in J) $ be a $ J $-indexed family of topological spaces and $ X $ any non-empty set. Let $ (f_{\alpha} \mid \alpha \in J) $ be a $ J $-indexed family of functions, where $ f_{\alpha}: Y_{\alpha} \to X $. What topology $ \tau $ can you put on $ X $ that will make all of the $ f_{\alpha} $’s continuous with respect to the $ \tau_{\alpha} $’s and $ \tau $?
Please help me with this. I think that I only need to put the indiscrete topology on $ X $, so that the only open sets in $ X $ are $ X $ itself and $ \varnothing $, and the inverse image of each of these under $ f_{\alpha} $ is open in $ Y_{\alpha} $.
| One can define a topology $ \tau $ on $ X $ as follows:
Declare a subset $ U $ of $ X $ to be $ \tau $-open if and only if $ {f_{\alpha}^{\leftarrow}}[U] \in \tau_{\alpha} $ for each $ \alpha \in J $.
Then $ \tau $ is the finest topology on $ X $ that makes $ f_{\alpha}: (Y_{\alpha},\tau_{\alpha}) \to (X,\tau) $ continuous for each $ \alpha \in J $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
diagonalizability of matrix If $A$ is invertible, $F$ is algebraically closed, and $A^n$ is diagonalizable for some $n$ that is not an integer multiple of the characteristic of $F$, then $A$ is diagonalizable.
My question is:
(1) Why the condition "$A$ is invertible" essential?
(2) In wikipedia, diagonalizable matrix , it says $A$ satisfies the polynomial $p(x)=(x^n- \lambda_1) \cdots (x^n- \lambda_k)$, is that means $p(x)$ is the minimal polynomail of $A^n$? If yes, why it can be written in the form of $(x^n- \lambda_1) \cdots (x^n- \lambda_k)$ instead of $(x- \lambda_1) \cdots (x- \lambda_k)$??
Thank you very much!
| If $A$ is invertible then $A^n$ is also invertible, so $0$ is not an eigenvalue of $A^n$.
$A^n$ is diagonalizable then its minimal polynomial $P$ is a product of distinct linear factors over $F$:
$$P(X)=(X-\lambda_1)\cdots(X-\lambda_k),\quad \lambda_i\neq\lambda_j$$
We know that $A^n$ is annihilated by $P$: $P(A^n)=0$ so $A$ is annihilated by the polynomial:
$$Q(X)=(X^n-\lambda_1)\cdots(X^n-\lambda_k)$$
and since $\lambda_i\neq0$ then each polynomial $(X^n-\lambda_i)$ has no multiple root (since its derivative is $nX^{n-1}$ and $n$ is not an integer multiple of the characteristic of $F$), then $Q$ is a product of distinct linear factors and $A$ is annihilated by $Q$. Hence $A$ is diagonalizable matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\sum \limits_{k=1}^{\infty} \frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} $ as a rational number. $$\sum \limits_{k=1}^{\infty} \frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} $$
I know from the ratio test it convergest, and I graph it on wolfram alpha and I suspect the sum is 2; however, I am having trouble with the manipulation of the fraction to show the rational number.
ps. When it says write as a rational number it means to write the value of $S_{\infty}$ or to rewrite the fraction?
| That denominator should suggest the possibility of splitting the general term into partial fractions and getting a telescoping series of the form
$$\sum_{k\ge 1}\left(\frac{A_k}{3^k-2^k}-\frac{A_{k+1}}{3^{k+1}-2^{k+1}}\right)\;,$$
where $A_k$ very likely depends on $k$. Note that if this works, the sum of the series will be
$$\frac{A_1}{3^1-2^1}=A_1\;.$$
Now
$$\frac{A_k}{3^k-2^k}-\frac{A_{k+1}}{3^{k+1}-2^{k+1}}=\frac{3^{k+1}A_k-3^kA_{k+1}-2^{k+1}A_k+2^kA_{k+1}}{(3^k-2^k)(3^{k+1}-2^{k+1})}\;,$$
so you want to choose $A_k$ and $A_{k+1}$ so that
$$3^{k+1}A_k-3^kA_{k+1}-2^{k+1}A_k+2^kA_{k+1}=6^k\;.$$
The obvious things to try are $A_k=2^k$, which makes the last two terms cancel out to leave $3^{k+1}2^k-3^k2^{k+1}=6^k(3-2)=6^k$, and $A_k=3^k$, which makes the first two terms cancel out and leaves $6^k(3-2)=6^k$; both work.
However, summing $$\sum_{k\ge 1}\left(\frac{A_k}{3^k-2^k}-\frac{A_{k+1}}{3^{k+1}-2^{k+1}}\right)\tag{1}$$ to $$\frac{A_1}{3^1-2^1}=A_1$$ is valid only if
$$\lim_{k\to\infty}\frac{A_k}{3^k-2^k}=0\;,$$
since the $n$-th partial sum of $(1)$ is
$$A_1-\frac{A_{n+1}}{3^{n+1}-2^{n+1}}\;.$$
Checking the two possibilities, we see that
$$\lim_{k\to\infty}\frac{2^k}{3^k-2^k}=0,\quad\text{but}\quad\lim_{k\to\infty}\frac{3^k}{3^k-2^k}=1\;,$$
so we must choose $A_k=2^k$, and the sum of the series is indeed $A_1=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Eigenvalues - Linear algebra If $c$ is eigenvalue for matrix B.
How can I prove than $c^k$ is eigenvalue for matrix $B^k$?
am not sure what I should try here...will appreciate your help
| Induction...?
$$B(v)=cv\Longrightarrow B^k(v)=B(B^{k-1}v)\stackrel{\text{Ind. hypothesis}}=B(c^{k-1}v)=c^{k-1}Bv=c^{k-1}cv=c^kv$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Joint density with continuous and binary random variable Assume $X\in\mathbb{R}$, $Y\in\{0,1\}$ are two random variables. What allows us to claim that $$f_{X}(x) = f_{XY}(x,1) + f_{XY}(x,0)$$
where $f_X(x)$ and $f_{XY}(x,y)$ are densities.
| $$P(X\le x)=P(X\le x\mid Y=1)P(Y=1)+P(X\le x\mid Y=0)P(Y=0)\\
=P(X\le x, Y=1)+P(X\le x,Y=0)\\
f_X(x)=\frac{dP(X\le x)}{dx}=f_{XY}(x,1)+f_{XY}(x,0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Normal $T\in B(H)$ has a nontrivial invariant subspace I am wondering if the following is true:
Every normal $T\in B(H)$ has a nontrivial invariant subspace if $\dim(H)>1$?
| Let $ T \in B(\mathcal{H}) $ be a normal operator. Let $ \sigma(T) $ denote the spectrum of $ T $. We then have two cases to consider: (i) $ \sigma(T) $ is a singleton set, and (ii) $ \sigma(T) $ contains at least two points.
Case (i): Suppose that $ \sigma(T) = \{ \lambda \} $ for some $ \lambda \in \mathbb{C} $. Let $ \text{id}_{\lambda} $ and $ 1_{\lambda} $ denote, respectively, the identity function on $ \{ \lambda \} $ and the constant function on $ \{ \lambda \} $ with value $ 1 $. By the Continuous Functional Calculus, we can apply these two functions to $ T $. As $ \text{id}_{\lambda} = \lambda \cdot 1_{\lambda} $, we obtain $ T = \lambda I $, where $ I: \mathcal{H} \to \mathcal{H} $ is the identity operator. Clearly, $ I $ has non-trivial subspaces (this is only true if we assume that $ \dim(\mathcal{H}) > 1 $), so $ T $ has non-trivial subspaces as well.
Case (ii): Suppose that $ \sigma(T) $ contains two distinct points $ \lambda_{1} $ and $ \lambda_{2} $ (note that this is not possible if $ \dim(\mathcal{H}) = 1 $). Let $ U_{1} $ and $ U_{2} $ be disjoint open neighborhoods (contained in $ \sigma(T) $) of $ \lambda_{1} $ and $ \lambda_{2} $ respectively. If $ \mathbf{E} $ denotes the resolution of the identity corresponding to $ T $, then we have non-zero projection operators $ P_{1} := \mathbf{E}(U_{1}) $ and $ P_{2} := \mathbf{E}(U_{2}) $ satisfying the following two properties:
*
*$ P_{1} P_{2} = \mathbf{0}_{B(\mathcal{H})} = P_{2} P_{1} $ and
*$ T P_{1} = P_{1} T $ and $ T P_{2} = P_{2} T $, i.e., $ P_{1} $ and $ P_{2} $ commute with $ T $.
Property (1) says that $ P_{1} $ is not the identity operator; otherwise $ P_{1} P_{2} = P_{2} \neq \mathbf{0}_{B(\mathcal{H})} $, which is a contradiction. Next, Property (2) says
$$
T[{P_{1}}[\mathcal{H}]] = {P_{1}}[T[\mathcal{H}]] \subseteq {P_{1}}[\mathcal{H}],
$$
which shows that $ {P_{1}}[\mathcal{H}] $ is a non-trivial invariant subspace of $ T $.
Conclusion: Every normal operator $ T \in B(\mathcal{H}) $ has a non-trivial invariant subspace if $ \dim(\mathcal{H}) > 1 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How do you solve linear congruences with two variables. For example: Solving for $x$ and $y$ given the following linear congruences.
$x + 2y \equiv 3 \pmod9\,$, $3x + y \equiv 2 \pmod9$
So far, I've tried taking the difference of the two congruences.
Since $x + 2y \equiv 3 \pmod9 \Rightarrow x + 2y = 3 + 9k\,$, and $3x + y \equiv 2 \pmod9 \Rightarrow 3x + y = 2 + 9l$.
$$(x + 2y = 3 + 9k) - (-3x + y = 2 + 9l) \Rightarrow -2x + y = 1 + 9(k - l)$$.
So now, are you supposed to solve this like a normal linear Diophantine equation?
| The CRT is used solve systems of congruences of the form $\rm x\equiv a_i\bmod m_{\,i}$ for distinct moduli $\rm m_{\,i}$; in our situation, there is only one variable and only one moduli, but different linear congruences, so this is not the sort of problem where CRT applies. Rather, this is linear algebra.
Instead, you are working with a $2\times2$ linear system over a given modulus, $9$. Here, the first two elementary methods of solving linear systems apply: substitution and elimination. The difference, however, is that we cannot generally divide by anything sharing divisors with $9$, i.e. multiples of $3$. And if, in your quest to eliminate variables, you multiply by things not coprime to the modulus, you can end up adding extraneous non-solutions, so it can be dangerous if you're not careful.
Let's use substitution. The congruences here are
$$\begin{cases}\rm x+2y\equiv 3 \mod 9, \\ \rm 3x+y\equiv2 \mod 9.\end{cases}$$
The first congruence gives $\rm x\equiv 3-2y$; plug this into the second to obtain
$$\rm 3x+y\equiv 3(3-2y)+y\equiv -5y\equiv2\mod 9.$$
Now $-5$ is coprime to $9$ so we can divide by it, i.e. multiply by its reciprocal mod $9$. In this case the reciprocal is $-5^{-1}\equiv-2\equiv7\bmod 9$, so the solution for $\rm y$ is $\rm y\equiv7\cdot2\equiv5\bmod 9$. To find $\rm x$, plug in $\rm y\equiv5$ into the congruences, obtaining $\rm x+10\equiv3$ and $\rm 3x+5\equiv2$. The first gives $\rm x\equiv2$, so that we have the unique solution $\rm (x,y)\equiv(2,5)$. However the second gives $\rm 3x\equiv-3\bmod9$, which, after dividing, gives $\rm x\equiv-1\equiv2\bmod3$ so that $\rm x\in\{2,5,7\}\bmod 9$; this doesn't change the fact that $(2,5)$ is the unique solution to the system, but it does illustrate that dividing by things that are not coprime to the modulus can introduce unwanted, fake solutions.
