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In a polynomial of $n$ degree, what numbers can fill the $n$? Until now, I've seen that the $n$ could be filled with the set $\mathbb{N}_0$ and $-\infty$ but I still didn't see mentions on other sets of numbers. As I thought that having 0 and $-\infty$ as degrees of a polynomial were unusual, I started to think if it would be possible for other numbers to also fill the gap.
| The degree of a polynomial can only take the values that you've specified. For that, let's revisit the definition of a polynomial. Personally, I was taught that a polynomial (in one variable) is an algebraic expression which can be written in the form $$a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$$ where $n$ is a non-negative integer. From this definition, as the degree is the same as $n$, therefore, the degree is also a non-negative integer (I'll not go into the degree of a zero polynomial, which is best left undefined).
On the other hand, Wikipedia says that
A polynomial is an expression of finite length constructed from
variables (also called indeterminates) and constants, using only the
operations of addition, subtraction, multiplication, and non-negative
integer exponents.
Clearly, there is no non-integer expenentiation, nor any division by a variable, so again it is clear that the degree is a non-negative integer.
| {
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How do you take the product of Bernoulli distribution? I have a prior distribution,
$$p(\boldsymbol\theta|\pi)=\prod\limits_{i=1}^K p(\theta_i|\pi).$$
$\theta_i$ can equal $0$ or $1$, so I am using a Bernoulli distribtion so that
$$p(\boldsymbol\theta|\pi)=\prod\limits_{i=1}^K \pi^{\theta_i}(1-\pi)^{1-\theta_i}.$$
I then want to add this distribution onto my marginal likelihood to make up my posterior. Should I solve it as
$$p(\boldsymbol\theta|\pi)=\pi^{K\theta_i}(1-\pi)^{K(1-\theta_i)} \, \, ?$$
But then is the product of bernoulli distributions the binomial distribution?
Then should my answer be
$$p(\boldsymbol\theta|\pi)=\left(\begin{array}c K\\ t \end{array}\right)\pi^{t}(1-\pi)^{K-t)} $$
where $K$ is the maximum number of $\theta_i$'s allowed, and $t=\{0, 1\}$ , (i.e. $t=0\, \, \text{or}\, \, 1$)?
What form do I add this prior to my likelihood?
| The equation you have can be represented as follows:
$$p(\boldsymbol x|\theta)=\prod\limits_{i=1}^K \theta^{x_i}(1-\theta)^{1-x_i}=\theta^{\sum_i x_i}(1-\theta)^{K-\sum_i x_i}$$
We have the Bayes rule
$$p(\theta|x)=\frac{p(x|\theta)p(\theta)}{p(x)}$$
as $\theta$ is known, we have the joint density $p(x,\theta)=p(\theta,x)$ which specifies all the information we need.
| {
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Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$. I encountered a problem in a book that was designed for IMO trainees. The problem had something to do with divisibility.
Prove that if $n$ is a positive integer then $2^{3n}-1$ is divisible by $7$.
Can somebody give me a hint on this problem. I know that it can be done via the principle of mathematical induction, but I am looking for some other way (that is if there is some other way)
| Hint: Note that $8 \equiv 1~~~(\text{mod } 7)$.
So,
$$2^{3n}=(2^3)^n=8^n\equiv \ldots~~~(\text{mod } 7)=\ldots~~~(\text{mod } 7)$$
Try to fill in the gaps!
Solution:
Note that $8 \equiv 1~~(\text{mod } 7)$. This means that $8$ leaves a remainder of $1$ when divided by $7$. Now assuming that you are aware of some basic modular arithmetic,
$$2^{3n}=(2^3)^n=8^n\equiv 1^n ~~(\text{mod } 7)=1~~(\text{mod } 7)$$
Now if $2^{3n}\equiv 1~~(\text{mod } 7)$ then it follows that,
$$2^{3n}-1=8^n-1\equiv (1-1)~~(\text{mod } 7)~\equiv 0~~(\text{mod } 7)\\ \implies 2^{3n}-1\equiv 0~~(\text{mod } 7)$$
Or in other words, $2^{3n}-1$ leaves no remainder when divided by $7$ (i.e. $2^{3n}-1$ is divisible by $7$). As desired
| {
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Convergence of a sequence of non-negative real numbers $x_n$ given that $x_{n+1} \leq x_n + 1/n^2$. Let $x_n$ be a sequence of the type described above. It is not monotonic in general, so boundedness won't help. So, it seems as if I should show it's Cauchy. A wrong way to do this would be as follows (I'm on a mobile device, so I can't type absolute values. Bear with me.) $$\left|x_{n+1} - x_n\right| \leq \frac{1}{n^2}.$$ So, we have $$
\left|x_m -x_n\right| \leq \sum_{k=n}^{m} \frac{1}{k^2}$$ which is itself Cauchy, etc., etc. But, of course, I can't just use absolute values like that. One thing I have shown is that $x_n$ is bounded. Inductively, one may show $$ \limsup_{n \to \infty} x_n \leq x_k + \sum_{k=n}^{\infty} \frac{1}{k^2},$$ although I'm not sure this helps or matters at all. Thanks in advance.
Disclaimer I've noticed that asking a large number of questions in quick succession on this site is often frowned upon, especially when little or no effort has been given by the asker. However, I am preparing for a large test in a few days and will be sifting through dozens of problems. Therefore, I may post a couple a day. I will only do so when I have made some initial, meaningful progress. Thanks.
| Note that
$$
\lim_{n\to\infty}\sum_{k=n}^\infty\frac1{k^2}=0
$$
and for $m\gt n$,
$$
a_m\le a_n+\sum_{k=n}^\infty\frac1{k^2}
$$
First, take the $\limsup\limits_{m\to\infty}$:
$$
\limsup_{m\to\infty}a_m\le a_n+\sum_{k=n}^\infty\frac1{k^2}
$$
which must be non-negative. Then take the $\liminf\limits_{n\to\infty}$:
$$
\limsup_{m\to\infty}a_m\le\liminf_{n\to\infty}a_n
$$
Thus, the limit exists.
| {
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A puzzle with powers and tetration mod n A friend recently asked me if I could solve these three problems:
(a) Prove that the sequence $ 1^1, 2^2, 3^3, \dots \pmod{3}$ in other words $\{n^n \pmod{3} \}$ is periodic, and find the length of the period.
(b) Prove that the sequence $1, 2^2, 3^{3^3},\dots \pmod{4}$ i.e. $\{\ ^nn \pmod{4}\}$ is periodic and find the length of the period.
(c) Prove that the sequence $1, 2^2, 3^{3^3},\dots \pmod{5}$ i.e. $\{\ ^nn \pmod{5}\}$ is periodic and find the length of the period.
The first two were not terribly difficult (but might be useful exercises in Fermat's little theorm), but the third one is causing me problems, since my methods are leading to rather a lot of
individual cases, and I would be interested to see if anyone here can find a neater way to solve it.
(In (c), I have evaluated the first 15 terms, and not found a period yet - unless I have made a mistake.)
| For $k\ge 1$ we have
$$\begin{align*}
{^{k+2}n}\bmod5&=(n\bmod5)^{^{k+1}n}\bmod5=(n\bmod5)^{^{k+1}n\bmod4}\bmod5\\
{^{k+1}n}\bmod4&=(n\bmod4)^{^kn}=\begin{cases}
0,&\text{if }n\bmod2=0\\
n\bmod4,&\text{otherwise}\;,
\end{cases}
\end{align*}$$
so
$${^{k+2}n}\bmod5=\begin{cases}
(n\bmod5)^0=1,&\text{if }n\bmod2=0\\
(n\bmod5)^{n\bmod4},&\text{otherwise}\;.
\end{cases}$$
In particular,
$${^nn}\bmod5=\begin{cases}
(n\bmod5)^0,&\text{if }n\bmod2=0\\
(n\bmod5)^{n\bmod4},&\text{otherwise}\;.
\end{cases}$$
for $n\ge3$, where $0^0\bmod5$ is understood to be $0$. Clearly the period starting at $n=3$ is a multiple of $20$; actual calculation shows that it is $20$, and that we cannot include the first two terms::
$$\begin{array}{r|c|c}
n:&&&&&&&&&1&2\\
{^nn}\bmod5:&&&&&&&&&1&4\\ \hline
n:&3&4&5&6&7&8&9&10&11&12\\
{^nn}\bmod5:&2&1&0&1&3&1&4&0&1&1\\ \hline
n:&13&14&15&16&17&18&19&20&21&22&\\
{^nn}\bmod5:&3&1&0&1&2&1&4&0&1&1
\end{array}$$
(Added: And no, I’d not seen that Qiaochu had already answered the question until after I posted.)
| {
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Theorems with an extraordinary exception or a small number of sporadic exceptions The Whitney graph isomorphism theorem gives an example of an extraordinary exception: a very general statement holds except for one very specific case.
Another example is the classification theorem for finite simple groups: a very general statement holds except for very few (26) sporadic cases.
I am looking for more of this kind of theorems-with-not-so-many-sporadic-exceptions
(added:) where the exceptions don't come in a row and/or
in the beginning - but are scattered truly sporadically.
(A late thanks to Asaf!)
| How about the Big Picard theorem? http://en.wikipedia.org/wiki/Picard_theorem
If a function $f:\mathbb{C}\to \mathbb{C}$ is analytic and has an essential singularity at $z_0\in \mathbb{C}$, then in any open set containing $z_0$, $f(z)$ takes on all possible complex values, with at most one possible exception, infinitely often.
| {
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Is a right inverse of a surjective linear map linear? On a finite dimensional vector space, the answer is yes (because surjective linear map must be an isomorphism). Does this extend to infinite dimensional vector space? In other words, for any linear surjection $T:V\rightarrow V$, AC guarantees the existence of right inverse $R:V\rightarrow V$. Must $R$ be linear?
How about $T:V\rightarrow W$ linear surjection in general?
| No. Let $V = \text{span}(e_1, e_2, ...)$ and let $T : V \to V$ be given by $T e_1 = 0, T e_i = e_{i-1}$. A right inverse $S$ for $T$ necessarily sends $v = \sum c_i e_i$ to $\sum c_i e_{i+1} + c_v e_1$ but $c_v$ may be an arbitrary function of $v$.
| {
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Limit of $(x_n)$ with $0
Let $0 < x_1 < 1$ and $x_{n + 1} = x_n - x_n^{n + 1}$ for $n \geqslant 1$.
Prove that the limit exists and find the limit in terms of $x_1$.
I have proved the existence but cannot manage the other part.
Thanks for any help.
| Note that $x_2=x_1(1-x_1)$ with $0\lt x_1\lt1$ hence $0\lt x_2\lt1/4$, that $(x_n)_{n\geqslant1}$ is decreasing, in particular $(x_n)_{n\geqslant1}$ converges to some value $\ell(x_1)$ in $[0,x_1)$, and that $x_n\leqslant x_2$ for every $n\geqslant2$. Hence, for every $n\geqslant2$, $x_{n+1}\geqslant x_n-x_2^{n+1}$, which implies that $x_n\geqslant x_2-x_2^3-\cdots-x_2^n$ for every $n\geqslant2$ and that $\ell(x_1)\geqslant x_2-x_2^3/(1-x_2)=x_2(1-x_2-x_2^2)/(1-x_2)$ hence $\ell(x_1)\gt0$.
Auto-quote:
To get an exact value of the limit $\ell(x_1)$ as an explicit function of $x_1$ is more difficult and, to venture a guess, probably not doable.
| {
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$\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to $U_n$. I've tried, but I can't solve the question. Please help me prove that:
$\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to $U_n$.
| (If you know about ring theory.) Since $\mathbb Z_n$ is an abelian group, we can consider its endomorphism ring (where addition is component-wise and multiplication is given by composition). This endomorphism ring is simply $\mathbb Z_n$, since the endomorphism is completely determined by its action on a generator, and a generator can go to any element of $\mathbb Z_n$. Therefore, the automorphism group $\mathrm{Aut}(\mathbb Z_n)$ is the group of units in $\mathbb Z_n$, which is $U_n=U(\mathbb Z_n)$.
| {
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Integer combination i want write a module to find the integer combination for a multi variable fomula. For example
$8x + 9y \le 124$
The module will return all possible positive integer for $x$ and $y$.Eg. $x=2$, $y=12$.
It does not necessary be exactly $124$, could be any number less or equal to $124$. Must be as close as possible to $124$ if no exact solution could be found.
I do not want to solve with brute force as the number of variable could be any...$(5,10,100,...n)$
Any algorithm could solve this?
| Considering the equal case, if you analyze some of the data points that produce integer solution to the original inequality:
$$...,(-34,44), (-25,36), (-16,28), (-7,20),(2,12),(11,4),...$$
you can see that:
$$x_{i}=x_{i-1}+9$$
to get $y$ values, re-write the inequality as follows (and use the equal part):
$$y = \frac{124-8x}{9}$$
using the above equations you can find $(x,y)$ values in a given range of $x$ values without brute force.
| {
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Optimization problem involving step function I've got to optimize the following function with respect to $\phi$:
$q(\phi, x) = \frac{1}{n} \sum_{i=1}^{n}{H(y_i)}$
where
$y_i = k - \phi l - x_i$
and $H(.)$ denotes the Heaviside function. $k$ and $l$ are constants, and $x$ follows either (1) a continuous uniform distribution or (2) a normal distribution. This is part of a quite standard programming problem but I'm a little stuck with finding the optimal $\phi$
I'm sure this is a totally simple question but I can't quite figure it out... any help is greatly appreciated...
| The step functions all add up in the same direction, since $\phi$ has the same sign in all $y_i$ – thus $q$ is minimal for all $\phi$ such that all step functions are $0$, which occurs for $\phi\lessgtr(k-x)/l$, where the inequality is $\lt$ or $\gt$ and $x$ is the greatest or least of the $x_i$, depending on whether $l$ is negative or positive, respectively.
| {
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natural solutions for $9m+9n=mn$ and $9m+9n=2m^2n^2$ Please help me find the natural solutions for $9m+9n=mn$ and $9m+9n=2m^2n^2$ where m and n are relatively prime.
I tried solving the first equation in the following way: $9m+9n=mn \rightarrow (9-n)m+9n=0 $
$\rightarrow m=-\frac{9n}{9-n}$
Thanks in advance.
| $$mn=9n+9m \Rightarrow (m-9)(n-9)=81$$
This equation is very easy to solve, just keep in mind that even if $m,n$ are positive, $m-9,n-9$ could be negative. But there are only 6 ways of writing 81 as the product of two integers.
The second one is trickier, but if $mn >9$ then it is easy to prove that
$$2m^2n^2> 18mn > 9m+9n $$
Added Also, since $9|2m^2n^2$ it follows that $3|mn$. Combining this with $mn \leq 9$ and $m|9n, n|9m$ solves immediately the equation.
P.S. Your approach also works, if you do Polynomial long division you will get $\frac{9n}{n-9}=9 +\frac{81}{n-9}$. Thus $n-9$ is a divisor of $81$.
P.P.S. Alternately, for the second equation, if you use $2\sqrt{mn} \leq m+n$ you get
$$18 \sqrt{mn} \leq 9(m+n)=2m^2n^2$$
Thus $$(mn)^3 \geq 81$$ which implies $mn=0 \text{ or } mn \geq 5$.
| {
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If $T$ is bounded and $F$ has finite rank, what is the spectrum of $T+F$? Suppose that $T$ is a bounded operator with finite spectrum. What happens with the spectrum of $T+F$, where $F$ has finite rank? Is it possible that $\sigma(T+F)$ has non-empty interior? Is it always at most countable?
Update: If $\sigma(T)=\{0\}$ then $0$ is in the essential spectrum of $T$ ($T$ is not invertible in the Calkin algebra), hence for any compact $K$, $\sigma_{ess}(T+K)=\sigma_{ess}(T)=\{0\}$. For operators such that the essential spectrum is $\{0\}$, it is known that their spectrum is either finite or consists of a sequence converging to $0$. I think it should be the same for operators with finite spectrum, but I cannot find a proof or reference.
| It is always true if $T$ is self-adjoint. Here is a theorem that you might be interested:
If $T$ is self-adjoint, a complex number is in the spectrum of $T$ but not in its essential spectrum iff it is an isolated eigenvalue of $T$ of finite multipliticity.
The result can be found on page 32 of Analytic K-homology by Nigel Higson and John Roe.
| {
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Finding a paper by John von Neumann written in 1951 There's a 1951 article by John von Neumann, Various techniques used in connection with random digits, which I would really like to read. It is widely cited, but I can't seem to find an actual copy of the paper, be it free or paying.
