问:
How could I write the following expression without expanding it?
in a induction proof I stumbled over the following situation:

$$\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\frac{1}{n^{2n}}\left(-m\right)^k\left(n^2+nm+n+m\right)^{n-k}$$

The degree of the numerator is always smaller as the degree of the denominator.. so this limit goes to zero for every piece of the sum..

(I expanded it for some terms to check if this statement is indeed true or atleast should be)

How could I rewrite the sum, so that its clean and visually pleasing to see that all the pieces of the sum go to zero?

Thanks alot :)

Edit:

I came from here:

$\lim\limits_{n\longrightarrow\infty}\left(1+\frac{m+1}{n}\right)^n=\lim\limits_{n\longrightarrow\infty}\left(\frac{(n+1)(n+m)}{n^2}-\frac{m}{n^2}\right)^n=\lim\limits_{n\longrightarrow\infty}\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)-\frac{m}{n^2}\right)^n$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=0}\binom{n}{k}\left(-\frac{m}{n^2}\right)^k\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^{n-k}$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\left(-\frac{m}{n^2}\right)^k\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^{n-k}+\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^n$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\left(-\frac{m}{n^2}\right)^k\left(\frac{n^2+nm+n+m}{n^2}\right)^{n-k}+\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^n$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\frac{1}{n^{2n}}\left(-m\right)^k\left(n^2+nm+n+m\right)^{n-k}+\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^n$

答:
The problem with the argument "all the pieces of the sum go to zero" is that as $n$ increases, you keep getting new pieces, and you never stop getting new pieces.

I think you would need to prove that the whole sum goes to zero, not just individual pieces of it.

What if you back up to the last expression before you applied the binomial theorem and try something like this instead?

$$
\left(1+\frac1n\right)\left(1+\frac mn\right) - \frac{m}{n^2}
= \left(1+\frac1n\right)\left(1+\frac mn\right)
  \left(1 -  \frac{m}{\left(1+\frac1n\right)\left(1+\frac mn\right)n^2} \right).
$$

Now if you can prove that

$$
\lim_{n\rightarrow\infty}
\left(1 -  \frac{m}{\left(1+\frac1n\right)\left(1+\frac mn\right)n^2} \right)^n
= 0,
$$

I think the rest of the proof follows easily.


答:
I'm not entirely sure but it looks like you're making the following commonly made mistake. Consider the limit
$$\lim_{n\to\infty}\left(\frac1n+\frac1n+\cdots+\frac1n\right) $$
where the sum has $n$ terms. We have $\lim_{n\to\infty}(1/n)=0$ so each term has limit $0$. Then
$$\lim_{n\to\infty}\left(\frac1n+\frac1n+\cdots+\frac1n\right)=0+0+\cdots+0=0$$
But this is wrong. The sum is just $n\cdot\frac1n=1$, so it's limit is $1$. You can't generally do this unless the number of terms is constant, i.e. it doesn't depend on $n$. On the other hand, in this example and in your problem we have dependence on $n$.

ContentLicense: CC BY-SA 4.0

Title: How could I write the following expression without expanding it?

Body:
in a induction proof I stumbled over the following situation:

$$\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\frac{1}{n^{2n}}\left(-m\right)^k\left(n^2+nm+n+m\right)^{n-k}$$

The degree of the numerator is always smaller as the degree of the denominator.. so this limit goes to zero for every piece of the sum..

(I expanded it for some terms to check if this statement is indeed true or atleast should be)

How could I rewrite the sum, so that its clean and visually pleasing to see that all the pieces of the sum go to zero?

Thanks alot :)

Edit:

I came from here:

$\lim\limits_{n\longrightarrow\infty}\left(1+\frac{m+1}{n}\right)^n=\lim\limits_{n\longrightarrow\infty}\left(\frac{(n+1)(n+m)}{n^2}-\frac{m}{n^2}\right)^n=\lim\limits_{n\longrightarrow\infty}\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)-\frac{m}{n^2}\right)^n$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=0}\binom{n}{k}\left(-\frac{m}{n^2}\right)^k\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^{n-k}$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\left(-\frac{m}{n^2}\right)^k\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^{n-k}+\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^n$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\left(-\frac{m}{n^2}\right)^k\left(\frac{n^2+nm+n+m}{n^2}\right)^{n-k}+\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^n$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\frac{1}{n^{2n}}\left(-m\right)^k\left(n^2+nm+n+m\right)^{n-k}+\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^n$


问:
Prove that "the set of all real numbers defined by Dedekind cuts" and "the set of all finite lengths on an infinite straight line" are isomorphic
I apologise for this question to be quite long... it looks wordy since I'm trying to express the process of my thinking.

I'm recently using the appendix at the end of chapter 1 in Rudin's Principle of Mathematical Analysis to study the construction of $\mathbb{R}$ from $\mathbb{Q}$.

I fall into a somewhat circular argument $\color{red}{(\ast)}$ when I try to visualize cuts as finite lengths on an infinite straight line (by finite length I mean the segment from a reference point "0" to some point on the line).

I'm trying to visualize cuts as lengths on the line because I think "length" is the intuitive concept that is central to the story: people in the old days first found out that there exist some certain lengths that could not be represented by the elements of $\mathbb{Q}$ (namely the diagonals of squares), then this became the reason for cuts to come.

So, the first question I asked myself is:

How could one be sure that, after defining the cuts, every finite length on the line is indeed included in $\mathbb{R}$?

My answer to this is: I can now pick any one point on the line, denote it as $x$, and I can associate a set $r$ which consists of all the rational points to the left of the point $x$ so that $r$ is a cut. And I'm now representing $x$ by $r \in \mathbb{R}$.

Then, I asked myself the second question:

How could I know that such a cut $r$ is actually unique to my length $x$? That is, how could I ensure that my cut $r$ will not represent $x$ anymore even if I move my point $x$ to the right on the line for just a tiny tiny bit?

For this question, let's denote the original $x$ as $x_0$ and the $x$ after moving as $x_1$. My idea goes like: if we can always find a rational number between $x_0$ and $x_1$ (no matter how close they are), then $x_1$ must be represented by a different cut (let's denote it as $s$) because by the definition of cut, $r$ is now a proper subset of $s$.

[Please note also that I did not justify my premise above using the property that $\mathbb{Q}$ is dense in $\mathbb{R}$, because I still don't know if $x_0$ and $x_1$ are two distinct reals at the moment]

So, at this point, I made two important assumptions:

Any two points on the line must either be the same point or be different. So, as long as I move the point, it becomes a different one to the original.

The settings for the lengths on the line make them isomorphic to the cuts. i.e. assume that some proper, possibly customary, addition, multiplication and order which operates on the lengths are defined, and are preserved by some bijection between the set of all lengths and the the set of all cuts.

Using those two assumptions, we now have the two different points on the line representing by two different cuts.

For assumption 1, I think it is rather like an axiom.

And Finally this leads to the humble question I wrote in the title:

How do we prove assumption 2? That is, how can we construct an isomorphism between the set of all real numbers defined by (Dedekind's) cuts and the set of all finite lengths (including zero length) on an infinite straight line?

For now, there are two angles that I tried to solve this question in general:

As it is a fact that any two ordered fields with the least-upper-bound property are isomorphic, I was trying to prove that the set of all lengths on the line forms an ordered field with the least-upper-bound property under some good definitions. But I'm stucking somewhere in the middle.

Assume that there exist two points on the line $x_0$ and $x_1$ as described above where $x_1$ is to the right of $x_0$, and there exists some cut $r$ that catches them both. Then, the length between $x_0$ and $x_1$ must be finite (due to assumption 1), and I don't see a reason why any rational cuts wouldn't be surrounded by such "blank zone" or why those blank zones would have various lengths (I don't think my naive understanding to the concept of length would let me say more on these). By mathematical induction, there are infinitely many rationals between any two different rationals, and thus we would have some finite constant multiplied by infinity equals some finite quantity, which seems logically incorrect. But to me, such "proof by contradiction" is far not rigorous.

By the way, I have a feeling that constructing $\mathbb{R}$ from $\mathbb{Q}$ by means of Cauchy sequences might be a better approach to this question, I haven't read such construction yet because I believe that cuts should be equally good.

$\color{red}{(\ast)}$ : I think what is (probably) being circular here is as follows (and this is what really confuses me):

From the definition of cut, it is clear that any two distinct cuts cannot represent the same length on the line. But now I want to prove that any two distinct lengths on the line cannot be represented by the same cut. Symbolically, I want to prove that if the two lengths on the line are denoted as $x_0$ and $x_1$ where $x_0 \neq x_1$, and their associated cuts are denoted as $r_0$ and $r_1$ respectively, then $r_0 \neq r_1$. By the definition of the order defined on cuts, to prove $r_0 \neq r_1$ I need to show that one is the proper subset of another. This means that I need to find a rational in between. However, to conclude that there exists a rational between two cuts it requires the two cuts not to be equal in the first place.

Could anyone please tell me how to get rid of this circularity?

There may be ambiguities in multiple places in this question as I am still being confused about it right now... so anyone who notices why I'm confused is more than welcomed to write an answer even if the answer is unrelated to my arguments above :)

Thank you for your patience :)

答:
We can show that $(i)$ any two nontrivial connected ordered groups are isomorphic and $(ii)$ the set of Dedekind cuts, with the usual ordering and addition, forms a nontrivial connected ordered group. So we are forced to conclude that the Dedekind cut model faithfully captures the "Platonic line" as long as we grant that the Platonic line is connected and that we can add and subtract lengths appropriately.

To me, both connectedness and nontrivial groupiness are totally fundamental to my conception of the line, so this solves the problem completely.


问:
Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$
I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?

$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$$

ContentLicense: CC BY-SA 3.0

Body:
First, let's make a common denominator of $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta}$.

Camera? Action!
$$\require{cancel}\begin{aligned}\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta}&=\frac{\sin^4{\theta}+\cos^4{\theta}}{\sin^2{\theta}\cos^2{\theta}}\\&=\frac{\left(\sin^2{\theta}+\cos^2{\theta}\right)^2-2\sin^2{\theta}\cos^2{\theta}}{\sin^2{\theta}\cos^2{\theta}}\\&=\frac{1-2\sin^2{\theta}\cos^2{\theta}}{\sin^2{\theta}\cos^2{\theta}}\\&=\frac{1}{\sin^2{\theta}\cos^2{\theta}}-\frac{2\cancel{\sin^2{\theta}\cos^2{\theta}}}{\cancel{\sin^2{\theta}\cos^2{\theta}}}\\&=\frac{1}{\sin^2{\theta}\cos^2{\theta}}-2\\&=\csc^2{\theta}\sec^2{\theta}-2\\&=\sec^2{\theta}\csc^2{\theta}-2\end{aligned}$$


问:
Example of knot that "becomes unknotted in a different 3-manifold"?
Suppose I had a knot $K \subset \mathbb{R}^3$, and an open ball $U$ inside a $3$-manifold $M$. If I now consider $K \subset U$ (via some homeomorphism of $\mathbb{R}^3$ with $U$), could it be possible that $K$ is unknotted in $M$? That is to say, $K$ becomes isotopic to the unknot in $U$; somehow "using the additional geometry that $M$ provides to make unknotting easier"? If so, are there any easy examples I can write down?

答: No, this isn't possible.  Let $S$ be the boundary of the closure of $U$, which is a $2$-sphere in $M$.  If $K$ is unknotted, that means there is an embedded disk $D\subset M$ whose boundary is $K$.  We can isotope $D$ so that it intersects $S$ transversely.  The intersection $D\cap S$ is a disjoint union of circles.  Let's assume $D$ is chosen to minimize the number of circles in this intersection.  If there is a circle, then there is one that bounds a disk $D'\subset S$ whose interior contains none of these circles.  Compressing $D$ along $D'$ (and then throwing away any $2$-sphere components) yields a new disk whose boundary is $K$ and which intersects $S$ in fewer circles, which is a contradiction. Therefore, $D\subset U$, and $K$ is an "unknot in $\mathbb{R}^3$."


问:
Banach space - using continuity to show contradiction
Let $X=(C[-1,1], \| \cdot \|_\infty)$. Let $Y=\{f \in C[-1,1] : \int_{-1}^0 f = \int_0^1 f \}$ be a subspace.

Let $\phi(f)= \int_{-1}^0 f -\int_0^1 f $.

Then we have $Y=Ker(\phi)= \phi^{-1}\{0 \}$.

Suppose there exists $v \in x$ with $\| v\|_{\infty}=1$ such that $\phi(v)= \| \phi\|$. Then I want to show that this is a contradiction.

I tried to show the contradiction by using that $\phi$ is continuous.

i.e. I used a similar argument as Let $X=\{f\in C[0,1] :f(0)=0 \}$. Prove that there does not exist $f\in X$ such that $\|f \|_{\infty}\leq 1$ and $|T(f)|= 1$., by taking $r>0$. Then I get $0 = | \phi(t)| <2 -3r$, which didn't get anywhere.

Any help is appreciated!

Thanks!

答:
For every $f$ one has
$$
  |\phi(f)| \leq
  \left|\int_{-1}^0f(x)\,  dx\right| +   \left|\int_0^1f(x)\,  dx|\right| \leq  \|f\| + \|f\| = 2 \|f\|,
  $$
so $\|\phi\|\leq 2$.  Letting
$$
  f_n(x) = \left\{\matrix{
  1, & \text {if }  -1\leq x<-1/n,\hfill \cr
  -nx, & \text {if } -1/n\leq x\leq 1/n, \cr
  -1, & \text {if }  \phantom{- \ } 1/n\leq x\leq 1,\hfill \cr
  }\right.
  $$
we have $\|f_n\|=1$, and $\lim_{n\to\infty }\phi(f_n)=2$, so actually  $\|\phi\|=2$.

Given $f$ in $C[-1, 1]$ with   $\|f\|=1$, let
$$
  A=\int_{-1}^0f(x)\,  dx, \quad \text{and}\quad   B=\int_0^1f(x)\,  dx.
  $$
Observing that $-1\leq A,B\leq 1$, the only way one can get $\phi(f)=2$, that is,
$A-B=2$, is when $A=1$ and $B=-1$. In this case we have
$$
  0 = 1-A = \int_{-1}^0\,  dx - \int_{-1}^0f(x)\,  dx
  = \int_{-1}^0\, 1-f(x)\,  dx.
  $$
Since $1-f(x)$ is continuous and nonnegative, this implies that $f(x)=1$ on $[-1,0]$.  In a similar way it follows
that $f(x)=-1$ on $[0, 1]$, but this violates the continuity of $f$.  Therefore you never get $\phi(f)=2$.