Note that matrix multiplication makes sense taken modulo an integer. The potential issues arise when we want inverses. If a matrix inverse $\rm A^{-1}$ exists of an integer-entry matrix $\rm A$, and every denominator appearing in the resulting rationals is coprime to the modulus (equivalently: $\rm\det A$ is coprime to the modulus), then $\rm A^{-1}$ can be reduced under the modulus (rationals like $\rm a/b$ become as $\rm ab^{-1}\bmod m$; this sort of thing is valid because there is a "ring homomorphism ${\bf Z}_{(p)}\to{\bf Z}/p{\bf Z}$"), and the result will be an inverse of $\rm A$ with respect to the modulus, i.e. $\rm A\, A^{-1}\equiv I\bmod m$. This is why bikenaga's answer works, using matrices, and it makes things quite easier when applicable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
If $g$ is an element in an abelian group $G$ and $H\leqslant G$, must there exist an $n$ such that $g^n\in H$? Let $G$ be an abelian group and $H$ a subgroup of $G$. For each $g \in G$, does there always exist an integer $n$ such that $g^{n} \in H$?
| This is evident that if $[G:H]=n<\infty$ then $\forall g\in G, g^n\in H$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
solving a basic complex equation but using de Moivres theorem I have a question which should be super super easy!
If I was to solve $z^2 = 1+i$ how would I do this using de-moivres theorem?
I have the answer here in front of me so I don't want the answer, I just dont understand the method very well!
Any help would be appreciated! I haven't had much experience with complex numbers and have just started a complex analysis course.
Many thanks
| $$\forall w=x+iy\in\Bbb C:$$
$$w=|w|e^{i\phi}=|w|(\cos\phi+i\sin\phi)\;,\;\;\phi=\begin{cases}\arctan\frac{y}{x}+2k\pi &,\;\;y\neq 0\\{}\\2k\pi\end{cases}\;\;,\;\;\;k\in\Bbb Z$$
In our case:
$$w=1+i\Longrightarrow |w|=\sqrt 2\,\,,\,\,\arctan\frac{1}{1}=\frac{\pi}{4}+2k\pi\Longrightarrow$$
$$z^2=1+i=\sqrt 2\,e^{\frac{\pi i}{4}\left(1+8k\right)}\;\;,\;\;k=0,1\;\Longrightarrow z=\sqrt[4]2\, e^{\frac{\pi i}{8}(1+8k)}\;,\;\;k=0,1$$
Why do we restrict ourselves only to the above values of $\,k\,$ ? Because any other integer value will give one of these two different ones (on the trigonometric circle, say) !
Thus, the solutions are
$$k=0:\;\;\;\;z_0:=\sqrt[4]2\,e^{\frac{\pi i}{8}}=\sqrt[4]2\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)\\k=1:\;\;\;z_1:=\sqrt[4] 2\,e^{\frac{9\pi i }{8}}=\sqrt[4] 2\left(\cos\frac{9\pi}{8}+i\sin\frac{9\pi}{8}\right)$$
By the way, it is a nice exercise to show that $\,z_0=-z_1\,$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Differentiating piecewise functions. Say we have the piecewise function $f(x) = x^2$ on the interval $0 \le x < 4$; and it equals $x+1$ on the interval $ x \ge 4$. Why is it that, when I take the derivative, the intervals loose their equality and become strictly greater or strictly less than?
| Note that $\displaystyle \lim_{x\to 4^-}(f(x))=\lim_{x\to 4^-}(x^2)=16$ while $\displaystyle\lim_{x\to 4^+}(f(x))=f(4)=5.$ Therefore $f$ isn't continuous on $x=4$ and so it can't be differentiable there. Hence the domain of the derivative doesn't include $4$.
Around $0$ the left lateral derivative isn't even defined, therefore $f$ can't be differentiable there.
Since $f$ is clearly differentiable on the interior of its domain, it follows that the domain of $f'$ is the interior of $f$'s domain, that is $\textbf{]}0,4[\cup \textbf{]}4+\infty[=\textbf{]}0,+\infty[\setminus \{4\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $x$, $y$, and $z$ are not distinct if $x^2(z-y) + y^2(x-z) + z^2(y-x) = 0$. Suppose that $x^2(z-y) + y^2(x-z) + z^2(y-x) = 0$.
How can I show that $x$, $y$, and $z$ are not all distinct, that is, either $x=y$, $y=z$, or $x=z$?
| $x^2(z-y)+y^2(x-z)+z^2(y-x)=x^2(z-y)+x(y^2-z^2)+z^2y-y^2z=(z-y)(x^2-x(y+z)+zy)=(z-y)(x-y)(x-z)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Logarithm proof problem: $a^{\log_b c} = c^{\log_b a}$ I have been hit with a homework problem that I just have no idea how to approach. Any help from you all is very much appreciated. Here is the problem
Prove the equation: $a^{\log_b c} = c^{\log_b a}$
Any ideas?
| If you apply the logarithm with base $a$ to both sides you obtain,
$$\log_a a^{\log_b c} = \log_a c^{\log_b a}$$
$$\log_b c = \log_b a \log_a c$$
$$\frac{\log_b c}{\log_b a} = \log_a c$$
however this last equality is the change of base formula and hence is true. Reversing the steps leads to the desired equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
Flow of D.E what is the idea behind conjugacy? I got some kinda flow issue, ya know?
well enough with the bad jokes let A be a 2x2 matrix, T a change of Coordinate matrix, and $B=T^{-1}AT$ the canonical matrix ascoiated with A. Show that the function $h=T^{-1}: \mathbb R^{2} \to \mathbb R^{2}$, $h(x)=T^{-1}X$ is a conjugacy between the flows of the systems $X^{'}$= $AX$ and $Y^{'}$= $BY$
Firstly Is there any chance someone could quantify this in terms of algebraic set theory for me? i seem to half comprehend flow of dynamical systems and half understand there counter part in group theory. they feel so similar yet i can't seem to put the 2 together. it feels like flow of this system is a Monomorphism as its a homomorphism and it is a one-one mapping.
Second question how do i prove the above is a conjugacy between the flows, the method we were shown in class was very odd and didn't involve using T and $T^{-1}$i have asked a somewhat similar question and someone was kind enough to show me how to do it using this kind of idea but with numbers. i don't actually understand why this concept works.
All that i really understand is that if the same number of eigenvalues with positive sign are in each matrix there should be a mapping of the flow that exists between the two of them.
| Suppose $X' = AX$. Let $Y = T^{-1} X T$. Then $$ Y' = (T^{-1} X T)' = T^{-1} X' T = T^{-1} (AX) T = (T^{-1} A T) (T^{-1} X T) = B (T^{-1} X T) = BY$$
So the flows of $B$ are simply conjugates of the flows of $A$. Is this what you wanted to know?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $2 xy < x^2 + y^2$ for $x$ is not equal to $y$ Show that $2 xy < x^2 + y^2$ for $x$ is not equal to $y$.
| If $x\neq y$ then without loss of generality we can assume that $x>y$ $x-y>0\Rightarrow (x-y)^2>0\Rightarrow x^2-2xy+y^2>0\Rightarrow x^2+y^2>2xy$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Homogeneous equation I am trying to solve the following homogeneous equation:
Thanks for your tips
$xy^3y′=2(y^4+x^4)$
I think this isHomogeneous of order4
=> $xy^3dy/dx=2(y^4+x^4/1)$
=> $xy^3dy=2(x^4+y^4)dx$
=> $xy^3dy-2(x^4+y^4)dx=0$
I do not know how to continued
| Make the substitution $v=y^4$. Then by the chain rule we have $v'=4y^3 y'$. Now your DE turns into:
$$x \frac{v'}{4}=2(v+x^4)$$
Then can be simplified to:
$$v'-8\frac{v}{x}=8x^3$$
We first solve the homogeneous part:
$$v_h' -8\frac{v_h}{x}=0$$
This leads to $v_h=c\cdot x^8$ so that $v_h'=c\cdot 8x^7$. This is our homogeneous solution. Now to find a particular solution we guess it will look something like $$v_p=ax^4+bx^3+dx^2+3x+f$$
Filling this into the DE we get:
$$(4ax^3 +3bx^2+2dx+3)-\frac{8}{x}(ax^4+bx^3+dx^2+3x+f)=8x^3$$
the only terms with $x^3$ in it are the $4ax^3$ and the $-8ax^3$. Conclusion: $-4a=8$ so $a=-2$. All the other terms are zero. We now have our particular solution $v_p=-2x^4$. The general solution is the sum of the particular and homogeneous solution so
$$v_g=c\cdot x^8 -2x^4$$
Now backsubstitute $v=y^4$ or $y=v^{\frac{1}{4}}$ to obtain the final answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Time complexity of a modulo operation I am trying to prove that if $p$ is a decimal number having $m$ digits, then $p \bmod q$ can be performed in time $O(m)$ (at least theoretically), if $q$ is a prime number. How do I go about this?
A related question is asked here, but it is w.r.t to MATLAB, but mine is a general one.
The relevant text, that I am referring: chapter 32 -String Matching, PP:991, "Introduction to Algorithms" 3rd edition Cormen et al.
| I will assume that $q$ is not part of your input, but rather a constant. Then, Algorithm D in 4.3.1 of Knuth's book "The Art of Computer Programming" (Volume 2) performs any long division in $O(m)$ steps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Variance of the number of balls between two specified balls Question:
Assume we have 100 different balls numbered from 1 to 100, distributed in 100 different bins, each bin has 1 ball in it.
What is the variance of the number of balls in between ball #1 and ball #2?
What I did:
I defined $X_i$ as an indicator for ball $i$ - "Is it in between balls 1 and 2?" Also I thought of the question as this problem: "We have actually just 3 places to put the 98 remaining balls: before, after and between balls #1,2, so for each ball there is a probability of 1/3 to be in between. So by this we have $E[X_i]= $$1 \over 3$ . Now $X=\sum _{i=1} ^{98} X_i$. Since $X_i$ is a Bernoulli RV then: $V(X_i)=p(1-p)=$$2 \over 9$.
