问:
Consequence of definition of Boolean algebra
The following is from Introduction to Mathematical Logic, Sixth Edition by Mendelson, Elliott

Let $B$ be a nonempty set, $f$ a function from $B$ into $B,$ and $g$ a function from
$B^2$ into $B.$ Write $x'$ for $f(x)$ and $x \cap y$ for $g(x, y).$ Then $〈B, f, g〉$ is called a Boolean
algebra if $B$ contains at least two elements and the following conditions are
satisfied:

(1) $x \cap y = y \cap x$ for all $x$ and $y$ in $B.$

(2) $(x\cap y)\cap z = x \cap (y\cap z)$ for all $x,y,z$ in $B.$

(3) $x\cap y' = z\cap z'$ if and only if $x\cap y = x$ for all $x,y,z$ in $B.$

Question : Prove that $z\cap z' = w\cap w'$ for any $w$ and $z$ in $B.$

My attempt :

Let $w\in B.$

It appears that if $w=x$ and $w=y$ in $(3)$ then what I want to show reduces to showing that $w\cap w = w$ for any $w\in B.$

I tried assuming that $w\cap w \ne w$ to bring forth a contradiction but couldn't get anywhere.

Any hints?

答: Taking $x=y=z=w$ in $(3)$, we have $w\cap w'=w\cap w'$ iff $w\cap w=w$; since it’s certainly true that $w\cap w'=w\cap w'$, it must be true that $w\cap w=w$.


问:
Function satisfying the relation $f(x+y)=f(x)+f(y)-(e^{-x}-1)(e^{-y}-1)+1$
Let f be the differentiable function satisfying the relation  $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$; $\forall x,y \in R$ and $\mathop {\lim }\limits_{h \to 0} \frac{{f'\left( {1 + h} \right) + f\left( h \right) - {e^{ - 1}}}}{h}$ exist. The value of $\int\limits_0^1 {f\left( x \right)dx}  = \_\_\_\_\_\_\_$.

My approach is as follow

$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$

$\Rightarrow f\left( {x + y} \right) - f\left( x \right) = f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$

y=h

$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - h}} - 1} \right) + 1}}{h}$

How will I proceed from here

答:
If you define $g(x) = f(x) + e^{-x}$, then the equation becomes $g(x + y) = g(x) + g(y)$.

It is well-known that all continuous solutions to that equation are linear, i.e. there exists $c$ such that $f(x) = cx$.

The rest should be easy.

ContentLicense: CC BY-SA 4.0


问:
$i^{th}$ root(s) of unity
If we define $S:=\{z\in\mathbb{C}:z^n=1\}$ (i.e. the $n^{th}$ roots of unity), then $|S|=n$ (i.e. we have $n$ of them). We can even go as far as to say: $$S=\{z_k:k\in\mathbb{N}\cap[1,n]\}=\{e^{2\pi i/k}:k\in\mathbb{N}\cap[1,n]\}$$
Plotting these points on an Argand diagram shows the above point, namely that the plot looks symmetric across the real axis, and has rotational symmetry order $n$. It follows that, if we let $n\to\infty$, the plot of points tends towards being a circle centred on the origin, radius $1$.

But $z^i$, and even $z^{1/i}$, are both (possibly) well-defined ideas, so can we talk about the $i^{th}$ roots of unity? How many of them are there? What geometric interpretation is there when we talk about them, and can we analogously look at the $2i^{th}$ roots, etc.?

I know that $z^{1/i}:=e^{(1/i)\log{z}}$, so $z^{1/i}=1$ means that we search for solutions to $(1/i)\log{z}=0 \Rightarrow \log{z}=0$.

I guess a similar question is: what geometric/insightful interpretation is there to the fact that $i^i=e^{(-\pi+4k)/2}$ (for $k\in\mathbb{Z})$?

Edit: turns out that I can't multiply things by $0$ correctly... mistake fixed, thanks.

答:
Define $\lg(z)$ to be the standard branch cut of $\ln$, discontinuous on the negative reals. Now since $\exp$ is periodic with period $2\pi i$, $\exp(y)=z\implies y=\lg(z)+(2\pi i)n$ for some $n\in\mathbb{Z}$.

So if you would like $z^i$ to mean $\exp(i\cdot\lg(z))$, then $$\begin{align}z^i=1&\implies\exp(i\cdot\lg(z))=1\\
&\implies i\cdot\lg(z)=\lg(1)+(2\pi i)n\\
&\implies \lg(z)= 2\pi n\\
&\implies z=\exp(2\pi n)\\
\end{align}$$

So the collection of $i$th roots of unity is countably infinite, consisting of the following positive real numbers: $$\{\ldots,e^{-4\pi},e^{-2\pi},1,e^{2\pi},e^{4\pi},\ldots\}$$
Repeating this for general complex roots, the $w$th roots of unity are the countable collection $$\{\exp(2\pi i n/w)\mid n\in\mathbb{Z}\}$$ (which may be merely finite if $w$ is such that there is overlap in this collection.)

ContentLicense: CC BY-SA 3.0

Title: $i^{th}$ root(s) of unity


问:
What is the min amount of $n$ so that the following conclusion be true? $a^n | b^3 \to a^3 | b^2$
What is the min amount of $n$ so that the following conclusion be true?

$a^n | b^3 \to a^3 | b^2$

We know If $a^n|b^3$ then there exists a $k \in \mathbb{Z}$ so that $b^3=k.a^n$ so :

\begin{align}
b^3 =k.a^n\to b^2 =a^n.k.b^{-1}\to b^2 =a^3 . a^{n-3}.k.b^{-1}
\end{align}
So If we want $a^3 |b^2$ to be true, $a^{n-3}.k.b^{-1}$ must belong to $\mathbb{Z}$. We have to check for at least which $n$ this is true? for $n=4$ I found a counter example, but I couldn't find a clear way to find the min of $n$.

答:
You already found counterexample for $n = 4$. So it suffices to show that $n = 5$ works.

Suppose we have $a^5 \mid b^3$. Taking squares gives $a^{10} \mid b^6$.

In particular, it implies $a^9 \mid b^6$, which in turn implies $a^3 \mid b^2$.


问:
What's the probability that the news station predicted the outcome incorrectly?
I wrote this problem for a high school statistics activity on the sampling distribution of sample proportions. When I sat down to write solutions I realized that it was maybe too confusing to assign, and now I'm not confident in my solution. Is my attempt a reasonable solution?

Question:

Suppose that during an election, a news agency that conducts a poll of 105 random voters and finds a sample proportion of 54% in support of a ballot measure that requires a majority to pass. The agency publishes a prediction that the measure will pass. How likely is it that the news agency is incorrect?

Attempt at a solution:

Let $n = 105,$ $\hat p=0.54$, and we use $ \sigma = \sqrt{\dfrac{0.25}{105}} = 0.0488$ as the worst case, since $p$ is unknown.

Notice that 0.54 is $0.4 = 0.82 \sigma$ away from $0.50$, which is the threshold for a “majority”. The proportion of samples with sample proportions that have z-scores within $.82$ of $0$ is $normalcdf(-.82,.82,0,1) = 0.587$.

So, 58.7% of random samples will have sample proportions within $.82\sigma$ of the true population proportion. In order for the news agency to be incorrect, $ \hat p=0.54$ must be more than $0.82 \sigma$ above the true population proportion.

Because of the symmetry of the normal distribution, at most $(1 - 0.587)/2 = 20.65\%$ of random samples will be at least this far above true proportion. So, since the sample was random, there is at most a $20.65\%$ chance that the outcome of the election will disagree with the news agency’s prediction.

