问:
Spivak's "Calculus", Chapter 22 Problem 26 part d
The problem investigates when the infinite power tower $a^{a^{a^a\ldots}}$ converges for $a > 0$ by considering the convergence of the sequence $b_{n+1} = a^{b_n}$ and calls the limit when it exists $b$.

Part c asks us to show using problem 25 that if $b$ exists then $e^{-1} \leq b$. Problem 25 proves the general result that in a sequence formed by iteratively applying a function $f$ with continuous first derivative (in this case $f(x) = a^x$), that $|f'(b)| \leq 1$.

However applying this to the problem just yields $|ln(a) a^b| \leq 1$. In the previous parts we have already shown that the inequality holds when $1 \leq a \leq e^{1/e}$, so I only need to apply this inequality for the case $0 < a < 1$. However this doesn't show that $b \geq e^{-1}$ or that $b \leq e$ for $0 < a < 1$.

Edit: I realise now that since $b$ is a fixed point of the recurrence, we have $|ln(a)b| \leq 1$, but I still don't see how this leads to the desired inequality.

答:
I've figured it out now. We just continue from
$$ |b \ln(a)| \leq 1$$
$$ |\ln(a^b)| \leq 1 $$
$$ |\ln b| \leq 1$$
$$ e^{-1} \leq b \leq e $$


问:
Solve the following system of equations
Solve the following system of equations:
$\left\{\begin{matrix}
x^3(1-x)+y^3(1-y)=12xy+18\\
\left | 3x-2y+10 \right |+\left | 2x-3y \right |=10
\end{matrix}\right.$</p>