But I know that the correct answer is 549 $8 \over 9$. I know that I should somehow use the formula to the sum of variances, but somehow I don't get to the correct answer.
| Denote the number of balls and the the number of bins by $b$. Suppose the first ball lands in bin $X_1$ and second ball lands in the bin $X_2$. The number of balls that will land in between them equals $Z = |X_2 -X_1| - 1$. Clearly
$$
\Pr\left( X_1 = m_1, X_2 = m_2 \right) = \frac{1}{b\cdot (b-1)} [ m_1 \not= m_2 ]
$$
Thus:
$$\begin{eqnarray}
\Pr\left(Z = n\right)&=&\sum_{m_1}^b \sum_{m_2=1}^b \frac{1}{b(b-1)} [ m_1 \not=m_2, |m_1-m_2|=n+1] \\ &=& \sum_{m_1}^b \sum_{m_2=1}^b \frac{2}{b(b-1)} [ m_1 > m_2, m_1=n+1+m_2] \\
&=& \frac{b-n-1}{\binom{b}{2}} [ 0 \leqslant n < b-1 ]
\end{eqnarray}
$$
With this it is straightforward to find:
$$
\mathbb{E}\left(Z\right) = \sum_{n=0}^{b-2} n \frac{b-n-1}{\binom{b}{2}} = \frac{b-2}{3}
$$
$$
\mathbb{E}\left(Z^2\right) = \sum_{n=0}^{b-2} n^2 \frac{b-n-1}{\binom{b}{2}} = \frac{(b-2)(b-1)}{6}
$$
Thus the variance reads:
$$
\mathbb{Var}(Z) =\mathbb{E}(Z^2) - \mathbb{E}(Z)^2 = \frac{(b+1)(b-2)}{18} = 549 \frac{8}{9}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Entire function dominated by another entire function is a constant multiple These two questions I didn't even find the way to solve
So please if you can help
*
*Suppose $f (z)$ is entire with $|f(z)| \le |\exp(z)|$ for every $z$
I want to prove that $f(z) = k\exp(z)$ for some $|k| \le 1$
*Can a non constant entire function be bounded in half a plane? Prove if yes , example if not.
| Hint:
1) Liouville + $\,\displaystyle{\frac{f(z)}{e^z}}\,$ is analytic and bounded...
2) Develop the function in power series and extend analytically into the "other" halpf plane.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Is the set of discontinuity of $f$ countable? Suppose $f:[0,1]\rightarrow\mathbb{R}$ is a bounded function satisfying: for each $c\in [0,1]$ there exist the limits $\lim_{x\rightarrow c^+}f(x)$ and $\lim_{x\rightarrow c^-}f(x)$. Is true that the set of discontinuity of $f$ is countable?
| Hopefully this more tedious proof is more illustrative...
In light of the assumptions on $f$, there are two ways that $f$ can fail to be continuous: (1) the left and right hand limits differ, or (2) the limits are equal, but the function value differs from the limit (thanks to @GEdgar for pointing this out).
Let $\epsilon>0$ and $\Delta_\epsilon = \{ x |\, |\lim_{y \downarrow x}f(y) - \lim_{y \uparrow x}f(y)| \geq \epsilon \}$. If $\Delta_\epsilon$ is not finite, then since $[0,1]$ is compact, there exists an accumulation point $\hat{x} \in [0,1]$. Let $c_+ = \lim_{y \downarrow \hat{x}}f(y), c_- = \lim_{y \uparrow \hat{x}}f(y)$. (Note that it is possible that $c_- = c_+$.) By assumption, there exists some $\delta>0$ such that for $y \in (\hat{x}-\delta,\hat{x})$, $|f(y)-c_-| < \frac{\epsilon}{4}$ and for $y \in (\hat{x}, \hat{x}+\delta)$, $|f(y)-c_+| < \frac{\epsilon}{4}$. However, this implies that $\hat{x}$ is an isolated point of $\Delta_\epsilon$, which is a contradiction. Hence $\Delta_\epsilon$ is finite.
If we let $\Delta = \cup_n \Delta_{\frac{1}{n}}$, we see that $\Delta $ is at most countable.
@GEdgar has pointed out an omission in my proof: It is possible that the two limits coincide at a point, but the function is still discontinuous at that point, ie, $\Delta$ is not the entire set of discontinuities.
Let $\Gamma_\epsilon = \{ x |\, \lim_{y \downarrow x}f(y) = \lim_{y \uparrow x}f(y), \ |\lim_{y \downarrow x}f(y)-f(x) | \geq \epsilon \}$. Suppose, as above, that $\Gamma_\epsilon$ is not finite, and let $\hat{x}$ be an accumulation point. Let $c_+, c_-$ be the limits as above. Again, there exists a $\delta>0$ such that if $y \in (\hat{x}-\delta,\hat{x})$, $|f(y)-c_-| < \frac{\epsilon}{4}$, and similarly, if $y \in (\hat{x}, \hat{x}+\delta)$, $|f(y)-c_+| < \frac{\epsilon}{4}$. Consequently, $\hat{x}$ is isolated, hence a contradiction, and $\Gamma_\epsilon$ is finite.
If we let $\Gamma= \cup_n \Gamma_{\frac{1}{n}}$, we see that $\Gamma$ is at most countable.
Since the set of discontinuities is $\Delta \cup \Gamma$, we see that the set of discontinuities is at most countable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
How to prove that if group $G$ is abelian, $H = \{x \in G : x = x^{-1}\}$ is a subgroup? $G$ is abelian, $H = \{x \in G : x = x^{-1}\}$ is a subgroup?
I know to prove that a subset $H$ of group $G$ be a subgroup, one needs to (i) prove $\forall x,y \in H:x \circ y \in H$ and (ii) $\forall x \in H:x^{-1} \in H.$
| Unless you know $H$ is a nonempty subset of $G$, we need also to show that the identity $e \in G$ is also in $H$:
(o) Clearly, the identity $e \in G$ is its own inverse, hence $e = e^{-1} \in H$.
$(ii)\;$ $\forall x \in H$, $x\in H \implies x = x^{-1} \in H$. Hence we have closure under inverses.
$(i)$ $x\circ y \in H$?
$$\; x, y \in H, \implies x = x^{-1}, y = y^{-1}$$ $$ x \circ y = x^{-1}\circ y^{-1} = y^{-1}\circ x^{-1} \quad\quad\quad\quad\tag{G is abelian}$$
$$y^{-1}\circ x^{-1} = (x\circ y)^{-1} \implies x\circ y \in H$$
Hence we have closure under the group operation.
Therefore, $H \leq G$. That is, $H$ is a subgroup of $G$.
Added: Note that this problem is equivalent to the task of proving that if $G$ is an abelian group, and $H$ is a subset of $G$ such that $H = \{x \in G\mid x^2 = e\}$, then $H$ is a subgroup of $G$. Why? For $x \in G$ such that $x^2 = e$, note that $$x^2 = e \iff x^{-1}\circ x^2 = x^{-1} \circ e \iff x^{-1}\circ (x\circ x) = x^{-1}$$ $$\iff (x^{-1}\circ x) \circ x = x^{-1} \iff e\circ x = x^{-1} \iff x = x^{-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Cardinals of set operations without AC Given info: $|A|=\mathfrak{c}$ , $|B|=\aleph_0$ in ZF (no axiom of choice).
Prove: $|A\cup B|=\mathfrak{c}$
If $B \subset A\implies|A \backslash B|=\mathfrak{c}$?
I have found several places proving that for $|\mathbb{R} \backslash \mathbb{Q}|,$ but none of the solutions appears to work for arbitrary sets. Maybe one should prove it for $|\mathbb{R} \backslash \mathbb{Q}|$ and then show that it will work for arbitrary $A$ and $B$ as well?
Here Showing that $\mathbb{R}$ and $\mathbb{R}\backslash\mathbb{Q}$ are equinumerous using Cantor-Bernstein David has a very nice idea (constructing bijection), but it requires that infinite set has countably infinite subset, which again!? needs some choice axiom.
In some sources I even saw statements that these can't be proved in ZF.
From what my teacher said I percieved that solution has something to do with Cantor–Bernstein theorem and that knowing how to prove $\mathfrak{c}+\mathfrak{c}=\mathfrak{c}$ would help as well.
Thanks!
| Prove it first for disjoint $A$ and $B$, relaxing the condition on $B$ to $|B|\le \aleph_0$. You then recover the full statement by considering $A\cup B = A\cup(B\setminus A)$.
You can restrict your attention even further to, say $A=(0,1)$ and $B$ being a subset of the integers. Once you have proved it for that case, the definition of "same cardinality" guarantees that it will be true for every other choice of disjoint $A$ and $B$ of the appropriate cardinalities.
It is true without any choice axiom that a set of size continuum has a countably infinite subset. By definition, because it has size continuum, there's a bijection from $\mathbb R$, and the image of $\mathbb N$ under that bijection is a countably infinite subset.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that that $\lim_{n\to\infty}\sqrt[n]{\binom{2n}{n}} = 4$ I know that
$$
\lim_{n\to\infty}{{2n}\choose{n}}^\frac{1}{n} = 4
$$
but I have no Idea how to show that; I think it has something to do with reducing ${n}!$ to $n^n$ in the limit, but don't know how to get there. How might I prove that the limit is four?
| Hint: By induction, show that for $n\geq 2$ $$\frac{4^n}{n+1} < \binom{2n}{n} < 4^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 3
} |
Convergence of $\int_3^{\infty} \frac{1}{(\ln(x))^2(x-\ln(x))}$ Does this integral converge?
$$\int_3^{\infty} \frac{1}{(\ln(x))^2(x-\ln(x))}$$
I've been trying to solve this for the past 2 hours...literally. I know the answer is fairly simple, but I just can't think of it
| We have $$\dfrac{x}e > \ln(x) \,\,\, \forall x > 0$$
Hence,
$$x - \ln(x) > x - \dfrac{x}e = x \left(1 - \dfrac1e\right)$$
Hence, we get that
$$I = \int_3^{\infty} \dfrac{dx}{\ln^2(x)(x-\ln(x))} < \left(\dfrac{e}{e-1} \right) \times \int_3^{\infty} \dfrac{dx}{x \ln^2(x)} = \dfrac{e}{(e-1)\ln(3)}$$
EDIT
A way to approximate the integral is as follows.
$$\dfrac1{x-\ln(x)} = \dfrac1x \sum_{k=0}^{\infty} \left(\dfrac{\ln(x)}{x} \right)^k$$
Hence,
$$I = \int_3^{\infty} \dfrac{dx}{\ln^2(x)(x-\ln(x))} = \sum_{k=0}^{\infty} \int_3^{\infty} \dfrac{\ln^{k-2}(x)}{x^{k+1}} dx$$
Let us evaluate each term now.
\begin{align}
f_k & = \int_3^{\infty} \dfrac{\ln^{k-2}(x)}{x^{k+1}} dx\\
& = \int_{\ln(3)}^{\infty} \dfrac{t^{k-2}}{e^{kt}} dt\\
& = \int_{\ln(3)/k}^{\infty} \dfrac{y^{k-2}}{k^{k-2}e^y} \dfrac{dy}k\\
& = \dfrac1{k^{k-1}} \int_{\ln(3)/k}^{\infty} y^{k-2} e^{-y} dy\\
& = \dfrac{\Gamma(k-1,\ln(3)/k)}{k^{k-1}}
\end{align}
where $\Gamma(m,z)$ is the incomplete $\Gamma$ function and there are many ways to compute incomplete $\Gamma$ function to arbitrary accuracy. For example, as shown here. Hence, we get that
$$I = \sum_{k=0}^{\infty} \dfrac{\Gamma(k-1,\ln(3)/k)}{k^{k-1}}$$which is an exponentially converging series and truncating this will provide us arbitrarily accurate answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Can three distinct points in the plane always be separated into bounded regions by four lines? How can I show that for any three points in the plane, four lines can be drawn that separate the three points into distinct enclosed regions?
Can any six points be enclosed in distinct regions formed by five lines?
Clarifications:
Points are distinct, enclosed regions mean bounded regions.
