问:
What's the inverse Laplace transform of $\sqrt{s^2-s}$
I know the inverse Laplace transform of $s$ is the delta function $\delta^\prime (t),$ i.e. $\mathcal{L}^{-1}(s)=\delta^\prime(t).$

And I know actually $\mathcal{L}^{-1}(s^n)=\delta^{(n)}(t),$ where $\delta^{(n)}$ means the nth-derivative of $\delta.$

My question is that, what is the inverse Laplace transform of $\sqrt{s^2-s},$ i.e. what's $\mathcal{L}^{-1}(\sqrt{s^2-s})\,?$

My attempts:

$$\sqrt{s^2-s}=s\sqrt{1-\frac{1}{s}}=s[1-\frac{1}{2s}+o(1)\frac{1}{s}]=s-\frac{1}{2}+o(1)\,\ \ \ \ \ \ \ \ \ \ \cdot\cdot\cdot\cdot\cdot\cdot\,\ \ \ (A)\,,$$ where the second equality is by the Taylor theorem. (I don't know if the first equality is vailed in the complex space.)

Then, $$\mathcal{L}^{-1}(\sqrt{s^2-s})=\mathcal{L}^{-1}(s)-\mathcal{L}^{-1}(\frac{1}{2})+\mathcal{L}^{-1}[o(1)]=\delta^\prime(t)-\frac{1}{2}\delta(t)+\mathcal{L}^{-1}[o(1)]$$

On one hand, we know $$\forall n\ge 0, \mathcal{L}^{-1}[\frac{1}{s^{n+1}}]=\frac{t^n}{n!}.$$

On the other hand, we know $o(1)$ is nothing than a power series in $\frac{1}{s}.$

So it seems like we have solved the question. But the result looks a mess, and I am not really confident of the first equality in $(A)$. And the application of Taylor theorem in $(A)$ is not vailed for those $s$ with $\lvert s \rvert$ not large enough.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\root{z}}$ and $\ds{\root{z - 1}}$ are their Principal Branchs.

The big arc integration vanishes out as its radius $\ds{\to \infty}$ mainly due to the factor $\ds{\expo{ts}}$ with $\ds{t > 0}$.

There's not any pole inside the contour. The integration is reduced to the subtraction of integrals above and belows the above mentioned branch-cuts.

Namely,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{1^{+}\ -\ \infty\ic}^{1^{+}\ +\ \infty\ic}
\root{s}\root{s - 1}\expo{ts}{\dd s \over 2\pi\ic}
\,\right\vert_{\,t\ >\ 0}}
\\[5mm] = &\
-\int_{-\infty}^{0}
\root{-s}\expo{\ic\pi/2}\root{1 - s}\expo{\ic\pi/2}\,\expo{ts}{\dd s \over 2\pi\ic}
\\[2mm] &\
-\int_{0}^{1}
\root{s}\root{1 - s}\expo{\ic\pi/2}\,\expo{ts}
{\dd s \over 2\pi\ic}
\\[2mm] &\
-\int_{1}^{0}
\root{s}\root{1 - s}\expo{-\ic\pi/2}\,\expo{ts}
{\dd s \over 2\pi\ic}
\\[2mm] &\
-\int_{0}^{-\infty}
\root{-s}\expo{-\ic\pi/2}\root{1 - s}\expo{-\ic\pi/2}\,\expo{ts}{\dd s \over 2\pi\ic}
\\[5mm] = &\
-\,{1 \over \pi}\int_{0}^{1}\root{s - s^{2}}\expo{ts}\dd s
\\[5mm] = &\
-\,{1 \over 2\pi}\expo{t/2}
\int_{-1/2}^{1/2}\root{1 - \pars{2s}^{2}}\expo{ts}\dd s
\\[5mm] \stackrel{2s\ \mapsto\ s}{=}\,\,\, &\
-\,{1 \over 4\pi}\expo{t/2}\
\underbrace{\int_{-1}^{1}\root{1 - s^{2}}\expo{ts/2}\dd s}
_{\ds{{\pi \over t/2}\on{I}_{1}\pars{t/2}}}
\end{align}
$\ds{\on{I}_{\nu}}$ is a Modified Bessel Function. The last result is A & S Table Identity $\ds{{\bf\color{black}{9.6.18}}}$.
Finally,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{1^{+}\ -\ \infty\ic}^{1^{+}\ +\ \infty\ic}
\root{s}\root{s - 1}\expo{ts}{\dd s \over 2\pi\ic}
\,\right\vert_{\,t\ >\ 0}}
=
\bbx{-\,{1 \over 2}\,{\expo{t/2} \over t}
\on{I}_{1}\pars{t \over 2}} \\ &
\end{align}
The integral $\ds{\underline{vanishes\ out}}$ whenever $\ds{t < 0}$.


问:
The number of degree one and degree three vertices in a graph with no isolated vertices and with more vertices than edges
Someone asked me this a few days ago:

Let $G=(V, E)$ be a graph and denote the set of vertices with degree $i$ by $V_i$. If $|V_0|=0$ and $|V_3| > |V_1|$, prove that $|V| < |E|$.

Is there a more concise proof than the following one:

Consider the graph $G=(V, E)$ with a nonzero minimum degree and such that $|E| \leq |V|$. We will show that $|V_1| \geq |V_3|$.

If $|E|<|V|-1$ then $G$ has more than one component. We can use the following arguments for each component.

If $|E|=|V|$ then $G$ is a cycle. Thus $|V_1| = |V_3| = 0$.

If $|E|=|V|-1$ then $G$ is a tree. Note that if $v\in V_1$ then $v$ cannot be adjacent to another element of $V_1$ because this means the graph is not connected (the edge between these vertices contributes the entirety of each vertices's degrees, implying they can't be adjacent to any other vertices). Now consider the graph $G_2 =G \setminus \{ V_2\}$, where we delete from $G$ the vertices with degree 2 and the edges incident to these deleted vertices. If $v \in V_1$ were adjacent to a vertex in $V_2$ in $G$, then $v$ will be an isolated vertex in $G_2$ (degree 0). The remaining vertices in $V_1$ are adjacent to vertices in $V_3 \cup V_4 \cup \dots \cup V_{max \ degree}$. Thus these remaining vertices in $V_1$ (via the edges incident to these vertices) have a one-to-one correspondence to vertices in $V_3 \cup V_4 \cup \dots \cup V_{max \ degree}$. If we count the number of these edges (in $G_2$) incident to vertices in $V_1$ and add the number of isolated vertices in $G_2$, we have
$$|V_1| \geq  |V_3 \cup V_4 \cup \dots \cup V_{max \ degree}| \geq |V_3|.$$

After giving the previous proof to the person who asked me about it, I am wondering if there is a shorter proof for this? Maybe an algorithmic one or a probabilistic one. Or is there a shorter and more direct way to prove the statement?

答:
By Handshaking Lemma/Degree Sum Formula, the sum of degrees of all vertices is twice the number of edges.

Let $|V|$ be the total number of vertices. We have:

\begin{align}2|E| = \sum_{v\in V}\deg(v)=\sum_{i\ge 0} i|V_i|&= \sum_{i \ge 4}i|V_i|+3|V_3| + 2|V_2| + |V_1| \\&> 2\sum_{i\ge 4}|V_i|+2|V_3|+|V_1|+2|V_2|+|V_1|\\&=2\sum_{i\ge 1} |V_i| = 2|V|\end{align}

proving the result.

ContentLicense: CC BY-SA 4.0


问:
Cardinality of $GL_n(K)$ when $K$ is finite
I don't know how to do the last task of an exercise.

Let $K$ be a field, $G=GL_n(K)$ and $X=K^n\backslash\{0\}$.

First task: Show that $G \times X \to X$, $(A,x)\mapsto Ax$ defines an action of $G$ on $X$. Done.

Second task: find the stabilizer of $\displaystyle x=\left(\begin{eqnarray} 1\\ 0 \\ \vdots \\0 \end{eqnarray}\right)$. Done.

Third task: Show that there is only one orbit. Done.

Last task: Let $|K|<\infty$. Use the last two tasks and induction to find $|G|$.

I know that $|G|=[G:St_G(x)]\cdot|St_G(x)|$ and i know that $[G:St_G(x)]=|\bar{x}|=m^n-1$ but i can't figure out what $|St_G(x)|$ is.

What should i do next?
I appreciate every hint :)

答: I think we can close to the last part like this. You know that if $V=V_n(q)$ be a vector space on the field $F=GF(q)$ of dimension $n$ then we have vectors $e_i, ~1\leq i\leq n$ such that $$V=\langle e_1,e_2,...,e_n\rangle$$ So $|V|=q^n$. In fact, $V\cong F\times F\times...\times F$ $n$ times. And you know that $$GL_n(q)=\{T\mid T:V\to V\ ~\text{where} ~~T~~\text{are all invertible linear transformations}\}$$ Now take $T\in GL_n(q)$. What is the order of $T(e_1)$? Obviously it is $q^n-1$. What will be $T(e_2)$? It is clearly $q^n-q$. By this way we will have $T(e_n)=q^n-q^{n-1}$. This gives you $$|GL_n(q)|=(q^n-1)(q^n-q)...(q^n-q^{n-1})$$</p>

ContentLicense: CC BY-SA 3.0


问:
Given marginal probability distribution functions, does there always exsit a joint distribution which can produce these marginals?
For N variables $X_1,X_2,\ldots,X_N$, the PDFs of all $X_i$s are given. Can we prove a joint distribution always exists?

What if some consistent PDFs of combinations of these variables are given, i.e., one PDF does not contradict with another? By "consistency" and "contradiction", I meant you cannot produce different margins from any two given PDFs. For example, one can calculate the marginal distribution of $X_1$ from the marginal distribution of $X_1,X_2$. If the marginal distribution of $X_1,X_2$ on $X_1$ is different from the given PDF of $X_1$, then there is an inconsistency (contradiction).

答:
You might want to check "Characterizing a Joint Probability Distribution by Conditionals" by Andrew Gelman and T. P. Speed, Journal of the Royal Statistical Society. Series B (Methodological) Vol. 55, No. 1 (1993), pp. 185-188 (4 pages).

From the abstract: We derive conditions under which a set of conditional and marginal probability distributions will uniquely specify an all-positive joint distribution.


问:
A question about floor functions and series
The sequence $\{a_n\}^{∞}_{n=1} = \{2,3,5,6,7,8,10,...\}$
consists of all the positive integers that are not perfect squares.

Prove that $a_n= n+ [\sqrt{n} + \frac{1}{2}]$.

Well, I managed to prove that
$ m^2 < n+ [\sqrt{n} + \frac{1}{2}] < (m+1)^2 $,
where $[\sqrt{n} + \frac{1}{2}] = m$.

But is this enough to answer the question? Or do we also need to prove that $n+ [\sqrt{n} + \frac{1}{2}] $ can take all non-perfect square values. If yes, then how?

Any help is appreciated, thanks!

答:
The increment $$a_{n+1}-a_n=1+\left\lfloor\sqrt{n+1}+\dfrac12\right\rfloor-\left\lfloor\sqrt{n}+\dfrac12\right\rfloor$$

is $1$, or $2$ when $\sqrt{n+1}$ rounds differently than $\sqrt n$.


问:
What is the order of $\bar{2}$ in the multiplicative group $\mathbb{Z}_{221}^\times$?
What is the order of $\bar{2}$ in the multiplicative group $\mathbb{Z}_{221}^\times$?

I keep computing $2^0, 2^1,...$ until we get $1\ (\operatorname{mod} 221)$, but this will take forever! Is there an easier way to do this? A hint that was given is that $221= 13 \times 17$.

答:
While the essence of the answer is the same as in that by JW Tanner, I would like to offer a solution in a more algebraic language:

By the Chinese remainder theorem, we know that
$$ \mathbb{Z}_{221}\simeq \mathbb{Z}_{13}\times \mathbb{Z}_{17}$$
by the isomorphism
$$ x+221\mathbb{Z}\mapsto (x+13\mathbb{Z},x+17\mathbb{Z})$$
The order of $x$ and its image have to be the same.

Both $\mathbb{Z}_{13}$ and $\mathbb{Z}_{17}$ are fields, so their multiplicative groups are of order $12=2^2\cdot 3$ and $16=2^4$ respectively. Thus the order of $2$ in $\mathbb{Z}_{13}$ has to be in $\{2,3,4,6,12\}$ (because its order must divide the order of the multiplicative group, this is Lagrange's theorem). Just trying these possibilities you see that its order is $12$. Similarly the order of $2$ in  $\mathbb{Z}_{17}$ is in $\{2,4,8,16\}$. Trying the first three possibilities gives that it's $8$.

Now if we are looking for the smallest integer $n>0$ such that
$$(x+13\mathbb{Z},x+17\mathbb{Z})^n=(x^n+13\mathbb{Z},x^n+17\mathbb{Z})=(1,1)$$
it becomes clear that it must be the least common multiple of $12$ and $8$, which is $24$.


问:
Quick Rejog of Partial Fraction Decomposition:
I don't need the problem to be solved I just need to have the decomposed equation. I have the following equation: \begin{equation} I=\int \frac{dx}{x^2(x^2-16)}\end{equation} The method that I have to use is Partial Fractions Decomposition. My question is since there is an $x^2$ term I have to increase in linearity which I wonder if I did correctly? $$\bbox[border:2px solid black]
{\begin{equation}\frac{1}{x^2(x^2-16)}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{D}{x+4}+\frac{E}{x-4} \end{equation}}$$

答:
You only increase the degree of numerator with irreducible factors (e.g. $x^2+1$), not repeated roots. Suppose the root $a$ is repeated $r$ times, then the decomposition has
$$\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_r}{(x-a)^r}$$


问:
how to prove that two iterated integral are different
We have the following function: $$f(x,y)=
    \begin{cases}
    1, & 0 \leq x-y \leq 1\\
    -1, & 0 \leq y-x \leq 1 \\
0, & \text{other cases}
    \end{cases} $$
and I need to prove that $$\int_0^{\infty}\int_0^{\infty} f(x,y) dx dy \not = \int_0^{\infty} \int_0^{\infty} f(x,y) dydx $$
I have defined the following sets:

$D_1=\{(x,y) \in \mathbb{R}^2:  0 \leq x-y \leq 1 \}=\{(x,y) \in \mathbb{R}^2: x \leq y \leq 1+x \}$
$D_2=\{(x,y) \in \mathbb{R}^2:  0 \leq y-x \leq 1 \}=\{(x,y) \in \mathbb{R}^2: x \geq y \geq x-1 \}$.

