问:
According to the identity laws, if A ∪ ∅ = A, then how comes A - ∅ = A as well?
This identity law is in a textbook I'm using but this question is also in the textbook and the solution is true. How can they both be true?

答:
Think of what the notation means, I have yet not earned enough reputation to just comment. So I will post this is an answer.

$A\cup\emptyset$ is the set $\{x\in U \mid x\in A \lor x\in\emptyset \}$ where $U$ us the universe of discourse. Since by definition the empty set has no elements $x\in \emptyset$ would be a contradiction and due to Universal Bound Laws, $P \lor (C\land\lnot C) = P$ where $(C\land\lnot C)$ is a contradiction. Thus the set in question $\{x\in U \mid x\in A \lor x\in\emptyset \}$ reduces to $\{x\in U \mid x\in A \}$

On the other hand the set $A-\emptyset$ is the set $\{x\in U\mid x\in A \land \lnot(x\in\emptyset)\}$ here the later is that $x$ is not an element of the empty set, which is always true for any element and is thus a tautology Universal Bound Laws. More explicitly $P \land T = P$ where $P$ is some statement and $T$ is a tautology, since its always true it reduces to just the statement $P$. Applying that to the set in question $\{x\in U\mid x\in A \land \lnot(x\in\emptyset)\} = \{x\in U| x\in A\}$

Thus we can see that $A\cap\emptyset = \{x\in U\mid x\in A\} = A-\emptyset$

I hope this clarifies this, usually looking at the definitions and logical laws helps shed light on those questions.


问:
Convergence of $f_n(x)=\frac{x}{1+nx^2}$ and its derivative $f'_n(x)=\frac{1-nx^2}{(1+nx^2)^2}$ on $\mathbb{R}$
Consider the sequence $f_n(x)=\frac{x}{1+nx^2}$ on $x\in \mathbb{R}$. Also consider $f'_n(x)=\frac{1-nx^2}{(1+nx^2)^2}$ on $x \in \mathbb{R}$.

I want to check their uniform convergence.

Convergence of $f_n$:

$f_n \to 0$ pointwise on $\mathbb{R}$.

$f_n(x) \ge 0$ for all $x \in \mathbb{R}^+$ and $f_n$ is an odd function so obtaining a global maximum in $[0, \infty)$ will provide use with an upperbound of $|f_n(x)|$

$f_n'(x)=0 \iff x=0 \ \text{or} \ x=\frac{1}{\sqrt n}$

At $x=0$ $f_n(x)$ is the global minimum so $f_n(x)$ attains its global maximum  at $x=\frac{1}{\sqrt n}$ in $\mathbb{R}$

$\implies M_n:=\operatorname{lub}\{|f_n(x)-0|:x \in \mathbb{R}\}=\frac{2}{\sqrt n} \to 0$

$\implies f_n \to 0$ uniformly in $\mathbb{R}$.

Convergence of $f'_n$:

$f'_n(x) \to \begin{cases}
                 0 & x \ne 0 \\
                 1 & x=0
                \end{cases}$
pointwise in $\mathbb{R}$ which tells $f'_n$ cannot be uniformly convergent in $\mathbb{R}$ as the limit function is not continuous.

Critical points of $f'_n(x)$ are $x=0$ and $x=\sqrt \frac{3}{2n} \to 0$

$\implies f'_n$ is monotone and continuous in [1,b] for any $b>1$ for sufficiently large $n$ and converges pointwise to $0$.

I want to ensure $f'_n$ converges uniformly to $0$ on $[1,b]$. How should I proceed?

答: You have not obtained the point-wise limit of $f_n'$ correctly. The limit is $1$ for $x=0$ and $0$ for all other $x$. The convergence is uniform on any compact interval not containing  $0$: $|f_n'(x)| \leq \frac { 1+nM^{2}} {(1+\epsilon^{2}n)^{2}}$ if $M$ is  a bound for $|x|$ on the given  interval $I$ and $\epsilon =\inf \{|x|: x \in I\}$.

ContentLicense: CC BY-SA 4.0


问:
How do I average out to a particular number if I have to add another number to get that average?
So, lets say, that you have the following numbers (4, 15, 3, 22) and you have to add the fifth number (which I would call X) and they must average out to say 20.

What do you have to do?

I figured the formula is something like (4+15+3+22+X)/5=20.

But for some reason I can't solve it. It just doesnt click in. Can anyone tell me if at least I have the formula right?

答:
AD's answer (and the comments that follow) as well as comments on the original question cover the algebraic solution pretty well.  This question, however, is one of the archetypal "algebra" problems that, out of context, most people wouldn't use algebra to do.  Here's a non-algebraic line of thinking that leads to the answer.

You're going to have 5 numbers that have an average of 20.  That means that when you add up these 5 numbers, it's like adding up 5 20s, or multiplying $5\times20$, which gives 100.  So the sum of all five of the numbers is $5\times20=100$.

The numbers you have so far add up to $4+15+3+22=44$.  To end up with a total of 100 with one more number, that number has to be $100-44=56$.

These are the same operations that fall out of doing the algebra, used in the same way.  The difference is in using formal symbols versus an informal chain of reasoning.  If this problem is posed to an elementary school student who has appropriate knowledge of averages, they will likely be able to solve it by thinking along the lines of what I've shown.  If this problem is posed to a student who has done more symbolic manipulation and variable-use, they will much more likely write out an equation and formally solve it.

edit: For a bit more on what I'm talking about with regard to non-algebraic "algebra" word problems, see "What Should Not Be in the Algebra and Geometry Curricula of Average College-Bound Students?" by Zalman Usiskin (Mathematics Teacher 73 (September 1980), pdf), particularly the section titled "Deletion 1: The traditional word problems," starting on the page numbered 70 in the linked PDF.

ContentLicense: CC BY-SA 3.0


问:
Pointwise supremum of a convex function collection
In Hoang Tuy, Convex Analysis and Global Optimization, Kluwer, pag. 46, I read:

"A positive combination of finitely many proper convex functions on $R^n$ is convex. The upper envelope (pointwise supremum) of an arbitrary family of convex functions is convex".

In order to prove the second claim the author sets the pointwise supremum as:

$$
f(x) = \sup \{f_i(x) \mid i \in I\}
$$

Then

$$
\mathrm{epi} f = \bigcap_{i \in I} \mathrm{epi} f_i
$$

As  "the intersection of a family of convex sets is a convex set" the thesis follows.

The claim on the intersections raises some doubts for me.

If $(x,t)$ is in the epigraph of $f$,   $f(x) \leq t)$ and, as  $f(x)\geq f_i(x)$, also  $f_i(x) \leq t)$; therefore $(x,t)$ is in the epigraph of every $f_i$ and the intersection proposition follows.

Now, what if, for some (not all) $\hat{i}$, $f_{\hat{i}}(x^0)$ is not defined? The sup still applies to the other $f_i$ and so, in as far as the sup is finite, $f(x^0)$ is  defined. In this case when  $(x^0,t) \in  \mathrm{epi} f$ not $\in \mathrm{epi} f_{\hat{i}}$ too, since $x^0 \not \in \mathrm{dom}\, f_{\hat{i}}$.

So it is simple to say that $f$ is convex over $\bigcap_{i \in I} \mathrm{dom} f_i\subset \mathrm{dom} f$, because in this set every  $x  \in \mathrm{dom}\, f_{\hat{i}}$.

What can we say when   $x \in \mathrm{dom}\, f$ but $x \not \in \mathrm{dom}\, f_{\hat{i}}$ for some $i$?

答: I think it is either assumed that the $f_i$ are defined on the same domain $D$, or that (following a common convention) we set $f_i(x)=+\infty$ if $x \notin \mathrm{Dom}(f_i)$. You can easily check that under this convention, the extended $f_i$ still remain convex and the claim is true.</p>


问:
Find the Taylor and Laurent series with center at $z_{o}= 1$ of the function $\frac{\sinh(z)}{(z-1)^4}$
Let $$f(z)=\frac{\sinh(z)}{(z-1)^4}.$$

First I do the following:

$$\sinh(z)=\sinh((z-1)+1)=\sinh(z-1)\cosh(1)+\cosh(z-1)\sinh(1)$$

The expansions of $\sinh(z)$ and $\cosh(z)$ are
$$\begin{align*}
\sinh(z) &= z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots \\
\cosh(z) &= 1+\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots
\end{align*}$$

Then
$$\begin{align*}
\sinh(z-1) &= (z-1)+\frac{(z-1)^3}{3!}+\frac{(z-1)^5}{5!}+\cdots \\
\cosh(z-1) &= 1+\frac{(z-1)^2}{2!}+\frac{(z-1)^4}{4!}+\cdots
\end{align*}$$

Now let's see what
$$\begin{align*}
\frac{\cosh(1)\sinh(z-1)}{(z-1)^4}
&= \cosh(1) \biggl( \frac{1}{(z-1)^3}+\frac{1}{3!(z-1)}+\frac{(z-1)}{5!}+\cdots \biggr) \\
\frac{\sinh(1)\cosh(z-1)}{(z-1)^4}
&= \sinh(1) \biggl( \frac{1}{(z-1)^4}+\frac{1}{2!(z-1)^2}+\frac{1}{4!}+\cdots \biggr)
\end{align*}$$

Rewriting these last two equalities we have
$$\begin{align*}
\frac{\cosh(1)\sinh(z-1)}{(z-1)^4}
&= \sum_{n=1}^{\infty} \frac{\cosh(1)}{(2n-1)!}(z-1)^{2n-5} \\
\frac{\sinh(1)\cosh(z-1)}{(z-1)^4}
&= \sum_{n=0}^{\infty} \frac{\sinh(1)}{(2n)!}(z-1)^{2n-4}
\end{align*}$$

Hence the Taylor and Laurent series of $f(z)$ is
$$
f(z) = \sum_{n=1}^{\infty} \frac{\cosh(1)}{(2n-1)!}(z-1)^{2n-5} + \sum_{n=0}^{\infty} \frac{\sinh(1)}{(2n)!}(z-1)^{2n-4}
$$

Where we have
$$
a_{n} = \frac{\cosh(1)}{(2n-1)!}
\quad \text{and} \quad
b_{n} = \frac{\sinh(1)}{(2n)!}.
$$

But I don't know if it's okay. I feel a bit confused regarding these series. Could you tell me if I'm okay? or in another case, could you give me any suggestions?

答:
Your work is good but you can make life a bit easier starting with $z=t+1$ (this is what you implicitly did)
$$\frac{\sinh (z)}{(z-1)^4}=\frac{\sinh (t+1)}{t^4}$$ Expanding as you did
$$\sinh (t+1)=\sinh (1) \cosh (t)+\cosh (1) \sinh (t)$$ Now, using the expansions of $\cosh (t)$ and $\sinh (t)$ around $t=0$ you then have
$$\frac{\sinh (t+1)}{t^4}=\sum_{n=0}^\infty a_n\, t^{n-4}$$ where
$$a_{2n+1}=\frac {\cosh(1)}{(2n+1)!}\qquad \text{and} \qquad a_{2n}=\frac {\sinh(1)}{(2n)!}$$ which is your result  (just make $t=z-1$).


答:
Note that application of the AM-GM inequality gives immediately that

$$\frac{|x|}{1+nx^2}\le \frac{|x|}{2\sqrt{nx^2}}=\frac1{2\sqrt n}$$

Therefore, $f_n$ converges uniformly to $0$ for $x\in \mathbb{R}$.

Next, we have

$$f_n'(x)=\frac{1-nx^2}{(1+nx^2)^2}$$

Let $0<a<b$ and suppose $x\in [a,b]$.  Then for $n\ge 1$

$$|f_n'(x)|= \frac{|nx^2-1|}{(nx^2+1)^2}\le \frac{nb^2+1}{(na^2+1)^2}$$

Hence, $f_n'(x)$ converges uniformly to $0$ for $x\in [a,b]$ for $0<a<b$.  And inasmuch as $f_n'$ is an even function, $f_n'(x)$ converges on $(-b,-a)$ also.

Title: Convergence of $f_n(x)=\frac{x}{1+nx^2}$ and its derivative $f'_n(x)=\frac{1-nx^2}{(1+nx^2)^2}$ on $\mathbb{R}$

Body:
Consider the sequence $f_n(x)=\frac{x}{1+nx^2}$ on $x\in \mathbb{R}$. Also consider $f'_n(x)=\frac{1-nx^2}{(1+nx^2)^2}$ on $x \in \mathbb{R}$.

I want to check their uniform convergence.

Convergence of $f_n$:

$f_n \to 0$ pointwise on $\mathbb{R}$.

$f_n(x) \ge 0$ for all $x \in \mathbb{R}^+$ and $f_n$ is an odd function so obtaining a global maximum in $[0, \infty)$ will provide use with an upperbound of $|f_n(x)|$

$f_n'(x)=0 \iff x=0 \ \text{or} \ x=\frac{1}{\sqrt n}$

At $x=0$ $f_n(x)$ is the global minimum so $f_n(x)$ attains its global maximum  at $x=\frac{1}{\sqrt n}$ in $\mathbb{R}$

$\implies M_n:=\operatorname{lub}\{|f_n(x)-0|:x \in \mathbb{R}\}=\frac{2}{\sqrt n} \to 0$

$\implies f_n \to 0$ uniformly in $\mathbb{R}$.

Convergence of $f'_n$:

$f'_n(x) \to \begin{cases}
                 0 & x \ne 0 \\
                 1 & x=0
                \end{cases}$
pointwise in $\mathbb{R}$ which tells $f'_n$ cannot be uniformly convergent in $\mathbb{R}$ as the limit function is not continuous.

Critical points of $f'_n(x)$ are $x=0$ and $x=\sqrt \frac{3}{2n} \to 0$

$\implies f'_n$ is monotone and continuous in [1,b] for any $b>1$ for sufficiently large $n$ and converges pointwise to $0$.

I want to ensure $f'_n$ converges uniformly to $0$ on $[1,b]$. How should I proceed?


问:
Inequality-constrained quadratic program
I am trying to find the optimal solution to this:
$$\begin{aligned}
\min_{} \quad & x^TAx\\
\textrm{s.t.} \quad & c^Tx \ge 1\\
\end{aligned}$$
where $A \in \mathbb{S}_{+}^{n}$ ($n \times n$ symmetric positive semidefinite matrix) and $c \in \mathbb{R}^n$ is a nonzero vector.

I know I can use the first order optimality condition and would get $2Ax(y-x) \ge 0$ for all feasibly $y$. $x$ is only optimal when this inequality is true for all feasible $y$ but I'm not sure how to get the optimal $x$ knowing the inequality.

答: assuming $A$ is symmetric positive definite, it has a Cholesky decompostions $A= W^T W.$   Taking $y= Wx,$ the objective becomes $y^T y$ which is just its squared length. The constraint is $(c^T W^{-1}) y \geq 1$ where the point nearest the origin is a multiple of the vector $(c^T W^{-1})$ as a row, or ${W^{-1}}^T c$ as a column. Something like that.


