问:
Operator Exponential $e^A e^B = e^{A+B}$
This question comes from an exam in my functional analysis class.

Suppose $X$ is a Banach space, and $T \in B(X,X)$ is a bounded linear operator on $X$. For any non-negative integer $n$, let
$$S_n=\sum_{k=0}^n \frac{1}{k!} T^k$$
where $T^k$ is the composition of $T$ with itself $k$ times and $T^0=I$.

We can show that for any integer $k>0$, $\Vert T^k \Vert \le \Vert T \Vert ^k$. Then we can show that $S_n \in B(X,X)$ and there is some $S \in B(X,X)$ such that $S_n \to S$. We write $S = e^T$ for this operator.

Finally, We were asked to show that $e^T$ has an inverse, and that it is $e^{-T}$. My thought is to prove the following claim first: if $A,B \in B(X,X)$ and $AB = BA$, then $e^A  e^B = e^{A+B}$. If the claim is true, it follows that $e^T e^{-T}=I$.

The claim can be proven provided that the product series can be computed with the Cauchy rule.

$$e^Ae^B=\sum_{i=0}^{\infty}\frac{A^i}{i!}\sum_{j=0}^{\infty}\frac{B^j}{j!}=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{A^lB^{k-l}}{l!(k-l)!}$$
$$=\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{l=0}^{k}\frac{k!}{l!(k-l)!}A^lB^{k-l}= \sum_{k=0}^{\infty}\frac{1}{k!}(A+B)^k= e^{A+B}$$

But why can the product series be summed in the Cauchy way? I know for real-number series, by Cauchy's theorem, if $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ are absolutely convergent to $A$ and $B$, respectively, then we can add $a_i b_j$ in any way, and the resulting series will converge to $AB$. Does this proposition still hold for commutable operators? (It would be greatly appreciated if ideas of proof or reference is suggested.)

答:
Technically, the post itself only asks about showing that the inverse of $e^{S}$ is $e^{-S}$. First, you can check that the power series $$e^{tS}=\sum\limits_{n=0}^\infty\frac{t^n}{n!}S^n$$ makes sense for all $t\in\mathbb{R}, S\in\mathcal{L}(E).$ Differentiating in $t$ gives $$\frac{d}{dt}e^{tS}=Se^{tS}=e^{tS}S.$$ Next, differentiate $e^{(s+t)S}e^{-tS}$ in $t$. You'll get that it's zero, so it's constant in $t$, and evaluation at $t=0$ gives $$e^{(s+t)S}e^{-tS}=e^{sS}.$$ Evaluate this at $s=0$ to get that $e^{tS}e^{-tS}=I$, which provides the desired inversion property (take $t=1$).

Next, we'll answer the question in your title. Compute that $$\frac{d}{dt}(e^{t(S+T)}e^{-tT}e^{-tS})=e^{t(S+T)}Se^{-tT}e^{-tS}-e^{t(S+T)}e^{-tT}Se^{-tS}.$$ We claim this equals zero. Indeed, the fact that $ST=TS$ implies that $$e^{-tT}S=\sum\limits_{n=0}^\infty \frac{(-t)^n}{n!}T^nS=S\sum\limits_{n=0}^\infty \frac{(-t)^n}{n!}T^n=Se^{-tT},$$ from which it follows that $Se^{-tT}=e^{-tT}S$.

Hence, $$e^{t(S+T)}e^{-tT}e^{-tS}$$ is constant in $t$. Evaluating at $t=0$ gives that $$e^{t(S+T)}e^{-tT}e^{-tS}=I.$$ Next, multiply on the right by $e^{tS}$, then by $e^{tT}$, and use the inversion property that we established earlier. This shows that $$e^{t(S+T)}=e^{tS}e^{tT}.$$ Evaluating at $t=1$ gives the result that you wanted.

This latter argument can be done directly using the holomorphic functional calculus, as well.


问:
Is this set bounded? (Hilbert spaces)
If $H$ is a Hilbert space, and we have a function $g: H \rightarrow H$, I have to prove if $g$ is a contraction, then the set $G=\{x \in H : \exists \alpha \in (0,1) / x=\alpha g(x)\}$ is bounded.

Why must be $G$ bounded?

ContentLicense: CC BY-SA 4.0

Title: Is this set bounded? (Hilbert spaces)

Body:
Let $g$ satisfy there is $\rho$: $\|g(x)-g(y) \|\le \rho\|x-y\|$ for all $x,y$.
By
$$
x = \alpha g(x) = \alpha (g(x)-g(0) + g(0))
$$
we have
$$
\|x\| \le 1\cdot \rho \|x-0\| + 1\cdot\|g(0)\|
$$
and so $\|x\| \le \frac1{1-\rho}\|g(0)\|$


问:
Number of ways to sum $1$s, $2$s and $3$s, taking into account the order
Say I have a number $n$. I want to see how many different ways I can create this number, where the order of my operations matter as well. For example I would for $n=3$ get the solutions $1+1+1$, $1+2$, $2+1$ and $3$, i.e. $4$ different solutions. From what I can tell, similar questions on here only care about the solutions to $1a + 2b + 3c = n$, with no respect to order.

答:
If there are $a_n$ ways to form $n$, the recurrence is $a_n=a_{n-1}+a_{n-2}+a_{n-3}$ because you can either form $n-1$ and add a $1$, or ...  This is the recurrence for the Tribonacci numbers.  The sequence starting from $n=0$ starts
$$ 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609$$
and is given in OEIS A000073.  You can find much more on this site if you search or in the OEIS entry


问:
What is the purpose of the first test in an inductive proof?
Learning about proof by induction, I take it that the first step is always something like "test if the proposition holds for $n = \textrm{[the minimum value]}$"

Like this:

Prove that $1+2+3+...+n = \frac{n(n + 1)}{2}$ for all $n \ge 1$.

Test it for $n = 1$:

$$n = \frac{n(n + 1)}{2} \implies 1 = \frac{2}{2}\textrm{, which holds.}$$

* The rest of the proof goes here *

So, I do it all the time (like a standard). But I never really thought about why. Why do I do such test? I can see that if the test does not hold, it cannot be proven by induction, I guess. But is there another reason we do this?

答:
Take a look at this: We prove by induction that the number

$$8^n+73$$

Is divisible by $7$ for every natural number $n$. Suppose that $8^n+73$ is divisible by $7$. Then

$$8^{n+1}+73 =8\cdot 8^n+73 = 7\cdot 8^n +8^n +73.$$

$7\cdot8^n$ is divisible by $7$ and $8^n+73$ is divisible by $7$ by assumption, so their sum is divisible by $7$.

ContentLicense: CC BY-SA 3.0


问:
Is this derivation right?
I was toying around with integration by parts and I found this formula. It works for all examples I have tried but I just want someone besides me to check the derivation: $$f(x)+C=\int f'(x)dx=xf'(x)-\int f''(x)xdx=xf'(x)-\frac{x^{2}}{2}f''(x)+\int f'''(x)\frac{x^{2}}{2}dx=xf'(x)-\frac{x^{2}}{2}f''(x)+\frac{x^{3}}{6}f'''(x)-\int f''''(x)\frac{x^{3}}{6}=\sum_{n=1}^{\infty}\frac{f^{(n)}(x)x^{n}(-1)^{n-1}}{n!}$$
Now use this formula for the inegral of f(x) and take the derivative of the sum to get rid of the constant:
$$\sum_{n=1}^{\infty}\frac{\frac{d}{dx}f^{(n-1)}(x)x^{n}(-1)^{n-1}}{n!}=\sum_{n=1}^{\infty}\frac{f^{(n)}(x)x^{n}(-1)^{n-1}+nf^{(n-1)}(x)x^{n-1}(-1)^{n-1}}{n!}$$
so:
$$f(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}(x)x^{n}(-1)^{n-1}+nf^{(n-1)}(x)x^{n-1}(-1)^{n-1}}{n!}$$
You can also solve for the constant denoted $C[f(x)]$:
$$C[f(x)]=\sum_{n=1}^{\infty}\frac{f^{(n-1)}(x)x^{n-1}(-1)^{n}}{(n-1)!}$$ Take the limit as x approaches 0 and you can see that: $C[f(x)]=-f(0)$

答:
What you have found is basically an inverted form of the Taylor expansion. Instead of expanding $f(x)$ in terms of $f^{(n)}(0)$, you can get this expression by writing $f(0)$ as an expansion in terms of $f^{(n)}(x)$.

\begin{align}
    f(0)&=f(x)+f'(x)(0-x)+\frac{f''(x)}{2}(0-x)^2+...\\
    f(0)&=f(x)-f'(x)x+\frac{f''(x)}{2}x^2+...\\
    -f(x)&+f(0)=-f'(x)x+\frac{f''(x)}{2}x^2+...\\
    f(x)&-f(0)=f'(x)x-\frac{f''(x)}{2}x^2+...
\end{align}

Title: Is this derivation right?


问:
Does the limit exist? Limit going to $(0,0)$
I tried to solve an exercise

Does $\lim\limits_{x, y\to\ (0,0)} \frac{(e^{x+y} - 1)}{x^2+y^2}$ exist?

I was thinking about looking at $(x, 0), (0,y)$ and $x=y=t$ and say it doesn't exist but i was told its not a proof. So could you please let me know how to solve this and this kind of questions? Thank you

答: Write:  $\dfrac{e^{x+y}-1}{x^2+y^2} = \dfrac{e^{x+y}-1}{x+y}\cdot \dfrac{x+y}{x^2+y^2}$. So the limit depends on the ratio $\dfrac{x+y}{x^2+y^2}$ that has no limit !


问:
Singular points of Steiner surface

Find the singular points in $\mathbb{P}^{3}$ of the Steiner surface

$$V(X_{2}^{2} X_{3}^{2} + X_{1}^{2}X_{3}^{2} + X_{1}^{2}X_{2}^{2} - X_{0}X_{1}X_{2}X_{3}).$$

As I understand it, we have to solve $\frac{\partial f}{\partial X_i} = 0$. So we have $$\frac{\partial f}{\partial X_0} = -X_1 X_2 X_3,$$ $$\frac{\partial f}{\partial X_1} = 2X_1 X_{3}^{2} + 2X_1 X_{2}^{2} - X_0 X_2 X_3,$$ $$\frac{\partial f}{\partial X_2} = 2 X_2 X_{3}^{2} + 2X_{1}^{2} X_{2} - X_0 X_1 X_3,$$  $$\frac{\partial f}{\partial X_3} = 2X_{2}^{2} X_{3} + 2X_{1}^{2} X_3 - X_0 X_1 X_2.$$

Setting all of them equal to zero yields $P = (a : 0 : 0 : b)$ or $P = (a : b : 0 : 0)$ or $P = (a : 0 : b : 0)$, where $a, b$ are not both zero.

Is my solution correct?

Is there perhaps a more methodological way of solving this? The way I solved it was to play around with the given equations and see which points will satisfy the conditions.

答: One can exploit the symmetry of the defining equation in the variables with positive index. The first equation says that at least one of $X_1,X_2,X_3$ is zero. Suppose it's $X_1$ - then the other equations reduce to $-X_0X_2X_3$, $2X_2X_3^2$, and $2X_2^2X_3$, which implies one of $X_2$ and $X_3$ must be zero (assuming we're not in characteristic 2 - in characteristic 2, we need to examine the original equation after setting $X_1,X_0=0$ to reach the same conclusion). So the singular locus is the union of $V(X_i,X_j)$ for $i,j\in\{1,2,3\}$ distinct.

Title: Singular points of Steiner surface


问:
When dealing with finite groups, if $a$ and $b$ are elements of $G$ and if $ba=a^4b^3$, how do I prove that ${\rm ord}(a^4b)={\rm ord}(a^2b^3)$?
When dealing with finite groups, if $a$ and $b$ are elements of $G$ and if $ba=a^4b^3$, how do I prove that ${\rm ord}(a^4b)={\rm ord}(a^2b^3)$?

I know that ${\rm ord}(ab)={\rm ord}(ba)$, and ${\rm ord}(a)={\rm ord}(a^{-1})$.

I think the answer might have something to do with conjugates in groups - perhaps showing that $a^4b$ and $a^2b^3$ are conjugates so that they may have the same order?

答:
Unfortunately, $a^4b$ and $a^2b^3$ may not be conjugates. However they both look similar to $a^4b^3$.

We may write:

$$a^4b = a^4b^3b^{-2} = bab^{-2}, \ a^2b^3 = a^{-2}a^4b^3 = a^{-2}ba$$

Now we may derive the result using $\operatorname{ord}(ab) = \operatorname{ord}(ba)$ and $\operatorname{ord}(a) = \operatorname{ord}(a^{-1})$.


问:
find $\lim_{x \to \infty} \frac{\ln(x+1)}{x}$
I need to find: $$\lim_{x \to \infty} \frac{\ln(x+1)}{x}$$
without using L'Hospital, integrals or derivative!

I would like your help in order to know i'm not doing something invalid.

I was thinking of the following way:

$$
\lim_{x \to \infty } \frac{\ln(x+1)}{x}  =  \lim_{x \to \infty } \frac{1}{x} \ln(x+1)  =  \lim_{x \to \infty } \ln(x+1)^\frac{1}{x}
$$

and because $\frac{1}{x}$ tend to $0$ as $x$ approaches $\infty$, than $(x+1)^\frac{1}{x}$ tends to $1$ and $\ln(1) = 0$
and the limit is $0$.

But this way sounds off. I was trying also to express $x$ with $t$ and use $e$ (euler) somehow, but it didn't work.

Is my way valid? If not, do you have a better way without using L'Hospital's rule?

Thanks!

Edit:

While reading answers I was thinking of another way,

$$\lim_{x \to \infty} \frac{\ln(x+1)}{x}  =  \lim_{x \to \infty} \frac{\ln(x(1+\frac{1}{x}))}{x} = \lim_{x \to \infty} \frac{\ln(x)}{x} + \frac{\ln(1+\frac{1}{x})}{x} = $$ $$\lim_{x \to \infty} \ln(\sqrt[x]{x}) + \frac{\ln(1+\frac{1}{x})}{x}$$

now $\sqrt[x]{x}$ tends to $1$ as x approaches to $\infty$, hence $\ln(\sqrt[x]{x})$ tends to $0$ (is it valid to say so?)

and $\frac{\ln(1+\frac{1}{x})}{x}$ tends to $0$ is pretty obvious.

Am on missing something in this one?

答:
$\ln\nearrow\, $ thus for $\, x\gg 1\, $ then $\, x+1<2x\implies 0<\dfrac{\ln(x+1)}{x}<\dfrac{\ln(2x)}{x}=\underbrace{\dfrac{\ln(2)}{x}}_{\to 0}+\dfrac{\ln(x)}{x}$

So we have reduced the problem to studying $f(x)=\dfrac{\ln(x)}{x}$

$f'(x)=\dfrac{1-\ln(x)}{x^2}<0$ at infinity, so $f\searrow$ at infinity and in particular this means that $f$ is bounded by some $M$ at infinity.