So we have that $$f(x,y)=
    \begin{cases}
    1, & (x,y) \in D_1\\
    -1, & (x,y) \in D_2 \\
0, & \text{other cases}
    \end{cases} $$
I have done the following:
$A=\{(x,y) \in \mathbb{R}^2: x,y \geq 0\}$
$$\int_0^{\infty}\int_0^{\infty} f(x,y) dy dx=\int\int_{D_1 \cap A} f(x,y) dy dx + \int\int_{D_2 \cap A} f(x,y) dy dx=\int_0^{\infty}\int_y^{y+1}-1 dxdy+\int_0^{\infty}\int_{y-1}^{y}1 dxdy=$$ $$= \int_0^{\infty}-y+y+1 dy+\int_0^{\infty}y-1-ydy=0$$
$$\int_0^{\infty}\int_0^{\infty} f(x,y) dx dy =\int\int_{D_1 \cap A} f(x,y) dx dy + \int\int_{D_2 \cap A} f(x,y) dx dy= \int_0^{\infty}\int_{x}^{x-1}1 dydx+\int_0^{\infty}\int_{x+1}^{x}-1 dxdy=0$$
I don't know what's wrong.. I would be really thankful if you helped me \
Thank you

答:
Splitting into two areas where $f(x,y)=+1$ or $-1$ will not work, as they are both infinite in extent, leaving you with $\infty - \infty$, which is not defined.

Instead, doing the first integral first:
$$\int_0^{\infty}\int_0^{\infty} f(x,y) dx dy \\= \int_0^{1}\int_0^{\infty} f(x,y) dx dy + \int_1^{\infty}\int_0^{\infty} f(x,y) dx dy \\= \int_0^{1}\left(\int_0^{y} f(x,y) dx +\int_y^{y+1} f(x,y) dx\right) dy \\+ \int_1^{\infty}\left(\int_{y-1}^{y} f(x,y) dx +\int_y^{y+1} f(x,y) dx\right) dy \\ = \int_0^{1}\left(-y+1\right) dy + \int_1^{\infty}0\, dy \\= \frac12$$

The second integral will be similar but end up with $\int\limits_0^{1}\left(x-1\right) dx + \int\limits_1^{\infty}0 \,dx = -\dfrac12$, showing that the order of integration matters. The cause is concealed in the integration of $0$ over an infinite range.


问:
Is this proposition incomplete?
In mendelson logic textbook , there is this proposition,(I will replace some letters to make it easier to read)

"If $P$ is a tautology containing as statement letters $A_1,A_2,...,A_n$ and $Q$ arises from $T$ by substituting statement forms $B_1,B_2,...,B_n$ for $A_1,A_2,...,A_n$ ,respectively, then $Q$ is a tautology;that is ,substituting in a tautology yields a tautology"

For me , this seems incomplete.Shouldn't there be this line also that ,"$A_1 \equiv B_1 , A_2 \equiv B_2 , ... , A_n \equiv B_n$ "?

答:
No.

$p \lor \lnot p$ is a tautology. According to the proposition, also $(p \lor q) \lor \lnot (p \lor q)$ is.

But $p$ and $p \lor q$ are not equivalent.

Title: Is this proposition incomplete?


答:
Basically you're correct, but you only need two constants, not three.

As mentioned in the comments,

$$ \frac Ax + \frac{Bx+C}{x^2} = \frac{A+B}{x} + \frac C{x^2} = \frac{A'}{x} +\frac{C}{x^2}$$

where $A'=A+B$.

In other words, either the $\frac {A}x$ term is redundant or the $Bx$ term is redundant.

Title: Quick Rejog of Partial Fraction Decomposition:

Body:
I don't need the problem to be solved I just need to have the decomposed equation. I have the following equation: \begin{equation} I=\int \frac{dx}{x^2(x^2-16)}\end{equation} The method that I have to use is Partial Fractions Decomposition. My question is since there is an $x^2$ term I have to increase in linearity which I wonder if I did correctly? $$\bbox[border:2px solid black]
{\begin{equation}\frac{1}{x^2(x^2-16)}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{D}{x+4}+\frac{E}{x-4} \end{equation}}$$


问:
Prove or disprove statements about the greatest common divisor
Help with prove or disproving either of these statements would be really appreciated, one or the other is fine, I just need a start or a solution to one and I'm sure I could probably figure the other out.

(a) For all $a, b, c \in \mathbb{Z}$, $\gcd(ac, bc) = c \gcd(a, b)$

(b) For all $a, b, c \in \mathbb{Z}$, $\gcd(ab, c) = \gcd(a, b) \gcd(b, c)$

答:
Hint for (a): Note that $c|ac$ and $c|bc$.

Hint for (b): Look at special cases, $a=b$, $a=c$ etc


问:
Image of group homomorphism.
Consider the subgroup $\mathbb{U}_{100}$ of the multiplicative group $\mathbb{C}^\times$ consisting of all the 100-th roots of unity. Define a group homomorphism as:
$$
f:\mathbb{U}_{100}\to\mathbb{U}_{100},\space\space\space\space z\mapsto z^{70}
$$

What is the order of the image of $f$?

I know that the image of $f$ is a set of the elements $\mathbb{U}_{100}$ such that the set is an image under $f$ but I'm not quite sure what to do...

答:
The elements of $\mathbb U_{100}$ are $u^k$ where $u$ is the first $100$th root of unity.
$$(u^k)^{70}=(u^{10})^{7k}$$
and $u^{10}$ is the first tenth root of unity. Powering this by any exponent will only give other tenth roots of unity, so the image of $f$ has order $10$. (Note that $7$ is coprime to $10$.)

Title: Image of group homomorphism.


答:
Hint:

We see that
$$
f(\mathbb{U}_{100})=\{ z^{70} | z \in \mathbb{U}_{100} \}=\{ e^{ik\pi\frac{7}{5}} | k \in [0,99]\}=\left < e^{i\pi\frac{7}{5}}\right>.
$$
We see that $o \mid 10$ where $o$ is the order of $f(\mathbb{U}_{100})$. So $o \in \{1,2,5,10\}$. You can try each case and see that $o=10$.

Title: Image of group homomorphism.

Body:
Consider the subgroup $\mathbb{U}_{100}$ of the multiplicative group $\mathbb{C}^\times$ consisting of all the 100-th roots of unity. Define a group homomorphism as:
$$
f:\mathbb{U}_{100}\to\mathbb{U}_{100},\space\space\space\space z\mapsto z^{70}
$$

What is the order of the image of $f$?

I know that the image of $f$ is a set of the elements $\mathbb{U}_{100}$ such that the set is an image under $f$ but I'm not quite sure what to do...


问:
Number of ways to express $n$ as a sum of squares of integers
Let $n$ be a positive integer.

The question is, how many ways are there, to write $n$ as
$\sum_i a_i^2$, where $a_i$ are some positive integers.

The lenght of $a_i$ is irrelevant so that one could formulate this abstractly as follows.
Let $$P_n := \{ a = (a_i)_{i\in \mathbb{N}} \in \mathbb{N}^{(\mathbb{N})} | \sum_{i} a_i^2 = n \text{ and } a_j = 0 \implies a_k = 0 \ \forall \ k \geq j \}.$$ Then, what is the size of $P_n$?

(I write $\mathbb{N}^{(\mathbb{N})}$ for sequences with values in $\mathbb{N}$ that are eventually $0$).

So far, I have a recursive formula:

$P_n = \coprod_{x \leq \sqrt{n}} P_{n - x^2}$

With this, one could compute the exact value for a given number, but the number of terms gets large quickly. (I wrote a basic program but the results neither enable me to find a formula, nor to reach numbers greater than ~60).

If a closed formula for the size of $P_n$ seems to be difficult, I would also be interested in upper bounding it, such that the upper bound is not to far off.

I couldn't find anything in literature (e.g. by using Google) but maybe I am missing the correct words, so I would also appreciate any hints on that direction.

答:
I computed the first few terms and looked  this up in the  Encyclopedia of Integer Sequences: https://oeis.org/A006456 .  Two of the references there are:

N. Robbins, On compositions whose parts are polygonal numbers, Annales Univ. Sci. Budapest., Sect. Comp. 43 (2014) 239-243. See p. 242. (http://ac.inf.elte.hu/Vol_043_2014/239_43.pdf and

J. Bohman et al., Partitions in squares, Nordisk Tidskr. Informationsbehandling (BIT) 19 (1979), 297-301. (Annotated scanned copy), https://oeis.org/A001156/a001156.pdf)

and these both give the asymptotic relation $P_n \sim C \lambda^n$, where (following Bohman's notation)

$\lambda \approx 1.417742546$ is the reciprocal of the unique  positive root $\mu \approx 0.705$ of $x + x^4 + x^9 + x^{16} + \cdots = 1$

$C = (\mu + 4 mu^4 + 9 \mu^9 + \cdots) \ \approx 0.46542113$

This follows from standard results in the theory of linear recurrences.  The constant $\lambda$ is close to $\sqrt{2}$ so this agrees with saulspatz's numerical work.  I don't expect that there's an exact formula.


问:
Do there exist some non-constant holomorphic functions such that the sum of the modulus of them is a constant
Do there exist some non-constant holomorphic functions $f_1,f_2,\ldots,f_n$such that $$\sum_{k=1}^{n}\left|\,f_k\right|$$ is a constant? Can you give an example? Thanks very much</p>

答:
NO. Suppose $f, g$ are holomorphic functions on the unite disc.
$$
2\pi r M=2\pi r( |f(z_0)|+|g(z_0)|)=|\int_{|z-z_0|=r} fdz|+|\int_{|z-z_0|=r} gdz|\le \int_{|z-z_0|=r} (|f|+|g|)|dz|=2\pi r M
$$
so all equal sign hold, then $f, g$ are constants.</p>


答:
Hint:

The order of the image of a group homomorphism

is the order of the group divided by the order of the kernel.

Title: Image of group homomorphism.


问:
Maximum and minimum of implicit function with a parameter.
Consider an equation $\frac{1}{y} + a\log{y} = x$, where $a \in \mathbb{R}$ is a parameter. I want to find the value of $a$ such that maximum (minimum) of $y(x)$ is the largest (smallest).

The idea is simple, all I have to do is to find the maximum of $y(x)$ for a fixed $a$ and the maximalize over all $a$. Same for minimum However, I have some problems with this.

Using the implicit function theorem, I have to assume, $y(x)$ is not equal $\frac{1}{a}$. So assume for example that $y(x) > \frac{1}{a}$ and calculate the derivative: $y'(x) = \frac{1}{\frac{-1}{y^2}+\frac{a}{y}} = \frac{y^2}{ay - 1} > 0$. Similarly, if $y(x) < \frac{1}{a}$ then $y'(x) < 0$.

How can I interpret that? Does that mean that the minimum is equal to $\frac{1}{a}$?

答:
As you noted, it is not possible to globally define $y$ as a function of $x$, so it is not clear what is meant as the maximum of $y(x)$. If $y(x)$ is taken as the lower part of the curve in the picture below ($y < \frac 1a$), then the maximum is precisely $\frac 1a$. Conversely, if you considerer the upper aprt ($y > \frac 1a$), the minimum will be $\frac 1a$ and there will be no maximum.

However, this will not get you any maximal or minimal $a$.


问:
if $f(x)=x^2-2$ then real solutions of $f^n(x)=x$

if $f(x)=x^2-2$ where $x\in[-2,2]$ then find the number of real solutions $f^n(x)=x$  for some natural $n$ where $f^n(x)=f(f^{n-1}(x))$

I had no clue on how to start , a recurrence relation seemed unobvious.Hence i started with examples!

$f(x)=x\Rightarrow x^2-x-2=0$ has two solutions.

Now $f(f(x))=x\Rightarrow x^4-4x^2-x+2=(x-2)(x+1)(x^2+x-1)=0$ has $4$ solutions.

By the time it reached $f(f(f(x)))=x$ i lost my patience.I hence went to WA which says that there are 8 roots.I see the pattern it looks like number of roots is $2^n$......

Any hints??

I would love a solution without calculus but i am ok with it otherwise..

答:
HINT:

$$f(2\cos x) = 4\cos^2 x - 2 = 2\cos 2x$$

hence

$$f^{\circ n}(2\cos x) = 2\cos 2^n x$$


问:
Non-empty intersection of infinitely many subsets in Finite measure space
Question is: Let $(X,\sum,\mu)$ be a finite measure space. Prove that if $A_n \in \sum$ and for every $n$ we have $\mu(A_n)\geq\epsilon>0$ then there is $x \in X$ such that $x \in A_n$ for infinitely many $n$.

My attempt: If $A_n$ is infinite sequence of sets with property $\mu(A_n)\geq\epsilon$ then $\sum_{i=1}^\infty\mu(A_i)=\infty$.
But because of $\bigcup_{i=1}^\infty A_n \subseteq X$ , we have $\mu(\bigcup_{i=1}^\infty A_n) \leq \mu(X) < \infty $.
Since $\mu(\bigcup_{i=1}^\infty A_n)<\sum_{i=1}^\infty\mu(A_i)$ we know that $A_n$ is not sequence of disjoint sets.

From that moment i have a problem with proving that there exist infinite subsequence of $A_n$ with no empty intersection.

答:
Set $B_k := \bigcup_{n \geq k} A_n$, then $\mu(B_k) \geq \mu(A_k) \geq \epsilon$. Moreover, $B_k \downarrow B:= \bigcap_{k \geq 1} B_k$ (the arrow indicates that the sequence is decreasing, i.e. $B_{k+1} \subseteq B_k$ for all $k$). By the continuity of the (finite!) measure $\mu$,  we get

$$\mu(B) = \lim_{k \to \infty} \mu(B_k) \geq \epsilon>0.$$

In particular, $B \neq \emptyset$, and so we can pick $x \in B$. By the definition of $B$, this means that $x \in A_n$ for infinitely many $n \in \mathbb{N}$.


问:
$P(A\cup B\cup C)$ when $C$ is a subset of $B$
I'm trying to solve the following problem:

Calculate $P(A\cup B\cup C)$, where $P(A) = 0.2$, $P(B) = 0.5$ and $P(C) = 0.1$.

$A$ and $B$ are independent.
$A$ and $C$ are independent.
$C$ is a subset of $B$.

Formula Union of three events:
$$P(A\cup B\cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(C\cap B) + P(A\cap B\cap C)$$

$P(A\cap B) = 0.2 \cdot 0.5 = 0.1$

$P(A\cap C) = 0.2 \cdot 0.1 = 0.02$

$P(C\cap B) = P(C)$ because $C$ is a subset of $B$ meaning all of $C$ exists in $B$.

$$P(A\cup B\cup C) = 0.2 + 0.5 + 0.1 - 0.1 - 0.02 - 0.1 + (0.2 \cdot 0.5 \cdot 0.1) = 0.59$$

But this is not correct..