问:
An interpolation inequality using Holder inequality
If $f \in L^{p}\left(\mathbb{R}^{n}\right) \cap L^{q}\left(\mathbb{R}^{n}\right),$ where $p<q,$ prove that $\|f\|_{r} \leq\left(\|f\|_{p}\right)^{\frac{1 / r-1 / q}{1 / p-1 / q}}\left(\|f\|_{q}\right)^{\frac{1 / p-1 / r}{1 / p-1 / q}}$ for any $p<r<q$. (I have already shown $f \in L^{r}\left(\mathbb{R}^{n}\right)$)

My approach: We have to show that $(\int |f|^r)^{\left(\frac{1}{r}\left(\frac{1}{p}-\frac{1}{q}\right)\right)}\leq(\int |f|^p)^{\left(\frac{1}{p}\left(\frac{1}{r}-\frac{1}{q}\right)\right)} (\int |f|^q)^{\left(\frac{1}{q}\left(\frac{1}{p}-\frac{1}{r}\right)\right)}$.
But I'm unable to understand how to pick quantities the reciprocals of which sum to one,  which would enable me to use Holder.

答: Write $|f|^{r}$ as $|f|^{a} |f|^{r-a}$ where $a= \frac {pq (q-r)} {q-p}$ and apply Holder's inequality with exponents $\frac p  a$ and $ \frac  q {r-a}$.


问:
Showing that posterior is also gamma
Suppose the prior for $\mu$ is $\sim gamma(k, \lambda)$, so

$f_{\mu}(m) = C m^{k - 1} \exp(-\lambda m)$ when $(m \geq 0)$

Further, suppose the data $X_1, \ldots, X_n$ given $\mu$ is iid Poisson with parameter $\mu$, so for each $[X_j \mid \mu = m] \sim [X \mid \mu = m]$,

$f_{X \mid \mu = m}(i) = \exp(-m) m^i / i!$ when $(i \in \{0, 1, \ldots\})$
Show that the posterior for $\mu$ given the data is also Gamma.

I know that the posterior = prior * Likelihood but am not sure how to recognize the Gamma.

答:
I know that the posterior = prior * Likelihood but am not sure how to recognize the Gamma

First, observe that the posterior is not exactly what you wrote but it is $\propto$, say it is prior $\times$ L $\times$ constant

Second, your L is the product of your $f$

$$L= C e^{-m n}m ^{\Sigma_k i_k}$$

In the constant C are included all the quantities that do not depend on $m$

Thus the posterior is

$$C m^{(k+\Sigma_k i_k)-1}e^{-(\lambda+n)m}$$

Here the variable is $m \geq 0$ and the posterior is in the same type of your prior, thus it is still gamma

Title: Showing that posterior is also gamma


问:
Polynomial close to separable polynomial in Gauss norm is separable itself.
Let $(K,|\cdot |)$ be a complete non-archimedean valuation field. Let $f(x)\in K[x]$ be a separable polynomial of degree $n$. Let $||\cdot||$ be the Gauss norm associated with $|\cdot|$, i.e. $||f||=\max_{0\leq i\leq n}\{a_i\}$, where $a_i$ is the coefficient of degree $i$ term in $f$. I want to prove that there exists a $\delta>0$, such that for every polynomial $g(x)\in K[x]$ of degree $n$, $||f-g||<\delta$ implies $g$ is separable.

In the context the condition that $K$ is complete and non-archimedean is given, but I am also wondering are these two conditions necessary for this particular result?

Any help is much appreciated.

答:
Hint: Let's look at $n=2$ first. As we all learned in school (just did not phrase it that way), the polynomial $ax^2+bx+c$ is separable if and only if $b^2-4ac\neq0$. Now if that is the case, and we change $a,b,c$ "just a little", then $b^2-4ac$ will still be $\neq 0$.

Now that thing $b^2-4ac$ for quadratic polynomials is called the discriminant. Look up if such a thing exists for general polynomials, and what properties it has.


问:
An identity for elementary automorphisms of a semisimple Lie algebra
$\newcommand{\ad}{\text{ad}}$ $\newcommand{\Ad}{\text{Ad}}$
Let $\mathfrak{g}$ be a semisimple Lie algebra and let $X \in \mathfrak{g}$ be nilpotent. Then $\ad X$ is nilpotent and $\exp \ad X$ acts on $\mathfrak{g}$ via the adjoint representation of $G$ (is this correct here? I am not sure if I may be interpreting it incorrectly, since there is in fact no mention to a Lie/algebraic group $G$).
I want to show that, for any $Y \in \mathfrak{g}$,
$$ \ad \left( \exp(\ad X) Y \right) = \exp (\ad X) \ad Y \exp(-\ad X).$$
I know that $\Ad(\exp X) = \exp (\ad X)$ for any $X \in \mathfrak{g}$ (in particular, $\exp$ and $\ad$ do not 'commute').

If I start from the left-hand side, I get
$$ \ad \left( \exp(\ad X) Y \right) = \ad \left( \Ad(\exp X) Y \right) = \ad (\exp X) \ad Y \ad (\exp -X).$$
If I start from the right-hand side, I get
$$ \exp (\ad X) \ad Y \exp(-\ad X) = \Ad(\exp (\ad X)) (\ad Y) = \exp (\ad (\ad X)) (\ad Y).$$
Neither of these give the identity, although they are almost there.

答:
Keep in mind $\Phi=\mathrm{Ad}\exp X=\exp\mathrm{ad}X$ is a lie algebra homomorphism, so

$$ \mathrm{ad}(\Phi Y)Z=[\Phi Y,Z]=\Phi[Y,\Phi^{-1}Z]=\Phi\circ\mathrm{ad}Y\circ\Phi^{-1} $$

which is what you wanted. In short,

$$ \mathrm{ad}((\exp\mathrm{ad}X)Y)=(\mathrm{Ad}\exp\mathrm{ad}X)Y. $$


问:
Ordered set notation in a simple case
apologies in advance for the noob question. (Really rusty when it comes to order theory notation.)

Suppose I have a set $\Omega := \{\alpha,\beta_1,...,\beta_n\} $ (for some fixed $n\in\mathbb{N}$). I would like to succinctly express the following:

$\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad $ "$C$ is a subset of elements from $\Omega$ ordered in the following way:

$\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad$ 1. If $\alpha$ is in $C$, then it is placed first.

$\quad \quad \quad\quad\quad \quad\quad\quad\quad\quad\quad$ 2. All $\beta$'s in $C$ are placed in increasing order of their indeces."

I hope this is sufficiently precise; to help illustrate, here are a couple of examples of what I have in mind:

$\quad \bullet\ $  $C = (\omega, \beta_1,\beta_3)$

$\quad \bullet\ $  $C = (\beta_2,\beta_3)$

If there is no succinct, mathematical way to state this, then no worries! Thanks in any case.

答: The set $\Omega = \{\alpha, \beta_1, \beta_2, \ldots, \beta_n\}$ is ordered by $\leqslant$ where $\alpha \leqslant \beta_1 \leqslant \beta_2 \leqslant \ldots \leqslant \beta_n$.The sequence $C$ is a subset of $\Omega$ where the elements appear in the order given by $\leqslant$. For example $C = (\alpha, \beta_2)$ or $C = (\beta_1, \beta_3, \beta_6)$.

Title: Ordered set notation in a simple case


问:
How to solve this coupled linear ODE
I've tried using mathematica and for some reason it's not working
I think it's a pretty standard problem but for some reason I'm having a hard time
This is essentially a variant of the Rabbi problem I'll really appreciate the help :)
I will appreciate both explanation on how to solve this using mathematica or a manual solution
\begin{align}
i\hbar \frac{d\alpha}{dt}&=\frac{\hbar\omega_0}{2}\alpha+\frac{dE}{2}e^{i\omega t}\beta\\
i\hbar \frac{d\beta}{dt}&=-\frac{\hbar\omega_0}{2}\beta+\frac{dE}{2}e^{-i\omega t}\alpha
\end{align}

答:
I'll begin by recasting this problem in a more concretely physical way. Introducing $\Psi=(\alpha,\beta)^\top$, we have

$$i\hbar\frac{d\Psi}{dt}=i\hbar \frac{d}{dt}\begin{pmatrix} \alpha\\ \beta\end{pmatrix}=\begin{pmatrix}\hbar \omega_0/2 & e^{i\omega t}dE/2\\ e^{-i\omega t}dE/2 & -\hbar\omega_0/2\end{pmatrix}\begin{pmatrix} \alpha\\ \beta\end{pmatrix}=H(t)\Psi$$
which is the time-dependent Schrodinger equation with Hamiltonian
$$\displaystyle H(t)
=
\begin{pmatrix}\hbar \omega_0/2 & e^{i\omega t}dE/2\\ e^{-i\omega t}dE/2 & -\hbar\omega_0/2\end{pmatrix}.$$
The Hamiltonian can be expressed in terms of Pauli matrices as

\begin{align}
H(t)&=\frac{\hbar\omega_0}{2}\begin{pmatrix} 1& 0 \\ 0 & -1\end{pmatrix}+\frac{dE}{2}\cos\omega t\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}-\frac{dE}{2}\sin\omega t\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\\
&=\frac{\hbar \omega_0}{2}\sigma_z+\frac{dE}{2}(\cos(\omega t)\sigma_x-\sin(\omega t)\sigma_y)
\end{align}
Note that the $dE$ portion amounts to rotating the operator $\sigma_x$ around the $z$-axis by an angle $-\omega$. Explicitly,

\begin{align}
e^{i\omega t\sigma_z/2}\sigma_x e^{-i\omega t \sigma_z/2}
&=\begin{pmatrix} e^{i\omega t/2} & 0 \\ 0 & e^{-i \omega t/2}\end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}
\begin{pmatrix} e^{-i\omega t/2} & 0 \\ 0 & e^{i\omega t/2}\end{pmatrix}\\
&=\begin{pmatrix} 0 & e^{i\omega t} \\ e^{-i\omega t} & 0\end{pmatrix}
\end{align}
and therefore the Hamiltonian may be expressed as

$$H(t)=\frac{\hbar\omega_0}{2}+\frac{dE}{2}e^{i\omega t\sigma_z/2}\sigma_x e^{-i\omega t \sigma_z/2}$$
This amounts to an electron subject to an external magnetic field $\vec{B}=\frac12 (dE\cos\omega t,dE\sin\omega t,\hbar\omega_0)$ which rotates in some cone around the $z$-axis.

This suggests choosing a reference frame which rotates along with the magnetic field. To do this, we introduce $\Psi_r =e^{-i\omega t\sigma_z/2}\Psi$ ($r$ for rotated). Then the relevant Schrodinger equation is
\begin{align}
i\hbar \frac{d}{dt}\Psi_r
&=i\hbar \frac{d}{dt}e^{-i\omega t\sigma_z/2}\Psi\\
&=\frac{\hbar\omega}{2} \sigma_z \Psi_r+e^{-i\omega t\sigma_z/2}i\hbar \frac{d}{dt}\Psi\\
&=\frac{\hbar\omega}{2} \sigma_z \Psi_r+e^{-i\omega t\sigma_z/2}H(t)\Psi\\
&=\frac{\hbar\omega}{2} \sigma_z \Psi_r+e^{-i\omega t\sigma_z/2}H(t)e^{i\omega t\sigma_z/2}\Psi_r\\
&=H_r\Psi_r
\end{align}
where we now have the time-independent Hamiltonian $\displaystyle H_r=\frac{1}{2}\hbar(\omega+\omega_0) \sigma_z+\frac{dE}{2}\sigma_x$
in the rotating frame. This Hamiltonian is solvable by hand. To facilitate this, let $$ \Omega=\frac12\sqrt{(dE/\hbar)^2+(\omega+\omega_0)^2},\quad \tan\phi = \frac{dE}{\hbar(\omega+\omega_0)},$$
with which we may write $H_r=\hbar\Omega [\cos(\phi)\sigma_z+\sin (\phi)\sigma_x].$ The same trick as before means that we may diagonalize the Hamiltonian as
$$H_r = e^{-i\phi \sigma_y/2}\hbar \Omega \sigma_z e^{i\phi\sigma_y/2}.$$

We now solve the Schrodinger equation as $\Psi_r(t)=U_r(t)\Psi_r(t=0)$ where

\begin{align}
U_r(t)=e^{-i H_r t/\hbar}
&=e^{-i\phi \sigma_y/2}e^{-i \Omega \sigma_z t}e^{i\phi\sigma_y/2}\\
&=\begin{pmatrix} \cos\frac{\phi}{2} & -\sin \frac{\phi}{2}\\ \sin \frac{\phi}{2} & \cos\frac{\phi}{2}\end{pmatrix}
\begin{pmatrix}e^{-i B_r/\hbar} & 0 \\ 0 & e^{i B_r/\hbar}\end{pmatrix}
\begin{pmatrix} \cos\frac{\phi}{2} & \sin \frac{\phi}{2}\\ -\sin \frac{\phi}{2} & \cos\frac{\phi}{2}\end{pmatrix}\\
&=\begin{pmatrix}
\cos \Omega t -i \cos\phi \sin\Omega t &
-i \sin\phi \sin \Omega t\\
-i \sin\phi \sin\Omega t &
 \cos\Omega t+i \cos\phi \sin\Omega t
\end{pmatrix}
\end{align}
Returning to the original reference frame, the solution takes the form

$$\Psi(t)=e^{i\omega t\sigma_z/2}\Psi_r(t)=e^{i\omega t\sigma_z/2}U_r(t)\Psi_r(0)=U(t)\Psi(0)$$
where the time-evolution operator is now
\begin{align}
U(t)
&= e^{i\omega t\sigma_z/2}U_r(t)\\
&=
\begin{pmatrix} e^{i\omega t/2} & 0 \\ 0 & e^{-i\omega t/2}\end{pmatrix}
\begin{pmatrix}
\cos \Omega t -i \cos\phi \sin\Omega t &
-i \sin\phi \sin \Omega t\\
-i \sin\phi \sin\Omega t &
 \cos\Omega t+i \cos\phi \sin\Omega t
\end{pmatrix}\\
&=\begin{pmatrix}
e^{i\omega t/2}(\cos \Omega t -i \cos\phi \sin\Omega t) &
-i e^{i\omega t/2}\sin\phi \sin \Omega t\\
-i e^{-i\omega t/2}\sin\phi \sin\Omega t &
 e^{-i\omega t/2}(\cos\Omega t+i \cos\phi \sin\Omega t)
\end{pmatrix}\\
&=\begin{pmatrix}
e^{i\omega t/2}(\cos \Omega t -i \Omega^{-1}(\omega+\omega_0) \sin\Omega t) &
-i e^{i\omega t/2}\Omega^{-1}(dE/\hbar) \sin \Omega t\\
-i e^{-i\omega t/2}\Omega^{-1}(dE/\hbar) \sin\Omega t &
 e^{-i\omega t/2}(\cos\Omega t+i \Omega^{-1}(\omega+\omega_0) \sin\Omega t)
\end{pmatrix}
\end{align}
This represents the full closed-form solution. As a sanity check, if $dE=0$ then $\Omega=\frac12( \omega+\omega_0)$ and thus

$$U(t)\to
\begin{pmatrix}
e^{i\omega t/2}e^{-i(\omega+\omega_0)t/2} & 0\\ 0 &
 e^{-i\omega t/2}e^{-i(\omega-\omega_0)t/2}
\end{pmatrix}
=
\begin{pmatrix}
e^{-i\omega_0 t/2} & 0\\ 0 &
 e^{-i\omega_0 t/2}
\end{pmatrix}
=e^{-i\omega_0 t \sigma_z/2}
$$
which agrees with $H(t)= \dfrac{\hbar\omega_0}{2}\sigma_z$ in this limit.