Now we can use the multiplicative to additive property of logarithm,

$f(x)=\dfrac{2\ln(\sqrt{x})}{(\sqrt{x})^2}<\dfrac{2M}{\sqrt{x}}\to 0$


问:
Let $a$ be an algebraic number, prove that $ia$ is also algebraic
Of course this is a particular case of the closure under addition and multiplication of algebraic numbers (which proof would also be appreciated), but I would prefer a simple transformation on the already known polynomial giving $a$ as a root in order to obtain a polynomial giving $ia$ as a root. Thank you in advance for your help.

答:
Let $b = ia$.

Since $a$ is algebraic, there is $p(x) = \sum\limits_{k=0}^{\deg p} \alpha_k x^k \in \mathbb{Q}[x]$ such that $p(a) = 0$.

For $\ell = 0,\ldots,3$, let $p_\ell(x)$ be the polynomial $\sum\limits_{k=0}^{\left\lfloor\frac{\deg p - \ell}{4}\right\rfloor} \alpha_{4k+\ell} x^{4k+\ell} \in \mathbb{Q}[x]$.

It is easy to see

$p(x) = p_0(x) + p_1(x) + p_2(x) + p_3(x)$

$p_\ell(b) = p_\ell(ia) = i^\ell p_{\ell}(a)$

This implies

$$\begin{align}
&\;(p_0(b) - p_2(b)) - i (p_1(b) - p_3(b))\\ = &\; (p_0(a) + p_2(a)) + (p_1(a) + p_3(a))\\ = &\; p(a) = 0\end{align}$$
Multiply by $(p_0(b) - p_2(b)) + i (p_1(b) - p_3(b))$, we find $b$ is a root of $q(x) \in \mathbb{Q}[x]$ where
$$q(x) \stackrel{def}{=} (p_0(x) - p_2(x))^2 + (p_1(x) - p_3(x))^2$$
As a result, $b = ia$ is also algebraic.


问:
Find the sum using properties of binomial coefficients
I need to find the following sum using properties of binomial coefficients:$$\sum\limits_{k=0}^{n} k^3 {n\choose k}^2.$$ I transformed this sum first in:$n^2\sum\limits_{k=1}^{n} k{n-1\choose k-1}^2$.
Then after substitution $j=k-1$ I get $n^2\sum\limits_{j=0}^{n-1}(j+1){n-1\choose j}^2$.
Then, after separating this in two sums for first one I get: $n^2\sum\limits_{j=0}^{n-1}j {n-1\choose j}^2$. And for the second $\sum\limits_{j=0}^{n-1}{n-1 \choose j}^2$. I am not sure how to solve these two. Probably is has something to do with Vandermonde convolution, but honestly, I don't know to use it properly.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Lets $\ds{{\cal I}_{m} \equiv
\sum_{k = 0}^{n}k^{m}{n \choose k}^{2}\,,\qquad
{\cal I}_{3}:\ {\Large ?}}$.

\begin{align}
{\cal I}_{3} & =
\bbox[5px,#ffd]{\sum_{k = 0}^{n}k^{3}{n \choose k}^{2}} =
\sum_{k = 0}^{n}\pars{n - k}^{3}{n \choose n - k}^{2}
\\[5mm] = &
\sum_{k = 0}^{n}\pars{n^{3} - 3n^{2}k + 3nk^{2} - k^{3}}{n \choose k}^{2}
\\[5mm] \implies {\cal I}_{3} & =
{1 \over 2}\,n^{3}\,{\cal I}_{0} -
{3 \over 2}\,n^{2}\,{\cal I}_{1} + {3 \over 2}\,n
\,{\cal I}_{2}\,,\qquad
\pars{\substack{\ds{Similarly,}
\\[1mm]
\ds{{\cal{I}_{1} = {1 \over 2}\,n\,{\cal I}_{0}}}}}
\\[5mm] \implies {\cal I}_{3} & =
-\,{1 \over 4}\,n^{3}\,{\cal I}_{0} + {3 \over 2}\,n
\,{\cal I}_{2}
\end{align}

\begin{align}
\sum_{k = 0}^{n}{n \choose k}^{2}x^{k} & =
\sum_{k = 0}^{n}{n \choose k}x^{k}{n \choose n - k}
\\[5mm] & =
\sum_{k = 0}^{n}{n \choose k}x^{k}\bracks{z^{n - k}}
\pars{1 + z}^{n}
\\[5mm] & =
\bracks{z^{n}}
\pars{1 + z}^{n}\sum_{k = 0}^{n}
{n \choose k}\pars{xz}^{k}
\\[5mm] & =
\bracks{z^{n}}\pars{1 + z}^{n}\pars{1 + xz}^{n}
\end{align}

$$
{\cal I}_{0} = \bracks{z^{n}}\pars{1 + z}^{2n} = \bbx{2n \choose n} \\
$$

\begin{align}
{\cal I}_{2} & =
\sum_{k = 0}^{n}k^{2}{n \choose k}^{2} =
\bracks{z^{n}}
\pars{1 + z}^{n}\,
\left.\partiald{}{x}\pars{x\,\partiald{}{x}}
\pars{1 + xz}^{n}\,\right\vert_{\, x\ =\ 1}
\\[5mm] & =
n\bracks{z^{n - 1}}
\pars{1 + z}^{n}\,
\left.\partiald{\bracks{x\pars{1 + xz}^{n - 1}}}{x}
\,\right\vert_{\, x\ =\ 1}
\\[5mm] & =
n\bracks{z^{n - 1}}
\pars{1 + z}^{n}\,
\bracks{\pars{1 + z}^{n - 1} + \pars{n - 1}z\pars{1 + z}^{n - 2}}
\\[5mm] & =
n{2n - 1 \choose n - 1} + n\pars{n - 1}{2n - 2 \choose n - 2}
\end{align}

\begin{align}
{\cal I}_{3} & =
\bbox[5px,#ffd]{\sum_{k = 0}^{n}k^{3}{n \choose k}^{2}}
\\[5mm] & =
-\,{1 \over 4}\,n^{3}{2n \choose n} + {3 \over 2}\,n
\bracks{n{2n - 1 \choose n - 1} + n\pars{n - 1}
{2n - 2 \choose n - 2}}
\end{align}
It can be simplified as
\begin{align}
{\cal I}_{3} & =
\bbox[5px,#ffd]{\sum_{k = 0}^{n}k^{3}{n \choose k}^{2}} =
\bbx{{1 \over 2}\,n^{2}\pars{n + 1}{2n - 2 \choose n - 1}} \\ &
\end{align}


问:
If $p \in\mathbb{R}^n$, $r > 0$, and $U = \{x \in\mathbb{R}^n \mid\|p-x\| > r\}$, then $U$ is an open subset of $\mathbb{R}^n$
Prove or give a counterexample: if $p \in \mathbb{R}^n$, $r > 0$, and $U = \{x \in \mathbb{R}^n \; | \; ||p − x|| > r\}$, then $U$ is an open subset of $\mathbb{R}^n$.

I think that it should be open but I can't figure out how to prove it.

The definition of open I am  using is for every $x \in U$, there exists a $\delta > 0$ such that  $B(x, \delta)$ is a subset of $U$.

答: Let's define $f: \mathbb{R}^n \to \mathbb{R}$ such that $f(x)=|x-p|$. Since $f$ is continuous, preimages of open sets are open. Since $(r, \infty )$ is open, $f^{-1}(r, \infty )$ is too. This is exactly the set in the question.


问:
Find the area of quadrilateral $ABCD$ using the hexagon properties.
Let $ABCDEF$ be a convex hexagon with the following properties.

(a) $AC$ and $AE$ trisect $\angle BAF$.

(b) $BE || CD$ and $CF ||DE$.

(c) $AB = 2AC = 4AE = 8AF$.

Suppose that quadrilaterals $ACDE$ and $ADEF$ have area $2014$ and $1400$, respectively. Find the area of quadrilateral $ABCD$.

The diagram was hard to make in geogebra. And the diagram I have attached might not be true for general case.

Not much progress, I got that triangles $AFE, AEC$ and $ACB$ are similar to one another, each being twice as large as the preceding one.

答:
To calculate areas, first we prove that $AD$ , $CF$ and $EB$ are concurrent. And to do that, first we state and prove a lemma about an exciting property of trapezoids.

Lemma: The line passing through the intersection of diagonals and the intersection of extensions of legs of a trapezoid, passes through midpoints of the bases of the trapezoid.

In the figure below, legs $EX$ and $CY$ of the trapezoid $XECY$ meet at $A$ and diagonals of the trapezoid intersect at $Z$ . $AZ$ intersects base $XY$ at $M$ and base $EC$ at $N$ . We show that $XM=YM$ and $EN=CN$ .

Triangles $XEC$ and $YEC$ share the same base ($EC$) and the same hight (the distance between the parallel bases), so they have equal areas: $ S_{XEC} = S_{YEC} $ . We can write this equality as:
$$ S_{ZXE} + S_{ZEC} = S_{ZYC} + S_{ZEC} $$
Therefore:
$$ S_{ZXE} = S_{ZYC} $$
Drawing $ZS$ and $ZT$ perpendicular to $EX$ and $CY$ , respectively, we have:
$$ ZS.EX = ZT.CY$$
$$ \frac{ZS}{ZT} = \frac{CY}{EX} $$
Now, noting that $XY$ and $EC$ are parallel and extensions of $EX$ and $CY$ meet at $A$ , we have:
$$ \frac{CY}{EX} = \frac{YA}{XA} $$
Therefore:
$$ \frac{ZS}{ZT} = \frac{YA}{XA} $$
$$ ZS.XA = ZT.YA $$
$$ S_{AZX} = S_{AZY} $$
Or, drawing $XH$ and $YK$ perpendicular to $AZ$ , we can write:
$$ XH.AZ = YK.AZ $$
$$ XH = YK $$
As a result, the right triangles $MHX$ and $MKY$ are equal. Therefore
$$ XM = YM $$
And it follows easily that $EN = CN$ .
That proves our lemma.

Next, we get back to our problem and show that $AD$ , $CF$ and $EB$ are concurrent.

First, note that the triangles $AFE$ , $AEC$ and $ACB$ are similar because they have equal angles at $A$ and equal ratios of sides:
$$ \widehat{FAE} = \widehat{EAC} = \widehat{CAB} $$
$$ \frac{AF}{AE} = \frac{AE}{AC} = \frac{AC}{AB} = \frac{1}{2} $$
These similarities have an important result, which we will used later on:
$$ S_{ACB} = 4S_{AEC} = 16S_{AFE} \qquad(1) $$

Triangles $AFE$ and $AEC$ are similar and they are placed such that they share the edge $AE$ , constructing the quadrilateral $AFEC$.

Similarly, triangles $AEC$ and $ACB$ are similar and they are placed such that they share the edge $AC$ , constructing the quadrilateral $AECB$.

The quadrilaterals $AFEC$ and $AECB$ are similar. Now, consider their respective diagonals, $FC$ and $EB$ , which intersect $AE$ and $AC$ at $X$ and $Y$ , respectively. Because of the similarity of the shapes it is easy to show that the triangles $AFX$ and $AEY$ are similar. As a result:
$$ \frac{AX}{AY} = \frac{AF}{AE} = \frac{1}{2} $$
Recalling that $\frac{AE}{AC}$ is also $\frac{1}{2}$ , we conclude that $XY$ is parallel to $EC$ .

We can now use the lemma we proved at the beginning: $XECY$ is a trapezoid. Its parallel bases are $XY$ and $EC$ . Its legs $EX$ and $CY$ meet at $A$ , and its diagonals intersect at $Z$ . $AZ$ intersects $EC$ at $N$ . We now know that $EN=CN$ .

Next, we consider that point D is at the intersection of two lines drawn from $E$ and $C$ , and parallel to $EB$ and $FC$ , respectively. This means that the quadrilateral $CZED$ , which is formed by these two pairs of parallel lines, is a parallelogram.

Now, we know that diagonals of a parallelogram bisect each other. We have also shown that the extension of $AZ$ bisects $EC$. Therefore, diagonal $DZ$ of the parallelogram should necessarily overlap $AN$ , as they both pass through points $N$ and $Z$ . This concludes our proof of the claim that $AD$ passes through $Z$ , the intersection of $CF$ and $EB$ .

We are now able to proceed with the calculation of the area of the quadrilateral $ABCD$ :
$$ S_{ABCD} = S_{ACB} + S_{ANC} + S_{NDC} $$
The area of $ANC$ is half of the area of $AEC$ , because $NC$ is half of $EC$. The areas of $AFE$ , $AEC$ and $ACB$ are related as we showed in equation (1). The area of $NDC$ is a quarter of the area of the parallelogram $CZED$. Therefore, if we have the areas of $AFE$ and the parallelogram $CZED$, we will be able to calculate the area of the shape as
$$ S_{ABCD} = (16+2)S_{AFE} + \frac{1}{4}S_{CZED} \qquad(2) $$
Note that we have two unknowns ($S_{AFE}$ and $S_{CZED}$) and we are given two pieces of information in the problem: areas of the quadrilaterals $ACDE$ and $ADEF$ . This allows us to build a system of two equations with two unknowns:
$$ S_{ACDE} = S_{AEC} + S_{EDC} = 4S_{AFE} + \frac{1}{2}S_{CZED}$$
$$ S_{ADEF} = S_{AFE} + S_{AEN} + S_{EDN} = 3S_{AFE} + \frac{1}{4}S_{CZED} $$
Note that because $N$ is the midpoint of $EC$ , we have $S_{AEN} = \frac{1}{2}S_{AEC}$ . The interested reader can solve the above system of equations for $S_{AFE}$ and $S_{CZED}$ and plug the results into equation (2) to calculate the requested area.


问:
Open, closed and continuous
I have some troubles to understanding something:

We were asked to find a function that is open and continuous but not closed and actually I found such a function, but our tutor gave us this example

$ e^x$ since this function is continuous as a mapping $$\exp: \mathbb{R} \rightarrow \mathbb{R}_{>0}$$ and the inverse function is continuous too, so this function is also open.

and the counterexpample was, that $\mathbb{R}$ is closed and is mapped to $\mathbb{R}_{>0}$ which is open.

then i thought, hey if the inverse function is continuous that means that the fiber of closed sets have to be closed, but $\ln^{-1}(\mathbb{R})$ is not closed at all. where am i wrong?

答:
If you consider $\def\R{\mathbb{R}}\def\exp{\mathrm{exp}}\exp\colon\R\to\R$, then the example is correct: the map is not closed, because

$$\exp(\R)=\R_{>0}$$

which is not closed in $\R$. On the other hand, the image of an open set in $\R$ is open in $\R_{>0}$, so also open in $\R$. Hence the function is open but not closed.

