问:
$\sum_{n=0}^\infty z^n = \prod_{m=0}^\infty \left(1+z^{2^m}\right)$
When reading Iwaniec and Kowalski's Analytic Number Theory, I came across the following "identity" on page 11 (the Amazon link has a free book preview which includes page 11):

$$\sum_{n=0}^\infty z^n = \prod_{m=1}^\infty \left(1+z^{2^m}\right).$$

However, I don't think this is true, as there will never be odd exponent terms in the product expansion. I think the identity should, instead, be $$\sum_{n=0}^\infty z^n = \prod_{m=0}^\infty \left(1+z^{2^m}\right).$$

Here's what I believe to be a proof of the equality between the sum and product.

Let $s_N=\sum_{n=0}^Nz^n$ and $p_N=\prod_{n=0}^N\left(1+z^{2^n}\right)$. I claim, first, that $p_N=s_{2^{N+1}-1}$ for all $N\geq0$, which I will prove by induction.

For $N=0$, we have $p_0=1+z=s_1=s_{2^{0+1}-1}$. Now, suppose that this is true for $N$, we will show that this implies the same for $N+1$, that is, $p_{N+1}=s_{2^{N+2}-1}$.

If $p_N=s_{2^{N+1}-1}$, then $$p_{N+1}=s_{2^{N+1}-1}\left(1+z^{2^{N+1}}\right)=\left(1+z+\cdots+z^{2^{N+1}-1}\right)\left(1+z^{2^{N+1}}\right)=\\\left(1+z+\cdots+z^{2^{N+1}-1}\right)\cdot1+\left(1+z+\cdots+z^{2^{N+1}-1}\right)\cdot z^{2^{N+1}}=\\1+z+\cdots+z^{2^{N+1}-1}+z^{2^{N+1}}+\cdots+z^{2^{N+1}+2^{N+1}-1}=\\1+z+\cdots+z^{2^{N+1}-1}+z^{2^{N+1}}+\cdots+z^{2^{N+2}-1}=s_{2^{N+2}-1},$$ as required.

Therefore, since $|p_N-s_{2^{N+1}-1}|=0<\varepsilon$ for any $\varepsilon>0$ and all $N\geq0$, the identity is proven. That is, the sum converges for a particular $z$ if and only if the product also converges for that $z$, and, in that case, they both converge to the same value.

Am I correct that the identity in Iwaniec/Kowalski wrong, and, if so, does my attempted proof of a corrected version work?

Body:
Another proof of $$\prod_{n\ge0}(1+z^{2^n})=\sum_{n\ge0}z^n=\frac1{1-z}$$ is as follows.
Define the pair of functions $f,g$ as
$$\begin{align}
f(z)&=\prod_{n\ge0}(1-z^{2^n})\\
g(z)&=\prod_{n\ge0}(1+z^{2^n}),
\end{align}$$
for $|z|<1$. Then we see that
$$\begin{align}
f(z)g(z)&=\prod_{n\ge0}(1-z^{2^n})(1+z^{2^n})\\
&=\prod_{n\ge0}(1-z^{2\cdot2^n})\tag 1\\
&=\prod_{n\ge0}(1-(z^2)^{2^n})\\
&=f(z^2).
\end{align}$$
However, note that
$$f(z^2)=\prod_{n\ge0}(1-z^{2\cdot2^n})=\prod_{n\ge0}(1-z^{2^{n+1}})=\prod_{n\ge1}(1-z^{2^n})=\frac{1}{1-z}\prod_{n\ge0}(1-z^{2^n})=\frac{f(z)}{1-z}.$$
Thus, from $(1)$, $$f(z)g(z)=\frac{f(z)}{1-z}.$$
That is,
$$g(z)=\prod_{n\ge0}(1+z^{2^n})=\frac{1}{1-z}.$$


问:
Let $z$ be the complex number $7-3i$, find a complex number $w$ such that $zw$ is a negative real number
I feel a bit bad asking this question but here it goes:

My initial idea was to rotate the complex number $z$ onto the negative real axis and then compute what complex number was used to do that but that feels a bit too long-winded for this type of problem, I was wondering if anyone has a shortcut.

答:
Let $z^\ast$ be the conjugate of $z$.

$$zz^\ast=|z|^2$$

Let $a$ be any negative number.
$$z(az^\ast)=a|z|^2$$

ContentLicense: CC BY-SA 4.0


问:
Bernstein polynomial: function to approximate
I am studying the Bernstein polynomial with Matlab. I wish to code the example in the wikipedia page of Bernstein polynomial, but I can not figure out the explicit form of the red function (see figure).

答:
Is impossible figure out the explicit formula of the red function using only a picture of its graph. However, this description says:

This animation was created with Maple 9.5 and Jasc Animation Shop. It shows Bernstein polynomials approximating $x \sin(\frac 72\pi {\sqrt {x}})$ (chosen for purely aesthetic reasons).

Although, it appears that the true formula is $-x \sin(\frac 72\pi {\sqrt {x}})$.


问:
limits of marginal probability of polar coordinate probability
I'm self-studying probability and came across a problem of finding the marginal probability $f_\theta$ of a pdf in polar coordinate $f_{r,\theta}$. To find the $f_r$, I know we do $\int^{2\pi}_{0} f_{r,\theta} d\theta$, but how about $f_\theta$? Is it from 0 to 1? Or 0 to infinity?

EDIT: the original problem is
find $f_\theta = \int_{0}^{?} \frac{r}{2\pi\sigma^2}exp(\frac{-r^2}{2\sigma^2})dr$

答: From the looks of it, $r$ is Rayleigh distributed and $\phi$ is uniform over $[0,2\pi]$. The $(r,\phi)$ pair corresponds to the magnitude and the phase of the vector $(x,y)$, where $x$ and $y$ are independent standard Normal variables. In that case, since the support of $x$ and $y$ is $(-\infty, +\infty)$, then the support of $r$ is $[0, +\infty)$.


问:
Suppose $x$ and $y$ are two real numbers such that $x + y = 6$ and $x^2 + y^2 = 43$. Find the value of $x^3 + y^3$.
I tried quadratic and substituting and cubing the equations but it doesn't work and I'm pretty sure i did not make any careless mistake.</p>

答:
HINT

Based on @CalvinLin's comment, we can proceed as follows.

Notice that $x^{3} + y^{3} = (x+y)(x^{2} - xy + y^{2}) = 6(43 - xy)$. Moreover, we also have that
\begin{align*}
\begin{cases}
(x+y)^{2} = 36\\\\
(x+y)^{2} = 43 + 2xy
\end{cases}
\end{align*}

Can you take it from here?


答:
Suppose $(X,Y)\sim f_{XY}$ where $f_{XY}:\mathbb{R}^2 \longrightarrow (0,\infty)$ is any pdf supported on all of $\mathbb{R}^2$.

Set $X=R\cos(\Theta)$ and $Y=R\sin(\Theta)$.

Then $(R,\Theta)\sim f_{R \Theta}$ where $f_{R \Theta}$ is the density supported on $[0,\infty)\times [0,2\pi)$ given by $$f_{R \Theta}(r,\theta)=f_{XY}\Big(r\cos(\theta),r \sin(\theta)\Big)\Bigg|\frac{\partial\big(r\cos(\theta),r \sin(\theta)\big)}{\partial (r,\theta)}\Bigg|=rf_{XY}\Big(r\cos(\theta),r\sin(\theta)\Big)$$ In order to obtain the marginal density $f_{\Theta}$ you simply "integrate away" the $r$ component and get $$f_{\Theta}(\theta)=\int_0^{\infty}f_{R\Theta}(r,\theta)dr=\int_0^{\infty}rf_{XY}\Big(r\cos(\theta),r\sin(\theta)\Big)dr$$ In your example $f_{XY}(x,y)=\frac{1}{2\pi \sigma^2}e^{-\frac{x^2+y^2}{2 \sigma^2}}$ so $f_{R \Theta}(r,\theta)=\frac{r}{2\pi \sigma^2}e^{-\frac{r^2}{2 \sigma^2}}$ and for $\theta \in [0,2 \pi)$ $$f_{\Theta}(\theta)=\int_0^{\infty}\frac{r}{2\pi \sigma^2}e^{-\frac{r^2}{2 \sigma^2}}dr=\frac{1}{2\pi}$$ i.e. $\Theta \sim \mathcal{U}[0,2\pi)$. If we replaced the upper bound of $\infty$ with any positive constant $\alpha$ notice that $$\int_0^{\alpha}\frac{r}{2\pi \sigma^2}e^{-\frac{r^2}{2\sigma^2}}dr=\frac{1}{2\pi}\Bigg(1-e^{-\frac{\alpha}{2\sigma^2}}\Bigg)$$ Integrating the right hand side from $\theta=0$ to $\theta= 2\pi$ would yield $1-e^{-\frac{\alpha}{2\sigma^2}}\neq 1$ so $f_{\Theta}(\theta)$ wouldn't even be a density. The upper bound has to be $\infty$.

Title: limits of marginal probability of polar coordinate probability

Body:
I'm self-studying probability and came across a problem of finding the marginal probability $f_\theta$ of a pdf in polar coordinate $f_{r,\theta}$. To find the $f_r$, I know we do $\int^{2\pi}_{0} f_{r,\theta} d\theta$, but how about $f_\theta$? Is it from 0 to 1? Or 0 to infinity?

EDIT: the original problem is
find $f_\theta = \int_{0}^{?} \frac{r}{2\pi\sigma^2}exp(\frac{-r^2}{2\sigma^2})dr$


问:
Dart expected number of throws problem
To win a game, you need to hit the bullseye $n$ times in a row (if easier, let $n$ be $3$). The probability of hitting the bullseye is $p$ for each throw, so it's independent (if easier, let $p$ be $0.6$).
What is the expected number of throws it takes you to win?

The difficulty I couldnt get past was the fact that it has to be in a row, so you could for example hit, hit, miss and then still have to hit three times in a row again, so every miss kind of resets you. If it helps, I wrote a short program which estimates the number of throws by simulating the game a $100000$ times, and the expected score for $n=3$ and $p=0.6$ turned out to be about $9.07$, but I need to get the exact value for any probability $p$, not just an estimate.

答:
Let $k_n$ be the expected number of throws to get $n$ bullseyes in a row. Then

With probability $1-p$ we miss, so waste the first throw.

With probability $p(1-p)$ we score then miss, so waste the first two throws

$\vdots$

With probability $p^{n-1}(1-p)$ we score $n-1$ then miss the $n$-th, so waste the first $n$ throws.

With probability $p^n$ we score $n$ in a row.

So putting that all together, we get
$$k_n=p^nn+\sum_{i=1}^np^{i-1}(1-p)(k_n+i)=\left(1-p^n\right)k_n+\sum_{i=1}^{n+1}p^{i-1}\implies\boxed{k_n=\frac{1-p^n}{p^n(1-p)}}.$$

Alternatively, note that it takes $k_{n-1}$ throws in expectation to get $n-1$ in a row. Once we get there, either we toss a bullseye and win (with probability $p$), or we miss and start over (with probability $1-p$). So
$$k_n=k_{n-1}+p+(1-p)(k_n+1)\implies k_n=\frac{k_{n-1}+1}{p},$$
and we are done by induction, as $k_1=\frac{1}{p}$.

Just for the record, here's a fun martingale solution.

Suppose at time $k$, a new gambler $g_k$ enters the casino and bets £$1$ that throw $k$ will hit the bullseye. If you miss, then $g_k$ loses so leaves the game; else $g_k$ wins £$p^{-1}$. In that case, $g_k$ then bets all their money that throw $k+1$ will hit the bullseye, and so on. This continues until either you miss, or you hit $n$ bullseyes in a row and $g_k$ wins £$p^{-n}$.

If $M_k$ is the total winnings for the casino at time $k$, then it's clear that $M_k$ is martingale. Let $\tau$ be the time at which you hit $n$ consecutive bullseyes. Clearly $\mathbb E\left[\lvert M_{n+1}-M_n\rvert\right]$ is bounded, so we can apply the optional stopping theorem:
$$0=\mathbb E[M_0]=\mathbb E[M_\tau]=\mathbb E\left[\tau-p^{-1}-p^{-2}-\dots-p^{-n}\right]\implies\mathbb E[\tau]=\frac{1-p^n}{p^n(1-p)}.$$

Title: Dart expected number of throws problem


答:
The probability density function $f_{r,\theta}$ is normalized, that is
$ \int_0^{2\pi} \int_0^\infty f_{r,\theta} rdr d\theta = 1 $.

We can solve the $f_\theta$ by calculating

$f_\theta = \int_0^\infty \frac{r}{2\pi\sigma^2} exp(- \frac{r^2}{2\sigma^2}) dr $

$ = \frac{1}{2\pi} \int_0^\infty \frac{1}{2\sigma^2} exp( - \frac{r^2}{2\sigma^2}) dr^2 $

$ = -\frac{1}{2\pi} \int_0^\infty exp(- \frac{r^2}{2\sigma^2}) d (-\frac{r^2}{2\sigma^2}) $

$ = -\frac{1}{2\pi}  exp(- \frac{r^2}{2\sigma^2}) |_0^\infty $

$ = 0-(-\frac{1}{2\pi})   $

$ = \frac{1}{2\pi}$

Hence,

$f_\theta  = \frac{1}{2\pi}$.

It is clear that the integral of probability density function $f_\theta$ over the domain $ \theta \in [0,2\pi] $should be 1, such that

$\int_0^{2\pi}  f_\theta d\theta = \frac{\theta}{2\pi} |_0^{2\pi} = 1 - 0 = 1 $


Title: Suppose $x$ and $y$ are two real numbers such that $x + y = 6$ and $x^2 + y^2 = 43$. Find the value of $x^3 + y^3$.