答:
Printout from a recent release of Minitab statistical software. I entered 57 Yes's out of the 105 interviews,
this gives the estimate $\hat p = 542857,$ not $\hat p = 0.54.$ Why not give the actual number in favor?

```<code>Test and CI for One Proportion

Test of p = 0.5 vs p > 0.5

                                              Exact
Sample   X    N  Sample p  95% Lower Bound  P-Value
1       57  105  0.542857         0.457999    0.218
</code>```

P-value exceeds 5% = 0.05, so you cannot reject $H_0.$
The actual percentage of the population in favor might be as low as 45.8%.

With so small a sample, it is not surprising
that the poll does not give useful results.
In election polling, it takes about $n=600$ subjects to get minimally useful results. This sample size gives a 2-sided confidence
interval with a margin of sampling error about $\pm 4\%.$ (For $\pm 2\%$ use $n = 2500.)$

Body:
Printout from a recent release of Minitab statistical software. I entered 57 Yes's out of the 105 interviews,
this gives the estimate $\hat p = 542857,$ not $\hat p = 0.54.$ Why not give the actual number in favor?

Test and CI for One Proportion

Test of p = 0.5 vs p > 0.5

                                              Exact
Sample   X    N  Sample p  95% Lower Bound  P-Value
1       57  105  0.542857         0.457999    0.218

P-value exceeds 5% = 0.05, so you cannot reject $H_0.$
The actual percentage of the population in favor might be as low as 45.8%.

With so small a sample, it is not surprising
that the poll does not give useful results.
In election polling, it takes about $n=600$ subjects to get minimally useful results. This sample size gives a 2-sided confidence
interval with a margin of sampling error about $\pm 4\%.$ (For $\pm 2\%$ use $n = 2500.)$


问:
How to solve $\int_{0}^{x} \frac{y^{m}}{(1+y/p)^{2m+2}} \, \mathrm{d}y$?
$$\int_{0}^{x} \frac{y^{m}}{(1+y/p)^{2m+2}} \, \mathrm{d}y$$

where $m$ and $p$ are constant. I tried to solve it using LIATE rule but getting stuck. Any help will be highly appreciated.

答:
1. If $m$ is a non-negativeinteger, then by writing

\begin{align*}
\frac{y^m}{(1 + y/p)^{2m+2}}
&= \frac{p^m (1 + (y/p) - 1)^m}{(1 + y/p)^{2m+2}} \\
&= p^m \sum_{k=0}^{m} (-1)^k \binom{m}{k} \frac{1}{(1 + y/p)^{m+k+2}},
\end{align*}

we get

\begin{align*}
\int \frac{y^m}{(1 + y/p)^{2m+2}} \, \mathrm{d}y
&= p^{m+1} \sum_{k=0}^{m} \frac{(-1)^{k+1}}{m+k+1} \binom{m}{k} \frac{1}{(1 + y/p)^{m+k+1}}.
\end{align*}

2. Alternatively, apply the integration by part $m$ times:

\begin{align*}
&\int \frac{y^m}{(1 + y/p)^{2m+2}} \, \mathrm{d}y \\
&= -\frac{p}{2m+1} \frac{y^m}{(1 + y/p)^{2m+1}} + \frac{pm}{2m+1} \int \frac{y^{m-1}}{(1 + y/p)^{2m+1}} \, \mathrm{d}y \\
&\hspace{4em}\vdots \\
&= - \sum_{k=1}^{m+1} p^k  \frac{\prod_{j=1}^{k-1} (m+1-j)}{\prod_{j=1}^{k} (2m+2-j)} \frac{y^{m+1-k}}{(1 + y/p)^{2m+2-k}} \\
&= - \sum_{k=1}^{m+1} p^k  \frac{m!/(m-k+1)!}{(2m+1)!/(2m-k+1)!} \frac{y^{m+1-k}}{(1 + y/p)^{2m+2-k}}
\end{align*}

3. Yet another way is to write $q = -1/p$, and then starting from

$$ \int \frac{1}{(1 - qy)^{m+2}} \, \mathrm{d}y = \frac{1}{m+1} \frac{1}{q(1 - qy)^{m+1}}, $$

apply the $m$-fold differentiate with respect to $q$ to both sides:

\begin{align*}
\frac{(2m+1)!}{(m+1)!} \int \frac{y^m}{(1 - qy)^{2m+2}} \, \mathrm{d}y
&= \frac{\partial^m}{\partial q^m} \int \frac{1}{(1 - qy)^{m+2}} \, \mathrm{d}y \\
&= \frac{1}{m+1} \frac{\partial^m}{\partial q^m} \left( \frac{1}{q(1 - qy)^{m+1}} \right)
\end{align*}

The last line is not hard to compute using the general Leibniz rule.

In case $m$ is no longer an integer, I suspect that we will need special functions such as incomplete beta function.


问:
Equation of latus rectum of parabola
If parabola $\left ( y+1 \right )^{2}=k\left ( x-2 \right )$ passes through a point (1,-2), then equation of its latus rectum and directrix ?
My attempt

Is this correct answer?
because this answer is not mention in given options.

答:
The parabola is $$(y+1)^2=(2-x)~~~~(1)$$
The most general form of parabpla is $$L_1^2=4AL_2$$ if $L_1$ and $L_2$are non-parallel. If these two are parallel and normalised $(L=\frac{ax+by+c}{\sqrt{a^2+b^2}}),$ Then length of its latus rectum is $4A$, the Eq. of directrix is $L_2=-A$, Eq. of its latus rectum is $L_2=A$. Eq. of its axis is $L_1=0$ tangent at vertex is $L_2=0$, focus is given by $(L_2=A, L_1=0)$ vertex is $(L_2=0, L_1=0).$

Eq. of directrix for (1) is $2-x=-1/4 \implies x=9/4$ and Eq. of latus rectum is $2-x=1/4 \implies x=7/4$

Title: Equation of latus rectum of parabola


答:
Continuing from your method $$f'(x)=\lim_{h\to 0} \frac{f(h)+1 -(e^{-x}-1)(e^{-h}-1))}{h}$$ for derivative to exist $f(0)=-1$ hence by L-Hospitals rule $$f'(x)=\lim_{h\to 0}\frac{f(h)+1}{h}-\lim_{h\to 0}(e^{-x}-1)\frac{(e^{-h}-1)}{h}$$
$$f'(x)=f'(0)+e^{-x}-1=e^{-x}+C\Rightarrow f(x)=-e^{-x}+Cx$$
Now for the given limit to exist $f'(1)+f(0)=e^{-1}$...hence find $C$...

Title: Function satisfying the relation $f(x+y)=f(x)+f(y)-(e^{-x}-1)(e^{-y}-1)+1$

Body:
Let f be the differentiable function satisfying the relation  $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$; $\forall x,y \in R$ and $\mathop {\lim }\limits_{h \to 0} \frac{{f'\left( {1 + h} \right) + f\left( h \right) - {e^{ - 1}}}}{h}$ exist. The value of $\int\limits_0^1 {f\left( x \right)dx}  = \_\_\_\_\_\_\_$.

My approach is as follow

$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$

$\Rightarrow f\left( {x + y} \right) - f\left( x \right) = f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$

y=h

$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - h}} - 1} \right) + 1}}{h}$

How will I proceed from here


问:
Proving $\binom{2n}{n} \geq \frac{2^{2n}}{n+1}$ combinatorially
On trying to answer this question, with duplicate here, I was wondering if there is a combinatorial interpretation of the expression.

 I tried something like
$$n\binom{2n}{n} \geq 2^{2n}-\binom{2n}{n}=|A_n|$$ where $A_n=\{x\in \{0,1\}^{2n}:\text{number of 1's}\neq \text{number of 0's}\}$ and I could not find a meaningful injection from $A_n$ to $[n]\times \binom{[2n]}{n}$

One of the things I try was noticing that at some point the symbol with more appearances passes $n$ from left to right (i.e., there exists $k$ such that $\max \{|x_1\cdots x_k|_0,|x_1\cdots x_k|_1\}>n$, and so I would swap all of the symbols but is not injective.

I guess the opposite problem (i.e., finding a surjection) helps but seems harder. Also, the difference is not in the OEIS.

答:
In
Two bijections for the area of Dyck paths, E. Pergola and Area of Catalan paths, W. Woan
one can read that if $D$ is a Dyck path and $a(D)$ is the area under the path, then $$A_n=\sum _{D\text{ Dyck path length 2n}}a(D)=\frac{1}{2}\left (4^n-\binom{2n+1}{n}\right ).$$
check the oeis  of the sequence $2A_n$.

But by giving full area to all Dyck paths, one can bound this by $\frac{n^2}{2}\times C_n=\frac{n^2}{2(n+1)}\binom{2n}{n}$ so
$$2A_n=2^{2n}-\binom{2n+1}{n}\leq \frac{n^2}{n+1}\binom{2n}{n},$$
so
$$2^{2n}\leq \binom{2n}{n}\left (\frac{n^2}{n+1}+\frac{2n+1}{n+1}\right )=\binom{2n}{n}(n+1).$$


问:
Parametric equations for given line
How would you find the parametric equations for:

1) a line through $(3,1)$ and $(-5,4)$.

2) a segment joining $(1,1)$ and $(2,3)$.

Can anyone show me the steps of doing it cause the way my textbook shows it, it has gotten me in a loop.

答:
1) The point-slope form for the equation of a line: $y-y_0=m(x-x_0)$. So an equation for the line is $y=\frac{4-1}{-5-3}(x-3)+1$. A parametric equation here would be $$y=\frac{4-1}{-5-3}(t-3)+1$$$$x=t$$
2) You could use the same technique and restrict $t$, or you could restrict $t$ using another method. First, an equation for the line is $y=2(x-1)+1$. We want $1\le x\le2$, so we need a bounded function that goes from $1$ at $-\infty$ to 2 at $\infty$. One function that is similar is the tanh (hyperbolic tangent: $\displaystyle\frac{e^x-e^{-x}}{e^{-x}+e^x}$. The tanh function goes from $-1$ at $-\infty$ to $1$ at $\infty$, so $\frac12\tanh x + \frac32$ goes from $1$ to $2$, so this is something we can use.

The unrestricted parametric plot is:$$y=2(t-1)+1$$$$x=t$$
After substituting $\frac12\tanh t + \frac32$ for $t$, we get what we want:

$$y = 2(\frac12\tanh t + \frac32-1)+1$$
$$x = \frac12\tanh t + \frac32$$

Title: Parametric equations for given line


问:
Show that there exists $(k, n) \in \mathbb{N}^2$ such that $P \cap [\![ n, +\infty [\![ = k \mathbb{N} \cap [\![ n, +\infty [\![$
The exercise

Let $P \subseteq \mathbb{N}$ non-empty such that $\forall a, b \in P, a+b \in P$. Show that there exists $(k, n) \in \mathbb{N}^2$ such that $P \cap [\![   n, +\infty  [\![ = k \mathbb{N} \cap [\![   n, +\infty  [\![$.

My try

We easily get: $\forall (a_1, \dots a_p) \in P, \forall (l_1, \dots, l_p) \in \mathbb{N}, l_1 a_1+\dots+l_pa_p \in P$.

The case $P = \{ 0 \}$ is trivial, so let's assume $P - \{ 0 \}$ is non-empty and denote $m$ its minimum. I'm not sure this approach will work because if I take as $P$ the subset of finite sums of prime numbers, I can't even figure the values of $k$ and $n$.

I think there is a link with the gcd of some elements of $P$, but which ones? The "smallest" ones?

Thanks for the help.

答:
The answer was given by Greg Martin in a comment, let me write it down.

Let $k = \inf \{d \in \mathbb{N}^*:\ \exists p_1,...,p_n: \gcd(p_1,...,p_m)=d\}$.

We have $P \subset k\mathbb{N}$. Conversely, we can also find $n$ large enough so that $ k\mathbb{N} \cap [\![n,\infty[\![ \subset P \cap [\![n,\infty[\![$.

You can see at the following link a proof that a solution to the coin problem exists. Namely, with $p_1,...,p_m$ such that $\gcd(p_1,...,p_m)=k$, there exists $n$ such that for all $q \ge n$ with $k | q$, there exist $a_1,...,a_m \in \mathbb{N}$ such that $\sum \limits_{i=1}^m a_ip_i = kq$. Since $P$ is stable under addition, $\sum \limits_{i=1}^m a_ip_i \in P$, so $k \mathbb{N} \cap [\![n,\infty[\![ \subset P \cap [\![n, \infty[\![$.

Combining both gives you $P \cap [\![n, \infty[\![ = k\mathbb{N} \cap [\![n,\infty[\![$.


问:
Does it look like an expansion of $\log$?
For a real number $\lambda>0$ and a positive integer $N$, consider
$$
\sum_{\ell=1}^{\infty}\frac{(-\lambda)^{\ell}}{\ell}\cdot\frac{N!}{(N+\ell)!}.
$$
Could we simplify it further as a function of $\lambda$? For example, it equals to
$$
\sum_{\ell=1}^{\infty}\frac{N!}{(N+\ell)!}\int_{0}^{-\lambda}v^{\ell-1}dv=N!\int_{0}^{-\lambda}\sum_{\ell=1}^{\infty}\frac{v^{\ell-1}}{(N+\ell)!}dv.
$$
Please give me some help or hints. Thanks a lot.

答:
$$F_N=\sum_{\ell=1}^{\infty}\frac{(-\lambda)^{\ell}}{\ell}\cdot\frac{N!}{(N+\ell)!}$$ has the typical shape of a generalized hypergeometric function.

In fact, it is
$$F_N=-\frac{\lambda }{N+1}\,\, _2F_2(1,1;2,N+2;-\lambda )$$ Even for small values of $N$, the results are not the most pleasant
$$F_0= -\log (\lambda )-\Gamma (0,\lambda )-\gamma$$
$$F_1=\frac{(1-\gamma)  \lambda +e^{-\lambda }-\lambda  (\log (\lambda )-\Gamma
   (0,\lambda ))-1}{\lambda }$$


问:
If 0 is an equilibrium point and an attractor, then 0 is also stable?
I need to show that, given $x'=f(x)$ with $f$ Lipschitz, $f(0)=0$, and that $\varphi(t,x_0) \rightarrow 0$ as $t \rightarrow \infty$ for all $x_0 \in \mathbb{R}$, then $0$ is stable. I feel like you need to relate $\varphi$ to $f$ somehow but I'm very hazy on how flows actually work, so I feel pretty stuck with this.

答:
Well, think of the phase portrait. The only way for the origin to be unstable is if some solution moves away from it, and for systems in $\mathbb{R}$, such an outward-moving solution will be a barrier for other solutions to approach the origin from that direction.

Note, by the way, that this statement is false for systems in $\mathbb{R}^2$ (or higher dimension), as shown for example by the system
$$
\dot x = \frac{x^2 (y-x)+y^5}{1+(x^2+y^2)^2}
,\qquad
\dot y = \frac{y^2 (y-2x)}{1+(x^2+y^2)^2}
;
$$
It can be shown that all of its solutions tend to $(0,0)$ as $t \to +\infty$, and that the origin is unstable anyway.
(See phase portrait plotted on Wolfram Alpha to get an idea of how this is possible.)


问:
Calculating the limit of the following series
I want to prove that $$\lim_{x\to1^-}(1-x)\sum_{n=1}^{\infty}(-1)^{n-1}\frac{nx^n}{1-x^{2n}} = \frac14$$

So far I tried to manipulate the series for instance using $$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{nx^n}{1-x^{2n}} =  -\sum_{n=1}^{\infty}(-1)^{n}nx^n\sum_{m=0}^{\infty}\left(x^{2n}\right)^m$$ since $x < 1$. Interchanging the two sums (not sure if allowed) I obtained, assuming I did not make mistakes, the sum $$\sum_{m=0}^{\infty}\frac{x^{2m+1}}{(1+x^{2m+1})^2}$$
I am unable to continue from this point. Perhaps my work isn't actually useful at all. Can you help me?

答:
The sum in question belongs more properly to the theory of theta functions and elliptic integrals and this is an approach which makes use of standard results from this theory.

Let's put $q=-x$ so that the expression under limit becomes $$-(1+q)\sum_{n=1}^{\infty} \frac{nq^n} {1-q^{2n}}$$ The sum above can be written as $$\sum_{n\geq 1}\left(\frac{nq^{n}}{1-q^{n}}-\frac{nq^{2n}}{1-q^{2n}}\right)$$ which in terms of Ramanujan function $$P(q)=1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n} $$ becomes $$\frac{P(q^2)-P(q)}{24}$$ and it follows that the original expression under limit is $$(1-x)\cdot\frac{P(-x)-P(x^2)}{24}$$ Let's replace this variable $x$ again by $q$ (for convention)  and the limit we seek is $$\lim_{q\to 1^-}(1-q)\cdot\frac{P(-q)-P(q^2)} {24}$$ Treating $q$ as the nome we use the following standard results
\begin{align}
q&=\exp\left(-\pi\frac{K'} {K} \right)\notag\\
P(-q) &=\left(\frac{2K}{\pi}\right)^2\left(\frac{6E}{K}+4k^2-5\right)\notag\\
P(q^2) &=\left(\frac{2K}{\pi}\right)^2\left(\frac{3E}{K}+k^2-2\right)\notag
\end{align}
(for proofs see this post) and the expression under limit equals $$(1-e^{-\pi K'/K}) \frac{1}{8}\left(\frac{2K}{\pi}\right)^2\left(\frac{E} {K} - k'^2\right)$$ where moduli $k, k'$ and elliptic integrals $K, K'$ correspond to nome $q$. As $q\to 1^-$ we have $k\to 1^-,k'\to 0^+$ and $K\to\infty, K'\to\pi/2,E\to 1$ and the desired limit equals the limit of $$\frac{\pi K'} {K} \cdot\frac{K} {2\pi^2}\cdot(E-k'^2K)$$ This works out to be $1/4$ if we can show that $k'^2K\to 0$. This is an easy consequence of the asymptotic $$K=\log(4/k')+o(1)$$ as $k\to 1^-$ (for proof see this post).


