问:
Can we always numerically integrate?
I posted this question in physics SE, as few suggested me to post it here: when I watch online lectures (usually of physics) most lecturers say that if we cant solve a integral exactly we can always numerically integrate it. (E.g. in central force problem the orbit where we find $\theta(r)$ or $r(\theta)$ is pretty difficult to compute and professor said that if needed we can always integrate it numerically if we cannot do it exactly)

So is it always possible to numerically integrate? Is there any example in physics where it not possible to integrate numerically or any integral which cannot be even integrated numerically? Is there any limitation of numerical integration?

答:
Suppose the integrand $f$ is Riemann integrable on $[a,b]$.  Then the Riemann sums converge to $\int_a^b f(x)\;dx$.  If you know a bound $M$ for $|f(x)|$ on $[a,b]$ you can rigorously prove a bound on the error for that approximation.  That error bound goes to zero as the number of terms goes to infinity.  So you CAN numerically integrate in this setting.

If you know $f$ is $n$ times differentiable, there are numerical integral calculations that converge more rapidly.  For example, Simpson's rule.

But of course there are functions (according to mathematicians) that are not Riemann integrable.  Do any of them come up in physics?  You will have to ask physicists about that.  (Mandelbrot reported that his early papers on fractals were routinely rejected by physics journals as "unphysical".  But that is no longer the case today.)

For example, in quantum physics you deal with Hilbert spaces like $L^2$, where there are lots of non-Riemann-integrable functions.

Mathematical physicists are often writing down integrals with delta functions in there, but pretending they are actual functions.  Will they try to compute these numerically?

Physicists write down divergent integrals like this:
$$
\delta(x-\alpha) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{ip(x-\alpha)} \; dp .
$$
Will they try to compute these numerically?

Title: Can we always numerically integrate?


问:
Packing circles on a line
On today's TopCoder Single-Round Match, the following question was posed (the post-contest write-up hasn't arrived yet, and their explanations often leave much to be desired anyway, so I thought I'd ask here):

Given a maximum of 8 marbles and their radii, how would you put them next to each other on a line so that the distance between the lowest point on the leftmost marble and the lowest point on the rightmost marble is as small as possible?

8! is small enough number for brute forcing, so we can certainly try all permutations. However, could someone explain to me, preferably in diagrams, how to calculate that distance value given a configuration?

Also, any kind of background information would be appreciated.

答:
If I understand well, the centers of the marbles are on the line.

In that case, we can fix a coordinate system such that the $x$-axis is the line, and the center $C_1$ of the first marble is the origin. Then, its lowest point is $P=(0,-r_1)$. Calculate the coordinates of the centers of the next circles:
$$C_2=(r_1+r_2,0),\ C_3=(r_1+2r_2+r_3,0),\ \dots,\ \\C_n=(r_1+2r_2+\ldots+2r_{n-1}+r_n,\ 0)$$
The lowest point of the last circle is $Q=(r_1+2r_2+..+2r_{n-1}+r_n,-r_n)$. Now we can use the Pythagorean theorem to calculate the distance $PQ$:
$$PQ^2=(r_1+2r_2+..+2r_{n-1}+r_n)^2+(r_1-r_n)^2=
\\=\left(2\sum_{k=1}^n r_k-(r_1+r_n)\right)^2+(r_1-r_n)^2\,.$$
It follows that the distance is independent of the order of the intermediate marbles, only the first and the last one counts. Hopefully from here you can calculate the minimum of this expression.

ContentLicense: CC BY-SA 3.0

Title: Packing circles on a line


问:
Express solutions of equation $ \tan x= x $ in closed form
I know that the equation
$$ \tan(x) = x $$
can be solved using numerical methods, but I’m looking for a closed form of the solutions. In my opinion, having only numerical solutions means that we don’t know the problem, and sooner or later, we’ll be able to find a closed-form solution or at least a power-series solution.

I’m looking for an explicit form of a sequence $ (x_{n})_{n \in \mathbb{Z}} $ in $ \mathbb{R} $ such that

$ 0 < x_{0} < \dfrac{\pi}{2} $,

$ \tan(x_{n}) = x_{n} $ for all $ n \in \mathbb{Z} $, and

$ (2 n - 1) \dfrac{\pi}{2} < x_{n} < (2 n + 1) \dfrac{\pi}{2} $ for all $ n \in \mathbb{Z} \setminus \{ 0 \} $.

The existence of such a sequence follows from the continuity of $ x \mapsto \tan(x) - x $ over its domain. Its uniqueness follows from the monotonicity of $ \tan(x) - x $ over the intervals
$$
\left( (2 n - 1) \frac{\pi}{2},(2 n + 1) \frac{\pi}{2} \right), \quad n \in \mathbb{Z}.
$$

答:
Here is a simple, yet fairly accurate close-form solution to the equation $\tan x=x$

$$x_k = \frac{(1+2k)\pi}2 - \frac2{(1+2k)\pi}$$

with $k=1,2,3\>...$ for all the positive roots. Due to symmetry, the negative roots are $-x_k$.

ContentLicense: CC BY-SA 4.0


问:
$A^3$ is similar to $B^3$ but $A^2$ is not similar to $B^2$
Find an example of $A,B$ such that $A^3$ is similar to $B^3$, and $A^2$ is not similar to $B^2$, and $A$ is not similar to $B$.

I only have some clues that $p_{A^2}(x) \neq p_{B^2}(x)$ and $p_{A}(x) \neq p_{B}(x)$

I first wanted to start from $A^3 = S^{-1} B^3 S$

Then let $S$ and $B$ to be some arbitrary matrices, where $S$ is invertiable, so that I could obtain $A^3$ which is similar to $B^3$. Thus, $A = \sqrt[\leftroot{-3}\uproot{3}3]{A^3}$
However, I realize I don't know how to find square root or cube root of matrices.

答: HINT: Can you think of an example where both cubes are the zero matrix but one square is not?</p>


问:
How to prove this function is increasing $\frac{e^x - x -1}{x^2}$
How to prove that the function $\displaystyle h(x) = \frac{e^x - x - 1}{x^2} = \sum_{k=2}^{\infty} \frac{x^{k-2}}{k!}$ is increasing?

Edit for context: I was asking because this function appears in Wainwright's high dimensional statistics textbook, as part of proving a Bernstein type probability inequality for random variables that are bounded on one side.

答:
The derivative
$$h’(c)=\sum_{k=0}^\infty \frac{k+1}{(k+3)!}c^k$$
is clearly positive for $x>0$.
Combine two Summands
$$\frac{k+1}{(k+3)!}x^k+\frac{k+2}{(k+4)!}x^{k*1}
=\frac}(k+1)(k+4)+(k+2)x}{(k+4)!}x^k
$$where $k$ is even. This is positive at least for $x>-2$. We conclude that $ h$ is increasing on $[-2,\infty)$.

We also  have
$$h‘(x)=\frac{(e^x-1)x^2-(e^x-x-1)2x}{x^4}
=\frac{(x^2-2x)e^x+x^2+2x}{x^4}
=\frac{((x-1)^2-1)e^x+(x+1)^2-1}{x^4}$$
In which form we readily see that $h’$ is positive for $x<-2$.


问:
Existence of subgroup of Abel group
$G$ is a Abel group, the group of order is $n$. If there is an integer $m$ that is a factor of $n$, then there exists a subgroup $H$ of $G$ which order is $m$ .How to prove that?

答:
By the fundamental theorem of finite abelian groups, $G$ is a product of cyclic groups.

But cyclic groups have the property that there is a subgroup of order $d$ for any divisor of the order of the group.

The result now follows easily.

Incidentally, this is summed up by saying that finite abelian groups are "CLT groups".  That's they satisfy the converse to Lagrange's theorem.

The smallest group which isn't CLT is $A_4$ (no subgroup of order $6$).

Title: Existence of subgroup of Abel group


答:
Let $\omega:=e^{\frac{2\pi i}{3}}$.

Then
$$
A=\begin{pmatrix}
\omega & 0\\
0 &\omega
\end{pmatrix}
\text{  and  }
B=\begin{pmatrix}
\omega & 0\\
0 &1\\
\end{pmatrix}
$$
will do the trick.

Title: $A^3$ is similar to $B^3$ but $A^2$ is not similar to $B^2$

Body:
Find an example of $A,B$ such that $A^3$ is similar to $B^3$, and $A^2$ is not similar to $B^2$, and $A$ is not similar to $B$.

I only have some clues that $p_{A^2}(x) \neq p_{B^2}(x)$ and $p_{A}(x) \neq p_{B}(x)$

I first wanted to start from $A^3 = S^{-1} B^3 S$

Then let $S$ and $B$ to be some arbitrary matrices, where $S$ is invertiable, so that I could obtain $A^3$ which is similar to $B^3$. Thus, $A = \sqrt[\leftroot{-3}\uproot{3}3]{A^3}$
However, I realize I don't know how to find square root or cube root of matrices.


问:
$A$ is nilpotent $4\times 4$ matrix and $\ker A^2 \ne \ker A^3$, Prove/Disprove the following
I'm stuck in this question and pretty lost here.

Question : if $A$ is $4\times 4$ nilpotent matrix such that $\ker A^2 \ne \ker A^3$. Prove or disprove the following :

a) Jordan form of $A$ has only one block

b) Jordan form of $A$ has a block with dimension $\ge 3\times3$

I'm not looking for answers, I want a hint/explanation which information $\ker A^2 \ne \ker A^3$ gives us.

by definition it means that exist $ u \in \mathbb{R^4}$ such that $u \in \ker A^3 \wedge u \notin \ker A^2$ (Because $\ker A^2 ⊆ \ker A^3$)

But I still don't know how to proceed... Thanks in advance

Body:
Not sure if I understand it correctly but it can be both.

a) $J$ can have just one block: $$J = \pmatrix{0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0} $$  then if $u=(1,1,1,0)$ we have $A^3u =0$ but $A^2u\ne 0$.

b) $J$ can have block $3\times 3$: $$J = \pmatrix{0&1&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0}   $$
then if $u= (0,0,1,0)$ we have $A^3u=0$ and $A^2u \ne 0$.


问:
How to use the fact that $\mathbb P_{x}$ is the image of $\mu_{0}$ under a translation for canonical Brownian motion.
For the canonical Brownian motion process on $(C(\mathbb R_{+}), \mathcal{B}, \mu_{0},(X_{t})_{t\geq 0})$ I want to show that. Define for $x \in \mathbb R$, the measure $\mathbb P_{x}$ as the image of $\mu_{0}$ under $C(\mathbb R_{+})\to C(\mathbb R_{+}), \; \gamma \mapsto \gamma +x$. Note that $X_{t}(\gamma)= \gamma(t)$ for $\gamma \in C(\mathbb R_{+})$

I want to show that:

$\mathbb P_{x}(X_{t_{0}}\in A_{0},...,X_{t_{n}}\in A_{n})=\delta_{x}(A_{0})\int P_{t_{1}}(x,dx_{1})....P_{t_{n}-t_{n-1}}(x_{n-1},dx_{n})1_{A_{1}}(x_{1})...1_{A_{n}}(x_{n})\;\; (**)$

for $0=t_{0} < ... < t_{n}$ and $A_{0},...,A_{n} \in \mathcal{B}(\mathbb R)$

What has been proven until this point:

$\mu_{0}(X_{t_{0}}\in A_{0},...,X_{t_{n}}\in A_{n})=\mu_{0}(\{ \gamma \in C(\mathbb R_{+}): \gamma(t_{i})\in A_{i}\; \forall i = 0,...,n\})\\=\delta_{0}(A_{0}) \int P_{t_{1}}(x,dx_{1})....P_{t_{n}-t_{n-1}}(x_{n-1},dx_{n})1_{A_{1}}(x_{1})...1_{A_{n}}(x_{n})\;\ \; \; (*)$

I am struggling to use the $\mu_{0}$ under the translation by $x$.

My attempt:

$\mathbb P_{x}(X_{t_{0}}\in A_{0},...,X_{t_{n}}\in A_{n})=\mu_{0}(\{\gamma \in C(\mathbb R_{+}): \gamma(t_{i})\in A_{i}-x\; \;\;\forall i = 0,...,n\})$ and by $(*)$ we obtain:

$\mu_{0}(\{\gamma \in C(\mathbb R_{+}): \gamma(t_{i})\in A_{i}-x\; \;\;\forall i = 0,...,n\})=\delta_{0}(A_{0}-x) \int P_{t_{1}}(x,dx_{1})....P_{t_{n}-t_{n-1}}(x_{n-1},dx_{n})1_{A_{1}-x}(x_{1})...1_{A_{n}-x}(x_{n})\\=\delta_{x}(A_{0}) \int P_{t_{1}}(x,dx_{1})....P_{t_{n}-t_{n-1}}(x_{n-1},dx_{n})1_{A_{1}}(x_{1}+x)...1_{A_{n}}(x_{n}+x)$

I then thought about substituting $y_{n}=x_{n}+x\implies dy_{n}=dx_{n} $ but realized this does not help either. Any ideas or tips where I am going wrong?

$\textbf{Explanation of my problem:}$ I want to be able to justify why I can conclude $(**)$ from $(*)$. It is clear that I have to work with the image of the measure $\mu_{0}$ under a translation, however, I am struggling as you can see in my above attempt.

答:
The substitution you suggested actually helps. After a slight change of notation,
$$
\mu_0(A_0\times A_1\times \cdots\times A_n)=\delta_0(A_0)\int_{A_1} dx_1\cdots\int_{A_n}dx_n\prod_{j=1}^n P_{t_j-t_{j-1}}(x_{j-1},x_j),
$$
where
$$
P_{t}(x,z)=\frac{1}{\sqrt{2\pi t}}\exp\left(\frac{(x-z)^2}{2t}\right).
$$
Using the substitution, $y_j=x_j+x$, and setting $B_j=A_j-x$,
\begin{align}
\mu_0(B_0\times B_1\times \cdots\times B_n)&=\delta_0(B_0)\int_{B_1} dx_1\cdots\int_{B_n}dx_n\prod_{j=1}^n P_{t_j-t_{j-1}}(x_{j-1},x_j) \\
&=\delta_x(A_0)\int_{A_1} dy_1\cdots\int_{A_n}dy_n\prod_{j=1}^n P_{t_j-t_{j-1}}(y_{j-1},y_j).
\end{align}


问:
Help me prove this inequality :
How would I go about proving this?

$$ \displaystyle\sum_{r=1}^{n} \left( 1 + \dfrac{1}{2r} \right)^{2r} \leq n \displaystyle\sum_{r=0}^{n+1} \displaystyle\binom{n+1}{r} \left( \dfrac{1}{n+1} \right)^{r}$$

Thank you! I've tried so many things. I've tried finding a series I could compare one of the series to but nada, I tried to change the LHS to a geometric series but that didn't work out, please could someone give me a little hint? Thank you!

