问:
Lindeberg-Feller CLT for weighted sumation.
Problem : Let $\{a_n\}_{n\ge 1}$ be an increasing positive sequence of integers $s.t.$ the following condition holds $$\frac{a_n^2}{a_1^2+\cdots+a_n^2} \rightarrow 0.$$
Assume $\{X_n\}_{n \ge 1}$ are $iid$ sequence of random variables with mean $0$ and variance $1$. TO SHOW
$$\frac{a_1X_1 + a_2X_2 + \cdots + a_nX_n}{\sqrt{a_1^2+\cdots+a_n^2}} \rightarrow N(0,1)$$

Direction: Well I feel we need to use the Linderberg's condition for CLT. By the definition of triangular arrows define $X_{n,k} = \frac{a_kX_k}{\sqrt{a_1^2+\cdots+a_n^2}}$, where $k \le n$. The Lindeberg condtion states $\sum_{k=1}^n\mathbf{E}[X_{n,k}^2\mathbf{1}_{X_{n,k} > \delta}] \rightarrow 0$ as $n \rightarrow \infty$ for any arbit $\delta$ implies the result we set out to show.

Thus trying to show the condition holds:
$$\mathbf{E}[X_{n,k}^2\mathbf{1}_{X_{n,k} > \delta}]$$
$$= \frac{a^2_k}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_k}}\right] $$
$$\leq \frac{a^2_n}{a_1^2+\cdots+a_n^2}\mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$$
But now summing over $n$ terms give me
$$\frac{na^2_n}{a_1^2+\cdots+a_n^2}\mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$$
I can conclude $\mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$ converges to $0$ because of finite variance and the condtion stated. However there remains a $\frac{na^2_n}{a_1^2+\cdots+a_n^2}$ term which I am pretty sure I cant show it goes to $0$.

Any help is appreciated

Edit : I was wondering if it could be shown that $\frac{na^2_n}{a_1^2+\cdots+a_n^2}$ this could be shown to be a constant as $\frac{na^2_n}{a_1^2+\cdots+a_n^2} \rightarrow 0$ for increasing sequences of $(a_n)$

答:
The bound
$$
\frac{a^2_k}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_k}}\right]\leqslant \frac{a^2_n}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]
$$
is a bit too crude; instead, use
$$
\frac{a^2_k}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_k}}\right]\leqslant \frac{\color{red}{a^2_k}}{a_1^2+\cdots+a_n^2} \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right].
$$
The last term does not dependent of $k$ and the summation over $k$ leaves us only with $ \mathbf{E} \left[X_1^2\mathbf{1}_{|X_1| > \delta\frac{\sqrt{a_1^2+\cdots+a_n^2}}{a_n}}\right]$. We can then conclude by dominated convergence.


问:
Compute norm for a continuous function.
Given that $f$ is a function from $[0, 1]$ such that $$f(x)=x-x^2.$$Compute the $L^{\infty}$ norm of the function $f(x)$.
My attempt:

I know that
$$\|f\|_{\infty }= \inf\{C\geq 0:|f(x)|\leq C{ \ \text{ for almost every }}x\}.$$
Since $f'(x)=1-2x$ means that $f$ attains its maximum value at $\frac12$ in $[0, 1]$ and maximum value is $$f(1/2)=1/4.$$ Can I say that $\|f\|_{\infty}=\frac14$?

答: Yes, that is correct. You can reach the same conclusion without differentiation:\begin{align}x-x^2&=\frac14-\frac14+x-x^2\\&=\frac14-\left(x-\frac12\right)^2\\&\leqslant\frac14,\end{align}and we have $x-x^2=\frac14$ when $x=\frac12$. Besides, you always have $x-x^2\geqslant0$.

ContentLicense: CC BY-SA 4.0

Title: Compute norm for a continuous function.


问:
Is this $\theta$ measured through vertical axis or from initial position?

Two equal particles $A$ and $B$ are attached to the ends of a light rigid rod of length
$2a$. The rod is supported in the position of unstable equilibrium with $A$ in contact
with a smooth horizontal table and $B$ above $A$ making an acute angle $\alpha$ with the
vertical axis. Find the velocity of $A$ when the rod has turned through angle $θ$ and
thrust on the table.

Is this $\theta$ measured through vertical axis or from initial position? Can anyone help me to understand the problem?

答:
Hint.

The gravity center maintains the $x$ coordinate constant along the movement. The initial potential energy is equal to the kinetic energy when the rod thrust on the table. Also there exists a holonomic constraint which is $y = a\sin\theta(t)$


问:
Proving two random vectors are following the same distribution using characteristic function?
Suppose we have two centered Gaussian random vectors $X$ and $Y$ I.I.D.

Find the distribution of $X_a=X\cos(a)+Y\sin(a)$ and $Y_a=-X\sin(a) + Y\cos(a)$ with $a$ being in $[0,2\pi)$.

What I tried to do is write the characteristic function of $X_a$ and prove that it's something obvious (like maybe the same law as $X$).

\begin{align}
\phi_{X_a}(u)&=E[\exp(i\langle u,X\cos(a)+Y\sin(a)\rangle)]\\
&=E[\exp(i\langle u,X\cos(a)\rangle)]\cdot E[\exp(i\langle u,Y\sin(a)\rangle)]
\end{align}

but I got stuck right there and didn't know how to proceed. Any hints?

答: If $a$ is a constant, then $X_a = X \cos(a) + Y \sin(a)$ is a linear combination of two independent zero-mean Gaussian R.V.s, which is a zero-mean Gaussian R.V. with the variance $\sigma_{X_a}^2 = \cos^2(a) \sigma_X^2 + \sin^2(a) \sigma_Y^2$. The same goes for $Y_a$. If $X,Y\sim\mathcal{N}(0,1)$, then $X_a,Y_a\sim\mathcal{N}(0,1)$ too.


问:
If $x(t)=\cos(2t)$ and $y(t)=\sin(2t)$, find $\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}$
So I was given the following prompt:

A curve in the $xy$-plane is defined by the parametric equations $x(t)=\cos(2t)$ and $y(t)=\sin(2t)$ for $t\ge0$. Find $\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$.

I believe I was able to find an answer for this, I'm just looking for some clarification as to whether or not this answer might actually be correct or not.

I was able to come up with an answer of $2$, after I found that square-root equation to be equal to the following: $\sqrt{(2\cos(2x))^2+(-2\sin(2x))^2}$, and I evaluated from there.

Any help/clarification would be appreciated!

答:
What you find is correct.

Using
$$
x'(t)=-2\sin(2t)\quad y'(t)=2\cos(2t)
$$
and $\sin^2x+\cos^2x=1$, you conclude that
$$
\sqrt{(x'(t))^2+(y'(t))^2}=2
$$


问:
How is this tuple-valued function written in arrow notation?
Let $f: \mathbb{R} \to \mathbb{R}$ and $r: \mathbb{R} \to \mathbb{N}$,
then suppose $T(x) = ( f(x), r(x) ) : x \in \mathbb{R}$.

How is the function $T$ written in arrow notation? Is
$$
T : \mathbb{R} \to \mathbb{R} \times \mathbb{N}
$$
correct?

答:
If $f : \mathbb{R} \to \mathbb{R}$, $r : \mathbb{R} \to \mathbb{N}$ and $T(x) = (f(x), r(x))$, then $T$'s arrow notation definition is
$$
T : \mathbb{R} \to \mathbb{R} \times \mathbb{N}.
$$


问:
Show that equation has no roots $ \in \mathbb{Z} $
Given $ P(x) := 7x^5-23x^2 +2x -8,$ show that $ P(x) $ has no roots in $ \mathbb{Z}. $

My (primitive solution);

Assume that it exists atleast one root in $ \mathbb{Z}. $ We can then write $ 7x^5 -23x^2+2x = x(7x^4-23x+2)=8.$

$ 8 $ is then a product of 2 factors, and both must divide $8.$ We have the set of $ D_8 = \{\pm 1,\pm2,\pm4,\pm8 \}$. So we can test for every $ x \in D_8 $ and see that none of them yields 8.

Questions;

Is this a solid proof and are there any other way to show it? And does "Integral root" mean a root in $ \mathbb{Z} $?

答:
A common technique used to prove there are no integer solutions is modular arithmetic.

Here for instance working modulo $3$:

$p(x)\equiv 7x^5-23x^2+2x-8\equiv x^5+x^2+2x+1\equiv x+x^2+2x+1\equiv x^2+1\pmod 3$

And we can use that there is no square $x^2\equiv 2\pmod 3$


问:
Spectral Theory for "Unrepresented" C*-algebras
If you Google something like "Borel functional calculus" or "spectral decomposition" you get plenty of results for operators on Hilbert spaces. What about an (unrepresented) $\mathrm{C}^*$-algebra $A$?

Apologies if this question seems a bit dumb: I am a little unsteady after learning some things about unitary inequivalence of Hilbert space representations of $\mathrm{C}*$-algebras.

We have an isometric *-isomorphism (direct sum of GNS of pure states) of $A$ with $\pi(A)\subset B(\mathsf{H})$. I am happy to use the Gelfand Naimark *-isomorphism to say that when a finite-spectrum self-adjoint element of $A$, say $f$, we can look at it in $\pi(A)$, and then give it a spectral decomposition back in $A$. Briefly: $\pi(a)$ is a self-adjoint operator with finite spectrum (the spectrum of an element is a unitary invariant), and so it has a spectral decomposition:

$$\pi(f)=\sum_{f_i\in\sigma(f)}f_i\,p_i,$$

and the $(p_i)_{i}$ form a partition of unity. I can take the inverse isomorphism then and so $f$ has a spectral decomposition back in $A$:

$$f=\sum_{f_i\in\sigma(f)}f_i\,\pi^{-1}(p_i).$$

From Murphy:

Let $u$ be a normal operator on a Hilbert space $\mathsf{H}$. Then there is a unique spectral measure relative to $(\sigma(u),\mathsf{H})$ such that $u=\int z\,dE$ where $z$ is the inclusion of $\sigma(u)$ in $\mathbb{C}$

Question 1: So the question is, if I have a self-adjoint bounded operator $f\in A$, no longer with a finite spectrum, do I still have a spectral decomposition back in $A$?

$$f=\pi^{-1}(\pi(f))=\pi^{-1}\left(\int_{\sigma(\pi(f))}z\,dE\right)\overset{?}{=}\int_{\sigma(f)}z\,\pi^{-1}(dE).$$

If Question 2 is false, what I really want is the following. The spectrum of $f$ (not e.g. the point spectrum but the full spectrum) is the same in every faithful representation ($\sigma(f)=\sigma(\pi(f))$). In $\pi(f)$ I can certainly use Borel functional calculus to talk about for a Borel set $E\subset \sigma(f)$, the projection $\mathbf{1}_{E}(\pi(f))\in B(\mathsf{H})$, can I safely define:
$$\mathbf{1}_E(f):=\pi^{-1}(\mathbf{1}_E(\pi(f))).$$

In other words:

Question 2: Do I have Borel functional calculus on 'unrepresented' $A$?

答: In general, you don't have a spectral decomposition in $A$ because the projection $1_E(\pi(x))$ does not necessarily belong to $\pi(A)$. For example, if $X$ is a connected compact space, the only projections in $C(X)$ are $0$ and $1$, and the only elements you can write as limits of linear combinations of $0$ and $1$ are multiples of the identity.


问:
Diffeomorphism preserves dimension
I read from Milnor's book $\textit{Topology from the Differentiable Viewpoint}$ this assertion

"If $f$ is a diffeomorphism between opensets $U\subset R^k$ and $V\subset R^l$, then k must equal l, and the linear mapping

$$df_x:R^k\rightarrow R^l$$ must be nonsingular."

The proof was: The composition $f^{-1}\circ f$ is the identity map of U; hence $d(f^{-1})_v\circ df_x$ is the identity map of $R^k$. Similarly $df_x \circ d(f^{-1})_v$ is the identity map of $R^l$. Thus $df_x$ has a two-sided inverse, and it follows that $k=l$.

My question is as follows:

I don't understand why the fact that $df_x$ has a two-sided inverse implies $k=1$. The way I would prove it is as follows: Instead of proving $df_x$ has a two-sided inverse, I would say one-sided inverse will suffice. Since $d(f^{-1})_v\circ df_x$, it follows that $df_x$ must be an isomorphism(Structure preserving comes from linearity. And bijectivity comes from the identity map.). Therefore $k=l$.

Thanks so much for your help!

答: It is not true that if a linear map $f:V\to W$ has a one sided inverse, then $\dim V=\dim W$ nor that $f$ is an isomorphism.</p>

ContentLicense: CC BY-SA 3.0

Title: Diffeomorphism preserves dimension


问:
Analytic Function such that $f(2z) = 4\,f(z)$

Let $f \colon \mathbb{C} \longrightarrow \mathbb{C}$ such that $f(2z) = 4\,f(z)$, for all $z \in \mathbb{C}$. Given that $f(0) = 0$ and $f(1) = 1 + 2i$, find $f(\sqrt{2} + i\,\sqrt{2})$.

My attempt was to write $f(z) = \sum a_n\,z^n$ and plug in the conditions, i.e.,
$$\sum a_n\,2^n\,z^n = \sum 4\,a_n\,z^n$$ which gives $a_n\,2^n = 4\,a_n$, but this didn't help at all...

How would I start this question? Any help is appreciated.

答: $a_n\,2^n = 4\,a_n$ helps a lot, actually. It tells you almost all you need to know. For which $n$ is $a_n\neq0$ at all possible?


问:
$\ln(a+b)\leq\ln(a)+\ln(b)$
Suppose $a,b>0$ are integers. Do we then have that
$$
\ln(a+b)\leq \ln(a)+\ln(b)?
$$

I think, since this is equivalent to $a+b\leq ab\Leftrightarrow 1\leq b-\frac{b}{a}$

it holds for $1<b<a$ only and, moroever, it is a strict inequality for these values.

答:
Suppose $a$ and $b$ are positive integers. Then:

\begin{align}
\ln(a+b)\leq \ln(a)+\ln(b) \\
\\
(*)\quad \iff a+b \leq ab \quad\\
\\
\iff 1 \leq b - \frac{b}{a} = b\left(1-\frac{1}{a}\right),\\
\\
\iff b \geq 2\ \text{ and }\ a \geq 2.
\end{align}

The $\impliedby$ at $(*)$ is justified by the fact that $\ln$ is an increasing function.
The $\implies$ at $(*)$ is justified by the fact that $\exp$ is an increasing function.

