问:
Total derivative of function with not continuous derivatives
I was wondering if the following function is totally differentiable in $(0, 0)$:
$$f(x, y) = \frac{x^2y}{x^2+y^2}$$ and $f(0,0) = 0$. I know that the partial derivatives are
$$f_x(x, y)= \frac{2xy^3}{(x^2+y^2)y^2}, \quad f_y(x, y) = \frac{x^4-x^2y^2}{(x^2+y^2)^2}.$$
Is it correct that both of these partial derivatives are not continuous in $(0,0)$? For $f_x(x,y)$, one can consider the limits $\lim_{h\to 0}f_x(h, h)=1/2$ and $\lim_{h\to 0}f_x(h, 0) = 0$, so this derivative can't be continuous, similarly for $f_y(x,y)$, which also can't be continuous, and in summary, $f$ can't be totally differentiable in $(0,0)$.

All I'm wondering is if my thoughts are correct here.
Thanks!

答:
We should find the value of partial derivatives at the point by the definition (via limits).

We have $f(x,0) = 0$ and $f(0,0) = 0$. Then,
$$f'_x(0,0) = \lim_{x\to 0}\frac{f(x,0) - f(0,0)}{x-0} = 0$$
Similarly, we get
$$f'_y(0,0) = \lim_{y\to 0}\frac{f(0,y) - f(0,0)}{y-0} = 0$$
Hence we need to check if the following equality is true:
$$\lim_{x\to0 \\ y\to0} \frac{f(x,y) - f(0,0) - 0 - 0}{\sqrt{x^2 + y^2}} \overset ? = 0$$
which is equivalent to
$$\lim_{x\to0 \\ y\to0} \frac{\frac{x^2y}{x^2+y^2}}{\sqrt{x^2+y^2}}= \lim_{x\to0 \\ y\to0} \frac{x^2y}{(x^2+y^2)^{3/2}} \overset ?= 0$$
Can you take it from here?


问:
$a+bi\mid n \implies a^2 + b^2\mid n$ if $(a,b)=1$
This seems very simple, but I have not been able to come up with an answer to it.

Suppose $a+bi \mid n$ in $\mathbb{Z}[i]$, where $a,b \in \mathbb{Z}$ with $(a,b)=1$. Prove that $a^2 + b^2 \mid n$ in $\mathbb{Z}$.

My attempt was: $a+bi \mid n \implies \overline{a+bi}\mid n$ (otherwise $n$ is not an integer), and if we have that $(a+bi,a-bi)=1$ (or a unit) in $\mathbb{Z}[i]$, then by unique factorization this implies that $(a+bi)(a-bi)\mid n$. However, I found it difficult to prove $(a,b)=1$ in $\mathbb{Z} \implies$ $(a+bi,a-bi)=1$ in $\mathbb{Z}[i]$.

Does anyone have a nice solution to this? Any solution is welcome.

For additional context, this question came up as I was trying to prove that $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$ when $a,b$ are coprime in the integers. A way of doing this using natural injections is given in Quotient ring of Gaussian integers, where there are many very good answers. One way is to use the first isomorphism theorem with the canonical map $\mathbb Z \to \mathbb Z[i]/(a - i b)$, and I need to prove that the kernel of that map is $\langle a^2+b^2 \rangle$. In every answer I saw on this, the solution skips over that step, making me suspect that there's something very simple I'm missing.

答: Some $c,\,d\in\Bbb Z$ satisfy $(a+bi)(c+di)=n$. For this to be real, $c+di=k(a-bi)$ for some $k\in\Bbb Q$, whence $k(a^2+b^2)=n$. Now we just need $k\in\Bbb Z$, which follows from $(a,\,b)=1$.

ContentLicense: CC BY-SA 4.0


问:
simple probability question picking a pen out of 4 and out of 3 colors
I have this simple problem which I cannot seem to solve , it is in an introduction to probability chapter.

so we have 4 pens , and 3 colors.. 2 blue , 1 red 1 black we pick them randomly one by one and with no return until we get a black pen EDIT- the question is that we need to write the probability mass function (the simple ones like in those tables) like find all the probabilities for $P(X=i,Y=j)$ hope it is understandable it is hard to translate this

X - counts the number of pens we picked , Y - counts the number of colors

so since we have 4 pens $X∈{{(1,2,3,4)}}$ and 3 colors so $Y∈{(1,2,3)}$

$P(X=1,Y=2)=P(X=1,Y=3)=P(X=2,Y=1)=P(X=2,Y=3)=P(X=3,Y=1)=P(X=4,Y=2)=P(X=4,Y=1)=0$

because these cannot happen.

now I think that I should use Hypergeometric distribution but it only works on some of them..

works on $P(X=3,Y=3)=$ $\frac {{1 \choose 1}*{1 \choose 1}*{2 \choose 1}} {{4 \choose 1}{3 \choose1}}$
I will get $\frac 2{12}$  which is right according to the book

${1 \choose 1}$ for 1 black pen out of 1 , ${1 \choose 1}$ 1 red pen out of 1 , ${2 \choose 1}$ 1 blue out of 2 , and for ${4 \choose 1}{3 \choose1}$ its because we need 1 pen out of 4 and 1 color out of 3 but I am not certain if it worked randomly or not

$P(X=2,Y=2)=$ $\frac {{1 \choose 1}*{1 \choose 1}+{1 \choose 1}*{2 \choose 1}} {{4 \choose 1}{3 \choose1}}$I get $\frac 3{12}$ which is also right according to the book

${1 \choose 1}*{1 \choose 1}$ since we need 2 pens and 2 colors we can have 2 options or 1 black and 1 red which is this option ,and I added ${1 \choose 1}*{2 \choose 1}$ because it can be one of the blue pens.

but this method doesnt work for (atleast for what I tried) on $P(X=1,Y=1)$ , $P(X=3,Y=2)$ and for $P(X=4,Y=3)$.

appreciate any help and tips, please the explanation is more important than actually solving this is just from a random practice book I have and I know that the question is probably very simple for most of the people here but I am new to this topic so I would like to know the approach to this question more than the final answer..
*** the answers for what is left : $P(X=1,Y=1)= \frac 1{4}$ , $P(X=3,Y=2)= \frac 1{12}$  and $P(X=4,Y=3)= \frac 1{4}$
thank you.

EDIT2 - by the end of every $P(X=i,Y=j)$ we will always have a black pen that is why some of them are $P(X=x,Y=y)=0$

EDIT3 - to make it clear for the $P(X=i,Y=j)$ as stated in the question X tells us how many pens were picked until we got the black one and Y is the number of colors picked so for example $P(X=1,Y=1)$ means 1 black pen and 1 color $P(X=2,Y=2)$ means 1 black pen and 1 more pen which we don't know its color , but we do know that we have 2 colors (Y=2 so 1 black and 1 can be red or blue)

答:
Here is a way to approach it. I will take a few cases to explain.

First $P(X = 3, Y = 3)$. So the third pen is black and you get pens of all $3$ colors in $3$ picks. I am denoting Black with capital $B$ and Blue with small $b$. So the possibilities are $(R, b, B)$ and $(b,R,B)$.

$P(X=3,Y=3) = \frac{1}{4} \times \frac{2}{3} \times \frac{1}{2} + \frac{2}{4} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$

or you could write it as $P(X=3,Y=3) = \displaystyle \frac{{2 \choose 1}{1 \choose 1}}{4 \choose 2} \times \frac{{1 \choose 1}}{2 \choose 1}$

Not let's take the case of $P(X=1,Y=1)$ that you were getting wrong answer for. The only way to get it is the first one is black out of $4$ pens.

So $P(X=1,Y=1) = \frac{1}{4}$

Next take $P(X=3,Y=2)$. What are the possibilities? Only $(b,b,B)$

So $P(X=3,Y=2) = \frac{2}{4} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{12}$

or you could write it as $P(X=3,Y=2) = \displaystyle \frac{2 \choose 2}{4 \choose 2} \times \frac{1 \choose 1}{2 \choose 1}$

Now for $P(X=4,Y=3)$, possibilities are $(b,b,R,B), (b,R,b,B), (R,b,b,B)$.

So $P(X=4,Y=3) = \displaystyle \frac{3 \choose 3}{4 \choose 3} \times \frac{1 \choose 1}{1 \choose 1} = \frac{1}{4}$.


问:
Geometrical Interpertation of Cauchy's Mean Value Theorem
Cauchy MVT: If functions f and g are both continuous on the closed interval [a,b], and differentiable on the open interval (a, b), then there exists some c ∈ (a,b), such that

$$\frac{f'(c)}{g'(c)}= \frac{f(b) - f(a)}{g(b)-g(a)}$$

Lately, after I proved the CMVT, I was trying to intuitively understand (using geometry of course) the meaning of CMVT by comparing it to the MVT; for, I know that the CMVT is just an extension of the MVT, such that the only difference is that $$g(x)=x$$ for the MVT. Yet, even though this is self-evident, I could not inhabit an intuition through a geometric representation of it; for, in all demonstrations of the MVT I have viewed, I only see one function $$f(x)$$ in the geometrical representation, which thusly implies there exists no $$g(x)$$ - not to my perspective at least.So, I searched for another demonstration of the CMVT and I found something related to parametric curves as follows:

And,  since I have not encountered parametric curves, I could not fully grasp this demonstration. So, can anyone help me with my confusions over a graphical demonstration of the CMVT. (Note: If a explanation of this intuition requires a knowledge of parametric curves etc..., feel free to include it in the answer).

答:
This likely won't be helpful to someone who's not familiar with parametric curves, but it did help me improve my geometric understanding of the Cauchy MVT.

In the wiki article on the Cauchy MVT, $h(x)=f(x)-rg(x)$ is defined so that $h(b) = h(a)$, so that Rolle's theorem can be applied to $h$.  Since I like linear algebra, it helped me to realize that they choose $r$ such that the vector $\left[-r, 1\right]$ is orthogonal to the vector $\left[g(b), f(b)\right] - \left[g(a), f(a)\right]$, which is the vector between the two endpoints.

The, you can think of $h$ as a dot product, $h(x) = \left[g(x), f(x)\right] \cdot \left[-r, 1\right]$.  So this is proportional to the projection onto $\left[-r, 1\right]$, which is the component of $\left[g(x), f(x)\right]$ orthogonal to the vector between the two endpoints -- i.e., the "distance away from" the vector between the endpoints.

From Rolle's theorem we know that $h'(x)$ is 0 at some point $c$.  This corresponds to the distance of $\left[g(x), f(x)\right]$ from the vector between the endpoints being stationary.  This happens when the (vector-valued) derivative is either the 0 vector, or is orthogonal to $\left[-r, 1\right]$.

ContentLicense: CC BY-SA 3.0


问:
Does $f, f' \in L^1([0, \infty))$ imply that $\lim_{x \to \infty} xf(x) = 0$?
Does $\int_0^\infty |f(x)| \, dx$ and $\int_0^\infty |f'(x)| \, dx$ being finite imply that $\lim_{x \to \infty} xf(x) = 0$? </p>

<p>(Context: I am working through an analytic number theory textbook. In a lemma, the book assumes the conditions on $f$ and $f'$. The term $xf(x)$ appears from an integration by parts, but is tossed out, which must mean that $\lim_{x\to\infty} xf(x) = 0$. However, I've tried to prove this and was not successful. I also cannot think of any counterexamples.)</p>

答:
To estimate $xf(x)$ we consider $$\int^{a}_{0} f(x)dx=xf(x)|^{a}_{0}-\int^{a}_{0}xf'(x)$$ We know the first term is finite since $f\in L^{1}$, so if $\lim_{a\rightarrow \infty}af(a)=0$ we must require $\int^{\infty}_{0}xf'(x)$ be finite as well. To be more precise we need $$\int^{\infty}_{0}f(x)dx=-\int^{\infty}_{0}xf'(x)$$because the middle term goes to $0$. </p>

<p>However, it is not clear to me how to find a counter example using the above criterion. The condition that $xf'(x)$ must be integrable is a quite strong one, but seems to be okay for all elementary test functions I come up with. On the other hand I also do not know how to prove it using $f,f\in L^{1}$. Hope this 25 cents might be helpful. </p>

Body:
To estimate $xf(x)$ we consider $$\int^{a}_{0} f(x)dx=xf(x)|^{a}_{0}-\int^{a}_{0}xf'(x)$$ We know the first term is finite since $f\in L^{1}$, so if $\lim_{a\rightarrow \infty}af(a)=0$ we must require $\int^{\infty}_{0}xf'(x)$ be finite as well. To be more precise we need $$\int^{\infty}_{0}f(x)dx=-\int^{\infty}_{0}xf'(x)$$because the middle term goes to $0$.

However, it is not clear to me how to find a counter example using the above criterion. The condition that $xf'(x)$ must be integrable is a quite strong one, but seems to be okay for all elementary test functions I come up with. On the other hand I also do not know how to prove it using $f,f\in L^{1}$. Hope this 25 cents might be helpful.


问:
How do unitary preserve inner products? Why is $\langle U x, U y \rangle = \langle x, U^\dagger U y \rangle$?
On many links online [1], [2], [3], etc. they mention the following:

$U$ is a unitary matrix, i.e. $UU^\dagger = U^\dagger U = I$, where $U^\dagger$ is the Hermitian conjugate of $U$.

Then,
$$\langle U x, U y \rangle = \langle x, U^\dagger U y \rangle$$

Why is this so?

答:
Since $\langle u,v\rangle=u^\dagger v$, we generally have
$$\langle Tu,\,v\rangle=(Tu)^\dagger v=u^\dagger T^\dagger v=\langle u,\,T^\dagger v\rangle$$
for any vectors $u,v$ and any linear map $T$.


问:
How are the hyperoperations defined?
For the first few operations such as addition and multiplication they follow rules such as
$$A(x,y+1)=A(x,y)+1$$
$$A(x,0)=x$$
For multiplication
$$M(x,y+1)=M(x,y)+x$$
$$M(x,1)=x$$
So for a general hyper operation you would think that
$$\Pi_n(x,y+1)=\Pi_{n-1}(\Pi_n(x,y),x)$$
But then this doesnt work for tetration because
$$\Pi_4(x,y+1)=\Pi_{3}(\Pi_4(x,y),x)$$
$$^{y+1}x =(^{y}x)^x$$
But this implies
$$^{y}x =(x)^{x^y}$$
Because exponents multiply
When it should be
$$^{y}x =x^{x^{x^{x^{x^{{...}}}}}}$$
What should be the correct formula for the nth hyperoperation.
Thanks.