答: Let's divide both parts by $n$. Then LHS becomes an arithmetic average of $\left(1+\frac{1}{2r}\right)^{  2r }$, $r=1..n$. The RHS becomes (binomial coefficients) $\left(1+\frac{1}{n+1}\right)^{ n+1 }$. The function $x\to \left(1+\frac{1}{x}\right)^{ x }$ seems to be concave(need to check it, though), hence the average is not greater than the value in the middle of the segment.</p>

Title: Help me prove this inequality :


问:
Why $y^2\ge x^4$ can be written as $|y|\ge x^2$?
I'm studying the domain of $f(x,y)=\sqrt{y^2-x^4}$

答:
As $y^2 \geq 0$ and $x^{4} \geq 0$, we may write
$$\sqrt{y^2} \geq \sqrt{x^4},$$
which indicates that

$$\lvert y\rvert \geq x^{2}$$


问:
In how many ways can I select 13 cards from a standard deck of 52 cards so that 5 of those cards are of the same suit?
My professor says the solution is $47 \choose 8$, but I don't understand how and why. Could you please explain the solution or say whether it is correct or not?
Thanks in advance!

答: Neither can I. Say $A$ is a set of selection of 13 cards with exactly 5 diamonds and similary for $B,C$ and $D$. Then $$|A|= |B|= |C|= |D| = {13\choose 5}\cdot {39\choose 8}$$  and $$|A\cap B | = |A\cap C|=...= |C\cap D| = {13\choose 5}\cdot {13\choose 5} \cdot  {26\choose 3}$$ Then by PIE we have $$|A\cup B\cup C\cup D| = 4\cdot  {13\choose 5}\cdot {39\choose 8}     - {4\choose 2} \cdot {13\choose 5}\cdot {13\choose 5} \cdot  {26\choose 3}$$


Title: Why $y^2\ge x^4$ can be written as $|y|\ge x^2$?

Body: I'm studying the domain of $f(x,y)=\sqrt{y^2-x^4}$

Body:
Note that
$$
y^2\geq x^4 \iff 0\leq y^2 - x^4=(|y|-x^2)(|y|+x^2)\iff|y|\geq x^2.
$$


答:
This is an answer for an IMV reasonable interpretation.

Let $H$ denote the selections that contain exactly $5$ hearts and let there be similar definitions for $D$, $S$ and $C$.

Then to be found is $|H\cup D\cup S\cup C|$.

Applying inclusion/exclusion and symmetry we find that this equals:$$4|H|-6|H\cap D|=4\binom{13}5\binom{39}{8}-6\binom{13}5^2\binom{26}3$$

This gives you the number of selections characterized by the fact that at least one of the suites is represented by exactly $5$ cards.

Title: In how many ways can I select 13 cards from a standard deck of 52 cards so that 5 of those cards are of the same suit?

Body:
My professor says the solution is $47 \choose 8$, but I don't understand how and why. Could you please explain the solution or say whether it is correct or not?
Thanks in advance!


问:
Subgroups of nth root of unity over an arbitrary field
Let $F$ be an arbitrary field and consider $K_n:=\{f\in F|f^n=1\}$, the nth root of unity in $F$ and I was asked to verify the following property:

Property I wanted to verify: Let $m=|K_n|$ then for every divisor $d|m$ there exists at most one subgroup of $K_n$ which has the cardinality of $d$.

I wanted to verify this because this characterises cyclic property of a finite Abelian group as I described here.

My progress thus far:

I figured there was a way using Theorem 1.1 in KConrad's note. This is because Thm 1.1 shows any finite subgroup of $F^{\times}$ are precisely roots to $X^l-1$ for some integer $l$ then the rest is clear enough.

However if I were to follow this method it would be clear straightaway that $K_n$ is cyclic and thus I thought this way was a bit of a waste and not intended for the original exercise.

Hence my question is if there exists some more straightforward way of showing $K_n$ satisfies the property I described above?

Many thanks in advance!

答: Your question is strange because you mentioned most of the ideas. If a subgroup $H_d\subset F^\times$ has $d$ elements then all its elements are roots of $x^d-1$, there are at most $d$ such roots (here we are using that $F$ is an integral domain) thus $H_d$ is precisely the roots of $x^d-1$. If $H_{p^k}$ exists (it exists in $\overline{F}$ iff $char(F)\ne p$)  then any element of $H_{p^k}-H_{p^{k-1}}$ has order $p^k$ so $H_{p^k}$ is cyclic. If $H_n$ exists then it contains $ \prod_{p^k \| n} H_{p^k}$ which contains an element of order $n$ so $H_n$ is cyclic. Knowing that those groups are cyclic makes the reasoning more intuitive. There is no easier way to solve those questions.


问:
Expectation of submartingale always increasing?
In a proof (upcrossing inequality) our professor used something, which i'm not sure i understand.

So let's say we have a $(\mathcal F_n)$-submartingale $(X_n)_{n\geq0}$ with $X_0 = 0$.

He then said that $E[X_n] \geq 0$ $\forall n \geq 1$, but how can you see this? I know that   $E[X_{n+1} | \mathcal F_n] \geq X_n $, but how can i go from the conditional expectation to the "usual" expected value?

答:
You can use the law of total expectation:

$$\mathbb{E}[X_{n+1}] = \mathbb{E}[\mathbb{E}[X_{n+1} \: | \: \mathcal{F}_n]] \geq \mathbb{E}[X_n]$$


问:
If the trace of the product of two matrices is zero,is the product of the matrices zero?
Does $\:\operatorname{ Tr}(XY) = 0 \Rightarrow XY=O $, for all $X,Y \succeq O$ ?

答:
Hint: If $X = \sum_{i}x_ix_i^*$ and $Y = \sum_{j}y_jy_j^*$, then
$$
XY = \sum_{i,j} (x_i^*y_j)x_iy_j^*, \quad \operatorname{tr}(XY) = \sum_{i,j} |x_i^*y_j|^2.
$$


问:
If $\sum_{n=1}^\infty | x'(x_n) | < \infty$ for all $x' \in X'$ then $\sup_{\|x'\| \leq 1} \sum_{n=1}^\infty | x'(x_n) | < \infty$.
I have this question to solve, but I have no idea how to solve:

a) Let $X$ a normed space and $\{ x_n \}_{n \in \mathbb{N}} \subset X$ a sequence with $\sum_{n=1}^\infty | x'(x_n) | < \infty$ for all $x' \in X'$. Show that $\sup_{\|x'\| \leq 1} \sum_{n=1}^\infty | x'(x_n) | < \infty$.

b) Let $K \subset C([a,b])$ be $K = \{ f \in C([a,b]) : \int_{a}^{(a+b)/2} f(x) \mathrm{d}x = \int_{(a+b)/2}^{a} f(x) \mathrm{d}x \}$. Show that $K$ is a closed vector subspace and find $x' \in C([a,b])'$ so that $\ker (x')=K$.

Thanks for any help :)

答:
(b) To ease discussion, without loss of generality, we assume that
$a=-1$ and $b=1$. Let $K=\{f\in C[-1,1]\mid\int_{-1}^{0}f=\int_{0}^{1}f\}$.
Let $K^{\bot}=\{\mu\in(C[-1,1])^{\ast}\mid\mu(f)=0 \mbox{ for all } f\in K\}.$

We assert that:

(i) $K$ is a a closed subspace of $C[-1,1]$, where $C[-1,1]$ is
equipped with the supremum norm.

(ii) $\mu\in K$ iff $\mu$ is a finite signed measure on $([-1,1],\mathcal{B}([-1,1]))$
and there exists $c\in\mathbb{R}$ such that $\mu(A)=c\cdot\lambda(A\cap[-1,0])-c\cdot\lambda(A\cap[0,1])$,
where $\lambda$ is the usual Lebesgue mearsure, or in short $\frac{d\mu}{d\lambda}(x)=h_{c}(x),$
where $h_{c}=c\cdot1_{[-1,0]}-c\cdot1_{[0,1]}.$

Proof: (i) Clearly $K$ is a vector subspace of $C[-1,1].$ Let $(f_{n})$
be a sequence in $K$ and suppose that $||f_{n}-f||\rightarrow0$
for some $f\in C[-1,1]$. We have that
\begin{eqnarray*}
 &  & \int_{-1}^{0}f\\
 & = & \lim_{n}\int_{-1}^{0}f_{n}\\
 & = & \lim_{n}\int_{0}^{1}f_{n}\\
 & = & \int_{0}^{1}f.
\end{eqnarray*}
Therefore, $f\in K$. Hence, $K$ is closed.

(ii) Define a map $\theta:C[-1,1]\rightarrow\mathbb{R}$ by $\theta(f)=\int_{-1}^{0}f-\int_{0}^{1}f$.
Define $\iota:[-1,1]\rightarrow\mathbb{R}$ by $\iota(x)=x$. Observe
that for each $f\in C[-1,1]$, $f+\theta(f)\iota\in K.$

Let $\mu\in K^{\bot}$. By Riesz Representation Theorem, $\mu$ is
a finite signed measure on $([-1,1],\mathcal{B}([-1,1]))$ and for
any $f\in C[-1,1]$, $\mu\left(f+\theta(f)\iota\right)=0$, i.e.,
$$
\int_{-1}^{1}fd\mu=-\theta(f)\mu(\iota).
$$

Let $a,b$ be arbitrary such that $-1\leq a<b\leq0$. Choose a sufficiently
large $N\in\mathbb{N}$. For $n\geq N$, define
$$
f_{n}=\begin{cases}
0, & \mbox{if }x\in[-1,a]\\
n(x-a) & \mbox{if }x\in(a,a+\frac{1}{n})\\
1, & \mbox{if }x\in[a+\frac{1}{n},b]\\
1-n(x-b), & \mbox{if }x\in(b,b+\frac{1}{n})\\
0, & \mbox{if }x\in[b+\frac{1}{n},1]
\end{cases}.
$$
Clearly $f_{n}\in C[-1,1]$ and $f_{n}\rightarrow1_{(a,b]}$ pointwisely
as $n\rightarrow\infty$. (Note that the graph of $f_{n}$ is a trapezium
with two sharp edges). Hence, $\int_{-1}^{1}f_{n}d\mu=-\theta(f_{n})\mu(\iota).$
Note that $|f_{n}|\leq1$ and $1$ is $\mu$-integrable. By Dominated
Convergence Theorem and letting $n\rightarrow\infty$, we obtain $\mu\left((a,b]\right)=c(b-a)$,
where $c=-\mu(\iota)$. Let $\mathcal{P}=\{(a,b]\mid-1\leq a<b\leq0]\}$
and $\mathcal{L}=\{A\in\mathcal{B}((-1,0])\mid\mu(A)=c\cdot\lambda(A)\}$.
Clearly $\mathcal{P}$ is a $\pi$-class and $\mathcal{L}$ is a $\lambda$-class
and $\mathcal{P}\subseteq\mathcal{L}$. By Dynkin's $\pi$-$\lambda$
theorem, we have $\sigma(\mathcal{P})\subseteq\mathcal{L}$ and hence
$\mathcal{L}=\mathcal{B}((-1,0])$. It can be proved similarly that
$\mu(\{-1\})=0$. Now, it follows that if $A\in\mathcal{B}([-1,0])$,
then $\mu(A)=c\cdot\lambda(A).$ Arguing in the same way, we can also
prove that for any $0\leq a<b\leq1$, $\mu\left([a,b)\right)=-c(b-a)$.
By invoking Dynkin's theorem again and observing that $\mu(\{1\})=0$,
we have: $\mu(A)=-c\cdot\lambda(A)$ for any $A\in B([0,1])$. Finally,
if $A\in B([-1,1])$, we have (observe that $\mu(\{0\})=0$)
\begin{eqnarray*}
\mu(A) & = & \mu\left(A\cap[-1,0]\right)+\mu\left(A\cap[0,1]\right)\\
 & = & c\cdot\lambda\left(A\cap[-1,0]\right)-c\cdot\lambda\left(A\cap[0,1]\right).
\end{eqnarray*}

Conversely, if $\mu$ is a finite signed measure and there exists
$c\in\mathbb{R}$ such that $\mu(A)=c\cdot\lambda\left(A\cap[-1,0]\right)-c\cdot\lambda\left(A\cap[0,1]\right),$
it is trivial to verify that $\mu\in K^{\bot}.$


问:
Solve $S= (a+b)^2 -3ab \geqslant 1$ for $a>0$ and $b>0$
I am trying to solve
$S= (a+b)^2 -3ab \geqslant 1$ for $a>0$ and $b>0$.

After dividing both sides by $ab$, I get $a^2+b^2 \geqslant 2ab$, which means $(a-b)^2  \geqslant 0$ which is always true.

This should also mean that $S=1$ if $a=b$, which, however, does not always hold when I take specific numeric examples. Can anyone tell me why, or where I went wrong?

答:
If $a,b$ are integers, then $$ S = a^2-ab+b^2  = {1\over 2}\Big((a-b)^2+a^2+b^2\Big)\geq {1\over 2}(0+1+1) = 1$$

If they are not, then the statment is not true, take $a=b$ and let $a\to 0$.


问:
All 2 dimensional points with cosine similarity 1
I'm reading this data analytics book, and I came across the following problem:

Consider cosine similarity for two-dimensional non-negative feature data

Find all points that have a cosine similarity of 1 with (1, 1).

The solution is this equation:
$\frac{x + y}{\sqrt{2 \cdot (x^2 + y^2)}} = 1$.

But I don't understand how we got this equation. As far as I know the cosine similarity takes 2 vectors and maps them onto the real numbers. The formula for the cosine similarity is $cos(x,y)= \frac{x \cdot y}{\sqrt{\sum x_i^2 \cdot \sum y_i^2}}$. How do we get $\frac{x + y}{\sqrt{2 \cdot (x^2 + y^2)}} = 1$ out of this? I don't see how that's possible.

答:
If $a$ and $b$ are two vectors, cosine similarity is defined by $\cos(a,b)=\frac{a\cdot b}{||a||\times||b||}$.
The formula you gave
$$
\cos(a,b)= \frac{a \cdot b}{\sqrt{\sum a_i \cdot \sum b_i}}
$$
is missing a power 2. The correct one is
$$
\cos(a,b)= \frac{a \cdot b}{\sqrt{\sum a_i^2 \cdot \sum b_i^2}}
$$
So in the case of two dimmensional vectors with $a=(1,1)$ and $b=(x,y)$ you directly get the answer you are looking for.


问:
Given the convergence of a sequence show the average of the series of that sequence also converges to the same limit
The title is pretty ambiguous but I'll try to clarify it here, given the following sequences:

$$a_n \rightarrow l$$
$$b_n = \frac{\sum_{k=1}^{n}{a_k}}{n}$$

I want to show $b_n \rightarrow l$.