Thank you.
| Okay, I think this works. By scaling and rotation, we can assume that two of the points are $(0,0)$ and $(0,1)$. Then the other point is $(x,y)$. Now the problem can be solved if the third point is $(1,0)$, with something like
Now if $x\ne 0$, the linear transformation $A=\pmatrix{x&0\\y&1}$ maps the point $(0,1)$ to $(x,y)$ and fixes the other two points, and also maps each green line to some new line, so $A$ applied to each line gives you four lines which enclose the points $(0,1), (0,0)$ and $(x,y)$.
If the third point is collinear with the other two points then it is easy to come up with the four lines that work.
Just make a cone that contains the two top points and another which contains the two bottom points. Then only the middle point will be in the intersection of the cones.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/320980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Give the combinatorial proof of the identity $\sum\limits_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$ Given the identity
$$\sum_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$$
Need to give a combinatorial proof
a) in terms of subsets
b) by interpreting the parts in terms of compositions of integers
I should not use induction or any other ways...
Please help.
| $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Leftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{\sum_{j = 0}^{n}{k - 1 + j \choose k - 1}} & =
\sum_{j = 0}^{n}{k - 1 + j \choose j} =
\sum_{j = 0}^{n}{-k + 1 - j + j - 1 \choose j}\pars{-1}^{j} =
\sum_{j = 0}^{n}{-k \choose j}\pars{-1}^{j}
\\[3mm] & =
\sum_{j = -\infty}^{n}{-k \choose j}\pars{-1}^{j} =
\sum_{j = -n}^{\infty}{-k \choose -j}\pars{-1}^{-j} =
\sum_{j = 0}^{\infty}{-k \choose n - j}\pars{-1}^{j + n} =
\\[3mm] & =
\pars{-1}^{n}\sum_{j = 0}^{\infty}\pars{-1}^{j}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{-k} \over z^{n - j + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm] & =
\pars{-1}^{n}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{-k} \over z^{n + 1}}\sum_{j = 0}^{\infty}\pars{-z}^{j}
\,{\dd z \over 2\pi\ic} =
\pars{-1}^{n}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{-k - 1} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm] & =
\pars{-1}^{n}{-k - 1 \choose n} =
\pars{-1}^{n}{k + 1 + n - 1 \choose n}\pars{-1}^{n} =
{n + k \choose n} = \color{#f00}{n + k \choose k}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Expected value uniform decreasing function We are given a function $f(n,k)$ as
for(i=0;i < k;i++)
n = rand(n);
return n;
rand is defined as a random number generator that uniformly generates values in the range $[0,n)$. It returns a value strictly less than $n$; also $\operatorname{rand}(0)=0$.
What is the expected value of our function $f(n,k)$ given $n$ and $k$?
| If the random numbers are not restricted to integers but are uniformly generated on the entire interval $[0,n)$, the expected value is halved in each iteration, and thus by linearity of expectation the expected value of $f(n,k)$ is $2^{-k}n$.
If $n$ and the random numbers are restricted to integers, the expected value decreases from $a_i$ to $(a_i-1)/2$ in each iteration, so we need to shift by $1$ to obtain the recurrence $a_{i+1}+1=(a_i+1)/2$, so in this case the expected value of $f(n,k)$ is $2^{-k}(n+1)-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
What are some open research problems in Stochastic Processes? I was wondering, what are some of the open problems in the domain of Stochastic Processes. By Stochastic Processes.
Any examples or recent papers or similar would be appreciated.
The motivation for this question is that I was studying stochastics from a higher level (i mean, brownian motion and martingales and stuff; beyond the undergrad markov chains and memoryless properties) and was wondering what are the questions that still lie unanswered in this field?
| Various academics have lists on their website
*
*Richard Weber, University of Cambridge (operations research)
*David Aldous, University of California, Berkeley with updates from Thomas M. Liggett
*Krzysztof Burdzy, University of Washington
*Hermann Thorisson, University of Iceland
Other resources that might be of interest
*
*The journal Queueing Systems published a Special Issue on Open Problems in 2011. James Cruise maintains an ungated copy of the papers along with research progress on the highlighted problems.
*John Kingman published a 2009 paper titled The first Erlang century—and the next where open problems and future reserch were discussed
*Jewgeni H. Dshalalow edited a book in 1985 titled Advances in queueing: theory, methods, and open problems
*Lyons, Russell, Robin Pemantle, and Yuval Peres. Unsolved problems concerning random walks on trees Classical and modern branching processes. Springer New York, 1997. 223-237.
*Questions tagged open problem and probability on mathoverflow
*Chapter 23 Open Problems from Levin, Peres, and Wilmer's book Markov Chains and Mixing Times
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 1,
"answer_id": 0
} |
Solving equation $\log_y(\log_y(x))= \log_n(x)$ for $n$ I'm just wondering, if I log a constant twice with the same base $y$,
$$\log_y(\log_y(x))= \log_n(x)$$
Can it be equivalent to logging the same constant with base $n$? If yes, what is variable $n$ equivalent to?
| Yes you can do that with the initial conditions for a logarithm satisfying.
The conditions for log(x) to the base n are: x > 0, n > 0 and n != 1.
so you should be careful with the domain that you are choosing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Universal algorithm to estimate probability of drawing certain combination of coloured balls I am writing AI for a board game and would be happy for some guidance in creating the function to calculate probabilities.
So, there's a pool of N coloured balls. Up to 7 colours, the quantity of balls in each colour is given (zero or more).
We are about to draw M balls from there. Question is, what is the probability of drawing certain combination of colours (ie, 3 red, 2 blue and 4 white).
All the quantities in pool and drawn are arbitrary, passed to a function as array.
If that would make the algorithm simpler, it is possible use recursion, ie
P = something * P([rest of pool], [rest of drawn balls])
I got as far as this:
N - total number of balls in the pool, quantity of each color is n1, n2, n3, etc.
M - total balls drawn. Quantity of each color is m1, m2, m3, etc.
The probability, of course, is a fraction, where the denominator is all the possible combinations of M bals drawn, and numerator is the number of valid combinations we are interested in.
The total number of combinations how M unique balls can be drawn from the pool of N unique balls, can be calculated like this:
$$\begin{equation}C = \frac{N!}{(N - M)!}\end{equation}$$
At the numerator, I have to calculate the product of all $$\frac{n_i!}{(n_i - m_i)!}$$, as this reflects in how many ways each colour can be picked mi times, from the pool where there are ni balls in that colour.
So far, I have this:
$$\frac{\frac{n_1!}{(n_1 - m_1)!}\cdot\frac{n_2!}{(n_2 - m_2)!}\cdot\frac{n_3!}{(n_3 - m_3)!}\cdot\ldots}{\frac{N!}{(N - M)!}}$$
Now, this number has to be multiplied by something, because the balls can come at any order of colours they want. And I am not sure how to solve this.
Next, I will have to calculate the probability of drawing certain combination of M balls, when total Q balls are drawn, but that is even more complex, so I have to solve the first step first.
| We use your notation. Imagine that we have put identity numbers on all the balls.
There are $\binom{N}{M}$ ways to choose $M$ balls from $N$. Note that the binomial coefficient counts the number of possible "hands" of $M$ balls. Order is irrelevant.
In case you are unfamiliar with binomial coefficients, $\binom{n}{k}$, pronounced, in English, "$n$ choose $k$," is the number of ways to choose $k$ objects from $n$ objects. By definition, $\binom{n}{0}=1$ for all $n$, and $\binom{n}{k}=0$ if $k\gt n$.
It turns out that if $0\le k\le n$, then $\binom{n}{k}=\frac{n!}{k!(n-k)!}$.
Now we turn to our problem. We want to find the number of ways to choose $m_1$ objects of colour $1$, $m_2$ objects of colour $2$, and so on, where $m_1+m_2+\cdots+m_k=M$.
There are $\binom{n_1}{m_1}$ ways to choose $m_1$ objects of colour $1$. For each of these ways, there are $\binom{n_2}{m_2}$ ways to choose $m_2$ objects of colour $2$, and so on. So the total number of ways to choose $m_1$ of colour $1$, $m_2$ of colour $2$, and so on is $\binom{n_1}{m_1}\binom{n_2}{m_2}\cdots \binom{n_k}{m_k}$. Thus the required probability is
$$\frac{\binom{n_1}{m_1}\binom{n_2}{m_2}\cdots \binom{n_k}{m_k}}{\binom{N}{M}}.\tag{$1$}$$
As to efficient ways of computing this, the subject has been studied a fair bit, particularly for the case $k=2$. One minor suggestion is to use the following recurrence:
$$\binom{b}{a}=\frac{b}{a}\binom{b-1}{a-1}.$$
Remark: The above analysis is close to the one you produced. The difference is that in Formula $(1)$, your denominator gets divided by $M!$, and your numerator gets divided by $m_1!m_2!\cdots m_k!$, precisely to deal with the order issues that you identified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Expectation of a stochastic exponential In class a while ago we used the following simplification:
$$ \mathbb E \left[ \exp\left(\langle \boldsymbol a,\mathbf W_t\rangle \right) \right] \quad =\quad \exp\left(\frac12 |\boldsymbol a|^2 t\right) $$
with $\boldsymbol a$ a constant $n$-dim vector, $\mathbf W_t$ an $n$-dim Brownian Motion.
I can recognize the quadratic variation on the right hand side and came up with the following (a bit hand wavy):
$$ \exp\left(\langle \boldsymbol a,\mathbf W_t\rangle \right) = \exp\left(\int_0^t d(\langle \boldsymbol a,\mathbf W_s\rangle) \right) \stackrel{\text{ito}}{=} \exp\left(\int_0^t \langle \boldsymbol a,d\mathbf W_s\rangle \right)\exp\left(\frac12 |\boldsymbol a|^2 t\right) $$
and then we'd need the expected value of the first term to be equal to $1$. But I don't see why this holds. It also looks fairly similar to the mgf of a normal distribution but again I don't see the connection clearly.
| Let $W_t = (W_t^1,\ldots,W_t^n)$ a $n$-dimensional Brownian motion. Then the processes $(W_t^j)_{t \geq 0}$ are independent 1-dimensional Brownian motions ($j=1,\ldots,n$). Thus
$$\mathbb{E}\big(e^{\langle a,W_t \rangle} \big) = \prod_{j=1}^n \mathbb{E}\big(e^{a_j \cdot W_t^j} \big) \stackrel{W_t^j \sim N(0,t)}{=} \prod_{j=1}^n e^{\frac{1}{2} a_j^2 \cdot t} = e^{\frac{1}{2} |a|^2 \cdot t}$$
where we used $\mathbb{E}(e^{a \cdot X}) = e^{\frac{1}{2}a^2}$ for $X \sim N(0,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can a continuous function map closed sets to open sets (and vice versa)? Definition of continuity: A function $f: X \to Y$ (where $X$ and $Y$ are topological spaces) is continuous if and only if for any open subset $V$ of $Y$, the preimage $f^{-1}(V)$ is open in $X$.