答:
What should be the correct formula for the nth hyperoperation?

Well, there is no answer because both of them are currently called Hyperoperations.
Your definition gives what are called lower-hyperoperations.

The recursive relation for lower h.os is the one you give:
$$h_{n+1}(x,y+1)=h_n(h_{n+1}(x,y),x)$$

As you can read on Wikipedia, this is only one of the possible definitions and probably not the most natural one.

The "natural hyperoperations", as I call them, or Goodstein's hyperoperations satisfy instead this relation:
$$H_{n+1}(x,y+1)=H_n(x,H_{n+1}(x,y))$$

Goodstein himself, who proposed the term tetration, pentation , ecc, defines them in this way in R. L. Goodstein (Dec 1947). "Transfinite Ordinals in Recursive Number Theory". Journal of Symbolic Logic. 12 (4): 123–129.

It's easy to prove that both definitions agree on the first 3 operations, and diverges at the fourth.

Is this the correct definition? We can't know unless we specify what
we mean with correct. It's not the oldest definition. For sure it's
the most natural and, I claim, the most fundamental one.

Title: How are the hyperoperations defined?


问:
Variance of AR(2) stationary process
Given $AR(2)$ stationary process
$$ y_t = 2 + 0.6y_{t-1} - 0.08y_{t-2} +u_t$$
where $u_t$ white noise from $N(0,4)$

Find $Var(y_t)$

My problem:

When I take the variances of left and right I have a kind of recursive covariance of $Cov(y_{t-1}, y_{t-2})$

答:
Indeed, you will have two unknown variables, so you need to write two equations.

Let $Cov(y_t, y_{t+k}) = \gamma_k$.

Then $$Var(y_t) = \gamma_0 = 0.6^2 Var(y_{t-1}) + 0.08^2 Var(y_{t-2}) + 2\cdot 0.6 \cdot 0.08 Cov(y_{t-1}, y_{t-2}) = $$ $$ = 0.6^2 \gamma_0  + 0.08^2 \gamma_0  + 2\cdot 0.6 \cdot 0.08 \gamma_1 $$
$$ Cov(y_t, y_{t+1}) = \gamma_1 = 0.021\gamma_0  - 0.0018 \gamma_1 - 0.014 \gamma_1 - 0.00012 \gamma_0 $$
After solving the system of two equations you should obtain $\gamma_0, \gamma_1$.

Note that $\gamma_0 = Cov(y_t, y_{t+0}) = Var(y_t)$

Title: Variance of AR(2) stationary process


问:
AXB = 0. A, B positive semi-definite. X in range of A $\implies$ X = 0?
I am reading a paper concerning the equation AXB = 0. A and B are positive semi-definite. The authors state that the columns of X belong to the range of A and deduce that X = 0. I am struggling to see how this was deduced, could anyone provide any hints?

Edit: A $\neq$ 0, B$\neq$ 0. Furthermore, $X$ can be expressed as $X = S + U$ where the columns of U belong to the null space of $A$.

Edit: The link to the paper is:

https://www.researchgate.net/publication/221345012_Learning_Output_Kernels_with_Block_Coordinate_Descent.

I am specifically concerned with the proof of Theorem 3.3.

答: This is false. Consider e.g. $A=X=I$ and $B=0$. There are perhaps some missing conditions that you have overlooked.


问:
Suppose that $G$ is nonabelian. Must $|\mathrm{Out}(G)| = |\mathrm{Aut}(G)|/|\mathrm{Inn}(G)|$?
I would say not.  Suppose that $G = D_4$.  By the article I found, $D_4$ has 4 outer automorphisms.  I understand how reflection comes to play, but $|\mathrm{Out}(G)| \neq |\mathrm{Aut}(G)|/|\mathrm{Inn}(G)|$.  Is this true?</p>

答: $\operatorname{Out}(G)$ is defined as the quotient $\operatorname{Aut}(G)/\operatorname{Inn}(G)$, so $$\left|\operatorname{Out}(G)\right|=\left[\operatorname{Aut}(G):\operatorname{Inn}(G)\right]=\frac{\left|\operatorname{Aut}(G)\right|}{\left|\operatorname{Inn}(G)\right|}$$ no matter what $G$ is.  Your confusion seems to be that you think that $\operatorname{Out}(G)$ denotes the set of otuer automorphisms of $G$, which is not true.


问:
I need a function whose integral ratio for f(2x)/f(x) remains stable
This is a very difficult problem to describe, but in essence its very simple, so I'll make an example. If we take the function

$f(x)=x$

its integral (as function of how far we want to take the integral) is

$g(a)=\int_{0}^{a} x \,dx= \frac{a^2}{2} $

now we want to test the following:

If we take the ratio of g(2x)/g(x), will it be the same ratio for every x?

so basically:

$\frac{g(2x)}{g(x)}=\frac{\frac{(2a)^2}{2}}{\frac{a^2}{2}}=\frac{(2a)^2}{a^2}=\frac{4*a^2}{a2}=4$

Which means no matter what x we plug into $g(2x)/g(x)$, it will be 4. But what if I want to replace $f(x)=x$ with another function, for a specific value of g(2x)/g(x), like for example 3? I've tried using $f(x)=bx$, but no matter which b I plug in, it will always be 4 (because of the square cube law). And if I use some other formula $g(2x)/g(x)$ will become unstable (different ratios for different x), So heres my question, what is the simplest formula I could use instead of $f(x)=x$ that would remain stable and would allow me to change $g(2x)/g(x)$?

答:
You could consider a function of the form

$$f(x)=x^b,\, b \neq -1.$$

This gives

$$g(x)=\frac{x^{b+1}}{b+1}$$

and

$$\frac{g(2x)}{g(x)}=\frac{2^{b+1}x^{b+1}}{b+1}/\frac{x^{b+1}}{b+1}=2^{b+1}.$$

That means you have to choose $b=\log_2 r -1$ to get a desired ratio $r=\frac{g(2x)}{g(x)}$.

This gives you a solution for any positive $r$, for $r=4$ you get $b=1$, as your own example shows.

Now, $f(x)=x^b$ will generally not be defined for non-positive $x$ if $b$ is not an integer. If this is a problem for you, you can simply define $f(-x) = f(x)\; \forall x \ge 0$ to get  function that is defined on all the real numbers.

$r=0$ is obviously impossible.

For negative $r$ I'm not sure if such a function can be found.


问:
Diagonalizability in $\mathbb{R}$ and $\mathbb{C}$
Give an example of a matrix $A\in M_{n\times n}(\mathbb{R})$ that is not diagonalizable, but A is diagonalizable viewed as a matrix over the field of complex numbers $\mathbb{C}.$</p>

答: The characteristic polynomial $x^2+1$ of $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is irreducible over $\mathbb R[x]$ but not over $\mathbb C[x].$</p>


问:
$\lim_{x\to{0^+}}{\sin(\log(x))}$
How can I solve this: $\lim_{x\to{0^+}}{\sin(\log(x))}$ ?

Since $\log(x)\to -\infty$ I don’t know how to evaluate the limit of $\sin(x)$

答: The sequences $x_n=e^{-(2n+1/2)\pi},\,x_n=e^{-(2n-1/2)\pi}$ have respective $x\to0^+$ limits for $\sin\ln x$ of $-1,\,1$, so the proposed limit doesn't exist.

Title: $\lim_{x\to{0^+}}{\sin(\log(x))}$


问:
if function is continuous and real she must be Inverse?
$R=${$f:[0,1]-> \Bbb R$}

R is all the real and continuous functions from [0,1] to $R$,i want to check if there is a additive inverse number relative to +,so is it possible to say that $f(x)+f(x)^-=0\in R$?

Maybe I did not explain myself correctly,I'm trying to prove:

$R=${$f:[0,1]-> \Bbb R$}

(all the continuous and real function )is ring

So the first step is prove $(R,+)$ is abelian group

And that's why I have to prove $f(x)+fx)^-=0$
to get additive inverse number.

答:
In fact $(R,+)$ is abelian group.

First you should prove

$(i)\forall f,g\in R$, $f+g\in R$

Given $f,g\in R$ then $f+g:[0,1] \to \mathbb{R}$ and since $f,g$ are real then the sum too is.Since $f,g$ are continuous the sum is continuous (is a easy result from calculus )
Then $f+g\in R$

$(ii)\forall f\in R \exists -f\in R:f+(-f)=0$

Given $f\in R$ exists $-f\in R$( because if $f$ is continuous then $-f$ is continuous)where $-f$ is such that
$f+(-f)=0$

$(iii)\text{Exists a function $h$ such that for every function $f\in R$, $f+h=f$} $

Since zero function is continous and real then we denote by $0$ the zero function
$0=0(x)$ is such that for any $f\in R$ $(f+0)(x)=f(x)+0=f(x)=f$

$(iv)\text{$\forall f,g\in R$, $f+g=g+f$}$

Since the set of functions under the sum is connmutative then
any subset too is. Since the set of continuous and real functions is a subset too is connmutative.

$(v) \forall f,g,h\in R, f+(g+h)=(f+g)+h$

It holds since the set of functions under the sum is connmutative and since $R$ is a subset then $R$ inherit that propery

Then $(R,+)$ is abelian group


问:
Does projection operation preserve perpendicularity?
Let $C$ be a nonempty convex closed subset of $R^n$, and $\pi_C$ be an orthogonal projection mapping from $R^n$ onto $C$. Suppose that $y \in R^n$ is perpendicular to $x \in R^n$. Is it possible to prove/disprove that $y$ is also perpendicular to $\pi_C (x)$ (i.e. the orthogonal projection of $x$ onto $C$)?

答:
No, for example, take standard projections along last coordinate, that is $$P(x) = (x_{1},...,x_{n-1},0)$$
where $x=(x_{1},x_{2},...,x_{n})$.

If x and y are perpendicular then $\sum_{k=1}^{n} x_{k}y_{k} = 0$ and this does not imply that $\sum_{k=1}^{n-1} x_{k}y_{k} = 0$


问:
Taylor series remainder term to yield $\ln4$ with error terms less than $2^{-10}$
I need to derive the taylor series for $\ln(1+x)$ about $1$ then create an inequality that gives the number of terms that must be taken to yield $\ln4$ with error terms less that $2^{-10}$. I got the taylor series to be $$f(x) = ln(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-1)^n}{n \cdot 2^n}$$ I was wondering if someone could help me with the the error term, I think the answer is suppose to be $n > 2^{10}-1$ but I cant seem to figure out how to get there. thanks

答:
The remainder term for the partial sum of the first $n$ terms of
$$
\ln(1+x)=\sum_{k=1}^\infty\frac{(-x)^k}{k},
$$
apart from the use of advanced Taylor remainder formulas, can be bounded by a geometric series as
$$
\left|\sum_{k=n+1}^\infty\frac{(-x)^k}{k}\right|\le\frac{|x|^{n+1}}{n+1}\frac1{1-|x|}
$$

This obviously does not work for $x=1$.

Using the formula $\ln4 = -2\ln(1-\frac12)$ with $x=-\frac12$ would require
$$
\frac{2^{1-n}}{n+1}\le 2^{-10}\iff 2^{11}\le(n+1)2^n
$$
which is surely the case for $n=11$, but remains true down to $n=8$.

Using the formula $\ln4=2\bigl(\ln( 1+\frac13)-\ln(1-\frac13)\bigr)$ with $x=\pm\frac13$ naively would require half the error for each series, thus
$$
\frac{3^{-n}}{2(n+1)}\le 2^{-11}\iff 2^{10}=2·8^{3}\le(n+1)9^{n/2}
$$
which is surely given for $n=6$, but also works for $n=5$. Note that both series have the same terms up to the sign.

In fact, this last formula has cancelling terms, so the series difference can be written as
$$
\ln 4=2\ln2 = \sum_{k=0}^\infty \frac4{(2k+1)3^{2k+1}}
$$
so that the remainder for the partial sum of the first $n$ terms has the bound
$$
\left|\sum_{k=n}^\infty \frac4{(2k+1)3^{2k+1}}\right|
\le\frac4{(2n+1)3^{2n+1}}\frac98=\frac1{2(2n+1)3^{2n-1}}
$$
So for the demanded accuracy
$$
\frac1{2(2n+1)3^{2n-1}}\le 2^{-10}\iff 3·8^3\le (2n+1)9^{n}
$$
which is satisfied for $n=3$.
$$
\ln 4\approx\frac43\left(1+\frac1{3·9}+\frac1{5·9^2}\right)=\frac43·\frac{421}{405}=1.386008....
$$
compared to $\ln4=1.386294...$


问:
Hermiticity/self-adjointness of the Laplacian operator
I'm trying to understand the concept of hermicity/self-adjointness but even after reading through some threads here, I just don't get it (first problem being that I don't understand the difference between the two, if there is any). But anyway, let's look e. g. at the one-dimensional eigenvalue problem of the Laplace operator $$\Delta f(x)=\frac{d^2}{dx^2} f(x)=cf(x)$$ where $c$ is a constant and the domain $\Omega$ for $x$ is e. g. $[x_0;x_1]$. From what I've read, the Laplace operator is self-adjoint/hermitian for boundary constraints $f(x_0)=f(x_1)=0$ (though I don't know why that is). Could someone expain why? (And maybe also showcase a set of BCs for which the Laplacian operator is not self-adjoint/hermitian?)

答:
The Laplacian is only partially defined on $L^2$ and it is unbounded. Usually, the Laplacian is defined on the dense subspace of compactly supported smooth functions and is extended to a self-adjoint closure (still partially defined only).