My method (which I think is pretty dodgy) is as follows:

$$|b_n - l| = |\frac{\sum_{k=1}^{n}{a_k}}{n} - l| \le \frac{\sum_{k=1}^{n}{|a_k-l|}}{n}$$

From the convergence of $a_n$ I have that $$\forall \epsilon > 0 ,\exists N_A,\forall n \ge N_A, |a_n-l|<\epsilon$$

Listing the terms of our new series $\ge b_n$:

$$\frac{\sum_{k=1}^{n}{|a_k-l|}}{n} = \frac{|a_1-l|}{n} + \frac{|a_2-l|}{n}+ ...+\frac{|a_{N_A}-l|}{n}+...+\frac{|a_n-l|}{n}$$

Here is where I am struggling, I need to find a value for $n$ such that the first $N_A -1$ terms are sufficiently small, but I can't seem to do so. The terms after $N_A$ are fine as they are less than $\epsilon / n$ from  above.

Am I anywhere on the right tracks? Or is my approach completely incorrect? I understand I could try to show the sequence of partial sums converges as well, would that method yield better results?

答:
Hint:
You're on the right track. Observe that, if you denote $S_{N_A}=\sum_{k=1}^{N_A}|a_k-l|$, this sum is fixed w.r.t. $n$, so we may choose $n$ so that
$$\frac{\sum_{k=1}^{N_A}|a_k-l|}n<N_A\, \varepsilon\iff n>\frac{\sum_{k=1}^{N_A}|a_k-l|}{N_A\,\varepsilon}.$$


问:
Find $n$th number which does not contain the digit $c$
So, this problem is a competitive programming problem, but I find it to be quite mathematical. We write down the numbers which do not contain the digit c. Find an algorithm which finds the nth number written down. I have found a very interesting algorithm, but I do not understand why it works: if $c \ne 0$, we tranform n - 1 (since 0 is part of the series) into base 9. Now we add 1 to every digit bigger or equal than c. And this is the answer, but in base 10. Hopefully, you can see that such a question is quite Maths-related. Could you explain why this algorithm works?

答:
Hint: You described a strictly increasing bijection $f: N^9 \rightarrow N_c$ where $N_9$ are the integers expressed in base 9 and $ N^c $ is the set of integers which do not contain the digit $c$.

Corollary: Accounting for $0$, the $n$th non-negative integer that doesn't contain a $c$ is $f(n-1_9)$.


问:
Show that $cd[(c+d)^2-3] > 0$ where $c > 0$ and $d > 0$
Show that $cd[(c+d)^2-3] > 0$ where $c > 0$ and $d > 0$

Can anyone help me with how to proceed?

答:
As others have pointed out it's false if $c,d \in \mathbb R^+$.

If $c,d \in \mathbb N$ then $cd[(c+d)^2-3] \ge cd[(1+1)^2-3]=cd >0 $.


问:
Show that there doesn't exist $f \in X$ with $\|f\|_{\infty} = 1$ such that $d \left (f,Y \right ) \geq 1.$

Let $X = \left \{f \in C[0,1]\ \big |\ f(0) = 0 \right \}$ and $Y = \left \{f \in C[0,1]\ \bigg |\ \displaystyle {\int_{0}^{1} f(t)\ dt = 0} \right \}.$ Show that there does not exist any $f \in X$ with $\|f\|_{\infty} = 1$ such that $d \left (f,Y \right ) \geq 1.$

If for any $f \in X$ with $\|f\|_{\infty} = 1$ we can able to find $g_f \in Y$ such that $\left \|f - g_f \right \|_{\infty} \lt 1$ then we are through. But for a given $f \in X$ how to find such a $g_f\ $? Any help in this regard will be appreciated.

Thanks for your time.

答:
Let $g:X \to Y$ be the mapping $g(f) = g_f := f - \int_0^1 f$.  Then

$$d(f,Y) \le d(f,g_f) = ||f - g_f||_\infty = \left|\int_0^1 f \right| \le \int_0^1 ||f||_\infty = 1.$$

To see the strict inequality, we observe that the second '≤' above is actually a strict inequality, since $f$ is a continuous function defined on $[0,1]$ starting at the origin.


问:
$\{\overline{0}\}^{\tau_{1}}$ and $\{\overline{0}\}^{\tau_{2}}$, $\{\overline{n}\}^{\tau_{1}}$ and $\{\overline{n}\}^{\tau_{2}}$
Be $\tau_{1}$ and $\tau_{2}$ topologies over $\mathbb{N}$ defined by:

$\tau_{1}=\{\{m\in \mathbb{N}:m<n\}:n\in \mathbb{N}\}\cup \{\mathbb{N}\}\}$ and $\tau_{2}=\{A\subseteq \mathbb{N}: 0\in A\}\cup \{\emptyset\}$.

Definition. The closing $\overline{A}$ is defined by the intersection of all closed ones contained in $A$. In other words the $\overline{A}$ is smaller set contained in $A$.

Would you like to know who these sets are? Is it correct what I did?
\begin{equation}
\{\overline{0}\}^{\tau_{1}},\{\overline{0}\}^{\tau_{2}}, \{\overline{n}\}^{\tau_{1}},                    \{\overline{n}\}^{\tau_{2}}
\end{equation}
com $n\in \mathbb{N}-\{0\}$.

Before, who are the closed in the $\tau_{1}$ topology and $\tau_{2}$?.

closed $\tau_1=\{\emptyset, \mathbb{N}, \mathbb{N}-\{0\}, \mathbb{N}-\{0,1\}, \mathbb{N}-\{0,1,2\}, \cdots\}$

closed $\tau_2=\{\emptyset, \mathbb{N},\mathbb{N}-\{0\}, \mathbb{N}-\{0,1,18,78\}, \cdots\}$

Can someone help me? Thank you.

答:
Every (non-empty) open set in both $\tau_1$ and $\tau_2$ contains $0$ so the closure of $\{0\}$ in both is $\Bbb N$.

If $n \neq 0$, in $\tau_1$ the closure is $\{m\mid m \ge n\}$ (which is closed and all whose members have minimal neighbourhoods that contain $n$). $\{n\}$ is closed in $\tau_2$ (as its complement contains $0$ and so is open) and so equals its own closure.


问:
Finding eigenvalues and eigenfunctions of differential equation
I need to write the following d.e. in Sturm-Liouville form and find the eigenvalues and eigenfunctions.

$$\frac {\mathrm d ^2 y}{\mathrm d x^2} + 7 \frac {\mathrm d y} {\mathrm d x} + (e^{3x} + \lambda)y = 0$$

where $y(0)=y(1)=0$.

I have written it as $-(p(x)y')' + q(x)y = \lambda r(x) y$ with $p(x) = e^{7x}$ , $q(x) = -e^{10x}$ and $r(x) = e^{7x}$.

I am not sure how to find the eigenvalues though. I know that the boundary conditions are relevant but I'm not sure where to start in terms of finding the eigenfunctions. How do I know by looking at the d.e. when it is possible to find them?

答:
First let's get rid of the first derivative term. Let $y(x)=e^{-7x/2}g(x)$. Then
$$\frac{dy}{dx}=e^{-7x/2}\frac{dg}{dx}-\frac72e^{-7x/2}g$$
$$\frac{d^2y}{dx^2}=e^{-7x/2}\frac{d^2g}{dx^2}-7e^{-7x/2}\frac{dg}{dx}+\frac{49}4e^{-7x/2}g$$
So
$$\frac{d^2g}{dx^2}+\left(e^{3x}+\lambda-\frac{49}4\right)g=0$$
Now we can wipe out that butt-ugly $e^{3x}$ business. Let $u=e^{3x/2}$. Then
$$\frac{dg}{dx}=\frac{dg}{du}\frac{du}{dx}=\frac32u\frac{dg}{du}$$
$$\frac{d^2g}{du^2}=\frac32u\left(\frac32\frac{dg}{du}+\frac32u\frac{d^2g}{du^2}\right)$$
So
$$u^2\frac{d^2g}{du^2}+u\frac{dg}{du}+\left(\frac49u^2+\frac49\lambda-\frac{49}9\right)g=0$$
This is Bessel's differential equation with general solution
$$g(u)=C_1J_{\sqrt{49-4\lambda}/3}\left(\frac23u\right)+C_2Y_{\sqrt{49-\lambda}/3}\left(\frac23u\right)$$
That brings us back to
$$y(x)=e^{-7x/2}\left(C_1J_{\sqrt{49-4\lambda}/3}\left(\frac23e^{3x/2}\right)+C_2Y_{\sqrt{49-\lambda}/3}\left(\frac23e^{3x/2}\right)\right)$$
Now you have to find $\lambda$ such that $y(0)=y(1)=0$ for $C_1$ and $C_2$ not both zero.


问:
What is derivative of Matrix with respect to a Matrix?
I want to calculate the derivative of dot product of two Matrices of not the same order.

$X = \begin{bmatrix}x_{11} & x_{12} & x_{13}\\x_{21} & x_{22} & x_{23}\\x_{31} & x_{32} & x_{32}\end{bmatrix}$

$y= \begin{bmatrix}y_{11} & y_{12}\\y_{21} & y_{22}\\y_{31} & y_{32}\end{bmatrix}$

Problem is I can't figure out What does it mean to derivative of matrix with respect of matrix individual elements.
I tried to use the sum notation to calculate derivative of a single element of the resultant matrix.

$c_{i,j} = \sum_{k=1}^na_{i,k}\cdot b_{k,j}$

$\frac{\partial (X y)_{11}}{\partial X} =
\begin{bmatrix}y_{11} & y_{12} & y_{21}\\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$

and the other partial Derivatives are similar to this.
I want to know that what is

$\frac{\partial Xy}{\partial X} = ?$

I can't figure out how to get this when the element derivative itself is a matrix.
and the matrices as inputs are not even of the same order.

答: Since $$Xy = \mathrm{vec}(Xy) = \mathrm{vec}(IXy) = (y\otimes I)'\mathrm{vec}(X)$$ take the derivative wrt $\mathrm{vec}(X)$ to obtain $y\otimes I$. This is consistent with the comment of Ben Grossmann as it is the "vectorization" of said fourth order tensor.


问:
$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$

Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)

My approach:

Solving the quadratic, I get:

$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$

Case 1: If the $b^2 - 4ac$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.

Case 2: If $b^2 - 4ac$ is not a perfect square, $b =  \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$

Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)

So the equation is not quadratic and $c = 0$.

$a + bp^\frac{1}{3} = 0$

$-\dfrac{a}{b} = p^\frac{1}{3}$

This is a contradiction and hence $b = 0$ and $a = 0$

Is there any other way to solve this?

答:
Here's another way. If $a+bq+cq^2=0$, then $a=-(bq+cq^2)$, hence $a^2=b^2q^2+2bcq^3+c^2q^4$. So if $q^3=p$, then we have two equations:

$$\begin{align}
a+bq+cq^2&=0\\
(a^2-2bcp)-c^2pq-b^2q^2&=0
\end{align}$$

Multiplying both sides of the first equation by $b^2$ and the both side of the second equation by $cp$, and then adding the resulting equations, we have

$$(ab^2+a^2c-2bc^2p)+(b^3-c^3p)q=0$$

Now if $q$ is irrational (i.e., if $p$ is not a perfect cube) and the other variables are rational, then we must have $ab^2+a^2c-2bc^2p=b^3-c^3p=0$. But $b^3-c^3p=0$ for a non-cube $p$ implies $b=c=0$, in which case the original equation, $a+bq+cq^2=0$, implies $a=0$.


问:
Integral of a vector field
I'm trying to evaluate the following integral:

$ \int_C(y+\sin x) dx +(z^2+\cos y)dy+(x^3)dz$

Where $C$ is the curve: $c(t) = (\sin t, \cos t, \sin 2t) $. Note that
  $C$ lies on the surface $z= 2xy$.

Question:
Could I just use the parameterization of the surface S $\phi(u,v)=(u, v,2uv) $ with $-1\leq  u\leq 1 $ and $-1\leq v \leq 1$ and use Stokes' Theorem to evaluate $ \int_S \nabla \times  F\cdot dS$ which should be equal to the required integral?

My only trepidation about the above method is that $c(t)$ may not necessarily be on the boundary of $S$, which would be required to use Stokes' Theorem.

答:
The answer is Yes.

Your parametrization is correct.

Now you need to compute the surface normal
$$
\nu = \frac{\partial \phi}{\partial u}\times \frac{\partial \phi}{\partial v}.
$$
Applying the Stokes theorem then the surface integral becomes:
$$
\oint_C F\cdot dr =  \iint_S \nabla \times F\cdot dS = \iint_D \nabla \times F\big(x(u,v),y(u,v),z(u,v)\big) \cdot \nu/\|\nu\| \,du dv,
$$
where $D = \{u^2 + v^2 \leq 1\}$.

$C$ which is giving by:
$$
\begin{aligned}
&x = \sin t
\\
&y = \cos t
\\
&z = \sin (2t)
\end{aligned}
$$ assuming $t\in [0,2\pi)$, $C$ lies on $z = 2xy$ simply because the parametrization satisfies the equation of the surface:
$$
z = \sin(2 t) = 2\sin t\cos t = 2xy,
$$
therefore, all the points on $C$, must be on $S$.

The boundary of $S$ is artificial chosen by cut the unbounded surface $z = 2xy$ using the cylinder $x^2+y^2 = 1$ (notice the parametrization of the curve also satisfies this), the intersection curve of $z = 2xy$ and $x^2+y^2 = 1$ is $C$.

Title: Integral of a vector field


问:
Chain rule with constraints
I am self-learning about the partial derivative with constraint in the online course Multivariable Calculus of MIT OpenCourseware and I'm struggling at understanding the exercise below (as shown by the image).

My question is that:  In this problem,   $f(x,y,z)$ and  $g(x,y)$ is basically the same thing.

But why the gradient of  $ f(x,y,z) $ has 3 components when the gradient vector of $ g(x,y)$ only has 2 components ?

Moreover, why the gradient of $ f(x,y,z)$ has 3 components when it is radically a function of $(x,y)$ ?

Thank you very much for your help!!!

答:
$g$ and $f$ are not the same. And the way how the gradient of $g$ is found is not an explicit application of the chain rule as one could expect from the title of the chapter.

So, I give here the solution to the problem by a direct application of the chain rule.

