问:
Exponential generating function in the case of a singular symbol
Suppose I have the following set {p, q, s, 0, 1, 2, 3} where we have $l_n$ is the number of strings with the symbols in this set of length $n$. However, there is exactly one number in each string that is a digit.

To find the generating function for this I started off by treating the alphabets as a ternary string and the digits 1 unit. Essentially we have our exponential generating function $F(x) = x\sum _{n=0}^{\infty }\:3^nx^n\frac{1}{n!}$. But I feel like my approach so far is wrong (it seems too simple) and this for the case where we only have 1 digit. I don't really know how to factor in the other 3 digits.

答:
There are $3^{n-1}$ strings of length $n-1$ with no digit; every string of length $n$ with exactly one digit is uniquely obtained by choosing one of these $3^{n-1}$ strings, choosing one of the $4$ digits, and inserting it in one of the $n$ possible positions in the chosen string. For $n\ge 1$ this can be done in $4n\cdot 3^{n-1}$ ways, so $\ell_0=0$, and $\ell_n=4n\cdot 3^{n-1}$ for $n\ge 1$. The exponential generating function must therefore be

$$\begin{align*}
\sum_{n\ge 0}\ell_n\frac{x^n}{n!}&=\sum_{n\ge 1}\frac{4n}{n!}3^{n-1}x^n\\
&=4x\sum_{n\ge 1}\frac{3^{n-1}x^{n-1}}{(n-1)!}\\
&=4x\sum_{n\ge 0}\frac{3^nx^n}{n!}\,,
\end{align*}$$

for which you should fairly easily be able to find a closed form.


问:
Evaluate this limit $ \lim_{x\to\infty}\left (\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x}$
Please help to evaluate this limit

$$ \lim_{x\to\infty} \left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x},$$

where $0 \leq a$ and $a \not= 1$.

I tried to logarithm from both sides, and apply taylor series but so far without success.

答:
Case 1: If $a>1$ then from  $\text{exp}^{\large{\log{u(x)}}}=u(x)$ we deduce

\begin{align}
\left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x} &=
\exp\left(\frac{\log\left(\frac{1}{x}\frac{a^x - 1}{a - 1}\right)}x\right)\\&=
\exp\left(\frac{\log\left(\frac 1x\right)+\log\left(\frac{a^x - 1}{a - 1}\right)}x\right)\\&=
\exp\left(\frac{-\log(x)+\log(a^x-1)-\log(a-1)}x\right)\\&=
\exp\left(\frac{-\log(x)+\log\big[a^x\left(1-a^{-x}\right)\big]-\log(a-1)}x\right)\\&=
\exp\left(\frac{-\log (x)+\log(1-a^{-x})-\log(a-1)}x+\log(a)\right)
\end{align}

Sending $x\to \infty$ will make the middle terms vanish since

$$\lim_{x\to\infty}\frac{\log(x)}{x}=0$$

Therefore

$$\lim_{x\to+\infty}\left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x}=\exp\bigl(\log(a)\bigr)=a.$$

Case 2: If $0< a <1$ then we first write

$$\frac{a^x - 1}{a - 1} = \frac{1-a^x}{1-a}$$

where
\begin{align}
\log\left(\frac{1}{x}\frac{1-a^x}{1-a}\right)&=
\log\left(\frac{1}{x}\right)+\log\left(\frac{1- a^x}{1-a}\right)\\&=
-\log(x)+\log\left(1- a^x\right)-\log\left({1-a}\right)
\end{align}

and

$$\lim_{x\to\infty}\frac{-\log(x)+\log\left(1- a^x\right)-\log\left({1-a}\right)}{x}=0$$

thus

$$\lim_{x\to+\infty}\left(\frac{1}{x}\frac{1-a^x}{1-a} \right)^\frac{1}{x}=\exp\bigl(0)=1.$$

ContentLicense: CC BY-SA 4.0


问:
How to understand the limit placed on integration variable when doing trig substitution?
For the calculation of  $\int \frac{1}{x^2-1}{d}x$,  trig substitution method  is to set $\sec\theta=x$.Doing this,  a limit is placed on  x$\in(-\infty,-1]\cup  [1, +\infty) $ . The integration result is $\frac{1}{2}$ $ln|\frac{x-1}{ x+1}$|+C with domain x$\in(-\infty,-1)\cup (-1,1)\cup (1, +\infty) $  , different from the x$\in(-\infty,-1]\cup  [1, +\infty) $.

Another example, if   same trig substitution $\sec\theta=x$ is used to calculate $\int (x^2-1){d}x$, the result is $\frac{1}{3}x^3-x+C$. In this case, the domain of the resulting function is $(-\infty,+\infty)$, again different from x$\in(-\infty,-1]\cup  [1, +\infty) $.

Here is my question, why are we allowed to do the trig substitution in these cases without any special cautions? It does not seem to cause any problems, but why?

答:
Here are the plots of $f(x) = \frac{1}{x^2-1}$ in blue and $g(x) = \frac{1}{2} \log |\frac{x-1}{x+1}|$ in orange.

Notice that $f$ is discontinuous at $|x| = 1$, thus the antiderivative $g$ is also discontinuous there.  The apparent continuity in the picture is a plotting artifact; in actuality, the portion of the curve on $|x| < 1$ is not connected to either of the curves in the regions $x > 1$ or $x < -1$.

Also notice that $g$ is well defined on the set $|x| \ne 1$, not just $|x| > 1$ as you suggest as a result of the substitution $x = \sec \theta$.  Why should this even work?  After all, if $|x| < 1$, there is no real-valued angle such that $\sec \theta = x$.  And while this is correct, what it doesn't acknowledge is that when $|x| < 1$, we could also choose the substitution $$x = \cos \varphi, \quad dx = -\sin \varphi \, d\varphi,$$ hence $$\int f(x) \, dx = \int -\frac{1}{1 - \cos^2 \varphi} \cdot (-\sin \varphi) \, d\varphi = \int \csc \varphi \, d\varphi.$$  And this is the same result as if we used the substitution $x = \sec \theta$ when $|x| > 1$.  One substitution (cosine) handles the $|x| < 1$ case, and the other (secant) handles the $|x| > 1$ case.  Or if you want to do away with trigonometric substitutions entirely, we would simply observe
$$f(x) = \frac{1}{(x-1)(x+1)} = \frac{1}{2}\left( \frac{1}{x-1} - \frac{1}{x+1} \right),$$ hence $$\int f(x) \, dx = \frac{1}{2} \left( \int \frac{dx}{x-1} - \int \frac{dx}{x+1} \right) = \frac{\log |x-1| - \log |x + 1|}{2} + C = \frac{1}{2} \log \left| \frac{x-1}{x+1} \right| + C.$$

The key insight here is to keep track of any points of discontinuity of the integrand, how these are mapped under substitution, and what happens after the substitution is undone.  In the other case, where we have no discontinuities, we can also use either substitution and again we recover the same antiderivative.


问:
Calculate the number of elements of order 2 and order 4 in each of $\mathbb{Z}_{16}, \mathbb{Z}_{8}\bigoplus\mathbb{Z}_{2}$ etc.
I'm trying to understand how to computer the order of an element in a finite abelian group in the form of $$\mathbb{Z}_{8}\bigoplus\mathbb{Z}_{2}$$ and I am just not getting it.  In particular I stumbled across this question:

Calculate the number of elements of order $2$ and order $4$ in each of: $\mathbb{Z}_{16}, \mathbb{Z}_{8}\bigoplus\mathbb{Z}_{2}, \mathbb{Z}_{4}\bigoplus\mathbb{Z}_{4}, \mathbb{Z}_{4}\bigoplus\mathbb{Z}_{2}\bigoplus\mathbb{Z}_{2}$

I even know the answer:

$\mathbb{Z}_{16}$ has 1 element of order 2 and 2 elements of order 4.

$\mathbb{Z}_{8}\bigoplus\mathbb{Z}_{2}$ has 3 elements of order 2 and 4 elements of order 4.

$\mathbb{Z}_{4}\bigoplus\mathbb{Z}_{4}$ has 3 elements of order 2 and 12 elements of order 4.

$\mathbb{Z}_{4}\bigoplus\mathbb{Z}_{2}\bigoplus\mathbb{Z}_{2}$ has 7 elements of order 2 and 8 elements of order 4.

Can someone please explain this in a simple way?  I've reread the appropriate sections in my textbook time and again and looked for explanations on the net without avail.

Thank you.

答:
This is a fairly simple way

For $\mathbb{Z} _8 \times \mathbb {Z} _2$.

Note $2(a,b)=0$ iff $2a=0$ and $2b=0$, then $a\in\{0,4\}  $ and $b\in \mathbb {Z} _2$. So there are $2\cdot 2=4$ possibilities for $|(a,b)|=2 $, but $|(0,0)|=1<2 $. Thus there are $4-1=3$ elements of order $2$ in $\mathbb{Z} _8 \times \mathbb {Z} _2$.

Similarly,  $4(a,b)=0$ iff $4a=0$ and $4b=0$, then $a\in\{0,2,4,6\}  $ and $b\in \mathbb {Z} _2$. So there are $4\cdot 2=8$ possibilities for $|(a,b)|=4 $, but we have found $4$ elements not having order $4$, which means that $8-4=4$ elements have order $4$.

ContentLicense: CC BY-SA 3.0


问:
stochastic integral equation
For $0 \leq t \leq T$, define

$$Z_t:=\exp {\left\lbrace \int_0^t X_sdW_s - \frac 12 \int_0^t X_s^2ds \right\rbrace }$$

Show that this process satisfies the stochastic integral equation

$$Z_t=1+\int_0^tZ_s X_s dW_s , \qquad 0≤t≤T$$

答:
here is solution with details :
indeed
 For $0 \leq t \leq T$ let $$Z_t:=\exp {Y_t }$$
with $$Y_t:= \int_0^t X_sdW_s - \frac 12 \int_0^t X_s^2ds  $$
let $ f \in C^2(R) $ such that $ Z_t=f(Y_t) $
so by Ito's formula  we get :
$$ f(Y_t)=f(Y_0)+\int_0^t \frac{\partial f}{\partial x}(Y_s)dY_s+ \frac{1}{2}\int_0^t \frac{\partial^2f}{\partial x^2}(Y_s)d<Y>_s $$
to calculate the quadratic variation of process Y

we have  $$\mathrm dY_t = -\frac12 X^2_t\mathrm dt + X_t\mathrm dW_t.$$ then
 $$d<Y>_s=X^2_sd<W>_s=X^2_sds  $$ and we have also $$ Z_s=f(Y_s)=exp(Y_s)=\frac{\partial f}{\partial y}(Y_s)=\frac{\partial^2f}{\partial y^2}(Y_s) $$
$$ f(Y_0)=exp(Y_0)=exp{\left\lbrace \int_0^0 X_sdW_s - \frac 12 \int_0^0 X_s^2ds\right\rbrace }=exp(0)=1  $$ then we can write :
$$Z_t=f(Y_t)=1+\int_0^t Z_sdY_s+ \frac{1}{2}\int_0^t Z_sX^2_sds $$
or $$ \mathrm dY_s = -\frac12 X^2_s\mathrm ds + X_s\mathrm dW_s $$
then $$ Z_t=f(Y_t)=1+\int_0^t Z_s(-\frac12 X^2_s\mathrm ds + X_s\mathrm dW_s) + \frac{1}{2}\int_0^t Z_sX^2_sds $$

hence $$Z_t=1+\int_0^tZ_s X_s dW_s , \qquad 0≤t≤T$$

Title: stochastic integral equation


问:
Why $E[\int_0^t e^{W_t}dW_t] = 0$?
I keep reading about one proof I found on internet and I see that they state that $$E[\int_0^t e^{W_s} dW_s] =0 $$

where $W_t$ is Wiener process.

Their justification is the following:

Note that by definition of the stochastic integral, we have
$\mathbb{E}\left[\int_0^Te^{B_t}dB_t\right]=0$. This follows from the
fact that the function $f : (t,\omega) \mapsto \exp(B(t,\omega))$
belong to $H^2[0,T]$ where $H^2$ is the space of progressively
measurable function such that : \begin{equation} \mathbb{E}\int_0^T
> f^2(t,\omega)dt < \infty \end{equation}

And I don't quite get why this solve the issue. Could you please explain to me why, or show another, maybe simpler way how it can be proved?

答: This is just a consequence of the basic property of Ito's integral, for any "nice function" $f$, we define that $$\int_0^T f(W_t)\, dW_t$$ to be $\lim_{\Delta t \to 0} \sum_{i=0}^n f(W_{t_i})\cdot(W_{t_{i+1}} - W_{t_i}) $ where $\{t_i\}$ is a equi-distance partition of $[0,T]$. Now use the fact that $f(W_{t_i})$ and $W_{t_{i+1}} - W_{t_i}$ are independent conditional on $\mathcal{F}_t$ (the natural filtration generated by the Brownian motion at time $t$) and take the expectation. The last sentence should be translated to $$\mathbb{E}[f(W_{t_i})\cdot(W_{t_{i+1}} - W_{t_i}) \mid \mathcal{F}_t] = f(W_{t_i})\,\mathbb{E}[W_{t_{i+1}} - W_{t_i}] = 0$$

Title: Why $E[\int_0^t e^{W_t}dW_t] = 0$?