问:
formal definition of limit to prove floor function limit
$$\lim_{ x\to1−} ([|x|] − 2) = −2 $$

I'm trying to understand the proof of the following limit using a formal definition.

I know that;

$\lim_{x\to c−}f(x)=L$,means that given any $ϵ>0$, there exists $δ>0$ such that for all $x<c$, if $|x−c|<δ$, then $|f(x)−L|<ϵ$.
but I'm not sure how I can handle the rest

答:
Hint: for a floor function, take note of the following ($n \in \mathbb{N}$):

$\lim_{x \rightarrow n^{-}}\lfloor x \rfloor = n-1$

and on the other hand,

$\lim_{x \rightarrow n^{+}}\lfloor x \rfloor = n$.

Now, think of this from the perspective of the $\delta$-$\epsilon$-definition for a specific $\delta$. Then consider $|x-1|<\delta$. What follows from this?


问:
Basic question about $\sup_{x\neq 0}{} \frac{\|Ax\|}{\|x\|} = \sup_{\|x\| = 1}{\|Ax\|} $, $x \in\mathbb{R}^n$
I am having trouble with understanding the following definition while studying some basic things related with matrix norms:

For every matrix $A\in M_n(\mathbb{R})$

$$\sup_{x\neq 0}{} \frac{\|Ax\|}{\|x\|} = \sup_{\|x\| = 1}{\|Ax\|},\; x \in\mathbb{R}^n$$

Why we are taking $\|x\| = 1$?  Is there any proof of above statement?

Added: Why to take sup in the definition of norm?

I need help to understand above mentioned fact.

Thank you very much.

答:
$$\frac{ \|Ax\| }{\|x\|} = \left|\left| A \frac{x}{\|x\|} \right|\right|$$
Running through all $x \neq 0$ is equivalent to running through all $y:= \frac{x}{\|x\|} $ with $\| y \|=1$.</p>


问:
Laplace Equation -> Boundary Conditions
Given $\nabla^2 \phi=0$,

With B.C.

$\Gamma_1 =-V$   (Left side of rectangle)

$\Gamma_2 = V$   (right side of rectangle)

$\Gamma_0 = 0$   (top and bottom sides of rectangle)

Can Separation of Variables be applied to this problem?  Will the superposition principle have to be applied to the second ODE because there are two non-zero BC?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\underline{\tt Domain}:\ {\large\pars{0,b}\times\pars{0,h}}}$

It's always better, at the very beginning, to take advantage of:

the homogeneous boundary conditions : It's
$\ds{\underline{\mbox{equivalent to}\ variable\ separation}}$ and it's less "painful".

the problem symmetries: The solution must changes its sign when $\ds{x \mapsto b - x}$ and it's invariant when
$\ds{y \mapsto h - y}$.

The above statements lead to
$\ds{\phi\pars{x,y} =
\sum_{n = 0}^{\infty}\on{a}_{n}\pars{x}\sin\pars{k_{n}y}}$ where $\ds{k_{n} \equiv \pars{2n + 1}\,{\pi \over h}}$ and
$\ds{\on{a}_{n}\pars{x} = -\on{a}_{n}\pars{b -x}}$.
Note that, with $\ds{m, n \in \mathbb{N}_{\,\geq\ 1}}$,
$$
\left\{\begin{array}{rcl}
\ds{\int_{0}^{h}\sin\pars{k_{m}y}\sin\pars{k_{n}y}\,\dd y}
& \ds{=} &
\ds{{1 \over 2}\,h\,\delta_{mn}}
\\
\ds{\int_{0}^{h}\sin\pars{k_{n}y}\,\dd y}
& \ds{=} &
\ds{{2 \over \pars{2n + 1}\pi}\,h\,\delta_{mn}}
\end{array}\right.
\label{1}\tag{1}
$$

$\ds{\phi\pars{x,y}}$ must satify the differential equation:
$$
\sum_{n = 0}^{\infty}
\bracks{\pars{\totald[2]{}{x} - k_{n}^{2}}\on{a}_{n}\pars{x}}\sin\pars{k_{n}y} = 0
\label{2}\tag{2}
$$

Use (\ref{1}) and (\ref{2}). I get
\begin{align}
&\pars{\totald[2]{}{x} - k_{n}^{2}}\on{a}_{n}\pars{x} = 0
\implies
\on{a}_{n}\pars{x} =
c_{n}\sinh\pars{k_{n}\bracks{x - {b \over 2}}}
\end{align}
The general solution is reduced to:
$$
\phi\pars{x,y} =
\sum_{n = 0}^{\infty}
c_{n}\sinh\pars{k_{n}\bracks{x - {b \over 2}}}\sin\pars{k_{n}y}
$$

The boundary conditions yield
\begin{align}
&V = \sum_{n = 0}^{\infty}
c_{n}\sinh\pars{k_{n}\bracks{b \over 2}}\sin\pars{k_{n}y}
\\[5mm] \implies &\
\overbrace{V\,{2 \over h}\int_{0}^{h}\sin\pars{k_{n}y}
\dd y}^{\ds{4V \over \pars{2n + 1}\pi}}
\\[2mm] = &\
\sum_{m = 0}^{\infty}
c_{m}\sinh\pars{k_{m}\,{b \over 2}}\
\underbrace{\bracks{{2 \over h}
\int_{0}^{h}\sin\pars{k_{n}y}\sin\pars{k_{m}y}\dd y}}
_{\ds{\delta_{nm}}}
\\[5mm] &\ \implies
c_{n} =
{4V \over \pars{2n + 1}\pi\sinh\pars{k_{n}b/2}}
\end{align}

Finally,
$$
\bbx{\phi\pars{x,y} =
{4V \over \pi}\sum_{n = 0}^{\infty}
{\sinh\pars{k_{n}\bracks{x - b/2}}\sin\pars{k_{n}y} \over
\pars{2n + 1}\sinh\pars{k_{n}b/2}}} \\
$$

Title: Laplace Equation -> Boundary Conditions


问:
Verify algebraically that the equation $\frac{\cos(x)}{\sec(x)\sin(x)}=\csc(x)-\sin(x)$ is an identity
I am stuck when I get to this point $\frac{\cos^2(x)}{\sin(x)}$.
Am I on the right track?

Verify algebraically that the equation is an identity:

$$\frac{\cos(x)}{\sec(x)\sin(x)}=\csc(x)-\sin(x)$$

My work:

$$\frac{\cos(x)}{\frac{1}{\cos(x)}\cdot\sin(x)}=\frac{\cos(x)}{\frac{\sin(x)}{\cos(x)}}=\frac{\cos(x)}{1}\cdot\frac{\cos(x)}{\sin(x)}=\frac{\cos^2(x)}{\sin(x)}$$

答:
First, we'll expand $\sec{x} = \frac{1}{\cos{x}}$.

Let's begin!
$$\require{cancel}\begin{aligned}\frac{\cos{x}}{\sec{x}\sin{x}}&=\frac{\cos{x}}{\frac{1}{\cos{x}}\cdot\sin{x}}\\&=\frac{\cos{x}}{\frac{\sin{x}}{\cos{x}}}\\&=\cos{x}\cdot\frac{\cos{x}}{\sin{x}}\\&=\frac{\cos^2{x}}{\sin{x}}\\&=\frac{1-\sin^2{x}}{\sin{x}}\\&=\frac{1}{\sin{x}}-\frac{\cancelto{\sin{x}}{\sin^2{x}}}{\cancelto{1}{\sin{x}}}\\&=\csc{x}-\sin{x}\end{aligned}$$
I hope this helps.


问:
$\lim_{(x,y,z)\rightarrow(0,0,0)}((\frac{xyz}{x^2+y^2+z^2})^{(x+y)})$
I'm quite stumped. The answer to this question is it does not exist, but I've tried using $y=x$, $z=x$ and $y=x^2$ and $z=x^2$, or even $y=\sqrt{x}$ and $z=\sqrt{x}$, but they all give me the answer $1$. How can I find two diferent pathways to prove it does not exist?

答: First take limit along the line x=y=z and we get the limit to be 1. Next take limit along the y-z plane , that is when x=0. The limit is 0. Since along two different paths we get two different limits, we conclude that the limit doesn't exist.</p>


问:
How do I find the function represented by the following series?
$$\sum_{k=0}^{\infty}\frac{2(-1)^k x^{k+2}}{k!}$$

I'm supposed to find the function represented by the power series but I'm really stumped... (x is raised to the k+2 power)

答: Factor out $2x^2$ and rewrite the rest as a series of the form $$\sum_{k=0}^\infty \frac{z^k}{k!} = e^z.$$


问:
Calculating permutations if the sequences have to be in ascending order?
How would you go about calculating the number of permutations in ascending order.

Obviously if you had (a set of) 3 numbers you have $ 3! $ permutations:

(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1)

But only one of these is in ascending order (1,2,3).

Consider the lottery - picking 6 numbers from 49. Normally the order for this doesn't matter you just calculate '49 choose 6' but if the numbers have to be in ascending order how would you calculate that?

答: Hint: A combination is a permutation where order does not matter.


答:
If you’re allowed to use the fact that $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ for all x, then you can write

\begin{align*}
\sum_{k=0}^{\infty}\frac{2(-1)^k x^k+2}{k!} &= \sum_{k=0}^{\infty}2\cdot\frac{(-x)^k+1}{k!}\\
&= 2\sum_{k=0}^{\infty}\left(\frac{(-x)^k}{k!}+\frac{1}{k!}\right)\\
&= 2\sum_{k=0}^{\infty}\frac{(-x)^k}{k!}+2\sum_{k=0}^{\infty}\frac{1}{k!}\\
&= 2e^{-x}+2e
\end{align*}

Title: How do I find the function represented by the following series?

Body:
$$\sum_{k=0}^{\infty}\frac{2(-1)^k x^{k+2}}{k!}$$

I'm supposed to find the function represented by the power series but I'm really stumped... (x is raised to the k+2 power)


问:
Why is the $n$-dimensional Lebesgue measure by default not complete?
Since the Lebesgue measure is complete, why is the product of Lebesgue measures not necessarily complete? Would anyone be able to provide some intuition or an example?</p>

答:
Let $\mu$ be the Lebesgue measure on $\mathbb{R}$.  The product measure $\mu \times \mu$ is defined on the $\sigma$-algebra generated by rectangles of the form $A \times B$ where both $A$ and $B$ are Lebesgue measurable subsets of $\mathbb{R}$.

Now suppose that $A \subseteq \mathbb{R}$ has measure $0$, and $B \subseteq \mathbb{R}$ is measurable.  Then $(\mu \times \mu)(A \times B) = 0$, but an arbitrary subset of $A \times B$ does not have to be in the $\sigma$-algebra generated by measurable rectangles.

For a concrete example, consider the case when $A = \{0\}$ and $B = [0, 1]$.  Let $V \subseteq [0, 1]$ be the Vitali set (or any other non-measurable set you like).  Then $(\mu \times \mu)(A \times B) = (\mu \times \mu)(\{0\} \times [0, 1]) = 0$, and $\{0\} \times V \subseteq \{0\} \times [0, 1]$, but $\{0\} \times V$ is not in the $\sigma$-algebra generated by measurable rectangles.

By the way, I think it's more common to use the phrase "$2$-dimensional Lebesgue measure" to refer to the completion of $\mu \times \mu$, rather than to mean $\mu \times \mu$ itself.  With that naming convention, $n$-dimensional Lebesgue measure is complete, by definition.


问:
Direct formula for area of a triangle formed by three lines, given their equations in the cartesian plane.
I read this formula in some book but it didn't provide a proof so I thought someone on this website could figure it out. What it says is:
If we consider 3 non-concurrent, non parallel lines represented by the equations :
$$a_1x+b_1y+c_1=0$$
$$a_2x+b_2y+c_2=0$$
$$a_3x+b_3y+c_3=0$$
Then the area of the triangle that these lines will enclose is given by the magnitude of :
$$\frac{det\begin{bmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{bmatrix}^2}{2C_1C_2C_3}$$
Where $C_1,C_2,C_3$ are the co-factors of $c_1,c_2,c_3$ respectively in the above matrix.

What I'm wondering is, where did this come from? And why isn't it famous? Earlier we had to calculate areas by finding the vertices and all but this does it in a minute or so and thus deserves more familiarity.

答:
Let   a1x + b1y + c1 = O be the equation of the first side.( We will represent this by L1=O )

Similarly   a2x + b2y + c2 = O
and   a3x + b3y + c3 = O be the equation of 2nd and 3rd side respectively ,represented by L2 and L3 .

Let (x1 , y1) , (x2 , y2) , (x3, y3) be the vertices of the Triangle

So The Area of the  Triangle will be the Determinant $\Delta$o

\begin{align*}
  \begin{vmatrix}
        x_1 & y_1 & 1 \\
        x_2 & y_2 & 1 \\
        x_3 & y_3 & 1 \\
    \end{vmatrix}.
\end{align*}
Consider another Determinant $\Delta$1
\begin{align*}
  \begin{vmatrix}
        a_1 & a_2 & a_3 \\
        b_1 & b_2 & b_3 \\
        c_1 & c_2 & c_3
    \end{vmatrix}.
\end{align*}

Let Determinant $\Delta$2 be $\Delta$o $\Delta$1
which is equal to the Determinant
\begin{align*}
  \begin{vmatrix}
        a_1x_1+b_1y_1+c_1 & a_2x_1+b_2y_1+c_2 & a_3x_1+b_3y_1+c_3 \\
        a_1x_2+b_1y_2+c_1 & a_2x_2+b_2x_2+c_2 & a_3x_2+b_3y_2+c_3 \\
        a_1x_3+b_1y_3+c_3 & a_2x_3+b_3x_3+c_3 & a_3x_3+b_3x_3+c_3
    \end{vmatrix}.
\end{align*}

But (x1 ,y1) lies on L2 and L3

So a1x2 + b1y2+ c1 = O
and  a1x3 + b1y3+ c1 = O
Similarly

a2x1 + b2y1+ c2 = O
a2x3 + b2y3+ c2 = O
and
a3x1 + b3y1+ c3 = O
, a3x2 + b3y2+ c3 = O

So the Determinant reduces to

\begin{align*}
  \begin{vmatrix}
        a_1x_1+b_1y_1+c_1 & 0 & 0 \\
        0 & a_2x_2+b_2x_2+c_2 & 0 \\
        0 & 0 & a_3x_3+b_3x_3+c_3
    \end{vmatrix}.
\end{align*}
So $\Delta$2 = L1(x1,y1)L2(x2,y2)L3(x3,y3)

Now we shall prove that

a1x1+b1y1+c1 = K1(Let) = $\Delta$1/a3b2-a2b3

Now we observe  that (x1,y1) is the solution of the system of equations

a1x1+b1y1+c1-K1 = O
a2x1+b2y1+c2=O
a3x1+b3y1+c3=O

So
\begin{align*}
  \begin{vmatrix}
        a_1 & a_2 & a_3 \\
        b_1 & b_2 & b_3 \\
        c_1-K_1 & c_2 & c_3
    \end{vmatrix}.
\end{align*}
=0=$\Delta$1 - $\Delta$3
Where $\Delta$3 =
\begin{align*}
  \begin
{vmatrix}
        a_1 & a_2 & a_3 \\
        b_1 & b_2 & b_3 \\
        K & 0 & 0
    \end{vmatrix}.
\end{align*}

So $\Delta$3 = K1(a3b2-2b3)

So 0 = $\Delta$1 - K1(a3b2-2b3)

Therefore K1= $\Delta$1/a3b2-a2b3

So K1 =L1(x1,y1) = $\Delta$1/a3b2-a2b3

Similarly  K2 =L2(x2,y2) = $\Delta$1/a3b1-a1b3

and

K3 =L3(x3,y3) = $\Delta$1/a1b2-a2b1

Since C1= a3b2-a2b3
and C2= a3b1-a1b3
and C3= a1b2-a2b1

So $\Delta$o$\Delta$1=L1(x1,y1) L2(x2,y2) L3(x3,y3)= K1K2K3 = $\Delta$3/C1C2C3

And Therefore the area of the triangle is

$\Delta$2/C1C2C3


问:
What categorical limits and colimits does $\pi_1$ preserve?
$\pi_1$ is a functor from the category of pointed topological spaces to the category of groups.