The situation is as follows

$$\mathbb{R}^2\stackrel{h}{\longrightarrow}\mathbb{R}^3\stackrel{f}{\longrightarrow}\mathbb{R}$$
with

$$h(x,y)=\begin{pmatrix}x\\y\\z(x,y)\end{pmatrix}=\begin{pmatrix}x\\y\\y^2-x^3\end{pmatrix} \text{ and } g=f\circ h$$

Now, the chain rule gives

$$\nabla g(1,2) = \nabla f(1,2,z(1,2))D_h(1,2)$$ where
$$D_h(x,y) =\begin{pmatrix}1 & 0 \\ 0 & 1 \\\partial_x z(x,y) & \partial_y z(x,y) \end{pmatrix}=\begin{pmatrix}1 & 0 \\ 0 & 1 \\-3x^2 & 2y \end{pmatrix}$$

Hence,

$$\nabla g(1,2) = \begin{pmatrix}1 & -2 & 3\end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & 1 \\-3 & 4 \end{pmatrix} = \begin{pmatrix}-8 & 10\end{pmatrix}$$

Title: Chain rule with constraints


问:
$m^*(E_j) = m^*(F_j) \implies m^*(\bigcup_{j=1}^\infty E_j) = m^*(\bigcup_{j=1}^\infty F_j)$
Define $m^*(E) = \inf \{m(U): U \text{ open and }E \subseteq U\},$ where $m$ is the Lebesgue measure on $\mathbb R^n$.

Suppose $\{E_j\},\{F_j\} \subseteq \mathbb R^n$ are increasing sequences of sets satisfying $F_j$ is a $G_\delta$ set, $E_j \subseteq F_j$ and $m^*(E_j) = m(F_j)$ for all $j \in \mathbb N$. Here, a $G_\delta$ set is defined to be a countable intersection of open sets. I know that
$$m^*(\bigcup_{j=1}^\infty  E_j) \leq m^*(\bigcup_{j=1}^\infty  F_j)$$ by the monotonicity of $m^*$.
Is it also true that $$m^*(\bigcup_{j=1}^\infty  E_j) \geq m^*(\bigcup_{j=1}^\infty  F_j)?$$
I think this is true but am unsure of a proof.

答:
Since each $F_j$ is measurable,
$$
m^*\!\left(\bigcup_{j\ge 1}E_j \right)\ge \limsup_{k\to\infty}m^*(E_k)=\limsup_{k\to\infty}m(F_k)=m\!\left(\bigcup_{j\ge 1}F_j \right).
$$


问:
If $g(s) = \int_0^{\infty}\left( f(t)-1\right)t^{s-1} dt$ and $f(t) = \frac{1}{t}f(\frac{1}{t})$ , then $g(s)=g(1-s)$ for all $s$
Let f be a very nice positive real valued-function and $g(s) =  \int_0^{\infty}\left( f(t)-1\right)t^{s-1} dt$ and $f(t) = \frac{1}{t}f(\frac{1}{t})  \hspace{7mm} (\text{  for all } t>0 )$

Then, show that $g(s)=g(1-s)$ for all $s$.

(* The 'very nice' means that we don't need to consider integrable, differentiable and  $ \int_0^{\infty}$ is also well-defined, f has a rapidly decaying at $\infty$ , etc...)

[My attempt]

First, I make the change of variable, $t \mapsto \frac{1}{x}$ , $dt= -\frac{1}{x^2}dx$.

$$g(s) =  \int_0^{\infty}\left( f(t)-1\right)t^{s-1} dt = \int_0^{\infty}\left( f(\frac{1}{x})-1\right)x^{1-s} x^{-2}dx = \int_0^{\infty}\left( f(\frac{1}{x})-1\right)x^{-1-s}dx$$
$$ = \int_0^{\infty}\left( xf(x)-1\right)x^{-1-s}dx = \int_0^{\infty}\left( f(x)-x^{-1}\right)x^{-s}dx  $$

However, $$ g(1-s) =  \int_0^{\infty}\left( f(x)-1\right)x^{-s} dx$$

I'm stuck here. How to solve this ?

Thank you for your attention.

答:
As pointed out in the comments: This is false, to see why:
$$g(s) =  \int_0^{\infty}\left( f(t)-1\right)t^{s-1} dt = \int_0^{1}\left( f(t)-1\right)t^{s-1} dt  +\int_1^{\infty}\left( f(t)-1\right)t^{s-1}dt .$$

Make the subsitutiong $t\mapsto 1/t$ in the second intgeral.

Use the functional equation of $f.$

Repeat for $g(1-s).$

Compute $g(1-s) - g(s).$


问:
Periodic functions counter example

Okay I was given this problem. I concluded that it's false, since if we move the sin(x) to the left side of the equation, we will have a periodic expression that and no matter what the function f does it will remain periodic..
Am I missing something? and how would you prove it?

答:
That is completely right. $f(\cos x) - \sin(x)$ is a $2\pi$-periodic function (no matter what $f$ is), and $\operatorname{sign}(x)$ is not periodic.

It would also suffice to set $x= -\pi$ and $x= \pi$ to obtain a contradiction:
$$
f(\cos(-\pi)) = \sin(-\pi) + \operatorname{sign}(-\pi) \implies f(-1) = 0 + (-1) \\
f(\cos(\pi)) = \sin(\pi) + \operatorname{sign}(\pi) \implies  f(-1) = 0 + 1 \\
$$

Title: Periodic functions counter example


问:
about graph of zero set of $f(x, y, z)=\left(x^{2}+y^{2}+z^{2}+R^{2}-r^{2}\right)^2-4 R^{2}\left(x^{2}+y^{2}\right) $
Let $f:\mathbb{R^3} \to \mathbb{R}$
$$f(x, y, z)=\left(x^{2}+y^{2}+z^{2}+R^{2}-r^{2}\right)^2-4 R^{2}\left(x^{2}+y^{2}\right)
$$
and $R,r \in \mathbb{R} , 0<r<R$

now define :

$Z=\{ (x,y,z) \in \mathbb{R^3}  | f(x,y,z)=0\}$

what is shape of $Z$ in $\mathbb{R^3}$ ? how we can draw $Z$ in $\mathbb{R^3}$  ?

first of all we must compute $Z$ :

$\left(x^{2}+y^{2}+z^{2}+R^{2}-r^{2}\right)^2-4 R^{2}\left(x^{2}+y^{2}\right)=0\Rightarrow $

$z^2=\pm2R\sqrt{x^2+y^2}-(\sqrt{x^2+y^2})^2-(R^{2}-r^{2})$

note that in the other hand we have $A \subset Z$ such that ,

$A=\{ (x,y,z) \in \mathbb{R^3}  | x^2+y^2=R^2,z^2=r^2\}$

so $Z$ must  include  two circles with radios $R$ and $z=\pm r$. now for example how we can draw $Z$ for $R=2, r=1$ or what is graph of following equation ? is this torus ?

$$z^2=\pm2  \sqrt{x^2+y^2}-(\sqrt{x^2+y^2})^2-(2^{2}-1^{2})$$

答:
It is a torus

Section with plane $y=0$ are the two circles
$$x^2+z^2-2 R x=R^2-r^2 ;\; x^2+z^2+2 R x=R^2-r^2$$
the rotation of $2\pi$ of one of the two around $z$-axis gives the surface.

$$...$$


问:
Is the value of $\pi$ in 2d the same in 3d?
I am starting with my question with the note "Assume no math skills". Given that, all down votes are welcomed. (At the expense of better understanding of course!)

Given my first question: What is meant by the perimeter of a Sector

Why is the value of $\pi$ not exactly $3$? why is it $3.14$.......... or a fraction $\frac{22}{7}$?

Is the value of $\pi$ of $3.14$... or $\frac{22}{7}$ the same as for $3$ dimensions?

答:
There are several ways to define $\pi$, but whichever you choose it is a number that happens to be irrational (it's not equal to any fraction, although some fractions are close).  It does represent the ratio between a circle's circumference and diameter for any  circle (but not for spheres, squares, or other shapes).

With regards to your first question, a sector is a part of a circle, like a slice of pie (the food).  Its perimeter consists of a round bit on the outside, and two straight bits toward the center.


问:
Show that the inverse of a triangular matrix is triangular
Consider the system of $d$ equations contained in the rows of $R\bar{x} = \bar{e_k}$ for the $d × d$ upper-triangular matrix $R$, where $\bar{e_k}$ is a $d$-dimensional column vector with a single value of $1$ in the $kth$ entry and $0$ in all other entries. Discuss why solving for $\bar{x} = [x_1 \cdots x_d]^T$ is simple in this case by solving for the variables in the order $x_d, x_{d−1}, \dots, x_1$. Furthermore, discuss why the solution for $R\bar{x} = \bar{e_k}$ must satisfy $x_i = 0$ for $i > k$. Why is the solution $x$ equal to the $kth$ column of the inverse of $R$? Discuss why the inverse of $R$ is also upper-triangular.

I am confused by the first part of the question and don't know where to start. For the last two questions, my answer for Why is the solution $x$ equal to the $kth$ column of the inverse of $R$? is that solving the equation $\bar{x}=R^{-1}\bar{e_k}$, all the columns of $R^{-1}$ would be multiplied by the scalars $ith$ entries $0$ of $\bar{e_k}$ except for the $kth$ entry of $\bar{e_k}$ where it is equal to $1$. For the last part Discuss why the inverse of R is also upper-triangular. The polynomials and inverses of triangular matrices are also triangular matrices of the same type.

答: It is all about definitions. Any matrix is equivalent to its reduced row echelon form. In order to guarantee that $R$ is invertible, none of its diagonal entries can be zero (See Exercise 1.6.9 in Hoffman and Kunze). Now performing elementary row operations (Only operation types 1 and 2. In other words, only scaling and shearing, no switching is needed) on $R$ will convert $R$ to $I$. Since each elementary row operation works on the upper part of $R$, it is equivalent to multiply an upper elementary matrix on the left side of $R$. Hence, $E_s \cdots E_1 R=I$, which implies $R^{-1}=E_s\cdots E_1$. The product of upper-triangular matrices is still upper-triangular.


问:
Rotating (circular shifting) a binary number - what is the rotational distance between the minimum and maximum?
I started continuing thoughts from What exactly happens, when I do a bit rotation?.
Let us take a binary number $B$ of length $L$ with $N$ ones, for example $L=8$, $N=5$ and $B=10110101=181$, the minimum (obtainable by rotating $B$) is $B_{min}=01011011=91$ and the maximum is $B_{max}=11011010=218$. The (left-)rotational distance $d(B_{min},L)=r$ is the required amount of left rotates from the minimum to the maximum. In this example $d(B_{min},L)=r=3$, since we obtain the maximum $11011010$ by three left rotates ($r=3$) of the minimum $01011011$.

What is the general formula of calculating the rotational distance $d(B_{min},L)$?

What we know so far is that $B_{max}=(B_{min}\cdot2^r)\bmod{(2^L-1)}$ which means in our example $218=(91\cdot2^3)\bmod(255)$ and thus $r$ might be obtained by the discrete logarithm, which is not so manageable.

Hopefully there is way to calculate $r$ explicitely (avoiding trial-&-error).

Maybe we can show that $L\mid N\cdot r+1$

答:
In order that the divisibility $L\mid N\cdot r+1$ is given, we must show that:

$\gcd(B_{min},2^L-1)\mid B_{max}$

$2^L-1\mid2^rB_{min}-B_{max}$

which leads us to the diophantine equation that must have an integer solution $a,b$:

$(2^L-1)\cdot b=2^{\frac{a\cdot L-1}{N}}\cdot B_{min}-B_{max}$

In our example $a=b=2$ provide a solution: $(2^8-1)\cdot b=2^{\frac{a\cdot 8-1}{5}}\cdot91-218$.


问:
Converse of density theorem for $L^1(\Omega)$
We know that if A is dense set in $L^1(\Omega)$. Then if $$\int_{\Omega}fg=0$$
for all $g\in A$ then it is easy to say $f=0$ a.e.

But I am interested to check whether converse also true or false.

A be any subset of $L^1(\Omega )$ with following property:

If $f\in L^{\infty }(\Omega )$ such that $\int fg=0$ for any $g\in A$ then $f=0$.

Then question is that can we prove span of  $A$ is dense in $L^1(\Omega)$?

Any Help or hint will be appreciated.

Edit: As s. harp suggested $A$ need not be dense. But span of $A$ is dense .

答:
The closest thing you get to your result as suggested in the comment is the following:

Definition: A subset $A$ in a normed space is called total if the smallest subspace containing $A$ is dense in X. You can take $X = L^{1} (\Omega)$ so that $X^{*} = L^{\infty}(\Omega)$.

Theorem: Prove that $A$ is total if and only if for $f \in X^{*}$ and $\langle f, a \rangle = f(a) = 0$ for each $a \in A$ implies $f =0$.

Proof: Let $M = \overline{ \text{Span} (A)}$. Then $M$ is a closed subspace of a normed linear space $X$. If $A$ is not total, then $M \neq X$. Hence, there exists $x_{0} $ not in $M$. Then by Hahn-Banach Theorem (one of its corollaries), there exists a functional $f \in X^{*}$ such that $f(M) =0$ and $f(x_{0}) = 1 \neq 0$. This proves that if for $f \in X^{*}$ and $f(a) = 0$ for each $a \in A$ implies $f =0$ implies that $A$ must be total.

Now suppose that $A$ is total. Then $X = \overline{ \text{Span} (A)}$. It follows if $x \in X$ then either $x$ is in $A$ or there exists a sequence such that $x_{n} \in A$ such that $x_{n} \to x$ and then by continuity of $f$ together with the fact that $A$ is total imply that if for $f \in X^{*}$ and $\langle f, a \rangle = \int_{\Omega} fa = f(a) = 0$ for each $a \in A$ implies $f =0$.


问:
Does having multiple limit values at a point imply essential discontinuity?
In Complex Analysis, do "jump discontinuities" exist?

If I find that a function of $z$ approaches two different values as z is approached from two different directions, can I immediately conclude that the point is an essential singularity? Or do other possibilities exist?

答: Assuming the singularity is isolated, yes. However, take $f(z)=\log z$ (principal branch) and consider what happens when $z \to -1$ along paths in the upper and lower halfplanes respectively.


问:
Determinant of matrix in index notation
The determinant of the $3\times 3$ square matrix $A=[a_{ij}]$   in index form  is given by
$$\mathrm{det}(A)=\epsilon_{ijk}a_{1i}a_{2j}a_{3k}$$

Wikipedia suggests that I can write it as
$$\mathrm{det}(A)=\frac{1}{3!}\epsilon_{ijk}\epsilon_{pqr}a_{ip}a_{jq}a_{kr}$$

using two epsilon symbols. But I don't understand How this supposed to happen. Can someone help me with this?

