答:
I am afraid I can't afford the time to write down a detailed solution but do allow me to provide you with a sketch of the natural line of argumentation for establishing this result. Consider an arbitrary (does not even need to be necessarily commutative) ring $A$ together with the notations:

$\mathscr{Id}_{\mathrm{b}}(A)$ for the set of all bilateral ideals of $A$.  We shall also employ the symbol $I \leqslant_{\mathrm{b}} A$ to denote the fact that $I \in \mathscr{Id}_{\mathrm{b}}(A)$ is a bilateral ideal of $A$.

$\mathscr{Spec}(A)$ for the prime spectrum of $A$, i.e. the set of all prime ideals of $A$.

$\mathscr{V}(X)=\left\{P \in \mathscr{Spec}(A) \mid X \subseteq P \right\}$ for any subset $X \subseteq A$.

$\mathrm{Id}\left(\mathscr{X}\right)=\begin{cases}\bigcap \mathscr{X}, &\mathscr{X} \neq \varnothing \\ A, &\mathscr{X}=\varnothing\end{cases},$
for any subset $\mathscr{X} \subseteq \mathscr{Spec}(A)$.

$\mathscr{R}=\left\{I \in \mathscr{Id}_{\mathrm{b}}(A) \mid \sqrt{I}=I\right\}$ for the set of all semiprime (this is the general terminology) ideals, or radical ideals (as they are called in the commutative setting).

$\mathscr{Za}_{\mathrm{f}}(A)=\left\{\mathscr{V}(X)\right\}_{X \subseteq A}$ for the set of all Zariski closed subsets of the prime spectrum.

$G.H=\left\langle GH \right\rangle$ as the subgroup product (the subgroup generated by the subset product $GH=\{xy\}_{\substack{x \in G\\ y \in H}}$) for any two additive subgroups $G, H \leqslant A$. When $I \leqslant_{\mathrm{s}}A$ is a left ideal and $J \leqslant_{\mathrm{d}}A$ a right one we have in particular that $I.J \leqslant_{\mathrm{b}}A$ is a bilateral ideal.

One proceeds by first setting up the following:

Proposition 1. For any bilateral ideal $I \leqslant_{\mathrm{b}} A$ we have $\mathrm{Id}\left(\mathscr{V}(I)\right)=\sqrt{I}$ (the radical of $I$ which is defined in the general case of not necessarily commutative rings and reduces to the well-known construction in the commutative case). Similarly, for any subset $\mathscr{X} \subseteq \mathscr{Spec}(A)$ we have $\overline{\mathscr{X}}=\mathscr{V}\left(\mathrm{Id}\left(\mathscr{X}\right)\right)$.

and continues with the following:

Corollary. The maps given by:
\begin{align*}
\mathscr{R} &\to \mathscr{Za}_{\mathrm{f}}(A)\\
I &\mapsto \mathscr{V}(I)\\
\mathscr{Za}_{\mathrm{f}}(A) &\to \mathscr{R}\\
\mathscr{F} &\mapsto \mathrm{Id}\left(\mathscr{F}\right)
\end{align*}
are mutually inverse antiisotonic maps when ordering both domains by inclusion.

From the corollary we infer in particular that any closed subset $\mathscr{F} \subseteq \mathscr{Spec}(A)$ must be of the form $\mathscr{F}=\mathscr{V}(I)$ with $I \in \mathscr{R}$ being a semiprime (radical) ideal. With these observations in place, it is easy to establish:

Proposition 2. $\mathscr{F} \in \mathscr{Za}_{\mathrm{f}}(A)$ is irreducible if and only $\mathrm{Id}\left(\mathscr{F}\right) \in \mathscr{Spec}(A)$ is prime.

Proof. For the direct implication, consider the irreducible closed subset $\mathscr{F} \subseteq \mathrm{Spec}(A)$ and set $P=\mathrm{Id}\left(\mathscr{F}\right)$ such that $P \in \mathscr{R}$ is radical and $\mathscr{F}=\mathscr{V}(P)$ by virtue of the above corollary. Since $\mathscr{F}$ is irreducible it is nonempty and therefore $P \neq A$, hence $P$ is a proper bilateral ideal (bear in mind that $\mathscr{V}(A)=\varnothing$). Consider now arbitrary $I, J \leqslant_{\mathrm{b}} A$ such that $I.J \subseteq P$. This entails $\mathscr{V}(I.J)=\mathscr{V}(I) \cup \mathscr{V}(J) \supseteq \mathscr{V}(P)$ and by virtue of irreducibility further leads to $\mathscr{V}(P) \subseteq \mathscr{V}(I)$ or $\mathscr{V}(P) \subseteq \mathscr{V}(J)$. By applying the $\mathrm{Id}$ operator and reversing the inclusions we infer $P=\sqrt{P} \supseteq \sqrt{I} \supseteq I$ or $P=\sqrt{P} \supseteq \sqrt{J} \supseteq J$. This means that $I \subseteq P$ or $J \subseteq P$ and thus proves that $P$ is prime.

For the converse implication it suffices to show that $\mathscr{V}(P)$ is irreducible whenever $P$ is prime. Obviously we have $P \in \mathscr{V}(P) \neq \varnothing$. Consider now two relatively closed subsets $\mathscr{F}, \mathscr{G} \subseteq \mathscr{V}(P)$ such that $\mathscr{V}(P)=\mathscr{F} \cup \mathscr{G}$. As subsets closed relatively to an absolutely closed subset, $\mathscr{F}$ and $\mathscr{G}$  are in their own turn absolutely (Zariski) closed and thus we can write $\mathscr{F}=\mathscr{V}(I)$ respectively $\mathscr{G}=\mathscr{V}(J)$ for some bilateral ideals $I, J \leqslant_{\mathrm{b}}A$. We have that $\mathscr{V}(I) \cup \mathscr{V}(J)=\mathscr{V}(I.J)=\mathscr{V}(P)$ whence $P \in \mathscr{V}(I.J)$. This leads successively to $I.J \subseteq P$ - therefore either $I \subseteq P$ or $J \subseteq P$ as $P$ is prime - and signifies that $\mathscr{V}(P) \subseteq \mathscr{V}(I)=\mathscr{F}$ or $\mathscr{V}(P) \subseteq \mathscr{V}(J)=\mathscr{G}$. We conclude by noticing that we are led to either $\mathscr{V}(P)=\mathscr{F}$ or $\mathscr{V}(P)=\mathscr{G}$ which entails the irreducibility of $\mathscr{V}(P)$. $\Box$

Title: Necessary condition to be closed Irreducible subspace

Body:
Let $X=\text{Spec}(R)$ where $R$ is a commutative ring with unity. If $Y\subset X$ is a closed subspace that is irreducible, then $Y = V(P)$ for some $P\in X$.

Proof. Since $Y$ is closed, $Y = V(I)$ for some ideal $I\subset R$. Need to show $I$ is prime ideal. Let $f,g\notin I$. Then consider $D(f)\cap Y$ and $D(g)\cap Y$. Assume they are both nonempty. Then they are nonempty open sets in $Y$ and as $Y$ is irreducible, $D(f)\cap D(g)\cap Y = D(fg)\cap Y$ is nonempty. But if $P\in D(fg)\cap Y$ then $I\subset P$ and $fg\notin P$ which implies $fg\notin I$.

The problem is I don't know how prove  $D(f)\cap Y$ is nonempty. I don't know if it's helpful information but I found that such $f,g$ are non-units.
But still don't know. Could you give any hint?


问:
What is the trick for the Taylor series of $\dfrac{e^x}{1+x}$
Thanks for above comments. Post here is higly edited:

I know that :
$$e^x = \sum_{k=0}^n \dfrac{x^k}{k!} + o(x^n)$$
$$\dfrac{1}{1+x} = \sum_{i=0}^n (-1)^i x^i + o(x^n)$$

Thus:
$$ \dfrac{e^x}{1+x} = \sum_{k=0}^n \dfrac{x^k}{k!}\sum_{i=0}^n (-1)^i x^i  + o(x^n)$$

Using the Cauchy product thanks to the comments and answers of this question:
$$ \dfrac{e^x}{1+x} = \sum_{k=0}^n c_i x^k + o(x^n)$$
with $$c_i = \sum_{i=0}^{k} \dfrac{(-1)^{k-i}}{i!}$$.

Is this okay ? Is it a definitive form or can I go further in the developpement ?

答:
Thanks for above comments. I edit :

I know that :
$$e^x = \sum_{k=0}^n \dfrac{x^k}{k!} + o(x^n)$$
$$\dfrac{1}{1+x} = \sum_{i=0}^n (-1)^i x^i + o(x^n)$$

Thus:
$$ \dfrac{e^x}{1+x} = \sum_{k=0}^n \dfrac{x^k}{k!}\sum_{i=0}^n (-1)^i x^i  + o(x^n)$$
Using the Cauchy product:
$$ \dfrac{e^x}{1+x} = \sum_{k=0}^n c_i x^k + o(x^n)$$
with $$c_i = \sum_{i=0}^{k} \dfrac{(-1)^{k-i}}{i!}$$.

Is this okay ? Is it a definitive form or can I go further in the developpement ?

ContentLicense: CC BY-SA 4.0


问:
If $x\equiv y \pmod{\gcd(a,b)}$, show that there is a unique $z\pmod{\text{lcm}(a,b)}$ with $z\equiv x\pmod a$ and $z\equiv y\pmod b$

If $x\equiv y \pmod{\gcd(a,b)}$, show that there is a unique $z\pmod{\text{lcm}(a,b)}$ with $z\equiv x\pmod{a}$ and $z\equiv y\pmod{b}$

What I have so far:

Let $z \equiv x\pmod{\frac{a}{\gcd(a,b)}}$ and $
z \equiv y\pmod b
 $. Then by the chinese remainder theorem there is a unique $z\pmod{\text{lcm}(a,b)}$ which satisfies this...

Is this the right approach here? I can't figure out how to get from $$z \equiv x\pmod{\frac{a}{\gcd(a,b)}}$$ what I need.

答:
Put $d=\gcd(a,b)$ and $\delta=x\bmod d=y\bmod d$ (here "mod" is the remainder operation). Then the numbers $x'=x-\delta$, $y'=y-\delta$ are both divisible by$~d$. In terms of a new variable $z'=z-\delta$ we need to solve the system
$$
  \begin{align}z'&\equiv x'\pmod a,\\z'&\equiv y'\pmod b.\end{align}
$$
Since $x',y',a,b$ are all divisible by $d$, any solution $z'$ will have to be as well; therefore we can divide everything by$~d$, and the system is equivalent to
$$
  \begin{align}\frac{z'}d&\equiv \frac{x'}d\pmod{\frac ad},\\
               \frac{z'}d&\equiv \frac{y'}d\pmod{\frac bd}.\end{align}
$$
Here the moduli $\frac ad,\frac bd$ are relatively prime, so by the Chinese remainder theorem there is a solution $\frac{z'}d\in\mathbf Z$, and it is unique modulo $\frac ad\times\frac bd$. Then the solutions for $z'$ will then form a single class modulo $\frac ad\times\frac bd\times d=\frac{ab}d=\operatorname{lcm}(a,b)$, and so will the solutions for $z=z'+\delta$.</p>

ContentLicense: CC BY-SA 3.0


答:
The secret hier is to use different summation indices in the sums:
$$\begin{align}
 \dfrac{e^x}{1+x} &= \sum_{i=0}^\infty \dfrac{x^i}{i!}\sum_{j=0}^\infty (-1)^jx^j\\
&=\sum_{i=0}^\infty\sum_{j=0}^\infty(-1)^j \frac{x^{i+j}}{i!}\\
&=\sum_{n=0}^\infty x^n\sum_{i=0}^n\frac{(-1)^{n-i}}{i!}.
\end{align}$$

Title: What is the trick for the Taylor series of $\dfrac{e^x}{1+x}$

Body:
Thanks for above comments. Post here is higly edited:

I know that :
$$e^x = \sum_{k=0}^n \dfrac{x^k}{k!} + o(x^n)$$
$$\dfrac{1}{1+x} = \sum_{i=0}^n (-1)^i x^i + o(x^n)$$

Thus:
$$ \dfrac{e^x}{1+x} = \sum_{k=0}^n \dfrac{x^k}{k!}\sum_{i=0}^n (-1)^i x^i  + o(x^n)$$

Using the Cauchy product thanks to the comments and answers of this question:
$$ \dfrac{e^x}{1+x} = \sum_{k=0}^n c_i x^k + o(x^n)$$
with $$c_i = \sum_{i=0}^{k} \dfrac{(-1)^{k-i}}{i!}$$.

Is this okay ? Is it a definitive form or can I go further in the developpement ?


问:
If limit of two sequences converges, do the sequences converge too?
We assume that $a_n\neq b_n$ and $a_n\neq \frac{1}{b_n}$, and we know $\frac{a_n}{b_n}$ converges to the limit $1$ (in other words $\lim_{n\to\infty}\big(\frac{a_n}{b_n}\big)=1$).

This means that $\frac{\lim_{n\to\infty}(a_n)}{\lim_{n\to\infty}(b_n)}=1$ and thus $\lim_{n\to\infty}(a_n)=\lim_{n\to\infty}(b_n)$.

Can we now assume that both of these sequences converge, in particular to a same limit $L$? Can we say $$\exists L\in\mathbb{R}\Rightarrow \lim_{n\to\infty}(a_n)=L=\lim_{n\to\infty}(b_n)?$$

答: I dont think they even need to converge take $a_n = n +1 $ and $b_n = n $


问:
Jensen's Formula application $f(z)$ has no zeros on $|z|=r, r<1$
I am reading the Balazard Saias and Yor paper where he defines,
$s=s(z)=\frac{1}{1-z}$ and $$f(z)=(s-1)\zeta(s)$$ where $\zeta$ is the Riemann zeta function.Then by Jensen's formula we have for $f(0)\neq 0$ and $r<1$,
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}log\ |f(re^{i\theta})|d\theta=log\ |f(0)|+\sum_{|\alpha|<r,f(\alpha)=0} log \ \frac{r}{|\alpha|}$$

For applying Jensen's formula we need that $f(z)$ has no zeros on $|z|=r$.

Question Show that $f(z)$ has no zeros on $|z|=r$, $r<1$.