It's clear that this functor preserves products, since $\pi_1(X\times Y)=\pi_1(X)\times\pi_1(Y)$, and a special case of Van Kampen's theorem states that $\pi_1$ preserves coproducts, since $\pi_1(X\vee Y)=\pi_1(X)*\pi_1(Y)$.

It doesn't seem like too much trouble to show that $\pi_1$ preserves pullbacks, and Van Kampen's theorem helps us with pushouts up to some degree of niceness concerning our spaces.

How many other limits or colimits does $\pi_1$ preserve? I'm not sure I know where to start if I want to talk about equalizers and coequalizers!

If we could show how $\pi_1$ preserves products and equalizers could we conclude that $\pi_1$ preseved all limits (since $Top$ and $Grp$ are both complete and cocomplete)? Could we say the same thing about coproducts?

Are there any obvious counter-examples?

I've had a good look through Hatcher's 'Algebraic Topology' and a quick breeze through JP May's 'Concise Course', but I'm not finding all the answers I'm after.

Any words on this topic would be appreciated :)

答:
In addition to the other answers,
$π_1$ also preserves something for homotopy inverse limits over $ω$ (or any other ordinal sequence).

Let
$$X \to \dots \to X_2 \to X_1 \to X_0$$
denote an inverse system of pointed spaces with $X$ being the homotopy limit of the inverse system.
Then the natural map
$$π_1(X) \longrightarrow \varprojlim_{n < ω} π_1(X_n)$$
is surjective.


问:
$f:\mathbb{R}\mapsto \mathbb{R}$ continuous and $f_n(x)=f(nx)$ is equicontinuous then...
This is problem 9 from page 264 of Pugh's Real Mathematical Analysis. If $f:\mathbb{R}\mapsto \mathbb{R}$ continuous and the sequence of functions $f_n(x)=f(nx)$ is equicontinuous then what can be said about $f$?

My gut tells me that $f$ must be a constant function, and I've tried to prove it by showing that $f'$ must be zero everywhere, but I haven't been successful. Could someone help me prove this, or tell me what is actually going on here?

答:
Your gut is right: $f$ must be constant. Fix $x_0,y_0\in\mathbb R$ and take any $\varepsilon>0$. By equicontinuity, there must exist some $\delta>0$ such that
\begin{align*}
\left.\begin{array}{l}x,y\in\mathbb R\\|x-y|<\delta\end{array}\right\}\quad\text{imply that}\quad|f(nx)-f(ny)|<\varepsilon\text{ for all $n\in\mathbb N$.}
\end{align*}
Now take $n\in\mathbb N$ so large that $|x_0-y_0|<n\delta$. Since $$\left|\frac{x_0}{n}-\frac{y_0}{n}\right|<\delta,$$ it follows that $$|f(x_0)-f(y_0)|=\left|f\left(n\frac{x_0}{n}\right)-f\left(n\frac{y_0}{n}\right)\right|<\varepsilon.$$
Since $\varepsilon>0$ can be arbitrarily small, one can conclude that $f(x_0)=f(y_0)$.


问:
Integration problem yields two solutions. But are they the same?
I have an integration problem: $\int\frac{1}{x\sqrt {x^{12}-1}}dx$.

My attempt was to let $u=\sqrt {x^{12}-1}$. Solving it yields:

$$\frac{1}{6}\int\frac{1}{u^2+1}du$$
$$=\frac{1}{6}\tan^{-1}(u)+C$$
$$=\frac{1}{6}\tan^{-1}\sqrt {x^{12}-1}+C$$

However, my TA substituted $u=\frac{1}{x}$, and subsequently $v=u^6$. Solving it yields:

$$-\frac{1}{6}\int\frac{1}{\sqrt {1-v^2}}dv$$
$$=-\frac{1}{6}\sin^{-1}(v)+C$$
$$=-\frac{1}{6}\sin^{-1}\frac{1}{x^6}+C$$

I do not understand how the two answers are different but mathematically are correct. When I tried to draw a right-angled triangle using the first solution, I would have:

opposite = $\sqrt{x^{12}-1}$

adjacent = $1$

hypotenuse = $x^6$

So from the above, $\sin\theta=\frac{\sqrt{x^{12}-1}}{x^6}$ and $\theta=\arcsin\frac{\sqrt{x^{12}-1}}{x^6}$, which is not the same as $-\arcsin\frac{1}{x^6}$

May I know what is wrong with my deduction above?

Note: I also tried to differentiate both solutions. The first solution will yield back to the integrand. But the second solution does not. But strangely, if I use any arbitrary limits on the two solutions, both give me the same answer.

Thank you in advance.

答:
According to your consideration,

$\cos\theta = \frac{1}{x^6} \Rightarrow \theta = \cos^{-1}(\frac1{x^6})$

And $\cos^{-1}x = \frac\pi2-\sin^{-1}x$.

So, $\boxed{\theta = \frac\pi2 - \sin^{-1}\left(\frac{1}{x^6}\right)}$.

Now, $\frac{1}{6}\cdot\frac\pi2$ is just a constant and you have the required result as $\boxed{- \frac16\sin^{-1}\left(\frac{1}{x^6}\right) +k}$


答:
The graph of $\frac{1}{6}\tan^{-1}\sqrt{x^{12}-1}$ and $-\frac{1}{6}\sin^{-1}\frac{1}{x^6}$ looks like this:

The red line is $-\frac{1}{6}\sin^{-1}\frac{1}{x^6}$ and the other $\frac{1}{6}\tan^{-1}\sqrt{x^{12}-1}$. They are just the same, but what makes the difference is the $+C$ constant. This is why the constant of integration is important. Both expressions differ only by a constant.

Title: Integration problem yields two solutions. But are they the same?

Body:
I have an integration problem: $\int\frac{1}{x\sqrt {x^{12}-1}}dx$.

My attempt was to let $u=\sqrt {x^{12}-1}$. Solving it yields:

$$\frac{1}{6}\int\frac{1}{u^2+1}du$$
$$=\frac{1}{6}\tan^{-1}(u)+C$$
$$=\frac{1}{6}\tan^{-1}\sqrt {x^{12}-1}+C$$

However, my TA substituted $u=\frac{1}{x}$, and subsequently $v=u^6$. Solving it yields:

$$-\frac{1}{6}\int\frac{1}{\sqrt {1-v^2}}dv$$
$$=-\frac{1}{6}\sin^{-1}(v)+C$$
$$=-\frac{1}{6}\sin^{-1}\frac{1}{x^6}+C$$

I do not understand how the two answers are different but mathematically are correct. When I tried to draw a right-angled triangle using the first solution, I would have:

opposite = $\sqrt{x^{12}-1}$

adjacent = $1$

hypotenuse = $x^6$

So from the above, $\sin\theta=\frac{\sqrt{x^{12}-1}}{x^6}$ and $\theta=\arcsin\frac{\sqrt{x^{12}-1}}{x^6}$, which is not the same as $-\arcsin\frac{1}{x^6}$

May I know what is wrong with my deduction above?

Note: I also tried to differentiate both solutions. The first solution will yield back to the integrand. But the second solution does not. But strangely, if I use any arbitrary limits on the two solutions, both give me the same answer.

Thank you in advance.


问:
On the Hilbert Curve
(1)A mapping  \begin{equation}{f}\colon{X}\rightarrow{Y}\end{equation} is said to be a homeomorphism of it is (1) f is a bijection (2) f is continuous (3)the inverse mapping f inverse is also continuous.
https://en.wikipedia.org/wiki/Homeomorphism#Definition

The bijection \begin{equation}R^2\leftrightarrow R\end{equation} between the unit square and the Hilbert curve should be a homeomorphism unless the definition includes differentiability as an essential criterion[or perhaps if the inverse function is not continuous].
Requesting a clarification to this issue.

答: Recall that a homeomorphism should preserve topological properties. If you remove one point $p$ from $\mathbb{R}$, then a homeomorphism $\phi: \mathbb{R} \to \mathbb{R}^2$ should restrict to a homeomorphism $$\phi:\mathbb{R} \setminus \{p\} \to \mathbb{R}^2 \setminus \{\phi(p)\}.$$ This cannot be since $\mathbb{R} \setminus \{p\}$ is disconnected whereas $\mathbb{R}^2 \setminus \{\phi(p)\}$ is connected. Moreover, the hilbert curve is not bijective since it is not necessarily injective.

Title: On the Hilbert Curve


问:
Proving $6 \sec\phi \tan\phi = \frac{3}{1-\sin\phi} - \frac{3}{1+\sin \phi}$

$$6 \sec\phi \tan\phi = \frac{3}{1-\sin\phi} - \frac{3}{1+\sin \phi}$$

I can't seem to figure out how to prove this.

Whenever I try to prove the left side, I end up with $\frac{6\sin\theta}{\cos\theta}$, which I think might be right.

As for the right side, I get confused with the denominators and what to do with them. I know if I square root  $1-\sin\phi$, I'll get a Pythagorean identity, but then I don't know where to go from there.

Please help me with a step-by-step guide. I really want to learn how to do this.

答:
You can do this either from LHS to RHS or from RHS to LHS.

Solution 1: LHS $\rightarrow$ RHS
$$\require{cancel}\begin{aligned}6\sec\phi\tan\phi&=6\frac{1}{\cos\phi}\frac{\sin\phi}{\cos\phi}\\&=\frac{6\sin\phi}{\cos^2\phi}\\&=\frac{3\sin\phi+3\sin\phi}{\left(1-\sin\phi\right)\left(1+\sin\phi\right)}\\&=\frac{3\left(1+\sin\phi\right)-3\left(1-\sin\phi\right)}{\left(1-\sin\phi\right)\left(1+\sin\phi\right)}\\&=\frac{3\cancel{\left(1+\sin\phi\right)}}{\left(1-\sin\phi\right)\cancel{\left(1+\sin\phi\right)}}-\frac{3\cancel{\left(1-\sin\phi\right)}}{\cancel{\left(1-\sin\phi\right)}\left(1+\sin\phi\right)}\\&=\frac{3}{1-\sin\phi}-\frac{3}{1+\sin\phi}\end{aligned}$$
Solution 2: RHS $\rightarrow$ LHS
$$\begin{aligned}\frac{3}{1-\sin\phi}-\frac{3}{1+\sin\phi}&=\frac{3\left(1+\sin\phi\right)-3\left(1-\sin\phi\right)}{\left(1-\sin\phi\right)\left(1+\sin\phi\right)}\\&=\frac{\cancel3+3\sin\phi\cancel{-3}+3\sin\phi}{\left(1-\sin\phi\right)\left(1+\sin\phi\right)}\\&=\frac{6\sin\phi}{1-\sin^2\phi}\\&=\frac{6\sin\phi}{\cos^2\phi}\\&=\frac{6\sin\phi}{\cos\phi\cos\phi}\\&=6\sec\phi\tan\phi\end{aligned}$$
I hope this helps.


问:
About Lusin's condition (N)
We say that $f:[0,1]\to \mathbb{R}$ satisfies Lusin's condition (N) provided
$$m(f(B))=0 \quad\mbox{whenever}\quad B\subseteq [0,1] \mbox{ with }m(B)=0$$
where $m$ stands for the Lebesgue measure on $\mathbb{R}$.

I found in here the following definition.

We say that $f:[0,1]\to X$ satisfies Lusin's condition (N) provided
$$\mathcal{H}^1(f(B))=0 \quad\mbox{whenever}\quad B\subseteq [0,1] \mbox{ with }m(B)=0$$
where $X$ is a metric space  and $\mathcal{H}^1$ stands for 1-dimensional Hausdorff measure.

What comes to my mind is the possible formulation of condition (N) for the function $f:[0,1]\to X$ but this time our $X$ is a Hausdorff locally convex topological vector space. I don't know if such formulation exists. If such formulation exists, I would be greatful if you can provide me such definition.

答:
Lusin's condition (N) deals exclusively with the measure space structure. A measurable map $f:X\to Y$ between measure spaces $X=(X,\Sigma_X,\mu_X)$ and $Y=(Y,\Sigma_Y,\mu_Y)$ satisfies the condition (N) if
$$ \mu_X(B)=0\implies \mu_Y(f(B))=0\tag{N}$$

The rest concerns not (N) itself, but rather the availability of measures on our spaces. On any metric space we have the family of Hausdorff measures $\mathcal H^s$. So one can choose (as the authors you quoted did) to use $\mathcal H^1$ on the target space when the domain is one-dimensions. They would probably use $\mathcal H^n$ on the target space if the domain was $n$-dimensional.

If your locally convex space is metrizable, you can use a metric to define $\mathcal H^1$ and hence, the  property (N). However, the property will depend on the choice of the metric, even if it is required to be translation-invariant. Indeed, if $d$ is one such metric, than $d^{\alpha}$ is another (for any $\alpha\in (0,1)$). The Hausdorff measure $\mathcal H^1$ with respect to $d^\alpha$ is equivalent to the Hausdorff measure $\mathcal H^{\alpha}$ with respect to $d$.  This measure is quite different from $\mathcal H^{1}$ with respect to $d$; the ideals of null sets are not the same.