Title: $\ln(a+b)\leq\ln(a)+\ln(b)$


问:
Calculate Probability of Bachier model hitting upper barrier
Let $B_t$ be a standard brownian motion and $\mu, \sigma \in \mathbb{R}$ with $\sigma \neq 0$. Now let $X_t = \sigma B_t + \mu t$ be the Bachier model with votality $\sigma$ and drift $\mu$. Further let $a,b > 0$ and define the stopping time $\tau = \text{inf} \{t \geq 0 : X_t = a$ or $X_t = -b\}$.

As part of an exercise I need to prove that:
\begin{equation}
\mathbb{P}(X_\tau=a) = \frac{\text{exp}(-\frac{2\mu}{{\sigma}^2})-1}{\text{exp}(-\frac{2\mu}{{\sigma}^2}(a+b))-1}
\end{equation}
I am not sure how to tackle this problem. We did prove in our lecture that if we define the function $g: \mathbb{R} \to \mathbb{R}$ as:
\begin{equation}
g(x) = \frac{\text{exp}(-\frac{2\mu }{{\sigma}^2}x)-\text{exp}(\frac{2\mu }{{\sigma}^2}b)}{\text{exp}(-\frac{2\mu}{{\sigma}^2}a)-\text{exp}(\frac{2\mu }{{\sigma}^2}b)}
\end{equation}
then $g(X_t)$ is a martingale satisfying $g(a) = 1$ and $g(-b) = 0$, but I don't see how this helps. All this tells me is
\begin{equation}
\mathbb{P}(X_\tau=a) = \mathbb{P}(g(X_{\tau})=1)
\end{equation}
Any help or hints would be appreciated.

答:
As you note, $g(X_t)$ is a martingale; therefore (and more useful) the stopped process $Y_t:=g(X_{t\wedge \tau})$ is a bounded martingale, with $Y_0=g(0)$. Therefore
$$
g(0)=E[Y_0] = E[Y_\tau]= g(a)P[X_\tau = a]+g(-b)P[X_\tau=-b].
$$
(This requires the previously garnered knowledge (or the stipulation) that $P[\tau<\infty] = 1$.)


问:
How to solve this differential equation via power series
Studying Quantum Mechanics, I encountered the following differential equation:

how to solve $$y''-x^2y=0$$
Griffiths, the author of the book, just said that its solutions are
$$y=Ae^{x^2/2}+Be^{-x^2/2}$$
I can immediatelly see that this is true, but I do not know how to get to this solution, and I cannot find it solved anywhere else.
I tried to use the power series method:
$$y=\sum_{n=0}^{\infty}c_n x^n$$
So the equation becomes:
$$\sum_{n=0}^{\infty}c_{n+2}x^n(n+1)(n+2)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
Here I need to get both sums to the same power of n, so I removed the first two terms from the first sum, so that I can later change the index:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}c_{n+2}x^n(n+1)(n+2)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
now I change the index in the fist sum:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}c_{n+4}x^{n+2}(n+4)n(n+3)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
Now I factor out the sum:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}x^{n+2}(c_{n+4}(n+4)(n+3)-c_n)=0$$
Now all three terms must be zero at the same time, so $c_2=c_3=0$, and:
$$c_{n+4}(n+4)(n+3)-c_n=0$$
$$c_{n+4}=\frac{c_n}{(n+4)(n+3)}$$
The first terms of the series are:
$c_4=\frac{c_0}{4\times 3}$, $c_8=\frac{c_4}{8\times 7}=\frac{c_0}{8\times 7\times 4\times 3}$
However, I don't really know how to proceed from here. I tried putting the coefficients into the sum for $y$, but I can't really come close to the answer. Could you please guide me as how to advance from here?

答:
HINT

Perhaps easier to write
$$
\begin{split}
c_n
 &= \frac{c_{n-4}}{n(n-1)}\\
 &= \frac{c_{n-8}}{n(n-1)(n-4)(n-5)} \\
 &= \frac{c_{n-12}}{n(n-1)(n-4)(n-5)(n-8)(n-9)}
\end{split}
$$
So you can try to write a closed form for $c_n$ and you have 4 series driven by $c_0,c_1,c_2,c_3$ but $c_2=c_3=0$ makes life much simpler. You can finish it that way.

A different approach is to note that if differentiating twice adds two powers of $x$ in the front, perhaps differentiating once would add one, getting $y'=xy$ which is separable and implies
$$
\frac{dy}{y} = xdx \iff \ln y = \frac{x^2}{2}+C
$$
which is equivalent to one family of your solutions.


问:
Find all the functions $f:\left[0,1\right]\rightarrow\mathbb{R}$ so that ${x}'f\left(x\right)\geq 2xf\left(x^{2}\right)$

Find all the functions $f:\left [ 0, 1 \right ]\rightarrow\mathbb{R}$ on the interval $\left [ 0, 1 \right ]$ so that
$${x}'f\left ( x \right )\geq {\left ( x^{2} \right )}'f\left ( x^{2} \right ),$$
in other words, that $f\left ( x \right )\geq 2xf\left ( x^{2} \right ).$

I simplify the original problem, I think we need to find the increasing function $f\left ( x \right )= {g}'\left ( x \right )$ since $x\geq x^{2}$ but are these all ? And how to find at least one function $g\left ( x \right )$ like that.. is it necessary to add the condition that the function $f$ is continuous ? Thanks for helping.

答:
The only continuous function on $[0,1]$ such that $f(x)\ge 2xf(x^2)$ is identically zero.

Proof.

We can immediately note that $f(0)\ge 0$ and $f(1)\le 0$.

Suppose $f(x)<0$ for some $x\in [0,1]$. By continuity we can choose $x\in (0,1)$.
Then $f(x^2)\le \frac{f(x)}{2x}$  and so $f$ is negative on $x,x^2,x^4,...$ where the arguments are tending to $0$. Furthermore, $f$ is strictly monotonic decreasing on this sequence as soon as the argument becomes less than a half.
Therefore, by continuity $f(0)<0$, a contradiction.

Now suppose $f(x)>0$ for some $x\in [0,1]$. By continuity we can choose $x\in (0,1)$.
Then $f(\sqrt x)\ge 2x{f(x)}$  and so $f$ is positive on $x,x^{1/2},x^{1/4},...$ where the arguments are tending to $1$. Furthermore, $f$ is strictly monotonic increasing on this sequence as soon as the argument becomes more than a half.
Therefore, by continuity $f(1)>0$, a contradiction.


答: It is a standard fact from elementary set theory that a map $f$ is a bijection $\iff$ $f$ has a two-sided inverse.  Remember, $df_x$ is a linear map, and therefore, an isomorphism.  This shows $k = l$ since isomorphic vector spaces have equal dimension.</p>

Title: Diffeomorphism preserves dimension

Body:
I read from Milnor's book $\textit{Topology from the Differentiable Viewpoint}$ this assertion

"If $f$ is a diffeomorphism between opensets $U\subset R^k$ and $V\subset R^l$, then k must equal l, and the linear mapping

$$df_x:R^k\rightarrow R^l$$ must be nonsingular."

The proof was: The composition $f^{-1}\circ f$ is the identity map of U; hence $d(f^{-1})_v\circ df_x$ is the identity map of $R^k$. Similarly $df_x \circ d(f^{-1})_v$ is the identity map of $R^l$. Thus $df_x$ has a two-sided inverse, and it follows that $k=l$.

My question is as follows:

I don't understand why the fact that $df_x$ has a two-sided inverse implies $k=1$. The way I would prove it is as follows: Instead of proving $df_x$ has a two-sided inverse, I would say one-sided inverse will suffice. Since $d(f^{-1})_v\circ df_x$, it follows that $df_x$ must be an isomorphism(Structure preserving comes from linearity. And bijectivity comes from the identity map.). Therefore $k=l$.

Thanks so much for your help!


答: $$a+b\le ab\iff 1\le(a-1)(b-1),$$ which is wrong for $a=1$ or $b=1$ and works for other naturals.

Title: $\ln(a+b)\leq\ln(a)+\ln(b)$

Body:
Suppose $a,b>0$ are integers. Do we then have that
$$
\ln(a+b)\leq \ln(a)+\ln(b)?
$$

I think, since this is equivalent to $a+b\leq ab\Leftrightarrow 1\leq b-\frac{b}{a}$

it holds for $1<b<a$ only and, moroever, it is a strict inequality for these values.


问:
$ \int_{\partial B_2(0)} \frac{3z+1}{z(z-1)(z-3)}dz $
How do i calculate $$\int_{\partial B_2(0)} \frac{3z+1}{z(z-1)(z-3)}dz$$ $   z  \in C$

I have tried partial fraction but i dont get it...

Body: $$\int_{\partial B_2(0)} \frac{3z+1}{z(z-1)(z-3)}dz=2\pi i\bigg[\text{Res}\bigg(\frac{3z+1}{z(z-1)(z-3)},z=0\bigg)+\text{Res}\bigg(\frac{3z+1}{z(z-1)(z-3)},z=1\bigg)\bigg]=2\pi i(\frac13-2)=\frac{10\pi i}{3}.$$


问:
about closed set of $\mathbf{A}_{k}^{1}$
Theorem : $A$ regular function is continuous, when $k$ is identified with $\mathbf{A}_{k}^{1}$ in its Zariski topology.

PROOF. It is enough to show that $f^{-1}$ of a closed set is closed. A closed set of $\mathbf{A}_{k}^{1}$ is a $ \color{red} {finite}$ set of points, so it is sufficient to show that $f^{-1}(a)=$ $\{P \in Y \mid f(P)=a\}$ is closed for any $a \in k .$ This can be checked locally: a subset $Z$ of a topological space $Y$ is closed if and only if $Y$ can be covered by open subsets $U$ such that $Z \cap U$ is closed in $U$ for each $U .$ So let $U$ be an open set on which $f$ can be represented as $g / h$, with $g, h \in A,$ and $h$ nowhere 0 on $U .$ Then $f^{-1}(a) \cap U=\{P \in U \mid g(P) / h(P)=a\} .$ But $g(P) / h(P)=$ $a$ if and only if $(g-a h)(P)=0 .$ So $f^{-1}(a) \cap U=Z(g-a h) \cap U$ which is closed. Hence $f^{-1}(a)$ is closed in $Y$

why a closed set of $\mathbf{A}_{k}^{1}$ is  $ \color{red} {finite}$ ? is every finite set  in $\mathbf{A}_{k}^{n}$ ,close ? also why a subset $Z$ of a topological space $Y$ is closed if and only if $Y$ can be covered by open subsets $U$ such that $Z \cap U$ is closed in $U$ for each $U$ ?

答:
Because Zariski topology on $\Bbb A^1_k$ is the cofinite topology.

Yes, finite subsets are closed in $\Bbb A^n_k$ because Zariski topology is T1.

If $\mathcal U\subseteq \tau$, $\bigcup \mathcal U=Y$ and $Z\cap U$ is closed in $U$ for all $U\in \mathcal U$, then $$Z=Y\setminus\left(\left(\bigcup_{U\in\mathcal U}U\right)\setminus Z\right)=Y\setminus\bigcup_{U\in\mathcal U}(U\setminus Z)$$ We've established that $U\setminus Z$ is open in $U$. Since $U$ is open, this means that $U\setminus Z$ is open $Y$. Therefore $Z$ is closed.


问:
How do we show that $\sum_{n=1}^{\infty}\frac{n \log (n+1)}{1+n^3x^2}$ converges?
Which criterion do we use to show the convergence of $$\sum_{n=1}^{\infty}\frac{n \log (n+1)}{1+n^3x^2}$$ ?

Do we maybe use that $\log (n+1)<n$ and so we get $\frac{n\log (n+1)}{1+n^3x^2}<\frac{n^2}{1+n^3x^2}$ ?

Since $1+n^3x^2>n^3x^2$ we get $$\frac{n\log (n+1)}{1+n^3x^2}<\frac{n^2}{n^3x^2}=\frac{1}{nx^2}$$ But this one diverges, right?

So do we have to find an other uppoer bound? Or do we use an other criterion?

答:
You can use that $\log (n+1) \leqslant n^\alpha$ for any $\alpha \in (0,1)$ so you have estimation

$$\frac{n\log (n+1)}{1+n^3x^2}<\frac{n^{1+\alpha}}{n^3x^2}=\frac{1}{n^{2-\alpha}x^2}$$

Addition.

As in comments arise question about continuity and differentiability, then let me say, that here key is domain with respect we are speaking. Function $f_n(x)=\frac{n\ln (n+1)}{1+n^3x^2}$ is defined on $\mathbb{R}$ (for simplicity I considering $\ln$). Same with $f'_n(x)=-\frac{2n^4x \ln (1+n)}{(1+n^3x^2)^2}$. Suppose we take any $x_0>0$, then exists $x_0 \gt a \gt 0$ such that $x_0 \in [a,b]$ and using same technique  as above gives us uniform convergence for $\sum\limits_{n=1}^{\infty}f'_n(x)$ on $[a,b]$, so differentiability of $f(x)=\sum\limits_{n=1}^{\infty}f_n(x)$ (so continuity) in $x_0$.


问:
Picking pawns from pouch
You have a pouch containing 4 chess pawns. Your prior knowledge is that the pouch does
not contain exactly 2 black and 2 white pieces. All other proportions of colours are equally
probable (e.g. 3 black and 1 white is equally probable as 4 black pieces).

The first piece drawn from the bag is black. What is the probability that the next piece drawn is also black?

I tried solving this problem in the following way:

The states can be represented by the number of black pieces in the pouch and is then written as:
$$B = \{0,1,3,4\}$$
, with probability distribution
$$P = \{\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\}$$.
The possible states of a measurement are picking up a black piece or picking up a white one. We denote these states as
$$C = \{b, w\}$$
The first measurement has a probability distribution of
$$Q = \{q_{b}, q_{w}\} = \{\frac{1}{2},\frac{1}{2}\}$$
where we are equally likely to pick up a white or a black piece. This is due to the weighted sum of black and white pieces in the possible states being equal. Let's say we pick a black piece.
Now reevaluate the probabilities of the states $B$. This can be done by counting in how many ways we could have picked a black piece from each state and weighting the probabilities by that number.
$$P|b = \{0,\frac{1}{8},\frac{3}{8},\frac{4}{8}\}$$
From these probabilities we can once again do a weighted sum of black and white pieces. It's a bit more complicated this time so I'll write it out.
$$b_{sum} = 0\cdot-1+\frac{1}{8}\cdot 0 +\frac{3}{8}\cdot 2 +\frac{4}{8}\cdot 3 = \frac{18}{8} = \frac{9}{4}$$
$$w_{sum} = 0\cdot4+\frac{1}{8}\cdot 3 +\frac{3}{8}\cdot 1 +\frac{4}{8}\cdot 0 = \frac{6}{8} = \frac{3}{4}$$
$$Q|b = \{q_b|b, q_w|b\} = \{\frac{b_{sum}}{b_{sum} + w_{sum}}, \frac{w_{sum}}{b_{sum} + w_{sum}}\} = \{\frac{3}{4}, \frac{1}{4}\}$$

This result was so unintuitive to me that I had to doubt the logic of my reasoning. I tried redoing the calculations with a bag of pieces where the state of two black and two white was allowed and got a similar result of black being much more likely as the second pick if it was the first. Is this correct? How can it be understood in relation to 4 uncorrelated chess pawns in a bag where each measurement would have a probability of $\frac{1}{2}$ to be true?