Attempt
$$f(z)=(s-1)\zeta(s)$$
$$f(z)= \frac{z}{1-z}\zeta(\frac{1}{1-z})$$
Since $|z|=r$ so write $z=re^{i\theta}$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1}{1-re^{i\theta}})    $$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1}{1-rcos\theta-irsin\theta})    $$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1-rcos\theta+irsin\theta}{(1-rcos\theta)^2+r^2sin^2\theta})    $$
$$\zeta(\frac{1}{1-z})= \zeta(\frac{1-rcos\theta+irsin\theta}{r^2-2rcos\theta+1})    $$
How to show that $f(re^{i\theta})\neq 0$ on $|z|=r, r<1$?.

答:
Jensen's theorem is valid even if $f$ has zeroes on the circle $|z|=r$ (in practice since such circles are discrete, one avoids them easily in the usual applications of Jensen which is to count zeros in open discs)

The proof is standard assuming the result first without zeroes; then the number of zeroes on the circle $|z|=r$ is finite ($f$ is analytic in a neighborhood of $|z|=r$ in Jensen), hence let them be $a_1,..a_n, |a_k|=r$ counted with multiplicity.

Hence $f=(z-a_1)..(z-a_n)g$ with $g$ same zeroes as $f$ except those on the circle and $\log |f|=\log |g|+\log |z-a_1|+...\log |z-a_k|$

But now applying Jensen for $g$, using the easy to prove fact that $\int_0^{2\pi}\log |r(e^{i\theta}-e^{i\theta_0})|d\theta=2\pi \log r$, the fact that $|\log f(0)|=n\log r+|\log g(0)|$ and the observation that $\log r/|a_k|=0$ gives the result.


问:
Showing $\sum_{j=1}^n p_j\log_2\frac1{r_j}-\sum_{j=1}^n p_j\log_2\frac1{p_j}=-\sum_{j=1}^np_j\log_2\frac{r_j}{p_j}$
How is
$$
\sum_{j=1}^{n} p_j \cdot \log_2\frac1{r_j} - \sum_{j=1}^{n} p_j \cdot \log_2\frac1{p_j}
$$
the same as
$$
-\sum_{j=1}^{n} p_j \cdot \log_2 \frac{r_j}{p_j}
$$
?

(It is for proving that the kullbach-lieber divergence is nonnegative and first I have to rewrite it)

答:
Just looking at a single value of $j$ and using properties of logarithms:

$p\log\frac1r-p\log\frac1p=p\log\frac{p}{r}=-p\log\frac{r}{p}$


问:
Lemma/Proposition/Theorem, which one should we pick?
This is something that confuses me.
I have read a few mathematical texts and they often seem to use Lemma/Proposition/Theorem if they have a particular statement.

Now, which one to use? A lemma can be something you need to prove a more important theorem, but then what about Fatou's Lemma?

When to pick Proposition or Theorem?

答:
I'm by no means knowledgeable on this topic, but this article seems to have useful definitions.

What is the difference between a theorem, a lemma, and a corollary?

Quote:

Posted by DAVE RICHESON.

I prepared the following handout for my Discrete Mathematics class
(here’s a pdf version).

Definition — a precise and unambiguous description of the meaning of a
mathematical term.  It characterizes the meaning of a word by giving
all the properties and only those properties that must be true.

Theorem — a mathematical statement that is proved using rigorous
mathematical reasoning.  In a mathematical paper, the term theorem is
often reserved for the most important results.

Lemma — a minor result whose sole purpose is to help in proving a
theorem.  It is a stepping stone on the path to proving a theorem.
Very occasionally lemmas can take on a life of their own (Zorn’s
lemma, Urysohn’s lemma, Burnside’s lemma, Sperner’s lemma).

Corollary — a result in which the (usually short) proof relies heavily
on a given theorem (we often say that “this is a corollary of Theorem
A”).

Proposition — a proved and often interesting result, but generally
less important than a theorem.

Conjecture — a statement that is unproved, but is believed to be true
(Collatz conjecture, Goldbach conjecture, twin prime conjecture).

Claim — an assertion that is then proved.  It is often used like an
informal lemma.

Axiom/Postulate — a statement that is assumed to be true without
proof. These are the basic building blocks from which all theorems are
proved (Euclid’s five postulates, Zermelo-Fraenkel axioms, Peano
axioms).

Identity — a mathematical expression giving the equality of two (often
variable) quantities (trigonometric identities, Euler’s identity).

Paradox — a statement that can be shown, using a given set of axioms
and definitions, to be both true and false. Paradoxes are often used
to show the inconsistencies in a flawed theory (Russell’s paradox).
The term paradox is often used informally to describe a surprising or
counterintuitive result that follows from a given set of rules
(Banach-Tarski paradox, Alabama paradox, Gabriel’s horn).

ContentLicense: CC BY-SA 2.5


问:
I would like to know an intuitive way to understand a Cauchy sequence and the Cauchy criterion.
My understanding from the definition in my book (Rudin) is this.

A seq. $\{p_n\}$ in a metric space $X$ (I only really know $\mathbb R^k$) is said to be a Cauchy sequence if for any given $\epsilon > 0$, $\exists N\in \mathbb N$ such that $\forall n,m\ge N$, $d(p_n,p_m)<\epsilon$.

(1) I see it as, given any tiny value $\epsilon$, we can find a natural number $N$ large enough so that the distance between $p_n$ and $p_m$ is less than $\epsilon$. Am I right ?

The reason I'm asking this is because I was trying to understand the proof of how
$$\sum a_nb_n$$
can converge, and the book said this

$$\left\lvert \sum_{n=p}^{q}a_nb_n\right\rvert \leq \epsilon$$

satisfies the Cauchy criterion and therefore it converges.

I read other questions and answers about the Cauchy sequence, but it didn't really help me…

Can someone explain me what's going on?

Edit:

Suppose

a) the partial sums of $A_n = \Sigma a_n$ form a bounded sequence

b) $b_0 \geq b_1 \geq \dotsb$

c) $\lim_{b \to \infty} b_n = 0$

Using the partial summation formula, algebraically the equation in the bottom is proved

$$\left\lvert \sum_{n=p}^{q}a_nb_n \right\rvert \leq \epsilon$$

Algebraically I had no problem, but I don't know why this proves convergence. I thought to show that a sequence is Cauchy, we need to find the distance between two terms in a sequence. That's where I'm confused.

答:
You are right about you understanding of Cauchy sequence.

Cauchy criterion means that if a real (or complex) sequence is Cauchy, then it is convergent. In this case the sequence is

$ S_{m} = \sum_{n=0}^{m}a_{n}b_{n} $.

Now what is the meaning of $S_{m}$ being Cauchy? And no doubt you would know that the convergence of the series is defined as the convergence of the sequence $S_{m}$.


问:
If two matrices are similar are their submatrices again similar?
I am struggling a bit with the following: if we have two similar matrices and we delete one column and one row in both of them are they again similar?

Thanks!

答: Try to think about $\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ and  $\begin{pmatrix} b & 0 \\ 0 & a \end{pmatrix}$.


问:
Let $x_0$ be an extreme point of $f$ and $g$. Is $x_0$ an extreme point of $\max\{f,g\}$?

Let $x_0$ be an local extreme point of $f$ and $g$. Is $x_0$ an local extreme point of $\max\{f,g\}$?

Let $x_0$ is the local maximum point of $f$ and $g$, it is OK! What may happen if $x_0$ is the local minimum point of $f$, and the local maximum point of $g$?

答:
As written, $f,g$ are not required to be continuous.
Let $$f(x)=\begin{cases}2&x=x_0\\0&\text{otherwise}\end{cases}$$
$$g(x)=\begin{cases}1&x<x_0\\0&x=x_0\\3&x>x_0\end{cases}$$
Then $f$ has a local maximum and $g$ a local minimum at $x_0$. However,
$$\max\{f,g\}(x)=\begin{cases}1&x<x_0\\2&x=x_0\\3&x>x_0\end{cases}$$
does not have an extremum at $x_0$.


答:
There is no simpler form. The number

$$n!\sum_{k=0}^n\frac{(-1)^k}{k!}$$ is known as the subfactorial of $n$.


问:
Among real numbers, find the real solutions of: $x^2(y^2+z)=zy^2$, $y^2(z^2+x)=xz^2$, $z^2(x^2+y)=yx^2$
Among real numbers, find the real solutions of the system: $x^2(y^2+z)=zy^2$, $y^2(z^2+x)=xz^2$, $z^2(x^2+y)=yx^2$.

I tried solving it as follows:

$x^2y^2+x^2z=zy^2$

$x^2y^2=z(y^2-x^2)$

$z=\frac{x^2y^2}{y^2-x^2}$

$x(z^2-y^2)=y^2z^2$

$x=\frac{y^2z^2}{z^2-y^2}$

$y=\frac{x^2z^2}{x^2-z^2}$

And here is where I got stuck. Could you please help me finish off the question?

答:
Substituting $z$ we obtain two equations in $x$ and $y$,
which we can solve. Then we need to consider the case $x^2=y^2$ separately.

In summary, we have only the following solutions (even over the complex numbers):

$$
(x,y,z)=(0,0,z),(0,y,0),(x,0,0).
$$
So two of the variables equal to zero, and the third one arbitrary.

More precisely, the two equations are, assuming $x^2-y^2$ nonzero and substituting $z=x^2y^2(y^2-x^2)$
$$
(x^2y^2 - x^2 - xy^2 + y^2)(x^2 - xy^2 - y^2)xy^2=0,
$$
and
$$
(x^2y^2 + x^2y + x^2 - y^2)(x^2y - x^2 + y^2)x^2y=0.
$$


问:
Overlap between vector subspaces
Given two vector subspaces $U, V$ in $\mathbb{R}^n$ with $\dim(U), \dim(V) < n$, is there a quantity that expresses the amount of "overlap" between the two spaces in a meaningful way? The quantity should increase when the overlaps between individual vectors from $U, V$ increase.

EDIT, MORE DETAIL:

Let us say that I know $U$ exactly; it is spanned by some eigenvectors of a matrix $M$ I have. However $M$ might be hard to diagonalize, so I try to approximate by some algorithm I have. This algorithm generates $V$, and I hope I will find the vectors that span $U$ in there, or vectors that have large overlaps with the aforementioned eigenvectors.

Let us for simplicity say that $U$ and $V$ have the same dimension, say, 4. By using the vectors in $V$ I find that the best overlaps I can get with the eigenvectors I'm seeking are 0.99, 0.98, 0.90, and 0.80. Overlap between vectors is the scaled inner product

$$\frac{\langle u | v \rangle^2}{\langle u | u \rangle \langle v | v \rangle}$$

In this case none of the overlaps are 1, so wouldn't that make the dimension of the intersection zero? If so then I need some other quantity than the one suggested by @JLinsta and @Aldoggen that somehow takes the overlaps above into consideration... If such a quantity exists :P

答:
Regarding the new problem, I do not know of a generalization of individual overlaps, but a simple ad hoc solution exists: gather your overlaps into a vector $o=(0.99, 0.98, 0.90, 0.80)$, take the norm, and normalize with the norm of $(1,1,1,1)$. This normalized norm I will call the summary $s$. It is a number between 0 and 1, and increases as the individual overlaps increase.

You want the summary $s$ to only be close to 1 when all overlaps are close to 1, so you could take the $p$-norm

$$\lVert x\rVert_p = \left({\sum_{i=1}^k}x_i^p\right)^\frac1p$$

with $p$ a small number, for example $p=0.2$. If $p<1$ this formula is not actually a norm, but it's still useful. With $p=0.2$ and $o$ as above, we find $s\approx0.914$. I also made this interactive desmos example.

An alternative is to go with my answer for the original question: the intersection $W$ of $U$ and $V$ is again a subspace of $\mathbb R^n$, so you could look at its dimension. If you want to scale this quantity, you can notice that the dimension of $W$ will be at most $\min(\dim U, \dim V)$, so you can divide by that to get the formula which JLinsta mentioned in the comments:
$$\frac{\dim W}{\min(\dim U, \dim V)}.$$

Even if none of the overlaps are 1, the approximating eigenvectors can still generate exactly the same vector space as the eigenvectors of $M$ itself, in which case the above quantity will be 1. For example, the real plane is generated by the standard basis, but also by the standard basis rotated by one degree.

Title: Overlap between vector subspaces


答:
Take any sequence $c_n$ that converges to $1$.

Now the ratio of the sequences $a_n:=b_nc_n$ and $b_n$ obviously tends to $1$ regardless the convergence of $a_n$ and $b_n$.

But if they do converge, the converge to the same limit.

Title: If limit of two sequences converges, do the sequences converge too?

Body:
We assume that $a_n\neq b_n$ and $a_n\neq \frac{1}{b_n}$, and we know $\frac{a_n}{b_n}$ converges to the limit $1$ (in other words $\lim_{n\to\infty}\big(\frac{a_n}{b_n}\big)=1$).

This means that $\frac{\lim_{n\to\infty}(a_n)}{\lim_{n\to\infty}(b_n)}=1$ and thus $\lim_{n\to\infty}(a_n)=\lim_{n\to\infty}(b_n)$.

Can we now assume that both of these sequences converge, in particular to a same limit $L$? Can we say $$\exists L\in\mathbb{R}\Rightarrow \lim_{n\to\infty}(a_n)=L=\lim_{n\to\infty}(b_n)?$$


问:
Intuitive explanation of Cauchy's Integral Formula in Complex Analysis
There is a theorem that states that if $f$ is analytic in a domain $D$, and the closed disc {$  z:|z-\alpha|\leq r$} contained in $D$, and $C$ denotes the disc's boundary followed in the positive direction, then for every $z$ in the disc we can write:
$$f(z)=\frac{1}{2\pi i}\int\frac{f(\zeta)}{\zeta-z}d\zeta$$

My question is:
What is the intuitive explanation of this formula? (For example, but not necessary, geometrically.)

(Just to clarify - I know the proof of this theorem, I'm just trying to understand where does this exact formula come from.)

答:
Perhaps you can see the Cauchy's integral formula as the "Winding number" of a curve $C$ around any complex number $a$.