Now, if $U$ is a closed subset of $X$ (meaning that the complement of $U$, $U^c$ is open and it contains all of its cluster points) and $f(U)$ (the image of $U$ under $f$) $= V$ is open in $Y$, then if $f$ is continuous, $f^{-1}(V) = f^{-1}(f(U)) = U$ is open. So if $U$ is closed then this leads to a contradiction. Conversely, if $U$ is open in $X$ and $f(U)=V$ is closed in $Y$, then $V^c$ is open. However, the complement of the preimage $f^{-1}(V)$ is closed since $(f^{-1}(V))^c = U$ which is open; which again leads to a contradiction. If there is anyone who has some valid counterexamples I'd be eager to see them.
| Let $X$ be your favorite topological space, perhaps $\mathbb{R}$, and $Y$ be the space with just a single point.Then the only function $f:X\to Y$ is continuous, and, just like with Brian M. Scott's example (indeed, $Y$ has the discrete topology here), every set maps to a set which is both open and closed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Why is the quotient map $SL_n(\mathbb{Z})$ to $SL_n(\mathbb{Z}/p\mathbb Z)$ is surjective? Recall that $SL_n(\mathbb{Z})$ is the special linear group, $n\geq 2$, and let $q\geq 2$ be any integer. We have a natural quotient map $$\pi: SL_n(\mathbb{Z})\to SL_n(\mathbb{Z}/q).$$ I remember that this map is surjective (is it correct?). It seems the Chinese Remainder Theorem might be helpful, but I forgot how to prove it.
Can anyone give some tips?
| The result is true for $n\geq 1$ and any integer $q\geq 1$.
The group $SL_n(\mathbb{Z}/q\mathbb{Z})$ is generated by the elementary (transvection) matrices.
It is easily seen that every elementary matrix is in the image of $\pi$, as the image of an elementary matrix in $SL_n(\mathbb{Z})$.
So $\pi$ is surjective.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Proving a set is a subset of a union with itself and another set Let A and B be sets. Show that
$$\space A \subseteq (A \cup B)$$
My proof is as follows, but I don't feel confident that what I've done is correct. Any input is much appreciated.
$$x \in A \cup B \equiv x \in A \wedge x \in B \equiv x \in A$$
Therefore, A is a subset of the union of A and B.
| You're going in the wrong direction. Don't assume $x\in A\cup B$, assume $x\in A$: you want to show that each element in $A$ is also an element of $A\cup B$ (not vice versa). Also, $x\in A\cup B\iff x\in A\textrm{ OR (not and) }x\in B$.
Let $x\in A$, and note that $x\in A\cup B\implies x\in A$ or $x\in B$. But, as $x\in A$, $x\in A\cup B$. Therefore, each element of $A$ is an element of $A\cup B$, so that $A\subseteq A\cup B$.
However, you should note that your proof is the correct proof of $A\cap B\subseteq A$ if you replace your $A\cup B$ with $A\cap B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Existence of an improper integral without the existence of a limit Can a continuous non-decreasing function exist for all $x \in [0, \infty)$ with $\int_0^\infty f \, dx$ existing, but the $\lim_{x \to \infty} f$ not existing? If it does, what does it look like? I feel that if the limit does not exist, how can the improper integral exist?
| The limit must exist, and it must be $0$.
If $f$ is not bounded above, the integral clearly doesn't exist. If $f(x)$ is bounded above, then since $f$ is non-decreasing, $\lim_{x\to\infty} f(x)$ exists. One can show that if that limit is non-zero, then the integral doesn't converge.
The assumption of continuity is not necessary. But we did use strongly the hypothesis that the function is non-decreasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Alternating Series Using Other Roots of Unity $\sum (-1)^n b_n$ is representative of an alternating series.
We look at whether $\sum b_n$ converges and if $b_{n+1}<b_n$ $\forall n\in \mathbb{Z}$.
What if our alternating series is of the form $\sum z^n b_n$ where $z$ is any primitive root of unity. Do the same tests still apply?
Another question: $\sum 1/n$ diverges but $\sum (-1)^n 1/n$ converges.
Is there a $p \in \mathbb{Z}$ where $\sum z^n 1/n$ converges where $z$ is a primitive $p^{th}$ root of unity,
but diverges when $z$ is a primitive $(p+1)^{th}$ root of unity
| In Dirichlet's test you can replace $(-1)^n$ in the alternating series test by any sequence with bounded partial sums, and thus as a special case also by $z^n$ where $z$ is a root of unity.
Thus the answers to your questions are "yes" and "no", respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/321937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is there an exponential bound for $(1+p)^{-n}$ when $p$ is small? Is there an exponential upper bound for $(1+p)^{-n}$ when $0<p<1$? Similarly a exponential lower bound for $(1+p)^n$ will also be good. Do you know of any resources where one can pick up bounds as such quickly? Thanks.
| When $np \ll 1, (1\pm p)^n \approx 1 \pm np$ with the accuracy improving as $np$ gets smaller. The error is of order $\frac {(np)^2}2$. When you say exponential bound, do you want less than exponential growth? It certainly doesn't fall exponentially
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Adjacency graph of cutting plane is a bipartite graph Each time you draw a line on a plane, you are cutting it in half. Suppose you keep doing this without drawing a line parallel to a previous one. An adjacency graph can be constructed to represent this where each node represents an undivided portion of the plane, and edges exist between portions that share a boundary edge (as created by the lines).
I want to prove that the adjacency graph is bipartite regardless of what lines are drawn.
I know that a graph is bipartite if it is 2-colourable and this could be used to prove that the adjacency graph is 2-colourable but I'm stuck with proving how a graph is 2-colourable by induction.
Any help is appreciated!!
Thanks.
EDIT
Here's an example
| Theorem: A graph with atleast one edge is 2-chromatic if and only if it has no circuit of odd length.
Reference: Page 168 Theorem 8-2 in Graph Theory by narsingn Deo.
Your adjacency graph doesnt have any circuit with odd degree.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If every subsequence is convergent, prove that the sequence is convergent
If every subsequence of a given sequence of real numbers is convergent, prove that the sequence is convergent.
Help me please. I could not understand how to solve this question.
| Your question is trivial unless you change the "If" to "If and only if". I'll prove the "if and any if" version.
(Trivial direction) Any sequence is a subsequence of itself, so if all subsequences of a given sequence converge, so does the original sequence.
(Nontrivial direction) Suppose $(x_n)$ converges to $L$. Let $(x_{n_i})_{i \geq 1}$ be a subsequence of $(x_n)$. Let $\epsilon > 0$. Since $(x_n)$ converges to $L$, there exists $N \equiv N(\epsilon) \in \mathbb{N}^+$ with the property that if $i \in \mathbb{N}^+$ with $i \geq N$, then $|x_i-L|<\epsilon$. This is from the definition of convergence of a sequence. Now $n_i \geq i$, so if $j\geq i$, then $n_j \geq n_i \geq i$, and $|x_{n_j}-L| < \epsilon$. So the subsequence $(x_{n_i})_{i \geq 1}$ converges to $L$.
There is nothing special about $\mathbb{R}$ in the proof. It would work the same in any metric space (I'm rusty when it comes to non-metric spaces).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate $\int_0^1 \frac{\ln(x+1)}{x^2+1} dx$ This problem appears at the end of Trig substitution section of Calculus by Larson. I tried using trig substitution but it was a bootless attempt
$$\int_0^1 \frac{\ln(x+1)}{x^2+1} dx$$
| First make the substitution $x=\tan t$ to find
$$I=\int_0^1 dx\,\frac{\ln(x+1)}{x^2+1}=\int_0^{\pi/4} dt\,\ln(1+\tan t).$$
Now a substitution $u=\frac{\pi}{4}-t$ gives that
$$I=\int_0^{\pi/4} du\,\ln\left(\frac{2\cos u}{\cos u+\sin u}\right).$$
If you add these, you get
$$2I=\int_0^{\pi/4} dt\,\ln\left(\frac{\sin t+\cos t}{\cos t}\cdot\frac{2\cos t}{\cos t+\sin t}\right)=\frac{\pi}{4}\ln 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
If we have $f$ is one-to-one, why can we conclude that $n\mathbb{Z}_m=\mathbb{Z}_m? $ Suppose $m,n \in \mathbb{Z},m,n\geq1.$ Define a map $$f:\mathbb{Z}_m \rightarrow \mathbb{Z}_m$$ where $[x] \rightarrow [nx]$ If we have $f$ is one-to-one, why can we conclude that $n\mathbb{Z}_m=\mathbb{Z}_m? $
| Any one to one function $f$ from a set $A$ with $m$ elements to a set $B$ with $m$ elements is onto. And any onto function is one to one.
This does not hold for infinite sets of the same cardinality.
The proof for finite sets is a matter of counting. Since $f$ is one to one, the values of $f$ at the $m$ elements of $A$ are all different. So $f$ takes on $m$ distinct values. Since $B$ only has $m$ elements, the values of $f$ must include all the elements of $B$.
Think of it this way. Let $A$ be a set of $m$ women, and let $B$ be a set of $m$ men. Each woman $a$ chooses a man $f(a)$, with the rule that no two women can choose the same man. (That says $f$ is one to one.) Will there be a man who remains unchosen? Certainly not. Thus the function $f$ is onto.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $G$ with normal $H_1,\ldots,H_n$ such that $G=H_1\cdots H_n$ and $H_i\cap H_j=\{e\}$ for $i\neq j$, but $G\not\cong H_1\times\cdots\times H_n$ How can I find a group $G$ with normal subgroups $H_1,\ldots,H_n$ such that $G=H_1H_2\cdots H_n$ and $H_i\cap H_j=\{e\}$ for all $i\neq j$, but $G\not\cong H_1\times H_2\times\cdots\times H_n$.
I'm thinking about $Q_8$. Is that right? Thank you very much!
| $Q_{8}$ is not ok, as in it two non-trivial normal subgroups intersect nontrivially.
Take instead the Klein $4$-group $V = \{e, a_1, a_2, a_3\}$, and $H_i = \langle a_i \rangle = \{ 1, a_i \}$ for $i = 1, 2, 3$.
You have indeed $G = H_1 H_2 H_3$ (actually, two factors already suffice) and $H_i \cap H_j = \{ e \}$ for $i \ne j$, but $H_1 \times H_2 \times H_3$ has order $8$, not $4$ like $V$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
For which $\alpha$ does the series $\sum_{n=1}^\infty n^\alpha x^{2n} (1-x)^2$ converges uniformly at $[0,1]$. $$\sum_{n=1}^\infty n^\alpha x^{2n} (1-x)^2$$
Find for which $\alpha$ this series converges uniformly at $[0,1]$.
As $(1-x)^2$ is not dependent of $n$, I thought about rewriting it as:
$$(1-x)^2 \sum_{n=1}^\infty n^\alpha x^{2n} $$
And then you can use the M-test, with $n^\alpha$. Which would make me conclude that it converges if $\alpha < -1$.
I was thinking that it may also converge at $\alpha \in [-1,1)$. If I take calculate the supremum of $f_n(x)$ I find that this is at $\frac{n}{n+1}$, which gives: $$f_n(\frac{n}{n+1})=n^\alpha (\frac{n}{n+1})^{2n}(1-\frac{n}{n+1})^2= n^\alpha ((1+ \frac{1}{n})^{n})^{-2}(\frac{1}{n+1})^2\leq n^\alpha \frac{1}{(n+1)^2}\leq n^{(\alpha-2)}$$.
This implies it also converges at $\alpha\in[−1,1)$. The thing I don't understand is that this would mean that the factor $(1-x)^2$ makes it converge at $\alpha\in[-1,1)$. But intuitively I would say that $\sum_{n=1}^\infty n^\alpha x^{2n}$ and $\sum_{n=1}^\infty n^\alpha x^{2n} (1-x)^2$ should converge for the same $\alpha$.