Body: I tried quadratic and substituting and cubing the equations but it doesn't work and I'm pretty sure i did not make any careless mistake.</p>

Body:
you don't need to factor $x^n + y^n$

from $xy = -\frac{7}{2}$ and $x+y = 6$ we know $x,y$ are the roots of
$$ t^2 -6t - \frac{7}{2}  $$

This means that we have a recurrence
$$  T_n = x^n + y^n  $$
which obeys
$$  T_{n+2} = 6 T_{n+1}  + \frac{7}{2} T_n $$

With $T_1 = 6, T_2= 43 $ we get
$$  T_3 = 6 \cdot 43 +  \frac{7}{2} \cdot 6 =  279 $$

the sequence being
$$ 6, 43, 279, \frac{3649}{2}, \frac{23847}{2},  \frac{311707}{4},  \frac{2037171}{4}, \frac{26628001}{8}, $$


问:
How to formulate a mapping function in set theory
I wanted to prove two sets are equal. For that I assume that there is a mapping function that maps from domain set to range set. Let the range set be one among the two sets which are equal. Also consider the universal set as domain set. So if we have same function and same domain in two cases,will the resultant range set also be equal? Also how do I formulate such a mapping function?</p>

答:
The easier and MOSTLY used method to show 2 sets are equal is to prove that $x\in A$ iff $x\in B$.

Now why exactly is your method of proving that 2 sets are equal generally harder?

Well let me phrase it in a concise way

2 sets $A$ and $B$ are equal iff

There exists a bijection(denoted by $f$) such that $f = I$. Where $I$ is the identity map (or equivalently $f(x) = x$)

This is more difficult to prove since you have to either construct a bijection which is also the identity, or prove it exists. Now proving there exists a bijection is hard enough, and for 2 general sets is VERY difficult since we would need to know how to compute $f$ and on top of that proving that out of the set of functions which is a bijection the identity (i.e $f$ s.t $f(x) = x$) also belongs to that set.


Body:
$258 = 6\cdot 43 =(x+y)(x^{2}+y^{2})=x^{3}+x^{2}y+xy^{2}+y^{3}$.

$216 = 6^{3}=(x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$

Subtracting these equations we get:

$42 = -2(x^{2}y+xy^{2}) \implies -21 = x^{2}y+xy^{2}$ and by substitution we have

$x^{3}-21+y^{3} = 258 \implies x^{3} +y^{3} = 279.$


问:
Introductory problem to understand Schroeder-Bernstein Theorem

Consider $A=[0,1]$ and $B=[0,1)$ and injections $f(x) = \frac13x$ from $A$ to $B$ and $g(x) =x$ from $B$ to $A$. Construct a bijection $h$ from $A$ to $B$ such that on some points of $A$, $h(x) = f(x)$ and for the other points $h(x) = g^{-1}(x).$

It seems that $f(x) = \frac13x$ is a bijection from $A \to B$ for the points $[0, \frac13]? $ If I understood the question correctly (I doubt I did...) they would want me to construct a bijection for the rest of the points? I’m reading ”topology through inquiry” and this was a problem from the book. Also I’m not sure if they mean the inverse function by $g^{-1}(x)$ or the preimage?

答:
I'm answering this question because it's been a while since I studied the Schroeder-Bernstein Theorem, it should be a fun review for me.

Here is a constructive proof of the Schroeder-Bernstein Theorem, page 1-9.

Given injective funtions $f:A \rightarrow B$ and  $g:B\rightarrow A$, in your case $f(x)=\frac{1}{3} x$ and $g(x)=x$ with $A=[0,1]$ and $B=[0,1)$.

Word for word from the document: let $B_1=B\backslash f(A)=[0,1)\backslash[0,\frac{1}{3}]= \left( \frac{1}{3},1 \right).$

If $B_k \subset B$ is defined for some $k \in \mathbb{N}$, let $A_k=g(B_k)$ and $B_{k+1}=f(A_k)$, thus defining the sets inductively for all $k \in \mathbb{N}$.

Define $\tilde{A}= \cup_{k \in \mathbb{N}} A_k$

Then the sought after bijective function is

$$ h(x) =
\begin{cases}
g^{-1}(x),  & x \in \tilde{A} \\
f(x), & x \in A\backslash \tilde{A}
\end{cases}$$

$A_1 = g(B_1) = \left( \frac{1}{3},1\right)$

$B_2=f(A_1)=\left(\frac{1}{9},\frac{1}{3}\right)$

$A_2=g(B_2)= \left(\frac{1}{9},\frac{1}{3}\right)$

$B_3 =f(A_2)=\left(\frac{1}{27},\frac{1}{9}\right)$

$A_3 = g(B_3)= \left(\frac{1}{27},\frac{1}{9}\right)$

$\vdots$

$B_k=\left(\frac{1}{3^k},\frac{1}{3^{k-1}}\right)$

$A_k = B_k = \left(\frac{1}{3^k},\frac{1}{3^{k-1}}\right)$ inductively. From which $\tilde{A} = \cdots \cup \left(\frac{1}{3^k},\frac{1}{3^{k-1}}\right)\cup \cdots \cup \left(\frac{1}{3^3},\frac{1}{3^2}\right) \cup \left(\frac{1}{3^2},\frac{1}{3}\right) \cup \left(\frac{1}{3},1\right)=(0,1)\backslash \left\lbrace\frac{1}{3^k}\Bigg|k\geq 1\right\rbrace$ and

$$ h(x) =
\begin{cases}
x,  & x \in (0,1)\backslash \left\lbrace\frac{1}{3^k}\Bigg|k\geq 1\right\rbrace \\
\frac{1}{3}x, & x \in \left\lbrace\frac{1}{3^k}\Bigg|k\geq0\right\rbrace\cup\{0\}
\end{cases}$$

To illustrate the bijection under $h$

$$0 \mapsto 0$$

$$\frac{1}{3^k} \mapsto \frac{1}{3^{k+1}} \text{for $k\geq 0$}$$

while $h$ fixes all other elements in $[0,1]$.

For more details on why $h(x)$ in general using this method is always well-defined, injective and surjective is described in the remainder of the proof.

Hope this helps!


问:
Prove specially rearranged alternating harmonic series converges to $\frac 12 \ln{\frac{4p}{q}}$
By Leibnitz's test the alternating series is convergent.

$\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac 1n
=\frac 12 \ln{\frac{4×1}{1}}$
\begin{align}
& \left(1-\frac 12- \frac 14\right)+\left(\frac 13-\frac 16-\frac 18\right)+...\\
& =\frac 12 \ln {\frac{4\cdot 1}{2}}
\end{align}
Let $\sum \sigma$ is a rearrangement of the alternating series.$\sum t$ is obtained by grouping it's terms in such  a manner that each group contains $p$ positive terms and $q$ negative terms. Prove that $\sum t=\frac 12\ln {\frac{4p}{q}}$

Body:
$t_{(p+q)n}$
\begin{align}
& =\left(1+\frac 13+...+\frac{1}{2p-1}-\frac 12-\frac 14-...-\frac{1}{2q}\right)+\left(\frac{1}{2p+1}+...+\frac{1}{4p-1}-\frac{1}{2q+2}-...-\frac{1}{4q}\right)+...+\left(\frac{1}{2p(n-1)+1}+...+\frac{1}{2pn-1}-\frac{1}{2q(n-1)+2}-...-\frac{1}{2qn}\right)\\
&=\left(1+\frac 13+...+\frac{1}{2pn-1}\right)-\frac 12\left(1+\frac 12+\frac 13+...+\frac{1}{qn}\right)\\
&=\left(1+\frac 12+\frac 13+...+\frac{1}{2pn}\right)-\left(\frac 12+\frac 14+...+\frac{1}{2pn}\right)-\frac 12(\gamma_{qn}+\ln{qn})\\
 &=(\gamma_{2pn}+\ln{2pn})-\frac 12(\gamma_{pn}+\ln{pn})-\frac 12(\gamma_{qn}+\ln{qn}) \\
& =\left(\frac 12\ln 2^2+\frac 12\ln\frac pq\right)+\left(\gamma_{2pn}-\frac 12\gamma_{pn}-\frac 12\gamma_{q n}\right)
 \end{align}
$\therefore \displaystyle \lim_{n \to \infty}t_{(p+q)n}=\frac 12\ln{\frac{4p}{q}} $


问:
Under what conditions is $\frac{dF(x)}{dx}\Big|_{x-x_0} = \frac{dF(x-x_0)}{dx}\Big |_x$?
I have a very complicated function for which I need to evaluate

$$ \frac{dF(x)}{dx}\Bigg|_{x-x_0}.$$

How can I tell when this will be equivalent to evaluating

$$ \frac{dF(x-x_0)}{dx}\Bigg|_x ?$$

Under what conditions on $F$ is this equivalent? Would this condition change for higher derivatives?

Body:
Assuming $g(x) := F(x-x_{0})$ is well-defined and differentiable, we have:
$$\frac{dF(x)}{dx}\bigg{|}_{x-x_{0}} = \lim_{y \to 0}\frac{F(y+(x-x_{0})) - F(x-x_{0})}{y} = \lim_{y\to 0} \frac{F((y+x)-x_{0})-F(x-x_{0})}{y} = \lim_{y\to 0}\frac{g(y+x)-g(x)}{y} = \frac{dg(x)}{dx} = \frac{dF(x-x_{0})}{dx}\bigg{|}_{x}$$


问:
Product of matrices; MAPLE giving a strange answer
Either my brain is seriously fried up right now or the computer is wrong.

If I have a matrix $\begin{bmatrix}
4 & -2\\
2 & -1 \\
0 &  0
\end{bmatrix}$ multiply by its transpose $\begin{bmatrix}
4 &2  &0 \\
-2&-1 &0
\end{bmatrix}$, I should get a $3 \times 3$ $\begin{bmatrix}
20 &10  &0 \\
10 &5  &0 \\
 0&0  &0
\end{bmatrix}$

For some reason Maple is giving me a $2 \times 2$ $\begin{bmatrix}
20 &10 \\
10 &5   \\
  \end{bmatrix}$

Why did they delete the last row and column of 0s? You can't do that

答:
Do you try the result with one row of os or try some square matrix with one row of os?

Maybe is the storage of maple?

ContentLicense: CC BY-SA 3.0


问:
Why use the term "arbitrary"?
At the beginning of proofs, mathematicians will often write something like:

Let $x$ be an arbitrary integer

I understand that the point of this statement is to get the reader of the proof to think about "a new symbol $x$ which represents one of the integers, although the specific integer is unspecified". Because the specific integer is unspecified, nothing will be assumed about the object represented by $x$ throughout the proof, except that it is an  integer. So the proof will apply for all integers.

The above is discussed here: Question about how to interpret arbitrary elements

My question is "Why is the term arbitrary used to invoke the above ideas?". This is my guess.

The term "arbitrary" generally means "done without any reasoning or logic". So when a mathematician says "Let $x$ be an arbitrary integer", they are saying "Let $x$ be an integer, where the integer is chosen without any reasoning or logic". Because the integer is chosen withou any reasoning, it is unknown / unspecified. So a reader  will then think about "a new symbol $x$ which represents one of the integers, although the specific integer is unspecified".

Am I correct in understanding why the term "arbitrary" is used?

Thank you for your time and please feel free to elaborate.

答:
Yes, "arbitrary" is just a context clue to the reader. It reminds the reader that when we introduce $x$, we're not assuming any particular features about it beyond those explicitly stated (e.g. that $x$ is an integer). At the formal level this is totally unnecessary, but for readers new to proofs it can help the intuition.

(Incidentally, note that one should not use "random" in this context!)

Title: Why use the term "arbitrary"?


问:
How many axiomatizations of set theory are there?
https://en.wikipedia.org/wiki/Skolem%27s_paradox

In reading about Skolem's paradox on wikipedia, I encountered the claim that every consistent countable axiomatisation of set theory in first order logic has a countable model. That got me thinking that it's strange that set theory could have more than one axiomatisation at all, countably modeled or otherwise. The only set theory I typically hear about is ZFC, and then I recalled NBG set theory, but that is an extension of ZFC, not a separate axiomatisation. Or is NBG considered a separate axiomatisation?

Moreover, I had thought that set theory required second order logic, since it involves quantifying over both sets and relations between sets. How could it have an axiomatisation in first-order logic at all, let alone more than one?

Finally, if indeed set theory (and n.b. that they didn't clarify which set theory, so I would take it to mean ZFC) has more than one axiomatisation in first-order logic, does it have finitely many, denumerably many, or perhaps even non-denumerably many?

Thank you for any advice.

答:
You can indeed get the same theory from a different set of axioms. A bit more precisely, for a set of sentences $\Gamma$, let the deductive closure of $\Gamma$ be the set $$Ded(\Gamma):=\{\varphi:\Gamma\vdash\varphi\}$$ of sentences which $\Gamma$ proves (I'm tacitly fixing a particular language to work in here, e.g. the language of set theory). Then we can have $Ded(\Gamma_1)=Ded(\Gamma_2)$ even if $\Gamma_1\not=\Gamma_2$.

For example, given $\Gamma$ consider $$\Gamma_{double}:=\{\varphi\wedge\varphi:\varphi\in\Gamma\}.$$ Obviously $Ded(\Gamma)=Ded(\Gamma_{double})$, but in general $\Gamma$ will be a different set of sentences than $\Gamma_{double}$. More interestingly, we can show the following:

Every finitely axiomatizable theory is axiomatizable by a single sentence. That is, if $\Gamma$ is finite, then there is some single sentence $\gamma$ such that $Ded(\Gamma)=Ded(\{\gamma\})$. There's actually less to this than meets the eye: just let $\gamma$ be the conjunction of the finitely many sentences in $\Gamma$.

Every computably enumerably axiomatizable theory is in fact computably axiomatizable, and indeed primitive recursively axiomatizable and more - this is Craig's trick, and is often useful in the context of incompleteness problems.

And getting back to $\mathsf{ZFC}$ in particular, it's easy to show that the Separation scheme is in fact redundant: we can prove each instance of Separation from the remaining axioms of $\mathsf{ZFC}$. Similarly, we can replace Choice with Zorn's Lemma, the Well-Ordering Principle, or various other statements. So there are lots of natural alternative axiomatizations of $\mathsf{ZFC}$.

(As to how many axiomatizations $\mathsf{ZFC}$ has, it's a good exercise to show that there are in fact continuum-many ways to axiomatize $\mathsf{ZFC}$. In fact, there are continuum-many ways to do this non-redundantly, but that takes more work. Even constructing a single non-redundant axiomatization is nontrivial - see here.)