For a partially defined operator $A: D(A)\subseteq L^2 \to L^2$, it is self-adjoint by definition if and only if $D(A) = D(A^*)$ and for every $f,g\in D(A)$ it follows $$\int f Ag = \int Af g. $$

I will explain the symmetry for functions in one variable for sake of simplicity. Consider $\Delta: C_c^\infty(a, b) \subseteq L^2(a, b) \to L^2(a, b)$ defined by $\Delta f = f''$ and $f,g\in C_c^\infty(a, b)$. By integration by parts, it follows
$$ \langle f, \Delta g \rangle = \int_a^b f g'' = \underbrace{[f g']_a^b}_{=0} - \int_a^b f' g' = -\underbrace{[f' g]_a^b}_{=0} + \int_a^b f'' g = \langle \Delta f, g \rangle. $$
That is, $\Delta$ is symmetric, and thus, we have $D(\Delta) \subseteq D(\Delta^*)$. But, for example $D(\Delta^*)$ also contains $C^2(a,b)$ and is larger than $D(\Delta)$. Thus, $\Delta$ is not self-adjoint. However, as $-\Delta$ is a non-negative (or positive semidefinite), there are many ways to extend $-\Delta$ to a self-adjoint operator.


问:
Wrong/naive inductive hypothesis and tower of hanoi as an example
I am reading about how a wrong formulation of the tower of Hanoi and the inductive hypothesis can lead to a dead-end.

The example I am reading states the following:

The task is to move N discs from a specific pole to another
specific pole. Assume there are poles $A$, $B$ and $C$.   The base case is that when the number of discs is 0 then no steps are needed
to complete the tasks.      For the inductive step assume that we
can move $n$ discs from pole $A$ to pole $B$ and we are required to
show how to move $n + 1$ discs from $A$ to $B$

Then it highlights that this definition is a dead-end since the only 2 ways to use the induction hypothesis as set don't lead anywhere.

Specifically the options are:

Move the top $n$ discs from pole $A$ to $B$. After this point all possibilities of using the induction hypothesis have been exhausted
since $n$ discs are on pole $B$ and we do not have hypothesis about
moving discs from that pole.

Move the smallest disc from pole $A$ to $C$. Then move the remaining $n$ discs from $A$ to $B$. Once again we have exhausted all
possibilities of using the induction hypothesis, because $n$ discs are
now on pole $B$, and we have no hypothesis about moving discs from
this pole.

The reasoning for $1$ is clear to me. If we move the top $n$ discs i.e. the $n$ smaller discs to pole $B$ then at some point we would have to move them again since we would need to place the $nth + 1$ remaining largest disc bellow them. And there is no inductive hypothesis for that so I think I get it.

I need help with the second part. How I understand it is we move the smallest disc to pole $C$. Then we use the inductive hypothesis to move the remaining $n$ larger discs from pole $A$ to $B$. (At this part I am not sure how this could happen if pole $C$ is occupied by the smallest disc but I guess it is part of the logic of using induction in proof.)

Then at that state it seems to me that the only thing pending would be to move the smallest disc from $C$ to $B$ and finish the task.

Why does it state that we would need to move the $n$ discs from pole $B$ and that is not possible since we don't have an induction hypothesis?

Am I misunderstanding something about statements on induction here?

答:
You are right to be concerned about the second part.

Your statement of the only thing pending is correct. However, the second part is nonsense since the only sensible inductive hypothesis concerns 'moving n discs from pole A to pole B, assuming that 3 poles are available'.

An added observation

Looking at your queries in the various comments I feel it might be useful to point out that proof by induction and proof by minimal counterexample are logically equivalent but that some of the things you are questioning seem so much clearer w.r.t. the minimal counterexample method.

If you are trying to prove a result then

(1) Assume it is false.

(2) Consider a counterexample which in some well-defined sense is minimal.

(3) Then try to prove the result by whatever means you wish but where, whenever you need to, you can assume the result for anything 'smaller'.

As an example of the benefits of this way of thinking you can see from this that distinctions such as 'strong' and 'weak' induction are just a distracting irrelevance.


问:
$\mathcal{E}'(\mathbb{R^n}) \subset \mathcal{S}'(\mathbb{R^n})$?
Can I have pls the proof of why is the space of distribution with compact support $\mathcal{E}'(\mathbb{R^d})$ subset in the set of tempered distribution $\mathcal{S}'(\mathbb{R^n})$?

答:
By definition of $\mathcal{S}$ we have $\mathcal{S} \subset C^\infty = \mathcal{E}$.

Next, if $\varphi_k \to 0$ in $\mathcal{S}$ then by definition of convergence in $\mathcal{S}$, for every multi-index $\alpha$ and $\beta$, $\sup |x^\alpha \partial^\beta \varphi_k|\to 0.$ Especially, taking $\alpha=(0,\ldots,0),$ one has $\sup |\partial^\beta\varphi_k|\to 0$ for all multi-index $\beta$, which means that $\varphi_k\to 0$ in $\mathcal{E}.$ Thus, if $\varphi_k \to 0$ in $\mathcal{S}$ then $\varphi_k\to 0$ in $\mathcal{E}.$

Take $u\in\mathcal{E}'$ and $\varphi_k\to0$ in $\mathcal{S}$. Since $\mathcal{S}\subset\mathcal{E}=C^\infty,$ the pairing $\langle u, \varphi_k \rangle$ is defined. And since $\varphi_k\to 0$ in $\mathcal{S}$ implies that $\varphi_k\to 0$ in $\mathcal{E}$ we have $\langle u, \varphi_k \rangle \to 0$. Thus, $u$ is continuous also over $\mathcal{S}$. This shows that $\mathcal{E}' \subset \mathcal{S}'.$


问:
If a binary quadratic form primitively represents $n$ and $n^3$, must it be the identity form?
I've been searching numerically, and whenever a positive definite binary quadratic form can primitively represent an integer $n>1$ and $n^3$, it turns out to be equivalent to the identity form in that class group.

Is there some elementary way to see that this is only possible for identity forms?

Here is what initially gave me hope that it could be possible with other forms.

For any binary quadratic form $A x^2 + B xy + C y^2$, if it can represent integers $r$, $s$, $t$ then it can also represent $rst$.

This can be seen as follows:

For any form with coefficients $(A,B,C)$, in the class group its inverse is $(A,-B,C)$.

Both a form and its inverse can represent the same numbers (if $(x,y)$ with the former, use $(x,-y)$ with the later).

If a form can represent $r$ with $(x_1,y_1)$, and $s$ with $(x_2,y_2)$ then the identity form can represent $rs$ with some integers $(u,v)$ due to form composition:

$$
\begin{aligned}
rs &= (A x_1^2 + B x_1 y_1 + C y_1^2)(A x_2^2 + B x_2 y_2 + C y_2^2) \\
   &= (A x_1^2 + B x_1 y_1 + C y_1^2)(A x_2^2 - B x_2 (-y_2) + C (-y_2)^2) \\
   &= I(u,v) \\
\end{aligned}
$$

Futhermore, if the form can represent $t$ with $(x_3,y_3)$, then it can represent $rst$ by some integers $(x_4,y_4)$ due to form composition.

$$rst = I(u,v)(A x_3^2 + B x_3 y_3 + C y_3^2) = (A x_4^2 + B x_4 y_4 + C y_4^2)$$

It is definitely possible to primitively represent cubes, for example

```<code>a) 2 x^2 - xy + 6 y^2 = 3^3  with x=3,y=-1
b) 2 x^2 - xy + 6 y^2 = 8^3  with x=2,y=-9
</code>```

But in example 'a', it is not possible for that form to represent 3. While in example 'b', the form can represent 8, but not primitively (x=2,y=0).

This seems to be the case for any form not equivalent to an identity form.

While for an identity form, I've been able to find a lot of examples that work.

```<code>    x^2 + xy + 12 y^2 = 18     with x=2, y=1, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 18^3   with x=-76, y=7, gcd(x,y)=1

    x^2 + xy + 12 y^2 = 118    with x=2, y=3, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 118^3  with x=964, y=207, gcd(x,y)=1

    x^2 + xy + 12 y^2 = 63     with x=3, y=2, gcd(x,y)=1
    x^2 + xy + 12 y^2 = 63^3   with x=-501, y=2, gcd(x,y)=1
</code>```

Can one prove that if a form can primitively represent $n$ and $n^3$, that is must be equivalent to the identity form?

UPDATE:

A related question On products of ternary quadratic forms $\prod_{i=1}^3 (ax_i^2+by_i^2+cz_i^2) = ax_0^2+by_0^2+cz_0^2$ shows a possible solution for any form with B=0 (all that is left is to analyze if the result is a primitive representation). Such forms are in the same equivalence class as its inverse in the class group, and are called "ambiguous forms".

Below Will Jagy gave an explicit solution for an ambiguous form which is not equivalent to the identity form, showing that the result can be a primitive representation.

The connection to what I tried is that the identity is always an ambiguous form. So that likely refines the question to:

If a form can primitively represents $n$ and $n^3$, is the form necessarily an ambiguous form?

答:
Too big for a comment,

this is mostly observations, perhaps incorrect, that may help you find an answer that makes sense to you.
Perhaps I'm missing the point. I'm not sure what you mean by "identity form" even after trying to look it up.

If we allow  $B=2$, one of your "forms" appears to be

$$(x_1+y_1)^2(x_2 -y_2)^2=x_3^4-2x_3^2y_3^2+y_3^4=(x_3^2-y_3^2)^2
$$

Take a look at Euclid's formula where your $(x,y)$ are represented by $(m,k).\quad$

$$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$

The only restrictions are $m> k>0\quad GCD(m,k)=1\space$ for non-trivial primitives.
The $A$-value takes the form $(2n+3);\quad$ the $B$-value takes the form $4n;\quad $ the $C$ value takes the form $4n+1.\quad$ Any of these can be cubes. Even $C\in\{5^3, 13^3, 17^3, 53^3, 241^3,\cdots\}$

Your form appears to the square of side-$A$ of a Pythagorean triple. and there are easy ways of finding $m,k$ values if any of $A,B,C$ are given.


问:
Volume of a set in phase space. How many dimensions?
Suppose I have a $6N$ dimensional space with points looking like this:

$$(r_x^{(1)},r_y^{(1)},r_z^{(1)}, p_x^{(1)}, p_y^{(1)}, p_z^{(1)},...,r_x^{(N)},r_y^{(N)},r_z^{(N)}, p_x^{(N)}, p_y^{(N)}, p_z^{(N)})$$

This is a phase space for a system of $N$ particles. In this space, I define a set by the following conditions:

$$\sum_{k=1}^{N} \left( \left[ p_{x}^{(k)} \right] ^2 +\left[ p_{y}^{(k)}  \right]^2 +\left[ p_{z}^{(k)} \right]^2 \right) =const>0 .$$

$$0\le r_{x}^{(k)} \le l ; \quad 0\le r_{y}^{(k)} \le l ; \quad 0\le r_{z}^{(k)} \le l ; \quad k=1,2,...,N; \quad l=const>0.$$

What is the dimension of this object? How to calculate its "volume"? I was thinking that the first equation specifies a sphere in $3N$ dimensions, which itself is a $3N-1$ dimensional object, but now I'm confused on how to treat the remaining conditions.

PS. Would it be helpful to first consider a similar problem, but with the first condition replaced by

$$0<\sum_{k=1}^{N} \left( \left[ p_{x}^{(k)} \right] ^2 +\left[ p_{y}^{(k)}  \right]^2 +\left[ p_{z}^{(k)} \right]^2 \right) \le const?$$

I've seen such procedures done in some textbooks but I don't really understand why not work with the problem directly.

Update: What would be the units of this set be? I'm assuming the $r$'s have units of length and $p$'s of momentum.

答:
I think it'll be helpful to first think of 1 dimensional physical space which for your case of $N$ non interacting particles is actually $2N$ dimensions (position and momentum). In this space, the conditions read:
$$p_i = c_i,\ r_i \in[0,l]$$
So the dimension of this space is simply $N$ (since $p$ is constant i.e. zero dimensional). Now let's move to $2D$ physical space ($4N$ phase space). This time we have:
$$p_{i1}^2 + p_{i2}^2 = c_i^2,\ r_i \in[0,l]$$
We only have on constraint, so this is a $4N-1$ dimensional space. What is this space? if we consider one particle and ignore one of the $r_i$'s, this is a hollow cylinder in $3D$ space, embedded in a $4D$ space. Now let's return to physical $3D$ space, It should be obvious that the dimension of the space is $6N-1$.</p>


答: No. Consider $x=(1,1)$ and $y=(-1,1)$ which has $x \cdot y = 0$ and take $\pi_c(x)$ the projection onto the first coordinate. We can now see that $\pi_c(x)=(1,0)$ so $\pi_c(x) \cdot y = -1 \neq 0$.

Title: Does projection operation preserve perpendicularity?

Body: Let $C$ be a nonempty convex closed subset of $R^n$, and $\pi_C$ be an orthogonal projection mapping from $R^n$ onto $C$. Suppose that $y \in R^n$ is perpendicular to $x \in R^n$. Is it possible to prove/disprove that $y$ is also perpendicular to $\pi_C (x)$ (i.e. the orthogonal projection of $x$ onto $C$)?


问:
Prove that limit doesn't exists at all
I have a function $f(x)=\frac{1}{x}+\sin(\frac{1}{x})$ which I have proved that $\lim_{x \to0^+}f(x)=\infty$, and I need to prove that $\lim_{x \to0^+}f'(x) \ne -\infty$.

I have got that $f'(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$

I am trying to prove that:

$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$$

Does not equal to $-\infty$ (or to prove that it doesn't exists at all).

I have tried to assume by contradiction that
$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\infty$$

So there exists $\delta_{1}>0$ such that for every $0<x<\delta_{1}$ we say that $$f(x)<\square $$

We choose $\delta=\delta_{1}$, and let $0<x<\delta$ so:

$$f(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\frac{1}{x^2}(1+\cos(\frac{1}{x}))\le -\frac{2}{x^2}<\frac{2}{\delta^2}=k$$

I do not know what to choose instead of $\square$, and if what I did is correct.

Appreciate your help!

I have also tried to think about the continuity of $f$, and saying that it is not bounded so it can't be continuous, so the limit doesn't exists.

Thanks a lot!

答:
Part 1:

You can try making it a bit "easier" by substitution.