And how could I prove that this series diverges for $\alpha \geq 1$ ?
Edit: If set $\alpha=1$, then I get $$\sum_{n=1}^\infty n x^{2n} (1-x)^2$$
I know the supremum is at $\frac{n}{n+1}$, so this gives $f_n(\frac{n}{n+1})=((1+ \frac{1}{n})^{n})^{-2}\frac{n}{(n+1)^2}$. So if $n\to\infty$, then this goes also to zero. Therefore I would conclude that it converges $\alpha\leq 1$. Is this correct ?
| Let's denote by
$$f_n(x)=n^\alpha x^{2n} (1-x)^2.$$
First, we search for which values of $\alpha$ we have the normal convergence of the series that implies the uniform convergence. The supremum of $f_n$ is attained at $x_n=\frac{n}{n+1}$ so
$$||f_n||_{\infty}=f_n(x_n)=n^\alpha (1+\frac{1}{n})^{-2n}(\frac{1}{n+1})^2\sim e^{-2}n^{\alpha-2},$$
hence the series $\sum_n ||f_n||_{\infty}$ converges for $\alpha<1$. Thus we have the uniform convergence for $\alpha<1.$
Now, what about $\alpha\geq 1$? We have uniform convergence if
$$\lim_n\sup_{x\in[0,1]}|\sum_{k=n+1}f_k(x)|=0.$$
I'll explain that if the series is of positive terms, then the uniform and normal convergence are same. Indeed, in this case we have
$$\lim_n\sup_{x\in[0,1]}|\sum_{k=n+1}f_k(x)|\geq\lim_n\sum_{k=n+1}f_k(x_k)=\lim_n\sum_{k=n+1}e^{-2}k^{\alpha-2}=+\infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Principal $n$th root of a complex number This is really two questions.
*
*Is there a definition of the principal $n$th root of a complex number? I can't find it anywhere.
*Presumably, the usual definition is $[r\exp(i\theta)]^{1/n} = r^{1/n}\exp(i\theta/n)$ for $\theta \in [0,2\pi)$, but I have yet to see this anywhere. Is this because it has bad properties? For instance, according to this definition is it true that for all complex $z$ it holds that $(z^{1/a})^b = (z^b)^{1/a}$?
| There really is not a coherent notion of "principal" nth root of a complex number, because of the inherent and inescapable ambiguities.
For example, we could declare that the principal nth root of a positive real is the positive real root (this part is fine), but then the hitch comes in extending this definition to include all or nearly all complex numbers. For example, we could try to require continuity, but if we go around 0 clockwise, versus counter-clockwise, we'd obtain two different nth roots for number we've "analytically continued" to. A/the traditional "solution" (which is not a real solution) is to "slit" the complex plane along the negative real axis to "prevent" such comparisons. And some random choice about whether the negative real axis is lumped together with one side or the other.
But even avoiding that ambiguity leaves us with a root-taking function that substantially fails to be a group homomorphism, that is, fails to have the property that the nth root of a product is the product of the nth roots.
The expression in terms of radius and argument "solves" the problem by not really giving a well-defined function on complex numbers, but only well-defined on an infinite-fold (ramified) covering of the complex plane... basically giving oneself a logarithm from which to make nth roots. But logarithms cannot be defined as single-valued functions on the complex plane, either, for similar reasons. Partially defined in artificial ways, yes, but then losing the fundamental property that log of a product is sum of the logs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Sequences and Languages Let $U$ be the following language. A string $s$ is in $U$ if it can be written as:
$s = 1^{a_1}01^{a_2}0 ... 1^{a_n}01^b$,
where $a_1,..., a_n$ are positive integers such that there is a 0-1 sequence $x_1, ..., x_n$ with $x_1a_1 + ... + x_na_n = b$. Show that $U \in P$.
Not sure how to even approach the problem. Any help would be appreciated. I dont want the answer specifically, but any hints or help would be great.
Thanks
| Hint for constructing a pushdown automaton recognising your language $U$ by empty stack:
[I didn't re-read the question before writing this, so I forgot the $a_i$'s had to be positive. You have to make a slight modification, probably adding two extra states, because of that, but the basic idea is still the same.]
Use three states apart from the initial state: two ($p_0$ and $p_1$) to indicate whether the current $x_i$ is $0$ or $1$, and one ($b$) to indicate that we are guessing we're up to the $1^b$ part. We only need one stack symbol, which we use to count the $1$s we read in $p_1$ and then check them off against the $1$s we read in $b$.
The fact that context-free languages are in $P$ is well-known, but I suppose if you haven't covered it in your course and this is homework, I guess you will need to convert either the pushdown automaton or the grammar into a polynomial-time Turing machine (which is easy to do).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Totient function and Euler's Theorem Given $\big(m, n\big) = 1$, Prove that
$$m^{\varphi(n)} + n^{\varphi(m)} \equiv 1 \pmod{mn}$$
I have tried saying
$$\text{let }(a, mn) = 1$$
$$a^{\varphi(mn)} \equiv 1 \pmod{mn}$$
$$a^{\varphi(m)\varphi(n)} \equiv 1 \pmod{mn}$$
$$(a^{\varphi(m)})^{\varphi(n)} \equiv 1 \pmod{mn}$$
but I can't see where to go from here. I'm trying to somehow split the $a^{\varphi(mn)}$ into an addition so I can turn it into $m^{\varphi(n)} + n^{\varphi(m)} \equiv 1 \pmod{mn}$.
| Hint What is $m^{\varphi(n)} + n^{\varphi(m)} \pmod{m}$? What about $\pmod n$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If a group $G$ has only finitely many subgroups, then show that $G$ is a finite group. If a group $G$ has only finitely many subgroups, then show that $G$ is a finite group. I have no idea on how to start this question. Can anyone guide me?
| Suppose $G$ were infinite. If $G$ contains an element of infinite order, then _.
Otherwise, every element of $G$ has finite order, and if $x_1, x_2, \ldots, x_n$ are any finite set of elements of $G$, then the subgroups $\langle x_i \rangle$ cover only finitely many elements of $G$. Therefore __.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Injectivity is a local property Let $R$ be a commutative noetherian ring, and let $M$ be an $R$-module. How can I show that if any localization $M_p$ at a prime ideal $p$ of the ring $R$ is injective over $R_p$, then $M$ is injective?
| Baer's criterion shows that it suffices to show that $\hom(B,M) \to \hom(A,M)$ is surjective for $B=R$ and $A=$ an ideal, in particular both are finitely presented. But then $\hom$ commutes with localization and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How do I show the equivalence of the two forms of the Anderson-Darling test statistic? It's stated in many places regarding the Anderson-Darling test statistic, which is defined as
$$n\int_{-\infty}^\infty \frac{(F_n(x) - F(x))^2}{F(x)(1 - F(x))}dF(x)$$
that this is functionally equivalent to the statistic
$$A^2 = -n - S$$
where
$$S = \sum_{k=1}^n\frac{2k-1}{n}\left(\ln F(Y_k) + \ln(1 - F(Y_{n+1-k}))\right)$$
Note that $F_n(x)$ is the empirical distribution function and $F(x)$ is the distribution to which we are comparing the sample. $Y_k$ is the $k^{th}$ ranked element in the sample.
I even went so far as to read the original 1954 paper by Anderson and Darling and I have yet to discover how this equivalence was computed - these authors merely stated the equivalence too. I've tried writing out the numerator inside the integral and splitting into 3 integrals - I was only able to simplify one of them.
I have an inkling that maybe the Probability Integral Transformation should be applied, but I'm not really sure how. I'd really appreciate if anyone could give any pointers.
| I think you can prove it by dividing the integral into (n+1) integrals on the intervals $[Y_k; Y_{k+1})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Choose 38 different natural numbers less than 1000, Prove among these there exists at least two whose difference is at most 26. Choose any 38 different natural numbers less than 1000.
Prove that among the selected numbers there exists at least two whose difference is at most 26.
I think I need to use pigeon hole principle, not sure where to even begin.
| Pigeonhole-principle is a good idea.
Hint: Think about partitioning $\{1,2,\ldots,999\}$ into subsets $\{1,2,\ldots,27\}$, $\{28,29,\ldots, 54\}$, $\{55,56,\ldots,81\}$, ..., $\{\ldots,998,999\}$ of size $27$ each.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Counting problem - How many times an inequality holds? Let $k>2$ be a natural number and let $b$ be a non-negative real number.
Now, for each $n$, $m \in \{ 1, 2, ... k \}$, consider the following inequalities:
$$ mb < k - n $$
We have $k^2$ inequalities. How can I count the couples of $n$ and $m$ such that the inequality (for $n$ and $m$ fixed) holds?
Clearly, an algorithm would easily give an answer. But I was wondering if there is a "more-mathematical" solution.
| Presumably $b$ is given. We can note that the right side ranges from $0$ to $k-1$, so define $p=k-n \in [0,k-1]$ and ask about $mb \lt p$. For a given $p$, the number of allowable $m$ is $\lfloor \frac {p-1}b \rfloor$ So we are asking for $\sum_{p=1}^{k-1}\lfloor \frac {p-1}b \rfloor$ Let $q=\lfloor \frac {k-2}b\rfloor$ Then $$\sum_{p=1}^{k-1}\lfloor \frac {p-1}b \rfloor=b\sum_{i=0}^{q-1}i+q(k-1-bq)=\frac{bq(q-1)}2+q(k-1-bq)$$ because the left sum starts with $b\ \ 0$'s, $b\ \ 1$'s, on to $b \ \ q-1$'s and finish with $q$'s.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/322975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
example of a set that is closed and bounded but not compact
Find an example of a subset $S$ of a metric space such that $S$ is closed and bounded but not compact.
One such example that comes from analysis is probably a closed and bounded set in $C[0,1]$. I attempt to construct my own example to see if it works.
Is $\{ \frac{1}{n} | n \in \mathbb{N} \}$ endowed with discrete topology a set that is closed and bounded but not compact? My guess is that it is indeed an example of closed and bounded does not imply compact.
Every element is less than or equal to $1$, and it is closed as a whole set. If we let $\mathcal{A}$ be a covering of the set that consists of singletons in $\{ \frac{1}{n} \}$ so that any finite subcover
$\{ \frac{1}{n_j} |j =1,...,k \quad \text{and} \quad n_j \in \mathbb{N} \}$ will not cover $\{\frac{1}{n}\}$, because if we take $n = \max \{{n_j}\}, \frac{1}{n+1}$ is not in the finite subcover.
Thanks in advance for pointing out any mistake.
| The "closed" ball $\lVert x \rVert \leq 1$ in any infinite dimensional Banach space is closed and bounded but not compact. It is closed because any point outside it is contained in a small open ball disjoint from the first one, by the triangle inequality. That is, if $\lVert y \rVert = 1 + 2 \delta,$ then the sets $\lVert x \rVert \leq 1$ and $\lVert x - y \rVert < \delta$ are disjoint. Hence the complement of the "closed" unit ball is open and the "closed" unit ball really is closed. But not compact if not in finite dimensions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 9,
"answer_id": 7
} |
Express $\sin 4\theta$ by formulae involving $\sin$ and $\cos$ and its powers. I have an assignment question that says "Express $\sin 4\theta$ by formulae involving $\sin$ and $\cos$ and its powers."
I'm told that $\sin 2\theta = 2 \sin\theta \cos\theta$ but I don't know how this was found.