Conclusion: If the target space lacks a canonical choice of a measure, there is no canonical notion of condition (N) for maps into it.

Title: About Lusin's condition (N)


问:
Vertex and edge deletion
I was given this example. I am wondering how does this work? The question says that G -v where v is a subset of V. From my understanding, this means that we remove perform a vertex deletion on graph G. I do not understand the first two and last two graphs. Those four graphs clearly performed a vertex and edge deletion. In addition, does have 2 same graphs mean anything? Thank you

答:
When the vertices of the "center square" are removed, we obtain the following graphs:

which are isomorphic to the left two and the right two graphs respectively.

The 2 same graphs appear since the original graph is (rotationally) symmetric: removing the leftmost vertex and removing the rightmost vertex will generate the middle two graphs, which are isomorphic, but are distinct graphs.

The set $\{G-v \mid v\in V\}$ must have the same number of elements as $V$; if the "same" graphs are omitted, this rule will be violated.

Title: Vertex and edge deletion


问:
Convergence of $\operatorname{Binomial}(1/\sqrt n, n)$
Let $Z_n \sim \operatorname{Binomial}(1/\sqrt n, n)$. Is the following expression asymptotically normally distributed for some normalization rate $r_n$?

$$r_n\left(Z_n - \mathbb E Z_n \right )$$

It's easy to show that $\mathbb E Z_n = \sqrt n$. The typical rate is $\sqrt n$, but a rate other than that might be required here.

What I've tried

Computing the moment generating function of the desired expression:

$$\mathbb E \exp \left ( \lambda r_n\left(Z_n - \mathbb E Z_n \right )\right ) = e^{-\lambda r_n \sqrt n}\left( 1 + \frac {r_n} {n} \left( e^{\lambda r_n /n} - 1 \right )\right)^n$$
where the terms in parens on the RHS grows exponentially. I dont think there is any rate $r_n$ such that the expression under consideration is asymptotically normal.

答:
The log-moment generating function under consideration is given by:

$$-\lambda r_n n \theta_n + n \log(1-\theta_n + \theta_n e^{\lambda r_n})$$

Taking a taylor expansion of the exponential we have the following inside the argument of the logarithm:

$$1 - \theta_n + \theta_n ( 1 + \lambda r_n + (\lambda r_n)^2/2 + o(r_n^2)) = 1 +  \theta_n ( \lambda r_n + (\lambda r_n)^2/2 + o(r_n^2))$$
Now returning to our log-mgf and taking a taylor expansion of the $\log(1 + y)$:

$$-\lambda r_n n \theta_n + n\left(\theta_n\left[\lambda r_n + (\lambda r_n)^2/2 + o(r_n^2)\right] - y^2/2 + y^3/3 + o(y^3)\right) =
n\left(\theta_n \left[(\lambda r_n)^2/2 + o(r_n^2))\right] - y^2/2 + y^3/3 + o(y^3)\right)$$

Notice that $y^2 = o(\theta_n^2 r_n^2)$ and that $n \, \theta_n = \sqrt n$ so we'll choose the norming rate $r_n = n^{-1/4}$ so that the expression converges to a non-zero value. Now, the expression reduces to:

$$\lambda^2 /2 + o(1)$$

Which is the log-mgf of a normal with mean 0 and variance 1. Computer simulations verify this:

```<code>n = 10000
outs = [np.power(n, 1/4) * ((np.random.rand(n) < 1/np.sqrt(n)).sum()/np.sqrt(n) - 1) for i in range(100000)]
plt.hist(outs, bins=30, alpha= 0.5, density=True)
plt.hist(np.random.normal(size=10000), alpha= 0.5, density=True)
plt.show()
</code>```

And as expected it converges faster than a binomial (whose norming rate $r_n=1$)

```<code>plt.hist((np.random.binomial(n, 0.5, size=len(outs)) - n/2)/np.sqrt(n/4), bins=30, density=True, alpha=0.5, label='bin')
plt.legend()
</code>```

Body:
The log-moment generating function under consideration is given by:

$$-\lambda r_n n \theta_n + n \log(1-\theta_n + \theta_n e^{\lambda r_n})$$

Taking a taylor expansion of the exponential we have the following inside the argument of the logarithm:

$$1 - \theta_n + \theta_n ( 1 + \lambda r_n + (\lambda r_n)^2/2 + o(r_n^2)) = 1 +  \theta_n ( \lambda r_n + (\lambda r_n)^2/2 + o(r_n^2))$$
Now returning to our log-mgf and taking a taylor expansion of the $\log(1 + y)$:

$$-\lambda r_n n \theta_n + n\left(\theta_n\left[\lambda r_n + (\lambda r_n)^2/2 + o(r_n^2)\right] - y^2/2 + y^3/3 + o(y^3)\right) =
n\left(\theta_n \left[(\lambda r_n)^2/2 + o(r_n^2))\right] - y^2/2 + y^3/3 + o(y^3)\right)$$

Notice that $y^2 = o(\theta_n^2 r_n^2)$ and that $n \, \theta_n = \sqrt n$ so we'll choose the norming rate $r_n = n^{-1/4}$ so that the expression converges to a non-zero value. Now, the expression reduces to:

$$\lambda^2 /2 + o(1)$$

Which is the log-mgf of a normal with mean 0 and variance 1. Computer simulations verify this:

n = 10000
outs = [np.power(n, 1/4) * ((np.random.rand(n) < 1/np.sqrt(n)).sum()/np.sqrt(n) - 1) for i in range(100000)]
plt.hist(outs, bins=30, alpha= 0.5, density=True)
plt.hist(np.random.normal(size=10000), alpha= 0.5, density=True)
plt.show()

And as expected it converges faster than a binomial (whose norming rate $r_n=1$)

plt.hist((np.random.binomial(n, 0.5, size=len(outs)) - n/2)/np.sqrt(n/4), bins=30, density=True, alpha=0.5, label='bin')
plt.legend()


问:
Does $\lim_{n\to\infty} \frac{2^{n}}{n!} = 0$ hold true?
I was solving a problem in real analysis where the author used the below to support his argument.

$$\lim_{n\to\infty} \frac{2^{n}}{n!} = 0$$

How does the above limit hold ?

答: $2^n=2*2*2*2*...$ and $n!=1*2*3*4*5...$. It should be clear that the denominator get multiplied by a higher number whereas the numerator only receives another $2$. If the numerator and the denominator would have a boxing mach the numerator would be a rookie and the denominator Mike Tyson. A more formal approach would be induction, if needed. Side note, you could replace the $2$ in the numerator by a $3$ or $4$ or even $2020$. Even then the denominator would win. (Use same method of induction)


问:
How to solve recurrence relation $a_k = ba_{k−1} + cr^k$, assuming $b \neq r$
Solve for $a_k$ in terms of $a_0$ and the other parameters in the following recurrence relation:

$a_k = ba_{k−1} + cr^k$, assuming $b \neq r$.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{b \not= r,\quad a_{k} = ba_{k − 1} + cr^{k}}$.

\begin{align}
\mbox{Lets}\quad d_{k} & \equiv {a_{k} \over r^{k}} \implies d_{k} =
{b \over r}\,d_{k - 1} + c
\\[2mm] \implies
d_{k} - {rc \over r - b}& =
{b \over r}\pars{d_{k - 1} - {rc \over r - b}}
\end{align}

\begin{align}
\implies &
d_{k} - {rc \over r - b} = \pars{b \over r}^{2}\pars{d_{k - 2} - {rc \over r - b}}
\\[2mm] & = \cdots =
\pars{b \over r}^{k}\pars{d_{0} - {rc \over r - b}}
\\[5mm] \implies &
d_{k} = {rc \over r - b} +
\pars{b \over r}^{k}\pars{d_{0} - {rc \over r - b}}
\\[5mm] \implies &
\bbx{a_{k} = {c \over r - b}\,r^{k + 1}\ +\
\pars{a_{0} - {rc \over r - b}}b^{k}} \\ &
\end{align}


答: Observe that: $\dfrac{2^n}{n!}= \dfrac{2\cdot 2\cdot 2\cdot...2}{1\cdot 2\cdot 3\cdot...n} < \dfrac{2}{n}, n \ge 5$. Thus for any $\epsilon > 0$, choose $N \in \mathbb{N}$ such that $N > \text{max}\left(\dfrac{2}{\epsilon},5\right)$. Then if $n \ge N \implies \dfrac{2^n}{n!} < \epsilon \implies \text{limit} = 0$.

Title: Does $\lim_{n\to\infty} \frac{2^{n}}{n!} = 0$ hold true?

Body:
I was solving a problem in real analysis where the author used the below to support his argument.

$$\lim_{n\to\infty} \frac{2^{n}}{n!} = 0$$

How does the above limit hold ?


答:
Since $n! = 1\cdot2\cdot3...\cdot n \ge 1\cdot 2\cdot3^{n-2}$ for $n \ge 3$, $$0<\frac{2^n}{n!} \le \frac{2^n}{2\cdot3^{n-2}}$$

Then since $$\lim_{n \to \infty}\frac{2^n}{2\cdot3^{n-2}} = 0$$

the limit in question is also $0$.


问:
The orthogonal complement of the space of row-null and column-null matrices
I propose the following lemma and its proof. It is related to row-null and column-null matrices - i.e. matrices whose rows and columns both sum to zero. Could you please give your opinion on the plausibility of the lemma, and the validity of the proof?

Lemma: Let $Z\in\text{GL}(n,\mathbb{R})$ be a general $n\times n$ real matrix, and let $Y\in\mathcal{S}(n,\mathbb{R})$, where $\mathcal{S}(n,\mathbb{R})$ is the space of row-null column-null $n\times n$ real matrices. Then $\text{Tr}(ZY)=0$ for all $Y$ in $\mathcal{S}(n,\mathbb{R})$ if and only if $Z$ has the form $$Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)$$.

Proof:
Consider the space of row-null and column-null matrices

$$\mathcal{S}(n,\mathbb{R})= \left\{ Y_{ij}\in GL(n,\mathbb{R}):\sum_{i}Y_{ij}=0,\sum_{j}Y_{ij}=0 \right\} $$

Its dimension is
$$\text{dim}(S(n,\mathbb{R}))=N^{2}-2N+1$$
since the row-nullness and column-nullness are defined by $2N$ equations, only $2N-1$ of which are linearly independent.
Consider the following space

$$\mathcal{G}(n,\mathbb{R})=\left\{ Z_{ij}\in GL(n,\mathbb{R}):Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)\right\}$$

Its dimension is

$$\text{dim}(\mathcal{G}(n,\mathbb{R}))=2N-1$$
where $N-1$ is the contribution from the antisymmetric part and $N$ is from the symmetric part.

Assume $Y\in\mathcal{S}$ and $Z\in\mathcal{G}$, then the Frobenius inner product of two such elements is
$$
\text{Tr}(ZY) =\sum_{ij}\left[\left(p_{j}-p_{i}\right)Y_{ji}+\left(q_{j}+q_{i}\right)Y_{ji}\right]
$$
$$
=\sum_{j}(q_{j}+p_{j})\sum_{i}Y_{ji}+\sum_{i}(q_{i}-p_{i})\sum_{j}Y_{ji}=0
$$
Since $\text{dim}(\mathcal{G})+\text{dim}(\mathcal{S})=\text{dim}(GL)$ and $\mathcal{G}\perp\mathcal{S}$, then $\mathcal{G}$ and $\mathcal{S}$ must be complementary in $GL$. Therfore, if $Y$ is orthogonal to all the matrices in $\mathcal{S}$, it must lie in $\mathcal{G}$.

PS: How can I get the curly brackets {} to render in latex mode?

答:
I've checked your proof. It is correct.

I want to say obvious thing: I know two methods for checking the proof is correct.

1. Check the answer, if you can get it from something else.

1'. If you can't get the answer independently, check some of it's properties.

2. Check the proof line by line.

2'. Divide your proof on parts and check each of them, using one of the above.

Applying second method is obvious. If you want, to apply first, you can try to prove the following.

1. Every symmetric matrix, that is orthogonal to S(n,R), is of the form $q_i+q_j$.

2. Every symmetric matrix, that is orthogonal to S(n,R), is of the form $q_i+q_j$.

This can be done by using equations of the form Tr(ZY)=0 with matrices $Y\in S(n,\mathbb{R})$ of the form $Y_{ij}=a_i b_j$, where a and b are vectors of the form vectors with one coordinate equal to 1, another --- equal to -1, all other coordinates equal to 0 (like $(0,\dots,0,1,0,\dots,0,-1,0,\dots,0)$).

ContentLicense: CC BY-SA 2.5


答:
First, note that $\lim_{n\to\infty} \frac{2}{n} = 0$ and then use that for large $n$ we have $$0 <\frac{2^n}{n!} = \frac{2}{n}\cdot \frac{2^{n-1}}{(n-1)!} < \frac{2}{n}$$

and the result follows by the squeeze theorem.


问:
Prove that the limit of a sequence of conformal mappings is also conformal
Let $\mathbb{D}$ denote the open unit disc in the complex plane, and let $\{D_n\}$ be a descending sequence of simply connected domains in $\mathbb{D}$, i.e. $D_{n+1}\subset D_{n}$ for all $n\geq 1$. Assume that $D=\text{Int}(\bigcap_{n}D_n)$ is non-empty, and $0\in D$.

For every $n\geq 1$, let $g_n:\mathbb{D}\to D_n$ be a conformal mapping such that $g_n(0)=0$, and assume that the sequence $\{g_n \}$ converges normally on $\mathbb{D}$ to an analytic function $g$.

What I would like to show is that $g$ is a conformal mapping from $\mathbb{D}$ to $D$.

By invoking Hurwitz's Theorem, we know that $g$ is either univalent or constant, so as long as we can prove that $g$ is non-constant, we know that it's a conformal mapping onto $g(\mathbb{D})$. However neither do I know how to show that $g$ is non-constant, nor do I know how to show that its image is exactly equal to $D$ (provided that it's non-constant). Again, by Hurwitz's Theorem, I can manage to show that $g(\mathbb{D})\subset D$ if it's non-constant. However I don't know how to show the other direction of inclusion.

Any help is appreciated.