Is there a general strategy to find copies of relatively old papers like this one?
EDIT: I've searched quite a lot before posting this question and fond the following reference:
Journal of Research of the National Bureau of Standards, Appl. Math. Series (1951), 3, 36-38
Unfortunately, my library doesn't have it, and it is not in NIST's online archive (neither at http://www.nist.gov/nvl/journal-of-research-past-issues.cfm nor at http://nistdigitalarchives.contentdm.oclc.org/cdm/nistjournalofresearchbyvolume/collection/p13011coll6)
| One of the citations gives the bibliographic info,
von Neumann J, Various Techniques Used in Connection with Random Digits, Notes by G E
Forsythe, National Bureau of Standards Applied Math Series, 12 (1951) pp 36-38. Reprinted in von
Neumann's Collected Works, 5 (1963), Pergamon Press pp 768-770.
That should be enough information for any librarian to find you a copy.
| {
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about the differentiability : the general case Let $U$ be an open set in $\mathbb{R}^{n}$ and $f :U \subset \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ be a given function. We say that $f$ is differentiable at $x_{0}\in U$ if the partial derivatives of $f$ existi at $x_{0}$ and if
$$\displaystyle \lim_{x \rightarrow x_{0}} \frac{\|f(x)-f(x_{0})-T(x-x_{0})\|}{\|x-x_{0}\|}=0$$
where $T=Df(x_{0})$ is the $ m \times n$ matrix with elements $\displaystyle \frac{\partial f_{i}}{\partial x_{j}}$ evaluated at $x_{0}$ and the $T(x-x_{0})$ means the product of $T$ with $x-x_{0}$ (regarded as a column matrix). We call $T$ the derivative of $f$ at $x_{0}$.
Now I consider a particular case($m=n=2$) $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$
Following the definition I obtain :
$$\displaystyle f(a,b)-f(a_{0}, b_{0})- \begin{pmatrix}\frac{\partial f_{1}}{\partial a} & \frac{\partial f_{1}}{\partial b}\\ \frac{\partial f_{2}}{\partial a} & \frac{\partial f_{2}}{\partial b} \end{pmatrix} \begin{pmatrix}a-a_{0} \\ b-b_{0} \end{pmatrix} = f(a,b)-f(a_{0},b_{0})- \begin{pmatrix}\frac{\partial f_{1}}{\partial a} \cdot(a-a_{0})+\frac{\partial f_{1}}{\partial b}\cdot(b-b_{0})\\ \frac{\partial f_{2}}{\partial a}\cdot(a-a_{0})+\frac{\partial f_{2}}{\partial b}\cdot(b-b_{0}) \end{pmatrix}$$ where $f(a,b)=(f_{1}(a,b),f_{2}(a,b))$.
My question is :
How can I compute this limit because the last element is a matrix and first two aren't. And why $f(x)-f(x_{0})-T(x-x_{0})$ I have to put into $\| \|$ ? I have an idea why I have to put into norm but I'm not sure and can you give an concrete example how I compute the limit - for example $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2, f(a,b)=(a^2+b^2, a^2+b^2)$ when $ (a,b) \rightarrow (1,1)$.
Thanks :)
| Let me compute for the example, $f(x,y)=(x^2+y^2, x+y)$. We write $f_1(x,y)=x^2+y^2$ and $f_2(x,y)=x+y$. Then
$$
\frac{\partial f_1}{\partial x}(x,y) = 2x,\quad
\frac{\partial f_1}{\partial y}(x,y) = 2y,\quad
\frac{\partial f_2}{\partial x}(x,y) = 1,\quad
\frac{\partial f_2}{\partial y}(x,y) = 1.
$$
Since these four functions are continuous then $f$ is differentiable. THEN $T$ is already the Jacobian matrix:
$$
T=\begin{pmatrix} 2x& 2y\\ 1&1\end{pmatrix}.
$$
Finally
$$
f(x,y)-f(x_0,y_0)-T\begin{pmatrix}x-x_0\\y-y_0\end{pmatrix}=
\begin{pmatrix}x^2+y^2-x_0^2-y_0^2-2x_0(x-x_0)-2y_0(y-y_0) \\ x+y-x_0-y_0 -(x-x_0+y-y_0)\end{pmatrix},
$$
taking limits
$$
\lim_{(x,y)\to(x_0,y_0)}
\frac{(x-x_0)^2+(y-y_0)^2}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=
\lim_{(x,y)\to(x_0,y_0)}
\sqrt{(x-x_0)^2+(y-y_0)^2}=0.
$$
| {
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How many numbers less than $x$ are co-prime to $x$ Is there a fast way , or a direct function to give the count of numbers less than $x$ and co-prime to $x$ .
for example if $x$ = 3 ; then $n = 2$ and if $x$ = 8 ; then $n = 4$.
| Yes there is. First of all, you have to prime factorize your $x$, any put it in exponential form. Suppose you have the number $x = 50$. The prime factorization is $5^2 * 2^1$. Now take each number seperately. Take the bases and subtract 1 from all of them. $5-1=4$. $2-1=1$. Now evaluate the each base/exponent combination after subtracting the exponent by 1. $5^{2-1}=5$. $2^{1-1}=1$. Now multiply your answers. $(4)(1)(5)(1) = 20$.
If you want to try out your examples, $3=3^1$. So your answer would be $(3-1)(3^{1-1}) = (2)(1) = 2.$ For your other example, $8=2^3. n=(2-1)(2^{3-1})=(1)(4)=4$.
I guess in general, prime factorize your $x$ to $a_1^{b_1}*a_2^{b_2}* ... * a_n^{b_n}$ and your answer will be $((a_1-1)(a_1^{b_1-1}))*((a_2-1)(a_2^{b_2-1}))*...*(a_n-1)(a_n^{b_n-1})$
| {
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Nonsingularity of Euclidean distance matrix Let $x_1, \dots, x_k \in \mathbb{R}^n$ be distinct points and let $A$ be the matrix defined by $A_{ij} = d(x_i, x_j)$, where $d$ is the Euclidean distance. Is $A$ always nonsingular?
I have a feeling this should be well known (or, at least a reference should exists), on the other hand, this fact fails for general metrics (take e.g. path metric on the cycle $C_4$)
edit: changed number of points from $n$ to general $k$
| I think it should be possible to show that your distance matrix is always nonsingular by showing that it is always a Euclidean distance matrix (in the usual sense of the term) for a non-degenerate set of points. I don't give a full proof but sketch some ideas that I think can be fleshed out into a proof.
Two relevant papers on Euclidean distance matrices are Discussion of a Set of Points in Terms of Their Mutual Distances by Young and Householder and Metric Spaces and Positive Definite Functions by Schoenberg. They show that an $n\times n$ matrix $A$ is a Euclidean distance matrix if and only if $x^\top Ax\le0$ for all $x$ with $e^\top x=0$ (where $e$ is the vector with $1$ in each component) and that the affine dimension of the points is $n$ if and only if the inequality is strict.
It follows that a Euclidean distance matrix can only be singular if the affine dimension of the points is less than $n$: If the affine dimension is $n$, there cannot be an eigenvalue $0$, since there is a positive eigenvalue (since $e^\top Ae\gt0$), and the span of these two eigenspaces would non-trivially intersect the space $e^\top x=0$, contradicting the negative definiteness of $A$ on that space.
To use all this for your case, one could try to show that a distance matrix in your sense is always a Euclidean distance matrix in the usual sense for points with affine dimension $n$. I think this could be done by continuously varying the exponent $\alpha$ in $A_{ij}=d(x_i,x_j)^\alpha$ from $1$ to $2$ and showing a) that there is always a direction in which the points can move such that $A$ remains their distance matrix with the changing exponent and b) that this movement necessarily causes them to have affine dimension $n$.
To get a feel for how this might work, consider a square: The movement would bend the square into a tetrahedron. The proof would need to account for the fact that this seems to hold only for $\alpha\lt2$; you can see from the example of three points in a line that they can be bent to accommodate $\alpha\lt2$ but not $\alpha\gt2$.
| {
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What is the best way to solve a problem given remainders and divisors? $x$ is a positive integer less than $100$. When $x$ is divided by $5$, its remainder is $4$. When $x$ is divided by $23$, its remainder is $7$. What is $x$ ?
| So, $x=5y+4$ and $x=23z+7$ for some integers $y,z$
So, $5y+4=23z+7=>5y+20=23z+23$ adding 16 to either sides,
$5(y+4)=23(z+1)=>y+4=\frac{23(z+1)}{5}$, so,$5|(z+1)$ as $y+4$ is integer and $(5,23)=1$
$=>z+1=5w$ for some integer $w$.
$x=23(5w-1)+7=115w-16$
As $0<x<100$ so,$x=99$ putting $w=1$
Alternatively, We have $5y=23z+3$
Using convergent property of continued fractions,
(i)$\frac{23}{5}=4+\frac{3}{5}=4+\frac{1}{\frac{5}{3}}=4+\frac{1}{1+\frac{2}{3}}$
$=4+\frac{1}{1+\frac{1}{\frac{3}{2}}}=4+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}$
So, the last but one convergent is $4+\frac{1}{1+\frac{1}{1}}=\frac{9}{2}$
$=>23\cdot2-5\cdot9=1$
$5y=23z+3(23\cdot2-5\cdot9)=>5(y+27)=23(z+6)=>5|(z+6)=>5|(z+1)$
(ii)$\frac{23}{5}=4+\frac{3}{5}=4+\frac{1}{\frac{5}{3}}=4+\frac{1}{1+\frac{2}{3}}$
$=4+\frac{1}{1+\frac{1}{\frac{3}{2}}}=4+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}$
$=4+\frac{1}{1+\frac{1}{1+\frac{1}{1+1}}}$
So, the last convergent is $4+\frac{1}{1+\frac{1}{1+1}}=\frac{14}{3}$
$=>23\cdot3-5\cdot14=-1$
$5y=23z-3(23\cdot3-5\cdot14)=5(y-42)=23(z-9)=>5|(z-9)=>5|(z+1)$
| {
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Infinite descent This Wikipedia article of Infinite Descent says:
We have
$ 3 \mid a_1^2+b_1^2 \,$.
This is only true if both $a_1$ and $b_1$ are divisible by $3$.
But how can this be proved?
| Suppose $a_1 = 3 q_1 + r_1$ and $b_1 = 3 q_2 + r_2$, where $r_1$ and $r_2$ is either $-1$, $0$ or $1$. Then
$$
a_1^2 + b_1^2 = 3 \left( 3 q_1^2 + 3 q_2^2 + 2 q_1 r_1 + 2 q_2 r_2 \right) + r_1^2 + r_2^2
$$
For $a_1^2 + b_1^2$ to be divisible by $3$, we should have $r_1^2 + r_2^2 = 0$, since $0\leqslant r_1^2+r_2^2 < 3$. Enumerating 9 cases, only $r_1 = r_2 = 0$ assure the divisibility.
This is essentially the answer anon gave in his comment.
| {
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Linear algebraic dynamical system that has complex entries in the matrix Suppose that there is a dynamical system that has the form of $\mathbb{x}_{k+1} = A\mathbb{x}_k$.
Suppose that one eigenvalue of $A$ matrix is complex number, the form of $a-bi$.
We then convert $\mathbb{x}_k = P\mathbb{y}_k$ where matrix $P$ is the matrix of corresponding eigenvector of $a-bi$ eigenvalue.
Then we would be able to write the dynamical system as the following: $\mathbb{y}_{k+1} = C\mathbb{y}_k$ where $C$ is \begin{bmatrix}a & -b \\b & a \end{bmatrix}
and then $A = PCP^{-1}$.
The question is, why does the eigenvalue of the matrix C equal to the eigenvalue of the matrix A?
| Note that if $Av=\lambda v$ then $Cw=\lambda w$ where $w=P^{-1}v$.
| {
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Dedekind Cuts in Rudin' analysis - Step 4 This question refers to the construction of $\mathbb{R}$ from $\mathbb{Q}$ using Dedekind cuts, as presented in Rudin's "Principles of Mathematical Analysis" pp. 17-21.
More specifically, in the last paragraph of step 4, Rudin says that for $\alpha$ a fixed cut, and given $v \in 0^*$, setting $w=- v / 2$, there exists an integer $n$ such that $nw \in \alpha$ but $(n+1)w$ is not inside $\alpha$. Rudin says that this depends on the Archimedean property of the rationals, however he has not proved it.
Could somebody prove the existence of the integer $n$?
| If $v\in 0^*$, then $v<0$, so $w>0$. Let $\gamma=\sup \alpha$. The Archimedean property says that for any $\gamma$ there is some integer $m$ such that $mw\geq\gamma$. Since $\mathbb{N}$ is well-ordered, there is a smallest $m$ such that $mw\geq\gamma$, call this $n+1$. This means that $nw<\gamma$, and hence $nw\in \alpha$. But also $(n+1)w\notin \alpha$ since it is larger than $\gamma$ (OK, so there might be some subtleties with endpoints you should think about, but essentially this is the idea).
| {
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$H\vartriangleleft G$ and $|H|\not\equiv 1 (\mathrm{mod} \ p)$ then $H\cap C_{G}(P)\neq1$
Let $G$, a finite group, has $H$ as a proper normal subgroup and let $P$ be an arbitrary $p$-subgroup of $G$ ($p$ is a prime). Then $$|H|\not\equiv 1 (\mathrm{mod} \ p)\Longrightarrow H\cap C_{G}(P)\neq1$$
What I have done:
I can see the subsequent well-known theorem is an especial case of the above problem:
Let $G$ is a finite non trivial $p$-group and $H\vartriangleleft G$. Then if $H\neq1$ so $H\cap Z(G)\neq1$.
So I assume that $G$ acts on $H$ by conjugation and therefore $$|H|=1+\sum_{x\in H-\{1\}}|\mathrm{Orbit}_G(x)|$$ $|H|\not\equiv 1 (\mathrm{mod} \ p)$ means to me that there is $x_0\in H$ such that $p\nmid|\mathrm{Orbit}_G(x_0)|$. Am I doing right? Thanks.
This problem can be applied nicely in the following fact:
Let $p$ is an odd prime and $q$ is a prime such that $q^2\leqslant p$. Then $\mathrm{Sym}(p)$ cannot have a normal subgroup of order $q^2$.
| Let $P$ act on $H$; the number of fixed points is the number of elements in $C_G(P)\cap H$. Now use the easy fact that the number of fixed points is congruent to $|H|\pmod{p}$.
| {
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Solving this linear system based on the combustion of methane, has no constants I recently discovered that I could solve a chemical reaction using a linear system. So I thought I would try something simple like the combustion of methane.
where x y z and w are the moles of each molecule
x $CH_4$ + y $O_2$ = z $H_2$O + w C$O_2$
the linear system for this would be:
x = w
2y = z + 2w
4x = 2z
I got as far as y = z and y = 2w but without any constants, I am stumped. Can anyone help me? I was assuming that elimination and substitution would suffice, but I must be wrong.
| Each of your equations can be rearranged to give
$$x-w=0$$
$$2y-z-2w=0$$
$$4x-2z=0$$
Your idea of using Gaussian elimination is then a very good one. Or you can solve directly. From the first equation, you get $$x=w.$$ From the third equation, you get $$z=\frac{x}{2}=\frac{w}{2}.$$ Then from the second equation you get $$y=\frac{z}{2}+w=\frac{5w}{4}.$$
In each subsequent equation, we have used previous information and any value for $w$ will suffice ($w$ is called a free variable).
Note that this makes sense as $w$ is a particular number of type atoms and changing it will change how the equation is to be balanced.
| {
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Why are two vectors that are parallel equivalent? Why are two parallel vectors with the same magnitude equivalent?
Why is their start point irrelevant?
How can a vector starting at $\,(0, -10)\,$ going to $\,(10, 0)\,$ be the same as
a vector starting at $\,(10, 10)\,$ and going to $\,(20, 20)\,$?
| Components of a vector determine the length and the direction of the vector, but not it's basepoint. Therefore 2 vectors are equivalent if and only if they have the same components, but this gives the idea that any 2 parallel vectors are equal though they have different base points. To avoid this confusion, we take that all vectors are based on the origin unless stated.
| {
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Inequality with two absolute values I'm new here, and I was wondering if any of you could help me out with this little problem that is already getting on my nerves since I've been trying to solve it for hours.