$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{cos(\frac{1}{x})}{x^2}\underset{t=\frac{1}{x}}{\Rightarrow}$$
$$\lim_{t\to\infty}-t^{2}\left(1+\cos t\right)$$
Now it should be easier to negate by definition, as you can easily see that no matter how big $t$ is, and how small $f'$ is for those big $t$'s, you can choose $t<t_1$ such that $f'(t_1)=0$

If you'll still be struggling, make sure to leave a comment and I will help you further.

Part 2:

Assuming, in contradiction that $\lim_{t\to\infty}-t^{2}\left(1+\cos t\right)=-\infty$

Then there exists $M_{1}<0$ such that for every $t<M_{1}$ we have:$$f\left(t\right)<-1$$
Choosing $M_{1}>t_{1}=\lfloor M_{1}\rfloor\cdot2\pi-\pi$ , we get: $$f\left(t\right)=0>-1$$ Contradiction!

Title: Prove that limit doesn't exists at all


问:
Lebesgue theorem - calculate the limit of the integral
$$\mbox{Let } f:X\to[0,\infty] \mbox{ and} \int fd\mu=c \mbox{ and we know that } 0<c<\infty \mbox{ and } 0<\alpha<\infty$$
$$\lim_{n\to\infty} {\int {n\ln\left[1+\left[{{f(x)}\over{n}}\right]^{\alpha}\right]}}d\mu(x) $$ The instruction says to consider cases separately. $$\mbox{ So I calculated that for } \alpha=1 \to \lim_{n\to\infty} {\int {n\ln\left[1+{{f(x)}\over{n}}\right]}}d\mu(x) = c$$ But I have a problem with calculating this integral for: $$0<\alpha<1 \mbox{ and } 1<\alpha<\infty$$Does anyone know how to solve this for the other two cases?

答:
You would want to apply the Dominated Convergence Theorem, in fact
$$n\ln{\left[1+{\frac{f(x)}{n}}\right]^{\alpha}} = \alpha n \ln\left[1+{\frac{f(x)}{n}}\right] \le \alpha n {\frac{f(x)}{n}} = \alpha f(x)$$
and it is integrable since
$$\int |\alpha f(x)|d\mu = \alpha \int f(x) = \alpha c < +\infty .$$

So now we apply the DCT to say
$$\lim_{n\to\infty} {\int {n\ln\left[1+{{f(x)}\over{n}}\right]^{\alpha}}}d\mu(x) = {\int \lim_{n\to\infty}{n\ln\left[1+{{f(x)}\over{n}}\right]^{\alpha}}}d\mu(x)$$$$ = {\int \lim_{n\to\infty}{\alpha \ln\left[1+{{f(x)}\over{n}}\right]^n dx= \int \alpha \ln\Big[e^{f(x)}}}\Big]d\mu(x) = $$
$$= \int \alpha f(x) dx = \alpha \int f(x) dx = \alpha c.$$


问:
Probability that $n$ random walks (1D) intersect a single point
I am struggling finding a relation to determine the probability $n$ random walks (1D) intersect in a single point at step $s$. In the method below my attempts. My method is somewhat intuitive based. I am looking for more rigorous proof.

note:
This question arises from someone who claims that a matching cumulative digit sum of: $\pi$, $e$ and $\varphi$ (golden ratio) is unique and "cosmological" [1]. I tend to disprove it. This digit sum can be seen as a random walk if the constants are normal (every digit occurs with same frequency).

Method:

For every step $s$ on the random walk we can determine the probability density function if we know the standard deviation on every step $s$.

The standard deviation of a single step can be calculated it's a discrete uniform distribution "equally likely outcomes", where $q$ is the number of outcomes e.g. the number of digits $[0,1,2,3,4,5,6,7,8,9]$, $q=10$:

$$\sigma=\sqrt{\frac{q^{2}-1}{12}} $$

All the (1D) random walks start in the origin for this example. The standard deviation will grow with every step $s$, the variance is proportional to the number of steps [2].

$$Var(s)=s \cdot \sigma^{2}$$

$$\sigma(s)=\sqrt{s} \cdot \sigma$$

While the bins grow rapid I assume a normal approximation of the Binomial distribution.

$$f(x)=\int_{-\infty}^{\infty} {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}}dx $$

The probability that $n$ random walks intercept in a single point is (not sure):

$$p(s)=\int_{-\infty}^{\infty} \left[ {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}} \right]^n dx $$

With help of Wolfram Alpha [3] the solution is found for $n=3$ meaning the probability of $3$ point intersecting random walks.

$$p(s)=\frac{1}{2 \sqrt{3} \ \pi \ \sigma^{2}} \cdot \frac{1}{s}$$

The total probability $p(n)$ is proportional to the reciprocal sum of $s$. So the total probability is proportional to the harmonic series:

$$p(n)=\frac{1}{2 \sqrt{3} \ \pi \ \sigma^{2}} \cdot \sum_{s=1}^{\infty}\frac{1}{s}$$

This series diverges, meaning there are infinate point intersections of $n=3$ random walks. So a matching cumulative digit sum of $\pi$, $e$ and $\varphi$ is not unique, probability $\sim 8 \%$ for the first 1200 digits (see graph).