答:
Perhaps asking Mathematica(WolframAlpha gives the answer as well) to solve it:

```<code>Solve[{x^3 (1 - x) + y^3 (1 - y) == 12 x y + 18,
       Abs[3 x - 2 y + 10] + Abs[2 x - 3 y] == 10}, {x, y}, Reals]
</code>```

immediately gives:

$$\left\{\left\{x\to -\sqrt{3},y\to \sqrt{3}\right\}\right\}$$

And there is this nice plot of the two curves:

P.S. I will probably(if I find a reason to) add an analytic answer later.

ContentLicense: CC BY-SA 3.0

Title: Solve the following system of equations

Body:
Perhaps asking Mathematica(WolframAlpha gives the answer as well) to solve it:

Solve[{x^3 (1 - x) + y^3 (1 - y) == 12 x y + 18,
       Abs[3 x - 2 y + 10] + Abs[2 x - 3 y] == 10}, {x, y}, Reals]

immediately gives:

$$\left\{\left\{x\to -\sqrt{3},y\to \sqrt{3}\right\}\right\}$$

And there is this nice plot of the two curves:

P.S. I will probably(if I find a reason to) add an analytic answer later.


问:
What does getting a number on 2 dice means?
In chapter 13 (beginning of probability theory) of the book "challenge and thrill of pre college mathematics", after theorem 3 (i.e. for any 2 events E and F, P(E union F)=P(E)+P(F)+P(E intersection F)), there is an illustration in which "P(throwing either an even number or a sum of 2,3,4,with two dice)" has to be calculated.

The illustration is :
E=event{2,4,6},
F=event{2,3, 4}.
Then E U F={2,3,4,6},P(E U F) =11/36.
P(E) =9/36, P(F) =6/36, P(EF) =P(2, 4) =4/36.
P(E) +P(F) -P(EF) =(9+6-4) /36=11/36=P(E U F).

My questions are,
1.) what does it mean "to throw an even number on 2 dice"?(should the sum of 2 numbers on 2 dice be even?)

2.) "sum of 2,3,4" means all possible sums from 2,3,4(i.e. 2+3=5,3+3=6,3+4=7,4+4=8,3+3=2+3+4=9 etc.),or sum of numbers on the dice should be 2,3 or 4?

答:
This is what your textbook means:

Let's throw 2 fair dice:

Event $E$: the sum of the "2 dice's face up" is even

Event $F$: the sum "2 dice's face up" is 2,3 or 4

Find

$$\mathbb{P}[E \cup F]=\mathbb{P}[E ]+\mathbb{P}[ F]-\mathbb{P}[E \cap F]=\frac{1}{2}+\frac{6}{36}-\frac{4}{36}=\frac{5}{9}$$

ContentLicense: CC BY-SA 4.0


问:
Distribution of product of normal random variables
Suppose $Z$ is a $p\times n$ matrix and $v$ is a $n\times 1$ vector. Both have entries that are independent, standard normals (in particular $Z$ and $v$ themselves are independent). What is the distribution of the vector $w=Zv$?

I had read that it has the distribution of $$w \sim \chi_n \cdot \chi_p \cdot U_p$$ where $\chi^2_k$ is a chi-squared variable with $k$ degrees of freedom and $U$ is a vector uniform on the surface of a $p-1$ dimensional sphere. I wasn't sure how this was obtained.

答:
First, the distribution of $v \in N(0, I_n)$ is spherically symmetric. For suppose $A^T A = I_n$. Then $(Av)_i = \sum_j A_{ij} v_j$ is a sum of independent normal random variables, so it's normal with variance $\sum_j A_{ij}^2 = 1$. It follows that $v/|v| \sim U_n$. Moreover, $|v| \sim \chi_n$ by definition, so $v \sim \chi_n \cdot U_n$. For essentially the same reason, $w/|w| \sim U_p$, so all that remains is to show $|w| \sim \chi_n \cdot \chi_p$.

Let $Z^i$ be the $i$th row of $Z$. For any particular value of $v$, we may pick an orthogonal matrix $A$ depending on $v$ such that $Av = |v|e_1$. Let $Y^i := AZ^i \sim N(0, I_n)$, which remains spherically symmetric for the same reasons as before. Thus

\begin{align*}
w_i &= \langle Z^i, v\rangle = \langle AZ^i, Av\rangle \\
&= \langle Y^i, |v|e_1\rangle \\
&= |v|Y_i^1 \sim \chi_n \cdot N(0, 1).
\end{align*}

Hence
\begin{align*}
|w| &= \left(\sum_i w_i^2\right)^{1/2} \\
&= \left(\sum_i |v|^2 (Y_i^1)^2\right)^{1/2} \\
&= |v| \left(\sum_i (Y_i^1)^2\right)^{1/2} \sim \chi_n \cdot \chi_p.
\end{align*}


问:
Range of numbers not divisible by $2$ or $3$ or $5$
How many numbers lie in the range of $1331$ and $3113$ (Endpoints Inclusive) such that they are not divisible by $2$ or $3$ or $5$?

I came across this Number System question and was not able to figure out the way to get the answer. At first I thought to separate out the numbers which are divisible by the LCM of $2$, $3$ & $5$ but this will still leave some numbers which will be exclusively divisible by any one of $2$ or $3$ or $5$.
Another method I found out to solve these type of questions is by using the Euler's Totient method but I am not aware of this.
So can someone please explain a general way to solve these types of question?

General question can be : How many numbers lie in the range of $A$ and $B$ such that they are not divisible by $x$ or $y$ or $z$ and so on?

答:
By nos I mean numbers.

For total nos between $1331$ and $3113  = 3113-1331 -1 = 1781$ (excluding end nos)

Of $1781$ nos , the nos divisible by $2  = 1781 +1 = 891$ (as the starting no is even so no of even nos is greater than no of odd nos  )

now nos which aren't divisible by $2 = 890$.

Of $890$ nos, nos which are divisible by $5$ but don't end in $0 =  1781/10= 178$
(as the nos ending in $0$ are already eliminated)

(for every $10$ nos there is one no divisible by $5$ and not ending in $0$)

so now, nos not divisible $2$ or $5 =  890 -178 = 712 $

now for  nos divisible by $3$,  take the nos from $1338$  to $3108$ which has $1770$  nos
for every $30$ nos , there are $10$ nos divisible by $3$ of which $5$ are also divisible by $2$  and one is divisible by $5$ but not ends in $0$.  So for every $30$ nos ,  $4$ nos are divisible by $3$ but not by $2$ or $5$

so of $1770$ nos there are $59$ sets of $30$ nos . hence $4 * 59 = 236 $ nos are divisible by $3$ but not by $2$ or $5$.

so nos not divisible by $2$ or $3$ or $5$ in given range  $712 - 236 = 476$

but we haven't taken  $3$ multiples in range $1331$ to $1338$ and $3108$ to $3113$

nos multiple of $3$ in this range is $= 1334$  (even and hence eliminated) , $1335$ (divisible by $5$ and ends in $5$) ,  $3111$(not eliminated)
so $476 - 1 = 475$ .

but i have excluded end nos  so adding $2$

$ = 475 + 2 = 477 $.


问:
Fast way to put boundaries of integration on the 3-dimensional integral

Calculate the triple integral
$$
\iiint_{G}(x+z) \,d x d y d z, \text { domain } G\text { is bounded by planes }\\ x+y=1, x-y=1
,x+z=1, z=0, x=0
$$

My question is, what is the easiest and most efficient way to put boundaries on such integral? For integral, that I provided, I built a 3D representation, but from it it very difficult to understand, how $x,y,z$ are changed. And especially it hard to do with only 2D representation.

答:
In an exam, we need minimal visualization of the domain. I would immediately try to apply Fubini, so the question is how to "split the space" in a simplest way.

Observe that some equations are involving only $x,y$, so we will consider them first:
$$
x+y =1\ ,\ x-y=1\ ,\ x=0\ .
$$
They are delimiting in the $xOy$-plane a (solid) triangle $\Delta$ with vertices $(0,1)$, $(0,-1)$, and $(1,1)$. So $G$ contains only points $(x,y,z)$ with $(x,y)$ in $\Delta$. Now we also consider the delimiting planes "in $z$". One of them is $z=0$, so we either take points in the half-space $z\ge 0$, or points in the half-space $z\le 0$. To see further, consider also the plane $x+z=1$. It passes through points $(0,y,1)$ (and $(1,y,0)$), so we are in the "upper" half-space $z\ge 0$.

So it remains to integrate.

For each $(x,y)\in\Delta$, the possible values of $z$ are those between $z=0$ and $z=(1-x)$, so we integrate on the segment $[0,1-x]$. The integral $J$ to be computed is thus:
$$
\begin{aligned}
J
&=
\iiint_G(x+z)\; dx\; dy\; dz\\
&=
\iint_{(x,y)\in\Delta}\; dx\; dy\int_{[0,1-x]}(x+z)\; dz\\
&=
\iint_\Delta\; dx\; dy\;\left[xz+\frac 12z^2\right]_{[0,1-x]}\\
&=
\iint_\Delta\; dx\; dy\;\cdot\; \left(x(1-x)+\frac 12(1-x)^2\right)\\
&=
\int_{x\in[0,1]}dx\;\cdot\; \left(x(1-x)+\frac 12(1-x)^2\right)
\int_{y\in[-(1-x),\ (1-x)]}\; dy
\\
&=
\int_{x\in[0,1]}dx\;\cdot\; \left(x(1-x)+\frac 12(1-x)^2\right)
\cdot 2(1-x)
\\
&=
\int_{x\in[0,1]}dx\;\cdot\; \left((1-x)x+\frac 12x^2\right)
\cdot 2x
\\
&=
\int_0^1(2x^2(1-x)+x^3)\; dx
=
\int_0^1(2x^2-x^3)\; dx
\\
&=\frac 23-\frac 14=\frac 1{12}(8-3)=\frac 5{12}\ .
\end{aligned}
$$

We can alternatively compute the integral,
after we visualize it as the pyramid with vertex $(1,0,0)\in Ox$ and with basis the rectangle with vertices  in the $yOz$-plane $(0,1,1)$, $(0,-1,1)$, $(0,1,0)$, $(0,-1,0)$.
So it is $\{0\}\times[-1,1]\times[0,1]$.

We let $x$ run from $0$ to $1$, then the corresponding $x$-section in $G$ is the rectangle $G_x=\{x\}\times[-1+x,1-x]\times[0,1-x]$. So

$$
\begin{aligned}
J
&=
\iiint_G(x+z)\; dx\; dy\; dz\\
&=
\int_{x\in[0,1]}dx\int_{y\in[-(1-x),\ (1+x)]}dy\int_{z\in[0,1-x]}(x+z)\; dz\ .
\end{aligned}
$$
This is the same, but maybe the visualization is simpler.


问:
Summation of function subtraction with a limit as $\epsilon \rightarrow 0$
I got stuck in a problem in the middle of my calculations of integrals and sums.

$$\lim_{\epsilon \rightarrow 0} \sum_{n=1}^\infty f(n-\epsilon)-f(n+\epsilon)=0$$

where $f$ is continuous on all of the values in the summation

I am not sure for the general case of any $f$, but I needed verification for $f(n) = -\frac{1}{n}$ will be gratefull for an explanation for the any function

First I tried to verify for no sum and here what I got;

$$\forall_{n \in \mathbb{R}} \;\lim_{\epsilon \rightarrow 0} \frac{1}{n+\epsilon} - \frac{1}{n-\epsilon} =
\frac{1}{n} - \frac{1}{n} = 0$$

$$\lim_{\epsilon \rightarrow 0} \sum_{n=1}^\infty \frac{1}{n+\epsilon} - \frac{1}{n - \epsilon}= \lim_{\epsilon \rightarrow 0} \sum_{n=1}^\infty\frac{-2\epsilon}{(n+\epsilon)(n-\epsilon)} = 0$$

Is this correct? And what about any other function

答:
For this $f(x)=1/x$ you have $$\left|\sum_{n=1}^\infty f(n-\epsilon)-f(n+\epsilon)\right|=2\epsilon\left|\sum_{n=1}^\infty \frac{1}{n^2-\epsilon^2}\right|$$
and one way to finish: for any $\epsilon<1/2$ the sum in the modulus bars is positive and bounded uniformly in $\epsilon$ by $\sum_{n=1}^\infty \frac1{n^2-1/2}$ which is convergent, so by squeeze rule the limit is $0$.

The result is not true in general without something about $f$ becoming smaller as $n\to\infty$. For example if $f(x)=x$ then
$$ f(n-\epsilon) - f(n+\epsilon) = n-\epsilon - n - \epsilon = -2\epsilon$$
which means $\sum_{n=1}^\infty f(n-\epsilon)-f(n+\epsilon)=-\infty$ for all $\epsilon>0$.

It's e.g. worse if $f(x)=x^2$, $f(n-\epsilon)-f(n+\epsilon) = -4\epsilon n$, and its better if $f(x)=\sqrt x$, $f(n-\epsilon)-f(n+\epsilon)=\sqrt{n-\epsilon}-\sqrt{n+\epsilon}=\frac{-2\epsilon}{\sqrt{n-\epsilon}+\sqrt{n+\epsilon}}\sim \frac{-2\epsilon}{\sqrt n}$, but not good enough.


问:
An inequality in the proof of Lebesgue Dominated Convergence Theorem in Royden's book.
In the proof of the Lebesgue Dominated Convergence Theorem, Royden used the following equality,

$$\liminf\int_X[g - f_n] \, d\mu = \int_X g \, d\mu - \limsup \int_X f_n \, d\mu$$

How do you justify this statement?
This seems very strange, isn't this equality equivalent to $\liminf \int_X f_n = \limsup \int_X f_n$?

Attached is the original texts. (p.378, Real Analysis by Royden 4th edition)

答: If $a=\int gd\mu$ and $a_n=\int f_nd \mu$ then $\lim \inf \, [a-a_n]=a-\lim \inf (-a_n)=a-\lim\sup a_n$.


问:
$\Delta_{0}$ formulas
I am working through the Jech Set Theory book, and at the moment I am stuck at his definition of the $\Delta_{0}$ formulas:

A formula of set theory is a $\Delta_{0}$ formula if:

(i) it has no quantifiers, or

(ii) it is $\varphi \wedge \psi$, $\varphi \vee \psi$, $\neg \varphi$, $\varphi\rightarrow\psi$ or $\varphi\leftrightarrow\psi$, with $\varphi$ and $\psi$ $\Delta_{0}$ formulas, or

(iii) it is $(\exists x\in y)\varphi$ or $(\forall x\in y)\varphi$ where $\varphi$ a $\Delta_{0}$ formula.

I'm not really sure if I understand right $\Delta_{0}$ formula is.

In class we wrote the same, only in the first point $\varphi$ is $\Delta_{0}$ if it is atomic: $x\in y$, $x=y$.

Is a $\Delta_{0}$ formula hence only the combination of $x\in y$ or $x=y$ with (ii)?

Afterwards we have in the notes: $x=\mathscr{P}(y)$ is not $\Delta_{0}$.

Is that because $x=\mathscr{P}(y)\leftrightarrow (\forall u\in x)(u\subseteq y)\wedge (\forall u)(u\subseteq y\rightarrow u\in x)$ and then the part $\forall u$ is not writable like (i)?

Also some other papers about this definition would be helpful! Thanks, Luca

答:
The definitions are the same. This is because omitting quantifiers effectively reduces the language to propositional calculus.

Namely, given atomic formulae, if we only apply (ii), we get exactly the quantifier-free formulae.

So the (i) of Jech basically amounts to what one can achieve with the (i) and (ii) from your class, without using (iii). As such, it is not minimal, but it does avoid tedious applications of (ii), when the actual interest in $\Delta_0$ formulae lies in the use of the restricted quantifications.

Regarding $x = \mathscr P(y)$, you are correct. More intuitively, since the definition of $\mathscr P(y)$ requires an unrestricted universal quantifier, it cannot be a $\Delta_0$ formula.

Unfortunately, Jech is the only source I have actually studied on the subject, so I can't help you with other references. I hope that the above clarified the definition a bit for you, though.

Title: $\Delta_{0}$ formulas


问:
Is there a function $f\in L_{1}[0,1]$ but $ \frac{d}{dx}\Big(\int_{0}^{x}\frac{f(s)}{(x-s)^{\alpha}}ds\Big)\not\in L_{1}[0,1]? $
Let $\alpha\in (0,1).$ For a given $f\in L_{1}[0,1]$ consider
$$
\phi(x)=\int_{0}^{x}\frac{f(s)}{(x-s)^{\alpha}}ds, \,\,\,\,\,x\in [0,1].
$$
It is clear that if $\phi\in AC[0,1]$ then $\phi$ is differentiable a.e. and $\phi'\in L_{1}[0,1].$
However, there is a function $f\in L_{1}[0,1]$ such that  $\phi\not\in AC[0,1].$ I am interested in the following question. Is there a function $f:[0,1]\rightarrow R$ such that $f\in L_{1}[0,1]$ but
$\phi'\not\in L_{1}[0,1]?$

答:
$$
\int_{0}^{x}\frac{f(s)}{(x-s)^{\alpha}}ds=\Gamma(1-\alpha)\frac{d^{(\alpha-1)}}{dx^{(\alpha-1)}}f(x)$$
$\Gamma$ is the Gamma function.

$\frac{d^\nu}{dx^\nu}$ denotes the fractionnal derivative of non-integer degree $\nu$ or the fractionnal antiderivative of non-integer degree $-\nu$.

Thus this question is related to fractional calculus : https://en.wikipedia.org/wiki/Fractional_calculus

There is an extensive literature about fractional calculus. At lower level, an article for the general public : https://fr.scribd.com/doc/14686539/The-Fractional-Derivation-La-derivation-fractionnaire

Also you can find a wide documentation about the above intgral referenced as Riemann-Liouville integral and/or Riemann-Liouville's transform.


问:
Question about the proof of the fact that subsequential limits are closed

Theorem: For any sequence $\{p_n\}$ in a metric space $X$, the set $E$ of all subsequential limits of $\{p_n\}$ is closed relative to $X$.

I am not sure I follow the(?) proof of this theorem in my book so I'll just rewrite it in my words.

Let $q$ be a limit point of $E$. Then if $q \in E$, we are done. $\color{red}{\text{To that end we want to show some $\{p_{n_i}\}$ converges to $q$}}$.

Choose $n_1 \in \mathbb N$ s.t. $p_{n_1}$ is an element of $\{p_n\}$ with $p_{n_1} \ne q.$ Set $\delta = d(p_{n_1}, q)$.

Assume we have $p_{n_1}, \ldots p_{n_i}$ with $n_1 < \ldots n_i$ s.t. $p_{n_k} \ne q$ and $d(p_{n_k}, q) < \frac{\delta}{k}$ for any $k \le i.$

Since $q$ is a limit point, every neighborhood of $q$ contains a point $q \ne x \in E$ meaning $x \in N_{\text{any radius}}(q)$. In particular, $x \in N_{\frac{\delta}{2(i+1)}}(q)$ meaning $d(x, q) < \frac{\delta}{2(i + 1)}$. Since $x$ is a limit of some subsequence of $\{p_n\}$, past some $N \in\mathbb N$, for all $\epsilon > 0$, each element of that subsequence is closer than a distance of $\epsilon$ to $x$. In particular, that is true for $\epsilon = \frac{\delta}{2(i + 1)}$. So, for $n_{i + 1} > n_i > N$, we have $d(p_{n_{i + 1}}, x) < \frac{\delta}{2(i + 1)}$. Then $d(p_{n_{i + 1}}, q) \le d(p_{n_{i + 1}}, x) + d(x, q) < 2\frac{\delta}{2(i + 1)} = \frac{\delta}{i + 1}$. Now we have a subsequence $\{p_{n_i}\}$ with $d(p_{n_i}, q) < \frac{\delta}{i}$ for any $i \ge 2.$ From here, it's not hard to show $p_{n_i} \to p.$

I have two questions. Is my understanding of the proof correct? Also, regarding the part in red of the proof above, why does $p_{n_i} \to q$ imply $q \in E$?

答:
Your proof is essentially correct, but I think some points could be improved. Here is a suggestion:

Let $n^{(q)}_1 = 1$ and $\delta = d(p_1,q) + 1 > 0$. [It is irrelevant whether or not $p_1 \ne q$.]

Assume we have $p_{n^{(q)}_1}, \ldots p_{n^{(q)}_i}$ with $n^{(q)}_1 < \ldots < n^{(q)}_i$ such that $d(p_{n^{(q)}_k}, q) < \frac{\delta}{k}$ for any $k \le i.$ [It is irrelevant whether or not $p_{n^{(q)}_k} \ne q$.] Since $q$ is a limit point, every neighborhood of $q$ contains a point $x \in E \setminus \{q\}$. [Again it is irrelevant whether or not $x \ne q$.] In particular, we find $x \in E$ such that $d(x, q) < \epsilon := \frac{\delta}{2(i + 1)}$. Since $x$ is the limit of some subsequence $(p_{n^{(x)}_r})$ of $(p_n)$, for some $R \in \mathbb N$ each element $p_{n^{(x)}_r}$ of that subsequence with $r \ge R$ is closer than $\epsilon$ to $x$. Pick $r(i) \ge R$ such that $n^{(x)}_{r(i)} > n^{(q)}_i$ and let $n^{(q)}_{i+1} = n^{(x)}_{r(i)}$. Then we have $d(p_{n^{(q)}_{i + 1}}, x) < \frac{\delta}{2(i + 1)}$ and $d(p^{(q)}_{n_{i + 1}}, q) \le d(p^{(q)}_{n_{i + 1}}, x) + d(x, q) < 2\frac{\delta}{2(i + 1)} = \frac{\delta}{i + 1}$.