答:
The 4 composition of the pouch are equiprobably thus this is not a parameter we have to consider (anyway nothing changes if you consider it)

You are requested to calculate

$$\mathbb{P}[BB|\text{First Black}]=\frac{\mathbb{P}[BB]}{\mathbb{P}[\text{First Black}]}=\frac{0+0+\frac{3}{4}\cdot\frac{2}{3}+1}{0+\frac{1}{4}+\frac{3}{4}+1}=\frac{3}{4}$$

Title: Picking pawns from pouch


问:
Is every matrix in $M_n(\mathbb{C})$ a sum of symplectic matrices?
A matrix $A \in GL(2n)$ is called symplectic iff $A^{t}JA = J$ where $J = \begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix}$. Is it true that every matrix in $M_{2n}(\mathbb{C})$ can be written as a sum of symplectic matrices? Any hints on how to prove this?

答: The result is indeed proved in the paper Every $2n$-by-$2n$ complex matrix is a sum of three symplectic matrices, in Linear Algebra and its Applications no. 517 (2017). For $n=1$ one can improve the result, i..e., that every complex matrix of size $2$ is the sum of two symplectic matrices. There are examples, which show that some matrices need at least three summands. So the fatcor $3$ is best possible in general.


问:
BlackWhite mathematics

I got this question:

In the famous Atlantis city BlackWhite the houses have five levels and each level is painted either by black or by white paint, but two adjacent floors can’t be both black or they would anger the gods.

(It is allowed to have adjacent white stories.)

a) How many different ways are there to paint a house in BlackWhite?

b) Try and find the different permutations for a certain number of levels per house.

Do you spot a pattern?

c) Explain why this pattern exists.