问:
How a continuous map from a compact m-manifold into $\mathbb{R^n}$ is an open map?
In the proof of Theorem 36.2 Munkres' Topology it claims implicitly that because X is compact thus f is an open map.

To prove that F is an imbedding we need only to show that F is injective *(because X is compact) *.

I believe by compactness it implies continuity of the inverse map to make it a homeomorphism. But how it is true?

答:
Hints:

If $F$ is a continuous map from a compact space to a Hausdorff space, then $F$ is a closed map.

An injective closed map is an open map onto its image.

The idea for the first part is the same as in the proof of Theorem 26.6 in Munkres.


问:
An estimator for a Cumulative distribution function
The salaries in a company distribute in a cumulative distribution function,

λ is a constant.

Am I correct to understand that η represents the minimum salary that is taken into account, and that the probability for every single salary is in relation to it? i.e the bigger the minimum salary, the smaller the probability becomes for every salary in the function.

How can I calculate an estimator for η by moments and with n values?

Thank you

答:
In this post I am going to assume that $\eta$ and $\lambda$ are both positive constants.

You are correct in saying that the smallest possible observable salary is  $\eta$, but larger values of $\eta$ don't necessarily yield smaller probabilities. For example, taking $\eta=6$ and $\lambda=9$ gives $$P(\eta+1<Y<\eta+2)=\int_{\eta+1}^{\eta+2}f_{Y}(y)dy\approx 0.174$$ Increasing $\eta$ to $7$ while keeping $\lambda$ fixed at $9$ yields a slightly larger probability of $0.196$. You will notice upon inspection of the pdf $f_{Y}=F_{Y}^{\prime}$ that salaries are skewed to the right are thus are concentrated around the minimum salary of $\eta$.

For part $(b)$, consider a simple random sample $Y_1=y_1, \ldots ,Y_n=y_n$ of $n$ independent salaries in the company. The likelihood function is $$\mathcal{L}(\eta)=f_{Y}(y_1)\times \dots \times f_{Y}(y_n)=\frac{\lambda^n \eta^{n\lambda}}{(y_1 \times \dots \times y_n)^{\lambda+1}}$$ whenever $0<\eta \leq \min\Big\{y_1,\ldots ,y_n\Big\}$ and $0$ elsewhere. Since $\mathcal{L}$ is increasing in $\eta$ we see that the value of $\eta$ which maxmimizes $\mathcal{L}$ is $\hat{\eta}=\min\Big\{y_1,\ldots ,y_n\Big\}$. This is your MLE and, in my opinion, a natural estimator for the minimum salary $\eta$.


问:
How to perform integration for questions involving cumulative distribution functions?
From a textbook, the following example is shown. I do not understand why $2(x-1)$ integrates to become $(x-1)^2$. How come it isn't $x^2-2x$? Is it something to do with the constant since $x^2-2x (+1)$ would then equal $(x-1)^2$. Also, isn't this a definite integral with limits $k$ and $1$ so isn't the constant not included? Any clarification would be appreciated.

答:
The antiderivatives $(x-1)^2 +C$ and $x^2-2x+C$ are both correct, they're equivalent.

When you expand the square, you get $x^2 - 2x + 1 + C$, which is the same as $x^2 + 2x + C_2$ where $C_2 = C + 1$. The constants are arbitrary so these are two ways of addressing the same set of functions.

The fact that this is a definite integral isn't a concern because both functions satisfy $F'(x) = 2(x-1)$, so the fundamental theorem of calculus applies equally to both. Equivalently you can see this as the fact that any constant will cancel out when you do $F(k) - F(1)$.


答:
You can use $x^2-2x$ if you wish; you'll get the same answer. Since it is a definite integral, the constant will cancel.

$$[x^2-2x]_{x=1}^k = k^2 - 2k - (1^2 - 2) = k^2 - 2k + 1 = (k-1)^2.$$

Title: How to perform integration for questions involving cumulative distribution functions?

Body: From a textbook, the following example is shown. I do not understand why $2(x-1)$ integrates to become $(x-1)^2$. How come it isn't $x^2-2x$? Is it something to do with the constant since $x^2-2x (+1)$ would then equal $(x-1)^2$. Also, isn't this a definite integral with limits $k$ and $1$ so isn't the constant not included? Any clarification would be appreciated.


问:
How many generator blocks can I fit inside a $3\times3\times3$ cube?
In my Minecraft world I have a $3\times3\times3$ cube of space which I want to fill with $1\times1\times1$ generator and wire blocks.

I can install a single outlet as part of the room's wall; it takes up none of the $3\times3\times3$ space.

Each generator block must be connected to the outlet through wires, so for each generator there is always a path of adjacent wire blocks that goes from that generator to the outlet. Diagonal neighbors are not adjacent in this case. The outlet can only be touching one block, it can't be installed on the edge between blocks.

I can choose where to place the generators, wires, and outlet. I want to fit as many generators as I can.

I can see a way to get $18$ powered generators, if the middle layer is $9$ wire blocks and the top and bottom layers are $9$ and $9$ generator blocks. But I can't tell if this is the maximum possible.

答:
You can solve the problem via integer linear programming as follows.  Let $[n]=\{1,\dots,n\}$, and for each cell $(i,j,k)\in [n] \times [n] \times [n]$, let $N_{i,j,k}$ be the set of neighboring cells.  Let binary decision variable $g_{i,j,k}$ indicate whether cell $(i,j,k)$ is a generator, and let binary decision variable $w_{i,j,k,d}$ indicate whether cell $(i,j,k)$ is a wire block at distance $d$ from the outlet.  The problem is to maximize $\sum_{i,j,k} g_{i,j,k}$ subject to
\begin{align}
g_{i,j,k} + \sum_{i,j,k,d} w_{i,j,k,d} &\le 1 &&\text{for all $i,j,k$} \tag1 \\
g_{i,j,k} &\le \sum_{(\bar{i},\bar{j},\bar{k})\in N_{i,j,k}} \sum_d w_{\bar{i},\bar{j},\bar{k},d} &&\text{for all $i,j,k$} \tag2 \\
w_{i,j,k,d} &\le \sum_{(\bar{i},\bar{j},\bar{k})\in N_{i,j,k}} w_{\bar{i},\bar{j},\bar{k},d-1} &&\text{for all $i,j,k,d$} \tag3 \\
\sum_{i,j} w_{i,j,k,0} &= \begin{cases}1 &\text{if $k=1$}\\0 &\text{otherwise}\end{cases}  &&\text{for all $k$}\tag4
\end{align}
Constraint $(1)$ prevents a cell from simultaneously being a generator and a wire block.
Constraint $(2)$ enforces a generator to be adjacent to a wire block.
Constraint $(3)$ enforces a wire block at distance $d$ to be adjacent to a wire block at distance $d-1$.
Constraint $(4)$ enforces the unique outlet to be along the $k=1$ wall (without loss of generality).

For $n=3$, the maximum does turn out to be $18$.


问:
Closed form for $\prod_{n=0}^\infty (1-z^{2^n})$
Is there a closed form for the product $$f(z) = \prod_{n=0}^\infty 1-z^{2^n}$$
either as a formal power series or as an analytic function in the disk $|z| < 1$? It's not hard to see that Taylor series coefficients of $f$ about 0 are all $\pm 1$:
$$f(z) = 1 - z - z^2 + z^3 - z^4 + z^5 + z^6 - z^7 + \dotsb$$
and that they form a pattern in blocks, so to speak:
$$+- \quad\to\quad +--+ \quad\to\quad +--+-++- \quad\to\quad \dotsb$$
But I don't know much else. I came across this infinite product when I incorrectly transcribed an exercise from a book, wasting a good deal of time before returning to the book to ask it, "are you sure?". (The exercise was to prove that $\prod_{n=0}^\infty 1 + z^{2^n} = (1-z)^{-1}$, which is easy.)</p>

答:
Converting my comment from 2016 into an answer, with elaborations:

The coefficients of $z^n$ in $f(z)$ is $-1$ to the power of the number of $1$s in the binary expansion of $n$; this is a slight variant of the Thue-Morse sequence with $1$ and $-1$ instead of $0$ and $1$ (A106400 on the OEIS). The coefficient is $1$ if $n$ is "evil" (has an even number of $1$s; A001969 on the OEIS) and $-1$ if $n$ is "odious" (has an odd number of $1$s; A000069 on the OEIS). The reciprocal

$$\frac{1}{f(z)} = \prod_{n \ge 0} \frac{1}{1 - z^{2^{n}}}$$

is the generating function for the sequence $c_n$ counting the number of partitions of $n$ into powers of $2$, which is A018819 on the OEIS. $c_{2n}$ is A000123 on the OEIS because $c_{2n+1} = c_{2n}$.

According to Kachi and Tzermias' On the $m$-ary partition numbers this problem was first studied by Euler. Mahler showed in On a special functional equation that

$$c_n = e^{O(1)} 2^{ -{k \choose 2}} \frac{n^k}{k!}$$

where $k$ is the unique positive integer satisfying $2^{k-1} k \le n < 2^k (k+1)$. This gives $k \approx \frac{W(n \log 2)}{\log 2}$ where $W(x) \sim \log x$ is the Lambert W function, which lets us write down an asymptotic for the logarithm of $c_n$ the leading term of which is

$$\log c_n \sim \frac{(\log n)^2}{2 \log 2}.$$

de Brujin's On Mahler's partition problem analyzes the asymptotics more precisely.

These asymptotics imply that $c_n$ grows faster than polynomially but slower than exponentially and so rule out $f(z)$ having a particularly simple closed form; for example it follows that $f(z)$ cannot be rational (although this has an easier proof) or more generally meromorphic or algebraic.


问:
How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$?
The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both side and drawing triangles. However, the worked solutions does a simplification from LHS to RHS instead, namely $$\cos\arcsin\left(\frac{3}{5}\right)\cos\arctan\left(\frac{7}{24}\right)-\sin\arcsin\left(\frac{3}{5}\right)\sin\arctan\left(\frac{7}{24}\right)=\frac{3}{5}$$which I don't understand. Can someone please explain?

答:
By drawing triangles it should be easy to see that \begin{align*}
\cos \arcsin \left(\frac{3}{5}\right)&=\frac{4}{5}\\
\cos \arctan \left(\frac{7}{24}\right)&=\frac{24}{25}\\
\sin \arcsin\left(\frac{3}{5}\right)&=\frac{3}{5}\\
\sin \arctan\left(\frac{7}{24}\right)&=\frac{7}{25}
\end{align*}
(note the Pythagorean triples (3,4,5), and (24,7,25).) Putting this all together: $$\cos \arcsin \left(\frac{3}{5}\right)\cos \arctan \left(\frac{7}{24}\right)-\sin \arcsin\left(\frac{3}{5}\right)\sin \arctan\left(\frac{7}{24}\right)=\frac{4}{5}\cdot \frac{24}{25}-\frac{3}{5}\cdot \frac{7}{25}=\frac{3}{5}$$


问:
Determining whether an operator is positive or not
Let $T=\int_a^bT_w dw$, where each $T_w$ is a strictly positive, self-adjoint, compact operators acting on a separable Hilbert space $X$. Then, I wonder whether T is also a strictly positive operator. That is, its eigenvalues are all strictly positive. I am naturally guessing that the answer is yes, but I cannot prove it. Above all, I do not know how rigorously one can define the operator $T$ given by an integral of operators, which I think should be the starting point of the proof. I am thinking that the other steps would be just simply straightforward if only it is defined rightly. Also, any reference related to this kind of material would be appreciated. Thanks.

答:
I will assume that the integral is defined to be the Bochner integral.

An operator $T \in B(H)$ is strictly positive iff for all $x \in H$, we have $\langle Tx, x \rangle > 0$. We note that for any $x \in H$, we have
$$
\langle Tx,x \rangle = \left\langle \left(\int_a^b T_w dw\right)x,x\right\rangle
= \int_a^b \langle T_wx,x \rangle\,dw.
$$
Because each $T_w$ is strictly positive, we have $\langle T_wx,x \rangle > 0$ for all $w$. By the properties of Lebesgue integration, we can conclude that $\int_a^b \langle T_wx,x \rangle\,dw > 0$. So, it indeed holds that $\langle Tx, x \rangle$ is positive for all $x \in H$.

So, $T$ is necessarily a strictly positive operator.


问:
Degree of $ \mathbb{Q}(\sqrt[80]{43},\sqrt[84]{43})/\mathbb{Q}$
I have to calculate the degree of the field extension $\mathbb{Q}(\sqrt[80]{43},\sqrt[84]{43})/\mathbb{Q}$.

I think the formulas $X^{80}-43$ ie. $X^{84}-43$ solve the equations. But I don't know how I can use this information to my advantage

答:
If $L$ is the field in question, we see that $L$ contains $$43^{\frac{c}{80} + \frac{d}{84}} = 43^{\frac{80c + 84d}{80\cdot 84}}$$ for any integers $c,d$. In particular, $L$ contains $43^{4/(80\cdot 84)} = 43^{1/1680}$. It is now clear that $L = \mathbb Q(43^{1/1680})$, since $43^{1/80}, 43^{1/84}$ are both powers of $43^{1/1680}$. Since $X^{1680}-43$ is irreducible (by Eisenstein's criterion, for example), the degree of this extension is $1680 = \text{lcm}(80,84)$.

More generally, if we had $\mathbb Q(43^{1/a_1},\dots, 43^{1/a_n})$, we would find via essentially the exact same argument that the degree of this extension is $\text{lcm}(a_1,\dots,a_n)$.


问:
Density and distributions of those numerically or analytically KNOWN solutions of Riemann $\zeta(1/2 + r i)=0?$
We know the conjecture about the Riemann hypothesis is about the nontrivial zeros are on
$$(1/2 + r i)$$
for some $r \in \mathbb{R}$ of Riemann zeta function.

My question is how much is known about the density and the distributions of those numerically or analytically KNOWN solutions of
$$\zeta(1/2 + r i)=0?$$

I found a related post but it was about 8 years ago, so maybe we have a better update?