Title: How to solve this coupled linear ODE


问:
Is $|\mathbb{R}^{\omega_1}| > \aleph_1$?
Intuatively I'd guess no, since $|\omega_1^{\omega_1}| = \aleph_1$.

But then there's cardinal exponentiation $\aleph_1^{\aleph_1}\geq2^{\aleph_1}=\aleph_2$.

A formal explanation would be appreciated.

答:
Write $\Bbb R=2^{\aleph_0}$, so $$|\Bbb R|^{\aleph_1}=(2^{\aleph_0})^{\aleph_1}=2^{\aleph_0\cdot\aleph_1}=2^{\aleph_1}>\aleph_1,$$
where I used that for infinite cardinals $\kappa,\lambda$ we have $\kappa\cdot\lambda=\max\{\kappa,\lambda\}$.

Also note that $2^{\aleph_1}=\aleph_2$ is independent of $\mathsf{ZFC}$, but nonetheless we have $2^{\aleph_1}>\aleph_1$ by Cantor's theorem.


问:
solution to a root inequality
I have the inequality
$$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc} \geq \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab}.$$I tried to do $u=a^2+b^2+c^2$ and $v=ab+ac+bc$ and $x=a^2+2bc$,  $y=b^2+2ac$,  $z=c^2+2ab$ ...but I did not find any solution. Any help is appreciated.</p>

答:
I will assume that $a, b, c \geq 0$. Change to cylindrical coordinates with axis in the direction of $(1, 1, 1)^T$ as follows:
$$\left(\begin{matrix} a \\ b \\ c \end{matrix}\right) = \frac{z}{\sqrt{3}}\left(\begin{matrix} 1 \\ 1 \\ 1\end{matrix}\right) + \rho\left[\frac{\cos\phi}{\sqrt{6}}\left(\begin{matrix} 1 \\ 1 \\ -2\end{matrix}\right) + \frac{\sin\phi}{\sqrt{2}}\left(\begin{matrix} 1 \\ -1 \\ 0\end{matrix}\right)\right]$$
The inequality then becomes:
$$\sum_{j=1}^3 \sqrt{z^2+\rho^2\cos(2\phi+\frac{2\pi}{3}j)} \leq \sqrt{z^2+\rho^2}+2\sqrt{z^2-\frac{1}{2}\rho^2}$$
The left side depends on $\phi$, while the right side does not, and we have equality when $\phi = \frac{k\pi}{3}$ with $k$ an integer. Evaluating the partial derivative with respect to $\phi$ of the left side shows the existence of critical points at $\phi = \frac{k\pi}{6}$ for integer $k$. Plugging these in to the inequality shows that equality holds when $\phi = \frac{k\pi}{3}$, while strict inequality holds when $\phi = \frac{(2k+1)\pi}{6}$ (unless $\rho = 0$).

We first examine the case when $z^2 \geq \rho^2$. Consider the function
$$f(x_1, x_2, x_3) = \sum_{i=1}^3\sqrt{z^2 + \rho^2x_i}$$
subject to the constraints
$$g(x_1,x_2,x_3)=x_1 + x_2 + x_3 = 0$$
$$h(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2=\frac{3}{2}$$
The method of Lagrange multipliers dictates that local maxima will only be found where $\nabla f + \lambda \nabla g + \mu \nabla h = 0$. Then for each $x_i$ we have
$$\rho^2(z^2+\rho^2x_i)^{-1/2}+\lambda+2\mu x_i=0$$
Let $\psi(x) = \rho^2(z^2+\rho^2x)^{-1/2}+\lambda+2\mu x$. Since $\frac{d^2 \psi}{d x^2} = \frac{3}{4}\rho^6(z^2+\rho^2x)^{-5/2} > 0$, $\psi$ can have at most two roots, so local maxima can only be found when two of the $x_i$ are equal. This corresponds to $\phi = \frac{k\pi}{6}$ for integer $k$. Since we have equality at $\phi = \frac{k\pi}{3}$ and strict inequality at $\phi = \frac{(2k+1)\pi}{6}$, the critical points at $\phi=\frac{k\pi}{3}$ are maxima and the critical points at $\phi = \frac{(2k+1)\pi}{6}$ are local minima. Since the inequality holds at the maxima, it holds for all $\phi$.

Now consider the case when $z^2 < \rho^2$. We know that the term $z^2 - \frac{1}{2}\rho^2$ in the right-hand side is always well-defined, so $\frac{1}{2}\rho^2 \leq z^2 < \rho^2$. The left-hand side is then undefined within a neighborhood of $\phi = \frac{(2k+1)\pi}{6}$. Since $\phi = \frac{k\pi}{3}$ are maxima for $z^2 \geq \rho^2$, by continuity with respect to $z$ and the fact that critical points can only occur at $\frac{k\pi}{6}$, it follows that $\phi = \frac{k\pi}{3}$ are maxima for all $z^2 \geq \frac{1}{2}\rho^2$. Then since the inequality holds at the maxima, it holds for all $\phi$ such that it is well-defined, and thus for all $a, b, c \geq 0$. $\square$

Title: solution to a root inequality


问:
$\langle v, v \rangle \geq 0$ meaning in inner product space.
In Linear Algebra Done Right: https://zhangyk8.github.io/teaching/file_spring2018/linear_algebra_done_right.pdf

Inner product space is defined as a vector space V together with an inner product $V \times V \to \mathbb{F}$ where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$.
The inner product must satisfy $\langle v, v \rangle \geq 0$.

How would it be defined if $\langle v, v \rangle \in \mathbb{C}$?

In the definition of norm, they defined norm as $||v|| = \sqrt{\langle v, v \rangle}$. From another source, norm is defined as a nonnegative real value function, that implies $\langle v, v \rangle$ must be nonnegative real.

Additionally, they defined a positive operator as a linear map from $V$ to $V$ that is self-adjoint and $\langle Tv, v \rangle \geq 0 \text{ } \forall v \in V$

How would it be defined if $\langle Tv, v \rangle \in \mathbb{C}$?

Thank you!

答:
This is sometimes an annoying convention, but for complex numbers when we write (for instance) $z\ge 0$, we really mean that $z$ is actually a positive real number. In this case, note that the standard Hermitian inner product on $\Bbb{C}^n$ given by $\langle z,w\rangle =\sum_{i=1}^n \overline{z}^iw^i$ has
$$
\langle z,z\rangle =\sum_{i=1}^n \overline{z}^iz^i=\sum_{i=1}^n \lvert z\rvert^2\in \Bbb{R}.
$$


答:
By definition it means that for every $v\in V$, $\langle v,v\rangle\in\Bbb{R}$ and $\langle v,v\rangle\geq 0$; or simply, because the set $[0,\infty)$ is a subset of $\Bbb{C}$, we can just say $\langle v,v\rangle\in [0,\infty)$. It is common to avoid saying all of this, and just write $\langle v,v\rangle\geq 0$.

Here's another long-winded way of saying it. The inner product on a complex vector space is a function $\langle\cdot,\cdot\rangle:V\times V\to \Bbb{C}$. From this function we can define the "quadratic function" $Q:V\to\Bbb{C}$ defined as $Q(v):=\langle v,v\rangle$. Now although $Q$ has the complex numbers $\Bbb{C}$ as the target space, what the axiom is requiring is that the image of the function $Q$ is actually a subset of $[0,\infty)\subset \Bbb{C}$.

It's the same story with $\langle T(v),v\rangle$.

Title: $\langle v, v \rangle \geq 0$ meaning in inner product space.

Body:
In Linear Algebra Done Right: https://zhangyk8.github.io/teaching/file_spring2018/linear_algebra_done_right.pdf

Inner product space is defined as a vector space V together with an inner product $V \times V \to \mathbb{F}$ where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$.
The inner product must satisfy $\langle v, v \rangle \geq 0$.

How would it be defined if $\langle v, v \rangle \in \mathbb{C}$?

In the definition of norm, they defined norm as $||v|| = \sqrt{\langle v, v \rangle}$. From another source, norm is defined as a nonnegative real value function, that implies $\langle v, v \rangle$ must be nonnegative real.

Additionally, they defined a positive operator as a linear map from $V$ to $V$ that is self-adjoint and $\langle Tv, v \rangle \geq 0 \text{ } \forall v \in V$

How would it be defined if $\langle Tv, v \rangle \in \mathbb{C}$?

Thank you!


问:
Are Radon measures on Polish spaces $\sigma$-finite?
If $\Omega$ is a Polish space and $\mu$ is a Radon measure on $\Omega$ (i.e. an inner-regular Borel measure), is $\mu$ $\sigma$-finite?

I know that Radon measures in general need not be $\sigma$-finite, and $\sigma$-finite measures need not be Radon. The standard counterexample to the former is an uncountable set with the discrete topology and counting measure (which is Radon, but not $\sigma$-finite), and one counterexample to the latter is $\Omega = \mathbb R$ with $\mu$ the counting measure on $\mathbb Q$ (which is $\sigma$-finite, but not Radon, or even Borel).

But what if $\Omega$ is Polish? If $Q \subset \Omega$ is countable and dense, since $\mu$ is Borel, every $q \in Q$ has an open neighborhood $U_q \ni q$ for which $\mu(U_q) < \infty$. But a priori, there's no guarantee that $W:=\bigcup_{q \in Q} U_q = \Omega$. We know that by density of $Q$ in $\Omega$, we have that $W^c$ has empty interior, but $W^c$ could still have infinite measure in principle (e.g. let $\mu = \lambda^2 + \lambda^1$ on $\mathbb R^2$, where $\lambda^2$ is the $2$-dimensional Lebesgue measure on $\mathbb R^2$ and $\lambda^1$ is the $1$-dimensional Lebesgue measure on the $x$-axis, and take $Q = \left\{(p,q) \in \mathbb Q^2 : p \neq 0\right\}$). And taking the closures $\overline U_q$ might not work because if $\Omega$ is an infinite-dimensional Banach space, for example, $\overline U_q$ need not be compact, so $\mu\left(\overline U_q\right) = \infty$ is possible.

I'm really not sure one way or the other about the answer to this question. I can't think of a counterexample, but I can't think of a proof, either. Anything I'm not thinking of?

答:
For every point $x$ of the space $X$ pick a neighbourhood $U_x$ with $\mu(U_x)<\infty$. Since Polish spaces are separable they are also Lindelöf, hence you can find a countable $I\subset X$ such that $\{U_y\mid y\in I\}$ still covers $X$, and this cover witnesses that $\mu$ is $\sigma$-finite.

Also note that in hereditarily Lindelöf spaces (so separable metric spaces for example) every Radon measure is not only $\sigma$-finite, but also moderated, which is a slightly stronger property.


问:
Proving two countably infinite sets have a one-to-one correspondence.
Let $A$ and $B$ be two sets. Suppose that $A$ and $B$ are both countably infinite sets. Prove that there is a one-to-one correspondence between $A$ & $B$.

Here is the proof I have. I tried going about the proof by first show $f: A \rightarrow N$ has a one-to-one correspondence and $g: B \rightarrow N$ as a one-to-one correspondence as well. Not sure if this was the right thing to do. Then I did theta: $A \rightarrow B$. Not sure if I defined theta right and accurately showed that there is a one-to-one correspondence between $A$ and $B$.

Proof: Suppose that $A$ and $B$ are both countably infinite sets. As a result of the two sets being countably infinite, we know set $A$ and $B$ can be put in a one-to-one correspondence with the set of positive integers. Thus we need to find a one-to-one correspondence between $N$ to $A$ and $N$ to $B$.

Lets define $$f: A \rightarrow N \,|\, f(n)= 2n, \,\, \forall n\in \{A\}$$

Then suppose there is a a,b in the set $A$ such that $f(a)=f(b)$ $\rightarrow$ $2a = 2b$ $\rightarrow a=b$.
Therefore, there is a one-to-one correspondence.

Similarly, define
$$g: B \rightarrow N \,|\, f(m) = 2m-1, \,\, \forall m \in \{B\}$$

Then suppose there is a $c,d$ in the $B$ such that $g(c)=g(d)$ $\rightarrow 2c-1 =2d-1$ $\rightarrow 2c=2d$ $\rightarrow c=d$

Therefore, there is a one-to-one correspondence.

The $|A|=|B|=|N|$ $(N \in \{\mathbb{N}\})$

Now, let $\theta: A \rightarrow B$ via $\theta(2n)=2n-1$ for all $n$ in the set $N$

Suppose $x,y$ in the set $N$ such that $\theta(x)=\theta(y)$.
$\theta(x) = \theta(y)$ $\rightarrow 2x-1 =2y-1 \rightarrow 2x=2y \rightarrow x=y$

Hence there is a one-to-one correspondence between the two countably infinite sets $A$ & $B$.

答:
Assuming countably infinite means there is a bijection between the $\mathbb N$ and the set.

So if $A$ and $B$ are both countably infinite then we know there must exist two bijections

$\phi: \mathbb N \to A$, and $\psi: \mathbb N\to B$.

Then I claim that $f = \psi\circ \phi^{-1}$ is a well defined bijection.

Can you prove that?

Just use the definitions.  First prove $\phi^{-1}:B\to \mathbb N$ defined by: for any $a\in A$ then $\phi^{-1}(a)$ will be equal to the unique $n\in \mathbb N$ so that $\phi(n)= a$; is well defined.

Then prove $\phi^{-1}$ is a bijection.

Then prove the composition of bijections is a bijection.

Then prove $\psi\circ \phi^{-1}$  is surjective and one-to-one.

......

Step 1:  $\phi^{-1}: \mathbb A\to \mathbb N$ is well defined.

$\phi:\mathbb N\to A$ is surjective. SO for ever $a\in A$ there is an $n\in \mathbb N$ so that $\phi(n) = a$.  And as $\phi$ is one-to-one, there is only one distinct $n$ so that $\phi(n) = a$.  So for every $a$ we can unique identify a distinct and unique $n_a$ so that $\phi(n_a) = a$ and we can define a function so that $\phi^{-1}(n_a) =a$ for each $a \in A$.  SO $\phi^{-1}:A\to \mathbb N$ is well defined.