答: This just averages $3!$ copies of the first formula. For example,$$\epsilon_{ijk}a_{1i}a_{2j}a_{3k}=\epsilon_{ijk}a_{2j}a_{1i}a_{3k}=\epsilon_{jik}a_{2i}a_{1j}a_{3k}=\sigma_{213}\epsilon_{ijk}a_{2i}a_{1j}a_{3k}.$$where $\sigma_{213}=-1=-(-1)^p,\,2\nmid p$ is a symbol I've invented to represent a coefficient caused by $213$ being an odd permutation. (The second $=$ above results from relabelling the indices.) More generally, if a permutation $\sigma$ introduces coefficient $\epsilon_\sigma$,$$\epsilon_{ijk}a_{1i}a_{2j}a_{3k}=\epsilon_\sigma\prod_{l=1}^3a_{(ijk)_l\sigma_l}.$$In fact, $\sigma_{213}$ is just $\epsilon_{213}$, but if I'd called the coefficient $\epsilon_{213}$ up front that would have been more confusing, so I've had to delay the usual notation $\epsilon_\sigma$. We can rewrite my last equation as$$\epsilon_{ijk}a_{1i}a_{2j}a_{3k}=\epsilon_{ijk}\epsilon_{pqr}a_{ip}a_{jq}a_{kr},$$provided this is interpreted as using one choice of $p,\,q,\,r$ rather than summing over them. Since that's not the usual reading of this notation, when we do use Einstein summation we instead have$$\epsilon_{ijk}a_{1i}a_{2j}a_{3k}=\frac{1}{3!}\epsilon_{ijk}\epsilon_{pqr}a_{ip}a_{jq}a_{kr}.$$Incidentally, this points us to another identity, $\epsilon_{ijk}\epsilon_{ijk}=3!$. This is obvious because there are $3!$ nonzero terms, each $(\pm1)^2=1$.

Title: Determinant of matrix in index notation


问:
What is the essence of ``the naturality of the cap product''?
Associated to a continuous map $f : X → Y$, there are natural pushforward and pullback maps on homology and cohomology, respectively, denoted $f_∗ : H_∗(X) → H_∗(Y)$ and $f^*: H^* (Y) → H^* (X)$.
These are related by the projection formula, also called ``the naturality of the cap product'':
$$f_∗(f^∗ c \cap σ) = c \cap f_∗ σ.$$

My questions are that:

what is the essence of this ``the naturality of the cap product''?

what are the uses of this ``the naturality of the cap product''?

Suppose it can be used in the proof of Alexander-Lefschetz-Poincaré Duality. What are the essence behind?

答:
One way to think about the cap product, which is a pairing
$$ \cap\colon H_{k+l}(X)\otimes H^k(X) \rightarrow H_l(X) $$
is via its adjoint which is a homomorphism
$$ \alpha_X \colon H_{k+l}(X)\rightarrow \hom(H^k(X),H_l(X)). $$
Notice that both sides of the equation are covariant functors in $X$ and naturality of the cup product precisely means that $\alpha_X$ is a natural transformation.

A very important application of this naturality statement is the following: Let
$$f\colon M \rightarrow N$$ be a continous map of degree $d$ between closed, connected and oriented manifolds of dimension $n$. The Poincaré duality isomorphism is defined as the "capping with the fundamental class of a manifold":
$$ PD_M = \_\cap [M]\colon H^k(M) \rightarrow H_{n-k}(M).$$
Let us apply naturality of the cap product to our map $f$.
For $\phi\in H^k(N)$:
$$f_*(f^*(\phi)\cap [M]) = \phi \cap d\cdot [N] =d\cdot (\phi\cap [N]). $$
If $f^!\colon H_k(N)\rightarrow H_k(M)$ denotes the "Umkehr map" $PD_M\circ f^* \circ PD^{-1}_N$, the the above shows that $f_* \circ f^! \colon H_k(N)\rightarrow H_k(N)$ is multiplication by $d$, which can be useful. In particular, if $d=1$, then $f_*$ is split surjective. Also, using the Theorem of Whitehead, a map of degree $1$ from a sphere to a closed and simply connected manifold is a homotopy equivalence.


问:
Find all of the Eigenvalues of T
Let there be $V$, vector space from a finite dimenstion contains all real numbers $\mathbb{R}$

$T:V$$\rightarrow $ $V$ linear transformation such that:

$T^2$ = $T$

$Im(T)$ $\neq$ $V$

$Im(T)$ $\neq$ {$\vec{0}$}

Find all of the Eigenvalues of $T$

答: For each $v\in V$, $v=\bigl(v-T(v)\bigr)+T(v)$. Now, not that$$T\bigl(v-T(v)\bigr)=T(v)-T^2(v)=0$$and that therefore $v-T(v)$ belongs to the eigenspace associated with the eigenvalue $0$. Besides,$$T\bigl(T(v)\bigr)=T(v),$$and so $T(v)$ belongs to the eigenspace associated with the eigenvalue $1$. So, $V$ is the direct sum of the eigenspace associated with the eigenvalue $0$ with the eigenspace associated with the eigenvalue $1$, and therefore the only possible eigenvalues are $0$ and $1$. But if $0$ was not an eigenvalue, then $T$ would be the identity map, in which case $\operatorname{Im}(T)=V$. And if $1$ was not an eigenvalue, then $T$ would be the null map, in which case $\operatorname{Im}(T)=\left\{\vec0\right\}$. So, the set of all eigenvalues is $\{0,1\}$.

Title: Find all of the Eigenvalues of T


问:
What is a practicized (or suitable) label for morphisms that satisfy the condition mentioned in the body of this question?
Let it be that $\mathfrak A$ and $\mathfrak B$ denote $\mathcal L$-models for some language $\mathcal L$ and that $A$ and $B$ denote their domains/universes.

A function $h:A\to B$ can be recognized as a morphism $\mathfrak A\to\mathfrak B$ if it satisfies the conditions:

$h(c^{\mathfrak A})=c^{\mathfrak B}$ for constants symbols.

$h\circ f^{\mathfrak A}=f^{\mathfrak B}\circ h^n$ for function symbols $f$.

$(a_1,\dots,a_n)\in r^{\mathfrak A}\implies (h(a_1),\dots,h(a_n))\in r^{\mathfrak B}$ for relation symbols $r$.

In both cases $n$ denotes the arity of the symbol.

If moreover $h$ is injective and satisfies the stronger condition:

$$(a_1,\dots,a_n)\in r^{\mathfrak A}\iff (h(a_1),\dots,h(a_n))\in r^{\mathfrak B}\text{ for relation symbols }r\tag1$$then $h$ is an embedding.

My question:

Is there some suitable terminology for a morphism $h$ that satisfies $(1)$?

I would like to say things as: "a morphism $h$ is an embedding if it is injective and ...."

Edit:

The answer of Alex is fine and educating. I have found a more direct answer to my question later in some script. Homomorphisms that satisfy $(1)$ were called strong homomorphisms there. So a homomorphism is an embedding iff it is strong and injective. Also I was told there that strong homomorphisms correspond one-to-one with congruences (hence quotients).

答:
The following terminology is commonly used. For an $L$-formula $\varphi(x_1,\dots,x_n)$, $L$-structures $A$ and $B$, and a function $h\colon A\to B$:

$h$ preserves $\varphi$ means that for all $a_1,\dots,a_n\in A$, if $A\models \varphi(a_1,\dots,a_n)$, then $B\models \varphi(h(a_1),\dots,h(a_n))$.

$h$ reflects $\varphi$ means that for all $a_1,\dots,a_n\in A$, if $B\models \varphi(h(a_1),\dots,h(a_n))$, then $A\models \varphi(a_1,\dots,a_n)$.

This terminology can also be applied to relation symbols, where we identify an $n$-ary relation symbol $R$ with the atomic formula $R(x_1,\dots,x_n)$.

You can also apply it to function symbols (and constant symbols, which are are just $0$-ary function symbols), by identifying an $n$-ary function symbol $f$ with the atomic formula $f(x_1,\dots,x_n) = y$. But since function symbols have a different character than relation symbols, I usually prefer to use the equivalent definition suggested in your question: $h$ preserves (or respects) $f$ if $h\circ f^A = f^B\circ h^n$.

Using this terminology, a homomorphism is a function which preserves the symbols in the language. An embedding is a homomorphism which is injective and reflects the relation symbols in the language.

A more elegant characterization is that a homomorphism is a function which preserves all atomic formulas, and an embedding is a function which preserves and reflects all atomic formulas.


问:
A Connected Graph Theorem
In our early days of calculus, we said that a function is continuous if you could draw its graph without having to lift your pen.
This is related to two concepts: that of the graph being closed, and that of it being connected.
Indeed, a powerful theorem I've recently learned is that a linear function (and more generally, of any function if the range is compact) between Banach spaces has a closed graph in the product topology if and only if the function which corresponds to that graph is continuous.

Now, because every normed vector space is path connected, so must be its continuous image, and hence its graph, if I'm not mistaken.
However, much less clear, is whether or not the converse holds:

If the graph of a function (either, linear between Banach spaces, or perhaps more generally) is path connected, does this imply that the function must have been continuous?

I see similar results for when the domain of the function has the property that bounded sequences contain convergent sub-sequences (example 1 and example 2),
but that is of course not going to be true of the domain is an infinite dimensional normed vector space.

Thanks very much.

答:
Generally this is false. Consider any metric space $X$ and let $f:X\to[0,1]$ be any discontinuous map. Define

$$h_f:X \times[0,1]\to[0,1]$$
$$h_f(x,t)=tf(x)$$

Clearly $h_f$ is discontinuous because $f$ is. But the graph of $h_f$ is path connected which I leave as an exercise. This shows that the statement is false even in low dimensional and compact case (e.g. $X=[0,1]$).

In the case of normed spaces (Banach or not) and linear maps this is even simpler. Let $f:V\to W$ be linear and $(v,f(v))\in Gr(f)$. Then there is a path

$$\lambda:[0,1]\to Gr(f)$$
$$\lambda(t)=(tv, f(tv))=(tv,tf(v))$$

which connects $(v,f(v))$ with $(0,0)$. Thus the graph of any linear map is path connected. But in the infinite dimension it is well known that there are discontinuous linear maps.

Title: A Connected Graph Theorem


答:
I'm not exactly sure what you mean by the value of $ \pi $ for 3 dimensional applications. $ \pi $ is a constant value - it is always equal to $ 3.1415 \ldots $.

Of course formulas don't always translate the same from 2 dimensional to 3 dimensional. For example, the area of a circle is $ \pi r^2 $ but the volume contained within a sphere is $ \frac{4\pi}{3} r^3 $. You could almost say that $ \pi $ in 2 dimensional geometry is analogous to $ \frac{4\pi}{3} $ but then this causes fault in other places with $ \pi $, such as surface area.

Title: Is the value of $\pi$ in 2d the same in 3d?

Body:
I am starting with my question with the note "Assume no math skills". Given that, all down votes are welcomed. (At the expense of better understanding of course!)

Given my first question: What is meant by the perimeter of a Sector

Why is the value of $\pi$ not exactly $3$? why is it $3.14$.......... or a fraction $\frac{22}{7}$?

Is the value of $\pi$ of $3.14$... or $\frac{22}{7}$ the same as for $3$ dimensions?


问:
Limits of trigonometric functions

I'm stuck on this problem, it seems that I have tried everything in my arsenal but I'm still unable to solve it.. how would you calculate this limit?

I should have added that we are not allowed to use L'Hopital just yet.

Title: Limits of trigonometric functions

Body:
Hint :

$$ \dfrac{\cos 3x-\cos 2x}{x^2}=\dfrac{(1-\cos 2x) - (1-\cos3x)}{x^2}$$


问:
Prove that $\displaystyle\left|\int_L \frac{z^3}{z^2+1} dz\right| \le \frac{9\pi}{8}$ where $L=\big(z: |z|=3, Re(z) \ge 0\big)$
I tried solving this by using the fact that the length of L is half the circumference of a circle with radius 3, which would mean that
if
$$\left|\frac{z^3}{z^2+1}\right| \le \frac{3}{8}$$ this problem is solved.
however, It appears to me as though this cannot be true:
$$\left|\frac{z^3}{z^2+1}\right| = \frac{\left|z\right|^3}{|z^2+1|} \ge \frac{27}{10}$$
using the triangle inequallity.
Is there something wrong with the way I'm going about this or is the statement really false?

答:
I believe the inequality you stated is not true. Firstly, if $|z|=3$, by the reverse triangle inequality we get that $|z^2+1|=|z^2-(-1)| \geq | |z^2|-1| = |3^2-1| = 8$. Taking reciprocals gives us that $\dfrac{1}{|z^2+1|} \leq \dfrac{1}{8}$.

If we multiply both sides of the inequality by $|z^3|$, we obtain for $|z|=3$ that
$$\left|\frac{z^3}{z^2+1}\right|\leq \dfrac{27}{8}$$

Now if we apply standard integral estimates, we get $\displaystyle\left|\int_L \dfrac{z^3}{z^2+1}dz\right| \leq \dfrac{27}{8}\cdot\operatorname{length}(L)$. As you said, the length of $L$ is half the circumference of a circle with radius $3$. That is, $\operatorname{length}(L) = \dfrac{6\pi}{2} = 3\pi$.

Thus, $$\displaystyle\left|\int_L \dfrac{z^3}{z^2+1}dz\right| \leq \dfrac{27}{8}\cdot3\pi= \dfrac{81\pi}{8} $$


答:
$Im(T) ≠ V$ says that the transformation is not "full rank"
At least one eigenvalue is $0.$

$Im(T) ≠ \{0\}$ nor is the transformation a zero matrix that flattens everything to the zero vector.

$T^2 = T.$ The eigenvalues will be roots of this equation.

Title: Find all of the Eigenvalues of T

Body:
Let there be $V$, vector space from a finite dimenstion contains all real numbers $\mathbb{R}$

$T:V$$\rightarrow $ $V$ linear transformation such that:

$T^2$ = $T$

$Im(T)$ $\neq$ $V$

$Im(T)$ $\neq$ {$\vec{0}$}

Find all of the Eigenvalues of $T$


Title: Limits of trigonometric functions

Body:
I'm stuck on this problem, it seems that I have tried everything in my arsenal but I'm still unable to solve it.. how would you calculate this limit?

I should have added that we are not allowed to use L'Hopital just yet.

Body: Another option uses $\cos(3x)-\cos(2x)=-2\sin(5x/2)\sin(x/2)$, so the limit is $-2(5/2)(1/2)=-5/2$.


答: $T$ satisfies the polynomial $f(x) = x(x -1)$. The minimal polynomial must divide $f$ and so the possibilities are $x, x - 1, x(x-1)$. By Cayley-Hamilton theorem the first two conditions imply that $T = 0$ or $T = I$, but both of these are ruled out by the hypothesis on the image of $T$. Hence the minimal polynomial must be $f(x) = x(x-1)$. The eigenvalues are the roots of minimal polynomial.

Title: Find all of the Eigenvalues of T


答:
You can just be more explicit and do the parametrization $z = 3e^{it}, -\pi/2\leq t \leq\pi/2:$

$$\left|\int_L\dfrac{z^3}{z^2+1}dz\right| = \left|\int_{-\pi/2}^{\pi/2}\dfrac{81e^{3it}i}{9e^{2it}+1}dt\right|\leq\dfrac{81}{8}\int_{-\pi/2}^{\pi/2}dt=\dfrac{81}{8}\pi$$
so the naive triangle-inequality bound gives something different than what you asserted.