Mean density of the nontrivial zeros of the Riemann zeta function

答:
In my humble opinion, a key paper is the one published in year $2014$ by G.Franca and A.LeClair. In particular, they provide a very good and simple approximation (equation $(229)$ in the linked paper).
$$\Im\left(r _{n}\right) \sim \frac{2 \pi  \left(n-\frac{11}{8}\right)}{W\left(\frac{n-\frac{11}{8}}{e}\right)}$$ where $W(.)$ is Lambert function;

Repeating some of their calculations for $n=10^k$, we have
$$\left(
\begin{array}{ccc}
k & \text{approximation} & \text{solution} \\
 1 & 50.233653 & 49.773832 \\
 2 & 235.98727 & 236.52423 \\
 3 & 1419.5178 & 1419.4225 \\
 4 & 9877.6296 & 9877.7827 \\
 5 & 74920.891 & 74920.827 \\
 6 & 600269.64 & 600269.68
\end{array}
\right)$$


问:
How to find the order of the group of rigid motions of platonic solids in $\mathbb{R}^3$?
The following appear as exercises in Dummit and Foote's Algebra (Section $1.2$ - Dihedral Groups):

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a tetrahedron. Show that $|G| = 12$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a cube. Show that $|G| = 24$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of an octahedron. Show that $|G| = 24$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of a dodecahedron. Show that $|G| = 60$

Let $G$ be the group of rigid motions in $\mathbb{R}^3$ of an icosahedron. Show that $|G| = 60$

From this answer I figured that rigid motions are orientation-preserving isometries, i.e. reflections are not allowed.

So, for a tetrahedron, I thought of symmetry axes passing through a vertex and the centroid of the opposite face. There are four such axes (let's call them $A,B,C,D$). Along every axis, we can define $1_i, r_i, r_i^2$ as three rotations with $r_i^3= 1$, the identity element ($i=A,B,C,D$). Since there are four such axes, $|G| = 3\times 4 = 12$. Is this fine, or am I missing something? I'm slightly concerned about the fact that $1_A,1_B,1_C,1_D$ may all possibly be the same (since they are identity transformations), and that I am over-counting?

Minor question (detour): Are identity transformations corresponding to different axes different, or the same?

For the cube, I did the following:

For every pair of opposite faces, we have an axis of symmetry. There are $3$ such pairs, hence $3$ such axes (say $A,B,C,D$). About each axis we define $1,r_i,r_i^2,r_i^3$ with $r_i^4 = 1$ where $i=A,B,C,D$.

There are four body diagonals (say $E,F,G,H$), and about each diagonal (symmetry axis) we define $1,r_j,r_j^2$ with $r_j^3= 1$ where $j=E,F,G,H$.

In view of the above calculations, we have $|G| = 3\times 4 + 4\times 3 = 24$.

Using this method becomes difficult hereafter, for larger solids. It is not easy to identify all the axes of symmetry by hand. Moreover, the only group I have learned about in some detail at this point is $D_{2n}$, so please do not give solutions such as "the required group $G$ is isomorphic to a known and well-studied group $X$, and we know $|X| = ?$ so $|G| = ?$"

I think it boils down to having a good way to count all the distinct rigid motions. Could someone help me with this?

I came across James Ha's solutions here, but I don't understand how the solutions presented in the PDF are equivalent to mine even for the tetrahedron and cube cases. It'd be nice if someone can help me see the equivalence, and also tell me how to proceed with the other platonic solids! Thanks a lot!

答:
To add some elaboration to existing answers, and additional comments:

As orangeskid mentions, you can infer the size of the symmetry group from the number of transformations between two edges. Here is a way to see this more clearly:

Consider directed edges on the polyhedron, which consist of a vertex and an edge emanating from that vertex (or equivalently, an edge with one of its endpoints distinguished). If we have $e$ edges, then we have $2e$ of these directed edges. Because we're using Platonic solids, every one of these can be taken to any other (this follows pretty easily from most definitions of the Platonic solids, but should be pretty intuitive).

But once we know that one directed edge $(v_1,e_1)$ goes to another directed edge $(v_2,e_2)$, we have completely specified the rotation: once we move $v_1$ to $v_2$, we've constrained the possible rotations to a single axis around which things can turn (since we have a point that is now immobile), and only one of those ways to rotate it will move $e_1$ to $e_2$.

In particular, this means that a rotation is uniquely specified by where it takes a single directed edge; since each of the $2e$ possibilities gives a unique rotation, there must be $2e$ possible rotations total.

(If we permit orientation-reversing transformations, there are twice as many; for every way to take a directed edge to another, we get a second transformation fixing that directed edge by reflecting about it.)

As for the identity transformations fixing an axis, these are all the same identity transformation; they leave the shape unchanged.

To more clearly spell out the types of (orientation-preserving) rotations possible for each possible platonic solid:

For every platonic solid, the possible rotations are either a nontrivial rotation about a vertex, a $180^\circ$ rotation about an edge, a nontrivial rotation about a face, or the identity transformation.

For the tetrahedron, faces are opposite vertices, so there are $4\cdot (3-1)$ nontrivial vertex/face rotations, $1$ identity, and $3$ edge-flips ($6$ edges, but two used per flip), for a total of $12$.

For the cube, there are $8\cdot (3-1)/2$ vertex rotations, $6\cdot(4-1)/2$ face rotations, $12/2$ edge flips, and $1$ identity, for a total of $24$.

For the octahedron, there are $6\cdot(4-1)/2$ vertex rotations, $8\cdot (3-1)/2$ face rotations, $12/2$ edge flips, and $1$ identity, for a total of $24$.

For the dodecahedron, there are $20\cdot(3-1)/2$ vertex rotations, $12\cdot(5-1)/2$ face rotations, $30/2$ edge flips, and $1$ identity, for a total of $60$.

For the icosahedron, there are $12\cdot(5-1)/2$ vertex rotations, $20\cdot(3-1)/2$ face rotations, $30/2$ edge flips, and $1$ identity, for a total of $60$.


问:
Gaussian integral in multiple dimensions
I would like to solve $f_k=\int_{-\infty}^{\infty}{x_k}e^{-\frac{1}{2}\sum_{i}^{n}\sum_{j}^{n}A_{ij}(x_{i}-C)(x_{j}-C)}d^{n}x$ for real constant C.
I feel like this should be obvious but I'm unable to come up with a solution or find the answer anywhere. Thanks in advance.

答:
First, let us recall that:
$$\int_{-\infty}^{\infty}dx e^{-\frac{1}{2}\alpha x^{2}} = \sqrt{\frac{2\pi}{\alpha}}$$
where $\alpha > 0$. This is the usual Gaussian integral in one dimension. Now, let me write:
$$\langle x, A x\rangle = x^{T}Ax = \sum_{i,j=1}^{n}x_{i}A_{ij}x_{j}$$
for simplicity. In order to the $n$-dimensional Gaussian integral to be well-defined, we must demand $A$ to be positive definite. In this case, there exists a unitary matrix $V$ such that:
$$V^{-1}AV = D = \mbox{diag}\{\alpha_{1},...,\alpha_{n}\}$$
where $\alpha_{1},...,\alpha_{n}$ are the eigenvalues of $A$, which are all strictly positive since $A$ is positive definite. Now, set $y := Vx$. Then, we have:
$$\langle x, Ax \rangle = \langle y, D y \rangle = \sum_{i=1}^{n}\alpha_{i}y_{i}^{2}$$
and, also $dx = |\det V| dy = dy$, since $|\det V| = 1$ because $V$ is unitary. Thus:
$$\int_{\mathbb{R}^{n}}dx e^{-\frac{1}{2}\langle x, Ax \rangle} = \int dy e^{-\frac{1}{2}\sum_{i=1}^{n}\alpha_{i}y_{i}^{2}} = \prod_{i=1}^{n}\int_{-\infty}^{\infty}dy_{i} e^{-\frac{1}{2}\alpha_{i}y_{i}^{2}} = \prod_{i=1}^{n}\sqrt{\frac{2\pi}{\alpha_{i}}} =\bigg{(}\sqrt{\frac{2\pi}{\alpha_{i}}}\bigg{)}^{n} = \frac{(2\pi)^{\frac{n}{2}}}{{\det A}^{\frac{1}{2}}}$$
where I used the one-dimensional Gaussian integral result stated above and the fact that $\det A = \prod_{i=1}^{n}\alpha_{i}$.

We can proceed in a similar way to evaluate the Laplace transform:
$$g(p) := \int_{\mathbb{R}^{n}}dx e^{-\frac{1}{2}\langle x, Ax \rangle}e^{\langle p, x\rangle}$$
with $p \in \mathbb{R}^{n}$. Using the same transformation as before and noticing that:
$$\langle p, x \rangle = \langle p, V^{-1}y\rangle = \langle Vp, y\rangle = \langle \zeta, y\rangle$$
with $\zeta := Vp$, we can obtain:
$$g(p) = \prod_{i=1}^{n}\int_{-\infty}^{\infty} dy_{i} e^{-\frac{1}{2}\alpha_{i}y_{i}^{2}}e^{\zeta_{i}y_{i}} = \frac{(2\pi)^{\frac{n}{2}}}{(\det A)^{\frac{1}{2}}}e^{\frac{1}{2}\langle p, A^{-1}p\rangle}$$

Finally, if you want to evaluate the object:
$$f_{k} = \langle x_{k}\rangle := \int_{\mathbb{R}^{n}}x_{k}e^{-\frac{1}{2}\langle x, Ax\rangle}$$
then the usual trick is to use:
$$f_{k} = \frac{\partial g(p)}{\partial p_{k}}\bigg{|}_{p=0} = \bigg{[}\frac{\partial}{\partial p_{k}}\bigg{(}\frac{(2\pi)^{\frac{n}{2}}}{(\det A)^{\frac{1}{2}}}e^{\frac{1}{2}\langle p, A^{-1}p\rangle}\bigg{)}\bigg{]}\bigg{|}_{p=0}$$

I'll leave the details to you!


问:
If for every $h>0$, $M_1\leq hM_2+M_0/h$ then $M_1^2\leq 4M_0 M_2$?
This is the last step of my proof and I'm not sure how to prove it. Can anybody help with this please?</p>

Body: Simple. Rewrite it: $f(h) = M_2h^2 - hM_1 + M_0 \ge 0, \forall h > 0\implies f(\frac{M_1}{2M_2}) \ge 0 \implies M_2\cdot (\dfrac{M_1}{2M_2})^2- \dfrac{M_1^2}{2M_2}+M_0\ge 0 \implies \dfrac{M_1^2}{4M_2}-\dfrac{M_1^2}{2M_2}+M_0\ge 0\implies M_0 - \dfrac{M_1^2}{4M_2} \ge 0\implies 4M_0M_2 - M_1^2 \ge 0\implies 4M_0M_2 \ge M_1^2. $


问:
Equality for the Gradient
We have that $f : \mathbb{R}^2 \mapsto \mathbb{R}, f \in C^2$ and $h= \nabla f  = \left(\frac{\partial f}{\partial x_1 },\frac{\partial f}{\partial x_2 }  \right)$, $x=(x_1,x_2)$.

Now the proposition I try to show says that

$$\int_0^1 \! \langle \nabla f(x \cdot t),x \rangle\,dt \,=
\int_0^{x_1} \! h_1(t,0) ,dt \,
+\int_0^{x_2} \! h_2(x_1,t) \,dt \,$$

I know that $\langle \nabla f(x \cdot t),x \rangle=d f(tx) \cdot x$ but it doesn't seem to help, maybe you have to make a clever substitution? (Because the range of integration changes). Thanks in advance.

答:
Before you read this answer, fetch a piece of paper and draw the following three points on it: $(0,0)$, $(x_{1},0)$ and $(x_{1},x_{2})$. These are the corners of a right-angled triangle whose hypotenuse I'd like to call $\gamma$ whose side on the $x_1$-axis I call $\gamma_{1}$ and whose parallel to the $x_2$-axis I call $\gamma_2$.

More formally, let $\gamma: [0,1] \to \mathbb{R}^2$ be the path $t \mapsto tx$. Similarly, let $\gamma_{1} : [0,1] \to \mathbb{R}^2$ be the path $t \mapsto (tx_{1},0)$ and $\gamma_{2} : [0,1] \to \mathbb{R}^2$ be the path $t \mapsto (x_{1}, tx_{2})$.

The integral on the left hand side can be written as
$$\int_{0}^{1} df(\gamma(t))\cdot\dot{\gamma}(t)\,dt = \int_{0}^{1} \frac{d}{dt}(f \circ \gamma)(t)\,dt = f(\gamma(1)) - f(\gamma(0)) = f(x_1, x_2) - f(0,0).$$

Similarly, after some simple manipulations the right hand side is equal to
$$\int_{0}^{1} \frac{d}{dt} (f \circ \gamma_{1})(t)\,dt + \int_{0}^{1} \frac{d}{dt} (f \circ \gamma_2)(t)\,dt = \left( f(\gamma_{1}(1)) - f(\gamma_{1}(0))\right) + \left(f(\gamma_2 (1)) - f(\gamma_2(0))\right)$$
and as $\gamma_1 (1) = (x_1,0) = \gamma_2 (0)$ two terms cancel out and what remains is $f(x_{1},x_{2}) - f(0,0)$.

Thus the left hand side and the right hand side are equal.

Title: Equality for the Gradient


问:
Find for all value of constant $a>0$; the interval of convergence of the power series $\sum_{n=0}^{\infty}\frac{1}{1+a^{n}}x^{n}$
Find for all value of constant $a>0$; the interval of convergence of the power series
$\sum_{n=0}^{\infty}\frac{1}{1+a^{n}}x^{n}$.

What I have tried is; if we let $b_{n}=\frac{1}{1+a^{n}}x^{n}$ so we get by the Ratio test

$\lim_{n \to \infty}\frac{\left | b_{n+1} \right |}{\left | b_{n} \right |}=\lim_{n \to \infty} \left | x \right |\frac{\frac{1}{a^n} + 1}{\frac{1}{a^n} + a}=\left\{\begin{matrix}
\left | x \right | \frac{1}{a}& \mathrm{if} \ \frac{1}{a}<1\\
\left | x \right | & \mathrm{if} \ \frac{1}{a}\geq 1
\end{matrix}\right.$

So the radius of convergence we get

$r=\left\{\begin{matrix}
a & \mathrm{if} \  \frac{1}{a}<1 \\
1 & \mathrm{if} \ \frac{1}{a}\geq 1
\end{matrix}\right.$

Now what do I do?