问:
What is $\lim\limits_{n→∞}(\frac{n-x}{n+x})^{n^2}$?
What is $$\lim_{n\rightarrow\infty}\left(\frac{n-x}{n+x}\right)^{n^2},$$ where $x$ is a real number. Mathematica tells me the limit is $0$ when I put an exact value for $x$ in (Mathematica is inconclusive if I don't substitute for $x$), but using $f(n)=(n-x)^{n^2}$ and $g(n)=(n+x)^{n^2}$, then $$\lim_{n\rightarrow\infty}f(n)=\lim_{n\rightarrow\infty}g(n)=\infty,$$ and by L'Hopital's rule we have
$$\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} = \lim_{n\rightarrow\infty}\frac{(s-z)^{-1+s^2} (s+z)^{1-s^2} (s+2 (s-z) \text{Log}[s-z])}{s+2 (s+z) \text{Log}[s+z]} = 1.$$

I'm not sure which to believe - Mathematica or L'Hopital.

答:
I think this goes to zero for positive $x$.  Consider the log of the expression:

$$n^2 \log{\left(\frac{1-(x/n)}{1+(x/n)}\right)} = n^2 \left[ \log{\left(1-\frac{x}{n}\right)}-\log{\left(1+\frac{x}{n}\right)}\right]$$

For fixed $x>0$, the value of $x/n$ is small compared with $1$, so Taylor expand and get

$$n^2 \left ( -\frac{x}{n} - \frac{x}{n}\right) = -2 n x$$

which clearly $\to -\infty$ as $n \to \infty$.  Taking exponentials, $e^{-\infty} = 0$, loosely speaking.

For $x<0$, this diverges.  For $x=0$, the limit is $1$.


问:
What is reasoning behind statement that $\log \frac{3}{2} < (\frac{3}{2} - 1)$?
In his paper, "Euler and the Partial Sums of the Prime Harmonic Series" (pdf), Paul Pollacks makes a small statement at the top of page 4:

$$ \log \frac{3}{2} < (\frac{3}{2} - 1) = \frac12 $$

I can see numerically this is true, but I can't see how he makes the inequality $\log \frac{3}{2} < (\frac{3}{2} - 1)$.

I have looked through references of power series approximations but can't see anything that matches.

答:
For any $x > 1$ we have
$$
\log(x) < x - 1.
$$
For a geometric proof, note that both functions vanish at $x = 1$ and that the graph of $\log(x)$ is concave down (so is trapped below the tangent line as you move to the right).


问:
Heat equation: Energy $\int_{\Omega}u(t, x) dx$ is decreasing with positive initial data and $u=0$ on the boundary
From David Borthwick - Introduction to Partial Differential Equations, Exercise 6.3b (paraphrased),

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain with piecewise $C^1$ boundary.
Suppose $u(t,x)$ satisfies the heat equation
$$u_t - \Delta u = 0$$
$$u(0,x) > 0 \quad\text{ for } x\in \Omega$$ $$u(t,x) = 0 \quad\text{ for } x\in \partial\,\Omega$$
Define the total thermal energy at time $t$ by

$$U(t) = \int_{\Omega}u(t,x)dx$$

Show that $U(t)$ is decreasing.

My attempts so far:

Using the divergence theorem, it can start like
$$U'(t) = \int_{\Omega} u_t(t, x) dx = \int_{\Omega} \Delta u(t, x) dx =\int_{\partial\Omega} \nabla u(t, x) \cdot \mathbf{n} \,dx = \,...?$$

By the maximum principle, it should end like
$$... = \int_{?} -u(t,x) (\,?\,) dx \leq \int_{?} -\min_{[0,t]\times \bar{\Omega}} u(t,x) (\,?\,) dx \leq 0$$
since $u(t,x)\geq 0$ for $(t,x) \in (\{0\}\times\Omega) \cup ([0,t]\times \partial \Omega )$

For the missing steps in the middle, I get stuck because I can no longer "integrate by parts" to bring $\nabla u$ "up" to $u$. Also, the negative sign seems to be crucial to prove $U$ is decreasing, but I have no ideas where it could come from.

Alternatively observe $u_t$ also satisfies a heat equation. Maybe we can use the maximum principle on $u_t$ directly, but we don't know $u_t(0,x)$ for $x\in \Omega$.

Are there more things to observe or are these the right tracks? Preferably only elementary results related to heat equation would be used. Any help is appreciated.

答: Have you tried the Hopf's lemma? (link: https://en.wikipedia.org/wiki/Hopf_lemma)</p>


问:
Can always transform coordinates into a space where an arbitrary curve will appear as a straight line?
I was just thinking about how a if you've got a circle, in cartesian coordinates that is given by the equation $x^2+y^2=k^2$. You can then move over to polar coordinates and this same circle when represented under polar coordinates, looks like a straight line.

I was just wondering if it was always the case that for any arbitrary curve, there always existed some way to transform the coordinates such that you could view that curve as a straight line?

答:
$\newcommand{\R}{\mathbb{R}}$

Locally, any curve in $\R^2$ is a graph of some function $f \colon \R \to \R$, i.e., as the set $\Gamma = \{ (x,f(x)): x \in \R \}$. To make it straight, one only needs a very simple change of coordinates:
$$
\Phi(x,y) = (x,y-f(x)), \quad \Phi^{-1}(x,y) = (x,y+f(x)).
$$
It's simple to check that $\Phi$ transform $\Gamma$ into the line $(x,0)$.

Of course it only works locally, and the circle is a good reason why.


问:
$L^p(\mu)$ space relations
Let $(X,M, \mu)$ be a measure space. Suppose $1 \leq p < r \leq \infty$.

Suppose $f \in L^p (\mu) \cap L^r(\mu)$. Show that $f\in L^s(\mu)$ for $p \leq s \leq r$.

Attempt:

Suppose $f \in L^p (\mu) \cap L^r(\mu)$. Then $\int |f| ^p d\mu < \infty$ and $\int |f|^r d\mu < \infty$.

Isn't it obvious that $\int |f|^s d\mu < \infty$ already?

I'm bit confused on what to show.

Any help will be appreciated!

Thank you for your help!

答:
Can also be done from the Hölder inequality, in the form:
If $1/\alpha+ 1/\beta = 1$, then

$$
\int |FG|\;d\mu \le
\left(\int |F|^\alpha\;d\mu\right)^{1/\alpha}\;\left(\int |G|^\beta\;d\mu\right)^{1/\beta}
\tag1$$

Given $p<s<r$ and $f \in L_p \cap L_r$ as in the problem, let
$$
\alpha = \frac{r-p}{r-s},\quad \beta = \frac{r-p}{s-p}
\quad\text{so that}\quad \frac{1}{\alpha}+\frac{1}{\beta} = 1\quad\text{and}\quad
\frac{p}{\alpha} + \frac{r}{\beta} = s,
$$
and let
$$
F = |f|^{p/\alpha},\qquad G = |f|^{r/\beta}
$$
Plug into $(1)$ to get
$$
\int|f|^s\;d\mu \le \left(\int |f|^p\;d\mu\right)^{(r-s)/(r-p)}
\;\left(\int |f|^r\;d\mu\right)^{(s-p)/(r-p)} < \infty.
$$

We will need a different (easier) argument in case $r = \infty$.

Title: $L^p(\mu)$ space relations


问:
Connected countable dense homogeneous space is homogeneous.
For a topological space $X$,

$\mathcal{H}(X)$ denotes the set of all autohomeomorphism of $X$. Note that, with the composition of functions,  $(\mathcal{H}(X),\circ)$ is a group with identity $\mbox{Id}_{X}$.

If $x\in X$, let $\mathcal{O}(x,X)=\{h(x):h\in\mathcal{H}(X)\}$, we call this set the orbit of $x$ (under homeomorphism). Note that $x=\mbox{Id}_{X}(x)\in \mathcal{O}(x,X)$.

Note that for each $x\in X$, $\mathcal{O}(x,X)$ is fixed under autohomeomorphisms of $X$ (that is, for each $h\in\mathcal{H}(X)$, $h[\mathcal{O}(x,X)]=\mathcal{O}(x,X)$)

A topological space $X$ is called countable dense homogeneous if

$X$ is separable, and

for any pair $D, E$ of countable dense subsets there exists $h\in\mathcal{H}(X)$ such that $h[D]=E$.

A topological space $X$ is homogeneous if for every $x, y \in X$ there is $h\in \mathcal{H}(X)$ such that $h(x) = y$.

The motivation for this question is to demonstrate the following claim

Every connected countable dense homogeneous space is homogeneous.

For this, we follow the following preliminary steps

For every $x\in X$, $\overline{\mathcal{O}(x,X)}$ is a clopen subset of $X$.

If $x,y\in X$ are such that $\overline{\mathcal{O}(x,X)}\cap \overline{\mathcal{O}(y,X)}\not=\emptyset$, then $\overline{\mathcal{O}(x,X)}=\overline{\mathcal{O}(y,X)}$.

Let $x\in X$, then $\overline{\mathcal{O}(x,X)}$ is a countable dense homogeneous space.

Let $x\in X$, then $\overline{\mathcal{O}(x,X)}=\mathcal{O}(x,X)$.

Finally, let $X$ be a connected countable dense homogeneous space and let $x,y\in X$. Note that $\mathcal{O}(x, X)$ and $\mathcal{O}(y, X)$ are clopen, also are non-empty sets. Then $\mathcal{O}(x, X)=\mathcal{O}(y, X)=X$. Therefore there is $h\in\mathcal{H}(X)$ such that $h(x)=y$.

Unfortunately I just couldn't prove step 4, someone has a suggestion on how to prove that, or some other new suggestion to directly prove the claim.

答: For the case of a first countable space we have that:


问:
Suppose $\lim_{n\rightarrow \infty}f_n(x)=0$ for all $x$ and $f_n$ non-increasing and continuous. Is $\lim_{n\rightarrow \infty}f_n(n)=0$?
Suppose $\lim_{n\rightarrow \infty}f_n(x)=0$ for all $x$ and $f_n$ non-increasing and continuous.  Is $\lim_{n\rightarrow \infty}f_n(n)=0$?

If $\lim_{n\rightarrow \infty}f_n(x)=0$, then
$$
(\forall \epsilon_1)(\exists N_1\in \mathbb{N})(\forall n)(n\geq N_1\implies |f_n(x)|<\epsilon_1)
$$
We want to show
$$
(\forall \epsilon_2)(\exists N_2\in \mathbb{N})(\forall n)(n\geq N_2\implies |f_n(n)|<\epsilon_2)
$$

I belive the above claim is true but the notation involving $n$ is too complicated for me right now.

答: $f_n(x)=-\frac  x n$ is a counter-example. Howver, if you assume that $f_n$'s are non-negative then the conclusion is true since $0 \leq f_n(n) \leq f_n(1)$ for all $n \geq 1$.


问:
Why could we use the complex basis for the cotangent space and what does Cauchy Integral Formula in this basis say?
I previously studied some differential forms on real manifolds from Introduction to Smooth Manifolds by Lee and I just started reading Principles of Algebraic Geometry by Griffith & Harris. However, I don't get what do they mean by a complex basis for the cotangent space. From the way I understand it, saying $C^n\cong R^{2n}$ means identifying the cotangent of the two. If this is the case, what do they mean by a complex linear combination of two real vectors?

Another question seems to be related to this question: where does the extra term in the equation below come from? The undergraduate version of Cauchy Integral Formula I learnt has only the first term on the right side of the equation.

答:
Introduction

I do not have a deep understanding of differential forms, and know almost nothing about multiple complex variables. But I think the concepts needed to answer your question won't rely on either; I don't think any major concepts in this exposition will change for $n>1$.

Given any smooth function $\mathbf{f}:\mathbb{R}^{2}\to\mathbb{R}^{2}$,
we may associate $\mathbf{f}\left(x,y\right)=\left\langle u\left(x,y\right),v\left(x,y\right)\right\rangle $
with $g:\mathbb{C}\to\mathbb{C}$ as $g\left(x+iy\right)=u\left(x,y\right)+iv\left(x,y\right)$. I wish I could have two large columns to easily present the parallels
between the real and complex perspectives. With the space limitations
of Mathematics StackExchange, I'm forced to go back and forth.

Cotangent Space

Decomposing a Real Form

We might write something like

\begin{align*}
\mathrm{d}\mathbf{f} & =\left\langle \mathrm{d}u,\mathrm{d}v\right\rangle \\
 & =\left\langle \dfrac{\partial u}{\partial x}\mathrm{d}x+\dfrac{\partial u}{\partial y}\mathrm{d}y,\dfrac{\partial v}{\partial x}\mathrm{d}x+\dfrac{\partial v}{\partial y}\mathrm{d}y\right\rangle \\
 & =\dfrac{\partial\left\langle u,v\right\rangle }{\partial x}\mathrm{d}x+\dfrac{\partial\left\langle u,v\right\rangle }{\partial y}\mathrm{d}y\\
 & =\dfrac{\partial u}{\partial x}\left\langle \mathrm{d}x,0\right\rangle +\dfrac{\partial v}{\partial x}\left\langle 0,\mathrm{d}x\right\rangle +\dfrac{\partial u}{\partial y}\left\langle \mathrm{d}y,0\right\rangle +\dfrac{\partial v}{\partial y}\left\langle 0,\mathrm{d}y\right\rangle \text{.}
\end{align*}

Decomposing a Complex Form

Analogously, we have

\begin{align*}
\mathrm{d}g & =\dfrac{\partial g}{\partial x}\mathrm{d}x+\dfrac{\partial g}{\partial y}\mathrm{d}y\\
 & =\dfrac{\partial\left(\mathfrak{R}\circ g+i\mathfrak{I}\circ g\right)}{\partial x}\mathrm{d}x+\dfrac{\partial\left(\mathfrak{R}\circ g+i\mathfrak{I}\circ g\right)}{\partial y}\mathrm{d}y\\
 & =\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}+i\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}\right)\mathrm{d}x+\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}+i\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}\right)\mathrm{d}y\\
 & =\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}\mathrm{d}x+\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}i\mathrm{d}x+\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\mathrm{d}y+\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}i\mathrm{d}y\text{.}
\end{align*}
Above, I have used $\mathfrak{R}$ and $\mathfrak{I}$ for the real
and imaginary parts.