Studying for my next test on inequalities with absolute values, I found this one:
$$ |x-3|-|x-4|<x $$
(I precisely found the above inequality on this website, here to be precise, but, the problem is that when I try to solve it, my answer won't be $(-1,+\infty)$, but $(1,7)$. I took the same inequalities that the asker's teacher had given to him and then put them on a number line, and my result was definitely not $(-1,+\infty)$
Here are the inequalities:
$$ x−3 < x−4 +x $$
$$ x−3 < −(x−4) +x $$
$$ −(x−3)<−(x−4)+x $$
And here are my answers respectively:
$$ x>1, \quad x>-1, \quad x<7 $$
I will really appreciate if anyone could help me out, because I'm already stressed dealing with this problem, that by the way, it is not demanding that I solve it, but you know, why not?
| $|x-3|-|x-4|< x$, I write in $|x-3|< x+|x-4|$
but remember: $|r|< s \implies -s < r < s$.
So I write the equation in the form: $-x-|x-4| < x-3 < x+|x-4|$.
From this inequality I obtain 2 equations:
(a) $-x-|x-4| < x-3$
(b) $x-3 < x+|x-4|$.
Remember too : $|r| > s \implies r > s \text{ or } r < -s $.
So this concept I will apply to equation (a) and equation (b).
From equation (a):
$-x-|x-4| < x-3$ I write so absolute value is in one side: -|x-4| < 2x-3 then I multiply by -1 : |x-4| > -2x+3 . Now I write the equation in the form of: |r| > s----> r > s OR r < -s .As a result we obtain 2 additional equation: x-4 > -2x+3 OR x-4 < 2x-3 . From the first one: 3x > 7 --->x > 7/3 or (7/3,∞) . From the second inequality: -1 < x or (-1,∞) .The first and second inequality are OR function then: (7/3,∞) U (-1,∞) is: (-1,∞)
Now the equation (b): x-3 < x+|x-4| we write to put absolute value in one side: $-3 < |x-4|$ or $|x-4| > -3$ . To satisfy this inequality $x$ will take any value positive or negative. As a result we can write the result the value of $x$ for equation (b) as: $(-∞,+∞)$
The final result from the equation (a) AND (b) will be the intersection of their value: $(a)∩(b)$ or $(-1,∞)∩(-∞,∞)$ and find the final result for $x: (-1,∞)$ that satisfy the inequality $|x-3|-|x-4| < x$
| {
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Existence of an Infinite Length Path I came across the following simple definition
A path $\gamma$ in $\mathbb{R}^n$ that connects the point $a \in \mathbb{R}^n$ to the point $b \in \mathbb{R}^n$, is a continuous $\gamma : [0, 1] \to \mathbb{R}^n$ such that $\gamma(0) = a$ and $\gamma(1) = b$. We denote by $\ell(\gamma)$ the (Euclidean) length of $\gamma$. $\ell(\gamma)$ is always defined and is either a non-negative realnumber or $\infty$.
However, I cannot seem to think of a path, defined in this manner (specifically, where the domain is compact), the length of which is infinite.
Can anyone provide an example ?
| Yes, for instance Koch snowflake is a such example. Let's do the following construction:
*
*Start with the segment $A_0 = [0,1]$.
*Subdivise $A_0$ in three equal pieces.
*Replace the middle third by an equilateral triangle with base $[\frac13, \frac23]$.
*Suppress the base of that triangle. You get the path $A_1$ something which looks like a saw teeth.
*By induction, to construct $A_{n+1}$ replace each of the $4^n$ segments of $A_n$ by an equilateral triangle of base this segment, and then remove the segment.
*The limit object is a path (whose image is compact set) joining $0$ and $1$.
The length of $A_n$ is $\left( \frac43 \right)^n$ tends to $+\infty$ as $n \to +\infty$.
The Hausdorff dimensions is $\frac{\ln 4}{\ln 3} \approx 1.26$
The following picture summarize the construction.
http://www.cl.cam.ac.uk/~dao29/tmp/koch/geometric-construction.png
| {
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Generating integer solutions to $4mn - m^2 + n^2 = ±1$ How can I generate positive integer solutions to $m$ and $n$ that satisfy the equation:
$4mn - m^2 + n^2 = ±1$,
subject to the constraints that $m$ and $n$ are coprime, $m-n$ is odd and $m > n$.
| Hint: Completing the square yields an equation of the form: $$x^2-Dy^2=\pm 1$$
for a particular $D$.
There's actually a simple recursion that generates all solutions.
Let $a_0=0$, $a_1=1$, and $a_{k+2}=4a_{k+1}+a_{k}$. Then the general solution is $(m,n)=(a_{k+1},a_{k})$.
This gives the positive solutions. The solutions with $n$ negative are of the form $(m,n)=(a_k,-a_{k+1})$.
There are no solutions with $m$ negative and $m> n$
| {
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Sufficiency to prove the convergence of a sequence using even and odd terms Given a sequence $a_{n}$, if I know that the sequence of even terms converges to the same limit as the subsequence of odd terms:
$$\lim_{n\rightarrow\infty} a_{2n}=\lim_{n\to\infty} a_{2n-1}=L$$
Is this sufficient to prove that the $\lim_{n\to\infty}a_{n}=L$?
If so, how can I make this more rigorous? Is there a theorem I can state that covers this case?
| If you are familiar with subsequences, you can easily prove as follows. Let $a_{n_k}$ be the subsequence which converges to $\limsup a_n$. it is obviously convergent and contain infinitely many odds or infinitely many evens, or both. Hence, $\limsup a_n = L$. The same holds for $\liminf a_n$, hence the limit of the whole sequence exists and equals $L$.
| {
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What is exactly "Algebraic Dynamics"? Could somebody please give a big picture description of what exactly is the object of study in the area of Algebraic Dynamics? Is it related to Dynamical Systems? If yes in what sense? Also, what is the main mathematical discipline underpinning Algebraic Dynamics? Is it algebraic geometry, differential geometry e.t.c.?
| The Wiki article states that it is a combination of dynamical systems and number theory. I know it's a redirect, but WP's information on this point is probably reliable enough :)
(Are you checking here because you are not comfortable with WP info? It is a serious question which I'm curious about.)
| {
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Is independence preserved by conditioning? $X_1$ and $X_2$ are independent. $Y_1|X_1\sim\mathrm{Ber}\left(X_1\right)$, $Y_2|X_2\sim\mathrm{Ber}\left(X_2\right)$. Are $Y_1$ and $Y_2$ necessarily independent? (Assume $\mathrm{P}\left(0<X_1<1\right)=1$, $\mathrm{P}\left(0<X_2<1\right)=1$)
| No. Let $X_1,X_2$ be independent uniform (0,1) random variables, and then define $$Y_1=1_{(X_1\geq X_2)}\,\mbox{ and }\, Y_2=1_{(X_2> X_1)}.$$
| {
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Cauchy nets in a metric space Say that a net $a_i$ in a metric space is cauchy if for every $\epsilon > 0$ there exists $I$ such that for all $i, j \geq I$ one has $d(a_i,a_j) \leq \epsilon$. If the metric space is complete, does it hold (and in either case why) that every cauchy net converges?
| Consider a Cauchy net:
$$\forall \lambda,\lambda'\geq\lambda_n:\quad d(x_\lambda,x_\lambda')<\frac{1}{n}$$
Extract a Cauchy sequence:
$$x_n:=x_{\lambda(n)}\quad\lambda(n):=\lambda_1\wedge\ldots\wedge\lambda_n$$
Apply completeness:
$$d(x_\lambda,x)\leq d(x_\lambda,x_{n_0})+d(x_{n_0},x)<\frac{N}{2}+\frac{N}{2}\leq\epsilon$$
where to choose the meet $n_0:=N\wedge n(N)$ with $N:=\lceil\frac{\epsilon}{2}\rceil$
| {
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Cardinality of the set of all pairs of integers The set $S$ of all pairs of integers can be represented as $\{i \ | \ i \in \mathbb{Z} \} \times \{j\ | \ j \in \mathbb{Z}\}$. In other words, all coordinates on the cartesian plane where $x, y$ are integers.
I also know that a set is countable when $|S|\leq |\mathbb{N}^+|$. I attempted to map out a bijective function, $f : \mathbb{N}^+ \rightarrow S$.
$1 \rightarrow (1,1) \\
2 \rightarrow (1,2)\\
3 \rightarrow (1,3) \\
\quad \vdots
$
I determined from this that the natural numbers can only keep up with $(1,*)$. But there is the ordered pairs where $x=2,3,4,\cdots$ not to mention the negative integers. In other words, $|S|\geq |\mathbb{N}^+|$ and therefore $S$ is not countably infinite.
Is this correct? (I don't think it is... Something to do with my understanding of infinite sets)
| Define $\sigma: \Bbb Z \times \Bbb Z \to \Bbb Z \times \Bbb Z$ by
$$
\sigma(m,n) = \left\{\begin{array}{lr}
(1,-1), & \text{for } (m,n) = (0,0)\\
(m, n+1), & \text{for } m \gt 0 \, \land \, -m \le n \lt m\\
(m-1, n), & \text{for } n \gt 0 \, \land \, -n \lt m \le n\\
(m, n-1), & \text{for } m \lt 0 \, \land \, m \lt n \le -m\\
(m+1, n), & \text{for } n \lt 0 \, \land \, n \le m \lt -n-1\\
(m+2,n-1), & \text{for } n \lt 0 \, \land \, m = -n-1
\end{array}\right\}
$$
Exercise: Show that $n \mapsto \sigma^n(0,0)$ is a bijective mapping between $\{0,1,2,3,...\}$ and $\Bbb Z \times\Bbb Z$.
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Solving improper integrals and u-substitution on infinite series convergent tests This is the question:
Use the integral test to determine the convergence of $\sum_{n=1}^{\infty}\frac{1}{1+2n}$.
I started by writing:
$$\int_1^\infty\frac{1}{1+2x}dx=\lim_{a \rightarrow \infty}\left(\int_1^a\frac{1}{1+2x}dx\right)$$
I then decided to use u-substitution with $u=1+2n$ to solve the improper integral. I got the answer wrong and resorted to my answer book and this is where they went after setting $u=1+2n$:
$$\lim_{a \rightarrow \infty}\left(\frac{1}{2}\int_{3}^{1+2a}\frac{1}{u}du\right)$$
And the answer goes on...
What I can't figure out is where the $\frac{1}{2}$ came from when the u-substitution began and also, why the lower bound of the integral was changed to 3.
Can someone tell me?
| $$u=1+2x\Longrightarrow du=2dx\Longrightarrow dx=\frac{1}{2}du$$
Remember, not only you substitute the variable and nothing more: you also have to change the $\,dx\,$ and the integral's limits:
$$u=1+2x\,\,,\,\text{so}\,\, x=1\Longrightarrow u=1+2\cdot 1 =3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\mathbb{Z}$ has no torsion? What does is mean to say that $\mathbb{Z}$ has no torsion?
This is an important fact for any course?
Thanks, I heard that in my field theory course, but I don't know what it is.
| @d555, you might want to know that the notion of torison is extremely important, for example if $A$ is finitely generated and abelian group, then it can be written as the direct sum of its torsion subgroup $T(A)$ and a torsion-free subgroup (but this is not true for all infinitely generated abelian groups). $T(A)$ is uniquely determined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trying to find angle at which projectile was fired. So let's say I have a parabolic function that describes the displacement of some projectile without air resistance. Let's say it's
$$y=-4.9x^2+V_0x.$$
I want to know at what angle the projectile was fired. I notice that
$$\tan \theta_0=f'(x_0)$$ so the angle should be
$$\theta_0 = \arctan(f'(0)).$$ or
$$\theta_0 = \arctan(V_0).$$
Is this correct? I can't work out why it wouldn't be, but it doesn't look right when I plot curves.
| From what you are saying, it looks like you are thinking of $f(x)$ as a real valued function, and in effect you are only considering linear motion. (This would be fine, if the cannon were firing straight up or straight down or straight up.)
However, you are interested in knowing the interesting two-dimensional trajectories, and this means that $V_0$ is a 2-D vector pointing in the direction of the initial trajectory, and that the coefficient of $x^2$ actually accompanies a vector which is pointing straight down (to show the contribution of gravity.)
Once you find what $V_0$ is, you can then compute its angle with respect to horizontal, and have your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Finding the value of y in terms of x. Is it possible to get the value of $y$ in terms of $x$ from the below equation? If so please give give me a clue how to do that :)
$$y \sqrt{y^2 + 1} + \ln\left(y + \sqrt{y^2 + 1}\right) = \frac{a}{x^2}.$$
| Since $\ \rm{asinh}(y)=\ln\left(y + \sqrt{y^2 + 1}\right)\ $ let's set $\ y:=\sinh(u)\ $ and rewrite your equation as :
$$\sinh(u) \sqrt{\sinh(u)^2 + 1} + u = \frac{a}{x^2}$$
$$\sinh(u) \cosh(u) + u = \frac{a}{x^2}$$
$$\sinh(2u) + 2u = 2\frac{a}{x^2}$$
After that I fear you'll have to solve this numerically (to get $u$ in function of $x$).
I don't see something simpler sorry...
To solve $\ \sinh(w) + w = r\ $ numerically you may :
*
*use iterations (Newton-Raphson) : $\ \displaystyle w_{n+1}=w_n-\frac {\sinh(w_n)+w_n-r}{\cosh(w_n)+1}$
(starting with $w_0=\frac r2$)
*use reversion of series to get $w=w(r)$ :
$w(r)= \frac 12 r - \frac 1{96}r^3 + \frac 1{1920}r^5 - \frac{43}{1290240}r^7 + \frac {223}{92897280}r^9 - \frac{60623}{326998425600}r^{11} + \frac{764783}{51011754393600}r^{13} - \frac {107351407}{85699747381248000}r^{15} + \mathrm{O}\bigl(r^{17}\bigr)$
*perhaps that other methods exist in the litterature...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ without changing into polar coordinates? How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ other than changing into polar coordinates? It is possible to prove it using infinite series?
| $$ \int_{-\infty} ^\infty e^{-x^2} \; dx =I$$
$$ \int_{-\infty}^\infty e^{-y^2} \; dx =I$$
$$\int_{-\infty}^\infty e^{-x^2}\times e^{-y^2} \; dx =I^2$$
$$\int_{-\infty}^\infty e^{-(x^2+y^2)}\; dx =I^2$$
Polar Coordinates: $x^2+y^2=R^2,-\pi\leqslant \theta\leqslant +\pi$
$$\int_{-\pi}^{+\pi} \int_{0}^ \infty e^{-(R^2)}\; R\times dR \times d(\theta) =I^2$$
$$2\times \pi \int_{0}^ \infty e^{-(R^2)}\; R\times dR =I^2$$
$$-R^2=P\implies -2Rd(R)=d(P)\implies Rd(R)=-d(P)/2$$
$$2\times \pi \int_{-\infty}^ 0 e^{(P)}d(P)/2$$
$$\pi=I^2\Longrightarrow I=\sqrt{\pi}$$
since $e^{x^2}$ is even $\implies$ ans=${\sqrt{\pi}/2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Chess tournament, graph
Problem. In chess tournament each player, from all $n$ players, played one game with every another player. Prove that it is possible to number all players with numbers from $1$ to $n$ in such way that every player has unique number and none of them lost the game against player with number greater by $1$.
My first association was that this tournament can be represented by directed clique and for example edge $\langle a,b \rangle$ means that player with number $a$ won the game against player with number $b$. Then what I need to prove is that vertices in every directed clique with $n$ vertices can be numbered in such way that there in no edge $\langle i+1, \ i \rangle$ for all $i=1,2,...,n-1$.
But it seems very hard, I don't know how to approach.
Or maybe it isn't the problem from graph theory, I don't know.
| You can prove this by induction. The root is - according to taste - a game with just one or two players.
So let's assume we have found a numbering for $n$ players and along comes the $n+1^{st}$ player and plays against every other player.
If he looses against the $n^{th}$ player, we give him the number $n+1$. If he wins against the $n^{th}$ player but also against the $1^{st}$ player, we give hm the number $1$ and move everybody else one up.
So assume he wins against the $n^{th}$ player and looses against the $1^{st}$ player. Then there is a number $k$ such that he wins against player $k$ and looses against player $k-1$. Then we give him the number $k$ and move everybody with a number $>k$ one up.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Is any Mersenne number $M_p$ divisible by $p+2$? More precisely, does there exist a natural number $p$ such that $(2^p-1)/(p+2)$ is also a natural number? It seems to me that this is a really simple problem (with the answer "no"), but I couldn't find anything on the web. There are some facts known about division by $p+1$, but nothing useful for $p+2$.
| Another proof is as follows. First for $p=2$, it follows because $2^p-1=3<4=p+2$.