For more information, please see : en.wikipedia.org/wiki/Winding_number

OwnerDisplayName: anonymous


问:
Why is the endomorphism ring of $\mathbb{Z}\times\mathbb{Z}$ noncommutative?
So I hear that the endomorphism ring of an abelian group is not always commutative. In particular, I'm looking at the abelian group $A=\mathbb{Z}\times\mathbb{Z}$, and considering $\text{End } A$. I can't find a counterexample to show that $\text{End } A$ is not commutative. Does anyone know of two such endomorphisms that don't commute?</p>

答:
I know other people have given good answers but here's an explicit example:
$\phi : (a,b) \mapsto (b,a)$ and $\theta: (a,b) \mapsto (a+a,b)$
both are homomorphisms from $A$ to $A$, but the order you apply them matters.</p>


问:
Find the polynomial $f(x)$ which have the following property
Find the polynomial $p(x)=x^2+px+q$ for which $\max\{\:|p(x)|\::\:x\in[-1,1]\:\}$ is minimal.

This is the 2nd exercise from a test I gave, and I didn't know how to resolve it. Any good explanations will be appreciated. Thanks!

答:
Here are some hints:

$p''>0$ implies that $\max\{p(x):x\in[-1,1]\}=\max\{p(1),p(-1)\}$,

$p(1)=1+p+q$ and $p(-1)=1-p+q$,

Figure out which of those two values is bigger for each $p$ and $q$.

EDIT: I misread the question, so that the first point is wrong but can be fixed looking at $p(v_x)$ where $v_x$ is the $x$-coordinate of the vertex $v_x=\frac{-p}{2}$.


答:
We have
$$\sum_{j=0}^{n}p_j\log_2(1/r_j)-\sum_{j=0}^{n}p_j\log_2(1/p_j)=\sum_{j=0}^{n}p_j(\log_2(1/r_j)-\log_2(1/p_j))$$
Two basic properties of logarithms is that
$$\log_n(a)-\log_n(b)=\log_n(a/b)$$
And
$$\log(a/b)=-\log(b/a)$$
Substituting these into the last sum gives the desired answer.

Title: Showing $\sum_{j=1}^n p_j\log_2\frac1{r_j}-\sum_{j=1}^n p_j\log_2\frac1{p_j}=-\sum_{j=1}^np_j\log_2\frac{r_j}{p_j}$

Body:
How is
$$
\sum_{j=1}^{n} p_j \cdot \log_2\frac1{r_j} - \sum_{j=1}^{n} p_j \cdot \log_2\frac1{p_j}
$$
the same as
$$
-\sum_{j=1}^{n} p_j \cdot \log_2 \frac{r_j}{p_j}
$$
?

(It is for proving that the kullbach-lieber divergence is nonnegative and first I have to rewrite it)


问:
Prove that $ \overrightarrow{a}\left(t\right)||\overrightarrow{x}\left(t\right) $ in circular motion
This may sound like a physics problem, but I got it in my homework in differential equations course, so Im asking here.

A particle moves on the standard sphere $ \mathbb{S}^{n-1}\subset\mathbb{R}^{n} $. I have to prove that $ \overrightarrow{v}\left(t\right)\perp\overrightarrow{x}\left(t\right) $, and given that $ |\overrightarrow{v}|=\text{const} $ I have to prove that $ \overrightarrow{a}\left(t\right)||\overrightarrow{x}\left(t\right) $.

The first part wasnt hard, if we note that $ 1=\overrightarrow{x}\cdot\overrightarrow{x} $ then $ 0=\left(\overrightarrow{x}\cdot\overrightarrow{x}\right)'=\overrightarrow{v}\cdot\overrightarrow{x}+\overrightarrow{x}\cdot\overrightarrow{v} $ and thus $ \overrightarrow{x}\cdot\overrightarrow{v}=0 $.

But Im not sure about the second part. If $|v|=const $, how can I show that $ a||x $ ?

I tried to write:

$ 0=\frac{d}{dt}\left(|v|^{2}\right)=\frac{d}{dt}\left(\sum_{k=1}^{n-1}v_{k}^{2}\left(t\right)\right)=\sum_{k=1}^{n-1}2v_{k}\left(t\right)a_{k}\left(t\right)=2\overrightarrow{v}\cdot\overrightarrow{a} $

Which shows that $ \overrightarrow{v}\perp\overrightarrow{a} $, But for general $ \mathbb{S}^{n-1} $ it does not promise that $ a||x $ as far as I understand.

Another way that I tried is to write:

$ 0=\left(\overrightarrow{x}\cdot\overrightarrow{v}\right)'=\overrightarrow{x}\cdot\overrightarrow{a}+|v|^{2} $

and thus $ \overrightarrow{x}\cdot\overrightarrow{a}=-|v|^{2} $
But Im not sure how to proceed or if it gives me anything...

Also, Im trying not to make any physical assumptions because its a mathenatical course, so a mathematical solution would be very helpful. Thanks in advance.

答:
You can re-parametrize any (sufficiently smooth) curve on the sphere so that its speed is constant, this is just a parametrization with some multiple of the curve length.

You only get the claim under additional assumptions, like if the particle moves on the sphere mechanically without the influence of any outside forces, only the holonomic constraint forces. Then the mechanical equation reads
$$
\ddot x =\lambda\nabla h(x)
$$
where $\lambda$ can depend on the time and $h(x)=0$ describes the constraint, here for example $h(x)=\frac12(\|x\|^2-1)$. That gives immediately $\nabla h(x)=x$ and thus $a=\ddot x$ parallel to $x$.

One can refine this by computing the time derivatives of $h(x(t))$ to get
$$
0=x·\dot x\\
0=\|\dot x\|^2+x·\ddot x=\|\dot x\|^2+\lambda\\
\dot\lambda=-2\dot x·\ddot x=-2\lambda\dot x·x=0.
$$
This gives more specifically that $\lambda=-\|v\|^2$ is constant, $\|v\|=c_v$, and $a=-c_v^2x$ which is of course still parallel to $x$.


问:
Boundedness in $L^2$ implies convergence in $L^2$ for martingales.
a. Show that if $X = (X_n)_n$ is an $L^2$ martingale bounded in $L^{2}$ then there exists a positive r.v. $U \in L^{2}$ that satisfies :
$$
P\left(\forall n \geq 1,\left|X_{n}\right| \leq U\right)=1
$$

b. Deduce from a) or otherwise that $X$ converges in $L^2$.

What I could do so far : since $(X^2_n)_n$ is a submartingale then $\sup_nEX^2_n = \sum_nEX^2_n < \infty$ and thus : $EX^2_n \to 0$ so $X_n$ converges to $0$ in $L^2$.

my questions :

is my work for b, correct?

how to show a) and can it actually be used to deduce b) ?

Thanks!

答:
Your work for $b)$ is not correct as Kavi Rama Murthy already mentioned in the comments. The statement $\sup_n EX_n^2=\sum_n EX_n^2$ is NOT true in general. Here is a solution for $a)$:

The r.v. $U$ you are looking for is given by $U:=\sup_n |X_n|$. So the $|X_n|\leq U$-part is trivial and you just have to show $U\in L^2$. This can be done by using the monotone convergence theorem for the sequence $U_n:=\sup_{m\leq n} |X_m|$,  Doob's maximal inequality and the fact that $M:=\sup_n EX_n^2<\infty$ by assumption
$$EU^2=E(\lim_nU_n)^2=\lim_nEU_n^2\leq\lim_n4 EX_n^2\leq4M<\infty.
$$

To your $b)$-part: From $a)$ you can conclude that $X^2$ is a uniformly integrable submartingale. Now use Doob's martingale convergence theorem to conclude that it converges a.s. an in $L^1$ to $X_\infty^2$. By Vitali's convergence theorem you can now conclude that $X$ converges in $L^2$. (The last part you can also be done different depending on which version of the theorems you are using.)


问:
Finite rank nilpotent operators
Let $H$ be a separable complex Hilbert space

Let $\mathcal{F(H)} =\{T \in \mathcal{B(H)}: \text{rank}(T) < \infty \}$ ,

and $\mathcal{N(H)} = \{ T \in \mathcal{F(H)}: T^k=0 \text{ for some }  k \in \mathbb{N} \}$

Find the norm closure $\overline{\mathcal{N(H)}}$.

Ideas:

I know that $\overline{\mathcal{F(H)}} = \mathcal{K(H)}$, where $\mathcal{K(H)}$ is the set of compact operators.

So the closure will be some compact operators.

And $\mathcal{N(H)} \subset \mathcal{K(H)} $.

Also, through some research, I've found that the quasinilpotnets are limits of nilpotent.

So I'm suspecting the that the norm close of finite rank nilpotents will be compact and quasinilpotents.

Any help will be appreciated!

Thank you in advance!

答:
Let $T$ be quasi-nilpotent and compact. Then $T\subset \overline {\mathcal N(H)}$. (This is the proof of Douglas mentioned in the Hilbert space problems paper by Halmos).

First, the spectrum is upper semicontinuous: For every $\epsilon>0$ there is $\delta>0$ such that $\sigma(S) \subset B_\epsilon(0)$ for all $S$ with $\|S-T\|<\delta$. (This can be seen as follows: If $|\lambda|\ge\epsilon$ then $\lambda I-T$ is continuously invertible by $\sigma(T)=\{0\}$, and so is $\lambda I-S$ for $S$ close to $T$.)

Take $\epsilon>0$ and let $\delta>0$ be as above.
Since $T$ is compact and $H$ is Hilbert, there is a finite-rank operator $S$ with $\|T-S\|<\delta$.
Since $S$ has finite-rank, we can triangularize $S=D+N$, where $D$ is diagonal and $N$ is nilpotent:
Write $S(x) = \sum_{i=1}^n \langle x,e_i\rangle f_i$, then build the complex Schur form of the matrix representation of $S$ on $span(e_1\dots e_n,f_1\dots f_n)$. Note that the resulting coordinate transform is unitary.
Since the diagonal $D$ contains the non-zero eigenvalus of $S$, it follows $\|D\|<\epsilon$, where we used again that the underlying basis is orthonormal. And $\|T-N\|< \|T-S+D\|<\epsilon+\delta$.

This proves one inclusion. I do not know how to prove the reverse.

Title: Finite rank nilpotent operators


问:
Calculate the ratio variance between two estimators
Let $X_1,X_2,...,X_n \sim Unif[0,\beta]$ and $\hat{\beta_1} = (n+1)X_{min}$ and $\hat{\beta_2} = (n+1)/n X_{max}$. Compute $\frac{var[\hat{\beta_1}]}{var[\hat{\beta_2}]}$. What I have done is the following:

$$
\begin{align*}
  \frac{var[\hat{\beta_1}]}{var[\hat{\beta_2}]} & = \frac{var[(n+1)X_{min}]}{var[\frac{n+1}{n}X_{max}]} \\
  & = \frac{var[(n+1) n \left(1 - \frac{x}{\beta} \right)^{n-1} \frac{1}{\beta}]}{var[\frac{n+1}{n} \frac{n}{\beta^n} x^{n-1}]} \\
  & = \frac{\int _0^{\beta \:}\:x^2\:\left(n+1\right)\:n\:\left(1\:-\:\frac{x}{\beta \:}\:\right)^{n-1}\:\frac{1}{\beta }dx-\beta ^2}{\int _0^{\beta \:}\:x^2\:\frac{n+1}{n}\frac{n}{\beta ^n}x^{n-1}dx-\beta ^2} \\
  & = \frac{\frac{\beta^2n^2 + \beta^2n}{n+2} - \beta^2n}{\frac{\beta^2(n+1)}{n+2}-\beta^2} \\
  & = n
\end{align*}
$$

However, I am not sure my final answer is correct and if it is what do I conclude? That the ratio variance between two estimators is equal to the amount of samples?

答:
Your final answer is not quite correct (somehow both the variances you found look negative):

Both estimators are unbiased

$X_{\min}$ and $X_{\max}$ have the same variance of $\frac{n}{(n+1)^2(n+2)}\beta^2$, which can be found from a Beta distribution

so $(n+1)X_{\min}$ has a variance of $\frac{n}{(n+2)}\beta^2$

and $\frac{n+1}{n}X_{\max}$ has a variance of $\frac{1}{n(n+2)}\beta^2$

giving a ratio of the variances of $n^2$

So the ratio of variances between the two estimators is equal to the square of the sample size.  Using the estimator based on the sample maximum looks much better, which makes sense as the maximum is the minimal sufficient statistic here.


问:
Solving cosh/sinh equation
I struggle with equations involving exponentials... In particular, I don't find how to solve this one analytically. I found graphically that there is exactly one solution when $a \geq 1$ but I cannot solve that analytically...

Any help is greatly appreciated. The goal is to find for which value(s) of $a$ the following equations admits exactly one solution. Any help is greatly appreciated...

$$
x (\mathrm{exp}({\frac{2x}{a^2}}) + 1) = \mathrm{exp}({\frac{2x}{a^2}}) - 1
$$

答:
Say $a\neq 0$.

$x=0$ is a solution. So you should make it to be the only one. The equation can be written, with $z=\frac{2x}{a^2}$, as $$\frac{a^2}{2}z=\tanh \frac z 2$$ Now note that $\frac{d}{dz}\tanh \frac z 2=\frac12\mbox{sech}^2 \frac{z}{2}\le\frac12$. On the other hand $\tanh \frac z 2 \leq1$ so if $\frac{a^2}2< \frac12$, then there will be three roots, otherwise you will have only one root which is $x=0$.

Title: Solving cosh/sinh equation


问:
Baby Rudin theorem 2.41
I am reading Baby Rudin chapter 2 and came across some question on Theorem 2.41.
When Rudin tries to prove that every infinite subset of $E$ has a limit points in $E$ implies $E$ is closed, he first supposes $E$ is not closed. Then there exists $x_0$ in $\mathbb{R}^k$ which is a limit point of $E$ but not a point in $E$. Then he constructs the set $S$. But later, he introduces $y$ and applies triangular inequality. I do not understand what $y$ is, and also, how he gets the last inequality of greater than or equal to $\frac12|x_0-y|$?

For reference, here is the proof given by Rudin:

2.41 $\ \ $ Theorem $\ \ $ If a set $E$ in ${\bf R}^k$ has one of the following three properties, then it has the other two:

$\quad(a)\ \ $ $E$ is closed and bounded.

  $\quad(b)\ \ $ $E$ is compact.

  $\quad(c)\ \ $ Every infinite subset of $E$ has a limit point in $E$.

Proof $\ \ $ If $(a)$ holds, then $E\subset I$ for some $k$-cell $I$, and $(b)$ follows from Theorems $2.40$ and $2.35$. Theorem $2.37$ shows that $(b)$ implies $(c)$. It remains to be shown that $(c)$ implies $(a)$.