a) and b) are quite simple -- a) requires a simple tree diagram and b) requires a little bit of thinking:

```<code>1)      B               W                   2
        |              / \
        |             /   \
        |            /     \
        |           /       \
        |          |         \
        |          |          \
        |          |           \
        |          |            |
2)      W          B            W           3
       / \         |            |
      /   \        |           / \
     |     |       |          /   \
     |     |       |         /     \
3)   B     W       W        B       W       5
     |    / \     / \       |      / \
4)   W   B   W   B   W      W     B   W     8
    / \  |  / \  |  / \    / \    |  / \
5) B   W W B   W W B   W  B   W   W B   W   13
</code>```

You can trace the lines for a), and find the pattern for b): 2, 3, 5, 8, 13.
It's the Fibonacci sequence!
You can generalise how many possibilities there are per story if s is the number of stories with the Binet formula:
$$\frac {(\phi)^{s+2} - (\frac {-1}{\phi})^{s+2}}{\sqrt 5}$$
But I still can't see why this house problem, along with other problems I've seen like the rabbit problem, or Dudeney's cows all link to the Fibonacci sequence!

答:
We are going to define three sequences, $(B_n), (W_n), (C_n)$, counting respectively the number of ways to colour the $n$ first level with the last one being coloured in Black, in White, or as we want.

We clearly have $B_n + W_n = C_n$

Now, notice that we have some equalities:

$B_n = W_{n-1}$ (because you must have a White level before a Black one, and you can always a Black one on top of a White one).

$W_n = B_{n-1} + W_{n-1}$ (simply counting them by splitting your $n$-levels colouring in two groups, depending on whether the $n-1$-th level was Black or White)

$W_n = C_{n-1}$ This is because you have no restriction on what comes before or after a White level.

Combining those equations, we get (using them in order):

$C_n = B_n + W_n = W_{n-1} + W_n = W_{n-1} + (B_{n-1} + W_{n-1}) = W_{n-1} + (B_{n-1} + C_{n-2}) = C_{n-1} + C_{n-2}$

This is the equation defining the Fibonnaci sequence. It is easy to compute $C_0, C_1$ and see how they are related !

Title: BlackWhite mathematics


答:
As it was pointed out in comments, you can use a better bound for the logarithm:

$$
\dfrac{n \log(n+1)}{1+n^3 x^2} \leq \frac{n \sqrt{n}}{n^3 x^2}= \frac{1}{x^2} \frac{1}{n^{3/2}}
$$

which now yields a convergent series.

Title: How do we show that $\sum_{n=1}^{\infty}\frac{n \log (n+1)}{1+n^3x^2}$ converges?

Body:
Which criterion do we use to show the convergence of $$\sum_{n=1}^{\infty}\frac{n \log (n+1)}{1+n^3x^2}$$ ?

Do we maybe use that $\log (n+1)<n$ and so we get $\frac{n\log (n+1)}{1+n^3x^2}<\frac{n^2}{1+n^3x^2}$ ?

Since $1+n^3x^2>n^3x^2$ we get $$\frac{n\log (n+1)}{1+n^3x^2}<\frac{n^2}{n^3x^2}=\frac{1}{nx^2}$$ But this one diverges, right?

So do we have to find an other uppoer bound? Or do we use an other criterion?


问:
Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$
We have a parabola and a circle with the following equations and their graph placed at the end of my question.

Parabola: $y^2 = 4x -4$

Circle: $(x-2)^2 + y^2 = 9$

My goal was to calculate their intersection points so I substituted $y^2$ from the parabola equation into the circle equation and I got

$(x-2)^2 + (4x-4)=9 \implies x^2 - 4x + 4 + (4x - 4) = 9 \implies x^2 = 9 \implies x = \pm3$

$x=3$ is the only correct solution but why is $x=-3$ produced as an extra invalid solution?

What is the exact mathematical explanation behind this? Why substituting one equation into the other has produced extra answers?

update

When I calculate $x$ from the parabola equation and substitute it in the circle equation, I don't get any extra answers for $y$:

$y^2=4x-4 \implies y^2 +4 = 4x \implies x = \frac{y^2}{4} + 1$

$(x-2)^2 +(4x-4)=9 \implies ((\frac{y^2}{4} + 1) - 2)^2 + (4x - 4)=9 \implies y^4 +8y - 128 = 0 \implies y^2=8,-16$

$y^2 = -16$ cannot be true so $y^2 = 8 \implies y=\pm 2\sqrt{2}$ and these are correct answers for $y$.

2nd update

I made a mistake in the calculation in the previous update although the final solutions where correct. I write the correct calculation:

$(x-2)^2 +y^2=9 \implies
((\frac{y^2}{4} + 1) - 2)^2 + y^2=9 \implies
(\frac{y^2}{4} - 1)^2 + y^2=9 \implies
(\frac{y^4}{16} - \frac{y^2}{2} + 1) + y^2=9 \implies
\frac{y^4}{16} + \frac{y^2}{2} + 1=9 \implies
(\frac{y^2}{4} +  1)^2=9 \implies
(\frac{y^2}{4} +  1)=\pm3 \implies
\frac{y^2}{4} =2,-4 \implies
y^2=8,-16$

答:
I suppose that we are working on the real number system and not consider complex numbers but I think my reasoning works in complex numbers too. We want to solve a system of equations containing two equations. I move all terms of each equation to one side and name them $A(x,y)$ and $B(x,y)$

$\begin{cases}
      A(x,y)=y^2 -4x +4=0\\
      B(x,y)=y^2 +(x-2)^2-9=0\\
    \end{cases}$

We want to find tuples like $(x', y')$ such that $A(x',y')=0$ and $B(x',y')=0$ simultaneously.

If we calculate $y^2$ from $A(x,y)=0$ and substitute it into $B(x,y)$ we arrive at a third equation $C(x)=0$

$C(x) = 4x-4 +(x-2)^2-9 = 0 \implies C(x) = x^2 -9=0$

And when we solve $C(x) = 0$ we get $x^2=9 \implies x=\pm 3$ but any point $(-3,y)$ with $y\in\mathbb{R}$ does not satisfy $\begin{cases}
      A(x,y)=0\\
      B(x,y)=0\\
    \end{cases}$
and the solving procedure has produced extraneous solutions. The reason is this the line of reasoning is not reversible.

$\begin{cases}
      A(x,y)=0\\
      B(x,y)=0\\
    \end{cases} \overset{1}{\implies} C(x)=0 \overset{2}{\iff} x=\pm 3$

The $\overset{1}{\implies}$ is not reversible. In this case when there exists $(x_0,y_0)$ as a solution to the system we will have:

$\begin{cases}
      A(x_0,y_0)=0\\
      B(x_0,y_0)=0\\
    \end{cases} \overset{3}\implies C(x_0)=0 \overset{4}\iff x_0=\pm 3$

but if there exits $(x_1,y_1)$ as a solution to $C(x) = 0$ we will have

$C(x_1)=0 \;\not\!\!\!\implies \begin{cases}
      A(x_1,y_1)=0\\
      B(x_1,y_1)=0\\
    \end{cases}$

When we combine the equations we loose information and cannot retrive the system $\begin{cases}
      A(x_1,y_1)=0\\
      B(x_1,y_1)=0\\
    \end{cases}$ from $C(x)=0$
so every solution to the system is a solution to $C(x)=0$ but we cannot say every solution to $C(x)=0$ must be a solution to the system and extraneous solutions might have been produced.


问:
Solutions of $y'-2y=1$
I claim the solution of $y'-2y=1$, is $$\phi(x)=e^{2x}\int_{x_0}^xe^{-2t}dt +ce^{2x}=e^{2x}\left(-\frac{1}{2}e^{-2x}\right)+ce^{2x}=-\frac{1}{2}+ce^{2x}$$

where $x_0$ is a fixed point on the interval where the function $b(x)=1$ is continuous.

Apparently the correct solution is $-\frac{1}{2}+ce^{-2x}$.

A second question is, why can I evaluate the integral above and ignore the term which involves $\frac{1}{2}e^{-2x_0}$ after applying FTC? Somehow it vanishes, and this integral is basically just the primitive evaluated at $x$.

答:
We can check whether your solution works:
\begin{align}
\phi(x)&=e^{2x}\int_{x_0}^xe^{-2t}dt +ce^{2x}\\
\phi'(x) &=2e^{2x}\int_{x_0}^xe^{-2t}dt + e^{2x}e^{-2x} + 2ce^{2x}=2e^{2x}\int_{x_0}^xe^{-2t}dt+1+2ce^{2x}.
\end{align}

Now,
\begin{align}
\phi'(x) - 2\phi(x) &= 2ce^{2x}=2e^{2x}\int_{x_0}^xe^{-2t}dt+1+2ce^{2x} - 2e^{2x}\int_{x_0}^xe^{-2t}dt +-2ce^{2x} = 1\\
\end{align}
which means that your solution is correct.  Now, to simplify $\phi(x)$:
\begin{align}
\phi(x)&=e^{2x}\int_{x_0}^xe^{-2t}dt +ce^{2x}\\
&=-\frac{e^{2x}}{2}\left(e^{-2t}\right)\Bigg|_{x_{0}}^{x} + ce^{2x}\\
&=-\frac{e^{2x}}{2}\left(e^{-2x} - e^{-2x_{0}}\right) + ce^{2x}\\
&=-\frac{1}{2}+\frac{e^{-2x_{0}}}{2}e^{2x} + ce^{2x}\\
&=-\frac{1}{2} + \left(\frac{e^{-2x_{0}}}{2} + c\right)e^{2x}\\
&=-\frac{1}{2} + Ce^{2x},
\end{align}
where $C = \frac{e^{-2x_{0}}}{2} + c$ is a constant.  So it seems like there is a typo in the answer you were given, and the reason that you can seemingly ignore the term with $\frac{1}{2}e^{-2x_0}$ is because it gets absorbed into the integration constant.

Title: Solutions of $y'-2y=1$


问:
Example to show that a bimonotonic function on a cartesian product of closed intervals on $\mathbb{R}$ need not be bounded.
I've come to a problem in a multivariate calculus book that I'm having trouble with.

Let $I$ and $J$ be closed and bounded intervals in $\mathbb{R}$. Give an example to show that a bimonotonic
function on $I \times J$ need not be bounded.

Here a function $f : \mathbb{R}^{2} \rightarrow \mathbb{R}$ is bimonotonically increasing if :
\begin{equation}
x_{1} \leq x_{2} \text{ and } y_{1} \leq y_{2} \Rightarrow f(x_{1},y_{2})+f(x_{2},y_{1}) \leq f(x_{1},y_{1}) + f(x_{2},y_{2})
\end{equation}
And a function is bimonotonically decreasing if :
\begin{equation}
x_{1} \leq x_{2} \text{ and } y_{1} \leq y_{2} \Rightarrow f(x_{1},y_{2})+f(x_{2},y_{1}) \geq f(x_{1},y_{1}) + f(x_{2},y_{2})
\end{equation}
A function is called bimonotonic if it is bimonotonically increasing or bimonotonically decreasing.

Can anyone come up with an example ?

答:
Here's an easy example:

Let $f': I \rightarrow \mathbb{R}$ be an unbounded function on the closed interval $I$, and let $$ f(x,y) := f'(x) \quad \forall x\in I, y \in J\,, $$
for some closed interval $J$.

Then $f$ is unbounded and bimonotonically increasing (and decreasing), because $$ f(x_1,y_2) + f(x_2,y_1)
 = f'(x_1) + f'(x_2) = f(x_1,y_1) + f(x_2,y_2) \,.$$


问:
Find \limsup_{n \rightarrow \infty}A_n and \liminf_{n \rightarrow \infty}A_n
Noy sure my understanding is right here. The $\limsup = A \cup B, \; \liminf=A\cap B$ because given any $n$, $A \cup B$ exists. Not sure how to explain $\liminf$ exactly.

答:
Hint.

Look for the definitions of $\liminf$ and $\limsup$ for sets. Then study the following sets
$$
\bigcap_{n\ge m}A_n,\quad
\bigcup_{n\ge m}A_n, \quad
\bigcup_{m\ge 1}\bigcap_{n\ge m}A_n,\quad
\bigcap_{m\ge 1}\bigcup_{n\ge m}A_n
$$


问:
Equality of integrations
I have a function $f(x)$ positive for $x>0$ and $f(0)=0$. I also have functions $g,h,k$ and a constant $a$ s.t. for all $t$:

$$\int_0^t f(t-x) g(x)dx=- k(t)\int_0^t f(t-x) h(x)dx$$

So I propose that, if $k(t)\neq 0$, this is equal to

$$\int_0^t f(t-x) \left(\dfrac{g(x)}{k(t)}+h(x) \right) dx=0$$

Once $f(t-x)\geq 0$, so I must have $\dfrac{g(x)}{k(t)}+h(x)\equiv 0$ for all $x\leq t$. In particular, $g(t)=-h(t)k(t)$ for all $t$.

I did not see error on this calculation. However, it seems a little naive, so I 'd like to understand if this calculaton is correct. Thank you!

答:
Just consider the integral $\int_{-1}^1 x^2 \cdot x \,dx$. You have that $x^2 \ge 0$, the integral is zero, but $g(x)= x$ is not identically zero.

In this case, maybe you can get something out of differentiating both sides on the relation...

$$
\frac{d}{dt} \int_0^t f(t-x) g(x) dx = f(t-t)g(t) + \int_0^t f'(t-x)g(x) dx= \int_0^t f'(t-x)g(x) dx
$$

$$
\frac{d}{dt} \left(-k(t) \int_0^t f(t-x) h(x) dx\right) = -k'(t) \int_0^t  f(t-x) h(x) dx -k(t) \int_0^t f'(t-x) h(x) dx
$$

Title: Equality of integrations


答:
Check that

$y(x)=A e^{-x^2/2}$ is solution of $$\frac{y''(x)}{y(x)}=x^2-1$$
and $y(x)=B e^{x^2/2}$ is solution of $$\frac{y''(x)}{y(x)}=x^2+1$$
It is in this sense Griffith has claimed that $$\psi(x)=A e^{-x^2/2}+ B e^{x^2/2}~~~~(1)$$ is only APPROXIMATE solution of $$\psi''(x)-x^2\psi(x)=0~~~~~~~(2)$$

So please note that (1) is not the exact solution of (2) instead it is an APPROXIMATE solution of (2).

Title: How to solve this differential equation via power series

Body:
Studying Quantum Mechanics, I encountered the following differential equation:

how to solve $$y''-x^2y=0$$
Griffiths, the author of the book, just said that its solutions are
$$y=Ae^{x^2/2}+Be^{-x^2/2}$$
I can immediatelly see that this is true, but I do not know how to get to this solution, and I cannot find it solved anywhere else.
I tried to use the power series method:
$$y=\sum_{n=0}^{\infty}c_n x^n$$
So the equation becomes:
$$\sum_{n=0}^{\infty}c_{n+2}x^n(n+1)(n+2)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
Here I need to get both sums to the same power of n, so I removed the first two terms from the first sum, so that I can later change the index:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}c_{n+2}x^n(n+1)(n+2)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
now I change the index in the fist sum:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}c_{n+4}x^{n+2}(n+4)n(n+3)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
Now I factor out the sum:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}x^{n+2}(c_{n+4}(n+4)(n+3)-c_n)=0$$
Now all three terms must be zero at the same time, so $c_2=c_3=0$, and:
$$c_{n+4}(n+4)(n+3)-c_n=0$$
$$c_{n+4}=\frac{c_n}{(n+4)(n+3)}$$
The first terms of the series are:
$c_4=\frac{c_0}{4\times 3}$, $c_8=\frac{c_4}{8\times 7}=\frac{c_0}{8\times 7\times 4\times 3}$
However, I don't really know how to proceed from here. I tried putting the coefficients into the sum for $y$, but I can't really come close to the answer. Could you please guide me as how to advance from here?


问:
False Interpretation of a logical statement
I have to find a false Interpretation of $$(\forall x)[P(x) \to Q(x)] \land (\exists x)(P(x) \lor Q(x)) \to (\exists x)(P(x) \land Q(x))$$

Isn't it true that $(\forall x)[P(x) \to Q(x)]$ means $\exists x(P(x) \land Q(x))$ ?

Thank you for your time.

答:
No, because if $Q$ is a tautology (meaning that $Q(x)$ is always true, for example $Q(x) \equiv (x = x)$  then $P(x) \to Q(x)$ is always a true implication.

This, by the way, shows you that your initial logical formula is false: indeed, take $Q$ any tautology (always True), $P$ any contradiction (always False), and show that this satisfies the left part of your implication but not the right part.


问:
Modulo of factorial
How do I solve for M modulo 2021 where M is the product of all numbers relatively prime to 2021 that are less than 2021.
I know Wilson's Theorem but that's applicable for modulo p where p is prime. Should I use CRT by breaking 2021 into 43 and 47 but even that seems intractable to me. Any help is appreciated.</p>

答:
All of these numbers will have multiplicative inverses (mod 2021), so they all cancel, with the exception of those numbers which are their own inverses. So $M$ is the product of the residues whose square is $1$.

