问:
Can $\omega_2$ be the order type of a countable union of countable sets of ordinals?
Working in ZF(wihtout choice), according to this answer:  $\omega_2$ is not the countable union of countable sets.

Can $\omega_2$ be the order type of a countable union of countable sets of ordinals?

答: The property of being a countable union of countable sets is preserved by bijections (just take the images of the sets under the bijection).  So, since $\omega_2$ is not a countable union of countable sets, neither is any set of cardinality $\aleph_2$.


问:
Convergence of fixed points
Problem: Let $f_t$ and $f_0$ be continuous functions where $f_t,f_0:[0,\infty) \to [0,\infty)$ and $t \in [0,1]$. Suppose that $f_0$ has a unique fixed point $x_0$ and $f_t$ has a unique fixed point $x_t$. If $f_t \to f_0$ uniformly on $[0,\infty)$ as $t \to 0$, is it true that $x_t \to x_0$?

My attempt: Define $f_t = tx$ then each $f_t$ has a unique fixed point at $0$. However, $f_1$ has infinitely many fixed points. This is where I'm having troubles with. Is there any hint or suggestion?

答: First you should note that for convergence you only need almost all elements of the series not all, that is your f_1 should not give you any troubles. I would suggest that you do not try to define a specific function right away, but get a good grasp of how a function must look like and draw one. Therefore note that the fixed points are precisely the intersections with the line f(x)=x and that a counterexample for f_0 must have infinite points that are almost fixed points. At the end, if you want an explicit set of functions you should be able to define them easily.</p>

ContentLicense: CC BY-SA 4.0

Title: Convergence of fixed points


问:
Proving $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

Prove the distributive property for sets:

$A \cup (B \cap C) =  (A \cup B) \cap (A \cup C)$

I'm not good with proofs but my understanding is that I have to prove 2 things:

(1)  $A \cup (B \cap C) \subset (A \cup B) \cap (A \cap C)$

(2)  $A \cap (B \cap C) \supset (A \cup B) \cap (A \cup C)$

This is what I have done so far:

Part (1)

If $x\in A$, then $x \in (A \cup B)$ and $x \in (A \cup C)$.

$\therefore x \in (A \cup B) \cap (A \cap C)$

If $x \in (B \cap C)$ then $x \in (A  \cup B)$ and $x \in (A \cup C)$ because $x \in B$ and $x \in C$.

$\therefore x \in (A \cup B) \cap (A \cap C)$

$\therefore A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)$

Part (2)

Now we have to prove the reverse inequality:  $(A \cup B) \cap (A \cap C)$.  Then $x \in A \cup B$ and $x \in (A \cup C)$

If $x \in A$, then $x \in A \cup (B \cap C)$

This is where I am up to.  I wanted to know whether my approach is correct and if I did part (1) correctly.  I'm stuck on part (2) and don't know how to proceed.  I'd appreciate any help.

Thank you!!

答:
A simple way is to do a calculational proof, starting at the most complex side, to calculate which elements are in $(A \cup B) \cap (A \cup C)$: for all $x$,
$$
\begin{align}
& x \in (A \cup B) \cap (A \cup C) \\
\equiv & \;\;\;\;\;\text{"definition of $\cap$; definition of $\cup$, twice"} \\
& (x \in A \lor x \in B) \land (x \in A \lor x \in C) \\
(*) \; \equiv & \;\;\;\;\;\text{"logic: simplify by 'factoring out' $x \in A$, using the fact that $\lor$ distributes over $\land$"} \\
& x \in A \lor (x \in B \land x \in C) \\
\equiv & \;\;\;\;\;\text{"definition of $\cap$; definition of $\cup$"} \\
& x \in A \cup (B \cap C) \\
\end{align}
$$
Now by extensionality (i.e., equal sets have the same elements) it follows that $$(0) \;\;\; A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$

Note how we didn't need to prove two separate $\subseteq$ cases.  Also note that the key step $(*)$ uses the logical law $$(1) \;\;\; P \lor (Q \land R) \;\equiv\; (P \lor Q) \land (P \lor R)$$  We see that $(0)$ and $(1)$ have the same structure.  Since $(1)$ is in the logic domain, it is more generally useful than $(0)$, which is only applicable when dealing with sets.

ContentLicense: CC BY-SA 3.0


问:
Prove that the series $\sum\limits_{n=2}^{\infty}(-1)^n\frac{\ln(n)}{n^x}$ converges to a positive real number for all $x > 0$
Prove that the series $\sum\limits_{n=2}^{\infty} (-1)^n\frac{\ln(n)}{n^x}=\frac{\ln(2)}{2^x} - \frac{\ln(3)}{3^x} + \frac{\ln(4)}{4^x} - \frac{\ln(5)}{5^x} + ...$ converges to a positive real number for all $x > 0$.