If you consider $\exp\colon\R\to\R_{>0}$, then of course its image is both open and closed in the codomain, being the whole of it. As already observed, this is a homeomorphism, so it's obviously open and closed.

Title: Open, closed and continuous


问:
Wire basket shape for maximum Moment of Inertia
Attempted to find the meridian shape for optimum moment  of inertia with a minimum mass of wire constraint and like to share in the iso-perimetric problem in the plane.

Assuming constant density per unit length

$$ \int y^2 ds - \lambda^2 \int ds= \int( y^2-\lambda^2) ds ,\; {ds}/{dx}=\sqrt{1+y^{'2}}\tag1$$

Lagrangian is

$$ ( y^2-\lambda^2) \sqrt{1+y^{'2}}$$

Using Euler-Lagrange equations on the Lagrangian we obtain

$$( y^2-\lambda^2) \sqrt{1+y^{'2}} -y'\cdot  ( y^2-\lambda^2)\dfrac{y'}{\sqrt{1+y^{'2}}} = c^2 $$

$$\dfrac{y^2-\lambda^2}{\sqrt{1+y^{'2}}} = c^2 \tag2 $$

To eliminate $c^2$ raising to second order with one differentiation with respect to arc

$$\dfrac{1}{c^2}
 =\dfrac{\sec \phi}{y^2- \lambda^2} = \dfrac{ \sec \phi \tan \phi\;k(s)}{ 2 y \sin \phi } $$

$$ \text{curvature} \;k(s)= \dfrac{d \phi}{ds} , \quad  k(s)= \dfrac{ 2 y \cos ^2\phi}{c^2} \tag3 $$

changing  $y=r$ as symbol for radius in cylindrical coordinates and since $ \sin \phi = \dfrac{dr}{ds},$

$$\sec^2 \phi \sin \phi\; d\phi=\sec \phi \tan \phi\; d\phi=\dfrac{2 r dr}{c^2} \tag4 $$

Integrating

$$ \sec\phi =\dfrac{r^2}{c^2}+\text{const.} \tag 5 $$

and  initial condition  $ \phi=0, r = r_1$

$$ \sec \phi -1 = \dfrac{r^2-r_1^2}{c^2} \tag 6 $$

which is the ODE for meridians of the basket. For the given graph initial conditions are at vertex $ \phi_{i}=0, r_i=0.8 $. Maximum curvature  at vertex is $ k_{max}=\dfrac{2r_1}{c^2}$.

Please comment  about correctness and or whether the curve is already known.

答:
From the above maximum curvature for other $c$ values are

$$ k_{max_1}=2 r_1/c^2$$

Other curves can be generated and basket profiles are shown plotted:


答: I think you can consider the following function instaed: $$\pi:\mathbb R^2\to\mathbb R\\\\ \pi\left((x,y)\right)= x$$</p>

Title: Open, closed and continuous

Body:
I have some troubles to understanding something:

We were asked to find a function that is open and continuous but not closed and actually I found such a function, but our tutor gave us this example

$ e^x$ since this function is continuous as a mapping $$\exp: \mathbb{R} \rightarrow \mathbb{R}_{>0}$$ and the inverse function is continuous too, so this function is also open.

and the counterexpample was, that $\mathbb{R}$ is closed and is mapped to $\mathbb{R}_{>0}$ which is open.

then i thought, hey if the inverse function is continuous that means that the fiber of closed sets have to be closed, but $\ln^{-1}(\mathbb{R})$ is not closed at all. where am i wrong?


问:
$0^2{{n}\choose{0}} + 1^2{{n}\choose{1}} + 2^2 {{n}\choose{2}} + ... +n^2{{n}\choose{n}} = n (1+n)* 2^{n-2}, n \in \mathbb{N} \setminus \{0\}$ proof.
I need to pove that:

$$0^2{{n}\choose{0}} + 1^2{{n}\choose{1}} + 2^2 {{n}\choose{2}} + ... +n^2{{n}\choose{n}} = n (1+n)* 2^{n-2}, n \in \mathbb{N} \setminus \{0\}$$

I think that the left hand side looks like binominal theorem. The problem is that in that case the sequence:
$$0^2+1^2+2^2+...n^2$$
would be therefore equal to the multiplication of growing powers on one element and decreasing powers of other element. That is impossible since powers are constant and equal.

And I can't do my induction proof without any transformation of that expression since adding $1$ to $n$ would make that structure in its current form irregular - I wouldn't be able to extract the seqence that is my thesis.

Therefore I don't know how to transform that expression so that it can be proven by induction (as I assume).

答:
Let

$$f(x)=\sum_{k=0}^nx^k{n\choose k}=(1+x)^{n}$$

Differentiating we get that

$$f'(x)=\sum_{k=0}^nkx^{k-1}{n\choose k}=n(1+x)^{n-1}$$

and so

$$xf'(x)=\sum_{k=0}^nkx^{k}{n\choose k}=nx(1+x)^{n-1}$$

differentiating again, we get that

$$(xf'(x))'=\sum_{k=0}^nk^2x^{k-1}{n\choose k}=n(1+x)^{n-1}+n(n-1)x(1+x)^{n-2}$$

plugging in $x=1$ we get that

\begin{align*}
\sum_{k=0}^nk^2{n\choose k}&=2^{n-1}n+n(n-1)2^{n-2}\\
&=n2^{n-2}(2+n-1)\\
&=n(1+n)2^{n-2}
\end{align*}

and we are done


问:
Minimum Value of Sum Absolute Functions
I have this problem

So the minimum value of the |5x-1| is the 0 because |-a|=a and then every negative value of the 5x-1 is positive and greater than 0?

Title: Minimum Value of Sum Absolute Functions

Body:
Note that $\,|A|+|B|+|C|\ge|A+B+C|=5|x-1|\ge0$. Particularly, $\,5|x-1|=0\,$ for $\,x=1$.

Further
$$A(1,y)=1\cdot y+3\cdot1-y-3=0$$
$$B(1,y)=4\cdot 1+y-4-1\cdot y=0$$
$$C(1,y)=2-2\cdot1=0$$
Therefore
$$|A(1,y)|+|B(1,y)|+|C(1,y)|=0$$
So, it is proved that your function takes a mininum ($0$) for $\,x=1$.


问:
To what extent is the Singular Value Decomposition unique?
In Adam Koranyi's article "Around the finite dimensioal spectral theorem",
in Theorem 1 he says that there exist unique orthogonal decompositions.

What is meant here by unique?

We know that the Polar Decomposition and the SVD are equivalent, but the polar decomposition is not unique unless the operator is invertible, therefore the SVD is not unique.

What is the difference between these uniquenesses?

答: The singular values are unique however the columns of $U$ and $V$ are only unique up to the complex sign (because of how Gram-Schmidt works you're only guaranteed that).


问:
Solving second order linear ordinary differential equation - cauchy problem
I would like to solve the following Cauchy problem:

$$
\left\{
\begin{array}{rcl}
y'' + 3y' + 2y &=& x^2 + 1,\\
y(2)&=&1,\\
y'(2)&=& 4.
\end{array}
\right.
$$

I found the characteristic equation: $r^2 + 3r + 2 = 0$, which has as roots $r_1 = -1$ and $r_2 = -2$.

The solutions of the homogeneous equation will be of type $y = Ae^{-x} + Be^{-2x}$.

To find a solution of the general equation, I used:
$$y = \lambda x^2 + \mu$$
$$y' = 2\lambda x$$
$$y'' = 2\lambda$$

So a particular solution of the general equation would something like:
$$ 2\lambda + 3(2\lambda x) + 2(\lambda x^2 + \mu) = x^2 + 1$$
and here is where I got stuck. I don't if the polynomial I used as "candidate" solution ($y = \lambda x^2 + \mu$) is incorrect. Does the "proposed" polynomial must have the exact terms or the one I proposed is correct? Could someone give a hint?
Thank you

答:
By using Annihilator's Method,, we have the ODE can be written like
$$(D^2+3D+2)[y]=x^2+1.$$
By other hand, it is straightforward verify that $D^3[x^2+1]=0$, thus, applying $D^3$ in the previous operator, we obtain:
$$D^3(D^2+3D+2)[y]=0.$$
The roots for the auxiliar equation are $-1,,2,0,0,0$, thus,
the "general solution" for the previous equation is
$$y=c_1e^{-x}+c_2e^{-2x}+c_3+c_4x+c_5x^2.$$
The first couple of terms are exactly the terms of the original homogeneous equation, the remain terms are the terms for a "candidate" of a particular solution, thus,
$$y_p=A+Bx+Cx^2.$$
Thus, applying this particular solution into the ODE we obtain
$$(2C+3B+2A)+(6C+2B)x+2Cx2\equiv 1+x^2$$
$$\Downarrow$$
$$A=\frac{9}{4},\quad B=-\frac{3}{2}\quad\mbox{e}\quad C=\frac{1}{2}.$$
Finally,
$$y_p(x)=\frac{9}{4}-\frac{3}{2}x+\frac{1}{2}x^2.$$
After the Initial Value substituition, we obtain the solution

$$\boxed{y(x)=\frac{13}{4}e^{2-x}-\frac{13}{4}e^{4-2x}+\frac{9}{4}-\frac{3}{2}x+\frac{1}{2}x^2.}$$


问:
Centralizers and conjugacy classes

Let $G$ be a finite group with order 168 such that for each element $g \in G$ wih order $7$, the centralizer $C_G (g)$ is equal to $\langle g \rangle$. Compute the number of conjugacy classes of elements $g$ of order $7$ and find the size of the conjugacy class for such an element.

I tried to begin with the class equation, but didn't have any luck. I don't know if I will need to use the definition of a Sylow subgroup at some point here.

I would appreciate any help.

答:
Let $G$ act on $G$ by conjugation.  Then each orbit of an element is its conjugacy class, and the size of the orbit is the index of the element's stabilizer.  We're given that the stabilizer (which is the element's centralizer) has size $7$ so it has index $24$.  Thus, each conjugacy class of an element of order $7$                                                                                has size $24$.

The $7$-Sylow subgroups of $G$ are not normal.  If one were, its non-trivial elements would have a conjugacy class of size no more than $6$.

Thus, $G$ must have $8$ $7$-Sylow subgroups (because no other divisor $d$ of $168$ satisfies $d \equiv 1 \pmod 7$).  Two distinct $7$-Sylow subgroups of $G$ must have trivial intersection, which means $G$ has $48$ distinct elements of order $7$.  Therefore, there are $2$ such conjugacy classes.

Title: Centralizers and conjugacy classes


问:
How do you solve $y'=\sin(y)$?

How do you solve $y'=\sin(y)$?

I ended up with
$$-\ln(\csc(y)+\cot(y)) = x+C$$

But then I simplified it to
$$e^{-x+C} = \frac1{\sin(y)} + \frac{\cos(y)}{\sin(y)}$$

Where do I go from here?

答:
Notice that for any integer $k$, the constant function $y(x)=\pi k$ is a solution because

$$\sin\left(y(x)\right)=\sin(\pi k)=0=y'(x)$$

To find the others, we may assume that $y(x)$ is not of the form $\pi k$. Then $\sin(y(x))$ is not identically zero, so

$$\csc(y(x))y'(x)=1$$

Antiderivatives of $\csc(y(x))y'(x)$ and $1$ are given by $-\tanh^{-1}\left[\cos(y(x))\right]$ and $x$, so these two must differ by a constant, say $c$.

$$-\tanh^{-1}\left[\cos(y(x))\right]=x+c$$

We can now solve equation for $y(x)$. As always, $c$ was redefined to absorb the minus sign.

\begin{align*}
\tanh^{-1}\left[\cos(y(x))\right] &= c-x\\
\cos(y(x)) &= \tanh(c-x)\\
y(x) &= 2\pi k\pm\cos^{-1}\left[\tanh(c-x)\right]
\end{align*}

In the last line, we made use of the not-so-easy-to-prove fact that for $|y|\leq 1$,

$$y=\cos(x)\iff x=2\pi k\pm\cos^{-1}(y)$$

In summary, the solutions are

$$y(x)=\pi k$$

and

$$y(x) = 2\pi k\pm\cos^{-1}\left[\tanh(c-x)\right]$$

Title: How do you solve $y'=\sin(y)$?


问:
How many 7 letter 'words' can be made from the English alphabet which contains at least 3 vowels?
Just a little bit unsure about this question:

How many 7 letter 'words' can be made from the English alphabet which contains at least 3 vowels?

And by 'words' I mean just a string of alphabet - does not have to be a real world in dictionary. I do know that you need to use Inclusion/Exclusion principle which says:

$|A \cup B| = |A| + |B| - |A\cap B|$

Let U be the set of all words containing 7 letters (26 letters to the power 7)
The only thing I'm not sure is what would be the sets A and B? Or do I need a third term C?

答:
I'm going to assume you are talking about the english language with letters, $L = \{ a, b, c, ... x, y, z \}$ ($|L| = 26$) and vowels, $V = \{a, e, i, o, u \}$ ($|V| = 5$). Let $C = L \setminus V$ be the set of consonants ($|C| = 21$).

Let's say you had no restrictions. You have $n$ positions ($n \ge 1$), each position has an element selected from $L$. Thus you have $|L|^n$ possible arrangements.

If you require at least $m$ vowels in your string ($m \le n$), then you will have to consider $n - m + 1$ cases.

Consider $m = n$. There is one case to consider, all of the characters are vowels and thus the total number of arrangements is $|V|^n$.

Consider $m = n - 1$. There are two cases that you will consider. The previous case and the one where all but one position is a vowel. There are $C(n,1)$ ways to position the consonant in the string. So you have $|V|^{n-1} |C| C(n,1)$ arrangements.

Consider $m = n - 2$. There are three cases. The previous two cases and the one where all but two positions are a vowel. So you have $ |V|^{n-2} |C|^2 C(n, 2)^2$.

Thus, consider $m = n - k$. There are $k + 1$ cases. The previous $k$ cases and the one where all but $k$ positions are a vowel. You have $|V|^{n-k} |C|^k C(n, k)^k$.

To account for everything, you have $\sum_{i = m}^{n}{ |V|^{i} |C|^{n-i} C(n, n-i)^{n-i} }$.


问:
Problem with Discrete Parseval's Theorem
I think I must be missing something obvious, but I can't for the life of me see what it is. The discrete version of Parseval's theorem can be written like this:

$\sum_{n=0}^{N-1} |x[n]|^2 = \frac{1}{N}\sum_{n=0}^{N-1} |X[k]|^2 $

Now, say you've got some function in time, like $x = \sin(\omega t)$. Depending on how large your value of $N$, the LHS could be arbitrarily large.

The FT of this function is a delta function at $\omega$. Everywhere else is zero, so your RHS is given by $1/N$. So how on earth does one side equal the other?! It seems to me like you've got a summation one one side and an average value on the other...

I must be missing something obvious!