Question

Does anyone know the general formula for the probability $p(n)$ that $n$ random walks (1D) intersect a single point?

```<code>import numpy as np

#Elements of digits [0,1,2,3,4,5,6,7,8,9] rescaled to fit random walk
array=[-9,-7,-5,-3,-1,1,3,5,7,9]

#steps, in single random walk, x walks to intercept, number of trial to find intercept
steps=2500
xwalks=3
trials=1500

#Set output array to zero
count=np.full([steps],0)

for n in range(trials):

    #Identify initial array, set total array to zero
    w0=np.random.choice(array,steps)
    w0=np.cumsum(w0)
    total=np.full([steps],0)

    #Select x random walk check for intercept
    for m in range(xwalks-1):

        #Next current random walk
        w=np.random.choice(array,steps)
        w=np.cumsum(w)

        #Compare previous and current random walk
        eq=np.equal(w0,w)
        eq=eq.astype(int)

        #Count intercepts
        total=total+eq

        #Set current walk to previous
        w0=w

    #Sum all interceptions for all trials
    count=count+np.where(total==(xwalks-1),1,0)

#Print output
print(count)
print(np.sum(count))
print(np.sum(count)/trials)
</code>```

答:
The probability $n$ random walks starting in origin (1D) intersect a single point can be calculated with:

$$p(s)=\int_{-\infty}^{\infty} \left[ {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}} \right]^n dx$$

Solutions have been found with Wolfram Alpha 1. This solution can be written as a function of the number of steps $s$. So the probability $p$ at $s$ steps can be calculated with a empirical found formula:

$$p(s)= \frac{1}{g(n) \cdot  \sigma ^{(n-1)}  } \cdot s^{-\frac{1}{2}(n-1)} $$

Where the function for $g(n)$ is found as:

$$g(n)=\sqrt { n} \cdot (2 \pi)^{\frac{1}{2}(n-1)}$$

The total probability over all steps $s$ is only dependent upon the $p$-series formed by $s$. This summation can be calculated with the Riemann zeta function:

$$\zeta(\small{\frac{1}{2}(n-1)} \normalsize) =\sum\limits_{s=1}^{\infty}s^{- \frac{1}{2}(n-1)}$$

$$p(n)=\frac{1}{g(n) \cdot \sigma^{(n-1)}} \cdot \sum\limits_{s=1}^{\infty}s^{- \frac{1}{2}(n-1)} $$

Graph information

Below a table and a plot for the probabilities of $p(n)$ point coinciding random walks (1D). Details:

The standard deviation is chosen: $10$ equal likely outcomes per step:
$$\sigma=\sqrt{\frac{99}{12}}$$

For infinate random walks $n\rightarrow \infty$ the $p$-series converges to $1$. This probability is plotted as: $p_{\infty}(n)$.

$p(n)$ can be graphed as an continuous function.

Values $n<3$ have not been graphed while Riemann zeta $\zeta(<1)$ are resulting in negative probabilities.

Observations:

$n=3$ random walks have infinate single point intersections. $p$-series: $\sum\limits_{s=1}^{\infty}s^{-1}=\infty$, Riemann zeta: $\zeta(1)=\infty$ both have infinity probability $p(n)\rightarrow \infty$.

$n=2$ random walks have infinate single point intersections for $p$-series: $\sum\limits_{s=1}^{\infty} s^{-\frac{1}{2}}=\infty$ , interpretation Riemann zeta is unclear: $\zeta(\frac{1}{2})=-1.46035...$ resulting in a negative probability $p(2)$.

It is possible that $n\geq4$ random walks never point intersect. For $n=4$ only $\sim 0.35 \%$ of them will point intersect (with 10 equal likely outcomes per step).

The probability $p(n)$ plot of $n$ random walks (1D) intersecting a single point is directly related to the $p$-series. A relation with Riemann zeta is unclear while the number of walks $n=a+ib$ would be a complex number.

I learned this has to do with: recurrence and transience 2.

Any more rigorous proofs or information is welcome please.


问:
Distributional derivative of a square integrable function and dual of Sobolev space
This is a simple question: Let $f\in L^2(\mathbb{R})$, $f'$ be its distributional derivative, then is $f'$an element of $H^{-1}(\mathbb{R})$, the dual of Sobolev space $H^1(\mathbb{R})$? Also, if I take an element of $H^{-1}(\mathbb{R})$, say $q$, then is it true that there exists an element $Q\in L^2(\mathbb{R})$ such that $Q'=q$ in distributional sense? Why?</p>

答:
I am gonna work with the fact that $H^{-1}(\mathbb{R})=(H^1(\mathbb{R}))^\star$ and $H^1(\mathbb{R})=H_0^1(\mathbb{R})$. By definition $$\langle f',\varphi\rangle=-\int_\mathbb{R} f\varphi',\ \forall\ \varphi\in C_c^\infty(\mathbb{R})$$

Because $C_c^\infty$ is dense in $H_0^1$ we can conclude by the last integral and the Lebesgue theorem that for $v\in H_0^1$, $\langle f',v\rangle$ is well defined by the formula  $$\langle f',v\rangle=\lim\langle f',\varphi_n\rangle$$
where $\varphi_n\in C_c^\infty$ converges to $v$ in $H_0^1$. Thereofore, $f'$ defines a bounded linear functional, i.e. $f'\in H^{-1}$.

For your second question, take a look in Brezis proposition 8.14.


问:
Cauchy-Schwarz Inequality with a twist
Let u and v be two vectors in $\mathbb{R}^2$ .

The Cauchy-Schwarz inequality states that $|u · v| ≤ |u||v|$

We are able to transform the above inequality so that it also shows us that

$|u+v|≤|u|+|v|$

But I cannot find a way to show that $\mathbf{|u-v|≤|u|+|v|}$

even though I now this has to be true. Any ideas?

答:
$$|u+v| \le |u| + |v|$$
is shown already, so let's use this.
$$|u-v| = |u +(-v)| \le |u| + |-v| = |u| + |v|$$
and we're done.

Title: Cauchy-Schwarz Inequality with a twist


问:
How does an absolute value operator interact with partial derivatives?
I wish to take the partial derivative with respect to $y$ of the following expression:

$$g(x,y)=\log{\left|\frac{\partial}{\partial x}f(x,y)\right|}$$

where $f(x,y):\mathbb{R}^2\rightarrow\mathbb{R}$ is some function which depends on $x$ and $y$, and $\frac{\partial}{\partial x}$ denotes the partial derivative with respect to $x$. How does the absolute value operator $|\cdot|$ interact with the partial derivative $\frac{\partial}{\partial y}$ if I want to evaluate $\frac{\partial}{\partial y}g(x,y)$?

I would greatly appreciate any suggestions or references on how to address such situations, also in more general cases.

答:
Assuming that $\frac{\partial }{\partial x}f(x,y)\ne 0$ everywhere, by the chain rule, you need the existence of partial derivative with respect to $y$ for the map
$$
y\mapsto |\frac{\partial }{\partial x}f(x,y)|
$$


问:
$I = \int_0^\infty y^2 [ (y^2 +1)/\sqrt{y^4 + 2 y^2 } - 1] dy$
I tried with contour integral:

$$I = \frac{1}{2}\int_{-\infty}^{\infty } dz z^2\left(\dfrac{z^2 +1}{\sqrt{z^4 + 2 z^2 }} - 1\right)$$

The contour can be deformed into the upper half plane. But there is a cut from $\sqrt{2} i $ up to $\infty i $.
The integral along the cut is no simpler than the original integral.

How to proceed? The answer is $\sqrt{2}/3$.

答:
Let $$I_n=\int_0^n \frac{y(y^2+1)}{\sqrt{y^2+2}}dy-\int_0^ny^2 dy $$

In the first integral, substitute $y^2+2 =t^2 \implies 2y dy =2t dt $.

$$I_n= \int_{\sqrt 2}^{\sqrt{n^2+2}}( t^2-1) dt -\int_0^n y^2 dy \\
=\frac{(n^2+2)^{3/2} -2\sqrt 2}{3}+\sqrt 2-\sqrt{n^2+2}-\frac{n^3}{3}$$

Since your integrand is even, $$I=\lim_{n\to \infty} I_n =\frac{\sqrt 2}{3} +\frac 13 \lim_{n\to\infty} (n^2+2)^{3/2} -n^3 -3\sqrt{n^2+2}\\
$$ All it remains now is to show that the latter limit is $0$. It can be written as $$(n^2-1)\sqrt{n^2+2} -n^3 \\ = (n^3-n) \sqrt{1+\frac{2}{n^2}} -n^3  \\ \to (n^3-n)\bigg( 1+\frac{1}{n^2} +\text{smaller terms}\bigg) -n^3 \\ =n^3-n+n-n^3 +\frac 1n \to 0$$


问:
Why second kind radial Mathieu function is not included in the elliptical membrane problem?
I am reading this paper:
it solves a vibrational elliptical membrane problem.

In equation 13 and 14, they assumed a solution without the second kind modified Mathieu function:

their explanation is:

Second  radial $Ne_m$ and $No_m$ functions are  excluded  because  the  wave solutions
  and their derivatives must to be continuous at $\xi=0$. See figure:

I don't quite understand this interpretation, why continuous at $\xi=0$ need to exclude the second kind, how can I see it from the graph? Please give me a clear and simple explanation...

If it is a circular membrane, the radial solution will be the Bessel function of first kind, and the second kind is excluded because it diverges at origin. This is very intuitive and understandable. The analogy to the elliptical case can't holds, because from the graph, it is finite value.

答:
It is somewhat technical, and due to elliptic cylindrical coordinates system, in which occurs a singularity in the interfocal line  (between the two focus of the ellipse) for the gradient along radial as well as angular direction. If the domain includes this interfocal-line then solution needs to be of the form of a product of a radial first-kind and angular first kind, other solutions will exhibit a discontinuity in derivatives when crossing the interfocal line. I name this condition, a regular gradient condition anywhere in the space of coordinate system. This kind of condition will appear whenever the gradient show a singularity in the space of coordinates

Henri Lévêque


问:
Work to provide explanation on the definition of the area of a Jordan-measurable set
The problem is as follows:

Given this theorem:

Let $D$ be bounded & Jordan-measurable set

Let $f$ be a bounded function on $D$

And $f$ is continuous except for a set of zero area

Then the conclusion is that the $\int\int _D f$ exists

and $\int\int_D f$ = $\int\int_R F$ where $F$ takes values $f(p)$ if
  the point $p$ is in D, and $0$ otherwise.

Goal:  To use this theorem to explain why for a Jordan-measurable set, I can say that the area $A(D) = \int\int _D 1 dA$

So far, I have several ideas coming to mind, but I don't see how my ideas have anything to do with the explanation that they are looking for >"<

My thoughts:

(1) A region $D$ is Jordan-measurable if the region has definite area

(2) In other context, $D$ is Jordan-measurable means that I can find a polygon which is totally inside $D$, and also another polygon such that this polygon contains the previous polygon, thus this "new" polygon contains some boundary points of $D$

(3) Another idea about $D$ is that $bdy(D)$ has zero area

(4) From the theorem and using the fact that the function $f = 1$ clearly is bounded on $D$ and continuous except for a set of zero area, then the double integral of $f=1$ over region $D$ exists and can also be transfered to a double integral of $F$ over rectangle $R$, by how F is defined in the theorem.

But I don't see the connection.

Well, consider only $A(D) = \int\int _D 1 dA$, then my understanding is that for $dA = dxdy$ or $dA = dydx$, when I do the double integral over region $D$ with respect to this change of area, what I'm doing is to "cover" region $D$ by two rectangles, one with respect to $x$ (call it U) and one with respect to $y$ (call it V).

So I think the idea of (2) comes here since I'm trying to find such 2 polygons.  (3) also holds trivially.  Together with how $F$ is defined, which is basically accepting the function value of point $p$ if it's in the set $D$, and discard the function value of $p$ if it isn't in the set.  Then, as conclusion, the area A(D) is the sum of the two rectangles U and V?

Would someone please help me on this question? I'm a bit lost >_>

Thank you in advance.

答:
I'm not exactly sure what the actual question is, but I'll try to give an answer nevertheless.

First, a matter of notation. Usually, an integral of $f$ over some set $M$ (no matter in how many dimensions) is written simply as $\int_M f\,d\mu$, where $\mu$ indicates the measure that is to be used. For pure riemann integrals, just writing $\int_M f$ suffices, because there really one is one possible measure. I'm guessing that $\int \int_M f\,dA$ is simply supposed to mean $\int_M f$, i.e. the integral of $f$ over $M$.

Since the question doesn't include a definition of $A(D)$ other than $A(D) = \int_D 1$, I also asumme that what's wanted here is an explanation why that integral yields the area of $D$, rather than a formal proof (which would first require a definition of what the area of $D$ is supposed to be).

Remember that the definite integral of a positive function $f$ in one dimension between $a$ and $b$ is simply the area under the curve $f(x)$ between the two lines $x=a$ and $x=b$. In two dimensions, the same holds, but you can pick more complicated shapes. In other words, for a positive $f$ and $D \subset \mathbb{R}^2$, the integral $\int_D f$ is the volume of the set $$
  S := \left\{(x,y,z) \in \mathbb{R}^3 \,:\, (x,y) \in D, 0 \leq z \leq f(x,y)\right\} \text{.}
$$

Now, if $f(x)$ is constant on $D$, i.e. $f(x)=c$, then $V$ is simply $D \times [0,c]$ (For example, if $D$ is a circle, $V$ is a disc with height $c$). In that case, you thus have that $$
  A_3(S) = A_2(D) \cdot c \text{.}
$$
where $A_3$ measures volumes in $\mathbb{R}^3$ and $A_2$ measures areas in $\mathbb{R}^2$.

If in particular $c=1$, i.e. if $f(x)=1$, then $A_3(S) = A_2(D)$. Since $A_3(S) = \int_D 1$ it follows that $$
  A_2(D) = \int_D 1 \text{.}
$$

I hope all this formalism doesn't distract from the simply geometric idea behind this. If you take some shape, and look at the volume between that shape at the constant function $1$, then this volume is the area of the shape times 1.


问:
$\int \frac{1}{x(x+1)...(x+n)}dx$
I have not managed to evaluate the integral. I feel the idea is just split the integral into a sum of integrals

$$\int \frac{1}{x(x+1)...(x+n)}dx = \int \left(\frac{A_1}{x} + \frac{A_2}{x+1} +... +\frac{A_n}{x+n} \right) dx$$

So, technically, the main problem is to determine the coefficients in the numerators of these fractions. Hhm, it has occurred to me that I might bypass the system of $n$ equations just putting $x = 0$, $x = - 1$,..., $x = - n$ after multipling the whole expression by $x(x+1)...(x+n)$. The idea can be of help to solve the risen problem.

Another idea is just to decompose the fraction by representing the numerator of the initial fraction, which is 1, as an expression considered to be suitable to attain the objective. For example,

$$ 1 = x + 1 - (x + 2) + (x + 3) - (x + 4) +... $$

I might attempt to adopt the idea but there is a nice trick that solves the problem? I came across the integral when searching for the competition problems on integration.

When trying to apply "brutal force" method, I got many brackets and equations afterwards. Hence, there must be more cunning technique.

答:
Your idea is correct.

This is the easiest case where you can do partial fraction decomposition since the polynomial in the denominator splits as distinct linear factors. In this case, you can apply the Heaviside cover-up method.

Title: $\int \frac{1}{x(x+1)...(x+n)}dx$


问:
Why are the values of $\sin (\theta)$ and $\csc(\theta)$ positive in the second quadrant?
Why is the value of $\sin (\theta)$ and $\csc(\theta)$ positive in the second quadrant?
Similarly why the value of other trigonometric ratios so in other quadrants?

答: The reason is that the sine is the projection of the point  on the trigonometric circle with curvilinear abscissa $\theta$. As to the cosecant, it is the reciprocal of the sine.


答: Write $$g(x,y)=\log|f_x|.$$ Then you have that $$g_y=\frac{1}{|f_x|}\frac{f_x}{|f_x|}f_{xy}=\frac{f_{xy}}{f_x}.$$

Title: How does an absolute value operator interact with partial derivatives?

Body:
I wish to take the partial derivative with respect to $y$ of the following expression:

$$g(x,y)=\log{\left|\frac{\partial}{\partial x}f(x,y)\right|}$$

where $f(x,y):\mathbb{R}^2\rightarrow\mathbb{R}$ is some function which depends on $x$ and $y$, and $\frac{\partial}{\partial x}$ denotes the partial derivative with respect to $x$. How does the absolute value operator $|\cdot|$ interact with the partial derivative $\frac{\partial}{\partial y}$ if I want to evaluate $\frac{\partial}{\partial y}g(x,y)$?

I would greatly appreciate any suggestions or references on how to address such situations, also in more general cases.


答:
First of all note that the cosecant is just $\frac{1}{\sin\theta}$, so they are both positive or both negative.

In the second you can view the sin of $\theta$ as the value of the vertical segment that is perpendicular to the $X$ axis drawn from the point where the radios of angle $\theta$ cuts the unit circle, as the figure shows:

Thus, sine is positive in the first two quadrants, but negative in the last two.

Title: Why are the values of $\sin (\theta)$ and $\csc(\theta)$ positive in the second quadrant?

Body:
Why is the value of $\sin (\theta)$ and $\csc(\theta)$ positive in the second quadrant?
Similarly why the value of other trigonometric ratios so in other quadrants?


问:
Conditional Probability Problem

An insurance company examines its pool of auto insurance customers and gathers the following information:

(i)  All customers insure at least 1 car

(ii)  64% of all customers insure more than one car

(iii)  20% of the customers insure a sports car

(iv)  Of those customers who insure more than one car, 15% insure a sports car.

What is the probability that a randomly selected customer insures exactly one car, and that car is not a sports car?

Let's use the following variable definitions:

O= owns 1 car, O' = owns more than 1 car

S= sports car, S' = Not sports car.

N() = Cardinal Number of a set

From statements (i)-(iii), we get the following:  $N(O') = 64, N(O) = 36, N(S) = 20$

From statement (iv):  $\Pr(S \mid O')=15$

We are asked to find $\Pr(S' \mid O)$

By definition:

$Pr(S' \mid O) = \cfrac{\Pr(S' \cap O)}{\Pr(O)}=\cfrac{\Pr(S' \cap O)}{N(O)}\tag{1}$

Pr()=N() since this is a uniform distribution--I interpret this when it says "randomly selected"

Next, I did:

$N(S)=N(S \cap O') + N(S \cap O)\tag{2}$

$0.2 = .64*.15 + .36 * x$

$x=0.28$, but we want 1-x because we want S' in $\Pr(S' \cap O)$ which equals 0.72.

See diagram below:

So plugging back into (1):

$Pr(S' \mid O) = \cfrac{0.71*.36}{.36}=.71$

but the answer is .26.

I mainly wanted to know why equation (2) is wrong.  I know that is the crux of the problem.  Why can't I use that equation in this case?

Any help is appreciated.  Thank you.

答:
Your problem is very simple. The question asks for the probability $\mathrm P(S'\cap O)$ not $\mathrm P(S'|O)$.

Two things: I don't like the notation where probabilities are greater than 1. Divide them by a hundred or at least put percentage symbols. Also, your notation is rather convoluted. There's no need to define a counting function and a probability function when, as you know, both are identical here.

Title: Conditional Probability Problem


问:
Tangent line to the curve
Write the equations off the tangent line and normal line to the curve:
$$x = e^t\cos(t)$$
$$y = e^t\sin(t)$$
at the point A(1,0).
Do I take $x(t_0) = 0$ and $y(t_0) = 1$?
I solved it this way and I got the tangent line:
$$(x-e\cos(1))/(e(\sin(1)-\cos(1))) = (y-0)/1$$
Is this correct or do I have to find $t_0$ in another way?

答:
First you have to find the parameter $t$ for which
$$c(t)=e^t\cdot\begin{pmatrix}\cos(t)\\\sin(t)\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}.$$
An easy calculation shows that $t=0$.  As
$$\dot c(t)=e^t\cdot\begin{pmatrix}\cos(t)-\sin(t)\\\sin(t)+\cos(t)\end{pmatrix}$$
the tangent vector in $t=0$ is
$$\dot c(0)=e^0\cdot\begin{pmatrix}1+0\\0+1\end{pmatrix}
=\begin{pmatrix}1\\1\end{pmatrix}$$
the tangent's equation is
$$\begin{pmatrix}x\\y\end{pmatrix}
=\begin{pmatrix}1\\0\end{pmatrix}+s\begin{pmatrix}1\\1\end{pmatrix}
$$
or $y=x-1$.

Title: Tangent line to the curve


问:
Find residues from different Laurent series
The link below shows the working steps of finding two different Laurent series with different valid annulus region.

Finding the Laurent series of $f(z)=1/((z-1)(z-2))$

Is it possible to find residues from two different Laurent series?

答:
If you want to find the residue of $f:\mathbb{C}\to\mathbb{C}$ at $z_0$ by looking at the coefficients of some expansion, you need the Laurent series of $f$ in a punctured disc $D(z_0,r)\setminus\{z_0\}$.

For example, if you consider the function $f(z)=\frac{1}{z}$, its residue at $z=1$ is $0$, not $1$.


问:
Is $\sqrt[3]{x^3}$ monomial?
I have been confused by different interpretations of polynomial and monomial. Is $\sqrt[3]{x^3}$ really a monomial?

答:
$\sqrt[3]3$ equals $x$, $\xi x$ or $\xi^2 x$, where $\xi$ is a primitive 3rd root of unity.