This question arose from this one:

Relatively simple proof that real-valued Dirichlet eta function is increasing, and also a query of the function at / about $x = 0$

Update:

There is a number $\alpha \in (1,2)$ such that for each $x > \alpha,$ the absolute sequence $(a_n)_{n\geq2}  = \left(\frac{\ln(n)}{n^x}\right)_{n\geq2} \ $is strictly decreasing. So for $x > \alpha$, the alternating series converges to a positive number.

$x \leq \alpha, \ $ in particular when $x << 1$ is more difficult.

I investigated the series for different value of $x$ below $0$ and it seems that the obvious approach is to show that $\exists$ an odd number $j$ such that $\frac{\ln(2)}{2^x} - \frac{\ln(3)}{3^x} + \frac{\ln(4)}{4^x} - \frac{\ln(5)}{5^x} + ... + \frac{\ln(j-1)}{(j-1)^x} - \frac{\ln(j)}{j^x} > 0,$ and that the absolute sequence $\left(\frac{\ln(n)}{n^x}\right)_{n\geq j}$ is strictly decreasing. I think that, even for the lowest value of $j$ for a given $x$, the absolute sequence will be decreasing. However, this may need to be proven as well.

$x = \frac{12}{20} \implies j = 51$

You can certainly check with a program, terms for "large enough $x$ " i.e. for $x \in (0.5, \alpha)$, but for very small positive $x$ you cannot check with a program. So we probably have to prove it using by rearranging the first $j$ terms and using inequalities to show that the sum of the first $j$ terms is $>0$. And yes, you also need to prove that the absolute sequence is strictly decreasing after the $j-th$ term also, but I don't think this will be as difficult as proving the sum of the first $j$ terms is $>0$.

I have also done a table showing x and values of $j$ where the sum is first positive.

\begin{array}{|c|c|c|c|}
\hline
x& j\\ \hline
 1&5 \\ \hline
 \frac{19}{20}&7 \\ \hline
 \frac{18}{20}&7 \\ \hline
 \frac{17}{20}&9 \\ \hline
 \frac{16}{20}&11 \\ \hline
 \frac{15}{20}&15 \\ \hline
 \frac{14}{20}&21 \\ \hline
 \frac{13}{20}&31 \\ \hline
 \frac{12}{20}&51 \\ \hline
 \frac{11}{20}&89 \\ \hline
 \frac{10}{20}&181 \\ \hline
 \frac{9}{20}&441 \\ \hline
 \frac{8}{20}&1395 \\ \hline
 \frac{7}{20}&6483 \\ \hline
 \frac{6}{20}&54805 \\ \hline
\end{array}

I made this table by manually inputting values into WA, so it may in fact be incorrect. A Python / C++ program or something may ensure more accurate results?

The bottom line so far: we need to prove $ \exists j \in \mathbb{N}$ for every $x$ with $0< x <\alpha$.

答:
I doubt there is any formula telling immediately that it is non-negative. What I mean is that sometimes you need numerical checks to know the sign of a given series.

For $s > 0$

$$\sum_{n=1}^\infty (-1)^{n+1} n^{-s}=
\sum_{n=1}^\infty (\sum_{m=1}^n (-1)^{m+1} ) (n^{-s}-(n+1)^{-s})=\sum_{n=1}^\infty \frac{1+(-1)^{n+1}}{2} (n^{-s}-(n+1)^{-s})$$
$$=\frac12+\frac{s}2 \sum_{n=1}^\infty (-1)^{n+1} \int_n^{n+1} t^{-s-1}dt$$
The latter converges for $s >-1$ and defines the analytic continuation as well as of the derivative.

The uniform convergence for $s\ge 0$ means that you can easily prove, with numerical checks, that $$\eta'(s)=\frac{1}2 \sum_{n=1}^\infty (-1)^{n+1} \int_n^{n+1} t^{-s-1}dt-\frac{s}2 \sum_{n=1}^\infty (-1)^{n+1} \int_n^{n+1}\log(t) t^{-s-1}dt$$ is non-negative on small enough intervals. Doing so with enough intervals you will cover $[0,T]$. Next for $T$ large enough it is non-negative on $[T,\infty)$  because $\eta'(s)=\log(2)2^{-s}+O(3^{-s})$.


问:
Find the polar decomposition
Find the left and right polar decomposition of the matrix:
\begin{align*}
A = \left[\begin{matrix} 1 & 0 \\ 1 & 1\end{matrix}\right]
\end{align*}

So a polar decomposition would be of the form $A = U\sqrt{A^\dagger A} = \sqrt{AA^\dagger}\,U$ for some unitary $U$.