问:
Show Sequence of Functions Converges Uniformly
I'm working on a Real Analysis book and am stuck on the following problem:

Let $f$ be continuous on $\mathbb R$ and let $$f_n(x)=\frac1n\sum_{k=0}^{n-1}f\left(x+\frac kn\right).$$ Prove that $f_n(x)$ converges uniformly to a limit on every finite interval $[a,b]$.

How would I rigorously solve this?

答:
The candidate for the limit is $g(x) := \displaystyle{\int_x^{x+1}} f(t)dt$.

For any $x$, you can bound the error between the integral and its approximation with the Riemann sums in $f_n$:

\begin{align*}
|g(x)-f_n(x)| & = \Big| \sum \limits_{k=0}^{n-1} \displaystyle{\int_{x+\frac{k}{n}}^{x+\frac{k+1}{n}}} f(t) - f\big(x+\frac{k}{n}\big)dt\Big| \\
& \le \sum \limits_{k=0}^{n-1} \displaystyle{\int_{x+\frac{k}{n}}^{x+\frac{k+1}{n}}} \big|f(t) - f\big(x+\frac{k}{n}\big)\big|dt
\end{align*}

$f$ is continuous on $[a,b]$, so it is uniformly continuous on that interval. For $\varepsilon>0$, we know there exists $\eta>0$ such that $x,y \in [a,b], |x-y|<\eta \implies |f(x)-f(y)| \le \varepsilon$. Now if you take $n$ large enough to have $\frac{1}{n} \le \eta$, for $k \in [\![0,n-1]\!]$ and $t \in \big[x+\frac{k}{n}, x+\frac{k+1}{n}\big]$, $\big|t-x-\frac{k}{n}\big| \le \eta$ so $\big|f(t)-f\big(x+\frac{k}{n}\big)\big| \le \varepsilon$.

Hence $|g(x)-f_n(x)| \le \sum \limits_{k=0}^{n-1} \displaystyle{\int_{x+\frac{k}{n}}^{x+\frac{k+1}{n}}}  \varepsilon dt = \varepsilon$, for all $x \in [a,b]$, for $n$ large enough.


问:
Angles and ellipse (proof)
First of all, sorry for my poor English! Can you please help me? I'm trying to prove that, given a point P at an ellipse.

Please help me prove that the angles are equal.

Thanks!

答:
You haven't stated the problem. You want to prove that a light ray from one focus reflects off the ellipse back to the other focus. You don't tell us what you know. Personally, I would represent the ellipse as a level set of the function
$$f(\mathbf x) = \|\mathbf x-F_1\| + \|\mathbf x-F_2\|$$ and use the fact that $\nabla f$ is the normal vector.

You could also write down the equation
$$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$
and use implicit differentiation and lots of slopes.

Title: Angles and ellipse (proof)


问:
Is $C(\mathbb{Z}, \mathbb{R})$ seperable?
During this pandemic I was self studying metric spaces and I came across this problem as exercise in a book.

Let $\mathbb{Z}$ (Integers)  be a discrete metric space where each set is open and $C(\mathbb{Z}, \mathbb{R})$ be a metric space  of all continous bounded functions from integers to real numbers under the norm $||f|| = sup{f}$. Show that $C(\mathbb{Z}, \mathbb{R})$ is not separable.

I know that each function from integers to real number is continous since each inverse image is subset of integers which is open.

I know that if we want show a metric space is not seperable then for every countable subset of metric space there is a point which is not limit point of that subset. I am not able to solve this problem with this approach. Is there any other approach? I'm not getting how can we use countability of subset to show it is not dense.

答:
For each subset $A$ of  $\mathbb N$ define $f_A(n)=1$ if $ n \in A$ and $0$ otherwise. Then $\|f_A-f_B\|=1$ whenever $A \neq B$. Since there are uncoutbaly many subsets of $\mathbb N$ it follows that $C(\mathbb Z, \mathbb R)$ is not separable.

[Suppose there is a countable dense set $(f_k)$. Then the ball of radius $1/2$ around $f_A$ must contain $f_k$  for at least one $k$. Call this $f_{k(A)}$. Verify that $ A \to k(A)$ is a one-to-one function from the family of all subsets of $\mathbb N$ into the set of positive integers. This is a contradiction].


问:
Find the product and coproduct of the category of Set with a given set
I am learning Category theory and I've found a problem :

Let $S$ be a fixed set. Define a category $\textbf{Set}_S$ , where collection of object is a set map $ f: X \rightarrow S$. Let $f':X' \to S$ be another object.
For two objects $f: X \rightarrow S, f^{\prime}: X^{\prime} \rightarrow S, \text { a morphism } h: f \rightarrow f^{\prime} \text { is a map } h: X \rightarrow X^{\prime}$  such that $f=f^{\prime} \circ h $. The composition of two morphisms in $\textbf{Set}_{S}$  is defined in the obvious way.
Describe the product and coproduct of $n$ objects
$$ f_{1}: X_{1} \rightarrow S, f_{2}: X_{2} \rightarrow S, \ldots, f_{n}: X_{n} \rightarrow S.$$

My idea for product is just the Cartesian product $X_1\times \cdots \times X_n$ with a map $T^{(i)}:X_1\times \cdots \times X_n \to X_i$. But it seems not to be true as $f_{i}$ is defined from  $X_{i}$ to $S$ instead of form $S$ to $X_i$.

Also, for coproduct, should it be the disjoint union of $X_1,\cdots,X_n$? I got the commutative diagram

By the way, I don't know where the morphism $h$ should be used.

答:
First of all, since the objects of $\textsf{Set}_S$ are maps with $S$ as codomain, it is non-sense to say that the product in this category is the cartesian product. The product of $f_1,\dots,f_n$ is a set-map $f_1 \times \cdots \times f_n$ from some set $X$ to $S$ together with morphisms $p_i : f_1 \times \cdots \times f_n \to f_i$ (that is, $p_i$ is a set-map $X \to X_i$ for which $f_1 \times \cdots \times f_n = f_i \circ p_i$) such that for any object $f : Y \to S$ and morphisms $g_i : f \to f_i$ (that is, $g_i$ is a set-map $Y \to X_i$ for which $f = f_i \circ g_i$) there is a unique morphism $g : f \to f_1 \times \cdots \times f_n$ (again, $g$ is a set-map $Y \to X$ for which $y = (f_1 \times \cdots \times f_n) \circ g$) such that $p_ig = g_i$, where $p_ig$ denotes the composition in $\textsf{Set}_S$.

So, take $X = \{(x_1,\dots,x_n) \in X_1 \times \cdots \times X_n : f_1(x_1) = \cdots = f_n(x_n)\}$, and $p_i$ the restriction on $X$ of the $i$-th coordinate projection $X_1 \times \cdots \times X_n \to X_i$ (what is $f_1 \times \cdots \times f_n$ then?).


问:
Showing or refuting that $[K(\alpha):K]=n$ and $p(\alpha)=0$, $deg(p(x))=n$ allows to conclude that $p(x)$ is the minimal polynomial.
solving an exercise (specifically problem 14.2.27 of dummit) the following question arose.

Let $K$ a field and $K(\alpha)/K$ extension field with $\alpha\not\in K$ and $ [K(\alpha):K]=n$ and $deg(p(x))=n$ and $p(\alpha)=0$
Is it enough to conclude that $p(x)$ is the minimal polynomial of $\alpha$ over $K$?

Thanks ins advance.

答: If $p(x)$ is monic, then yes. Observe that $p(\alpha)=0$ implies that the minimal polynomial of $\alpha$, $\min(K,\alpha)$, divides $p(x)$, but since they are of the same degree, $p(x)$ is just a scalar multiple of $\min(K,\alpha)$.


问:
Special function related to a nonlinear ODE
I am interested in finding a special function solution of the following ODE

$ (r S)^{\prime \prime} = - 2 r S^2 $
with initial conditions
$ S(0)=1, S^\prime(0) = 0 $.

The method of Frobenius gives the infinite series

$
S=
1 -\frac{{{r}^{2}}}{3}+\frac{{{r}^{4}}}{15}-\frac{11 {{r}^{6}}}{945}+\frac{16 {{r}^{8}}}{8505}-\frac{97 {{r}^{10}}}{334125} +\frac{914 {{r}^{12}}}{21049875} +H.O.T
$

and the following recursion among the coefficients:

$ a_n= - \frac{2}{n (n+1)}\sum\limits_{j=0}^{n-2}{\left. a_j a_{n-j-2} \right.}$

The question is: is this function related to the generalized hypergeometric functions?

If yes - what is the relationship?

答:
Let $U=rS$ ,

Then $U''=-\dfrac{2U^2}{r}$ with initial conditions $U(0)=0$ , $U'(0)=1$

Which reduces to an Emden–Fowler equation.


问:
Necessary and sufficient condition for an Euler trail between two vertices
Graph theory question one of those obvious math proofs so its going to be a pain to prove.

Show that G=(V,E) has an Euler trail between (different) vertices u and v if and only if G is connect and all vertices except u and v are of even degree.

By obviousness if there all other vertices are of even degree then you can make a circuit from v to a1 to a2 to an eventually you will come back to u or v if you continue doing this you eventually will arrive at u or v with no way to leave thus Euler trail.

But how would you actually prove this?

答:
Method 1:

Hint: Use the extremal principle. Consider the longest path.

Method 2:

Hint: Induct on the number of paths.

Method 3:

Hint: Connect $u$ with $v$, and apply the proof of Euler circuit.


答:
Note that we may write

\begin{align*}
F_N(\lambda) := \sum_{\ell=1}^{\infty} \frac{N!}{(N+\ell)!} \frac{(-\lambda)^{\ell}}{\ell}
&= \int_{0}^{-\lambda}\sum_{k=0}^{\infty}\frac{N! k!}{(N+1+k)!} \frac{v^k}{k!} \, \mathrm{d}v \\
&= \int_{0}^{-\lambda} \int_{0}^{1} (1 - t)^N e^{vt} \, \mathrm{d}t \, \mathrm{d}v \\
&= - \int_{0}^{1} (1 - t)^N \frac{1 - e^{-\lambda t}}{t} \, \mathrm{d}t.
\end{align*}

This allows us to reduce each $F_N(\lambda)$ to an elementary expression plus some special functions, such as an exponential integral or incomplete gamma function.

If your primary goal is to study the asymptotic behavior of this function, then the integral representation might be more useful. For instance, we can show that $F_N(\lambda) \sim -\log \lambda$ as $\lambda \to \infty$ for each fixed $N$.

Addendum. If $\lambda = N/a$ for some $a > 0$, then as $N\to\infty$

\begin{align*}
F_N(N/a)
&= - \int_{0}^{1} (1 - t)^N \frac{1 - e^{-(N/a)t}}{t} \, \mathrm{d}t \\
&= - \int_{0}^{N/a} \left( 1 - \frac{au}{N} \right)^N \frac{1 - e^{-u}}{u} \, \mathrm{d}u \tag{$t = au/N$} \\
\xrightarrow[\text{as } N\to\infty]{\text{DCT}}&\quad - \int_{0}^{\infty} e^{-au} \left( \frac{1 - e^{-u}}{u} \right) \, \mathrm{d}u \\
&= \log \left( \frac{a}{a+1} \right),
\end{align*}

where the last line follows from the Frullani integral.

Title: Does it look like an expansion of $\log$?

Body:
For a real number $\lambda>0$ and a positive integer $N$, consider
$$
\sum_{\ell=1}^{\infty}\frac{(-\lambda)^{\ell}}{\ell}\cdot\frac{N!}{(N+\ell)!}.
$$
Could we simplify it further as a function of $\lambda$? For example, it equals to
$$
\sum_{\ell=1}^{\infty}\frac{N!}{(N+\ell)!}\int_{0}^{-\lambda}v^{\ell-1}dv=N!\int_{0}^{-\lambda}\sum_{\ell=1}^{\infty}\frac{v^{\ell-1}}{(N+\ell)!}dv.
$$
Please give me some help or hints. Thanks a lot.


问:
Multiples of an effective degree 2 divisor on a hyperelliptic curve.
Let $L$ denote the Riemann-Roch space, and $l$ its dimension.

We are given the following definition:
A curve $C$ of genus $g$ is hyperelliptic if and only if there is a degree 2 divisor $D$ such that $l(D)\ge2$

In that case, without loss
of generality we may assume $D$ is effective and that $L(D) = (1,x)$ where $(...)$ denotes span and $x$ is nonconstant.

I need to prove that $L(nD)=(1,x,x^2,...,x^n)$ for $n\le g-1$. The $\supseteq$ direction is easy.
Using ultrametric absolute values, I've shown that $1,x,...,x^n$ are linearly independent. It remains to show that $l(nD)\le n+1$, which is what I'm having trouble with. I went to my professor and was given a hint to use Riemann-Roch Theorem, but I can't figure out how to obtain that bound. Here's what I have.

Let $W$ be a canonical divisor. Then $l(nD)-l(W-nD)=2n-g+1\le n+1$ for $g\ge n$. Since $l(W-nD)\ge0$, I don't have an upper bound on $l(nD)$. If someone can help, that would be great. Thanks.

I've shown that $(g-1)D$ is canonical, but I can't use that since my proof involves the assumption that $l(nD)=n+1$.

答:
In the end, I think this comes down to a pigeonhole principle argument. First, note that (I think) you've shown that $x^n \in L(nD) \setminus L((n-1)D)$ for $n \leq g$ (see the discussion in the comments). Thus
$$
\ell(nD) - \ell((n-1)D) \geq 1
$$
for all $n \geq 1$, i.e., each time we increase $n$, the dimension has to go up by at least $1$. Now since $\deg(gD) = 2g > 2g - 2$, then $\deg(K - gD) < 0$, so
$$
\require{cancel}
\ell(gD) - \cancelto{0}{\ell(K - gD)} = 1 - g + \deg(gD) = 1 - g + 2g = g+1
$$
by Riemann-Roch. Here's a table of the values $\ell(nD)$.
$$
\begin{array}{c|c|c|c|c|c}
\ell(D) & \ell(2D) & \ell(3D) & \cdots & \ell((g-1)D) & \ell(gD)\\ \hline
2 & ? & ? & \cdots & ? & g+1
\end{array}
$$
Thus $\ell(2D) < \ell(3D) < \cdots < \ell((g-1)D)$ is a strictly increasing sequence of $g-2$ values chosen from the set $\{3, 4, \ldots, g\}$. But $\{3, 4, \ldots, g\}$ only contains $g-2$ distinct values, so there's only way this can occur: we must have
$$
\ell(2D) = 3, \ell(3D) = 4, \ldots, \ell((g-1)D) = g \, ,
$$
as desired.


问:
Complex Potential
An incompressible inviscid fluid fills a long circular cylinder of radius a. Show that if there is a line vortex of circulation $2\pi \Gamma$ parallel to the axis of the cylinder and at a distance $b (< a)$ from it, it will revolve around the axis with a constant speed $(\Gamma b)/(a^2−b^2)$.

So i know that we have a vortex and the complex potential is
$-(i\Gamma/2\pi)*\ln(z)$

since we are given speed that is $w'(z)$, right? so how would we go about showing all this. where would we start? would we start by integrating $w'(z)$? or could we start with a vortex pair within the cylinder?

答:
Orient the coordinate system so that the vortex inside the cylinder is at the point $z_b =b +i\cdot 0$ on the real axis.  To satisfy the no flux condition at the surface of the cylinder the velocity field inside the cylinder can be obtained using the method of images with another vortex of equal and opposite strength placed at the point $z_c =c + i \cdot 0$ where $bc = a^2$.

There is no self-induced motion of the vortex at $z_b$.  The motion of this vortex is due to the velocity field produced by the image vortex alone with potential

$$f_c(z) = \frac{i2 \pi\Gamma}{2\pi}\log (z - z_c),$$

and complex velocity

$$u-iv =-\frac{d f_c}{dz} = -i \Gamma \frac{1}{z - z_c}$$

The velocity at the location $z_b$ is

$$u(z_b) - i v(z_b) = -i \Gamma \frac{1}{z_b - z_c} = i\frac{\Gamma}{c - b} = i\frac{\Gamma b}{bc - b^2} = i \frac{\Gamma b}{a^2- b^2}$$

Hence the vortex at $z_b$ moves with speed given by the modulus

$$|u(z_b) - i v(z_b)| = \frac{\Gamma b}{a^2 - b^2}$$

Title: Complex Potential


问:
If $X, A_1$ are independent, and $X, A_2$ are independent, then are $X, A_1 \cdot A_2$ independent?
There are RVs $A_1, ..., A_n$ and $X$.

$A_1$ and $X$ are independent,

$A_2$ and $X$ are independent,

...

$A_n$ and $X$ are independent.