Step 2: $\phi^{-1}: A\to \mathbb N$ is a bijection.

a) $\phi^{-1}$ is surjective. Pf:  $\phi:\mathbb N \to A$ so for any $n\in \mathbb N$ there is an $a\in A$ so that $\phi(n) = a$.  So by definition $\phi^{-1}(a) = n$.  SO for every $n$ there is an $a\in A$ so that $\phi^{-1}(a) = n$.  SO $\phi^{-1}$ is surjective.

b) $\phi^{-1}$ is one to one.  Pf: Suppose $\phi^{-1}(a) =\phi^{-1}(\alpha)$.  Label that $n =\phi^{-1}(a) =\phi^{-1}(\alpha))$.  Then by definition $\phi(n) =a$ and $\phi(n) =\alpha)$.  But $\phi$ is a function.  It maps every value of $n$ to one (and only one) value of $A$.  So we must have $a =\alpha$.  So $\phi^{-1}$ is one to one.

So $\phi^{-1}$ is a bijection.

(Note: this argument and proof would work for any bijection between any two sets.  All bijections will have an inverse function and the inverse will be a bijection.

Step 3:  Prove $\psi\circ \phi^{-1}:\mathbb A \to B$ is a bijection.

a) $\psi\circ \phi^{-1}$ is surjective: Pf: Suppose $b\in A$.  Then as $\psi$ is surjective there is an $n_b\in \mathbb N$ so that $\psi(n_b) = b$.

And $\phi^{-1}$ is surjective so there is an $a \in A$ so the $\phi^{-1}(a) = n_b$.

So $\psi(\phi^{-1}(a)) = \psi(n_b) = b$. So there is an $a\in A$ so that $\psi\circ\phi^{-1}(a) = b$.  So $\psi\circ\phi^{-1}$ is surjective.

b) $\psi\circ\phi^{-1}$ is one to one.  Suppose $\psi\circ\phi^{-1}(a)=\psi\circ\phi^{-1}(\alpha) = b$.

Let $\phi^{-1}(\alpha) = n$ and $\phi^{-1}(a)= m$.  Then $\psi(n)=\psi(\phi^{-1}(\alpha)) = b$ and $\psi(m)=\psi(\phi^{-1}(a)) = b$.  But $\psi$ is one to one so $n=m$ and $\phi^{-1}(\alpha) = \phi^{-1}(a)$.

But $\phi^{-1}$ is one to one so $\alpha = a$.  so if $\psi\circ\phi^{-1}(a)=\psi\circ\phi^{-1}(\alpha)\implies a=\alpha$ so $\psi\circ\phi^{-1}$ is one to one.

SO $\psi\circ\phi^{-1}$ is a bijection between $A$ and $B$.

Not... all this would hold for any two sets of the same (any) cardinality.

If $|M| = |N|$ and $|N| = |C|$ there always exists a bijection between $M$ and $C$ and $|M| = |C|$>


问:
Does $\mathfrak{(p<q)\land(r<s)}$ imply $\mathfrak{p^r<q^s}$?

Does $\mathfrak{(p<q)\land(r<s)}$ imply $\mathfrak{p^r<q^s}$, where $\mathfrak{ p,q,r,s}$ are cardinal numbers?

Is it possible to prove in $\mathsf{ZFC}$ that there is a counterexample?

答:
The inequality does not hold in general. First, note that $2^\kappa=\kappa^\kappa$ for any infinite $\kappa$. Second, it is consistent (using the technique of forcing) that $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$, even though $\aleph_0<\aleph_1$. This gives us that it is consistent that $\aleph_0<\aleph_1$ and yet $\aleph_0^{\aleph_0}=\aleph_1^{\aleph_1}$.

For a (combinatorially more involved) $\mathsf{ZFC}$ example, $\beth_\omega^{\aleph_0}=\beth_\omega^{\aleph_1}$, so $\beth_{\omega}^{\aleph_0}=(\beth_\omega^+)^{\aleph_1}$. (For a similar computation, see Jech's exercise 5.18, here or here.)


答:
There is already a good answer. This can be looked upon differently.

Since $A$ and $B$ are countable, let us list their elements, i.e. $A=\{a_1, a_2, \cdots\}$ and $B=\{b_1, b_2, \cdots\}$ (all distinct elements). Define the one-to-one correspondences $f:A\rightarrow \mathbb{N}$ by $f(a_i)=i$ and $g:B \rightarrow \mathbb{N}$ by $g(b_i)=i$. Then $g^{-1} \circ f$ is the required one-to-one correspondence from $A$ to $B$.

But if you list the elements as, $A=\{\cdots, a_{-2}, a_{-1}, a_0, a_1, a_2, \cdots\}$, similar formula for $B$. The one-to-one correspondence would be $f:A \rightarrow \mathbb{N}$ define by $f(a_i)=2i$, if $i\geq 0$; $f(a_i)= -(2i+1)$ if $i<0$. Similarly define $g:B \rightarrow \mathbb{N}$. $g^{-1}\circ f$ is the one-to-one correspondence as before.

Title: Proving two countably infinite sets have a one-to-one correspondence.

Body:
Let $A$ and $B$ be two sets. Suppose that $A$ and $B$ are both countably infinite sets. Prove that there is a one-to-one correspondence between $A$ & $B$.

Here is the proof I have. I tried going about the proof by first show $f: A \rightarrow N$ has a one-to-one correspondence and $g: B \rightarrow N$ as a one-to-one correspondence as well. Not sure if this was the right thing to do. Then I did theta: $A \rightarrow B$. Not sure if I defined theta right and accurately showed that there is a one-to-one correspondence between $A$ and $B$.

Proof: Suppose that $A$ and $B$ are both countably infinite sets. As a result of the two sets being countably infinite, we know set $A$ and $B$ can be put in a one-to-one correspondence with the set of positive integers. Thus we need to find a one-to-one correspondence between $N$ to $A$ and $N$ to $B$.

Lets define $$f: A \rightarrow N \,|\, f(n)= 2n, \,\, \forall n\in \{A\}$$

Then suppose there is a a,b in the set $A$ such that $f(a)=f(b)$ $\rightarrow$ $2a = 2b$ $\rightarrow a=b$.
Therefore, there is a one-to-one correspondence.

Similarly, define
$$g: B \rightarrow N \,|\, f(m) = 2m-1, \,\, \forall m \in \{B\}$$

Then suppose there is a $c,d$ in the $B$ such that $g(c)=g(d)$ $\rightarrow 2c-1 =2d-1$ $\rightarrow 2c=2d$ $\rightarrow c=d$

Therefore, there is a one-to-one correspondence.

The $|A|=|B|=|N|$ $(N \in \{\mathbb{N}\})$

Now, let $\theta: A \rightarrow B$ via $\theta(2n)=2n-1$ for all $n$ in the set $N$

Suppose $x,y$ in the set $N$ such that $\theta(x)=\theta(y)$.
$\theta(x) = \theta(y)$ $\rightarrow 2x-1 =2y-1 \rightarrow 2x=2y \rightarrow x=y$

Hence there is a one-to-one correspondence between the two countably infinite sets $A$ & $B$.


问:
Modular quadratic equation question- Where did I go wrong?
\begin{gather}
\frac{N^2+N}{2} \equiv 0 \pmod 4 \\
N^2+N\equiv 0\\
(N+\frac{1}{2})^2-\frac{1}{4}\equiv 0\\
4(N+\frac{1}{2})^2-1\equiv 0\\
2^2(N+\frac{1}{2})^2\equiv 1\\
(2N+1)^2\equiv 1 \\
2N+1 \equiv 1\\
2N=0\\
N=2,4,6,8...\\
2N+1 \equiv 3\\
2N=2\\
N=1,3,5,7...\\
\end{gather}
In other words I've shown $\forall N >0 \to \frac{N^2+N}{2} \equiv 0 \pmod 4$, which is false e.g. for $N=3$,  $\frac{N^2+N}{2}$ has a remainder of 2

I'd like a correct solution as well. Thanks.

答:
In arithmetic mod $4$, you cannot divide by $4$ or $2$. It is like dividing by $0$ in normal arithmetic; you can write the expression but there is no number that is equal to it.

If you divide and multiply by zero you can prove that all numbers are equal. This is of course inconsistent with our normal arithmetic, and also with modular arithmetics. The reason is that you cannot divide by zero.

If you allow division by zero you get the uninteresting arithmetic where there is only a single number.

Note that the second line is a legal equation, and its solutions, by trial and error, are $3$ and $4$.

Your example is interesting in the way it disguises $0$ as $4$, and also as $2$, since $2=\sqrt{4}=\sqrt{0}=0$.

I hadn't seen an example like this using modular arithmetic.


问:
What is the smallest solution to $x = 2 \operatorname{mod} (6y+1)$, $x = 3 \operatorname{mod} (12y+1)$, $x= 3 \operatorname{mod} (18y+1)$?
I'd like to solve a system of modular equations, but it's a somewhat unusual system. Specifically, what is the smallest natural number $x$ for which $x = 2\pmod{6y+1}$, and $x=3\pmod{12y+1}$, and $1 \pmod{18y+1}$ for some natural number $y$?

If $y=1$, then WolframAlpha tells me that $x=1654$.  But is it possible that a bigger choice of $y$ could yield a smaller value of $x$?  This came up in the context of Godel coding, by the way.

Body:
\begin{align}
   x &\equiv 2\pmod{ 6y+1} \\
   x &\equiv 3\pmod{12y+1} \\
   x &\equiv 1\pmod{18y+1}
\end{align}

Note
\begin{align}
   2( 6y+1) - 1(12y+1) &= 1 \\
   3(12y+1) - 2(18y+1) &= 1 \\
  -2(18y+1) + 3( 6y+1) &= 1 \\
\end{align}

So the numbers $6y+1, 12y+1$, and $18y+1$ are pairwise prime.

\begin{align}
   (12y+1)(18y+1)
      &\equiv 216y^2+30y+1 \pmod{6y+1} \\
      &\equiv (216y^2+30y+1)\pmod{6y+1} \\
      &\equiv (36y-1)(6y+1) + 2 \pmod{6y+1} \\
      &\equiv 2 \pmod{6y+1} \\
      &\equiv 0 \pmod{12y+1} \\
      &\equiv 0 \pmod{18y+1}
\end{align}

\begin{align}
   -12(6y+1)(18y+1)
      &\equiv (-1296y^2-288y-12) \pmod{12y+1} \\
      &\equiv (-108y - 15)(12y + 1) + 3 \pmod{12y+1} \\
      &\equiv 0 \pmod{6y+1} \\
      &\equiv 3 \pmod{12y+1} \\
      &\equiv 0 \pmod{18y+1}
\end{align}

\begin{align}
   (9y+5)(6y+1)(12y+1)
      &\equiv 648y^3 + 522y^2 + 99y + 5 \\
      &\equiv (36y^2 + 27y + 4)(18y + 1) + 1 \pmod{18y+1} \\
      &\equiv 0 \pmod{6y+1} \\
      &\equiv 0 \pmod{12y+1} \\
      &\equiv 1 \pmod{18y+1}
\end{align}

\begin{array}{c}
   (12y+1)(18y+1)-12(6y+1)(18y+1)+(9y+5)(6y+1)(12y+1) \\
   = 648y^3-558y^2-159y-6
\end{array}

So, a solution is $$648y^3-558y^2-159y-6 \pmod{(6y+1)(12y+1)(18y+1)}$$


答:
What you did incorrectly is basically explained in Peter's answer, and several question comments. Note, though, the more general issue when manipulating congruences, since you are really dealing with integers, is that involving "fractions" of the form $\frac{p}{q}$ where $\gcd(p,q) = 1$, really means the division by $q$ is multiplying by its modulo multiplicative inverse, i.e., $q^{-1}$. However, only integers which are relatively prime to the modulus have inverses. Since $\gcd(4, 4) = 4 \neq 1$ and $\gcd(4, 2) = 2 \neq 1$, neither $2$ nor $4$ have an inverse in modulo $4$, so while working in that modulus you can't, in general, just "divide" by either value, e.g., $\frac{1}{2}$ and $\frac{1}{4}$ don't make sense.

As for how to correctly solve your congruence equation, note for any solution $N$, there's some integer $j$ where

$$\begin{equation}\begin{aligned}
& \frac{N^2 + N}{2} \equiv 0 \pmod{4} \iff \frac{N^2 + N}{2} = 4j \iff \\
& N^2 + N = 8j \iff N^2 + N \equiv 0 \pmod{8}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Due to $N$ and $N + 1$ being relatively prime to each other, with one being odd and one being even, then since $8 = 2^3$ has only prime factors of $2$, this then gives

$$\begin{equation}\begin{aligned}
& N(N + 1) \equiv 0 \pmod{8} \implies \\
& (N \equiv 0 \pmod{8}) \lor (N + 1 \equiv 0 \pmod{8}) \implies \\
& N \equiv 0, 7 \pmod{8}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

This means the allowed values for $N$ are $N = 8k$ and $N = 8k + 7$ for any $k \in \mathbb{Z}$.

Title: Modular quadratic equation question- Where did I go wrong?

Body:
\begin{gather}
\frac{N^2+N}{2} \equiv 0 \pmod 4 \\
N^2+N\equiv 0\\
(N+\frac{1}{2})^2-\frac{1}{4}\equiv 0\\
4(N+\frac{1}{2})^2-1\equiv 0\\
2^2(N+\frac{1}{2})^2\equiv 1\\
(2N+1)^2\equiv 1 \\
2N+1 \equiv 1\\
2N=0\\
N=2,4,6,8...\\
2N+1 \equiv 3\\
2N=2\\
N=1,3,5,7...\\
\end{gather}
In other words I've shown $\forall N >0 \to \frac{N^2+N}{2} \equiv 0 \pmod 4$, which is false e.g. for $N=3$,  $\frac{N^2+N}{2}$ has a remainder of 2

I'd like a correct solution as well. Thanks.


问:
Critical points: $u = x - 2y + 2z$ if $x^2 + y^2 + z^2 = 9$
Find the critical points $P$ of $u =x-2y-2z$ given that $x^2 + y^2 + z^2 =9$ and classify if it is maximum, minimum or neither

I've find that $ x = 1 $ or $x=-1$, $y=-z =2$ or $y = -z = -2$ using $ x=^+/_- \sqrt{9 -y^2 - x^2}$. But, it looks like I will have so much work to calculate the determinant of the hessian matrix to decide if it is maximum, minimum or neither for each possible point. Is there a easiest way to do this (I've tried spherical coordinates but doesn't look like a good idea too)? Maybe I am using the hardest path, since my second order derivatives dont have a good form.

Thanks!