Grateful for any help!

答:
You don't need any fancy formula like that. If $C\subseteq B$, then $B\cup C=B$, and $A\cup B\cup C= A\cup B$. Therefore, you only have to compute $P(A\cup B)$.

(You can use your formula for this one, but $A$ and $B$ are already independent, therefore...)


问:
Can't that every standard positive real number is limited be proved?
I'm reading a short section about internal set theory(see here), in which

$x$ is limited in case for some standard $r$ we have $|x| ≤ r$.

while the predicate “standard” is not defined.
I'm interpreting an element $x \in A$ is standard, if it dosn't belong to $^*A \setminus A$.

One exercise in it asks:

Can one prove that every standard positive real number is limited?

I think not. Because either $\{x \in {\bf{R}} : x >0 \land x \text{ is standard} \}$ or $\{x \in {\bf{R}} : x \text{ is limited} \}$ are  "illegal set formations" in the lauguage of set theory. so there's no way to tell whether an element in one set necessarily belongs to the other.

On the other hand,  I'm doubtful about such reasoning, since if $x$ is a standard positive real number, so is $x+1$. Because $x  < x+1$, we have every standard positive real number must be limited.

What's wrong?

答:
There is a difference between saying that the set of limited numbers exists, and that the set of standard numbers exists, and that every standard number is limited.

To see the full meaning of this, let me give you a close example. In $\Bbb R$ we cannot define $\Bbb Z$, but we can define each of the integers, and therefore we can define every natural number if we wish to define it. Still we cannot define $\Bbb N$ itself.

Similarly here, we cannot define the set of standard reals, nor the set of limited reals. But we can prove that every standard real is limited. Simply by arguing that if $r$ is standard then $|r|$ is also standard, and therefore $|r|\leq|r|$ as wanted.


问:
MGF Expect Random
If I have a two norm distributed RV, i.e. $L \sim N(m,s^2)$ and $M \sim N(0,1)$, with coo coefficient $ \operatorname{Corr}(L,M) =\rho$, what is then:

$$ E[e^L M^2] = \mathop{?} $$

答: I don't think there's enough information to know this expectation. For example, suppose that $m = 0$, $s = 1$, and $\rho = 0$. This would be equally consistent with either i) $A$ and $B$ being independent or ii) $B = AW$ where $W$ is the Rademacher distribution (see Wikipedia). These result in different values for the sought expectation. Specifically in case ii) the expectation will be larger because whenever $A$ is large $B^2$ is also large.

Title: MGF Expect Random


问:
How many scalar multiplications are possible on a $\mathbb{Q}$-Vector Spaces?
Let $V$ be a $\mathbb{Q}$ Vector-Space. Then from the definition of Vector-Spaces, we have $1.x=x$ for all $x \in V$. From this one can show $n.x=x+x+...+x(\text{n times})$ for all $n \in \mathbb{Z}$.

So any scalar multiplication defined on a $\mathbb{Q}$-Vector Space, when restricted to $\mathbb{Z}$ becomes identical. Now the question is, is there only one possible way to define scalar multiplication on a $\mathbb{Q}$-Vector Space?

This can be generalized to ask

How many scalar multiplications are possible on $\mathbb{R}$-Vector Space and $\mathbb{C}$-vector space?

答:
Hint: If $q = m/n \in \mathbb Q$, then $n\cdot(q\cdot v)=m\cdot v$.

For the general case, if $V$ is a vector space over a field $F$ and $\phi$ is an automorphism of $F$, then $\lambda \circ v := \phi(\lambda) v$ defines a scalar multiplication on $V$. If $\phi$ is not the identity, then this scalar multiplication is not the original one.

The argument for $\mathbb Q$ boils down to the fact that $\mathbb Q$ has only one automorphism. Nor does $\mathbb R$ (see here).
On the other hand, complex conjugation is a nontrivial automorphism of $\mathbb C$. This gives an indication of how to handle the general case.


问:
Intuition behind the difference between absolute continuity and $\mu(A) = 0 \implies \mathcal H^{d-1}(A)<+\infty$
In the book Optimal Transport for Applied Mathematicians, the author states that the condition that a measure $\mu \in \mathcal P(\mathbb R^d)$ is absolutely continuous relative to the Lebesgue measure is a stronger than "$\mu(A) = 0$ for any $A \subset \mathbb R^d$ such that $\mathcal H^{d-1}(A)<+\infty$".

Now, I can intuitively understand what absolutely continuity of a measure implies, for example, but I do not have any intuition when it comes to the "$\mu(A) = 0$ for any $A \subset \mathbb R^d$ such that $\mathcal H^{d-1}(A)<+\infty$".

What would be the intuition behind such condition? Can anyone provide examples where $\mu$ is not absolutely continuous, but satisfies this second condition?

答:
Yes, in this context you can think about the Lebesgue measure as a sort of characterization of the (let me use this word a bit informally) dimension of the sets. All the Borel sets $A\subset\mathbb{R}^d$ of dimension $d$ verify $\mathcal{L}^d(A)>0$, while if $\mathcal{L}^d(A)=0$ then dimension of $A$ is strictly smaller than $d$. Thus, for $\mu$ to be absolutely continuous w.r.t. $\mathcal{L}^d$ it should vanish also in sets of dimension strictly smaller than $d$. On the other hand, the condition $\mu(A)=0$ for any $A\subset\mathbb{R}^d$ such that $\mathcal{H}^{d-1}(A)<+\infty$ leaves room to define $\mu$ so that it does not vanish for Borel sets $A$ of dimension $d-1$ but such that $\mathcal{H}^{d-1}(A)=+\infty$.

Example: A line in $\mathbb{R}^2$. Consider $d=2$ and $B=\{(x,y)\in\mathbb{R}^2:y=x\}$, hence $\mathcal{L}^2(B)=0$ but $\mathcal{H}^{1}(B)=+\infty$. Then you can define $\mu$ so that $\mu(A)=0$ for any $A\subset\mathbb{R}^d$ such that $\mathcal{H}^{d-1}(A)<+\infty$ but $\mu(B)>0$. Therefore it is not absolutely continuous w.r.t. $\mathcal{L}^2$.


问:
Superadditive games and the worth of the grand coalition
I'm studying a first course in Game Theory, I don't understand why in superadditive games, the grand coalition has the highest payoff among all coalitional structures.

If by contraddiction we assume that in a superadditive game $(G,v)$ there exist $S\subset N$ such that $v(S)>v(N)$, then by superadditivity $v(N)\geq v(S)+v(N - S)>v(N)+v(N-S).$

Is there something that force $v(N-S)$ to be positive or null?

答: By convention, $v(S)\ge0$ for all subsets $S$ of $N$.


问:
why $\sup$ come before $(a_n + b_n) ?$
Taken from Rudin real and complex analysis  book     chapter $1$ excercise

let $\{a_n\}$ and $\{b_n\}$ be  sequence  in $[-\infty,\infty]$.Prove the following assertion

$b).\lim_{n \to \infty} \sup(a_n +b_n) \le  \lim_{\to \infty} \sup a_n + \lim_{n\to \infty} \sup b_n$

My attempt : $\sup a_n$ is upper bound  of  $a_n$  and $\sup b_n$ is a upper bound of  $b_n$.

Now $ a_n \le \sup  a_n \tag1$

and  $b_n\le \sup b_n\tag2 $

From $(1)$ and $(2)$  we can say that

$a_n + b_n \le \sup a_n  + \sup b_n$

Put  limit both side

$\lim_{n\to \infty} a_n + b_n \le \lim_{n \to \infty}  \sup a_n  + \lim_{n \to \infty} \sup b_n$

My doubt : why  $\sup$  come  before  $(a_n + b_n) ?$

答:
If you take the supremum of a sequence you get a fixed value, so it doesn't make much sense to take the limit after it. I think you're confusing $\limsup$ and $\sup$.

They're not the same thing, since $\sup\limits_{n\in\Bbb N}\{a_n\}$ is the biggest value you can approach with the sequence (could be $+\infty$ if not bounded), while $\limsup\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\sup\limits_{k\ge n}\{a_k\}$, which means that you consider $a_1,a_2,a_3,\ldots$, you take the biggest value approachable and you add it to your collection; but then you consider $a_2,a_3,\ldots$, you take the biggest value approachable and you add it to your collection; now consider $a_3,\ldots$ and etc. You are collecting the biggest value approachable of each $\{a_k\}_{k\ge n}$ for each $n\in\Bbb N$. Then you take the limit, which you can do because the way you collection is constructed is as a non-increasing sequence (you have to allow $+\infty$ as a value). Doing this you're making sure that you're getting either $+\infty$ if not bounded, either a value that your sequence doesn't stop taking or approaching to.

This is the key difference. For example: If you consider the sequence $0,0,0,0,0,1,0,0,0,0,0,\ldots$ then the supremum is $1$, but that value just appears once, so it would be nice if there was something that makes sure we get a value that keeps appearing or a value that our sequence keeps approaching to. That's a job for $\limsup$, since eventually it gets rid of that single instance of $1$ and give us $0$, which makes sense. Another example could be the sequence $1+\frac{1}{n}$: If you consider $\sup$ you get the biggest term, which is $2$, the first one. Considering $\limsup$ you get the actual limit if the sequence, which is $1$. You can prove that if the limit of a sequence exists, then $\limsup$ and $\liminf$ give the same value as the limit.

Now, for the result you're asking about:

$a_k\le\sup\limits_{k\ge n}\{a_k\}\;\forall k\ge n$ and $b_k\le\sup\limits_{k\ge n}\{b_k\}\;\forall k\ge n$ implies $a_k+b_k\le\sup\limits_{k\ge n}\{a_k\}+\sup\limits_{k\ge n}\{b_k\}\;\forall k\ge n$, which implies $\sup\limits_{k\ge n}\{a_k+b_k\}\le\sup\limits_{k\ge n}\{a_k\}+\sup\limits_{k\ge n}\{b_k\}\;\forall k\ge n$.

Taking limit on both sides you get $\limsup\limits_{n\to\infty}(a_n+b_n)=\lim\limits_{n\to\infty}\sup\limits_{k\ge n}\{a_k+b_k\}\le\lim\limits_{n\to\infty}(\sup\limits_{k\ge n}\{a_k\}+\sup\limits_{k\ge n}\{b_k\})\le\lim\limits_{n\to\infty}\sup\limits_{k\ge n}\{a_k\}+\lim\limits_{n\to\infty}\sup\limits_{k\ge n}\{b_k\}=\limsup\limits_{n\to\infty}a_n+\limsup\limits_{n\to\infty}b_n$

Title: why $\sup$ come before $(a_n + b_n) ?$


问:
Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$

I couldn't do this question and hence I looked at the solution which goes as follows:

if $x=0$ then $x^2+2xy=0\ne5$ hence $x\ne0$

I state that $y=lx$. Hence the system becomes:

$x^2(1+2l)=5$, $x^2(l^2-3l)=-2$ which becomes $5l^2-11l+2=0$, hence $l=2$ or $l=\frac{1}{5}$.

Hence the possible solutions are:

$(1, 2), (-1,-2), (\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}), (-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}})$.

My question is why was I supposed to think of substituting $y=lx$? Why was this supposed to be intuitive and is there a more intuitive approach?

答:
Alternative approach.

Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$

$3x^2 + 6xy = 15~$ and $~2y^2 - 6xy = -4.$

Adding gives $3x^2 + 2y^2 = 11.$

Then, you have that $5 - x^2 = 2xy.$

Therefore, $25 - 10x^2 + x^4 = 4x^2y^2 = (2x^2)(11 - 3x^2) = 22x^2 - 6x^4.$

Therefore, $7x^4 - 32x^2 + 25 = 0.$

Applying the quadratic equation against $x^2$ gives that $x^2 \in \{1, \frac{25}{7}\}.$

Using that $2y^2 = 11 - 3x^2$ gives that
$y^2 \in \{4, \frac{1}{7}\}.$

Therefore, there are only 8 possible solutions to the equations:

$(x,y) \in \{(\pm 1, \pm 2), (\pm \frac{5}{\sqrt{7}}, \pm \frac{1}{\sqrt{7}})\}.$

Checking each of these 8 possibilities against the original equation gives final answer of:

$$(x,y) \in \left\{(1, 2), (-1, -2),
\left(\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right),
\left(-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}}\right)\right\}.$$


答:
I find it helpful to look at the prime factor decomposition of the numbers $a,b$ and $c$. For two numbers $x,y$ the $\gcd(x,y)$ is just the maximal set of common prime factors.

So for (a) you it is clear that the maximal sets of prime factors of $ac$ and $bc$ contain  both the prime factors of $c$. So if you take the prime factors of $c$ out, both sets remain maximal.

For (b) I would suggest to play around with some examples.

Title: Prove or disprove statements about the greatest common divisor

Body:
Help with prove or disproving either of these statements would be really appreciated, one or the other is fine, I just need a start or a solution to one and I'm sure I could probably figure the other out.

(a) For all $a, b, c \in \mathbb{Z}$, $\gcd(ac, bc) = c \gcd(a, b)$

(b) For all $a, b, c \in \mathbb{Z}$, $\gcd(ab, c) = \gcd(a, b) \gcd(b, c)$


问:
What's the inverse Laplace transform of $\sqrt{s^2-s}$
I know the inverse Laplace transform of $s$ is the delta function $\delta^\prime (t),$ i.e. $\mathcal{L}^{-1}(s)=\delta^\prime(t).$

And I know actually $\mathcal{L}^{-1}(s^n)=\delta^{(n)}(t),$ where $\delta^{(n)}$ means the nth-derivative of $\delta.$

My question is that, what is the inverse Laplace transform of $\sqrt{s^2-s},$ i.e. what's $\mathcal{L}^{-1}(\sqrt{s^2-s})\,?$

My attempts:

$$\sqrt{s^2-s}=s\sqrt{1-\frac{1}{s}}=s[1-\frac{1}{2s}+o(1)\frac{1}{s}]=s-\frac{1}{2}+o(1)\,\ \ \ \ \ \ \ \ \ \ \cdot\cdot\cdot\cdot\cdot\cdot\,\ \ \ (A)\,,$$ where the second equality is by the Taylor theorem. (I don't know if the first equality is vailed in the complex space.)