  $\qquad$ If $E$ is not bounded, then $E$ contains points ${\bf x}_n$ with $$|{\bf x}_n|>n\qquad(n=1,2,3,...).$$ The set $S$ consisting of three points ${\bf x}_n$ is infinite and clearly has no limit point in ${\bf R}^k$, hence has none in $E$. Thus $(c)$ implies that $E$ is bounded.

  $\qquad$ If $E$ is not closed, then there is a point ${\bf x}_0\in {\bf R}^k$ which is a limit point of $E$ but not a point of $E$. For $n=1,2,3,...,$ there are points ${\bf x}_n\in E$ such that $|{\bf x}_n-{\bf x}_0|<1/n$. Let $S$ be the set of these points ${\bf x}_n$ Then $S$ is infinite (otherwise $|{\bf x}_n-{\bf x}_0|$ would have a constant positive value, for infinitely many $n$), $S$ has ${\bf x}_0$ as a limit point, and $S$ has no other limit point in ${\bf R}^k$. For if $\mathbf y\in{\bf R}^k$, ${\bf y}\neq{\bf x}_0$, then $$\begin{align}|{\bf x}_n-{\bf y}| & \geq |{\bf x}_0-{\bf y}| - |{\bf x}_n-{\bf x}_0| \\ & \geq |{\bf x}_0-{\bf y}|-\dfrac1n\geq \dfrac12|{\bf x}_0-{\bf y}|\end{align}$$ for all but finitely many $n$; this shows that $\bf y$ is not a limit point of $S$ (Theorem $2.20$).

  $\qquad$ Thus $S$ has no limit point in $E$; hence $E$ must be closed if $(c)$ holds.

$\qquad$ We should remark, at this point, that $(b)$ and $(c)$ are equivalent in any metric space (Exercise $26$) but that $(a)$ does not, in general, imply $(b)$ and $(c)$. Examples are furnished by Exercise $16$ and by the space ${\scr L}^2$, which is discussed in Chap. 11.

答:
In the proof, Rudin shows that $S$ has a limit point $x_0$. However, at the start of the proof we find that $x_0 \in \mathbb{R}^k - E$ and as $S \subset E$ we have found an infinite subset of $E$ which doesn't have a limit point in $E$. Well, not quite yet. It might be the case that $S$ has another limit point $y$ which is not equal to $x_0$ and is in $E$. So we must first exclude this possibility to be able to conclude that we have found an infinite subset $S \subset E$ which has no limit point in $E$. The string of inequalities that follow in Rudin's proof is to show that $x_n$ can't be simultaneously getting close to $x_0$ and $y$. I suppose the hard part is the last part:

$$\vert x_0 - y \vert - \frac{1}{n} \geq \frac{1}{2} \vert x_0 - y \vert.$$

It is important to read what follows: "for all but finitely many $n$". If $n$ is large enough (and therefore, $\frac{1}{n}$ is very small) we would essentially have

$$\vert x_0 - y \vert \geq \frac{1}{2} \vert x_0 - y \vert.$$

Which should be very clear to why it is true. In fact, we find that for $n \geq \frac{2}{\vert x_0 - y \vert}$ (well defined as $x_0 \neq y$) the first inequality holds and for $n < \frac{2}{\vert x_0 - y \vert}$ (which are finitely many) it doesn't hold. So what exactly have we shown about $y$?  We have shown that $\vert x_n - y \vert$ isn't getting smaller and smaller as it should be if $y$ were a limit point of $S$. So $S$ doesn't have any other limit point besides $x_0$ and as $x_0 \notin E$ we have constructed an infinite set $S \subset E$ which has no limit point in $E$. However, this is a contradiction as we assumed c) to be true to prove a). Therefore, $E$ must be closed.

Hope this helps!

Title: Baby Rudin theorem 2.41


问:
Find a function describing distance between a moving point and a fixed point
In 2D space, a point moves at a constant velocity (vx, vy) starting from position (x1, y1). A second, stationary point is located at (x2, y2).
What is the function describing distance between the moving and the stationary point over time t?

```<code>distance(t) = some expression over x1, x2, y1, y2, vx and vy
</code>```

答: The euclidean distance at time $t$ is given by:$$d(\pmatrix {x_1\\y_1}+t\pmatrix{v_x\\v_y},\pmatrix{x_2\\y_2})=\sqrt{(x_1-x_2+tv_x)^2+(y_1-y_2+tv_y)^2}$$


答:
Ok, this is embarrassingly easy.

```<code>distance(t) = sqrt((x1 - x2 + vx * t) ^ 2, (y1 - y2 + vy * t) ^ 2)
</code>```

I should have spent more time over it before posting...

Title: Find a function describing distance between a moving point and a fixed point