We can find all such $x$ using the CRT to be $1, 988, 1033, 2020$. So their product $M = 1$.

Title: Modulo of factorial


问:
Norm of $L^2$ functional
Consider the functional $T: L^2((0,1)) \rightarrow \mathbb{R}$
$$
T(f) = \int_0^1 f(x) (1-x) dx.
$$
Calculate $\lVert T \rVert$.

I can show that
$$
|T(f)| \leq \int_0^1 |f(x) (1-x)| dx \leq \left( \int_0^1 |f(x)|^2 dx \right)^{1/2}
\left( \int_0^1 (1-x)^2 dx \right)^{1/2}= \lVert f \rVert_2 \sqrt{\frac{1}{3}}
$$
using Hölder inequality. Which means $\lVert T \rVert \leq \sqrt{\frac{1}{3}}$.

Now I need to find $f \in L^2(0,1)$, $\lVert f \rVert \leq 1$ (or a sequence) such that
$\lVert T(f) \rVert = \sqrt{\frac{1}{3}}$. How can I do that?

Thanks.

答: In both inequalities in the question we have inequality if $f(x)=\alpha(1-x)$ for $\alpha\in\mathbb{R}$. In particular, if $\alpha=\pm\sqrt{3}$, then $\|f\|_2=1$ and $\|Tf\|_2=\frac1{\sqrt{3}}$.

Title: Norm of $L^2$ functional


答:
You seem to be drawing without replacement.

Let us consider three possible prior distributions for starting with $b_4=0,1,2,3,4$ black balls in a bag:

$p_1(b_4) = \left(\frac14,\frac14,0,\frac14,\frac14\right)$ - your problem

$p_2(b_4) = \left(\frac15,\frac15,\frac15,\frac15,\frac15\right)$ - allowing a $2+2$ bag

$p_3(b_4) = \left(\frac1{16},\frac4{16},\frac6{16},\frac4{16},\frac1{16}\right)$ - binomial if each piece independent

You then happen to draw a Black ball (event $B_1$), which has probability $\mathbb P(B_1 \mid b_4)= \frac{b_4}4$.  Your posterior probabilities are then proportional to $\left(0,\frac1{16},0,\frac3{16},\frac4{16}\right)$ or $\left(0,\frac1{20},\frac2{20},\frac3{20},\frac4{20}\right)$ or $\left(0,\frac4{64},\frac{12}{64},\frac{12}{64},\frac{4}{64}\right)$ which normalised by the probability of $\mathbb P(B_1)=\frac12$ of having drawn a Black ball so they sum to $1$, gives

$p_1(b_4 \mid B_1) = \left(0,\frac1{8},0,\frac3{8},\frac4{8}\right)$ - your problem

$p_2(b_4 \mid B_1) = \left(0,\frac1{10},\frac2{10},\frac3{10},\frac4{10}\right)$ - allowing a $2+2$ bag

$p_3(b_4 \mid B_1) = \left(0,\frac1{8},\frac{3}{8},\frac{3}{8},\frac{1}{8}\right)$ - if each piece independent (in effect binomial)

The probability of drawing a second Black piece (event $B_2$) given the first was Black is then

$\mathbb P_1(B_2 \mid B_1) = \frac1{8}\times \frac03+0\times \frac13 +\frac3{8}\times \frac23+\frac4{8}\times \frac33 = \frac34$ - your problem

$\mathbb P_2(B_2 \mid B_1) = \frac1{10}\times \frac03+\frac2{10}\times \frac13 +\frac3{10}\times \frac23+\frac4{10}\times \frac33 = \frac23$ - allowing a $2+2$ bag

$\mathbb P_3(B_2 \mid B_1) = \frac1{8}\times \frac03+\frac3{8}\times \frac13 +\frac3{8}\times \frac23+\frac1{8}\times \frac33 = \frac12$ - if each piece independent

These look intuitively sensible results, as the first problem makes the first draw to be informative about the rest of the bag because the prior distribution is not what you would have if the pieces are independent.

Title: Picking pawns from pouch

Body:
You have a pouch containing 4 chess pawns. Your prior knowledge is that the pouch does
not contain exactly 2 black and 2 white pieces. All other proportions of colours are equally
probable (e.g. 3 black and 1 white is equally probable as 4 black pieces).

The first piece drawn from the bag is black. What is the probability that the next piece drawn is also black?

I tried solving this problem in the following way:

The states can be represented by the number of black pieces in the pouch and is then written as:
$$B = \{0,1,3,4\}$$
, with probability distribution
$$P = \{\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\}$$.
The possible states of a measurement are picking up a black piece or picking up a white one. We denote these states as
$$C = \{b, w\}$$
The first measurement has a probability distribution of
$$Q = \{q_{b}, q_{w}\} = \{\frac{1}{2},\frac{1}{2}\}$$
where we are equally likely to pick up a white or a black piece. This is due to the weighted sum of black and white pieces in the possible states being equal. Let's say we pick a black piece.
Now reevaluate the probabilities of the states $B$. This can be done by counting in how many ways we could have picked a black piece from each state and weighting the probabilities by that number.
$$P|b = \{0,\frac{1}{8},\frac{3}{8},\frac{4}{8}\}$$
From these probabilities we can once again do a weighted sum of black and white pieces. It's a bit more complicated this time so I'll write it out.
$$b_{sum} = 0\cdot-1+\frac{1}{8}\cdot 0 +\frac{3}{8}\cdot 2 +\frac{4}{8}\cdot 3 = \frac{18}{8} = \frac{9}{4}$$
$$w_{sum} = 0\cdot4+\frac{1}{8}\cdot 3 +\frac{3}{8}\cdot 1 +\frac{4}{8}\cdot 0 = \frac{6}{8} = \frac{3}{4}$$
$$Q|b = \{q_b|b, q_w|b\} = \{\frac{b_{sum}}{b_{sum} + w_{sum}}, \frac{w_{sum}}{b_{sum} + w_{sum}}\} = \{\frac{3}{4}, \frac{1}{4}\}$$

This result was so unintuitive to me that I had to doubt the logic of my reasoning. I tried redoing the calculations with a bag of pieces where the state of two black and two white was allowed and got a similar result of black being much more likely as the second pick if it was the first. Is this correct? How can it be understood in relation to 4 uncorrelated chess pawns in a bag where each measurement would have a probability of $\frac{1}{2}$ to be true?


答:
There is something wrong somewhere since, if
$$y=A e^{\frac{x^2}{2}}+B e^{-\frac{x^2}{2}}$$ then
$$y''-x^2 y=A e^{\frac{x^2}{2}}-B e^{-\frac{x^2}{2}}$$ which is obviously not $0$.

If fact, the solution of the given differential equation is much more complex
$$y=c_1 D_{-\frac{1}{2}}\left(\sqrt{2} x\right)+c_2 D_{-\frac{1}{2}}\left(i \sqrt{2}
   x\right)$$ where appear the parabolic cylinder functions.
It can also write
$$y=c_1 e^{-\frac{x^2}{2}} H_{-\frac{1}{2}}(x)+c_2 e^{\frac{x^2}{2}} H_{-\frac{1}{2}}(ix)$$ where appear Hermite polynomials.

So, the series solution is really the best way (at least, for the time being).

Edit

Coefficients $c_m$ are given by
$$c_m=\frac{2^{-m} \Gamma \left(\frac{3}{4}\right) \left(k_1+k_2 (-1)^m+\left(k_1-k_2\right)
   \sin \left(\frac{\pi  m}{2}\right)+\left(k_1+k_2\right) \cos \left(\frac{\pi
   m}{2}\right)\right)}{\Gamma \left(\frac{m}{4}+1\right) \Gamma
   \left(\frac{m+3}{4}\right)}$$ which make
$$y=\frac{2 \pi  \sqrt{x}}{\Gamma \left(\frac{1}{4}\right)}\left(\left(k_1-k_2\right) I_{\frac{1}{4}}\left(\frac{x^2}{2}\right)+\left(k_1+k_2\right)
   I_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)\right)$$


问:
If $F$ is a finite field of size $q=p^n$ and $b\in K$ is algebraic over $F$ then $b^{q^{m}} = b$ for some $m > 0$
I want to apply a similar type of argument to show that $\alpha^{q} = \alpha$ for all $\alpha\in F$.  When we know that the characteristic of $F$ is $p$ and $|F| = q = p^{n}$.  But I dont know how to incorporate the condition that $b\in K$ is algebraic over $F$.  I know there is some $f\in F[t]$ such that $f(b) = 0$.  Also, $F(b)/F$ is finite... I don't know how to proceed. Any suggestions?  </p>

答: If $b=0$ then you're done. Otherwise it is a unit and the multiplicative order of a unit $b$ in the group of units divides the order of said group. Given that $F(b)$ is a vector space over a field of size $q$, we find that $|F(b)|=q^m$ for $m=\dim_{\,F}F(b)$. What is the size of the group of units in $F(b)$ then?</p>


问:
How to show that $4n^2 +20n\log(n)+2000n+4 \log(n)+1111$ is in $\Theta(n^2)$?
I am practising these types of questions for more understanding. But I am not getting how to show it is in $\Theta(n^2)$. I tried solving to show it is in ${O}(n^2)$ but I am not sure my approach to this problem is correct or not. Can you please help me to understand?

I did a few observations as $\log(n) \leqslant n$ for all $n > 0$ so we can replace $\log(n)$ with $n$. Also $n \leqslant n^2$ for all $n \geqslant 1$. So this way I'll get

$$4n^2 + 20n\log(n) + 2000n + 4 \log(n) + 1111 \leqslant 4n^2 +20 (n^2)+ 2000n + 4(n) +1111,$$ for $n>0$.

With another estimation, I've considered $$n \leqslant n^2$$ for $n > 0$, so replacing $n$ with $n^2$ yields to

$$4n^2 + 20n^2 + 2000n + 4n +1111 \leqslant 4n^2 + 20n^2 + 2000n^2 + 4n^2 + 1111,$$ for $n>0$,
which is less or equal to
$$4n^2 + 20n^2 + 2000n^2 + 4n^2 + 1111n^2,$$  since $1111 \leqslant 1111 n^2$  for   $n > 0$.

This yields to $3136n^2$ and if I pick $n_0$ for $0$ my $C$ would become $3136$, hence $g(n)$ is in ${O}(f(n))$ as $g(n) \leqslant C \cdot f(n)$ by letting $C=3136$ and $n_0 \geqslant 0$.

Is this way correct? Also, can you help me to get big-$\Omega$ equation for the given $f(n)$?

答:
Yes, your reasoning is correct. In summary, you have obtained that
$$4n^2\leqslant 4n^2 + 20n\log n+ 2000n+ 4 \log n+ 1111 \leqslant 3136n^2,$$
for all $n\geqslant 1$, which is precisely what it means to be $\Theta(n^2)$ (as $n\to\infty$).

For big-$\Omega$ (assuming you are referring to Knuth's definition), we just need to show that for all $n\geqslant A$ (for some fixed $A$), the function $f(n) = 4n^2 + 20n\log n+ 2000n+ 4 \log n+ 1111 $ is larger than some fixed multiple of $n^2$.

But this is obviously true for $n\geqslant 1$ since
$$4n^2 + \underbrace{20n\log n+ 2000n+ 4 \log n+ 1111}_{>0} \geqslant 4n^2 = \Omega(n^2).$$


问:
max of absolute value of log convex, concave or neither?
$$f(x)=\exp\left\{\max_{i=1,\ldots,n}\left|\log(a_i^Tx)\right| \right\}$$
where $a_i \in \mathbb{R}^m$ and $\operatorname{dom} f = \{x \in \mathbb{R}^m_+ \mid a_i^Tx>0, i = 1,\ldots,n \}$

I know that $|\log(a_i^Tx)|$ is neither convex or concave but is this also the case for $f$? If so how can I show that?

答:
Since $\exp$ is a monotone transformation, $\exp(\max_i y_i) = \max_i \exp(y_i)$ for any vector $y$. Therefore:

$$f(x) = \exp\left\{\max_{i=1,\ldots,n}\left|\log(a_i^Tx)\right| \right\} = \max_{i=1,\ldots,n} \exp\left\{\left|\log(a_i^Tx)\right| \right\}$$

Since the maximum of convex functions is convex, it suffices to check if $g_i(x) = \exp\left\{\left|\log(x)\right| \right\}$ is convex. You can either plot a graph to convince yourself it's convex, or you can formally prove it.


问:
Ring homomorphism $f: \Bbb{Z}_p[X] \to \Bbb{Z}_p^{n\times n}$. Prove it is never bijective and find the amount of elements in Im$(f)$.
Let $f: \mathbb{Z}_p[X] \to \mathbb{Z}_p^{n \times n}$
with $n \geq 2$ be a ring homomorphism with $f(1) = I_n, f(X) = M$.

a) Prove it is never injective and never surjective.

b) Find the minimal polynomial and the amount of elements $ \operatorname{Im}(f)$ with $p=3, n=6$ and with $M$ given:
$f(X)= M  = \left( \begin{matrix} 2 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} \right).$

a) I first proved that $f$ is fully determined by $f(X)$.
$f(p(X)) = f(\sum_{i=1}^n a_iX^i) = \sum_{i=1}^n f(a_i)f(X^i) =  \sum_{i=1}^n a_if(X)^i$, so we see that indeed $f$ is determined by the value of $f(X)$. (These operations are allowed since $f$ is a ring homomorphism).

Then I tried proving that it is never injective.
Since $M \in \mathbb{Z}_p^{n \times n}$, I know that there exists a $q \in \mathbb{N}$ such that $M^q = 0$ with $q$ a divisor of $p$.
Consider $p(X) = X^i$, then $f(p(X)) = p(f(X)) = p(0) = 0$, which means that $p(X) \in \text{ker}(f)$ and since the kernel is not equal to $\{0\}$ we know it is not injective.

I am not sure how to prove that it is never surjective.

b) I think that the minimal polynomial is equal to $\phi_M(X) = (X-2)^2(X-1)$ since there are 2 Jordan blocks for eigenvalue 2 and 1 for eigenvalue 1.

I don't really know how to calculate the amount of elements that $ \operatorname{Im}(f)$ has.

答:
For a): Your proof of injectivity is not correct. For example if $M$ is the identity matrix, then $M^k=I\ne0$ for all $k$. One way would be to argue via cardinalities: $\Bbb Z_p[X]$ is infinite while $\Bbb Z_p^{n\times n}$ is finite, so ....

A bit cleaner/more general argument would be to look at the dimension. $\Bbb Z_p^{n\times n}$ has dimension $n^2$, hence the set $\{I,M,\dots,M^{n^2}\}$ is linearly dependent, so there are coefficients $a_0,\dots,a_{n^2}$, not all equal to $0$, such that $$a_0I+\dots+a_{n^2}M^{n^2}=0$$ Now see how you can make a non-zero polynomial $p$ out of this such that $p(M)=0$.

Hint for the surjective part: $\Bbb Z_p[X]$ is commutative while $\Bbb Z_p^{n\times n}$ is not.

For b): For the minimal polynomial you need to look at the size of the largest Jordan block for each eigenvalue and not the number of Jordan blocks.

Assume that we found the minimal polynomial $\phi$ and say it is of degree $d$. We know that $\ker f=\langle\phi\rangle$, hence $$\operatorname{im} f\cong \Bbb Z_p[X]/\langle\phi\rangle$$
In general for a field $K$ and a polynomial $p$ over $K$ of degree $n$ we have that $K[X]/\langle p\rangle$ is a $K$-vector space of dimension $n$ (if you don't know this try to prove this, e.g. by explicitely writing down a basis). So in this case we know that $\operatorname{im} f$ is a $\Bbb Z_p$-vector space of dimension $d$. What does this tell us about the cardinality?


问:
Subspaces for the set of polynomials with the property $p(1)=p(2)=p(3)=0$

Let $P$ be the set of all polynomials. Let $A \subset P$ such that for every polynomial the following holds $p(1)=p(2)=p(3)=0$. Give an example of a subspace $V \subset A$ and $A \subset W$ such that $\{0\} \ne V \ne A$ and $A \ne W \ne P$.

The last conditions seem very confusing to me. What's with all the, not equal restrictions?

For the first part I would need to find a subset $V$ of $A$ for which the following properties hold.

$0 \in V$

If $u,v \in V$ then $u+v \in V$

If $u \in V$ then $cu\in V$, when $c\in \mathbb{R}$

If $P$ is the set of all polynomials and $A$ is a subset of $P$ could I take $V$ to be for example the set of polynomials of let's say degree $3$ for which the condition holds? Any hints on this would be appreciated...

答:
Note that the a polynomial of degree $3$ such that the condition holds is of the form
$$c(x-1)(x-2)(x-3)$$
with $c\not=0$ which is not a linear space because it does not contain the zero polynomial.

Indeed, as  $$ you can take the set of polynomials of degree $\leq 3$ for which the condition holds, that is the one generated by the polynomial $(x-1)(x-2)(x-3)$:
$$V=\{c(x-1)(x-2)(x-3): c\in\mathbb{R}\}.$$
Evidently $V\subset A$, and $V\not=A$ because $x(x-1)(x-2)(x-3)\in A\setminus V$.

On the other hand, as $W$ you can take the set of all polynomials of any degree such that $p(1)=0$. We have that $A\subset W$ and $W\not=A$ because $(x-1)\in W\setminus A$.