Then, $$\mathcal{L}^{-1}(\sqrt{s^2-s})=\mathcal{L}^{-1}(s)-\mathcal{L}^{-1}(\frac{1}{2})+\mathcal{L}^{-1}[o(1)]=\delta^\prime(t)-\frac{1}{2}\delta(t)+\mathcal{L}^{-1}[o(1)]$$

On one hand, we know $$\forall n\ge 0, \mathcal{L}^{-1}[\frac{1}{s^{n+1}}]=\frac{t^n}{n!}.$$

On the other hand, we know $o(1)$ is nothing than a power series in $\frac{1}{s}.$

So it seems like we have solved the question. But the result looks a mess, and I am not really confident of the first equality in $(A)$. And the application of Taylor theorem in $(A)$ is not vailed for those $s$ with $\lvert s \rvert$ not large enough.

答:
From MacLaurin expansion we have
$$s\sqrt{1-\frac{1}{s}}=\sum _{n=0}^{\infty } -\frac{s^{2-n} \Gamma \left(n-\frac{3}{2}\right)}{2 \sqrt{\pi } \;\Gamma (n)}$$
$$\mathcal{L}^{-1}\left(s\sqrt{1-\frac{1}{s}}\right)=-\sum _{n=0}^{\infty }\frac{\mathcal{L}^{-1}\left(s^{2-n}\right) \Gamma \left(n-\frac{3}{2}\right)}{2 \sqrt{\pi } \;\Gamma (n)}=\ldots$$
Knowing that
$$\mathcal{L}_s^{-1}\left(s^{2-n}\right)(t)=\frac{t^{n-3}}{\Gamma (n-2)};\;n>2$$
we have
$$\ldots=-\sum _{n=0}^{\infty }\frac{\frac{t^{n-3}}{\Gamma (n-2)} \Gamma \left(n-\frac{3}{2}\right)}{2 \sqrt{\pi } \;\Gamma (n)}=-\frac{e^{t/2} I_1\left(\frac{t}{2}\right)}{2 t}$$
where $I_1(x)$ is the modified Bessel function of the first kind.


问:
What is the density of $(W, Z)$ where $W = \frac{X}{X+Y}$ and $ Z = \frac{Y}{X+Y}$?
I have tried to use the method involving the Jacobian matrix but I am struggling to find the inverse of the function $f(x,y) = (\frac{x}{x+y}, \frac{y}{x+y})$.

We have that $X,Y \geq 0$.

答:
There is no inverse of $f$, for the two components of the result are related by $W+Z=1$. $f(1,2)=f(2,4)$, for example. It suffices to find the density of one of the two coordinates (e.g. by finding the density of $(X/(X+Y),Y)$) and then complementing to find the density of the other.

If $(W,Y)=(X/(X+Y),Y)$ then $YW/(1-W)=X$, i.e. $(X,Y)=(YW/(1-W),Y)$.


答:
In any group we have $\langle g^k \rangle = \langle g^d \rangle$ for $d=\gcd(k,n)$.
One inclusion is obvious. The other follows from Bézout's identity.

Let $g$ be a primitive $100$th root of unity. Then $\mathbb{U}_{100} = \langle g \rangle$ and so the image of $f$ is $\langle g^{70} \rangle = \langle g^{10} \rangle$ since $10=\gcd(70,100)$.
The order of $g^{10}$ is $100/10=10$ and so the image has order $10$.

Title: Image of group homomorphism.


问:
Supremum and Expectation
Let $\Omega$ be a finite set, $X$ be a random variable on $(\Omega, 2^{\Omega})$, $\mathcal{M}$ be a set of probability measure on $(\Omega, 2^{\Omega})$ with $\forall \omega , P(\{\omega\})>0$, and $\mathcal{F}$ be a $\sigma$ -algebra on $\Omega$.

I want to prove or disprove that $\displaystyle \sup_{Q\in \mathcal{M}}E^Q[\sup_{Q'\in \mathcal{M}}E^{Q'}[X|\mathcal{F}]]=\sup_{Q\in\mathcal{M}}E^Q[X]$.

My Attempt

By definition, for any $Q\in\mathcal{M}$,

$\sup_{Q'}E^{Q'}[X|\mathcal{F}] \geq E^Q[X|\mathcal{F}]$.

By taking the expectation w.r.t. $Q$ of both sides, we have

$E^{Q}[\sup_{Q'}[X|\mathcal{F}]] \geq E^Q[E^Q[X|\mathcal{F}]]=E^Q[X]$

By taking $\sup$ of both sides,

$\sup_{Q}E^Q[\sup_{Q'}E^{Q'}[X|\mathcal{F}]]\geq\sup_{Q}E^Q[X]$

Now we need to prove the $\leq$ part of the inequality. By definition of sup, we have

$\forall \epsilon >0, \exists Q(\epsilon)\in\mathcal{M}$ s. t.$ \sup_{Q'}E^{Q'}[X|\mathcal{F}]\geq E^{Q(\epsilon)}[X|\mathcal{F}]\geq \sup_{Q'}E^{Q'}[X|\mathcal{F}]-\epsilon$

By the squeezing theorem, we have $E^{Q(\epsilon)}[X|\mathcal{F}] \to \sup_{Q}E^Q[X|\mathcal{F}](\epsilon \to 0).$

By taking the expectation w.r.t $Q(\epsilon)$ of $E^{Q(\epsilon)}[X|\mathcal{F}]\geq \sup_{Q'}E^{Q'}[X|\mathcal{F}]-\epsilon$, we have

$E^{Q(\epsilon)}[E^{Q(\epsilon)}[X|\mathcal{F}]]\geq E^{Q(\epsilon)}[\sup_{Q'}E^{Q'}[X|\mathcal{F}]]-\epsilon$.

By using $\sup_{Q}E^{Q}[E^{Q}[X|\mathcal{F}]] \geq E^{Q(\epsilon)}[E^{Q(\epsilon)}[X|\mathcal{F}]]$, we have

$\sup_{Q}E^{Q}[E^{Q}[X|\mathcal{F}]] \geq E^{Q(\epsilon)}[\sup_{Q'}E^{Q'}[X|\mathcal{F}]]-\epsilon$

Now I am stuck here. I wanted to take $\epsilon \to 0$, but what we have is $E^{Q(\epsilon)}[X|\mathcal{F}] \to \sup_{Q}E^Q[X|\mathcal{F}](\epsilon \to 0)$, which is not useful here.

I feel like there is a counterexample but haven't come up with it by now.

Any help? As a reference, here is a definition of conditional expectation I am using.

Definition of conditional expectation

Let $\{B_i\}$ be atoms of $\mathcal{F}$, and $P\in \mathcal{M}$. For a random variable $X$, $E[X|\mathcal{F}]:\Omega \to \mathbb{R}$ is defined by

$E^P[X|\mathcal{F}](\omega) = \frac{E^P[X,B_i]}{P(B_i)}$, where $B_i$ is an atom that has $\omega$.

答:
Let us write

$$ S = \sup_{Q} \mathbb{E}^{Q}\biggl[\sup_{Q'}\mathbb{E}^{Q'}[X\mid\mathcal{F}]\biggr] $$

for simplicity. Then we already know that $ S \geq \sup_Q \mathbb{E}^{Q}[X] $, and so, it suffices to prove the reverse inequality.

Now, let $\omega_0 \in \Omega$ be such that $X(\omega_0) = \max X$. We also write $M = \max \lvert X \rvert$. Then for each $\epsilon > 0$, there exists $Q_{\epsilon} \in \mathcal{M}$ such that $Q_{\epsilon}(\{\omega_0\}) > 1-\epsilon$, and so,

\begin{align*}
\mathbb{E}^{Q_{\epsilon}}[X]
&\geq \mathbb{E}^{Q_{\epsilon}}[X \mathbf{1}_{\{\omega_0\}} - M \mathbf{1}_{\Omega\setminus\{\omega_0\}}] \\
&= X(\omega_0)(1-\epsilon) - M\epsilon \\
&= \max X - (X(\omega_0) + M)\epsilon \\
&\geq S - (X(\omega_0) + M)\epsilon.
\end{align*}

Letting $\epsilon \downarrow 0$, this proves that

$$ \sup_Q \mathbb{E}^{Q}[X] \geq S $$

and therefore the equality holds. This proof also hints that the supremum is essentially "achieved" at the point mass concentrated at the maximum point of $X$.

Title: Supremum and Expectation


问:
Show that the unit circle minus a point is connected (without using path connectedness)
I am trying to show that the closed unit interval $I = [0,1]$ is not homeomorphic to $S^1$ using the notion of connectedness (and not the notion of path connectedness).

A proof here shows that the two are not homeomorphic because, if $f: I \to S^1$ were a homeomorphism, then a contradiction arises when we restrict the domain of $f$ to $I \setminus \{a\}$. The contradiction comes from the fact that $I \setminus \{a\}$ is disconnected whereas $S^1 \setminus \{p\}$ is connected, for any point $a \in (0,1)$ (that is, any interior point $a$) and $f(a) = p$.

But I can't figure out how to show that $S^1 \setminus \{p\}$ is connected using just the definition of connectedness.

The definition of connectedness that I have is the following, from Willard:

A space $X$ is disconnected iff there are disjoint nonempty open sets
$H$ and $K$ in $X$ such that $X = H \cup K$. When no such
disconnection exists, $X$ is connected.

Equivalently, a disconnection could be made from two disjoint, nonempty closed sets (sometimes this construction seems easier, depending on the topology).

Again, equivalently, we can say that a space $X$ is disconnected iff there exists a proper clopen (both closed and open) subset of $X$.

I've tried to show that, if we assume $S^1 \setminus \{p\}$ is disconnected then we get some contradiction, but I haven't had any luck.

答: Suppose $f:I\rightarrow \mathbb{S}^1$ is a homeomorphism. Pick a point $x\in I\backslash \{0,1\}$. Denote the point this point is mapped to by N. We get that $I\backslash \{ x\}$ is homeomorphic to $\mathbb{S}^1 \backslash \{N\}$. However, via sterographic projection, $\mathbb{S}^1\backslash \{N\}$ is homeomorphic to $\mathbb{R}$. Therefore $\mathbb{S}^1\backslash \{N\}$ is connected. Since $I\backslash \{x\}$ is disconnected, $I$ and $\mathbb{S}^1$ cannot be homeomorphic.


问:
What function is denoted by $\mathbb{I}(x=y)$?
I am reading a paper that uses the function $\mathbb{I}(x=0)$. I assume this is a common function, but I haven't seen it before.

From the context it seems like it returns $0$ when the argument is true and $1$ when it is false, but I suspect that it is the opposite and that there is a typo in the paper. Just want to make sure.

答:
$\mathbb I$ is an indicator function. It returns $1$ if the predicate inside is true and $0$ if false.

Sometimes this is written $[P]$, the Iverson bracket.


问:
Quick Rejog of Partial Fraction Decomposition:
I don't need the problem to be solved I just need to have the decomposed equation. I have the following equation: \begin{equation} I=\int \frac{dx}{x^2(x^2-16)}\end{equation} The method that I have to use is Partial Fractions Decomposition. My question is since there is an $x^2$ term I have to increase in linearity which I wonder if I did correctly? $$\bbox[border:2px solid black]
{\begin{equation}\frac{1}{x^2(x^2-16)}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{D}{x+4}+\frac{E}{x-4} \end{equation}}$$

答:
One can easily observe what redundant is as follows
$$\frac{1}{x^2(x^2-16)}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{D}{x+4}+\frac{E}{x-4}$$
$$=\frac{A}{x}+\frac{B}{x}+\frac{C}{x^2}+\frac{D}{x+4}+\frac{E}{x-4}$$
Thus either $\frac Ax$ or $\frac{bx}{x^2}=\frac Bx$ is redundant


问:
Uniform convergence in real valued injective functions sequence
If $f_{n}$ is a sequence of real valued injective functions that converges uniformily to $f$, then is it necessary that $f$ be injective? Thanks in advance!

答:
Easy counter-example :

Pick $f_n(x):= \frac{x}{n+1}$ for $x\in[0,1],n\in\mathbb{N}$

$\frac{1}{n+1}\ge\lvert{f_n-0}\rvert$ which means that $\frac{1}{n+1}\ge {sup_{x\in[0,1]}\lvert{f_n-0}\rvert}$, which means that $f_n$ converges uniformly towards the trivial function, yet $f_n$ is clearly injective while the zero function clearly isn't


问:
Confused about the relation between $\ell^p$ and $L^p$
I know from Wikipedia that $\ell^p(S)$ is a special case of an $L^p$ space, namely it is the space $L^p(S, \mu)$, where $\mu$ is the counting measure on $S$.

There are also the $\ell^{\infty}$ and $L^{\infty}$ spaces. $\ell^{\infty}$ is defined as the space of bounded sequences with the sup norm. I read, also on Wikipedia, that this space $\ell^{\infty}$ is a special case of an $L^{\infty}$ space, namely it is the space $L^p(\mathbb N, \mu)$, where $\mu$ is the counting measure on $\mathbb N$.

One can consider a slight generalization of $\ell^{\infty}$: For any set $S$, consider the space $\ell^{\infty}(S)$ of all bounded functions $S\to k$ ($k$ either $\mathbb R$ or $\mathbb C$) equipped with the sup norm. Question: Can one also consider this space as a special case of $L^{\infty}(\cdots)$? It seems, Wikipedia doesn't answer this question, in fact, the Wikipedia page on "L-infinity" doesn't even consider the space $\ell^{\infty}(S)$.

答: Sure, on any set $S$ you have a counting measure $\#$. Then there are no nonempty null sets and thus the essential supremum (as in the definition of $L^\infty(S,\#)$) is just a supremum. In particular $\ell^\infty(S)=L^\infty(S,\#)$.


问:
Product of $\cos(a^kx)$ where $k=1,2,3,\ \dots$ and $a$ is a natural number
I have quite frequently encountered the relation, $$\cos(x)\cos (2x)\cos(4x)\dots \cos(2^{n-1}x)= \frac{\sin(2^nx)}{2^n\sin(x)}$$but is there any explicit formula for the series, $$\cos(x)\cos (3x)\cos(9x)\dots \cos(3^{n-1}x)$$I have tried hard to come up with a relation, but in vain. If there is such a formula, I would really be interested in how to prove it, that is you can give me a hint.

答:
While there isn't a directly comparable formula, by looking at various forms of the identity you give we can find similar expressions.

The usual derivation goes through the doubling formula for $\sin x$: starting from $\sin(2x)=2\sin(x)\cos(x)$ or, put another way, $\frac{\sin(2x)}{\sin(x)}=2\cos(x)$, we can look at $\frac{\sin(4x)}{\sin(x)}=\frac{\sin(4x)}{\sin(2x)}\cdot\frac{\sin(2x)}{\sin(x)}$ $= \left(2\cos(2x)\right)\cdot\left(2\cos(x)\right)$, $\frac{\sin(8x)}{\sin(x)}=\frac{\sin(8x)}{\sin(4x)}\cdot\frac{\sin(4x)}{\sin(x)}$, etc.