Does it mean that $A_1 \cdot ... \cdot A_n$ and $X$ are also independent?

It seems that this can be proved by induction. But I can't figure out how to prove it for case $n = 2$.

答: In two independent tosses of fair coin consider the event $A$ of the first toss being Head, the event $B$  of the second toss being Heads and the event $C$ of the two outcomes being both Heads or both Tails. Then $I_A$ is independent of $I_B$, $I_A$ is independent of $I_C$ but it is not true that $I_A$ is independent of $I_BI_C=I_{B \cap C}$


问:
Simplify $\mathbb{E}\left[ -\log \left( \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{z-\mu}{\sigma}\right)^{2}} \right) \right]$
How to simplify the following expected value?

\begin{align}
&= \mathbb{E}[ -\log f(x)] \\
&= \mathbb{E}\left[ -\log \left( \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{z-\mu}{\sigma}\right)^{2}} \right) \right]\\
&= ?
\end{align}

Is the next step

$$\mathbb{E}\left[ -\log\frac{1}{\sigma \sqrt{2 \pi}}- \left( -\frac{1}{2}\left(\frac{z-\mu}{\sigma}\right)^{2} \right) \right] ?$$
and from there, how to reduce further to non-expectation form?

答: @Joitandr's approach is overkill, given what you've already done. By the linearity of expectation, the result is $\log(\sigma\sqrt{2\pi})+\frac{1}{2\sigma^2}\underbrace{\Bbb E[(z-\mu)^2]}_{\sigma^2}=\log(\sigma\sqrt{2e\pi})$.


问:
Determining Which Statistician Has The Greatest Posterior Degree of Belief
I am given that we have data $s$ and that there are $\textbf{Two}$ statisticians analyzing these data using the same sampling model but with different priors, and they are asked to assess a hypothesis $H_0$.

Both statisticians report a Bayes factor in favor of $H_0$ equal to $100$

$\textbf{Statistician I}$ assigned a prior probability $\frac{1}{2}$ to $H_0$

$\textbf{statistician II}$ assigned prior probability $\frac{1}{4}$ to $H_0$

How do we determine which statistician has the greatest posterior degree of belief in $H_0$ being true?

答:
How do we determine which statistician has the greatest posterior degree of belief in $H_0$ being true?

having the same Bayes Factor, the posterior degree of belief in $H_0$ is greater for Stat I  as Stat II gives a bigger prior probability to $H_1$

The decision in favour of $H_0$ is governed by the following rule:

$$\frac{P(\theta=\theta_0)}{P(\theta=\theta_1)}\times \underbrace{\frac{p(\mathbf{x}|\theta_0)}{p(\mathbf{x}|\theta_1)}}_{\text{Bayes Factor}}>1$$

the likelihood ratio is called "Bayes Factor"

Thus for Stat 1 the decision rule is 1:1 while for Stat II the decision rule is 1:3. This means that the degree of belief in $H_0$ is greater for Stat  I who beliefs the two hypotheses are equal in probability. Stat II beliefs that $H_1$ is more probabily. Decision could change after seeing the data, but as per the fact that BF is the same for both...


问:
How to find the following integral: $ \iint\limits_D\max\left\{\sin x,\sin y\right\}dxdy,\ D=\left\{0\le x\le\pi,\ 0\le y\le\pi\right\} $

How to evaluate the following integral:
$$
\iint\limits_D\max\left\{\sin x,\sin y\right\}dxdy,\ \ \text{where}\ D=\left\{0\le x\le\pi,\ 0\le y\le\pi\right\}
$$

To be honest, I don't understand how to approach this problem. In my opinion, $\sin y$ is always greater than $\sin x$, since $x=\sin y$ lies higher than $y=\sin x$ for any given point. But it doesn't seem to be true. Could anyone give me a hint about how I should evaluate this integral?

答:
Personally, I believe you have to integrate over $z = \sin{x}$ and $z = \sin{y}$, like as you said, it won't make sense for $x = \sin{y}$ and $y = \sin{x}$.

Perhaps a graph could help you visualize since I belive that's the part you're having problem with. Here's what I plotted on GeoGebra:

Now I believe you can take care of the integration.


问:
"Simplest" symbolic equation with "most complex" symbolic answer
The below was inspired by this question, in which the OP was surprised that the "simple" equation $x^3 + x = 1$ could have such a "complex" solution.

In rough form, the question here is a bit more general:

What is an example of the "simplest" symbolic equation whose solution is "most complex"?

That rough statement is of course inadequate to constrain or compare candidate answers.  Thus, to be specific, here are the conditions:

The equation involves a single variable, say $x$, and any number of constants.

The "complexity" of the problem equation and of the symbolic solution is the number of bytes in its expression (for instance in a MathJax representation).

The global criterion to be optimized is the ratio of the number of bytes in the solution to the number of bytes in the problem equation.

Both the problem and solution must be represented in their simplest canonical form.  For instance, one must use $\sin x$ rather than its infinite series representation.

All standard special functions and symbols (factorial !, power ^, etc.) are allowed, where the complexity is in the number of bytes in their canonical MathJax representation.

Both the problem and the solution must be symbolic, not numerical.  (Otherwise simple problem equations such as $x^2 = 2$, which have an "infinitely complex" numerical solution, would trivially optimize the criterion.)

There must be a true, single closed-form solution.  (Thus one cannot include, for example, $x^{9!} = 2$ and get a large list of solutions.)

The solution is not a simple numerical calculation, for instance $x = 9 \uparrow \uparrow \uparrow$ (using Knuth's uparrow notation) or $x = (9!)^{9!}$, etc.  As illustrated by the linked question above, it is rather the algebraic complexity of a solution.

One can get lost in nit picking (e.g., deciding which is to be preferred, \sqrt{x} versus x^{1/2}, or a/b versus \frac{a}{b}, and such).  I suspect there are some dramatic examples where such minutia are irrelevant.

答:
Problem: express $\sin(2\pi/17)$ in algebraic terms.

Solution: Let $\epsilon=\sqrt{17+\sqrt{17}}$.

Let $\epsilon'=\sqrt{17-\sqrt{17}}$.

Let $\delta=\sqrt{17}-1$.

Let $\alpha=\sqrt{34+6\sqrt{17}+\sqrt2\delta\epsilon'-8\sqrt2\epsilon}$.

Then $\sin(2\pi/17)={1\over16}\sqrt2\sqrt{4\epsilon'^2-2\sqrt2\delta\epsilon'+8\sqrt2\epsilon-(\sqrt2\delta+2\epsilon')\alpha}$

Note: it's even worse for $\sin(2\pi/257)$, and far worse for $\sin(2\pi/65537)$. For the first, see https://math.stackexchange.com/q/517218 – for the second, you're on your own.


问:
A question related to $S^{\perp}$ and closure of span of $S$
This question was asked in my linear algebra quiz previous year exam and I was unable to solve it.

Let V be an inner ( in question it's written integer , but i think he means inner) product space and S be a subset of V. Let $\bar S$denote the closure of S in V with respect to topology induced by the metric given by inner product. Which of the following statements are  true?

A $ S $=$ (S^{\bot})^{\bot}$

B $ \overline S$= $(S^{\perp})^{\perp}$

C $\overline {\text{span}(S)}$=$(S^{\bot})^{\bot}$

D $ S^{\bot} $=$ ((S^{\bot})^{\bot})^{\bot}$

I was completely blank on how can I approach this problem although I have studied linear algebra carefully. Can you please tell on how I should approach the problem .

Edit : I tried It again . I marked A ,D but answer is C,D. If A is false I don't see why D must be true. So, I think I am missing some concepts.

答:
The condition that $S$ is a subset (and is not, for example, given to be a subspace of $V$) is rather odd, and makes some of this problem different from standard linear algebra. In general, in any inner product space, $S^\perp$ is a closed subspace, so the right sides of each of A, B, C, and D are closed subspaces.

However, for each of $A$, $B$, and $D$ (the original version of $D$, and not the current one), it is possible for the left side not to be a closed subspace. (The example given by Peter Franek of when $S$ is a one-element set is likely the simplest.) In general, though, $\operatorname{span}(S)$ should be a subspace, and $\overline{\operatorname{span}(S)}$ should be a closed subspace, so it should seem plausible that C could be true. It in fact is, as it essentially only deals with subspaces --- see if you can prove this.


问:
Linear independence, invertibility and the determinant
I'm trying the following proof:

Suppose that $A$ is an $nxn$ matrix. Show that if $A$ has linearly dependent rows, then $\det(A) = 0$.

My attempt:

I considered two approaches, namely, a direct proof vs proof by contrapositive. Off the bat, I realized that proof by contrapositive (i.e. $\det(A) \neq 0 \implies$ invertibility $\implies$ linearly independent columns) didn't seem to work. I decided to use a direct proof, as follows:

Suppose $A_1, ..., A_n$ are linearly dependent rows of $A$. By definition, there are scalars $c_1, ..., c_n$ s.t. $c_1A_1 + ... + c_nA_n = 0$, where not all $c_i$ are $0$ for $i = 1, ..., n$. Let $c_j$ be the non-zero coefficient with corresponding row $A_j$ where $j \in {1, ..., n}$, then:

$$A_j = -\frac{c_1}{c_j}A_1 - ... - \frac{c_n}{c_j}A_n$$

Let $B$ be the matrix obtained after performing the series of operations above on row $A_j$, then $A_j = (0\space ... \space 0)$. Also, by $n$-linearity, $\det(B) = \det(A)$. Moreover:

$$\det(B) = \sum_{\sigma} (\operatorname{sgn} \sigma) A_{1\sigma_1} \cdots A_{n\sigma_n}$$

Notice that each product in the summand above contains an entry from row $A_j$. Hence, $\det(B) = \det(A) = 0$

QED.

I was curious if my proof above is correct. It was a little confusing at points and I'm not sure if showing the last 'observation' as a proof is necessary. Any feedback is much appreciated.

Also, out of sheer curiosity, would proof by contrapositive work?

Edit: Sorry for a few typos.

答:
Generally, your proof looks good. There are some indexing issues in your $A_j=-c_1/c_jA_1+\cdots$ line, but I assume these are all typos, and you probably want to say something like "fix some $1\leq j\leq n$ for which $c_j\neq 0$" instead of "let $c_j$ be the non-zero coefficient," as there could be multiple nonzero $c_j$. Otherwise, everything looks good.

Whether the last observation (that a matrix with an entirely $0$ row has determinant $0$) is necessary depends on the amount of rigor you're looking for (e.g. if this is for a class where proof-writing is a focus, etc.), and I don't think there's a one-size-fits-all answer. There's certainly no harm in including it, especially if the goal is for your solution to be as clear as possible, but it might end up being cumbersome if this proof is part of some longer document or solution. In most scenarios, I'd imagine it would be fine to just state that a matrix with an entirely $0$ row has $0$ determinant, and maybe cite "expansion by minors" or "the permutation sum formula" or something like that.

Re: proof by contradiction: this sort of depends on what you're assuming. There are so many different definitions of the determinant, and it takes a good bit of work to determine that they're all equivalent. For the definition based on volumes of parallelepipeds, for example, the implication $\det(A)\neq 0\implies \text{invertibility}$ is not difficult, but it's not obvious given the "sum of permutations" definition you've cited in your proof. If all you have is that definition (or expansion by minors or something), you can prove invertibility by defining the adjugate matrix (essentially explicitly calculating the inverse), but this is a lot more work. The implication $\text{invertibility}\implies\text{independent rows}$ should not be too hard, though.


问:
How to show: if $b \mid a$ and $c \mid a$ and $\mathrm{gcd}(b,c) = 1$, then $bc \mid a$?
A little stumped on this problem, any help would be greatly appreciated.

Show that for all $a,b,c \in \mathbb{Z}$, if $b \mid a$ and $c \mid a$ and $\mathrm{gcd}(b,c) = 1$, then $bc \mid a$.

答:
If the highest power of prime $p$ in $a,b,c$ are $r_a,r_b,r_c$ respectively,

$r_a$ must be  $\ge$ max$(r_b,r_c)$

$\implies $lcm $(b,c)$ divides $a$

If gcd$(b,c)=1,$lcm $(b,c)=b\cdot c$


答:
Denote
$$f(x,y) = \max \{\sin x, \sin y\}.$$ $f$ being the max of two continuous maps is continuous and integrable on the compact $D$.

$D$ is the disjoint union of the four regions

$$\begin{cases}
D_1 &= \{ (x,y) \in D \mid x \lt y\, , x \lt \pi -y\}\\
D_2 &= \{ (x,y) \in D \mid x \lt y\, , x \gt \pi -y\}\\
D_3 &= \{ (x,y) \in D \mid x \gt y\, , x \gt \pi -y\}\\
D_4 &= \{ (x,y) \in D \mid x \gt y\, , x \lt \pi -y\}\\
\end{cases}$$ and a null set.

Moreover $f(x, y) = f(y,x)= f(\pi -x,y) = f( x, \pi -y)$ for all $(x,y) \in D$. Therefore

$$
\iint\limits_{D_i} f(x,y)dxdy = \iint\limits_{D_j} f(x,y)dxdy = 2\iint\limits_{0 \lt x \lt \pi/2\, ,0 \lt y \lt x} f(x,y)dxdy$$ for $(i,j) \in \{1, \dots, 4\}^2$ and

$$I=
\iint\limits_{D} f(x,y)dxdy = 8\iint\limits_{E} f(x,y)dxdy$$
where $E =\{0 \lt x \lt \pi/2\, ,0 \lt y \lt x\}$

As $x \mapsto \sin x$ is increasing on $[0, \pi/2]$, $\sin x \ge \sin y$ for $(x,y) \in E$ and

$$\begin{aligned}
I &= 8 \iint\limits_{E} f(x,y)\ dxdy = 8 \iint\limits_{E} \sin x \ dxdy\\
&= 8\int_0^{\pi/2} \int_0^x \sin x \ dy dx = 8 \int_0^{\pi/2} x \sin x \ dx\\
&=8
\end{aligned}$$

Note: I think you're confusing in your question single variate maps like $y = \sin x$ and bivariate ones like $f$. Clearly $\sin y$ is not always greater than $\sin x$ on $D$. Just take $(x,y)=(\pi/2, 0)$ to convince yourself.

Title: How to find the following integral: $ \iint\limits_D\max\left\{\sin x,\sin y\right\}dxdy,\ D=\left\{0\le x\le\pi,\ 0\le y\le\pi\right\} $

Body:
How to evaluate the following integral:
$$
\iint\limits_D\max\left\{\sin x,\sin y\right\}dxdy,\ \ \text{where}\ D=\left\{0\le x\le\pi,\ 0\le y\le\pi\right\}
$$

To be honest, I don't understand how to approach this problem. In my opinion, $\sin y$ is always greater than $\sin x$, since $x=\sin y$ lies higher than $y=\sin x$ for any given point. But it doesn't seem to be true. Could anyone give me a hint about how I should evaluate this integral?


问:
How do sides of a constant area quadrilateral rotate?
A quadrilateral of constant area having  three fixed length sides $(a,c,b)$ has fourth side length variable. End coordinates of fixed side length  $c$ are $(0,0),(c,0)$.

Find relation between angles $ (x,y).$

答:
Let $S$ be the constant area of the quadrilateral.  We consider the right trapezoid obtained by projecting the vertices of the variable edge on the horizontal line.  We have then