For odd primes $p$, there is a theorem (see Mersenne Prime for proof) that states that every factor of $2^p-1$ is of the form $2kp+1$ for some integer $k\geq0$. Thus we want a $p$ for which $p+2=2kp+1$. There is no such prime because for $p>2$, we have $1<p+2<2p+1$ thus requiring $0<k<1$, a contradiction to the theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Random Sequence Generator function I want to find out a function or algorithm, whichever is suitable, which can provide me a random sequence. Like
Input: 3
Output: {1,2,3} or {1,3,2} or {2,1,3} or {2,3,1} or {3,1,3} or {3,2,1}
Same as if I will enter a number N, output will be a random permutation of the set {1,2,...N}
How can a I write this type of algorithm. Actually I want to find out the logic behind it.
Edit: I don't want to use any buffer to to save anything.
| The first $O(n)$ shuffle or random permutation generator was published by Richard Durstenfeld in 1964. This algorithm came to the notice of programmers because it was included in Knuth's TAOCP, Vol 2, 1969, page 125, as Algorithm P. A succinct statement of the algorithm is:
\begin{equation}
\text{for }\ k \leftarrow n,n-1,\ldots,2\ \text{ do }
\begin{cases}
\text{Choose at random a number}\ \ r \in [1,k] \\ \text{Interchange }\pi[r] \text{ and }\pi[k],\
\end{cases}
\end{equation}
where $\pi[1\ldots n]$ is the array to be shuffled or randomly permuted.
See my notes for more information here:RPGLab
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove the boundedness of the solutions of a nonlinear differential equation I have the following differential equation:
$$
\ddot{x} = -\log(x) - 1
$$
and I need to prove that every solution of this equation is a bounded function. From the phase plane portrait, it is obvious that this is true:
How can I construct a formal proof for this?
| $x_1=x$, $x_2=\dot{x}$
$\dot{x}_1=x_2$
$\dot{x}_2=-\log(x_1)-1$
we know that to have the solution for the ODE we need $x_1>0$
consider the following function
$$V=(x_1\log(x_1)+0.5*x_2^2)$$
the derivative of this function along the system trajectories is
$$\dot{V}=x_2\log(x_1)+x_2+x_2(-\log(x_1)-1)=0$$
Therefore, the solutions are bounded!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Possible Duplicate:
Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
Summation of natural number set with power of $m$
How to get to the formula for the sum of squares of first n numbers?
how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Let,
$T_{2}(n)=1^2+2^2+3^2+\cdots+n^2$
$T_{2}(n)=(1^2+n^2)+(2^2+(n-1)^2)+\cdots$
$T_{2}(n)=((n+1)^2-2(1)(n))+((n+1)^2-2(2)(n-1))+\cdots$
| The claim is that
$\sum_{i = 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$
We will verify this by induction.
Clearly $n = 1$ holds.
Suppose the formula holds for $n$. Lets verify it holds for $n + 1$.
$$\sum_{i = 1}^{n + 1} i^2 = \sum_{i = 1}^n i^2 + (n + 1)^2 = \frac{n(n + 1)(2n + 1)}{6} + (n + 1)^2
\\
= \frac{n(n + 1)(2n + 1)}{6} + \frac{6(n^2 + 2n + 1)}{6}
\\
= \frac{2n^3 + 3n^2 + n}{6} + \frac{6n^2 + 12 n + 6}{6}
\\
= \frac{2n^3 + 9n^2 + 13n + 6}{6}$$
If you factor you get
$$= \frac{(n + 1)(n + 2)(2n + 3)}{6} = \frac{(n + 1)((n + 1) + 1)(2(n + 1) + 1)}{6}$$
The result follows for $n + 1$. So by induction the formula holds.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility
*
*$9^n$ $-$ $2^n$ is divisible by 7.
*$4^n$ $-$ $1$ is divisible by 3.
*$9^n$ $-$ $4^n$ is divisible by 5.
Can these be generalized as
$a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer?
But why is $a^n$ $-$ $b^n$$ = (a-b)N$ ?
I also see that $6^n$ $- 5n + 4$ is divisible by $5$ which is $6-5+4$ and $7^n$$+3n + 8$ is divisible by $9$ which is $7+3+8=18=9\cdot2$.
Are they just a coincidence or is there a theory behind?
Is it about modular arithmetic?
| Another proof:
Denote $r=b/a$. We know that the sum of a geometric progression of the type $1+r+r^2+\ldots+r^{n-1}$ is equal to $\frac{1-r^n}{1-r}$. Thus, we have
\begin{align}
1-r^n&=(1-r)(1+r+r^2+\ldots+r^{n-1}),\quad\text{substituting $r=b/a$ gives:}\\
a^n-b^n &= (a-b)\color{red}{(a^{n-1}+a^{n-2}b+\ldots+b^{n-1})}\\
a^n-b^n &= (a-b)N
\end{align}
The last step follows since $a,b$ are integers and a polynomial expression of the type in $\color{red}{red}$ font is also an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 8,
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Limsups of nets The limsup on sequences of extended real numbers is usually taken to be either of these two things, which are equivalent:
*
*the sup of all subsequential limits.
*The limit of the sup of the tail ends of the sequence.
For the situation with nets, the same arguments guarantee the existence of the above quantities 1. and 2, as long as you understand that a subnet of a net must be precomposed with an increasing function that is also cofinal. Also, one has that 1. $\leq$ 2., and I just can't see the reverse inequality. Don't forget that one cannot imitate the argument for sequences because the following fact fails:
Given a directed set, how do I construct a net that converges to $0$.
| As far as I can say, the more usual definition of limit superior of a net is the one using limit of suprema of tails:
$$\limsup x_d = \lim_{d\in D} \sup_{e\ge d} x_e = \inf_{d\in D} \sup_{e\ge d} x_e.$$
But you would get an equivalent definition, if you defined $\limsup x_d$ as the largest cluster point of the net. This definition corresponds (in a sense) to the definition with subsequential limits, since a real number is a cluster point of a net if and only if there is a subnet converging to this number.
I think that it is relatively easy to see that $\limsup x_d$ is a cluster point of the net $(x_d)_{d\in D}$.
To see that for every cluster point $x$ we have $x\le\limsup x_d$ it suffices to notice that, for any given $\varepsilon>0$ and $d\in D$, the interval $(x-\varepsilon,x+\varepsilon)$ must contain some element $x_e$ for $e\ge d$. Hence we get
$$
\begin{align*}
x-\varepsilon &\le \sup_{e\ge d} x_e\\
x-\varepsilon &\le \lim_{d\in D} \sup_{e\ge d} x_e.
\end{align*}$$
and, since $\varepsilon>0$ is arbitrary, we get
$$x\le \lim \sup_{e\ge d} x_e.$$
Thus the limit superior is indeed the maximal cluster point.
So the only thing missing is to show that cluster points are precisely the limits of subnets - this is a standard result, which you can find in many textbooks.
Some references for limit superior of a net are given in the Wikipedia article and in my answer here.
Perhaps some details given in my notes here can be useful, too. (The notes are still unfinished.) I should mention, that I pay more attention there to the notion of limit superior along a filter (you can find this in literature defined for filter base, which leads basically to the same thing). The limit superior of a net can be considered a special case, if we use the section filter; which is the filter generated by the base $\mathcal B(D)=\{D_a; a\in D\}$, where $D_a$ is the upper section $D_a=\{d\in D; d\ge a\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Fréchet derivative I have been doing some self study in multivariable calculus.
I get the geometric idea behind looking at a derivative as a linear function, but analytically how does one prove this?
I mean if $f'(c)$ is the derivative of a function between two Euclidean spaces at a point $c$ in the domain... then is it possible to explicitly prove that $$f'(c)[ah_1+ bh_2] = af'(c)[h_1] + bf'(c)[h_2]\ ?$$ I tried but I guess I am missing something simple.
Also how does the expression
$f(c+h)-f(c)=f'(c)h + o(\|h\|)$
translate into saying that $f'(c)$ is linear?
| It is not true that $f'(c_1+c_2)=f'(c_1)+f'(c_2)$ in general.
However, the derivative $f'(c)$ is the matrix of the differential $df_c$ and the expression which you write shows why $df_c(h) = f'(c)h$ is linear in the $h$-variable. It is simply matrix multiplication and all matrix multiplication maps are linear. The subtle question is if $f'(c)$ exists for a given $f$ and $c$.
Recall that $f: \mathbb{R} \rightarrow \mathbb{R}$ has a derivative $f'(a)$ at $x=a$ if
$$ f'(a) = \lim_{ h \rightarrow 0} \frac{f(a+h)-f(a)}{h} $$
Alternatively, we can express the condition above as
$$ \lim_{ h \rightarrow 0} \frac{f(a+h)-f(a)-f'(a)h}{h} =0.$$
This gives an implicit definition for $f'(a)$. To generalize this to higher dimensions we have to replace $h$ with its length $||h||$ since there is no way to divide by a vector in general. Recall that $v \in \mathbb{R}^n$ has $||v|| = \sqrt{v \cdot v}$. Consider $F: U \subseteq \mathbb{R}^m \rightarrow \mathbb{R}^n$ if $dF_{a}: \mathbb{R}^m \rightarrow \mathbb{R}^n$ is a linear transformation such that
$$ \lim_{ h \rightarrow 0} \frac{F(a+h)-F(a)-dF_a(h)}{||h||} =0 $$
then we say that $F$ is differentiable at $a$ with differential $dF_a$. The matrix of the linear transformation $dF_{a}: \mathbb{R}^m \rightarrow \mathbb{R}^n$ is called the Jacobian matrix $F'(a) \in \mathbb{R}^{m \times n}$ or simply the derivative of $F$ at $a$. It follows that the components of the Jacobian matrix are partial derivatives of the component functions of $F = (F_1,F_2, \dots , F_n)$
$$ J_F = \left[ \begin{array}{cccc}
\partial_1 F_1 & \partial_2 F_1 & \cdots & \partial_n F_1 \\
\partial_1 F_2 & \partial_2 F_2 & \cdots & \partial_n F_2 \\
\vdots & \vdots & \vdots & \vdots \\
\partial_1 F_m & \partial_2 F_m & \cdots & \partial_n F_m \\
\end{array} \right] = \bigl[\partial_1F \ | \ \partial_2F \ | \cdots | \ \partial_nF\ \bigr] =
\left[ \begin{array}{c}
(\nabla F_1)^T \\ \hline
(\nabla F_2)^T \\ \hline
\vdots \\ \hline
(\nabla F_m)^T
\end{array} \right].
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Module Theory for the Working Student Question:
What level of familiarity and comfort with modules should someone looking to work through Hatcher's Algebraic Topology possess?
Motivation:
I am taking my first graduate course in Algebraic Topology this coming October. The general outline of the course is as follows:
*
*Homotopy, homotopy invariance, mapping cones, mapping cylinders
*Fibrations, cofibrations, homotopy groups, long-exact sequences
*Classifying spaces of groups
*Freudenthal, Hurewicz, and Whitehead theorems
*Eilenberg-MacLane spaces and Postnikov towers.
I have spent a good deal of this summer trying to fortify and expand the foundations of my mathematical knowledge. In particular, I have been reviewing basic point-set and algebraic topology and a bit of abstract algebra. My knowledge of module theory is a bit lacking, though. I've only covered the basics of the following topics: submodules, algebras, torsion modules, quotient modules, module homomorphisms, finitely generated modules, direct sums, free modules, and a little bit about $\text{Hom}$ and exact sequences, so I have a working familirity with these ideas. As I do not have much time left before the beginning of the semester, I am trying to make my studying as economical as possible. So perhaps a more targeted question is:
What results and topics in module theory should every student in Algebraic Topology know?
| Looks like you'll definitely want to know the homological aspects: projective, injective and flat modules. There are lots of different characterizations for these which are useful to know. (Sorry, don't know the insides of D&F very well, perhaps it is covered.)
The Fundamental theorem of finitely generated modules over a principal ideal domain is a pretty good one to know too, since it kind of bundles up the entire theory of linear algebra.
(How much algebraic geometry do you know?)
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "12",
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Evaluating $\int_0^{\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx$ I need to solve
$$
\int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx
$$
I tried to use symmetric properties of the trigonometric functions as is commonly used to compute
$$
\int_0^{\Large\frac\pi2}\ln\sin x\ dx = -\frac{\pi}{2}\ln2
$$
but never succeeded. (see this for example)
| Rewrite the integral as
$$
\int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx=\int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{\sqrt{1-\sin^2 x}}}{\sin x}\cdot\cos x\ dx.
$$
Set $t=\sin x\ \color{red}{\Rightarrow}\ dt=\cos x\ dx$, then we obtain
\begin{align}
\int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx&=\frac12\int_0^1\frac{\ln t\ \ln(1-t^2)}{t}\ dt\\
&=-\frac12\int_0^1\ln t\sum_{n=1}^\infty\frac{t^{2n}}{nt}\ dt\tag1\\
&=-\frac12\sum_{n=1}^\infty\frac{1}{n}\int_0^1t^{2n-1}\ln t\ dt\\
&=\frac12\sum_{n=1}^\infty\frac{1}{n}\cdot\frac{1}{(2n)^2}\tag2\\
&=\large\color{blue}{\frac{\zeta(3)}{8}}.
\end{align}
Notes :
$[1]\ \ $Use Maclaurin series for natural logarithm: $\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}\ $ for $|x|<1$.
$[2]\ \ $$\displaystyle\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}\ $ for $ n=0,1,2,\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
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What am I losing if I decide to perform all math by computer? I solve mathematical problems every day, some by hand and some with the computer. I wonder: What will I lose if I start doing the mathematical problems only by computer? I've read this text and the author says that as techonology progress happens, we should change our focus to things that are more important. (You can see his suggestions on what is more important, in the suggested text.)
I'm in search of something a little more deep than the conventional cataclysmic hypothesis: "What will you do when a EMP hit us?!" and also considering an environment without exams which also exclude the "Computers are not allowed in exams" answer.
This is of crucial importance to me because as I'm building some mathematical knowledge, I want to have a safe house in the future. I'm also open to references on this, I've asked something similar before and got a few indications.
| Yes you can solve most of the problems by computer but you will lose your critical thinking ability, you are going to be an operator not a creator eventually!
| {
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"timestamp": "2023-03-29T00:00:00",
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Which separation axiom? Let $X$ be a topological space. Assume that for all $x_1,x_2 \in X$ there exist open neighbourhoods $U_i$ of $x_i$ such that $U_1 \cap U_2 = \emptyset$. Such a space, as we all know, is called Hausdorff. What would we call a space, and which separation axioms would the space satisfy, if $\overline{U_1} \cap \overline{U_2} = \emptyset$ for all $x_i \in X$?
| Such a space is known as $T_{2\frac{1}{2}}$ or Urysohn according to Wikipedia.
| {
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Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$ I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can be found in that pdf but I will write here to be more explicitly.
Exercise 1.3.4(a) Let $a,b,c$ be positive real numbers. Prove that
$$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4.$$
(b) For real numbers $a,b,c \gt0$ and $n \leq3$ prove that:
$$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$
| Write $$\frac ab+\frac ab+\frac bc\geq \frac{3a}{\sqrt[3]{abc}}$$ by AM-GM.