Now we have a subsequence $(p_{n^{(q)}_i})$ with $d(p_{n^{(q)}_i}, q) < \frac{\delta}{i}$ for all $i$. Therefore $p_{n^{(q)}_i} \to q$. Hence by definition $q \in E$.


答:
I think the authors are subtly trying to point out that the probability of throwing two dice both even is the same as throwing two dice that then sum to $2,4$ or $6$, which is easy to see by any mapping of:

$$22,24,26,42,44,46,62,64,66$$

onto $$11,13,22,31,15,24,33,42,51$$

such as $(x, y)\to (x+\min(x,y), y+\min(x,y))$.

Title: What does getting a number on 2 dice means?

Body:
In chapter 13 (beginning of probability theory) of the book "challenge and thrill of pre college mathematics", after theorem 3 (i.e. for any 2 events E and F, P(E union F)=P(E)+P(F)+P(E intersection F)), there is an illustration in which "P(throwing either an even number or a sum of 2,3,4,with two dice)" has to be calculated.

The illustration is :
E=event{2,4,6},
F=event{2,3, 4}.
Then E U F={2,3,4,6},P(E U F) =11/36.
P(E) =9/36, P(F) =6/36, P(EF) =P(2, 4) =4/36.
P(E) +P(F) -P(EF) =(9+6-4) /36=11/36=P(E U F).

My questions are,
1.) what does it mean "to throw an even number on 2 dice"?(should the sum of 2 numbers on 2 dice be even?)

2.) "sum of 2,3,4" means all possible sums from 2,3,4(i.e. 2+3=5,3+3=6,3+4=7,4+4=8,3+3=2+3+4=9 etc.),or sum of numbers on the dice should be 2,3 or 4?


问:
Show that if the matrix A has the form $\begin{pmatrix}a &b\\ b& -a\end{pmatrix}$, so $ a^2+b^2= 1 $
I need to prove that if matrix A has the form $\begin{pmatrix}a &b\\ -b& a\end{pmatrix}$, or    $\begin{pmatrix}a &b\\ b& -a\end{pmatrix}$

so $ a^2+b^2= 1 $

I picked a matrix $\begin{pmatrix}a &b\\ c& d\end{pmatrix}$, and I proved for this one.Then I changed $c = b$ and $d = -a$, but I'm not sure if this is correct. I'm not getting the correct answer, I didn't prove it.

答:
Let $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}.$ Then we have: $A$ is orthogonal if $AA^T=I$. Hence

$$AA^T=I \iff a^2+b^2=c^2+d^2=1, ac+bd=0.$$


问:
In Calculus, how can a function have several different, yet equal, derivatives?
I've been pondering this question all night as I work through some problems, and after a very thorough search, I haven't found anything completely related to my question. I guess i'm also curious how some derivatives are simplified as well, because in some cases I just can't see the breakdown. Here is an example:

$f(x) = \dfrac{x^2-6x+12}{x-4}$ is the function I was differentiating. Here is what I got:

$f '(x) = \dfrac{x^2-8x+12}{(x-4)^2}$ which checks using desmos graphing utility.

Now, when I checked my textbook(and Symbolab) they got:

$f '(x) = 1 - \dfrac{4}{(x-4)^2}$ which also checks on desmos.

To me, these derivatives look nothing alike, so how can they both be the equal to the derivative of the original function? Both methods used the quotient rule, yet yield very different results. Is one of these "better" than the other? I know that it is easier to find critical numbers with a more simplified derivative, but IMO the derivative I found seems easier to set equal to zero than the derivative found in my book.I also wasn't able to figure out how the second derivative was simplified, so I stuck with mine.

I'm obviously new to Calculus and i'm trying to understand the nuances of derivatives. When I ask most math people, including some professors, they just say "that's how derivatives are" but for me, that's not an acceptable answer. If someone can help me understand this, I would appreciate it.

答:
There is perhaps a very different approach to solve this question

Here lets suppose your function =y=x^2-6x+12/x-4
take log both sides

log y=log(x^2-6x+12/x-4)
apply the property of log

logy=log(x^2-6x+12)-log(x-4)
differentiate both sides with respect to x
1/y dy/dx=2x-6/x^2-6x+12-(1/x-4)

so dy/dx=y x 2x-6/x^2-6x+12-(1/x-4)
hence your final derivative is

(x^2-6x+12/x-4) .{2x-6/x^2-6x+12-(1/x-4)}


问:
Function spaces $L^p$ and $C^k$ inclusion?
Are there any inclusions in terms of set-theory (or estimates for corresponding norms) for $L^p$ and $C^k$ function spaces?

So, is there any $p, k$ so that $L^p  \subset  C^k$ or $C^k  \subset  L^p$ ?

答:
No

First, you have the problem that $L^p$ is defined as a set of equivalence classes, which $C^{k}$ is not (as @Cameron Williams) pointed out.

Moreover, $f(x)=x$ is in $C^k$ for any $k\in\mathbb{N}$, but not in $L^p(\mathbb{R})$.

Finally, $f(x)=1\{x=0\}$ is in $L^p(\mathbb{R})$, but not $C^k$.


问:
Solve $\int \frac{\arctan(x)}{x^3} dx$ without partial fractions
So I'm trying to solve for the indefinite integral $$\int \frac{\arctan(x)}{x^3}\, dx$$

Here is what I tried:
Use integration by parts:
$$\int v\,du = vu - \int u\,dv$$
Let $v = \arctan(x)\\$
Let $dv = \frac{dx}{1 + x^2}.\\$
Let $du = x^3\, dx.\\$
Let $u = -\frac{1}{2x^2}.\\$

$$ -\frac{\arctan(x)}{2x^2} + \frac{1}{2}\int \frac{1}{x^2} \cdot \frac{1}{1 + x^2} \,dx$$

Now I'm sort of stuck, and since we haven't been taught partial fractions, I'm not sure how to proceed. Would someone mind explaining how to approach this? Thanks in advance!

Body:
Hint. One may just write
$$
\frac{1}{x^2(1 + x^2)}=\frac{(1+x^2)-x^2}{x^2(1 + x^2)}=\frac{1}{x^2}-\frac{1}{1 + x^2}.
$$


问:
Directly proportional to which?
This might be a simple question, but the exact wording of this statement is important. If I know that the relationship between two variables is this:

$$\gamma \propto  \frac{1}{\sqrt{k}}$$

Is the appropriate way to write that in a sentence: 'y is inversely proportional to k' or should it be 'y is inversely proportional to the square root of k'.

If somebody could explain why whichever is correct is correct, that would be great.

答: I would take "$y$ is inversely proportional to $k$" to mean $y=c/k$ for some constant $c$, so for your situation I'd definitely say "$y$ is inversely proportional to the square root of $k$."</p>

Title: Directly proportional to which?


问:
Normal distributed $\log x$
Assuming $\log x$ is normally distributed. How do I get the distribution of $x$? Also how can I get the standard deviation of $x$ assuming I know some asymmetric error on $x$ like e.g. Table 1 in https://iopscience.iop.org/article/10.3847/1538-4357/aad90c/pdf

答:
If

$$Y=\log X\sim N(\mu;\sigma^2)$$

Then

$$X\sim \text{Log-Normal}$$

to find the pdf of $X$ simply use the fundamental transformation theorem (EDIT) that is the following

$$f_X(x)=f_Y(g^{-1}(x))|\frac{d}{dx}g^ {-1 }(y)|$$

Where $Y=g^{-1}(X)=\log X$

Thus

$$f_X(x)=\frac{1}{\sigma x\sqrt{2\pi}}e^{-(\log x-\mu)^2/(2\sigma^2)}$$

$x >0$

Here read more on Log-normal distribution

Title: Normal distributed $\log x$


问:
given continuous function $f:\mathbb{R}^n\to\mathbb{R}$ prove $\exists\delta>0$ s.t. $x\in B(c;\delta)$ $f(x)>m$
So for a homework assignment for a real analysis class I'm given the following:

Given a continuous function $f:\mathbb{R}^n\to\mathbb{R}$ and a point $c\in\mathbb{R}^n$.

Then assume $0<m<f(c)$ with $m\in\mathbb{R}$. Show $\exists\delta>0$ such that for all $x\in B(c;\delta)$ the following holds: $f(x)>m$.

I worked out the following: Since $f$ is continuous, there must be a $\delta$ s.t $|x-c|<\delta$ for every $\epsilon$ s.t. $|f(x)-f(c)|<\epsilon$, so, using the triangle inequality, I see that: $\epsilon>|f(x)-f(c)|\geq|f(x)|-|f(c)|>|f(x)|-m>0$, meaning that $|f(x)|>m$ should hold then. I think it's alright what I've done, yet I can't shake that I'm still missing something, mainly because of the absolute value of $f(x)$. I would gladly take any help that I can get, thank you in advance.

答: If $|f(x) - a| < \varepsilon$ then $f(x) > a - \varepsilon$ so you can choose $\varepsilon = f(c) - m$ and then continue your proof


问:
Understanding the textbook of Humphreys Lie algebras
I'm studying Humphreys Introduction to Lie Algebras and Representation Theory.
I do not understand the second paragraph in page 26.

Given a representation $\phi : L \to \mathfrak gl (V)$, the associative algebra (with 1) generated by $\phi(L)$ in End$V$ leaves invariant precisely the same subspaces as $L$. Therefore, all the usual results (e.g. Jordan-Holder Theorem) for modules over associative rings hold for $L$ as well.

My Question is,

How the associative algebra generated by $\phi(L)$ leaves L invariant?

From 1, how can we deduce that all the results for modules over associative rings hold for L as well?

Any help would be appreciated. Thanks.

答:
Let $A$ be that associative algebra. It is, as a set, the set of all linear combinations of $\operatorname{Id}$ (which leave invariant every subspace of $V$) and endomorphisms of $V$ of the type $\phi(X)$, with $X\in L$. So, if $W\subset V$ is a subspace which is invariant with respect to the action of $L$ on $V$, then it is invariant with respect to the action of any element of $A$.

Since $A$ is an associative algebra, every theorem about representations of associative algebras applies. And the invariant subspaces are the same in both cases. So, studying theorems about invariant spaces with respect to the action of $L$ is the same thing as studying invariant subspaces with respect to the action of $A$, to which you can apply those theorems.


答: If $Y:=\ln X$ has CDF $F_Y$, $X$ has CDF $F_X(x):=P(X\le x)=P(Y\le\ln x)=F_Y(\ln x)$, so its PDF is $f_X(x):=\frac{d}{dx}F_X(x)=\tfrac1xF_Y^\prime(\ln X)$ by the chain rule. In terms of $Y$'s PDF $f_Y:=F_Y^\prime$, $f_X(x):=\tfrac1xf_Y(\ln x)$. This gives the distribution discussed here.

Title: Normal distributed $\log x$

Body: Assuming $\log x$ is normally distributed. How do I get the distribution of $x$? Also how can I get the standard deviation of $x$ assuming I know some asymmetric error on $x$ like e.g. Table 1 in https://iopscience.iop.org/article/10.3847/1538-4357/aad90c/pdf


问:
Winning probability to choose a chip of a box.
The Player got 10 blue and 10 red chips, and he was offered a game where he should distribute those chips into 2 similar boxes. (All chips should be placed in one of the boxes.) Then, another person randomly chooses a box and then takes out one random chip from this box (unless the box is empty). The player wins if a red chip is drawn and lose otherwise. What's the maximum probability of the player's winning outcome assuming that he distributes chips the right way?

I have been trying this problem for the past 1 week and not able to proceed further.
The best I have reached is:

P(getting R) = P(getting Red | in chosen Box 1)*P(choosing Box 1) + P(getting Red | in chosen Box 2)*P(choosing Box 2)

But not able to think of a method to maximise the probability.
Can anyone please help me with this?

答:
It is more or less clear, that the distribution
$$
\color{red}{1} +
\color{blue}{0}
\qquad\text{ and }\qquad
\color{red}{9} +
\color{blue}{10}
$$
in the first box, and respectively in the second box is optimal.
(Or switch boxes.)

The probability of winning is then
$$
\frac 12\cdot 1+\frac 12\cdot \frac 9{19}=\frac{14}{19}\ .
$$
This is the maximal value among all possible ways to split the $20$ chips,
$$
\color{red}{p} +
\color{blue}{q}
\qquad\text{ and }\qquad
\color{red}{(10-p)} +
\color{blue}{(10-q)}
$$
where the probability is
$$
\frac 12\cdot \frac {p}{p+q}
+
\frac 12\cdot \frac {10-p}{20-p-q}\ , \qquad p+q\not\in\{0, 20\}\ .
$$

For the sake of completeness, i'll add a...

Proof: Here is a quick estimation.
If $p=0$, we obtain a probability $\le 1/2$, this is not optimal. So
we may and do assume by symmetry that $1\le p\le 5$.

The difference $\displaystyle
\frac 12\left(\frac {p}{p+q}
+
\frac {10-p}{20-p-q}\right)-\frac {14}{19}
$
has positive denominator, and the numerator is
$$
-5p^2 + 5p + 9pq + 14q^2 - 185 q
\ .
$$
As a function in $q$,
the part $9pq + 14q^2 - 185 q$ is zero in $q=0$, then
it is monotone decreasing on $J:=[0, (185-9p)/2]\supseteq[0, (185-45)/14/2]=[0,5]$, then it increases on an interval of the same length as $J$, "following" $J$, so the value in $10$ is zero or less. This shows that an optimal value is obtained for $q=0$. It remains to maximize $-5p^2+5p$ on $[1,5]$. Of course, $p=1$ gives the maximum.

$\square$


问:
Why is this map an identification?
Let us define $\Delta^n=\{(t_0,...,t_n) \in \mathbb{R}^{n+1} \mbox{ such that } t_i \geq 0 \mbox{ and } \sum_i t_i=1\}$ for $n \geq 0$. If $q(\mu_0,...,\mu_{n-1},t)=(t,(1-t)\mu_0,...,(1-t)\mu_{n-1})$ is a map $\Delta^{n-1}\times [0,1] \to \Delta^n$, why is it an identification?

Id est, why $A \subset \Delta^n$ is open if and only if $q^{-1}(A)$ is open?

答:
It's a map between compact Hausdorff spaces, so it's a closed continuous map, and in particular a quotient map (aka an identification map).

Short proof: if $A \subseteq \Delta^n$ is open, so is $q^{-1}[A]$ as $q$ is continuous ( I think that's easy enough to see), and if $q^{-1}[A]$ is open, $(\Delta^{n-1}\times [0,1]) \setminus q^{-1}[A]$ is closed, hence compact and so $q[(\Delta^{n-1}\times [0,1]) \setminus q^{-1}[A]]$ is compact (by continuity of $q$), hence closed and the complement of that set is precisely $A$.

Title: Why is this map an identification?


问:
Minimal residue making congruence system solvable (CRT)

A number when divided by $902$, $802$ and $702$ leaves remainder of $602$, $502$ and $402$ respectively. What would be the minimum remainder of when the number is divided by $2005$?

My attempt:

For the given number $n$, $n + 300$ is divisible by $902$, $802$ and $702$. Hence $n + 300$ can be written as $\mathrm{lcm}(902,802,702)\cdot K$ where $K$ is the integer and we can keep varying $K$ from $1$ onwards to manually check the minimum remainder. However this is not a great approach. Any simplistic approach to this problem would be appreciated.

答:
We seek the least $\,r\in \Bbb N\,$ such that the following congruence system is solvable

$$\begin{align}
&x\equiv -300\pmod{702}\\
&x\equiv -300\pmod{802}\\
&x\equiv -300\pmod{902}\\
&x\equiv\ \  \ r\quad \pmod{2005}
\end{align}\qquad$$

By the CRT solvability criterion it has a solution $\!\iff\!$ each pair has a solution, and a pair  $\,x\equiv a\pmod{\!m},\ x\equiv b\pmod{\!n}\,$ is solvable $\!\iff\!$ they are consistent mod the gcd $\,d=(m,n),\,$  i.e. $\,a\equiv x\equiv b\pmod{\! d},\,$ i.e. $\,\color{#c00}{d\mid a-b}.\,$ The first three are consistent since they have obvious solution $\,x = -300\,$ so it remains to check that the last is consistent with the first three, which is true for the first and third because their modulus is coprime to $2005.\,$ For the second we require $\,(2005,802)=\color{#c00}{401\mid r+300}.\,$ Thus $\,r=101\,$ is clearly the least $\,r\in\Bbb N\,$ making it solvable.


问:
Two questions about implications between $\mathsf{DC}, \mathsf{BPI}$ and $\mathsf{AC}_\omega$
Does the implication $\mathsf{DC} \implies \mathsf{BPI}$ hold?

And does the implication  $\mathsf{BPI} \implies \mathsf{AC}_\omega$ hold?

I checked with Howard/Rubin's "Consequences of the Axiom of Choice" in part V where they list relations between forms, in particular, I could not find any mention of either of the above implications on page 324 and 326 respectively.

答: Both implications are false, as can be verified at the Consequences website using the "list of all models with specified characteristics" feature (DC is form 43, BPI is 14, AC$_\omega$ is 8).</p>


问:
Integration with respect to finite variation processes
Hey can anyone explain me (or recommend a book) how we construct stochastic integral of (not necessarily continuous) process $H$ with respect to process with finite variation (also not continuous)?

答:
The stochastic integral is usually constructed for adapted càglàd (continuous on the left, limit on the right) processes as integrands and semimartingales $\big($which are càdlàg (continuous on the right, limit on the left) processes and include the FV-processes$\big)$ as integrator by using predictable simple processes and the property that the predictable simple processes are dense under the ucp-topology in the set of adapted càglàd processes. From this you can extend it to a bigger class of integrands. But all this you can read for example in the book of Protter ("Stochastic Integration and Differential Equations").

By the way, for integrators which are FV-processes you can also directly use the Lebesgue-Stieltjes integration path-by-path. A theorem states (also in the book I recomended) that the Lebesgue-Stieltjes integral and the stochastic integral are indistinguishable in that case.


问:
Local minima of the modulus of a square-free complex polynomial is always zero?
I came across the following questions in my research:

Let P(z) be a complex polynomials which has only simple roots.
Consdier the two real-variable function h(x,y)=|P(x+yi)|. Then
is it true that h has no positive local minima?

If P(x) is real polynomial, then it's not very difficult to see that
it is true. It's a straightforward application of Rolle's Theorem

答:
Suppose that there is $(x_0,y_0) \in \mathbb R^2$ and an open neigborhood $U$ of $(x_0,y_0)$ such that