I'm trying to take the square root by finding the spectral decomposition $\lbrace|\lambda\rangle\rbrace$ of $A^\dagger A$ and using:
\begin{align*}
\sqrt{A^\dagger A} = \sum_{\lambda}\sqrt{\lambda}|\lambda \rangle \langle \lambda |
\end{align*}
but since we have repeated eigenvalues I'm not sure how to proceed

答:
A brute force (numerical) method for obtaining the right polar decomposition has been presented in
this answer at MSE.
When applied to the OP's problem (Test3) we get an outcome like this:
$$
\begin{bmatrix}
    1.000000 &  0.000000 \\
    1.000000 &  1.000000
\end{bmatrix} = \\
\begin{bmatrix}
    0.894427 & -0.447214 \\
    0.447214 &  0.894427
\end{bmatrix}
\begin{bmatrix}
    1.341641 &  0.447214 \\
    0.447214 &  0.894427
\end{bmatrix}
$$
However, when searching for the number $0.447214$ on the internet, we find that it's approximately equal to
What is 1 over square root of 5?.
So let's do an educated guess:
$$
\begin{bmatrix}
 1 & 0 \\
 1 & 1
\end{bmatrix} =
\begin{bmatrix}
 2/\sqrt{5} & -1/\sqrt{5} \\
 1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
\begin{bmatrix}
 3/\sqrt{5} & 1/\sqrt{5} \\
 1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
$$
Finding the left polar decomposition is left as an exercise to the reader.

EDIT.

However, the square root of a real-valued general symmetric positive definite $2 \times 2$ matrix can be calculated exactly:
$$
\begin{bmatrix} a & b \\ b & c \end{bmatrix}^2 = \begin{bmatrix} A & B \\ B & C \end{bmatrix} \quad \Longrightarrow \\
\begin{bmatrix} a & b \\ b & c \end{bmatrix}\begin{bmatrix} a & b \\ b & c \end{bmatrix} =
\begin{bmatrix} a^2+b^2 & b(a+c) \\ b(a+c) & b^2+c^2 \end{bmatrix} = \begin{bmatrix} A & B \\ B & C \end{bmatrix}
$$
It follows that:
$$
\begin{cases} a^2+b^2 = A \\ b(a+c) = B \\ b^2+c^2 = C \end{cases} \quad \Longrightarrow \quad
\begin{cases} a^2 = A - b^2 \\ b^2(a^2+2ac+c^2) = B^2 \\ c^2 = C - b^2 \end{cases}
$$
Using the determinants gives an additional equation:
$$
(ac-b^2)^2 = AC-B^2 \quad \Longrightarrow \quad ac = \sqrt{AC-B^2}+b^2
$$
Making smart combinations already gives the end-result:
$$
b^2(a^2+2ac+c^2) = B^2 \quad \Longrightarrow \\ b^2([A - b^2]+2[\sqrt{AC-B^2}+b^2]+[C - b^2]) = B^2
\quad \Longrightarrow \\ b^2(A+2\sqrt{AC-B^2}+C) = B^2 \quad \Longrightarrow \\ b = \frac{B}{\sqrt{A+2\sqrt{AC-B^2}+C}}
$$
Together with
$$
a = \sqrt{A-b^2} \quad ; \quad c = \sqrt{C-b^2}
$$
So here comes a much simpler solution. First form the transpose times the original matrix:
$$
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^T\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} =
\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} =
\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} A & B \\ B & C \end{bmatrix} =
\begin{bmatrix} a & b \\ b & c \end{bmatrix}^2
$$
With the above formulas:
$$
b = \frac{B}{\sqrt{A+2\sqrt{AC-B^2}+C}} = \frac{1}{\sqrt{2 + 2\sqrt{1} + 1}} = 1/\sqrt{5} \\
a = \sqrt{A-b^2} = \sqrt{2 - 1/5} = 3/\sqrt{5} \\
c = \sqrt{C-b^2} = \sqrt{1 - 1/5} = 2/\sqrt{5}
$$
So we have, indeed:
$$
\sqrt{\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^T\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}} =
\begin{bmatrix} 3/\sqrt{5} & 1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}
$$
At last, the orthogonal matrix is found with:
$$
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}
\begin{bmatrix} 3/\sqrt{5} & 1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}^{-1} =
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}
\begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ -1/\sqrt{5} & 3/\sqrt{5} \end{bmatrix} / 1 =
\begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}
$$
Which leads to the same conclusion:
$$
\begin{bmatrix}
 1 & 0 \\
 1 & 1
\end{bmatrix} =
\begin{bmatrix}
 2/\sqrt{5} & -1/\sqrt{5} \\
 1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
\begin{bmatrix}
 3/\sqrt{5} & 1/\sqrt{5} \\
 1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
$$

Title: Find the polar decomposition


问:
What is entire function and how can I show that it is not continuous?
Given the function
$$
f(z)=\frac{e^{z}}{z^{2}-z-2}
$$
Is this function entire? If not, for what value(s) of $z$, will this function cease to be analytic?