You get $$\operatorname{LHS} \geq \frac{a+b+c}{\sqrt[3]{abc}}+n\left(\frac{\sqrt[3]{abc}}{a+b+c}\right).$$
Set $$z:=\frac{a+b+c}{\sqrt[3]{abc}}$$ and then notice that for $n\leq 3$, $$z+\frac{3n}{z}\geq 3+n.$$ Indeed the minimum is reached for $z=\sqrt{3n}\leq 3$; since $z\geq 3$, the minimum is reached in fact for $z=3$.
| {
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How to construct a one-to one correspondence between$\left [ 0,1 \right ]\bigcup \left [ 2,3 \right ]\bigcup ..$ and $\left [ 0,1 \right ]$ How can I construct a one-to one correspondence between the Set $\left [ 0,1 \right ]\bigcup \left [ 2,3 \right ]\bigcup\left [ 4,5 \right ] ... $ and the set $\left [ 0,1 \right ]$ I know that they have the same cardinality
| For $x\in{(k,k+1)}$ with $k\geq 2$ and $k$ even define $f(x)=\frac{1}{2x-k}$; for $x\in(0,1]$ define $f(x)=\frac{x+1}{2}$; set $f(0)=0$. Now it remains to map $A=\{2,3,4,5,..\}$ bijectively to $\{1/2,1/4,1/6,1/8..,\}$ to do this define $f(x)=\frac{1}{2(x-1)}$ on $A$. This should give you the desired bijection. Please, let me know if I have made any silly error this time.
| {
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Closed form representation of an irrational number Can an arbitrary non-terminating and non-repeating decimal be represented in any other way? For example if I construct such a number like 0.1 01 001 0001 ... (which is irrational by definition), can it be represented in a closed form using algebraic operators? Can it have any other representation for that matter?
| Since $0.1 = \frac{1}{10}$, $0.001 = \frac{1}{10^3}$, $0.0000001 = \frac{1}{10^6}$. Making a guess that $n$-th term is $10^{-n(n+1)/2}$ the sum, representing the irrational number becomes
$$
0.1010010001\ldots = \sum_{k=0}^\infty \frac{1}{10^{\frac{k(k+1)}{2}}} = \left.\frac{1}{2 q^{1/4}} \theta_2\left(0, q\right)-1\right|_{q=\frac{1}{\sqrt{10}}}
$$
where $\theta_2(u,q) =2 q^{1/4} \sum_{n=0}^\infty q^{n(n+1)} \cos((2n+1)u)$ is the elliptic theta function.
| {
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Prime elements in $\mathbb{Z}/n\mathbb{Z}$ I'm tring to determine the prime elements in the ring $\mathbb{Z}/n\mathbb{Z}$.
| Every ideal of $\mathbb{Z}/n\mathbb{Z}$ is of the form $m\mathbb{Z}/n\mathbb{Z}$, where $m$ is a divisor of $n$. And its residue ring is isomorphic to $\mathbb{Z}/m\mathbb{Z}$.
Hence a prime ideal of $\mathbb{Z}/n\mathbb{Z}$ is of the form $p\mathbb{Z}/n\mathbb{Z}$, where $p$ is a prime divisor of $n$.
Hence every prime element of $\mathbb{Z}/n\mathbb{Z}$ is of the form $p$ (mod $n$), where $p$ is a prime divisor of $n$.
| {
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Future Lifetime Distribution Suppose the future lifetime of someone aged $20$ denoted by $T_{20}$ is subject to the force of mortality $\mu_x = \frac{1}{100-x}$ for $x< 100$. What is $\text{Var}[\min(T_{20},50)]$?
So we have: $$E[\min(T_x,50)|T_{20} > 50] = 50$$ $$\text{Var}[\min(T_{20},50)|T_{20} > 50] = ?$$ $$E[\min(T_x,50)|T_{20} \leq 50] = 25$$ $$\text{Var}[\min(T_{20},50)|T_{20} \leq 50] = 50^{2}/12$$
What is the second line?
Also I know that we use the "expectation-variance" formula to calculate $\text{Var}[\min(T_{20},50)]$.
| Hint: The random variable $\min(T_{20},50)|T_{20} > 50)$ doesn't vary much!
If you then want to use a formula, call the above random variable $Y$. We want $E((Y-50)^2)$. How did you decide earlier that $E(Y)=50$?
| {
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What branch of the Math can help me with this? I would love to focus on the branches of the Math that can help me with:
*
*generation of entropy, i suppose that most of the works are based on statistic since even a big part of the cryptographic world starts from this and is tested with the help of the statistic. But i do not wont to generate confusion, with entropy i mean to fake randomness
*creating structures programmatically and via a procedural way, like the classic Voronoi for example, but in N dimensions and with boundaries, for example generating a structure, a building, a road, an entire city, you get the point.
What is the big topic of the Math that can help me with this?
| 1) Dynamical systems, more precisely: Ergodic theory. See Introduction to ergodic theory by Ya. G. Sinai or Ya. B. Pesin's books.
2) Fractal geometry. You can refer to Fractal Geometry by Kenneth Falconer.
You can also see Chaos and Fractals: New Frontiers of Science (more elementary) by Heinz-Otto Peitgen, Hartmut Jürgens and Dietmar Saupe.
For programming, see here, here or here.
Notice that these two branches are strongly linked.
| {
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Intuition behind gradient VS curvature In Newton's method, one computes the gradient of a cost function, (the 'slope') as well as its hessian matrix, (ie, second derivative of the cost function, or 'curvature'). I understand the intuition, that the less 'curved' the cost landscape is at some specific weight, the bigger the step we want to take on the landscape terrain. (This is why it is somewhat superior to simply gradient ascent/descent).
Here is the equation:
$$
\mathbf {w_{n+1} = w_n - \alpha\frac{\delta J(w)}{\delta w}\begin{bmatrix}\frac{\delta^2J(w)}{\delta w^2}\end{bmatrix}^{-1}}
$$
What I am having a hard time visualizing, is what is the intuition behind 'curvature'? I get that is the 'curviness of a function', but isnt that the slope? Obviously it is not, that is what the gradient measures, so what is the intuition behind curvature?
Thanks!
| Intuitively, curvature is how fast the slope is changing: a greater rate of change of the slope means it is more curved.
So it is related to the derivative of the slope, i.e. the derivative of the derivative or the second derivative.
| {
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Proving that one quantity *eventually* surpasses the other I want to prove the following; $$\forall t>0\ \ \forall m\in \mathbb{N} \ \ \exists N \in \mathbb{N} \ \ \forall n\geq N: \ (1+t)^n > n^m.$$
For readers who hate quantifiers, here's the version in words: "$(1+t)^n$ eventually surpasses $n^m$ for some $t>0$ and $m\in \mathbb{N}$".
Though this sounds simple enough, I couldn't manage to prove it using only very elementary statements about the real numbers, like the archimedian property etc. (so no derivates and so on involved).
My questions are:
1) Is there a simple (as described above) proof for this ? (If there isn't, I would also be happy with a proof using more of the analysis machinery.)
2) Is there a way to express the least $N$, which satisfies the above, in a closed form ?
3) I have somewhere heard of a theorem, that two real convex functions can have at most two intersection points (and the above statement seems closely related to this theorem), so I would be very happy, if someone could also give me a reference for this theorem.
| 1) Yes. The inequality is equivalent to $((1+t)^{\frac{1}{m}})^n>n$ by taking the $\frac{1}{m}$th power of both sides which reduces to your original problem to the problem with $m=1$ since $(1+t)^{\frac{1}{m}}$ is of the form $1+u$ with $u$ a strictly positive real.
So we'll assume $m=1$. We can expand $(1+t)^n$ using the binomial theorem as (assuming $n>2$)
$$(1+t)^n=1+nt+\frac{n(n-1)}{2}t^2+...$$
So in particular we get $(1+t)^n>1+\frac{n(n-1)}{2}t^2$ if $n>2$, so if we pick $N$ such that $\frac{N-1}{2}t^2>1$ and $N>2$ (which we can do by the Archimedean property), we find that $(1+t)^n>n$ for all $n\geq N$.
2) The Lambert W function should be able to do that, yes.
| {
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Does $(\mathbf A+\epsilon \mathbf I)^{-1}$ always exist? Why? Does $(\mathbf A+\epsilon \mathbf I)^{-1}$ always exist, given that $\mathbf A$ is a square and positive (and possibly singular) matrix and $\epsilon$ is a small positive number? I want to use this to regularize a sample covariance matrix ($\mathbf A = \Sigma$) in practice, so that I can compute the inverse, which I need to calculate a Mahalanobis distance between two samples. In practice, my covariance matrix is often singular. I know the term $(\mathbf A+\epsilon \mathbf I)^{-1}$ often appears in the context of least squares problems involving Tikhonov regularization (ridge regression). However, I've never seen a statement, proof, or reference which says that the expression is always invertible.
Can any of you help me with a proof or reference?
| Yes, if $\mathbf A$ is any $n \times n$ matrix, then $\mathbf A+\epsilon \mathbf I$ is invertible for sufficiently small $\epsilon > 0$. This is because $\det (\mathbf A + \epsilon \mathbf I)$ is a polynomial in $\epsilon$ of degree $n$, and so it has a finite number of zeroes.
| {
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Physical meaning of spline interpolation I remember that when I took my Numerical Analysis class, the professor said the spline interpolation take its name from a kind of wood sticks used to draw curved lines. Also Wikipedia say that the name is due to those elastic rulers:
Elastic rulers that were bent to pass through a number of predefined
points (the "knots") were used for making technical drawings for
shipbuilding and construction by hand
I now wonder if the spline are only inspired to those rulers or if they moreover precisely follow the physical laws that govern the bending of particular wood sticks (and if it is not the case, I ask if anyone know of any alternative physical interpretation of them).
| The equations of cubic splines are derived from the physical laws that govern bending of thin beams. For example, see http://stem2.org/je/cs.pdf.
The spline equation is an approximate solution of the minimum energy bending equation, valid when the amount of bending is small.
Generally, in computer-aided geometric design, minimising some sort of "energy" function is often used as a way to smooth the shape of a curve or surface.
| {
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What's the meaning of $C$-embedded? What's the meaning of $C$-embedded? It is a topological notion. Thanks ahead.
| A set $A \subset X$ ($X$ is a topological space) is $C$-embedded in $X$ iff every real-valued continuous function $f$ defined on $A$ has a continuous extension $g$ from $X$ to $\mathbb{R}$ (so $g(x) = f(x)$ for all $x \in A$).
A related notion of $C^{\ast}$-embedded exist where continuous real-valued functions are (in both cases) replaced by bounded real-valued continuous functions.
The Tietze theorem basically says that a closed subset $A$ of a normal space $X$ is $C$ and $C^{\ast}$-embedded in it.
| {
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solving for a coefficent term of factored polynomial.
Given: the coefficent of $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$ is $48,$ find the value of the constant $a.$
I expanded it and got
$64-64\,x-144\,{x}^{2}+320\,{x}^{3}-260\,{x}^{4}+108\,{x}^{5}-23\,{x}^{
6}+2\,{x}^{7}+64\,a{x}^{2}-192\,a{x}^{3}+240\,a{x}^{4}-160\,a{x}^{5}+
60\,a{x}^{6}-12\,a{x}^{7}+a{x}^{8}
$
because of the given info $48x^2=64x^2-144x^2$ solve for $a,$ $a=3$.
Correct?
P.S. is there an easier method other than expanding the terms?
I have tried using the bionomal expansion; however, one needs still to multiply the terms. Expand $(2-x)^6$ which is not very fast.
| It would be much easier to just compute the coefficient at $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$. You can begin by computing:
$$ (2-x)^6
= 64 - 6 \cdot 2^5 x + 15 \cdot 2^4 x^2 + x^3 \cdot (...)
= 64 - 192 x + 240 x^2 + x^3 \cdot (...) $$
Now, multiply this by $(1+2x+ax^2)$. Again, you're only interested in the term at $x^2$, so you can spare yourself much effort by just computing this coefficient: to get $x^2$ in the product, you need to take
$64, \ -192 x, \ 240 x^2$ from the first polynomial, and
$ax^2,\ 2x, 1$ from the second one (respectively).
$$(1+2x+ax^2)(2-x)^6 = (...)\cdot 1 + (...) \cdot x
+ (64a - 2\cdot 192 + 240 )\cdot x^2
+ x^3 \cdot(...) $$
Now, you get the equation:
$$ 64a - 2\cdot 192 + 240 = 48 $$
whose solution is indeed $a = 3$.
As an afterthought: there is another solution, although it might be an overkill. Use that the term at $x^2$ in polynomial $p$ is $p''(0)/2$. Your polynomial is:
$$ p(x) = (1+2x+ax^2)(2-x)^6$$
so you can compute easily enough:
$$ p'(x) = (2+2ax)(2-x)^6 + 6(1+2x+ax^2)(2-x)^5 $$
and then:
$$ p''(x) = 2a(2-x)^6 + 2 \cdot 6(2+2ax)(2-x)^5 + 30(1+2x+ax^2)(2-x)^4 $$
You can now plug in $x=0$:
$$ p''(0) = 2a \cdot 2^6 + 2 \cdot 6 \cdot 2 \cdot 2^5 + 30 \cdot 2^4 $$
On the other hand, you have
$$p''(0) = 2 \cdot 48$$
These two formulas for $p''(0)$ let you write down an equation for $a$.
| {
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Image of a morphism According to Wikipedia, image of a morphism $\phi:X\rightarrow Y$ in a category is a monomorphism $i:I\rightarrow Y$ satisfying the following conditions:
*
*There is a morphism $\alpha:X\rightarrow I$ such that $i\circ\alpha=\phi$.
*If $j:J\rightarrow Y$ is a monomorphism and $\beta:X\rightarrow J$ is a morphism such that $\beta\circ j=\phi$, then there exists a unique morphism $\gamma:I\rightarrow J$ such that $\beta=\gamma\circ\alpha$ and $j\circ\gamma=i$.
It is easy to see that $\alpha$ is unique. Intuitively, such $\alpha$ should be an epimorphism, but I can't seem to show it. Is it true?
| The term "image" suggests that this concept is modeled on the image of a map, a morphism in the category of sets. In that case, $I$ can be any set equipotent with the image (in the conventional sense) of $\phi$, and $\alpha$ is generally neither unique, nor an epimorphism (a surjective map). For uniqueness, note that you can compose $\alpha$ with any permutation of $I$ to obtain another suitable $\alpha$ for a given $I$.
| {
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How to check if a point is inside a rectangle? There is a point $(x,y)$, and a rectangle $a(x_1,y_1),b(x_2,y_2),c(x_3,y_3),d(x_4,y_4)$, how can one check if the point inside the rectangle?
| Given how much attention this post has gotten and how long ago it was asked, I'm surprised that no one here mentioned the following method.
A rectangle is the image of the unit square under an affine map. Simply apply the inverse of this affine map to the point in question, and then check if the result is in the unit square or not.
To make things clear, consider the following image,
where the vectors in the pictures are $\mathbf{u} = c - d$, $\mathbf{v} = a - d$, and $\mathbf{w} = d$.
Since the legs of a rectangle are perpendicular, the matrix $\begin{bmatrix}\mathbf{u} & \mathbf{v}\end{bmatrix}$ is orthogonal and so we even have a simple formula for the inverse:
$$\begin{bmatrix}\mathbf{u} & \mathbf{v}\end{bmatrix}^{-1} = \begin{bmatrix}\mathbf{u}^T/||u||^2 \\ \mathbf{v}^T/||v||^2\end{bmatrix}.$$
If you want to check many points for the same rectangle, this matrix can be easily precomputed and stored, so that you perform the (typically more expensive) division operations only once at the start. Then you only need to do a few multiplications, additions, and subtractions for each point you are testing.
This method also applies more generally to checking whether a point is in a parallelogram, though in the parallelogram case the matrix inverse does not take such a simple form.
| {
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Determining whether or not spaces are separable I've been going over practice problems, and I ran into this one. I was wondering if anyone could help me out with the following problem.
Let $X$ be a metric space of all bounded sequences $(a_n) \subset \mathbb{R}$ with the metric defined by
$$d( (a_n), (b_n)) = \sup\{ |a_n - b_n| : n = 1, 2, \ldots \}.$$
Let $Y \subset X$ be the subspace of all sequences converging to zero.
Determine whether or not $X$ and $Y$ are separable.
Thanks in advance!
| Suppose that $A=\{\alpha_n:n\in\Bbb N\}$ is a countable subset of $X$, where $\alpha_n$ is the sequence $\langle a_{n,k}:k\in\Bbb N\rangle$. Note that for any $x\in\Bbb R$ there is always a $y\in[-1,1]$ such that $|x-y|\ge 1$. Thus, we can construct a sequence $\beta=\langle b_k:k\in\Bbb N\rangle$ such that $b_k\in[-1,1]$ and $|b_k-a_{k,k}|\ge 1$ for each $k\in\Bbb N$. Clearly $\beta\in X$, but for each $n\in\Bbb N$ we have $\sup_{k\in\Bbb N}|b_k-a_{n,k}|\ge|b_n-a_{n,n}|\ge 1$, so the distance between $\beta$ and $\alpha_n$ is at least $1$. Thus, $A$ is not dense in $X$, and $X$ is not separable.