$$
S = \frac{1}{2}[(a\sin x +b\sin y)(c -a\cos x +b\cos y) +a^2\sin x \cos x -b^2\sin y\cos y],
$$
hence the relation
$$
a\sin x +b\sin y = \frac{2S}{c}.
$$


问:
Boundedness of sequence of functions with continuity and nonnegativity with convergent improper integral
Let $(f_{n})_{n\in\mathbb{N}}$ be a sequence of continuous nonnegative functions in $[a,\infty)$ for some $a\in\mathbb{R}$. Moreover, we assume that
$$\int_{a}^{\infty}f_{n}(t)dt\to 0 \text{ as }n\to\infty$$
Can we prove that $(f_{n})_{n\in\mathbb{N}}$ is uniformly bounded with respect to $t\in[a,\infty)$ and $n\in\mathbb{N}$?
In other words, I want to show that
$$\exists M>0 \,\forall n\in\mathbb{N} \,\forall t\in[a,\infty)\, |f_{n}(t)|\leq M$$

This is my attempt so far:

First, I use continuity of $f_{n}$ and assume the conclusion does not hold for the sake of contradiction. Then, I know that there exists a sequence $(f_{n}(t_{n}))_{n\in\mathbb{N}}$ such that
$$\forall n\in\mathbb{N}\text{ (large enough) }, \exists \eta_{n}>0 \forall t\in(t_{n}-\eta_{n},t_{n}+\eta_{n}), f_{n}(t)\geq \frac{1}{2}$$
Therefore, for $n$ sufficiently large, we deduce that
$$\eta_{n}\leq  \int_{t_{n}-\eta_{n}}^{t_{n}+\eta_{n}}f(t)dt\leq \int_{a}^{\infty}f_{n}(t)dt\to0\text{ as }n\to\infty$$

The problem is that I cannot take $\eta_{n}$ uniformly to ensure $\inf\limits_{n\in\mathbb{N}}\eta_{n}>0$. Any help will much appreciated! Thank you!

答:
Let $f_n(x)=n^{3}(x-n)$ for $n \leq x \leq n+\frac 1 {n^{2}}$, $f_n(x)=n^{3}(n+\frac 2 {n^{2}}-x)$ for $ n+\frac1 {n^{2}} \leq x \leq  n+\frac 2 {n^{2}}$ and $0$ elsewhere. Verify that this is a counter-example.

[$f(n+\frac 1  {n^{2}}) =n \to \infty$]


问:
Totally busted euler trail.
New to graph theory proofs would like some help.

Let $G_{1}$ and $G_{2}$ be Eulerian graphs . let $v_{1}$ be a vertex in $G_{1}$ and $v_{2}$ be a vertex in $G_{2}$  jion $v_{1}$ and $v_{2}$ by a single edge.

1)Does the graph have a Euler trail?
EDIT:

I think i finally get this part start at $v_{1}$ make a Euler circiut back to $v_{1}$ ignoring the new path to $v_{2}$ now delete these edges and cross form $v_{1}$ to $v_{2}$ now complete that euler circiut jion the ends together of these 2. Now as shown this is actually a Eulerian Trail. Which is where my confusion arose form.

2) is the resulting graph eulerian? why or why not?

The requirements for a graph to be Eulerian is that every vertex is of even degree, and clearly this was true in $G_{1}$ and $G_{2}$ as they where Eulerian thus all vertex of $G_{1}$ and $G_{2}$ where even. even +1 is not even hence there The new Graph is not Eulerian. ($ v_{1}$ and $v_{2}$ must both be of odd degree)

答:
1) Yes. Take the Eulerian circuit, and delete that edge. You get a eulerian path.

2) No. Use the classification that a graph is Eulerian if and only if every vertex has even degree.

Title: Totally busted euler trail.


答:
Counterexample

Set $f_n(x)=n$ for $x\in [0,\frac{1}{2n}]$ and $f(x)=-2n^2x+2n$ for $x\in [\frac{1}{2n}, \frac{1}{n}]$ and $f$ is zero for $x\geq \frac{1}{n}$. Then every $f_n$ is continuous (in $C([0,\infty))$) and the sequence of the integrals goes to $0$ but clearly $f_n$ is not uniformly bounded

Title: Boundedness of sequence of functions with continuity and nonnegativity with convergent improper integral

Body:
Let $(f_{n})_{n\in\mathbb{N}}$ be a sequence of continuous nonnegative functions in $[a,\infty)$ for some $a\in\mathbb{R}$. Moreover, we assume that
$$\int_{a}^{\infty}f_{n}(t)dt\to 0 \text{ as }n\to\infty$$
Can we prove that $(f_{n})_{n\in\mathbb{N}}$ is uniformly bounded with respect to $t\in[a,\infty)$ and $n\in\mathbb{N}$?
In other words, I want to show that
$$\exists M>0 \,\forall n\in\mathbb{N} \,\forall t\in[a,\infty)\, |f_{n}(t)|\leq M$$

This is my attempt so far:

First, I use continuity of $f_{n}$ and assume the conclusion does not hold for the sake of contradiction. Then, I know that there exists a sequence $(f_{n}(t_{n}))_{n\in\mathbb{N}}$ such that
$$\forall n\in\mathbb{N}\text{ (large enough) }, \exists \eta_{n}>0 \forall t\in(t_{n}-\eta_{n},t_{n}+\eta_{n}), f_{n}(t)\geq \frac{1}{2}$$
Therefore, for $n$ sufficiently large, we deduce that
$$\eta_{n}\leq  \int_{t_{n}-\eta_{n}}^{t_{n}+\eta_{n}}f(t)dt\leq \int_{a}^{\infty}f_{n}(t)dt\to0\text{ as }n\to\infty$$

The problem is that I cannot take $\eta_{n}$ uniformly to ensure $\inf\limits_{n\in\mathbb{N}}\eta_{n}>0$. Any help will much appreciated! Thank you!


问:
How to find the angle formed by a cevian in a right triangle given a condition?
The problem is as follows:

In a right triangle $\triangle ABC$, right at $B$, it is traced the
height $BH$, in the triangle $BHC$, then is traced the interior cevian
$HM$ in such a way that $MC = AB$. Find the measure of the angle
$\angle MHC$ if it holds that $HC = BH + 2AH$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\frac{53^{\circ}}{2}\\
2.&\frac{37^{\circ}}{2}\\
3.&53^{\circ}\\
4.&37^{\circ}\\
4.&30^{\circ}\\
\end{array}$

In this particular problem. I'm stuck at the very beginning. What's exactly the right drawing or interpretation for this?. Can someone help me here?. Can this be solved using congruency?. Which among the identities of congruency of triangles would apply?.

Please include a drawing in the answer because I don't know which sort of appropiate drawing would be for this particular problem. Can it be solved using only euclidean geometry postulates?.

Typically I would put or place an effort in the solution of this problem, but in this particular subject I'm stuck at the very beginning because I don't know how to translate what it is written in words into the problem with accuracy so that the problem can be solved, thus I really need help.

If it helps, I found this problem in a puzzles book which goes by the name Reason and logic which is from 2000s and includes topics probably borrowed from Martin Gardner's 50's book and Moise Geometry's textbook from the 70's.

I really wish this can suffice to give information. So far, as I indicated in the problem my best guess is that the problem uses triangle congruency, but I don't know how to make the right sketch for making this to happen.

Upon on my progress I could spot that there are going to be formed two congruent triangles and if this condition is used then I believe the problem can be solved, hence what might be needed is to establish a relationship between the altitude from $M$ to $HC$ would allow the answer. Such approach is mentioned in the answer given below.

Therefore, since I'm not good in this topic, it would help me a lot if the answer would include a step by step solution so I can understand.

答:
By the chord-chord power theorem
with respect to the point $H$,
\begin{align}
|BH|^2&=|AH|\cdot|CH|
\tag{1}\label{1}
.
\end{align}

Given that

\begin{align}
|CH|&=|BH|+2\cdot|AH|
\tag{2}\label{2}
,
\end{align}

we have a quadratic equation

\begin{align}
\left(\frac{|BH|}{|AH|}\right)^2
&= \frac{|BH|}{|AH|}+2
\tag{3}\label{3}
,
\end{align}

hence
\begin{align}
\frac{|BH|}{|AH|}
&=2=\tan\alpha
\tag{4}\label{4}
,\\
\cos\alpha&=\tfrac{\sqrt5}5
,\quad
\sin\alpha =\tfrac{2\sqrt5}5
\tag{5}\label{5}
.
\end{align}

Let $|OA|=|OC|=R=\tfrac12$.
Then

\begin{align}
c&=|AB|=|CM|=\cos\alpha=\tfrac{\sqrt5}5
\tag{6}\label{6}
,\\
|AH|=|MP|&=c\cos\alpha=\cos^2\alpha
=\tfrac15
\tag{7}\label{7}
,\\
|BH|=|CP|
&=|AH|\tan\alpha
=\tfrac25
\tag{8}\label{8}
,\\
|CH|&=\tfrac45
\tag{9}\label{9}
,\\
|HP|&=
|CH|-|CP|=\tfrac25
\tag{10}\label{10}
,\\
\tan x&=\frac{|MP|}{|HP|}
=\tfrac12
\tag{11}\label{11}
,\\
x&=\arctan \tfrac12
\approx 26.565^\circ
\tag{12}\label{12}
.
\end{align}


问:
Let $G \le S_n$ ,show that if $G$ has an odd permutation then half of the elements in $G$ are even permutation and the other half are odd permutation.
Assume $G \le S_n$ ,show that if $G$ has an odd permutation then half of the elements in $G$ are even permutation and the other half are odd permutation.   Moreover conclude that $G$ has a subgroup of index $2$.  I've seen the following related proof:

Define a map  $\text{sgn}:S_n→\left\{−1,1\right\}$ which is given by:

$$  \begin{align} \text{sgn(σ)} = \left\{ \begin{array}{cc}                  1 & \hspace{5mm} \text{if σ is a product of an even number of >transpositions} \\                 -1 & \hspace{5mm} \text{if σ is a product of an odd number of transpositions} \\                 \end{array} \right. \end{align}$$

Then the kernel of the mapping would be the alternating group $A_n$,by the isomorphism theorem $S_n/A_n≅\left\{−1,1\right\}$.

Now by the Lagrange's  theorem  it  follows  that  $|A_n|= \frac{|S_n|}{2}.$

This  means  that  exactly  half  of  the  permutations  in    Sn    are  even  and  so  exactly  half  are  also  odd.

But  I  don't  see  how  to  conclude  that  for    $G \le  S_n$  with  the  given  assumptions  the  claim  that  half  of the  elements  in  $G$  are  even  permutation  and  the  other  half  are  odd  permutation  does  hold. Maybe after  this  it  looks  straight  forward  but  can  someone $2$? show  me why $G$ has a subgroup of index

答: Hint. Restrict $\mathrm{sgn}$ to a homomorphism $\mathrm{sgn}:G\to\{\pm1\}$.


问:
How to decide on the standard error for difference of standard deviations

In this problem the null hypothesis is that $\sigma_1 = \sigma_2$ then why in the standard error a common sigma value is not used? It can he found as $\sigma = ({n_{1}s_1^{2}+ n_{2}s_{2}^2})/({n_{1}+n_{2}})$
And also while testing for difference of sample means i.e $H_0: \mu_1=\mu_2$, does it always mean that $\sigma_1=\sigma_2$? Or does that happen only when the samples are drawn from the same population? Basically what’s the difference from the null hypothesis that $H_0: \mu_1=\mu_2$ and the case when samples are drawn from same population? Is the testing done for the significant difference of population mean or sample mean? also in difference of proportions, under the hypothesis $P_1=P_1$ the standard error also has one term $P_1=P_2=P$ why doesn’t that happen here for sigma? Or let me put it this way, under the null hypothesis $P_1=P_1$ the standard error also has one term $H_0: P_1=P_2=P$ Is this $P$ same as the one calculated from $P=(n_1p_1+n_2p_2)/(n_1+n_2)$ where $p_i$ is the sample proportion.

EDIT: I referred another text and here it says, under null hypothesis $H_0: \sigma_1=\sigma_2$ and continues to say that, i.e sample standard deviations don’t differ significantly. But are not $\sigma_1$ and $\sigma_2$ population standard deviations?

答:
The provided solution is complete garbage.  Disregard it entirely.  The test statistic they use is not normally distributed under the assumption of the null hypothesis.  I don't know where this text comes from, but such an elementary mistake discredits the whole text.

To test the equality of variances (and in turn, standard deviations), the usual test is the $F$-ratio test for normally distributed populations, and Levene's test when departure from normality is considered.  But under no circumstances does that solution make any sense whatsoever.  It looks like the author of the solution simply treated the sample standard deviations like sample means, and assumed something like Welch's $t$-test would be an appropriate test statistic.  But even then, there are mistakes:  first, $z$ is not appropriate because the true population standard deviation is unknown.  Second, there's an extra factor of $2$ in the denominators.  Third, there is a sign error in the last expression where the author uses $-$ instead of $+$.  But this is all irrelevant because as I have already stated, the solution is garbage.


问:
Let M, K, and L be points on line (AB), (BC), and (CA), respectively. Find the maximum area of smallest the three triangles MAL, KBM, and LCK?
Let M, K, and L be points on line (AB), (BC), and (CA), respectively. Find the maximum area of smallest the three triangles  MAL,  KBM, and  LCK in respect to ABC?

I try to guess the answer and then prove it, however it doesn't work to randomly guess.
I draw some pictures, but it hardly tells me anything

答:
We shall denote the area of a triangle $ABC$ with $S_{ABC}$. We shall prove that $\min(S_{MAL}, S_{KBM}, S_{LCK}) \leq \frac{1}{4}S_{ABC}$.

Without loss of generality assume that $S_{MAL}=\max(S_{MAL}, S_{KBM}, S_{LCK})$. Clearly if $\frac{S_{MAL}}{S_{ABC}} \leq \frac{1}{4}$, we are done. Otherwise consider $\frac{S_{MAL}}{S_{ABC}}=\frac{AM}{AB} \cdot \frac{AL}{AC}>\frac{1}{4}$. Let $\frac{AM}{AB}=x, \frac{AL}{AC}=y, 0 \leq x, y \leq 1$. Then $xy>\frac{1}{4}$. Also, we have $\frac{MB}{AB}=1-x, \frac{LC}{AC}=1-y$. Clearly if any of $K, L, M$ are equal to $A, B$, or $C$, then $\min(S_{MAL}, S_{KBM}, S_{LCK})=0 \leq \frac{1}{4}S_{ABC}$. Thus $0<x, y<1$.

Let $K'$ be on $BC$ s.t. $\frac{BK'}{CK'}=\frac{1-y}{1-x}$. Then $$\frac{S_{K'BM}}{S_{ABC}}=\frac{BK'}{BC} \cdot \frac{BM}{AB}=\frac{(1-x)BK'}{BC}=\frac{(1-y)CK'}{BC}=\frac{CK'}{BC} \cdot \frac{LC}{AC}=\frac{S_{LCK'}}{S_{ABC}}$$

If $BK \leq BK'$, then $$\frac{S_{KBM}}{S_{ABC}}=\frac{BK}{BC} \cdot \frac{BM}{AB} \leq \frac{BK'}{BC} \cdot \frac{BM}{AB}=\frac{S_{K'BM}}{S_{ABC}}$$

Otherwise $BK>BK'$ so $CK<CK'$, so $$\frac{S_{LCK}}{S_{ABC}}=\frac{CK}{BC} \cdot \frac{LC}{AC}<\frac{CK'}{BC} \cdot \frac{LC}{AC}=\frac{S_{LCK'}}{S_{ABC}}=\frac{S_{K'BM}}{S_{ABC}}$$

In any case, we have $$\min(\frac{S_{KBM}}{S_{ABC}}, \frac{S_{LCK'}}{S_{ABC}}) \leq \frac{S_{K'BM}}{S_{ABC}}=\frac{(1-x)BK'}{BC}=\frac{(1-x)(1-y)}{1-x+1-y}$$

It thus suffices to prove that $\frac{(1-x)(1-y)}{1-x+1-y} \leq \frac{1}{4}$, or equivalently, $\frac{1}{1-x}+\frac{1}{1-y} \geq 4$.

Note that $x>xy>\frac{1}{4}$ so $4x-1>0$. Thus

\begin{align}
\frac{1}{1-x}+\frac{1}{1-y}-4 & =\frac{1}{1-x}+\frac{4x}{4x-4xy}-4 \\
& >\frac{1}{1-x}+\frac{4x}{4x-1}-4 \\
& =\frac{(4x-1)+4x(1-x)-4(1-x)(4x-1)}{(1-x)(4x-1)} \\
& =\frac{12x^2-12x+3}{(1-x)(4x-1)} \\
& =\frac{3(2x-1)^2}{(1-x)(4x-1)} \\
& \geq 0
\end{align}

Therefore $\min(S_{MAL}, S_{KBM}, S_{LCK}) \leq \frac{1}{4}S_{ABC}$. Furthermore, equality holds when $M, K, L$ are the midpoints of $AB, BC, AC$ respectively, so $\max(\min(S_{MAL}, S_{KBM}, S_{LCK}))=\frac{1}{4}S_{ABC}$.


问:
Calculating a transition probability in FSM
I have a Finite State Machine represented in following form:

```<code>[ 0, 1, 0, 0, 0, 1,]
[ 4, 0, 0, 2, 3, 0,]
[ 0, 0, 0, 0, 0, 0,]
[ 0, 0, 0, 0, 0, 0,]
[ 0, 0, 0, 0, 0, 0 ]
</code>```

Where each row represents a state (first row is s0, second one is s1, etc) and the values represent the probability of transitioning to the next state. For instance, s0 has 50% chance of going to s1 and 50% chance of going to s5.

How can I calculate probability of going from s0 to s3? If these states did not have a loopback (you can go from s0 to s1 and then back to s0 according to this matrix) answer would be trivial, but I am bit confused about how this loop or feedback changes the probabilities.

答:
It looks as if $s_2,s_3,s_4$ all absorb and that $s_2$ is inaccessible.

Let $p$ be the probability you want.  Then

$$p = \frac12\times\left(\frac49\times p +\frac29 \times 1 + \frac39 \times 0\right)+\frac12 \times 0$$  which is easy to solve for $p$


问:
Schwartz Inequality and the Norm Convergence Theorem
I was trying the show that if $f\in PC(2\pi)$ and $x_0 \in [-\pi,\pi]$ then the Fourier series for $f$ can be integrated term by term as
$$\int_{x_0}^{x}f(t)dt = \frac{1}{2}a_0(x-x_0) + \sum_{n=1}^{\infty} \int_{x_0}^{x}(a_ncos(nt)b_nsen(nt))dt$$

The problem came with the suggestion that I should consider using the Schwarz Inequality and the Norm Convergence Theorem.

I was able to conclude that
$$\left| \int_{x_0}^{x} f(t) dt\right| \le ||f||_2|x-x_0|^2 \le ||f||_2 \sqrt{2\pi}$$  using the suggested Inequality, but I haven't been able to move forward from here.

Thanks in advance.

答:
You forgot the integral on  the right side.

Hints: Since $\sum |a_n|^{2} $ and $\sum |b_n|^{2} $ are finite the suggested inequality shows that the series on RHS is uniformly convergent.

Deonte the sum by $g$. LHS and $g$ are both periodic continuous functions. To show that they are equal it is enough to show that they have the same Fourier coefficients. This can shown by an integration by parts.


问:
Find The Posterior of $(\mu, \frac{1}{\sigma^2})$
I am given the following sample:

$6.56, 6.39, 3.30, 3.03, 5.31, 5.62, 5.10, 2.45, 8.24, 3.71,\\
4.14, 2.80, 7.43, 6.82, 4.75, 4.09, 7.95, 5.84, 8.44, 9.36$

which is a sample from a $N(μ, σ^2)$ distribution.

I am also give that:

The prior $μ|σ^2 \stackrel{}{\sim} N(3, 4σ^2)$

$σ^2 \stackrel{}{\sim} Gamma(1, 1)$

I am being asked to determine the posterior of $\large(\mu, \frac{1}{\sigma^2})$. How do I find that? Any help is much appreciated!

答:
Let's set $\theta_1=\mu$ and $\theta_2=\frac{1}{\sigma^2}$ so that $\theta_2$ indicates the precision of a normal distribution.

Your model is the following

$$p(x|\theta_1,\theta_2)=\sqrt{\frac{\theta_2}{2\pi}}e^{-\frac{\theta_2}{2}(x-\theta_1)^2}$$

I think that the Gamma(1;1) is the prior distribution of $\theta_2$ thus the prior

$$h(\theta_1,\theta_2)=\sqrt{\frac{1}{2\pi}}\theta_2^{\frac{1}{2}}\cdot e^{-\frac{\theta_2}{2}(\theta_1-3.4)^2}\cdot e^{-\theta_2}$$

is a Normal Gamma distribution with the following parameters_

$\alpha=1$

$\beta=1$

$\lambda=1$

$\mu=3.4$

Now, as usual, multiplying the prior and the likelihood, after some algebraic manipulations, you will find that the posterior distribution is still a Normal Gamma with the following parameters:

$\mu_{post}=\frac{3.4+n\overline{X}_n}{n-1}$

$\lambda_{post}=n+1$

$\alpha_{post}=1+\frac{n}{2}$

$\beta_{post}=1+\frac{1}{2}\Sigma_i(X_i-\overline{X}_n)^2+\frac{n(\overline{X}_n-3.4)^2}{2(n+1)}$

Concluding, after seeing the sample you posted, the posterior is

$$ \bbox[5px,border:2px solid red]
{
h(\theta_1,\theta_2|\mathbf{x})=\sqrt{\frac{21}{2\pi}}\frac{43.14^{11}}{10!}\cdot \theta_2^{10.5}\cdot e^{-10.5\theta_2(\theta_1-8.70)^2}\cdot e^{-43.14\theta_2}
\
}
$$


问:
What is the min amount of $n$ so that the following conclusion be true? $a^n | b^3 \to a^3 | b^2$
What is the min amount of $n$ so that the following conclusion be true?

$a^n | b^3 \to a^3 | b^2$

We know If $a^n|b^3$ then there exists a $k \in \mathbb{Z}$ so that $b^3=k.a^n$ so :

\begin{align}
b^3 =k.a^n\to b^2 =a^n.k.b^{-1}\to b^2 =a^3 . a^{n-3}.k.b^{-1}
\end{align}
So If we want $a^3 |b^2$ to be true, $a^{n-3}.k.b^{-1}$ must belong to $\mathbb{Z}$. We have to check for at least which $n$ this is true? for $n=4$ I found a counter example, but I couldn't find a clear way to find the min of $n$.

答:
Recall if $\,q\in\Bbb Q,\, k\in\Bbb N_{>0}\ \ \color{#c00}{\rm then}\ \ q^k\in\Bbb Z\!\color{#c00}{\iff}\! q\in\Bbb Z,\,$ by the Rational Root Test. Using this we can power the quotients to put them over a common denominator, then the rest is straightforward.

Hence $\ \dfrac{b^{\large 3}}{a^{\large n}}\in \Bbb Z\,\Rightarrow\, \dfrac{b^{\large 2}}{a^{\large 3}}\in \Bbb Z\ $ is $\rm\color{#c00}{equivalent}$ to below (with $\,k=3,\,k=n)$

$\qquad\ \ \dfrac{b^{\large \color{#0a0}9}}{a^{\large 3n}}\in \Bbb Z\,\Rightarrow\, \dfrac{b^{\large \color{#0a0}{2n}}}{a^{\large 3n}}\in \Bbb Z,\,$ clearly true if $\,\color{#0a0}{9\le 2n},\,$ else false (e.g. let $\,a\!=\!c^{\large 3},\,b\!=\!c^{\large n},\,c>1)$


问:
Probability of same feature inside a population

Let us have a population of many units, each unit has $n$ features, with $p$ of them are set to $1$ and other to $0$.

Question:
What is the probability, that if O take 2 random different units and compare one random feature, the value of both will be 1?

My naive answer:

Since probability of a feature for one unit to be 1 is p/n, the probability of 2 units to have it is
$$
(p/n)^2
$$
I want to ask, if this calculation correct, or it is naive and I must involve further distributions, like binominal, whatever but I do not understand how to apply them here.

Yes, i assume the features are independent, that is why i calculate like that.
Also it is known Hardy–Weinberg principle

https://en.wikipedia.org/wiki/Hardy%E2%80%93Weinberg_principle

what tells the same.

答:
You seem to be assuming independence between the features of different units and that the features are equally likely.

If that is true then your calculation is correct. If not then not, and you would need to know more than the question states.


问:
E measurable with m(E) < $\infty$?
Suppose that $E$ is measurable with $m(E)$ $<$  $\infty$.

ii) Show that $\displaystyle \ \ \int_E 2f\,\,\,$ $=$  $2$$\displaystyle \ \ \int_E f\,\,\,$ if $f$ is bounded and measurable.

I told my TA about it in my self study class and he said do not worry about it. We are not going to go into this topic where we would need to solve these types of problems. $ugh$.

Here is what I know from reading:

If $E$ is an element in $M$ we say $E$ is Lebesgue measurable. $M$ will be called Lebesgue measure.

The proof for ii) is very $vague$. I wanted to see the proof of this because I did not learn it. I am new to this and was wondering how we would prove it to be bounded and measurable. This proof might help later with other proofs (even though my TA said do not look at it. $ugh$). That is why I would like to see this.

答:
For 1), start by writing $f$ as a simple function — that is, write down what it means (decomposition of $f$). Then, observe that $\int_E f=\int_\Omega \mathbf{1}_E f$ (where $\Omega$ is the whole space), and $\mathbf{1}_E f$ is also simple (as $E$ is measurable).

For 2), use the theorem relating any function bounded and measurable to simple fucntions (as a limit), and 1).

Edit:
For 2):
Since $f$ is bounded and measurable, there exists a sequence $(f_n)_{n\in\mathbb{N}}$ of simple measurable functions uniformly converging to $f$.

Fix any $\varepsilon > 0$. Then there exists $N\in\mathbb{N}$ such that for all $n\geq N$, for all $x\in\Omega$, $|f_n(x) - f(x)|\leq \varepsilon^\prime\stackrel{\rm{}def}{=}\frac{\varepsilon}{m(E)}$.
We then have that, $\forall n\geq N$,

$\left| \int_E fd\mu - \int_E f_nd\mu \right| \le \int_E \left|f - f_n\right|d\mu \le \int_E \varepsilon^\prime d\mu = \varepsilon^\prime m(E) =\varepsilon$ ($\dagger$)
and similarly
$\left| \int_E (2f)d\mu - \int_E (2f_n)d\mu \right| \leq 2\varepsilon$ ($\ddagger$)

i.e $\int_E (2f)d\mu < \infty$, $\int_E f d\mu < \infty$ and, more importantly,

$\int_E (2f)d\mu \stackrel{(\ddagger)}{=} \displaystyle\lim_{n\to\infty} \int_E (2f_n)d\mu \displaystyle\stackrel{(1)}{=} 2\displaystyle\lim_{n\to\infty} \int_E f_n d\mu \stackrel{(\dagger)}{=} 2\int_E fd\mu$

Title: E measurable with m(E) < $\infty$?