答:
Pick $c>0, D_c(0) \subset D \subset D_n$. Hence $g_n^{-1}(z)$ maps $D_c(0) \to \mathbb D=D_1(0)$ or $g_n^{-1}(cz)$ maps the unit disc to itself and preserves zero, so by Schwarz Lemma its derivative at $0$ is at most $1$ in absolute value; but this means $|g_n^{-1}(0)'| \le 1/c$ or $|g_n'(0)| \ge c$ hence $|g'(0)| \ge c>0$ so $g$ is non-constant

Let $G=g(\mathbb D) \subset D \subset D_n$ open because $g$ is holomorphic nonconstant; considering the restrictions of $g_n^{-1}$ to $G$ they form a normal family being uniformly bounded and we will show that any limit point of it is $g^{-1}$ which imnplies that $g_n^{-1} \to g^{-1}$ normally on $G$; note first that if $g_{n_k}^{-1} \to h$ we have that $|g_{n_k}^{-1}(0)'|=|1/g'_{n_k}(0)| \to |1/g'(0)| >0$ by the above, hence $h$ is not constant so it is univalent.

Fix $w \in \mathbb D$; if $\epsilon >0$ $g_n(z)-g(w)$ has a zero $|z_n-w| <\epsilon/2$ for large enough $n$ hence $g_{n_k}(z_{n_k})=g(w)$ for $n_k$ large enough, or $g_{n_k}^{-1}(g(w))=z_{n_k}$ while also $|h(g(w))-g_{n_k}^{-1}(g(w))| < \epsilon/2$ again for large enough $k$ by the defintion of $h$. Remembering that $|z_{n_k}-w| <\epsilon/2$ also, by the triangle inequality we get that $|h(g(w))-w| < \epsilon$ hence $h(g(w))=w$ for all $w \in \mathbb D$ and the above claim has been proved.

But now this means that $h=g^{-1}$ is a univalent function mapping $G$ onto $\mathbb D$ and of course if $G \ne D$ we can get a subsequence of $g_n^{-1}$ normally convergent to some $h_1$ on $D$; by definition $h=h_1$ on $G$, hence $h_1$ is also univalent by Hurwitz as it is nonconstant and $h_1(D) \subset \mathbb D$ which is a contradiction since the restriction of $h_1$ to $G$ already covers the whole $\mathbb D$ so $h_1$ cannot be univalent. This means $G=D$ and we are done!


问:
Proving $\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$

Prove this trigonometric identity:

$$\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$$

I've simplified it until

$$\frac{2\cos^2\theta}{1-\sin\theta}$$

but couldn't get $2+2\tan\theta$ from it.

答:
First, let's make a common denominator of $\frac{1+\cos\theta}{\sec\theta-\tan\theta}+\frac{\cos\theta-1}{\sec\theta+\tan\theta}$.

Ready, set, go!
$$\require{cancel}\begin{aligned}\color{blue}{\frac{1+\cos\theta}{\sec\theta-\tan\theta}}+\color{red}{\frac{\cos\theta-1}{\sec\theta+\tan\theta}}&=\frac{\color{blue}{\left(1+\cos\theta\right)\left(\sec\theta+\tan\theta\right)}+\color{red}{\left(\cos\theta-1\right)\left(\sec\theta-\tan\theta\right)}}{\color{blue}{\left(\sec\theta+\tan\theta\right)}\color{red}{\left(\sec\theta-\tan\theta\right)}}\\&=\frac{\color{blue}{\cancel{\sec x}+\tan x+\cos{\theta}\sec{\theta}\cancel{+\cos{\theta}}\tan{\theta}}+\color{red}{\cos{\theta}\sec{\theta}\cancel{-\cos{\theta}\tan{\theta}}\cancel{-\sec x}+\tan x}}{\color{blue}{\left(\sec x+\tan x\right)}\color{red}{\left(\sec x-\tan x\right)}}\\&=\frac{2\tan x+2\cos{\theta}\sec{\theta}}{\sec^{2}x-\tan^{2}x}&&\text{The denominator is equal to 1.}\\&=\frac{2\tan x+2\cos{\theta}\sec{\theta}}{1}\\&=2\tan x+2\cos{\theta}\sec{\theta}\\&=2\tan\theta+2\cancel{\cos\theta}\cdot\frac{1}{\cancel{\cos\theta}}\\&=2+2\tan\theta\end{aligned}$$
I hope this helps.


问:
Calculating generalized inverse of singular matrix using singular value decomposition (SVD)
I became confused about how singular value decomposition can be used to find generalized inverse of singular matrix.

Specifically, I am dealing with the matrix $G=\begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & \sqrt{2} & \sqrt{2} & 0 \\ \sqrt{2} & 0 & 0 & \sqrt{2} \end{pmatrix}$. Calculating $A=G^{T}G$, $A=\begin{pmatrix} 3 & 0 & 1 & 2 \\ 0 & 3 & 2 & 1 \\ 1 & 2 & 3 & 0 \\ 2 & 1 & 0 & 3 \end{pmatrix}$ and its eigenvalues are $6,4,2,0$ so the matrix $A$ is singular. However, my textbook describes that it is possible to calculate the generalized inverse of $G$ using the decomposition called Lanczos decomposition, namely $G=U_{p}\Lambda_{p} V_p^{T}$. It is written that the generalized inverse is then $G^{-p}=V_{p} \Lambda_{p}^{-1} U_{p}^{T}$.

Now, matrices $U_{p}$, $\Lambda_{p}$, and $V_{p}^{T}$ are defined as follows: $\Lambda_{p}$ is the diagonal matrix where its diagonal components are $p$ nonzero eigenvalues of $G$ (in this case, $p=3$ because $rank(G)=3$). Also, it is written that $U_{p}$ and $V_{p}$ can be attained by calculating singular value decomposition of $GG^{T}$ and $G^{T}G$, respectively.

I understand $\Lambda_{p}$ part, but I want to know how to exactly calculate $U_{p}$ and $V_{p}$. I did an experiment using NumPy module but I think I am misunderstanding something because result is different from desired one.