Now, it turns out that $\frac{\sin(3x)}{\sin(x)}$ is also a polynomial in $\cos(x)$; specifically, if $P(t)=4t^2-1$, then $\frac{sin(3x)}{\sin(x)}=P(\cos(x))$, so we can continue expanding out in very similar ways and get the equation:
$$\frac{\sin(3^nx)}{\sin(x)}=\prod_{j=0}^{n-1}P\left(\cos(3^jx)\right)$$
In fact, it turns out that for all $k$ we have $\sin(kx)=\sin(x)\cdot P_k(\cos(x))$ for some polynomial $P_k$; specifically, $P_k$ is the Chebyshev polynomial of the second kind $U_{k-1}(t)$. This means that we can generalize the formula above as:
$$\frac{\sin(k^nx)}{\sin(x)}=\prod_{j=0}^{n-1}U_{k-1}\left(\cos(k^jx)\right)$$.

It turns out that the polynomial $U_1(x)=2x$, so the case $k=2$ is exactly the formula that you gave.

There's another angle (ahem) on the formula, too: it expresses the uniqueness of binary expansions!  To see this, let's let $z=e^{ix}$ and, for simplicity's sake, I'll write $z'=e^{-ix}=z^{-1}$. Then $\cos(x)=\frac12(z+z')$ $=z'\cdot\frac12(z^2+1)$ and $\sin(x)=\frac1{2i}(z-z')$ $=z'\cdot\frac1{2i}(z^2-1)$. Now I'll write $w=z^2$, again for convenience's sake: we can write $\sin(2^nx)=z'^{2^n}\cdot\frac1{2i}(w^{2^n}-1)$, so $\dfrac{\sin(2^nx)}{\sin(x)}=z'^{2^n-1}\dfrac{w^{2^n}-1}{w-1}$. Now, the term in $w$ on the right hand side is just the sum $\sum_{j=0}^{2^n-1}w^j=1+w+w^2+\ldots+w^{2^n-1}$. On the other hand, $2\cos(x)=z'(1+w)$, $2\cos(2x)=z'^2(1+w^2)$, etc. So the product $\prod_{j=0}^{n-1}2\cos(2^jx)$ is $\prod_{j=0}^{n-1}z'^{2^i}(1+w^{2^j})$; the product of the $z'^{2^j}$ terms is just $z'^{2^n-1}$, and what we wind up with is the identity
$$\prod_{j=0}^{n-1}\left(1+w^{2^j}\right)=\sum_{j=0}^{2^n-1}w^j=\frac{w^{2^n}-1}{w-1}$$
Now, the coefficient of $w^k$ on the left hand side of this identity is the number of ways of writing $k$ as a sum of different terms of the form $2^i$, and this identity says that this sum is $1$ for every $k$; in other words, binary expansions are unique!

We can go the other direction from this now, and try to do the same thing with ternary expansions: the generating-function version of the statement 'every number has a unique ternary expansion' is the identity $\prod_{j=0}^{n-1}(1+w^{3^j}+w^{2\cdot3^j})$ $=\sum_{j=0}^{3^n-1}w^j$ $=\dfrac{w^{3^n}-1}{w-1}$. We can think of the right hand side of this as being related to $\dfrac{\sin(3^nx)}{\sin(x)}$ and try to write the left hand side in similar fashion. Since $1+w^{3^j}+w^{2\cdot 3^j} = w^{3^j}\cdot(w^{-(3^j)}+1+w^{3^j})$, these terms should be expressible in the form $a+b\cos(2\cdot3^jx)$ for some $a$ and $b$. I'm a bit too lazy to do the algebra on this right now, but I'm 99% confident that if you go all the way through this you'll find that it's exactly equivalent to the identity in terms of the product of $P(\cos(3^jx))$ that I derived from the triplication formula above.

This formulation also suggests why there's no 'nice' formula for the product you're after: The '$w$-analog' of $\prod_{j=0}^{n-1}\cos(3^jx)$ is $\prod_{j=0}^{n-1}(1+w^{3^j})$; expanding this out into a polynomial gives $\sum_{j\in C_n}w^j$, where $C_n$ is the 'Cantor set' of numbers whose ternary expansion is at most n digits and has only zeroes and ones in it. It's (hopefully) not too surprising that there's no 'nice' way of writing this set other than that characterization.


问:
Why does zero correlation not imply independence?
Although independence implies zero correlation, zero correlation does not necessarily imply independence.

While I understand the concept, I can't imagine a real world situation with zero correlation that did not also have independence.

Can someone please give me an example so I can better understand this phenomenon?

答:
Think about this problem visually.
If the joint distribution is symmetric to the origin, then the correlation could be $0$.
But the joint distribution may not be uniform in the domain.
e.g.
$$p(x,y) = \frac{|x| + |y|}{4}, \text{ for }x,y \in [-1,1].$$
Then the marginal densities are $p(x) = |x|/2 + 1/4$, and $p(y) = |y|/2 + 1/4$.
Obviously, these two variables are not independent since they tend to appear at the corners. Strictly speaking, $p(x,y) \neq p(x)p(y)$.
Brighter color means higher probability.

Next, it is obvious that the correlation is $0$ as $\text{Cov}(X,Y)= E(XY) -EX\cdot EY = 0$.

Another example is $p(x,y) = 1/2$, for $|x|+|y| < 1$, which is a uniform distribution over the diamond domain. Then $p(x) = 1 - |x|$, and $p(y) = 1 - |y|$. The reasoning is similar to the above example.


问:
Prove that $x^2y^2(4-x^2)(4-y^2) \leq16$ for all $|x|\leq2$ and $|y| \leq 2$
Prove that $x^2y^2(4-x^2)(4-y^2) \leq16$ for all $|x|\leq2$ and $|y| \leq 2$

My attempt:

If $x^2(4-x^2) \leq 4$ for $|x|\leq2$ and $y^2(4-y^2) \leq 4$ for $|y|\leq2$ then the inequality holds.

Define $h(x) = x^2(4-x^2)-4$. $h'(x) = 8x-4x^3$, hence we get critical points $x=0,\pm \sqrt{2}$. By the second derivative test we find that $0$ is a maximum in the $|x| \leq 2$ range  . Hence:

$$x^2(4-x^2)-4 \leq 0$$
$$x^2(4-x^2)\leq 4$$

A similar argument for the $y$ based function, and hence the top inequality holds.

Is this attempt correct?

Body:
Note

$$x^2y^2(4-x^2)(4-y^2)=[4-(2-x^2)^2][4-(2-y^2)^2]\le4\cdot4=16
$$


Body:
Prove that $x^2y^2(4-x^2)(4-y^2) \leq16$ for all $|x|\leq2$ and $|y| \leq 2$

My attempt:

If $x^2(4-x^2) \leq 4$ for $|x|\leq2$ and $y^2(4-y^2) \leq 4$ for $|y|\leq2$ then the inequality holds.

Define $h(x) = x^2(4-x^2)-4$. $h'(x) = 8x-4x^3$, hence we get critical points $x=0,\pm \sqrt{2}$. By the second derivative test we find that $0$ is a maximum in the $|x| \leq 2$ range  . Hence:

$$x^2(4-x^2)-4 \leq 0$$
$$x^2(4-x^2)\leq 4$$

A similar argument for the $y$ based function, and hence the top inequality holds.

Is this attempt correct?

Body:
$$
(x^2-2)^2 \geq0
\Rightarrow (x^2)^2-4x^2+4 \geq0
\Rightarrow (x^2)^2-4x^2 \geq-4
\Rightarrow 4x^2-(x^2)^2 \leq4 \\
\Rightarrow x^2(4-x^2) \leq4
$$
Similarely $y^2(4-y^2) \leq4$.

When $0<y<x$ and $0<z<t$, then $0<yz<xt$. Now we have $x^2(4-x^2) \leq4$ and $y^2(4-y^2) \leq4$, and hence $x^2y^2(4-x^2)(4-y^2) \leq16$.


问:
Why do you multiply by the integrating factor when solving differential equations?
Differential equations of the form
$$
\frac{dy}{dx}+p(x)y=q(x)
$$
can be solved by multiplying both sides of the equation by the 'integrating factor' $e^{\int p(x) \, dx}$. This transforms the equation to
\begin{align}
&e^{\int p(x) \, dx}\frac{dy}{dx}+e^{\int p(x) \, dx}p(x)y = e^{\int p(x) \, dx}q(x) \\
\implies &\frac{d}{dx}(e^{\int p(x) \, dx}y) = e^{\int p(x) \, dx}q(x) \, \tag{*}\label{*}.
\end{align}
This makes the equation much easier to solve, since we can integrate both sides of $\eqref{*}$ and proceed from there. However, while it is not too difficult to verify that
$$
\frac{d}{dx}(e^{\int p(x) \, dx}y)= e^{\int p(x) \, dx}\frac{dy}{dx}+e^{\int p(x) \, dx}p(x)y \, ,
$$
it still seems that the 'multiply by $e^{\int p(x) \, dx}$' trick has been pulled out of a hat. In this post, I present a way of arriving at the above method in a way that seems less ad hoc.

答:
The integrating factor method is based upon what might be termed the 'reverse product rule'. First, note that any differential equation of the form
$$
f(x)\frac{dy}{dx} + f'(x)y=g(x)
$$
may be rewritten as
$$
\frac{d}{dx}(f(x)y)=g(x)  \, .
$$
For example,
\begin{align}
&\sin(x)\frac{dy}{dx} + \cos (x)y = x^2 \\
\implies &\frac{d}{dx}(\sin(x)y) = x^2 \\
\implies &\sin(x)y = \int x^2 \, dx \\
\implies &\sin(x)y = \frac{x^3}{3}+C \\
\implies &y = \frac{x^3}{3\sin(x)} + \frac{C}{\sin (x)} \, .
\end{align}
So when we have a differential equation on the form
$$
\frac{dy}{dx}+p(x)y=q(x) \tag{1}\label{1} \, ,
$$
we want to convert it to
$$
f(x)\frac{dy}{dx} + f'(x)y=g(x) \, , \tag{2}\label{2}
$$
since $\eqref{2}$ boils down to
$$
\frac{d}{dx}(f(x)y)=g(x) \, .
$$
This is where the integrating factor steps in. Let $I(x)$ be the integrating factor, the multiplier that helps us get from $\eqref{1}$ to $\eqref{2}$. If we multiply $\eqref{1}$ by $I(x)$, we get
$$
I(x)\frac{dy}{dx}+I(x)p(x)y=I(x)q(x) \, .
$$
If the integrating factor is of the correct form, then we should now be able to rewrite the LHS as
$$
\frac{d}{dx}(I(x)y) \, .
$$
This tells the condition that $I(x)$ must satisfy: we require that $\frac{d}{dx}(I(x)y)$ equals $I(x)\frac{dy}{dx}+I(x)p(x)y$. To reiterate, multiplying a differential equation by the integrating factor should allow us to rewrite the LHS as the derivative of the product of two functions. This means that
$$
I(x)\frac{dy}{dx} + I'(x)y = I(x)\frac{dy}{dx}+I(x)p(x)y \, .
$$
This condition boils down to $I'(x)=I(x)p(x)$. Hence,
\begin{align}
&p(x) = \frac{I'(x)}{I(x)} \\
\implies & \int p(x) = \ln|I(x)| \\
\implies & I(x) = e^{\int p(x) \, dx} \, .
\end{align}
This solution can be verified by plugging $I(x)=e^{\int p(x) \, dx}$ into the equation
$$
I(x)\frac{dy}{dx}+I(x)p(x)y=I(x)q(x) \, .
$$
For ease of notation, let $P$ be an antiderivative of $p$, so that $I(x)=e^{\int p(x) \, dx}=e^{P(x)}$. The above equation is thus
$$
e^{P(x)}\frac{dy}{dx}+e^{P(x)}p(x)y = e^{P(x)}q(x) \, .
$$
We have fulfilled our objective of arriving at an equation of the form
$$
f(x)\frac{dy}{dx} + f'(x)y=g(x) \, ,
$$
with $f(x)=e^{P(x)}$ and $g(x)=e^{P(x)}q(x)$. Therefore, our work is done.


问:
Relation between lubs of a set and its subset
If $A\subset B$ and $B$ is bounded above, show that $\operatorname{lub}A \leq \operatorname{lub}B$.

This seems very obvious but I am not able to write a proper solution for this. Does it really require a proof or we can logically conclude this? Kindly help.

答:
Yes it requires a proof, by logically concluding.

Let $x$ be the least upper bound of $B$, then it's an upper bound for $A\subseteq B$, hence $\sup A\le x$.


问:
Test series for divergence or convergence
I have to find whether the infinite series $$\sum_{n=0}^\infty \left(\frac{n}{b(n-3)}\right)^n, \quad\text{for }b>0$$ diverges or find for which values of $b$ it converges to some finite value. So far I've tried to apply the ratio test but it didn't seem to work. Any advice on how to solve the problem is greatly appreciated.

答:
You have
$$\left(\frac{n}{b(n-3)}\right)^{n} = \frac{1}{b^n}\left(1-\frac{3}{n}\right)^{-n}$$

and $$\lim_{n \rightarrow +\infty} \left(1-\frac{3}{n}\right)^{-n} = e^3$$

so $$\left(\frac{n}{b(n-3)}\right)^{n} \sim \frac{e^3}{b^n}$$

So the series converges iff $$\frac{1}{b}<1$$
i.e. iff $b>1$.


问:
The probability of waiting time in a Poisson Process
Let $T_n$ be the time of the $n^{th}$ arrival time in a Poisson process N with rate $\lambda$.

I want to define  something called an excess lifetime process {E(t), t $\geq$ 0} where E(t) = $T_{N(t)+1}$ - t

E(t) can be understood as the time one must wait from time t until the arrival of the next event.

I have a problem deriving the following cdf for E(t),

P(E(t) > x) = $e^{\lambda(x+t)}$ + $\int_0^t$P(E(t-u)>x)($\lambda$$e^{-\lambda e}$)du

I am attempting to condition the process to $T_1$ so that it makes the equation easier, but I am still lost on how to get the integral part of the probability equation.

How do i derive the cdf of E(t)?

答:
I am not sure if I understand your question correctly. Are you trying to derive that specific equation, or are you trying to derive the probability distribution of delays between events in a Poisson process?

If the latter, then I can copy in mine here:

The Poisson Process

A stochastic process that produces independent spikes at an average rate $\lambda$, can be viewed as there being a probability $p$ every discrete time interval $\Delta t = \frac{1}{n}$ that a spike is created, found as
\begin{equation}
    p = \frac{\lambda}{n},
\end{equation}
where $n$ is time resolution, i.e. the number of intervals per unit of time.