答:
The discrete fourier transform is usually defined as $$
  X_k = \sum_{n=0}^{N-1} x_ne^{-i2\pi \frac{n}{N}k} \text{,}
$$
i.e. without a normalizing factor. Now assume a signal is $T$-periodic, and do the transform for $N=T,2T,3T,\ldots$. You get the sequence $X_k,2X_k,3X_k,\ldots$ for the individual $X_k$. In Parseval's theorem, those are squared though, so on the RHS without the $\frac{1}{N}$ you'd get $R,4R,8R,\ldots$. On the LHS, however, you get $L,2L,3L,\ldots$ since there's no square-of-a-sum involved there, you just square the individual coefficients. For the equality to hold, you thus have to account for this effect, which is where the $\frac{1}{N}$ comes from.

Your argument doesn't hold because by saying "The FT is a delta function" you're glossing over the fact that this $\delta$ is scaled linearly as $N$ increases. Squaring makes that quadratic, and multiplying with $\frac{1}{N}$ makes it linear again. Just as the RHS ist.


Title: $0^2{{n}\choose{0}} + 1^2{{n}\choose{1}} + 2^2 {{n}\choose{2}} + ... +n^2{{n}\choose{n}} = n (1+n)* 2^{n-2}, n \in \mathbb{N} \setminus \{0\}$ proof.

Body:
I need to pove that:

$$0^2{{n}\choose{0}} + 1^2{{n}\choose{1}} + 2^2 {{n}\choose{2}} + ... +n^2{{n}\choose{n}} = n (1+n)* 2^{n-2}, n \in \mathbb{N} \setminus \{0\}$$

I think that the left hand side looks like binominal theorem. The problem is that in that case the sequence:
$$0^2+1^2+2^2+...n^2$$
would be therefore equal to the multiplication of growing powers on one element and decreasing powers of other element. That is impossible since powers are constant and equal.

And I can't do my induction proof without any transformation of that expression since adding $1$ to $n$ would make that structure in its current form irregular - I wouldn't be able to extract the seqence that is my thesis.

Therefore I don't know how to transform that expression so that it can be proven by induction (as I assume).

Body:
Let $n \in \mathbb{N}^*$, and $S_n = \displaystyle \sum_{k=0}^{n} k^2 \binom{n}{k} = \sum_{k=0}^{n} k(k-1) \binom{n}{k} +\sum_{k=0}^{n} k \binom{n}{k} $

You can rapidly check that $\displaystyle k \binom{n}{k} = n \binom{n-1}{k-1}$ and $ \displaystyle k (k-1) \binom{n}{k} = n(n-1) \binom{n-2}{k-2} $.

So $\displaystyle \sum_{k=0}^n k \binom{n}{k} = n \sum_{k=1}^n\binom{n-1}{k-1} = n \sum_{k=0}^{n-1} \binom{n-1}{k} = n (1+1)^{n-1} = n 2^{n-1} $

Also $ \displaystyle \sum_{k=0}^n k(k-1) \binom{n}{k} = n(n-1) \sum_{k=2}^n\binom{n-2}{k-2} = n(n-1) \sum_{k=0}^{n-2} \binom{n-2}{k} = n(n-1) 2^{n-2}$.

Thus $S_n = n2^{n-1} + n(n-1)2^{n-2} = 2n 2^{n-2}+n(n-1)2^{n-2} = n(n+1)2^{n-2} $


问:
Is the $L^1$-norm of the FT of $(x+i)^n/(x-i)^{n+2}$ bounded as a sequence in $n$?
Let the function $f_n\in L^1(\mathbb R)\cap C_0(\mathbb R)$ be defined for $n\in\mathbb N$ by
$$f_n(x):=\left(\frac{x+i}{x-i}\right)^n\frac{1}{(x-i)^2}\,.$$
Then its Fourier transform is always of the form
$$\hat{f_n}(x)=q_n(x)e^{-x}\theta(x)\,,$$
for a polynomial $q_n$ of degree at most $n+1$ and $\theta$ the Heaviside step function. As Jason pointed out in the comments, $q_n(x)=-\sqrt{2\pi}\int_0^x L_n(2t)dt$, with $L_n$ the $n$th Laguerre polynomial.

Clearly, $\|\hat{f_n}\|_1<\infty$. However, is $\|\hat{f_n}\|_1$ bounded as a function in $n$?

Why I need this:

I'm studying regularity properties of functions on $\mathbb R$ obtained via the Cayley transform from functions on the circle. I found out that the problem of integrability of the Fourier transform of a large class of functions can be reduced this very concrete question.

What I've tried:

I tried to use the general estimate $\|\hat{f_n}\|_1\leq c(\|f_n\|_2+\|f_n'\|_2)$, but unfortunately $\|f_n'\|_2$ blows up as $n\to\infty$. I thought about using complex analysis, like e.g. contour integration, but I am not very well versed in that subject.
I also used Wolframalpha to determine the polynomials $q_n$, and subsequently determining $\|\hat{f_n}\|_1$ for small $n$. I got (up to a possible factor of $\sqrt{2\pi}$):

$\|\hat{f_0}\|_1=1$

$\|\hat{f_1}\|_1=1.20728$

$\|\hat{f_2}\|_1=1.37548$

$\|\hat{f_3}\|_1=1.52031$

$\|\hat{f_4}\|_1=1.64932$

before my Wolframalpha computation time was exceeded. It looks as though this sequence might stay below 2 (or some larger number) but I couldn't prove this.

答:
This may be coming late to the game at this point, but here's how we can work through the details to show that the $L^1$ norms of the $\hat{f_n}$ are unbounded. I won't get into a justification of the formula for the $q_n$ in terms of the Laguerre polynomials. And I want to note that I'm drawing heavily on Szegö's Orthogonal Polynomials, and will be mirroring the notation there quite a bit.

Step 1: Converting to generalized Laguerre polynomials

To begin, let's start with the formula for the Laguerre polynomails themselves: we have
$$
L_n(x) = \sum_{k = 0}^{n}
\binom{n}{k} \frac{(-1)^k}{k!} x^k.
$$
Using the relationship $q_n(x) = -\sqrt{2 \pi} \int_{0}^{x} L_n(2t) \, dt$,
but dropping the constant factor of $-\sqrt{2 \pi}$ for simplicity, we get
$$
q_n(x) = \sum_{k = 0}^{n}
\binom{n}{k} \frac{(-1)^k}{(k+1)!} 2^k x^{k+1}.
$$
With a little extra algebra, we can express the $q_n$ in terms of the generalized Laguerre polynomials $L_n^{(\alpha)}$ as follows:
$$
q_n(x) = \frac{x}{n+1} L_n^{(1)}(2x).
$$

Step 2: Asymptotics

Now we can directly use the Plancherel-Rotach asymptotics for the $L_n^{(\alpha)}$ as developed in Orthogonal Polynomials. These asymptotics give an approximation for $L_n^{(\alpha)}(x)$ for $x_0 \leq x \leq (4 - \eta)n$, where we can think of $x_0$ and $\eta$ as fixed, small positive numbers. To re-state these asymptotics, we let $l_n = (4n + 4)^{1/2}$ and also introduce auxiliary variables $\xi$ and $\phi$, related to $x$ and to each other by the relationship $x = \xi^2 = l_n^2 \cos^2 \phi$. Then
\begin{align*}
e^{-x/2} L_n^{(1)}(x)
&= e^{-\xi^2 / 2} L_n^{(1)}(\xi^2) \\
&= (-1)^n (\pi \sin \phi)^{-1/2} \xi^{-3/2} n^{1/4} \\
& \qquad \times \left\{
\sin \left(
(n + 1)
(\sin 2 \phi - 2 \phi)
+ \frac{3 \pi}{4}
\right)
+ O(\xi^{-1} l_n^{-1}).
\right\}
\end{align*}
For the $q_n$, we therefore obtain
\begin{align*}
e^{-x/2} q_n(x/2)
&= \frac{x}{2n + 2} e^{-x/2} L_n^{(1)}(x) \\
&= 2 l_n^{-2} (-1)^n (\pi \sin \phi)^{-1/2} \xi^{1/2} n^{1/4} \\
& \qquad \times \left\{
\sin \left(
(n + 1)
(\sin 2 \phi - 2 \phi)
+ \frac{3 \pi}{4}
\right)
+ O(\xi^{-1} l_n^{-1})
\right\} \\
&= g_n(x/2) + R_n(x/2),
\end{align*}
where again $x = \xi^2 = l_n^2 \cos^2 \phi$ and the range of $x$-values where this holds is $x_0 \leq x \leq (4 - \eta) n$.

Now we turn to estimating
$$
\int_{x_0 / 2}^{(2 - \eta/2) n} | \hat{f_n}(y) | \, dy
= \frac{1}{2} \int_{x_0}^{(4 - \eta) n} | e^{-x/2} q_n(x/2) | \, dx
$$
from below.

Step 3: The remainder term

We begin with the remainder term $R_n$. Using $C$ to denote a constant that may vary from line to line but which is independent of $n$, we have via the change of variable $t = \cos \phi = \xi / l_n$:
\begin{align*}
I_R
&= \int_{x_0}^{(4 - \eta)n} | R_n(x/2) | \, dx \\
&\leq C l_n^{-3} n^{1/4}
\int_{x_0}^{(4 - \eta)n} (\sin \phi)^{-1/2} \xi^{-1/2} \, dx \\
&\leq C l_n^{-3} n^{1/4}
\int_0^1 (1 - t^2)^{-1/4} (l_n t)^{-1/2} \cdot 2 l_n^2 t \, dt \\
&= C l_n^{-3/2} n^{1/4}
\int_0^1 (1 - t^2)^{-1/4} t^{1/2} \, dt \\
&= C l_n^{-3/2} n^{1/4}.
\end{align*}
So $I_R = O(n^{-1/2})$.

Step 4a: The main term - changes of variable

The main term with $g_n$ is more involved. As a first step, we carry out a change of variable to $\phi$ in the integral:
\begin{align*}
I_g
&= \int_{x_0}^{(4 - \eta)n} | g_n(x/2) | \, dx \\
&= C l_n^{-2} n^{1/4} \int_{x_0}^{(4 - \eta)n}
(\sin \phi)^{-1/2} \xi^{1/2}
\left|
\sin \left(
(n + 1)
(\sin 2 \phi - 2 \phi)
+ \frac{3 \pi}{4}
\right)
\right|
\, dx \\
&= C l_n^{1/2} n^{1/4} \int_{a_n}^{b_n}
(\sin \phi)^{1/2} (\cos \phi)^{3/2}
\left|
\sin \left(
(n + 1)
(\sin 2 \phi - 2 \phi)
+ \frac{3 \pi}{4}
\right)
\right|
\, d\phi,
\end{align*}
where $a_n = \cos^{-1} (\sqrt{(4 - \eta)n}/l_n)$ and $b_n = \cos^{-1} (\sqrt{x_0}/l_n)$. Note that as $n \to \infty$, we have $a_n \to \cos^{-1} (\sqrt{1 - \eta/4})$ and $b_n \to \pi / 2$.

We can additionally change variable via $s = \sin 2 \phi - 2 \phi$; this transformation doesn't permit an entirely closed-form expression for the new integrand, but we can still write
$$
I_g
= C l_n^{1/2} n^{1/4} \int_{c_n}^{d_n}
\tilde{g}(s)
\left|
\sin \left(
(n + 1) s
+ \frac{3 \pi}{4}
\right)
\right|
\, ds
$$
for an appropriate function $\tilde{g}$ and numbers $c_n$ and $d_n$, where $c_n \to -\pi$ and $d_n \to d = \sqrt{\eta (1 - \eta/4)} - 2 \cos^{-1} (\sqrt{1 - \eta/4}) < 0$.

Step 4b: The main term - Féjer's lemma

We can now apply Fejér's lemma to understand the limiting behavior of the remaining integral. This lemma states that if $h_1 \in L^1([0, \pi])$, and if $h_2 \in L^{\infty}(\mathbb{R})$ is $\pi$-periodic, then
$$
\lim_{n \to \infty} \int_{0}^{\pi} h_1(s) h_2(ns) \, ds
= \left(
\frac{1}{\pi} \int_{0}^{\pi} h_2(x) \, ds
\right)
\left(
\int_{0}^{\pi} h_1(s) \, ds
\right).
$$
A straightforward corollary is that if $h_1^{(n)} \to h_1$ in $L^1([0, \pi])$, then
$$
\lim_{n \to \infty} \int_{0}^{\pi} h_1^{(n)}(s) h_2(ns) \, ds
= \left(
\frac{1}{\pi} \int_{0}^{\pi} h_2(s) \, ds
\right)
\left(
\int_{0}^{\pi} h_1(s) \, ds
\right).
$$

In our problem, the role of $h_2$ is taken by $| \sin (\cdot + 3\pi / 4) |$, and
$$
h_1^{(n)}
= \chi_{[c_n, d_n]} \tilde{g}
\to \chi_{[-\pi, d]} \tilde{g}
= h_1.
$$
We can therefore conclude that
\begin{align*}
\lim_{n \to \infty}
&\int_{c_n}^{d_n}
\tilde{g}(s)
\left|
\sin \left(
(n + 1) s
+ \frac{3 \pi}{4}
\right)
\right|
\, ds \\
&= \frac{2}{\pi}
\int_{-\pi}^{d} \tilde{g}(s) \, ds \\
&= \frac{2}{\pi}
\int_{\cos^{-1} (\sqrt{1 - \eta/4})}^{\pi / 2}
(\sin \phi)^{1/2} (\cos \phi)^{3/2}
\, d\phi \\
&> 0.
\end{align*}

So concluding with the main term, we can say that $I_g \geq C n^{1/2}$ for sufficiently large $n$, where $C > 0$ does not depend on $n$.

The lower bound for $I_g$ and the upper bound for $I_R$ now imply that the norms $\| \hat{f_n} \|_{1}$ are unbounded.


问:
Double Expectation Problem: Finding E[N] of coin flipped until 3 heads appear in a row. N=number of flips
A coin having a probability p of coming up heads is successively flipped until three heads appear in a row. Let N denote the number of flips and find E[N]

This is what I did:

答:
If probability of getting a head in a flip is $p$, probability of getting a tail is $(1-p)$, not $0.5$ as in your working.