$Y$, on the other hand, is separable. Let $D$ be the set of sequences of rational numbers that are $0$ from some point on, i.e., that have only finitely many non-zero terms. Show that $D$ is both countable and dense in $Y$. For the latter, start with any $\langle a_k:k\in\Bbb N\rangle\in Y$ and any $\epsilon>0$ and construct an element of $D$ that is less than $\epsilon$ away from $\langle a_k:k\in\Bbb N\rangle\in Y$.
| {
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Integer solutions of $p^2 + xp - 6y = \pm1$ Given a prime $p$, how can we find positive integer solutions $(x,y)$ of the equation:
$$p^2 + xp - 6y = \pm1$$
| If $p=2$ or $p=3$, you can't.
Otherwise, the extended Euclid algorithm produces a solution $(x_0,y_0)$ of $x_0 p - 6 y_0 = \pm 1$ (in fact, one for $+1$ and one for $-1$).
Then $x=x_0+6k-p$ and $y=y_0+pk$ is a solution of $p^2+xp-6y=\pm 1$. Since both $x$ and $y$ grow with $k$, we find infinitely many positive solutions, one for eack $k\ge\max\{\lfloor {p-x_0\over6}\rfloor, \lfloor {-y_0\over p}\rfloor\}+1$.
| {
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Are these two quotient rings of $\Bbb Z[x]$ isomorphic? Are the rings $\mathbb{Z}[x]/(x^2+7)$ and $\mathbb{Z}[x]/(2x^2+7)$ isomorphic?
Attempted Solution:
My guess is that they are not isomorphic. I am having trouble demonstrating this. Any hints, as to how i should approach this?
| Suppose they are isomorphic. Then
$\left(\mathbb{Z}[x]/(x^2+7)\right)/(2) \cong \left(\mathbb{Z}[x]/(2x^2+7)\right)/(2)$. Ravi helpfully pointed out that considering ideals in either ring in terms of $x$ will often give us different ideals, but we do not suffer from this problem when using ideals generated by integers such as $(2)$.
Letting overlines represent the canonical homomorphism from $\mathbb{Z}[x]$ to $\mathbb{F}_2[x]$, applying the correspondence theorem for rings on the left side tells us that $$\left(\mathbb{Z}[x]/(x^2+7)\right)/(2) \cong \left(\mathbb{Z}[x]/(2)\right)/(\overline{x^2 + 7}) \cong \mathbb{F}_2[x]/(x^2 + 1)\,.$$
Similarly, on the right, we get
$$\left(\mathbb{Z}[x]/(2x^2+7)\right)/(2) \cong \left(\mathbb{Z}[x]/(2)\right)/(\overline{2x^2 + 7}) \cong \mathbb{F}_2[x]/(1) = \{0\}\,.$$
One can verify the former ring contains four distinct elements. The latter is trivial, which is absurd.
EDIT: To further bolster intuition for an approach like this, see Bill's mention of it in his first comment on his own post.
| {
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Find $\lim\limits_{n\to+\infty}(u_n\sqrt{n})$ Let ${u_n}$ be a sequence defined by $u_o=a \in [0,2), u_n=\frac{u_{n-1}^2-1}{n} $ for all $n \in \mathbb N^*$
Find $\lim\limits_{n\to+\infty}{(u_n\sqrt{n})}$
I try with Cesaro, find $\lim\limits_{n\to+\infty}(\frac{1}{u_n^2}-\frac{1}{u_{n-1}^2})$ then we get $\lim\limits_{n\to+\infty}(u_n^2n)$
But I can't find $\lim\limits_{n\to+\infty}(\frac{1}{u_n^2}-\frac{1}{u_{n-1}^2})$
| If ever $u_N\le 0$, then all $-1/n\le u_n\le0$ for all $n>N$, hence $u_n\sqrt n\to 0$.
Therefore we may assume for the rest of the argument that $u_n>0$ for all $n$.
Let $e_n = n+2-u_n$. Then $0<e_0<2$. Using the recursion formula for $e_n$ show that the assumption that $e_n\le2$ for all $n$ leads to $e_n\ge2^n e_0$. Therefore $e_n>2$ for some $n$, i.e. $u_n<n$ for some $n$.
Let $q_n = {u_n\over n}$ for $n\ge 1$. We have seen that $0<q_n<1$ for big $n$. Find the recursion formula for $q_n$ and show that $q_n< q_{n-1}^2$ for big $n$ and therefore $q_n<\frac1n$ for some $n$. But then $u_{n+1}<0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How to prove that the Torus and a Ball have the same Cardinality How to prove that the Torus and a Ball have the same Cardinality ?
The proof is using the Cantor Berenstein Theorem.
I now that they are subsets of $\mathbb{R}^{3}$ so I can write $\leq \aleph$ but I do not know how to prove $\geqslant \aleph$.
Thanks
| Hint:
Show that a circle is equipotent with $[0,2\pi)$ by fixing a base point, and sending each point on the circle to its angle, where $0$ is the base point.
Since both the torus and the ball contain a circle, both have at least $\aleph$ many elements.
| {
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If $G$ is a finite group and $g \in G$, then $O(\langle g\rangle)$ is a divisor of $O(G)$ Does this result mean:
*
*Given any finite group, if we are able to find a cyclic group out of it (subgroup), then the order of the cyclic group will be a divisor of the original group.
If I am right in interpreting it, can one suggest an example of highlighting this? And also make me understand the possible practical uses of this result. It surely looks interesting
Thanks
Soham
| You have a finite group $G$ and you take any element $g\in G$. Then $\langle g \rangle$ is a subgroup of $G$. Then, as mentioned in the comment by anon, you can apply Lagrange's theorem to get the conclusion that you want.
As an example of this, you could consider the symmetric group $S_5$. You pick a random element $\sigma \in S_5$, for example $\sigma = (1, 2, 4)$. Then you get the subgroup
$$
\langle \sigma\rangle = \{(1,2,4), (1, 4, 2), (1) \}.
$$
Hence the order of $\langle \sigma\rangle$ is $3$, and indeed 3 is a divisor in $O(S_5) = 5! = 120$.
You ask in the comment above about an example with a subgroup of the complex numbers. Consider $z = e^{\frac{2\pi i}{15}}$. Then you have the group $G = \langle z\rangle$ (under multiplication). This group has order $15$. Can you find/write down the elements?)Now take $w= e^{\frac{2\pi i}{5}}$. Then $\langle w\rangle$ is a subgroup of $G$ of order ... ( I will let you think about that).
As an application of this someone else might have something helpful to say.
| {
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Is $x^y$ - $a^b$ divisible by $z$, where $y$ is large? The exact problem I'm looking at is:
Is $4^{1536} - 9^{4824}$ divisible by $35$?
But in general, how do you determine divisibility if the exponents are large?
| You can use binomial theorem to break up the individual bases into multiples of the divisor and then you can expand binomially to check divisibilty. This is in a general case of doing such a problem . There may be more methods of doing such a problem.
| {
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"source": "stackexchange",
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Every point closed $\stackrel{?}{\Rightarrow}$ space is Hausdorff If a topological space is Hausdorff, then every point is closed.
Is the converse true?
Edited: Let $G$ be a topological group and $H$ the intersection of all
neighborhoods of zero. Since every coset of $H$ is closed, every point
of $G/H$ will be closed. Why does that make $G/H$ Hausdorff?
| Notice that $H$ is a closed normal subgroup of $G$, for that see e.g. this for proof that it is a closed subgroup (equal to $\operatorname{cl} \{e\}$), and for normality just notice that conjugation preserves the neighbourhoods of identity (as a set), so it does preserve intersection as well.
From that we see that $G/H$ is a topological group.
It is a known fact that for topological groups, $T_0$ implies completely regular Hausdorff. Every point being closed is equivalent to $T_1$, from which $T_{3\frac {1}{2}}$, so in particular $T_2$, follows.
A proof can be found in many places, e.g. Engelking's General Topology iirc.
A short one for closed $\{e\}\implies T_2$: notice that Hausdorffness is equivalent to the diagonal being closed. But the diagonal is the preimage of identity by the map $(x,y)\mapsto xy^{-1}$.
In general, we do not have the implication, as shown by e.g. the cofinite topology on an infinite space.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Finding disjoint neighborhoods of two points in $\Bbb R$ Let $x$ and $y$ be unique real numbers. How do you prove that there exists a neighborhood $P$ of $x$ and a neighborhood $Q$ of $y$ such that $P \cap Q = \emptyset$?
| Hint: open intervals of length $e$ centered at $x$ and $y$ are such neighborhoods, and if $e$ is small enough, they will not intersect. How small does $e$ have to be for this to happen?
| {
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find the eigenvalues of a linear operator T Let $A$ be $m*m$ and B be $n*n$ complex matrices, and consider the linear operator $T$ on the space $C^{m*n}$ of all $m*n$ complex matrices defined by $T(M) = AMB$.
-Show how to construct an eigenvector for $T$ out of a pair of column vectors $X, Y$, where $X$ is an eigenvector for $A$ and $Y$ is an eigenvector for $B^t$.
-Determine the eigenvalues of $T$ in terms of those of $A$ and $B$
-Determine the trace of this operator
| There is a standard way to do this kind of exercise. Firstly assume that $ A$ and $ B$ are diagonal. Then a short calculation shows that the eigenvalues are as given in previous solutions, i.e., the pairwise products of those of these matrices. The result then holds for diagonalisable matrices by a suitable choice of bases. The easiest way to get the final version is to use the fact that the diagonalisable matrices are dense in all matrices and employ a continuity argument involving the characteristic polynomials.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's the answer of (10+13) ≡? As to Modulus operation I only have seen this form:
(x + y) mod z ≡ K
So I can't understand the question, by the way the answers are :
a) 8 (mod 12)
b) 9 (mod 12)
c) 10 (mod 12)
d) 11 (mod 12)
e) None of the above
| The OP may be taking a Computer Science course, in which if $b$ is a positive integer, then $a\bmod{b}$ is the remainder when $a$ is divided by $b$. In that case $\bmod$ is a binary operator. That is different from the $x\equiv y\pmod{m}$ of number theory, which is a ternary relation (or, for fixed $m$, a binary relation).
Calculating $(10+13)\bmod{12}$ is straightforward. Find $10+13$, and calculate the remainder on division by $12$. We get $(10+13)\bmod{12}=11$. So $(10+13)\bmod{12} \equiv 11\pmod{12}$.
Remark: It seems unusual to use the binary operator and the ternary relation in a single short expression.
| {
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Determinant of matrices along a line between two given matrices The question, with no simplifications or motivation:
Let $A$ and $B$ be square matrices of the same size (with real or complex coefficients). What is the most reasonable formula one can find for the determinant $$\det((1-t)A + tB)$$ as a function of $t \in [0,1]$? If no reasonable formula exists, what can we say about these determinants?
So we're taking a line between two matrices $A$ and $B$, and computing the determinant along this line. When $A$ and $B$ are diagonal, say $$A = \operatorname{diag}(a_1,\ldots,a_n), B = \operatorname{diag}(b_1,\ldots,b_n),$$ then we can compute this directly:
$$\begin{aligned}
\det((1-t)A + tB) &= \det \operatorname{diag}((1-t)a_1 + tb_1, \ldots, (1-t)a_n + tb_n) \\
&= \prod_{j=1}^n ((1-t)a_j + tb_j).
\end{aligned}$$
I'm not sure if this can be further simplified, but I'm sure someone can push things at least a tiny bit further than I have.
I'm most curious about the case where $A = I$ and each $(1-t)A + tB$ is assumed to be invertible. Here's what I know in this case: writing
$$D(t) = \det((1-t)I - tB),$$
we can compute that
$$ \dot{D}(t) = D(t) c(t)$$
where
$$c(t) := \operatorname{trace}(((1-t)I + tB)^{-1}(B-I))$$
(a warning: I am not 100% sure this formula holds). Thus we can write
$$D(t) = \exp\left(\int_0^t c(\tau) \; d\tau\right)$$
since $D(0) = 1$.
I have no idea how to deal with the function $c(\tau)$ though. Any tips?
| I think I have an answer to the last case I mentioned ($A=I$, all $(1-t)I + tB$ invertible). The key is to write
$$\begin{aligned}
\int_0^t c(\tau) \; d\tau &= \operatorname{trace} \int_0^t ((1-\tau)I + \tau B)^{-1} (B-I) \; d\tau \\
&= \operatorname{trace} \log ((1-t)I + tB)
\end{aligned}
$$
using that $\frac{d}{dt}((1-t)I + tB) = B-I$ and $\frac{d}{dt}\log(A(t)) = A(t)^{-1} \frac{d}{dt} A'(t)$ (I think this is true!). Taking matrix logarithms here should be ok, since everything in sight is invertible. Then
$$
D(t) = e^{\operatorname{trace} \log((1-t)I + tB)}.
$$
Of course, this leaves one with the problem of computing a matrix logarithm, but in the case where $B$ is diagonal, this reduces to the first formula in my question. This could probably be generalised to the case of general invertible $A$ by computing $\dot{D}(t)$ as before. I'll do this tomorrow and make an edit with the results (or you can do it!)
| {
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Proof of the Hardy-Littlewood Tauberian theorem Can someone point me to a proof of the Hardy-Littlewood Tauberian theorem, that is suitable enough to be shown to high school students? (with knowledge of calculus, sequences and series of course)
| Have you looked at the presentation in Titchmarsh's Theory of Functions (Section 7.5)? The only non-elementary part of the argument is Weierstrass's approximation theorem, which you can probably assume as a fact. The preliminary material given also include an "easy" special case where the exposition certainly can be understood by someone with knowledge of calculus, sequences, and series.
| {
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Evaluating $ \lim\limits_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $ How would you evaluate the following series?
$$\lim_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $$
Thanks.
| We have the following important theorem,
Theorem: Let for the monotonic function $f$ ,$\int_{0}^\infty f(x)dx$ exists and we have $\lim_{x\to\infty}f(x)=0$ and $f(x)>0$ then
we have
$$\lim_{h\to0^+}h\sum_{v=0}^\infty f(vh)=\int_{0}^\infty f(x)dx$$
It is enough to take $h^{-1}=t$ and $f(x)=\frac{2}{1+x^2}$, then we get the desired result.
So we showed that
$$\lim_{t\to\infty}\left(\frac{2}{t}+\frac{2t}{t^2+2^2}+\cdots+\frac{2t}{t^2+n^2}+\cdots\right)=\pi$$
Let me show some additional infinite sum by this theorem
We show
$$\lim_{t\to
1^{-}}(1-t)^2\left(\frac{t}{1-t}+\frac{2t^2}{1-t^2}+\frac{3t^2}{1-t^2}+\cdots
\frac{nt^2}{1-t^2}+\cdots\right)=\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$
Now in previous theorem if we take $e^{-h}=t$ and $f(x)=\frac{xe^{-x}}{1-e^{-x}}$ then since
$$\int_0^\infty\frac{xe^{-x}}{1-e^{-x}}=\int_0^\infty x(\sum_{n=1}^\infty e^{-nx})dx=\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$
The previous theorem gives us an important relation of infinite sum about Euler constant
We show
$$\lim_{t\to
1^{-}}(1-t)\left(\frac{t}{1-t}+\frac{t^2}{1-t^2}+\frac{t^3}{1-t^3}+\cdots \frac{t^n}{1-t^n}+\cdots\right)-\log\frac{1}{1-t}=C=\text{Euler
constant}$$
By using the previous theorem we just need to take $f(x)=\frac{e^{-x}}{1+e^{-x}}$ and $e^{-h}=t$ and since it is known that $$\int_0^\infty e^{-x}\left(\frac{1}{1-e^{-x}}-\frac{1}{x}\right)dx=C=\text{Euler constant}$$
See this book for more examples of G.polya, problems and theorems in analysis , vol I Springer
| {
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is it true that the infinity norm can be bounded using the $L_2$ norm the following way? Let $v \in \mathbb{R}^k$, and let $A \in \mathbb{R}^{m \times k}$ and let $B \in \mathbb{R}^{m \times n}$ such that each column of $B$, $B_i$, has $$||B_i||_2 \le 1.$$
Is it true that:
*
*$||v A^{\top} B||_{\infty} \le ||v A^{\top}||_2$ ?
*If the spectral norm of $A$ is such that $||A||_{\mathrm{spectral}} \le 1$, is it true that $||v A^{\top}||_2 \le ||v||_2$ ?