Title: Prove that $\displaystyle\left|\int_L \frac{z^3}{z^2+1} dz\right| \le \frac{9\pi}{8}$ where $L=\big(z: |z|=3, Re(z) \ge 0\big)$

Body:
I tried solving this by using the fact that the length of L is half the circumference of a circle with radius 3, which would mean that
if
$$\left|\frac{z^3}{z^2+1}\right| \le \frac{3}{8}$$ this problem is solved.
however, It appears to me as though this cannot be true:
$$\left|\frac{z^3}{z^2+1}\right| = \frac{\left|z\right|^3}{|z^2+1|} \ge \frac{27}{10}$$
using the triangle inequallity.
Is there something wrong with the way I'm going about this or is the statement really false?


问:
The smallest nontrivial conjugacy class in $S_n$

Find the smallest nontrivial conjugacy class in $S_n$.

For small $n$, the answer can be found by counting the permutations of each possible cycle type. The result is:
$$\begin{array}{ccc}
n & \text{smallest nontrivial class(es)} & \text{size} \\
\hline
1 & \text{none} & \\
2 & \text{transpositions} & 1 \\
3 & 3\text{-cycles} & 2 \\
4 & \text{double transpositions} & 3 \\
5 & \text{transpositions} & 10 \\
6 & \text{transpositions, triple transpositions} & 15
\end{array}$$

For $n\geq 7$, I think the unique smallest nontrivial conjugacy class in $S_n$ is given by the transpositions.

The conjugacy classes in $S_n$ are given by the sets of permutations of the same cycle type. For the cycle type $\alpha = 1^{a_1} 2^{a_2} \ldots$, the size of the respective conjugacy class is known to be
$$N_\alpha = \frac{n!}{\prod_i (i^{a_i} a_i!)}.$$
In particular, the number of transpositions in $S_n$ is $\binom{n}{2}$.

In this way, it remains to show the following:

Let $n \geq 7$ and $\alpha$ be a partition of $n$ distinct from $1^n$ and $2^1 1^{n-2}$. Show that $N_\alpha > \binom{n}{2}$.

So far I can only come up with a lengthy and technical case-by-case study.
I'm looking for a more elegant proof of the statement.

答:
This is only a partial argument that solves some cases. The idea is simple: instead of focussing on a particular conjugacy class, we consider that of an appropriate power which has simpler cycle structure.

It follows from the fact that $\binom{n}{2}<\binom nl$ for all $2<l<n-2$ that for any permutation with total length $l$ (i.e. having $n-l$ fixed points), its conjugacy class is stricly bigger than that of the transpositions. Indeed, for every choice of $n-l$ points there are several permutations in that conjugacy class having these as fixed pionts. Also, for $l=n-2$ and $n\geq 5$, for every choice of two points there are several permutations in that conjugacy class having these points as fixed points. So there are only $2$ cases of interest: $l=n-1$ and $l=n$, i.e. $\sigma$ has one fixed point and none respectively.

In any case, if there are two distinct cycle lengths $2\leq k<K$, then we may consider the permutation $\sigma^k$ instead. This permutation is different from the identity, and has at least $2\leq k$ fixed points, so by the previous argument its conjugacy class contains at least $\binom n2$ elements. The map
$$\mathrm{Conj}(\sigma)\to\mathrm{Conj}(\sigma^k),~\theta\mapsto \theta^k$$
is onto and non injective whenever $k\geq 3$ or $k=2$ and there are at least two $2$-cycles in $\sigma$. Furthermore if $\sigma$ contains only one $2$ cycle, then there are obviously more than $\binom n2$ elements in the conjugacy class. So your claim is true whenever there $\sigma$ contains cycles of different length.

So there remains one case: $\sigma$ has length $n$ or $n-1$, and only one cycle length $k$.

If $\sigma$ has one fixed point, then the choice of fixed point accounts for $n$ choices already. However it might be easiest to resort back to the formula you mentioned at the beginning: we have $n=qk+1$ and the cardinality equals $\frac{n!}{k^q q!}$.


问:
Does $\int \tan^3x\sec^2x \space dx$ have 2 solutions?
So normally, you evaluate $\int\tan^3x\sec^2x \space dx$ by substituting $u = \tan x$ and $du = \sec^2x\space dx$ right? So,
$$\begin{equation}\begin{aligned}
\int\tan^3x\sec^2x \space dx &= \int u^3 \space du \\
  &= \frac{u^4}{4} + C\\
  &= \frac{tan^4x}{4} + C \\
\end{aligned}\end{equation}$$

But, I tried to solve it this way instead,
$$\begin{equation}\begin{aligned}
\int\tan^3x\sec^2x \space dx &= \int\tan^2x\tan x\sec x\sec x\space dx \\
  &= \int(\sec^2x-1)\tan x\sec x\sec x\space dx \\
  &= \int\sec^3x\tan x\sec x-\sec x\tan x\sec x\space dx \\
  &= \int\sec^3x\tan x\sec x-\sec x\tan x\sec x\space dx \\
  &= \int\sec^3x\tan x\sec x\space dx - \int\sec x\tan x\sec x\space dx \\
\end{aligned}\end{equation}$$

and then substituting $u = \sec x$, thus $du = \tan x\sec x\space dx$. So,
$$\begin{equation}\begin{aligned}
\int\sec^3x\tan x\sec x\space dx - \int\sec x\tan x\sec x\space dx &= \int u^3 du - \int u\space du\\
&= \frac{u^4}{4} - \frac{u^2}{2} + C \\
&= \frac{\sec^4x}{4} - \frac{\sec^2x}{2} + C \\
\end{aligned}\end{equation}$$

What?! Can someone explain to me where I made the mistake?

答:
There's no mistake. Both methods resulted in the same solution, up to a constant of integration. Indeed, notice that:

\begin{align*}
\frac{1}{4}\tan^4 x + C
&= \frac{1}{4}(\tan^2 x)^2 + C \\
&= \frac{1}{4}(\sec^2 x - 1)^2 + C \\
&= \frac{1}{4}(\sec^4 x - 2\sec^2 x + 1) + C \\
&= \frac{1}{4}\sec^4 x - \frac{1}{2}\sec^2 x + \frac{1}{4} + C \\
&= \frac{1}{4}\sec^4 x - \frac{1}{2}\sec^2 x + D \\
\end{align*}

where $D = \frac{1}{4} + C$ is an arbitrary constant of integration.


问:
Solve $y'' + 2y' + y = \begin{cases} 4e^t, & 0 \leq t < 1 \\ 0, &t \geq 1 \end{cases}$ where $y(0) = 0, y'(0) = 0$
Hi can you help me solve this?  I have no idea how to solve the initial problem by Laplace transform when it is a piecewise function.

Solve the initial value problem using Laplace transform on $$y'' + 2y' + y = \begin{cases} 4e^t, & 0 \leq t < 1 \\ 0, &t \geq 1 \end{cases}$$ where $y(0) = 0, y'(0) = 0$

答:
\begin{align}
(\mathcal Lf)(s) = {} & \int_0^\infty f(t) e^{-st} \, dt \\[8pt]
= {} & \int_0^1 4e^t e^{-st} \, dt \\[8pt]
& \text{and so on.}
\end{align}


问:
How do I show this inequality that $\frac{\ln({x+1})}{\ln({x})} \leq \frac{x}{x-1}$ for $x>1$?
How do I show this inequality that for $x>1$ $$\frac{\ln({x+1})}{\ln({x})} \leq \frac{x}{x-1}$$

So far I have tried to use the inequality $\ln({x+1}) < x$, but it is not good enough.

答:
(Fill in the gaps as needed. If you're stuck, show you work and explain what you've tried.)

Hint: Prove that $(x+1)^{x-1} < x^x $ for $ x > 1$.

Further hint: Direct application of weighted AM-GM.

Corollary: Inequality as stated, after taking logarithms.


问:
Proof that $x \in \Bbb C[x]/\langle p(x)\rangle$ has a square root
I'm looking for a "nice" proof of the following result.

Proposition: Let $p \in \Bbb C[x]$ be such that $p(0) \neq 0$. Prove that there exists a $q \in \Bbb C[x]$ such that
$$
q(x)^2 = x \pmod {p(x)}.
$$
In other words, $x + \langle p\rangle \in \Bbb C[x]/\langle p\rangle$ is a perfect square.

Here is a proof that I don't find particularly nice.

Let $V = \Bbb C[x]/\langle p\rangle$; note that $V$ is a finite-dimensional vector space. The multiplication operator $T:V \to V$ define by $[Tf](x) = xf(x)$ defines an invertible linear operator. By various means (cf. this post or this post), it can be shown that there exists an operator $S:V \to V$ for which $S = q(T)$ for some $q \in \Bbb C[x]$ and $S^2 = T$. It follows that $[q + \langle p\rangle]^2 = x + \langle p\rangle$.

This proof relies on linear-algebraic machinery, and in particular on Jordan normal form. Is there a nicer approach, perhaps one that is "purely algebraic"?

Here's a formal extension of Eric's "explicit" construction. With the application of the Chinese remainder theorem and the substitution $y = x-a$, the problem is reduced to that of showing that $y + a$ has a square root modulo $y^n$. We can see that such a square root exists as follows. Define
$$
q(y) = b_0 + b_1 y + \cdots + b_{n-1} y^{n-1}.
$$
We find that the coefficients of $q^2$ are given by the "Cauchy product", i.e.
$$
q(y)^2 = b_0^2 + (2b_0b_1) y + (2b_0b_2 + b_1^2)y^2 + \cdots + (2b_{n-1}b_0 + \cdots)y^{n-1} \pmod{y^n}.
$$
Define $a_j$ to be the coefficient of $y^j$ in the above product. We note that for $j \geq 1,$
$$
a_j = 2b_0b_j + f_j(b_0,\dots,b_{j-1}).
$$
If we take $b_0 = \sqrt{a} \neq 0$, we then get $a_0 = a$ as desired and find that the system
$$
a_1 = 1, \quad a_j = 0 \text{ for } j \geq 2,
$$
can be solved recursively by taking
$$
b_1 = \frac 1{2b_0}, \quad b_j = -\frac{f_j(b_0,\dots,b_{j-1})}{2b_0} \text{ for } j \geq 2.
$$
Thus, there indeed exists a polynomial $q(y)$ for which $[q(y)]^2 = x + a$, which was what we wanted.

答: Well, this isn't really any different from using Jordan normal form, but here's how I'd think about it.  By the Chinese remainder theorem, the quotient $\mathbb{C}[x]/(p(x))$ splits as a product of $\mathbb{C}[x]$-algebras of the form $\mathbb{C}[x]/((x-a)^n)$, where the $a$'s are the roots of $p$ and the $n$'s are their multiplicities.  So, it suffices to show that $x$ is a square in $\mathbb{C}[x]/((x-a)^n)$ if $a\neq 0$.  Or, setting $y=x-a$, it sufffices to show that $y+a$ is a square in $\mathbb{C}[y]/(y^n)$.  This follows immediately from the fact that $a$ is a square in $\mathbb{C}$ and Hensel's lemma.  Explicitly, you can construct a square root of $y+a$ by starting with $b\in\mathbb{C}$ that is a square root of $a$, then taking $\frac{y}{2b}+b$ to get a linear term of $y$ when you square it, then adding a quadratic term to make the quadratic term of the square vanish, and so on up to the $y^{n-1}$ term.


问:
Given $16$ pens to be distributed between $4$ people ,then find the number of distributions with the given assumptions.
Given $16$ pens to be distributed between $4$ people assuming the fourth  person gets at least $3$ pens and the three other ones should not get more than $5$ pencils,in how many ways it's possible?

We are looking for the integral solutions to the equation $b_1+b_2+b_3+b_3+b_4=16$ such that $b_4 \ge 3$ and $b_1,b_2,b_3 \le 5$,this is given by :

$$[x^{16}]\left[\sum_{3 \le k \le 16}^{}x^k \right]\left[\sum_{0 \le k \le 13}^{}x^k \right]^3$$
$$=[x^{13}]\left[\sum_{0 \le k \le 13}^{}x^k \right]\left[\sum_{0 \le k \le 13}^{}x^k \right]^3$$
$$=[x^{13}](1-x^{14})^4(1-x)^{-4}$$
$$=[x^{13}]\sum_{k=0}^{4}\binom{4}{k}(-1)^kx^{14k}\sum_{k \ge 0}^{}\binom{k+3}{k}x^{k}$$
$$=\binom{16}{13}$$

However the answer is not this,so where was I wrong?

答:
$b_1 + b_2 + b_3 + b_4=16$

As $b_4 \geq 3$, the number of ways to distribute $16$ pens will be equivalent to

$b_1 + b_2 + b_3 + b_4 = 13 \,$ without a restriction on $b_4$. The number of ways -

$\displaystyle  = {16 \choose 3}$

Now $b_1, b_2, b_3 \leq 5$ so we should exclude number of ways in which one of them is $\geq 6$ which will be given by -

$b_1 + b_2 + b_3 + b_4 = 7$

Number of ways $\displaystyle  = {3 \choose 1} {10 \choose 3} \,$ as there are ${3 \choose 1}$ ways to choose a number from $b_1, b_2, b_3$ that can be $\geq 6$.

But this will overcount cases where $2$ numbers are $6$ or greater. So we need to add those back.

$b_1 + b_2 + b_3 + b_4 = 1$ (two numbers are $6$ or more).

Number of ways $\displaystyle  = {3 \choose 2} {4 \choose 3} \,$ as there are ${3 \choose 2}$ ways to choose two numbers from $b_1, b_2, b_3$ that can be $\geq 6$.

Now $3$ numbers cannot be all $\geq 6$ at the same time.

So applying P.I.E, our final answer is

${16 \choose 3} - {3 \choose 1} {10 \choose 3} + {3 \choose 2} {4 \choose 3} = 212$


问:
Modifying distinct lists in GAP when their values are equal
I am making some functions in GAP and am having weird issues with lists and variables. It seems like whenever two variables have the same value at a given point, any changes made to one will change the other even though I have defined them separately.