If you take the ground field $\Bbb C$ or say $\Bbb R(\xi)$, then all solutions are terms with coefficient and variable $x$. One of it is a monomial: $x$.

Title: Is $\sqrt[3]{x^3}$ monomial?


答:
Assume by contradiction that for every $K<0$ there exists a $\delta>0$ such that for every $0<x<\delta$, we get $f'(x)<K$.

We choose $n \in \mathbb{N}$ such that $n>K$, and we choose $x_{n}= \frac{1}{\pi +2 \pi n}$.

Note that $0<\frac{1}{\pi +2 \pi n}<1$. (So in this case $\delta=1$)

And: $f'(x)=-(\pi +2\pi n)(\cos(\pi+2\pi n)+1)=0>K$

So $f'(x)>K$, in contradiction!

Is that correct?

Thanks!

Title: Prove that limit doesn't exists at all

Body:
I have a function $f(x)=\frac{1}{x}+\sin(\frac{1}{x})$ which I have proved that $\lim_{x \to0^+}f(x)=\infty$, and I need to prove that $\lim_{x \to0^+}f'(x) \ne -\infty$.

I have got that $f'(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$

I am trying to prove that:

$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$$

Does not equal to $-\infty$ (or to prove that it doesn't exists at all).

I have tried to assume by contradiction that
$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\infty$$

So there exists $\delta_{1}>0$ such that for every $0<x<\delta_{1}$ we say that $$f(x)<\square $$

We choose $\delta=\delta_{1}$, and let $0<x<\delta$ so:

$$f(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\frac{1}{x^2}(1+\cos(\frac{1}{x}))\le -\frac{2}{x^2}<\frac{2}{\delta^2}=k$$

I do not know what to choose instead of $\square$, and if what I did is correct.

Appreciate your help!

I have also tried to think about the continuity of $f$, and saying that it is not bounded so it can't be continuous, so the limit doesn't exists.

Thanks a lot!


问:
Prove/refute property about a constrained sequence of numbers
I need to prove or refute a property about a sequence of numbers.
Here is what is given to me:

Sequence ($a_1,a_2,...,a_k,a_{k+1},a_{k+2}$) containing $k+2$ numbers. Every number $0 < a_i \leq M, i=1,...,k+2$ for the same given constant $M$. Moreover, $\sum_{i=1}^{k+2} a_i = N$, for another given constant $N > 0$. The constants $N$ and $M$ are related by $kM \geq N, k > 1$.

Then, I need to prove or refute the following property: Can all consecutive pairs of numbers $a_i, a_{i+1}, i=1,...,k+1$ be defined such that $a_i + a_{i+1} > M$ ? Or does it lead to a violation of one of the given constraints?

I tried searching for something similar but in truth, I barely know what to search for. My tentative proofs do not really go anywhere meaningful, so I had to resort to more experienced math guys to help me with this. It has been a long time since I had to prove some property like this.

答:
Assuming that given $k$, $N$ is selected such that
$$\sum_{i=1}^{k+2} a_i = N,$$
I claim it is possible $ \iff k\ge 3$

First of all we see the cases $k=1,2$ to check it is impossible:

(i) if $k=1$ then
$$M< a_1+a_2 < N \le kM = M;$$
(ii) if $k=2$ then we only have $(a_1,a_2,a_3,a_4)$, and
$$a_1+a_2>M \mbox{ and } a_3+a_4 >M ,$$
so
$$N = \sum_{i=1}^{k+2} a_i >M + M =2M = kM .$$

Now let $k\ge 3$: let $\epsilon > 0$ (arbitrarily small) and such that $\epsilon < M$, then I pick:
$$a_i= M \mbox{ if } i \mbox{ is even; } a_i = M -\epsilon \mbox{ if } i \mbox{ is odd.}$$
Now we obseve that $a_i + a_{i+1} = 2M - \epsilon > M$ anf that $0 < a_i \le M$, so they are good, then:

(i) if $k$ is even
$$\sum_{i=1}^{k+2} a_i = N = (2M -\epsilon)\Big(\frac{k}{2} + 1\Big) = (k+2)M - \epsilon\Big(\frac{k}{2}+1\Big)$$
which is possible if I pick $\epsilon$ such that
$$kM \ge (k+2)M - \epsilon\Big(\frac{k}{2}+1\Big)$$
which implies that
$$\epsilon \ge \frac{2M}{\frac{k}{2}+1} .$$
Also we have that $M> \epsilon$, so we can pick such a $\epsilon$ only if
$$M> \frac{2M}{\frac{k}{2}+1},$$
which means that $k > 2$.

(ii) if $k$ is odd
$$\sum_{i=1}^{k+2} a_i = N = (2M-\epsilon)\Big(\frac{k+1}{2}\Big) + a_{k+2} = (k+1)M - \epsilon \Big(\frac{k+1}{2}\Big) + a_{k+2} =$$
$$= (k+2)M - \epsilon \Big(\frac{k+1}{2} +1 \Big)$$
and here too we see it is possible if $\epsilon$ is such that
$$\epsilon \ge \frac{2M}{\frac{k+1}{2}+1} .$$
This implies that it is tue if
$$M > \frac{2M}{\frac{k+1}{2}+1},$$
which means if $k>1$.

This concludes my claim.


答:
In the context of real numbers, yes, this is just $x$, which is a monomial. However--

It's not in the form of a monomial, but it is functionally equivalent to a monomial in the sense that its values agree everywhere with the monomial $x$.

So I suppose the real answer depends on what your definition of "is" is.

Title: Is $\sqrt[3]{x^3}$ monomial?

Body: I have been confused by different interpretations of polynomial and monomial. Is $\sqrt[3]{x^3}$ really a monomial?


问:
Help with: $\tan2θ+ 2\tanθ= 3\cotθ$ for $0\le θ\le 180$
This is a trigonometry question I have asked everyone I know and no one seems to be able to solve it . Any hits or steps on how to get to the answer I appreciate it! Thank you!</p>

Body:
$$\tan 2x + 2\tan x = 3\cot x$$
$$\frac{2\tan x}{1-\tan^2x} + 2\tan x = 3\cot x$$
$$\frac{2}{1-\tan^2x} + 2 = \frac3{\tan^2 x}$$
$$2 = \frac3{\tan^2 x}-\frac{2}{1-\tan^2x} $$

Let $tan^2 x = t$ and solve

$$2 = \frac3{t}-\frac{2}{1-t} $$
$$2(t)(1-t)=3-5t$$
$$7t-2t^2-3=0$$


问:
Representing a hexagon as an intersection of 6 halfplanes
how can I represent a hexagon (6-gon) as an intersection of 6 half planes?</p>

答:
Is this the type of answer you require?

$$y\le\sqrt3$$
$$y\ge-\sqrt3$$
$$y\ge \sqrt3(x-2)$$
$$y\le -\sqrt3(x-2)$$
$$y\le \sqrt3(x+2)$$
$$y\ge -\sqrt3(x+2)$$


答:
You may recall this image:

Now, the unit circle definition tells us that $\cos \theta$ and $\sin\theta$ are the $x$ and $y$ coordinates of the point whose radius makes angle $\theta$ with the positive $x$-axis.

Therefore, in the first quadrant ($0<\theta<90^\circ$), both $\cos\theta$ and $\sin\theta$ are positive, and in the second quadrant ($90^\circ<\theta<180^\circ$), $\cos\theta$ is negative and $\sin\theta$ is positive, and so on for third and fourth quadrants.

Define:
$$\sec\theta=\frac{1}{\cos\theta}\text{ and }\csc\theta=\frac{1}{\sin\theta}$$
Thus, the sign of $\sec\theta$ is the same as $\cos\theta$ and the sign of $\csc\theta$ is the same as $\sin\theta$.

Define:
$$\tan\theta=\frac{\sin\theta}{\cos\theta}\text{ and }\cot\theta=\frac{\cos\theta}{\sin\theta}$$
Thus, the signs of $\tan\theta$ and $\cot\theta$ are positive, if $\sin\theta$ and $\cos\theta$ have same signs, and negative, if different.

For example, for the second quadrant, $\cos\theta$ is negative and $\sin\theta$ is positive. So, $\sec\theta$ is negative and $\csc\theta$ is positive, and , $\tan\theta$ and $\cot\theta$ are negative.

Hope this helps. Ask anything if not clear :)


答:
Hint:

Set $t=\tan\theta$. Using the duplication formula for the tangent, yhe equation becomes
$$\frac{2t}{1-t^2}+2=\frac{2t(2-t^2)}{1-t^2}=\frac 3t\iff2t^2(2-t^2)=3(1-t^2)\iff 2t^4-7t^2+3=0.$$
Thus you obtain a biquadratic equation in $t$ that you can solve with the usual substitution $u=t^2$.

Title: Help with: $\tan2θ+ 2\tanθ= 3\cotθ$ for $0\le θ\le 180$

Body: This is a trigonometry question I have asked everyone I know and no one seems to be able to solve it . Any hits or steps on how to get to the answer I appreciate it! Thank you!</p>


问:
Show that f is uniformly continuous
I am having difficulty determining what value I should assign to $\delta$ for the following problem.  How do I determine what it should be?

Define $f:[3.4,5] \rightarrow \mathbb{R}$ by $f(x)=2/(x-3)$.  Show that $f$ is uniformly continuous on [3.4,5].

This is what I have so far.

Choose $\varepsilon >0,$ and $\delta=???$.  If $|x-y|< \delta$ and $x,y \in [3.4,5]$, then

$|f(x)-f(y)|=|2/(x-3)-2/(y-3)| = |2(y-x)/((x-3)(y-3))|<???$

答:
Here We have a very important result that ensures the following:

"If $f : K \rightarrow \mathbb{R}$ is a continuous function and K is compact, then $f$ is uniformly continuous."

In our case [3.4 , 5] is closed and bounded in the line hence is compact by Heine-Borel's Theorem, $f$ is continuous (note that denominator of the fraction never is zero so we have a quotient of continuous functions hence continuous) Hence the result above apllies and $f$ is uniformly continuous.

Title: Show that f is uniformly continuous


答:
That's because cartesian coordinate system is defined in that way. This needs explanation, I know.

We have generalised trigonometry, from only inputs between $0$ and $90$ to any real number $\theta$, by using a unit circle (Wikipedia link) on cartesian coordinate system with its centre on origin i.e. $(0,0)$.

To do this generalisation we define $\sin (\theta)$  as $y-axis$ of the the end of the point on radius of that unit circle (Khan Academy link).
Now it is by definition that in $2nd$ and $1st$ quadrant $y-coordinate$ is positive, so clearly it follows that $sine$ function must be positive in $2nd$ and $1st$quadrant as it is $y-axis$. Now since $\sin (\theta)$ is positive, so must be $\csc(\theta)$. As we have defined it as reciprocal of $\sin (\theta)$, now reciprocal  of any positive number must be a positive number($0$ is not positive).

Now we have similarly defined $\cos(\theta)$ as the $x-coordinate$ of the the end of the point on radius of that unit circle, similarly it must be positive in $1st$ and $4rth$ quadrant.
Following image may be helpful:

It is left as an exercise for you to find why other trigonometric functions ..... you understand what I mean(right)!


问:
if function is continuous and real she must be Inverse?
$R=${$f:[0,1]-> \Bbb R$}

R is all the real and continuous functions from [0,1] to $R$,i want to check if there is a additive inverse number relative to +,so is it possible to say that $f(x)+f(x)^-=0\in R$?

Maybe I did not explain myself correctly,I'm trying to prove:

$R=${$f:[0,1]-> \Bbb R$}

(all the continuous and real function )is ring

So the first step is prove $(R,+)$ is abelian group

And that's why I have to prove $f(x)+fx)^-=0$
to get additive inverse number.

答:
If you freeze $x$, $f(x),g(x),\cdots$ are reals and enjoy the properties of the Abelian group $\mathbb R,+$. These remain $\forall\,x$.

All you have to do is to prove closure, i.e. the sum of two continuous functions is a continuous function.


问:
Let $(G,*)$ be a group and $a \in G$.
Suppose that $|a|=n$ and $n=mk$ for some positive integers $m$ and $k$.
What is $|a^k|$?
Let $(G,*)$ be a group and $a \in G$.
Suppose that $|a|=n$ and $n=mk$ for some positive integers $m$ and $k$.
What is $|a^k|$?

attempt:
Let $a \in G$ such that $|a|=n$. Then,
\begin{equation*}
a^n = a^{mk} = (a^k)^m = e_G
\end{equation*}
where $e_G$ is the identity element of $G$. Thus, the order of $a^k$ is $m$.

In particularly, by theorem:
Let $(G,*)$ be a group, $a \in G$, and
$|a|=n$. Then, for all $k \in \Bbb N$,
$|a^k| = \frac{n}{\gcd(k,n)}$,
we have, $|a^k| = \frac{n}{\gcd(k,mk)} = \frac{n}{k} = m$.

Does it true? On the other hand, in the answer key says that the answer is
$\frac{n}{m}$. How to get this?

答:
In your first attempt, note that what you obtained, $(a^k)^m = e$ does not guarantee that $o(a^k) = m$. It tells you that
$$o(a^k) \mid m \implies o(a^k) \le m$$

To show that  $o(a^k) = m$, we need to also show that $o(a^k) \ge m$. To show this, suppose for a contradiction that $o(a^k) < m$. Then, $\exists t \in \mathbb{Z}^+$ such that $t < m$ and $(a^k)^t = e$. But then, we have
$(a^k)^t = a^{kt} = e$ for $kt < km$ (since $t < m$). But this means $o(a) < km = n$, contradiction.

Your second attempt is perfectly valid.


答:
I do not know if this will answer your question, but consider this if your question is about "different Laurent series for the same function".

Let $f(z)=\sum \limits_{n=-\infty}^{-1}{a_{n}}{z^n}+\sum \limits_{n=0}^{\infty}a_nz^n=\sum \limits_{n=1}^{\infty}\dfrac{a_{-n}}{z^n}+\sum \limits_{n=0}^{\infty}a_nz^n$

Define $g(z)$ as

\begin{align}
g(z)&=\sum \limits_{n=0}^{\infty}a_nz^n, \forall z, |z|\leqslant R.
\\
&=\sum \limits_{n=1}^{\infty}\dfrac{a_{-n}}{z^n}, \forall z, |z|>R
\end{align}

Clearly $g(z)$ is bounded and analytic on both $|z|\leqslant R$ and $|z|>R$. So is bounded on entire domain.

Since $\sum \limits_{n=-\infty}^{-1}a_nz^n=\sum \limits_{n=1}^{\infty}\dfrac{a_{-n}}{z^n}$ for $|z|=R$, $g(z)$ is analytic on entire domain through analytic continuation. Thus $g(z)$ is bounded entire function, and $g(z)=c$, or $a_n=0$ for all $n$ except $n=0$.

Then if $f(z)\equiv 0$, $a_n=0,$ for all $n$, which means Laurent Series is unique.

Title: Find residues from different Laurent series

Body:
The link below shows the working steps of finding two different Laurent series with different valid annulus region.

Finding the Laurent series of $f(z)=1/((z-1)(z-2))$

Is it possible to find residues from two different Laurent series?


问:
Show that $m\le\left(\frac{1}{b-a}\int_{a}^{b}f^{2}\left(x\right)dx\right)^{\frac{1}{2}}\le M$
Assume $f$ is a real-valued function which is integrable over the interval $I=[a,b]$ and for every $x \in [a,b]$ we have that $0\le m \le f(x) \le M$,show the following inequality does hold:
$$m\le\left(\frac{1}{b-a}\int_{a}^{b}f^{2}\left(x\right)dx\right)^{\frac{1}{2}}\le M$$

I know that for $M_I=\sup \{f(x):x \in [a,b]\}$ and $m_I=\inf \{f(x):x \in [a,b]\}$ we have that :

$$ U(P,f)=\sum_{i=0}^{n-1} M_i \Delta x_{i}\le M_I\sum_{i=0}^{n-1} \Delta x_{i} =M_I(b-a)$$

And
$$  m_I(b-a)=m_I\sum_{i=0}^{n-1} \Delta x_{i}  \le \sum_{i=0}^{n-1} m_i \Delta x_{i}=L(P,f)$$

Where $P=(a=x_0,x_1,...,x_n=b)$ is a partition of $[a,b]$,combining these two inequality gives:

$$m_I(b-a) \le L(P,f) \le  U(P,f)\le M_I(b-a)$$

And if the function $f$ is integrable over $[a,b]$,then $$m_I(b-a) \le \int_{a}^{b}f\left(x\right)dx\le M_I(b-a)$$

And if  for every $x \in [a,b]$ we have that $0\le m \le f(x) \le M$ then from the definition of supremum and infimum:

$$m \le \frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx\le M$$

But I don't know how the get the main inequality.

答:
Hint:

If you take the square, what you want to show is
$$
m^2\le\frac{1}{b-a}\int_a^b f^2\le M^2
$$

but $m^2\le f^2\le M^2$. Simply integrate it.

You should assume that $m,M\ge 0$. Otherwise, you need to replace them with $|m|$ and $|M|$.


答:
You are off to a good start. Whenever we have epsilon-delta proofs, we need to use a two step process. First we work to figure out what delta should be. Then we show it works.

Step 1 is to determine what delta should be. This means we start as you did by saying that we WANT: $$ \epsilon > |f(x) - f(y)| =  \left| \frac{2}{x-3} - \frac{2}{y-3} \right| =
\left| \frac{2(y-x)}{(x-3)(y-3)} \right| $$

Using this we can multiply and rewrite the inequality as $$
\epsilon \frac{\left|(x-3)(y-3)\right|}{2} > \left| y-x \right|$$

As $f$ is defined on the interval $[3.4, 5]$, we can see that $|(x-3)| < 5-3=2$ for all $x \in [3.4, 5]$. Therefore, we have $$ \epsilon \frac{\left|(x-3)(y-3)\right|}{2}  < \epsilon \frac{2\cdot2}{2} = 2\epsilon.$$

Now we proceed to step 2, where we choose $\delta = 2 \epsilon.$ We then proceed to show that this choice of delta will make the limit uniformly continuous.

(Proof) Let $\epsilon > 0$ then choose $\delta = 2\epsilon$ and let $x,y \in [3.4, 5]$ and $|x-y| < \delta.$ Then we have $$|x-y| < 2 \epsilon  \implies \frac{|x-y|}{2} < \epsilon  $$

Now we note that $$ \frac{|x-y|}{2} \geq \frac{2\cdot2}{2} \geq \frac{\left|(x-3)(y-3)\right|}{2} = |f(x) - f(y)|$$ as $x,y \in [3.4, 5]$. We can then rewrite the right side as in step 1 until we have $|f(x) - f(y)|$.

Putting all of the inequalities together yields $|f(x)-f(y)| < \epsilon$.

Title: Show that f is uniformly continuous

Body:
I am having difficulty determining what value I should assign to $\delta$ for the following problem.  How do I determine what it should be?

Define $f:[3.4,5] \rightarrow \mathbb{R}$ by $f(x)=2/(x-3)$.  Show that $f$ is uniformly continuous on [3.4,5].

This is what I have so far.

Choose $\varepsilon >0,$ and $\delta=???$.  If $|x-y|< \delta$ and $x,y \in [3.4,5]$, then

$|f(x)-f(y)|=|2/(x-3)-2/(y-3)| = |2(y-x)/((x-3)(y-3))|<???$


答:
For your second question, define
$$c(t)=\begin{pmatrix}
\frac{t+2}{3t+4}\\\frac{4t+3}{2t+1}
\end{pmatrix}.$$
Compute
$$\dot c(t)=-2\begin{pmatrix}
\frac{1}{(3t+4)^2}\\\frac{1}{(2t+1)^2}
\end{pmatrix}.$$
The tangent line passes through the origin iff $\dot c(t)$ is parallel to $c(t)$. From here you'll arrive in solving
$$(t+2)(3t+4)=(4t+3)(2t+1),$$
which has the solutions $t=\pm1$.

Title: Tangent line to the curve

Body:
Write the equations off the tangent line and normal line to the curve:
$$x = e^t\cos(t)$$
$$y = e^t\sin(t)$$
at the point A(1,0).
Do I take $x(t_0) = 0$ and $y(t_0) = 1$?
I solved it this way and I got the tangent line:
$$(x-e\cos(1))/(e(\sin(1)-\cos(1))) = (y-0)/1$$
Is this correct or do I have to find $t_0$ in another way?


问:
VCdim of a range space that consists of all open disks of the R^d for some d
Please help me prove this question:

For the range space $(\Bbb R^d, D_d)$ where $D_d$ consists all the open disks in $\Bbb R^d$, each disk in $D_d$ is in the form of: $D(p,r):=\{x\in\Bbb R^d | \Vert x-p\Vert_2<r\}$, for $p\in\Bbb R^d $ and $r>0$.

Show that $\operatorname{VCdim}(\Bbb R^d, D_d) \le d+2$.

答:
There are two ways to solve this (that I know of), one of them is explained here: https://cstheory.stackexchange.com/a/11089.
This proof is for for closed disks not open, but we can change this proof to consider two different cases, one of them is that there are points on the edge of the ball B (i.e. points such that $||x-p||_2=r$) and the other that there aren't such points, if there aren't any points on the edge their proof doesn't change, if there are we shrink $r$ a little bit such that the points on the edge are no longer in the ball but we also didn't remove any other points from B, and now we continue the proof without changes.

And a different way to proof this is this:
Let S be the largest set of points such that $2^{|S|}=|\Pi(S)|$. For some fixed p,r let $T=\{x|x\in S,||x-p||_2<r\}$ and let $S'$ be the minimum enclosing hypersphere (d-dimentional ball) of the points of $T$. A point $x\in S$ satisfies $||x-p||_2<r$ if and only if $x\in S'$. The minimum enclosing sphere of a set of points can be determined by at most d+1 points from the set i.e. there is a subset of size at most d+1 such that it's minimum enclosing sphere is the minimum enclosing sphere of all the points (see Lemma 3.7 from https://www.google.com/url?sa=t&source=web&rct=j&url=https://people.inf.ethz.ch/emo/DoctThesisFiles/fischer05.pdf&ved=2ahUKEwjs8ZnO16DuAhUWH-wKHVqSBKUQFjAAegQIARAB&usg=AOvVaw0lnz91H2v8Oqa1oPvgyk6Y)  . From that we get that there exists a subset of at most d+1 points from $T$ (and therefore also from S) that defines for all the rest of the points whether or not their distance from p is at most r. Meaning that:

$$2^{|S|}=|\Pi(S)|\leq \sum_{i=0}^{d+1}{\binom{|S|}{i}}$$
Now from the binomial theorem we get:
$$\sum_{i=0}^{|S|}{\binom{|S|}{i}}\leq \sum_{i=0}^{d+1}{\binom{|S|}{i}}$$

Meaning that $VCdim(R^d,D_d)=|S|\leq d+1$

Sorry for bad english.


问:
Points in space
Let $P =\left \{ A_{1}, A_{2} \cdots A_{n} : A \in \mathbb{R}^{3}\right \}$ (where $A_{0}=A_n, A_{1}=A_{n+1})$. By middling of a point $A_{i+1}$ we mean setting it's value to a rectangular projection of that point on the line $A_iA_{i+2}$. We middle every point of $P$ and start the process again ... ad infinitum. Decide whether all points of $P$ will be collinear after a finite amount of steps.

I came with this myself, please be kind.

答:
Convergence looks to be (it's not a mathematical proof!)

towards a point when points $A_i$ are smoothly spaced (i.e. with distances $A_kA_{k+1}$ of the same order of magnitude (Fig. 1).

towards a line, indeed, as you have assumed it, when there is large gap between two points. (Fig. 2).

Fig. 1: Viviani's curve, in black (intersection of a sphere and a certain cylinder)  shrinks into almost a dot after 300 iterations, with graduated colors, the last ones being more reddish. Only one out of ten of these curves have been plotted.

Fig; 2: The case of a helix with three periods (in black, with connected endpoints). Here,  after 500 steps, this still intermediate stage (in red) in the form of a bow with its string displays a trend to a limit which would be approximately the line segment connecting the initial endpoints.

Matlab program:

```<code>clear all;close all;hold on;axis equal;
% t=0:pi/20:6*pi;A=[cos(t);sin(t);t]; % helix
t=-2*pi:pi/20:2*pi; A=[1+cos(t);sin(t);2*sin(t/2)]; % Viviani's curve
n=size(A,2);
pf=300; % number of steps
A=[A,A(:,1:2)]; % enables cyclicity
plot3(A(1,:),A(2,:),A(3,:),'k','linewidth',2); % initial  "curve"
for p=1:pf;
    if mod(p,10)==0
        plot3(A(1,:),A(2,:),A(3,:),'color',[p/pf,0,1-p/pf]);
    end;
    for q=1:n
        v=A(:,q+2)-A(:,q);v=v/norm(v); % unit vector of [A_k, A_k+2]
        w=A(:,q+1)-A(:,q);
        B(:,q)=A(:,q)+(w'*v)*v; % orth. projection
    end;
    A=[B(:,1:n),B(:,1:2)];
end;
</code>```

Title: Points in space

Body:
Convergence looks to be (it's not a mathematical proof!)

towards a point when points $A_i$ are smoothly spaced (i.e. with distances $A_kA_{k+1}$ of the same order of magnitude (Fig. 1).

towards a line, indeed, as you have assumed it, when there is large gap between two points. (Fig. 2).

Fig. 1: Viviani's curve, in black (intersection of a sphere and a certain cylinder)  shrinks into almost a dot after 300 iterations, with graduated colors, the last ones being more reddish. Only one out of ten of these curves have been plotted.

Fig; 2: The case of a helix with three periods (in black, with connected endpoints). Here,  after 500 steps, this still intermediate stage (in red) in the form of a bow with its string displays a trend to a limit which would be approximately the line segment connecting the initial endpoints.

Matlab program:

clear all;close all;hold on;axis equal;
% t=0:pi/20:6*pi;A=[cos(t);sin(t);t]; % helix
t=-2*pi:pi/20:2*pi; A=[1+cos(t);sin(t);2*sin(t/2)]; % Viviani's curve
n=size(A,2);
pf=300; % number of steps
A=[A,A(:,1:2)]; % enables cyclicity
plot3(A(1,:),A(2,:),A(3,:),'k','linewidth',2); % initial  "curve"
for p=1:pf;
    if mod(p,10)==0
        plot3(A(1,:),A(2,:),A(3,:),'color',[p/pf,0,1-p/pf]);
    end;
    for q=1:n
        v=A(:,q+2)-A(:,q);v=v/norm(v); % unit vector of [A_k, A_k+2]
        w=A(:,q+1)-A(:,q);
        B(:,q)=A(:,q)+(w'*v)*v; % orth. projection
    end;
    A=[B(:,1:n),B(:,1:2)];
end;


问:
Confused with a proof about group solubility
I am reading about soluble groups in Ian Stewart's book about Galois theory (fourth edition)
and I came across a theorem which I can not understand the proof of. Down below is the theorem and the proof which can be found in the book mentioned before on page 163:

Theorem. Let $G$ be a group, $H$ a subgroup of $G$, and $N$ a normal subgroup of $G$.

(1) If $G$ is soluble, then $H$ is soluble.

(2)...

Proof. (1) Let $1 = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_r = G$
be a series for $G$ with abelian quotients $G_{i+1}/G$. Let $H_i = G_i \cap H$. Then $H$ has a series $1 = H_0 \triangleleft \cdots \triangleleft H_r = H.$
We show the quotents are abelian. Now (here is the part that I do not understand)

$\frac{ H_{i+1}}{H_i} = \frac{G_{i+1} \cap H}{G_i \cap H} = \frac{G_{i+1} \cap H}{G_i \cap (G_{i+1} \cap H)} \cong \frac{G_i(G_{i+1} \cap H)}{G_i}$

by the first isomorphism theorem. But this latter group is a subgroup of $G_{i+1}/G_i$ which is abelian. Hence $H_{i+1}/H_i$ is abelian for all $i$, and $H$ is soluble. $\square$

I understand everything in the proof up until the point when Steward mentions the first isomorphism theorem. The first thing that is confusing is the notation $G_i(G_{i+1} \cap H)$, what does this mean? Is this some sort of "group multiplication"? What's confusing me is that there's no symbol between $G_i$ and $(G_{i+1} \cap H)$. I have been self-studying some Galois theory and group theory and I have never come across this notation before.

The second thing that I can not grasp is how $\frac{G_{i+1} \cap H}{G_i \cap (G_{i+1} \cap H)}$ can be isomorphic to $\frac{G_i(G_{i+1} \cap H)}{G_i}$ by the first isomorphism theorem. The first isomorphism theorem states that if $\varphi: G \rightarrow H$ is a homomorphism between two groups then $G/\ker(\varphi) \cong$ im$(\varphi)$. To use this theorem, one first needs an isomorphism between two groups. However, in the proof, there is in my opinion, no clear isomorphism that is equivalent to $\varphi$ so I can not understand how one would use this theorem in this case.

Edit

Okay so I have been playing around and I think that I am on to something.
First, define the map $\varphi: G_{i+1} \cap H \rightarrow \frac{G_i(G_{i+1} \cap H)}{G_i}$ defined by $\varphi: g_h \mapsto (gg_h)G_i = g_hG_i$ where $g_h \in G_{i+1} \cap H$ and $g \in G_i$. We will now use $g$ and $g_h$ like we did here throughout this part.