I attach my Python code I used for computation.

```<code>import numpy as np
from numpy.linalg import svd

G=np.array([[1, 0, 1, 0],
            [0, 1, 0, 1],
            [0, np.sqrt(2), np.sqrt(2), 0],
            [np.sqrt(2), 0, 0, np.sqrt(2)]])

V, S, Vh=svd(G.T@G)
Vp=V[:, :3] # Eigenvectors corresponding to 3 nonzero eigenvalues

U, S, Uh=svd(G@G.T)
Up=U[:, :3] # Eigenvectors corresponding to 3 nonzero eigenvalues

D=np.sqrt(np.diag(S[:3])) # square root of eigenvalues of G.T@G
print(Up@D@Vp.T)
print(G)
</code>```

Two results must be equal according to the theory but it is not. Specifically, it seems that the first two rows of $U_{p}\Lambda_{p}V_{p}^{T}$ are swapped but I don't get why. I think this is because I incorrectly solved for $V_{p}$ and $U_{p}$. I want to know what am I misunderstanding.

答:
When you take the SVD of $GG^T = U_p\Lambda_p \Lambda_p^T U_p^T$, there is an arbitrary sign in the columns of $U_p$. Similarly, there is an arbitrary sign in the columns of $V_p$ when you take the SVD of $G^T G$.

When calculating $U$ and $V$ directly from the SVD of $G$, you can multiply corresponding columns of both $U$ and $V$ by an arbitrary sign, but you can't do it to a column of $U$ only or to a column of $V$ only. In your calculation, however, the signs of the columns of $U_p$ and $V_p$ are getting changed arbitrarily relative to one another, leading to an incorrect answer when you reconstruct G. The link between the columns of $U_p$ and the corresponding columns of $V_p$ with regard to their signs gets broken when you calculate them separately in this manner.

If you take your code and follow it up by computing $U$ and $V$ directly from the SVD of $G$, and compare them to the $U_p$ and $V_p$ you computed, you will see that this is exactly what happened. The third column of $U_p$ is the negative of the third column of $U$, but the first three columns of $V_p$ match those of $V$.


问:
Right tail of log-concave densities.
Consider any density $g$ with support $(a,b)$ and $-\infty<a<b<+\infty$.

Is it possible to have $g$ log-concave on $(a,b)$ even if $\lim_{q\uparrow b}g(q)=+\infty$?

An (1995, 1998) proves that the right tail of a log-concave density is at most exponential, but the proof seems to assume $(a,b)=\mathbb{R}$. It should hold for, say, $(a,b)=(0,1)$, right?

答: If$a,b$ are finite then random variables distributed according to $g$ are bounded. Bounded random variables are subgaussian. Subgaussian random variables are subexponential. So then using the claim cited in your question, yes it is possible.

Title: Right tail of log-concave densities.


答:
Hint:

$$a_k = ba_{k−1} + cr^k\\
a_{k+1} = ba_k+cr^{k+1}\\
\implies a_{k+1}-ba_k=r(a_k-ba_{k-1})\\
\implies a_{k+1}-ra_k=b(a_k-ra_{k-1})
$$

So both $a_{k+1}-ba_k$ and $a_{k+1}-ra_k$ are geometric sequences. Can you start from here?

If you still get stuck, take a look at this post.

Title: How to solve recurrence relation $a_k = ba_{k−1} + cr^k$, assuming $b \neq r$

Body:
Solve for $a_k$ in terms of $a_0$ and the other parameters in the following recurrence relation:

$a_k = ba_{k−1} + cr^k$, assuming $b \neq r$.


问:
How to approach an unusual quadratic question
Question: $a,b\in\mathbb{R}\mid a\neq 0$ When the polynomial $x^2-2ax+a^4$ is divided by $x+b$ the remainder is $1$. The polynomial $bx^2+x+1 $ has $ax-1$ as a factor. What is the value of $b$

My Attempt
$$\frac{x^2-2ax+a^4}{x+b}=... \frac{1}{x+b}\Rightarrow b^2+2ab+a^4=1$$
$$bx^2+x+1=(ax-1)(b/a-1)\Rightarrow a^2+a+b=0$$
How do you proceed from here? Am I approaching this problem correctly?

Body:
for $x=\frac{1}{a}$

$$\frac{1}{a^2}-\frac{2a}{a}+a^4=0$$

$$\frac{1}{a^2}-2+a^4=0$$

$$(\frac{1}{a}-a)^2+a^2=0$$

$$a\not=0, a=-1, a=1$$

Let $a=-1$ then $x^2+2x+1=0$

for $x=-b$

$$b^2-2b+1=1$$

$$b^2-2b=0$$

Let $a=1$ then $x^2-2x+1=0$

for $x=-b$

$$b^2+2b+1=1$$

$$b^2+2b=0$$


问:
Probability of picking out of equal number of red, blue, green marbles
A bag contains equal number of red, green, and yellow marbles. If Geeta pulls three marbles out of the bag, replacing each marble after she picks it, what is the probability that at least one will be red?

I know that we can solve this very easily using the probability of this not happening, for eg.
there is a 2/3 probability that the marble will not be red.
so the probability of all three marbles not being red is
2/3 * 2/3 * 2/3 = 8/27.

so the probability of atleast one marble is 1 - 8/27 = 19/27.

But I want to solve this problem by actually calculating the probability of this happening one by one, how do I do it?

I tried doing this like this but the answer came out incorrect:
1/3 + 1/3 * 1/3 + 1/3 * 1/3 * 1/3 = 13/27.
which is incorrect.

Requesting any help with this, Thank you.

答:
To do this the additive way, you'll need to calculate the probability of getting exactly one red marble, exactly two red marbles, and exactly three red marbles. These you can add up because they're disjoint sets of outcomes.

If the first marble is red, then the second and third aren't: $1/3 \times 2/3 \times 2/3 = 4/27$. Same for only the second marble red, and for only the third marble red, for a total of $12/27$.

Can you finish from here?


答:
From your equations we have $(a+b)^2 = (a^2)^2= a^4$.

Hence $1 = b^2 + 2ab + a^4 = (b+a)^2-a^2 + a^4 = 2a^4 - a^2$.

This is a quadratic equation in $a^2$, which can be solved to obtains value(s) of $a$ and $b$.

Title: How to approach an unusual quadratic question

Body:
Question: $a,b\in\mathbb{R}\mid a\neq 0$ When the polynomial $x^2-2ax+a^4$ is divided by $x+b$ the remainder is $1$. The polynomial $bx^2+x+1 $ has $ax-1$ as a factor. What is the value of $b$

My Attempt
$$\frac{x^2-2ax+a^4}{x+b}=... \frac{1}{x+b}\Rightarrow b^2+2ab+a^4=1$$
$$bx^2+x+1=(ax-1)(b/a-1)\Rightarrow a^2+a+b=0$$
How do you proceed from here? Am I approaching this problem correctly?


问:
Limit of $n^{\frac{1}{n}}$.
We have to prove that $n^{\frac{1}{n}}$ converges to $1$. I have proved it using the binomial theorem where we can substitute $(1+t)$ in place of $n$ and proceed forward. However along with the question another approach was mentioned where we can make use of the Monotone Convergence Theorem. For this approach, I proved that $a_n=\left(1+\frac{1}{n}\right)^n$ is increasing and using $a_n<a_{n+1}$ I proved that $n^{\frac{1}{n}}$ is decreasing after the third term. Also it is bounded as all terms are greater than $0$. So by MCT it should be convergent. But I am not able to evaluate the limit and am only able to prove its existence here. Please help.

答: $$a_n = n^{1/n} \implies \ln a_n = \frac{\ln n}{n} \implies \lim\ln a_n = 0 \implies \ln \lim a_n = 0 \implies \lim a_n = e^0 = 1$$

Title: Limit of $n^{\frac{1}{n}}$.


问:
Bounding the roots of the sum of two monic polynomials with real coefficients.
Let $P_1(z)$ and $P_2(z)$ be monic polynomials with real coefficients and roots $\{z_1^{(1)},z_1^{(2)},...\}$ and $\{z_2^{(1)},z_2^{(2)},...\}$, respectively. Are there any results relating the non-trivial roots $\{z^{(1)}, z^{(2)},...\}$ of
\begin{equation}
P(z) = \xi P_1(z) + P_2(z)
\end{equation}
to the roots of $P_1$ and $P_2$?

In particular, if $P_1, P_2$ and $P$ have then same number of complex valued roots, under what circumstances do we have (up to a relabelling)
\begin{equation}
\min \{\mathrm{Im} \, z_1^{(k)}, \mathrm{Im} \, z_2^{(k)} \} \leq \, \mathrm{Im} \, z^{(k)}
\leq \max \{\mathrm{Im} \, z_1^{(k)}, \mathrm{Im} \, z_2^{(k)} \}
\end{equation}
for all $k$ and for all $\xi \in [0,\infty)$?

Example:
Let $P_1(z) = z (z^2 + 1)$ and $P_2(z) = z^2 + r^2$. The polynomial $P(z) = \xi P_1(z) + P_2(z)$ satisfies the above property given $r\leq 1$.

答:
One way to approach this is to bound the roots of your polynomials in terms of the polynomial coefficients (which, in contrast, behave nicely under addition):

Let $(K, \vert \cdot \vert)$ be a valued field and let
$$f(X) = X^d +a_1 X^{d-1}+ \ldots + a_d = (X-b_1)(X-b_2)\cdots(X-b_d)$$
be a polynomial, where $a_1,\ldots, a_d,b_1,\ldots,b_d \in K$.  Let $C=2$ if the absolute value is archimedean and let $C=1$ if the absolute value is non-archimedean.  Then
$$C^{-d} \vert f \vert \leq \prod_{i=1}^d \max\{\vert b_i \vert, 1\} \leq C^d \vert f \vert.$$

(The proof is by induction on $d$.) In particular, this gives bounds on the absolute values of the roots (both above and below) in terms of the coefficients of $f$.  By adding two polynomials $f$ and $g$, we quickly obtain bounds for the roots of their sum (simply by replacing $\vert f \vert$ by $\vert f+g \vert$).


问:
How many non-isomorphic full binary trees are there?
A full binary tree is a rooted tree where every node has either zero children or two children.  My question is, how many non-isomorphic full binary trees are there, countably many or uncountably many?

Or more formally, what is the maximum cardinality a set can have if its elements are all full binary trees and any two elements are non-isomorphic?

答: There are $2^{\aleph_0}$ full binary trees up to isomorphism.  This is obviously an upper bound, since every full binary tree is countable.  To see that it can be achieved, given a binary tree $T$, consider the set $S\subseteq\mathbb{N}$ consisting of those $n$ such that every node in the $n$th level of $T$ has two children.  It is easy to see that every subset $S\subseteq\mathbb{N}$ which contains $0$ can be realized by some full binary tree $T$, and so this gives $2^{\aleph_0}$ different isomorphism types of full binary trees.


答: After you have shown that $L=\lim_{n\to \infty}n^{1/n}$ exists, and is positive (because every $n^{1/n}\ge 1$),  then $$0<L=\lim_{n\to\infty}(2n)^{1/2n}=\lim_{n\to\infty}2^{1/2n}(n^{1/n})^{1/2}=L^{1/2}$$ because $2^{1/2n}\to 1.$

Title: Limit of $n^{\frac{1}{n}}$.

Body: We have to prove that $n^{\frac{1}{n}}$ converges to $1$. I have proved it using the binomial theorem where we can substitute $(1+t)$ in place of $n$ and proceed forward. However along with the question another approach was mentioned where we can make use of the Monotone Convergence Theorem. For this approach, I proved that $a_n=\left(1+\frac{1}{n}\right)^n$ is increasing and using $a_n<a_{n+1}$ I proved that $n^{\frac{1}{n}}$ is decreasing after the third term. Also it is bounded as all terms are greater than $0$. So by MCT it should be convergent. But I am not able to evaluate the limit and am only able to prove its existence here. Please help.


答:
Let $$T=\{0^n: n\in\mathbb{N}\}\cup\{0^n1: n\in\mathbb{N}\}.$$ $T$ consists of a single branch with a bunch of length-$1$ "spines" sticking out.

Now given $A\subseteq\mathbb{N}$, let $$T_A=T\cup\{0^n10: n\in A\}\cup\{0^n11: n\in A\}.$$ That is, we stick a "v" onto each "spine" corresponding to an element of $A$. Each $T_A$ is a full binary tree in your sense and we have $$T_A\cong T_B\iff A=B.$$ So there are continuum-many isomorphism types of full binary trees.

Title: How many non-isomorphic full binary trees are there?

Body:
A full binary tree is a rooted tree where every node has either zero children or two children.  My question is, how many non-isomorphic full binary trees are there, countably many or uncountably many?

Or more formally, what is the maximum cardinality a set can have if its elements are all full binary trees and any two elements are non-isomorphic?


问:
Proving $\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+1}{n+1})}k>\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+3}{n+1})}k$
A few days ago I asked a question about an interesting property of the partial sums of the series $\sum\sin(nx)/n$.
Here's the link:
Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$

The proof given there left some details for me, in particular I have to prove that
$$
\sum_{k = 1}^{n}{1 \over k}
\sin\left(k\pi\frac{2m+1}{n+1}\right) >
\sum_{k=1}^{n}{1 \over k}\sin\left(k\pi\frac{2m+3}{n+1}\right)
$$
for every natural $n$ and $m$ such that $m = 0, 1, \ldots, \left\lfloor\left(n-1\right)/2\right\rfloor$. I don't think this is relatively simple; anyway I tried a few expansions with sum-to-product and viceversa formulas, but everything I tried leads to $$\sum_{k=1}^{n}\frac{\cos(2(m+1)x_k)\sin(x_k)}{k} < 0$$
where $x_k = k\pi/(n+1)$. From here I don't see any further semplification. Also I noted that $\sin(x_k)$ is always positive since $k$ is at most $n$, but we can't say much about $\cos$.
Then I tried to pair terms since for $k' = n+1-k$ we have $\sin(x_k) = \sin(x_k')$, which gives $$\sum_{k=1}^{n/2}\left[\frac{\cos(2(m+1)x_k)}{k}+\frac{\cos(2(m+1)x_{k'})}{n+1-k}\right]\sin(x_k) < 0$$ for $n$ even. Then I hoped that the term between square brackets were always negative, but it's not always the case... How can I do?

答:
You have already determined the local maxima of a partial sum $S_n(x)$ in $[0,\pi]$ are at
$x_{n,m} = \frac{2m+1}{n+1} \pi$ for $m = 0,1, \ldots ,\lfloor\frac{n-1}{2} \rfloor$.

We first show that the vertical distance between successive peaks is diminishing in that

$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right)>  S_n\left(\frac{2m+3}{n+1}\pi \right) - S_n\left(\frac{2m+5}{n+1}\pi \right) > \ldots $$

Note that $\frac{2m+3}{n+1}\pi = \frac{2m+1}{n+1}\pi + \frac{2\pi}{n+1}$ and

$$S_n'\left(x\right) - S_n'\left(x+ \frac{2\pi}{n+1} \right) = \frac{\sin \frac{n}{2}x \cos \frac{n+1}{2}x}{\sin \frac{x}{2}} - \frac{\sin \left(\frac{n}{2}x+ \frac{n\pi}{n+1} \right) \cos \left(\frac{n+1}{2}x+\pi\right)}{\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)} \\ =\frac{\sin \frac{n}{2}x \cos \frac{n+1}{2}x}{\sin \frac{x}{2}} - \frac{\sin \left(\frac{n}{2}x- \frac{\pi}{n+1} \right) \cos \frac{n+1}{2}x}{\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)} \\ = \frac{\cos \frac{n+1}{2}x}{\sin \frac{x}{2}\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)}\left[\sin \frac{x}{2}\sin \left(\frac{n}{2}x+ \frac{\pi}{n+1}\right) -\sin \frac{x}{2}\sin \left(\frac{n}{2}x- \frac{\pi}{n+1}\right)  \right]$$

Applying angle addition identities and  simplifying we eventually get

$$\tag{1}S_n'\left(x\right) - S_n'\left(x+ \frac{2\pi}{n+1} \right) = \frac{\sin \frac{\pi}{n+1}\sin (n+1)x}{\cos \frac{\pi}{n+1} - \cos \left(\frac{\pi}{n+1} +x \right)}$$

Now consider the points

$$\frac{2m+1}{n+1}\pi \leqslant  \frac{2m+2}{n+1}\pi-\frac{y}{n+1} \leqslant \frac{2m+2}{n+1} \leqslant \frac{2m+2}{n+1}\pi+\frac{y}{n+1} \leqslant \frac{2m+3}{n+1}\pi, \\ \frac{2m+3}{n+1}\pi \leqslant  \frac{2m+4}{n+1}\pi-\frac{y}{n+1} \leqslant \frac{2m+4}{n+1} \leqslant \frac{2m+4}{n+1}\pi+\frac{y}{n+1} \leqslant \frac{2m+5}{n+1}\pi$$

Using (1) with $x_- = \frac{2m+2}{n+1}\pi-\frac{y}{n+1}$, $x'_-= x_- + \frac{2\pi}{n+1} = \frac{2m+4}{n+1}\pi-\frac{y}{n+1}$, $x_+ = \frac{2m+2}{n+1}\pi+\frac{y}{n+1}$,  and$x'_+=x_+ + \frac{2\pi}{n+1} =  \frac{2m+4}{n+1}\pi+\frac{y}{n+1}$we get

$$G'(y) = \frac{d}{dy} \left[S_n \left(x_-) -  S_n \left(x'_-\right) \right] - S_n(x_+) - S_n(x'_+)\right] \\ = \frac{\sin \frac{\pi}{n+1} \sin y}{n+1}\left[\frac{1}{\cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi-y}{n+1}\right)}  - \frac{1}{\cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi+y}{n+1}\right)} \right] $$

Notice that $G(0) = 0$ and $G'(y) > 0$ for $0 < y < \pi$, so $G(\pi) > 0$ which implies

$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right)>  S_n\left(\frac{2m+3}{n+1}\pi \right) - S_n\left(\frac{2m+5}{n+1}\pi \right)$$

When $m = \lfloor \frac{n-1}{2}\rfloor -1 $ we have $\frac{2m+3}{n+1} < \pi < \frac{2m+5}{n+1} = \pi + \delta$. Hence,

$$S_n\left(\frac{2m+5}{n+1}\pi \right) = S_n(\pi + \delta) = S_n(\pi + \delta - 2\pi) = S_n(\delta - \pi)= - S_n(\pi - \delta) < 0,$$

Since $\frac{2m+3}{n+1}$ is the last relative extremum point before $\pi$, $S_n(\pi) = 0$, and $\frac{2m+3}{n+1} < \pi - \delta < \pi$,  it follows that

$$S_n\left(\frac{2m+3}{n+1}\pi \right) > 0 >  S_n\left(\frac{2m+5}{n+1}\pi \right)$$

Thus, for all $m = 0,1,\ldots, \lfloor\frac{n-1}{2}\rfloor -1$ we have

$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right) > 0$$


问:
Convergence in $C(X,Y)$ with graph topology
I can't figure out this point-set topology assertion. Suppose $X$ is paracompact (Hausdorff?) and $Y$ is metrizable with metric $d$. Then a basis of open sets for the graph topology $WO$ on $C(X,Y)$ (take this as the definition if you like) is given by open sets of the form $$U(f,\epsilon) := \{g \in C(X,Y) : d(f(x),g(x)) < \epsilon(x), \epsilon \in C(X,(0,\infty)).$$

The assertion is this:

$f_n \rightarrow f$ in $WO$ iff there is a compact set $K$ such that
  $f_n|_K \rightarrow f|_K$ uniformly and $f_n \equiv f$ off $K$ for all
  but finitely many $n$.

Certainly the condition implies WO convergence, but I can't figure out how to get a converse. Convergence in WO implies that we get uniform convergence on for any subset, but I can't make sense of how to get the other part. The source asserts that we can choose a sequence of points $x_{n_k}$ with no cluster such that $f(x_{n_k}) \neq f_{n_k}(x_{n_k})$. From here, the result is clear enough since we can use a partition of unity to build an $\epsilon$ that makes a problem.

But how do we pick that sequence? If $X$ is hemicompact then I can see how I might proceed, but otherwise I just don't know enough about the compact subsets of $X$. Can anyone who is better at point-set topology than me help out? The source of this statement is Michor's Manifolds of Differentiable Mappings, Lemma 3.4.

答:
@user-j

It is easy to check that Zach L.’s counterexample is OK.

Concerning your reference to [GG] (see the fragments from the pages 42,  43, 44), there is written

the sequence of mappings $f_n$ converge to $f$ (in the Whitney $C^k$ topology) iff there is a compact subset $K$ of $X$ such that $j^kf_n$ converges uniformly to $j^kf$ on $K$ and all but a finite number of the $f_n$'s equal $f$ off $K$. The "if" part is clear, and we shall prove the "only if" part by contradiction. Assume $f_n$ converges to $f$ and that there does not exist a compact set $K$ with the above property. Let $K_1, K_2,\dots$ be a sequence of compact subsets of $X$ such that $K_i\subset \operatorname{Int} (K_{i + 1})$ and $X =\bigcup_{i=1}^\infty K_i$.

For topological spaces the latter assumption is even stronger than hemicompactness. Indeed, according to exercise 3.4.E from [Eng] (referenced to [Are]):

A Hausdorff space $X$ is hemicompact if in the family of all compact subspaces of $X$ ordered by $\subset$ there exists a countable cofinal subfamily.

(a) Prove that every first-countable hemicompact space is locally compact.

(b) Give an example of a countable hemicompact space which is not a $k$-space.

(c) Show that in the realm of second-countable spaces hemicompactness is equivalent to local compactness.

(d) Prove that if the space $\Bbb R^X$ with the compact-open topology is first-countable and $X$ is a Tychonoff space, then $X$ is hemicompact.

References

[Are] Arens R., A topology for spaces of transformations, Ann. Math. 47 (1946) 480-495.

[Eng]  Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

[GG] Martin Golubitsky, Victor Guillemin, Stable Mappings and Their Singularities, Springer-Verlag, 1973.


问:
Find the congruence classes (mod $(3 + \sqrt{ −3})/2)$ in $Q[ \sqrt{ −3}]$.
Find the congruence classes (mod $(3 + \sqrt{ −3})/2)$ in $Q[ \sqrt{ −3}]$.

This one really has me stumped. How do I find a clear, number-theoretic approach towards a solution to this problem?

EDIT: I have solved this problem, but am looking for a solution that preferably doesn't really use the notions of rings and specifically fields/field extensions

Since -3 = 1 mod 4 the ring of integers in Q(√(-3)) is Z[ (-1+√(-3))/2 ]. (-1 + √(-3))/2 is a root of the irreducible polynomial f(X) = X^2 + X + 1 over Z, and is a cube root of 1, X^3 =1. Z[ (-1+√(-3))/2 ] is a Euclidean domain, and so a UFD.

Norm ((3 + √(-3))/2) = 3, so (3 + √(-3))/2 must be a prime { else if = QR then Norm(Q) Norm(R) = 3 --- doesn't make sense unless Q or R is unit}.

Since (3 + √(-3))/2 is a prime then Z[ (-1+√(-3))/2 ] mod (3 + √(-3))/2 is a field {a general fact}, containing Z mod (3 + √(-3))/2. Since (3 + √(-3))/2 divides 3, then Z mod (3 + √(-3))/2 is really the subfield Z mod 3. More precisely if A and B are in Z and A = B mod (3 + √(-3))/2 then A-B = (3 + √(-3))/2*Q where Q is an integer in Z[(-1+√(-3))/2]. Taking norms you have (A - B)^2 = 3*Norm(Q), so 3 divides A-B and A = B mod 3.

So Z[ (-1 + √(-3))/2] mod (3 + √(-3))/2 is the field extension of Z mod 3 gotten by adjoining a root of f(X) = X^2 + X + 1 to Z mod 3. But X^2 + X + 1 = (X-1)^2 mod 3, so f(X) splits completely mod 3, which means that there is no extension of Z mod 3, ie,         Z[ (-1 + √(-3))/2 ] mod (3 + √(-3))/2 = Z mod 3.