Expected number of flips to get first head -

$E({\small{H}}) = 1 + (1-p)E({\small{H}}) \implies E({\small{H}}) = \frac{1}{p}$

Expected number of flips to get $2$ heads in a row -

$E({\small{HH}}) = 1 + E({\small{H}}) + (1-p)E({\small{HH}}) \implies E({\small{HH}}) = \frac{1}{p} + \frac{1}{p^2}$

Expected number of flips to get $3$ heads in a row -

$E({\small{HHH}}) = 1 + E({\small{HH}}) + (1-p)E({\small{HHH}})$

$ \implies E({\small{HHH}}) = \frac{1}{p} + \frac{1}{p^2} + \frac{1}{p^3}$


问:
Chebyshev differentiation matrices and Galerkin method
I've been reading the "Spectral Methods in Matlab" by Lloyd N. Trefethen and I'm interested in solving PDE's with spectral methods. From what I understand this book gives us a method (at least until chapter 6 "Chebyshev Differentiation Matrices") that I take it to be as follows:

Say we're trying to solve $u_{xx}=f(x)$. Then, what we do is to let $u(x)$ be approximated by some polynomial given by $p(x)=\sum_j^ny_i \Pi_{k\neq j}\bigg(\frac{x-x_j}{x_j-x_k}\bigg)$. Now if we let the evaluation points $\{ x_i\}$ be the Chebyshev nodes we get rid of the Runge phenomena. Knowing this it is "easy" to calculate some matrix $D_N$ that will give us the action of the differential operator and we have then reduced our ptoblem to a linear system of equations, and the later interpolation of the calculated $u_i$ will give me a very good aproximation to the exact solution of $u$.

Now I sort of understand this. But what we're doing at the end of the day is calculating the values of $u_i$ at the points $\{x_i\}$ that guarantee a very good interpolation. And this is fine. But what does it have to do with the Galerkin method? What I understand from the Galerkin method is that we let $u(x)\approx\sum_n^N \tilde{u}_n\phi_n(x)$ for some family of $\phi_i$ such that they respect the boundary conditions of the problem at hand and those same functions are to be described by some orthogonal basis of polynomials up to degree $N$. Our task now is to invoke the minimization of the residual $R=\tilde u_{xx}-s$ and after some working out we are tasked of finding the coefficients $\tilde u_n$. This has a very apparent difference to the first method described where we were tasked to find the values of $u$ at specific points $x_i$ which were meticulously chosen to minimize the error in the interpolation. But since in class we were first introduced to the Galerkin method (and also Tau and Colocation) and afterwards the other method by Trefethen without any bridge to make any conection I'm struggling to discover what that connection is, or if there's any connection at all. Can anyone give a hand?

PS: This is the reference that was used to have us introduced to the Galerkin method https://arxiv.org/pdf/gr-qc/0609020.pdf

答:
The goal of all approximation methods is to find accurate weights $\tilde{u}$ to minimize a special norm or metric:

$u(x)\approx\sum_n^N \tilde{u}_n\phi_n(x)$.

The Galerkin method can be assigned to a class of error minimization methods which is optimal in the sense of an energy $L_2$ based norm. Popular methods:

Collocation methods or Interpolation

Minimization at specific points

Optimal in the sense of the Lebesgue constant $L_{\infty}$

Least Squares or Galerkin methods

Minimization considering an integral norm

Optimal in the sense $L_{2}$ or similar norm

So to speak: The Galerkin and Least Squares methods look at the error on a complete interval rather than at the error at specific singular point on this interval. You can think of it as minimizing the area under a curve. Note that the math behind these methods is not so important for a good understanding.

Regards


问:
Why $ \mathcal{F}_{0} \equiv\{A \times A': A \in \mathcal{A} \text { and } A' \in \mathcal{A}\} $ is not a $\sigma$-field?
I asked a related question yesterday:Extend $(A^{(1)}_1\times A^{(1)}_2)\cap(A^{(2)}_1\times A^{(2)}_2)=(A^{(1)}_1\cap A^{(2)}_1)\times(A^{(1)}_2\cap A^{(2)}_2)$ where I am confused and think

$$
\mathcal{F}_{0} \equiv\left\{A \times A^{\prime}: A \in \mathcal{A} \text { and } A^{\prime} \in \mathcal{A}^{\prime}\right\}
$$

should be a $\mathcal{F_0}$ is a $\sigma$-field,  if both $\mathcal{A}$ and $\mathcal{A'}$ are $\sigma-$field. My reasoning can be found in the link I attached but I was told this supposition is wrong because this class of set is not closed under countable unions.

I am very confused by intuition if both $\mathcal{A}$ and $\mathcal{A'}$ are $\sigma$-field, how can the collection of all product of sets end up not closed under union? I proved why the product is closed under intersection in the link provided which I hope is not wrong. But on the textbook I used, it didn't provide me with the formula for the union. Can someone provide some counter examples to show that $\mathcal{F_0}$ is not a $\sigma-$field?

答:
Consider $\mathcal{F}_{0} \equiv\left\{A \times A^{\prime}: A \in \mathcal{A} \text { and } A^{\prime} \in \mathcal{A}^{\prime}\right\}$.
The complement of an element of $\mathcal{F}_0$ need not be an element of $\mathcal{F}_0$.  The union of two elements of $\mathcal{F}_0$ need not be an element of $\mathcal{F}_0$.  Intersection is OK, however.

Surely you can work out a counterexample for the complement using the two-element set $\Omega = \{0,1\}$ with $A = A' = \{0\}$.


问:
Matrix Representation - Linear Operator
I have this question but I seem to be deadlocked in solving it:

Given any basis of the vector space $A = \mathbb{R}^{3\times 3}$, where the vector space $A$ is the space of all $3\times 3$ matrices.

If $B\in \mathbb{R}^{3\times 3}$ is a matrix and also defines a linear mapping:

$ S_B\colon A\rightarrow A$

$ ~~~~~~~\kappa \mapsto B\kappa B^* $

Choose a basis of the vector space A, and find a matrix representation of $S_B$ with respect to the chosen basis.

My approach:

I chose the usual standard basis $E = \{e_1, e_2, \dots e_9\}$ as:

$e_1 = \begin{pmatrix}
  1 & 0 & 0\\
  0 & 0 & 0\\
  0 & 0 & 0
\end{pmatrix}$

$e_2 = \begin{pmatrix}
  0 & 1 & 0\\
  0 & 0 & 0\\
  0 & 0 & 0
\end{pmatrix}$ , etc

Then proceeded to obtain the images of these bases:

$S_B(e_1) = \begin{pmatrix}
  \kappa & 0 & 0\\
  0 & 0 & 0\\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  1 & 0 & 0\\
  0 & 0 & 0\\
  0 & 0 & 0
\end{pmatrix} = \begin{pmatrix}
  \kappa & 0 & 0\\
  0 & 0 & 0\\
  0 & 0 & 0
\end{pmatrix}$

Then, $S_B(e_1) = \kappa e_1 + 0e_2 + 0e_3 + \dots + 0e_9 $

This column yields a $9\times 1$ and eventually all the bases would yield $9\times 9$ as a representative matrix. But the linear mapping is an operator since $S_B\colon A \rightarrow A$, which is $3\times 3$. Can anyone help unravel where I am missing it?

答:
An operator $\;T\;$ on a linear space $\; V\;$ of dimension $\;n\;$ is represented by an $\;n\times n\;$ matrix. In this case, $\;S_B\;$ is a linear operator on a $\;9\,-$ dimensional space, namely $\;\Bbb R^{3\times3}\;$ and it thus represented by an $\;9\times9\;$ matrix with respect to some given fixed basis of $\;\Bbb R^{3\times3}\;$.

I hope this addresses and solves your problem.

Title: Matrix Representation - Linear Operator


答: $\mathcal F_0$ isn't even closed under finite unions.  For example, let $\mathcal A$ and $\mathcal A'$ both be the Borel $\sigma$-algebra on $\mathbb{R}$.  Then $(0,1) \times (0,1)$ and $(1,2) \times (1,2)$ are both in $\mathcal F_0$, but $((0,1) \times (0,1)) \cup ((1,2) \times (1,2))$ is not a product of any two subsets of $\mathbb{R}$.

Title: Why $ \mathcal{F}_{0} \equiv\{A \times A': A \in \mathcal{A} \text { and } A' \in \mathcal{A}\} $ is not a $\sigma$-field?

Body:
I asked a related question yesterday:Extend $(A^{(1)}_1\times A^{(1)}_2)\cap(A^{(2)}_1\times A^{(2)}_2)=(A^{(1)}_1\cap A^{(2)}_1)\times(A^{(1)}_2\cap A^{(2)}_2)$ where I am confused and think

$$
\mathcal{F}_{0} \equiv\left\{A \times A^{\prime}: A \in \mathcal{A} \text { and } A^{\prime} \in \mathcal{A}^{\prime}\right\}
$$

should be a $\mathcal{F_0}$ is a $\sigma$-field,  if both $\mathcal{A}$ and $\mathcal{A'}$ are $\sigma-$field. My reasoning can be found in the link I attached but I was told this supposition is wrong because this class of set is not closed under countable unions.

I am very confused by intuition if both $\mathcal{A}$ and $\mathcal{A'}$ are $\sigma$-field, how can the collection of all product of sets end up not closed under union? I proved why the product is closed under intersection in the link provided which I hope is not wrong. But on the textbook I used, it didn't provide me with the formula for the union. Can someone provide some counter examples to show that $\mathcal{F_0}$ is not a $\sigma-$field?


问:
Measures on all subsets of $\aleph_0$
A theorem of Ulam says:

A finite measure $\mu$ defined on all subsets of a set of cardinality $\aleph_1$ must be $0$ for all subsets if it sends every $1$-element subset to $0$.

Will this statement hold if we substitute $\aleph_0$ for $\aleph_1$ and require the measure to be finitely additive only?  In particular, could we define a nontrivial finite measure on the set of all finite graphs that assigns each finite set a measure $0$?

答:
The finite additive measures are somehow easy. There exists a non-trivial finite additive measure on any infinite set, which is $0$ on any finite subset. What makes all the problems with measures is countable additivity.

The key to the solution is the word `ultrafilter'.

A filter on set $\aleph_0$ is a system of sets $\cal F$ (system of some `big' sets, in our case that will be sets of non-zero measure) which satisfies

if $A\in \cal F$, and $B\supseteq A$ then $B\in \cal F$,

if $A,B\in \cal F$ then $A\cap B\in \cal F$,

$\aleph_0\in \cal F$, and $\emptyset \neq \cal F$.

An example of such system is for example the system of all infinite subsets of $\aleph_0$.

A filter is an ultrafilter if it is maximal in respect to inclusion, i.e. we have to add one condition: `for every $A$, either $A \in \cal F$, or $\aleph_0 \setminus A \in \cal F$'. By axiom of choice (Zorn's lemma) we know that every filter can be extended to an ultrafilter.

So fix an ultrafilter $\cal F$ which extends the system of all infinite subsets of $\aleph_0$. Then we can finally define the measure as:
$$
\mu(A) = \cases{
1 & if $A\in \cal F$,\cr
0 & if $A\notin \cal F$.\cr
}
$$

You can check that this is a finitely additive measure. Well, all the dificult steps have been done in the ultrafilter notation which can be very confusing. If you want to understand ultrafilters better, you can solve the following exercise:

Given a finitely additive measure $\mu$ defined on all subsets of $X$, such that $\mu(X) = 1$, show that the systems of sets $\{ A : \mu(A) = 1 \}$, and $\{ A : \mu(A) \neq 0\}$ are filters on $X$.

There are also many other filters on $X$ that can be defined by the measure $\mu$, unless $\mu$ gives just values $0$, or $1$, in that case there is only one such filter, which is also an ultrafilter.

Title: Measures on all subsets of $\aleph_0$


问:
Center of mass of $4z^2=x^2+y^2$ within $x=0, y=0, z=0, z=3$
I have a solid limited by the cone: $4z^2=x^2+y^2$ and the following planes: $x=0, y=0, z=0, z=3$

This solid has a density = $1$

I already found the volume using the following triple integral ($r$ referring to radius):

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{6}\int_{\sqrt{r^2/4}}^{3} r dz dr d\theta =9\pi$$

From that, what would be the best way to find the moment of inertia and finally the center of mass / barycenter?

Thanks in advance

答:
Note that the volume in the first quadrant is one-quarter that of the whole cone, i.e. $\frac14\cdot \frac13\cdot 3 \cdot (\pi \cdot6^2)= 9\pi$. Then, the coordinates of its barycenter are

$$z_c =\frac1{9\pi}\int_0^3 \int_0^{2z} \int_0^{\frac\pi2}z \>r dr d\theta dz=\frac94
$$

$$x_c=y_c =\frac1{9\pi}\int_0^3 \int_0^{2z} \int_0^{\frac\pi2}
r\cos\theta \> rdr d\theta dz=\frac6\pi
$$


答:
We can model this problem using an absorbing Markov chain. The state space is $S=\{0,1,2,3\}$, where a state $i\in S$ corresponds to the number of consecutive heads flipped since the last tails (once $3$ consecutive heads are flipped, the process remains in that state). The initial distribution is $\mathbb P(X_0=0)=1$, and the transition probabilities are given by
$$
P=\begin{pmatrix}
1-p & p & 0 & 0\\
1-p & 0 & p & 0\\
1-p & 0& 0 & p\\
0 & 0 & 0 & 1
\end{pmatrix}.
$$
(That is, the $(i,j)$ entry of $P$ denotes $\mathbb P(X_{n+1}=j\mid X_n=i)$.)

Note that we can write $P$ in block form:
$$
P = \begin{pmatrix}
T & \mathbf T^0\\
\mathbf 0 & 1,
\end{pmatrix}
$$
where $T$ is the substochastic matrix consisting of transitions between transient states and $\mathbf T^0 + T\mathbf 1=\mathbf 1$. Let $p_N(k) := \mathbb P(N=k)$ for nonnegative integers $k$. $N$ has a discrete phase-type distribution, and it is well-known that
$$
p_N(k) = \tau T^{k-1}\mathbf T^0,\quad k=0,1,2,\ldots
$$
There isn't a nice closed form for powers of $T$, but we can still find the expectation. Consider the fundamental matrix
$$
R := \sum_{k=0}^\infty T^k.
$$
Since $T$ is a strictly substochastic matrix (i.e. there is a row which sums to less than one), this series converges, and in fact $R = (\mathbf I - T)^{-1}$, where $\mathbf I$ denotes the identity matrix. The expected number of stepts before being absorbed when starting in transient state $i$ is the $i^{\mathrm{th}}$ entry of $R\mathbf 1$. Since
$$
R =  (\mathbf I - T)^{-1} = \left(
\begin{array}{ccc}
 p & -p & 0 \\
 p-1 & 1 & -p \\
 p-1 & 0 & 1 \\
\end{array}
\right)^{-1} = \left(
\begin{array}{ccc}
 \frac{1}{p^3} & \frac{1}{p^2} & \frac{1}{p} \\
 \frac{1-p^2}{p^3} & \frac{1}{p^2} & \frac{1}{p} \\
 \frac{1-p}{p^3} & \frac{p-p^2}{p^3} & \frac{1}{p} \\
\end{array}
\right),
$$
it follows that
$$
\mathbb E[N] = p^{-3}(1+p+p^2).
$$

Title: Double Expectation Problem: Finding E[N] of coin flipped until 3 heads appear in a row. N=number of flips

Body:
A coin having a probability p of coming up heads is successively flipped until three heads appear in a row. Let N denote the number of flips and find E[N]

This is what I did:


问:
When is there an adjunction between two categories?
Call two categories $C$ and $D$ adjivalent if there is an adjunction
$$F\dashv U,\quad F\colon C\to D,\quad U\colon D\to C$$
between $C$ and $D$.