A New Real Basis

We have written $\mathrm{d}\mathbf{f}$ as a linear combination of
$\left\langle \mathrm{d}x,0\right\rangle $, $\left\langle 0,\mathrm{d}x\right\rangle $,
$\left\langle \mathrm{d}y,0\right\rangle $, $\left\langle 0,\mathrm{d}y\right\rangle $.
And we can rewrite $\mathrm{d}\mathbf{f}$ (or $2\mathrm{d}\mathbf{f}$)
in terms of a new basis: $\left\langle \mathrm{d}x,\mathrm{d}y\right\rangle $,
$\left\langle -\mathrm{d}y,\mathrm{d}x\right\rangle $, $\left\langle \mathrm{d}x,-\mathrm{d}y\right\rangle $,
$\left\langle \mathrm{d}y,\mathrm{d}x\right\rangle $. We have

\begin{align*}
2\mathrm{d}\mathbf{f} & =2\dfrac{\partial u}{\partial x}\left\langle \mathrm{d}x,0\right\rangle +2\dfrac{\partial v}{\partial x}\left\langle 0,\mathrm{d}x\right\rangle +2\dfrac{\partial u}{\partial y}\left\langle \mathrm{d}y,0\right\rangle +2\dfrac{\partial v}{\partial y}\left\langle 0,\mathrm{d}y\right\rangle \\
 & =\left(\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}\right)\left\langle \mathrm{d}x,\mathrm{d}y\right\rangle +\left(\dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial y}\right)\left\langle -\mathrm{d}y,\mathrm{d}x\right\rangle \\
 & \phantom{=}+\left(\dfrac{\partial u}{\partial x}-\dfrac{\partial v}{\partial y}\right)\left\langle \mathrm{d}x,-\mathrm{d}y\right\rangle +\left(\dfrac{\partial v}{\partial x}+\dfrac{\partial u}{\partial y}\right)\left\langle \mathrm{d}y,\mathrm{d}x\right\rangle \text{.}
\end{align*}

As an aside, note that the latter two coefficients would be $0$ if
the Cauchy-Riemann equations
are satisfied.

A New Complex Basis

We have written $\mathrm{d}g$ as a real linear combination
of $\mathrm{d}x$, $i\mathrm{d}x$, $\mathrm{d}y$, $i\mathrm{d}y$.
And we can rewrite $\mathrm{d}g$ (or $2\mathrm{d}g$) in terms of
a new basis: $\mathrm{d}x+i\mathrm{d}y$, $-\mathrm{d}y+i\mathrm{d}x$,
$\mathrm{d}x-i\mathrm{d}y$, $\mathrm{d}y+i\mathrm{d}x$. Note that
those expressions can be rewritten in a suggestive way as $\mathrm{d}x+i\mathrm{d}y$,
$i\left(\mathrm{d}x+i\mathrm{d}y\right)$, $\mathrm{d}x-i\mathrm{d}y$,
$i\left(\mathrm{d}x-i\mathrm{d}y\right)$. Inspired by $z=x+iy$ and $\overline{z}=x-iy$, we can make this even tidier by introducing the abbreviations $\mathrm{d}z=\mathrm{d}x+i\mathrm{d}y$ and $\mathrm{d}\overline{z}=\mathrm{d}x-i\mathrm{d}y$;
this reduces the basis to $\mathrm{d}z$, $i\mathrm{d}z$, $\mathrm{d}\overline{z}$,
$i\mathrm{d}\overline{z}$. We have

\begin{align*}
2\mathrm{d}g & =2\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}\mathrm{d}x+2\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}i\mathrm{d}x+2\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\mathrm{d}y+2\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}i\mathrm{d}y\\
 & =\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}\right)\left(\mathrm{d}x+i\mathrm{d}y\right)+\left(\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}-\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\left(-\mathrm{d}y+i\mathrm{d}x\right)\\
 & \phantom{=}+\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}-\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}\right)\left(\mathrm{d}x-i\mathrm{d}y\right)+\left(\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\left(\mathrm{d}y+i\mathrm{d}x\right)\\
 & =\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}\right)\mathrm{d}z+\left(\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}-\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\left(i\mathrm{d}z\right)\\
 & \phantom{=}+\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}-\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}\right)\mathrm{d}\overline{z}+\left(\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\left(i\mathrm{d}\overline{z}\right)\\
 & =\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}+i\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}-i\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\mathrm{d}z\\
 & \phantom{=}+\left(\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial x}+i\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial x}-\dfrac{\partial\left(\mathfrak{I}\circ g\right)}{\partial y}+i\dfrac{\partial\left(\mathfrak{R}\circ g\right)}{\partial y}\right)\mathrm{d}\overline{z}\\
 & =\left(\dfrac{\partial\left(\mathfrak{R}\circ g+i\mathfrak{I}\circ g\right)}{\partial x}+\dfrac{\partial\left(\mathfrak{I}\circ g-i\mathfrak{R}\circ g\right)}{\partial y}\right)\mathrm{d}z\\
 & \phantom{=}+\left(\dfrac{\partial\left(\mathfrak{R}\circ g+i\mathfrak{I}\circ g\right)}{\partial x}+\dfrac{\partial\left(i\mathfrak{R}\circ g-\mathfrak{I}\circ g\right)}{\partial y}\right)\mathrm{d}\overline{z}\\
 & =\left(\dfrac{\partial g}{\partial x}+\dfrac{\partial\left(-ig\right)}{\partial y}\right)\mathrm{d}z+\left(\dfrac{\partial g}{\partial x}+\dfrac{\partial\left(ig\right)}{\partial y}\right)\mathrm{d}\overline{z}\\
 & =\left(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\right)g\,\mathrm{d}z+\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)g\,\mathrm{d}\overline{z}\text{.}
\end{align*}
Note that $\mathrm{d}g=\dfrac{\partial g}{\partial x}\mathrm{d}x+\dfrac{\partial g}{\partial y}\mathrm{d}y$
started as a complex linear combination of $\mathrm{d}x$ and $\mathrm{d}y$,
and ended up as a complex linear combination of $\mathrm{d}z$ and
$\mathrm{d}\overline{z}$, even though we passed through a representation
as a real linear combination to show the parallel with $\mathbf{f}$.
Dividing through by $2$, we obtain $\mathrm{d}g=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\right)g\,\mathrm{d}z+\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)g\,\mathrm{d}\overline{z}$.
This form suggests defining the "Wirtinger derivatives"
: $\dfrac{\partial}{\partial z}=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\right)$
and $\dfrac{\partial}{\partial\overline{z}}=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)$
so that we may write the tidy $\boxed{\mathrm{d}g=\dfrac{\partial g}{\partial z}\mathrm{d}z+\dfrac{\partial g}{\partial\overline{z}}\mathrm{d}\overline{z}}$.
In this form, the Cauchy-Riemann equations amount to "$\dfrac{\partial g}{\partial\overline{z}}=0$ as a complex number".

Cauchy Integral Formula

As you noted, the Cauchy Integral Formula doesn't usually have that extra term, because it's not needed for holomorphic functions. This generalization for smooth functions is sometimes known as the Cauchy-Pompeiu formula. An exposition (starting from an understanding of the boxed equation above) can be found in C.K. Fong's notes for "Calculus of Differential Forms with Applications". Specifically, in "Section 2. Complex 2 - forms: Cauchy - Pompeiu's Formula" of "Chapter 3: Complex Differential
Forms". I'll summarize the argument here, translated into your book's variables.

Let $z$ be a particular complex number, and use $w$ for a varying
one. Consider $\dfrac{f\left(w\right)}{w-z}\mathrm{d}w$. Then a routine
calculation (using $\mathrm{d}\left(\dfrac{\mathrm{d}w}{w-z}\right)=0$)
shows that $\mathrm{d}\left(\dfrac{f\left(w\right)}{w-z}\mathrm{d}w\right)=\dfrac{\partial f(w)}{\partial\overline{w}}\dfrac{\mathrm{d}\overline{w}\wedge\mathrm{d}w}{w-z}$.

We can apply Green's Theorem/general Stokes' Theorem to the large disc
$\Delta$ by removing a small disc around the point
$z$ and letting the small radius go to zero, just as in the standard proof of the Cauchy Integral Formula
for a holomorphic $f$.

By the standard argument about what happens
to the integral of $\dfrac{f(w)\mathrm{d}w}{w-z}$ around a circle
whose radius approaches $0$, we end up with $\iint_{\Delta}\dfrac{\partial f(w)}{\partial\overline{w}}\dfrac{\mathrm{d}\overline{w}\wedge\mathrm{d}w}{w-z}=\int_{\partial\Delta}\dfrac{f(w)\mathrm{d}w}{w-z}-2\pi if\left(z\right)$.
This is equivalent to the formula in your book (make sure to note
that $\mathrm{d}\overline{w}\wedge\mathrm{d}w=-\mathrm{d}w\wedge\mathrm{d}\overline{w}$).

Note that only when $f$ is holomorphic could you note that $\dfrac{\partial f(w)}{\partial\overline{w}}=0$ and simplify.


问:
Greatest number of occurrences of the pattern 4213 in a permutation.
Statistic St000750 in the FindStat database is a map $\operatorname{St000750} \colon S_n \rightarrow \mathbb N_{\geq 0}$ given by

The number of occurrences of the pattern $4213$ in a permutation.

This is the number of length-4 substrings such that the first number is largest, the second is third-largest, the third is smallest, and the fourth is second-largest. For example, the permutation $\pi = 563214$ has $\operatorname{St000750}(\pi) = 6$ occurrences of this pattern:
$$5324, 5314, 5214, 6324, 6314, \text{and}\ 6214.$$

I'm interested in computing$$
   a(n) = \max \{ \operatorname{St000750}(\pi) : \pi \in S_n\}
$$
By brute-forcing $0 \leq n \leq 9$, I get that

$a(0) = a(1) = a(2) = a(3) = 0$,

$a(4) = 1$,

$a(5) = 3$,

$a(6) = 6$,

$a(7) = 13$,

$a(8) = 24$, and

$a(9) = 40$.

Moreover, this sequence does not appear in the On-Line Encyclopedia of Integer Sequences.

However, an analogous sequence for the pattern $132$ is A061061, with a simple formula: $A061061(n) = \max\{A061061(k) + k\binom{n-k}{2} : 1 \leq k < n\}$.

Is there a way to compute this sequence which is better than brute force? If not, how can you construct a better upper bound than $a(n) \leq \binom{n}{4}$?

答:
You can solve the problem via integer linear programming as follows.  Let binary decision variable $x_{p,v}$ indicate whether position $p\in\{1,\dots,n\}$ takes value $v\in\{1,\dots,n\}$.  Let integer decision variable $y_p$ be the value in position $p$.  Let $Q=\{q\subset \{1,\dots,n\}:|q|=4\}$ be the $4$-subsets of positions. For $q=\{q_1,q_2,q_3,q_4\}\in Q$, let binary decision variable $z_q$ indicate whether the values taken by the four positions in $q$ match the pattern 4213.  The problem is to maximize $\sum_{q\in Q} z_q$ subject to linear constraints
\begin{align}
\sum_{p=1}^n x_{p,v} &= 1 &&\text{for $v\in\{1,\dots,n\}$} \tag1 \\
\sum_{v=1}^n x_{p,v} &= 1 &&\text{for $p\in\{1,\dots,n\}$} \tag2 \\
\sum_{v=1}^n v\cdot x_{p,v} &= y_p &&\text{for $p\in\{1,\dots,n\}$} \tag3 \\
y_{q_3} + 1 - y_{q_2} &\le n(1-z_q) &&\text{for $q\in Q$} \tag4 \\
y_{q_2} + 1 - y_{q_4} &\le n(1-z_q) &&\text{for $q\in Q$} \tag5 \\
y_{q_4} + 1 - y_{q_1} &\le n(1-z_q) &&\text{for $q\in Q$} \tag6 \\
\end{align}
Constraint $(1)$ assigns each value to exactly one position.
Constraint $(2)$ assigns each position to exactly one value.
Constraint $(3)$ enforces $x_{p,v}=1 \iff y_p = v$.
Constraints $(4)$, $(5)$, and $(6)$ enforce $z_q=1 \implies y_{q_3} < y_{q_2} < y_{q_4} < y_{q_1}$, as required by pattern 4213.

I confirmed your values of $a(n)$, and the next three are $a(10)=62$, $a(11)=96$, and $a(12)=138$.

The linear programming relaxation appears to yield upper bound $\binom{n}{4}(1-1/n)$ for $n \ge 5$.


问:
Simplifying $a(a-2) = b(b+2)$
I reduced a number theory problem to finding all ordered pairs $(a,b)$ that satisfy the equation $a(a-2) = b(b+2)$ in a certain range. After thinking about this for a while, I figured that either $a = b + 2$, $a = -b$, $a = b = 0$ or $a = 2 \text{ and } b = -2$. It is easy to prove that these values for $a$ and $b$ will all satisfy the equation, but how would I solve this equation had I not come up with these solutions? And how do I know I did not miss one? In short, can this equation be solved (more) rigorously?