答:
A similar problem appears here:   find absolute maximum and minimum with explanation

Summarizing the method using a geometric interpretation, the function to be extremized can be considered as the description of a plane.  For $ \ x \  - \  2y \  + \  2z \ = \ 0 \ , $ we have a plane through the origin with normal vector $ \ \langle 1 \ ,  -2 \ , 2 \rangle \ ; $  all planes parallel to this are    $ \ x \  - \  2y \  + \  2z \ = \ d \ . $  There are two such planes that are just tangent to the sphere $ \ x^2 \ + \ y^2 \ + \ z^2 \ = \ 9  \ $ of radius $ \ 3 \ $ centered on the origin.  The distance of these planes from the origin is given by

$$ D \ \ = \ \ 3 \ \ = \ \ \frac{|d \ - \ 0 |}{\sqrt{1^2 \ + \ (-2)^2 \ + \ 2^2}} \ \ = \ \ \frac{|d|}{\sqrt{9}} \ \ , $$

hence, $ \ |d| \ = \ 3 \ · \ 3 \ = \ 9 \ $ .  The maximum and minimum values of  $ \ u \ = \ x \  - \  2y \  + \  2z \  $ are $ \ 9 \ $ and $ \ -9 \ , $ respectively.

Note that the Lagrange equations represented by $ \ \nabla u \ = \ \lambda ·\nabla g \ ,  $ with $ \ g(x,y,z) \ =  \ x^2 \ + \ y^2 \ + \ z^2 \ - \ 9 \ , $ are used to find the locations of points on the sphere where the normal vector to its surface is aligned with the normal vector for the plane (or, the gradient vector of the function $ \ g \ $ is a scalar multiple of the gradient vector of the function $ \ u \ $ ) .  The three equations are

$$ 1 \ \ = \ \ \lambda \ · \ 2x \ \ \ , \ \ \ -2 \ \ = \ \ \lambda \ · \ 2y \ \ \ , \ \ \ 2 \ \ = \ \ \lambda \ · \ 2z $$

$$ \Rightarrow \ \ \ \lambda \ \ = \ \ \frac{1}{2x} \ \ = \ \ \frac{-2}{2y} \ \ = \ \ \frac{2}{2z} \ \ \ \Rightarrow \ \ \ -2x \ \ = \ \ y \ \ = \ \ -z \ \ , $$

which leads us to the above result when inserted into the equation for the sphere.


问:
How do you set up a system of ODE's for this problem?
The problem is as follows:

Black and White balls are being created inside an arbitrary volume at rates of $Q_{B}$ and $Q_{W}$. They also disappear from the volume at rates $\lambda_{B}$ and $\lambda_{W}$. Along the surface of the volume are detection machines to which white and black balls attach. These machines are created with rate $Q_{D}$, they disappear(along with all the balls stored in them at any given time)with rate $\lambda_{D}$, and the white and black balls attach to a detection machine with rates $\beta_{B}$ and $\beta_{W}$. Once the number of white balls stored in a detection machine reaches $n$, that detection machine turns off and no more balls can go into it.
How do you go about setting up differential equations that describe this system?

Without the condition of a detection machine turning off after $n$ white balls attach, I'd use this system:

$\dot{B}=Q_{B}-\lambda_{B}B-\beta_{B}BD$

$\dot{W}=Q_{W}-\lambda_{W}W-\beta_{W}WD$

$\dot{D}=Q_{D}-\lambda_{D}D$

But I'm not sure how to rewrite the system to account for $n$. Let me know if you need clarification about any parts of the problem. Thank you!

答:
The formulation of your problem is a bit odd since you're talking about discrete events (n balls) but then you write down an ODE which is something you do in the macroscopic limit -- you represent your variables as continuous concentrations.

Apart from that, I would start tackling the question you pose by introducing two variables for the detector: the unoccupied detector and the occupied detector.

So something of this form:

$\dot{D_u} = Q_D - \lambda_D D_u - \left( 1 - D_o \right) D_u (B + W) $

$\dot{D_o} = \left( 1 - D_o \right) D_u (B + W)$

assuming that your concentrations live on the unit interval ($D_o \in [0,1]$).


问:
Inequality (Decay of Green function)
Let $d\ge 3$ be an integer and $m,r>0$. Then how would one show that
$$
\int_m^\infty (s^2-m^2)^{(d-3)/2}e^{-sr}ds \le \frac{C}{r^{(d-1)/2}} e^{-mr}
$$
for somce constant $C$ which may depend on $m$ but not $r$.

Background. The LHS is basically the Fourier transform $G(x)$ of $1/(k^2+m^2)$ at $x$ where $|x|=r$ (you can check this by taking $d=3$). It is known that $G(x)$ is bounded by the RHS (or at least, I have seen papers mentioning this), but how would one prove the aforementioned inequality.

答:
To make the story short, using quite obvious changes of variables, the integral is
$$I=\frac{2^{\frac{d}{2}-1} \left(\frac{m}{r}\right)^{\frac{d-2}{2}}}{\sqrt{\pi }} \,\Gamma \left(\frac{d-1}{2}\right)\,K_{\frac{d-2}{2}}(m r)$$ At least, when $m$ is large,we have
$$I=\frac{(2 m)^{\frac{d-3}{2}}\,\Gamma \left(\frac{d-1}{2}\right) e^{-mr}}{r^{\frac {d-1 } 2}}\Bigg[1+\frac{(d-1)(d-3)}{8 m r}+O\left(\frac{1}{m^2}\right) \Bigg]$$ If we neglect the high order terms, then you would like to show that
$$\frac{(2 m)^{\frac{d-3}{2}}\,\Gamma \left(\frac{d-1}{2}\right) e^{-mr}}{r^{\frac {d-1 } 2}} < \frac C {r^{\frac {d-1 } 2}} e^{-mr}$$ that is to say
$$(2 m)^{\frac{d-3}{2}}\,\Gamma \left(\frac{d-1}{2}\right) < C$$

Title: Inequality (Decay of Green function)


问:
Showing $a^2=b^2$ in a group with 8 elements isomorphic to Quaternion group
I am working in a group $G$ with 8 elements such that $a, b\in G$ and $ba=a^3b$. Moreover, I am given that both $b$ and $a$ have order 4 and $\langle a \rangle \neq \langle b \rangle$. I understand this group is isomorphic to Quaternion group, but this is not given and I am trying to prove this. I want to show $a^2=b^2$, but I am stuck.
Here is what I have done so far:

$$(ab)^2=abab=a(ba)b=aa^3bb=a^4b^2=b^2$$

Also from the equality $ba=a^3b$ I can derive many other equalities, but they do not seem to give me anything useful. For instance

$$aba=b \text{ or } a^3ba^3=b$$

Another observation I have made is that both $b^2$ and $a^2$ have order 2. But I am not sure if that is enough to conclude $a^2=b^2$. Please guide me in the right direction.

答:
We do not need to assume $\langle a\rangle\neq\langle b\rangle$.

$\;\;\;\;\;$ As a result of Cauchy's Theorem the $G$ group center $\mathcal Z(G)$ is non-trivial. However, both $a,b\notin\mathcal Z(G)$ or else $ba=a^3b$ would imply $a^2=e$ which contradicts the OP assumption $$\text{ord}(a)=\text{ord}(b)=4$$
$\;\;\;\;\;$ Therefore, $\frac{G}{\mathcal Z(G)}$ is not cyclic and thus $|\frac{G}{\mathcal Z(G)}|=4$ with $\text{ord}(g\cdot\mathcal Z(G))=2$ for all $g\in G$. Therefore, $a^2,b^2\in\mathcal Z(G)$ but $a^2,b^2\neq e$ and thus $a^2=b^2$ and $\mathcal Z(G)=\langle a^2\rangle=\langle b^2\rangle$.


问:
$f(x) = a_1^x + a_2 ^x +... + a_n^x$ is increasing and $a_1a_2...a_n = 1$
I have encountered this problem in an Olympiad material provided by my teacher and I do not know how to deal with it.
If $a_1, a_2, \cdots, a_n > 0$ and $a_1a_2...a_n = 1$, prove the function $f:[0, \infty] \to \mathbb{R}, f(x) = a_1^x + a_2 ^ x + ... + a_n ^ x$ is increasing. One of my friends managed to solve it using derivatives, but I would like an approach which does not involve calculus, since it is a 10th grade problem.

答: If $x>y$, then $$f(x)-f(y)=\sum_j{a_j^y(a_j^{x-y}-1)}$$  By AM-GM, $$\frac{1}{n}\sum_j{a_j^{x-y}}\geq\sqrt[n]{\prod_j{a_j^{x-y}}}=\left(\prod_j{a_j}\right)^{\frac{x-y}{n}}=1$$  Thus $$\sum_j{a_j^{x-y}}\geq n$$  Finally, let $a_m$ be the $a_j$ of smallest magnitude.  Then $$f(x)-f(y)=\sum_j{a_j^y(a_j^{x-y}-1)}\geq a_m^y\sum_j{(a_j^{x-y}-1)}=a_m^y\left(\left(\sum_j{a_j^{x-y}}\right)-n\right)\geq0$$


答:
Your first error is that you you scaled by $\,2,\,$ which is not invertible $\!\bmod 4,\,$ so this will not yield an equivalent congruence. Rather, it yields necessary but not sufficient conditions on roots (so possibly  extraneous roots). See here for more on the insufficiency of unidirectional inferences.

To get an equivalent congruence we need to scale the modulus too, since $\,4\mid a/2\iff 8\mid a,\,$ so

$$(n^2+n)/2\equiv 0\!\!\!\pmod{4}\iff n^2+n\equiv 0\!\!\!\pmod{8}\qquad$$

Now we can complete the square as you did, but since this too involves scaling the modulus, this will end up being fruitless, leading back to where we started, namely

$$\begin{align} n^2+n&\equiv 0\pmod{8}\\
\iff\ \ \ \ \ \ \ 4n^2+4n&\equiv 0\pmod{32}\\
\iff 4n^2+4n+1&\equiv 1\pmod{32}\\
\iff \ \ \ \ \ \, \color{#c00}{(2n+1)^2}&\:\color{#c00}{\equiv 1}\pmod{32}\\
\iff (2n)(2n+2)&\equiv 0\pmod{32}\\
\iff\ \ \ \ \ \ \  n(n+1)&\equiv 0\pmod{8}
\end{align}\qquad$$

where we solved $\,\color{#c00}{a^2\equiv 1}\,$ by factoring the difference of squares $\,0\equiv a^2-1\equiv (a-1)(a+1).\,$  You can't simply take square roots as you did, e.g. $\,x^2\equiv 1\pmod{8}\,$ has $4$ roots $\,x\equiv 1,3,5,7$.

Instead, by $\,n,\,n\!+\!1\,$ coprime, $\,8\mid n(n\!+\!1)\iff 8\mid n\,$ or $\,8\mid n\!+\!1,\,$ thus we conclude that  $\,n(n+1)/2\equiv 0\pmod{4}\iff n\equiv 0,7\pmod{8}$

Finally, beware that modular fractions are well-defined (uniquely exist) only when they are writable with denominator coprime to the modulus, when $\,a/b := ab^{-1}.\,$ For more on modular fractions see here and here.


问:
Show the sum of probabilities is less than or equal to twice the probability of the union
I can't solve this problem out of my class's textbook although it seemingly is very simple. Could someone solve this or point me in the correct direction please?

$Let(Ω,F,P)$ be a probability space and let $E_1,...,E_n ∈ F$ be mutually independent events. If $P(⋃^{n}_{i=1}E_i)≤1/2$, then

$$∑^{n}_{i=1}P(E_i)≤2P(⋃^{n}_{i=1}E_i)$$

This would be trivial if they were disjoint events, but unsure how to show this if they're only independent, although I can tell this is likely easy and I'm missing something terribly obvious.

My first thought was to use inclusion/exclusion but i cant get that to work.

答:
This is easily proved by induction on $n$. For $n=2$ we have $P(E_1)+P(E_2)\leq P(E_1 \cup E_2)+P(E_1 \cup E_2)=2P(E_1 \cup E_2)$.

Suppose that result is true for $n$ independent events and consider $n+1$ independent events: $E_1,E_2,...,E_{n+1}$. We have $$ \sum\limits_{i=1}^{n+1} P(E_i)= \sum\limits_{i=1}^{n} P(E_i)+P(E_{n+1})$$ $$\leq P(\bigcup_{i \leq n} E_i)+P(E_{n+1})\leq 2P(\bigcup_{i \leq {n}} E_i)+P( E_{n+1}) \leq 2P(\bigcup_{i \leq {n+1}} E_i)$$ The last step follows from the folowing :

If $A$ and $B$ are independent and $P(A\cup B) \leq \frac 1  2$ then  $2P(A)+P(B) \leq 2P(A\cup B)$.  [Use the fact that $P(A\cup  B)=P(A)+P(B)-P(A)P(B)$].


问:
Trace and eigenvalue decomposition, and convexity of a matrix function.
I am studying the convexity of a function with trace. It is described as follows:

$f(X)=tr(X^{-1})$ on $\boldsymbol{dom} ~f=\boldsymbol{S}_{++}^n$

Define $g(t)=f(Z+tV)$, where $Z\succ 0$ and $V \in \boldsymbol{S}^n$.

$g(t)= tr((Z+tV)^{-1}) \\
     ~~~~~~= tr(Z^{-1}(I+tZ^{-1/2}VZ^{-1/2})^{-1}) \\
     ~~~~~~= tr(Z^{-1}Q(I+t\Lambda)^{-1}Q^T) \\
     ~~~~~~= tr(Q^TZ^{-1}Q(I+t\Lambda)^{-1})) \\
     ~~~~~~= \sum_{i=1}^{n}(Q^TZ^{-1}Q)_{ii}(1+t\lambda_i)^{-1},$

where $Z^{-1/2}VZ^{-1/2}=Q\Lambda Q^T$ by eigenvalue decomposition.

In this derivation, I cannot understand why the last line $tr(Q^TZ^{-1}Q(I+t\Lambda)^{-1})) = \sum_{i=1}^{n}(Q^TZ^{-1}Q)_{ii}(1+t\lambda_i)^{-1}$ holds.
For me, it looks like the result that the trace of a matrix is a sum of its eigenvalues.

Is $Q^TZ^{-1}Q$ a diagonal matrix? If it is true how can we show that it is true. If it is not, how come the last line hold?

I am stuck to this problem for days, your help will be much appreciated.

答:
By the definition of the trace, and then using the fact that $I+t\Lambda$ is diagonal, we get
\begin{equation}\begin{aligned}
\text{tr}\left(Q^TZ^{-1}Q(I+t\Lambda)\right)&=\sum_{i=1}^n\sum_{j=1}^n\left(Q^TZ^{-1}Q\right)_{ij}\left(I+t\Lambda\right)_{ji}\\
&=\sum_{i=1}^n\sum_{j=1}^n\left(Q^TZ^{-1}Q\right)_{ij}\delta_{ji}\left(1+t\lambda_i\right)\\
&=\sum_{i=1}^n\left(Q^TZ^{-1}Q\right)_{ii}\left(1+t\lambda_i\right)
\end{aligned}\end{equation}


问:
Taylor Series Approximation: Squaring and Scaling
This was seen from a numerical analysis text/reference.