I used Wolfram Alpha to get the answer but this is what I could get :
$$
4\cos^3\theta\sin\theta- 4\cos\theta \sin^3\theta
$$
How can I solve this problem?
| That's a trig identity.
So...
$\sin{4\theta} = 2\sin{2\theta}cos{2\theta}$
Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Compact injections and equivalent seminorms Let $V$ and $H$ be two Banach spaces with norm $\lVert \cdot \rVert$ and $\lvert \cdot \rvert$ respectively such that $V$ embeds compactly into $H$. Let $p$ be a seminorm on $V$ such that $p(u) + \lvert u \rvert$ is a norm on $V$ that is equivalent to $\lVert \cdot \rVert$. Set $N = \{u \in V: p(u) = 0\}$. Prove that there does not exist a sequence $(u_n)$ in $V$ satisfying
*
*$\operatorname{dist}(u_n, N) = 1$ for all $n$
*$p(u_n) \to 0$.
Ideas: I have no reason why I should expect such a result, so I can't motivate it. Anyway, I want to claim that $u_n$ approach a limit $u$. Then hopefully $p(u_n) \rightarrow p(u) = 0$ so $u \in N$, contradicting $1 = \operatorname{dist}(u_n,N) \rightarrow \operatorname{dist}(u,N) = 0$. It would help greatly if the $(u_n)$ were bounded, since then the compact injection means that $u_n$ approach a limit in $\bar{V} \subset H$. Then somehow argue that the limit is actually in $V$?
| We will first show that, given $x\in X$, there exists $z\in N$ s.t. $\|x-z\|_X=d(x-z,N)$ and $p(x)=p(x-z)$. Let $\{z_n\}\subset N$ s.t. $\|x-z_n\|_X\to d(x,N)$. Then $\infty>\|x\|_X+\sup_{n}\|x-z_n\|_X>\sup_{m}\|z_m\|_X$. Since $N$ is finite dimensional, there exists a subsequence $z_{n_k}$ and $z\in N$ s.t. $z_{n_k}\to z$. Therefore, $\|x-z_{n_k}\|_X\to\|x-z\|_X=d(x,N)$. Moreover, $p(x-z)=p(x)$ because
\begin{align*}
p(x)&\leq p(x-z)+p(z)=p(x-z)\\
p(x-z)&\leq p(x)+p(-z)=p(x)
\end{align*}
and $d(x-z,N)=d(x,N)$ since $N$ is a subspace.
We now prove the main claim. AFSOC that for all $n\in\mathbb{N}$ there exists $x_n\in X$ s.t. $d(x_n,N)>np(x_n)\geq 0$. Normalizing $x_n$ by $d(x_n,N)$ we have that $1>np(x_n)$. For each $x_n$, let $z_n\in N$ be the $z$ found above. Then we have that $\|x_n-z_n\|=1$ for all $n\in\mathbb{N}$ so $\{x_n-z_n\}$ is a bounded sequence. By compactness of the embedding, $\{x_n-z_n\}$ has a subsequence, $\{x_{n_k}-z_{n_k}\}$ that converges to some $\hat{z}$. By continuity of $p$, we have that $p(x_{n_k}-z_{n_k})\to p(\hat{z})=0$. This implies that $\hat{z}\in N$. However, this implies that $d(x_{n_k}-z_{n_k},N)\leq\|(x_{n_k}-z_{n_k})-\hat{z}\|\to0$: a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $H \triangleleft G$ and $|G/H|=m$, show that $x^m \in H$ for $\forall x \in G$
If $H \triangleleft G$ and $|G/H|=m$, show that $x^m \in H$ for $\forall x \in G$.
My attempt is: since $|G/H|=\frac{|G|}{|H|}=m$, we have $x^{|G|}=x^{m|H|}=e$, then I stuck here. Can anyone guide me ?
| Since $(xH)^m=H$, it follows that $x^mH=H$, and thus that $x^m\in H$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What was the first bit of mathematics that made you realize that math is beautiful? (For children's book) I'm a children's book writer and illustrator, and I want to to create a book for young readers that exposes the beauty of mathematics. I recently read Paul Lockhart's essay "The Mathematician's Lament," and found that I, too, lament the uninspiring quality of my elementary math education.
I want to make a book that discredits the notion that math is merely a series of calculations, and inspires a sense of awe and genuine curiosity in young readers.
However, I myself am mathematically unsophisticated.
What was the first bit of mathematics that made you realize that math is beautiful?
For the purposes of this children's book, accessible answers would be appreciated.
| When I was a kid I realized that
$$0^2 + 1\ (\text{the first odd number}) = 1^2$$
$$1^2 + 3\ (\text{the second odd number}) = 2^2$$
$$2^2 + 5\ (\text{the third odd number}) = 3^2$$
and so on...
I checked it for A LOT of numbers :D
Years passed before someone taught me the basics of multiplication of polynomial and hence that $$(x + 1)^2 = x^2 + 2x + 1.$$
I know that this may sound stupid, but I was very young, and I had a great time filling pages with numbers to check my conjecture!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "662",
"answer_count": 162,
"answer_id": 4
} |
What was the first bit of mathematics that made you realize that math is beautiful? (For children's book) I'm a children's book writer and illustrator, and I want to to create a book for young readers that exposes the beauty of mathematics. I recently read Paul Lockhart's essay "The Mathematician's Lament," and found that I, too, lament the uninspiring quality of my elementary math education.
I want to make a book that discredits the notion that math is merely a series of calculations, and inspires a sense of awe and genuine curiosity in young readers.
However, I myself am mathematically unsophisticated.
What was the first bit of mathematics that made you realize that math is beautiful?
For the purposes of this children's book, accessible answers would be appreciated.
| If you've ever heard of $3,529,411,764,705,882$ being multiplied by $3/2$ to give $5,294,117,647,058,823$ (which is the same as the 3 being shifted to the back), you might consider including that in the book.
There are lots of other examples, like $285,714$ turning into $428,571$ (moving the 4 from back to front) when multiplied by $3/2$, or the front digit of $842,105,263,157,894,736$ moving to the back four times in a row when you divide it by $2$. (There's a leading zero in front of the last term, though.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "662",
"answer_count": 162,
"answer_id": 36
} |
What was the first bit of mathematics that made you realize that math is beautiful? (For children's book) I'm a children's book writer and illustrator, and I want to to create a book for young readers that exposes the beauty of mathematics. I recently read Paul Lockhart's essay "The Mathematician's Lament," and found that I, too, lament the uninspiring quality of my elementary math education.
I want to make a book that discredits the notion that math is merely a series of calculations, and inspires a sense of awe and genuine curiosity in young readers.
However, I myself am mathematically unsophisticated.
What was the first bit of mathematics that made you realize that math is beautiful?
For the purposes of this children's book, accessible answers would be appreciated.
| For me, it was the beauty of the number 1, how it can be multiplied with anything , and it won't change the number it is being multiplied with, also how it can be represented as any number divided by itself such as 4/4=1
I would also love to share this beautiful poem by Dave Feinberg that is titled "the square root of 3" and was also featured in a Harold and Kumar Movie, it renewed my love for math and is and always has been one of my favorite poems! :
I’m sure that I will always be
A lonely number like root three
The three is all that’s good and right,
Why must my three keep out of sight
Beneath the vicious square root sign,
I wish instead I were a nine
For nine could thwart this evil trick,
with just some quick arithmetic
I know I’ll never see the sun, as 1.7321
Such is my reality, a sad irrationality
When hark! What is this I see,
Another square root of a three
As quietly co-waltzing by,
Together now we multiply
To form a number we prefer,
Rejoicing as an integer
We break free from our mortal bonds
With the wave of magic wands
Our square root signs become unglued
Your love for me has been renewed
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "662",
"answer_count": 162,
"answer_id": 68
} |
What was the first bit of mathematics that made you realize that math is beautiful? (For children's book) I'm a children's book writer and illustrator, and I want to to create a book for young readers that exposes the beauty of mathematics. I recently read Paul Lockhart's essay "The Mathematician's Lament," and found that I, too, lament the uninspiring quality of my elementary math education.
I want to make a book that discredits the notion that math is merely a series of calculations, and inspires a sense of awe and genuine curiosity in young readers.
However, I myself am mathematically unsophisticated.
What was the first bit of mathematics that made you realize that math is beautiful?
For the purposes of this children's book, accessible answers would be appreciated.
| Maybe the fact that the homotopy category of a model category is equivalent to the full subcategory of fibrant-cofibrant objects with homotopy classes of morphisms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "662",
"answer_count": 162,
"answer_id": 100
} |
What was the first bit of mathematics that made you realize that math is beautiful? (For children's book) I'm a children's book writer and illustrator, and I want to to create a book for young readers that exposes the beauty of mathematics. I recently read Paul Lockhart's essay "The Mathematician's Lament," and found that I, too, lament the uninspiring quality of my elementary math education.
I want to make a book that discredits the notion that math is merely a series of calculations, and inspires a sense of awe and genuine curiosity in young readers.
However, I myself am mathematically unsophisticated.
What was the first bit of mathematics that made you realize that math is beautiful?
For the purposes of this children's book, accessible answers would be appreciated.
| I think one of my early favourite mathematical things was simply "proof by contradiction" -- any of them.
I think its appeal is that you nearly have proof by example, except that you're proving a negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "662",
"answer_count": 162,
"answer_id": 132
} |
How do we integrate, $\int \frac{1}{x+\frac{1}{x^2}}dx$? How do we integrate the following integral?
$$\int \frac{1}{x+\large\frac{1}{x^2}}\,dx\quad\text{where}\;\;x\ne-1$$
| The integral is equivalent to
$$\int dx \frac{x^2}{1+x^3} = \frac{1}{3} \int \frac{d(x^3)}{1+x^3} = \frac{1}{3} \log{(1+x^3)} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving a matrix is positive definite using Cholesky decomposition If you have a Hermitian matrix $C$ that you can rewrite using Cholesky decomposition, how can you use this to show that $C$ is also positive definite?
$C$ is positive definite if $x^\top C x > 0$ and $x$ is a vector.
| From Wikipedia:
If A can be written as LL* for some invertible L, lower triangular or
otherwise, then A is Hermitian and positive definite.
$A=LL^*\implies x^*Ax=(L^*x)^*(L^*x)\ge 0$
Since $L$ is invertible, $L^*x\ne 0$ unless $x=0$, so $x^*Ax>0\ \forall\ x\ne 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show $ \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{1}+a_{2}+\cdots+a_{n}}}=1 $ if $\displaystyle \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{n}}}=1 $ Let $\{a_{n}\}$ be a positive sequence with $\displaystyle \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{n}}}=1 $. How can we show that
$$ \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{1}+a_{2}+\cdots+a_{n}}}=1 $$
I am not sure the problem is true. If it is false, what is the counterexample?
| This is true.
1) Pick $\epsilon>0$. Then there exists $N$ such that
$$
\sqrt[n]{a_n}\leq 1+\epsilon\quad\Rightarrow\quad a_n\leq (1+\epsilon)^n\qquad\forall n\geq N.
$$
Now
$$
\sum_{k=1}^Ka_k =\sum_{k=1}^{N-1}a_k+\sum_{k=N}^Ka_k\leq C+\sum_{k=N}^K(1+\epsilon)^k=C+ (K-N+1)(1+\epsilon)^K\leq C+K(1+\epsilon)^K
$$
where $C=\sum_{k=1}^{N-1}a_k$ is fixed.