So the residue class representatives are just 0, 1 and 2.

答:
I know this question has already been answered, but I'd like to offer a graphical solution too for those you come in the future.

If you are familiar with the concept of a lattice, this problem becomes very easy.

In case you don't know what a lattice is, here is a quick description:
To create the lattice, first you have to graph the multiples of the integer in question. The x-axis is the $a$ value of $\alpha$ and the y-axis is the $b\sqrt{d}$ value of $\alpha$. To generate multiples of $\alpha$, you would multiply it by $x+yi$ where x and y are integers. If you connect the grid points or multiples you create a lattice.

For this problem, you only need 1 lattice grid, ie any 4 points or multiples that creates a grid. The next step is to find all quadratic integers in the field $Q[\sqrt{d}]$ (where d=-3 in our case) that are "inside" this grid or box (but not on it). The final step is to subtract the value of the nearest gridpoint from each of these quadratic integers, and you get your congruence classes.

In our case, $\alpha=\frac{3+1\sqrt{-3}}{2}$, so 4 easy points/multiples would be $(0,0)$, $(1.5,.5\sqrt{-3})$, $(1.5,-.5\sqrt{-3})$, and $(3,0)$. We can graph all quadratic integers of $Q[\sqrt{-3}]$ in that area and we see that there are only 2 eligible points, being $(1, 0\sqrt{-3})$ and $(2, 0\sqrt{-3})$. Subtracting this from the nearest lattice values (0 for the first one and 3 for the second) we get the values 1 and 2. So your final answer would be 0,1, and 2.

Here is a picture of a lattice that I found off the internet. The yellow points are multiples of your $\alpha$ (in the picture, its just a+bi), and the red points represent quadratic integers inside the lattice of that specific field


Title: Limit of $n^{\frac{1}{n}}$.

Body:
Easiest proof:

By Bernoulli,

$(1+n^{-1/2})^n
\ge 1+n^{1/2}
\gt n^{1/2}
$.

Raising to $2/n$ power,

$n^{1/n}
\lt (1+n^{-1/2})^2
=1+2n^{-1/2}+n^{-1}
\lt 1+3n^{-1/2}
\to 1
$.


问:
Can someone explain this “visual proof” of $(1 + 2 + 3 +…+ n)^2 = 1^3 + 2^3 + 3^3 +…+ n^3$?
I see a lot of people saying this visual proof is beautiful and I really want to be able to understand it. If anyone could help me out, I'd really appreciate it! Thanks in advance!

答:
The lower left figure (red and gray) has $(1+2+3+...+n)$ little cubes. (In the diagram, $n=6$.)

Going across the bottom row from left to right, we see that the figures are $1$ thick, $2$ thick, $...$, $n$ thick.

So this is $(1+2+3+...+n)$ little cubes, repeated $(1+2+3+...+n)$ times, which is $(1+2+3+...+n)^2$.

The top row is a bit easier to visualize: $1^3 + 2^3 + 3^3 + ... + n^3$.

There is the same number of each color of cube in the top and bottom row.

For $n=1$ just the gray cube is used: one gray cube on the bottom; one gray cube on the top.

For $n=2$ look at the top two layers of cubes in the two leftmost figures in the bottom row (gray and red cubes). Seven gray cubes and two red ones are reconfigured to make the $1\times 1 \times 1$ and $2\times 2 \times 2$ cubes in the top row.

For $n=3$ look at the top three layers of cubes in the three leftmost figures in the bottom row ($25$ gray, $5$ red, and $6$ yellow cubes).

And so forth.


问:
Solving the equation: $F''(r) + a(r)F'(r) + bF = 0$
How do I solve the following equation:

$$
F''(r) + \frac{1}{2}rF'(r) - \frac{1}{2}kF(r) = 0
$$

This equation is found when substituting the solution $u = t^{\frac{k}{2}}F(r)$ into the heat equation : $u_{xx} = u_t$.

The solution for $u$ is found by solving the characteristic equation:
$$
\frac{dx}{x} = \frac{dt}{2t} = \frac{du}{ku}
$$
Where I get
$$
r = xt^{\frac{-1}{2}} \\
v = ut^{\frac{-1}{2}} \\
\text{where   } v = F(r) \\
u = t^{\frac{k}{2}}F(xt^{\frac{-1}{2}})
$$

The general solution given is:
$$
F(r) = c_1U(k+\frac{1}{2}, 2^{\frac{-1}{2}}r) + c_2V(k+\frac{1}{2}, 2^{\frac{-1}{2}}r)
$$

Where $U(p,z)$ and $V(p,z)$ are parabolic cylinder functions.

How do I arrive at the given solution?

答:
There is at least anoother solution for $$F''(r) + \frac{1}{2}rF'(r) - \frac{1}{2}kF(r) = 0$$ which looks very close to
Hermite differential equation.

Let $F(r)=e^{-\frac{r^2}{4}}G(r)$ to make
$$2 G''(r)-r G'(r)-(k+1) G(r)=0$$which does not look very different but easier to integrate making $r=2 x $.

So, using the solution given in the linked page, you should end with
$$G(r)=c_1 H_{-(k+1)}\left(\frac{r}{2}\right)+c_2 \,
   _1F_1\left(\frac{k+1}{2};\frac{1}{2};\frac{r^2}{4}\right)$$ where appear Hermite polynomial and Kummer hypergeometric function.

Multiplied by $e^{-\frac{r^2}{4}}$, you arrive to the parabolic cylinder functions  (have a look here).


问:
Solving a Diophantine Equation dealing with powers of 2
Let x and y be positive integers with no prime factors larger than 5. Find all such $x$ and $y$ which satisfy $$x^2-y^2=2^k$$
where the range of k is $1\leq k\leq2019$.

I managed to solve the equation for general solutions where $(x, y)=(3 \times 2^z, 2^z)$ and $(x, y)=(5 \times 2^z, 3 \times 2^z)$ for non-negative integers for $z$. However, I still have trouble finding the range of values for $z$ that fit the given range of $k$.

答:
Given $x^2-y^2=2^k\implies x^2=y^2+2^k\quad$ we have the Pythagorean theorem $A^2+B^2=C^2\quad $where

$A=m^2-n^2\quad B=2mn\quad C=m^2+n^2\\\implies A=y, B=2^k, C=x$

Side-B can be any multiple of $4$ such as in the triples
$$(3,4,5)\qquad (15,8,17)\qquad (5,12,13), (35,12,37)\\ (12,16,20), (63,16,65)$$

$B=2^k\land 1\le n \le 2019\implies n\in\{2,3,4,\cdots,2019\}$ and $2^2\le B \le 2^{2019}.$
Notice that $B\not\in\{12,20,24,28,36...\}$ but this is not a problem.

Now we can find the triples for these values of $B$ by solving $B=2mn$ for $n$ and testing a defined range of $m$-values to find which yield integers. For example

\begin{equation}
B=2mn\implies n=\frac{B}{2m}\qquad\text{for}\qquad \bigg\lfloor \frac{1+\sqrt{2B+1}}{2}\bigg\rfloor \le m \le \frac{B}{2}\
\end{equation}
The lower limit ensures $m>n$ and the upper limit ensures $m\ge 2$
$$B=64\implies\qquad \bigg\lfloor \frac{1+\sqrt{128+1}}{2}\bigg\rfloor =6 \le m \le \frac{64}{2}=32\quad \\
\land \quad m\in\{8,16,32\}\implies n\in\{4,2,1\}$$
$$f(8,4)=(48,64,80)\qquad f(16,2)=(252,64,260)\qquad
 f(32,1)=(1023,64,1025)$$
Here we have
$(x,y,k)=(80,48,6),(260,252,6),(1025,1023,6)$

The only problem here is the restriction on prime factors greater than $5$. For the example given only $(80,48,6)$ works. The larger the value of k, the harder it will be to find triples that meet this criteria, expecially since $2^{2019}$ contains over $600$ digits.

A little more testing shows that the following Pythagorean triples meet the criteria

$(3,4,5)\quad (12,16,20)\quad (48,64,60)\quad (192,256,320)\quad  (768,1024,1280)\quad
(3072,4096,5120)\quad (12288,16384,20480)$

It looks like a pattern where each B-value is 4-times the one before it but it does not hold up with $B=65536$ so testing is required for each triple.


问:
Volume of a Cylinder With a Cylinder Cutout
I was considering a cylinder with a cylinder cut out through the middle of it (on the curved side, parallel to the base) inspired by a doughnut themed coffee mug I saw. Would there be any way to obtain a general formula for the volume of such a cylinder? The volume of the overall cylinder is simple of course ($\pi r^2h$), and most of the cutout can be calculated easily enough. However, the ends where the cutout meet the edge of the cylinder are giving me trouble. The only way I can think of to solve it is using a triple integral, and that seems very messy. Would there be any better way to do this? Here are some pictures to show the situation for added clarity.

The Entire Cylinder

The Simple Part of the Cutout

The Complicated Part of the Cutout

答:
Let's calculate the entire small radius cylinder. Easy for me to set it up that way. Let's call $x$ the horizontal direction perpendicular to both cylindrical axes, $y$ the vertical direction (along the axis of the big cylinder), and $z$ the direction along the axis of the small cylinder. Also, call the radius of the small cylinder $r$, and of the big cylinder $R$. Not divide the volume of the small cylinder into rectangles perpendicular to the $x$ axis. You can see that the width in the $y$ direction is $2\sqrt{r^2-x^2}$, and the height in the $z$ direction is $2\sqrt{R^2-x^2}$. The limits of integration are from $-r$ to $r$:
$$V=8\int_0^r\sqrt{r^2-x^2}\sqrt{R^2-x^2}dx$$
Make the substitution $x=r\sin\theta$, $dx=r\cos\theta d\theta$. Can you take it from here?


问:
For what values of a the function $y=x^6+ax^3-2x^3-2x^2+1$ is even
I want to know for what valuyes this function is even

I know that $f(x)=f(-x)$ to proove that function is even. how its helps me?$$y=x^6+ax^3-2x^3-2x^2+1$$
Thanks!

答:
Let us start by evaluationg $f(x)$ and $f(-x)$.

$f(x) = x^6 + ax^3 - 2x^3 -2x^2 +1$

and

$f(-x) = x^6 - ax^3 + 2x^3 -2x^2 +1$

If $f(x)$ should equal $f(-x)$ then $a = $...


答: $n^{1/n}=e^{\ln n^{1/n}}=e^{\ln n/n}$.  Now exp is continuous, and $\ln n/n\to0$.  So the limit is $e^0=1$.

Title: Limit of $n^{\frac{1}{n}}$.


问:
Conway's Functional Analysis V Prop. 4.7

Hi, for the proof of this proposition, why do we have $dim (X/M)=1$? In the forward direction, why does this imply there exists $x\in X$ such that $\| x+M\|=1$ and $f(x+M)= \|f\|$?

In the backwards direction, the reason why we get $\| x_0+M\|=1$, is it because that $dim (X/M)=1$ again?

Thanks!

答:
For any nonzero linear functional $x^*$, the quotient space $\dim X/\ker x^*$ is one-dimensional. This is straightforward: fix $z\in X\setminus\ker x^*$ with $x^*(z)=1$. Then, for any $x\in X$, you have $x-x^*(x)z\in \ker x^*$. So $X/\ker x^*=\mathbb C\, z+\ker x^*$ is one-dimensional.

If you have a linear functional $f$ on a one-dimensional space $\mathbb C\,v$, then $f(\lambda v)=\lambda f(v)$, and $$\|f\|=\sup\{|\lambda f(v)|/\|\lambda v\|:\ \lambda\in\mathbb C\}=\frac{|f(v)|}{\|v\|}=|f(v/\|v\|)|.$$ Now write $v'=v/\|v\|$ and $f(v')=e^{i\theta}\,|f(v')|$. Then $$\|f\|=|f(v')|=e^{-i\theta}\,f(v')=f(e^{-i\theta}v')$$ and $\|e^{-i\theta}v'\|=1$.

In the converse, suppose that $\|x_0-m\|<1$ for some $m\in M$. Then
$$
\frac{|x^*(x_0-m)|}{\|x_0-m\|}=\frac{\|x^*\|}{\|x_0-m\|}>\|x^*\|,
$$
a contradiction. So $\|x_0+M\|=\|x_0\|=1$.


问:
Prove that $\lim_{n \to \infty}\int_{A}f_{n}d\mu=\int_{A}fd\mu.$

Let $(f_{n})_{n=1}^{\infty}$ be a sequence of measurables functions non-negatives such that  $\displaystyle \lim_{n\to \infty}f_{n}(x)=f(x)$ almost everywhere and $$\lim_{n \to \infty} \int_{X}f_{n}d\mu =\int_{X}fd\mu<+\infty. $$
Prove that for all measurable set $E\subseteq X$, $$\lim_{n \to \infty}\int_{E}f_{n}d\mu=\int_{E}fd\mu.$$

My approach: Let $E\subseteq X$ measurable set, so we have
$$\lim_{n \to \infty}\int_{E}f_{n}d\mu\overbrace{=}^{definition}\lim_{n\to \infty}\int_{X} f_{n}\cdot \chi_{E}d\mu\overbrace{=}^{de: \chi_{E}}\lim_{n \to \infty} \int_{E}f_{n}d\mu\overbrace{=}^{hypothesis?}\int_{E}fd\mu $$

Is my proof correct?
I don't know how can I use the hypothesis $f_{n}\overset{a.e.x}{\to}f$
How can I conclude in my prove?

答:
Let $(X,\mathcal{F},\mu)$ be a measure space. Let $(f_{n})$ be a
sequence of real-valued integrable functions defined on $X$. Let
$f:X\rightarrow\mathbb{R}$ be an integrable function. Suppose that
$f_{n}\rightarrow f$ a.e., then $\lim_{n}\left(||f_{n}||-||f||-||f_{n}-f||\right)=0$,
where $||f||=\int|f|d\mu$ etc...

We go to prove the above assertion. Observe that $|f_{n}-f|\leq|f_{n}|+|f|$,
so $|f_{n}|-|f|-|f_{n}-f|\geq-2|f|.$ On the other hand, $|f_{n}|\leq|f_{n}-f|+|f|$
and hence $|f_{n}|-|f|-|f_{n}-f|\leq0\leq2|f|$. Combining, we have
$\Biggl||f_{n}|-|f|-|f_{n}-f|\Biggr|\leq2|f|$. Clearly, $|f_{n}|-|f|-|f_{n}-f|\rightarrow0$
a.e.. By Dominated Convergence Theorem, we have that $\int\left(|f_{n}|-|f|-|f_{n}-f|\right)d\mu\rightarrow 0$.
That is, $||f_{n}||-||f||-||f_{n}-f||\rightarrow0$.

Go back to our question. Since $f_{n}$ and $f$ are non-negative,
$||f_{n}||=\int f_{n}$ and $||f||=\int f$. It is given that $||f_{n}||\rightarrow||f||$.
Therefore, $||f_{n}-f||=\left(||f_{n}||-||f||\right)-\left(||f_{n}||-||f||-||f_{n}-f||\right)\rightarrow0$.
Finally, if $E\subseteq X$ is measurable, we have $\int_{E}|f_{n}-f|\leq||f_{n}-f||\rightarrow0$.
Therefore $\int_{E}f_{n}\rightarrow\int_{E}f$.


问:
Remainder of division when $p$ is not prime
Calculate the remainder of the division of $1160^{602}$ by $35$.

What I thought of doing right away is to factor out $602$ until I get to a point where I can apply Fermat's theorem. However, $35$ is not a prime number. How to proceed in this case?

答:
Let  $N \equiv 1160^{602} \pmod{35} \equiv 5^{602} \pmod {35}$.

Clearly $N \equiv 0 \pmod 5$

By Euler's theorem, $5^6 \equiv 1 \pmod 7$.

So $N \equiv 5^{600} \cdot 5^2 \pmod 7 \equiv 1^{100} \cdot25 \pmod 7 \equiv 4 \pmod 7$.

Ww now seek a multiple of $5$ that is congruent to $1$ modulo $7$.

$$5 \equiv 5 \pmod 7, \qquad 10 \equiv 3 \pmod 7, \qquad
    \color{red}{ 15\equiv 1 \pmod 7}$$

Next we seek a multiple of $7$ that is congruent to $1$ modulo $5$.

$$7 \equiv 2 \pmod 5, \qquad 14 \equiv 4 \pmod 5, \qquad
    \color{red}{21 \equiv 1 \pmod 5}$$

It follows that $N \equiv 4 \cdot 15 + 0 \cdot 21 \equiv 60 \equiv 25\pmod{35}$


问:
Inner product defined by summation
I am trying to prove that $V$ is a finite dimension space over C and $(u_1,...,u_n)$ is a basis for $V$. Show that $\langle\,,\rangle$ : $V × V → C,$ defined by $\langle\sum_{i=1}^{n} a_iu_i, \sum_{i=1}^{n} b_iu_i\rangle=\sum_{i=1}^{n} a_i\overline {b_i}$

I tried to show the 4 properties of an inner product:

for additivity: $\langle\sum_{i=1}^{n} a_iu_i+\sum_{i=1}^{n} c_iu_i,\sum_{i=1}^{n} b_iu_i\rangle$ = $\sum_{i=1}^{n} (a+c)_i\overline {b_i} = \sum_{i=1}^{n} a_i\overline {b_i}+\sum_{i=1}^{n} c_i\overline {b_i} = \langle\sum_{i=1}^{n} a_iu_i, \sum_{i=1}^{n} b_iu_i\rangle +\langle\sum_{i=1}^{n} c_iu_i, \sum_{i=1}^{n} b_iu_i\rangle$

Then for homogeneity and conjugate symmetry, it's the same way as above, it's just properties of summations and scalars

For positivity and defineness, $\langle\sum_{i=1}^{n} a_iu_i, \sum_{i=1}^{n} a_iu_i\rangle = \sum_{i=1}^{n} a_i\overline {a_i} = \sum_{i=1}^{n} |a_i|^2 \geq0$

and if  $\langle\sum_{i=1}^{n} a_iu_i, \sum_{i=1}^{n} a_iu_i\rangle = 0, \sum_{i=1}^{n} |a_i|^2 =0$ and since $|a_i|^2 \geq0$ , $a_i=0$ and each  $\sum_{i=1}^{n} a_iu_i =0$

My question is, is my outline correct? If so, why did my proof did not mention anything about a basis and linear independence which is mentioned in the assumption? Where should those conditions about a basis appear in the proof?

答: In your proof you assumed any vector can be written uniquely as $\sum a_i u_i$. For these you need the properties of a basis.

Title: Inner product defined by summation


问:
Markov chain: frog jumping on infinite ladder

A frog jumps on an infinite ladder. He goes up one step with probability $(1-p)$, and falls back down to the ground with probability $p$. If the frog has just fallen to the bottom, on average how many jumps will it take before he next reaches step $i$?

I showed that the frog's jumps form a Markov chain, with transition probability $p_{i,i+1}=(1-p)$ from step $i$ to step $i+1$, and probability $p_{i,0}=p$ from step $i$ to the ground. It then follows that the stationary distribution $(\pi_0,\pi_1,\dots)$ is geometric, with $\pi_i=(1-p)^{i}p$. Let $k_{i}^{j}$ denotes the mean time to go from step $i$ to step $j$ and $m_i = k_{i}^{i}$ is the mean return time from step $i$ to itself. Then, using the law of total expectation, we have \begin{align*}&m_i=(1-p)\left(1+k_{i+1}^{i}\right)+p\left(1+k_{0}^{i}\right), \text{ and} \\ &k_{i+1}^{i} = (1-p)\left(1+k_{i+2}^{i}\right)+p\left(1+k_{0}^{i}\right)\end{align*} Solving the latter recurrence relation, we get \begin{equation*}k_{j}^{i} = \frac{1}{p}\left(1+pk_{0}^{i}-\left(\frac{1}{1-p}\right)^j\right)\end{equation*} We also have the well-known result $m_i = \frac{1}{\pi_i}$, so substituting all of this into the first equation gives \begin{equation*}\frac{1}{(1-p)^{i}p} = (1-p)\left(1+\frac{1}{p}\left(1+pk_{0}^{i}-\left(\frac{1}{1-p}\right)^{i+1}\right)\right)+p\left(1+k_{0}^{i}\right)\end{equation*} so we can solve for the desired $k_{0}^{i}$ as \begin{equation*}k_{0}^{i} = \frac{2}{(1-p)^{i}p}-\frac{1}{p}\end{equation*}
However, the answer is apparently supposed to be \begin{equation*}\frac{1}{(1-p)^{i}p}-\frac{1}{p}\end{equation*} the explanation being that $m_i = k_{i}^{0}+k_{0}^{i}$, with $m_i = \frac{1}{(1-p)^{i}p}$ as above and $k_{i}^{0} = \frac{1}{p}$ by a property of the geometric distribution. So I can't figure out what I did wrong to get the $2$ in the numerator - any assistance would be much appreciated. Thanks in advance for reading this long post.

答:
If $k_j^i$ is the mean time to go from step $i$ to step $j$, then $k_i^i$ should equal $0$. If the frog is already on step $i$, then it wouldn't have to take any more jumps to reach that step. There's at least two ways to solve this correctly.

Approach $1$:
One thing you could do is completely ignore $m_i$ and solve for $k_0^i$ in $k_{j}^{i} = \frac{1}{p}\left(1+pk_{0}^{i}-\left(\frac{1}{1-p}\right)^j\right)$ assuming that $k_i^i = 0$. Plugging in $j = i$ and $k_i^i = 0$ yields that $$k_0^i = \frac{1}{(1-p)^ip}-\frac{1}{p}$$

Approach $2$: You're right that $m_i$ should be the mean return time from $i$ to itself, but the frog has to move first, so it isn't equal to $0$. Instead, it has to fall to the ground $\left(k_i^0 \text{ steps}\right)$ and then go back up to $i$ $\left(k_0^i\text{ steps}\right)$. This is why $m_i$ should equal $k_i^0 + k_0^i$. Since you know $k_i^0 = \frac{1}{p}$ by the geometric distribution and $m_i = \frac{1}{\pi_i} = \frac{1}{(1-p)^{i}p}$, $k_0^i$ would be equal to $\frac{1}{(1-p)^{i}p}-\frac{1}{p}$.


问:
How to prove this stronger form of Cauchy Goursat's theorem?

Let $W$ be an open subset in $C$ and $f$ be a continuous function on $
 W$. Let $a\in W $ and assume that $f : W \setminus \{a\} \to C$ is
complex differentiable function. Then, for any closed rectangle $R
\subseteq W$, we have

$\int_{\partial R} f(z) = 0$

If $a $ is not inside or on our rectangle, then the result simply follows from Cauchy Goursat's theorem. However, how to prove when $a$ is inside or on the rectangle $R$ ?

答:
Enclose $a$ with a small circle $C_t:|z-a|=t$ traversed in negative direction so that $\partial R\cup C_t$ encloses a domain where $f$ is holomorphic

(if you want you can add an edge $L$ joining $\partial R$ and $C_t$ so that $\partial R\cup L\cup C_t\cup L^-$ is the boundary of a simply connected domain)

then $$\int_{\partial R\cup C_t}f(z)dz=0$$
and $$\lim_{t\to 0}\int_{C_t}f(z)dz=0$$

since $f$ is continuous.


问:
Typesetting Multiline Inequalities
If we write
$$\begin{align}
a&<b\\
&=c\end{align}$$
does this mean
$$a<b=c$$
or
$$a<b; a=c$$
?
How do we write the other in multiline format?

(Edited to Add)
What about
$$\begin{align}
a&<b\\
&=c\\
&=d\\
&=f\end{align}$$
?

答: It means $a<b=c$. Basically, this notation is due to lack of space, so terms are separated into multi lines.

Title: Typesetting Multiline Inequalities


Title: For what values of a the function $y=x^6+ax^3-2x^3-2x^2+1$ is even

Body:
I want to know for what valuyes this function is even

I know that $f(x)=f(-x)$ to proove that function is even. how its helps me?$$y=x^6+ax^3-2x^3-2x^2+1$$
Thanks!

Body:
$$f(x)=f(-x)\\
x^6+ax^3-2x^3-2x^2+1=x^6-ax^3+2x^3-2x^2+1\\
(a-2)x^3=(2-a)x^3
$$
So for what values of $a$ is this equality satisfied?

$$a-2=2-a\iff 2a=4\iff a=2$$


答:
Generally,

\begin{align*}
a &< b \\
&= c
\end{align*}

is to be interpreted as $a < b = c$. This can be convenient for a series of derivations or inequalities, when putting them on the same line could be ugly or hard to follow. To pull an example from a homework of mine:

\begin{align*}
f(z)
&= \frac{1}{(z-ia)^2} \frac{1}{(2ia)^2} \frac{1}{(z+ia)^2}  \\
&= \frac{1}{(z-ia)^2}  \frac{1}{(2ia)^2} \ \sum_{k=0}^\infty \binom{2+k-1}{k} \left( - \frac{z-ia}{2ia} \right)^k  \\
&= \frac{1}{(z-ia)^2} \frac{1}{(2ia)^2} \sum_{k=0}^\infty \frac{(-1)^k (k+1)}{(2ia)^k} (z-ia)^k   \\
&=  \sum_{k=0}^\infty \frac{(-1)^k (k+1)}{(2ia)^{k+2}} (z-ia)^{k-2} \\
&=  \sum_{n=-2}^\infty \frac{(-1)^n (n+3)}{(2ia)^{n+4}} (z-ia)^{n}
\end{align*}

(Usually I accompany these with minor notes on the right side, or number the equal signs and then explain the equalities afterwards, it simply depends. The former looks ugly on MSE most of the time so I left it out.)

If you want to write $a < b; a = c$, then you could just use