Can one say something about when two categories are adjivalent? In particular, are there pairs of categories which aren't adjivalent?

答: It's very common that categories should admit an adjunction between them. In particular if $C$ has a terminal and $D$ an initial object, then take $F$ constant at the initial and $G$ constant at the terminal. On the other hand, any adjunction between $C$ and $D$ induces a homotopy equivalence between their nerves, which means that "adjivalent" categories are weakly equivalent in the Thomason model structure on categories, though the converse need not hold.


问:
Neumann Stability Analysis
$$
u_{j+1,m} = \frac{1}{2}\sigma(\sigma-1)u_{j,m+1}-(\sigma^1-1)u_{j,m}+\frac{1}{2}\sigma(\sigma+1)u_{j,m-1}\\
\sigma = \frac{c\Delta t}{\Delta x}\\
u_t+cu_x = 0
$$
I need find the Neumann Stability of the above scheme. My work is below:

$$
u_{j+1,m} = \lambda e^{ikx_m} \hspace{5mm} u_{j,m+1} = e^{ik(x_m+\Delta x)} \hspace{5mm} u_{j,m} = e^{ikx_m} \hspace{5mm} u_{j,m-1} = e^{ik(x_m-\Delta x)}\\
\lambda e^{ikx_m} = \frac{1}{2}\sigma(\sigma-1)e^{ik(x_m+\Delta x)}-(\sigma^1-1)e^{ikx_m}+\frac{1}{2}\sigma(\sigma+1)e^{ik(x_m-\Delta x)}\\
\lambda = \sigma^2\cos(k\Delta x)-i\sigma\sin(k\Delta x)+1-\sigma^2\\
\left|\lambda\right| \leq 1 \implies \left|\lambda\right|^2 \leq 1\\
\left|\lambda\right|^2 = (1-\sigma^2(1-\cos(k\Delta x)))^2+\sigma^2\sin^2(k\Delta x)
$$

I know the above work is correct but I have been struggling to get past this point to find the correct solution.

$$
\left|\lambda\right|^2 = 2\sigma^2\cos(k\Delta x) -2\sigma^4\cos(k\Delta x)+\sigma^4\cos^2(k\Delta x) -2\sigma^2+\sigma^4+1+\sigma^2\sin(k\Delta x)\leq 1
$$
I know
$$
-1 \leq \cos(k\Delta x) \leq 1 \hspace{5mm} \text{and} \hspace{5mm} 0 \leq \cos^2(k\Delta x) \leq 1
$$

Applying these conditions to the equation for $\left|\lambda\right|^2$ that I found gives me the final solution of a constant. This would show unconditional stability, but the correct answer should be conditional stability. The correct answer should also be the same as the CFL condition. Can someone please help show me where I keep going wrong?

答:
So let's start with
$$
\rho \equiv |\lambda|^2 = (1 - \sigma^2 + \sigma^2 \cos \alpha)^2 + \sigma^2 \sin^2 \alpha.
$$
I've denoted $\alpha = k \Delta x$. First, let's introduce $\mu = \cos \alpha$ and get rid of $\sin \alpha$:
$$
\rho = (1 - \sigma^2 + \sigma^2 \mu)^2 + \sigma^2 (1 - \mu^2).
$$
When $\alpha$ runs for all possible values, $\mu$ runs only from $-1$ to $1$. So basically we need to find for which $\sigma$ the following inequality holds
$$
\max_{\mu \in [-1, 1]} \rho \leq 1. \tag{*}
$$
Observe that $\rho$ is a quadratic polynomial in $\mu$:
$$
\rho(\mu) = (\sigma^2 - 1) \sigma^2 \mu^2 + 2(1-\sigma^2) \sigma^2 \mu +
(1 - \sigma^2)^2 + \sigma^2.
$$
so we could find the maximum pretty easy. The maximum could be attained only at $\mu = \pm 1$ or at the root of $\frac{d\rho}{d\mu} = 0$ if it lies in $\mu \in [-1, 1]$.
$$
\rho(\mu = 1) = 1, \qquad \rho(\mu = -1) = (1 - 2\sigma^2)^2.
$$
It is apparent now that $|\sigma| > 1$ violates (*), since $1 < \rho(\mu = -1) \leq \max_{\mu \in [-1, 1]} \rho(\mu)$. So the scheme is unstable when $|\sigma| > 1$.

The case $\sigma = \pm 1$ is also trivial: $\rho(\mu) = 1$. The scheme is stable.

Now let's consider $|\sigma| < 1$. Since $(\sigma^2 - 1) \sigma^2 < 0$ the $\rho(\mu)$ is a concave function. It's extremum is a maximum. It is located at
the root of $\rho'(\mu) = 0$
$$
2(\sigma^2 - 1)\sigma^2 \mu + 2 (1 - \sigma^2) \sigma^2 = 0.
$$
The maximum always is at $\mu = 1$, so
$$
\max_{\mu \in [-1, 1]} \rho(\mu) = \rho(\mu = 1) = 1.
$$
Summarizing, the scheme is stable for $|\sigma| \leq 1$.

Here's some plots of $\rho(\mu)$ for different $\sigma$:

Title: Neumann Stability Analysis


问:
Given the sequence $a_1=1$,$a_{n+1}=1+\frac{n}{a_n}$, does the sequence $a_n+n-a_n^2$ converges?

Given the sequence $a_1=1$,$a_{n+1}=1+\frac{n}{a_n}$, does the sequence $a_n+n-a_n^2$ converges? If it converges, find the limit.

Now, I proved
$\frac{1+\sqrt{1+4(n-1)}}{2}\leq a_n\leq \frac{1+\sqrt{1+4n}}{2}$,
and
$0\leq p_n=a_n+n-a_n^2\leq 1,p_{n+1}+\frac{n}{a_n^2}p_n=1$,which implies $\lim\limits_{n\rightarrow+\infty}p_{n+1}=\lim\limits_{n\rightarrow\infty}p_{n-1}$,
but I cannot prove $\lim\limits_{n\rightarrow+\infty}p_{n+1}=\lim\limits_{n\rightarrow\infty}p_{n}$ which means the squence $a_n+n-a_n^2$ limit equals to ${1\over2}$

so the question is: how can I prove $\lim\limits_{n\rightarrow+\infty}p_{n+1}=\lim\limits_{n\rightarrow\infty}p_{n}$ ?

答:
Link with the number of involutions in the symmetric goup $S_n$.
Let $I_0=1$. Let, for $n\geqslant 1$, $I_n$ be the number of involutions in the symmetric group $S_n$ (i.e $\ \sigma \in S_n \ | \ \sigma^2=I$). We can show that $\ I_1=1 \ $ , and $\ \forall n \in \mathbb N^{\star} \ , \ I_{n+1}=I_n+nI_{n-1}$. And, by induction, $\ \forall n\in \mathbb N \ , \ a_n=\frac{I_{n+1}}{I_n}\  $ .

L Moser and M Wyman, On solutions of x^d=1 in symmetric groups, Canad. J. Math., 7 (1955)  This paper can be found here . In this article, it is shown that exists $(\beta_n)_{n\in\mathbb N}$ such that $\ \forall p \in \mathbb N \ , \ a_n =\sqrt{n} +\displaystyle \sum_{k=0}^p \dfrac{\beta_k}{n^{k/2}}+{\rm o}\left(\dfrac{1}{n^{p/2}}\right)\ $. For example $\ a_n=\sqrt{n}+\dfrac{1}{2}-\dfrac{1}{8\sqrt{n}} +{\rm o}\left(\dfrac{1}{\sqrt{n}}\right)$. With that  result, it's easy to show that $\ \lim (a_n+n-a_n^2)=\dfrac{1}{2}$ .

If $\ b_0 \in \mathbb R_+^{\star}\ $ and $\ \forall n \in \mathbb N \ , \ b_{n+1}=1+\frac{n}{b_n} \ $, then, $(b_n)_{n\in\mathbb N}$ has the same asymptotic expansion: $\ \forall p \in \mathbb N \ , \ a_n =\sqrt{n} +\displaystyle \sum_{k=0}^p \dfrac{\beta_k}{n^{k/2}}+{\rm o}\left(\dfrac{1}{n^{p/2}}\right)\ $. (the same coefficients $\beta_k$ )


问:
Prove that $z^{\alpha} = \sum_{k=0}^{\infty} {\alpha \choose k} z_{0} ^{ \alpha - k}(z -z_{0})^k$
$$z^{\alpha} = \sum_{k=0}^{\infty} {\alpha \choose k} z_{0}  ^{ \alpha - k}(z -z_{0})^k$$

for $z,z_{0} > 0$ with $ |z - z_{0}| < |z_{0}|$

I assume that I should use differentiation

$\alpha \cdot z^{\alpha-1} = $ but I'm not sure that this is the right for to the solution and how does the RHS look after differentiating it. I'm thankful for every hint.

答:
It is sufficient to prove that $$z^\alpha=\sum_{k\ge0}{\alpha \choose k}(z-1)^k.\tag1$$
This is because
$$\begin{align}
z^\alpha&=z_0^\alpha\left(\frac{z}{z_0}\right)^\alpha \\
&=z_0^\alpha\sum_{k\ge0}{\alpha\choose k}\left(\frac{z}{z_0}-1\right)^k\\
&=\sum_{k\ge0}{\alpha\choose k}z_0^{\alpha-k}\left(z-z_0\right)^k.
\end{align}$$
Here is a proof for $(1)$. Let $$f(z)=(z+1)^\alpha.$$
We will write
$$f(z)=\sum_{k\ge0}\frac{f^{(k)}(0)}{k!}z^k.$$
Let $D$ be differentiation w.r.t. $z$, $D:\phi(z)\mapsto\phi'(z)$.

Then we have
$$\begin{align}
D^0f&=(z+1)^\alpha\\
D^1f&=\alpha (z+1)^{\alpha-1}\\
D^2f&=\alpha(\alpha-1)(z+1)^{\alpha-2}\\
D^3f&=\alpha(\alpha-1)(\alpha-2)(z+1)^{\alpha-3}\\
&...\\
D^kf&=p_k(\alpha)(z+1)^{\alpha-k},
\end{align}$$
where $$p_k(\alpha)=\prod_{j=0}^{k-1}(\alpha-j),$$
and $p_0(\alpha)=1$. Thus $f^{(k)}(0)=D^kf|_{z=0}=p_k(\alpha)$ and
$$f(z)=\sum_{k\ge0}\frac{p_k(\alpha)}{k!}z^k.$$
Recall the definition $${n\choose k}=\prod_{i=1}^{k}\frac{n-i+1}{i}=\frac{\prod_{i=1}^{k}(n-i+1)}{\prod_{i=1}^{k}i}=\frac{\prod_{i=0}^{k-1}(n-i)}{k!}=\frac{p_k(n)}{k!}.$$
Thus
$$\frac{p_k(\alpha)}{k!}={\alpha\choose k},$$
and $$(1+z)^\alpha=\sum_{k\ge0}{\alpha\choose k}z^k.$$
Replace $z$ with $z-1$, and get
$$z^\alpha=\sum_{k\ge0}{\alpha\choose k}(z-1)^k.$$
We are done.


问:
Applying Chebychev's Inequality
I am trying to figure out how to apply Chebychev's inequality on this example: Suppose $\bar{X}$ is the mean of $n=100$ observations from a population with mean $\mu$ and variance $\sigma^2 = 9$. What are the limits in which $\bar{X} - \mu$ will lie with probability of at least $0.90$.

I know Chebyshev's inequality is
$$P[g(Y) \ge r] \le \frac{\mathrm{E}[g(Y)]}{r}$$

I think $r$ can be $0.90$ in this case, so I would have
$$P[(\bar{X} - \mu) \ge 0.90]$$

I am not sure if I am on the right track, and I am unsure how to apply this inequality. What should I do from here?

答:
Tchebychev's inequality is
$$ \mathbb{P}(|Y-\mathbb{E}[Y]|\geqslant\varepsilon)\leqslant\frac{\mathbb{V}[Y]}{\varepsilon^2} $$
Since $\mathbb{E}[\overline{X}]=\mu$, we have
$$ \mathbb{P}(|\overline{X}-\mu|\geqslant\varepsilon)\leqslant\frac{\sigma^2}{n\varepsilon^2} $$
But $\frac{\sigma^2}{n\varepsilon^2}\leqslant 0.10\iff\varepsilon\geqslant\sqrt{\frac{0.10}{n}}\sigma$. With such a $\varepsilon$, you have
$$ \mathbb{P}(|\overline{X}-\mu|<\varepsilon)\geqslant 1-\frac{\sigma^2}{n\varepsilon^2}\geqslant 0.90 $$

Title: Applying Chebychev's Inequality


答:
$\bar{X}$ has mean $\mu$ and variance $\sigma^2/n$. Chebychev's inequality implies
$$P(|\bar{X}-\mu| \ge t) \le \frac{\sigma^2/n}{t^2}.$$
Set the right-hand side to $0.1$ and solve for $t$ to obtain the interval $[\mu - t, \mu + t]$.

Title: Applying Chebychev's Inequality

Body:
I am trying to figure out how to apply Chebychev's inequality on this example: Suppose $\bar{X}$ is the mean of $n=100$ observations from a population with mean $\mu$ and variance $\sigma^2 = 9$. What are the limits in which $\bar{X} - \mu$ will lie with probability of at least $0.90$.

I know Chebyshev's inequality is
$$P[g(Y) \ge r] \le \frac{\mathrm{E}[g(Y)]}{r}$$

I think $r$ can be $0.90$ in this case, so I would have
$$P[(\bar{X} - \mu) \ge 0.90]$$

I am not sure if I am on the right track, and I am unsure how to apply this inequality. What should I do from here?


问:
Proof that $x^n=y$ with accumulation point
given a sequence $(y_n)$ with $y_n \ge 0$ for all integers $n$ and $(x_n)=\sqrt[N]{y_n} $ with a constant value $N$ out of the natural numbers.

I want prove that $x^N=y$.

I know that $(y_n)$ is limited because $(y_n)$ converges and thus it follows by definition. From that I already concluded that $(x_n)$ is limited too and using Bolzano-Weierstraß Theorem $(x_n)$ has at least one accumulation point which is called $x$.
I thought that I can solve this problem by showing that $(x_n)$ converges to $\sqrt[N]{y}$ and from that one can conclude that $x^N=y$.
But I am stuck at this point. If it would be just the square root of $y$ it would be easy to show by definition of convergent sequences but here we have the Nth root for a given N.