Body:
In 2D space, a point moves at a constant velocity (vx, vy) starting from position (x1, y1). A second, stationary point is located at (x2, y2).
What is the function describing distance between the moving and the stationary point over time t?

```<code>distance(t) = some expression over x1, x2, y1, y2, vx and vy
</code>```

Body:
Ok, this is embarrassingly easy.

distance(t) = sqrt((x1 - x2 + vx * t) ^ 2, (y1 - y2 + vy * t) ^ 2)

I should have spent more time over it before posting...


问:
Continuous random vector with joint pdf (find marginal and validity)
Let (X,Y) be a continuous random vector with joint probability density function
$$f_{X,Y}(x,y)=\left\{\begin{matrix}
6xy & for \; 0<x<1, 0<y<\sqrt{x}\\
 0& otherwise
\end{matrix}\right.$$

Show that it is a valid joint pdf and that the marginal pdf's are given by
$$f_{X}(x)=\left\{\begin{matrix}
3x^2 & for \; 0<x<1\\
 0& otherwise
\end{matrix}\right.$$
and
$$f_{Y}(y)=\left\{\begin{matrix}
3y(1-y^4) & for \; 0<y<1\\
 0& otherwise
\end{matrix}\right.$$

My try

I would think that to show that it is indeed a valid joint pdf I would have to show the double integral over both variables from $-\infty$ to $\infty$ is equal to 1. Would this be it?

For the marginal pdf's I am thinking that I should take the integral from $0$ to $1$ with respect to the variable of the pdf that I am not interested in. I.e for pdf for $X$ i want to integrate over $y$. Would this be correct?
I need this for the next part to find $P(Y>X^2).

答:
in order to finally clarify the question, have a look at the following drawings

In the left picture, when you want to derive $f_X(x)$ you integrate $dy$, in fuction of $X$  thus your integral bounds are $y \in [0;\sqrt{x}]$

In the right picture, when you want to derivee $f_Y(y)$ you integrate in $dx$, in function of $Y$ thus your integration bouds are $x \in [y^2;1]$


问:
G-oriented Manifold
Let M be a compact manifold provided with an action of  a lie group G.

What is the meaning when we say  that M is G-oriented and if possible what is the intuition behind this notion?

Actually I'm reading an article on equivariant cohomology which say: "if M is G-oriented, integration over M gives us a map from $H^{-\infty}_G(M)$ (G-equivariant de Rham cohomology of M with generalized coefficients)to $C^{-\infty}(\mathfrak{g})^G$ (the subspace of generalized functions on $\mathfrak{g}$ which are invariant under the action of G endowed from the adjoint action)".

Thanks!

答:
We say that $M$ is $G$-oriented if $M$ is oriented and the $G$-action preserves the orientation of $M$.

If $M$ (not necessarily compact) is $G$-oriented, then we can talk about integrating (compactly supported) smooth equivariant form $\alpha\in\mathcal{A}^\infty_G(\mathfrak{g},M)=C^\infty(\mathfrak{g},\mathcal{A}(M))^G$
$$
\int_M\alpha\colon X\in\mathfrak{g}\mapsto\int_M\alpha(X)
$$
so get a map
$$
\int_M\colon \mathcal{H}^\infty_{G,c}(\mathfrak{g},M)\to C^\infty(M)^G
$$
where the differential defining the $G$-equivariant (compactly supported) cohomology group $\mathcal{H}^\infty_{G,c}$ is $(d_\mathfrak{g}\alpha)(X)=(d_M\alpha)(X)-i(X_M)(\alpha(X))$.  Now your article generalises this to distributions and do some more exotic things.

Title: G-oriented Manifold


问:
Regarding subgroups of $GL_{2}(\mathbb{F}_q)$ of order $p$, where $p$ is an odd prime dividing $q+1$.
The problem (this is not a homework problem, though it does feel like one):

Suppose p and q are primes such that $p \vert q+1$ and p is odd. Then all subgroups of $GL_{2}(\mathbb{F}_{q})$ of order p are conjugate to one another.

What I know so far:

First of all, $\vert GL_{2}(\mathbb{F}_{q})\vert = q(q+1)(q-1)^2$.

If $q=2$, then $p=3$ and  $\vert GL_{2}(\mathbb{F}_{q})\vert =6$. Then the only subgroup of order $p$ is the unique Sylow $3$-subgroup. Of course this subgroup is conjugate to itself.

So, now we look at the case where $q$ is an odd prime. Obviously $p \nmid q(q-1)^2$. But, $p$ may not be the largest power of $p$ dividing $q+1$. So, the subgroups of order $p$ are not necessarily Sylow $p$-subgroups. Of course, they must be conjugate to a subgroup of a Sylow $p$-subgroup.

What is the right way to complete this argument? I am hoping for an answer that pushes me in the right direction without giving it all away (at least for now).

Thanks for your time. My apologies for the very basic question.

Edit: Check the comments below Derek's hint for a solution. Please let me know if you spot any problems with my argument.

答:
The Sylow $p$-subgroups of ${\rm GL}_2(q)$ (for $p$ an odd prime dividing $q+1$) are cyclic. Using that fact, you should be able to prove using Sylow's Theorem that all subgroups of order $p$ are conjugate.

To see that the Sylow $p$-subgroups are cyclic, observe that ${\rm GL}_2(q)$ contains an element of order $q^2-1$ arising from multiplication by a multiplicative generator of the field ${\mathbb F}_{q^2}$ of order $q^2$.


问:
Locus of points where tangents drawn on hyperbola make angle 45
Find the locus of points where a pair tangents drawn on the hyperbola $x^2 - y^2 = a^2 $ enclose an angle of $45$ degrees. This is what I've done so far.

$\theta$ between tangents is $45$ so

$\lvert \frac{m_2-m_1}{1+m_1m_2)}\rvert = tan45 =1$

$\lvert m_2-m_1\rvert = \lvert 1+m_1m_2\rvert$

I think point-slope form of hyperbola tangent will also be helpful here but I'm not sure how to apply it:
$$y = mx \pm \sqrt{a^2m^2-b^2}$$

答:
Hint :

The equation for pair of tangents can be converted into a quadratic in $m$ :

$$(x^2-a^2)m^2-2xy \cdot m + y^2+b^2=0$$

Now use Vieta's formulas  on the obtained condition for $m$,
$$(m_1+m_2)^2-4m_1m_2=(m_1-m_2)^2=(1+m_1m_2)^2$$

to obtain the desired locus.


问:
Beta function's domain
Why is the Beta function only defined for x,y > 0 ?
$$B(x,y)=\int_0^1t^{x−1}(1−t)^{y−1}dt $$

答:
set them $=0$ and verify that

$$B(0;0)= \int_0^1 \frac{1}{t(1-t)}dt$$

diverges

The same happens for any values of $a,b\leq 0$

Title: Beta function's domain


问:
Solving the following ODE.
I came up with the following ODE:

$\frac{dx}{dt}+\frac{dy}{dt}=k_1(a-x-y)(b-x)+k_2(a-x-y)(c-y)$

where $k_1,k_2,a,b,c$ are constants. I know under changes of variable, the equation can be changed into the following form:

$\frac{dx}{dt}+\frac{dy}{dt}=k_1(a-x-y)x+k_2(a-x-y)y$

which is simpler. But I can't go any further. Hope someone could help. Thanks!

答:
Starting from your simplification:

$$\frac{dx}{dt}+\frac{dy}{dt}=k_1(a-x-y)x+k_2(a-x-y)y$$

$$\frac{dx}{dt}+\frac{dy}{dt}\;=\left[k_1x(a-x-y)+k_2y(a-x-y)\right]$$

$$\int\left(dx+dy\right)=\int\bigg(ak_1x+ak_2y-k_1x^2-k_2yx-k_1xy-k_2y^2\bigg)dt$$

$$x+y=\int\bigg(ak_1x+ak_2y-k_1x^2-k_2yx-k_1xy-k_2y^2\bigg)dt$$

We cant integrate  $x$ and $y$ wrt. $t$ because they implicitly depend on $t$.

Title: Solving the following ODE.


答:
Hint.

If $\|\vec v\| = c_0$ then $\|\vec v\|^2 = c_0^2$ and deriving $
\vec a\cdot \vec v = 0$ but if the movement is circular $\|\vec x\|= c_1$ or $\|\vec x\|^2= c_1^2$ and deriving $\vec v\cdot \vec x = 0$ so we arrived to

$$
\cases{\vec a\cdot \vec v = 0\\
\vec v\cdot \vec x = 0
}
$$

NOTE

Also from $\vec v\cdot \vec x = 0$ deriving again $\vec a\cdot \vec v + \|\vec v\|^2=0$ then $\vec a\cdot \vec v \le 0$

A general result can be obtained assuming the material point moves on the sphere without the influence of external forces. In this case the lagrangian can be written as

$$
L = \frac 12 m \|\dot{\vec x} \|^2-\lambda(\| \vec x \|^2-r^2)
$$

with $\lambda$ as a Lagrange multiplier so the movement equations follow

$$
m\ddot{\vec x}+\lambda\vec x = 0
$$

concluding that $\ddot{\vec x} = \vec a$ is parallel to $\vec x$ in the same way as @Lutz Lehmann's answer.

Title: Prove that $ \overrightarrow{a}\left(t\right)||\overrightarrow{x}\left(t\right) $ in circular motion

Body:
This may sound like a physics problem, but I got it in my homework in differential equations course, so Im asking here.

A particle moves on the standard sphere $ \mathbb{S}^{n-1}\subset\mathbb{R}^{n} $. I have to prove that $ \overrightarrow{v}\left(t\right)\perp\overrightarrow{x}\left(t\right) $, and given that $ |\overrightarrow{v}|=\text{const} $ I have to prove that $ \overrightarrow{a}\left(t\right)||\overrightarrow{x}\left(t\right) $.

The first part wasnt hard, if we note that $ 1=\overrightarrow{x}\cdot\overrightarrow{x} $ then $ 0=\left(\overrightarrow{x}\cdot\overrightarrow{x}\right)'=\overrightarrow{v}\cdot\overrightarrow{x}+\overrightarrow{x}\cdot\overrightarrow{v} $ and thus $ \overrightarrow{x}\cdot\overrightarrow{v}=0 $.

But Im not sure about the second part. If $|v|=const $, how can I show that $ a||x $ ?

I tried to write:

$ 0=\frac{d}{dt}\left(|v|^{2}\right)=\frac{d}{dt}\left(\sum_{k=1}^{n-1}v_{k}^{2}\left(t\right)\right)=\sum_{k=1}^{n-1}2v_{k}\left(t\right)a_{k}\left(t\right)=2\overrightarrow{v}\cdot\overrightarrow{a} $

Which shows that $ \overrightarrow{v}\perp\overrightarrow{a} $, But for general $ \mathbb{S}^{n-1} $ it does not promise that $ a||x $ as far as I understand.

Another way that I tried is to write:

$ 0=\left(\overrightarrow{x}\cdot\overrightarrow{v}\right)'=\overrightarrow{x}\cdot\overrightarrow{a}+|v|^{2} $

and thus $ \overrightarrow{x}\cdot\overrightarrow{a}=-|v|^{2} $
But Im not sure how to proceed or if it gives me anything...

Also, Im trying not to make any physical assumptions because its a mathenatical course, so a mathematical solution would be very helpful. Thanks in advance.


答:
You have

$$
B(x,y)\geq\int_0^{1/2}t^{x−1}(1−t)^{y−1}dt \geq \frac{1}{2^{y-1}} \int_0^{1/2}t^{x−1} \, dt.
$$

Now the last integral is finite if and only if $x>0$. Similarly for $y$.

Title: Beta function's domain

Body:
Why is the Beta function only defined for x,y > 0 ?
$$B(x,y)=\int_0^1t^{x−1}(1−t)^{y−1}dt $$


答:
Here's a start, but I must sleep now.

Maxima can occur at $-1$, $1$, and points where the polynomial has derivative $0$.

At $-1$, the polynomial has absolute value $|1 - p + q|$.

At $1$, it has absolute value $|1 + p + q|$.

$p'(x) = 2x + p$, which vanishes at $x = -p/2$ (assuming this is in range), at which point the absolute value is $|p ^ 2/4 - p^2/2 + q|$ = $|-p^2/4+q|$

Title: Find the polynomial $f(x)$ which have the following property

Body:
Find the polynomial $p(x)=x^2+px+q$ for which $\max\{\:|p(x)|\::\:x\in[-1,1]\:\}$ is minimal.

This is the 2nd exercise from a test I gave, and I didn't know how to resolve it. Any good explanations will be appreciated. Thanks!


问:
Integral using gamma function $\int_0^1 x^a(1-x)^b dx = \frac{\Gamma(1 + a) \Gamma(1 + b)}{\Gamma(2 + a + b)} $
Does anyone knows how to show

$$\int_0^1 x^a(1-x)^b dx = \frac{\Gamma(a+b-2)}{\Gamma( a -1)\Gamma(b-1)} $$
for $a,b \gt -1?$

Also why it doesn't work for $a$ or $b\lt -1$ ?

答:
The formula you wrote in the question is incorrect. In the title you have the correct one. To prove it, let's consider
\begin{align} \Gamma(a+1)\Gamma(b+1) &= \int_0^\infty t^a e^{-t}dt \int_0^\infty s^b e^{-s}ds = \\
&= \int_0^\infty  \Big( \int_0^\infty t^a s^b e^{-t-s}ds \Big) dt = \\
&= \int_0^\infty \Big( \int_t^\infty t^a (u-t)^b e^{-u}du \Big) dt = \\
&= \int_0^\infty \Big( \int_0^u t^a (u-t)^b e^{-u} dt \Big) du =  \\
&= \int_0^\infty \Big( \int_0^1 (ux)^a (u-ux)^b udx \Big) e^{-u}  du = \\
&= \int_0^\infty \Big( \int_0^1 x^a (1-x)^b dx \Big) u^{a+b+1} e^{-u} du = \\
&= \int_0^\infty u^{a+b+1} e^{-u} du \int_0^1 x^a (1-x)^b dx = \\
&= \Gamma(a+b+2)\int_0^1 x^a (1-x)^b dx  \end{align}

It doesn't work if $a<-1$ or $b<-1$ because this integral are not converegnt for such values of $a$ and $b$.


问:
Optimization approaches to solving PDEs
In modern numerical methods, a PDE is often recast into the form of a variational problem, which is sometimes equivalent to a minimization problem.
However in my courses on numerical analysis (say, finite element methods) the focus is not (apparently) on developing optimization techniques to minimize the arosen energy functional, but rather on approximating the variational problem on a smaller subspace.

Are there interesting approaches that focus on the minimization of the energy directly? Is research being done in this field, and could you maybe provide some reference?

答:
Many physical phenomena are a result of dynamic/static equilibrium energy/work conservation in physics/mathematical physics. A pde represents a phenomenon based on the laws of physics that  obey the phenomenon.

This answer would do little justice to so many phenomena in Science.

Kontorovich's text book in Variational methods is informative in geometry. A particular case from geometry in linear plate theory by Sophie Germaine:

$$\left(\dfrac {\partial^2 }{\partial x^2}+\dfrac {\partial^2 }{\partial y^2} \right)\left(\dfrac {\partial^2 w}{\partial x^2}+\dfrac {\partial^2 w}{\partial y^2} \right)= \dfrac{p}{D}$$

where $ \text{ bending deformation w, pressure p, D plate bending rigidity} $  are obtained by neglecting second order terms from

$$ A \Delta(e_i e_j) = B \Delta(e_i \kappa_j) = D \Delta(\kappa_i \kappa_j) $$

where $(A,B,D) $ are in-plane , cross coupling and stiffness coefficients  relating  double strain, cross strain-curvature and Gauss curvature  that are involved on  von Kármán's  equations of pure geometric non-linearity...

Navier-Stokes equation has not been generally solved using known mathematical functions.

Title: Optimization approaches to solving PDEs


问:
Can you turn an infinite direct sum into a finite one?
In a linear algebra book, the author defines, given a direct sum decomposition $V = U \oplus W$, the projection of $V$ onto $U$ with respect to this decomposition as $P_{U,W} \in \mathcal{L}(V)$ such that, given the unique decomposition $v = u+w$ where $u\in U, w \in W$, we have $P_{U,W}v=u$. Just to play around i decided to generalize it to arbitrary (potentially infinite) direct sums:

Suppose $$V = \bigoplus_{\alpha \in L}U_\alpha$$

(I defined this as each element of $V$ having uniquely as a sum $\sum u_\alpha$ where each $u_\alpha \in U_\alpha$ and only finitely many elements of this sum are non-zero. And defined infinite sums that have only finitely many non-zero elements as the corresponding finite sum of it's non-zero elements.)

Then we could define the projection onto one of those $U_\alpha$'s in the obvious way. However after thinking about this, i noticed that, at least if the sum is finite, this construction adds nothing new: We could have an equivalent decomposition that uses only 2 terms, for example if $V = U_1 \oplus$...$\oplus U_n$ then also $V=U_i \oplus \bigoplus_{k\neq i} U_k$. Can we do the same this if the direct sum is infinite?