答: Therefore, the series $\sum\limits_{n=0}^{\infty}\dfrac{1}{1+a^{n}}x^{n}$ is convergent for $|x| < a$ if $a>1$ and for $|x|<1$ in the case $a\leqslant{1}.$ Can you check convergence for $|x|=a, \;\; a>1$ and for $|x|=1 \text{ if } a\leqslant{1}?$</p>


问:
Odd perfect numbers and egyptian fraction conjecture
I have reached the conclusion that the non-existence of odd perfect numbers is related to the following

Conjecture Let it be $R=\{d_1,d_2,...,d_n\}$ the set of distinct proper divisors less than $\sqrt{O}$ of some odd positive integer $O$, such that $d_1<d_2<...<d_n$, and such that for some divisor $d_k\in R$ holds that $d_k\leq d_n<3d_k$. Then, it exist no solution to the egyptian fraction $$\frac{2}{d_k}+\frac{1}{d_1}+\frac{1}{d_2}+...=1$$

The process to reach the conjecture is the following (for you to check it is correct):

Let $P$ be some odd perfect number.

Let $R=\left\{ d_{1},d_{2},...,d_{n}\right\}$  be the set of proper divisors of $P$ less than $\sqrt{P}$ excluding $1$, and $T=\left\{ \frac{P}{d_{1}},\frac{P}{d_{2}},...\frac{P}{d_{n}}\right\}$  be the set of proper divisors of $P$ greater than $\sqrt{P}$ excluding $P$.

As $P$ is a perfect number,

$$1+d_{1}+d_{2}+...+d_{n}+\frac{P}{d_{n}}+...+\frac{P}{d_{2}}+\frac{P}{d_{1}}=P (1)$$

Operating,

$$1+d_{1}+d_{2}+...+d_{n}=P-\frac{P}{d_{1}}-\frac{P}{d_{2}}-...-\frac{P}{d_{n}}$$

$$1+d_{1}+d_{2}+...+d_{n}=P\left(1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}\right)$$

$$\frac{1+d_{1}+d_{2}+...+d_{n}}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}=P$$

As $1+d_{1}+d_{2}+...+d_{n}$ is an integer, it follows that $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ must be integer. Therefore, $P$ is a perfect number only if both $1+d_{1}+d_{2}+...+d_{n}$ and $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ are proper divisors of $P$.

It is trivial to show that one of the two expression is less than $\sqrt{P}$, and therefore belongs to $R$, and the other is greater than $\sqrt{P}$ and belongs to $T$.

It is clear that $1+d_{1}+d_{2}+...+d_{n}$ is greater than the greatest element of $R$, so subsequently we can state that

$$\left(\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}\right)=d_{k}\in R (2)$$

$$1+d_{1}+d_{2}+...+d_{n}=\frac{P}{d_{k}}\in T (3)$$

The proof showing that $d_k\leq d_n<3d_k$ is a bit long, so I skip it for the shake of briefness. And operating with $(2)$, the conjecture follows. I have found (thanks to some help provided here in MathStackExchange) examples of odd positive integers for which $(3)$ alone holds, but not any example of an odd positive integer for which $(2)$ holds with the restrictions exposed.

Any help on how to attack this conjecture would be welcomed. Otherwise, if someone is able to show any counterexample to the conjecture (that is, provide some odd positive integer for which $(1)$ holds), it would be more than welcomed to save efforts proving it. I have not seen this conjecture elsewhere, so it is possible that it is mistaken.

Thanks in advance!

Edit

I post the proof showing that $d_k\leq d_n<3d_k$, as it is fundamental for the conjecture and for the solution I have provided for verification.

From (3), and operating,

$$1+\sum_{j=1}^{n}d_{j}=\frac{P}{d_{k}}$$

$$P=d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)$$

As by definition we have that $d_{n}<\sqrt{P}$, just substituting we can set that

$$d_{n}<\sqrt{d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)}$$

$$d_{n}^{2}<d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)$$

$$\frac{d_{n}}{d_{k}}<\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}} (4)$$

Other hand, taking (1) and dividing by $\sqrt{P}$, we get that

$$\frac{1}{\sqrt{P}}+\frac{d_{1}}{\sqrt{P}}+\frac{d_{2}}{\sqrt{P}}+...+\frac{d_{n}}{\sqrt{P}}+\frac{P}{d_{n}\sqrt{P}}+...+\frac{P}{d_{2}\sqrt{P}}+\frac{P}{d_{1}\sqrt{P}}=\sqrt{P}$$

$$\frac{1}{\sqrt{P}}+\frac{d_{1}}{\sqrt{P}}+\frac{d_{2}}{\sqrt{P}}+...+\frac{d_{n}}{\sqrt{P}}+\frac{\sqrt{P}}{d_{n}}+...+\frac{\sqrt{P}}{d_{2}}+\frac{\sqrt{P}}{d_{1}}=\sqrt{P}$$

As $\forall d_{j}\,\frac{\sqrt{P}}{d_{j}}>1$, and $1+\sum_{j=1}^{n}d_{j}>\sqrt{P}$, then we get inmediately that $n<\sqrt{P}-1$; otherwise, the sum of all the terms would be greater than $\sqrt{P}$ and $P$ would not be a perfect number. Thus, we can affirm that

$$n\leq\sqrt{P}-2 (5)$$

As the minimum gap between consecutive elements of $R$ is $2$, the maximum sum of elements of $R$ with the minimum gap between them is

$$d_{n}+\left(d_{n}-2\right)+\left(d_{n}-4\right)+...+\left(d_{n}-2\left(n-1\right)\right)$$

Therefore, we can establish that

$$1+\sum_{j=1}^{n}d_{j}\leq d_{n}+\left(d_{n}-2\right)+\left(d_{n}-4\right)+...+\left(d_{n}-2\left(n-1\right)\right)$$

$$1+\sum_{j=1}^{n}d_{j}\leq nd_{n}-\left(\left(n-1\right)^{2}+\left(n-1\right)\right)$$

Subsequently, we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\frac{nd_{n}-\left(\left(n-1\right)^{2}+\left(n-1\right)\right)}{d_{n}}$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq n-\frac{\left(\left(n-1\right)^{2}+\left(n-1\right)\right)}{d_{n}}$$

Substituting $n$ by the inequality obtained in (5), we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\sqrt{P}-2-\frac{\left(\left(\sqrt{P}-2-1\right)^{2}+\left(\sqrt{P}-2-1\right)\right)}{d_{n}}$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\sqrt{P}-2-\frac{\left(\left(\sqrt{P}-3\right)^{2}+\left(\sqrt{P}-3\right)\right)}{d_{n}} (6)$$

Operating with the numerator of the third term of the right handside of (6), we get that

$$\left(\sqrt{P}-3\right)^{2}+\left(\sqrt{P}-3\right)=$$

$$P+9-6\sqrt{P}+\sqrt{P}-3=$$

$$P-5\sqrt{P}+6$$

As by definition $d_{n}<\sqrt{P}$, we can affirm that

$$\frac{P-5\sqrt{P}+6}{d_{n}}>\sqrt{P}-5+\frac{6}{\sqrt{P}}$$

Therefore, substituting at (6), we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<\sqrt{P}-2-\left(\sqrt{P}-5+\frac{6}{\sqrt{P}}\right)$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<3-\frac{6}{\sqrt{P}}$$

Subsequently,

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<3$$

$$1+\sum_{j=1}^{n}d_{j}<3d_{n}$$

Finally, substituting in (4), we get the desired result

$$\frac{d_{n}}{d_{k}}<3$$

答:
There are at least two (2) gaps in your reasoning, that I have been able to identify so far.

First, the fact that
$$\bigg(1 + d_1 + d_2 + \ldots + d_n\bigg)\Bigg(\dfrac{1}{1 - \dfrac{1}{d_1} - \dfrac{1}{d_2} - \ldots - \dfrac{1}{d_n}}\Bigg) = P$$
does NOT imply that
$$1 + d_1 + d_2 + \ldots + d_n = a \mid P$$
and
$$\dfrac{1}{1 - \dfrac{1}{d_1} - \dfrac{1}{d_2} - \ldots - \dfrac{1}{d_n}} = b \mid P.$$

What you do have is the biconditional
$$a \mid P \iff b \mid P$$
but NOT the conjunction
$$(a \mid P) \land (b \mid P).$$

Indeed, you have neither proved $a \mid P$ nor $b \mid P$ in the OP.

Second, the inequality
$$\frac{P - 5\sqrt{P} + 6}{d_n} \geq \sqrt{P} - 5 + \frac{6}{\sqrt{P}}$$
does NOT follow from
$$d_n < \sqrt{P}$$
basically because if we set $c = d_n, e = \sqrt{P}$ and $d = -5 < 0$, then
$$c < e \implies 1 < \frac{e}{c} \implies d > \frac{de}{c}.$$

So, your proof for the claimed upper bound $d_n < 3d_k$ is FLAWED.


答:
I'd say that the most general integral one usually considers for operators on a Hilbert space is the weak integral, in
the following sense: A function
$$
  w\in [a, b]\mapsto T_w\in B(H)
  $$
is weakly integrable if,  for every vectors $\xi $ and $\eta $ in $H$,  the map
$$
  w\mapsto \langle T_w(\xi ), \eta \rangle
  $$
is Lebesgue integrable on $[a, b]$.  In this case it may be shown that there exists a unique bounded operator $T$ such that
$$
  \int_a^b \langle T_w(\xi ), \eta \rangle \,  dw = \langle T\xi , \eta \rangle ,
  $$
for every $\xi $ and $\eta $ in $H$.

All of this immediately generalizes if the interval $[a,b]$ is replaced by  a general  measure space.

This integral is more general than both Bochner's and Pettis'.  As @Ben Grossmann noticed, a Bochner integral with a
compact integrand gives a compact operator, but this property is not shared by either the weak or the Pettis integral.

If $T_w$ is weakly integrable and each $T_w$ is strictly positive, then so is $\int_a^b T_w\, dw$ by the very same
argument as in @Ben's answer.

EDIT (Answering @Ben Grossmann's comment)

There is a bit of a conflict surrounding the word "weak" in operator theory, because what is universally known as the
weak operator topology (WOT) on $B(H)$ has little to do with its weak topology (W) as a Banach space.

In fact, $B(H)$ is a dual space, namely it is the dual of the space $L^1(H)$ of trace class operators, so $B(H)$ also
has a weak$^*$ topology (W$^*$) arising from the duality
$$
  (T,S)\in L^1(H)\times B(H)\mapsto  \text{tr}(TS)\in \mathbb C.
  $$

The functionals defining the WOT, namely
$$
  T\in B(H)\mapsto \langle T\xi , \eta \rangle \in \mathbb C,
  \tag 1
  $$
for $\xi ,\eta \in H$,
are in fact W$^*$-continuous, so the hierarchy is
$$
  \text{WOT} ⊆  \text{W}^* ⊆  \text{W},
  $$
(all inclusions being proper).

While  the expression "weak operator theory" motivated the terminology   "weak
integral" I used in my answer, a Banach space theorist would perhaps be more inclined to use a term involving the expression "weak$^*$".

Terminology apart, it is conceivable that a function $f$ from a measure space $(\Omega , \mu )$ to $B(H)$ is such that the integral
$$
  \int_\Omega \varphi (f(x))\,d\mu(x)
  $$
is well defined for every $\varphi $ in the dual space $B(H)'$, and yet $f$ is not Pettis integrable.

En passant,  in this case,  one can find a vector $\Lambda $ in the double dual space $B(H)''$ such that
$$
  \int_\Omega \varphi (f(x))\,d\mu(x) = \Lambda (\varphi ), \quad\forall \varphi \in B(H)'.
  $$
The lack of the Pettis integrability of $f$ will be precisely expressed by the fact that $\Lambda $ lies not in the canonical
copy of $B(H)$ within $B(H)''$.

The above  hypothetical function  $f$  will of course satisfy (1),  so it is weakly integrable (WOT) and its weak integral will be the operator
$T$ in $B(H)$ obtained as the image  of $\Lambda $ under the natural projection
$$
  B(H)'' = L^1(H)''' \to   L^1(H)' = B(H).
  $$

Concluding, I am a bit embarrassed not to be able to give an explicit example of a function $f$ with the above
properties but  I am basing my hopes of finding it on the fact that such non-Pettis integrable functions
exists in many non-reflexive spaces.  On the other hand, I might be totally wrong and, in this case, proving myself
wrong will surely lead to a worthwhile result!

Title: Determining whether an operator is positive or not

Body: Let $T=\int_a^bT_w dw$, where each $T_w$ is a strictly positive, self-adjoint, compact operators acting on a separable Hilbert space $X$. Then, I wonder whether T is also a strictly positive operator. That is, its eigenvalues are all strictly positive. I am naturally guessing that the answer is yes, but I cannot prove it. Above all, I do not know how rigorously one can define the operator $T$ given by an integral of operators, which I think should be the starting point of the proof. I am thinking that the other steps would be just simply straightforward if only it is defined rightly. Also, any reference related to this kind of material would be appreciated. Thanks.


问:
Probability of getting all face cards when splitting deck of cards in half
Question: What is the probability of getting all of the face cards in one pile when a standard deck of cards is split into two piles?

Additional info: Generally a deck of cards is split into two by putting cards in two different piles one card at a time in an alternating fashion until the entire deck is split into two even piles. A standard deck of cards contains 16 face cards out of 52 total cards. The order of the cards in either pile does not matter, all that matters is that one pile has all the face cards.

My approach and why I think it is wrong: I tried solving this problem by calculating the binomial probability ($_nC_xp^x(1-p)^{n-x}$), where $n=26$, $x=16$, and $p=\frac{16}{52}$, which results in $P(x)=0.00087$. I believe this is incorrect because it does not take into account the decreasing number of face cards and total cards as more are put into the two separate piles (correct me if I am wrong).