So
$$
\sqrt[K]{\sum_{k=1}^Ka_k }\leq (1+\epsilon)\left(C +\frac{K}{(1+\epsilon)^K}\right)^\frac{1}{K} \longrightarrow 1+\epsilon.
$$
This proves that
$$
\limsup \sqrt[K]{\sum_{k=1}^Ka_k }\leq 1+\epsilon\quad\forall\epsilon>0\quad\Rightarrow \quad\limsup \sqrt[K]{\sum_{k=1}^Ka_k }\leq 1.
$$
2) Take a subsequence such that
$$
\lim \sqrt[n_k]{a_{n_k}}=1.
$$
Pick $\epsilon>0$. There exists $K$ such that
$$
\sqrt[n_k]{a_{n_k}}\geq 1-\epsilon\quad\Rightarrow \quad a_{n_k}\geq (1-\epsilon)^{n_k}\quad\forall k\geq K.
$$
Now
$$
\sum_{n=1}^{n_k}a_n\geq a_{n_K}\geq (1-\epsilon)^{n_k}\quad\forall k\geq K.
$$
So
$$
\sqrt[n_K]{\sum_{n=1}^{n_k}a_n}\geq 1-\epsilon \quad\forall k\geq K.
$$
Hence
$$
\limsup \sqrt[N]{\sum_{n=1}^{N}a_n}\geq 1-\epsilon\quad\forall\epsilon>0\quad \limsup \sqrt[N]{\sum_{n=1}^{N}a_n}\geq 1.
$$
Both inequalities are now proven, so
$$
\limsup \sqrt[N]{\sum_{n=1}^{N}a_n}= 1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
The definition of continuous function in topology I am self-studying general topology, and I am curious about the definition of the continuous function. I know that the definition derives from calculus, but why do we define it like that?I mean what kind of property we want to preserve through continuous function?
| If you mean the definition that $f\colon X\to Y$ is continuous if $f^{-1}(U)$ is open for every open $U\subseteq Y$, then this is because this property is equivalent to continuity in metric spaces, but doesn't refer to the metric. This makes it a natural candidate for the general definition.
It is a little harder to motivate the definition in terms of the preservation of properties, because we have to do it backwards. We are saying that a function $f$ is continuous if $f^{-1}$ (thought of as a function from the set of subsets of $Y$ to the set of subsets of $X$) preserves the property of openness. We run into problems if we try to insist that $f(U)$ is open for every open $U\subseteq X$ (maps with this property are called open), because the image function from subsets of $X$ to subsets of $Y$, taking a subset $V$ to the subset $f(V)$, doesn't respect the operations of union and intersection in the same way that $f^{-1}$ does. For example, $f^{-1}(U\cap V)=f^{-1}(U)\cap f^{-1}(V)$, but it is not true in general that $f(U\cap V)=f(U)\cap f(V)$. As the definition of a topology involves certain intersections and unions preserving the property of openness, any "structure preserving map" must do so as well, which is why we consider $f^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
} |
How to prove a generalized Gauss sum formula I read the wikipedia article on quadratic Gauss sum. link
First let me write a definition of a generalized Gauss sum.
Let $G(a, c)= \sum_{n=0}^{c-1}\exp (\frac{an^2}{c})$, where $a$ and $c$ are relatively prime integers.
(Here is another question. Is the function $e(x)$ defined in the article equal to $\exp(2\pi i/x)$ or $\exp(\pi i/x)$?)
In the article, a formula is given according to values of $a$ and $c$.
For example, if $a$ is odd and $4|c$, then
$G(a, c)=(1+i)\epsilon_a^{-1} \sqrt{c} \big(\frac{c}{a}\big)$, where $\big(\frac{c}{a}\big)$ is the Jacobi symbol.
I would like to prove it but I don't know how to do it. Could you give me a guide?
| The quadratic Gauss sum is given by
\begin{eqnarray*}
G(s;k) = \sum_{x=0}^{k-1} e\left(\frac{sx^2}k\right),
\end{eqnarray*}
where $\displaystyle e(\alpha) = e^{2\pi \imath \alpha}$, $s$ is any integer coprime to $p$ and $k$ is a positive integer. The generalized Gauss sums is given by
\begin{eqnarray*}
G(a,b,c) = \sum_{x=0}^{|c|-1} e\left(\frac{ax^2+bx}c\right),
\end{eqnarray*}
where $ac \neq 0$ and $ac+b$ is even.
It is well known that
\begin{eqnarray*}
G(s;k) = \begin{cases}
\left(1+\imath^s\right)\left(\frac ks\right)\sqrt{k} &\mbox{ if } k \equiv 0 \mod 4\\
\left(\frac sk\right)\sqrt{k} &\mbox{ if } k \equiv 1 \mod 4\\
0 &\mbox{ if } k \equiv 2 \mod 4\\
\imath \left(\frac sk\right)\sqrt{k} &\mbox{ if } k \equiv 3 \mod 4
\end{cases}
\end{eqnarray*}
There are many proofs for the above formula: Gauss proved it using elementary methods, Dirichlet used a poisson summation formula, Cauchy used a transformation function for the classical theta function, etc... An elementary proof in the style of Gauss is available in the book Gauss and Jacobi Sums by Berndt, Evans and Williams and also the book Introduction to Number Theory by Nagell.
One method is to show, for $k$ odd, $|G(s;k)|^2 = k$, and then determining the sign of $G(s;k)$ will be the hard part. From here you can use reduction properties of the quadratic Gauss sum and the Chinese Remainder Theorem to prove the even cases.
There is no general formula for a generalized Gauss sum. You can find a reciprocity theorem for these sums in the book Gauss and Jacobi Sums as well, also in Introduction to Analytic Number Theory by Apostol.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry functions How do I verify the following identity:
$$\frac{1-(\sin x - \cos x)^2}{\sin x} = 2\cos x$$
I have done simpler problems but got stuck with this one.
Please help.
Tony
| $$\frac{1-(\sin x -\cos x)^2}{\sin x}=\frac{1-(\sin^2x+\cos^2x)+2\sin x\cos x}{\sin x}=\frac{2\cos x\sin x}{\sin x}=2\cos x$$
Here I used the identity: $\sin^2x+\cos^2x=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Intersection of a Collection of Sets I am trying to figure out how to write out the answer to this.
If I am given: $$A_i = \{i,i+1,i+2,...\}$$
And I am trying to find the intersection of a collection of those sets given by:
$$\bigcap_{i=1}^\infty A_i $$
First off, am I right in saying its the nothing since:
$$A_1 = \{ 1,2,3,...\}$$
$$A_2 = \{ 2,3,4,...\}$$
So, would the answer be the empty set?
| Yes, that's correct, the answer is the empty set. To explain this properly you want to take a number $n \in \mathbb N$ and explain why $n \notin \bigcap A_i$. If you do this with no conditions on $n$ you will have shown that the intersection doesn't contain any positive integers, hence it's empty.
To show $n \notin \bigcap A_i$ you must pick an $i$ such that $n \notin A_i$. I leave it to you to determine which $i$ to pick.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Discuss the eigenvalues and eigenvectors of $A=I+2vv^T$.. I need some help with this question: Let $v\in\mathbb R^n$. Discuss the eigenvalues and eigenvectors of $$A=I+2vv^T$$. Can anyone help me?
| The matrix $vv^T$ is real symmetric, so it's diagonalizable, then there's an invertible matrix $P$ and a diagonal matrix $D$ such that
$$vv^T=PDP^{-1}.$$
Since the rank of $vv^T$ is $0$ or $1$ then $D=diag(||v||^2,0,\ldots,0).$
Now, we have:
$$A=I+2vv^T=P(I+2D)P^{-1}$$
so $A$ is diagonalizable in the same basis of eigenvectors than $vv^T$ and has the eigenvalues $1+2||v||^2,1,\ldots,1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/323931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Complex representation of sinusoids question A sum of sinusoids defined as:
$$\tag1f(t) = \sum_{n=1}^{N}A\sin(2\pi tn) + B\cos(2\pi tn)$$ is said to be represented as:
$$\tag2f(t) = \sum_{n=-N}^{N}C\cdot e^{i2\pi tn}$$
which is derived from Euler's identity $e^{ix} = \cos(x) + i\sin(x)$ from which follows that:
$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$ and $$\sin(x)=\frac{e^{ix} - e^{-ix}}{2i}$$
therefore:
$$ \cos(x) + \sin(x) = \frac{e^{ix} + e^{-ix}}{2} + \frac{e^{ix} - e^{-ix}}{2i} $$
The thing is, I cannot see how this turns into $(2)$? The last statement, no matter how I try to manipulate it, I cant get it to the form $$\cos(x) + \sin(x) = Ce^{ix}$$ so that the summation works and it makes no sense for it to be so, since Euler's equation is using $i\sin(x)$ - clearly a different thing...
Further, the statement :
$$ |\cos(x) + i\sin(x)| = 1$$ is also confusing since ${\cos(x)^2 + i\sin(x)}^2$ is not the same thing as ${\cos(x)^2 + \sin(x)}^2$ and it appears people just "drop" the $i$, which is terribly confusing.
Could someone please explain what am I not understanding? Thanks a bunch!
| Using the identities:
$$\begin{aligned}
f(t)
&= \sum_{n=1}^N A \cos(2 \pi t n) + B \sin(2 \pi t n) \\
&= \sum_{n=1}^N \left[A\left(\frac{e^{i 2 \pi t n} + e^{-i 2 \pi t n}}{2} \right) +
B \left(\frac{e^{i 2 \pi t n} - e^{-i 2 \pi t n}}{2i} \right)\right] \\
&= \sum_{n=1}^N \left[\left(\frac{A}{2}+\frac{B}{2i}\right)e^{i 2 \pi t n} +
\left(\frac{A}{2}-\frac{B}{2i}\right)e^{-i 2 \pi t n}\right] \\
&=\sum_{n=1}^N \left(\frac{A}{2}+\frac{B}{2i}\right)e^{i 2 \pi t n}
+\sum_{n=1}^N \left(\frac{A}{2}-\frac{B}{2i}\right)e^{-i 2 \pi t n} \\
&=\sum_{n=1}^N \left(\frac{A}{2}+\frac{B}{2i}\right)e^{i 2 \pi t n}
+\sum_{n=-N}^{-1} \left(\frac{A}{2}-\frac{B}{2i}\right)e^{i 2 \pi t n}
\end{aligned}$$
So you are right about the fact that there is no single $C$ that fulfills the sum from $-N$ to $N$, unless $A=B=0$. Otherwise, define
$$
C_n =
\begin{cases}
\frac{A}{2}+\frac{B}{2i} \quad \text{if } n>0 \\
0 \qquad ~~~~~~~~\text{if } n=0 \\
\frac{A}{2}-\frac{B}{2i} \quad \text{if } n<0 \\
\end{cases}
$$
and you get something similar.
Regarding $|\cos(x)+i\sin(x)|=1$, note that the square of the modulus of a complex number $z$ is given by $|z|^2=z\bar{z}$, which in this case gives
$$\begin{aligned}
|\cos(x)+i\sin(x)|^2
&= \left(\cos(x)+i\sin(x)\right) \left(\cos(x)-i\sin(x)\right) \\
&=\cos^2(x) -i^2\sin^2(x) \\
&=\cos^2(x) + \sin^2(x) \\
&=1
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.