答:
To get a concrete picture in mind, think of the space $X$, the direct sum of countably many copies of $\mathbb R$. This is simply (isomorphic to) the space of all eventually null sequences of real numbers. It is clear that indeed any such eventually null sequence is uniquely the finite sum of its projections to the non-null coordinates.

Now, if you single out any particular coordinate position, say the first one, and eliminate it, then you still have the space of all eventually null sequences. In other words, you get $X$ again. A moment's thought should thus show that $X\cong \mathbb R \oplus X$.

In the most general case the same phenomenon is true. The proof can be given directly in terms of the condition you gave in terms of unique decomposition (it's a straightforward exercise). Much more generally though it is a categorical principle. The direct sums you are looking at are the categorical coproducts in the category of vector spaces. First, note that similarity of the property you are asking about to associativity. You are basically asking if the brackets can be moved around in the usual way. Your question is about what happens when there are infinitely many summands. The universal property of coproducts guarantees that such moving around of brackets results in isomorphic answers. No axiom of choice is involved.


答:
Without further restrictions on the motion, it will not be true in general that $\ \vec{a}\,\|
\,\vec{x}\ $. If $\ \vec{x}(t)=\frac{1}{\sqrt{2}}\big(\cos(t)\,\mathbf{i}+\sin(t)\,\mathbf{j}+\mathbf{k}\big)\ $ in $\ \mathbb{R}^3\ $, for instance, then the particle is moving on $\ \mathbb{S}^2\ $, $\ \vec{v}(t)=\frac{1}{\sqrt{2}}\big(-\sin(t)\,\mathbf{i}+\cos(t)\,\mathbf{j}\big)\ $, so $\ \left|\,\vec{v}(t)\,\right|=\frac{1}{\sqrt{2}}\ $ is constant, but $\ \vec{a}(t)=-\frac{1}{\sqrt{2}}\big(\cos(t)\,\mathbf{i}+\sin(t)\,\mathbf{j}\big)\ $ is not parallel to $\ \vec{x}(t)\ $.

Reply to question in comments by OP

If $\ \vec{a}(t)\,\|\,\vec{x}(t)\ $ for all $\ t\ $, then $\ \vec{a}(t)=\ddot{\vec{x}}(t)=\lambda(t)x(t)\ $ for some scalar function $\ \lambda(t)\ $ of $\ t\ $.  Let $\ \vec{u}_1, \vec{u}_2,\dots, \vec{u}_{n-2}\ $ be  a basis for the subspace of $\ \mathbb{R}^n\ $ perpendicular to $\ \vec{v}(0)\ $ and $\ \vec{x}(0)\ $. I'm assuming here that $\ \vec{v}(0)\ne\vec{0}\ $, so $\ \vec{v}(0)\ $ and $\ \vec{x}(0)\ $ must span a space of dimension $2$.

Let $\ y_j(t)=\vec{u}_j\cdot\vec{x}(t)\ $.  Then $\ y_j(t)\ $ must satisfy the linear differential equation $\ \ddot{y}_j(t)=\lambda(t)y_j(t)\ $, and the initial  conditions $\ y_j(0)=\dot{y}_j(0)=0\ $. It follows that $\ y_j(t)=0\ $ for all $\ t\ $, and hence $\ \vec{x}(t)\ $ must remain in the $2$-dimensional subspace spanned by $\ \vec{v}(0)\ $ and $\ \vec{x}(0)\ $.

Therefore, if $\ \vec{a}\,\|\,\vec{x}\ $, and $\ \left|\,\vec{v}\ \right|\ $ is a non-zero constant, then the only possible motions $\ \vec{x}\ $ can have must be of uniform speed around a circle centered on the origin.


问:
Finding the integers roots of 2 variables polynomial $x^2-xy+y^2-1$
I'm looking for a way to convince myself that the roots of the polynomial  $x^2-xy+y^2-1$ s.t $(x,y) \in \mathbb{Z}^2$ are bounded with $|x|,|y|\leq 1$.

Is there is a way to present the sets of roots as elipse?

答:
As
$$x^2-xy+y^2-1=\frac{(x-y)^2+x^2+y^2}2-1 $$
any root will have
$$ (x-y)^2+x^2+y^2=2$$
and for integers this means that one of the three squares is $0$ and the other two are $1$.


问:
Invariance of syndetic set in topological semigroup
Let $T$ be a topological semigroup.  $A\subseteq T$ is syndetic, if there is compact set $K\subseteq T$ with $T=KA$.

Let syndetic set $A\subseteq T$ and $g\in T$ be given. I think that $Ag^{-1}=\{y: yg\in A\}$ is syndetic. Indeed  since $A$ is syndetic, there is compact set $K$ with $T=KA$. For every $t\in T$, since $tg\in KA$, hence $t\in KAg^{-1}$. This implies that $T= KAg^{-1}$ i.e.

if $A$ is syndetic, then $B=Ag^{-1}$ is syndetic.

What can say about situation of $Ag$, is it syndetic?

Please help me to know it.

答: The answer is no.  Take the semigroup $T=\{0,1\}$, viewed as a subgroup of the multiplicative integers, and equipped with the discrete topology.  Then $A=\{1\}$ is syndetic (take $K=T$), but if $g=0$, then  $Ag$ is not.


问:
If $A = \{x_n:n\ge 1\}$ is totally bounded, then $(x_n)$ has a Cauchy subsequence (Carothers, Lemma 7.3)

Let $(x_n)$ be a sequence in $(M,d)$, and let $A = \{x_n:n\ge 1\}$ be the range of the sequence. If $A$ is totally bounded, then $(x_n)$ has a Cauchy subsequence.

First, here are the definitions/characterizations of totally bounded I know:

$A\subset (M,d)$ is said to be totally bounded if for all $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in M$ such that $A\subset\bigcup_{i=1}^n B(x_i,\epsilon)$. Equivalently, $A\subset (M,d)$ is said to be totally bounded if for all $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in A$ such that $A\subset\bigcup_{i=1}^n B(x_i,\epsilon)$.

$A$ is totally bounded iff for all $\epsilon>0$,  there are finitely many sets $A_1,A_2,...,A_n\subset A$ with $\text{diam}(A_i) < \epsilon$ for all $1\le i\le n$ such that $A\subset \bigcup_{i=1}^n A_i$. (The $A_i$'s might as well be subsets of $M$, we don't care - it is easy to see the equivalence of the two characterisations).

I need some help in understanding the proof:

If $A$ is finite, we are done. Is the following reasoning correct: If $A$ is finite, then $x_n$ is eventually constant. Eventually constant sequences are Cauchy themselves - and so have a Cauchy subsequence. $\color{red}{\text{(Resolved).}}$

Why is $A_1$ also totally bounded? We know that $A\subset \cup_{i=1}^n A_i$, where $A_1$ contains infinitely many points of $A$. I know that subsets of a totally bounded set are also totally bounded, but $A_1\subset A$ is not necessarily true! $\color{red}{\text{(Resolved).}}$

Why does the author write (How?) in the second last line. Isn't it obvious that I can pick $x_{n_k}\in A_k$ since all the $A_k$'s are non-empty? What is there to justify? $\color{red}{\text{(Resolved).}}$

$\text{diam}(A_k) < 1/k$ by construction, but why is $\text{diam}\{x_{n_j}:j\ge k\} \le \text{diam}(A_k)$? We need this to conclude that the chosen subsequence is indeed Cauchy. $\color{red}{\text{(Resolved).}}$

I could simply have selected $1/2^k$ instead of $1/k$ right (or any other monotone decreasing function of $k$)? I don't find the choice of $1/k$ special. $\color{red}{\text{(Resolved).}}$

Here's the proof attached for reference:

Thanks a lot for the help!

答:
$A_1 \subseteq A$ is true! $A$ is a union by finitely many sets of diameter $<1$ (we can take $\varepsilon=\frac12$ in the definition of being totally bounded and know there are $x_1, \ldots x_n$ so that $A \subseteq \bigcup_{i=1}^n B(x_i, \frac12)$, we just take $A_1  =B(x_i, \frac12) \cap A$ for the ball that contains infinitely many points of $A$. So $_1$ is still totally bounded and its diameter is at most the diameter of $B(x_i, \frac12)$ which is $\le 1$ by the triangle inequality..

The above process we can keep on doing, using recursion.

Picking $x_{n_k}$ as a subsequence needs that each $A_k$ has infinitely many elements, not just non-emptyness: for $x_{n_1}$ we use non-emptyness, but then we need some $x_n$ with $n > n_1$ to be in $A_2$ and this is what the infinity of points gives us. This keeps on: having $n_k$ we need elements with larger index than $n_k$ to be in $A_{k+1}$ to keep the subsequence going (which must have a monotonically increasing subset $n_1 < n_2 < \ldots$ as indices).

And yes any other decreasing sequence of diameters would have done just as well, as long as it goes below any given $\varepsilon>0$.


答:
Consider $f(x,y)=x^2-xy+y^2-1$. Then $\nabla f(x,y)=(2 x - y, 2 y - x)$. The bounding rectangle for the ellipse is found by solving these equations:

$f(x,y)=0$, $2x-y=0$; solve for $y$

$f(x,y)=0$, $2y-x=0$; solve for $x$

This gives a square centered at the origin: $[-L,L] \times [-L,L]$, where $L=\frac{2}{\sqrt3}<2$.

Title: Finding the integers roots of 2 variables polynomial $x^2-xy+y^2-1$

Body:
I'm looking for a way to convince myself that the roots of the polynomial  $x^2-xy+y^2-1$ s.t $(x,y) \in \mathbb{Z}^2$ are bounded with $|x|,|y|\leq 1$.

Is there is a way to present the sets of roots as elipse?


问:
Solve the system of equations: $32y+32x^3=6x+17$, $16z+32y^3=6y+9$, $8x+32z^3=6z+5$ where $x,y,z\in \mathbb{R}$
Solve the system of equations: $$\begin{cases} 32y+32x^3=6x+17 \\16z+32y^3=6y+9 \\8x+32z^3=6z+5 \end{cases}$$ where $x,y,z\in \mathbb{R}$ (Bulgaria 1960)

I attempted to solve this question as follows:

$32y+32x^3=6x+17$

$y=-x^3+\frac{6x}{32}+\frac{17}{32}$

$y-\frac{1}{2}=-x^3+\frac{3x}{16}+\frac{17}{32}-\frac{1}{2}$

Here it started getting very complex, and hence I don't think it can be solved this way. After this I tried doing something similar with the other two equations, but once again it was ending up way too complex. It is obvious that the solution is $x=y=z=\frac{1}{2}$, but I can't manage to prove it. Could you please explain to me how to solve this question?

答:
Just for simplicity's sake, make a change of variables $$(a,b,c) = (2x,2y,2z)$$
We have the system of equations
\begin{align}
16b + 4a^3 &= 3a+17 \\
8c + 4b^3 &= 3b+9 \\
4a + 4c^3 &= 3c+5 \\
\end{align}
or
\begin{align}
16(b-1)  &= - 4a^3+3a+1 \\
8(c-1)   &= - 4b^3+3b+1 \\
4(a-1)  &=-4c^3+ 3c+1 \\
\end{align}
or
\begin{align}
16(b-1)  &= - (a-1)(4a^2+4a+1) \tag{1}\\
8(c-1)   &= - (b-1)(4b^2+4b+1) \tag{2}\\
4(a-1)  &=-(c-1)(4c^2+4c+1) \tag{3}\\
\end{align}

Suppose $b >1$ then from $(1)$: $a<1$ $\implies$ from $(3)$: $c>1$ $\implies$ from $(2)$: $b>1$  $\implies$ contracdiction!

Same for  $b <1$, we have also a contradiction.

For $b=1$, we have $c =1$ and $a =1$.

Conclusion: the system of equations has a unique solution $(a,b,c) = (1,1,1)$ or $$(x,y,z) = \left(\frac{1}{2},\frac{1}{2},\frac{1}{2} \right)$$


问:
Convex functions and determinants
In my introductory class to calculus, we were given this statement and we were supposed to find a proof of it.

Function is convex on a closed interval $\iff $  if given three random numbers in the interval, which satisfy this inequality: $x_1<x_2<x_3$; determinant of $$ \left( \begin{matrix} x_1 & f(x_1) & 1 \\ x_2 & f(x_2) & 1 \\ x_3 & f(x_3) & 1 \end{matrix} \right) $$
is positive.

I do not know how to approach this problem;
I figured out that the determinant of the matrix is equal to
$$x_1(f(x_2)-f(x_3)) + x_2(f(x_3)-f(x_1)) + x_3(f(x_1)-f(x_2))$$
Still, I don't see how this helps me prove that this is always positive for all convex functions. I do not want anyone to solve this for me, I just think I need a nudge in the right direction.

答: This is the shoestring formula for the signed area of the "triangle" formed by the points $(x_i, f(x_i))$, $i = 1, 2, 3$. (I put "triangle" in quotes in case these points lie on a straight line). What does convexity of the function imply about this signed area?

Title: Convex functions and determinants


问:
Show that $\sum^\infty_{n=m}\frac{1}{n^2} = O(\frac{1}{m})$
I have been told it is very simple, but I cannot work it out. I was thinking I use that I know the left hand side converges?</p>

Body:
If $m\geq 2$, then
$$
\sum\limits_{n = m}^\infty  {\frac{1}{{n^2 }}}  \le \sum\limits_{n = m}^\infty  {\frac{1}{{n(n - 1)}}}  = \sum\limits_{n = m}^\infty  {\left( {\frac{1}{{n - 1}} - \frac{1}{n}} \right)}  = \frac{1}{{m - 1}} \leq \frac{2}{m}.
$$


问:
$x-\frac{1}{x} - \frac{1}{x-\frac{1}{x}} - \dots $ can we represent this infinite expression by an integral or a summation?
$$x-\frac{1}{x} - \frac{1}{x-\frac{1}{x}} - \frac{1} {x-\frac{1}{x} - \frac{1}{x-\frac{1}{x}}} - \dots$$
this expression is the result of recursively subtracting $1/x$ from $x$,

i.e.

$$x_{n+1} = x_n - 1/x_n$$

when executing the previous expression multiple times it does not converge, as the number of executions increase the value of $x$ tends to change slower and slower.

I wonder if this expression can be represented by some integral or summation

答: If $x_{n+1} = x_n - \frac{1}{x_n}$ were to converge to a limit $L$ for some value of $x_1$, taking limits in the recurrence relation we would have $L = L - \frac{1}{L}$ which is absurd.


问:
A few questions on nonstandard analysis
I know that nonstandard analysis is analysis plus the existence of infinitesimal numbers. Does it mean that nonstandard analysis is the same theory as $ZF+\exists$infinitesimal numbers?

From what I read about it on Wikipedia there seem to be a few different approaches to it. There is "Robinson's method" which is "based on studying models". There is "Nelson's method" which is "Internal set theory" which is an extension of ZF. Are they the same? Or are there theorems that are true in one but not in the other?

If IST is an extension of ZF how much stronger than ZF is the "theory of nonstandard analysis"?

答:
To answer your first question, one can indeed think of nonstandard analysis, in a first approximation, as entailing the existence of infinitesimal numbers as you write, but to be more precise one would have to elaborate further.  I like to think of it in terms of three approaches (other editors may disagree), as follows:

(1) the most straightforward approach (in my opinion) is through the construction of a proper field extension $\mathbb R^\star$ (notation varies, but let's stick with this one which is used in Keisler's book, see http://www.math.wisc.edu/~keisler/calc.html)  of the real field $\mathbb R$.  This field extension can be constructed in a way somewhat similar to constructing $\mathbb R$ out of the rationals $\mathbb Q$.  Namely, one starts with sequences of real numbers, introduces a suitable equivalence, and obtains the hyperreal field $\mathbb R^\star$ as the quotient.  This construction is called the ultrapower construction.  A serious undergraduate algebra course provides enough background to understand this construction; namely what is required is the existence of  a certain maximal ideal in a suitable ring.

(2) A more sophisticated route (and the one taken by Robinson in his 1966 book, see http://www.google.co.il/books?id=OkONWa4ToH4C&source=gbs_navlinks_s) is to invoke the compactness theorem from mathematical logic (more specifically, model theory) so as to prove the existence of such an $\mathbb R^\star$ (in Robinson's book this is denoted $^\star\mathbb R$).