There is a vague statement here: square the result repeatedly. The scaling is rather clear: divide by 2 until $|x|<1$. How about the squaring? I tried squaring the result by the number of times I scaled, and the result was way off.

Have you seen any proof or justification to this technique?

thanks

答:
Seems to be working for me. Just make sure you follow the directions exactly. The first sixteen terms of the Taylor expansion are $e^x=1+x+\frac{x^2}{2!}+...+\frac{x^{15}}{15!}$. So write $e^{30}$ as $(e^{30/2^5})^{2^5}$, because $2^5$ is the smallest power of 2 that makes the exponent less in absolute value than 1. Do the Taylor expansion for $e^{30/2^5}$, and then raise the result to the power of $2^5$. It matches $e^{30}$. I also tried $\exp -3$ and it worked.

```<code>taylor=function(x){
  sum=1
  for (i in 1:15) {
    sum=sum+x^i/factorial(i)
  }
  return(sum)
}
</code>```

-out

```<code>> taylor(30/2^5)^(2^5)
[1] 1.068647e+13
> exp(30)
[1] 1.068647e+13
> taylor(-3/2^2)^(2^2)
[1] 0.04978707
> exp(-3)
[1] 0.04978707
</code>```

Body:
Seems to be working for me. Just make sure you follow the directions exactly. The first sixteen terms of the Taylor expansion are $e^x=1+x+\frac{x^2}{2!}+...+\frac{x^{15}}{15!}$. So write $e^{30}$ as $(e^{30/2^5})^{2^5}$, because $2^5$ is the smallest power of 2 that makes the exponent less in absolute value than 1. Do the Taylor expansion for $e^{30/2^5}$, and then raise the result to the power of $2^5$. It matches $e^{30}$. I also tried $\exp -3$ and it worked.

taylor=function(x){
  sum=1
  for (i in 1:15) {
    sum=sum+x^i/factorial(i)
  }
  return(sum)
}

-out

> taylor(30/2^5)^(2^5)
[1] 1.068647e+13
> exp(30)
[1] 1.068647e+13
> taylor(-3/2^2)^(2^2)
[1] 0.04978707
> exp(-3)
[1] 0.04978707


答:
$(2N+1)^{2} \equiv 1  $

$2N+1 \equiv 1$

This step is wrong, for example: $(2*3+1)^2 \equiv 1$.


问:
Show that ($\mathscr{P}(\Bbb{R})$)' and $\Bbb{R}^\infty$ are isomorphic.
Apologies for any bad formatting.

This is number 35 from chapter 3.F of Axler's Linear Algebra Done Right, if you have the text to view.

I've been stuck on this one for quite some time:
obviously because the vector spaces are infinite dimensional, we cannot create the isomorphism using bases of the spaces. Also, because $\Bbb{R}^\infty$ contains infinite sequences, it is not clear to me how to create an injective function...

I also thought that maybe I could develop the isomorphism using finite dimensional subspaces of the respective spaces, which is relatively easy, but being able to abstract that iso for the infinite dimensional spaces is beyond me.

Any assistance/explanation would be greatly appreciated.

答: As with all algebraic duals, the correspondence is based on looking at how each functional behaves on a basis for the original vector space (in your case, polynomials). So let $\beta = \{1, t, t^2, \ldots\}$ be the standard monomial basis for the space of real polynomials. Then, by linearity, every linear functional $L \in (P(R))'$ is uniquely defined by its values on $\beta$, namely $L(1), L(t), L(t^2), \ldots$. Perhaps, more suggestively, let $L(1) = a_0, L(t) = a_1, \ldots, L(t^k) = a_k$. Then $L$ is determined by a set $(a_0, a_1, a_2, \ldots)$. Because it sounds like you are a new student, it might be helpful that I leave some remaining details for you to work out, namely making sure that this correspondence is both injective and surjective.


问:
Which example can show that the conclusion may fail if we remove the hypothesis $f(0)=0$?
Suppose $f: [0,1] \rightarrow \mathbb{R}$ is continuous. $f(0)=0$ and $f'(x)$ exists and finite for all $x \in (0,1)$. Suppose further $f'(x)$ is incresing on $(0,1)$. Show that $f(x)/x$ is also incresing on $(0,1)$.Give an example to show that this conclusion can fail if we remove the hypothesis that $f(0)=0$.

I have done the first part but can't think of any example which will show the conclusion can fail. I want a proper example. Thank you.

答:
If $f(x)/x$ is increasing on $(0, 1)$ then
$$
 \frac{f(x)}{x} \le \frac{f(1/2)}{1/2} \implies f(x) \le 2x f(\frac 12)
$$
for $0 < x < 1/2$. It follows that $f(0) \le 0$, so that is a necessary condition: The conclusion fails for all functions with $f(0) > 0$.

A concrete example is $f(x) = 1 + x^2$. $f'(x) = 2x$ is increasing, but $f(x)/x = x + 1/x$ is decreasing on $(0, 1)$.

The conclusion is true if $f(0) \le 0$: An increasing derivative means that $f$ is convex. Writing the convexity condition for $0 < x < y$ and using $f(0) \le 0$ shows that $f(x)/x \le f(y)/y$.


答: Take $f(x) = 3/4 + (x-1/2)^2$ for a counterexample.

Title: Which example can show that the conclusion may fail if we remove the hypothesis $f(0)=0$?

Body:
Suppose $f: [0,1] \rightarrow \mathbb{R}$ is continuous. $f(0)=0$ and $f'(x)$ exists and finite for all $x \in (0,1)$. Suppose further $f'(x)$ is incresing on $(0,1)$. Show that $f(x)/x$ is also incresing on $(0,1)$.Give an example to show that this conclusion can fail if we remove the hypothesis that $f(0)=0$.

I have done the first part but can't think of any example which will show the conclusion can fail. I want a proper example. Thank you.


问:
Understanding Kolmogorov Extension Theorem
I'm following these notes where the author proves the Kolmogorov Extension Theorem. In order to do so, he uses an auxiliary lemma to construct a sequence of compact sets $K_{n} \subset B_{n}$ such that $K_{n+1}\subset K_{n}\times \mathbb{R}$ and:
\begin{eqnarray}
\mu_{n}(K_{n}) \ge \frac{\epsilon}{2} \tag{1}\label{1}
\end{eqnarray}
Then, he states:

"Since $K_{n}$ is non-empty, we pick $(x_{1}^{n},...,x_{n}^{n})\in K_{n}$ [...]"

My problem is: where (\ref{1}) being used in the proof? Does it assure that each $K_{n} \neq \emptyset$? How come?

答: Yes, note that $\mu_n(K_n)\geq \epsilon/2>0$. On the other hand $\mu_n(\emptyset)=0$. Hence, $K_n\neq \emptyset$.


问:
According to the identity laws, if A ∪ ∅ = A, then how comes A - ∅ = A as well?
This identity law is in a textbook I'm using but this question is also in the textbook and the solution is true. How can they both be true?

答:
$A-\emptyset$  is , by definiton of the " difference" operation, the set of all $x$ such that

(1) $x$ belongs to $A$

and

(2) $x$ belongs to $\bar\emptyset$ ( for anyset $S$ , $\bar S$ is the complement of S).

But $\bar\emptyset = U $ ( for any object whatsoever is a non-member of the emptyset).

In virtue of another identity law : $A \cap U = A$.

So $A-\emptyset = A \cap U = A$


问:
$\mathbb{Q}[X,Y]/(Y^2-X^3)$ is not a UFD
I'm trying to show that $R = \mathbb{Q}[X,Y]/(Y^2-X^3)$ is not a UFD, but I got stuck.

To prove this, I could try to find two "different" factorisations for one element, but I am not familiar with this, so I tried to use a lemma and one of the previous exercises.
If my syllabus is right, in every UFD counts
$$ x \ \text{is irreducible} \quad \iff \quad x \ \text{is prime}$$
My syllabus also states that the elements $\bar{X}, \bar{Y} \in R$ are irreducible.
So I tried to show that at least one of the elements $\bar{X}, \bar{Y} \in R$ does not generate a prime ideal.

This would mean that I should find two polynomials $f,g \in \mathbb{Q}[X,Y]$, such that
$$\exists p \in \mathbb{Q}[X,Y], \quad fg - pX \in (Y^2-X^3)$$
and $$\forall q \in \mathbb{Q}[X,Y], \quad f-qX \notin (Y^2-X^3) \ \wedge \ g-qX \notin(Y^2-X^3)$$ or the same thing but then for $Y$.

I hope that you can tell me if this approach is correct, and provide me a hint. I'd appreciate it if you told me the solution, but please start your answer with a hint, clearly separated from the rest.

Body:
Counterexample

$(y-x)(y+x) = y^2-x^2 = x^3-x^2 = x^2(x-1)$.


问:
Proving Tightness of Markov's Inequality for Discrete Random Variable
For any value of $t \geq 1$, construct a non-negative random variable $X$ where $Pr[X \geq tE[X]] = \frac{1}{t}$. This shows that the Markov bound is tight. If we parameterize a discrete random variable by $t$ to get two values $\{0, t\}$ where $Pr[t] = \frac{1}{t^2}$ and $Pr[0] = 1 - \frac{1}{t^2}$ then we have proven the equation above. My question is how do you prove this? Working out the algebra, the equation holds, but how would you show your work?

This post here addresses the issue, but the random variable is different and the answer is essentially "no proof" which I am finding hard to believe.

答:
It may help to think of this as a counter-example: You are providing a counter-example to the statement that the Markov Inequality is not tight, e.g. the exists a $t \geq 1$ such that $Pr[X > tE[X]] = \frac{1}{t}$ for all non-negative random variables $X$.

Counter-examples can take a lot of thinking to produce, but often lead to very short proofs, since there is no requirement to show the thought process that lead you to discover the counter-example. However, providing this thought process can often be very enlightening for the reader.

In this case, my intuition would be that inequalities often attain equality at "extremes". Think of the AM-GM inequality where equality is attained when all variables are equal, or the fact that a bounded convex function attains its maximum at a boundary point. In this case the "extreme" is that all probability mass is distributed at the end-points of an interval $[0,s]$. You can also see that due to linearity of expectation, that you can assume without loss of generality that the interval endpoints can be 0 and 1.
Now you can parameterise all distribution on two values ${0,1}$ by the probability $Pr[1]=p$ and see if a particular choice of $p$ will lead to equality in the Markov Inequality.


问:
Finding $\frac{\mathrm d}{\mathrm dx}(6)$ by first principles
I had an exercise in my text book:

Find $\frac{\mathrm d}{\mathrm dx}(6)$ by first principles.

The answer that they gave was as follows:

$$\lim_{h\to 0} \frac{6-6}{h} = 0$$

However surely that answer would be undefined if we let $h$ tend towards $0$, as for finding all other limits like that we substitute $0$ in the place of $h$?

答:
In all cases in which the $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ exists, both the numerator and the denominator approach $0$.  And even if the limit does not exist the numerator and denominator both become $0$ when $h=0$.

But $\lim\limits_{h\to0}$ of anything depends only on the behavior of that function of $h$ when $h$ is not $0$.

These are the most important things you can learn about limits before getting into these problems about derivatives.  If these things were not as they are, the reason for learning about limits before going on to derivatives just wouldn't be there at all.


问:
Perpendicular line to a plane ( cube )

The Question

I'm trying to show that $A'C$ perpendicular on plane ($C'BD$). Found that $AB = 6$ and that $A'C'BD$ and $CC'BD$ are both a tetrahedron with the same base ( triangle $C'BD$). Still I can't explain why $A'C$ is perpendicular on plane ($C'BD$).

My Understanding

$AE=3\sqrt{6}$

$A'C\perp$(triangle $C' B)$?

I found that $AB=6$

答:
HINT:
$$
A'C'=A'B=A'D,
\quad
CC'=CB=CD
\quad\text{and}\quad
C'B=BD=DC'.
$$

Title: Perpendicular line to a plane ( cube )


问:
In how many ways can they rearrange such that each person is in a new seat and no 2 persons swap?
Beforehand, each person is assigned a seat (n seats). How many ways can they rearrange such that each person is in a new seat and no 2 people swap?

I'm guessing $(n-1)!$, but I have no idea on how to get the answer.

I was trying to get the answer to this problem:

A small class of nine boys are to change their seating arrangement by drawing their new
seat numbers from a box. After the seat change, what is the probability that there is only
one pair of boys who have switched seats with each other and only three boys who have
unchanged seats?

where I got:
$(9C3 * 6C2 *4!)/9!$ but I realized that none of the (4!) people should be on the same seat. And no pair of students swapped seats

The answer to this problem is $9C3*6C2*3!/9!$. I want to find a generalization, with n seats, how many ways can you arrange n people such that no 2 people swap and no one is in the same seat.

答:
I think there is a way to write a recurrence relation by the following reasoning: we are counting the permutations of $\left\{1 , \ldots  , n\right\}$ which decomposition in cycles contains only cycles having at least $3$ elements. Let ${X}_{n}$ be that number. We can start with ${X}_{1} = {X}_{2} = 0$ and ${X}_{3} = 2$. Conventionally, we take ${X}_{0} = 1$.

Let $n  \geqslant  3$. We choose $k  \geqslant  2$ elements in $\left\{1 , \ldots  , n-1\right\}$ to form a cycle with the element $n$. There are $\binom{n-1}{k}$ ways to find these elements and there are $k !$ ways to align them in a row to build the cycle with element $n$. For the remaining $n-k-1$ elements, we choose one of the
${X}_{n-k-1}$ valid permutations. Hence we obtain the relation

\begin{equation}
{X}_{n} = \sum _{k = 2}^{n-1} \binom{n-1}{k} \  k ! \  {X}_{n-k-1}
= (n-1)!\sum _{k = 2}^{n-1} \frac{{X}_{n-k-1}}{(n-k-1)!}
\end{equation}
This easily entails the relation $X_n = (n-1)X_{n-1} + (n-1)(n-2) X_{n-3}$.

Note that this order 3 relation can be shown directly by another way of counting the permutations:

Second reasoning We can separate the valid permutations between the permutations for which the number $n$ belongs to a 3-cycle (there are $(n-1)(n-2) X_{n-3}$ of them because there are $(n-1)(n-2)/2$ ways of choosing the two other elements of the 3-cycle and each choice gives two possible cycles) and the permutations for which the number $n$ belongs to a larger cycle. Those are obtained by inserting the number $n$ in one of the cycles of a valid permutation of the $n-1$ first elements. For a given such permutation, there are exactly $n-1$ places where to insert the $n$-th element, between one of the $n-1$ first numbers and its successor in a cycle. Hence there are $(n-1)X_{n-1}$ such permutations. This proves the order 3 relation.