答: The first half of the deck must contain $26$ of the $36$ non-face cards, or it must contain $16$ face cards and $10$ non-face cards.  $$\frac{\binom{36}{26}+\binom{36}{10}}{\binom{52}{26}}\approx1.025\times10^{-6}$$


问:
How to show this particular $4 \times 4$ matrix is positive definite?
I am preparing for an exam in Numerical Analysis, and I am solving some practice problems. The question is to show that the following matrix $A$ is positive definite. I would only have about 10 minutes to work on this problem, so I am trying to solve this as fast as possible.

We are also told that all the eigenvalues of $A$ are distinct (but this information may not be useful here; there is a part b to this question which might make use of this fact)

$$A = \begin{bmatrix}
1 & -1 & 2 & 0\\
-1 & 4 & -1 & 1\\
2 & -1 & 6 & -2\\
0 & 1 & -2 & 4
\end{bmatrix}$$

My Attempts: My first tought is to use Greshgorin's Circle Theorem to show that all the eigenvalues are positive. However, this does not work because  the first Greshgorin disk contains negative reals.

My second tought is to use Sylvester's Criterion. This is perhaps doable in under 10 minutes, but it is prone to mistakes (especially when going fast). I am also not sure if Sylvester's Criterion was taught in the class that this problem comes from.

答:
HINT:

The rows from $2$ to $4$ are diagonally dominant. The first one is not. But we can make it so by multiplying on both sides by the matrix $\operatorname{diag}(t, 1, 1, 1)$, where $t$ is large.

$\bf{Added:}$ Looks plausible, but the other rows are affected. Indeed, need $t> 3$, and then the second row is not dominant anymore. So the solution is not good. In fact, one can check that there is no way to transform our matrix with a diagonal matrix to make it diagonally dominant.

Maybe just showing directly that the determinant is positive. Since the principal minor $(2,3,4)$ is positive definite, being dominant, this would be good enough.

Using WolframAlpha, I got the Cholesky decomposition of the matrix

$$\left[\begin{matrix} 1 & -1 & 2 & 0  \\ -1& 4 & -1& 1\\ 2& -1& 6&-2\\ 0&1 &-2&4\end{matrix} \right]=\\=\left[\begin{matrix} 1 & 0 & 0 & 0  \\ -1& 1 & 0& 0\\ 2& 1/3 & 1&0\\ 0&1/3 &-7/5&1\end{matrix} \right]\left[\begin{matrix} 1 & 0& 0 & 0  \\ 0& 3 & 0& 0\\ 0& 0& 5/3&0\\ 0&0 &0&2/5\end{matrix} \right]\left[\begin{matrix} 1& -1& 2& 0\\ 0& 1& 1/3& 1/3\\ 0& 0& 1& -7/5\\0& 0& 0& 1\end{matrix} \right]$$

The diagonal part has moderate eigenvalues. Now, $A$ has a small eigenvalue, $\approx 0.02$, and this is possible since the upper diagonal part has a small singular value.


问:
Evaluating a triple summation ${\sum \sum \sum}_{1 \le i < j \le k \le 20} a_ia_ja_k$

If $$a_i= \begin{cases} 1 && 1 \le i \le 10 \\ 2 && 11 \le i \le 20 \end{cases}  $$

Find the value of $$\displaystyle {\sum \sum \sum}_{1 \le i < j \le k \le 20} a_ia_ja_k $$

I could only evaluate the summation when $1 \le i < j \le k \le 10$, by writing it as $\displaystyle {\sum _{i=1}^9\sum_{j=i+1}^{10} \sum_{k=j}^{10}} 1$ which came out to be $\displaystyle \binom{11}{3}$.

Should I make different cases like $i \le 10, 11 \le j \le k \le 20$ , etc.?

How do I evaluate it for the other cases? Is there any easier way to evaluate it without evaluating each sum one by one? Could there possibly be a combinatorial approach? I'd be delighted to see different approaches to this problem.

答:
Essentially you want to count how many triplets $(i,j,k)$ will lead to $(a_i,a_j,a_k)$ to be one of the follow four possibilities: $(1,1,1),(1,1,2),(1,2,2),(2,2,2)$ (think about why these are the only possible outputs for $(a_i,a_j,a_k)$.

Now, count how many triplets of $(i,j,k)$ under the constraint $1\le i <j \le k\le 20$ will give you each of those four outputs. In particular,

$(a_i,a_j,a_k)=(1,1,1)$ if we have $1\le i <j \le k\le 10$. Say there are $A$ many such $(i,j,k)$.

$(a_i,a_j,a_k)=(1,1,2)$ if we have $1\le i <j \le 10$, $11\le k \le 20$. Say there are $B$ many such $(i,j,k)$.

$(a_i,a_j,a_k)=(1,2,2)$ if we have $1\le i \le 10$, $11\le j\le k \le 20$. Say there are $C$ many such $(i,j,k)$.

$(a_i,a_j,a_k)=(2,2,2)$ if we have $11\le i < j\le k \le 20$. Say there are $D$ many such $(i,j,k)$.

Then your final result will be $(1\cdot 1\cdot 1)A + (1\cdot 1\cdot 2)B + (1\cdot 2\cdot 2)C + (2\cdot 2\cdot 2) D$.

This turns it into a counting problem to find $A,B,C,D$.

(To check your work, you should get $A = 165, B = 450 , C = 550, D = 165$. And this gives the correct final sum.)


问:
Sum of normalized B-splines
Let $N_j$, $j=-k,\ldots , m-1$ the normalized B-splines of the set of nodes $x_0, \ldots , x_m$ of degree $k$.

Show that $$\sum_{j=-k}^{m-1}N_j(x)=1 \ \text{ for all } x\in [x_0, x_m]$$

Do we use for that the below formula?
$$ N_{i,0,\tau }(u)=\begin{cases}1,&u\in \left[\tau _{i},\tau _{i+1}\right)\\0,&{\text{else}}\end{cases}
\\
N_{i,p,\tau }(u)={\frac {u-\tau _{i}}{\tau _{i+p}-\tau _{i}}}\,N_{i,p-1,\tau }(u)\;+\;{\frac {\tau _{i+p+1}-u}{\tau _{i+p+1}-\tau _{i+1}}}\,N_{i+1,p-1,\tau }(u) \text{ for } p>0$$

Or do we use an other formula to calculate the sum?

$$$$

EDIT:

In the notes I found also a formula with divided differences, which is
\begin{align*}&N_j(x)=(x_{j+k+1}-x_j)B_j(x) \\ \text{ with } \ B_j&=(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k+1}]\\ & =\frac{(\cdot -x)_+^k[x_{j+1}x_{j+2}\ldots x_{j+k+1}]-(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k}]}{x_{j+k+1}-x_j}\end{align*} where $(x)_+^k=\begin{cases}x^k & \text{ if } x\geq 0 \\ 0 & \text{ if } x<0\end{cases}$.

Then we have \begin{align*}&\sum_{j=-k}^{m-1}N_j(x)=\sum_{j=-k}^{m-1}(x_{j+k+1}-x_j)B_j(x)\\ & =\sum_{j=-k}^{m-1}(x_{j+k+1}-x_j)\frac{(\cdot -x)_+^k[x_{j+1}x_{j+2}\ldots x_{j+k+1}]-(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k}]}{x_{j+k+1}-x_j}\\ & =\sum_{j=-k}^{m-1}\left ((\cdot -x)_+^k[x_{j+1}x_{j+2}\ldots x_{j+k+1}]-(\cdot -x)_+^k[x_jx_{j+1}x_{j+2}\ldots x_{j+k}]\right )\end{align*}
How can we continue?

答:
Yes. you use those formulas. In fact, your course probably gave those formulas as the definitions of b-spline basis functions, so, if you don't know any other facts about them, there's no other place to start.

In case it's not obvious, you use induction on the degree, $p$.

Title: Sum of normalized B-splines


问:
Outward pointing vectors on the boundary of the interval $[a,b] \subset \mathbb{R}$
According to the Loring W. Tu's book "An Introduction to Manifolds", we have a standard oriented atlas on $[a,b]$, with the charts $([a,b[, \phi)$ and $(]a,b],\psi)$, where $\phi \colon [a,b[ \longrightarrow [0,b-a[$ is given by $\phi (t) = t-a$ and $\psi \colon ]a,b]\longrightarrow ]a-b,0]$ given by $\psi (t) = t-b$. Also, $\psi$ is considered an atlas, since for 1-dimensional manifolds with boundary, we allow both $\mathbb{H}^1 := \mathbb{R}_{\geq0}$ and $\mathbb{L}^1 := \mathbb{R}_{\leq0}$ as models, which allows us to call a closed interval in $\mathbb{R}$ an oriented 1-dimensional manifold with boundary. I show below a paragraph in the book in which the author discusses the outward-pointing vectors on $[a,b]$.

One of the things I'm possibly not understanding fully is why he says we have a standard coordinate $x$ in the interval (since we need two charts to cover it) and consequently, why he says that we have a vector field $\frac{d}{dx}$ on the manifold and an orientation form $dx$. But I think of that, maybe wrongfully so, as minor details which are left for the reader to fill.

The "main issue" I'm having with this example is the fact that, in this example, as the author states, the outward-pointing vector on $b$ is $\frac{d}{dx}$, which has a positive component associated with $\frac{d}{dx}$, seemingly contradicting a previous statement which he had made (shouldn’t it have a negative component?). I will leave a screenshot of the statement below. I tried to work out the details in this example, but I ran into the same problem.

Where am I think wrong? I appreciate all the help in advance.

答:
$x$ is really just the identity map on $[a,b]$. So $x\circ \phi^{-1} = \phi^{-1}, x \circ \psi^{-1} = \psi^{-1}, \phi \circ x^{-1} = \phi, \psi \circ x^{-1} = \psi$. All of these maps are smooth, which means that $x$ is an embedding of $[a,b]$ into $\Bbb R$. So, yes, it is a coordinate on $[a, b]$, and quite obviously, a natural one. The description given of $\frac d{dx}$ is accurate. At $b, \frac d{dx}$ does point outward, and at $a$, it is $-\frac d{dx}$ that points outward. What previous statement do you think this contradicts?

The only reason you need two charts here is that for technical reasons, the charts on manifolds with boundaries are into half-spaces, so a single chart can only handle one of the ends. For $1$-dimensional manifolds, there would be no problem with allowing a chart to be into a compact interval as well. But there is no reason to make this special exception to the normal definition, as it doesn't add anything that wasn't already available.

I think I see your problem now:  You are concerned about this:

In terms of local coordinates, a vector $X_p$ is outward-pointing if and only if $a^n(p) < 0$

where $p$ is on the boundary, and in local coordinates at $p, X_p = \sum_i a^i(p)\left.\frac{\partial}{\partial x^i}\right|_p$.

In this, they are specifically talking about a chart carrying $p$ to the boundary of $\Bbb H^n$, not $\Bbb L^n$. $a^n(p) < 0$ for outward pointing vectors because for points on the boundary of $\Bbb H^n$, it is $-\frac{\partial}{\partial x^n}$ that points out of $\Bbb H^n$. For $\Bbb L^n$, the opposite is true. $\frac{\partial}{\partial x^n}$ points outward and $-\frac{\partial}{\partial x^n}$ points inward.

By section 22.5, they are no longer talking about oriented manifolds, and have slipped back into describing unoriented manifolds, where it is sufficient to have charts into subsets of $\Bbb H^n$ exclusively. As such, they did not think about the need to have charts into $\Bbb L^n$ to have a compatible orientation.


问:
All Polynomials Such That All Roots of $f(x)$ are in $f(x^2)$
Let there be a polynomial, $f(x)$ and denote $g(x) = f(x^2)$.

Find all monic polynomials, f, with complex coefficients, such that for all $r_i$ satisfying $f(r_i) = 0$, $g(r_i) =0$ as well. (source: An Old (first written in 1985) Indian Math Book Filled With Interesting Problems)

What I've basically broken the problem down to is that if all roots of $f$ are in $g$, then $f$ must be divisible by $g$.

Starting off with degree $1$, it's pretty easy to see that the possibilities are $f(x) = x$ and $f(x) = x-1$.

For degree $2$, I used polynomial long division to get the following polynomials: $f(x) = x^2, x(x-1), x^2 - 1, x^2 + x + 1,$ and $(x-1)^2.$

Afterwards, I realized that if $p(x)$ and $q(x)$ both work, then so much $p(x) \cdot q(x)$.

Furthermore, I saw that all cyclotomic polynomials work.

Other than polynomial long division (which can get quite tedious and long), are there any easier ways to determine all functions $f$ that work for higher degrees (such as $3$ or $4$ or $5$)?

答:
"If all roots of $f$ are in $g$, then $f$ must be divisible by $g$" is not quite right; first, you mean $f$ must divide $g$, and second, it's possible that the multiplicities of the roots of $f$ are bigger than those of $g$. E.g. we could have $f(z) = z^2, g(z) = z$.

Anyway, the question is to find all monic polynomials $f$ such that the set of roots of $f$ is closed under the squaring operation $r \mapsto r^2$. $f$ can have the root $0$ with any multiplicity; we now restrict our attention to nonzero roots $r$ only. If $|r| \neq 1$ then repeated squaring $k \mapsto r^{2^k}$ rapidly sends the absolute value to either $0$ or $\infty$, and in particular would imply that $f$ has infinitely many roots, so all roots of $f$ must have absolute value exactly $1$. If $r = e^{2 \pi i t}$ then $r^{2^k} = e^{2^{k+1} \pi i t}$ which takes on infinitely many distinct values unless $t$ is rational.

So all of the roots of $f$ must be roots of unity, and moreover if a primitive $d^{th}$ root of unity $\zeta_d$ is a root then $\zeta_d^{2^k}$ must all be a root for all $k \ge 1$. So the answer is exactly the products of copies of $f(z) = z$ and polynomials of the form

$$f(z) = \prod_{k=0}^{n-1} (z - \zeta_d^{2^k})^{m_k}$$

where $\zeta_d$ is some primitive $d^{th}$ root of unity and $n$ is the smallest positive integer such that $2^n \equiv 2^i \bmod d$ for some $i < n$ and the multiplicities $m_k \ge 1$ are arbitrary positive integers. ($i = 0$ if $d$ is odd but not otherwise.)