Can someone give me a hint? :)

答:
You can go with $N$th root the same way as with square roots:

$$x - x_n = \sqrt[N]{y} - \sqrt[N]{y_n} = \frac{y - y_n}{\sqrt[N]{y}^{N - 1} + \sqrt[N]{y}^{N - 2}\sqrt[N]{y_n} + \ldots + \sqrt[N]{y_n}^{N - 1}}$$

Numerator converges to $0$ and, is $y \neq 0$, denominator is separated from $0$ (as it's at least $\sqrt[N]{y}^{N - 1}$), so fraction converges to $0$.

If $y = 0$, we should use definition of limit: choose $\epsilon$, find $n_0$ s.t. $y_n < \epsilon^N$ if $n > n_0$, and take the same $n_0$ for $x_n$: if $n > n_0$ then $x_n = \sqrt[N]{y_n} < \sqrt[N]{\epsilon^N} = \epsilon$.

Alternatively, it's not hard to show that $x \to x^N$ is continuous and bijective on any segment that doesn't include $0$, so inverse of it, $x \to \sqrt[N]{x}$ is continuous too (and we still need to separately show continuity in $0$ for even $N$).


答:
We obtain for $z_0\ne 0$ according to the binomial series expansion:
\begin{align*}
\color{blue}{z^{\alpha}}&=\left(z_0+(z-z_0)\right)^\alpha\\
&=z_0^{\alpha}\left(1+\frac{z-z_0}{z_0}\right)^{\alpha}\\
&=z_0^{\alpha}\sum_{k=0}^{\infty}\binom{\alpha}{k}\left(\frac{z-z_0}{z_0}\right)^k\\
&\,\,\color{blue}{=\sum_{k=0}^{\infty}\binom{\alpha}{k}\left(z-z_0\right)^{k}z_0^{\alpha-k}\qquad\qquad\left|\frac{z-z_0}{z_0}\right|<1}
\end{align*}

Title: Prove that $z^{\alpha} = \sum_{k=0}^{\infty} {\alpha \choose k} z_{0} ^{ \alpha - k}(z -z_{0})^k$

Body:
$$z^{\alpha} = \sum_{k=0}^{\infty} {\alpha \choose k} z_{0}  ^{ \alpha - k}(z -z_{0})^k$$

for $z,z_{0} > 0$ with $ |z - z_{0}| < |z_{0}|$

I assume that I should use differentiation

$\alpha \cdot z^{\alpha-1} = $ but I'm not sure that this is the right for to the solution and how does the RHS look after differentiating it. I'm thankful for every hint.


问:
Jacobian linearization: There is not an equilibrium point set
To apply Jacobian linearization procedure, we should find equilibrium points of the system. But, in some nonlinear systems, it is impossible to find equilibrium points set which make system's differential equations equal to zero. So, we have to handle the problem in another way.

I have this kind of nonlinear system and cannot find equilibrium points. What do you suggest me to linearize this nonlinear system with Jacobian matrices?

I tried to solve this problem with optimization methods in MATLAB via fmincon, but I think it is not a clear way.

My reference source is: http://www.sosmath.com/diffeq/system/nonlinear/linearization/linearization.html

Thanks a lot!

答:
For linearization, it is not necessarily that the linearization point is an equilibrium. For $x$ being the state and $u$ being the input signal, you have $$\dot{x} = f(x,u) = f(x_0,u_0)+A(x-x_0)+B(u-u_0)+H.O.T.$$ Here $A$ and $B$ are the linearization (Jacobian) matrices, $$A = \frac{\partial}{\partial x}f(x,u)|_{x=x_0,\, u=u_0}$$ and $$B = \frac{\partial}{\partial u}f(x,u)|_{x=x_0,\, u=u_0}.$$

If $x_0$, $u_0$ are an equilibrium, then you have $f(x_0,u_0) = 0$; otherwise, the constant term will remain.

For $x_0=0$ and $u_0=0$ we have the standard linearization: $\dot{x} \approx Ax + Bu$. However, if $x_0$, $u_0$ is not the origin, then you should work with the deviations $\tilde{x}=x-x_0$ and $\tilde{u}=u-u_0$.

Finally, you obtain $$\dot{\tilde x} = A \tilde x + B \tilde u + C,$$ where $C = f(x_0,u_0)$ is the constant term that you must take into account for control design.


问:
Suppose $\mu(E\cap I)\geq C\mu(I)$ for every interval $I\subseteq [0,1]$. Show that $\mu(E)=1$.
Suppose $E\subseteq [0,1].$

$C$ is some positive number and $\mu$ is Lebesgue measure.

My approach:

Assume $\mu(E)<\frac{q}{p}<1$.

We know $E$ can be approximated by finite summation of intervals. Intervals can be approximated by summation of intervals like $[\frac{i}{2^N},\frac{i+1}{2^N}]$. Therefore, $E$ can be approximated by finite summation of intervals like $[\frac{i}{2^N},\frac{i+1}{2^N}]$.

Choose a $N$ such that there is a $\delta$-approximation.

Suppose there are $n$ intervals used in the approximation. Since $n\cdot 2^{-N}\leq \frac{q}{p}+\delta$, we have $2^N-n\geq \frac{p-q-p\delta}{p}2^N$, which means there are at least about $\frac{p-q-p\delta}{p}2^N$ intervals remained blank.

Then, since it's a $\delta$-approximation, there is some interval on which $\mu(E\cap I)\leq \frac{\delta p}{p-q-p\delta}2^{-N}$.

Then, we have $\mu(E\cap I)/\mu(I)\leq \frac{\delta p}{p-q-p\delta}\to 0$ as $\delta\to 0$, which is a contradiction.

I think this is a correct proof, but is too ugly. Is there an elegant one?

答:
The premises imply that $E$ is measurable because $\mu (E)=\mu(E\cap I)$ exists when $I=[0,1].$

Let $E'=E\setminus \{0,1\}.$ I found it easier to show that $\mu(E')=1.$ Note that $\mu(E'\cap J)\ge C\mu(J)$ for every interval $J\subseteq (0,1).$

By contradiction, suppose $\mu(E')=1-r<1.$ Lebesgue measure is inner-regular: $\mu(E')=\sup \{\mu (B): B=\overline B\subseteq E'\}.$ So for any $d>0$ there exists $B_d=\overline B_d\subseteq E'$ with $$\mu(B_d)>\mu(E')-d=1-r-d.$$

Let $F_d= (0,1)\setminus  B_d.$ Then $F_d\supseteq (0,1)\setminus E'$ so $$\mu(F_d)\ge 1-\mu(E')=r.$$

Now $F_d$ is open in $\Bbb R$ and is a subset of $(0,1)$ so $F_d=\cup G_d$ where $G_d $ is a countable pairwise-disjoint family of open intervals, with $J\subseteq (0,1)$ for every $J\in G_d.$ Hence $$\mu (E'\cap  F_d)=\sum_{J\in G_d}\mu (E'\cap J)\ge$$ $$\ge\sum_{J\in G_d}C\mu(J)=$$ $$=C\sum_{J\in G_d}\mu(J)=C\mu(\cup G_d)=$$ $$=C\mu(F_d)\ge Cr.$$ Now since $B_d, F_d$ are disjoint and $B_d\cup F_d=(0,1),$ and since $B_d\subseteq E'$ we have $$1-r=\mu(E')= \mu (E'\cap B_d)+\mu (E'\cap F_d)=$$ $$=\mu(B_d)+\mu(E'\cap F_d)>$$ $$>(1-r-d)+Cr.$$  This holds for any $d>0$ so $$\forall d>0\,(\,1-r>(1-r-d)+Cr\,)$$ which implies $1-r\ge 1-r+Cr,$ which is absurd if $C$ and $r$ are positive.


问:
Converting DFA to Regex
I am trying to convert these finite automata to a regex. I tried using Arden's Theorem and state-elimination pattern. I cannot figure out how to do either as I keep getting stuck.
For Arden's Theorem, I am unable to solve the "system of equations"
and for state-elimination, I'm not sure how to remove each state and look at the pattern. This is the DFA:DFA</p>

答:
This one seems simple enough to do by inspection. The regular expression $(aa^*b+bb^*a+\epsilon)^*$ describes the inputs that leave the DFA in the initial state. The words accepted by the DFA are precisely those that take it at some point to $q_0$ and then end with either $aa^*$ or $bb^*$, so the language of the DFA is

$$(aa^*b+bb^*a+\epsilon)^*(aa^*+bb^*)\,.$$

Title: Converting DFA to Regex


问:
Let X be a random variable with density f(x) = e^−(x+2) for x > −2, and f(x)=0 otherwise. Find the moment generating function of X, E[X] and E[X^2].
As f(x)=e$^-$$^($$^x$$^+$$^2$$^)$ hence P(x)=-e$^-$$^($$^x$$^+$$^2$$^)$ hence shouldn't M(t) the integral from infinity to minus infinity of e$^t$$^x$.-e$^-$$^($$^x$$^+$$^2$$^)$, is this true? How do I find E[X] and E[X$^2$] from there?

答:
By definition, the moment-generating function of $X$ is
\begin{align}
M_X(t) &:= \mathbb E[e^{t X}]\\
&= \int_{\mathbb R} e^{t x}f_X(x)\ \mathsf dx\\
&= \int_{-2}^\infty e^{t x}e^{-(x+2)}\ \mathsf dx\\
&= e^{-2}\int_{-2}^\infty e^{-(1-t)x}\ \mathsf dx\\
&= \frac{e^{-2t}}{1-t},\quad t<1.
\end{align}
We can compute the $n^{\mathrm{th}}$ moment of $X$ by
$$
\mathbb E[X^n] = \lim_{t\to 0}\frac{\mathsf d^n}{\mathsf dt^n}M_X(t).
$$
In this case,
$$
\lim_{t\to 0}\frac{\mathsf d^n}{\mathsf dt^n}\left[\frac{e^{-2t}}{1-t}\right] = e^{-2}\cdot\Gamma(n+1, -2),
$$
where $\Gamma(a,z)$ denotes the incomplete gamma function:
$$
\Gamma(a,z) := \int_z^\infty s^{a-1}e^{-s}\ \mathsf ds.
$$
In particular for $n=1$ and $n=2$ this yields
\begin{align}
\mathbb E[X] &= -1\\
\mathbb E[X^2] &= 2.
\end{align}


问:
Hamiltonian for Geodesic Flow
I'm trying to prove that geodesic flow on the cotangent bundle $T^* M$ is generated by the Hamiltonian vector field $X_H$ where

$$H = \frac{1}{2}g^{ij}p_i p_j$$

but I am stuck. Could somebody show me how to complete the calculation, or where I've made a mistake? Cheers!

I know that vector field for geodesic flow is

$$X = p^i \partial/\partial x^i - \Gamma^i_{jk}p^j p^k \partial /\partial p^i$$

so I must verify that

$$X \ \lrcorner\  \omega = - dH$$

where $\omega = dp_i \wedge dx^i$. It's easy to check that

$$X \ \lrcorner\  \omega = - p^i dp_i - \Gamma^l_{jk}g_{li}p^j p^k dx^i$$

$$-dH = -g^{ij}p_j dp_i -\frac{1}{2}\partial_i g^{jk}p_j p_k dx^i$$

Here I am stuck. Using the explicit formula for the Christoffel symbols in terms of the metric doesn't seem to work! Have I done something wrong, or am I missing something?

N.B. I'm aware of solutions that do not use coordinates, but I'd like to understand this one which does!

答:
I've worked out my mistake - the vector field $X$ above is incorrect. The subtlety comes in raising and lowering the relevant indices.

Firstly we can all agree that the following vector field on the tangent bundle generates geodesic flow

$$Y = p^i \partial/\partial x^i - \Gamma^i_{jk} p^j p^k \partial/\partial p^i$$

where $(x^i, p^i)$ are coordinates on $TM$.

Now we want to use the tangent-cotangent isomorphism to determine a vector field on $T^*M$ which also gives geodesics after projection. Write out the ODEs for geodesic flow

$$\dot{x}^i = p^i $$
$$\dot{p}^i = -\Gamma^i_{jk}p^j p^k$$

Now define $p_i = g_{ij}p^j$ and calculate

\begin{align*}\dot{p}_i &= g_{ij,k}\dot{x}^k p^j + g_{ij}\dot{p}^j \\&= g_{ij,k}p^j p^k - g_{il}\Gamma^l_{jk}p^jp^k
\\ &= p^j p^k(g_{ij,k} - g_{ij,k} + \frac{1}{2}g_{jk,i}) \end{align*}

Thus the correct vector field for geodesic flow on $T^* M$ is in fact

\begin{align*}X &= g^{ij}p_j \frac{\partial}{\partial x^i}  + \frac{1}{2}\frac{\partial g_{jk}}{\partial x^i}  p^j p^k \frac{\partial }{\partial p_i}\\ &= g^{ij}p_j \frac{\partial}{\partial x^i}  - \frac{1}{2}\frac{\partial g^{jk}}{\partial x^i}  p_j p_k \frac{\partial }{\partial p_i} \end{align*}

by raising and lowering indices, and using the derivative of the inverse metric. Now this agrees with $-dH$ in the question, after contracting with $\omega$.

Title: Hamiltonian for Geodesic Flow


问:
Getting two very different answers for Maclaurin series for $xe^{x}$ - how is this possible?
I am currently trying to find the Maclaurin series for $xe^{x}$ in Calculus II. This is what I've done so far:

$$f(x) = xe^{x}$$
$$f'(x) = xe^{x} + e^{x}$$
$$f''(x) = xe^{x} + 2e^{x}$$
$$f'''(x) = xe^{x} + 3e^{x}$$
$$f^{[4]}(x) = xe^{x}+4e^{x}$$

and so on. Evaluating them at x = 0 gives me 0, 1, 2, 3, 4.... respectively.

Thus, I can say that:

$$xe^{x} = x + \frac{2x^{2}}{2!} + \frac{3x^{3}}{3!} + \frac{4x^{4}}{4!} ... \frac{nx^{n}}{n!}$$
And therefore:
$$xe^{x} = \sum_{n=0}^{\infty}\frac{nx^{n}}{n!}$$

However, there is more to it. I can rewrite the function:

$$xe^{x} = x + x^{2} + \frac{x^3}{2} + \frac{x^4}{6} ...$$

And then:
$$xe^{x} = \frac{x}{0!} + \frac{x^{2}}{1!} + \frac{x^{3}}{2!} + \frac{x^{4}}{3!} ... \frac{x^{n+1}}{n!}$$

Which means:
$$xe^{x} = \sum_{n=0}^{\infty}\frac{x^{n+1}}{n!}$$

How is this possible? How can $$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!} = \sum_{n=0}^{\infty}\frac{nx^{n}}{n!}$$?

Have I made an error somewhere? Or is this mathematically possible? Thank you.