If the process starts at time $t_0=0$, then the probability that the first spike is created after a time $t=N\Delta t$ is equal to the probability that it is not created in the $N$ first time intervals.
The probability that a spike is not created in any given interval is of course $(1-p)$, so the probability that it does not occur after $N$ intervals is therefore $(1-p)^N$, giving us the expression
\begin{equation}
    \text{Prob}(\text{spike after } t) = (1-p)^N.
\end{equation}
Substituting $N=\dfrac{t}{\Delta t}=n t$ and $p=\dfrac{\lambda}{n}$ gives us
\begin{equation}
    \text{Prob}(\text{spike after } t) = \Big(1- \frac{\lambda}{n}\Big)^{n t},
\end{equation}
and in the limit of perfect time resolution $n \to \infty$ we get
\begin{align}
    \text{Prob}(\text{spike after } t) & = \lim_{n \to \infty} \Big(1 - \frac{\lambda}{n}\Big)^{nt}
\end{align}
which we see is the definition of the exponential function, i.e.
\begin{equation}
    \boxed{\text{Prob}(\text{spike after } t) = e^{-\lambda t}.}
    \label{Poisson_exp}
\end{equation}

The probability that a spike occurs after a time $t$ is in other words exponentially decreasing as $t$ gets larger.

I hope this answers your question.


问:
Active transformation to switch basis vectors
I was wondering how I could construct a $2\times 2$ active transformation matrix which would switch the basis vectors $e_1$ and $e_2$.

Any help is appreciated.

答:
If I understand correctly, by $\{e_1,e_2\}$ you mean the standard basis $e_1=(1,0)$, $e_2=(0,1)$, and you want to find the matrix associated to a linear function $f:\mathbb{R}^2\to \mathbb{R}^2$ such that $f(e_1)=e_2$ and $f(e_2)=e_1$.

This is quite simple, since such linear function is unique and its associated matrix (with respect to this basis) is just $A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ (why?).
You can easily compute that this matrix "switches" $e_1$ and $e_2$.


答:
I'm using the root test, which you can read about if you want here: https://en.wikipedia.org/wiki/Root_test
.

Consider $\lim \limits_{n\rightarrow \infty}((\frac{n}{b(n-3)})^n)^{\frac{1}{n}}=\lim \limits_{n\rightarrow \infty}\frac{n}{b(n-3)}=\lim \limits_{n\rightarrow \infty}\frac{1}{b}=\frac{1}{b} $ by L'hopital's rule. By the root test, the series converges if this limit is less than $1$, so the series converges if $b>1$.

Hope this helps.

Title: Test series for divergence or convergence

Body: I have to find whether the infinite series $$\sum_{n=0}^\infty \left(\frac{n}{b(n-3)}\right)^n, \quad\text{for }b>0$$ diverges or find for which values of $b$ it converges to some finite value. So far I've tried to apply the ratio test but it didn't seem to work. Any advice on how to solve the problem is greatly appreciated.


问:
2 degree polynomial problem
I'm not sure how to title this but here's the problem;

Define $a,b,c$ so that $P(x) + Q(x) = 0$ with every value of $x$.

$P(x) = 5x^2 - ax + 4 - (bx^2 - 7x + 3)$ and $Q(x) = 8x^2 + x + c$.

So you need to find values for $a, b$ and $c$ which will make the sum of $P(x)$ and $Q(x)$ equal to zero, no matter what you will input as the value of $x$.

I've tried simplifying the sum of the polynomials but that's how far I'll get. I can deduce the correct answer by trying different values but i don't think it's how you are supposed to solve this problem.

How you should mathematically approach this problem? Can you create an equation out of this?

答:
In general what you want to do is group the like terms.
What does $P(x)+Q(x)$ look like?
$P(x)+Q(x)=5x^2-ax+4-bx^2+7x-3+8x^2+x+c$
now we can group the like terms to have that this is equal to:
$(5-b+8)x^2+(-a+7+1)x+(4-3+c)$
we want this to equal zero, so in particular we need $(5-b+8)=0$, $(-a+7+1)=0$ and $(4-3+c)=0$.

The same strategy is how you would do this in general.

Title: 2 degree polynomial problem


问:
Question about Laurent series and analytic functions.
I don't think I understand when a Laurent series vs Taylor series should be used. More importantly, I don't think I understand how a Taylor series can be analytic with a singularity. A few questions:

Can an analytic function (as one where the derivative, defined as a limit of ratios, existed) exist even if there's a singularity at a point?

So only analytic functions can be expressed as power series? Is that correct?

So in the example below, it has a singularity yes but it can be expressed as a Taylor series and not a Laurent series. Why is this?

If the radius R = 0, what does that mean? Does it converge at the singularity? What does that mean?

I guess my main question is: Can a function be analytic in a domain even if it has a singularity?

For example, I know this can be expressed as a Taylor series: $\frac{1}{3-z}$ around the point z = 1

$$\frac{1}{2 - (z - 1)} = \frac{1}{2} \cdot \frac{1}{1 - \frac{z - 1}{2}}$$
Note that the leading 1 in the denominator is in the right form so we can use:

$$\frac{1}{1-z} = 1 + z + z^2 + ...\sum_{n=0}^{\infty} z^n$$ and this converges if: \left | z \right | < 1

So even though this series has a singularity at z = 3, it can be expressed as a Taylor series. But is this Taylor series analytic everywhere? It's not at z=3 right?

The series above: $$\frac{1}{2} \cdot \frac{1}{1 - \frac{z - 1}{2}} = \frac{1}{2}( 1 + (\frac{z-1}{2}) + (\frac{z - 1}{2})^2 + ...$$ converges if $$\left | \frac{z-1}{2} \right | < 1$$
Here's a quote about Laurent series:

A “Laurent series” for a function is a generalization of the Taylor series, and like a Taylor series, it is essentially a power series. With a Laurent series, however, the powers can be negative. A major advantage of the Laurent series over the Taylor seriesis that Laurent series can be expanded around singular points for a given function, and can then be used to analyze the function in the neighborhoods of its singularities. This analysis will be needed when we develop “residue theory” for computing all sorts of weird integrals that can arise in applications.
By the way, a singular point for a function f is a point on the complex plane where f
might not be analytic

What does it mean to expand around singular points for a given function?

答:
There is a lot to address in this question, maybe more than ought to be addressed in one question in a forum like this. First of all, I will suggest that you read about Laurent series, analyticity and singularities in an introductory text on complex analysis. Some good alternatives are Theodore Gamelin's Complex Analysis and Saff and Snider's Fundamentals of Complex Analysis. That said, I will briefly address some of the points here:

A analytic function can have a limit at a singular point. This is what we call a removable singularity, and it means that the function can be redefined such that it is analytic in that point. In this case the Laurent series will equal the Taylor series. In addition to removable singularities, we distinguish between poles and essential singularities. These are defined through properties of the Laurent series around the point in question.

The Laurent series is, as you mention, a generalized Taylor series. It is generalized in the sense that we allow for terms with negative exponents. A Laurent series is therefore usually written as

$$\sum_{n = -\infty}^{\infty} a_n (z-z_0)^n,$$

which is a notation for the formal series

$$\sum_{n = 1}^{\infty} \frac{a_{-n}}{(z-z_0)^n} + \sum_{n=0}^{\infty} a_n (z-z_0)^n.$$

Obsere that the last parts looks exactly like a Taylor series. Whereas a Taylor series converges in a disc, a Laurent series converges in an annulus. An annulus is a region in the complex plane of the form $r < |z-z_0| < R$ (You should be able to see this from the expression above if you are familiar with power series). Note that we allow for $r=0$, where the annulus is a punctured disc, and $R= \infty$. This is why a Laurent series is defined around a singularity. $z_0$ would typically be a singularity and the Laurent series is a representation of the function around that singularity. Intuitively the terms with negative exponents "make up for" the singularity, and how many such terms are included in the series tells us what kind of singularity we are near. If there are none, it is a removable singulary. If there are finitely many but at least one, we call it a pole, whereas if there are infinitely many, we are dealing with an essential singularity.

Yes, analytic functions can be expressed locally as power series. In some textbooks this is in fact the definition of analyticity. Be aware that there is not just one Laurent series representation of an analytic function. As mentioned, the representation is local, and will depend on the domain you are studying.


问:
Prime numbers of certain form
Are there infinitely many prime numbers of the form $p^{2h}+p^h+1$, where $p$ is a prime and $h$ is a positive integer?

答:
The generalized Bunyakovsky conjecture implies that there are, as long as $x^{2h} + x^h + 1$ is irreducible (which I think is true if and only if $h$ is a power of $3$).

EDIT: Yes, that is the case.  $x^{2h}+x^h+1 = \dfrac{x^{3h}-1}{x^h - 1}$ is the product of the cyclotomic polynomials $C_d(x)$ where $d$ divides $3h$ but not $h$.  If $h$ is a power of $3$, the only such $d$ is $3h$ itself: $x^{2h} + x^h + 1 = C_{3h}(x)$ is irreducible.  If $h$ is divisible by some prime $q \ne 3$, then $d = 3h/q$ is another $d$, and $x^{2h}+x^h+1$ is reducible.

Title: Prime numbers of certain form


答:
Actually, having all the expression affected by an n power, you will probably have better results using the root test. This is:

$L = \sqrt[n]{|{a_n}|} = \frac{n}{b(n-3)}$.

As that is a quotient of same-degree polynomials, the limit to infinity is:

$L = \lim_{n\to\infty} \sqrt[n]{|{a_n}|} = \frac{1}{b}$

As you may know, $L < 1$ ensures convergence, and $L > 1$ ensures divergence, thus the first clear conclusion is that if $b > 1$, the series converges, and if $b < 1$ the series diverges

The case $b = 1$ must be analysed separately, for in that case the root test does not give any information on the series' convergence. However, we can see that if $b = 1$:

$\lim_{n\to\infty} {a_n} = \lim_{n\to\infty} (\frac{n}{n-3})^n = \lim_{n\to\infty} (1 + \frac{3}{n-3})^n = e^3$

As $e^3 \neq 0$, the series is not convergent when $b = 1$, thus converging only when $b > 1$


答:
You can use the root test
\begin{align*}
\limsup_{n \to \infty} \sqrt[n]{\left( \frac{n}{b(n-3)} \right)^n} &= \limsup_{n \to \infty} \frac{n}{b(n-3)}
\end{align*}
The series converges, when
$$ \limsup_{n \to \infty} \frac{n}{b(n-3)} > 1 $$
Therfore, for the series to converge, you have to choose $b$ as
$$ b > \lim_{n\to \infty} \frac{n}{n-3} = 1 $$


问:
$A$ closed subset of metric space. Then there exists a continuous function $f:A \to [0,1] \subset \mathbb R$ with $f^{-1}(0)=A$.
Suppose $A$ is a closed non-empty subset of a metrix space $(X,d)$. We define the distance from $x\in X$ to the set $A$ as:
$$ d(x,A):= \inf_{a\in A} d(x,a).$$
The problem I then solved is that it turns out that $x\mapsto d(x,A)$ is continuous. We derive some estimates using the triangle inequality and the property of the infimum being the greatest lower bound. We then show with the usual $\epsilon-\delta$ that whenever $x \to x_0$ we get  $f(x)\to f(x_0)$ for any $x_0 \in X$.

Now the exercises I am working on suggest that this problem can be used to tackle the following:

There exists a continuous function $f:X \to [0,1] \subset \mathbb R$ with $f^{-1}(0)=A$.

Now I know that continuity implies sequential continuity and that closed sets contain all their limit points. This means that for $(x_n)_{n \in \mathbb N}$ a sequence in $A$ with $x_n \to x$ we know $x \in X$, and also such a function $f$ has the property that $f(x_n) \to f(x)$. I am not entirely sure how the pieces fit together.

答: Define$$f(x)=\frac{d(x,A)}{1+d(x,A)}.$$Then $f$ is continuous, $f$ is a map from $X$ into $[0,1]$ and $f^{-1}(0)=A$ (since $A$ is closed; in general, $f^{-1}(0)=\overline A$).


问:
Product rule for limits of functions
When asking wolfram alpha to solve the following limit $$\lim_{x \to 0} \frac{3^x+\ln|x|}{x^5+x^4}$$ it uses the following step: $$\lim_{x \to 0} \frac{3^x+\ln|x|}{x^5+x^4} =\lim_{x \to 0}\frac{1}{x^5+x^4} \times \lim_{x \to 0}\left(3^x+\ln|x|\right).$$ Can you really do this, since the limit for $\frac{1}{x^4+x^5}\rightarrow \infty$ ?

I have read the the product rule for limits is "if the limit for $f(x)$ and $g(x)$ exists when $x$ goes to $a$ then $\lim_{x \to a}f(x)g(x)=(\lim_{x \to a}f(x))(\lim_{x \to a}g(x))$ and the other way around"

答:
Because both the limits tend to either $\infty$ or $-\infty$, that separation is allowed and raises no ambiguity. The ambiguity may only happen only if one of the indefinite limits (such as $\infty\times 0$) happens. An example is
$$
{\lim_{x\to 0} x\times {1\over x}
\\
\lim_{x\to 0} x^2\times {1\over x}
\\
\lim_{x\to 0} x\times {1\over x^2}
}
$$

Title: Product rule for limits of functions


答:
For each $a > 1$, let us define $F_a$ by

$$ F_a(x) = \prod_{n=1}^{\infty} \cos(x/a^n). $$

Then we may write

$$ \cos(x)\cos(ax)\cdots\cos(a^{n-1}x) = \frac{F_a(a^n x)}{F_a(x)}. $$

So the question boils down to identifying $F_a(x)$. Unfortunately, there is a compelling evidence that $F_a(x)$ reduces to an elementary, explicit function only when $a = 2$.

Indeed, note that $F_a(x)$ can be realized as the characteristic function

$$ F_a(x) = \mathbb{E}[e^{ix S_a}] = \int_{\mathbb{R}} e^{ixu} \, \mu_{S_a}(\mathrm{d}u) $$

of the random variable $S$ defined by

$$ S_a = \sum_{n=1}^{\infty} \frac{X_n}{a^n}, $$

where $X_1, X_2, \dots$ are independent Rademacher random variables (i.e., each $X_n$ takes each of the values $\pm 1$ with probability $\frac{1}{2}$) and $\mu_{S_a}$ denotes the law of $S_a$.