Thanks.
| 1) Cauchy-Schwarz says $$|(v A^T B)_i| = |v A^T B_i| \le \|v A^T\|_2 \|B_i\|_2 \le \|v A^T\|_2$$
2) Yes because the spectral norm is the operator norm corresponding to the $2$-norm on vectors, and $\|A\|_{\text{spectral}} = \|A^T\|_{\text{spectral}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Evaluate $\int\frac{dx}{\sin(x+a)\sin(x+b)}$ Please help me evaluate:
$$
\int\frac{dx}{\sin(x+a)\sin(x+b)}
$$
| The given integral is: $$\int\frac{dx}{\sin(x+a)\sin(x+b)}$$
The given integral can write:
$$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}$$
We substition
$$\frac{\sin(x+a)}{\sin(x+b)}=t$$
By the substition of the above have:
$$\frac{dx}{\sin^2(x+a)}=\frac{dt}{\sin(a-b)}$$
Now have:
$$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}=\frac{1}{\sin(a-b)}\int\frac{dt}{t}=\frac{1}{\sin(a-b)}\ln |t|=\frac{1}{\sin(a-b)}\ln|\frac{\sin(x+a)}{\sin(x+b)}|+C$$
| {
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calculate $\int_{-\infty}^{+\infty} \cos(at) e^{-bt^2} dt$ Could someone please help me to calculate the integral of:
$$\int_{-\infty}^{+\infty} \cos (at) e^{-bt^2} dt.$$
a and b both real, b>0.
I have tried integration by parts, but I can't seem to simplify it to anything useful. Essentially, I would like to arrive at something that looks like: 7.4.6 here:
textbook result
| Hint:
Use the fact that
$$\int_{-\infty}^\infty e^{iat- bt^2}\,dt = \sqrt{\frac{\pi}{b}} e^{-a^2/4b} $$
which is valid for $b>0$.
To derive this formula, complete the square in the exponent and then shift the integration contour a bit.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Representations of integers by a binary quadratic form Let $\mathfrak{F}$ be the set of binary quadratic forms over $\mathbb{Z}$.
Let $f(x, y) = ax^2 + bxy + cy^2 \in \mathfrak{F}$.
Let $\alpha = \left( \begin{array}{ccc}
p & q \\
r & s \end{array} \right)$ be an element of $SL_2(\mathbb{Z})$.
We write $f^\alpha(x, y) = f(px + qy, rx + sy)$.
Since $(f^\alpha)^\beta$ = $f^{\alpha\beta}$, $SL_2(\mathbb{Z})$ acts on $\mathfrak{F}$ from right.
Let $f, g \in \mathfrak{F}$.
If $f$ and $g$ belong to the same $SL_2(\mathbb{Z})$-orbit, we say $f$ and $g$ are equivalent.
Let $f = ax^2 + bxy + cy^2 \in \mathfrak{F}$.
We say $D = b^2 - 4ac$ is the discriminant of $f$.
Let $m$ be an integer.
If $m = ax^2 + bxy + cy^2$ has a solution in $\mathbb{Z}^2$, we say $m$ is represented by $ax^2 + bxy + cy^2$.
If $m = ax^2 + bxy + cy^2$ has a solution $(s, t)$ such that gcd$(s, t) = 1$,
we say $m$ is properly represented by $ax^2 + bxy + cy^2$.
Is the following proposition true?
If yes, how do we prove it?
Proposition
Let $ax^2 + bxy + cy^2 \in \mathfrak{F}$.
Suppose its discriminant is not a square.
Let $m$ be an integer.
Then $m$ is properly represented by $ax^2 + bxy + cy^2$ if and only if there exist integers $l, k$ such that $ax^2 + bxy + cy^2$ and $mx^2 + lxy + ky^2$ are equivalent.
| Lemma 1
Let $f = ax^2 + bxy + cy^2 \in \mathfrak{F}$.
Let $\alpha = \left( \begin{array}{ccc}
p & q \\
r & s \end{array} \right)$ be an element of $SL_2(\mathbb{Z})$.
Then $f^\alpha(x, y) = f(px + qy, rx + sy) = kx^2 + lxy + my^2$,
where
$k = ap^2 + bpr + cr^2$
$l = 2apq + b(ps + qr) + 2crs$
$m = aq^2 + bqs + cs^2$.
Proof:
Clear.
Proof of the proposition
Let $f(x, y) = ax^2 + bxy + cy^2$.
Suppose $m$ is properly represented by $f(x, y)$.
There exist integers $p, r$ such that gcd$(p, r) = 1$ and $m = f(p, r)$.
Since gcd$(p, r) = 1$, there exist integers $s, r$ such that $ps - rq = 1$.
By Lemma 1, $f(px + qy, rx + sy) = mx^2 + lxy + ky^2$,
where
$m = ap^2 + bpr + cr^2$
$l = 2apq + b(ps + qr) + 2crs$
$k = aq^2 + bqs + cs^2$.
Hence, $ax^2 + bxy + cy^2$ and $mx^2 + lxy + ky^2$ are equivalent.
Conversely suppose $ax^2 + bxy + cy^2$ and $mx^2 + lxy + ky^2$ are equivalent.
There exists integer $p, q, r, s$ such that $ps - rq = 1$ and
$f(px + qy, rx + sy) = mx^2 + lxy + ky^2$.
Letting $x = 1, y = 0$, we get $f(p, r) = m$.
Since $ps - rq = 1$, gcd$(p, r) = 1$.
Hence $m$ is properly represented by $ax^2 + bxy + cy^2$.
| {
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derive the formula for the left rectangle sum $f(x)=x^2+1$ from $0$ to $3$ Simply that, derive the formula for the left rectangle sum $f(x)=x^2+1$ from $0$ to $3$
This is when you use like rectangles and Riemann sums to approximate an integral. Not really sure what this means to derive the formula ?
| Can you follow the very similar example on this web site?
http://www2.seminolestate.edu/lvosbury/CalculusI_Folder/RiemannSumDemo.htm
It is important for you to learn what is going on here.
I would strongly recommend you use all three (left, right and midpoint) to find the integral.
Of course, you know what the answer should be by doing the integral.
Hint 1: Area = 12
Hint 2: See the left sum here:
http://www.wolframalpha.com/input/?i=INTEGRATE%5Bx%5E2%2B1%2C%7Bx%2C0%2C3%7D%5D&t=crmtb01
Please show your work if this is confusing.
HTH ~A
| {
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Sum of angles in $\mathbb{R}^n$ Given three vectors $v_1,v_2$ and $v_3$ in $\mathbb{R}^n$ with the standard scalar product the follwing is true
$$\angle(v_1,v_2)+\angle(v_2,v_3)\geq \angle(v_1,v_3).$$
It tried to substitute $\angle(v_1,v_2) = cos^{-1}\frac{v_1 \cdot v_2}{\Vert v_1 \Vert \Vert v_2 \Vert}$ but I could not show the resulting inequality. What is the name of the inequality and do you know reference that one can cite in an article?
| You can reduce the problem to $\mathbb{R}^3$, and there wlog v2=(0,0,1)
then one gets an easy to prove inequality if one writes everything in polar coordinates.
| {
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Example of a sequence with countable many cluster points Can someone give a concrete example of a sequence of reals that has countable infinite many cluster points ?
| First fix some bijection $f : \mathbb{N} \to \mathbb{N} \times \mathbb{N}$. For $n \in \mathbb{N}$ let $g(n)$ denote the first coordinate of $f(n)$ and let $h(n)$ denote the second corrdinate.
Then define a sequence $\{ x_n \}_{n=1}^\infty$ by $$x_n = g(n) + 2^{-h(n)}.$$
Then every natural number is a cluster point of this sequence.
For a more concrete example, consider the following sequence:
$$\begin{array}{c|c}
n & x_n \\
\hline
1 & 1 + 2^{-1} \\
2 & 1 + 2^{-2} \\
3 & 2 + 2^{-1} \\
4 & 1 + 2^{-3} \\
5 & 2 + 2^{-2} \\
6 & 3 + 2^{-1} \\
7 & 1 + 2^{-4} \\
8 & 2 + 2^{-3} \\
9 & 3 + 2^{-2} \\
10 & 4 + 2^{-1} \\
\vdots & \vdots
\end{array}
$$
(I'm too lazy to give an exact formula at the moment, but I think the idea is clear.)
| {
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A formal proof that a sum of infinite series is a series of a sum? I feel confused when dealing with ininities of any kind. E.g. the next equation is confusing me.
$$\displaystyle\sum^\infty_{n} (f_1(n) + f_2(n)) = \displaystyle\sum^\infty_{n_1=1} f_1(n_1) + \displaystyle\sum^\infty_{n_2=1} f_2(n_2)$$
How do people deal with such a special object as infinity with such an ease (it makes sense in general though)? Is there any axiomatic theory for that? I ask because I saw people relying on this when calculating the exact value of a series.
And last but not least: does my question sound tautological? Better delete it then.
| You have not quite stated a result fully. We state a result, and then write down the main elements of a proof. You should at least scan the proof, and then go to the final paragraph.
We will show that if $\sum_{i=1}^\infty f(i)$ and $\sum_{j=1}^\infty g(j)$ both exist, then so does $\sum_{k=1}^\infty (f(k)+g(k))$, and
$$\sum_{k=1}^\infty (f(k)+g(k))=\sum_{i=1}^\infty f(i)+\sum_{j=1}^\infty g(j).$$
Let $\sum_{i=1}^\infty f(i)=a$ and $\sum_{j=1}^\infty g(j)=b$.
We want to show that $\sum_{k=1}^\infty (f(k)+g(k))=a+b$.
Let $\epsilon \gt 0$. By the definition of convergence, there is an integer $K$ such that if $n \gt K$ then
$$\left|\sum_{i=1}^\infty f(i)-a\right|\lt \epsilon/2.$$
Similarly, there is an integer $L$ such that if $n \gt L$ then
$$\left|\sum_{j=1}^n g(j)-b\right|\lt \epsilon/2.$$
Let $M=\max(K,L)$. Then, from the two inequalities above, and the Triangle Inequality, it follows that if $n \gt M$, we have
$$\left|\sum_{k=1}^n (f(k)+g(k))-(a+b)\right|\lt \epsilon.$$
This completes the proof.
If you do not have experience with arguments like the one above, it may be difficult to understand. But there is one important thing you should notice. After the initial statement of the result, there is no more mention of "infinity." All subsequent work is with finite sums. The existence and value of an expression like $\sum_{n=1}^\infty h(n)$ is defined purely in terms of finite sums.
Remark: It is perfectly possible for the sum on your left-hand side to exist, while the sums on the right do not. A crude example would be $f(i)=1$ for all $i$, and $g(j)=-1$ for all $j$. But one can prove that if any two of the sums exist, then the third one does.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Locally integrable functions Formulation:
Let $v\in L^1_\text{loc}(\mathbb{R}^3)$ and $f \in H^1(\mathbb{R}^3)$ such that
\begin{equation}
\int f^2 v_+ = \int f^2 v_- = +\infty.
\end{equation}
Here, $v_- = \max(0,-f)$, $v_+ = \max(0,f)$, i.e., the negative and
positive parts of $v=v_+ - v_-$, respectively.
Question: Does $g\in H^1(\mathbb{R}^3)$ exist, such that
\begin{equation}
\int g^2 v_+ < \infty, \quad \int g^2 v_- = +\infty \quad ?
\end{equation}
Some thoughts:
Let $S_\pm$ be the supports of $v_\pm$, respectively.
One can easily find $g\in L^2$ such that the last equation holds, simply multiply $f$ with the characteristic function of $S_-$. The intuitive approach is then by some smoothing of this function by a mollifier, or using a bump function to force the support of $g$ away from $S_+$.
However, the supports of $S_\pm$ can be quite complicated: for example, fat Cantor-like sets. Thus, a bump function technique or a mollifier may "accidentally" fill out any of $S_\pm$.
My motivation:
The problem comes from my original research on the mathematical foundations of Density Functional Theory (DFT) in physics and chemistry. Here, $f^2$ is proportional to the probability density of finding an electron at a space point, and $v$ is the potential energy field of the environmentn. $\int f^2 v$ is the total potential energy for the system's state. The original $f$ gives a meaningless "$\infty-\infty$" result, but for certain reasons, we are out of the woods if there is some $other$ density $g$ with the prescribed property.
Edit:
Removed claim that $S_\pm$ must be unbounded. This does not follow from the stated assumptions.
| When $v \in L^1_{loc}(\mathbb{R}^3)$, then $v$ is the density of an absolutely continuous signed measure $\mu$. Take a Hahn decomposition of $\mathbb{R}^3$ in two measurable sets, so that $\mathbb{R}^3$ is disjoint union of say $P$ and $N$, $P$ is positive for $\mu$ and $N$ is negative for $\mu$, defined as $\mu$ is non-negative on measurable subsets of $P$ and non-positive on measurable subsets of $N$. This gives a Jordan decomposition in measures $\mu^+$ and $\mu^-$, concentrated on $P$ and $N$ respective. Now $\mu^+$ on a measurable set $A$ is $\int_A v_+$ and $\mu_-$ on a measurable set $A$ is $\int_A v_-$. When $f \in H(\mathbb{R}^3) = L^2(R^3)$ has $\int f^2 v_+ = \infty $ and $ \int f^2 v_- = \infty$, then $f \,1_N \in L^2(\mathbb{R}^3)$, $\int f^2\,1_N v_- = \infty $ and $\int (f \, 1_N)^2 v_+ = \int f^2 \, 1_N v_+ = 0$. The not-shown measures in the integrals is Lebesgue measure on $\mathbb{R}^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Random walking and the expected value I was asked this question at an interview, and I didn't know how to solve it. Was curious if anyone could help me.
Lets say we have a square, with vertex's 1234. I can randomly walk to each neighbouring vertex with equal probability. My goal is to start at '1', and get back to '1'. How many average walks will I take before I return back to 1?
| By symmetry, the unique invariant probability measure $\pi$ for this Markov chain is uniform on the four states. The expected return time is therefore $\mathbb{E}_1(T_1)=1/\pi(1)=4.$
This principle is easy to remember and can be used to solve other interesting problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $i^i$ is a real number According to WolframAlpha, $i^i=e^{-\pi/2}$ but I don't know how I can prove it.
| This would come right from Euler's formula. Let's derive it first.
There are many ways to derive it though, the Taylor series method being the most popular; here I’ll go through a different proof.
Let the polar form of the complex number be equal to $z$ .
$$\implies z = \cos x + i\sin x$$
Differentiating on both sides we get,
$$\implies \dfrac{dz}{dx} = -\sin x + i\cos x$$
$$\implies dz = (-\sin x + i\cos x)dx$$
Integrating on both sides,
$$\implies \displaystyle \int \frac{dz}{z} = i \int dx$$
$$\implies \log_e z = ix + K$$
Since $K = 0$, (Set $x = 0$ in the equation), we have,
$$\implies z = e^{ix}$$
$$\implies e^{ix} = \cos x + i\sin x$$
The most famous example of a completely real number being equal to a real raised to an imaginary is
$$\implies e^{i\pi} = -1$$
which is Euler’s identity.
To find $i$ to the power $i$ we would have to put $ x = \frac{\pi}2$ in Euler's formula.
We would get
$$e^{i\frac{\pi}2} = \cos \frac{\pi}2 + i\sin \frac{\pi}2$$
$$e^{i\frac{\pi}2} = i$$
$${(e^{i\frac{\pi}2})}^{i} = i^{i}$$
$$i^{i} = {e^{i^{2}\frac{\pi}2}} = {e^{-\frac{\pi}2}}$$
$$i^{i} = {e^{-\frac{\pi}2}} = 0.20787957635$$
This value of $i$ to the power $i$ is derived from the principal values of $\sin$ and $\cos$ which would satisfy this equation. There are infinite angles through which this can be evaluated; since $\sin$ and $\cos$ are periodic functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "83",
"answer_count": 6,
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Evaluating $\int_0^\infty\frac{\sin(x)}{x^2+1}\, dx$ I have seen $$\int_0^\infty \frac{\cos(x)}{x^2+1} \, dx=\frac{\pi}{2e}$$ evaluated in various ways.
It's rather popular when studying CA.
But, what about $$\int_0^\infty \frac{\sin(x)}{x^2+1} \, dx\,\,?$$
This appears to be trickier and more challenging.
I found that it has a closed form of
$$\cosh(1)\operatorname{Shi}(1)-\sinh(1)\text{Chi(1)}\,\,,\,\operatorname{Shi}(1)=\int_0^1 \frac{\sinh(x)}{x}dx\,\,,\,\, \text{Chi(1)}=\gamma+\int_0^1 \frac{\cosh(x)-1}{x} \, dx$$
which are the hyperbolic sine and cosine integrals, respectively.