问:
$\lim_{n\to\infty} \left(\frac{\left(1 + \frac{1}{n}\right)^n}{e}\right)^n$
Could anyone please help me find limit?
$$\lim_{n\to\infty} \left(\frac{\left(1 + \frac{1}{n}\right)^n}{e}\right)^n$$

Thank you

答:
Welcome to Math.SE! To guide you properly, it would be very helpful to understand the context of what you are asking and what you tried already.

To get you started, note that since $f(x) = \ln x$ is a continuous function for $x>0$, you have
$$
\lim_{n \to \infty} g(n)
 = \exp\left( \ln\left(\lim_{n \to \infty} g(n)\right)\right)
 = \exp\left( \lim_{n \to \infty} \ln(g(n))\right)
$$
and taking logarithm of the function you are taking the limit of will greatly simplify it.

Can you finish this now? If not, please update with details of what you tried and what is not working so we may guide you further.


问:
Prove $\frac{d}{dt} \Big|_{t=0}\mbox{tr}(e^{X+tY})=\mbox{tr}(e^XY)$
I’m asked to prove $\frac{d}{dt}\Big|_{t=0}\mbox{tr}(e^{X+tY})=\mbox{tr}(e^XY)$ for any $X,Y$ in $M_n(\mathbb{C})$. My attempt is to assume both $X$ and $Y$ are diagonalizable, and since the set of all diagonalizable matrices is dense in $M_n(\mathbb{C})$, if we can show this is true for diagonalizable matrices, then we are done. I expected this will somehow simplify the proof, but seems it does not work well unless I further assume $X,Y$ can be diagonalizable at the same time. Any suggestions on this?

答:
For ease of typing, define
$$\eqalign{
A&= A(t) = X+tY
\quad\implies\quad
 &dA = Y\,dt \\
 &&X = \lim_{t\to 0}\,A(t) \\
}$$
The Frobenius product notation for the trace will also prove convenient
$$A:B = {\rm Tr}(A^TB) = B:A$$
First, note that the Taylor series is valid for the matrix exponential
$$\eqalign{
e^A = \sum_{k=0}^\infty \frac{A^k}{k!}
\quad\implies\quad
de^A &= \sum_{k=0}^\infty\frac{1}{k!}\sum_{j=0}^{k-1} A^j\,dA\,A^{k-1-j} \\
}$$
Next, write the objective function and calculate its differential, gradient and derivative.
$$\eqalign{
\phi &= {\rm Tr}(Ie^A) \\
&= I:e^A \\
d\phi &= I:de^A \\
 &= I:\left[\sum_{k=0}^\infty\frac{1}{k!}\sum_{j=0}^{k-1} A^j\,dA\,A^{k-1-j}\right] \\
 &= \left[\sum_{k=0}^\infty\frac{1}{k!}\sum_{j=0}^{k-1} A^j\,I\,A^{k-1-j}\right]^T:dA \\
 &= \left[\sum_{k=0}^\infty\frac{1}{k!}\sum_{j=0}^{k-1} A^{k-1}\right]^T:dA \\
 &= \left[\sum_{k=0}^\infty\frac{A^k}{k!}\right]^T:dA
\qquad\implies \frac{\partial\phi}{\partial A} = \left[e^A\right]^T \\
 &= \left[e^A\right]^T:Y\,dt \\
\frac{d\phi}{dt} &= {\rm Tr}(e^AY) \\
}$$
Finally, take the limit as $t\to 0$.
$$\eqalign{
\lim_{t\to 0}\left(\frac{d\phi}{dt}\right)
 &= \lim_{t\to 0}\,{\rm Tr}\left(e^AY\right) \\
 &= {\rm Tr}\left(e^XY\right) \\
}$$

The rearrangement property of the Frobenius product was used in some of the steps above, e.g.
$$\eqalign{
A:BCD &= B^TA:CD \\
 &= AD^T:BC \\
 &= B^TAD^T:C \\
}$$
these follow directly from the cyclic property of the trace
$$
{\rm Tr}(ABCD)
 = {\rm Tr}(BCDA)
 = {\rm Tr}(CDAB) = etc.
$$

UPDATE

The above is an example of a formula (which can be found on page 12 of the Matrix Cookbook) which is valid for any analytic function $\;f(z)$
$$\eqalign{
\frac{\partial\,{\rm Tr}(f(A))}{\partial A} &= f^\prime(A)^T  \\
}$$
Let $f(z)=e^z$ then $f'(z)=e^z\;$ and the whole derivation reduces to
$$\eqalign{
\phi &= {\rm Tr}(e^A) \\
d\phi &= (e^A)^T:dA = {\rm Tr}(e^AY)\,dt \\
\frac{d\phi}{dt} &= {\rm Tr}(e^AY) \\
\lim_{t\to 0}\left(\frac{d\phi}{dt}\right) &= {\rm Tr}(e^XY) \\
}$$


问:
How to prove this set of congruences involving various moduli?
Consider:

$x\equiv 6\pmod {13}$

$x\equiv 0\pmod {43}$

$x\equiv 10^n \pmod {41}$.

For x and n nonnegative integers.

Using Wolphram I found a set of solutions:

$x=903+22919\cdot s$ is one solution

Another is:

$14319+22919\cdot s$

let’s call the generic solution:

$d+22919\cdot s=x$, with d an integer depending on n.

How can I prove that d is either $\equiv 0\pmod {215}$ or $\equiv 129\pmod {215}$ or in only one case $\equiv 43\pmod{215}$? Or is the statement false?

答:
If you first check the possibilities for $x\equiv 10^n \pmod {41}$ you will find that $x$ is congruent to one of $1,10,16,18,37$ modulo $41$.

You can now apply the Chinese Remainder Theorem to see that you require numbers of the form $43d+22919s$, where $43d\equiv 6 \pmod {13}$  and $43d\equiv 1,10,16,18,37 \pmod {41}$.

These conditions simplify to $d\equiv 8 \pmod {13}$  and $d\equiv 5,8,9,21,39 \pmod {41}$.

It should be straightforward now but ask if anything is unclear! The following answers should allow you to check if you have the right idea. These are the values for the $d$ in your post.

$$344,903,10965,14319,17114.$$


问:
How to represent a sequence of odd numbers given specific criterea
I'm trying to figure out how to represent a sequence of ODD numbers given the following conditions:

1) I know how many numbers are in the sequence (N).
2) I know the average of all the numbers in the sequence.
3) I know the sum of all the numbers in the sequence.

Example:  Given the sequence 21, 23, 25, 27, and 29 (N == 5), the average of the sequence is 25, and the sum of the numbers in the sequence is 125...  How do I express that sequence mathematically?  I need it to work for values of N that are much greater than 5.

N+1 would yield a sequence of 31, 33, 35, 37, 39, 41 with the average being 36 or (N*N).

答:
For a sequence of $N$ consecutive odd numbers,
$$
a, a + 2, a + 4, \dots, a + 2(N - 1),
$$
the average is
$$
\begin{align}
\frac{1}{N} &\left[ a + (a + 2) + (a + 4) + \cdots + (a + 2(N - 1)) \right] \\
&= \frac{1}{N} \left[ (a + a + a + \cdots + a) + (0 + 2 + 4 + \cdots + 2(N - 1) \right] \\
&= \frac{aN + N(N - 1)}{N} \\
&= a + N - 1.
\end{align}
$$
So, if you know $N$ and the average $m$, from the equation $m = a + N - 1$, you can find the first number in your sequence:
$$
a = m - N + 1.
$$
Now, count by twos.</p>


问:
How to get co-relate the curves
I come across a question where I need to co-relate the function $f(t)$ and $g(t)$. The answer is supposed to be $g(t)=f(\frac{t}{2}-\frac{3}{2})$. Can someone explain the steps involved to reach the expression from the plots shown below?

答:
We can think of the graph of $g(t)$ as being a transformation of the graph of $f(t)$ by a sequence of two steps:

Stretch the graph of $f(t)$ horizontally by a factor of two, so that the space between the vertical bars is two rather than one.  This gives us the graph of $f\left(\frac{t}{2}\right)$.

Shift the resulting graph to the right by $3$ so that the left corner point is at $x = 3$ instead of $x = 1$.  This gives us the graph of $f\left(\frac{t - 3}{2}\right)$.

We can simplify this transformation as $$f\left(\frac{t - 3}{2}\right) = f\left(\frac{t}{2} - \frac{3}{2}\right).$$

Title: How to get co-relate the curves


问:
$\mathbb Z[\sqrt{13}]$ is not a UFD.
I have seen complicated arguments to establish whether $\mathbb Z$ extended by this or that radical is a UFD.  They usually suppose that $u$ is a unit and then compute the reciprocal and then make arguments about each component being an integer.  Some of them go by defining a norm and proving things about that.  I'm a little confused about why these arguments are so elaborate when it seems to me that there is a very straight-forward argument in many cases.  My guess is that I'm overlooking some important caveat in one of my steps but I'm not sure where.

Take for example $\mathbb Z[\sqrt{7}]$.  To prove it's not a UFD I look at two factorizations of 6:

$$ 6=3\cdot 2 $$

$$ 6 = (\sqrt 7 - 1)(\sqrt 7 + 1) $$

I then argue that there is no unit (in fact no element at all) such that $3u=\sqrt 7-1$.  Since $u=a+b\sqrt 7$ then we would need

$$ 3a = -1 $$

but this is not true for any $a\in\mathbb Z$.  Of course the same argument could be used on any of $\pm 3,\pm 2$ on the left and $\pm (\sqrt 7-1),\pm (\sqrt 7+1)$ on the right.

Is there some logical error in saying that if $w+x\sqrt 7 = y+z\sqrt 7$ for integers $w,x,y,z$ then we must have $w=y$ and $x=z$?

答:
To repeat what KCd and Lubin had already said in a verbose way. While $2$ and $3$ are primes in $\mathbb{Z}$, but they are not primes in $\mathbb{Z}[\sqrt{7}]$ and hence not irreducible.

\begin{equation}
    2 = (3+\sqrt{7})(3-\sqrt{7})  \\
    3 = (\sqrt{7}+2)(\sqrt{7}-2)  \\
    \sqrt{7}+1 = (3-\sqrt{7})(5+2\sqrt{7})  \\
    \sqrt{7}-1 = -(3+\sqrt{7})(5-2\sqrt{7})
\end{equation}

Definitions: Let $R$ be an integral domain. Then an element $a \in R$ is said to be prime in $R$ if:

(a) $a \ne 0$ and $a$ is not a unit, and

(b) For all $b,c \in R$, $a \mid bc$ implies $a \mid b$ or $a \mid c$.

The element $a \in R$ is said to be irreducible (a weaker condition than primeness) if:

(a) $a \ne 0$ and $a$ is not a unit, and

(b) If $a = bc$ for some $b,c \in R$, then $b$ is a unit or $c$ is a unit.

If you are still unsure, you might want to look up the definition of unit (not unity). Very briefly, $a$ is a unit if and only if norm $N(a)=1$. BTW, $\mathbb{Z}[\sqrt{7}]$ is a Euclidean domain and hence a UFD.

Title: $\mathbb Z[\sqrt{13}]$ is not a UFD.


问:
Asymptotics of sum with delta function
How one should  compute the asymptotic of sum in these two form:

\begin{equation}
 \sum_{k,l,m,n=1}^\Lambda\delta_{k,l+m+n}\end{equation}
and
\begin{equation}
 \sum_{k,l,m,n=-\Lambda}^\Lambda\delta_{k,l+m+n},\end{equation}

where $\Lambda$ is some positive integer.

答:
The first sum is the number of choices of $l,\,m,\,n$ for which $l,\,m,\,n,\,l+m+n$ each range from $1$ to $\Lambda$. These comprise the lattice points in a pyramid; the asymptotic behaviour is of height $\Lambda$ and an isosceles right-angled base of leg length area $\Lambda$, $\tfrac13\cdot\Lambda\cdot\tfrac12\Lambda^2=\tfrac16\Lambda^3$.

The second sum admits a similar treatment. There are asymptotically $(2\Lambda)^3=8\Lambda^3$ choices of $l,\,m,\,n$, of which I'll show two thirds satisfy $|l+m+n|\le\Lambda$, making the sum's asymptotic behaviour $\tfrac{16}{3}\Lambda^3$. In particular, if $L,\,M,\,N$ are $U(-\Lambda,\,\Lambda)$ IIDs $L^\prime:=\tfrac{L+\Lambda}{2\Lambda}$ are $U(0,\,1)$ IIDs, and by the $n=3$ case of the Irwin–Hall distribution$$\begin{align}P(|L+M+N|\le\Lambda)&=P(1\le L^\prime+M^\prime+N^\prime\le2)\\&=\int_1^2\tfrac12(-2x^2+6x-3)dx\\&=\left[-\tfrac13x^3+\tfrac32x^2-\tfrac32x\right]_1^2\\&=\tfrac23.\end{align}$$

Title: Asymptotics of sum with delta function


问:
How to prove $\int_0^{2r} 2h \sqrt{r^2 -(x-r)^2}dx = \pi r^2 h$ without geometry?
I wanted to find the volume of a cylinder, radius r, height h by slicing it in to rectangles:

I placed the cylinder on the x-axis, one corner of the base diameter at (0,0) the opposite at (2r, 0). I have found that an area of a cross-section perpendicular to the x-axis is $A(x) = 2h \sqrt{r^2 -(x-r)^2}$ so:

$V=\int_0^{2r} 2h \sqrt{r^2 -(x-r)^2}dx$

I have tested this and it gives the volume correctly for various r and h. But how to show:

$\int_0^{2r} 2h \sqrt{r^2 -(x-r)^2}dx = \pi r^2 h$ without just saying "we know $V= \pi r^2 h$"?

This problem was my own device, perhaps it is not possible to do this.

答: I believe the difficulty comes down to finding $$\int\sqrt{a^2-x^2}\,dx$$ The standard technique is to substitute $x=a\sin\theta$, $dx=a\cos\theta\,d\theta$. Can you take it from there?</p>


问:
Pure State Definition Confusion
Let me start with some objects. Consider the $\mathrm{C}^*$-algebra $A$ defined by:
$$A=M_1(\mathbb{C})\oplus M_2(\mathbb{C})\subset B(\mathbb{C}^3).$$
Let $x=\mathbb{C}^3$ be given by $(e_1+e_2)/\sqrt{2}\,$  (the one dimensional factor acts on $e_1$).

We have a vector state given by $\rho_x(f)=\langle x,f(x)\rangle$, and

$$\rho_x\left((c_1)+\left(\begin{array}{cc}c_{11} & c_{12} \\ c_{21} & c_{22}\end{array}\right)\right)=\frac12 c_1+\frac{1}{2}c_{11}.$$

If I read something about quantum mechanics for such a state, as it is is given by a unit vector, it is called a pure state.

However this contradicts the answer to this question... and when I pick up Murphy he says that a pure state is such that if $\rho_0\leq \rho_x$ is a positive linear functional, then $\rho_0=t\rho_x$ for $t\in[0,1]$.

However for $\rho_x$ we have the linear functional:

$$\rho_0\left((c_1)+\left(\begin{array}{cc}c_{11} & c_{12} \\ c_{21} & c_{22}\end{array}\right)\right)=\frac12 c_1$$

is a bounded linear functional such that $\rho_x-\rho_{0}$ is a positive linear functional --- half the state given by the vector $e_2$, not $t\rho_x$... and so Murphy would say $\rho_x$ is not pure.

Can you help me with my confusion?

答:
The confusion is that the (introductory) quantum mechanical texts that I am reading are using the full $B(\mathsf{H})$ rather than closed self-adjoint subalgebras.

For example in full $B(\mathbb{C}^3)$, the state associated to the same vector is:

$$\rho_x\left([c_{ij}]_{i,j=1}^3\right)=\frac{c_{11}+c_{12}+c_{21}+c_{22}}{2},$$
a different beast to $\rho_x\in A^*$.

Title: Pure State Definition Confusion


问:
Prove that $C e^x$ is the only set of functions for which $f(x) = f'(x)$
I was wondering on the following and I probably know the answer already: NO.

Is there another number with similar properties as $e$? So that the derivative of $ e^x$ is the same as the function itself.

I can guess that it's probably not, because otherwise $e$ wouldn't be that special, but is there any proof of it?

答:
Ok, so all of the above answers either assume that the derivative of $e^x$ is itself, the derivative of $\log(x)$ is $1/x$, or some other theorem regarding Taylor series. Here is an argument which is, in some sense, from first principles.

Suppose you want to find a function which is it's own derivative. Suppose we have a real valued function $f(x)$ which is suitably well behaved such that every higher order "antiderivative" of $f$ to exist.

Let us denote the antiderivative of $f_0=f$ to be $f_1$, the antiderivative of $f_1$ to be $f_2$ and so on.

Suppose we consider the sum

\begin{equation}
E(x) = \sum_{i=0}^\infty f_i(x).
\end{equation}

Suppose we assume that the series is absolutely and uniformly convergent so that it can be differentiated term by term (this will be justified later on).

We end up with the equation

$$
\frac{d}{dx} E(x) = E(x) - f(x).
$$