$$0<h(x_0,y_0) \le h(x,y)$$

for all $(x,y) \in U$. With $z_0=x_0+iy_0$ we have

$$0<|P(z_0)| \le |P(z)|$$

for all $z \in U.$ Since $P$ is holomorphic on $U$, the minimum principle shows that $P$ is constant on $U$. The identity theorem shows then that $P$ is constant on $ \mathbb C$, hence $h$ is constant on $ \mathbb R^2.$


问:
For $z=x\sin (y/x)$ with $x=r+s$ and $y=r-3s$ find $\frac{\partial^2 z}{\partial r \partial s}$
For $z=x\sin (y/x)$ with $x=r+s$ and $y=r-3s$ find $$\frac{\partial^2 z}{\partial r \partial s}$$

My try:

$$\frac{\partial^2 z}{\partial r \partial s}=\frac{\partial }{\partial r }\frac{\partial z}{\partial s }$$
$$=\frac{\partial }{\partial r }(\frac{\partial z}{\partial x }\frac{\partial x}{\partial s }+\frac{\partial z}{\partial y }\frac{\partial y}{\partial s })$$
$$\frac{\partial }{\partial r }(\sin(y/x)+(-y/x-3)\cos(y/x))$$

$$=\color{red}{\frac{\partial }{\partial r }\sin(y/x)}+\color{blue}{\cos(y/x)\frac{\partial }{\partial r }(-y/x-3)}+\color{green}{(-y/x-3)\frac{\partial }{\partial r }\cos(y/x)}$$

$$=\color{red}{-y/x^2\cos(y/x)+1/x\cos(y/x)}$$$$+\color{blue}{(y/x^2-1/x)\cos(y/x)}$$$$+\color{green}{(y/x^2\sin(y/x)-1/x\sin(y/x))(-y/x-3)}$$

答:
A possibility since the $(x,y)\to(r,s)$ map is quite simple (and easily invertible)

is to express $z=(r+s)\sin(\frac{r-3s}{r+s})$ and calculate directly in $r,s$.

It is a bit messy because there are a lot of terms, but the chain rule goes more smoothly, anyway you get a lot of simplifications in the end, I arrived to:

$$\dfrac{\partial^2z}{\partial r\partial s}=16\,rs\,\dfrac{\sin\left(\frac{r-3s}{r+s}\right)}{(r+s)^3}=\dfrac{(3x+y)(x-y)}{x^3}\sin\left(\frac yx\right)$$


问:
How to find the Laurent expansion for $1/\cos(z)$
How to find the Laurent series for $1/\cos(z)$ in terms of $(z-\frac{\pi}{2})$ for all $z$ such that $0<|z-\frac{\pi}{2}|<1$</p>

答:
To get a recurrence for the series of the reciprocal, consider
$$
\begin{align}
1
&=\sin(x)\sum_{k=0}^\infty a_kx^{2k-1}\\
&=\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}\sum_{k=0}^\infty a_kx^{2k-1}
\end{align}
$$
and apply Cauchy products to get
$$
\sum_{k=0}^n(-1)^k\frac{a_{n-k}}{(2k+1)!}=\left\{\begin{array}{}
1&\text{if }n=0\\
0&\text{otherwise}
\end{array}\right.
$$
So that we get $a_0=1$ and
$$
a_n=\sum_{k=1}^n(-1)^{k-1}\frac{a_{n-k}}{(2k+1)!}
$$
which gives $\{a_k\}$ to be
$$
\left\{1,\frac16,\frac7{360},\frac{31}{15120},\dots\right\}
$$
Then we get
$$
\begin{align}
\frac1{\cos(x)}
&=-\frac1{\sin(x-\pi/2)}\\
&=-\sum_{k=0}^\infty a_k(x-\pi/2)^{2k-1}\\
&=-\frac1{x-\pi/2}-\frac{x-\pi/2}{6}-\frac{7(x-\pi/2)^3}{360}-\frac{31(x-\pi/2)^5}{15120}-\dots
\end{align}
$$</p>


问:
Probability St Petersburg Game
I am reading a book on probability and there is an interesting chapter on the St Petersburg game - where a coin is flipped until a head is landed and the prize is £2 if there is a head on the first throw, £4 is there is a head on the second flip, £8 for a head on the third etc. The idea of this game is it introduces a paradox since someone should be prepared to pay any amount to play since the expected yield is infinite.

However in the book it introduces the notion of changing the rewards to 2,4,6,8.... etc rather than the doubling in the original game. The book contends that even though the possible payouts increase without bound the calculation of the expected value yields a sensible answer of £4. It mentions that this is simple to calculate so omits the calculations. I have pondering this and cant see how the £4 has been arrived at - a entry price of £4 would be a fair price for the game but why? Any help greatly appreciated. I am sure I'm missing something to do with summing to infinity perhaps

答:
Let $x$ be the expected payout of the game.

You have a probability of $1/2$ of the game ending with a head on the first throw, in which case you receive $2$. If it lands tails (also probability $1/2$), it is now as if you start a new game where the payouts are $4,6,8,...$. This is identical to the original game, but all the payouts are exactly $2$ higher than before. So the expected payout of this new game is $x+2$.

The formula for the expected winnings is then:
$$x = \frac12\cdot 2 + \frac12(x+2)$$
which has the solution $x=4$.

Title: Probability St Petersburg Game


问:
How to show that the function $\log(1+\alpha t^2)$ is quasi convex?
I need to show that the function $f(t)=\log(1+\alpha t^2)$ is quasi convex. I checked its quasi convexity by plotting it using matlab as follows.

I really appreciate if you could please give me some directions to prove its quasi convexity.

答: Quasiconvexity of $f$ means that for all $a\in\mathbb R$, the set of all $t\in \mathbb R$ such that $f(t)\leq a$ is an interval. Can you compute that set ?


问:
How to find an exponential variable?
What is the way to solve a simple equation of this type?

$$x ^ 4 = 1.25$$

Update: I should have asked, how to calculate N'th root in a calculator?

Title: How to find an exponential variable?

Body:
$$x^4=a, a\in\mathbb R^{+}$$

$$\implies( x^2)^2-(\sqrt a)^2=0$$

$$\implies (x^2-\sqrt a)(x^2+\sqrt a)=0$$

$$\implies (x^2-(\sqrt[4]{a})^2)(x^2-(i\sqrt[4]{a})^2)=0$$

$$ \implies (x-\sqrt[4]{a})(x+\sqrt[4]{a})(x-i\sqrt[4]{a})(x+i\sqrt[4]{a})=0$$

I leave it here for you.


答:
The probability that the head shows up first in $n$-th throw is:
$$
p(n)=\frac1{2^n},
$$
and the corresponding expected value of the win:
$$
\sum_{n=1}^\infty 2n p(n)=4.
$$

Title: Probability St Petersburg Game

Body:
I am reading a book on probability and there is an interesting chapter on the St Petersburg game - where a coin is flipped until a head is landed and the prize is £2 if there is a head on the first throw, £4 is there is a head on the second flip, £8 for a head on the third etc. The idea of this game is it introduces a paradox since someone should be prepared to pay any amount to play since the expected yield is infinite.

However in the book it introduces the notion of changing the rewards to 2,4,6,8.... etc rather than the doubling in the original game. The book contends that even though the possible payouts increase without bound the calculation of the expected value yields a sensible answer of £4. It mentions that this is simple to calculate so omits the calculations. I have pondering this and cant see how the £4 has been arrived at - a entry price of £4 would be a fair price for the game but why? Any help greatly appreciated. I am sure I'm missing something to do with summing to infinity perhaps


问:
Square of a normal distribution
Let $Z$ have a normal distribution with mean $\mu$ and variance $1$. Show that $Z^2$ is a continuous random variable and find its p.d.f.

I really don't know what to do with this... I tried working out the CDF but it didn't get me anywhere.

答:
Note as we have $Z^2\ge 0$, we have $P(Z^2 \le x) = 0$ for $x < 0$. Now let $x \ge 0$, then
$$ F_{Z^2}(x) = P(Z^2 \le x) = P(Z \in [-\sqrt x,\sqrt x]) = F_Z(\sqrt x) - F_Z(-\sqrt x) $$
where $F_Z$ is the cdf of $Z$. Now take the derivative with respect to $x$, we get for the pdf that
\begin{align*}
  f_{Z^2}(x) &= \frac 1{2\sqrt x}f_Z(\sqrt x) + \frac 1{2\sqrt x} f_Z(-\sqrt x)\\
\end{align*}</p>

Title: Square of a normal distribution


问:
Prove the logarithmic spiral intersects all radius vectors at the same angle
I'm trying to prove that the logarithmic , the equation of which is $r = Ce^{k\theta}$, intersects all radius vectors at the same angle.

I'm assuming this relates to the radius-tangent angle that I will define as $\psi$. If i denote the respective radius-tangent angles by $\psi_1$ and $\psi_2$, then:

$$\alpha = \psi_1 - \psi_2$$

And

$$\tan\alpha = \frac{\tan\psi_1-\tan\psi_2}{1 + \tan\psi_1\tan\psi_2}$$

So I thought of rearranging the equation of $r$ to find $\theta$, such that $\frac{ln(r)}{k}+C = \theta$, though ?I'm not sure how to proceed from here and would really be grateful for your support.

答:
Quick way: Apply a similarity transform $r\mapsto e^\alpha r$, $\theta\mapsto\theta+\alpha/k$.

Slower way: Note that the angle between tangent and radial vector is $\cot^{-1}(\frac1r\frac{\mathrm{d}r}{\mathrm{d}\theta})$ since $\mathbf{e}_r,\frac1r\mathbf{e}_\theta$ is an orthonormal frame.


问:
Show that $\lim _{n \rightarrow+\infty} \sup _{x \in[0,1]}\left|f(x)-\mathcal{P}_{n}(x)\right|=0$
Let $f$ be a continuous function on $[0,1]$ and for any $n \geq 1$ let us define the polynomial function :
$$
\mathcal{P}_{n}(x)=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) x^{k}(1-x)^{n-k} f\left(\frac{k}{n}\right)
$$

For any $x \in[0,1]$ let $\left(\varepsilon_{n}(x)\right)_{n>1}$ be a sequence of i.i.d. Bernoulii r.v. of parameter $x$. Let then $S_{n}(x)=\varepsilon_{1}(x)+\cdots+\varepsilon_{n}(x)$.

$\mathcal{P}_{n}(x)$ can be interpreted as the following expectation : $$\mathbb{E}[f(\frac{S_n(x)}{n})]$$

By using Bienaymé-Chebychev's inequality we can show that :
$$
\forall \varepsilon>0, \quad \mathbb{P}\left(\left|\frac{S_{n}(x)}{n}-x\right| \geq \varepsilon\right) \leq \frac{1}{4 n \varepsilon^{2}}
$$

thus $\frac{S_{n}(x)}{n}$ converges in probability and thus in distribution to $x$, consequently $\mathcal{P}_{n}(x)$ converges to $E[f(x)] = f(x)$, I'm not sure how to establish the convergence with the supremum as stated in the title though, any help will be greatly appreciated.

答:
Denote $\zeta_n(x) = S_n(x)/n$. For any $\varepsilon>0$,
$$
\bigl|\mathrm{E}[f(\zeta_n(x)] - f(x)\bigr|\le \mathrm{E}\left[\bigl|f(\zeta_n(x))-f(x)\bigr| \right] \\= \mathrm{E}\left[\bigl|f(\zeta_n(x))-f(x)\bigr|\mathbf{1}_{|\zeta_n(x)-x|\le \varepsilon} \right] + \mathrm{E}\left[\bigl|f(\zeta_n(x))-f(x)\bigr|\mathbf{1}_{|\zeta_n(x)-x|> \varepsilon} \right]\\
\le \omega_f(\varepsilon) + 2\sup|f| \cdot \mathrm{P}(|\zeta_n(x)-x|> \varepsilon),
$$
where $\omega_f(\varepsilon) = \sup_{0\le s<t<s+\varepsilon\le 1}|f(t) - f(s)|$ is the modulus of coninuity of $f$.

I believe you can conclude from here, using Chebyshev's inequality.


答:
Let $s$ be the expected payout. Then
$$s=2\cdot\frac12 + 4\cdot\frac14 + 6\cdot\frac18 + 8\cdot\frac1{16}+\dots$$
Subtract
$$2 = 2\left(\frac12 + \frac14 + \frac18 + \frac1{16}+\dots\right)$$
$$s-2=2\cdot\frac14 + 4\cdot\frac18 + 6\cdot\frac1{16}+\dots$$
Double
$$2(s-2)=2\cdot\frac12 + 4\cdot\frac14 + 6\cdot\frac18+\dots=s$$
That is,
$$2s-4=s$$
so
$$s=4$$

Title: Probability St Petersburg Game


问:
Rational solutions of quadratic forms
Is there an algorithm or a method that one can use to determine whether an equation of the form $(\text{E})$:
$$ax^2+by^2+cz^2+dt^2=0$$
has a solution $(x,y,z,t)$ in whole numbers. In other words, given the tuple $(a,b,c,d)\in\mathbb{Z}^4$, how to determine whether there is a non-trivial tuple $(x,y,z,t)\in\mathbb{Z}^4$ that satisfies $(\text{E})$?

In case when one of $a,b,c$ or $d$ is zero, one can use Legendre's theorem on ternary quadratic forms. This paper shows the full process of how it's done. Is there anything similar for the case of four variables?

答:
Clearly, it is enough to decide whether their is rational solutions or not (multiply by the square of a common denominator to have integer solutions).

I assume that $a,b,c,d$ are  all nonzero (otherwise, you are reduced to known cases).

Multiplying by $a$ and replacing $x$ by $ax$, one may assume WLOG that $a=1$, so we are reduced to the equation $x^2+by^2+cz^2+dz^2=0$. (*)

This step is not necessary, but it allows to write down simpler conditions.

Hasse Minkowski says that (*) has a nonzero rational solution if and only if it has a nonzero solution over $\mathbb{R}$ and $\mathbb{Q}_p$ for all $p$.

Having a solution over the reals is equivalent to say that $b,c,d$ are not all $>0$.

For the $p$-adic case, it depends on the determinant and local Hasse invariants of the rational quadratic form $x^2+by^2+cz^2+dt^2$.

Here, the determinant is the square class of $bcd$, and if $p$ is prime , the local Hasse invariant is $(b,cd)_p(c,d)_p$.

I will below how to define $( , )_p$.

If $r,s$ are two non zero rationals, write $r=p^\alpha u, s=p^\beta v, p\nmid u, p\nmid v$.

Then $(r,s)_p=(-1)^{\alpha\beta\cdot\frac{p-1}{2}}\left(\dfrac{u}{p}\right)^\beta \left(\dfrac{v}{p}\right)^\alpha$, if $p\neq 2$, where $\left(\dfrac{\phantom{a}}{p}\right)$ is the Legendre symbol,
and $\displaystyle (r,s)_2=(-1)^{\frac{u-1}{2}\cdot \frac{v-1}{2}+\alpha\frac{v^2-1}{8}+\beta\frac{u^2-1}{8}}$

Note for later use that $(-1,-1)_p=1$ if $p\neq 2$ and $(-1,-1)_2=-1$.

Note also that if $p\neq 2$, and the $p$-adic valuations of $r,s$ are both zero,  $(r,s)_p=1.$

Recall also the following fact.

Fact. Let $r=p^\alpha u, p\nmid u$. Then $r$ is a square in $\mathbb{Q}_p$ if and only if :

$\alpha$ is even and $u$ is a square modulo $p$ if $p\neq 2$ (which can be decided using Legendre symbol)

$\alpha$ is even and $u\equiv 1 \ [8]$ if $p=2$.

If we translate Thm 6 of Chapter IV $\S$ 2 of Serre's "A course in arithmetic", we get that (*) has a nonzero solution over $\mathbb{Q}_p$ if and only if one of the following cases hold:

Case 1. $bcd$ is not a square modulo in $\mathbb{Q}_p^\times$

Case 2. $bcd$ is a square in $\mathbb{Q}_p^\times$ and $(b,cd)_p=(c,d)_p$ if $p\neq 2$

Case 3. $bcd$ is a square in  $\mathbb{Q}_2^\times$ and $(b,cd)_2=-(c,d)_2$

All these conditions amounts to computations of finitely many Legendre symbols, because if $p\nmid bcd$ and $p\neq 2$, then either $bcd$ is not a square in $\mathbb{Q}_p^\times $, or $bcd$ is a square in $\mathbb{Q}_p^\times$ but in this case both symbols $(b,cd)_p$ and $(c,d)_p$ are equal to $1$ !!

Hence for odd prime which are not divisors of $bcd$, you are automatically in one of the first two cases.

So you only have to test if you are in one of the three previous cases only for $p\mid 2bcd $. Note that if the $p$-adic valuation of $bcd$ is odd, $bcd$ is automatically not a square, so you can reduce to prime numbers $p$ such that $p\mid 2bcd$ and $v_p(bcd)$ is even !

Hence, you have an algorithm to decide the existence of a nontrivial solution over $\mathbb{Q}$ (note that the Legendre symbol coincide with the Jacobi symbol, so you can compute it without factoring you integers). However, it won't give you an explicit solution.

Example. Consider $3x^2+3\cdot 5y^2+7z^2-2\cdot 7\cdot 23 t^2=0$.

This is equivalent to consider $x^2+3^2 \cdot 5y^2+3\cdot 7z^2-2\cdot 3 \cdot 7\cdot 23 t^2=0$.

Here $b=3^2 \cdot 5, c=3\cdot 7, d=-2\cdot 3\cdot 7 \cdot 23,  cd=-2\cdot 3^2\cdot  7^2\cdot 23$ , and $bcd=-2\cdot 3^4\cdot 5\cdot 7^2\cdot 23$.

Since $b,c,d$ are not all positive, we have solutions over the reals.

For the $p$-adic case,  we just have to check two cases: $p=3,7$ since they are the only prime divisors of $bcd$ with an even valuation.

For $p=3$, we need to check that $-2\cdot 5\cdot 23$ is a square modulo $3$.
But $-2\cdot 5\cdot 23=1 \ [3]$, which is a square.

Now since $( r, s)_p$ only depends on the square classes of $r$ and $s$, we have
$(b,cd)_3= (5,-2\cdot 23)_3=(-1)^{\frac{5-1}{2}\frac{-46-1}{2}}=1$ and
$(c,d)_3=(3\cdot 7,-2\cdot 3\cdot 7\cdot 23)_3 =(-1)^{\frac{3-1}{2}}\left(\dfrac{7}{3}\right) \left(\dfrac{-2\cdot 7\cdot 23}{3}\right)=-1\cdot 1\cdot -1=1.$

Hence we are in Case 2.

For $p=7$,   we need to check that $-2\cdot 5\cdot 23$ is a square modulo $7$.
But $-2\cdot 5\cdot 23=1 \ [7]$, which is a square.

Since $7\nmid 5,2$ and $3$, $(b,cd)_7=1$. Now $(c,d)_7=(3\cdot 7,-2\cdot 3\cdot 7\cdot 23)_7=(-1)^{\frac{7-1}{2}}\left(\dfrac{3}{7}\right) \left(\dfrac{-2\cdot 3\cdot 23}{7}\right)=-1\cdot -1\cdot 1=1.$

Hence we are in Case 2.

All in all, the original equation must have a non trivial solution.

This is indeed the case since $3\cdot 2^2+3\cdot 5\cdot 3^2+7\cdot 5^2-2\cdot 7\cdot 23\cdot 1^2=0$.

Title: Rational solutions of quadratic forms


问:
Geometric interpretation of Laplace's formula for determinants
Coming from the geometric point of view, the determinant of an $n \times n$-Matrix computes the volume of an parallelepiped spanned by the columns of the matrix.

In context of this question, let the determinant be defined by Laplace's formula and let us assume that we use the Laplace expansion along the first column of the matrix.

In case of a $2\times 2$-matrix of the form $M=\begin{pmatrix} a & c\\ b & d\end{pmatrix}$, the expansion is quite intuitive: One takes $a$ - which can be seen as the projection of the vektor $\begin{pmatrix} a \\ b \end{pmatrix}$ onto the $e_1$-axis - and multiplies it with $d$, which can be seen as the projection of the vector $\begin{pmatrix} c \\ d \end{pmatrix}$ onto the $e_2$-axis. So it's the base $\times$ height-formula which would be correct if $b=0$. If $b \neq 0$, the error which is made is exactly $c \cdot d$, again by a base $\times$ height-formula, which one can check with the help of a figure by Salomon Golomb, which I found here:

Why determinant of a 2 by 2 matrix is the area of a parallelogram?

I hoped to transfer this principle to $3 \times 3$-matrices.