(3) Edward Nelson's approach, called IST (internal set theory) is a reformulation of Robinson's approach where, instead of extending the field $\mathbb R$, Nelson takes the "syntactic" route.  This means that the language of ordinary set theory is enriched by the addition of a unary predicate "Standard".  Then infinitesimals are found within the "ordinary" $\mathbb R$ itself; so to speak they have been there all along, it's just that we haven't noticed (because the ordinary syntax of set theory is insufficiently rich).  An infinitesimal number is NOT "Standard".

All three approaches are equivalent (at least at a basic level), so one proves the same theorems in all of them.  No new axioms are needed beyond those of ZFC.

A more detailed summary by Joel David Hamkins appears here.

Title: A few questions on nonstandard analysis


问:
Prove that a subclass of any set is also a set
I have been asked, while discussing the ZF subset axioms to prove that a subclass of any set is a set. I have put together the following proof and would be grateful if someone could confirm that it is correct.

By definition of classes, we know that classes and subclasses are given by predicates. Using the language of our axiom, whenever we have a set $A$ along with a predicate $\phi$, there exists a set $B$ which contains exactly the elements of $A$ which satisfy $\phi$. Hence if $A$ is a set then so must be the class $B=\{a\in A \ | \ \phi (a)\}$ which contains all elements of A which satisfy $\phi$.

答: This is exactly true because a class is a predicate and the axiom of separation holds by assumption: it tells us that the intersection of a set and a class is a set, in these terms.</p>


问:
Prove uniquness of solution for a simple ordinary differential equation
Consider the equation $ x'=f\left(x\right) $ Where $ x(t) $ is a fucntion of $ t $.

Assume $ x_0 $ is a root of $ f(x) $. That is $ f(x_0)=0 $, and assume that is the only root in some neighborhood of $ x_0 $.

So $x(t)=x_0 $ is a trivial solution for the differential equation. Now, assume in addition that $ f $ is continuous and $ f \neq 0 $ in the neighborhood $ \left(x_{0}-\varepsilon,x_{0}\right) $. (Without loss of generality assume $f<0 in that neighborhood).

So we can write the differential equation (locally) as:

$ \frac{x'}{f\left(x\right)}=1 $

And thus $ \intop_{x}^{x_{0}}\frac{ds}{f\left(s\right)}=\left(t_{0}-t\right) $

(where $ x_{0}-\varepsilon<x<x_{0} $). Now, my question is:

How can I prove that if $ \intop_{x}^{x_{0}}\frac{ds}{f\left(s\right)}=\infty $ then the solution is unique (meaning, the locally solution for the differential equation in the neighborhood $(x_0-\varepsilon,x_0) $ does not cross the trivial solution $ x(t)=x_0 $ and thus can be extended.

and if $ \intop_{x}^{x_{0}}\frac{ds}{f\left(s\right)}<\infty $ then there is no uniquness for the solution for a given initial value.

Im sorry if I wrote things that does now make sense, Im starting a new course in differential equation and I do not feel comfortable yet with all the terms.

Thanks in advance.

答:
Assume that you have a solution $x(t)$ with $x(t_0)\ne x_0$ and $x(t_1)=x_0$, $t_1$ minimal with that property. Then by the separation process, as you wrote
$$
t_1-t_0=\int_{x(t_0)}^{x_0}\frac{ds}{f(s)}
$$
Now the left side is finite, the right side diverges by assumption, which gives a contradiction. So there can not be such a $t_1$.


问:
Bijective function in a commutative diagram
Let $X,Y, Z$ be sets and $p_x:Z\rightarrow X$ and $p_y:Z \rightarrow Y$ so that for every Triple $(M, f_x, f_y)$ exactly one function $g:M\rightarrow Z$ exists, which makes following diagram commutative (Commutative $\Leftrightarrow f_x=g\circ p_x$ and $f_y=g\circ p_y$).

Show that for $(M, f_x, f_y)=(X\times Y, \pi_x, \pi_y)$, the received function $g:X\times Y \rightarrow Z$ is a Bijection.

$\pi_x(x,y) = x$ and $\pi_y(x,y) = y$

My attempt: Since the diagram is commutative, it follows...
$$x = \pi_x(x,y) = (p_x \circ g)(x,y) \Rightarrow x = (p_x \circ g)(x,y)$$
$$y = \pi_y(x,y) = (p_y \circ g)(x,y) \Rightarrow y = (p_y \circ g)(x,y)$$

So I think in order to get the desired values, $g$ must be...
$$g(x,y) = (p_x^{-1} \circ \pi_x)(x) \cap (p_y^{-1}\circ \pi_y)(y)$$

I'm unsure what to do next. I could also say...

$$ g^{-1}(x,y) = (\pi_x^{-1}\circ p_x)(x) \cap (\pi_y^{-1}\circ p_y)(y)$$

But since I don't know if the functions $\pi, p$ themselves are bijective, I cannot conclude that $g^{-1}$ exists

I'd like to have some tips on how to proceed.

答:
More specifically, we must have that the sets $p_X^{-1}(\pi_X(x,y))\cap p_Y^{-1}(\pi_Y(x,y))=p_X^{-1}(x)\cap p_Y^{-1}(y)$ contain exactly one element for each $x\in X,\ y\in Y$.

You just showed that $g(x,y)$ is such an element, and if it contained another one, we could modify $g$ for $x,y$ and the diagram would still commute, contradicting the uniqueness of $g$.

Explicitly, for a $z\in Z$, we will have $g^{-1}(z)=(p_X(z),\,p_Y(z))$.


问:
Finding Mean of Beta Distributed Continous Density function
Let $X ∈ (0,1)$ be beta distributed with β-parameter 10, so the density function is:

$p(x) = 10x^9$

How can i find the Mean ?

How can i find the Variance?

I'm not sure how properly find the values of this for a beta distribution.

I know how to find the Expected values and Variance for discrete values, and i generally i think i realize the formula for continous variables but with the beta distribution im not sure how the intervals should look on the integrals.

答:
$$E(X)=\int_0^1 10x^{10}dx$$

$$E(X^2)=\int_0^1 10x^{11}dx$$

$$V(X)=E(X^2)-E^2(X)$$

Your $X\sim Beta(10;1)$ thus the mean is $\frac{10}{11}$ and similarly for the variance you get $\frac{10}{11^2\cdot 12}$ but also direct calculation with the integral is easy


问:
If bitwise quadratic time is considered unfeasible, then what really is feasible for large numbers?
The naive algorithm for factoring a semi-prime $n$ runs $\mathcal{O}(\sqrt{n})$. Let's consider the bit-wise complexity of division to be $\mathcal{O}(b^2)$. That means the bitwise complexity of naive factoring is $\mathcal{O}(b^2\sqrt{n}) \in \mathcal{O}(\sqrt{n}^2\sqrt{n}) = \mathcal{O}(n^{1.5})$.

At first glance, this still seems pretty good. I understand that when $n$ is on the order of $2^{1000}$, even this under-quadratic complexity becomes unfeasible but then no matter what the runtime, you can always find an unfeasible $n$, e.g., if the runtime was $\mathcal{O}(\log_2(n))$, we could let $n = 2^{2^{1000}}$ and just like that even log time becomes unfeasible.

Could someone explain why then we often hear that factoring is an "exponentially" hard problem? Is that true? I mean, obviously it has been hard and things like RSA depend on it but is it really exponential? Or, is the matter really of picking the big enough number $n$?

If it is exponential how can I drive that in terms of big-Oh?

Edit: I understand that $f(b)=2^b$ is exponential thus when we increase the bits, $n=2^b$ grows exponentially. However, I am finding it hard to connect the dots between value versus length based complexity analysis. When we say factoring is $\mathcal{O}(\sqrt{n})$ we have assumed that the time complexity of division is constant therefore,

$$
\sum_{i=1}^{\sqrt{n}}\mathcal{O}(1) = \mathcal{O}(1)_1 + \mathcal{O}(1)_2 + ... + \mathcal{O}(1)_{\sqrt{n}} = \mathcal{O}(\sqrt{n}).
$$

Now I understand if we let go of the assumption that division is constant and plug the true complexity in the above set-up, we should get an exponential runtime. Say we take the Schoolbook long division then for "two b-digit numbers", division is $\mathcal{O}(b^2)$

$$
\sum_{i=1}^{\sqrt{n}}\mathcal{O}(b^2) = \mathcal{O}(b^2)_1 + \mathcal{O}(b^2)_2 + ... + \mathcal{O}(b^2)_{\sqrt{n}} = \mathcal{O}(b^2\sqrt{n}).
$$

Since $\mathcal{O}(\log_2(n)) = \mathcal{O}(b) \in \mathcal{O}(\sqrt{n})$, we can write this as $\mathcal{O}(n\sqrt{n}) = \mathcal{O}({n^{1.5}})$.

The answers say that since $n=2^b$, $\mathcal{O}(2^{1.5b})$ is exponential and when I think about it like that, it makes sense. However, the way we write $n$ in big-Oh shouldn't really matter because we have already accounted for the non-constant division cost so,

$$
n=2^b \Rightarrow \mathcal{O}(n^{1.5}) = \mathcal{O}(2^{1.5b})
$$

This feels somewhat rigging because if you give someone $g(x)=x$ and they say oh but I'll input $2^b, 2^{b+1}, 2^{b+3}, ...$ so your function $g$ is exponential! Well, what makes the function exponential is not really the growth of the function but the input given to function is being grown exponentially!

答:
Could someone explain why then we often hear that factoring is an "exponentially" hard problem?

Trial division is an exponential-time algorithm because runtime is measured as a function of the length of the input and not the value of the input bit-string. Because of this, you need $\mathcal O(2^{b/2})$ divisions to factor with trial division on a semi-prime of length $b$ bit. Clearly, this is an exponential-time algorithm. If you're still skeptical, compare this runtime to that of regular multiplication which runs in time $\mathcal O(b^2)$.

I mean, obviously it has been hard and things like RSA depend on it but is it really exponential?

Technically, no, because you have sub-exponential time factoring algorithms like the General Number Field Sieve. However, they are still sufficiently slow that a moderate increase in keysize keeps the required effort to be infeasibly high.

if the runtime was $\mathcal O(\log_2(n))$, we could let $n=2^{2^{1000}}$

Yes, that would work, but then you'd run into the situation where you need so absurdly large numbers that factoring the semiprime is faster than actually encrypting or decrypting data. Even if you picked a worse algorithm like $\mathcal O(\log_2^5(n))$, you still get into a situation where you need to pick keys that are like 1GB which is ... rather unpractical for most applications. In fact, quantum computers can factor in this kind of time and pqRSA (paper, slides) has done the math how large keys would need to be to offer reasonable security and they did arrive at about 1GB of key data.


问:
Suppose $p$ is prime and $m \gt 0$ and $n \gt 0$ and $m \equiv n \pmod{p-1}$. Prove that $a^m \equiv a^n \pmod{p}$ for all $a$.

Suppose $p$ is prime and $m \gt 0$ and $n \gt 0$ and $m \equiv n \pmod{p-1}$. Prove that $a^m \equiv a^n \pmod{p}$ for all $a$.

I know I have to apply Fermat's little theorem to prove this, but I don't know where to start.

答:
$$a^{m-n}=a^{k(p-1)}\equiv1\pmod p$$

$$\frac{a^m}{a^n}\equiv1\pmod p$$

$$a^m\equiv a^n \pmod p$$


问:
An example of infinitely differentiable function
The function $f:\mathbb{R}\to \mathbb{R}$ is given by the following formula: $$f(x)=\begin{cases}
\exp\left(-\dfrac{1}{(x-1)^2}-\dfrac{1}{(x+1)^2}\right), & |x|<1 \\
0, & |x|\geq 1
\end{cases}$$

Show that $f(x)$ is infinitely differentiable.

Proof: I was able to show that $f'(x)$ exists for any $x\in \mathbb{R}$ and $$f'(x)=\begin{cases}
\exp\left(-\dfrac{1}{(x-1)^2}-\dfrac{1}{(x+1)^2}\right)\times \left(\frac{2}{(x-1)^3}+\frac{2}{(x+1)^3}\right), & |x|<1 \\
0, & |x|\geq 1
\end{cases}$$

I was wondering is there any way to write explicit formula for $f^{(n)}(x)$.

I would be very thankful of someone can show the detailed proof of this problem.

答:
For the purpose of the desired claim, the explicit formula is not needed and rather it is sufficient to know (i.e., prove by induction) that
$$ f^{(n)}(x)=\begin{cases}\exp\left(-\frac1{(x-1)^2}-\frac1{(x+1)^2}\right)\times P_n(\frac1{x-1},\frac1{x+1}),&|x|<1\\0,&|x|\ge 1\end{cases}$$
where $P_n(x,y)$ is a polynomial. This is clear with $P_0(x,y)=1$ from the problem statement, and you already managed to show this for $n=1$ with $P_1(x,y)=2x^3+2y^3$. But we do not need to know $P_n$ explicitly, it suffices to know that it is a polynomial.


问:
Antiderivative of $\frac{1}{x^2-1}$ for $x>1$ or $x<-1$
We know that the antiderivative of $\frac{1}{x^{2} - 1}$ on $(-1,1)$ is $\operatorname{artanh}(x)$.  Is there a nice way of writing the antiderivative on the intervals $(-\infty,-1)$ and $(1,\infty)$?

I realize we could write $$\int_{a}^{x}\frac{1}{y^{2} - 1}\,dy$$ for $a>1$, for the antiderivative on $(1,\infty)$, for example, but I'm curious if there's an expression that doesn't include an integral, and WolframAlpha chokes when I ask it to compute this.

答:
For the sake of completeness, here are the two suggested ways of writing the antiderivative:

First Way:
$$\boxed{\int \frac{dx}{x^2-1} = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C.}$$
This holds because:
\begin{align}
\frac{d}{dx}\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| &=\frac{1}{2}\frac{d}{dx}\bigl(\ln|x-1| - \ln|x+1|\bigr)\\[5pt]
&=\frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)\\[5pt]
&=\frac{1}{x^2-1}.
\end{align}

Second Way:
$$\boxed{\int \frac{dx}{x^2-1} = \begin{cases}
-\operatorname{artanh}(x)+C, &|x|<1\\
-\operatorname{arcoth}(x)+C, &|x|>1.
\end{cases}}$$

This holds because:
$$-\operatorname{artahn}(x) = \frac{1}{2}\ln\left(\frac{1-x}{x+1}\right) = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| \quad\text{for } |x|<1$$ and
\begin{align}
-\operatorname{arcoth}(x) &= -\operatorname{artanh}\left(\frac{1}{x}\right)\\[5pt]
&= \frac{1}{2}\ln\left(\frac{1-\frac{1}{x}}{1+\frac{1}{x}}\right)\\[5pt]
&= \frac{1}{2}\ln\left(\frac{x-1}{x+1}\right)\\[5pt]
&=\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| \quad\text{for }|x|>1.
\end{align}


答:
The answer depends on $g$ in the following way:

Proposition. Let $A$ be syndetic. Then $Ag$ is syndetic iff $Tg = T$.

Indeed, if $Ag$ is syndetic, then $KAg = T$ for some compact $K$.

Then $KAg \subseteq Tg$, so that $T = Tg$.

Conversely, if $Tg = T$, and $KA = T$, then $KAg = Tg = T$.

Title: Invariance of syndetic set in topological semigroup

Body:
Let $T$ be a topological semigroup.  $A\subseteq T$ is syndetic, if there is compact set $K\subseteq T$ with $T=KA$.

Let syndetic set $A\subseteq T$ and $g\in T$ be given. I think that $Ag^{-1}=\{y: yg\in A\}$ is syndetic. Indeed  since $A$ is syndetic, there is compact set $K$ with $T=KA$. For every $t\in T$, since $tg\in KA$, hence $t\in KAg^{-1}$. This implies that $T= KAg^{-1}$ i.e.

if $A$ is syndetic, then $B=Ag^{-1}$ is syndetic.

What can say about situation of $Ag$, is it syndetic?

Please help me to know it.


问:
Prove that $0\le\int_0^1\log(u){\rm d}x+\frac1{2\pi^2}\int_0^1\frac1{u^2}\left(\frac{{\rm d}u}{{\rm d}x}\right)^2{\rm d}x$
I want to prove a special case of a functional inequality stated in a book. To be specific, let $u(x)$ being a positive, differentiable function on $[0,1]$ with unit mass (i.e., $\int_0^1 u(x){\rm d}x = 1$), then we want to show that