Edit:
Silly of me I did not recognize that my last two solutions were already covered by my first two.

Title: Simplifying $a(a-2) = b(b+2)$

Body:
Denote $c=b+2$. Then
$(a-1)^2=(c-1)^2$ etc.


答:
Summary: you were only off by a factor of four. You would be correct if you replaced $x$ with $4x$.

The EGF for choosing the digits is $4x^1$, since there can only be one digit, and there are $4$ choices for it.

The EGF for choosing the letters is $\sum_{n\ge 0} 3^nx^n/n!=e^{3x}$, since there can be any number $n$ of letters, and when there are $n$ letters, there are $3^n$ ways to choose the letters.

Therefore, the EGF for choosing the entire string (digits and numbers) is the product of these two EGF's.
$$
(4x)\cdot (e^{3x})
$$
To finish, you then need to extract the coefficient of $x^n$ and multiply by $n!$.

Title: Exponential generating function in the case of a singular symbol

Body:
Suppose I have the following set {p, q, s, 0, 1, 2, 3} where we have $l_n$ is the number of strings with the symbols in this set of length $n$. However, there is exactly one number in each string that is a digit.

To find the generating function for this I started off by treating the alphabets as a ternary string and the digits 1 unit. Essentially we have our exponential generating function $F(x) = x\sum _{n=0}^{\infty }\:3^nx^n\frac{1}{n!}$. But I feel like my approach so far is wrong (it seems too simple) and this for the case where we only have 1 digit. I don't really know how to factor in the other 3 digits.


问:
How does $\epsilon$-$\delta$ explain the idea of 'approach'?
If $\lim_{x\to x_0}f(x)=L$

I Informally the idea of limit is that as $x$ approaches $x_0$, the value that $f(x)$ approaches is its limit at $x_0$.

II The $\epsilon$-$\delta$ definition states that- $\forall \epsilon$, $\exists \delta$, such that:

$0<|x-x_0|<\delta\Rightarrow|f(x)-L|<\epsilon$. For any value of $\epsilon$ if I can provide a $\delta$, then $L$ is the limit of $f(x)$ at $x_0$.

In other words for any distance from $L$, that has been provided, I have to find a distance, from $x_0$, such that for all the values of $x$ closer to $x_0$, their respective $f(x)$s are closer to $L$.

In many of the calculus related videos that I watch, the informal definition (The approach definition) is used to prove theorems, and other things. Right now I'm only dealing with single variable calculus, so I can imagine for a 2D graph how values of $f(x)$ can approach $L$, as $x$ approaches $x_0$. But what about in a multi-variable setting.

So how does the $\epsilon$-$\delta$ definition explain the 'approach' idea, and how is the 'approach' idea of limit always true?

I've also read a statement that $\epsilon$-$\delta$ does not actually define the limit, but is instead just a way to prove a limit is true or false?

This is one statement in a proof I saw: If $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=f'(x)$, assuming $\frac{f(x+h)-f(x)}{h} = f'(x) + \sigma(h)$, where $\sigma(h)$ is a 'junk term' that approaches $0$ as $h$ approaches $0$. The proof was based on this assumption.

Mathematically when I apply limit on both sides I get $\lim_{h \to 0} \sigma(h)=0$. But I'm still skeptical if this (assuming function equal to limit + 'junk') can be done for any case.

答:
The way I look at it is this: there are two sort of Mathematical arguments you can make: informal arguments, where you don't care about sticking very close to definitions and instead make intuitive arguments - and formal arguments, where we care a lot about making sure we stick to definitions very carefully.

The "approach" idea of a limit you speak of is informal. It's not a definition. It's simply trying to show you what the essence of a limit is.

The ${\epsilon-\delta}$ definition is just that - a definition. This is rigorous. This is what you need to use in formal Mathematics to prove limits.

Trying to prove the two above ideas are the same doesn't really make sense. ${\epsilon-\delta}$ is a definition trying to formally define what is meant by a limit, and that "approach" idea is just giving you a taste of what the definition says.


问:
Why is - $\frac{8}{17}$ taken instead of $\frac{8}{17}$ when trying to find the $cos x$ from $\tan x = \frac{15}{8}$ and $\in [\pi, \dfrac{3\pi}{2}]$?
$\sec^2 (x) =  \tan^2 (x) + 1$

$\sec^2 (x) = \left(\dfrac{15}{8}\right)^2 + 1$

$\sec (x) = \pm \sqrt{\dfrac{289}{64}}$

$\sec (x) = - \dfrac{17}{8}$ because $x \in \left[\pi, \dfrac{3\pi}{2}\right]$ <--- This is where I don't understand why we take the negative.

答:
Because the primary trigonometric ratios ($\sin, \cos, \tan$) are only all non-negative in the first quadrant. In the other three quadrants, exactly one is non-negative.

Your angle is in the third quadrant, where only tangent is non-negative. You expect to get a negative cosine here. The secant is the reciprocal of the cosine, so it shares the same sign. That's why you take the negative value.

A useful memory aid here.


答:
Well, nothing is actually "moving".

But as you take values of $x$es that are "very near" $a$ then resulting values of $f(x)$es will in response be "very near" $L$.  And if you restrict or range of $x$es by much narrower and restrict you $x$es to be closer to $a$ that will result in the range of $f(x)$ values to be closer to $L$.

So $\lim_{x\to a} f(x) = L$ means for whatever target range we desire to hone in toward $f(x)\to L$ we'd like... that is to say if we want all of our $f(x)$ outputs to all be within $L' < f(x) < L''$ for values of $L'$ and $L''$ as close to $L$ as we practically like.  We can find cooresponding $a', a''$ so that whenever we choose $a' < x < a''$ then the result will be that $f(x)$ will be $L' < f(x) < L''$.

So how do we express that analytically?

Our target is that we want $L', L''$ to arbitrarily as close as a value of $\epsilon$ within $L$.  That is for example, if we want $L',L''$ to be within one-hundred billionth of a unit of $L$ that is $\epsilon = $ one-hundredth billion of a unit and our goal is that the $x$ we pick (that will be somewhere near but not equal to $a$) will result in $L - \epsilon = L -\frac 1{100000000000} < f(x) < L + \frac 1{100000000000} = L + \epsilon$.

Or in other word $|f(x) - L| < \epsilon$.

That's our target goal.  And we want to achieve that be picking $x$ close to $a$.

That is to say.... If $\lim_{x\to a} f(x) = L$ that says we can, by restricting our choices of $x$ to be very very very close to $a$ have the result that $|f(x)-L| < \epsilon$ and no matter how small we choose $\epsilon$ we can always reach this goal be pick $x$ closer and closer to $a$.

How close?

Well for every target distance of $\epsilon$ that we want $f(x)$ to be within $L$. there is $\delta$ that will be a sufficient requirement so that if $a' = a-\delta$ and $a'' = a+ \delta$ where $\delta$ that by choosing $x$ so that $a' = a-\delta < x <a+\delta < a''$ that guarantees that $L-\epsilon < f(x) < L + \epsilon$.

Now what if we can't guarantee that result?  Well... then we failed in our attempt of assuring we can hone in on $f(x)\to L$ by choosing $x\to a$.  If we fail we fail and we can't  do it. But to say that can force $f(x)$ to be close to $L$ by forcing $x$ to be close to $a$ means precisely:

For any target goal of getting $f(x)$ within a small $\epsilon$ range of $L$, we can find a small $\delta$ range (specifically with that $\epsilon$ in mind) so that choosing any $x$ within that $\delta$ near $a$, assures that the responding $f(x)$ will be within $\epsilon$ of $L$.

That is what a $\delta$-$\epsilon$ construction means.

......

Maybe a practical example:

Suppose want to show that $\lim_{x\to 2} x^2 = 4$.

So we want to show we can get $x^2$ close to $4$ but picking $x$ close to $2$.

How close?

Well we can get $x^2$ within $5$ units of $4$,  That is $-1=4-5 < x^2 < 4+5 = 9$ by choosing $x$ within $1$ unit of $2$.  That is if $1=2-1 < x < 2+1 = 3$ then $1^2 < x^2 < 9$ and we are within the target range.

Well, that's not very close.  Let's try to get closer.  Let try to get $x^2$ withing $1$ unit.  That is we want $3= 4- 1 < x^2 < 4+ 1$. And we can do that be taking $x$ closer to $2$.

How close?  Well if $x$ is within $\frac 14$ unit of $2$ then $1\frac 45 < x < 2\frac 15$ and $3 < 3.24=(1\frac 45)^2 < x^2 <(2\frac 15)^2 = 4.84 < 5$ and we are within range.

Still can we get closer?

Let's see if we can get $x^2$ within an $\epsilon = \frac 1{100,000,000,000}$ of $4$  That is can we get $3.99999999999 < x^2 < 4.000000000001$? by get $x$ close to $2$?

How close.

Well, if we choose $\delta = \frac 2{1,000,000,000,000}$ then  $2- \delta = 1.999999999998 < x < 2.000000000002=2+\delta$ will result in $3.999999999992000000000004 < x^2 < 4.000000000008000000000004$ and that is within our desired range.

So, so far, for every target goal of getting $x^2$ within some small $\epsilon$ of $4$ we've succeeding but taking $x$ within some some $\delta$ of $2$.

Can we do that for any $\epsilon$?

For any $\epsilon$ will we always be able to find a $\delta$ so that taking $x$ within $\delta$ of $2$ will guarantee that $x^2$ will be within $\epsilon$ of $4$?

Well.... gosh.... I sure hope so......  I mean.... I'm putting my reputation out there by claiming as $x\to 2$ we have $x^2 \to 4$.  I'd better be able to back that up.... so.... gosh, I'd better be able to claim that.... (guess I'm not going to sleep very well tonight if I can't....)

Title: How does $\epsilon$-$\delta$ explain the idea of 'approach'?

Body:
If $\lim_{x\to x_0}f(x)=L$

I Informally the idea of limit is that as $x$ approaches $x_0$, the value that $f(x)$ approaches is its limit at $x_0$.

II The $\epsilon$-$\delta$ definition states that- $\forall \epsilon$, $\exists \delta$, such that:

$0<|x-x_0|<\delta\Rightarrow|f(x)-L|<\epsilon$. For any value of $\epsilon$ if I can provide a $\delta$, then $L$ is the limit of $f(x)$ at $x_0$.

In other words for any distance from $L$, that has been provided, I have to find a distance, from $x_0$, such that for all the values of $x$ closer to $x_0$, their respective $f(x)$s are closer to $L$.

In many of the calculus related videos that I watch, the informal definition (The approach definition) is used to prove theorems, and other things. Right now I'm only dealing with single variable calculus, so I can imagine for a 2D graph how values of $f(x)$ can approach $L$, as $x$ approaches $x_0$. But what about in a multi-variable setting.

So how does the $\epsilon$-$\delta$ definition explain the 'approach' idea, and how is the 'approach' idea of limit always true?

I've also read a statement that $\epsilon$-$\delta$ does not actually define the limit, but is instead just a way to prove a limit is true or false?

This is one statement in a proof I saw: If $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=f'(x)$, assuming $\frac{f(x+h)-f(x)}{h} = f'(x) + \sigma(h)$, where $\sigma(h)$ is a 'junk term' that approaches $0$ as $h$ approaches $0$. The proof was based on this assumption.

Mathematically when I apply limit on both sides I get $\lim_{h \to 0} \sigma(h)=0$. But I'm still skeptical if this (assuming function equal to limit + 'junk') can be done for any case.


答:
Add $1$ to both sides to get
$$a^2 - 2a + 1 = b^2 + 2b + 1$$
i.e.
$$(a-1)^2 = (b+1)^2.$$

Hence $a-1 = b+1$ or $a-1 = -b-1$ and these are all solutions.

Title: Simplifying $a(a-2) = b(b+2)$

Body:
I reduced a number theory problem to finding all ordered pairs $(a,b)$ that satisfy the equation $a(a-2) = b(b+2)$ in a certain range. After thinking about this for a while, I figured that either $a = b + 2$, $a = -b$, $a = b = 0$ or $a = 2 \text{ and } b = -2$. It is easy to prove that these values for $a$ and $b$ will all satisfy the equation, but how would I solve this equation had I not come up with these solutions? And how do I know I did not miss one? In short, can this equation be solved (more) rigorously?

Edit:
Silly of me I did not recognize that my last two solutions were already covered by my first two.


问:
Question about the fundamental group of a space
From my understanding the fundamental group is the equivalence class of all homotopic loops. So the group consists of all homotopic closed loops, but if every loop is homotopic in the group, wouldn't that make the group trivial? Or am I missing something here?
Please explain in detail if you can what I'm getting wrong. Thank you in advance</p>

答: The fundamental group can be viewed as the quotient of the set of closed based loops by the based homotopy equivalence relation So, homotopic loops go to the same thing.


答: Consider the unit circle in the plane, centered at the origin.  A loop that goes around the circle once, is not equivalent to a loop that goes around the same circle twice, and none of these two are  equivalent to a loop that goes around three times.</p>

Title: Question about the fundamental group of a space

Body:
From my understanding the fundamental group is the equivalence class of all homotopic loops. So the group consists of all homotopic closed loops, but if every loop is homotopic in the group, wouldn't that make the group trivial? Or am I missing something here?
Please explain in detail if you can what I'm getting wrong. Thank you in advance</p>


答:
If all closed loops are homotopic to one another in your space, then the fundamental group of that space is trivial. The archetypal examples are the Euclidean spaces $\Bbb R^n$, where each closed loop is easily seen to be homotopic to, say, the constant loop at $0$. And since being homotopic is an equivalence relation, all loops are homotopic to one another. Thus we get $\pi_1(\Bbb R^n)\cong0$.