When $a = 2$, $\mu_{S_2}$ becomes absolutely continuous w.r.t. the Lebesgue measure. More precisely, the law of $S_2$ becomes a uniform distribution over $[-1, 1]$, hence giving

$$ F_a(x) = \int_{-1}^{1} \frac{1}{2} e^{ixu} \, \mathrm{d}u = \frac{e^{ix} - e^{-ix}}{2ix} = \operatorname{sinc}(x). $$

On the other hand, if $ a > 2 $ then $\mu_{S_a}$ is singular continuous. (It is connected to Cantor-like sets.) Since it sounds very unlikely that the inverse Fourier transform of an elementary function produces a singular continuous measure, I believe that $F_{a}$ is not an elementary function in this case.

When $1 < a < 2$, I suspect that $\mu_{S_a}$ is absolutely continuous w.r.t. the Lebesgue measure but the density function will be fractal-like. So by a similar reasoning, I believe that $F_{a}$ is not an elementary function in this case as well.

I added probability histograms for simulations of $S_{3/2}$, $S_{2}$, and $S_{3}$ for comparison:

Title: Product of $\cos(a^kx)$ where $k=1,2,3,\ \dots$ and $a$ is a natural number

Body: I have quite frequently encountered the relation, $$\cos(x)\cos (2x)\cos(4x)\dots \cos(2^{n-1}x)= \frac{\sin(2^nx)}{2^n\sin(x)}$$but is there any explicit formula for the series, $$\cos(x)\cos (3x)\cos(9x)\dots \cos(3^{n-1}x)$$I have tried hard to come up with a relation, but in vain. If there is such a formula, I would really be interested in how to prove it, that is you can give me a hint.


问:
A binomial identity from Mathematical Reflections
Here is the problem:

Let $m,n$ be positive integers with $n>m$. Prove that

$\displaystyle\sum_{k=0}^{\lfloor\frac{n+m}2\rfloor} (-1)^{k}\binom{n}{k}\binom{m+n-2k}{n-1}=\binom{n}{m+1}$

This problem is O243 of Mathematical Reflections. A solution had been published using complex integration (https://www.awesomemath.org/assets/PDFs/MR5sol(1).pdf). However, I would like to see a solution using the difference operator, if any, since the form of the summand brings this to mind.

答:
Here is an elementary proof, which I found by studying finite difference operations, but finally formulates somewhat smoother using generating functions. As for the difference operation approach, it easily shows that for the usual definition of binomial coefficients the left hand side is always $0$. Indeed with $\Delta$ the difference operation defined by $(\Delta f)(x)=f(x+1)-f(x)$ one has
$$
  (\Delta^n f)(x)=\sum_{k=0}^n(-1)^{n-k}\binom nkf(x+k)
$$
and the left hand side matches this at $x=0$ for $f:x\mapsto(-1)^n\binom{n+m-2x}{n-1}$, which with the usual defintion of binomial coefficients is a polynomial function of degree $n-1$, so $\Delta^n f=0$.

So one must assume the summation in the question is supposed to be truncated when $n+m-2k$ becomes negative. This can be obtained by rewriting the left hand side in the question to
$$
  \sum_{k=0}^{n} (-1)^{k}\binom{n}{k}\binom{m+n-2k}{m+1-2k},
$$
which makes the lower index negative whenever the upper index is, and with the usual definition of binomial coefficients a negative lower index makes them zero regardless. Now it turns out this is best interpreted using negative binomial powers, so let us make the upper index unconditionally negative by "negating the upper index" (making it $(m+1-2k)-(m+n-2k)-1=-n$):
$$
  \sum_{k=0}^{n} (-1)^{k}\binom{n}{k}(-1)^{m+1-2k}\binom{-n}{m+1-2k}.
$$
Now we see this is a certain convolution of the coefficients of $(1-X)^n$ and half of the coefficients of $(1-X)^{-n}=\sum_i(-1)^i\binom{-n}iX^i$, namely those whose parity is that of $m+1$. This is more conveniently described by making the left factor into a polynomial in $X^2$: the formula above describes the coefficient of $X^{m+1}$ in the power series (which turns out to be a polynomial)
$$
  (1-X^2)^n(1-X)^{-n}=((1-X)(1+X))^n(1-X)^{-n}=(1+X)^n.
$$
This coefficient is of course $\binom n{m+1}$.

Note that with the question reformulated properly, it works for any $n\in\mathbf N$ and $\def\Z{\mathbf Z}m\in\Z$.

Here's a description purely in terms of difference operators. The expression in the question naturally leads to considering the sequence $\binom k{n-1}_{k\in\mathbf N}$ and repeatedly taking differences of element two places apart. For simplicity extend the sequence by terms $0$ to the left so that it is indexed by $\Z$, and define the right-shift operator $R$ on sequences $f$ by $R(f)_i=f_{i-1}$ for all $i\in\Z$; using it define the "backward difference by $2$" operator $\nabla_2=I-R^2:f\mapsto(f_i-f_{i-2})_{i\in\Z}$. Now $I$ and $R$ commute, so applying the binomial formula to $(I-R^2)^n$ gives $\nabla_2^n(f)_k=\sum_{k=0}^n(-1)^n\binom nkf_{i-2k}$. In order to interpret the left hand side in the question as such an iterated difference, define sequences $C^{(k)}$ to be "column $k$" of Pascal's triangle shifted so as to start with a $1$, and $0$-extended to the left:
$$
  C^{(k)}_i=\begin{cases}\binom{k+i}k&\text{if $i\geq0$}\\
                         0&\text{if $i<0$;}\end{cases}
$$
then we are asked to prove that
$$
  \nabla_2^n(C^{(n-1)})_{m+1}=\binom n{m+1}.
$$
Now observe that with $\nabla_1=I-R$ one has $\nabla_1(C^{(k)})=C^{(k-1)}$ for all $k>0$, and $\nabla_1(C^{(0)})=D=(\delta_{i,0})_{i\in\Z}$, the sequence with a single nonzero term $1$ at index $0$ (this is where the truncation of negative index terms really kicks in). This suffices for computing $\nabla_1^n(C^{(n-1)})$, but we need the same with $\nabla_2$ instead of $\nabla_1$. Fortunately $\nabla_2=(1-R^2)=(1+R)(1-R)=(1+R)\nabla_1$, and everything commutes, so
$$
 \nabla_2^n(C^{(n-1)})=(1+R)^n(\nabla_1^n(C^{(n-1)}))
  =(1+R)^n(D)=\binom ni_{i\in\Z}.
$$
Evaluating this at index $i=m+1$ gives $\binom n{m+1}$, as desired.


问:
what additional conditions on $\varphi$ (if any) are sufficient to ensure that if $G$ is abelian, then so is $H$?
If $\varphi :G \to H$ is an isomorphism, prove that $G$ is abelian if and only if $H$ is abelian. If
$\varphi :G \to H$ is a homomorphism, what additional conditions on $\varphi$ (if any) are sufficient to
ensure that if $G$ is abelian, then so is $H$?

My   attempt :It is given that $G$ is abelian and  $\varphi$ is a isomorphism,then for every element $h \in H$, there exists $g \in G $ such that $ h = \varphi(g)$. Then consider any two elements $h_1, h_2 \in H$ and let
$g_1, g_2 \in G$ be such that $h_1 = \varphi(g_1)$ and $h_2 = \varphi(g_2)$. We have $h_1h_2 = \varphi(g_1)\varphi(g_2) =\varphi(g_1g_2) = \varphi(g_2g_1) = \varphi(g_2)\varphi(g_1) = h_2h_1$. So $H$
is abelian

My doubt:If $\varphi :G \to H$ is a homomorphism,what additional conditions on $\varphi$ (if any) are sufficient to ensure that if $G$ is abelian, then so is $H$?

答:
Note that in the proof you only used the fact that $\varphi$ is surjective. So this is sufficient.

In general, if $\varphi$ is any homomorphism and $G$ is abelian then it follows that the image $\varphi(G)$ is abelian as well. But if the image is not all $H$ (i.e if $\varphi$ is not surjective) then $H$ itself might not be abelian.


问:
How to prove this inequality on trace of inverse of a positive definite matrix?
I was stuck for a long time to prove the following trace inequality

\begin{equation}\label{ine1}
{\text {tr}} ({\bf A}_1^{-1}) \leq {\text {tr}} ({\bf A}^{-1}),  \tag{1}
\end{equation}

where ${\bf A} \in {\mathbb C}^{K \times K}$ is a positive definite matrix and ${\bf A}_1$ consists of the diagonal elements of ${\bf A}$, i.e., ${\bf A}_1 = {\text {diag}} ( {\text {diag}} ({\bf A}))$.
I have run a lot of simulation using Matlab and this inequality always holds.
(1) in the simple case with $K=2$ can be easily proven.
However, I can not prove it for the more general $K>2$ case.
Some of my efforts are as follows.

Denote ${\bf A}_1 = {\text {diag}} \{ a_1, \cdots, a_K \}$ and the eigenvalues of ${\bf A}$ by $ \{ \lambda_1, \cdots, \lambda_K \}$.
Then, it is obvious that

\begin{equation}\label{e1}
\sum_{k=1}^K a_k = \sum_{k=1}^K \lambda_k. \tag{2}
\end{equation}

In addition, using Hadamard's inequality, we have

\begin{equation}\label{ine2}
\prod_{k=1}^K a_k \geq \prod_{k=1}^K \lambda_k. \tag{3}
\end{equation}

Since (1) is equivalent to

\begin{equation}
\sum_{k=1}^K \frac{1}{a_k} \leq \sum_{k=1}^K \frac{1}{\lambda_k}, \tag{4}
\end{equation}

I tried to prove (4) using (2), (3), and the relationship between different means (Harmonic mean, etc.). But they didn't work.

Anyone providing the proof or relevant hints will be much appreciated. Thank you.

答:
The attempt at using Hadamard's Inequality is a nice idea since it is one of the original results in majorization.  In general, let  $\mathbf w\in \mathbb R_{\gt 0}^n$  have the $n$ components on the diagonal of $A$ and $\mathbf x\in \mathbb R_{\gt 0}^n$ have the $n$ eigenvalues of $A$.  Then

$\mathbf w \preceq \mathbf x$

where $\preceq$ denotes (strong) majorization

There are many ways to prove this result.  E.g. it is implied by
Maximize $\mathrm{tr}(Q^TCQ)$ subject to $Q^TQ=I$

now for $u\gt 0$ the mapping $u\mapsto u^{-1}$ is convex and for $\mathbf u\in \mathbb R_{\gt 0}^n$

$f(\mathbf u) = \sum_{k=1}^n u_k^{-1}$

is symmetric, hence Schur Convex.  Combining majorization with Schur Convexity gives

$\text{trace}\Big(\big(A_1\big)^{-1}\Big)$

$=\text{trace}\Big(\big(A\circ I\big)^{-1}\Big)$

$=f\Big(\mathbf w\Big)$

$\leq f\Big(\mathbf x\Big)$

$=\text{trace}\Big(\big(A\big)^{-1}\Big)$

which is the inequality OP has sought


问:
General Prediction Theory
Suppose that we have random variables $Y, X_1, \ldots, X_{p-1}$. Let $X = (X_1, X_2, \ldots, X_{p-1})^\top$. To predict $Y$ from the values of $X$, we use $f(X)$ a predictor of $Y$ such that the mean squared error is minimized, i.e., $$\min_f E [Y - f(X)]^2.$$

Theorem:
Let $\varrho (X) = E[Y|X].$ Then for any other predictor $f(X)$, if $$E[Y - \varrho (X)]^2 \leq E[Y - f(X)]^2$$
then $\varrho(X)$ is the best predictor of $Y.$

For predicting $Y = (Y_1, \ldots, Y_q)'$ from $(X_1, \ldots, X_{p-1})$. The predictor $f(X)$ is best if the the scalar is minimized i.e., $$\min E{[Y - f(X)]^\top [Y - f(X)]}.$$ With vector $Y$, what modifications do I have to make in order of the above theorem to hold? I need suggestions from learned members.

答:
If $Y$ is a vector with $q$ components then the predictor $f(X)$ must be a vector of $q$ component predictors, which you can write as
$$
f(X):=(f_1(X),f_2(X),\ldots, f_q(X)).\tag1
$$
Now use the fact that $v^\top v=\sum_\limits{i=1}^q (v_i)^2$ for any vector $v:=(v_1,\ldots,v_q)$ to write
$$
[Y-f(X)]^\top[Y-f(X)] = \sum_\limits{i=1}^q [Y_i-f_i(X)]^2\tag2
$$
so the expectation of (2) is
$$
E[Y-f(X)]^\top[Y-f(X)] = \sum_\limits{i=1}^q E\left([Y_i-f_i(X)]^2\right).\tag3
$$
Your goal is to minimize (3) over all $(f_1,\ldots,f_q)$. Since (3) is a sum of $q$ components, where component $i$ of the sum involves $Y_i$, $(X_1,\ldots,X_{p-1})$, and $f_i$, you can minimize separately each component of the sum, using the Theorem.

Title: General Prediction Theory


问:
Calculation of ordered pair $(x,y,z)$ in $x^2 = yz\;\;,y^2=zx\;\;,z^2 = xy$
(1) Total no. of integer ordered pair $(x,y,z)$ in $x^2 = yz\;\;,y^2=zx\;\;,z^2 = xy$

(2) Total no. of integer ordered pair $(x,y,z)$ in $x+yz = 1\;\;,y+zx = 1\;\;,z+xy = 1$

My Try:: (1) Clearly $ x = 0,y = 0,z = 0$ are the solution of given equation

and from three equation we observe that $x,y,z$ has same sign.

Now If $x\neq 0,y\neq 0$ and $z\neq 0,$ Then $x^2-y^2 =-z(x-y)\Leftrightarrow (x-y).(x+y+z)  =0$

Means either $x=y$ or $x+y+z = 0$

$\bullet$ If $x = y$, The put in $z^2 = xy=x^2=y^2\Leftrightarrow z = \pm x = \pm y$

means  $x = y =z$

So $(x,y,z) = (k,k,k)$ where $k\in \mathbb{Z}$

$\bullet$ If $x+y+z = 0$, Then put in third $z^2 = xy\Leftrightarrow x^2+y^2+xy = 0$

So $x^2+y^2+xy = x^2+y^2+(x+y)^2 = 0\Leftrightarrow x = 0,y = 0,x+y = 0$

So $(x,y,z) = (0,0,0)$

So Given equation has Infinite solution

My Question is I have Calculate Right or not

If not plz explain me.