Here's a simple example illustrating my main problem. Suppose I write the code:

```<code>list:=[1,2,3];
list2:=list;
Add(list2,4);
</code>```

Using this code, I would want to add 4 to list2, which would make list2=[1,2,3,4]. This part is fine. The unintended consequence here is that this also changes list to list=[1,2,3,4].

In this example, my question is: how do I change list2 without changing list?

答:
When assigning objects in GAP, the assignment is always a pointer to the objects. For objects that do not have the ability to change stored information that does not matter, but for lists or records it means that a change to the entries of the one object also changes the other.

To avoid this happening, there are two kinds of copy functions" ShallowCopy duplicates the top-level object, but changes the entries. StructuralCopy duplicates the whole structure (which is often wasting memory). If an object has no further substructures, both functions do the same.  For example:

```<code>gap> l:=[[1,[2]],[3],4];
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> a:=l;
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> b:=ShallowCopy(l);
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> c:=StructuralCopy(l);
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> l[1][1];
1
gap> l[1][1]:=5;l;
5
[ [ 5, [ 2 ] ], [ 3 ], 4 ]
gap> a;b;c;
[ [ 5, [ 2 ] ], [ 3 ], 4 ]
[ [ 5, [ 2 ] ], [ 3 ], 4 ]
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> l[4]:=6;l;
6
[ [ 5, [ 2 ] ], [ 3 ], 4, 6 ]
gap> a;b;c;
[ [ 5, [ 2 ] ], [ 3 ], 4, 6 ]
[ [ 5, [ 2 ] ], [ 3 ], 4 ]
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> Add(l[1][2],7);l;
[ [ 5, [ 2, 7 ] ], [ 3 ], 4, 6 ]
gap> a;b;c;
[ [ 5, [ 2, 7 ] ], [ 3 ], 4, 6 ]
[ [ 5, [ 2, 7 ] ], [ 3 ], 4 ]
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
</code>```

Body:
When assigning objects in GAP, the assignment is always a pointer to the objects. For objects that do not have the ability to change stored information that does not matter, but for lists or records it means that a change to the entries of the one object also changes the other.

To avoid this happening, there are two kinds of copy functions" ShallowCopy duplicates the top-level object, but changes the entries. StructuralCopy duplicates the whole structure (which is often wasting memory). If an object has no further substructures, both functions do the same.  For example:

gap> l:=[[1,[2]],[3],4];
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> a:=l;
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> b:=ShallowCopy(l);
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> c:=StructuralCopy(l);
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> l[1][1];
1
gap> l[1][1]:=5;l;
5
[ [ 5, [ 2 ] ], [ 3 ], 4 ]
gap> a;b;c;
[ [ 5, [ 2 ] ], [ 3 ], 4 ]
[ [ 5, [ 2 ] ], [ 3 ], 4 ]
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> l[4]:=6;l;
6
[ [ 5, [ 2 ] ], [ 3 ], 4, 6 ]
gap> a;b;c;
[ [ 5, [ 2 ] ], [ 3 ], 4, 6 ]
[ [ 5, [ 2 ] ], [ 3 ], 4 ]
[ [ 1, [ 2 ] ], [ 3 ], 4 ]
gap> Add(l[1][2],7);l;
[ [ 5, [ 2, 7 ] ], [ 3 ], 4, 6 ]
gap> a;b;c;
[ [ 5, [ 2, 7 ] ], [ 3 ], 4, 6 ]
[ [ 5, [ 2, 7 ] ], [ 3 ], 4 ]
[ [ 1, [ 2 ] ], [ 3 ], 4 ]


问:
Looking for better proof to Rutgers Algebra Qual Problem
On the August 2016 Rutgers Qualifying Exam appears the following problem:

Let $B$ be a symmetric nondegenerate form on a the 2 dimensional vector space over a field $\mathbb{F}_p$, where $p$ is an odd prime. Show that there is a vector $v$ such that $B(v,v)=1$.

The way I solved this (kinda brute force and did not use symmetric condition) was as follows:
Let $a=B(e_1,e_1), b=B(e_1,e_2), c=B(e_2,e_1), d=B(e_2,e_2)$ (where nondegeneracy gives $a,b=0$

Then if $v=xe_1+ye_2$, $B(v,v)=ax^2+(b+c)xy+dy^2$, so $B(v,v)=1$ can be restated as:

$(x+\frac{y(b+c)}{2a})^2 =\frac{1}{a}-\frac{d}{a}y^2$

Now set $\alpha=\frac{1}{a}, \beta=\frac{-d}{a}$ and note $\alpha,\beta\ne 0$

A solution $x,y$ exists in case $\alpha+\beta y^2$ is a quadratic residue mod $p$.

Now note that when $y$ varies from $0$ to $p-1$, $y^2$ repeats each value exactly once excepts $0$, so $\alpha+\beta y^2$ takes on $\frac{p+1}{2}$ values, but there are only $\frac{p-1}{2}$ non-quadriatic residues (mod p).

I don't like this solution because I never used the symmetry of the form, and don't think it was intended to be solved as a number theory problem.

Does anyone know of a more natural algebraic approach?

答:
More natural algebraic approach:

Suppose to the contrary that there exists no such vector $v$. For the vector space $V$ (over $F_p$) let $e_1$ and $e_2$ denote the basis vectors. Next, consider the cyclic subspaces generated by $e_1$ and $e_2$, denoted $<e_1>$ and $<e_2>$ respectively. The cardinalities of both of these subspaces is $p$ (obviously). Now, given the symmetric bilinear form $B$, suppose $B(v,v) \neq 1$ and $B(u,u) \neq 1$ for all $ v \in <e_1>$ and $ u \in <e_2>$. Since the field $F_p$ has $p$ elements and there are $p$ elements in both $<e_1>$ and $<e_2>$, there exists $ae_1, be_1 \in <e_1>$ and $ce_2, de_2 \in <e_2>$ such that $B(ae_1, ae_1)=B(be_1,be_1)$ and $B(ce_2,ce_2)=B(de_2,de_2)$ ($ a \neq b, c \neq d$) (by the pigeonhole principle.) Now, this is the case if and only if (1) $B(e_1(a-b), e_1(a-b))=0$ and  (2) $B(e_2(c-d), e_2(c-d))=0$. Given this, I now claim that $B(e_1(a-b) +e_2(c-d),x)=0$ for all $ x \in V$. Let $x \in V$. Then we have $x=\alpha e_1 + \beta e_2$, for $\alpha , \beta \in F_p$. Since $F_p$ is a field,  there exists $m_1$, $m_2$ such that $m_1 (a-b)=\alpha$ and $m_2(c-d)=\beta$. Now, multiply (1) and (2) by $m_1$ and $m_2$ respectively to get, (3) $B(e_1(a-b),\alpha e_1)$ and (4) $B(e_2(c-d), \beta e_2)$. Adding (3) and (4) we get $B(e_1(a-b) +e_2(c-d), \alpha e_1 + \beta e_2)=B(e_1(a-b) +e_2(c-d), x)=0$. This violates the non-degeneracy hypothesis.

Hope this works!


答:
You can show it using concavity of $\ln x$ as follows:

The given inequality is equivalent to

$$\ln x \geq \frac{x-1}{x}\ln(x+1)$$

Now, since $x>1$ we have

$$\frac{x-1}{x}\ln(x+1) + \frac 1x\ln 1 \stackrel{concavity}{\leq}\ln\left(\frac{x-1}{x}(x+1) +\frac 1x\cdot 1\right) =\ln x$$

Done.

Title: How do I show this inequality that $\frac{\ln({x+1})}{\ln({x})} \leq \frac{x}{x-1}$ for $x>1$?

Body:
How do I show this inequality that for $x>1$ $$\frac{\ln({x+1})}{\ln({x})} \leq \frac{x}{x-1}$$

So far I have tried to use the inequality $\ln({x+1}) < x$, but it is not good enough.


答:
You can get tricky with your applications of trig and algebra.

$\frac {\cos 3x - \cos 2x}{x^2}\\
\frac {4\cos^3 x - 3\cos x - 2 \cos^2 x + 1}{x^2}\\
\frac {(\cos x - 1)(4\cos^2 x + 2\cos x - 1)}{x^2}\\
\frac {(\cos x + 1)(\cos x - 1)(4\cos^2 x + 2\cos x - 1)}{(\cos + 1)x^2}\\
\frac {(\cos^2 x - 1)(4\cos^2 x + 2\cos x - 1)}{(\cos + 1)x^2}\\
\frac {(-\sin^2 x) (4\cos^2 x + 2\cos x - 1)}{(\cos + 1)x^2}\\
\lim_\limits{x\to 0} \frac {\sin x}{x} = 1\\
\lim_\limits{x\to 0} \frac {-(4\cos^2 x + 2\cos x - 1)}{(\cos + 1)x^2} = -\frac {5}{2}$

You could apply L'Hopital's rule twice...

$\lim_\limits{x\to 0}\frac {\cos 3x - \cos 2x}{x^2}\\
\lim_\limits{x\to 0}\frac {-3\sin 3x + 2\sin 2x}{2x}\\
\lim_\limits{x\to 0}\frac {-9\cos 3x + 4\cos 2x}{2} = -\frac {5}{2}$

Or, you could use a Taylor series.
$\cos x = 1 - \frac {x^2}{2} + o(x^4)$

$\lim_\limits{x\to 0}\frac {\cos 3x - \cos 2x}{x^2}\\
\lim_\limits{x\to 0}\frac {1- \frac {(3x)^2}{2} - 1+ \frac {(2x)^2}{2} + o(x^4)}{x^2}\\
\lim_\limits{x\to 0}-\frac {5}{2} + o(x^2) =-\frac {5}{2}$

Title: Limits of trigonometric functions


问:
Closed form for $n$-th derivative of exponential: $\exp\left(-\frac{\pi^2a^2}{x}\right)$
I need the closed-form for the $n$-th derivative ($n\geq0 $):

$$\frac{\partial^n}{\partial x^n}\exp\left(-\frac{\pi^2a^2}{x}\right)$$

Thanks!

By following the suggestion of Hermite polynomials:

$$H_n(x)=(-1)^ne^{x^2}\frac{\partial^n}{\partial x^n}e^{-x^2}$$

and doing the variable change $x=\pi a y^{-\frac{1}{2}}$, I obtain:

$$\frac{\partial^n}{\partial x^n}=-2\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n\frac{\partial^n}{\partial y^n}$$

and therefore

$$H_n(\pi a y^{-\frac{1}{2}})=(-1)^{n+1}e^{\frac{\pi^2a^2}{y}}2\left(\frac{\pi a}{y^{\frac{3}{2}}}\right)^n\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}$$

Finally

$$\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}=\frac{1}{2}e^{-\frac{\pi^2a^2}{y}}(-1)^{n+1}H_n(\pi a y^{-\frac{1}{2}})\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n$$

Is this correct?

答:
I obtained the following result using the Fourier transform:
$$\frac{\partial^n}{\partial x^n}\exp\left(-\frac{\pi^2a^2}{x}\right)=\\\frac12\pi^2\,a^2(-1)^nn!\ x^{-n-2}\left({_2F_3}\left(\frac{n}{2}+1,\frac{n}{2}+\frac{3}{2};\frac{3}{2},\frac{3}{2},2;\frac{a^4\,\pi^4}{4\,x^2}\right)\pi^2a^2(n+1)\\-{_2F_3}\left(\frac n2+\frac12,\frac n2+1;\frac12,1,\frac32;\frac{a^4\pi^4}{4\,x^2}\right)2\,x\right).$$

Probably, it can be simplified.


问:
A question regarding Runge-Kutta method of order 3.

Runge-Kutta method is defined by the iterations:

$u_{i+1}=u_i+ahf(x_i,u_i)+bhf(x_i+ch,u_i+dhf(x_
i,u_i))$

where $a,b,c,d$ are constants.
Given that the method converges of order 3.  Show that $a+b=1$, $cb=1/2$ , $db=1/2$

It is quite not useful i guess to use the definition of this method which appears in wikipedia page:
https://www.google.com/url.

How can one show this in another way, not based on the definition in the link above.I thought about using the Taylor expansion twice and compare coefficients but I am not sure how to do it correctly.

答:
[Note] : The idea that you should be aware of is that a $k^{th}$ order Runge-Kutta method scheme approximates $y(t_{i+1})$ by $y_{i+1}$ recursively with a global error of $O(h^{k})$ by using $k$ function evaluations. I will present you the derivation for Runge-Kutta method of order $2$ so that you get to derive to use similar argument to derive the third order Runge-Kutta method.

Therefore, we shall derive the Runge-Kutta method of order $2$ that is of the form :
$$
\left\{\begin{array}{l}
K_{1}=f\left(t_{i}, y_{i}\right) \\
K_{2}=f\left(t_{i}+\alpha h, y_{i}+\beta h K_{1}\right) \\
y_{i+1}=y_{i}+h\left(w_{1} K_{1}+w_{2} K_{2}\right)
\end{array}\right.
$$
Where $\alpha,\beta,w_{1}$, and $w_{2}$ are constants that we will determine.

The scheme is obtained by assuming that the exact value $y\left(t_{i}+h\right)$ satisfies the equation $(1)$. Then, $y\left(t_{i}\right)$ is approximated by $y_{i},$ which leads to equation $(2)$

\begin{align*}
y\left(t_{i}+h\right) &=y\left(t_{i}\right)+w_{1} h f\left(t_{i},y\left(t_{i}\right)\right)+w_{2} h f\left(t_{i}+\alpha h,y\left(t_{i}\right)+\beta h f\left(t_{i}, y\left(t_{i}\right)\right)\right)+O\left(h^{3}\right) \tag1\\
&=y_{i}+w_{1} h K_{1}+w_{2} h f\left(t_{i}+\alpha h, y_{i}+\beta hK_{1}\right)+O\left(h^{3}\right) \tag2
\end{align*}