However, there are plenty of spaces where not all loops are homotopic. The circle $S^1$, for one (it may be easier to visualize if you look at an annulus rather than a circle; the result is the same). There any two loops that are homotopic must necessarily go the same number of times in the same direction around the circle. A bit more calculation shows that the converse holds as well, so we get $\pi_1(S^1)\cong \Bbb Z$, with the isomorphism given by the winding number.


答:
Proposition 3.1 in the paper On Packing Densities of Permutations  by Albert, Atkinson, Handley, Holton, and Stromquist proves that the packing density of the $1342$ pattern (the limit of $a(n)/\binom n4$) is between $0.19657$ and $\frac29$, with an explicit upper bound of $\frac{n^2(n+1)^2}{108}$ on $a(n)$. This is also an upper bound for the $4213$ pattern, which is symmetric to it.  Theorem 4 in Sliacan and Stromquist's Improving bounds on packing densities of 4-point permutations improves the bounds on the packing density to the interval $$[0.198836597,0.198837287]$$ almost but not quite solving the problem. It doesn't look like anyone's finished it off since then.

I'm going to keep my earlier answer, which shows an upper bound of about $0.77\binom n4$, below, just because what else am I going to do with it.

The number of $213$ patterns (like $132$ patterns) in a permutation of length $n$ is at most $\frac{2 \sqrt 3 - 3}{6} n^3$, and we can use that to bound the number of $4213$ patterns.

Let's divide the possible $4213$ patterns into two types:

Ones in which the first number comes from the first quarter of the permutation, and the rest come from the last three quarters. There are $\frac n4 \binom{3n/4}{3}$ substrings of this form, but we'll bound the number of them that form $4213$ patterns.

All others. There are $\binom n4 - \frac n4 \binom{3n/4}{3}$ substrings of this form, and to get an upper bound we'll just assume that all of them form $4213$ patterns.

For the first type, if the substring forms a $4213$ pattern, its last three numbers form a $213$ pattern, so there can be at most $\frac{2\sqrt3 - 3}{6} (\frac{3n}{4})^3$ ways to choose those last three numbers; $\frac{2\sqrt3 - 3}{6} \frac n4 (\frac{3n}{4})^3$ patterns overall.

So this gives us an upper bound of $\binom n4 - \frac n4 \binom{3n/4}{3} + \frac{2\sqrt3 - 3}{6} \frac n4 (\frac{3n}{4})^3$. Asymptotically, this is $(\frac{27\sqrt3 - 22}{32} + o(1)) \binom n4$, or about $0.77\binom n4$.

Title: Greatest number of occurrences of the pattern 4213 in a permutation.

Body:
Statistic St000750 in the FindStat database is a map $\operatorname{St000750} \colon S_n \rightarrow \mathbb N_{\geq 0}$ given by

The number of occurrences of the pattern $4213$ in a permutation.

This is the number of length-4 substrings such that the first number is largest, the second is third-largest, the third is smallest, and the fourth is second-largest. For example, the permutation $\pi = 563214$ has $\operatorname{St000750}(\pi) = 6$ occurrences of this pattern:
$$5324, 5314, 5214, 6324, 6314, \text{and}\ 6214.$$

I'm interested in computing$$
   a(n) = \max \{ \operatorname{St000750}(\pi) : \pi \in S_n\}
$$
By brute-forcing $0 \leq n \leq 9$, I get that

$a(0) = a(1) = a(2) = a(3) = 0$,

$a(4) = 1$,

$a(5) = 3$,

$a(6) = 6$,

$a(7) = 13$,

$a(8) = 24$, and

$a(9) = 40$.

Moreover, this sequence does not appear in the On-Line Encyclopedia of Integer Sequences.

However, an analogous sequence for the pattern $132$ is A061061, with a simple formula: $A061061(n) = \max\{A061061(k) + k\binom{n-k}{2} : 1 \leq k < n\}$.

Is there a way to compute this sequence which is better than brute force? If not, how can you construct a better upper bound than $a(n) \leq \binom{n}{4}$?


问:
Is the series $ 1-1+1-1+1+\cdots = -1+1-1+1-1+1-\cdots $?
We have the series

$$ S = 1-1+1-1+1-1+1 \cdots $$

If we manipulate S, we get that
$$ 1 - S = S $$
$$ 1 = 2S $$
$$ \frac{1}{2} = S $$

Also, if we re-order S, we get that

$$ S = (1-1)+(1-1)+(1-1)\cdots $$

$$ = -1+1-1+1-1+1-1+1\cdots $$

So $$ -S = S + 1  $$

Solving for S we get that

$$ S = -\frac{1}{2} $$

My question is, is the new S re-ordered considered the same series as S? When we can say that two series are the same in that context? Thanks.

答:
First of all, this series is not convergent, so there's no such thing as sum of the series $S$.

Even if it was a convergent series, like for example
$$ 1 - \frac12 + \frac13 - \frac14 + \dots $$
in general, if you rearrange the terms, it is possible to obtain a different sum. For example, if you rearrange them as follows:
$$ 1 + \frac13 -\frac12 + \frac15 +\frac17-\frac14 + \frac19+\frac{1}{11}-\frac16 + \dots $$
(two positive terms then one negative, repeat) you will get a different sum.

Only if the series $\sum_{n=1}^\infty a_n$ is absolutely convergent, that is, the series $\sum_{n=1}^\infty |a_n|$ is convergent, then it is guaranteed that rearranging the terms will not change the sum. So, for example in the series
$$ 1-\frac14 + \frac19-\frac{1}{16}+\frac{1}{25} + \dots$$
you can rearrange terms whatever way you want, and the sum will remain the same.

The theorem it states that is called the Riemann series theorem.


答:
You can easily check that the function

$$f(x)= x-1-\log x$$

has a minimum equal to $0$ at $x=1$ through derivative's test , so for all other points $f(x) >0$

Title: What is reasoning behind statement that $\log \frac{3}{2} < (\frac{3}{2} - 1)$?

Body:
In his paper, "Euler and the Partial Sums of the Prime Harmonic Series" (pdf), Paul Pollacks makes a small statement at the top of page 4:

$$ \log \frac{3}{2} < (\frac{3}{2} - 1) = \frac12 $$

I can see numerically this is true, but I can't see how he makes the inequality $\log \frac{3}{2} < (\frac{3}{2} - 1)$.

I have looked through references of power series approximations but can't see anything that matches.


答:
Short not quite an answer. We don't explain "approaches", we do without it, by carefully defining something equivalent we can reason with.

When you think of limits in terms of "approaching" you run into philosophical questions about time passing or numbers being infinitely close together.

Mathematicians have decided to replace  vague notions of "infinitely close" by requiring "infinitely many" inequalities. That's what you do when you argue that "for every $\epsilon$ ...".

Added after the question was edited:

It's routine algebra to show that these two statements say the same thing about a function $f$ and a number $L$:
$$
\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = L
$$
and
$$
\lim_{h \to 0} \left( \frac{f(x+h) - f(x)}{h} - L \right) = 0 .
$$
If you write $\sigma(h)$ for the expression in the large parentheses in the second equation then showing that $L$ is the derivative of $f$ at $x$ is just the same as showing $\sigma$ has limit $0$ as $h \to 0$. It's an algebraic trick that's sometimes useful.


答:
You want $\lim f(n)$. Take logs and simplify:
$$
\ln f(n) = n^2 \ln \left(\frac{n-x}{n+x}\right)
         = n^2 \ln \left(1 - \frac{2x}{n+x}\right)
$$
and use the Taylor series
$$
\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \ldots
$$
to get
$$
\ln f(n) = -n^2 \left( \frac{2x}{n+x}
                      + \frac{1}{2} \left(\frac{2x}{n+x}\right)^2\ldots \right)
         \to -\infty
$$
and so $f(n) \to 0$.</p>

Title: What is $\lim\limits_{n→∞}(\frac{n-x}{n+x})^{n^2}$?

Body:
What is $$\lim_{n\rightarrow\infty}\left(\frac{n-x}{n+x}\right)^{n^2},$$ where $x$ is a real number. Mathematica tells me the limit is $0$ when I put an exact value for $x$ in (Mathematica is inconclusive if I don't substitute for $x$), but using $f(n)=(n-x)^{n^2}$ and $g(n)=(n+x)^{n^2}$, then $$\lim_{n\rightarrow\infty}f(n)=\lim_{n\rightarrow\infty}g(n)=\infty,$$ and by L'Hopital's rule we have
$$\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} = \lim_{n\rightarrow\infty}\frac{(s-z)^{-1+s^2} (s+z)^{1-s^2} (s+2 (s-z) \text{Log}[s-z])}{s+2 (s+z) \text{Log}[s+z]} = 1.$$

I'm not sure which to believe - Mathematica or L'Hopital.


问:
Natural deduction - help solving
I have this issue in natural deduction:

B -> (A & ~A) ├ ~B

I am having a really hard time understanding the rules in natural deduction - but this is what I got so far - can someone help me to the end?

B / premis

(A&~A) /premis

B-> ^ /1, 2  ^E

B /1

^ /3->E

~B /4,5 ->I

Thank you!

答:
Natural Deduction's rules of inference revolve around 'introducing' or 'eliminating' logical connectives.   So that is where we start.$\def\fitch#1#2{~~~~~\begin{array}{|l}#1\\\hline#2\end{array}}$

You are given a premise $B\to(A\land\lnot A)$ and wish to infer $\lnot B$.   Well there is a conditional and conjunction to eliminate in the premise, and a negation to introduce in the conclusion. Clearly these rules need to be understood (at least).

$$\dfrac{\phi\to\chi\quad\phi}{\chi}{~\small{\to}\mathsf e}\qquad\dfrac{\gamma\wedge\beta}{\gamma}{~\small{\land}\mathsf e}\qquad\dfrac{\gamma\wedge\beta}{\beta}{~\small{\land}\mathsf e}\qquad\dfrac{[\psi]\\~\vdots\\\bot}{\lnot\psi~}
{~\small\lnot\mathsf i}$$

So... to successfully apply negation introduction, an assumption must be made, under which a contradiction must be derived.   For this proof that assumption must be $B$, and this and the premise then allows conditional elimination to be applied, thus deriving $A\land\lnot A$.   Then from this... well...

$$\fitch{~~1.~~B\to(A\land\lnot A)\hspace{7ex}\textsf{Premise}}{\fitch{~~2.~~B\hspace{16ex}\textsf{Assumption}}{~~3.~~A\land\lnot A\hspace{10ex}{\to}\mathsf e~1,2\\~~4.~~A\hspace{16ex}{\land}\mathsf e~3\\~~5.~~\lnot A\hspace{14.5ex}{\land}\mathsf e~3\\~~~~\vdots\\~~\mathrm m.~~\bot}\\~~\mathrm n.~~\lnot B\hspace{17ex}\lnot\mathsf i~2{-}\mathrm m}$$

So all that is left is to derive that contradiction.

Title: Natural deduction - help solving


答:
The word "approach" does not serve well in a mathematical definition. I would argue that it is misleading to say that if $\lim_{x\to x_0} f(x) = L$ then $f(x)$ approaches $L$ as $x$ approaches $x_0.$

Let's define a function $f$ by
$$
f(x) = \begin{cases}
x \sin \frac1x & x \neq 0, \\
0 & x = 0.
\end{cases}
$$

You should be able to confirm by a $\delta$-$\epsilon$ proof that
$$
\lim_{x\to 0} f(x) = 0.
$$
For any $\epsilon,$ one choice of a suitable $\delta$ is to set $\delta = \epsilon.$
Just remember that $-1 \leq \sin\frac1x \leq 1$ and therefore
$\lvert x\sin\frac1x \rvert \leq \lvert x \rvert.$

But is is really OK to say that $f(x)$ approaches $0$ as $x$ approaches $0$?
The way I see it, if you decrease $x$ continously down toward $0,$
the value of $f(x)$ is sometimes approaching $0,$ but it infinitely often actually reaches $0$ and then starts going away from $0.$
It just keeps approaching, going away, approaching, going away, back and forth infinitely many times. It's not even possible to say whether $f(x)$ finally approaches $0$ from above or below, because there is no final approach.
There are just infinitely many oscillations of the function that inexorably get squeezed down to the limit $0$.

So I would argue that the purpose of $\delta$-$\epsilon$ is not to explain the "approach" idea, because the "approach" idea of the limit is not always true.

(That is, it is not always true unless you have a very technical definition of the word "approach" that means "satisfies the $\delta$-$\epsilon$ definition";
but that seems silly to me. We have a definition of "limit" expressed by $\delta$-$\epsilon$ without using the word "approach", so why introduce another word that we have to interpret with a special meaning that contradicts its plain English meaning?)


问:
Proof of Axiom of choice by well ordering
I am reading Folland's real analysis. In Prologue, proof of Axiom of choice using well ordering is as follows:

Let $X=\cup_{\alpha \in A}X_{\alpha}$. Pick a well ordering on $X$ and, for $\alpha \in A$, let $f(\alpha)$ be the minimal element of $X_{\alpha}$. Then $f\in \prod_{\alpha \in A}X_{\alpha}$.