The answer is much nicer if we ask for $f$ to have integer coefficients. Then if $f(z)$ has $\zeta_d$ as a root of multiplicity $m$ it must be divisible by the cyclotomic polynomial $\Phi_d(z)$ with multiplicity $m$. If $d$ is odd then the roots of $\Phi_d(z)$ are already closed under squaring; otherwise, if $d$ is even then squaring the roots of $\Phi_d(z)$ produces exactly the roots of $\Phi_{\frac{d}{2}}(z)$, so $f$ must also be divisible by $\Phi_{\frac{d}{2}}(z)$. Continuing in this way we conclude that $f$ is a product of copies of $z$ and cyclotomic polynomials $\Phi_{d_i}(z)$ with the property that the set of $d_i$ which occur is closed under division by $2$. This can be restated in a slightly nicer way as follows: if $d$ is odd then

$$\Phi_d(z) \Phi_{2d}(z) \Phi_{4d}(z) \dots \Phi_{2^k d}(z) = \Phi_d(z^{2^k})$$

so equivalently $f$ is a product of copies of $z$ and polynomials of the form $\Phi_d(z^{2^k})$ where $d$ is odd.

In particular, your claim that all cyclotomic polynomials work is not quite right, and $\Phi_2(z) = z + 1$ is the smallest counterexample.


问:
Concrete Example of Fatou's Lemma
I am not sure if I understand the $\inf$ meanings in Fatou's Lemma. The assertion states that:

$$\lim_{n \to \infty} \int_X \inf f_n d\mu \leq \lim \inf_{n \to \infty} \int_{X}  f_n d\mu$$

I would like to understand its meaning and see the use of the lemma through a simple example. So let the sequence of functions be $f_n(x) =  \big(1 + \frac{1}{n} \big)^n x$ for $n \geq 1$, where $n \in  \mathbb{N}$.

Obviously it is an increasing sequence that converges to $f(x) = e x$. Let us integrate over the $[0,1]$ interval. My questions are:

For the left hand side of the inequality we would need to integrate $\inf f_n$ and then evaluate its limit. This function would in our case be given by $x$ right?. If true, then the value of the integal is simply $\frac{1}{2}$. What do we do with the limit then?

And for the right hand side we have to evaluate $\lim \inf_{n \to \infty} \frac{1}{2} \big(1 + \frac{1}{n} \big)^n$. The smaller value $\forall n \geq 1$ is $1$. Again, what is the use of the limit here?

Thanks!

答: As people have pointed out in the comments, you are confusing $\liminf$ and the infimum notation. For a concrete example of strict inequality, consider $f_n = n\chi_{[0,1/n]}$ on $[0,1]$.

Title: Concrete Example of Fatou's Lemma


问:
Closed form for $n$th derivative of exponential of $f$
What is the closed form for:

$$\frac{\partial^n}{\partial x^n}\exp(f(x))=\exp(f(x))\cdot[????]$$

答:
I dont know how to get this result, but check wolfram's solution. It's exactly what you are looking for.

http://www.wolframalpha.com/input/?i=d%5En%2Fdx%5En%28e%5E%28f%28x%29%29%29


问:
Suppose $x$ and $y$ are two real numbers such that $x + y = 6$ and $x^2 + y^2 = 43$. Find the value of $x^3 + y^3$.
I tried quadratic and substituting and cubing the equations but it doesn't work and I'm pretty sure i did not make any careless mistake.</p>

答:
No tricks.  Just solve for $x$ and then $y$:

$x+y=6\iff y=6-x$

$x^2+y^2=43\implies x^2+(6-x)^2=43\iff 2x^2-12x-7=0$

$\iff x=3+\dfrac5{\sqrt2}$ or $x=3-\dfrac5{\sqrt2}$,

with $y=6-x=3-\dfrac5{\sqrt2}$ or $3+\dfrac5{\sqrt2}$, respectively.

Therefore, $x^3+y^3=\left(3+\dfrac5{\sqrt2}\right)^3+\left(3-\dfrac5{\sqrt2}\right)^3$

$=2\times3^3+2\times3\times3\times\left(\dfrac5{\sqrt2}\right)^2=279.$

(Odd powers of $\dfrac5{\sqrt2}$ cancel odd powers of $-\dfrac5{\sqrt2}$.)


问:
How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$?
The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both side and drawing triangles. However, the worked solutions does a simplification from LHS to RHS instead, namely $$\cos\arcsin\left(\frac{3}{5}\right)\cos\arctan\left(\frac{7}{24}\right)-\sin\arcsin\left(\frac{3}{5}\right)\sin\arctan\left(\frac{7}{24}\right)=\frac{3}{5}$$which I don't understand. Can someone please explain?

答:
The solution makes use of the Pythagorean identities.

I. $\arcsin(3/5)=\theta\implies \sin(\theta)=3/5$. Using the identity: $\cos^2(\theta)=1-\sin^2(\theta)\iff \cos(\theta)=\sqrt{1-\sin^2(\theta)}$. Hence, $\cos(\arcsin(3/5))=\sqrt{1-(\frac{3}{5})^2}=\sqrt{\frac{16}{25}}=4/5$.

II. $\arctan(7/24)=\theta\implies \tan(\theta)=7/24$. Using the identity $1+\tan^2(\theta)=\sec^2(\theta) \iff \cos(\theta)=\frac{1}{\sqrt{1+\tan^2(\theta)}}$. Hence, $\cos(\arctan(7/24))=\frac{1}{\sqrt{1+(\frac{7}{24})^2}}=24/25$.

III. $\sin(\arcsin(3/5))=3/5$ is trivial.

IV. $\arctan(7/24)=\theta \implies \tan(\theta)=7/24$ Using the identity $1+\cot^2(\theta)=\csc^2(\theta) \iff \sin(\theta)=\sqrt{\frac{\tan^2(\theta)}{1+\tan^2(\theta)}}$. Hence, $\sin(\arctan(7/24))=\sqrt{\frac{(\frac{7}{24})^2}{1+(\frac{7}{24})^2}}=7/25$.

Combining the 4 results we have: $I*II-III*IV=4/5*24/25-3/5*7/25=3/5$ yields the desired result.


问:
Combinations and Permutations in coin tossing
I understand the formulae for combinations and permutations and that for the binomial distribution. However, I'm confused about their application to coin tossing.

Consider three tosses. Outcomes with two heads are HHT, HTH and THH. So, there are three and that's what you get if you use the formula for combinations.

With combinations "the order is not important". How does that apply in this case? In what way are these three outcomes different if the order is not important?

What are the 6 permutations?

答:
Coin toss series can be viewed, depending on what you want to know, as either "combination" or "permutation" but in all cases "with repetition" (meaning same side can occur again and again).

*Probably the best page that summarizes the Combination vs Premutation with or without Repetition https://www.mathsisfun.com/combinatorics/combinations-permutations.html *

If the question is "How many ways a series of R coin tosses (N=2 sides) can go? Of these, how many will have 2 Heads in the row?", you are looking for Permutation with Repetition where "HHT" is different outcome from "THH".

Permutation with Repetition is the simplest of them all:

N to the power of R.

Example:
3 tosses of 2-sided coin is 2 to power of 3 or 8 Permutations possible.

In these, "at-least-2 Heads in a row" permutations are: HHH, HHT, THH - 3. Probability of "at least 2 heads in a row" is 3/8th (0.375)

If the question is "If you throw a 2-sided coin (N=2), R times, how many times can you get at least 2 heads?", you are looking for Combination (order is not important) with Repetition where "HHT" and "THH" are same outcomes (combination).

Combination with Repetition formula is the most complicated (and annoying to remember): (R+N-1)! / R!(N-1)!

For 3 2-sided coin tosses (R=3, N=2), Combination with Repetition: (3+2-1)! / 3!(2-1)! = 24 / 6 = 4

These are (because order is not important): HHH, HHT, HTT, TTT

If you are looking for "at least 2 Heads", 2 options match: HHH and HHT (order not important).

Probability of you getting at least 2 heads is 2 outcomes / 4 Combinations (with Repetition) = 0.5. Right?


问:
Does second countable implies countably many components?
I was proving that a manifold has countably many connected components and this question came up. The way I did the original question is to show that components in manifolds are open and then use the fact that second countability implies separable to show that there's only countably many components. The latter argument follows from the fact that, if there's uncountably many components, then there must be some component "left out" but since that component is open, we must get a contradiction.

However, in general, components are not open (in fact, they are always closed). So in the general case, I do not know how to prove that separability implies having a countable number of components or if second countability would imply such (since second countability is a stronger condition). Are those implications true and if not, what are the counter-examples?

答:
Let $\mathbb{R}$ have the usual topology and let $X$ be the set of irrational numbers with the subspace topology inherited from $\mathbb{R}$.

Since $\mathbb{R}$ is second countable, so is $X$.

But the components of $X$ are the singleton subsets of $X$, so $X$ has uncountably many components.


问:
exponential equation quick-select algorithm
i'm trying to proof that quickSelect algorithm RunTime complexity is: Theta(n)

With Akra-Buzzi method.

So i need to Solve this recursive function: T(n) = T(n/5) + T(7n/10) + n

i need to find p, such that: (1/5)^p + (7/10)^p = 1.

and i cant solve this exponential equation :/

答:
Consider that you look for the zero of function
$$f(p)=\left(\frac{1}{5}\right)^p+\left(\frac{7}{10}\right)^p-1$$ For simplicity of further notations, let $a=\frac{1}{5}$ and $b=\frac{7}{10}$.

We have $f(0)=1$ and $f(1)=-\frac{1}{10}$; so the solution is a bit smaller than $1$. Just start Newton method with $p_0=1$ and you will have as iterates
$$p_{n+1}=p_n-\frac{f(p_n)}{f'(p_n)}=p_n-\frac{a^{p_n}+b^{p_n}-1 }{a^p \log (a)+b^p \log (b) }$$ This would give
$$\left(
\begin{array}{cc}
n & p_n \\
 0 & 1.000000 \\
 1 & 0.825040 \\
 2 & 0.839659 \\
 3 & 0.839780
\end{array}
\right)$$


问:
A Urysohn Function on an Uncountable Fort Space
I'm struggling with a problem about determining if an Uncountable Fort Space is Tychonoff or not, at this moment I have proven that my space is regular Hausdorff but I can't find a Urysohn Function for the space in question.

Any hint or tip is welcome :)

答:
It’s a compact Hausdorff space, so it is not just Tikhonov but $T_4$ (normal plus $T_1)$.

However, we can also show the result directly. Let $X$ be the space, and let $p$ be the non-isolated point. The closed sets in $X$ are the sets containing $p$ and the finite subsets of $X\setminus\{p\}$.

Suppose first that $p\in F\subseteq X$ and $x_0\in X\setminus F$. Then $\{x_0\}$ is a clopen set, so the function

$$f:X\to[0,1]:x\mapsto\begin{cases}
0,&\text{if }x=x_0\\
1,&\text{otherwise}
\end{cases}$$

is continuous, $f(x_0)=0$, and $f[F]=\{1\}$.

Now suppose that $F$ is a non-empty finite subset of $X\setminus\{p\}$. Then $F$ is clopen, so the function

$$f:X\to[0,1]:x\mapsto\begin{cases}
1,&\text{if }x\in F\\
0,&\text{otherwise}
\end{cases}$$

is continuous, $f[F]=\{1\}$, and $f(x)=0$ for each $x\in X\setminus F$.


问:
Geometric yet rigorous motivation for the definition of the surface measure
Let $M \subset \mathbb{E}^n$ be a submanifold, $\dim M = m$. Then $M$ inherits Riemannian structure from the Euclidean space. Consider the $\sigma$-algebra induced by the Lebesgue measurable sets and the parametrizations.

The surface measure on $M$ is defined as:

$$\sigma(A) = \int_{r^{-1}(A)} \sqrt{\det[g_{ij}]} d\lambda_m ~~~~~~(*) $$

where $g_{ij} = \langle\frac{\partial r}{\partial x_i}\mid\frac{\partial r}{\partial x_j} \rangle$, i.e. $[g_{ij}] = (Dr)^T (Dr)$ is the Gram matrix of $Dr$ or, in other words, the determinant of the inherited Riemann metric tensor. For simplicity, we assume here that the set $A$ lies within the image of a single local parametrization $r$.

This definition was motivated in a rather sloppy way:

Take a parametrization $r$ around $p \in M$ such that $r(0) = p$. Take a small hypercube $K \subset \mathbb{R}^m$ having $0$ as one of its vertices. Then the image $r(K)$ will approximately be a parallelogram contained in $TM_p$, with its sides parallel to the vectors $\frac{\partial r}{\partial x_i} \in TM_p$, whose volume is $\sqrt{\det g_{ij}} \lambda_m(K)$. Therefore, we define the measure as above.

I'm not convinced by an argument like this. It's completely unclear that the error of such approximation will ever converge to 0. Keep in mind that naive triangulation doesn't work - recall the Schwarz lantern. Moreover, the approximation error is not quantified at all.

I do agree that we only need to argue that the equality holds on images of hypercubes, as they form a generating set for the $\sigma$-algebra, so by the $\pi-\lambda$ lemma we'll get uniqueness of the measure satisfying $(*)$ for images of hypercubes.

Are there any different ways in which the surface measure is unique? Are there any limit constructions which justify the name surface area for a 2-dimensional manifold in $\mathbb{E}^3$? Can we justify the geometric meaning of the surface measure in a more rigorous way?

答:
It is sad that such an important question has been left unanswered! So, let me attempt.   People already mentioned change of variables formula, Hausdorff measure $H^n$, but here is the connection. First, I must say that I presume that we agree that $H^n$ deserves to be what "surface area" should be. So, my goal is to show that the particular choice/definition of $\sigma$ works because $\sigma = H^n$. The reason for the particular definition of Jacobian comes from linear algebra and looing into how linear maps translate $m$-volume into a bigger Euclidean space.

Lemma: Given any $\lambda>1$, $M$ can be covered by (countably many) local parameterizations $f\colon \mathbb{R}^m \supset U \to M$ such that

$f$ is a $\lambda$-bilipschitz map with respect to both the Riemannian distance on $M$, or the Euclidean distance on $M$ inherited from the ambient space, and

$\lambda^{-1} \leq J(f,x) \leq \lambda$.