It's an odd function, so
$$\int_{-\infty}^\infty \frac{\sin(x)}{x^2+1} \, dx=0$$
But, does anyone know how the former case can be done? Thanks a bunch.
| Mellin transform of sine is, for $-1<\Re(s)<1$:
$$
G_1(s) = \mathcal{M}_s(\sin(x)) = \int_0^\infty x^{s-1}\sin(x) \mathrm{d} x =\Im \int_0^\infty x^{s-1}\mathrm{e}^{i x} \mathrm{d} x = \Im \left( i^s\int_0^\infty x^{s-1}\mathrm{e}^{-x} \mathrm{d} x \right)= \Gamma(s) \sin\left(\frac{\pi s}{2}\right) = 2^{s-1} \frac{\Gamma\left(\frac{s+1}{2}\right)}{\Gamma\left(1-\frac{s}{2}\right)} \sqrt{\pi}
$$
And Mellin transfom of $(1+x^2)^{-1}$ is, for $0<\Re(s)<2$:
$$
G_2(s) = \mathcal{M}_s\left(\frac{1}{1+x^2}\right) = \int_0^\infty \frac{x^{s-1}}{1+x^2}\mathrm{d} x \stackrel{x^2=u/(1-u)}{=} \frac{1}{2} \int_0^1 u^{s/2-1} (1-u)^{-s/2} \mathrm{d}u = \frac{1}{2} \operatorname{B}\left(\frac{s}{2},1-\frac{s}{2}\right) = \frac{1}{2} \Gamma\left(\frac{s}{2}\right) \Gamma\left(1-\frac{s}{2}\right) = \frac{\pi}{2} \frac{1}{\sin\left(\pi s/2\right)}
$$
Now to the original integral, for $0<\gamma<1$:
$$
\int_0^\infty \frac{\sin(x)}{1+x^2}\mathrm{d}x = \int_{\gamma-i \infty}^{\gamma+ i\infty} \mathrm{d} s\int_0^\infty \sin(x) \left( \frac{G_2(s)}{2 \pi i} x^{-s}\right) \mathrm{d}s = \frac{1}{2 \pi i} \int_{\gamma-i \infty}^{\gamma+i \infty} G_2(s) G_1(1-s) \mathrm{d}s =\\ \frac{1}{4 i} \int_{\gamma-i \infty}^{\gamma+i \infty} \Gamma(1-s) \cot\left(\frac{\pi s}{2}\right) \mathrm{d} s = \frac{2\pi i}{4 i} \sum_{n=1}^\infty \operatorname{Res}_{s=2n} \Gamma(1-s) \cot\left(\frac{\pi s}{2}\right) = \sum_{n=1}^\infty \frac{\psi(2n)}{\Gamma(2n)} = \sum_{n=1}^\infty \frac{1+(-1)^n}{2} \frac{\psi(n)}{\Gamma(n)}
$$
Since
$$
\sum_{n=1}^\infty z^n \frac{\psi(n)}{\Gamma(n)} = \mathrm{e}^z z \left(\Gamma(0,z) + \log(z)\right)
$$
Combining:
$$
\int_0^\infty \frac{\sin(x)}{1+x^2} \mathrm{d}x = \frac{\mathrm{e}}{2} \Gamma(0,1) - \frac{1}{2 \mathrm{e}} \Gamma(0,-1) - \frac{i \pi }{2 \mathrm{e}} = \frac{1}{2e} \operatorname{Ei}(1) - \frac{\mathrm{e}}{2} \operatorname{Ei}(-1)
$$
| {
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"url": "https://math.stackexchange.com/questions/191639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Help evaluating a limit I have the following limit:
$$\lim_{n\rightarrow\infty}e^{-\alpha\sqrt{n}}\sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose k}\sum_{m=0}^{n-1-k}\frac{(\alpha\sqrt{n})^m}{m!}$$
where $\alpha>0$.
Evaluating this in Mathematica suggests that this converges, but I don't know how to evaluate it. Any help would be appreciated.
| I would start even more simple-mindedly by replacing the inner sum
with its infinite $n$ value of $e^{\alpha \sqrt{n}}$.
This cancels out the outer expression, so we are left with
$\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose k}$.
Doing some manipulation,
$\sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose k}
= \sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose {n-1}}
= \sum_{k=n-1}^{2n-2}2^{-k-1} {{k}\choose {n-1}}
$.
As often happens, it is late and I am tired and not sure exectly what to do next,
so I'll leave it at this.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Showing $H=\langle a,b|a^2=b^3=1,(ab)^n=(ab^{-1}ab)^k\rangle$.
Let $G=\langle a,b|a^2=b^3=1,(ab)^n=(ab^{-1}ab)^k \rangle$. Prove that $G$ can be generated with $ab$ and $ab^{-1}ab$. And from there, $\langle(ab)^n\rangle\subset Z(G)$.
Problem wants $H=\langle ab,ab^{-1}ab \rangle$ to be $G$. Clearly, $H\leqslant G$ and after doing some handy calculation which takes time I've got:
*
*$ab^{-1}=(ab^{-1}ab)(ab)^{-1}\in H$
*$b=b^{-2}=(ab)^{-1}ab^{-1}\in H$
*$a=(ab)b^{-1}\in H$
So $G\leqslant H$ and therefore $G=H=\langle ab,ab^{-1}ab\rangle$.
For the second part, I should prove that $N=\langle(ab)^n\rangle\leqslant Z(G)$.
Please help me.
Thanks.
| The question is answered in the comments. However so that this question does not remain listed as unanswered forever, I will provide a solution. I will also give the details for the first part of the question.
Part 1 Show $G= \langle ab, ab^{-1}ab \rangle$
Let $H=\langle ab, ab^{-1}ab \rangle $. It is clear that $ H \leq G$. We will show that $G \leq H$ by showing that $a$ and $b$ are in $H$.
We make the following observations which we will use in the argument:
1) $ (ab)^{-1}=b^{-1}a^{-1} \in H $
2) $ a^{-1}=a $
3) $ b^{-2}=b $
Then $(ab^{-1}ab)(ab)^{-1}=ab^{-1} \in H$. Thus $ab^{-1} \cdot b^{-1}a^{-1} = ab^{-2}a = aba \in H$. Hence $(ab)^{-1} \cdot (aba) = a \in H$. Thus we have that $a \in H$ and then multiplying $ab$ with $a^{-1}$ on the left gives that also $b$ is in $H$. Hence $ G \leq H$ and we now can conclude $G=H$.
Part 2 Show $ \langle (ab)^{n} \rangle \leq Z(G)$
Since $G = \langle ab, ab^{-1}ab \rangle$ we will try to show $ab$ and $ab^{-1}ab$ commute with $(ab)^{n}$. Of course it is clear that $(ab)$ does.
Now consider $$(ab^{-1}ab)\cdot (ab)^{n} = ab^{-1}ab(ab^{-1}ab)^{k}$$ (using a relation given in the presentation for $G$ in the question).
$$=(ab^{-1}ab)^{k+1}=(ab^{-1}ab)^{k}ab^{-1}ab = (ab)^{n} \cdot ab^{-1}ab$$
Hence we see that $(ab)^{n}$ commutes with both of our generators of $G$ and thus $ \langle (ab)^{n} \rangle $ is central.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191791",
"timestamp": "2023-03-29T00:00:00",
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point outside a non-convex shape I have a non-convex shape (object) in black on the figure on the link. At the beginning, All red points are outside the shape. Next, I apply a random transformation on some points. This create a new shape (yellow).
What I want is to fill the outside of the shape with a specific color but because of the transformation, some red points are now inside the shape. I would like to know how to find a correct red point, i.e outside the shape to apply the filling algorithm.
Here is the link on the image.
http://postimage.org/image/sbjmgi1tr/
| In that case, use a point in polygon algorithm. Then you can test each point if it is in your new polygon or not.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing pass equivalence of cinquefoil knot According to C.C. Adams, The knot book, pp 224, "every knot is either pass equivalent to the trefoil knot or the unknot".
A pass move is the following:
Can someone show me how to show that the Cinquefoil knot is pass equivalent to unknot or trefoil? Been trying on paper but no luck. Don't see how pass moves apply here. Thanks.
| The general method is demonstrated in Kauffman's book On Knots. Put the knot into a "band position" So that the Seifert surface is illustrated as a disk with twisted and intertangled bands attached. Then the orientations match those of your figure. You can pass one band over another. Your knot is the braid closure of $\sigma_1^5$. The Seifert surface is two disks with 5 twisted bands between them. Start by stretching the disks apart.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question about a closed set Let $X = C([0; 1])$. For all $f, g \in X$, we define the metric $d$ by
$d(f; g) = \sup_x |f(x) - g(x)|$. Show that $S := \{ f\in X : f(0) = 0 \}$ is closed in $(X; d)$.
I am trying to show that $X \setminus S$ is open but I don't know where to start showing that.
I wanna add something more, I have not much knowledge about analysis and I am just self taught of it, what I have learnt so far is just some basic topology and open/closed sets.
| I'd try to show that $S$ contains all its limit points. To this end let $f$ be a limit point of $S$. Let $f_n$ be a sequence in $S$ converging to $f$ in the sup norm.
Now we show that $f$ is also in $S$:
By assumption, for $\varepsilon > 0$ you have that $\sup_{z \in [0,1]}|f_n(z) - f(z)| < \varepsilon$ for $n$ large enough. In particular, $|f_n(0) - f(0)| = | f(0)| < \varepsilon$ for $n$ large enough. Now let $\varepsilon \to 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Graph for which certain induced subgraphs are cycles Let us call a graph G $nice$ if for any vertex $v \in G$, the induced subgraph on the vertices adjacent to $v$ is exactly a cycle.
Is there anything that we can conclude about nice graphs? In particular, can we find a different (maybe simpler) but equivalent formulation for niceness?
| Wrong Answer
Given a finite connected "nice" graph, $G$, you can take all triples $\{a,b,c\}$ of nodes with $\{a,b\}$,$\{b,c\}$, and $\{a,c\}$ edges in the graph.
Take these as $2$-simplexes, and stitch them together in the obvious way.
The fact that $G$ is nice means that each edge must be on exactly two triangles. The fact that $G$ is nice also means that the interior of the union of the triangles that contain node $a$ will be homeomorphic to an open ball in $\mathbb R^2$.
So this all shows that stitching these together will yield $G$ as a triangulation of a compact $2$-manifold.
There is at least one "degenerate" case for which this is not true - the single-edge graph with two nodes. Depends on whether you consider a single node graph to be a cycle...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Quadratic equations that are unsolvable in any successive quadratic extensions of a field of characteristic 2
Show that for a field $L$ of characteristic $2$ there exist quadratic equations which cannot be solved by adjoining square roots of elements in the field $L$.
In $\mathbb{Z_2}$ adjoining all square roots we obtain again $\mathbb{Z_2}$, but $t^2+t+1$ does not have roots in this field. I don't know how to do the general case.
| If $L$ is a finite field of characteristic two, then consider the mapping
$$
p:L\rightarrow L, x\mapsto x+x^2.
$$
Because $F:x\mapsto x^2$ respects sums: $$F(x+y)=(x+y)^2=x^2+2xy+y^2=x^2+y^2=F(x)+F(y),$$ the mapping $p$ is a homomorphism of additive groups. We see that $x\in \mathrm{Ker}\ p$, if and only if $x=0$ or $x=1$. So $|\mathrm{Ker}\ p|=2$. Therefore (one of the basic isomorphism theorems) $|\mathrm{Im}\ p|=|L|/2$. In particular, the mapping $p$ is not onto.
Let $a\in L$ be such that it is not in the image of $p$. Then the quadratic equation
$$
x^2+x+a=0
$$
has no zeros in $L$.
Because the mapping $F$ is onto (its kernel is trivial), all the elements of $L$ have a square root in $L$. Thus adjoinin square roots of elements of $L$ won't allow us to find roots of the above equation.
In general the claim may not hold. By elementary Artin-Schreier theory we can actually show that quadratic polynomials of the prescribed type exist exactly, when the above mapping $p$
is not onto (irrespective of whether $L$ is finite or not). This is because, unless the quadratic $r(x)$ is of the form $x^2+a$ (when joining square roots, if needed, will help), its splitting field is separable, hence cyclic Galois of degree two. Thus the cited theorem of Artin-Schreier theory says that the splitting field of $r(x)$ can be gotten by joining a root of a polynomial of the form $x^2+x+a=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\sqrt{x-4} + 10 = \sqrt{x+4}$ Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<
| Square both sides, and you get
$$x - 4 + 20\sqrt{x - 4} + 100 = x + 4$$
This simplifies to
$$20\sqrt{x - 4} = -92$$
or just
$$\sqrt{x - 4} = -\frac{92}{20}$$
Since square roots of numbers are always nonnegative, this cannot have a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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limit at infinity $f(x)=x+ax \sin(x)$ Let $f:\Bbb R\rightarrow \Bbb R$ be defined by $f(x)= x+ ax\sin x$.
I would like to show that if $|a| < 1$, then $\lim\limits_{x\rightarrow\pm \infty}f(x)=\pm \infty$.
Thanks for your time.
| We start by looking at the case when $x$ is (large) positive. The idea is that if $|a|\lt 1$, then since $|\sin x|\lt 1$, the term $ax\sin x$, even if it happens to be negative, can't cancel out the large positiveness of the front term $x$. We now proceed more formally.
Note that $|\sin x|\le 1$ for all $x$, so $x|a\sin x| \le x|a|$, and therefore $x+ax\sin x\ge x-|a|x$. So our function is $\ge (1-|a|)x$ when $x$ is positive. Since $|a|\lt 1$, the number $1-|a|$ is a positive constant. But $(1-|a|)x$ can be made arbitrarily large by taking $x$ large enough.
Similarly, let $x$ be negative. Then $x+ax\sin x=x(1+a\sin x)$. But $1+a\sin x\ge 1-|a|$, and $(1-|a|)x$ can be made arbitrarily large negative by taking $x$ negative and of large enough absolute value.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The product of all elements in $G$ cannot belong to $H$
Let $G$ be a finite group and $H\leq G$ be a subgroup of order odd such that $[G:H]=2$. Therefore the product of all elements in $G$ cannot belong to $H$.
I assume $|H|=m$ so $|G|=2m$. Since $[G:H]=2$ so $H\trianglelefteq G$ and that; half of the elements of the group are in $H$. Any Hints? Thanks.
| Consider the image of the product under the quotient map $G\to G/H\cong C_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Internal Direct Sum Questiom I'm posed with the following problem. Given a vector space $\,V\,$ over a field (whose characteristic isn't $\,2$), we have a linear transformation from $\,V\,$ to itself.
We have subspaces
$$V_+=\{v\;:\; Tv=v\}\,\,,\,\, V_-=\{v\;:\; Tv=-v\}$$
I want to show that $\,V\,$ is the internal direct sum of these two subspaces. I have shown that they are disjoint, but can't seem to write an arbitrary element in V as the sum of respective elements of the subspaces...
Edit:forgot something important!
$$\, T^2=I\,$$
Sorry!
| Another approach:
Lemma: In arbitrary characteristic, if $V$ is a $K$-vector space and $P:V\to V$ is an endomorphism with $P^2=\lambda P$, $\lambda\ne 0$, then $V=\ker P\oplus\ker (\lambda I-P)$.
Proof: If $v\in \ker P\cap\ker (\lambda I-P)$, then $\lambda v = (\lambda I-P)v+Pv = 0$, hence $v=0$. Thus $\ker P\cap\ker (\lambda I-P)=0$.
For $v\in V$ let $u=\lambda^{-1}Pv$ and $w=v-u$. Then clearly $v=u+w$ and $(\lambda I-P)u =Pv-\lambda^{-1}P^2v = 0$ and $Pw =Pv-Pu = Pv-\lambda^{-1}P^2v=0$, i.e. $u\in\ker(\lambda I-P)$ and $w\in \ker P$. We conclude that $V=\ker P\oplus\ker(\lambda I-P)$.
Corollary: If $\operatorname{char} K\ne 2$, $V$ is a $K$-vector space and $T:V\to V$ is an endomorphism with $T^2=I$, then $V=T_+\oplus T_-$ (with $T_\pm$ defined as in the OP).
Proof: Let $P=T+I$. Then $P^2 = T^2+2T+I^2=I+2T+I=2P$. Since $\lambda:=2\ne0$, the lemma applies. Here $\ker P=T_-$ and $\ker (2 I-P)=\ker(I-T)=T_+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Subsets and Splits