My question is that why it is necessary to pick the minimal element of $X_{\alpha}$ so that we would need well ordering principle. Why can't we just choose an arbitrary element of $X_{\alpha}$ while we know that $X_{\alpha}$ is non-empty so it has at least one element.

I think I have not understood the axiom of choice well.

答: You can choose an arbitrary element of one $X_\alpha$ without using the axiom of choose. For that matter, you can choose one arbitrary element from any finite number of the sets $X_\alpha$ without using the axiom of choice. But to choose one member from each of an infinite family of sets you need either the axiom of choice or a specific algorithm for choosing them. Here the well-ordering on $X$ provides that algorithm: it allows to choose from $X_\alpha$ very specifically the minimal element of $X_\alpha$ with respect to the well-ordering of $X$. This is no longer an arbitrary choice: it is completely determined by the well-ordering.


问:
Given the octagon $ABCDEFGH$, with $\angle GHA=\angle ABC=\angle CDE=\angle EFG=120^o$, $AE\perp GC$ and $AO=OE=OG=OC$, find the area of the octagon
Given the octagon $ABCDEFGH$,  with $\angle GHA=\angle ABC=\angle CDE=\angle EFG=120^o$, $AE\perp GC$ and $AO=OE=OG=OC$ find the area of the octagon.

We have that $\triangle AGH, \triangle GFE, \triangle DEC, \triangle CBA$ isosceles. $ACEG$ is a square with side $2$. $\angle HGA=\angle HAG=\angle BAC=\angle BCA=\angle DCE=\angle DEC=\angle GEF=\angle EGF$.

Also we have that $HBDF$ is a square. Basically all you need to work out is $GH$, but I can't work it out. Could you please explain to me how to solve the question?

答:
With this symmetrical figure, focus on  $\triangle AOK$ , $\triangle ABK$,$\triangle AOB$.

For the simpliticy of calulation, first let us assume  BK=1 $\to$AB=2,AK=OK=$\sqrt 3$:

\begin{align} \frac{S_{ABCDEFGH}}{S_{ACEG}} =\frac{S_{\triangle AOB}}{S_{\triangle AOK}}=\frac{S_{\triangle AOK}-S_{\triangle ABK}}{S_{\triangle AOK}}=1-\frac{S_{\triangle ABK}}{S_{\triangle AOK}}=1-\frac{BK}{OK}=1-\frac{1}{\sqrt{3}}\end{align}

Now  scale back with $S_{ACEG}=4$, $S_{ABCDEFGH}$=
4$(1-\frac{1}{\sqrt{3}})$


问:
Find the n-th base 2 palindrome
So I want to compute the n-th base 2 palindrome, where n is a number less than or equal to 5000, and we can pick it.

So we know that in base 2 only odd numbers can be palindromes. Okay so that saves us some computation power, but still not enough.

My first thought was, to compute all the possible odd numbers in base 2 and check for every one of them if it is a palindrome. Then I would count in order how many palindromes there are, and when we reach the target $n$ we can print out the n-th palindrome. I think this is not enough, or my method of finding such numbers is bad. My question is, is there any other property of base 2 palindromes, besides that they are odd, that could help my search ? What kind of technique would you pick for search instead of a simple iteration (that's whagt I have done).
(I am coding in python.)

答: Yet another idea, use the fact that odd length palindromes need it's middle digit repeated an odd number of times total. Whereas, even length ones need all digits an even number of times. Also, half your search space: take the end two bits off, and bitxor with the Mersenne number($2^n-1$ aka base 2 repunit) of the length you were at, it's also a palindrome of length wanted.

Title: Find the n-th base 2 palindrome


问:
A smooth map $s\colon U\to E$ which is not a section where $\pi\colon E\to M$ is a vector bundle.
I have a silly question about an example of a smooth map $s\colon U\to E$ which is not a section on $U$ where  $\pi\colon E\to M$ is a vector bundle and $U$ is an open subspace of $M$. According to the definition, $s$ is going to be a section if it has a left inverse i.e.
$$\pi\circ s=id_U.$$
So, if I take $M=U=\mathbb{R}$, $E=\mathbb{R}^2$, and
$$s\colon\mathbb{R}\to \mathbb{R}^2\text{ with }s(t)=(\cos(t),\sin(t)),$$
then is it going to be an example of a smooth map that is not a section? What are the other examples? In other words, I want to understand the nature/motivation for sections.

答: A possible motivating example: take a smooth manifold $M$ and consider, for each $x\in M$ the tangent space $T_xM.$ Then, one defines the tangent bundle,  $TM$ so that $TM$ is itself a smooth manifold. It is also a vector bundle. Its elements are pairs $(x,v)$ where $x\in M$ and $v\in T_xM.$ The intuition for this is that we gather together the individual vector spaces $T_xM$ and combine them into a differentiable manifold. Then, the projection $\pi:TM\to M$ sends $(x,v)$ to $x$, as expected, and the existence of a section $s$ ensures that when we send a particular $x\in M$ into $TM$, it ends up in $T_xM$ (considered as a subspace of $TM).$ That is, we want to exclude maps that send $x$ to elements of the form $(y,v): y\neq x.$  This requirement is built into the definition of section for if a map $s$ sends $x$ to such a $(y,v)$ then $\pi\circ s(x)=y\neq x.$


问:
What is reasoning behind statement that $\log \frac{3}{2} < (\frac{3}{2} - 1)$?
In his paper, "Euler and the Partial Sums of the Prime Harmonic Series" (pdf), Paul Pollacks makes a small statement at the top of page 4:

$$ \log \frac{3}{2} < (\frac{3}{2} - 1) = \frac12 $$

I can see numerically this is true, but I can't see how he makes the inequality $\log \frac{3}{2} < (\frac{3}{2} - 1)$.

I have looked through references of power series approximations but can't see anything that matches.

答:
If $x = 1$ then $0 =\ln x = x-1$.

Would you accept that as $x, x-1$ and $\ln x$ get larger than $x$ and $x-1$ getter larger "faster" than $x$ does?

We can argue that with derivatives.

$\frac {dx}{dx} = \frac {d(x-1)}{dx} = 1$  but for $x > 1$ then $\frac {d\ln x}{dx} = \frac 1x < 1$.

So $\ln x$ increases more "slowly" than $x-1$.

So if they start on the racetrack at the same value at $x = 1$ (so $x-1 =0$ and $\ln x = 0$) but $x-1$ is always faster than $\ln x$ which just gets slower and slower, then $x-1$ immediate surpasses it.

Of course.... I have to take my roommate to the hospital as I just knocked his eye out but waving my hands about so wildly.)


问:
Parallel system when the time to repair of the components becomes very large.
A component has a mean to to failure $U$ and a mean time to repair $D$. Two of these components are combined into a parallel system. In other words, we need just one of the components to be working for the system to be working. A natural question is what the mean time to failure and mean time to repair for the system become. On page 614 of "Introduction to probability models" by Sheldon Ross (9th ed), this exact scenario is described (as a consequence of equations 9.32 and 9.33 we get the MTBF of the system ($\bar{U}$)), which is also calculated on that page for a parallel system:

$$\bar{U} = \frac{1-\left(\frac{D}{U+D}\right)^2}{\left(\frac{D}{U+D}\right)^2} \frac{D}{2}$$

Now, what happens when the mean time to repair of the components, $D$ becomes very large? When I calculate the limit of the above expression as $D \to \infty$, I surprisingly get $\bar{U} = U$.

This doesn't make sense since if we consider the time to repair of the components as exponential, we should see one failure of the system at the max of the times to repair of the two components. And the average of this time should be double the average of the time to failure of the components, not the same. What am I missing?

答:
It should not be surprising because most of the time the system is running one of the components is broken.  In that case the MTBF of the system is the same as the MTBF of one component.  In fact, most of the time the system is down completely.  You get one component repaired and it runs until it fails.  Then both are broken and the system is down.

As an example, imagine MTBF of a component is 1 hour and MTTR is 1000 hours.  You start with both working.  After about a half hour one of the two components fails.  After about another hour the other fails and the system is down.  You are down for about 1000 hours until one component is repaired.  Unless the other gets repaired in the next hour, you will only run for about an hour before the newly repaired one fails.  On average you are up for 2 hours out of every 1000, about 1 hour at a time.

Checking limiting cases like this is a good check of formulas.  If you made an error in the derivation, it often shows up in one of the limiting cases.  You can often find answers to the limiting cases without using your derivation, as I did here.


问:
Showing correct proof for finding equivalence classes for relation defined on $\mathbb{Z}$ by $aRb$ if $6 \mid (a^2 - b^2)$
For relation defined on $\mathbb Z$ by $aRb$ if $6 \mid (a^2 - b^2)$

I know that the equivalence classes are {0,1,2,3,4,5} but I'm unsure if my method of proof is valid. For the first part of the question, I proved it was an equivalence relation on $\mathbb{Z}$, and now I need to find and prove the equivalence classes.

My thoughts so far:

Let there be $a,r \in \mathbb{Z}$, so that $6 \mid (a^2-r^2)$.

I know by Euclidean Division that  $6 \mid (a^2-r^2) \equiv a^2=6q+r^2$ , $q \in \mathbb{Z}$

Then I also know that $0 \leq r^2 <6$

If $aRr$, the I know $[a]=[r]$ and also $a\in[r]$***

Then by transitivity, since I proved (in my homework) $bRa \land aRr \Longrightarrow bRr$ and therefore, $[b]=[a]=[r]$ and $a,b \in [r]$

***Can I state this? Or am I missing something key concept before I can say this?

Any feedback on my train of thoughts is appreciate. Pls let me know if you see any holes of logic that I need to fill.

答:
you can use $\pmod 2$ and $\pmod 3$ to describe things, but the result does not fit that well with $\pmod 6$

We always have $x^2 \equiv x \pmod 2,$ so $a^2 \equiv b^2 \pmod 2$ if and only if $a \equiv b \pmod 2.$   For this calculation, we are left with classses even and odd.

Next, squares are $0 \pmod 3$ or $1 \pmod 3.$  As with 2, what we can tell is whether a number is divisible by $3$ or not divisible by $3$.

The equivalence class could be summarized as YY, YN, NY, NN, where the first yes/no is divisbility by $2,$ the second letter is divisibility by $3$

two numbers are equivalent if and only if they have the same pair of letters


问:
Rudin and exponential function construction
I have been dusting the cobwebs off of my math knowledge and in working through Rudin's PMA, I have come to a rather small detail that I have been unable to supply an explanation for. In constructing the exponential and logarithmic functions in chapter 8, Rudin eventually writes that if $x>0$, $n$ is an integer, and $m$ is a positive integer, we have

$$ x^{\frac{1}{m}}=E\bigg(\frac{1}{m}L(x)\bigg) $$

I find myself unable to confidently supply the missing details using anything in the section on exponents and logarithms up to that point. Is it simply a consequence of the fact that Rudin has already constructed $E(p)=e^p$ for rational $p$ and defined $E(x)=e^x=\mathrm{sup}\ e^p$, where $x$ is real, and $p<x$, in which case I would suppose that $$E\bigg(\frac{1}{m}L(x)\bigg)=E\bigg(L(x)\frac{1}{m}\bigg)=(e^{ln(x)})^\frac{1}{m}=x^\frac{1}{m}$$ I feel as though I am overlooking something.

答:
Note that $E\left(\frac{1}{m}L(x)\right) > 0$ and $$\left(E\left(\frac{1}{m}L(x)\right)\right)^m = E\left(m\cdot \frac{1}{m}L(x)\right) = E(L(x)) = x$$
where the first equality follows from (29) with $z_1=\ldots=z_n = \frac{1}{m}L(x)$ and $n=m$.

It follows from the theorem-definition 1.21 that $$E\left(\frac{1}{m}L(x)\right) = x^{1/m}$$


问:
How to you evaluate ${\int \frac{1}{z\left(z-1\right)\left(z-3\right)}dz}$ using the Residue Theorem

How to you evaluate $${\int\limits_{|z|=2} \frac{1}{z\left(z-1\right)\left(z-3\right)}dz}$$ using the Residue Theorem.

This problem has been driving me crazy, I am supposed to apply the Residue formula for double poles, but I can't figure out how. I would appreciate help.

答:
You don't have a double pole.   You have simple poles at $0,1$ and $3$.  The residues are $1/3,-1/2$ and $1/6$, respectively.

If the contour encloses all three poles, you get $2\pi i (1/3-1/2+1/6)=0$.


答:
By Cauchy intgeral theorem:
$$\mathcal{I}:=\oint_{\lvert z\rvert =2}\frac{1}{z\left(z-1\right)\left(z-3\right)}dz$$ $$=2\pi i\left(\operatorname{Res}_{z=0}\left[\frac{1}{z\left(z-1\right)\left(z-3\right)}\right]+\operatorname{Res}_{z=1}\left[\frac{1}{z\left(z-1\right)\left(z-3\right)}\right]\right).$$

Evaluating these residues yields
$$\mathcal{I}=2\pi i(1/3-1/2)=-\frac{\pi i}{3}$$

Title: How to you evaluate ${\int \frac{1}{z\left(z-1\right)\left(z-3\right)}dz}$ using the Residue Theorem

Body:
How to you evaluate $${\int\limits_{|z|=2} \frac{1}{z\left(z-1\right)\left(z-3\right)}dz}$$ using the Residue Theorem.

This problem has been driving me crazy, I am supposed to apply the Residue formula for double poles, but I can't figure out how. I would appreciate help.