Thanks

答:
Hint to (2): subtract 2nd equation out 1st:
$$(x-y)(1-z)=0.\ \ \ (1)$$

Addition:

Similarly
$$(x-z)(1-y)=0,\ \ \ (2)$$
$$(z-y)(1-x)=0.\ \ \ (3)$$

If $z=1$, then $x+y=1, xy=0$ (from your $1$st and $3$rd equations) whence we obtain two solutions
$$z=1, x=1, y=0,$$
$$z=1, x=0, y=1.$$
Similarly for $y=1$ we get one more solution from ($2$):
$$y=1, x=1, z=0.$$

Further, let $x,y,z\ne 1$. Then $x=y=z$ and $x^2+x-1=0$ from your $1$st equation. Then we have two solutions:
$$x=y=z=\frac{-1+\sqrt{5}}{2}$$
and
$$x=y=z=\frac{-1-\sqrt{5}}{2}.$$


问:
Can the classifying space of a Lie group be also a Lie group?
How to show a classifying space of a Lie group is also a Lie group or not a Lie group?

For example, $U(1)$ is a Lie group, let us consider the following classifying spaces:

so $BU(1)=\mathbf{CP}^\infty$ can it be also be a Lie group, but how?

so $B^2 U(1)=B\mathbf{CP}^\infty$ can it be also be a Lie group, but how?

so $BSU(2)=\mathbf{HP}^\infty$ can it be also be a Lie group, but how?

答:
Let $G$ be a topological group with classifying space $BG$. Suppose that $BG$ supports a group structure. Then there are homotopy equivalences $G\simeq \Omega BG\simeq\Omega^2B^2G$, so $G$ is a double loop space, and in particular supports a homotopy abelian H-space multiplication. However:

Theorem: Let $X$ be a non contractible, connected, finite complex which is a homotopy commutative H-space, then $X$ has the homotopy type of a torus.

This is Theorem 1.1 in John Hubbuck's paper On Homotopy Commutative H-spaces Top. 8 (1969), 119-126.

Thus the only compact connected Lie groups whose classifying spaces could potentially support a group structure are the tori. Since $BT^k\simeq (BS^1)^k$, to consider these we should study $BS^1$.

Now, it's well-known that $BS^1\simeq K(\mathbb{Z},2)$, and that there are topological groups within this homotopy type. In the comments below I argue that there is in fact an infinite-dimensional Banach Lie group in the homotopy type of $BS^1$. I personally do not consider this a postive answer to your question, but you may be happy with it.

We can also non-compact, non-connected Lie groups. In the connected non-compact world there are examples like $\mathbb{R}^n$ and $T^k\times\mathbb{R}^n$ which meet our critera. However these are essentially the only examples, because in general a noncompact Lie group will deformation retract onto its maximal compact subgroup, and Hubbuck's theorem then gives the same no-go result as before.

In the compact non-connected case there are torsion abelian groups. These do have classifying spaces which are non-compact, non-Lie topological groups (i.e. $\mathbb{R}P^\infty$ and other infinite lens spaces).

Finally in the non-compact non-connected case we introduce the torsion free abelian groups $\mathbb{Z}^k$. These groups do provide a positive answer to your question, since, as I mentioned in the comments, you have $B\mathbb{Z}^k\simeq T^k$.


问:
Prove that $x^2y^2(4-x^2)(4-y^2) \leq16$ for all $|x|\leq2$ and $|y| \leq 2$
Prove that $x^2y^2(4-x^2)(4-y^2) \leq16$ for all $|x|\leq2$ and $|y| \leq 2$

My attempt:

If $x^2(4-x^2) \leq 4$ for $|x|\leq2$ and $y^2(4-y^2) \leq 4$ for $|y|\leq2$ then the inequality holds.

Define $h(x) = x^2(4-x^2)-4$. $h'(x) = 8x-4x^3$, hence we get critical points $x=0,\pm \sqrt{2}$. By the second derivative test we find that $0$ is a maximum in the $|x| \leq 2$ range  . Hence:

$$x^2(4-x^2)-4 \leq 0$$
$$x^2(4-x^2)\leq 4$$

A similar argument for the $y$ based function, and hence the top inequality holds.

Is this attempt correct?

答:
$$
\left(\frac x2\right)^2\left(\frac y2\right)^2\left(1-\left(\frac x2\right)^2\right)\left(1-\left(\frac y2\right)^2\right)\le \frac{1}{16}
$$

now make the substitutions

$$
\cases {
\frac x2=\sin a\\
\frac y2=\sin b
}
$$

then

$$
\left(\frac 12\sin(2a)\right)^2\left(\frac 12\sin(2b)\right)^2\le\frac{1}{16}
$$


问:
very ample divisors on curves using Riemann Roch Spaces
Let $C$ be a projective non-singular irreducible curve, let $D$ be a divisor on $C$. Suppose the Riemann-Roch Space is $L(D)=\langle f_1,...,f_n\rangle$.
Define $\phi_D:C\to \mathbb{P}^{n-1}$ by $\phi_D(P)=(f_1(P):...,f_n(P))$. We say that $D$ is very ample if $\phi_D$ is an embedding.

Suppose $P\neq Q\in C$, $\phi_D(P)=(1:0:...:0),\phi_D(Q)=(0:1:...:0)$. I need to show that if $D$ is very ample, then $L(D-P)\neq L(D)$. I'm not sure how to go about that.

Some literatures incorporate a base point free property to the definition of very ample, but I don't have such a definition in this case.

Since $f_1$ achieves minimum order at $P$ across all $f\in L(D)$, is it true that $f_1\notin L(D-P)$? How to show this?

It is clear that $f_1\notin L(D-P)\iff L(D-P)\neq L(D)$, but I'm not sure how to prove either side of this equivalence.

答:
Suppose that your complete linear system $|D|$ has base point, define the divisor
$$
F=\min_{E\in |D|} E
$$

Exercise: $|D-F|$ is base point free and $L(D)=L(D-F)$.

So we always can suppose that $|D|$ is base point free. In this case we can solve this with some propertys of the map $\phi_D$, for example $L(D-P)=L(D)$ implies that $P$ is base point of the complete linear system $|D|$.

Rmk: I think that for a good definition of very ample divisor it's necessary request that the complete linear system associated to this divisor is without base point.


问:
Difference between arrows in functions
I'd like you to explain to me the difference between the two arrows in functions
$\to$ and $\mapsto$
I know we use the first one to link the domain E and the range F while we use the second one to link the argument and the function. But I don't know why we can't just mix between them. Are there any reasons ?
Thank you in advance for your answers
Regards

答:
We write $f:E\to F$ to denote that $f$ is a function with domain $E$ and range $F$, and we can write the rule for $f$ as $f\mapsto f(x)$.  For example, we might define $f:\mathbb{R}\to\mathbb{R}$ by $x\mapsto 2x+1$.  It's just the convention, and it's useful to know, without needing any other context, that $\to$ is at the level of "types" and $\mapsto$ is at the level of "values."  Syntactically, in $x\mapsto E$ the left-hand side introduces a new variable $x$ that can be referenced in the expression $E$.  The $\to$ arrow does not introduce new variables like this, so it's a fundamentally different sort of thing.

(Something that can be useful with the $\mapsto$ notation is defining functions with types like $\mathbb{R}\to(\mathbb{R}\to\mathbb{R})$ with a rule like $x\mapsto(y\mapsto 2x+3y)$.  This is a function that takes a real number and produces a function from the real numbers to the real numbers.  A more complicated, but useful, example like this is the linear isomorphism that sends a vector space to its double dual, which might be written as $V\to ((V\to\mathbb{R})\to\mathbb{R})$ with the rule $v\mapsto (f\mapsto f(v))$.)

Title: Difference between arrows in functions


问:
Find PDF of sum of two independent random variables (exponential and uniform distribution)
Suppose that $X$ and $Y$ are independent and $Z=X+Y$. If the probability density functions of $X$ and $Y$ are:

$f_X(x)=e^{-x}, x>0$ ($0$ if otherwise),

$f_Y(x)=1/2, 0<y<2$ ($0$ if otherwise),

What's the PDF of $X+Y$?

I know the formula but I don't quite get the domains of $X$ and $Y$. Please help, thanks!

答:
The only thing that's tricky is that the ranges of $X$ and $Y$ are different; we can take care of that by breaking into cases.

Note that the ranges are such that $X>0$ and $Y\in(0,2)$. So, how can we make $X$ and $Y$ add up to a given value $z$?

First, consider the case where $z\leq 0$.  This has density $0$. (Why?)

Next, consider the case where $Z=z$, $z\geq 2$.  This happens by letting $Y$ take any value in $(0,2)$, and then using $X$ to "make up the rest".  So, you could model the density for a given $z\geq2$ by
$$
f_Z(z)=\int_0^2f_Y(y)f_X(z-y)\,dy.
$$

Finally, consider the case where $z\in (0,2)$. In this case, we are restricted in what values of $Y$ are possible; in particular, it must be the case that $Y=y\in(0,z)$. And then once again, we use $X$ to 'make up the rest'. So, for $z\in(0,2)$,
$$
f_Z(z)=\int_0^zf_Y(y)f_X(z-y)\,dy.
$$


问:
Show that the unique solution of $P_n(X) = 0$ is $x=i\frac{e^{2i\theta}+1}{e^{2i\theta}-1}=\frac{1}{\tan(\theta)}$
I need to how that when $ k \neq 0$, the unique solution of $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i} = 0$ is $ x=i\frac{e^{2i\theta}+1}{e^{2i\theta}-1}=\frac{1}{\tan(\theta)} $ with $\theta=\frac{k\pi}{2n+1} \in ]-\frac{\pi}{2},\frac{\pi}{2}[ \backslash {0}$.

In the previous questions I've shown that there is no solutions when $k=0$, that if $P_n(x) = 0$, then $x$ verify $(\frac{x+i}{x-i})^{2n+1} = 1$, and that $P_n(x) = 0$ when $\frac{x+i}{x-i} = \exp(\frac{2ik\pi}{2n+1})$ with $k$ between $-n$ and $n$.

I don't really know where to start with this question. Do i just need to replace x in the polynomial with what is said to be the unique solution ?

答:
We want the solutions to $$(x+i)^{2n+1} - (x-i)^{2n+1} =0.$$

Assume $x$ is real and $z=x+i = r e^{i\theta}$ so that $z^*=x-i=re^{-i\theta}$ is the conjugate of $z$.

Substituting and rearranging:

$$z^{2n+1} = {z^\ast}^{2n+1}$$

$$e^{i(2n+1)\theta}=e^{-i(2n+1)\theta}$$

$$e^{i2(2n+1)\theta}=1=e^{2\pi k i}$$

so $$\theta = \frac{k\pi}{2n+1}$$

but $z=x+i$ thus $\Im z \ne 0$, so $k\ne 0$ nor is $k$ a multiple of $2n+1.$

$$x=\Re \left(re^{i\theta}\right)=r\cos \theta, \quad { \text{and} }\quad
 1=\Im \left(re^{i\theta}\right)=r\sin \theta,$$

so $r = 1/\sin \theta$ and $$x=\frac{\cos \theta}{\sin \theta}=\cot \theta=\frac{1}{\tan \theta} =i \frac{e^{i\theta} + e^{-i\theta}}{e^{i\theta} - e^{-i\theta}}=i\frac{e^{2i\theta}+1}{e^{2i\theta}-1}.$$

One indexing is $k\in \{1,2,\cdots, 2n\}.$

The solution is not unique.  $P_n$ is a polynomial of order $2n$, and $P_n(x)=0$ has $2n$ solutions.

$$\begin{aligned} P_n(x)&= \frac{1}{2i}\left[(x+i)^{2n+1}-(x-i)^{2n+1}\right]\\&=\frac{1}{2i} \left[ \sum_{k=0}^{2n+1} \pmatrix{2n+1\\k}x^k i^{2n+1-k}-\sum_{k=0}^{2n+1} \pmatrix{2n+1\\k}x^k (-i)^{2n+1-k}\right]\\ &=\sum_{k=0}^{n} \pmatrix{2n+1\\2k}x^{2k} (-1)^{k+n}\\&=(2n+1)x^{2n}-\pmatrix{2n+1\\2n-2}x^{2n-2} + \cdots +(-1)^n.\end{aligned}$$

Please see a continuation of the discussion for further developments.


问:
Embedding of finite groups in small $\operatorname{GL}(n,F)$
How does we obtain a small (not necessarily smallest) value of $n$ such that a $p$-group of order $p^m$ can be embedded in $\operatorname{GL}(n,\mathbb{F}_p)$?

[For embedding of a finite group $G$ in $S_n$, we can compute a small value of $n$ by looking at the minimum index of a subgroup $H$ containing no proper normal subgroup of $G$. Here, this value of $n$ need be necessarily smallest.]

答:
Here's a theorem due to Minkowski-Schur-Blichtfeld which might be of some interest for you:

Theorem. Let $G$ be a subgroup of $GL_n(\mathbb Z)$ for some $n$. We suppose that $G$ is a $p$-group of order $p^m$. Then $m\leqslant N(p,n)$, with $$N(p,n)= \left[\frac{n}{p-1}\right] + \left[\frac{n}{p(p-1)}+ \right]\cdots $$
Conversely, for any prime $p$ and any $n>0$, there exists a subgroup $G$ of $GL_n(\mathbb Z)$ of order $p^{N(p,n)}$.

Another remark, due to Minkowski, is that if $G$ is a finite subgroup of $GL_n(\mathbb Z)$, then the natural restriction arrow $G \to GL_n(\mathbb F_p)$ is injective whenever $p\geqslant 3$. This enables to relate more precisely the last theorem to your question.


问:
Limit of $\frac{1}{\sqrt[n] n}$
I would like to show that the limit of the sequence $(\frac{1}{\sqrt[n] n})_{n=1}^{\infty}$ is equal to 1 using the definition of the limit of a sequence. (Another way to say this is that it converges to 1).

Definition: A sequence $(a_n)_{n=1}^{\infty}$ converges to a real number A iff for each $\epsilon > 0$ there is a positive integer N such that for all $n \geq N$ we have $|a_n - A| < \epsilon$.

Edit: Here is my work.

Let $\epsilon > 0$. There exists $N$ such that for $n \geq N$,

$$|\frac{1}{\sqrt[n] n}-1| = |\frac{n^{\frac{n-1}{n}}}{n} - 1| = |\frac{n^{\frac{n-1}{n}} - n}{n}| = \frac{n-n^{\frac{n-1}{n}}}{n}$$

Here is where I am stuck.

答:
I think the following method can help much. For each $n$, there exists an integer $k$ such that $2^{k-1}\le n\le2^k$ and we have either
$$
2^{k-1\over 2^{k-1}}\le n^{1\over n}\le 2^{k\over 2^k}
$$
or
$$
2^{k\over 2^{k}}\le n^{1\over n}\le 2^{k-1\over 2^{k-1}}
$$
Now prove $k\over 2^k$ tends to $0$.

Title: Limit of $\frac{1}{\sqrt[n] n}$