$$0 \leq \int_0^1 \log(u){\rm d}x + \frac{1}{2\pi^2}\int_0^1 \frac{1}{u^2}\left(\frac{{\rm d}u}{{\rm d}x}\right)^2{\rm d}x.$$
In a book, the author resorts to the construction of certain solutions of one-dimensional heat equation and used the eigen-expansion (hence the arguments are very lengthy and involved). Since I am only interested in this special inequality stated above, may I know whether there exists a simpler and more elegant proof ?

答: Here are some observations. Define the functional $$F(x,u,u')=\int_0^1\log u+\frac{u'^2}{2\pi^2u^2}\,dx$$ where $u$ satisfies the integral constraint $1=\int_0^1u(x)\,dx$ (so $u$ is a p.d.f.). We will assume that $F$ is finite over its stationary path. Consequently, define the functional to be minimised $$G(x,u,u')=\int_0^1\log u+\frac{u'^2}{2\pi^2u^2}+\lambda u\,dx$$ where $\lambda$ is a Lagrange multiplier. The Euler-Lagrange equation $G_{u'}'=G_u$ yields $$\frac d{dx}\frac{u'}{\pi^2u^2}=\lambda+\frac1u\implies uu''-2u'^2-\pi^2u^2(\lambda u+1)=0.$$ Let $v=u'^2$. This gives $v^*=2u'(u')^*=2u''$ where $v^*=dv/du$ so the differential equation becomes a first-order ODE $$v^*-\frac4uv=2\pi^2u(\lambda u+1).$$ It is easy to derive the solution $v=\pi^2(Cu^4-2\lambda u^3-u^2)$ for some constant $C$ which leads to $$u'=\pm\pi u\sqrt{Cu^2-2\lambda u-1}.$$ Separation of variables yields the solution $$\arctan\frac{\lambda u+1}{\sqrt{Cu^2-2\lambda u-1}}=\pi x+D$$ for some constant $D$ and after simplification, $$((\lambda^2+C)\cos^2(\pi x+D)-C)u^2+2\lambda u+1=0.$$ Solving the quadratic gives the stationary path $$u_s(x)=\frac{-\lambda-\sqrt{\lambda^2+C}\sin(\pi x+D)}{(\lambda^2+C)\cos^2(\pi x+D)-C}=-\frac1{\lambda+\sqrt{\lambda^2+C}\sin(\pi x+D)}$$ where the negative root is chosen to satisfy $u(x)>0$ on $[0,1]$ and $|F|<\infty$. Substituting this into the integral constraint gives $$1=\frac2{\pi\sqrt C}\left[\operatorname{arctanh}\frac{\sqrt{\lambda^2+C}+\lambda\tan(\pi x+D)/2}{\sqrt C}\right]_0^1$$ so that $$\sqrt{\lambda^2+C}=-\frac{\sqrt C}{\cos D\tanh(\pi\sqrt C/2)}.$$ Hence \begin{align}\min F(x,u,u')&=\int_0^1\log u_s+\frac12(Cu_s^2-2\lambda u_s-1)\,dx\\&=-\lambda-\frac12+\int_0^1\log u_s+\frac C2u_s^2\,dx\\&=-\frac{\lambda+1}{2}+\frac{\lambda\sqrt{\lambda^{2}+C}\cos D}{\pi\left(\left(\lambda^{2}+C\right)\cos^2D-C\right)}+\int_0^1\log u_s\,dx\\&=\frac{\Lambda-1}2+\frac{\Lambda\sinh\pi\sqrt C}{2\pi\sqrt C}-\int_0^1\log\left(\Lambda+\frac{\sqrt C\sin(\pi x+D)}{\cos D\tanh(\pi\sqrt C/2)}\right)\,dx\end{align} where $C,\Lambda:=-\lambda>0$.


问:
I would like to know an intuitive way to understand a Cauchy sequence and the Cauchy criterion.
My understanding from the definition in my book (Rudin) is this.

A seq. $\{p_n\}$ in a metric space $X$ (I only really know $\mathbb R^k$) is said to be a Cauchy sequence if for any given $\epsilon > 0$, $\exists N\in \mathbb N$ such that $\forall n,m\ge N$, $d(p_n,p_m)<\epsilon$.

(1) I see it as, given any tiny value $\epsilon$, we can find a natural number $N$ large enough so that the distance between $p_n$ and $p_m$ is less than $\epsilon$. Am I right ?

The reason I'm asking this is because I was trying to understand the proof of how
$$\sum a_nb_n$$
can converge, and the book said this

$$\left\lvert \sum_{n=p}^{q}a_nb_n\right\rvert \leq \epsilon$$

satisfies the Cauchy criterion and therefore it converges.

I read other questions and answers about the Cauchy sequence, but it didn't really help me…

Can someone explain me what's going on?

Edit:

Suppose

a) the partial sums of $A_n = \Sigma a_n$ form a bounded sequence

b) $b_0 \geq b_1 \geq \dotsb$

c) $\lim_{b \to \infty} b_n = 0$

Using the partial summation formula, algebraically the equation in the bottom is proved

$$\left\lvert \sum_{n=p}^{q}a_nb_n \right\rvert \leq \epsilon$$

Algebraically I had no problem, but I don't know why this proves convergence. I thought to show that a sequence is Cauchy, we need to find the distance between two terms in a sequence. That's where I'm confused.

答:
It's enough to consider the special case $\mathbb R$ for the purpose of understanding the gist of it. So, we'll assume everything happens in $\mathbb R$.

For a sequence $a_n$ to converge to a finite limit $L$, means that $$\forall \epsilon >0 \exists N s.t. \forall k>N: d(x_k,L)<\epsilon$$ which intuitively means that for any prescribed positive distance $\epsilon$, from some index onwards, all elements in the sequence are within that distance to the limit $L$.

Now, for the sequence to be Cauchy means that $$\forall \epsilon > 0\exists N s.t. \forall n,m>N:d(x_n,x_m)<\epsilon $$ which intuitively means that for any prescribed positive distance $\epsilon$ from some index onwards all elements in the sequence are no more than $\epsilon $ distance from each other.

Before we further examine the crucial differences let us remark that the oh so important property of $\mathbb R$ (and many other spaces) is that they are complete: A sequence converges if, and only if, it satisfies the Cauchy condition.

So, the conditions (in a complete space) actually mean the same thing. So, what are the differences? Well, the condition for convergence to a limit requires you to specify a limit. The Cauchy condition does not, and this is a great thing that occurs often when you want to show something converges but you have no idea what it might converge to. So you show it is Cauchy (in a complete space) and conclude it converges.

Intuitively, it is clear that any convergent sequences is Cauchy. This is so since if all elements from some index onwards are very close to $L$ then they must be very cose to each other as well. The converse though is not clear and not always true (as said above, it is the completeness property of the reals which is not a trivial matter). For instance, in $\mathbb Q$ with the standard metric, any sequence of rational numbers that converges in $\mathbb R$ to an irrational number is Cauchy but fails to converge in $\mathbb Q$.


问:
Probability distribution of a function of two random variables
I'm trying to find the probability distribution of a function $q(x, y)=x^2+y^2$, where $x$ and $y$ are normally distributed: $x$~$N[0,\sqrt{V_x}]$ and $y$~$N[0,\sqrt{V_y}]$.

I began by finding the probability distribution of $x^2$ and $y^2$, and this is the distribution I got: $$Z_1=g(x)=x^2$$ $$Z_2=h(y)=y^2$$ $$f_{Z_1}(z_1)=\frac{e^{-\frac{z_1}{2V_x}}}{\sqrt{2 \pi V_x}\cdot\sqrt{z_1}}$$ $$f_{Z_2}(z_2)=\frac{e^{-\frac{z_2}{2V_y}}}{\sqrt{2 \pi V_y}\cdot\sqrt{z_2}}$$

I then try to find the distribution of the sum of the two variables with newly found distributions, by using the convolution of the two PDFs: $$Z=q(x,y)=x^2+y^2$$ $$f_Z(z)=\int_{-\infty}^{\infty}f_{Z_1}(x)f_{Z_2}(z-x)dx$$

But I can't seem to be able to solve this integral. Am I missing something? Is there a way to compute this integral? Or is there some other way to find the distribution of the sum of the two?

答:
The density is $$f_Z(z)=\frac{d}{dz}\int_{-\sqrt{z}}^{\sqrt{z}}\int_{-\sqrt{z-x^2}}^{\sqrt{z-x^2}}f_X(x)f_Y(y)dydx=\frac{e^{a z}I_0\left(b z \right)}{c }$$
where:

$I_0(x)$ is the modified Bessel function of the first kind

$a=-\frac{1}{4}(\frac{1}{V_Y} + \frac{1}{V_X} )$

$b=\frac{1}{4}(\frac{1}{V_Y} - \frac{1}{V_X} )$

$c=2 \sqrt{V_XV_Y}$

You can find the Taylor series for the cdf by starting with the series for $I_0(x)$ and $e^{x}$, make the appropriate substitutions, integrating term by term:
$$F_Z(z)=\frac{1}c t + \frac{a}{2c}t^2 + \frac{2a^2 + b^2}{12c}t^3 + ...$$

Here is a graph of the cdf (blue) and the approximation (orange) above when $V_X=V_Y=1$. You need more terms in the approximation to get better accuracy for $z$ far from $0$.

The same approach works using the convolution formula. The density is

$$f_Z(z)=\int_{0}^{z}f_{Z_1}(x)f_{Z_2}(z-x)dx=\int_{0}^{z}\frac{e^{-\frac{x}{2V_X}}}{\sqrt{2\pi V_X}\sqrt{x}} \frac{e^{-\frac{z-x}{2V_Y}}}{\sqrt{2\pi V_Y}\sqrt{z-x}}dx$$

$$=\frac{e^{-\frac{z}{2 V_Y}}}{c \pi}\int_{0}^{z}\frac{e^{2 b x}}{\sqrt{x}\sqrt{z-x}}dx$$
Now, make the substitution $t=\frac{2x-z}{z}$ so that $x=\frac{z}{2}(1+t)$ and $dx=\frac{z}{2}dt$. The integral becomes

$$\frac{e^{-\frac{z}{2 V_Y}}}{c \pi}\int_{-1}^{1}\frac{e^{2 b \frac{z}{2}(1+t)}}{\sqrt{1+t}\sqrt{1-t}}dx
=\frac{e^{-\frac{z}{2 V_Y}}}{c \pi}\int_{-1}^{1}(1-t^2)^{-\frac{1}2}e^{2 b \frac{z}{2}(1+t)}dx$$
Now, use this formula for the modified Bessel function to find the integral
$$\int_{-1}^{1}(1-t^2)^{0-\frac{1}2}e^{2 b \frac{z}{2}(1+t)}dx=e^{b z}\pi I_0(b z)$$
After some simplification, you find
$$f_Z(z)=\frac{e^{a z}I_0\left(b z \right)}{c}$$


问:
Finding a vector in Euclidian space that minimizes a loss function subject to some constraints
I'm trying to solve the following minimization problem, and I'm sure there must be a standard methodology that I could use, but so far I couldn't find any good references. Please let me know if you have anything in mind that could help or any references that you think would be useful for tackling this problem.

Suppose you are given $K$ points, $p_i \in R^n$, for $i \in \{1,\ldots,K\}$. Assume also that we are given $K$ constants $\delta_i$, for $i \in \{1,\ldots,K\}$. We want to find the vector $x$ that minimizes:

$\min_{x \in R^n} \sum_{i=1,\ldots,K} || x - p_i ||^2$

subject the following $K$ constraints:

$\frac{ || x - p_i ||^2 } { \sum_{j=1,\ldots,K} ||x - p_j||^2} = \delta_i$

for all $i \in {1,\ldots,K}$.

Any help is extremely welcome!

Bruno

edit: also, we know that $\sum_{i=1,\ldots,K} \delta_i = 1$.

答:
Yes, this works. If $n \leq K-2,$ you have no guarantee of any legal solution, even when the $\delta_i$ sum to 1, as required. It may be that the sample points, your $v_j$ and $Y,$ were in a Euclidean space of much lower dimension, however, that does not guarantee you can repeat that piece of luck if the new $n$ in $\mathbf R^n$ is too small.

If $n = K -1,$ there should be a single feasible point, "near" the simplex with the $K$ points as vertices. No need (or ability) to minimize anything. Actually, unless the $\delta$'s are all equal, it appears there is a second feasible point far away. If all angles in the simplex are acute, there is a feasible point in its interior.

So, my advice is, figure out how to find a feasible point when $n=K-1.$ If circumstance forces $n \geq K,$ rotate so  the hyperplane containing all the $p_i$ becomes the hyperplane $x_1, x_2, \ldots, x_{K-1}, 0,0,\ldots,0,$ solve the problem there, then rotate back.

Meanwhile, I see nothing wrong with a numerical method for finding the single feasible point near the simplex when $n=K-1.$ Easier than finding the intersection of a large number of spheres and planes. Note that, when $n=K,$ the full set of all feasible points is either a straight line (if all $\delta_i$ are equal) or, in fact, an actual circle. Go figure. In either case, meeting the hyperplane that contains the $p_i$'s orthogonally.

For that matter, your easiest program is just to solve the problem in the original $v_i, Y$ location, that is, a numerical method that finds the point $Z$ near the $v_i$ simplex with the correct $\delta$'s. Then you can just map $Z$ along with the $v_i.$


答:
The rough intuition is that if we go far enough along the sequence we get to a point where it doesn't vary very much. And if that is the case it must stay within a narrow range of values.

If we can reduce the variation arbitrarily (choose $\epsilon$ as small as we like) by going far enough ($N$ terms), then we can narrow the range as much as we like, so that there is ultimately a single value - the limit.

The value of the criterion is that it proves there is a limit without needing to know what the limit is - just using the internal properties of the sequence.

Title: I would like to know an intuitive way to understand a Cauchy sequence and the Cauchy criterion.

Body:
My understanding from the definition in my book (Rudin) is this.

A seq. $\{p_n\}$ in a metric space $X$ (I only really know $\mathbb R^k$) is said to be a Cauchy sequence if for any given $\epsilon > 0$, $\exists N\in \mathbb N$ such that $\forall n,m\ge N$, $d(p_n,p_m)<\epsilon$.

(1) I see it as, given any tiny value $\epsilon$, we can find a natural number $N$ large enough so that the distance between $p_n$ and $p_m$ is less than $\epsilon$. Am I right ?

The reason I'm asking this is because I was trying to understand the proof of how
$$\sum a_nb_n$$
can converge, and the book said this

$$\left\lvert \sum_{n=p}^{q}a_nb_n\right\rvert \leq \epsilon$$

satisfies the Cauchy criterion and therefore it converges.

I read other questions and answers about the Cauchy sequence, but it didn't really help me…

Can someone explain me what's going on?

Edit:

Suppose

a) the partial sums of $A_n = \Sigma a_n$ form a bounded sequence

b) $b_0 \geq b_1 \geq \dotsb$

c) $\lim_{b \to \infty} b_n = 0$

Using the partial summation formula, algebraically the equation in the bottom is proved

$$\left\lvert \sum_{n=p}^{q}a_nb_n \right\rvert \leq \epsilon$$

Algebraically I had no problem, but I don't know why this proves convergence. I thought to show that a sequence is Cauchy, we need to find the distance between two terms in a sequence. That's where I'm confused.


问:
Finding a sequence satisfying this recurrence relation?
I just don't even know where to start with this,

Find a sequence $(x_n)$ satisfying the recurrence relation:

$2x_n$$_+$$_2$ = $3x_n$$_+$$_1$ + $8x_n$ + $3x_n$$_-$$_1$ Where n is a natural number and

$x_0$ = -1,
$x_1$ = 3 and
$x_2$ = 3

Thanks in advance!

答:
A general way to solve such is using generating functions. Define:

$\begin{align*}
   g(z)
     &= \sum_{n \ge 0} x_n z^n
\end{align*}$

Write your recurrence shifted (subtraction in indices gets messy), multiply by $z^n$, sum over $n \ge 0$ and recognize resulting sums:

$\begin{align*}
   2 x_{n + 3}
     &= 3 x_{n + 2} + 8 x_{n + 1} + 3 x_n \\
   2 \sum_{n \ge 0} x_{n + 3} z^n
     &= 3 \sum_{n \ge 0} x_{n + 2} z^n
          + 8 \sum_{n \ge 0} x_{n + 1} z^n
          + 3 \sum_{n \ge 0} x_n z^n \\
   2 \frac{g(z) - x_0 - x_1 z - x_2 z^2}{z^3}
     &= 3 \frac{g(z) - x_0 - x_1 z}{z^2}
          + 8 \frac{g(z) - x_0}{z}
          + 3 g(z)
\end{align*}$

Solve this for $g(z)$ (substitute the initial values), express as partial fractions:

$\begin{align*}
   g(z)
     &= \frac{1 - 5 z}{1 - 2 z - 3 z^2} \\
     &= \frac{1 - 5 z}{(1 + z) (1 - 3 z)} \\
     &= - \frac{3}{2 (1 + z)}
          + \frac{1}{2 (1 - 3 z)}
\end{align*}$

We want the coefficient of $z^n$ of this. But this is just two geometric series:

$\begin{align*}
   [z^n] g(z)
     &= - \frac{3}{2} (-1)^n + \frac{1}{2} \cdot 3^n \\
     &= \frac{3^n - (-1)^n}{2}
\end{align*}$