问:
Well-definedness of zeros and poles of rational functions in algebraic geometry
My question is about the well-definedness of zeros and poles of rational functions following Vakil's Rising Sea (November 18, 2017 Edition).

On a regular codimension $1$ point $p$, we define the number of zeros resp. poles of an element $f \in \operatorname{Quot}(\mathcal{O}_{X,p})$ via the discrete valuation given by the DVR $\mathcal{O}_{X,p}$ (p. 353). In algebraic geometry, one then usually considers some rational function $s$ and tries to analyze the number of zeros and poles of $s$. I'm confused in how this makes sense, i.e. why is $s$ naturally an element in $\mathcal{O}_{X,p}$?

Let me also recall quickly the definition of rational functions in Vakil's book. On p. 171 we define a rational function on a scheme $X$ as an equivalence class of pairs $(U, s \in \Gamma(U, \mathcal{O}_X))$ where $\operatorname{Ass}(X) \subseteq U$. Two pairs $(U,s)$ and $(U',s')$ are said to be equivalent if $s|_{U \cap U'} = s'|_{U \cap U'}$. I'm guessing that my confusion stems from an insufficient understanding of associated points.

So to summarize: Given a rational function $s \in \Gamma(U,\mathcal{O}_X)$ with $\operatorname{Ass}(X) \subseteq U$, why is $s$ naturally an element of $\operatorname{Quot}(\mathcal{O}_{X,p})$ for all regular codimension $1$ points $p$?

答:
If $p\in X$ is a regular codimension one point, then it is in the closure of exactly one associated point $\eta$ of $X$ (ref), and in fact that point is the generic point of the unique irreducible component it is on. As the reduced locus is open in a noetherian scheme (ref), we can pick an integral affine open neighborhood $\operatorname{Spec} A\subset X$ of $p$. Then $\mathcal{O}_{X,p}$ is a localization of $A$, and $\mathcal{O}_{X,\eta}$ is the quotient field of $A$, so $\operatorname{Frac}(\mathcal{O}_{X,p})=\mathcal{O}_{X,\eta}$.

Now suppose $U\subset X$ contains all the associated points of $X$. Let $p\in X$ be a regular codimension one point and $\eta$ the unique associated point of $X$ which is a generalization of $p$. Then $s\in\mathcal{O}_X(U)$ maps to $\mathcal{O}_{X,\eta}$ by restriction, and as $\mathcal{O}_{X,\eta}=\operatorname{Frac}(\mathcal{O}_{X,p})$ we've explained what you were interested in.

Let me note a couple of small imprecisions in your question, just in case these were contributing to some of your troubles.

... I'm confused in how this makes sense, i.e. why is $s$ naturally an element in $\mathcal{O}_{X,p}$?

You stated this incorrectly here, but correctly after the line break: $s$ is naturally an element of $\operatorname{Frac}(\mathcal{O}_{X,p})$, not $\mathcal{O}_{X,p}$.

... Given a rational function $s \in \Gamma(U,\mathcal{O}_X)$ ...

It would be more precise to say "a rational function represented by $s\in\Gamma(U,\mathcal{O}_X)$".


问:
Polynomial with small hessian determinant
Let $P:\mathbb R^2\to \mathbb R$ be a polynomial with no constant or linear terms, and with all other coefficients bounded by 1 in absolute value. Suppose its Hessian determinant is equal to $\epsilon Q(x,y)$ for some tiny $\epsilon$ and some other polynomial $Q$ with all coefficients bounded by 1 in absolute value. Then is it true that for some unit vector $u$, we have
$$
(\nabla\cdot u) P(x,y)=\epsilon^{1/2} C_d R(x,y)
$$
for some constant $C_d$ depending on the degree $d$ of $P$ only, and some polynomial $R$ with all coefficients bounded above by 1 in absolute value?

Edited: this question occurs in my research, where I need to consider the set where the Hessian determinant of a polynomial $P$ is small. There is one particular case which bothers me, namely, when $\det(D^2P)$ is of the order $O(\epsilon)$ in the entire $[0,1]^2$ but some coefficients of $P$ is large, say when
$$
P(x,y)=x^2+\epsilon^{1/2}y^2.
$$
In this case we see that $\partial_y P(x,y)$ is of the order $\epsilon^{1/2}$. However, as in the case
$$
P(x,y)=x^2+(2+\epsilon^{1/2})xy+y^2,
$$
all coefficients of $P$ are large, but taking $u=(1/\sqrt 2,1/\sqrt 2)$ also does the job. For a general polynomial with high order, this rotation may be difficult to find.

答:
I suspect what you want is false, the example would be
$$ P = x^2 + 2xy + y^2 $$
where the Hessian derivative is zero.

If this example does not finish things, you should explain why.


问:
Proof that $(\mathbb{R}, +) \simeq (\mathbb{R}_{>0}, \times)$
I am trying to prove that the additive group. of reals is isomorphic to the multiplicative group of positive reals. The obvious isomorphic is $f(x) = e^x$. I know this to be a homomorphism since $e^{x+y} = e^x e^x$ by the properties of the exponential function, but every proof that $f$ is bijective, including my own, draws on the properties of the logarithm function. For example, it takes the form "if $e^x = e^y$, then $\log e^x = \log e^y$, then $x = y$" to prove infectivity, and "given $y > 0$, $e^{\log(y)} = y$" to prove subjectivity. But this is circular, because invoking the existence of an inverse presupposes that $e$ is bijective, hence invertible.

One way to prove that $f$ is an isomorphism, surely, is to invoke the fact that $\log(x)$ is an inverse, so $f$ is invertible, hence bijective. But my question is, without invoking the existence of this function, is there another way to prove that $f$ is injective and surjective? Am I wrong that invoking the $\log$ function is circular?

答:
It is well known that  $e^x $ has an inverse, $\ln x $.  Thus the homomorphism is bijective.

You can differentiate  $e^x $ to get injectivity as well.  Since the derivative is always positive,  and the limits as $x\to\pm\infty $ are $0$ and $\infty $, it's surjective.


问:
Finding $n$, $n^n \equiv k \pmod m$
What is the general process of finding $n$ such that
$$n^n \equiv k \pmod m?$$

I tried working with an example, say $k=60$ and $m=120.$ Then, I know that $2,3,5 \mid n.$ So, Letting $n=30w,$ We have
$$30w^{30w} \equiv 30^{30w}(w)^{30w}$$
But continuing this thought leaves me an expression with $w^w$ again. The problem I feel is that the exponent of $n^n$ is varying, so I can't really deduce many properties about it.

答:
As stated in the comments, a solution does not always necessarily exist. However, we can always find a solution in the case of a prime modulus. I am copying my solution from this post here because that topic has been closed and the solution might not survive deletion of the thread:

Assume WLOG that $0\le a<p.$ If $p\mid a$ then we can choose $x=p.$ If $p\nmid a,$ then Fermat's little theorem says that $$a^{p-1} \equiv 1\pmod p.$$ Also, $$a^p \equiv a\pmod p.$$ Then for any integers $m,n,$
$$(a+pn)^{p+m(p-1)}\equiv a\pmod p.$$ So we just need to find integers $m,n$ such that $$a+pn=p+m(p-1).$$ This is equivalent to $$pn-(p-1)m=p-a.$$ Since $$p\cdot 1 + (p-1)\cdot (-1)=1,$$ we can let $n=p-a$ and $m=p-a.$ Indeed, this works because the base $a+p(p-a)$ is equal to the exponent $p+(p-a)(p-1).$ This can be checked by expansion.

Of course, this only constructs a solution and does not find all solutions. If you want a challenge, find infinitely many solutions using this line of reasoning and the classification of all solutions to a two-variable linear Diophantine equation.

Title: Finding $n$, $n^n \equiv k \pmod m$


问:
How do I prove that for every positive integer $ n$, there exist $2n$ consecutive positive integers, each of which is composite?
I honestly need help with this question. I'm not asking for the answer but can someone at least point in the right direction to prove this? I know a prime number is one whose factors are itself and 1 otherwise they are composite numbers. So I basically need to prove that there is a sequence of 2n consecutive positive integers containing no primes but I don't know where to start?  Do I define an integer x and multiply by 2? If so then I know that will always be composite. I'm afraid I'm going in the wrong direction. If anyone can just point me right I'd appreciate it.

EDIT TO SHOW WHAT I'VE DONE SO FAR:
Proof: Given $n>0$
For every number in the set $1,2,3,4,5\cdots n+1$, $x=(2n!)$

(Now I know that any number $2n!$ is a composite number but I don't know how to prove that formally)

答:
There are two simple ways.

One requires a rather strong theorem: If pi(n) is the number of primes ≤ n, then the limit of n / (pi(n) log n) is 1. So the average gap between two primes becomes larger and larger, therefore the maximum gap between primes becomes larger and larger, therefore there will eventually be two primes separated by 2n or more composite numbers.

One doesn't require much mathematics at all: The number (2n+1)! is the product of all integers from 1 to 2n+1, and therefore divisible by all integers from 1 to 2n+1. (2n+1)! + 2 is divisible by 2, (2n+1)! + 3 is divisible by 3 etc. until (2n+1)! + (2n+1) is divisible by 2n+1. These numbers are therefore all composite, and there are 2n of them.


问:
What is $\lim\limits_{n→∞}(\frac{n-x}{n+x})^{n^2}$?
What is $$\lim_{n\rightarrow\infty}\left(\frac{n-x}{n+x}\right)^{n^2},$$ where $x$ is a real number. Mathematica tells me the limit is $0$ when I put an exact value for $x$ in (Mathematica is inconclusive if I don't substitute for $x$), but using $f(n)=(n-x)^{n^2}$ and $g(n)=(n+x)^{n^2}$, then $$\lim_{n\rightarrow\infty}f(n)=\lim_{n\rightarrow\infty}g(n)=\infty,$$ and by L'Hopital's rule we have
$$\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} = \lim_{n\rightarrow\infty}\frac{(s-z)^{-1+s^2} (s+z)^{1-s^2} (s+2 (s-z) \text{Log}[s-z])}{s+2 (s+z) \text{Log}[s+z]} = 1.$$

I'm not sure which to believe - Mathematica or L'Hopital.

答:
Both, and more.

If $x=0$ the limit is trivially 1. Otherwise, if $x>0$

$$
\left(\frac{n-x}{n+x}\right)^{n^2} = \left(\frac{n+x-2x}{n+x}\right)^{n^2} = \left(1 - \frac{2x}{n+x}\right)^{n^2} = \left(1 - \frac{1}{n/\left(2x\right)+1/2}\right)^{n^2} = \left(1 - \frac{1}{\bar n}\right)^{(\bar n-1/2)^24x^2} = \left(1 - \frac{1}{\bar n}\right)^{(\bar n^2 -\bar n +1/4)4x^2}
$$

Splitting the final term into a product (on the basis that $a^{b+c+d} = a^ba^ca^d$) the limit $n\to\infty \iff \bar n \to \infty$ is $0\cdot$(something bounded) $= 0$. Another way of seeing this is considering the exponent to be $\Theta(n^2)$.

Finally, if $x<0$ the limit is $\infty$ since the term in the brackets is $(1+\frac 1 {\bar n})$ when the limit is $\bar n \to \infty$.

The basis of these observations is the limit $1<e = \lim_{n\to\infty}(1+\frac 1 n)^n$, and derivations thereof, like $(1-\frac 1n)^n\to\frac 1 e<1$, etc.


问:
Function of class $C^1$ implies locally Lipschitz
Let $A$ be open in $\mathbf{R}^{m} ;$ let $g: A \rightarrow \mathbf{R}^{n} .$ If $S$ is a subset of $A,$ we say that $g$ satisfies the Lipschitz condition on $S$ if the function
$$
\lambda(\mathbf{x}, \mathbf{y})=|g(\mathbf{x})-g(\mathbf{y})| /|\mathbf{x}-\mathbf{y}|
$$
is bounded for $\mathbf{x}, \mathbf{y}$ in $S$ and $\mathbf{x} \neq \mathbf{y}$. We say that $g$ is locally Lipschitz if each point of $A$ has a neighborhood on which $g$ satisfies the Lipschitz condition.

How do I go about showing that if $g$ is of class $C^{1}$, then $g$ is locally Lipschitz?

I know that the derivatives of $g$ are locally bounded and then I can apply the MVT, but am not sure how to formulate this. The $1-D$ case is easy, but I am struggling with the multidimensional case given.

答:
There is Taylor's theorem for multivariable functions, which gives a mean value estimate. But in this case, it is easy to derive the estimate from scratch. Let $M > 0$ and suppose that for all $x$ in some open convex set $U$, $||Dg(x)|| \leq M$. Let $x_1, x_2 \in U$. Define $\phi \colon [0, 1] \to \mathbb{R}^n$ by $\phi(t) = g(x_1 + t(x_2 - x_1))$. By the chain rule, $\phi'(t) = Dg(x_1 + t(x_2 - x_1))(x_2 - x_1)$. By the fundamental theorem of calculus (applied componentwise),
\begin{align}
||g(x_2) - g(x_1)|| &= ||\phi(1) - \phi(0)|| \\
&= ||\int_{0}^{1}Dg(x_1 + t(x_2 - x_1))(x_2 - x_1)\,dt|| \\
&\leq \int_{0}^{1}||Dg(x_1 + t(x_2 - x_1))(x_2 - x_1)||\,dt \\
&\leq \int_{0}^{1}||Dg(x_1 + t(x_2 - x_1))|| \cdot ||x_2 - x_1||\,dt \\
&\leq M||x_2 - x_1||.
\end{align}


问:
Why isn't $\mathbb{Z}$ an open set of the set of real numbers in the standard topology?
In class my professor said that the interior of $\mathbb{Z}$ is the empty set. That is because the interior is the union of all open sets contained in $A$. But why is it not an open set?