问:
Characterization of ${\rm PSU}(4,2)$ by its order $|{\rm PSU}(4,2)|$ and the number $n_5({\rm PSU}(4,2))$ of its Sylow $5$-subgroups?
We know that $|{\rm PSU}(4,2)|=25920=2^6\cdot3^4\cdot5$, and the number $n_5({\rm PSU}(4,2))$ of Sylow $5$-subgroups of ${\rm PSU}(4,2)$ is $1296=2^4\cdot3^4$.

In Theorem 3.2 of the paper Two new characterizations for sporadic simple groups, Khosravi and Khosravi prove that every sporadic simple group $S$ is uniquely determined by its order $|S|$ and the number of its Sylow $p$-subgroups $n_p(S)$, where $p$ is the largest element of $$\pi(S)=\{p\mid p~{\rm is~a~divisor~of}~|S|\}.$$ I just want to prove the same result for ${\rm PSU}(4,2)$. But I cannot follow the proof of Theorem 3.2. In particular, in the first paragraph of the proof of Theorem 3.2 it says that "... and so $|G/K|$ is a divisor of $n_p(G)$". I cannot catch the idea of this sentence. May someone give me a more detailed explanation？

答: Recalling the definition that $n_p(G)=|N_G(P)|$ where $P\in\operatorname{Syl}_p(G)$ we have $$N_G(P)/N_G(K)\cong G/K\implies \left|N_G(P)/N_G(K)\right|=|G/K|$$ and the result follows as $\left|N_G(P)/N_G(K)\right|$ divides $|N_G(P)|$.


问:
Is the set of irrational numbers countable when the set of algebraic numbers is involved?
The set of algebraic numbers (let's say $\mathcal{A}$ ) is countable. The set of irrational numbers (let's say $\mathcal{I}$ ) is uncountable. The set of algebraic numbers contains some of irrational numbers and some irrationals are not algebraic.

Now if we choose a subset $\mathcal{I_A}$ of all irrationals from the set of algebraic numbers, will it be countable?

If I consider that $\mathcal{I_A}\subset \mathcal{A}$, then it is countable [ $\because$ subset of a countable set is countable] but again $\mathcal{I_A}\subset \mathcal{I}$ [subsets of uncountable sets can be countable or uncountable].

Hence in this case is $\mathcal{I_A}$ countable? Can we establish a bijection $\mathbb{N}\to\mathcal{I_A}$ ?

Any help or explanation is valuable and highly appreciated.

答:
Yes, $\mathcal I_A$ is countable. If you want an injection $f:\mathcal I_A\to\Bbb N$, consider a bijection $g:\Bbb N\to A$ and call $$f(x)=\left\lvert\{n\in\Bbb N\,:\, g(n)\in\mathcal I_A\land 0\le n< g^{-1}(x)\}\right\rvert$$

The image of this function is an interval of $(\Bbb N,\le)$ containinig $0$. Its image is a proper subset of $\Bbb N$ if and only if $\mathcal I_A$ is finite (which, apparently, you are not excluding).


问:
Number with all 2 digits permutations from a set of digits
Lets say you are given a set of n consecutive digits , Ex {0,1,2,3}. You are required to form a number using the digits which abides by the following rule:

1.The number should cover all relations (ex 12 and 21) between any 2 digits from the set of numbers

For the given set, this number would be 0123130203210 or 0123021310320.

I was able to get this number manually by trial and error. Is there a deterministic way of achieving this number for any n consecutive digits?

The number will have all the 2 digit permutations only once.

0 -> [1,2,3] will yield 01 , 02 , 03

1 -> [0,2,3] will yield 10 , 12 , 13

2 -> [0,1,3] will yield 20 , 21 , 23

3 -> [0,1,2] will yield 30 , 31 , 32

We have to generate a number which will cover all the above relations.

答:
This problem is equivalent to finding a Euler path in a complete directed graph on n vertices. Using the example {0,1,2,3} you gave, the graph would look like this:

To find an Euler path you can use Hierholzer's algorithm, which in simple terms goes as follows:

Start at any vertex, follow a path of edges (the longer, the better) until you return to the beginning vertex, i.e. a cycle.

Repeat the above process until you've exhausted the edges adjacent to your beginning vertex.

If there are any unused edges, find a vertex on one of your cycles that has unused adjacent edges, use it as a starting vertex and repeat the above two steps.

Continue to do the above as long as there are any unused edges.

In the above graph, starting at 0, a simple way to start would be to traverse the outer edges in both directions, e.g. 012303210. At this point we're only missing the edges between 0 and 2, and the ones between 1 and 3. The former can be included by replacing any of the three 0's in our path with 020, or either of the two 2's with 202; similarly, the latter can be included by replacing either of the two 1's with 131, or either of the two 3's with 313. I'm choosing to replace the first 2, which yields 01202303210, and then the second 3, which yields 0120230313210.


问:
Show that if $x^*$ is optimal solution of P it is optimal solution of P'
Consider the problem  (P)
$\min f(x)$

s.t $g(x)\leq0$

$x\in X$

Where $f,g$ is convex and $X$ is convex. Suppose that $x^*$ is optimal solution of the problem above and $g(x^*)<0$ Show that $x^*$ is optimal solution of the problem below  (P')

$\min f(x)$

s.t $x\in X$

I've thought that $x^*$ is in the set of optimal solution of P and we can see that $P\subseteq P'$ and therefore $x^*\in P'$ and is optimal solution of $P'$ but I am not sure if this is the right way.

We've just started convex optimization

Any if anyone can give some geometric intuition it will be nice

答:
Let us assume the $x'$ minimize $(P')$.Looking at the interval $[x',x^*]\subseteq X$ because $X$ is convex. one can find $\lambda\in[0,1]$ such that $g(\lambda x'+(1-\lambda)x^*)\leq\lambda g(x')+(1-\lambda)g(x^*)\leq g(x^*)< 0$.\
Now looking at $f(\lambda x'+(1-\lambda)x^*)\leq\lambda f(x')+(1-\lambda)f(x^*)\leq f(x^*)$ and we got new minimizer for $(P)$ in contradiction that $x^*$ is the minimizer


问:
Difficult integral involving $\arctan x$

Question: Show that$$\int\limits_0^{\infty}\mathrm dx\,\frac {\arctan x\log(1+x^2)}{x(1+x^2)}=\frac {\pi}2\log^22.$$

I can't tell if I'm being an idiot, or if this is a lot more difficult than it looks. First, I tried integration by parts using the fact that $$\frac 1{x(1+x^2)}=\frac 1x-\frac x{1+x^2}.$$

But quickly I gave up as I wasn't sure what to do with the result. I then decided to make the substitution $t=\arctan x$ to get rid of the $1+x^2$ term in the denominator. Therefore$$\begin{align*}\mathfrak{I} & =\int\limits_0^{\pi/2}\mathrm dt\,t\cot t\log\sec^2t=-2\int\limits_0^{\pi/2}\mathrm dt\, t\cot t \log\cos t.\end{align*}$$

However, I'm not exactly sure what to do after this. Should I use integration by parts? Differentiation under the integral sign? I'm having trouble getting started with this integral. Any ideas?

Body:
\begin{align}J&=\int_0^\infty \frac{\ln(1+x^2)\arctan x}{x(1+x^2)}dx\\
&=\int_0^1 \frac{\ln(1+x^2)\arctan x}{x(1+x^2)}dx+\int_1^\infty \frac{\ln(1+y^2)\arctan y}{y(1+y^2)}dy\\
&\overset{x=\frac{1}{y}}=\int_0^1 \frac{\ln(1+x^2)\arctan x}{x(1+x^2)}dx+\int_0^1 \frac{x\ln\left(\frac{1+x^2}{x^2}\right)\arctan\left( \frac{1}{x}\right)}{1+x^2}dx\\
&=\int_0^1 \frac{\ln(1+x^2)\arctan x}{x}dx-2\int_0^1 \frac{x\ln(1+x^2)\arctan x}{1+x^2}dx+2\int_0^1 \frac{x\ln x\arctan x}{1+x^2}dx+\\
&\frac{\pi}{2}\int_0^1 \frac{x\ln(1+x^2)}{1+x^2}dx-\pi\int_0^1 \frac{x\ln x}{1+x^2}dx\\
&\overset{\text{IBP}}=\int_0^1 \frac{\ln(1+x^2)\arctan x}{x}dx-2\int_0^1 \frac{x\ln(1+x^2)\arctan x}{1+x^2}dx+\\&\Big[\ln(1+x^2)\ln x\arctan x\Big]_0^1-\int_0^1 \frac{\ln(1+x^2)\arctan x}{x}dx-\int_0^1 \frac{\ln(1+x^2)\ln x}{1+x^2}dx+\\&\frac{\pi}{2}\int_0^1 \frac{x\ln(1+x^2)}{1+x^2}dx-\pi\int_0^1 \frac{x\ln x}{1+x^2}dx\\
&=-2\int_0^1 \frac{x\ln(1+x^2)\arctan x}{1+x^2}dx-\int_0^1 \frac{\ln(1+x^2)\ln x}{1+x^2}dx+\frac{\pi}{2}\int_0^1 \frac{x\ln(1+x^2)}{1+x^2}dx-\\&\pi\int_0^1 \frac{x\ln x}{1+x^2}dx\\
K&=\int_0^1 \frac{2x\ln(1+x^2)\arctan x}{1+x^2}dx\\
&\overset{\text{IBP}}=\Big[\ln(1+x^2)^2\arctan x\Big]_0^1-K-\int_0^1 \frac{\ln^2(1+x^2)}{1+x^2}dx\\
K&=\frac{\pi\ln^2 2}{8}-\frac{1}{2}\int_0^1 \frac{\ln^2(1+x^2)}{1+x^2}dx\\
J&=-\frac{\pi\ln^2 2}{8}+\frac{1}{2}\int_0^1 \frac{\ln^2(1+x^2)}{1+x^2}dx-\int_0^1 \frac{\ln(1+x^2)\ln x}{1+x^2}dx+\frac{\pi}{2}\int_0^1 \frac{y\ln(1+y^2)}{1+y^2}dy-\\&\pi\int_0^1 \frac{y\ln y}{1+y^2}dy\\
&\overset{x=y^2}=-\frac{\pi\ln^2 2}{8}+\frac{1}{2}\int_0^1 \frac{\ln^2(1+x^2)}{1+x^2}dx-\int_0^1 \frac{\ln(1+x^2)\ln x}{1+x^2}dx+\frac{\pi}{4}\int_0^1 \frac{\ln(1+x)}{1+x}dx-\\&\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\
&=\frac{1}{2}\int_0^1 \frac{\ln^2(1+y^2)}{1+y^2}dy-\int_0^1 \frac{\ln(1+y^2)\ln y}{1+y^2}dy-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\
&\overset{y=\tan t}=2\int_0^{\frac{\pi}{4}}\ln^2\left(\cos t\right)dt-2\int_0^{\frac{\pi}{4}}\ln^2\left(\cos t\right)dt+2\int_0^{\frac{\pi}{4}}\ln\left(\cos t\right)\ln\left(\sin t\right)dt-\\&\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\
&=2\int_0^{\frac{\pi}{4}}\ln\left(\cos t\right)\ln\left(\sin t\right)dt-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\
&=\frac{1}{2}\left(\int_0^{\frac{\pi}{4}}\ln^2(\sin t\cos t)dt-\int_0^{\frac{\pi}{4}}\ln^2\left(\tan u\right)du\right)-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\
&\overset{u=2t}=\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin u}{2}\right)du-\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan u\right)du-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\
&=\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\sin u\right)du-\frac{\ln 2}{2}\int_0^{\frac{\pi}{2}}\ln\left(\sin u\right)du+\frac{\pi\ln^2 2}{8}-\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan u\right)du-\\&\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\
L&=\int_0^{\frac{\pi}{2}}\ln^2\left(\sin u\right)du\\
&=\frac{1}{4}\left(2\int_0^{\frac{\pi}{2}}\ln^2\left(\sin u\right)du+2\int_0^{\frac{\pi}{2}}\ln^2\left(\cos u\right)du\right)\\
&=\frac{1}{4}\left(\int_0^{\frac{\pi}{2}}\ln^2\left(\cos u\sin u\right)du+\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\right)\\
&\overset{t=2u}=\frac{1}{4}\left(\frac{1}{2}\int_0^{\pi}\ln^2\left(\frac{\sin t}{2}\right)dt+\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\right)\\
&=\frac{1}{8}\int_0^{\pi}\ln^2\left(\sin t\right)dt-\frac{\ln 2}{4}\int_0^{\pi}\ln\left(\sin t\right)dt+\frac{\pi\ln^2 2}{8}+\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\
&=\frac{1}{8}\left(L+\int_{\frac{\pi}{2}}^\pi\ln^2\left(\sin v\right)dv\right)-\frac{\ln 2}{4}\int_0^{\pi}\ln\left(\sin t\right)dt+\frac{\pi\ln^2 2}{8}+\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\
&\overset{t=\pi-v}=\frac{1}{4}L-\frac{\ln 2}{4}\int_0^{\pi}\ln\left(\sin t\right)dt+\frac{\pi\ln^2 2}{8}+\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\
L&=-\frac{\ln 2}{3}\int_0^{\pi}\ln\left(\sin t\right)dt+\frac{\pi\ln^2 2}{6}+\frac{1}{3}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\
&=-\frac{\ln 2}{3}\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt-\frac{\ln 2}{3}\int_{\frac{\pi}{2}}^{\pi}\ln\left(\sin v\right)dv+\frac{\pi\ln^2 2}{6}+\frac{1}{3}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\
&\overset{t=\pi-v}=-\frac{2\ln 2}{3}\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt+\frac{\pi\ln^2 2}{6}+\frac{1}{3}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\
\end{align}

Therefore,

\begin{align}J&=\frac{1}{6}\pi\ln^2 2-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx-\frac{2\ln 2}{3}\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt+\frac{1}{12}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan t\right)dt-\\&\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan t\right)dt\\
&=\frac{1}{6}\pi\ln^2 2-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx-\frac{2\ln 2}{3}\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt+\\&\frac{1}{12}\left(\int_0^{\frac{\pi}{4}}\ln^2\left(\tan t\right)dt+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln\left(\tan u\right)du\right)-\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan t\right)dt\\
&\overset{t=\frac{\pi}{2}-v}=\frac{1}{6}\pi\ln^2 2-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx-\frac{2\ln 2}{3}\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt-\frac{1}{3}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan t\right)dt\\
&=\frac{1}{6}\pi\ln^2 2-\frac{\pi}{4}\times -\frac{\pi^2}{12}-\frac{2\ln 2}{3}\times -\frac{\pi\ln 2}{2}-\frac{1}{3}\times \frac{\pi^3}{16}\\
&=\boxed{\frac{\pi\ln^2 2}{2}}
\end{align}

I assume that:

\begin{align}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan t\right)dt&=\frac{\pi^3}{16}\\
\int_0^1 \frac{\ln x}{1-x}dx&=-\zeta(2)=-\frac{\pi^2}{6}\\
\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt&=-\frac{\pi\ln 2}{2}\\
\int_0^1 \frac{\ln x}{1+x}dx&=\int_0^1 \frac{\ln x}{1-x}dx-\int_0^1 \frac{2y\ln y}{1-y^2}dy\\
&\overset{y=x^2}=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx=-\frac{\pi^2}{12}
\end{align}


问:
Is convolution area-preserving?
Consider two functions f(x) and g(x) with indefinite integrals ("area under the curve") Af and Ag. Does the convolution f * g preserve the area, i.e. is

Af *g = Af * Ag

i.e.