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


问:
Find all functions for which $f(f(x)) = x$ for all $x \in D$
I just came across a nice coincidence with regular polygons. To check if a polygon with $n$ sides can completely occupy a surface, it must hold that the integer multiple of the interior angle $\alpha = \pi - 2\pi/n$ gives exactly $2\pi$. Say:
\begin{align*}
k \cdot (\pi - 2\pi/n) = 2\pi
\end{align*}
By rearranging you then find:
\begin{align*}
k = \frac{2n}{n-2}, \hspace{30pt} n = \frac{2k}{k-2}
\end{align*}
It follows that this only works with triangles, squares, and hexagons. However, I do not want to go into that now. I noticed that the function
\begin{align*}
f(x) =\frac{2x}{x-2}
\end{align*}
has the strange property $f(x) = y$, $f(y) = x$. Means: $f(f(x)) =x \ $ or $\ f^{-1} = f$. Are there any other functions of this type? Ignoring the trivial answer $f(x) = x$.

答:
For instance, take any function of the form
$$f(x)=\frac{ax+b}{cx+d},$$
where $a,b,c,d\in\mathbb{R}$. Now:
$$y=\frac{ax+b}{cx+d} \implies x=\frac{b-dy}{cy-a}$$
So, any function of that family with $a=-d$ satisfies what you are looking for.

EDIT:
In fact, note that $f$ is an invertible function such that $f(x)=f^{-1}(x)$, for all $x\in D$. You can consider another family, which also includes the examples by DiegoMath. For any invertible function $g$ (probably, with some restrictions to be well defined) you have:
$$y=g^{-1}(f(g(x))) \implies x=g^{-1}(f(g(y))).$$
The example by DiegoMath takes $g(x)=e^x$.


答:
Yes, these two sums are equal. Remove the first (zero) term from the RHS sum so that your counter $n$ starts from 1, not from 0.

And then let $n-1 = k$. This will give you that they are indeed equal.

Title: Getting two very different answers for Maclaurin series for $xe^{x}$ - how is this possible?

Body:
I am currently trying to find the Maclaurin series for $xe^{x}$ in Calculus II. This is what I've done so far:

$$f(x) = xe^{x}$$
$$f'(x) = xe^{x} + e^{x}$$
$$f''(x) = xe^{x} + 2e^{x}$$
$$f'''(x) = xe^{x} + 3e^{x}$$
$$f^{[4]}(x) = xe^{x}+4e^{x}$$

and so on. Evaluating them at x = 0 gives me 0, 1, 2, 3, 4.... respectively.

Thus, I can say that:

$$xe^{x} = x + \frac{2x^{2}}{2!} + \frac{3x^{3}}{3!} + \frac{4x^{4}}{4!} ... \frac{nx^{n}}{n!}$$
And therefore:
$$xe^{x} = \sum_{n=0}^{\infty}\frac{nx^{n}}{n!}$$

However, there is more to it. I can rewrite the function:

$$xe^{x} = x + x^{2} + \frac{x^3}{2} + \frac{x^4}{6} ...$$

And then:
$$xe^{x} = \frac{x}{0!} + \frac{x^{2}}{1!} + \frac{x^{3}}{2!} + \frac{x^{4}}{3!} ... \frac{x^{n+1}}{n!}$$

Which means:
$$xe^{x} = \sum_{n=0}^{\infty}\frac{x^{n+1}}{n!}$$

How is this possible? How can $$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!} = \sum_{n=0}^{\infty}\frac{nx^{n}}{n!}$$?

Have I made an error somewhere? Or is this mathematically possible? Thank you.


答:
Good question!

You did not make any mistakes.

We may see that
$$f(x)=\sum_{n\ge0}\frac{n}{n!}x^n=0+\sum_{n\ge1}\frac{n}{n!}x^n.$$
We then see that $n/n!=1/(n-1)!$ when $n\ge1$. Thus
$$f(x)=\sum_{n\ge1}\frac{x^n}{(n-1)!}=\sum_{n\ge0}\frac{x^{n+1}}{n!}=xe^x.$$


问:
Analytic functions vanishing (sub)exponentially at infinity
Let $f$ be an analytic function in the upper complex half-plane and continuous up to the real axis, and let $a>0$. Suppose that the function
\begin{equation}
\zeta\in\mathbb{C}^+\rightarrow f(\zeta)\mathrm{e}^{-ia\zeta}\in\mathbb{C}
\end{equation}
is itself bounded. Intuitively, since the absolute value of the exponential grows as $|z|\to\infty$, this requires $f$ to decay at least exponentially, with exponent larger than $a$, at $|z|\to\infty$; for example, any function like $f(\zeta)=\mathrm{e}^{ib\zeta}$, $b>a$ will do the trick, as well as any combination of such functions.

I wonder whether the class of analytic, bounded functions in the half-plane satisfying this condition is in fact larger and/or can be characterized somehow.

答:
Holomorphic functions with controlled growth usually appear in the theory of integral transforms of generalized function. Consider, for example, the class of holomorphic functions bounded on the right half plane by an exponential function i.e. such that
$$
\mathscr{LH}_a\triangleq\big\{ f\text{ is holomorphic for }\Re\zeta>-a \text{ and } |f(\zeta)|\le Ce^{-L|\zeta|},\; \Re \zeta>0\big\}.\label{1}\tag{1}
$$
for some $L>0$ (assuming nothing on the regularity of the function $f$ for $\Im \zeta=0$).

It can be proved that ([2] p. 400 and p. 403) an analytic function $f$ belongs to $\mathscr{LH}_a$ if and only if it is the Laplace transform of a Laplace hyperfunction: and the class \eqref{1} up to a counterclockwise rotation of $\pi/2$ of the domain of definition of its members , strictly includes the class of holomorphic functions bounded in the upper half plane and continuous on the real axis, i.e. if $f$ is bounded on the upper half plane and continuous on the real axis, then $f(-i\zeta)\in\mathscr{LH}_0$.

Apart from this "modern" characterization on this class of function, Torsten Carleman used functions bounded on the upper and the lower half plane to define his generalized Fourier transform: his result are collected in the monograph [1].

References

[1] Thorsten Carleman, L’intégrale de Fourier et questions qui s’y rattachent (French), Publications Scientifiques de l'Institut Mittag-Leffler, 1, Uppsala. 119 p. (1944), MR0014165, Zbl 0060.25504.

[2] Eungu Lee and Dohan Kim, "Laplace hyperfunctions", Integral Transforms and Special Functions, 19:6, 399-407 (2008), MR2426730, Zbl 1186.46042.


问:
Laplace equation with weird boundary condition
So, guys, here's my problem. I have this differential equation

$$
U''_{xx}+U''_{yy}=0
$$

with these boundary conditions:

$$
U'_{y}(x,0)=0
$$
$$
U'_{y}(x,\pi)=0
$$
$$
U(0,y)=0
$$
$$
U(\pi,y)=1+\cos(2y)
$$

Now, I obtain this solution for the first three conditions:

$$
\sum_{n=1}^{\infty} K_n \cdot \sinh(nx) \cdot \cos(ny)
$$

And it has to verify the last condition, so

$$
U(\pi,y)=\sum_{n=1}^{\infty} K_n \cdot \sinh(n\pi) \cdot \cos(ny)=1+\cos(2y)
$$

So, $ K_n \cdot \sinh(n\pi) $ has to be the coefficient of the Fourier series of $ 1+\cos(2y) $. My problem is, how do I calculate $ K_n $?

I tried to obtain the constant by using the following formula

$$
K_n \cdot sinh(n\pi)=\frac{1}{\pi} \int_{-\pi}^{\pi} (1+ \cos(2y)) \cdot \cos(ny) \ dy
$$

But, according to Wolfram Alpha and my own results, it's equal to $0$.

Thank you very much, guys!

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

Lets $\ds{\on{U}_{y}\pars{x,y} \equiv
-\sum_{n = 1}^{\infty}\on{a}_{n}\pars{x}n\sin\pars{ny}}$ which already satisfies $\ds{\on{U}_{y}\pars{x,0} = \on{U}_{y}\pars{x,\pi} = 0}$.

Then, $\ds{\on{U}\pars{x,y} \equiv
\sum_{n = 1}^{\infty}\on{a}_{n}\pars{x}\cos\pars{ny} + \on{f}\pars{x}}$.

The above $\ds{\on{U}\pars{x,y}}$ expression must satisfy the Laplace differential equation:
$$
\sum_{n = 1}^{\infty}\bracks{\on{a}''_{n}\pars{x} - n^{2}\on{a}_{n}\pars{x}}\cos\pars{ny} + \on{f}''\pars{x} = 0
\label{1}\tag{1}
$$
Integrate both sides over $\ds{y \in \pars{0,\pi}}$. It yields $\ds{\on{f}''\pars{x} = 0 \implies \on{f}\pars{x} = bx + c}$ where $\ds{b\ \mbox{and}\ c}$ are constants.

Multiply both sides of (\ref{1}) by $\ds{\cos\pars{ny}}$ and integrate both sides over $\ds{y \in \pars{0,\pi}}$:
$$
\on{a}''_{n}\pars{x} - n^{2}\on{a}_{n}\pars{x} = 0
\implies
\on{a}_{n}\pars{x} = p_{n}\expo{nx} + q_{n}\expo{-nx}\,,\quad p_{n}, q_{n}:\ \mbox{constants}
$$

The general solution becomes
\begin{align}
\on{U}\pars{x,y} & =
\sum_{n = 1}^{\infty}
\pars{p_{n}\expo{nx} + q_{n}\expo{-nx}}\cos\pars{ny} + bx + c
\\[5mm]
0 & =
\sum_{n = 1}^{\infty}
\pars{p_{n} + q_{n}}\cos\pars{ny} + c \implies q_{n} = -p_{n}\,,\ c = 0
\\[5mm]
1 + \cos\pars{2y} & =
\sum_{n = 1}^{\infty}
\pars{p_{n}\expo{n\pi} + q_{n}\expo{-n\pi}}\cos\pars{ny} + b\pi + \overbrace{c}^{\ds{=\ 0}}
\\[2mm] \implies &
2p_{n}\sinh\pars{n\pi} = \delta_{n2}\,,\quad b\pi = 1 \implies b = {1 \over \pi}
\end{align}

Finally,
$$
\on{U}\pars{x,y} =
\bbx{{x \over \pi} + {\sinh\pars{2\pi x}\cos\pars{2y} \over \sinh\pars{2\pi}}} \\
$$


问:
A Bounded Continuous Function on a Circle
Suppose $f$ is continuous and:
$$
|f(e^{i \theta})| \leq M
$$

Suppose:
$$
|\int_{|z|=1}f(z)dz| = 2\pi M
$$
I am trying to show that $f(z) = c \bar{z}$ for some constant $c$ with $|c|=M$

答:
Let $\int_{|z|=1}f(z)dz=\alpha|\int_{|z|=1}f(z)dz|, |\alpha|=1$; also note that $dz=iz|dz|$ on the unit circle

$2\pi M=\int_{|z|=1}\alpha^{-1} f(z)dz= \int_{|z|=1}\alpha^{-1} if(z)z|dz| \le  \int_{|z|=1}|\alpha^{-1} if(z)z||dz| = \int_{|z|=1}|f(z)||dz| \le 2\pi M$ so we have equality (ae and then everywhere by continuity).

But $|dz|$ is a positive measure so this first means that $|f(z)|=M$ on the unit circle and then by taking real and imaginary parts that  $\Re \alpha^{-1} if(z)z=|f(z)|=M, \Im \alpha^{-1} if(z)z=0$ hence $f(z)=-i\alpha M/z=-i\alpha M \bar z$ so we are done!


问:
inner product space and polynomial
Let $V = \mathrm{span}\{1,x,x^2,x^3\}$ be a real inner product space with the
inner product defined by
$$
\langle f,g\rangle =\int\limits_{-1}^{1} fg
$$

Check that $T(f) = f(0)$ is a linear functional on this space and find the element of $V$ which represents it (corresponds to).

So how should I do it? From where to begin?

I am guessing the first part is easy because I have to prove that

$$T(f+g)=(f+g)(0)=f(0)+g(0)=T(f)+T(g)$$

and that

$$T(af)=af(0)=aT(f)$$

but what about this vector that I need to find?

I know that it has something to do with Gram-Schmidt but I do not know how.

答:
Hint:

Find an orthonormal basis for $\,V\,$ using Gram-Schmidt, say the basis $\,\{u_1,...,u_4\}\,$ you get by applying G-S to the given basis of $\,V\,$ . From here, you get that any element $\,w\in V\,$ has a representation (I'm copying (one of) the proof(s) of Riesz Representation Theorem):

$$w=\sum_{k=1}^4\langle w,u_i\rangle u_i\implies Tw=\sum_{k=1}^4\langle w,u_1\rangle Tu_i=\sum_{k=1}^4\langle w,T(u_i)\,u_i\rangle u_i$$

Well, define now

$$u:=\sum_{k=1}^4 T(u_i)\,u_i\;\ldots\ldots$$

Title: inner product space and polynomial


问:
Are indefinite integrals of "equivalent" functions always equal?
How is it possible to prove that the indefinite integrals of "equivalent" functions that cannot be defined at a particular point are equal to each other?

For example,

$$\int\dfrac {x^2}{x}dx=\int xdx $$

Or,

$$\int\dfrac {(x+1)^2}{x+1}dx=\int (x+1)dx $$

Obviously, the function $f(x)=\dfrac{x^2}{x}$ is not defined at the point $x = 0$, but $f(x)=x$ is defined  and likewise, the function $f(x)=\dfrac {(x+1)^2}{x+1}$ is not defined at the point $x = -1$, but $f(x)=x+1$ is defined.

How can I prove that, the indefinite integrals of these functions must be equal?

Thank you very much.

答:
When you are solving the indefinite integral you are looking for primitive functions (of the sub-integral function) in some interval $I$. You need to have an idea up-front what this interval is.

Now... if this interval includes the point(s) -1/0 the sub-integral function is not defined at that point, so really it makes no sense to look for a primitive function at that point (for one or both of the functions). You see if a function is not defined at a given point, it has no primitive too.

If on the other hand the interval does not include the point(s) -1/0, then of course you can cancel the sub-integral function/fraction and solve. So then the primitive function is if of course the same (plus some constant $C$).

So in general the two indefinite integrals are not quite the same.


问:
Why is the kronecker delta not defined for complex numbers?
In the wikipedia page for kronecker delta it says

The restriction to positive integers is common, but there is no reason
it cannot have negative integers as well as positive, or any discrete
rational numbers. (...) However, the Kronecker delta is not defined
for complex numbers.

This seems completely arbitrary. Why would the kronecker delta not be defined for complex numbers? Defining it over any set $X$ as $\delta: X^2 \to \{0,1\}$ with $\delta_{x,y} = 1 \iff x=y$ seems like a pretty natural generalization.

Is there any reason for it to say in wikipedia that it is not defined for complex numbers?

答: Math Wikipedia has a lot of small strange mistakes like this, more than I expected; I've corrected several over the last month or two. The Kronecker delta can be defined on an arbitrary set, including the real or the complex numbers but all sorts of others. I'll delete the sentence on Wikipedia.</p>