Now $\varphi$ is a surjective homomorphism so im$(\varphi) = \frac{G_i(G_{i+1} \cap H)}{G_i}$. Notice that

$\ker(\varphi) = \{g_h \; | \; g_hG_i = G_i\}.$

Since $g_h$, per definition, is an element of $G_{i+1} \cap H$
then $g_h$ must also be an element of $G_i$ and so $g_h \in G_i \cap (G_{i+1} \cap H)$. This shows that $\ker(\varphi) \subseteq G_i \cap (G_{i+1} \cap H).$ By a similar argument, one can show that $G_i \cap (G_{i+1} \cap H) \subseteq \ker(\varphi)$ which means that $\ker(\varphi) = G_i \cap (G_{i+1} \cap H)$. Now, by the first isomorphism theorem, we have our desired isomorphism. Is this correct or am I missing something?

答:
It is the set product, defined like so: for $K,L\le G$,

$$KL=\{k\ell\in G\mid k\in K,\ell\in L\}.$$


问:
Show that $H$ is a normal subgroup of $G$.

Let $N$ be a normal subgroup of $G$. Let $H$ be the set of all elements $h$ of $G$ such
that $hn = nh$ for all $n \in N$. Show that $H$ is a normal subgroup of $G$.

My attempt: I've been trying to do this exercise and so far I've been able to show that $H$ is a normal subgroup of $N$ (the proof is similar to showing that the center is a subgroup), but that $H$ is a normal subgroup of $N$ does not always imply that $H$ be a normal subgroup of $G$. How could I show in this case that $H$ is indeed a normal subgroup of $G$?

Body:
Let $h\in H, n\in N$ for every $g\in G$, $ghg^{-1}n(ghg^{-1})^{-1}$

$=ghg^{-1}ngh^{-1}g^{-1}$, since $N$ is normal, $g^{-1}ng\in N$, we deduce that
$hg^{-1}ngh^{-1}=g^{-1}ng$, this implies that

$ghg^{-1}n(ghg^{-1})^{-1}=gg^{-1}ngg^{-1}=n.$


答:
We have to prove that if $h \in H$ and $g \in G$, then $g^{-1}hg \in H$. Note that $x \in H$ iff $xnx^{-1} = n$ for every $n \in N$ (i.e., iff conjugation by $x$ fixes $N$). For any $g \in G$, $h \in H$ and $n \in N$, we have:
$$
\begin{align*}
(g^{-1}hg)n(g^{-1}hg)^{-1}
   &= (g^{-1}hg)n(g^{-1}h^{-1}g) \\
   &= g^{-1}(h(gng^{-1})h^{-1})g \\
   &= g^{-1}(gng^{-1})g \tag*{because $gng^{-1} \in N$ and $h \in H$} \\
   &= (g^{-1}g)n(g^{-1}g) \\
   &= n
\end{align*}
$$
So with $x = g^{-1}hg$, we have $xnx^{-1} = n$ for every $n \in N$, so that $x \in H$, which is what we had to prove.

Body:
Let $N$ be a normal subgroup of $G$. Let $H$ be the set of all elements $h$ of $G$ such
that $hn = nh$ for all $n \in N$. Show that $H$ is a normal subgroup of $G$.

My attempt: I've been trying to do this exercise and so far I've been able to show that $H$ is a normal subgroup of $N$ (the proof is similar to showing that the center is a subgroup), but that $H$ is a normal subgroup of $N$ does not always imply that $H$ be a normal subgroup of $G$. How could I show in this case that $H$ is indeed a normal subgroup of $G$?


问:
Proving $1^k+2^k+ \dots + n^k = O(n^{k+1})$
I have to prove that : $1^k+2^k+ \dots + n^k = O(n^{k+1})$, where $k \in \mathbb{Z}^+$ . I followed a technique different from the book , and I want to check if my way is correct. I approach the problem using Induction.

Let $P(n)$ be the statement "$1^k+2^k+ \dots + n^k = \sum_{i = 1}^ni^k =  O(n^{k+1})$ ".

Induction basis: I will show that $P(1)$ is true. That equals: $1^k = O(1^{k+1})$. In other words, I have to prove that $\lim_{x\to \infty}\frac{1^x}{1^{x+1}} = L \geq 0$.

$\lim_{x\to \infty}\frac{1^x}{1^{x+1}} = \lim_{x\to \infty}\frac{1^x}{1^{x}1^1} = 1 \in \mathbb{R}$ so the Induction basis is proved.

Induction step: Let $P(l)$ be true, for some $l>1$. I will prove that $P(l) \rightarrow P(l+1)$.

As $P(l)$ is true, that means that $\sum_{i=1}^{l} = O(l^{k+1})$. Let's prove that $P(l+1)$ is true. I want to prove that $\sum_{i=1}^{l+1} = O((l+1)^{k+1})$ is true, so: $1^k + 2^k + \dots + l^k + (l+1)^k = O((l+1)^{k+1})$

I have to prove that: $\lim_{x \to \infty}\frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^{x+1}} = L \geq 0$

$
\begin{align}
\\
\lim_{x \to \infty}\frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^{x+1}}
 &= \lim_{x \to \infty}\frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^x \times (l+1)} \\
& = \frac{1}{l+1} \lim_{x \to \infty} \frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^{x}} \\
& = \frac{1}{l+1} [\lim_{x \to \infty} (\frac{1}{l+1})^x+ (\frac{2}{l+1})^x + \dots (\frac{l}{l+1})^x + 1 ]
\end{align}$ .

As $l > 1 \rightarrow l+1 > 2$ the above limit equals $0 + 0 + \dots + 0 + 1 = 1$. So $\frac{1}{1+l} \in \mathbb{R}$, and thus the induction step is proven, and the proof is concluded. $\blacksquare$

Is this way correct?

答:
I suppose $ k $ is a fixed integer, that was the first thing you've said. So we wanna prove this : $$ \sum_{p=1}^{n}{p^{k}}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(n^{k+1}\right) $$

If we wanted to use induction, we must do it on $ k $, not on $ n $ which is the variable of the asymptotic developpement. I suggest to following method though :

Notice that, for any $ k\in\mathbb{N} $ : $$ \frac{1}{n^{k+1}}\sum_{p=1}^{n}{p^{k}}=\frac{1}{n}\sum_{p=1}^{n}{\left(\frac{p}{n}\right)^{k}}\underset{n\to +\infty}{\longrightarrow}\int_{0}^{1}{x^{k}\,\mathrm{d}x}=\frac{1}{k+1} $$

Thus : $$ \sum_{p=1}^{n}{p^{k}}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(n^{k+1}\right) $$

That is due to the fact that any convergent sequence is bounded, since $ \left(\frac{1}{n^{k+1}}\sum\limits_{p=1}^{n}{p^{k}}\right)_{n} $ converges, $ \frac{1}{n^{k+1}}\sum\limits_{p=1}^{n}{p^{k}}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(1\right) $, hence the result.


问:
What's the function before $f_0(x)$ in Fast-growing Hierarchy
I was wondering about a function related to the fast-growing hierarchy that would be before $f_0(x) = x + 1$, i.e a function $f_{-1}(x)$ such that:

$f_{-1}^x(x) = x+1$ where, for example $f_{-1}^3(x) = f_{-1}(f_{-1}(f_{-1}(x)))$

Edit: This function isn't technically in the fast-growing hierarchy, because they are functions of integers. This implies that $f_0(x)$ is the slowest function that we can ùake, but we're actually extending the domain right here.

Can we get a formula for such a function (if it exists) ? Can we find more than one ? Is there some good properties ?

Here some precisions and what I already know.

Actually, this is easy to find this solution:
$f_{-1}(x) = x + \frac{1}{\lfloor x \rfloor}$, but it's not (in this problem) a "good" function, i.e it's not continuous or differentiable. But $f_0(x), f_1(x) = 2x$ and $f_2(x) = x2^x$ have these properties. I actually ask the question for these ones.

$f^0(x) = x$ in general, this is coherent with some properties of the re-iterating functions $f^{-1}(f^1(x)) = x = f^{1-1}(x) = f^0(x)$. It also can be read as a f that isn't applied to x. But here the definition says that $f_{-1}^0(0) = 1$ (which should be false). I think a great domain for $f_{-1}(x)$ is $[1,\infty[$, otherwise there could be contradictions.

It seems like if $f_{-1}(x)$ can be of the form $x + \sum_{k=1}^{\infty}{\frac{c_k}{x^k}}$ with $c_k \in \mathbb{R}$ (and with a converging serie) then $f_{-1}^x(x) \approx x+1$ for some "big" $x$ (but actually the approximation is in general getting quiet good), i.e $\lim_{x\rightarrow\infty}{\frac{f_{-1}^x(x)}{x+1}} = 1$ But I wasn't able to demonstrate it, though it's intuitive.

Here is a possible approximation:

$x + \frac{1}{x} + \frac{1}{1.9364832724923593}(\frac{1}{x^2} - \frac{1}{x^3}) + \frac{1}{3.5300211407995237}(\frac{1}{x^4} - \frac{1}{x^5}) + \frac{1}{0.62770748195252}(\frac{1}{x^7} - \frac{1}{x^6})$

Thank you :)

答:
I've recently studied on internet iterated functions and how to solve functional equations like $f(f(x)) = e^x$, which was pretty interesting. There's a method that it is use, which is about finding a solution to the Abel functional-equations:

$\alpha(f(x)) = \alpha(x) + 1$

Because it will follow that

$\alpha(f^t(x)) = \alpha(x) + t
\\ \iff f^t(x) = \alpha^{-1}(\alpha(x) + t)$

Which gives a solution to all iterated functions of f!
We're searching a function $f$ such that $f^n(n) = n+1$, let's try to find it's solution to the Abel functional equation (which is how I solved the problem):

$\alpha(f^t(x)) = \alpha(x) + t
\\ \implies \alpha(f^t(t)) = \alpha(t) + t
\\ \iff \alpha(t+1) = \alpha(t) + t$

This now seems pretty easy, we first try to solve it for t as an integer.

$\alpha(t+n) = \alpha(t) + \sum_{k=0}^{n-1}{t+k}
\\ = \alpha(t) + \sum_{k=0}^{n-1}{t} + \sum_{k=0}^{n-1}{k}
\\ = \alpha(t) + tn + \frac{n(n-1)}{2}$

Choosing $t=0$ we get

$ \alpha(n) = \alpha(0) + \frac{n(n-1)}{2}$

This is not rigourous (as everything that we'll do next), but let's generalise to every real $x$ (just replace $n$ with $x$). We've just found a solution to our original functional equation, with $\alpha(0)$ a constant!

Now, let's try to go further and find $\alpha^{-1}(x)$

$\alpha(x) = \alpha(0) + \frac{x(x-1)}{2}
\\ \iff \alpha(x) = \frac{1}{2}x^2 - \frac{1}{2}x + \alpha(0)
\\ \iff x = \frac{1}{2}\alpha^{-1}(x)^2 - \frac{1}{2}\alpha^{-1}(x) + \alpha(0)
\\ \iff \frac{1}{2}\alpha^{-1}(x)^2 - \frac{1}{2}\alpha^{-1}(x) + \alpha(0) - x = 0$

After using the method for solving second degree equations and simplifying, we get:

$\alpha^{-1}(x) = \frac{1 \pm \sqrt{1 - 8\alpha(0) + 8x}}{2}$

Now we can perhaps find a solution by using this formula.

$f(x) = \alpha^{-1}(\alpha(x) + 1)
\\ = \alpha^{-1}(\frac{1}{2}x^2 - \frac{1}{2}x + \alpha(0) + 1)
\\ = \frac{1 \pm \sqrt{1 - 8\alpha(0) + 8(\frac{1}{2}x^2 - \frac{1}{2}x + \alpha(0) + 1)}}{2}$

And then after simplification, we get rid of $\alpha(0)$!

$f(x) = \frac{1 \pm \sqrt{4x^2 - 4x + 9}}{2}$

If we choose the sign to be positive, we get a function that actually works!
Well I didn't prove it (yet, but I don't now if it's difficult), but this is essential since I've made a lot of assumptions.

Also, I don't know if we can construct others solutions, which I think might be really interesting! I was also surprised that there was such a simple formula for the problem, I'm sure that somebody has found this before.

For answering the question, yes such a continuous and derivable function exists, and we have:
$f_{-1}(x) = \frac{1 + \sqrt{4x^2 - 4x + 9}}{2}$


问:
Asymptotic expansion derivation Airy function
How do I derive the asymptotic expansion of airy function. We have been using it in WKB approximation method while deriving connection. I tried to derive it by expanding the integral but didn't reach anywhere. In wiki it is written that it can be derived by using complex analysis but there isn't a derivation. How to do it?</p>

Body:
The expansions are (have a look here)

$$\log (\text{Ai}(x))=-\frac{2 }{3}x^{3/2}-\frac{1}{4} \log \left({16\pi ^2 x}\right)-\frac{5}{48x^{3/2} } +O\left(\frac{1}{x^2}\right)$$
$$\log (\text{Bi}(x))=\frac{2 }{3}x^{3/2}-\frac{1}{4} \log \left({\pi ^2 x}\right)+\frac{5}{48x^{3/2} }
  +O\left(\frac{1}{x^2}\right)$$

May be, you could prefer
$$\text{Ai}(x)=\frac{e^{-\frac{2 t^3}{3}}}{2 \sqrt{\pi t} }\sum_{n=0}^\infty\frac {a_n}{t^{3n}}\qquad \text{where}\qquad t=\sqrt x$$ where the $a_n$ form the sequence
$$\left\{1,-\frac{5}{48},\frac{385}{4608},-\frac{85085}{663552},\frac{37182145}{12740
   1984},-\frac{5391411025}{6115295232}\right\}$$
$$\text{Bi}(x)=\frac{e^{\frac{2 t^3}{3}}}{\sqrt{\pi t} }\sum_{n=0}^\infty\frac {b_n}{t^{3n}}\qquad \text{where}\qquad t=\sqrt x$$ where the $b_n$ form the sequence
$$\left\{1,\frac{5}{48},\frac{385}{4608},\frac{85085}{663552},\frac{37182145}{1274019
   84},\frac{5391411025}{6115295232}\right\}$$