∫ ( ∫f(x)g(t−x)dx ) dt = (∫f(x)dx) (∫g(x)dx) ?

I've learned that convolution in one space is multiplication in the dual space, or more precisely, that direct multiplication in one space corresponds to polynomial multiplication in the dual space. There is also a nice, short video where this intuition is explained, but I'm having difficulty transferring this to the question regarding areas and drawing the right conclusions.

To provide some practical context: Image formation in an optical system can be described as

Image = Source * PSF

with PSF being the point spread function describing the system's impulse response. The PSF is often approximated by a Gaussian and the convolution results in a Gaussian blur.

If the Gaussian is normalized to unit area (i.e. a normal distribution), does this equate to the fact that the integrated intensities of source and image are preserved? Meaning that the photons are merely "redistributed" (blurring process), but the total photon number is preserved (= energy conservation in an ideal, lossless system)?

Conversely, can I add e.g. a 10% loss of photons due to absorption by simply changing the normalization of the PSF from 1.0 to 0.9 (rescaling the Gaussian's normalization factor)?

答:
If by convolution you mean the standard

$$(f * g)(t) = \int f(x)g(t-x)dx$$

and by area

$$A_f = \int f(x)dx$$

then the question is whether the following is true

$$\int\left[\int f(x)g(t-x) dx\right]dt = \left[\int f(x)dx\right]\cdot \left[\int g(x)dx\right]$$

This is indeed true. By a simple change of variables

$$u = x$$
$$v = t - x$$

with Jacobian

$$\det \pmatrix{1 & -1 \\ 0 & 1} = 1$$

we get

$$\int\int f(x)g(t-x) dxdt  = \int\int f(u)g(v)dudv = \left[\int f(u)du\right]\cdot \left[\int g(v)dv\right]$$

provided that the conditions of the Fubini theorem are satisfied.

Title: Is convolution area-preserving?


问:
Find angle in triangle with incenter

In a triangle $ABC$ if $AD$, $BE$, $CF$ are the angle bisectors of $\angle A$, $\angle B$ $\angle C$ respectively with incenter $I$ and $\angle A=120^\circ$ Find $\angle DFI$

It's $30^\circ$. If $BE$ meet $DF$ at $G$ and $CF$ meet $DE$ at $H$ it seem that $\triangle GEI$ and $\triangle HFI$ is an isosceles triangle. And also $GA$ and $HA$ is the angle bisector of $\angle BAD$ and $\angle CAD$ respectively. I dun know how to prove all of these, I just put it in GeoGebra and see the accurate picture, all I could find is $\angle BIC=150^\circ$, $\angle BIF=30^\circ$ and $\angle DFI+\angle DEI=60^\circ$. Please help

答:
Construct an equilateral triangle $CFK$.

Since $\angle FAC + \angle FKC = 180^\circ$, $AFKC$ is a cyclic quadrilateral.

$\Rightarrow \angle FAK = \angle FCK = 60^\circ = \angle FAD$

$\Rightarrow AIDK$ is a straight line.

$\Rightarrow \angle DKC = \angle AFC = 180^\circ - \angle FAC - \angle ACF = 60^\circ - \frac{\angle C}{2} = \angle DCK$

$\Rightarrow DK = DC$

$\Rightarrow \triangle FDK \cong \triangle FDC$

$\Rightarrow \angle DFI = \frac{60 ^\circ}{2} = 30^\circ$

Title: Find angle in triangle with incenter


问:
how to solve this exponential inequality?
I'm studying for the admission test.
In particular, this exponential inequality is a little bit tricky.
this is the inequality:
$$
\left(\frac{2}{e}\right)^x > 1
$$
I've tried to use the change of basis formula to try to solve the logarithm to deduce the result. But it's not the optimal way.
Then, I've tried to rewrite 1 as (2/e)^0, and to solve the equation to find the critical point, but it's still not correct.
Can you help me?

Thanks in advance.

答:
The question is answered in the comment but it isn't clear.

Since $2/e$ is less than 1 therefore, if we go with $x>0$ in the comments for any $x$, it would yield a wrong result. For example $(2/e)^{50} = 2.17*10^{-7} \leq 1$

However, for $x < 0$ as a first step, should be a good answer, since the equation will be inversed making it $e/2$ which is number over 1 in the first place.

Start from here.

Continuation

I said start from here, when I gave the full answer without supporting it (proof). Therefore, please rewrite everything in an eloquent way.


问:
Necessary and sufficient condition for an Euler trail between two vertices
Graph theory question one of those obvious math proofs so its going to be a pain to prove.

Show that G=(V,E) has an Euler trail between (different) vertices u and v if and only if G is connect and all vertices except u and v are of even degree.

By obviousness if there all other vertices are of even degree then you can make a circuit from v to a1 to a2 to an eventually you will come back to u or v if you continue doing this you eventually will arrive at u or v with no way to leave thus Euler trail.

But how would you actually prove this?

答:
The way to proof goes is as follows. The book I am using gives a max-length contradiction argument to prove that if all vertices of a graph are even then Graph must have Eulerian Circuit.

Claim: The graph with all even vertices has a circuit.

By contradiction, you start by assuming that $T$ a trail of maximum length such that $T = (u=u_1,u_2....,u_k=v)$. Since whenever the vertex $v$ is encoutered on $T$ it has both an incoming and outgoing edge the last time it is encountered will lead it to have an odd degree. Thus either you have another vertex after $v$ in which case the trail you started with is not a maximum length trail or $u=v$ in which case you have a circuit $C$.

For sake of contradiction pick the largest circuit in the graph. Assuming that $C$ is not a Eulerian circuit consider the graph $G-C$. Since the circuit reduces the degree of every vertex by exactly two what $G-C$ gives you is a possibly disconnected graph with all vertices of degrees smaller by 2. Thus each component of $G-C$ has all even vertices and is connected to original circuit $C$ on a vertex say $x$ for component $G_x$. Using the same procedure as above you can construct an new Circuit $C'$ at $x$ inside the component $G_x$. By concatenating this new circuit at $x$ to the older circuit $C$  you get a larger circuit than before which is a contradiction with the fact that $C$ was supposed to be the largest circuit in the graph.Thus the graph with all even vertices will have an Eulerian circuit.

If you are not very fond of proof by contradiction you can think of this as recursively appending circuits of various components till all edges are part of the circuit.

More detailed by someone at university of texas


问:
How to find out (x2, y2) when I know the equation of a function, (x1, y1) and the distance between the two points?
I'm writing a program in which I have to draw something along a line multiple time and spaced evenly. I used two points [$(x_1,y_1)$ and $(x_3, y_3)$] to create a line. Then I found out the equation $y = mx+b$ of the line going through those two points. Now I want to draw multiple time between those two points with a distance $d$ between each drawing [first drawing is at $(x_1, y_1)$].

I know the squared distance between two points in a coordinate system is $d^2 = (x_1 - x_2)^2 + (y_1 -y_2)^2$. But I am not sure how to figure out $x_2$ and $y_2$ from there. is it even possible? is there another way to know $x_2$ and $y_2$ with all the other known information?

答:
$(x_2-x_1)^2+(y_2-y_1)^2=d^2$ is in fact the equation of a circle with radius $d$  and center on $(x_1, y_1)$, so to find $x_2$ and $y_2$ you have to find the intersection of line $y=mx+b$ with this circle , that is you have to solve following system of equation:

$\begin{cases}(x-x_1)^2+(y-y_1)^2=d^2\\ y=mx+b\end{cases}$

You have $x_1$, $y_1$ and $d$ so you can find x and y which in fact is $x_2$ and $y_2$


问:
uncoupled transport equation of (U) and another coupled linear pde, does $U$ decay?

Let $C = 1$ and $\gamma = 1$. and the initial conditions as $U(t=0) = e^{cos(2\pi x)}$ (periodic boundary) and $V(t=0) = 0$. Now I'm getting two numerical solutions. One models the $U$ as a transported wave moving to the right, the other models $U$ as a decaying wave over time. Which one is right?

答:
$$\frac{\partial U}{\partial t}+C\frac{\partial U}{\partial x}=0$$
There is no difficulty to analytically solve this PDE thanks to the method of characteristics or other method. The general solution is :
$$U(t,x)=F(x-Ct)$$
where $F$ is an arbitrary function.

Initial condition : $U(0,x)=e^{\cos(2\pi x)}=F(x-0t)$
$$F(x)=e^{\cos(2\pi x)}$$
Now the function $F$ is known. We put it into the above general solution where the variable is not $x$ but is $(x-Ct)$.
$$\boxed{U(x,t)=e^{\cos\left(2\pi (x-Ct)\right)}}$$
So the exact model for $U(t,x)$ is determined and this answers to your question.


答:
I think the easiest way is to write the line in vector form where $\vec{n} = \frac 1{\sqrt{1+m^2}}\begin{pmatrix}1 \\ m\end{pmatrix}$ is the normalized direction vector. You will find the $k$-th point on the line starting from $\begin{pmatrix}x_1 \\ y_1\end{pmatrix}$ by

$$\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}x_1 \\ y_1\end{pmatrix}+k\cdot d\cdot \vec{n} = \begin{pmatrix}x_1 \\ y_1\end{pmatrix}+k\frac{d}{\sqrt{1+m^2}}\begin{pmatrix}1 \\ m\end{pmatrix}$$

Here is a Desmos-graph which illustrates above formula. You may start with a horizontal line to see that the distance fits and then change the slope $m$ and animate $k$.

Title: How to find out (x2, y2) when I know the equation of a function, (x1, y1) and the distance between the two points?

Body:
I'm writing a program in which I have to draw something along a line multiple time and spaced evenly. I used two points [$(x_1,y_1)$ and $(x_3, y_3)$] to create a line. Then I found out the equation $y = mx+b$ of the line going through those two points. Now I want to draw multiple time between those two points with a distance $d$ between each drawing [first drawing is at $(x_1, y_1)$].

I know the squared distance between two points in a coordinate system is $d^2 = (x_1 - x_2)^2 + (y_1 -y_2)^2$. But I am not sure how to figure out $x_2$ and $y_2$ from there. is it even possible? is there another way to know $x_2$ and $y_2$ with all the other known information?


问:
Chain Rule for PDEs
I'm having a hard time seeing how to use the chain rule to prove these two statements and understanding how they work for PDEs in general

Given $D>0$

$u(x,t)$ satisfies $u_{t} = Du_{xx}$ iff $v(y,t) = u(\sqrt{D}y,t)$ satisfies $v_{t} = v_{yy}$

$u(x,t)$ satisfies $u_{t} = Du_{xx}$ iff $v(x,\tau) = u(x,\tau/D)$ satisfies $v_{\tau} = v_{xx}$

Any help would be greatly appreciated

答:
Example Solution

$v_t = \frac{\partial}{\partial t}(u(\sqrt{Dy},t))=\frac{\partial u(\sqrt{Dy},t)}{\partial t}*\frac{\partial}{\partial t}(u)=u_t(\sqrt{Dy},t)$

$v_{yy} = \frac{\partial}{\partial y}(\sqrt{D}\frac{\partial u}{\partial t})=Du_{xx}$

Title: Chain Rule for PDEs


问:
How to find the following integral: $ \iint\limits_D\max\left\{\sin x,\sin y\right\}dxdy,\ D=\left\{0\le x\le\pi,\ 0\le y\le\pi\right\} $

How to evaluate the following integral:
$$
\iint\limits_D\max\left\{\sin x,\sin y\right\}dxdy,\ \ \text{where}\ D=\left\{0\le x\le\pi,\ 0\le y\le\pi\right\}
$$

To be honest, I don't understand how to approach this problem. In my opinion, $\sin y$ is always greater than $\sin x$, since $x=\sin y$ lies higher than $y=\sin x$ for any given point. But it doesn't seem to be true. Could anyone give me a hint about how I should evaluate this integral?

答:
notice that since $x,y\in[0,\pi]$, $\sin x,\sin y\in[0,1]$ and that the sine function has a line of symmetry about $x=\pi/2$. Over the interval $[0,\pi/2]$ $\sin(x)$ is an increasing function and so if:
$$x>y,\sin(x)>\sin(y)\,\,\,x,y\in[0,\pi/2]$$
now repeat this process for all of the possible regions and then integrate the respective functions :)


问:
Reference: $d_r(\mathcal{C}_m)\geq \delta_{FR}^{r}(m)$ Generalized Feng-Rao distance is lower bound of generalized Hamming Weight
I am looking for a reference (article/book), where they give a proof for this claim:

$$d_r(\mathcal{C}_m)\geq \delta_{FR}^{r}(m)$$

Where $\mathcal{C}_m$ is the $m^{th}$ classical one point algebraic code, $d_r(\mathcal{C}_m)$ is the generalized Hamming weight of order $r$ for that code, and $\delta_{FR}^{r}(m)$ is the $r$ generalized Feng-Rao distance of $m$.

答:
For those interested, the best I could find is in the original paper, Theorem 3.14:

Petra Heijnen and Ruud Pellikaan. Generalized Hamming weights of q-ary Reed-Muller codes. 1998

https://www.win.tue.nl/~ruudp/paper/29.pdf