I considered the parallelogram spanned by the 2nd and the 3rd column of a $3 \times 3$-matrix. Keeping Laplace's formula in mind, the first entry of the first column is multiplied with the area of the projection of the parallelogram onto the $e_2$-$e_3$-plane, the second entry is multiplied with the area of the projection onto the $e_1$-$e_3$-plane and the third entry is multiplied with the area of the projection onto the $e_1$-$e_2$-plane. (That is, $\det(a,b,c)= c\cdot (a\times b)$).

I tried to give an geometric sense to this calculations mimicking what was done in Golomb's figure but I failed. Sure, there is a way to rearrange the three sets to conclude that actually the volume of the parallelepiped was computed but again I would not understand why.

Is the anyone who knows a way to interprete Laplace's formula geometrically? I think there might be a way which uses the projections of the parallelogram which I mentioned above...

I would be very thankful for all other (geometric) interpretations, too!

Ole

答:
The two answers above are excellent, and far beyond what I could hope to produce, but here's the "Laplace expansion of the determinant for dummies" version.

There are usually two starting definitions for the determinant:

As the signed volume of the parallelotope spanned by a tuple of vectors

As a multilinear alternating form

On a 2D plane, 2 vectors define a parallelogram. Moving to 3D and adding a third vector, the three vectors define a parallelepiped. Generalizing this, $n$ vectors in $n$-dimensional space define a so-called parallelotope. Just like a parallelepiped has a volume, so too does a parallelotope. A signed volume is a volume with a sign in front, depending on the "handedness" of the tuple of vectors.

The first definition is in my opinion the more natural way (certainly easier to think about, at least!). However, the second definition is useful because of the properties it states about the determinant. The two definitions are equivalent (one implies the other and vice-versa), but let's just see why the first implies the second.

First of all, a linear form is a linear map from a vector space to its underlying field (the field is the mathematical structure from which the coefficients come from). It only takes one vector as an argument. If we generalize this idea to more than one vector as input, with the function being linear in every argument, then we get what is called a multilinear form: takes in several vectors, outputs something from the underlying field (e.g. a real number) and is linear in every argument.

Adding the designation "alternating" before "multilinear form" also has multiple equivalent definitions, but the one which is interesting here is that inverting two vector arguments of a multilinear form makes it switch signs (e.g. $f(v_2, v_1) = -f(v_1, v_2)$).

Volume is multilinear: scaling one side of a parallelotope scales its volume by the same factor (try drawing a parallelogram or a parallelepiped and scaling one of its sides to see!). The volume obtained when adding a vector to one of its vectors is the same as the sum of volumes with the two vectors separate.

Signed volume is also alternating (that's what we mean by "signed").

One more property is necessary before illustrating the Laplace expansion: signed volume is invariant under shearing. A simple way to define shearing is as adding one of the arguments, scaled, to another argument: $\text{det}(v_1, v_2, v_3) = \text{det}(v_1, v_2 + \lambda v_1, v_3)$. Once again, to convince yourself, it's pretty useful to draw a parallelogram and try shearing it: in 2D, shearing doesn't change the height, and remember, the area of a parallelogram in 2D is the base times the height!

Finally, we can try to find out how the Laplace expansion works in 3D. We have the three properties we will need: the determinant is multilinear, alternating, and invariant under shearing. Usually, a matrix is considered to be a bunch of column vectors stuck together, but without loss of generality we can also consider it to be a bunch of row vectors, as this makes it visually a bit more similar to Gaussian elimination (a nice bonus to understand the link between the two!).

While the determinant is also invariant under transposition, this isn't immediately obvious, so let's not use this property, and instead remember that if we have the three vectors $v_1 = (a, b, c)$, $v_2 = (d, e, f)$ and $v_3 = (i, j, k)$, the matrix we write is $\begin{pmatrix} a & b & c \\ d & e & f \\ i & j & k \end{pmatrix}$.

$$\begin{vmatrix} a & b & c \\ d & e & f \\ i & j & k \end{vmatrix} = \begin{vmatrix} a & 0 & 0 \\ d & e & f \\ i & j & k \end{vmatrix} + \begin{vmatrix} 0 & b & 0 \\ d & e & f \\ i & j & k \end{vmatrix} + \begin{vmatrix} 0 & 0 & c \\ d & e & f \\ i & j & k \end{vmatrix}$$

Here, we decompose $(a, b, c)$ into its three components in the canonical basis: $(a, b, c) = (a, 0, 0) + (0, b, 0) + (0, 0, c)$.

$$ = a \begin{vmatrix} 1 & 0 & 0 \\ d & e & f \\ i & j & k \end{vmatrix} + b \begin{vmatrix} 0 & 1 & 0 \\ d & e & f \\ i & j & k \end{vmatrix} + c \begin{vmatrix} 0 & 0 & 1 \\ d & e & f \\ i & j & k \end{vmatrix}$$

Here, we made use of the multilinearity of the determinant.

$$ = a \begin{vmatrix} 1 & 0 & 0 \\ 0 & e & f \\ 0 & j & k \end{vmatrix} + b \begin{vmatrix} 0 & 1 & 0 \\ d & 0 & f \\ i & 0 & k \end{vmatrix} + c \begin{vmatrix} 0 & 0 & 1 \\ d & e & 0 \\ i & j & 0 \end{vmatrix}$$

This time, we used the invariance of the determinant under skew to eliminate the first component of the second and third vectors.

$$ = a \begin{vmatrix} e & f \\ j & k \end{vmatrix} - b \begin{vmatrix} d & i \\ f & k \end{vmatrix} + c \begin{vmatrix} d & e \\ i & j \end{vmatrix}$$

Finally, we noticed two things. First, having two coplanar vectors and one orthonormal vector means the volume is just that of the parallelogram spanned by the two colinear vectors. Second, to have positive sign for the determinant, $(0, 1, 0)$ should be on the second row instead of the first, which you can achieve with one transposition of the the first and second rows, so the second term has a negative sign. Same goes for $(0, 0, 1)$, which should be on the third row, which you can achieve with two transpositions to preserve the ordering of the second and third rows, so its sign is positive.


问:
How to find the expected number of insertions before a Hash table is full?
If I have a hash table with $m$ slots and each slot has $1/m$ probability of being picked for an insertion of an element. What is the expected number of insertions before all slots are full?

I would think that the expected number of insertions of elements is $m$ but not sure because an element can also hash to another element's slot? Reason for this thought is: $$\mathbb{E}[X]=\sum_x^{m}Pr[X=x] = \sum_x^{m}\frac{1}{m}=1$$
Where after $m$ insertions where each insertion has $1/m$ probability, the hash table is full. I don't care about what to do with the elements that collide. For simplicity sake we can just discard them.

答:
This question is well-known as Coupon collector problem, so that the expected number of insertions until all slots are full is:
$$\mathbb{E}[X]=m H_m
$$
where $H_m$ is the harmonic number:
$$
H_m=\sum_{k=1}^m\frac1k.
$$


问:
Find the last two digit of $5^{121}*3^{312}$
The answers key says by using
$$5^k\equiv 25 \pmod{100}, k>=2$$

and $$3^{40}\equiv 1 \pmod{100}$$
can have $$ 5^2 *3^4 \equiv25 \pmod{100},$$
and it follows that the last two digits of  $5^{143}*3^{312}$ are $25$.

Don't know why  $ 5^2 *3^4 \equiv25 \pmod{100}$ can applies that the last two digits of $5^{143}*3^{312}$ are $25$.

Also, the question is asking about $5^{121}$, why the last step relates to $5^{143}$

Body:
Clearer: use MDL: $\,\ \color{#c00}ab\bmod \color{#c00}ac\, =\, \color{#c00}a(b\bmod c)\ $ to factor out $\,\color{#c00}{a=5^2}$ from your mod, i.e.
$$\begin{align}
&\color{#c00}5^{\large \color{#c00}2+k}\,\cdot\,  3^{\large 2n}\ \:\!\bmod 4\cdot \color{#c00}{5^2}\\[.2em]
=\ & \color{#c00}{5^2} (5^{\large k}\cdot  3^{\large 2n}\ \  \bmod 4)\\[.2em]
=\ & 5^2 (1^{\large k} (-1)^{\large 2n}\!\bmod 4)\\[.2em]
=\  & 5^2 (1)
\end{align}\qquad\qquad$$


问:
Hausdorff distance union limit
Let $(M,d)$ be a compact metric space and $d_H(A,B)$ represent the Hausdorff distance between $A,B\subset M$. Let $A_n$ be a sequence of non-empty increasing compact subsets of $M$ (i.e. $A_n\subseteq A_m$ if $m>n$). Assume that $A=\text{cl}(\cup_nA_n)$ is non-empty, compact and bounded. Is it true that:

$$d_H(A_n,A)\xrightarrow{n\rightarrow \infty}0$$

I thought this was true but then I came across this document (which makes me think it isn't) and this question (which makes me think it is) and now I am unsure. If it is untrue what would be a good counterexample?

答:
True.

Let $A_n$ be an increasing sequence of nonempt compact sets.  Assume that $A := \text{cl}\;\bigcup_n A_n$ is compact.

For all $n$, we have $A_n \subseteq A$, so
$$
\forall x \in A_n\quad \inf\{d(x,y) : y \in A\} = 0
\tag1$$
Now let $\epsilon > 0$.  For each $n$, let
$$
B_n = \bigcup_{x \in A_n}\{y : d(y,x) < \epsilon\}
$$
Now $B_n$ is an open set, the sequence $B_n$ is increasing.  By the definition of "closure", $\bigcup_n B_n \supseteq A$.  Since $A$ is compact, there is a finite subcover; since $(B_n)$ is increasing, there exists $n_\epsilon$ so that $B_{n_\epsilon} \supseteq A$.  Then for all $n \ge n_\epsilon$ we also have $B_n \supseteq A$.  Thus:
$$
\forall n \ge n_\epsilon\quad \forall y \in A\quad\inf\{d(x,y): x \in A_n\} \le \epsilon .
\tag2$$
Combining this with $(1)$: for all $n \ge n_\epsilon$ we have
$d_H(A_n,A) \le \epsilon$.

That is true for all $\epsilon > 0$, so $\lim_n d_H(A_n,A) = 0$.

Title: Hausdorff distance union limit


问:
The way to deduce the following equations.
Given the $5$ equations.

$a_1=a_2+a_5 \tag{1}$

$a_4=a_3+a_5 \tag{2}$

$b=a_1+a_2 \tag{3}$

$b=a_3+\frac{a_4}{2} \tag{4}$

$\frac{a_5}{2}+a_1=a_3 \tag{5}$

I want to deduce the following.

$a_1=\frac{9}{16}*b\tag{6}$

$a_2=\frac{7}{16}*b\tag{7}$

$a_3=\frac{5}{8}*b\tag{8}$

$a_4=\frac{3}{4}*b\tag{9}$

$a_5=\frac{1}{8}*b\tag{10}$

Can anyone tell some hint(s)? so that I can deduce it in my own.

答:
Hint 1: As the comments have stated, one approach is to write the system as a matrix and then find the RREF. The matrix is given by $$A = \left(
\begin{array}{cccccc}
 1 & -1 & 0 & 0 & -1 & 0 \\
 0 & 0 & 1 & -1 & 1 & 0 \\
 1 & 1 & 0 & 0 & 0 & b \\
 0 & 0 & 1 & \frac{1}{2} & 0 & b \\
 1 & 0 & -1 & 0 & \frac{1}{2} & 0 \\
\end{array}
\right)$$

You already have the result in your question, so know what the RREF should produce.

Hint 2: A second approach is to eliminate equations.


问:
First order higher degree ODE
Considering the ODE
$$\left(\dfrac{dy}{dx}\right)^n+C_1y=C_2$$
where, $C_1$ and $C_2$ are constants.

Is there a way to find an analytical or approximate analytical solution to this ODE in terms of $n$?

答:
There is a analytical solution to this ODE:
\begin{align*}
\left(\dfrac{dy}{dx}\right)^{n}+C_{1}y & =C_{2}
\end{align*}
If we solve for $\dfrac{dy}{dx}$:
\begin{align*}
\left(\dfrac{dy}{dx}\right)^{n} & =C_{2}-C_{1}y\\
\dfrac{dy}{dx} & =\left(C_{2}-C_{1}y\right)^{\frac{1}{n}}
\end{align*}
Integrating:
\begin{align*}
\int\left(C_{2}-C_{1}y\right)^{-\frac{1}{n}}dy & =\int dx\\
-\frac{1}{C_{1}}\int-C_{1}\left(C_{2}-C_{1}y\right)^{-\frac{1}{n}}dy & =\int dx\\
\begin{cases}
\frac{n}{n-1}\left(C_{2}-C_{1}y\right)^{\frac{n}{n-1}} & n>1\\
\log\left(C_{2}-C_{1}y\right) & n=1
\end{cases} & =x+k
\end{align*}
You can finally solve for $y(x)$:
\begin{align*}
y\left(x\right) & =\begin{cases}
\frac{1}{C_{1}}\left(C_{2}-\left(\frac{n-1}{n}\left(x+k\right)\right)^{\frac{n-1}{n}}\right) & n>1\\
\frac{1}{C_{1}}\left(C_{2}-e^{x+k}\right) & n=1
\end{cases}
\end{align*}

Title: First order higher degree ODE


问:
Proving $f(u) = \lambda u \iff f(\bar u) = \bar \lambda \bar u$
I'm having trouble understand the following proposition:

Let $f: \mathbb C^n \to \mathbb C^n$ be an endomorphism and let $\lambda$ be an eigenvalue of $f$, then:

$\bar \lambda$ is also an eigenvalue of $f$

$f(u) = \lambda u \iff f(\bar u) = \bar \lambda \bar u$

If $p$ is the characteristic polynomial of $f$ then if $\lambda$ is a root, $\bar \lambda$ is also a root, that I understand. What I don't understand is proposition number (2). Why is that true?

答:
Counterexample: let  $A= diag (0,i)$

(a $2 \times 2$ diagonal matrix)

and $f(u):=Au$ for $u \in \mathbb C^2$.

The eigenvalues auf $f$ are $0$ and $i$. But $ \bar i =-i$ is not an eigenvalue of $f$.


问:
Finding the entire length of the cardioid $r = 1-\cos\theta$
I'm trying to find the entire length of the cardioid $r = 1-\cos\theta$.

Although, I'm not sure how to compute the definite integral for $a$ and $b$, given that the equation is for the length of the arc is this:

$$\int^{\theta}_{a}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\space d\theta$$

Here is my working out:

$\sqrt{(1-\cos\theta)^2-\sin\theta} = 1-2\cos \left(θ\right)+\cos ^2\left(θ\right)-\sin ^2\left(θ\right)$

Hence:

$$\int^{\theta}_{a}\sqrt{\left(\cos \left(θ\right)-1\right)^2-\sin ^2\left(θ\right)}\space d\theta$$

Though, where would I begin with computing the following integral? I appreciate the support.

答:
To find the full length of the cardioid you need to integrate from $0$ to $2\pi$:
$$
L=\int_0^{2\pi}\sqrt{(1-\cos\theta)^2\color{red}+(\sin\theta)^2}\,d\theta=
\int_0^{2\pi}\sqrt{2(1-\cos\theta)}\,d\theta=\int_0^{2\pi}2\sin\frac\theta2\,d\theta.
$$

Can you take it from here?


问:
Prove that if $p_1,\dots,p_k$ are distinct odd primes then 1 has $2^k$ square roots $\mod m$ where $m$ is the product of the primes.
I think I am most of the way through this proof but I am stuck. Here was my approach: I looked at the square roots of $1$ mod $105$, and noticed that each one corresponded to one less than an integer multiple of a subset of the prime factors of $105$, i.e. we have for instance $71=24\cdot3-1$ or $29=2\cdot(3\cdot 5)-1$ and so on. I conjectured that given a subset of the prime factors of $m$, I could construct a unique square root of $1$ mod $m$, which would show that there are (at least) $2^k$ of them.

So I did that and I think that it worked. Pick some set of the prime factors of $m$, say $p_{a_1},\dots,p_{a_j}$. Claim: we can find a corresponding $n$ such that $n\prod_{i=1}^jp_{a_i}-1$ is a square root of $1$ mod $m$.

Proof: We are looking for a solution to

$$\left(n\prod_{i=1}^jp_{a_i}-1\right)^2\equiv 1 \pmod m$$
which may be rewritten as

$$\left(\prod_{i=1}^jp_{a_i}\right)^2n^2 - 2\prod_{i=1}^jp_{a_i}n\equiv 0 \pmod m$$

Now since $\prod_{i=1}^jp_{a_i}|m$ we may write this as $$\prod_{i=1}^jp_{a_i}n^2-2n \equiv 0\mod \frac{m}{\prod_{i=1}^jp_{a_i}}$$

or

$$n\left(\prod_{i=1}^jp_{a_i}n-2\right) \equiv 0 \mod {\frac{m}{\prod_{i=1}^jp_{a_i}}}$$

This has solutions when $n\equiv0\mod \frac{m}{\prod_{i=1}^jp_{a_i}}$ or when $\prod_{i=1}^jp_{a_i}n-2\equiv 0 \mod \frac{m}{\prod_{i=1}^jp_{a_i}}$. In the first case we end up with the same root for any choice of $p_{a_i}$'s so that's sort of a degenerate one. But in the second case we have $\prod_{i=1}^jp_{a_i}n\equiv 2 \mod \frac{m}{\prod_{i=1}^jp_{a_i}}$, which has a unique and nonzero solution for $n$. Plugging that $n$ back into $n\prod_{i=1}^jp_{a_i}-1$ thus gives a square root of $1$ mod $m$ which corresponds uniquely to your choice of prime factors, thus showing that there are (at least) $2^k-1$ square roots of 1. Then we add 1 in which corresponds to the empty set of prime factors and we get the full $2^k$.

My problem is ruling out the possibility of more square roots of 1, i.e. if $x^2\equiv1\mod m$ I need to show that it corresponds uniquely to something of the form $n\prod_{i=1}^jp_{a_i}-1$. This is proving to be difficult if not impossible.

I put a lot of work into going the one way and I really don't want it all to be for nothing. Can anyone see a way to go the other direction?

答:
Note $\  p_1\!\cdots p_k\mid x^2\!-1 = (x\!-\!1)(x\!+\!1) \!\iff\! p_i \mid x\!-\!1\,\ {\rm or}\,\ p\mid x\!+\!1\!\iff\! x\equiv \pm1 \pmod{ p_i}$

Therefore there are precisely $\,2^k\,$ solutions $\ x\equiv\, (\pm1,\ldots,\pm1)\ \pmod{p_1,\ldots,p_k}$

Your method amounts to bijecting each solution with the subset of primes where $\,x\equiv -1,\,$ or, equivalently, the product of primes from that subset. So the solutions are in $1$-$1$ correspondence with all subsets of $\,\{p_1,\ldots,p_k\}\,$ or, equivalently, all factors of $\,p_1\!\cdots p_k.$


问:
Probability question - (Probably) Bayes' Rule and Total Probability Theorem
I just took a probability final exam and was fairly confident in my solution of 28/31, but I wanted to be sure... because according to http://www.stat.tamu.edu/~derya/stat211/SummerII02/Final.Summer02.doc which has it as the second question, the answer is .6627. What's discerning is that they have the decimal equivalent of 28/31 as one of their answers which makes it seem like they know something I don't...

"Seventy percent of all cattle are treated by an injected vaccine to combat a serious disease. The probability of recovery from the disease is 1 in 20 if untreated and 1 in 5 if treated. Given that an infected cow has recovered, what is the probability that the cow received the preventive vaccine?"

Here's my solution: Let A be the event a cow recovered, let B be the event a cow received the vaccine.

We are given:

```<code>P(A|B) = 1/5

P(A|~B) = 1/20

P(B) = 7/10
</code>```

We want to find P(B|A), so use Bayes' rule and the total probability theorem to find

P(B|A) = P(A|B) x P(B) / (P(A|B) x P(B) + P(A|~B) x P(~B) ).

Plugging in the values from what's given above, we get (.2 x .7) / (.2 x .7 + .05 x .3) which gives 28/31.

If I'm wrong, I'd love to be pointed in the right direction haha

Thank you!

答: I get .68, probably rounding but what happened was you fell off right at start and the checkers accepted your term definitions which are incorrect.  A=1/4 B=7/10 and P(A/~B)=1/10</p>


问:
Probability with multiple dice rolls
A friend of mine recently created a game where your movements on the plateau are determined by series of dice throws. The rules are simple: you have exactly $6$ dice that you should throw simultaneously. However, by trading a specific resource (not discussed here), you can choose to throw again any dice you want (except the dice you previously chose to keep). Let's clarify this with an example.

Consider this: I want $6 \times 6$ (the maximal output).

Throw $1$, output $6, 6, 2, 1, 4, 5 \implies$ I keep the two $6$ and throw again the $4$ other dice.

Throw $2$, output $6, 5, 3, 4 \implies$ I keep the single $6$ and I guess you figured it out.

Now, for the question, I would like to compute the probability of successfully outputting $6$ times $6$ with at most n throws. The blunt answer is to considered all the favorable "paths" to my output, for example with $6$ times $6$ in $3$ throws: $0$ first throw, $0$ second throw, $6$ third throw, etc., which is $36$ paths I reckon.

Needless to say, the process of computing such probabilities for a high number of throws is quite the hassle. Obviously we can consider that outputting $6\times6$ in at most $3$ throws is equivalent to doing so in exactly $1$ throw OR in exactly $2$ throw OR in exactly $3$ throw which is convenient because you have an iterative process. However, there is still $21$ unique paths for successfully outputting $6\times6$ in exactly $3$ throws.

Is there a crafty way to compute said probability that I'm not envisioning?

答:
I'm making this an answer for completeness.

It's actually quite easy. Consider each die separately. We want each die to roll $6$ at least once in $n$ throws. So, for a given die, the probability of getting a $6$ is $1−\left(\frac56\right)^n$. And, we want each dice to be a $6$, so our final answer is $$\left(1−\left(\frac56\right)^n\right)^6$$

Title: Probability with multiple dice rolls


问:
If $A$ is a bounded operator on a Banach space $X$, is there a topological decomposition $X=\mathcal N(A)\oplus_1X_2$?
Let $X,Y$ be Banach spaces and $A\in\mathfrak L(X,Y)$.