Can someone explain (without using metric definitions because we have not covered that)?

答:
Standard topology on $\mathbb{R}$ is generated by the open intervals $(a, b)$ for $a, b\in\mathbb{R}$. Therefore

$$
A = \bigcup_{k\in\mathbb{Z}} (k, k+1)
$$

is an open set. However, $A$ is the complement of $\mathbb{Z}$. Therefore, $\mathbb{Z}$ is closed. Finally, the only clopen sets in the standard topology are $\emptyset$ and $\mathbb{R}$ (a hallmark of a connected space). Therefore, $\mathbb{Z}$ is not open.


答: If you take an open interval centered around an integer $n$, it looks like $(n-\varepsilon, n + \varepsilon)$ for some real number $\varepsilon$. But $(n-\varepsilon, n + \varepsilon)$ will never be contained in $\mathbf{Z}$ for any $\varepsilon$; there will always be a non-integer real number in $(n-\varepsilon, n + \varepsilon)$. So $n$, which was arbitrarily chosen, cannot be in the interior of $\mathbf{Z}$.

Title: Why isn't $\mathbb{Z}$ an open set of the set of real numbers in the standard topology?

Body:
In class my professor said that the interior of $\mathbb{Z}$ is the empty set. That is because the interior is the union of all open sets contained in $A$. But why is it not an open set?

Can someone explain (without using metric definitions because we have not covered that)?


问:
Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$
I am trying to evaluate

$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$

My solution is as follow

$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$

$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$

The solution is correct  but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.

Is there any generalized method

答:
You can crudely bound $1+\frac{1}{2} + \cdots + \frac{1}{n}$ by
$$\sum_{k=1}^n \frac{1}{k} \le \sum_{k=1}^{\lceil\sqrt{n}\rceil} \frac{1}{k} + \sum_{k=\lceil \sqrt{n} \rceil}^n \frac{1}{k} \le \lceil \sqrt{n}\rceil \cdot 1 + n \cdot \frac{1}{\sqrt{n}} \le 3\sqrt{n},$$
so that when you divide this by $n$, the expression is bounded by $3 \frac{\sqrt{n}}{n} \to 0$.


答: If you know that a base for the standard topology on $\mathbb R$ is the collection of all intervals of the form $(a,b): a<b$ then it's easy to show that $\mathbb Z$ is not open, for if so, then there would be a basis element $(a,b)$ of the aforementioned type such that $(a,b)\subseteq \{0\}$. Using the definition of open set, this would mean that there are $a<c<d<b$ such that $(c,d)\subseteq (a,b)$ and now $c=0=d$ which is impossible since $c<d.$


答:
Your solution is incorrect. The good solution is that $\sum_{k=1}^n \frac1k \sim \log n,$ so that

$$ 1\leq T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + . + \frac{1}{n}}}{{{n^2}}}} \right)^n} \leq \mathop {\lim }\limits_{n \to \infty }\left(1 + \frac{\log n}{n^2} \right)^n$$

For any $k,$ there exists an $N,$ such that $\frac{\log n}{n^2} \leq \frac{1}{k n}$ for $n>N,$ so

$$\mathop {\lim }\limits_{n \to \infty }\left(1 + \frac{\log n}{n^2} \right)^n \leq \exp(\frac1k).$$ It follows that $T=1.$

Title: Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$

Body:
I am trying to evaluate

$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$

My solution is as follow

$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$

$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$

The solution is correct  but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.

Is there any generalized method


问:
A group is abelian if and only if $\varphi(x) = x^{-1}$ is an automorphism.
I am trying to prove that given $G$ a group, $\varphi: G \to G $ given by $\varphi(x) = x^{-1}$ is an automorphism if and only if $G$ is abelian.

I've already shown that $\varphi$ is bijective, irrespective of whether $G$ is abelian, so the only additional distinguishing factor of an automorphism is the homomorphism property. I believe I've found a way to prove this in a way where the steps are reversible. My attempt is:

Suppose $\varphi$ is an automorphism. Then $\varphi(xy) = \varphi(x) \varphi(y)$ for all $x,y \in G$, so $(xy)^{-1} = x^{-1} y^{-1} = (yx)^{-1}$. By uniqueness of inverses, this implies that $xy = yx$, so $G$ is abelian. All of these steps are reversible. If $G$ is abelian, given $x,y \in G$, $xy = yx$. Then $(xy)^{-1} = (yx)^{-1}$, but $(yx)^{-1} = x^{-1} y^{-1} = \varphi(x) \varphi(y)$, so $\varphi(xy) = \varphi(x) \varphi(y)$, and $\varphi$ is an automorphism.

Have I made a mistake?

答: You are correct. It's good :) :) :)</p>

Body: You are correct. It's good :) :) :)


答:
Your doubt is right - it is not possible to split limit in variable amount of summands, as shows example $1=\lim 1=\lim\left(\frac{1}{n}+\cdots + \frac{1}{n}\right)=\lim \frac{1}{n} +\cdots + \lim\frac{1}{n} = 0$.

On another way you can use representation

$$\sum_{i=1}^{n}\frac{1}{i}  = \ln n + \gamma + \varepsilon_{n}= \ln n +O(1), n\rightarrow \infty$$
where $ \varepsilon_{n} \sim \frac{1}{2n}, n\rightarrow \infty$ based on Euler–Mascheroni constant $\gamma$. This also gives you $\frac{{1 + \frac{1}{2} + \frac{1}{3} + . + \frac{1}{n}}}{n} \to 0, n\rightarrow \infty$.


答:
If
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + . + \frac{1}{n}}}{{{n^2}}}} \right)^n}$$
then
$$\log T = \mathop {\lim }\limits_{n \to \infty } n\log{\left( {1 + \frac{H_n}{{{n^2}}}} \right)}\le \mathop {\lim }\limits_{n \to \infty } \frac{H_n}{n}.$$
It can be easily shown that this limit tends to $0$ (for example by Cesaro-Stolz.) So
$$\log T \le 0 \Rightarrow \log T = 0 \Rightarrow T=1.$$

Addendum: Showing that $\mathop {\lim }\limits_{n \to \infty } \frac{H_n}{n}\to0$.

Cesaro-Stolz:

We may set $a_n=H_n$ and $b_n={n}$. (We can apply Cesaro-Stolz since $(n)_{n\ge 1}$ is strictly monotone and divergent. The limit
$$\lim_{n\to\infty}\frac{H_{n+1}-H_n}{n+1-n}=\lim_{n\to\infty}\frac{1}{n+1},$$
exists so
$$\lim_{n\to\infty}\frac{H_n}{n}=\lim_{n\to\infty}\frac{1}{n}=0.$$

Bounding $H_n$

In this question a user proved $H_n<\log(n)+1.$ That means that
$$\lim_{n\to\infty}\frac{H_n}{n}\le\lim_{n\to\infty}\frac{\log(n)+1}{n}=0.$$
Since our oringal limit is non-negative we can conclude $\lim_{n\to\infty}\frac{H_n}{n}=0$.

l'hopital

One may also stick to the classic l'hopital. Differentiating numerator and denominator yields
$$\mathop {\lim }\limits_{n \to \infty } \frac{H_n}{n}=\mathop {\lim }\limits_{n \to \infty } \frac{\pi^2/6-H_n^{(2)}}{1}.$$
Since $H_n^{(2)}\to\pi^2/6$ for $n\to\infty$, the overall limit equals zero too.


问:
Is it true that a smooth, harmonic functions with compactly supported gradient is trivial?
Let $u \in C^{\infty}(\mathbb{R}^n)$ be such that $\nabla u \in C^{\infty}_c(\mathbb{R}^3; \mathbb{R}^3)$ and $\Delta u=0$. I wish to prove or disprove that $u = c$ for some $c \in \mathbb{R}^n$; note the statement would be true with $c=0$ (by Liouville classical theorem) if $u$ itself is compactly supported.

I had the idea of exploiting a representation formula like:

$$ u(x) = \frac{1}{n \omega_n} \int_{\mathbb{R}^3} \nabla u(y) \cdot \frac{x-y}{|x-y|^n} dy $$

together with an integration by parts in order to get the Laplacian. Unfortunately, the latter is only valid if $u \in W^{1,1}(\mathbb{R}^n)$ and here we don't have the hypothesis $u \in L^1(\mathbb{R}^n)$. I am stuck here, but at same time I cannot produce a counterexample.

答:
Differentiating the equation we note
$$0=\partial_i\Delta u = \Delta \partial_i u$$ which means that the components of $\nabla u$ are themselves harmonic. Since they are compactly supported, they vanish identically, and so $u$ is constant.


问:
If abelian $P\in{\rm Syl}_p(G)$, and $H\le P$ and $H^g\le P$, show $g\in G$ is the left product of an element of $N_G(P)$ with an element of $C_G(H)$

If abelian $P \in{\rm Syl}_p(G)$, and $H \leq P$ and $H^g \leq P$, show $g \in G$ is the left product of an element of $N_G(P)$ with an element of $C_G(H)$

First, I set $g=nc$, where I hope to find that $n \in N_G(P)$ and $c \in C_G(H)$. I am trying to do this by contradiction, assuming either $n \notin N_G(P)$ (which implies $n \notin N_G(H) )$ or $c \notin C_G(H)$. In the first cast ($n \notin N_G(P)$, $c \in C_G(H)$), we have that $H^g = (H^n)^c \neq H^c = H$. However, $g$ is arbitrary (as far as I know) in $G$, so I can not use that $P$ is abelian to find that $H^g = H$. I know that $P$ must be the unique Sylow $p$-subgroup but I can't see how that will help me.

I am not sure where to go from here.

答:
Note. I had run into an issue when writing this, then mistakenly assumed $g$ fixed $H$ set-wise and that got me the right expression and a couple of comments below. When I fixed that error, I could not get the expression of $g$ as a product in the order in which I interpret the original question to be requiring: the product of an element in the normalizer times an element in the centralizer. Derek Holt has given a counterexample to show that it may be impossible to achieve that; since he refers to my answer, I'm adding this note and undeleting.

Following on Derek Holt’s suggestion: note that $H\leq P$, and $H\leq P^{g^{-1}}$. Since $P$ is abelian, $P\leq C_G(H)$, and since $P^{g^{-1}}$ is also abelian, $P^{g^{-=1}}\leq C_G(H)$.

Now note that $P^{g^{-1}}$ and $P$ are both Sylow $p$-subgroups of $C_G(H)$ (you should prove this). Therefore, they are conjugate in $C_G(H)$; that is, there exists $c\in C_G(H)$ such that $P^{g^{-1}c}=P$. Therefore, $g^{-1}c\in N_G(P)$.

But that gives the wrong product order: $g^{-1}c\in N_G(P)$, so $c^{-1}g=n$ for some $N\in N_G(P)$, $g = cn$.


问:
What conditions do I need for this calculation to work? (Product rule differentiation of bilinear form)
Let $b(\cdot,\cdot):X \times X \to \mathbb{R}$ be a bilinear mapping where $X$ is a Banach space.

Consider functions $f:[0,T]\to X$ and $g:[0,T] \to X$. What assumptions do I need on $b$ so that this is valid:

$$\lim_{h \to 0}\frac{b(f(t+h),g(t+h))-b(f(t),g(t))}{h} = \lim_{h \to 0}\frac{b(f(t+h),g(t+h))\pm b(f(t+h), g(t)) -b(f(t),g(t))}{h} = \lim_{h \to 0}\frac{b(f(t+h),g(t+h)-g(t))+b(f(t+h)-f(t),g(t))}{h}$$
$$=\lim_{h \to 0}\frac{b(f(t+h),g(t+h)-g(t))}{h}+\lim_{h \to 0}\frac{b(f(t+h)-f(t),g(t))}{h}=b(\lim_{h \to 0}f(t+h),\lim_{h \to 0}\frac{g(t+h)-g(t)}{h})+b(\lim_{h \to 0}\frac{f(t+h)-f(t)}{h},g(t))$$
$$=b(f(t), g'(t)) + b(f'(t), g(t))$$

Obviously $f$ and $g$ need to be differentiable in time. So say $f, g \in C^1([0,T];X)$. But not sure how to formulate the condition on $b$. It should be something like "$b$ is continuous"..

答: You need that the bilinear form is a bounded operator (i.e. continuous). If it is such, you can freely pass limits through. </p>


答:
We want to show that $$\displaystyle \lim_{n \to \infty} \dfrac{H_n}{n}=0$$.

We have $\log n = \displaystyle \int_{1}^{n} \dfrac{1}{t} dt$. Also, $H_n = \displaystyle \sum_{r=1}^{n}\dfrac{1}{r} = \dfrac{1}{n}\sum_{r=1}^{n}\dfrac{1}{\frac{r}{n}}$

Thus, by comparison with the Riemann sums of $\log n$, we see that $\log n < H_n < 1 + \log n$.

So we get that, $$\dfrac{\log n}{n} < \dfrac{H_n}{n} < \dfrac{1+\log n}{n}$$

But both the left and right sides of the above inequality tend to $0$ as $n \to \infty$, and hence, by the Sandwich theorem (or Squeeze principle), $$\lim_{n \to \infty} \dfrac{H_n}{n} = 0$$