Clearly, the same reasoning shows that if we generalize and we count instead the number $Y_n$ of permutations of $\{1,\cdots, n\}$ where all the cycles have a length $\ge p$, then we have the relation

\begin{equation}
Y_n = (n-1) Y_{n-1} + (n-1)\cdots (n-p+1) Y_{n-p}
\end{equation}


问:
Will $P$ be necessarily bounded?

Let $\mathcal H$ be a Hilbert space. Let $P : \mathcal H \longrightarrow \mathcal H$ be a self-adjoint idempotent linear operator. Can $P$ be necessarily bounded?

I can't prove boundedness of $P$ from these assumptions. Can anybody please help me in this regard?

Thanks for your time.

答: Here is an extended hint. You're given a lot of assumptions to work with here; idempotent, meaning $P^2 = P$, and self adjoint, meaning $\langle Pa, b\rangle = \langle a, Pb\rangle$ for any vectors $a$, $b$. Imagine combining these assumptions - say, consider $\langle P^2a, b\rangle$. On the one hand, this should be equal to $\langle Pa, b\rangle$ by the idempotence. On the other, it should also be equal to $\langle PPa, b\rangle = \langle Pa, Pb \rangle$ by the self-adjointness. If you choose $a = b$, then the latter can be related to $\| Pa \|$. Finally, relate the latter to the former, and try using the cauchy schwartz inequality. Hope this helps!

Title: Will $P$ be necessarily bounded?


问:
What does this definition of diagonalizable matrices mean?
In Serge Lang’s Linear Algebra, I don’t understand what this means on page 93:

If $A$ is an $n\times n$ matrix in $K$, we say that $A$ can be diagonalized (in $K$) if the linear map on $K^n$ represented by $A$ can be diagonalized.

Okay, so is this just referring to the map $L:K^n\to K^n$ such that $L($X$) = AX$? If so, for this matrix to be diagonalizable, must $A$ be a diagonal matrix, thus it would have been a lot simpler to just say that $A$ needs to be diagonal in the first place? I just don’t get the way it’s defined at all.

答:
An example is sometimes better than a whole theory. Take
$$A =\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} $$

The map $L$ associated to $A$ is the map $L : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined for all $(x,y) \in \mathbb{R}^2$ by
$$L(x,y)=\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y  \end{pmatrix}, \quad \text{i.e.} \quad L(x,y) =(x+y,2y)$$

Of course $A$ is not a diagonal matrix. However, if you consider $e_1=(1,0)$ and $e_2=(1,1)$, then $(e_1,e_2)$ is a basis of $\mathbb{R}^2$, and you get $L(e_1)=e_1$ and $L(e_2)=2e_2$, so the matrix of $L$ is the basis $(e_1,e_2)$ is given by
$$\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} $$

This matrix is diagonal, which means that $L$ (or equivalently $A$) is diagonalizable (even if $A$ is not diagonal).


问:
Is the tensor product of the cone of a morphism $f$ with the cone of the morphism $1-f$ isomorphic to zero in any tensor-triangulated category?
I think, if my computation ain't wrong, that if I look at a morphism $f:R\to R$ as a morphism of complexes, in the triangulated (derived, homotopy) category of complexes of $R$-modules, one has that the complex $C(f)\otimes C(\mathrm{id}_R-f)$ is the complex $\cdots\to 0\to R\overset{\mathrm{id}}\to R\to 0\to \cdots$ (in degrees $-1$ and $0$), which is isomorphic (quasi-isomorphic, homotopically equivalent) to zero.

My question is: is this true in a more general context? More precisely, let $(\mathbb{K},\otimes,1)$ be a tensor-triangulated category, $f:1\to 1$ a morphism in $\mathbb{K}$, and consider the completions to distinguished triangles of the morphisms $f$ and $\mathrm{id}-f$
$$1\overset{f}\longrightarrow 1\longrightarrow C\longrightarrow T1$$
and $$1\overset{\mathrm{id}-f}\longrightarrow 1\longrightarrow C'\longrightarrow T1$$
Does it follows that $C\otimes C'\cong 0$?

答:
Yes, it's true generally.

$\require{AMScd}$

Tensoring the triangle
$$1\xrightarrow{1-f}1\rightarrow C'\rightarrow T1$$
with $C$, to get
$$C\xrightarrow{1-f}C\rightarrow C\otimes C'\rightarrow T1,$$
shows that proving that $C\otimes C'=0$ is equivalent to proving that
$C\xrightarrow{1-f}C$ is an isomorphism.

Tensoring the triangle
$$1\xrightarrow{f}1\xrightarrow{\alpha} C\xrightarrow{\beta} T1$$
with $1\xrightarrow{1-f}1$ gives a morphism of triangles
\begin{CD}
  1@>f>>1@>\alpha>>C@>\beta>>T1\\
  @VV1-fV@VV1-fV@VV1-fV@VV1-fV\\
  1@>f>>1@>\alpha>>C@>\beta>>T1.\\
\end{CD}

But also the diagram
\begin{CD}
  1@>f>>1@>\alpha>>C@>\beta>>T1\\
  @VVfV@VVfV@VV0V@VVfV\\
  1@>f>>1@>\alpha>>C@>\beta>>T1.\\
\end{CD}
is a morphism of triangles, since $f\beta$ and $\alpha f$ are compositions of consecutive arrows in triangles, and are therefore zero.

Adding the vertical arrows,
\begin{CD}
  1@>f>>1@>\alpha>>C@>\beta>>T1\\
  @VV1V@VV1V@VV1-fV@VV1V\\
  1@>f>>1@>\alpha>>C@>\beta>>T1.\\
\end{CD}
is a morphism of triangles, and since the first two vertical arrows are
isomorphisms, by the axioms of a triangulated category the third
vertical arrow $C\xrightarrow{1-f}C$ is also an isomorphism, as
required.


问:
question on a notation in Big Rudin
I am wondering why Rudin used the following notation in his "Real and Complex Analysis". It is in Definition 8.7, as following.

If $(X, \mathscr{S}, \mu)$ and $(Y, \mathscr{T}, \lambda)$ are two $\sigma$-finite measure spaces, and $Q\in \mathscr{S}\times \mathscr{T}$, then define $(\mu\times \lambda)(Q)=\int_{X}{\lambda(Q_x) d\mu(x)}=\int_{Y}{\mu(Q^y) d\lambda(y)}$.

My question is why $\mu$ and $\lambda$ depend on $x$ and $y$ respectively? Aren't they fixed once the measure spaces $(X, \mathscr{S}, \mu)$ and $(Y, \mathscr{T}, \lambda)$ are given? What is the meaning of $\mu(x)$ and $\lambda(y)$ or the author wants to emphasize here?

Thanks.

答:
There is no meaning to $\mu(x)$ (and Rudin does not use this notation). On the other hand the integral of a function $f$ with respect to a measure $\mu$ is denoted in a variety of ways, and amongst these,
$$
\int f \mathrm{d}\mu=\mu(f)=\int f(x) \mathrm{d}\mu(x)=\int f(x) \mu(\mathrm{d}x).
$$</p>

Title: question on a notation in Big Rudin


问:
solve the differential equation $y'=ye^{-y'}$
how to solve differential equation like this

$y'=ye^{-y'}$

I have no idea to solve problem like this

is there any special function or special way to solve ?

答: Hint: When $f(y,y')=0$ and you can write $y$ with the respect to $y'$, i.e.;$$y=g(y')$$ then differentiate form both sides with repect to $x$ and then out $p=y'$. You will find a separable OE: $$pdx=g'(p)dp$$ I think it is better to indicate the solution as a parametric general solution: $$x=\int \frac{g'(p)}pdp,~~~y=g(p)$$</p>


答:
The empty set doesn't have anything in it.

So if you add everything in the empty set (nothing) you'd end up with the same. If you subtract everything in the emptyset (nothing) you'd end up with the same.

Imagine you have a set: $\text{Everything you own}$ and we added to it $\text{All the leprechauns in Antarctica}$.  As there are no leprechauns in Antarctica we didn't add anything.  The result is $\text{Everything you own}$.

Now imagine wetook $\text{Everything you own}$ and subtracted $\text{all the leprechauns in Antarctica}$.  Well you don't own any leprechauns in Antarctica (no one does) so you are left with... $\text{Everything you own}$.

More formally:

$A \cup \emptyset =\{x$ so that $x \in A$ or $x \in \emptyset\}$. As there is nothing in $\emptyset$ then $x\in \emptyset$ is impossible.  So $A \cup \emptyset =\{x$ so that $x \in A$ or $x \in \emptyset\}= \{x\in A\} = A$.

And $A\setminus \emptyset = \{x\in A| x\not \in \emptyset\}$.  Now as absolutely no $x\in A$ are in $\emptyset$ then every $x \in A$ is not in $\emptyset$.  SO $A \setminus \emptyset = \{x \in A|x\not \in \emptyset\} = \{x\in A\} = A$.

Title: According to the identity laws, if A ∪ ∅ = A, then how comes A - ∅ = A as well?

Body: This identity law is in a textbook I'm using but this question is also in the textbook and the solution is true. How can they both be true?


问:
Gradient of 2-norm squared
Could someone please provide a proof for why the gradient of the squared $2$-norm of $x$ is equal to $2x$?

$$\nabla\|x\|_2^2 = 2x$$

答:
I'm not sure if this is rigorous enough to count as a proof, but an elegant way to obtain derivatives of vector expressions is to use matrix differential calculus.

Let $y = \lVert x \rVert_2^2 = x^{T} x$ with $x \in \mathbb{R}^{n}$.
Using the product rule, the differential of $y$ is
$$
dy = dx^{T} x + x^{T} dx = 2 x^{T} dx
$$

We can then set
$$
dy = \frac{dy}{dx} dx = (\nabla_{x} y)^{T} dx = 2x^{T} dx
$$
where $dy/dx \in \mathbb{R}^{1 \times n}$ is called the derivative (a linear operator) and $\nabla_{x} y \in \mathbb{R}^{n}$ is called the gradient (a vector).

Now we can see $\nabla_{x} y = 2 x$.

If $x$ is complex, the complex derivative does not exist because $z \mapsto |z|^{2}$ is not a holomorphic function.

We can, however, instead consider the real derivatives with respect to the two components of $x$.
Let $x = u + i v$.
With this definition, $y$ is a real function of $u, v \in \mathbb{R}^{n}$ defined by
$$
y = x^* x = (u + i v)^* (u + i v) = u^T u - v^T v
$$
Taking the differential
$$
dy = 2 u^T du - 2 v^T dv = \frac{\partial y}{\partial u} du + \frac{\partial y}{\partial v} dv
$$
and therefore
$$
\nabla_{u} y = 2 u \enspace , \qquad \nabla_{v} y = -2 v
$$

For an introduction to matrix differential calculus, see the lecture of Geoff Gordon on YouTube or the paper on matrix derivatives of Mike Giles.

Title: Gradient of 2-norm squared


问:
Measure Algebra Question
Background:

Let $G$ be a locally compact group, and define $M(G)$ to be the vector space of all regular complex measures defined on $G$, normed by total variation.

For $\mu,\nu\in M(G)$, define the product $\mu * \nu$ as follows:

Define $\varphi\in C_{0}(X)^{*}$ by $$\varphi(f) = \int_{G}\int_{G}f(xy)d\mu(x)d\nu(y)\text{, for all }f\in C_{0}(X)$$

By the Riesz-Representation Theorem, there is a unique element (call it $\mu*\nu$) of $M(G)$ such that $\varphi(f) = \int_{G}fd(\mu *\nu)$.

This turns $M(G)$ into an algebra.

Proposition:

I'm reading a book which says that it is an immediate consequence of Fubini's Theorem that

$$\int_{G}\int_{G}f(xy)d\mu(x)d\nu(y) = \int_{G}\int_{G}f(xy)d\nu(y)d\mu(x)$$

My Question:

In order to apply Fubini's Theorem, we need
$$\int_{G\times G}|f(xy)|d(\mu\times\nu) < \infty$$

which doesn't seem to follow from $f\in C_{0}(X)$.

Counterexample:

Take $G = \mathbb{Z}$, $f\in C_{0}(G)$ defined by $f(n) = \frac{1}{|n|}$. Then $\int_{G}f(x)d\mu(x) = \sum_{n=-\infty}^{\infty}\frac{1}{|n|} = \infty$.

From here it follows that for fixed $y\in G$, we have $\int_{G}f(xy)d\mu(x) = \infty$ so obviously the hypothesis of Fubini's Theorem is not satisfied.

But in this case, $f$ is non-negative, so the desired integral equation instead follows from Tonelli's Theorem, so all is not lost.

So does the Proposition follow from a different argument? or is an additional hypothesis on $G$ required?

答: I don't know if this is the answer, but it just occured to me that for $\mu\in M(G)$ to have finite norm, it must have finite total variation, which I think would be sufficient to guarantee the hypothesis of Fubini's Theorem.  Maybe someone with more knowledge than I can confirm this?</p>

Title: Measure Algebra Question


问:
Show that $G\left(n,p\right)e^{-\frac{1}{12np\left(1-p\right)}}<{n\choose pn} < G\left(n, p\right)$
The problem:

Using the inequalities

$\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}e^{\left(\frac{1}{12n}-\frac{1}{360n^{3}}\right)} < n! < \sqrt{2\pi n}\left(\frac{n}{e}\right)^ne^{\frac{1}{12n}}$

show that, for a constant $p$ such that $0<p<1$, where $pn$ is an integer,

$G\left(n, p\right)=\frac{2^{nH\left(p\right)}}{\sqrt{2\pi n p\left(1-p\right)}}$, and $H\left(p\right)$ is the binary entropy function, show that

$G\left(n,p\right)e^{-\frac{1}{12np\left(1-p\right)}}<{n\choose pn} < G\left(n, p\right)$

My work so far:

First I have noted that

${n\choose pn} = \frac{n!}{\left(pn\right)!\left(\left(1-p\right)n\right)!}$

$ \approx \frac{\sqrt{2\pi n} \left( \frac{n}{e} \right)^{n}}{\left(pn\right)!\left(\left(1-p\right)n\right)!}$

$\approx \frac{ \sqrt{2\pi \left(1-p\right)n}\left( \frac{\left(1-p\right)n}{e}\right)^{\left(1-p\right)n}}{2\pi \sqrt{\left(1-2p\right)}\left(\frac{pn}{e}\right)^{pn} \left(\frac{\left(1-2p\right)n}{e}\right)^{\left(1-2p\right)n}} e^\left({\frac{1}{12\left(1-p\right)n}}-\frac{1}{12pn+1}-\frac{1}{12\left(1-2p\right)n+1}\right)$

$\leq 2^{nH\left(p\right)} \frac{e^{\frac{1}{12n}}}{\sqrt{2\pi np\left(1-p\right)}}$

and from here I am not sure where to go. Any hints or help would be greatly appreciated.