Under which conditions can we show that

$X=\underbrace{\mathcal N(A)}_{=:\:X_1}\oplus_1\underbrace{(1-\pi)X}_{=:\:X_2}$ (topological direct sum) for some projection $\pi\in\mathfrak L(X)$;

$X_2$ is complete;

$\left.A\right|_{X_2}$ is bijective?

Note that if $X$ is a Hilbert space, all claims are easily established, since we can take $\pi$ as the orthogonal projection of $X$ onto $X_1$. In that case, $X_2=X_1^\perp$.

The problem with the Banach space case is that the annihilator $X_1^\perp$ is a space of functionals (which can be identified with vectors of $X$ by Riesz' representation theorem in the Hilbert space case).

答:
Any closed subspace $X_1$ of $X$ is the null space for some bounded  linear map, e.g. the quotient map $$X\to X/X_1.$$

So the question is really whether or not every closed subspace is complemented.

A famous Theorem by Lindenstrauss and Tzafriri [1] asserts that every Banach space which is not isomorphic to a Hilbert space has a non complemented closed subspace.

On the other hand, should there be a closed subspace $X_2$ such that $X$ is the algebraic direct sum of $N(A)$ and $X_2$, then the projections onto both $N(A)$ and $X_2$ are bounded, thanks to the closed graph Theorem, and the restriction of $A$ to $X_2$ is injective.

Finally, if $A$ is surjective, then $A|_{X_2}$ is bijective, hence also an isomorphism, by the open mapping Theorem.

[1] Lindenstrauss, J., Tzafriri, L. On the complemented subspaces problem. Israel J. Math. 9, 263–269 (1971).


问:
Proof of an equation in a Poisson Process
In a Poisson Process with parameter $\lambda$, if $X_n$ is the time of the rth arrival.

Prove:

$$\sum^\infty_{n=1}fx_n(x)=\lambda\quad\text{for all}\ x\gt0$$

I’m new to Statistics and the Poisson Process in general. I know that $N(0) = 0$, that the number of arrivals in any interval $a\gt0$ has a Poisson$(\lambda a)$ distribution and that $N(t)$ increments are independent but I don’t see how any of that helps me to prove/show the given equation. I would appreciate any and all help.

答: $$\sum\limits^\infty_{n=1}f_{X_n}(x) \\= \sum\limits^\infty_{n=1} \dfrac{\lambda^{n-1} t^{n-1}e^{-\lambda t}}{(n-1)!}\lambda \\=  e^{-\lambda t}\lambda\sum\limits^\infty_{n=1} \dfrac{(\lambda t)^{n-1} }{(n-1)!} \\=  e^{-\lambda t}\lambda\sum\limits^\infty_{m=0} \dfrac{(\lambda t)^{m} }{m!} \\= e^{-\lambda t}\lambda e^{\lambda t} \\ = \lambda.$$


问:
How to get ${n \choose 0}^2+{n \choose 1}^2+{n \choose 2}^2+\cdots+{n \choose n}^2 = {x \choose y}$
I found this in my test book, any hints? Given $${n \choose 0}^2+{n \choose 1}^2+{n \choose 2}^2+\cdots+{n \choose n}^2 = {x \choose y}$$ Then find the value of x and y in n. According to the answer provided on last pages of that book, it's $2n \choose n$. What i dont understand is how to get the answer (there's no explanation written there). All i said is thank you so much for every activity to me and i'm sorry for  my bad english (english is not my native language) and my messy post.</p>

答: Hint: compare the coefficient of $x^n$ on both sides of $(1+x)^{2n} = ((1+x)^n))^2$.</p>


问:
Calculating Overtake Time for Two Cars
Yes I've seen several similar questions but they all seem to not have the same provided information as what I'm looking for. Given my very rusty math skills I've been unable to convert the answers I've found there to my use.

Car A is traveling at 20mph. Car B is traveling at 40mph and is 1 mile behind car A. How long will it take for Car B to overtake Car A?

I'm really looking for the correct formula here. So I could reword this to look more like:

Car A is traveling at a mph. Car B is traveling at b mph and is d miles behind car A. How much time (t) in hours will it take Car B to overtake Car A?

Thanks ahead of time. I've been trying to turn this into a formula and have been humbled by my inability to wrap my ahead around what seems to me to be a simple question.

答: Start by snowing B is approaching A at b - a mph.</p>

Title: Calculating Overtake Time for Two Cars


问:
Range of numbers not divisible by $2$ or $3$ or $5$
How many numbers lie in the range of $1331$ and $3113$ (Endpoints Inclusive) such that they are not divisible by $2$ or $3$ or $5$?

I came across this Number System question and was not able to figure out the way to get the answer. At first I thought to separate out the numbers which are divisible by the LCM of $2$, $3$ & $5$ but this will still leave some numbers which will be exclusively divisible by any one of $2$ or $3$ or $5$.
Another method I found out to solve these type of questions is by using the Euler's Totient method but I am not aware of this.
So can someone please explain a general way to solve these types of question?

General question can be : How many numbers lie in the range of $A$ and $B$ such that they are not divisible by $x$ or $y$ or $z$ and so on?

答:
This is a classic application of the Inclusion-Exclusion Principle.

I will write $a|b$ as a shorthand for "$a$ divides $b$". You want to know how many numbers (between 1331 and 3113) are not divisible by 2, 3, or 5. First I will count the number divisible by either 2, 3, or 5. So let $1331 \le n \le 3113$ (the question does not specify whether the endpoints are included, but at this point it doesn't matter).

We want to count the number of $n$ satisfying:

$2|n$ OR $3|n$ OR $5|n$.

You count 891, 594, 356. This is the first term in the inclusion-exclusion formula. The answer is not just $891+594+356$, because we have double-counted those numbers in the overlap - namely, those numbers such that:

($2|n$ AND $3|n$), OR ($2|n$ AND $5|n$), OR ($3|n$ AND $5|n$)

This group can be simplified to:

$6|n$ OR $10|n$ OR $15|n$

There are 297, 178, 119 in each group. So to remove the double-counted elements, we should take them away from our sum, right?

Well, $(891+594+356) - (297 + 178 + 119)$ is still not the answer because we have removed too many items. To cut a verbose story short, consider those numbers divisible by 2, 3 and 5 (i.e. by 30). Well, We counted them three times in the first counting, and removed them three times in the second counting.

So we have to add them back in, and there are 59 such numbers. So the final answer is

$$891+594+356 - 297-178-119 + 59 = 1306$$

So your final answer is 477, if "between" includes the endpoints for you.

The general question

If $x, y, z$ are coprime numbers, then the general question is exactly the same as above. You just do
$$(\lfloor B/x \rfloor - \lfloor A/x \rfloor) + (\lfloor B/y \rfloor - \lfloor A/y \rfloor) + \ldots - (\lfloor B/xy \rfloor - \lfloor A/xy \rfloor) - \ldots + (\lfloor B/xyz \rfloor - \lfloor A/xyz \rfloor) .$$

If they aren't coprime, then i t's slightly different - suppose for example $x=6$, $y=9$, $z=12$. Then "$6|n$ AND $9|n$" is equivalent to $18|n$, not to $54|n$ as you might naively guess.


问:
The lowest weight of a representation is $w_0 \lambda$
There is the well-known fact that if $\lambda$ is the highest weight of $V$ a finite-dimensional irreducible $\mathfrak{g}$-module, then $w_0 \lambda$ is the lowest weight (here $\mathfrak{g}$ is any complex semisimple Lie algebra), where $w_0$ is the longest element of the Weyl group.
Why is this true?

Related questions have been asked on MSE, but not answering why, only giving the statement.
It has almost been answered in this MO post but not concretely for me.

Specifically, this is the proof I am reading:

The set of weights of $V$, denote by $\chi(V)$, are invariant under the Weyl group action.
Let $Q_+$ be the semigroup generated by $\Phi^+$.
Since $w_0 \cdot Q_+ = -Q_+$, we have $w_0 (\lambda) \prec \mu$ for all $\mu \in \chi(V)$, which implies the assertion.

Why is the bolded statement correct? I'm not even sure it is, just graphically looking at the picture of the weights of the representation of $\mathfrak{sl}_3$ with highest weight $\lambda = \varpi_1 + 2 \varpi_2$ below.

The weight marked with the open circle at $2\pi/3$ representing $s_2 s_1 (\lambda - \alpha_2)$ does not graphically appear to be a higher weight than $w_0(\lambda)$.
What am I misunderstanding here/how do I reason the above statement algebraically?

答:
Perhaps a picture says more than words here.

If $x = s_2 s_1(\lambda - \alpha_2)$, then the green region contains all weights which are $\preceq x$. The green region is the region $x - \mathbb{R}_{\geq 0}\alpha_1 - \mathbb{R}_{\geq 0}\alpha_2$: every point that can be written as $x$ plus some strictly negative linear combination of simple roots, or equivalently $x$ plus some strictly negative linear combination of simple roots. (Beware: this order is usually defined by $y \preceq x$ if and only if $y \in x - \mathbb{N}\alpha_1 - \mathbb{N} \alpha_2$, so even though it looks like $0 \preceq \lambda$ from the region-description, actually $0$ and $\lambda$ are incomparable).

Algebraically, we have that $y \preceq x$ if and only if $y + q = x$ for some $x \in Q^+$. Applying $w_0$ to both sides gives that $w_0 y = -w_0 q + w_0 x$, and since $- w_0 q \in Q^+$ we have $w_0 x \preceq w_0 y$.


问:
Winning probability to choose a chip of a box.
The Player got 10 blue and 10 red chips, and he was offered a game where he should distribute those chips into 2 similar boxes. (All chips should be placed in one of the boxes.) Then, another person randomly chooses a box and then takes out one random chip from this box (unless the box is empty). The player wins if a red chip is drawn and lose otherwise. What's the maximum probability of the player's winning outcome assuming that he distributes chips the right way?

I have been trying this problem for the past 1 week and not able to proceed further.
The best I have reached is:

P(getting R) = P(getting Red | in chosen Box 1)*P(choosing Box 1) + P(getting Red | in chosen Box 2)*P(choosing Box 2)

But not able to think of a method to maximise the probability.
Can anyone please help me with this?

答:
Another derivation.

Without loss of generality we may assume that the left box contains $n\le10$ chips with $p$ chips from them being red ($0\le p\le n$).

Then the probability to win
$$
\frac12\left(\frac pn+\frac{10-p}{20-n}\right)=\frac{5n+p\,(10-n)}{n\,(20-n)}
$$
for $n<10$ is obviously maximized by the choice of the largest possible $p$, i.e. $p=n$, so that the expression to maximize is:
$$
\frac{5n+n\,(10-n)}{n\,(20-n)}=\frac{n(15-n)}{n(20-n)},
$$
which is clearly minimized by the choice of the smallest possible $n$ except for $0$, i.e. $n=1$.

That means that one red chip should be placed in one box, and the other 19 chips in the other box.


答: To elaborate on Ma Ming's answer, note that every cyclic group $C$ of even order $2b$ has a unique element of order $2$, namely the $b$-th power of a generator, which defines an order-$2$ subgroup in $C$.  Your question concerns the group $\mathbb{Z}_m^*$ which is the product of the groups $\mathbb{Z}_{p_i}^*$, each of even order. The subgroup of $\mathbb{Z}_m^*$ given by the product of the order-$2$ subgroups as above is precisely the collection of the "square roots" you are interested in. Since there are $k$ of these order-$2$ subgroups, the order of the product is $2^k$.</p>

Title: Prove that if $p_1,\dots,p_k$ are distinct odd primes then 1 has $2^k$ square roots $\mod m$ where $m$ is the product of the primes.

Body:
I think I am most of the way through this proof but I am stuck. Here was my approach: I looked at the square roots of $1$ mod $105$, and noticed that each one corresponded to one less than an integer multiple of a subset of the prime factors of $105$, i.e. we have for instance $71=24\cdot3-1$ or $29=2\cdot(3\cdot 5)-1$ and so on. I conjectured that given a subset of the prime factors of $m$, I could construct a unique square root of $1$ mod $m$, which would show that there are (at least) $2^k$ of them.

So I did that and I think that it worked. Pick some set of the prime factors of $m$, say $p_{a_1},\dots,p_{a_j}$. Claim: we can find a corresponding $n$ such that $n\prod_{i=1}^jp_{a_i}-1$ is a square root of $1$ mod $m$.

Proof: We are looking for a solution to

$$\left(n\prod_{i=1}^jp_{a_i}-1\right)^2\equiv 1 \pmod m$$
which may be rewritten as

$$\left(\prod_{i=1}^jp_{a_i}\right)^2n^2 - 2\prod_{i=1}^jp_{a_i}n\equiv 0 \pmod m$$

Now since $\prod_{i=1}^jp_{a_i}|m$ we may write this as $$\prod_{i=1}^jp_{a_i}n^2-2n \equiv 0\mod \frac{m}{\prod_{i=1}^jp_{a_i}}$$

or

$$n\left(\prod_{i=1}^jp_{a_i}n-2\right) \equiv 0 \mod {\frac{m}{\prod_{i=1}^jp_{a_i}}}$$

This has solutions when $n\equiv0\mod \frac{m}{\prod_{i=1}^jp_{a_i}}$ or when $\prod_{i=1}^jp_{a_i}n-2\equiv 0 \mod \frac{m}{\prod_{i=1}^jp_{a_i}}$. In the first case we end up with the same root for any choice of $p_{a_i}$'s so that's sort of a degenerate one. But in the second case we have $\prod_{i=1}^jp_{a_i}n\equiv 2 \mod \frac{m}{\prod_{i=1}^jp_{a_i}}$, which has a unique and nonzero solution for $n$. Plugging that $n$ back into $n\prod_{i=1}^jp_{a_i}-1$ thus gives a square root of $1$ mod $m$ which corresponds uniquely to your choice of prime factors, thus showing that there are (at least) $2^k-1$ square roots of 1. Then we add 1 in which corresponds to the empty set of prime factors and we get the full $2^k$.