Since $E$ is our guess answer for the question, we take our $f$ to be identically zero. This gives us that $f_1(x) = c_1$, $f_2(x) = c_1x+c_2$ and so on. But note here that the constants $c_1,c_2,\ldots$ should be chosen such that the series defining $E$ should converge absolutely and uniformly. Equivalently we can choose $c_i$ such that $C=\sum c_i < \infty$. Therefore we end up with the expression,

$$
E(x) = C + C\sum_{i=1}^\infty \frac{x^i}{i!}.
$$

This seems to answer the question but there is a slight caveat here.

Suppose $E$ is a function which is its own derivative, why should it be the sum of some $f_i$'s as given above?

Well, this question is answered if we can characterize the starting function $f$ in terms of $E$, which can be seen from the equation $f(x) = E(x) - E'(x)$. In fact this forces the choice of $f$ to be zero! Now, the claim is proved.


问:
How can we show that the equality doesn't occur?
Let $f:[0,1] \to [1,2]$, continuous on $[0,1]$ with $1<f(1)<2$.

Show that $$\int_0^1\left (f^2(x)-2f(x)\right )\, dx>-1.$$

We have that $$(f(x)-1)^2\geq 0\implies \int_0^1 (f(x)-1)^2\, dx\geq 0 \Rightarrow \int_0^1\left (f^2(x)-2f(x)\right )\, dx\geq -1.$$ Till here we have $\geq$ instead of $>$. How can we show that the equality doesn't occur?

Does the same holds also if $f$ is Riemann integrable?

答:
The equality occurs if and only if
$$\int_0^1 (f(x)-1)^2\, dx = 0 \iff \forall x\in(0,1), f(x)=1 \tag{1} $$
But because $f(1)>1$ and $f(x)$ continue, so there exists an interval $(1-\epsilon,1)$ near $1$ such that $\forall x \in (1-\epsilon,1), f(x)>1$.
Hence, the condition (1) can't be satisfied. That's why the equality doesn't occur.


问:
Question about when to use inclusion-exclusion principle.
I was solving a question that says that we're spreading $N^2$ balls into a grid $N \times N$, that has $N^2$ $1\times 1$ squares.

I had these two questions one after another and I failed the second one:

In how many ways we can spread the balls such that we get exactly two empty columns and in the rest of the little squares there's atleast one ball.
 So I just went and did $N \choose 2$${N^2-2N + 2N +1} \choose2N$, and got my answer.

2) In how many ways we can spread the balls such that we have exactly 2 empty columns?

And here I did the same thing I chose $2$ to be empty, and spread the balls in the rest of the empty squares.

When I went to look for the answer, this was answered using inclusion-exclusion principle, and I realized I have a big problem with understanding when should I really think about inclusion-exclusion, It's still my first practice exam, but I think it's a very big problem that I couldn't find the difference in this question and know in which way to answer.

 I would appreciate any tips about this topic from experienced people, that would really help me improve my understanding of this subject.

Thanks in advance to everyone.

答:
In the first problem you have two empty columns, and each of the remaining $N(N-2)$ squares is non-empty. This makes it a simple stars and bars problem: after you put one ball on each of those squares, you can distribute the remaining $2N$ balls to them arbitrarily, and this can be done in $\binom{N(N-2)+2N-1}{2N}$ ways.

In the second problem you must have exactly $2$ columns that are completely empty, so you must have at least one ball in each of the other $N-2$ columns, but any (or all) of those columns can be partly empty. It’s harder to count these arrangments. You might at first think that there are

$$\binom{N(N-2)+N^2-1}{N^2}=\binom{2N^2-2N-1}{N^2}$$

of them, since you’re distributing $N^2$ balls amongst $N(N-2)$ squares, but that includes all of the arrangements that leave one or more of the $N-2$ columns completely empty, and we don’t want to count those.

We can start with $\binom{2N^2-2N-1}{N^2}$ as a first approximation, but then we have to subtract the number of arrangements that leave one of those $N-2$ columns empty. For each of the $N-2$ columns there are

$$\binom{N(N-3)+N^2-1}{N^2}=\binom{2N^2-3N-1}{N^2}$$

ways to distribute the $N^2$ balls to the remaining $N-3$ columns, so a better approximation is

$$\binom{2N^2-2N-1}{N^2}-(N-2)\binom{N^2-3N-1}{N^2}\,.$$

Unfortunately, we’ve now subtracted too much: any arrangement of the balls that leaves $2$ of the $N-2$ columns completely empty has been counted once in the first term and subtracted twice in the second term, once for each of the $2$ extra empty columns, so it’s been counted altogether $-1$ times. We need to add it back in. There are $\binom{N-2}2$ pairs of columns that are not supposed to be completely empty, and for each pair there are

$$\binom{N(N-4)+N^2-1}{N^2}=\binom{2N^2-4N-1}{N^2}$$

arrangements that leave that pair completely empty, so the next correction gives us

$$\binom{2N^2-2N-1}{N^2}-(N-2)\binom{N^2-3N-1}{N^2}+\binom{N-2}2\binom{N^2-4N-1}{N^2}$$

as a third approximation. The full inclusion-exclusion computation leaves us with

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

ways to distribute the $N^2$ balls to the other $N-2$ columns so that each of them gets at least one ball, and since there are $\binom{N}2$ pairs of columns to be the empty ones, the final answer is

$$\binom{N}2\sum_{k=0}^{N-2}(-1)^k\binom{N-2}k\binom{2N^2-(k+2)N-1}{N^2}\,.$$


问:
Is there a closed expression for this one?
Has anyone ever seen a closed expression (and its sequence) for the following one:

$$\partial_x^{n} \, \text{arctanh}^k(x)|_{x=0}\,\,\,\,(1)$$

Provided $k,n \in \mathbb{N_0}$ and $x \in \mathbb{R}$.

When $k=1$, things become easier:

$$d_x^{n} \, \displaystyle\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}|_{x=0}, |x| < 1 $$

When $k=1$, it gives the following sequence:

$1,0,2,0,24,0,720,0,40320,0, \ldots$

Which clearly is:

$0!, 0,2!,0,4!,0,6!,0,8!,0 \ldots$

However, when $k=2$:

$0,0,2,0,16,0,368,0,16896,0,1297152, \ldots$

Which becomes hard to find a pattern.

But, for any natural $k$, is there another way to rewrite (1) and then find its sequence? Any suggestion on how to proceed? I don't think raising the McLaurin series to the $k$-th power and then expanding it is a good idea.

Thanks

答:
First approach. If $k>n$, the answer is $0$ since the Taylor series about $0$ begins with $x^k$. Suppose that $n\geq k$. By the Cauchy formula, the quantity in question is
\begin{align*}
\frac{{n!}}{{2\pi i}}\oint_{(0 + )} {\frac{{(\tanh ^{ - 1} z)^k }}{{z^{n + 1} }}dz}&  = \frac{{n!}}{{2\pi i}}\oint_{(0 + )} {\frac{{s^k }}{{(\tanh s)^{n + 1} }}\frac{1}{{\cosh ^2 s}}ds}
\\ & = \frac{{n!}}{{2\pi i}}\oint_{(0 + )} {\left( {\frac{{2s}}{{e^{2s}  - 1}}} \right)^{n + 1} e^{2s} \left( {\frac{{e^{2s}  + 1}}{2}} \right)^{n - 1} \frac{{ds}}{{s^{n - k + 1} }}} .
\end{align*}
Now, in terms of the generalised Bernoulli- and Euler polynomials,
$$
\left( {\frac{{2s}}{{e^{2s}  - 1}}} \right)^{n + 1} e^{2s}  = \sum\limits_{j = 0}^\infty  {2^j B_j^{(n + 1)} (1)\frac{{s^j }}{{j!}}}
$$
and
$$
\left( {\frac{{e^{2s}  + 1}}{2}} \right)^{n - 1}  = \sum\limits_{j = 0}^\infty  {2^j E_j^{(1 - n)} (0)\frac{{s^j }}{{j!}}} ,
$$
near $s=0$. Thus, our integral becomes
$$
\frac{{n!}}{{2\pi i}}\sum\limits_{j = 0}^\infty  {\left[ {\frac{{2^j }}{{j!}}\sum\limits_{m = 0}^j {\binom{j}{m}B_m^{(n + 1)} (1)E_{j - m}^{(1 - n)} (0)} } \right]\oint_{(0 + )} {\frac{{ds}}{{s^{n - k - j + 1} }}} } \\ = \frac{{2^{n - k} n!}}{{(n - k)!}}\sum\limits_{m = 0}^{n - k} {\binom{n-k}{m}B_m^{(n + 1)} (1)E_{n - k - m}^{(1 - n)} (0)} .
$$
In terms of the Stirling numbers of the first kind,
$$
B_m^{(n + 1)} (1) = \frac{{n - m}}{n}B_m^{(n)}  = \binom{n}{m}^{ - 1} s(n,n - m).
$$
The generalised Euler numbers may be expressed in terms of the Stirling numbers of the second kind as follows.
\begin{align*}
&\left( {\frac{{e^{2s}  + 1}}{2}} \right)^{n - 1}  = \left( {1 + \frac{{e^{2s}  - 1}}{2}} \right)^{n - 1}  = \sum\limits_{k = 0}^{n - 1} {\binom{n-1}{k}\frac{1}{{2^k }}(e^{2s}  - 1)^k }
\\ & = \sum\limits_{j = 0}^{n - 1} {\binom{n-1}{k}\frac{{k!}}{{2^k }}\sum\limits_{i = k - 1}^\infty  {2^j S(j,k)\frac{{s^j }}{{j!}}} }  = \sum\limits_{j = 0}^\infty  {\left[ {2^j \sum\limits_{k = 0}^j {\binom{n-1}{k}\frac{{k!}}{{2^k }}S(j,k)} } \right]\frac{{s^j }}{{j!}}} ,
\end{align*}
i.e.,
$$
E_j^{(1 - n)} (0) = \sum\limits_{k = 0}^j {
   \binom{n - 1}{k} \frac{{k!}}{{2^k }}S(j,k)} .
$$
Thus, after some simplification, the final result is that for $n\geq k$,
$$\boxed{
\left[ {\frac{{d^n }}{{dx^n }}(\tanh ^{ - 1} x)^k } \right]_{x = 0} = \\ =  2^{n - k} k!\sum\limits_{m = 0}^{n - k} {\sum\limits_{r = 0}^{n - k - m} {\binom{n - m}{k} \binom{n - 1}{r}\frac{{r!}}{{2^r }}s(n,n - m)S(n - k - m,r)} } .}
$$

Second approach. We expand $(\tanh ^{ - 1} x)^k$ about $x=0$ using the Stirling numbers of the first kind:
\begin{align*}
(\tanh ^{ - 1} x)^k & = \frac{1}{{2^k }}(\log (1 + x) - \log (1 - x))^k
\\ & = ( - 1)^k \frac{{k!}}{{2^k }}\sum\limits_{m = 0}^k {( - 1)^m \frac{{\log ^m (1 + x)}}{{m!}}\frac{{\log ^{k - m} (1 - x)}}{{(k - m)!}}}
\\ & = ( - 1)^k \frac{{k!}}{{2^k }}\sum\limits_{m = 0}^k {( - 1)^m \left( {\sum\limits_{j = 0}^\infty  {\frac{{s(j,m)}}{{j!}}x^j } } \right)\left( {\sum\limits_{j = 0}^\infty  {( - 1)^j \frac{{s(j,k - m)}}{{j!}}x^j } } \right)}
\\ &
 = ( - 1)^k \frac{{k!}}{{2^k }}\sum\limits_{m = 0}^k {( - 1)^m \sum\limits_{j = 0}^\infty  {\left[ {\sum\limits_{r = 0}^j {( - 1)^{j - r} \binom{j}{r}s(r,m)s(j - r,k - m)} } \right]\frac{{x^j }}{{j!}}} }
\\ & = \sum\limits_{j = 0}^\infty  {\left[ {\frac{{k!}}{{2^k }}\sum\limits_{r = 0}^j {\sum\limits_{m = 0}^r {( - 1)^{m + r} \binom{j}{r}\left| {s(r,m)s(j - r,k - m)} \right|} } } \right]\frac{{x^j }}{{j!}}} .
\end{align*}
Therefore, by the Taylor formula,
$$\boxed{
\left[ {\frac{{d^n }}{{dx^n }}(\tanh ^{ - 1} x)^k } \right]_{x = 0}  = \frac{{k!}}{{2^k }}\sum\limits_{r = 0}^n {\sum\limits_{m = 0}^r {( - 1)^{m + r} \binom{n}{r}\left| {s(r,m)s(n - r,k - m)} \right|} } .}
$$


问:
Simple question about absolute value
hi guys i was wondering about if

$|x| > z > 0 $

but

$|x - y | < | z - y|$

does it imply

$|x| < |z| => |x| < z$

and there a contradiction? :)

答: If $x=-2$, $y=-3$, $z=1$, then $$\begin{cases}0<1<\lvert -2\rvert\\ \lvert -2-(-3)\rvert<\lvert 1-(-3)\rvert\end{cases}$$

Title: Simple question about absolute value


问:
How to Kill Homotopy Groups Using Framed Cobordism?
Let $M$ be an orientable manifold (with or without boundary), $N$ a framed submanifold in the interior of $M$ and assume (if necessary) that $\dim N<(\dim M)/2$.

1. If some low-dimensional homotopy group $\pi_i(N)$ of $N$ is nonzero as well as its image in $\pi_i(M)$,

is it possible to find a $N'$ framed cobordant to $N$ in $M$ such that $\pi_i(N)$ (or at least its image in $\pi_i(M)$) is zero?

In other words, can I use framed cobordisms and framed surgeries in $M$ to transform $N$ to something $k$-connected?

2. Also, could you recommend a book which could give me more intuition on this kind of questions?

I just have read Milnor's Topology from Differentiable Viewpoint..

答:
The answer to "can I use framed cobordisms and framed surgeries in M to transform N to something k-connected" is no and I'll provide an answer below.  In fact, what I'll show is the following:

There are $N$ and $M$ as in the problem statement, with $\pi_2(N)\rightarrow \pi_2(M)$ non-trivial, but for which there is no closed manifold $X$ meeting all of the following requirements:

X is framed cobordant to $N$ in $M$.

The induced map $\pi_2(X)\rightarrow \pi_2(M)$ is the $0$-map.

$\pi_1(X)$ is finite.

I don't see how to get rid of condition 3, but there may be a way.  Getting rid of it would answer your first question negatively.  Also, no where in my proof do I use the fact that $X$ is framed cobordant to $N$ in $M$.  I use that it is framed, and that it is cobordant to $N$.

Consider $N:=\mathbb{C}P^2 \times \{p\}\subseteq M:=\mathbb{C}P^2\times S^n$ for $n > 4$.  Due to the product structure, $N$ is framed, and further, the inclusion $i:N\rightarrow M$ induces an isomorphism on $\pi_2$.  I'll show there is no closed $4$-manifold meeting all three of the above requirements.

Suppose for a contradiction that such an $X$, with inclusion map $j$, exists.

Lemma 1:  The space $X$ must be orientable with non-trivial first Pontrjagin class.

Proof:  We have $j^\ast(TM) = TX\oplus \nu$ where the normal bundle $\nu$ is trivial since $X$ is framed.  Then $w_1(TX) = w_1(j^\ast(TM)) = j^\ast(w_1(TM)) = 0$, so $X$ is orientable.

Now, as $X$ is cobordant to $\mathbb{C}P^2$, the Pontrjagin numbers must agree. Since $p_1(\mathbb{C}P^2) \neq 0$, it now follows that the same is true of $X$.  $\square$

Let $\pi:\overline{X}\rightarrow X$ denote the universal cover of $X$.

Lemma 2:  The first Pontrjagin class of $\overline{X}$ is non-trivial.

Proof:  The projection $\pi:\overline{X}\rightarrow X$ induces an isomorphism $\pi^\ast(TX)\cong T\overline{X}$.  Since $X$ is orientable and has finite fundamental group (say, of order $s$), it follows that the induced map $\mathbb{Z}\cong H^4(X)\rightarrow H^4(\overline{x})\cong \mathbb{Z}$ is multiplication by $s$.  Thus, $p_1(T\overline{X}) = \pi^\ast( p_1(TX)) = sp_1(X)\neq 0$.  $\square$

Lemma 2 above is clearly incompatible with the following lemma, which gives the contradiction to the existence of $X$.

Lemma 3:  The manifold $\overline{X}$ is stably parallelizable, so, in particular, the first Pontrjagin class of $\overline{X}$ is trivial.

Proof:  The map $j\circ \pi$ is trivial on $\pi_2$ since $j$ is trivial on $\pi_2$ by assumption.
Since the maps $\pi_2(\overline{X})\rightarrow H_2(\overline{X})$ and $\pi_2(\mathbb{C}P^2)\rightarrow H_2(\mathbb{C}P^2)$ are isomorphisms (from the Hurewicz theorem), we deduce that the map $H_2(\overline{X})\rightarrow H_2(M)$ is trivial.  Dualizing, we see the map on $H^2$ is also trivial.

It follows from my answer to this MSE question that the map $j\circ \pi$ lifts to a map $f:\overline{X}\rightarrow S^5\times S^n$.  This lifted map is homotopically trivial by cellular approximation, so it follows that $j\circ \pi$ is homotopically trivial.

Pulling back a bundle along a homotopically trivial map gives a trivial bundle, so we see $$(j\circ \pi)^\ast (TM) = \pi^\ast(j^\ast(TM)) = \pi^\ast(TX \oplus \nu) = T\overline{X}\oplus \pi^\ast(\nu)$$ is trivial.  On the other hand, the normal bundle $\nu$ is trivial since $X$ is framed and pulling back a trivial bundle gives a trivial bundle.$\square$.


问:
Element-wise multiplication or Hadamard product
$a_{ij}$ and $b_{ij}$ elements of matrix A and B (same dimensions).