答:
I put $q=1-p$. First, I shall use the inequalities
$$
\sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n  < n! < \sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n e^{\frac{1}{{12n}}}
$$
which hold for all positive integer $n$. We obtain
$$
\binom{n}{pn} > \frac{{\sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n }}{{\sqrt {2\pi pn} \left( {\frac{{pn}}{e}} \right)^{pn} e^{\frac{1}{{12pn}}} \sqrt {2\pi qn} \left( {\frac{{qn}}{e}} \right)^{qn} e^{\frac{1}{{12qn}}} }} = \frac{{2^{nH(p)} }}{{\sqrt {2\pi pqn} }}e^{ - \frac{1}{{12npq}}} .
$$
To prove the upper bound, I shall use the inequalities
$$
\sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n e^{\frac{1}{{12n+1}}} < n! < \sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n e^{\frac{1}{{12n}}}
$$
which hold for all positive integer $n$ (cf. H. Robbins, A remark on Stirling's formula, Amer. Math. Monthly 62 (1955), pp. 26–29.). We deduce
$$
\binom{n}{pn} < \frac{{\sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n e^{\frac{1}{{12n}}} }}{{\sqrt {2\pi pn} \left( {\frac{{pn}}{e}} \right)^{pn} e^{\frac{1}{{12np + 1}}}\sqrt {2\pi qn} \left( {\frac{{qn}}{e}} \right)^{qn} e^{\frac{1}{{12nq + 1}}}}} = \frac{{2^{nH(p)} }}{{\sqrt {2\pi pqn} }}e^{\frac{1}{{12n}} - \frac{1}{{12np + 1}} - \frac{1}{{12nq + 1}}} .
$$
Finally, note that
$$
\frac{1}{{12n}} - \frac{1}{{12pn + 1}} - \frac{1}{{12qn + 1}} =  - \frac{{144n^2 (pq + 1) + 12n - 1}}{{12n(12np + 1)(12nq + 1)}} < 0.
$$


问:
What would be a good error correction code for recovering from 50% data corruption?
For my use case it's quite likely that much of the data will be corrupted in such a way that I do not know which bits are corrupted and which bits aren't. As in, half or so of the data may be completely randomized. Hamming codes look good but I'm not sure how to extend it to more than 1 bit of correction. Reed-Solomon codes are also very good, but I don't know how to apply it without knowing which bits are corrupted. What would be a good coding scheme for this use case?</p>

答:
If by half data are corrupted, you mean that the binary probability of error is equal to $p = 1/2$, then you are in trouble. For a BSC (Binary Symmetric Channel), the channel capacity per symbol is equal to:

$$C_S = 1 + p \, log_2 p + (1-p)\,log_2(1-p)$$

And then the capacity is equal to $C_S = 0$ for $p = 1/2$.

This implies that no code can correct such a randomized channel.

In a comment, you detail that the data are corrupted by blocks (this should have been told in the post itself). For example, one half is highly corrupted, on half is correct. In such an extreme situation, the best if to make a block transmission, with a CRC for each block, to detect which blocks are corrupted, and which are not. Then data blocks detected as corrupted (highly incorrect) must be thrown. To get the whole information, you have to perform some kinds of data block repetition.

The important aspect is that if a given block is highly corrupted, you can use the above capacity analysis to decide that the best thing we can do with it is to throw it.

In a real transmission system, with a return channel, this is managed by retransmitting the erroneous blocks (ARQ process).


问:
Show that there is $x \in S^1$ s.t. $f(x)=f(-x)$ for $f \in C^0(S^1, \mathbb{R})$

Let $f: S^1 \subset \mathbb{R}^2 \to \mathbb{R}$ be a continuous function on the unit circle $S^1$.

Show that there is $x \in S^1$ s.t. $f(x)=f(-x)$

Intuitively this is quite clear since I can imagine if this does not hold, we should have sort of a spiral that is not continuous at one point. But I'm not sure how to show this exactly. Any ideas?

答:
Hint :

The function $g :[0,\pi] \rightarrow \mathbb{R}$ defined by
$$g(t)=f(e^{it})-f(-e^{it})$$

is continuous and satisfies $g(0)=-g(\pi)$. Then apply IVT to $g$.


问:
What is the difference between $\mathbb{R}^+$ and $\mathbb{R}^*$?
I know that both of them contain all positive numbers from $\mathbb{R}$ but one notation contains $0$ too. I don't know which one.

Thanks in advance.

答: $\Bbb R^+ = \{r\in \Bbb R\mid r>0\}$ (open half space) and $\Bbb R^* = \Bbb R\setminus\{0\}$ (carrier set of multiplicative group of field $\Bbb R$).


问:
find normal subgroup of symmetric group
Symmetric group $S_3=\{(),(1,2),(2,3),(1,3),(1,2,3),(1,3,2)\}$, I understand that $H=\{e,(1,2,3),(1,3,2)\}$ is the normal subgroup of S3 ($H\lhd S_3$) because:
$$ gH=Hg, \forall g\in S_3$$
e.g. let $g=(1,2)$ we have
$$gH=\{(1,2),(1,2)(1,2,3),(1,2)(1,3,2) \}=\{(1,2),(2,3),(1,3)\}$$
$$Hg=\{(1,2),(1,2,3)(1,2),(1,3,2)1,2) \}=\{(1,2),(1,3),(2,3)\}$$
or let $g=(1,2,3)$ we have
$$gH=\{(1,2,3),(1,3,2),()\}$$
$$Hg=\{(1,2,3),(1,3,2),()\}$$
Q1: how one can find such $H\lhd S_3$ (proper, non-trivial normal subgroup of maximum order) by a step-by-tep procedure? Finding the normal subgroup of S4 or large symmetric groups seems tedious.

Q2: Is there a way to find the order of the group without constructing it?

答:
As we know that if $H$ is a $p-$subgroup of a group $G$ then $H$ is a normal subgroup iff $H$ is unique in $G$.
Since we see that $H=\{I,(1 2 3),(1 3 2)\}$ is a sylow $3-$subgroup and is unique in $G$, then $H$ is a normal subgroup of $G$.

Title: find normal subgroup of symmetric group


问:
Matrix norm induced from vector explanation
Can someone explain as thoroughly as possible what a matrix induced norm is? A concrete example would for this norm would help a lot...

$A \in C^{n \times n}, v \in C^n$

$|| A || = \displaystyle\sup_{v \neq 0}\frac{||Av||}{||v||}$

Thank you.

答:
Hints: $Av$ is another vector when $v$ is a vector. Assuming one puts the Euclidean norm on $\mathbb{C}^{n},$ the norm of a vector $u \in \mathbb{C}^{n}$ is given by $\| u \| = \sqrt{\sum_{i=1}^{n} |u_{i}|^{2}}$, where the $u_{i}$ are the entries of $u$ (though other norms are possible). Armed with this information, to calculate the norm of a matrix is a matter of following the definitions. The sup is the least upper bound. It may help to note
that the ratio $\frac{\|Av\|}{\|v\|}$ is unchanged if we multiply $v$ by a non-zero scalar, so it suffices to consider vectors with $\| v \| = 1.$</p>


问:
Does a second-countable, locally compact, Hausdorff space admit a countable basis of pre-compact open sets?
Let $X$ be a such space. I know that for a Hausdorff space, being locally compact is equivalent to having a basis of pre-compact open sets. But how do I prove that $X$ can be covered by countably many such sets, by using second-countability of X? Thank you.

答:
If we have a countable base $\mathcal{B}$ for $X$ then any other base $\mathcal{C}$ of $X$ has a countable subfamily that is also a base. See here e.g.

We can apply this to the base of pre-compact sets that exists in a locally compact Hausdorff space.

Finally note that a base is in particular a cover of $X$ (for every $x$ there is at least one base element containing it, as $X$ itself is open and so a union of base sets).

Final remark: the closures of the cover elements of course still cover $X$ and show that $X$ is then $\sigma$-compact.


问:
Continuity of $x^2y^2/(x^2+xy+y^2)$ at $(0,0)$
I want to check whether the function $f(x,y)=\frac{x^2y^2}{x^2+xy+y^2}$ with $f(0,0)=0$ is continuous at $(0,0)$.

But due to term $xy$ in addition to squares in the denominator, I am unable to proceed.

Can one give some hint for it?

I tried to find partial derivatives.

Both partial derivatives are $0$ on $x$ as well as $y$ axis.

But at other points, we get as: $f_x=\frac{x^2y^3+2xy^4}{(x^2+xy+y^2)^2}$; I couldn't get any direction to see whether this partial derivative w.r.t. $x$ is bounded near $(0,0)$.

答:
Hint: Show that $|f(x,y)| \leq |x||y| $.

[$|x||y|-xy \leq 2|x|y| \leq x^{2}+y^{2}$ by AM-GM inequality].


问:
Proving trig derivatives without limit definition
How do I go about deriving say, $\cfrac{d}{dx}[\sin x]=\cos x$ without using the limit definition of a derivative? I believe with that approach I'd have to know how to do Taylor Series expansions, and so I haven't gone that far yet.

So far the best argument I can come up with is that the rate of change at the minimum and maximums, k, of $\sin x$ is $0$, yet it is also the case that $\cos k=0$. This isn't really a derivation, though.

Any simple ways of doing this without needing calc 2 and up?

答:
If you can buy yourself just a teeny bit of multivariate calculus, you can get this with no limits. All we need are three facts about the function $t\mapsto (\cos(t),\sin(t))$. For brevity and to remind ourselves we can't use any other special properties of the trig functions we'll just write these as $(c(t),s(t))$:

It moves around the unit circle - the Pythagorean identity $c^2+s^2=1$

It moves at unit speed - so $\dot{c}^2+\dot{s}^2=1$

It moves counterclockwise.

The tiny bit of multivariate calc is in (2). We know the length of the circular arc subtended by an angle, measured in radians, is the circle's radius times the angle; that's what defines radians in the first place. So with unit radius we just have $L(t)=t$ and $\frac{dL}{dt}=1$. And $dL$ is the hypotenuse of the right triangle with sides $dx$ and $dy$ so $1=\frac{dL}{dt}^2=\frac{dx}{dt}^2+\frac{dy}{dt}^2=\dot{c}^2+\dot{s}^2$.

By (1) and the chain rule (and dividing by two), $c\dot{c}+s\dot{s}=0$.

We solve for $\dot{c}$ and substitute into (2), yielding $(\frac{-s\dot{s}}{c})^2+\dot{s}^2=1$ and $(s^2+c^2)\dot{s}^2=c^2$. By (2) we simplify that to $\dot{s}^2=c^2$.

Clockwise or counterclockwise just tells us whether it's the negative or positive  square root. CCW means the signs of $\dot{s}$ and $c$ agree: when we're on the right-half plane we're moving up, on the left we're moving down. So we take the positive square root, $\dot{s}=c$. Using $c\dot{c}+s\dot{s}=0$ we also know $\dot{c}=-s$.

Actually, it's enough to know it's CCW at just one point. We know $\cos(0)>0$, and for small $t$, $\sin(t)\geq \sin(0)$ so $\sin'(0)\geq 0$.

In terms of a little more multivariable calc, a constant radius means our position and velocity vectors are perpendicular; their dot product is just the chain rule result $c\dot{c}+s\dot{s}=0$. Remembering how to come up with a perpendicular vector, you can skip the algebra step. I have to give credit to Michael O’Connor for the above way of putting it with less multivariate calc.


问:
Why decimal form of $\frac{A}{2^m\times5^n\times3^p\times7^q\cdots}$ has both repeating and non repeating numbers?
I read in my book that, decimal form of the fraction $\cfrac{A}{\underbrace{2^m\times5^n}_{\text{at least one}}\times\underbrace{3^p\times7^q}_{\text{at least one}}\times\cdots}\quad(A\in\mathbb{N})$, has both repeating numbers and non repeating numbers. I don't understand the reason of that. I know the decimal form of $\cfrac{A}{2^m\times5^n}$ is determinable because we can rewrite it as $A\times(0.5)^m\times(0.2)^n$. hence with other prime factors in denominator like $3^p$ and $7^q$ decimal is not determinable so it has repeating numbers. but for $\cfrac{A}{{2^m\times5^n}\times{3^p\times7^q}\times\cdots}$ which is a combination of two forms I don't know how to show it mathematically that contain both repeating and non repeating numbers in decimal form. I can rewrite it as :
$$A\times(0.5)^m\times(0.2)^n\times\frac{1}{3^p\times 7^q\cdots}$$
It is product of determinable and repeating decimals but it doesn't help me.

答:
I assume you're talking about the fact that every rational number, in decimal form, has an initial segment followed by an infinitely repeating segment. For instance, $\frac{11}{90}=0.1\bar{2}=0.1222\cdots$.

Consider a rational number $a/b$. We can factor it as $10^{-k}c/d$ where $d$ is coprime to $10$, so without loss of generality $b$ is coprime to $10$. For simplicity, I will also assume $a/b>0$ (a little bit of modification is necessary for negatives). Furthermore, we can subtract out the integer part to assume $a<b$.

The powers of ten must eventually repeat mod $b$, so there is an $n$ for which $10^n-1=b\bar{b}$. If we multiply and divide $a/b$ by $\bar{b}$ it is amenable to the geometric sum formula:

$$ \frac{a}{b}=\frac{a\bar{b}}{b\bar{b}}=\frac{a\bar{b}}{10^n-1}=\frac{1}{10^n}\frac{a\bar{b}}{1-\frac{1}{10^n}}=\frac{a\bar{b}}{10^n}+\frac{a\bar{b}}{10^{2n}}+\cdots $$

Since $a\bar{b}<10^n$, if we substitute $a\bar{b}$s decimal representation into the above we see that we get the decimal digits of $a\bar{b}$ (interpreted as having $n$ digits, with leading $0$s if necessary) repeat over and over again.

Adding back in the integer part we subtracted off can only affect the first so many digits, and similarly multiplying the whole shebang by $10^{-k}$ only shifts everything.


问:
Cloudy and Sunny Days
In Freedonia, every day is either cloudy or sunny (not both). If it's sunny on any given day, then the probability that the next day will be sunny is $\frac 34$. If it's cloudy on any given day, then the probability that the next day will be cloudy is $\frac 23$.

In the long run, what fraction of days are sunny?

How am I to solve this? My initial thought of 9/17 (obtained by $\frac{\frac{3}{4}}{\frac{3}{4}+\frac{2}{3}}$) was incorrect.

答:
Use the law of total probability:

$$P(S)=P(S_2\mid S_1)P(S)+ (1-P(C_1 \mid C_2))(1-P(S))$$
where $P(S)$ is the probability of a sunny day, $P(S_2|S_1)$ is the probability of a sunny day following a sunny day and $P(C_2|C_1)$ is the probability of a cloudy day following a cloudy day.

\begin{gather}
P(S)=3P(S)/4+ (1-2/3)(1-P(S))\\
P(S)/4=1/3 -P(S)/3\\
7P(S)/12=1/3\\
P(S)=4/7\\
\end{gather}

Title: Cloudy and Sunny Days