```<code><span class="math-container">\begin{align*}
a &< b \\
a &= c
\end{align*}</span>
</code>```

which yields

\begin{align*}
a &< b \\
a &= c
\end{align*}

Perhaps throw a curly brace on the left-hand side to indicate they should considered together or a group?

Title: Typesetting Multiline Inequalities

Body:
If we write
$$\begin{align}
a&<b\\
&=c\end{align}$$
does this mean
$$a<b=c$$
or
$$a<b; a=c$$
?
How do we write the other in multiline format?

(Edited to Add)
What about
$$\begin{align}
a&<b\\
&=c\\
&=d\\
&=f\end{align}$$
?

Body:
Generally,

\begin{align*}
a &< b \\
&= c
\end{align*}

is to be interpreted as $a < b = c$. This can be convenient for a series of derivations or inequalities, when putting them on the same line could be ugly or hard to follow. To pull an example from a homework of mine:

\begin{align*}
f(z)
&= \frac{1}{(z-ia)^2} \frac{1}{(2ia)^2} \frac{1}{(z+ia)^2}  \\
&= \frac{1}{(z-ia)^2}  \frac{1}{(2ia)^2} \ \sum_{k=0}^\infty \binom{2+k-1}{k} \left( - \frac{z-ia}{2ia} \right)^k  \\
&= \frac{1}{(z-ia)^2} \frac{1}{(2ia)^2} \sum_{k=0}^\infty \frac{(-1)^k (k+1)}{(2ia)^k} (z-ia)^k   \\
&=  \sum_{k=0}^\infty \frac{(-1)^k (k+1)}{(2ia)^{k+2}} (z-ia)^{k-2} \\
&=  \sum_{n=-2}^\infty \frac{(-1)^n (n+3)}{(2ia)^{n+4}} (z-ia)^{n}
\end{align*}

(Usually I accompany these with minor notes on the right side, or number the equal signs and then explain the equalities afterwards, it simply depends. The former looks ugly on MSE most of the time so I left it out.)

If you want to write $a < b; a = c$, then you could just use

\begin{align*}
a &< b \\
a &= c
\end{align*}

which yields

\begin{align*}
a &< b \\
a &= c
\end{align*}

Perhaps throw a curly brace on the left-hand side to indicate they should considered together or a group?