My problem is ruling out the possibility of more square roots of 1, i.e. if $x^2\equiv1\mod m$ I need to show that it corresponds uniquely to something of the form $n\prod_{i=1}^jp_{a_i}-1$. This is proving to be difficult if not impossible.

I put a lot of work into going the one way and I really don't want it all to be for nothing. Can anyone see a way to go the other direction?


问:
Probability St Petersburg Game
I am reading a book on probability and there is an interesting chapter on the St Petersburg game - where a coin is flipped until a head is landed and the prize is £2 if there is a head on the first throw, £4 is there is a head on the second flip, £8 for a head on the third etc. The idea of this game is it introduces a paradox since someone should be prepared to pay any amount to play since the expected yield is infinite.

However in the book it introduces the notion of changing the rewards to 2,4,6,8.... etc rather than the doubling in the original game. The book contends that even though the possible payouts increase without bound the calculation of the expected value yields a sensible answer of £4. It mentions that this is simple to calculate so omits the calculations. I have pondering this and cant see how the £4 has been arrived at - a entry price of £4 would be a fair price for the game but why? Any help greatly appreciated. I am sure I'm missing something to do with summing to infinity perhaps

答:
If you are familiar with the series
$\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots,$
you may recall that it adds up to $1$:
$$ 1 = \frac12 + \frac14 + \frac18 + \frac1{16} + \cdots, $$
which is just a number-heavy way to say that the probability is $1$ that the coin eventually will come up heads. (Probability $\frac12$ on the first throw, $\frac14$ that it takes exactly two throws, $\frac18$ that it takes exactly three throws, etc.)

If we multiply all the terms by $2$ we get twice as much for the sum:
\begin{align}
 2 &= 1 + \frac12 + \frac14 + \frac18 + \cdots, \\
 4 &= 2 + 1 + \frac12 + \frac14 + \cdots.
\end{align}

If we divide all the terms by $2$ we get half as much:
\begin{align}
\frac12 & = \frac14 + \frac18 + \frac1{16} + \frac1{32} + \cdots,\\
\frac14 & = \frac18 + \frac1{16} + \frac1{32} + \frac1{64} + \cdots,\\
\frac18 & = \frac1{16} + \frac1{32} + \frac1{64} + \frac1{128} + \cdots,\\
\end{align}
and so forth.

Rearranging terms in a series can alter the sum if some terms are positive and some are negative; but in the series you are looking at, all terms are positive, so we can write:
\begin{align}
2\cdot\frac12 + 4\cdot\frac14 &+ 6\cdot\frac18 + 8\cdot\frac1{16} + \cdots\\
&= 1 + 2\cdot\frac12 + 3\cdot\frac14 + 4\cdot\frac18 + \cdots\\
&= 1 + (1+1)\cdot\frac12 + (1+1+1)\cdot\frac14 + (1+1+1+1)\cdot\frac18 + \cdots\\
&= 1 + \frac12 + \frac14 + \frac18 + \cdots\\
&\phantom{{}= 1} + \frac12 + \frac14 + \frac18 + \cdots\\
&\phantom{{}= 1 + \frac12} + \frac14 + \frac18 + \cdots\\
&\phantom{{}= 1 + \frac12 + \frac14} + \frac18 + \cdots\\
&\phantom{{}= 1 + \frac12 + \frac14 + \frac18} + \cdots\\
&= 2 \\
&\phantom{{}= 2} + 1\\
&\phantom{{}= 2 + 1} + \frac12\\
&\phantom{{}= 2 + 1 + \frac12} + \frac14\\
&\phantom{{}= 2 + 1 + \frac12 + \frac14} + \cdots\\
&= 4.
\end{align}

Title: Probability St Petersburg Game


问:
Showing that the log is the inverse of the exponential
I know this question may seem easy however I am not "meant" to know any facts about log other than it is defined by $L=\int_{1}^{x}\frac{1}{t}dt$ and that the exponential function is given by the usual sum.

To start with I composed $L \circ E $ which fell out to get $x$ which is what I wanted. However I am struggling with the other way: $E \circ L $. So far I have that:

$$E \circ L =\sum_{n=0}^{\infty} \frac{(\int_{1}^{x}\frac{1}{t}dt)^n}{n!}$$

But I cannot see any obvious way to show this is equal to $x$ for $x\gt0$. Can anyone help me out?

答:
Same idea as @macton and @CameronWilliams.

We will define $$F=E\circ L=\sum_{n=0}^{\infty}\frac{1}{n!}\left[\int_{1}^{x}\frac{1}{t}\,\mathrm{d}t\right]^n.$$
We want to take the derivative of $F(x)$. Note that the Mc-Laurin series of the exponential function converges (absolutely) for all $x\in\mathbb{R}$. That means that we can interchange derivative and summation.
$$\frac{\mathrm{d}}{\mathrm{d}x}F(x)=\sum_{n=0}^{\infty}\frac{1}{n!}\frac{\mathrm{d}}{\mathrm{d}x}\int_{1}^{x}\frac{1}{t}\,\mathrm{d}t=\frac{1}{x}\sum_{n=0}^{\infty}\frac{1}{n!}\left[\int_{1}^{x}\frac{1}{t}\,\mathrm{d}t\right]^n.$$
We derived to the last step by using the chain rule and the fundamental theorem of calculus. Note that this series is equal to $F(x)$. So we may rewrite
$$F'(x)=\frac{1}{x}F(x).$$
This leads to an easy differential equation, with the solution $F(x)=x$.
That means that
$$E\circ L=x,$$
so we have proven that $L$ is the inverse function of $E$.


问:
Show that $f_n(x) = \frac{\sin{(nx+3)}}{\sqrt{n+1}}$ converges pointwise
I need to show that this sequence function converges pointwise with the help of ($\delta-\epsilon$) definition.

I tried to first split this sequence by sandwich theorem then I am stuck at applying pointwise convergence definition that how should I choose $\epsilon$ and $N$.

答:
$|f_n(x)| \le \frac{1}{\sqrt{n+1}}$ for all $x$ and all $n$. Hence $(f_n)$ converges uniformly (and hence pointwise) on $ \mathbb  R$ to $0$.

Can you proceed ?


问:
$ A \subseteq \mathbb{N} $ is infinite set, Show existence of strictly monotonic increasing sequence...
Problem: If $ A \subseteq \mathbb{N} $ is an infinite set then there exists a strictly monotonic increasing sequence $ (n_k)_{k=1}^\infty $ s.t. $ A = \{ n_k | k \in \mathbb{N} \} $

My attempt of  proof:

Perform the following algorithm,

step 1: Define $ n_1 = minA $

step 2: $ n_2 = min( A\setminus \{ n_1 \}) $

step 3: $ n_3 = min(A\setminus \{ n_1 , n_2 \}) $

$ \vdots$

step j: $ n_j = min(A\setminus \{ n_1 , n_2, ...,n_{j-1} \}) $

We'll perform $ |A|  $ steps and define each time  $ n_i $ for all $ 1 \leq i \leq |A| $.

Noticing that $ n_1 < n_2 < ... < n_{ |A| } $ ( we have a strictly monotonic sequence ) and that $ n_i \in A $ for all $ 1 \leq i \leq |A| $ so we're finished.

More compact attempt of proof:

Define $ n_1 = minA $ . For all $ n_i \in A $ s.t. $ 2 \leq i \leq |A| $, we'll define  $ n_i = min(A\setminus \{ n_1 , n_2, ...,n_{i-1} \}) $

Noticing that $ n_1 < n_2 < ... < n_{ |A| } $ and that $ n_i \in A $ for all $ 1 \leq i \leq |A| $ so we're finished.

I don't know if I'm correct since $ A $ is an infinite set and I haven't seen proof by algorithms in which the iteration is continuing infinitely ( there are $ |A| $ iterations in the algorithm above ), neither I have seen the usage of indexing on an infinite set ( for example, I have wrote: " for all $ 1 \leq i \leq |A| $ , but $ |A| $ is not a finite number since $ A $ is infinite set " ). Are my proofs correct? if not, what is the problem with them?

答: You should not be using $|A|$ anywhere since $|A|=\infty$. However your algorithm for constructing the sequence is fine. After describing it (step j) you can claim the sequence is monotonic increasing by construction. Then you use the infinite size of $A$ to ensure that there always exists a $n_{j+1} > n_j$ because the set $A\setminus \{ n_1 , n_2, ...,n_{j} \}$ is also infinite.


问:
Confusion on the proof of Haim Brezis Proposition 1.9 (application of Hahn-Banach Geometric Form)
I am reading Haim Brezis's functional analysis, and I am confused by on of his proof.

Let $(V, \|\ \cdot\ \|)$ be a Banach space on $\mathbb{C}$ and let $V^{*}$ be its dual, endowed with the dual norm $$\|f\|_{V^{*}}=\|f\|_{op}=\sup_{\|x\|_{V}\leq 1}|\langle x,f \rangle_{V, V^{*}}|.$$ For $M\subset V$, we define the annihilator of $M$ as $$M^{\bot}:=\{f\in V^{*}:\langle x,f\rangle_{V,V^{*}}=0\ \ \text{for all}\ \ x\in M\}\subset V^{*}.$$ and similarly for $N\subset V^{*}$, we define $$^{\bot}N:=\{x\in V:\langle x,f\rangle_{V,V^{*}}=0\ \ \text{for all}\ \ f\in N\}=\bigcap_{f\in N}\ker(f)\subset V.$$

Proposition 1.9 states the following:

Let $M\subset V$ be a linear subspace. Then, $$^{\bot}(M^{\bot})=\overline{span\ M}.$$

I have proved $\overline{span\ M}\subset ^{\bot}(M^{\bot})$, and the other direction $\overline{span\ M}\supset ^{\bot}(M^{\bot}$ is an application of Hahn-Banach theorem geometric form. The proof goes as follows:

Suppose by contradiction that there exists some $x_{0}\in ^{\bot}(M^{\bot}$ such that $x_{0}\notin \overline{span\ M}.$ It is clear that $\{x_{0}\}$ and $\overline{span\ M}$ then has empty intersection, both convex, and the first one is compact, the second one is closed.

Then, by Hahn-Banach theorem, second geometric form, there is a closed hyperplane that strictly separates $B=\{x_{0}\}$ and $A=\overline{span\  M}$. That is, there exists some $f\in V^{*}$ and $\alpha\in\mathbb{R}$ such that $$\langle x,f\rangle_{V,V^{*}}<\alpha<\langle x_{0}, f\rangle_{V,V^{*}}\ \ \text{for all}\ \ x\in \overline{span\ M}.$$

So far so good. But then, Brezis claimed that

Since $M$ is a linear space, it follows that $\langle x, f\rangle_{V,V^{*}}=0$ for all $x\in M$.  Why this is true? I thought about this for a while but don't follow why linear space implies this result...

The following is the statement of Hahn-Banach 2nd geometric form:

Let $A,B\subset V$ be two nonempty convex subsets such that $A\cap B=\varnothing$. Assume that $A$ is closed and $B$ is compact. Then, there exists a closed hyperplane that strictly separates $A$ and $B$. That is, there exists some $f\in V^{*}$ and $\alpha\in\mathbb{R}$ such that $$\langle x,f\rangle<\alpha\ \ \text{for all}\ \ x\in A\ \ \text{and}\ \ \langle x,f\rangle>\alpha\ \ \text{for all}\ \ x\in B.$$

答:
Replace $x$ by $\beta x$ in your inequality, with arbitrary $\beta$.

More abstractly, the image of $f$ is a subspace of $\mathbb{R}$ because $f$ is linear, and you already know it's not all of $\mathbb{R}$ from your inequality.


答:
Hint: Write it as
$${n \choose 0} {n \choose n} + {n \choose 1} {n \choose n-1} + \dots + {n \choose n} {n \choose 0},$$
and use the obvious combinatorial bijection.</p>

Title: How to get ${n \choose 0}^2+{n \choose 1}^2+{n \choose 2}^2+\cdots+{n \choose n}^2 = {x \choose y}$

Body: I found this in my test book, any hints? Given $${n \choose 0}^2+{n \choose 1}^2+{n \choose 2}^2+\cdots+{n \choose n}^2 = {x \choose y}$$ Then find the value of x and y in n. According to the answer provided on last pages of that book, it's $2n \choose n$. What i dont understand is how to get the answer (there's no explanation written there). All i said is thank you so much for every activity to me and i'm sorry for  my bad english (english is not my native language) and my messy post.</p>


问:
Pumping lemma definition!
I was studing Pumping lemma and I saw wondering and confusing definition,"Pumping length".

Here is my problems:

Question 1:

Is pumping length = number of states in DFA?

Question 2:

If any string can be divided into $\{xy^iz$ | $ i≥ 0\}$ , why $\vert xy\vert ≤$ Pumping length ?

I know Pigeonhole principle and how it used to epxlain we have a loop in DFA.

But I dont know how to use Pigeonhole principle to proof $\vert xy\vert ≤$ pumping length .

I tried but gather nothing:

I also checked this links but nothing was clear to me:

First link

Second one

Thanks for any kind of help ;)

答:
On your first question: This one is a bit ill-posed because you don't specify what (D)FA you are talking about. The PL only says that for every regular $L$, there exists some integer $p$ such that all words in $L$ of length at least $p$ can be pumped, i.e. it doesn't uniquely specify such a $p$ (of course one could fix that by requiring $p$ to be minimal).

If we take a look at the proof of the PL, we see that $p$ is chosen to be the number of states of a DFA recognizing $L$, so in that sense the answer to your question is yes, but as I laid out, the "pumping length" is really referring just to the specifically chosen $p$.

On the second question: This is usually stated as $|xy| \leq p$.
The idea of the PL is that the run of our DFA on some pumpable word can be segmented into three parts: One that leads to a loop ($x$), then the loop itself ($y$), and finally the part that leads from that loop to an accepting state ($z$). The proof argues that such a decomposition exists for all words of length at least $p$ by (as you said) employing the pidgeonhole principle.
Now, this same argument tells us that a loop must already occur after at most $p$ symbols and hence, $|xy|$ is at most $p$.

I want to close with saying that the PL can be completely stated without the term "pumping length" (see e.g. here on wikipedia) and if it causes you any trouble, you can always just disregard it.

Title: Pumping lemma definition!