where $\alpha, \beta, w_{1},$ and $w_{2}$ relations are obtained by comparing $(2)$ with the Taylor series expansion of $y(t)$ at the point $t_{i}$ for
$t=t_{i}+h$ up to second order.
$$
\begin{align*}
y\left(t_{i}+h\right) &=y\left(t_{i}\right)+h y^{\prime}\left(t_{i}\right)+0.5 h^{2} y^{\prime \prime}\left(t_{i}\right)+O\left(h^{3}\right) \\
y^{\prime}\left(t_{i}\right) &=f\left(t_{i}, y\left(t_{i}\right)\right) \\
y^{\prime \prime}(t) &=\left(y^{\prime}(t)\right)^{\prime}=\frac{d}{d t} f(t, y(t))=f_{t}(t, y(t))+y^{\prime}(t) f_{y}(t, y(t)) \quad \text {(chain rule)} \\
&=f_{t}(t, y(t))+f(t, y(t)) f_{y}(t, y(t)) \\
y^{\prime \prime}\left(t_{i}\right) &=f_{t}\left(t_{i}, y\left(t_{i}\right)\right)+f\left(t_{i}, y\left(t_{i}\right)\right) f_{y}\left(t_{i}, y\left(t_{i}\right)\right) \\
\Longrightarrow\left(t_{i}+h\right) &=y\left(t_{i}\right)+h f\left(t_{i}, y\left(t_{i}\right)\right)+\frac{h^{2}}{2}\left[f_{t}\left(t_{i}, y\left(t_{i}\right)\right)+f\left(t_{i}, y\left(t_{i}\right)\right) f_{y}\left(t_{i}, y\left(t_{i}\right)\right)\right]+O\left(h^{3}\right) \\
&=y_{i}+h K_{1}+\frac{h^{2}}{2}\left[f_{t}\left(t_{i}, y_{i}\right)+K_{1} f_{y}\left(t_{i}, y_{i}\right)\right]+O\left(h^{3}\right)\tag3
\end{align*}
$$
[Note] : Till now we can't just compare $(3)$ and $(2)$ because of the term $f(t_{i}+\alpha h,y_{i}+\beta h K_{1})$. So using Taylor series expansion
for the function $\phi(h):=f(t_{i}+\alpha h,y_{i}+\beta h K_{1})$ centered at zero up to first order, and then replacing it in $(2)$ we get :
$$
\begin{align*}
\phi(h) &=\phi(0)+h \phi^{\prime}(0)+O\left(h^{2}\right)=f\left(t_{i}, y_{i}\right)+h\left(\frac{d}{d h} f\left(t_{i}+\alpha h, y_{i}+\beta h K_{1}\right)\right)(0)+O\left(h^{2}\right) \\
&=f\left(t_{i}, y_{i}\right)+h\left(\alpha f_{t}\left(t_{i}+\alpha h, y_{i}+\beta h K_{1}\right)+\beta K_{1} f_{y}\left(t_{i}+\alpha h, y_{i}+\beta h K_{1}\right)\right)(0)+O\left(h^{2}\right) \\
&=K_{1}+\alpha h f_{t}\left(t_{i}, y_{i}\right)+\beta h K_{1} f_{y}\left(t_{i}, y_{i}\right)+O\left(h^{2}\right) \\
\Longrightarrow y\left(t_{i}+h\right) &=y_{i}+w_{1} h K_{1}+w_{2} h\left[K_{1}+\alpha h f_{t}\left(t_{i}, y_{i}\right)+\beta h K_{1} f_{y}\left(t_{i}, y_{i}\right)+O\left(h^{2}\right)\right]+O\left(h^{3}\right) \\
&=y_{i}+\left(w_{1}+w_{2}\right) h K_{1}+h^{2} w_{2}\left[\alpha f_{t}\left(t_{i}, y_{i}\right)+\beta K_{1} f_{y}\left(t_{i}, y_{i}\right)\right]+O\left(h^{3}\right)\tag4
\end{align*}
$$
By comparing equation (4) to (3), we get:
$$
\left\{\begin{array}{ll}
w_{1}+w_{2}=1 & \\
w_{2} \alpha=\frac{1}{2} & \text { where } \\
w_{2} \beta=\frac{1}{2}
\end{array} \quad\left\{\begin{array}{l}
K_{1}=f\left(t_{i}, y_{i}\right) \\
K_{2}=f\left(t_{i}+\alpha h, y_{i}+\beta h K_{1}\right) \\
y_{i+1}=y_{i}+h\left(w_{1} K_{1}+w_{2} K_{2}\right)
\end{array}\right.\right.
$$
As desired.

[Final Note] : Since we have $3$ equations and $4$ unknowns, then we can obtain infinitely many second order RK schemes. For instance let $w_{1}=w_{2}=\frac{1}{2}$, then $\alpha=\beta=1$ and then you would have deduced a special type of RK2 scheme :)

Now try deriving the RK3 scheme using this approach and you should get the result you want.


答: Since $\text{Im } T \neq V \land \text{Im } T \neq \lbrace 0 \rbrace$, we can conclude that T has a nontrivial kernel, but also a nontrivial Image. Suppose $v \in V$ is an eigenvector of T with the eigenvalue $\lambda \in \Bbb R$. Then $$T(v) = \lambda v$$ but further since T is a projection $$T \circ T(v) = T(\lambda v) = T(v)$$ and since T is linear we get that $$\lambda T(v) = T(v)$$ which is true for $\lambda \in \lbrace 0,1 \rbrace$ since for the case $\lambda=0$ we get $v \in \ker T$.

Title: Find all of the Eigenvalues of T


问:
Riemannian exponential map on symmetric spaces of non-compact type
Let $S$ be a symmetric space of non-compact type, specifically let $S= SL(n)/SO(n)$ for some $n\geq 2$.  I know that such spaces are geodesically complete, simply connected, and have non-positive curvature.  Therefore, by the Cartan-Hadamard Theorem, the Riemannian exponential map $Exp_p$ at any point $p \in S$ is a global diffeomorphism to the appropriately dimensioned Euclidean space.

Edit: Is it correct that $Exp_e(v)=\pi\circ \exp$ where $\exp$ is the matrix exponential and $\pi:SL(n)\rightarrow SL(n)/SO(n)=S$?

答:
For a linear group like $SL(n)$, the Lie theory exponential agrees with the matrix exponential $\mathrm{exp}$. However, the Riemannian exponential map doesn't always agree with the Lie theoretic exponential map. For example, if you choose a nonzero nilpotent matrix $N$, then $c(t) = \pi \circ \exp(tN)$ is not a geodesic in the associated symmetric space. Fortunately it is easy to answer your question in terms of something called the "Cartan decomposition."

When you pick a point in a symmetric space $G/K$, you can define a Cartan decomposition of the Lie algebra $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$. Here $\mathfrak{k}$ is the Lie algebra of $K$, and this decomposition is not a Lie algebra decomposition, rather it is a vector space decomposition with the bracket satisfying
$$ [\mathfrak{k},\mathfrak{k}] \subset \mathfrak{k}, \, [\mathfrak{k},\mathfrak{p}] \subset \mathfrak{p}, \, [\mathfrak{p},\mathfrak{p}] \subset \mathfrak{k} .$$

In your example, the natural basepoint to choose is the identity coset $[SO(n)]$. Then $\mathfrak{k} = \mathfrak{so}(n)$ is the Lie algebra of skew-symmetric matrices, and $\mathfrak{p}$ is the vector space of symmetric traceless matrices.

Theorem. For any $X \in \mathfrak{p}$, $c(t) = \pi \circ \exp(tX)$ is a geodesic. Morever, every geodesic through the basepoint arises this way.

So in your example, the geodesics through $[SO(n)]$ are exactly $\pi \circ \exp (tX)$ for $X$ a symmetric traceless matrix.


问:
If $X$ and $Y$ are independent random variables, are $X^2$ and $Y^2$ also independent?
If $X$ and $Y$ are independent random variables, are $X^2$ and $Y^2$ also independent?

Can this be proven?

答:
To prove that 2 (discrete) random variables are independent, it boils down to showing that $P(A=a, B=b) = P(A=a)P(B=b)$ all the time.

If they are not independent, then we must find a counter example.

If we suspect that they are independent, then we want to show that:

(Case when $x\neq 0, y \neq 0$)

P( X = x^2, Y = y^2 ) = P( X = x OR X = -x, Y = y OR Y = -y)

= P( X = x, Y = y) +  P( X = -x, Y = y ) + ...

= ?

= ? P(X = x^2 ) P(Y= y^2)

Can you take it from here?

Must this equality be true? How are we using the independence of $X, Y$?

What about the case when $ x =0$?
What if the random variables are continuous, or a mix?

More generally, we say that if $X, Y$ are independent, then for measurable functions $f, g$, we get that $f(X), g(Y)$ is also independent.

The proof is immediate from understanding measurable functions.

If you haven't learn about measurable functions, then just think of them "reasonable, nicely behaving functions" like $ f(x) = x^2$. The proof follows by a similar approach as above.


问:
Show that for any $Z\subseteq Y$, $Z$ is closed in $X$
Be $X$ a topological space and $Y \subseteq X$ closed in $X$. Suppose $Y$, with the subspace-induced topology, is a discrete space. Show that for any $Z\subseteq Y$, $Z$ is closed in $X$.

I wonder if my solution is correct.

If $Y$ is a discrete space so any subset of $Y$ belongs to the $Y$ topology.

(i) Can I say that if $Y$, with the subspace-induced topology, is a discrete space then $X$ is a discrete topological space due to this?

If yes, I can say that if $Y$ is closed then $Y$ is not in the $X$ topology, as $Z$ is a subset of $Y$, it means that $Z$ is not an element of the $X$ topology, so it is closed.

答: Answer to $(i)$: No, in general you cannot say this. Pick $X=[0,2]$ and $Y=\{1\}$. You easily see that $X$ is not a discrete space.


问:
testing if the infinite series are converges or diverges
here are two problems on testing if the infinite series are conv. or div.

$\sum_{n=1}^\infty \frac{{{7\sqrt[6]{n^{13}}+2n}}}{{\sqrt[3]{27n^{9}-10n+16}}}$

$\sum_{n=2}^\infty \frac{1}{\ln(n!)}$

the first one I couldn't do :(

but the second one here is what I did using the ratio test I got that

$$\frac{\ln(n!)}{\ln((n+1)!)}=\frac{\ln(n)+\cdots+ \ln(2)}{\ln(n+1)+\cdots +\ln(2)}<1$$

which I thought would be effective but it isn't :(

答:
Both are divergent. In the first one, the term is $\sim 7n^{13/6}/3n^3= \frac 73 n^{-5/6}$,
as one know $\sum n^{-p}$ is divergent when $p<1$. In the second one, the term is greater
than $\frac{1}{n\ln n}=\frac{1}{\ln n^n}$, you see $\int_2^\infty \frac{1}{x\ln x}dx$ is divergent so by the integral test $\sum  \frac{1}{n\ln n}$ is divergent, so $\sum \frac{1}{\ln n!}$ is more divergent.


问:
Reference for a proof of the Hahn-Mazurkiewicz theorem
The Hahn-Mazurkiewicz theorem states that a space $X$ is a Peano Space if and only if $X$ is compact, connected, locally connected, and metrizable. If anybody knows a book with a proof, please let me know. Thanks.

P.S. (added by t.b.) A Peano space is a topological space which is the continuous image of the unit interval.

答: Read the section on Peano spaces in General Topology by Stephen Willard.


问:
Differential equation (2nd order) with divergent coefficients.
I have this equation:

$$x(x-1)y''+6x^2y'+3y=0$$

I try to get the series for the solution around $x=0$, using Frobenius (however it's written). the first solution must be of the form:

$$y_1=\sum_{n=0}^\infty c_nx^{n+1}$$

If I try to get coefficients, I will get a divergent series of coefficients, in which I can't see any sense. The first one is $c_1=\frac{3}{2}c_0$, I calculated the first 1000 by computer and the never go down, although I think you can deduce that from the expression for the coefficients:
$$a_{n+2}=\frac{[(n+2)(n+1)+3]a_{n+1}+6(n+1)a_{n}}{(n+3)(n+2)}$$
Since the upper thing goes "like" a factorial with the $a_{n-1,2}$.

So... how would you get the solutions by series for this? What does it mean that the coefficients are divergent?

答:
Without looking at the actual form of your specific coefficients, the fact that the coefficients of a power series go to infinity is not a "curse" — it is possible, albeit it implies that the radius of convergence will be at most $1$ (if it is $0$, then it is a problem).

For instance, consider $a_n = n$: the radius of convergence is $R=1$, as $\sum_{n=1}^\infty nx^n$ is well-defined for all $|x|<1$ (a characterization is actually $R=\sup \{r \geq 0: a_n r^n \to 0\}$).

Even more striking, $a_n = 2^n$: the radius of convergence is $R=\frac{1}{2}$.


答:
In fact, we have: For any Borel functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$ , if $X,Y$ are independent, then $f(X),g(Y)$ are
also independent.

The proof is just few lines: Observe that $\sigma(f(X)):=\{f(X)^{-1}(B)\mid B\in\mathcal{B}(\mathbb{R})\}=\{X^{-1}(f^{-1}(B))\mid B\in\mathcal{B}(\mathbb{R})\}\subseteq\sigma(X)$.
Similary $\sigma(g(Y))\subseteq\sigma(Y)$. It is given that for any
$A\in\sigma(X)$ and $B\in\sigma(Y)$, $P(A\cap B)=P(A)P(B)$. Therefore,
it is trivial that for any $A\in\sigma(f(X))$ and $B\in\sigma(f(Y))$,
$P(A\cap B)=P(A)P(B)$.

Body:
If $X$ and $Y$ are independent random variables, are $X^2$ and $Y^2$ also independent?

Can this be proven?


问:
Energy norm/ Proof of the axiom of the norm
Let $u\in H^1_0(a,b)$(Sobolev space), $p\in C^1[a,b], r\in C[a,b]$, with the latter two being strictly positive.

It is known that: $\Vert u \Vert_{\mathscr A } = \left[\mathscr{A}(u,u)\right]^{\frac{1}{2}}=\int_a^b\left[ p(x)\left(\frac{\partial u}{\partial x}\right)^2 + r(x)\left(u(x) \right)^2\,dx  \right]^{\frac{1}{2}}$. This is the definition of the energy norm.

I would like to prove that:
If $\Vert u \Vert_{\mathscr A } = 0\Rightarrow u = 0 $ in $H^1_0(a,b)$

答: If $\|u\|_\mathscr{A}=0$, we must have that the integrand $(p(u')^2+ru^2)^{1/2}=0$ almost everywhere. This further implies that $p(u')^2+ru^2=0$ a.e. Since $p,r>0,$ we must have that each individual term in that sum is $0$ a.e., and in particular $ru^2=0$ a.e. Once again referring to the sign of $r$, we conclude finally that $u=0$ a.e.


答:
For $x>1$
$$x\ln x -(x-1)\ln(x+1)
=\int_1^x \int_t^\infty \frac{u-1}{u(u+1)^2}\>dudt\ge0$$
Rearrange the inequality to obtain

$$\frac{\ln({x+1})}{\ln x} \leq \frac{x}{x-1}$$


答:
The technical way to prove this...

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and suppose $X, Y: (\Omega, \mathcal{F}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$, where $\mathcal{B}(\mathbb{R})$ is the collection of Borel sets of $\mathbb{R}$, and $(\Omega, \mathcal{F})$ denotes the measurable space consisting of the sample space $\Omega$ with a $\sigma$-algebra of subsets of $\Omega$ given by $\mathcal{F}$.

Let $g: \mathbb{R} \to \mathbb{R}$ be given by $g(x) = x^2$. Because $g$ is continuous over $\mathbb{R}$, we have that $g: (\mathbb{R}, \mathcal{B}(\mathbb{R})) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$.

Thus, for any $A, B \in \mathcal{B}(\mathbb{R})$, $g^{-1}(A) \in \mathcal{B}(\mathbb{R})$ and $g^{-1}(B) \in \mathcal{B}(\mathbb{R})$. By definition of $X$ and $Y$ being independent, it follows that $X^{-1}(g^{-1}(A))$ and $Y^{-1}(g^{-1}(B))$ are independent. Thus, $(g \circ X)^{-1}(A)$ and $(g \circ Y)^{-1}(B)$ are independent for any $A, B \in \mathcal{B}(\mathbb{R})$.

It follows that $g(X) = X^2$ and $g(Y) = Y^2$ are independent random variables.

Title: If $X$ and $Y$ are independent random variables, are $X^2$ and $Y^2$ also independent?