We may write $A$ as union of $A_i$ each contained in image of one of these local parameterizations. For each $A_i$, from (1),
$$
H^n(A_i) \leq \lambda^n H^n(f^{-1}(A_i)),
$$
But on $\mathbb{R}^n$, $H^n=\mathcal{L}^n$, so,
$$
H^n(f^{-1}(A_i)) = \mathcal{L}^n(f^{-1}(A_i)) = \int_{f^{-1}(A_i)} 1 dx \leq \lambda \int_{f^{-1}(A_i)} J(f,x) dx = \lambda \sigma(A_i).
$$
Here we used the fact that $\sigma$ has the same expression in any parameteriation, which is true because of the change of variables formula (for $C^1$ diffeomorphisms between subsets of Euclidean spaces).

So,
$$
H^n(A_i) \leq \lambda^2 \sigma(A_i)
$$
Similar argument gives
$$
 H^n(A_i) \geq \lambda^{-2} \sigma(A_i).
$$
By additivity, we get same estimates for $A$.
Since $\lambda>1$ was arbitrary, we get
$$
 H^n(A) = \sigma(A).
$$


问:
What if function has cartesian product as domain but actual function only uses one variable?
If a function's domain is $ℤ \times ℤ$ and codomain is $ℤ$ but the actual function only uses one variable, is it a valid function?

eg.

$f(x) = x + 3$

Can I just put an integer $x$ value in, which will provide a value, even if $y$ is not used in the function itself?

答:
If the domain of $f$ in your example is $\mathbb{Z}\times\mathbb{Z}$, then $f(x)=x+3$ does not make sense, since $x$ is an element of $\mathbb{Z}\times\mathbb{Z}$ and $3$ is an element of $\mathbb{Z}$ (how can you add an element of $\mathbb{Z}\times\mathbb{Z}$ to an element of $\mathbb{Z}$?).

However, if you mean $f(x,y)=x+3$, then in this case it is understood that $x$ and $y$ are both elements of $\mathbb{Z}$, and that the input is $(x,y)$, an element of $\mathbb{Z}\times\mathbb{Z}$. In this case, $f(x,y)=x+3$ is well-defined, even though the function assignment doesn't use the variable $y$. (Think about a constant function $f(x)=3$, for example - it doesn't use the variable $x$, but is still well-defined.)


答:
This approach is kind of bruteforce, but I could do it successfully using only pen and paper.

First, we'll consider the case $1 \le i \le 10$. We get a 2-dimensional matrix for the values that the addend can assume:

When you're doing this on paper, there's no need to write out all terms once you get the pattern, and an ellipsis works. Notice that For a given $i$, we will sum the lower triangle of this matrix from the index $i+1$. Therefore, as an example, for i=5, the sum will look like this:

The sum for $i$ is given as
$$S_i = \sum_{t=1}^{10-i}t + 20(10-i) + 220$$
Summing this from 1 to 10, we get
$$S_1 = \sum_{n=1}^9 \frac{n(n+1)}{2} + 20 \sum_{n=1}^9 n + 220 \times 10$$

We'll now consider $11 \le i \le 20$. In this case, the lower triangle assumes a value of 8, and the same pattern follows. The sum in this case is
$$S_2 = 8\sum_{n=1}^9 \frac{n(n+1)}{2}$$

Adding $S_1$ and $S_2$, we get
$$S = 9\sum_{n=1}^9 \frac{n(n+1)}{2} + 20 \sum_{n=1}^9 n + 2200$$

Evaluating this gives us $\boxed{S = 4585}$.

As an addendum, here's a simple python program I wrote to check the sum, and the additions that I made.

```<code>sum = 0
for i in range(1, 21):
    for j in range(i+1, 21):
        for k in range(j, 21):
            addend = 1
            if i > 10:
                addend *= 2
            if j > 10:
                addend *= 2
            if k > 10:
                addend *= 2
            sum += addend
            print(addend, end=" ")
        print("")

print(sum)
</code>```

Title: Evaluating a triple summation ${\sum \sum \sum}_{1 \le i < j \le k \le 20} a_ia_ja_k$

Body:
If $$a_i= \begin{cases} 1 && 1 \le i \le 10 \\ 2 && 11 \le i \le 20 \end{cases}  $$

Find the value of $$\displaystyle {\sum \sum \sum}_{1 \le i < j \le k \le 20} a_ia_ja_k $$

I could only evaluate the summation when $1 \le i < j \le k \le 10$, by writing it as $\displaystyle {\sum _{i=1}^9\sum_{j=i+1}^{10} \sum_{k=j}^{10}} 1$ which came out to be $\displaystyle \binom{11}{3}$.

Should I make different cases like $i \le 10, 11 \le j \le k \le 20$ , etc.?

How do I evaluate it for the other cases? Is there any easier way to evaluate it without evaluating each sum one by one? Could there possibly be a combinatorial approach? I'd be delighted to see different approaches to this problem.

Body:
This approach is kind of bruteforce, but I could do it successfully using only pen and paper.

First, we'll consider the case $1 \le i \le 10$. We get a 2-dimensional matrix for the values that the addend can assume:

When you're doing this on paper, there's no need to write out all terms once you get the pattern, and an ellipsis works. Notice that For a given $i$, we will sum the lower triangle of this matrix from the index $i+1$. Therefore, as an example, for i=5, the sum will look like this:

The sum for $i$ is given as
$$S_i = \sum_{t=1}^{10-i}t + 20(10-i) + 220$$
Summing this from 1 to 10, we get
$$S_1 = \sum_{n=1}^9 \frac{n(n+1)}{2} + 20 \sum_{n=1}^9 n + 220 \times 10$$

We'll now consider $11 \le i \le 20$. In this case, the lower triangle assumes a value of 8, and the same pattern follows. The sum in this case is
$$S_2 = 8\sum_{n=1}^9 \frac{n(n+1)}{2}$$

Adding $S_1$ and $S_2$, we get
$$S = 9\sum_{n=1}^9 \frac{n(n+1)}{2} + 20 \sum_{n=1}^9 n + 2200$$

Evaluating this gives us $\boxed{S = 4585}$.

As an addendum, here's a simple python program I wrote to check the sum, and the additions that I made.

sum = 0
for i in range(1, 21):
    for j in range(i+1, 21):
        for k in range(j, 21):
            addend = 1
            if i > 10:
                addend *= 2
            if j > 10:
                addend *= 2
            if k > 10:
                addend *= 2
            sum += addend
            print(addend, end=" ")
        print("")

print(sum)


答:
@2rd_7's suggestion to calculate the pivots seems quick and straightforward. You just need to keep subtracting out multiples of the remaining rows (i.e., add the first row to the second, subtract 2 times the first row from the third, etc.) until you get an upper triangluar matrix:
$U = \begin{bmatrix} 1 & -1 & 2 & 0\\ 0 & 3 & 1 &1\\ 0 & 0 & 3 & -1\\ 0 & 0 & 0 & 26/9\end{bmatrix}$.

The pivots are now on the diagonal and are all positive, so the matrix is positive definite.

Title: How to show this particular $4 \times 4$ matrix is positive definite?

Body:
I am preparing for an exam in Numerical Analysis, and I am solving some practice problems. The question is to show that the following matrix $A$ is positive definite. I would only have about 10 minutes to work on this problem, so I am trying to solve this as fast as possible.

We are also told that all the eigenvalues of $A$ are distinct (but this information may not be useful here; there is a part b to this question which might make use of this fact)

$$A = \begin{bmatrix}
1 & -1 & 2 & 0\\
-1 & 4 & -1 & 1\\
2 & -1 & 6 & -2\\
0 & 1 & -2 & 4
\end{bmatrix}$$

My Attempts: My first tought is to use Greshgorin's Circle Theorem to show that all the eigenvalues are positive. However, this does not work because  the first Greshgorin disk contains negative reals.

My second tought is to use Sylvester's Criterion. This is perhaps doable in under 10 minutes, but it is prone to mistakes (especially when going fast). I am also not sure if Sylvester's Criterion was taught in the class that this problem comes from.


答:
There are 16 face cards so the remaining 10 cards in the half deck will have to be non-face cards. There are $\binom{36}{10}$ ways to choose 10 non-face cards from all 36 and $\binom{52}{26}$ ways to choose 26 cards from 52 in total. Therefore, the probability is as follows:

$$ \frac{\binom{36}{10}}{\binom{52}{26}} \approx 5.126 x 10^{-7}$$

Or you can think about it in terms of the other half without any face cards. You'll need to choose 26 non-face cards from the 36:

$$ \frac{\binom{36}{26}}{\binom{52}{26}} \approx 5.126 x 10^{-7}$$

You get the same result.

Title: Probability of getting all face cards when splitting deck of cards in half

Body:
Question: What is the probability of getting all of the face cards in one pile when a standard deck of cards is split into two piles?

Additional info: Generally a deck of cards is split into two by putting cards in two different piles one card at a time in an alternating fashion until the entire deck is split into two even piles. A standard deck of cards contains 16 face cards out of 52 total cards. The order of the cards in either pile does not matter, all that matters is that one pile has all the face cards.

My approach and why I think it is wrong: I tried solving this problem by calculating the binomial probability ($_nC_xp^x(1-p)^{n-x}$), where $n=26$, $x=16$, and $p=\frac{16}{52}$, which results in $P(x)=0.00087$. I believe this is incorrect because it does not take into account the decreasing number of face cards and total cards as more are put into the two separate piles (correct me if I am wrong).


问:
Proof of non-strictly convexity of $l_1$ and $l_{\infty}$
Given the Banach space $l_p = \left\{f\in R^n : \|f\| \leq \infty\right\}$ for $1\leq p \leq \infty$, we define the following norms

$\|f\|_p = \left(\displaystyle\sum_{j=0}^{\infty}
   |f_j|^p\,w_j\right)^{\frac{1}{p}},\quad p\geq 1,\,w_j>0,\,\forall
   j\in\mathbb N$

$\|f\|_{\infty} = \underset{j\in\mathbb N}{\sup} |f_j|$

I want to prove that the spaces $l_1$ and $l_{\infty}$ are not strictly convex.

For $l_1$, I'm trying to find two different sequences $\left\{a_n\right\}_{n=1}^{\infty}$ and $\left\{b_n\right\}_{n=1}^{\infty}$ such that

$$\displaystyle\sum_{j=0}^{\infty} |a_n| + \displaystyle\sum_{j=0}^{\infty} |b_n| = \displaystyle\sum_{j=0}^{\infty} |a_n+b_n|$$

Respectively, for $l_{\infty}$, I have to find two sequences such that

$$\underset{j\in\mathbb N}{\sup} |a_j|+ \underset{j\in\mathbb N}{\sup} |b_j| = \underset{j\in\mathbb N}{\sup} |a_j+b_j|$$

are these argument correct?

If they are, then I think that I can prove it for $l_{\infty}$. One just has to take the sequences:

$$\left\{a_n\right\}_{n=1}^{\infty}=\left\{\frac{1}{2^{n}}\right\}_{n=1}^{\infty} $$
$$\left\{b_n\right\}_{n=1}^{\infty}=\left\{\frac{1}{2n}\right\}_{n=1}^{\infty}$$

Since then we have:

$$\underset{j\in\mathbb N}{\sup} |a_j|+ \underset{j\in\mathbb N}{\sup} |b_j| = \frac{1}{2} + \frac{1}{2} = 1 = \underset{j\in\mathbb N}{\sup} |a_j+b_j|$$

答:
I think

$$\displaystyle\sum_{j=0}^{\infty} |a_n| + \displaystyle\sum_{j=0}^{\infty} |b_n| = \displaystyle\sum_{j=0}^{\infty} |a_n+b_n|$$

is not correct for all situations.

Consider the following
$$\left\{a_n\right\}_{n=0}^{\infty}=\left\{-\frac{1}{2^{n}}\right\}_{n=0}^{\infty} $$
$$\left\{b_n\right\}_{n=0}^{\infty}=\left\{\frac{1}{2^{n+1}}\right\}_{n=0}^{\infty}$$

$$\sum\limits_{j=0}^{\infty}|-\frac{1}{2^n}|+\sum\limits_{j=0}^{\infty}|\frac{1}{2^{n+1}}| > \sum\limits_{j=0}^{\infty}|-\frac{1}{2^n}+\frac{1}{2^{n+1}}|=\sum\limits_{j=0}^{\infty}|-\frac{1}{2^{n+1}}|=\sum\limits_{j=0}^{\infty}|\frac{1}{2^{n+1}}|$$


答: The provided solution first employs the angle addition identity $$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta,$$ with the choices $$\alpha = \arcsin \frac{3}{5}, \quad \beta = \arctan \frac{7}{24}.$$  That is to say, $$\sin \alpha = \frac{3}{5}, \quad \tan \beta = \frac{7}{24}.$$  In order to proceed with the evaluation, we now recall $$\sin^2 \theta + \cos^2 \theta = 1, \\ \tan^2 \theta + 1 = \sec^2 \theta.$$  So $$\frac{9}{25} = \sin^2 \alpha = 1 - \cos^2 \alpha,$$ hence $$\cos \alpha = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}.$$  Similarly, $$\frac{49}{24^2} = \tan^2 \beta = \sec^2 \beta - 1,$$ or $$\cos \beta = \frac{24}{25}.$$  It follows that $\sin \beta = \cos \beta \tan \beta = \frac{7}{25}$ and $$\cos(\alpha + \beta) = \frac{4}{5} \frac{24}{25} - \frac{3}{5} \frac{7}{25} = \frac{3}{5}.$$  This solution assumes that $-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2}$ and  $-\frac{\pi}{2} < \arctan x < \frac{\pi}{2}$.

Title: How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$?

Body: The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both side and drawing triangles. However, the worked solutions does a simplification from LHS to RHS instead, namely $$\cos\arcsin\left(\frac{3}{5}\right)\cos\arctan\left(\frac{7}{24}\right)-\sin\arcsin\left(\frac{3}{5}\right)\sin\arctan\left(\frac{7}{24}\right)=\frac{3}{5}$$which I don't understand. Can someone please explain?