I want to multiply the matrices element-wise so the resulting matrix $s$ have the same dimensions as A and B. Is this the correct way to express the mathematical operation?
And are they the same?

\begin{equation}
    s = \sum_{i}^n \sum_{j}^n w_{ij} a_{ij}
    \label{eq:observert}
\end{equation}

\begin{equation}
    s = w_{ij} \odot a_{ij}
    \label{eq:hadamard}
\end{equation}

答:
The first identity gives you a scalar, not a matrix

The Hadamard product of matrices $A$ and $B$ is usually denoted as $A\odot B$ and the element of $A\odot B$ is given by
$$
(A\odot B)_{ij}=(A)_{ij}(B)_{ij}
$$


答:
This limit is equivalent to
$$ \lim_{x\to 0} \left(\frac{\left(1 + x\right)^{\frac1x}}{e}\right)^{\frac1x} $$
$$ L = \lim_{x\to 0} \left(\frac{e^{\frac{\ln(1+x)}{x}}}{e}\right)^{\frac1x} $$
$$ = \lim_{x\to 0} \left(e^{\frac{\ln(1+x)}{x}-1} \right)^{\frac1x}$$
$$ = \lim_{x\to 0} \left(e^{1-\frac{x}{2}+\frac{x^2}{3} +\cdots -1} \right)^{\frac1x} $$
$$ = \lim_{x\to 0} \left(e^{-\frac{x}{2}+\frac{x^2}{3} +\cdots} \right)^{\frac1x} $$
$$ = \lim_{x\to 0} \left(e^{-\frac{1}{2}+\frac{x}{3} +\cdots } \right) $$
$$ = e^{-\frac12} $$
$$ \boxed{ \therefore \lim_{n\to\infty} \left(\frac{\left(1 + \frac{1}{n}\right)^n}{e}\right)^n = \lim_{x\to 0} \left(\frac{\left(1 + x\right)^{\frac1x}}{e}\right)^{\frac1x} = \color{red}{\dfrac{1}{\sqrt{e}}}}$$

Body:
Could anyone please help me find limit?
$$\lim_{n\to\infty} \left(\frac{\left(1 + \frac{1}{n}\right)^n}{e}\right)^n$$

Thank you


答:
$$\lim_{n\to\infty} \left(\frac{\left(1 + \frac{1}{n}\right)^n}{e}\right)^n$$

Let $\lim_{n\to\infty}\frac{\left(1 + \frac{1}{n}\right)^n}{e}=1+t\implies t= \lim_{n\to\infty}\frac{\left(1 + \frac{1}{n}\right)^n-e}{e}$

$$L=\lim_{n\to\infty}e^{n\frac{\left(1 + \frac{1}{n}\right)^n-e}{e}}$$

The series expansion of $\left(1 + \frac{1}{n}\right)^n$ when $n\to\infty$ is $e-\frac{e}{2n}+O(n^2)$

$$L=\lim_{n\to\infty}e^{n\frac{\left(1 + \frac{1}{n}\right)^n-e}{e}}=\lim_{n\to\infty}e^{n\frac{-e}{2ne}}=\frac{1}{\sqrt e}$$

The reason for introducing $t$ is to convert the function to the form $\lim_{f\to\infty}(1+f)^{1/f}$


答: No. $|x-y|<|z-y|$ tells us that $y$ is closer to $x$ than it is to $z$. If $x>0$, this means simply that $y>\frac12(x+z)$, since $\frac12(x+z)$ is halfway between $x$ and $z$. Otherwise, $x<-z<0$, and it tells us similarly that $y<\frac12(x-z)$.

Title: Simple question about absolute value

Body:
hi guys i was wondering about if

$|x| > z > 0 $

but

$|x - y | < | z - y|$

does it imply

$|x| < |z| => |x| < z$

and there a contradiction? :)


问:
Why we consider counterimages to define continuos, measurable and measure-preserving functions?
I will specify the question. In a conference by Arnold (on continued fractions, you can find it on YouTube) he says that the fact that we consider preimages when defining measure preserving maps is something deep, which links to controvariant and covariant concepts in mathematics. What does this mean? I know the definition of measure preserving map, and I know Ergodic Theory and all the consequences, but is there something more deep about why we have to take the preimage and not the image when defining measure preserving functions?

We have a similar phenomena in topology and measure theory, when we define continuos and measurable maps: to say that a map preserve some kind of structure, we ask something involving the preimage. Is there something deep also in those definitions? In topology for example we have that the continuous  image of a compact or of a connected is compact or connected, but the preimage of an open or a closed is an open or a closed (in topology we can define open and closed maps, but they are less used that continuous maps)

I have never asked myself this question before because in topology the definition of continuous map arises very easily from the epsilon-delta definition in metric space, and in measure theory the definition of measurable map it works well with integrals. But in Ergodic Theory if I were asked to chose the definition of measure preserving I would have said that the measure of the image is equale to the one of the starting set.

Are there similar example in Maths of this phenomena?

答:
In short because for two subset$A,B$ of a space $X$, $f^{-1}(A \cap B)=f^{-1}(A) \cap f^{-1}(B)$ but there are case where $f(A \cap B) \ne f(A) \cap f(B)$ (it is a good exercice to show the first claim and find counterexample of the secound).

So the preimage of a $\sigma$-algebra is a $\sigma$-algebra but not the image.


问:
convergence in measure does not imply weak convergence
Suppose $\sup_n\|f_n\|_1<\infty$ and $f_n\rightarrow f$ a.e.. However it is not necessary that $f_n\rightarrow f$ weakly in $L^1$.

Can someone raise an example?

Thanks in advance.

答: Try $f_n = n \, \chi_{(0,1/n)}$. The key to weak convergence in $L^1$ is uniform integrability. This does not hold for the above $f_n$.</p>


问:
Proof of divisibility by full induction
How can I prove by full induction that $n^{2(p+1)}-1$ is divisible by $n+1$?

Title: Proof of divisibility by full induction

Body:
Hint:

$n^2-1=(n+1)(n-1)\qquad$

Consider $n^{2(p+1)}-n^{2p}$


答:
Hint:

If $n^x\equiv1\bmod{n+1}$, and $n^2\equiv1\bmod{n+1}$, then $n^{x+2}\equiv1\bmod{n+1}$.

Title: Proof of divisibility by full induction

Body: How can I prove by full induction that $n^{2(p+1)}-1$ is divisible by $n+1$?


问:
Eliminate $\theta$ from the equation $\sec \theta + \tan \theta = a+\sqrt b,$
Eliminate $\theta$ from the equation $$\sec \theta + \tan \theta = a+\sqrt b,$$where $~a,b, \sec \theta \in \mathbb{Q},$ and $\sqrt b \notin \mathbb{Q}.$

My attempt: By the given condition we must have $~\tan \theta \notin \mathbb{Q}.$ Then we can have $$\tan \theta = x+y,~~ \text{ where} ~ x \in \mathbb{Q},~ y \notin \mathbb{Q}.$$ But after this I am not able to proceed suitably to eliminate $\theta$. I did square both side, took bar both side, but failed.

Please help me to solve this.

答:
You can use the identity $\tan(\theta) = \pm\sqrt{\sec^{2}(\theta)-1}$

Then you get $$\sec(\theta) + \sqrt{\sec^{2}(\theta)-1} = a + \sqrt{b}$$

Since $\sec(\theta)$ is in $Q$ and $b \in Q$ the negative version of the tangent identity is ruled out, then $\sec(\theta) = a$. So the equation reduces to

$$a + \sqrt{a^{2}-1} = a + \sqrt{b}$$

So

$$b = a^{2}-1$$


问:
Why is the subspace topology same as the Order topology?
Hi i am reading Topology by Munkres and there in the proof of theorem 27.1 it is written that

Given $a<b$, let $\mathcal A$ be a covering of $[a,b]$ by sets open in $[a,b]$ in the subspace topology(which is the same as the order topology).

I can't understand why is the subspace topology same as the order topology in this case? Is this always true? Btw i know how subspace topology and order topologies are defined. If $(X,\tau)$ is a topological space and $A$ be any subset of X then we define $\tau_A$ to be the set with subsets as : $\tau_A=\{U\cap A | U\in \tau\}$. And order topology is defined in a different way.So how can they be same here? For reference i am attaching the screenshot where i have highlighted the part where this is mentioned.

答:
In a L.O.T.S. (Linearly Ordered Topological Space) $(X,<),$ if $Y\subset X$ then the subspace topology $T_S$ on $Y$ may be strictly stronger than the $<$-order topology $T_<$ on Y. We always have $T_<\subseteq T_S.$

E.g. if $X=\Bbb R$ with the usual order and $Y=\{0\}\cup (1,2).$ For $Y$ is order-isomorphic to $[1,2)$ and the order-topology on $[1,2)$ has no isolated points, but in the subspace topology, the set $\{0\}=Y\cap (-1,1)$ is open.

A necessary & sufficient condition for $T_S\ne T_<$ is that there exists $y\in Y$ and $x\in X\setminus Y$ such that

$(i).\, x<y$ and the set $B=\{y'\in Y: y'<y\}$ is not empty and has no largest member, and $\forall y'\in B\,(y'<x),$... OR

$(ii).\, y<x$ and the set $C=\{y'\in Y:y'>y\}$ is not empty and has no smallest member, and $\forall y'\in C\,(y'>x).$

Some sufficient conditions for $T_S=T_<$ are

$(i').\, Y$ is order-dense in $X,$ that is, if $x_1,x_2\in X$ with $x_1<x_2$ then $Y\cap (x_1,x_2)\ne \emptyset,$... OR

$(ii').\, Y$ is convex in $X,$ that is, if $y_1,y_2\in Y$ with $y_1<y_2$ and $y_1<x<y_2$ then $x\in Y.$

Observe that $(ii')$ holds if $Y=[a,b]=\{x\in X:a\le x\le b\}.$

In some cases where $T_<\ne T_S,$ the subspace topology may be generated by a different linear order $<'$ on $Y.$ For example, if $X=\Bbb R$ and $Y= [0,1)\cup [2,3)$ then $Y$ with the subspace topology is homeomorphic to  the ordered space $(0,1]\cup [2,3).$ But being a L.O.T.S. is not a hereditary property in general. No linear order on the real subspace $\{0\}\cup (1,2)$ can generate its subspace topology, but I would not call this obvious.


问:
Calculate the value of $\sum_{(x,y,z) \in S} g(x,y,z).$
Let S be a triplet set of integers $(x, y, z)$ satisfying the equation $∣x∣ + ∣y∣ + ∣z∣ = 2021$. If we define $g (x, y, z) = x + y + z$ for every real number $x, y$, and $z$,
calculate the value of $\sum_{(x,y,z) \in S}  g(x,y,z).$

Previously I have looked for the x y and z values but the results do not match, is this problem using number theory?

答:
A MASSIVE HINT

@Lulu has pointed out that solutions occur in pairs $(x,y,z)$ and $(-x,-y,-z)$.

Consider for example the pair of solutions $(3,-4,2014)$ and  $(-3,4,-2014)$. What is the value of $(3-4+2014)+(-3+4-2014)$?

Do you now see what the point of the hint was?


问:
Representation of Hilbert functionals in terms of dual basis
Let $\mathscr{H}$ be a complex separable Hilbert space with (orthonormal) Hilbert basis $\{e_{n}\}_{n\in \mathbb{N}}$. By Riesz Representation Theorem, if $f: \mathscr{H}\to \mathbb{C}$ is a given functional, there exists $y \in \mathscr{H}$ such that:
$$f(x) = \langle y,x\rangle$$
for every $x \in \mathscr{H}$, where $||f|| = ||y||$. Thus, if $\mathscr{H}^{*}$ is the dual space of $\mathscr{H}$, we established a conjugate linear map $T:\mathscr{H}\to \mathscr{H}^{*}$ such that $\mathscr{H}\ni y \mapsto Ty := f$, so that $\mathscr{H}^{*}\cong \mathscr{H}$. Now, $(\forall n \in \mathbb{N})$ let $\epsilon_{n}$ be defined by $\epsilon_{n} := Te_{n}$.

Question: How can I prove that $\{\epsilon_{n}\}_{n\in \mathbb{N}}$ is an orthonormal basis on $\mathscr{H}^{*}$ so that every $f \in \mathscr{H}^{*}$ can be written as:
$$f = \sum_{n \in \mathbb{N}} f_{n}\epsilon_{n}$$
where $(\forall n\in \mathbb{N})$ $f_{n} \in \mathbb{C}$?

答:
In order to talk about orthonormal bases in $H^*$ one first needs that space to have an inner product. This is not specified in the question, but assuming it is the natural one defined by $$\langle \phi,\psi\rangle_{H^*}:=\langle T^{-1}\psi,T^{-1}\phi\rangle_H$$ then it follows that $\epsilon_n$ are orthonormal: $$\langle\epsilon_n,\epsilon_m\rangle_{H^*}=\langle e_m,e_n\rangle_H=\delta_{nm}.$$

Suppose $f\in\mathrm{Span}(\epsilon_n)^\perp$, that is $\langle \epsilon_n,f\rangle_{H^*}=0$ for all $n$. Then, taking $x=T^{-1}f$, $$\forall n\in\mathbb{N},\quad\langle x,e_n\rangle_H=0,\qquad\therefore x=0,\quad \therefore f=0$$ This implies that $\epsilon_n$ is an orthonormal basis for $H^*$. Moreover, for any functional, $f=\sum_n\langle\epsilon_n,f\rangle_{H^*}\epsilon_n$, by the standard theory of Hilbert spaces.


问:
Why orientation of plane region boundary implies surface boundary?
For parametrized surfaces $\Phi:D\to \mathbb R^3,$ we assume that $\Phi$ is one to one on the whole of $D$. Then the image of $D$ is $S$ and we define the image of $\partial D$ (the boundary of $D$) is  $\partial S$ (But I'm sure I'm missing something, eg., we know boundary of sphere (as surface $\mathbb R^3$) is empty).

Question: (1)If we assume that $\partial D$ is oriented as in Green's theorem, Can we get an orientation for $\partial S$ and $S$ itself is oriented so $\Phi_u \times \Phi_v$ is in the direction of the normal.  (2) Is the above definition of boundary  of surface  is correct in $\mathbb R^3$? If so, then I think  then $\partial S \neq \emptyset$

If $D$ is a  plane  region, with boundary curve $C,$ the positive orientation of $C$ is given by  the vector $k\times v_{out},$ where $v_{out}$ is the normal vector field pointing outward along curve.

As you walk along $C$ in the direction of positive orientation, the region $D$ is on your left.

Edit: Assume that $D$ is  plane region of elementary type (as in Greens theorem)

答:
I think what's going on is that you are thinking about a somewhat restricted version of Stokes Theorem, which I'll refer to as the "singly parameterized Stokes Theorem", in which one assumes that the surface $S$ is restricted to be a parameterized surface, with parameterization $\Phi : D \to S$ as stated in your question. In that case all I can say is that the answers to your two questions are "Yes" and that $\partial S$ is indeed nonempty, for the reasons that you suggest.

However, there is a more general version of Stokes Theorem where $S$ is not assumed to be "singly parameterized", and where $\partial S$ is allowed to be empty. In this more general situation there is no guarantee that $S$ is orientable (the Möbius band is a nonorientable, for example), and therefore the more general version has an extra hypothesis requiring that $S$ be orientable, and another hypothesis that $S$ be compact. The proof of this general version of Stokes Theorem does make use of the multiple coordinate systems needed to cover an arbitrary oriented surface $S$.

For example this more general version of Stokes Theorem does indeed apply to the case that $S \subset \mathbb R^3$ is the unit 2-sphere (which means the boundary of the unit 3-ball in $\mathbb R^3$). Of course we know intuitively that the unit 2-sphere is not a "singly parameterized" surface. For example spherical coordinates do not apply to the north or south pole, so we have to add a "north polar" coordinate system and a "south polar" coordinate system, which gives a covering of the unit 2-sphere by three coordinate systems. You can actually do better and cover the unit 2-sphere with only two coordinate systems. But there does not exist any single coordinate system that covers the whole 2-sphere in a one-to-one fashion.

It is difficult to explain the details of coordinate systems, orientations, and boundaries without some more advanced knowledge of manifolds, which one eventually learns about in a course on differential geometry or differential topology.