I am struggling to understand this question. How can I convert that equation to $A+iB$ form? How can I show that it is discontinuous?

答: Hint: Check when $z^2 - z - 2=0$ for $z\in \mathbb{C}$


问:
Solving for (or approximating) $x$ in $\frac{x-\sin x}{x^{0.4}} = c$
I'm an engineer and I'm deducting a formula, and I found a BIG problem: I can't isolate $x$.

The formula is:
$$\frac{x-\sin x}{x^{0.4}} = c$$
where $c$ is a constant.

How to find the value for $x$, or at least a good approximation?

答:
The leading term of $x-\sin x$ is $\frac{x^3}6$. This would result in a first approximation of $x=(6c)^{1/2.6}$ for smallish values of $c$.

For larger values of $c$ one would use that $|\sin x|\le  1$ to get approximately $x=c^{1/0.6}$.


问:
Self dual posets product

Let $P,Q$ be partially ordered sets such that $P\times Q$ is self dual.
Self dual means $(P\times Q)^*=P\times Q$ where $P^*$ is the dual of $P$.
Does that mean $P,Q$ are self dual?

My professor hinted that the answer is negative, that in particular there exist non self dual $P,Q: P\times Q$ is self dual. Why is that true? Is there a counterexample of two such sets?

答: Consider the partial orders $\langle\Bbb N,\le\rangle$ and $\langle\Bbb N,\ge\rangle$. Neither is self-dual — each is the dual of the other, in fact — but their product is self-dual.

Title: Self dual posets product


答:
I would use Newton-Raphson, with $c^{\frac{5}{3}}$ as initial guess. If $c$ is close to $0$ (e.g. in $(-1, 1)$), you could use $\sqrt[2.6]{6c}$ instead.

The function oscillates around $x^{\frac{3}{5}}$ and is nice and smooth. This should work fine.

I tested it for $c= 10$ and within 3 iterations the error was tiny. Depending on the required accuracy, you can easily add more iterations.

Title: Solving for (or approximating) $x$ in $\frac{x-\sin x}{x^{0.4}} = c$

Body:
I'm an engineer and I'm deducting a formula, and I found a BIG problem: I can't isolate $x$.

The formula is:
$$\frac{x-\sin x}{x^{0.4}} = c$$
where $c$ is a constant.

How to find the value for $x$, or at least a good approximation?


答:
I wonder in what context you came across this problem.

Entire functions are differentiable everywhere in the complex plane and therefore are continuous everywhere in the plane.

Functions are not considered holomorphic (i.e. not differentiable in some open neighborhood) at points where their value is $\infty.$

The function $z\mapsto e^z$ is nowhere $0.$ But the denominator $z^2-z-2$ is $0$ at two points in the plane (and both of those are real numbers). At a point where the numerator is a holomorphic function and is not $0$ and the denominator is a holomorphic function and is $0,$ the value of the function is $\infty.$ If such points exist, the function is not entire.

There is seldom any reason to look at anything in $\text{“}A+iB\text{ form”}$ in this subject.

Title: What is entire function and how can I show that it is not continuous?

Body:
Given the function
$$
f(z)=\frac{e^{z}}{z^{2}-z-2}
$$
Is this function entire? If not, for what value(s) of $z$, will this function cease to be analytic?

I am struggling to understand this question. How can I convert that equation to $A+iB$ form? How can I show that it is discontinuous?


问:
Using various inequalities to prove this martingale inequality?
I’ve got another martingale inequality that I would be grateful for a kickstart on.

Suppose $X_t$ is a local martingale such that $|X_t|$ and $\langle X_t\rangle\leq c$ $\forall$ $t\geq 0$ and for some constant
$c\in\mathbb R$. I need to show that $$\mathbb{E}\left(\sup_{t\geq
 0}X_t\right)^4\leq361\mathbb{E}\langle X\rangle_\infty^2.$$

How should I approach this? I asked a question of similar nature previously, so I suspect I will need Cauchy-Schwarz's, Doob's, and Grönwall's inequalities here. However, in the previous question, $X_t$ was given in SDE form, so applying the inequalities made sense — here, all I know is that it is a martingale, so I'm not too sure where to start. Please guide me, thank you!

答:
I am assuming that $X$ is a continuous local martingale such that $X_0=0$ and that $\langle X \rangle$ is the quadratic variation process of $X.$

Applying Ito's formula with $f(x)=x^4$ we have
$$|X_t|^4 = 4 \int_0^t X_t^3 dX_t + 6 \int_0^t X_t^2 d\langle X \rangle_t. \tag{1}$$

Since $X$ is bounded, $E \int_0^t|X_s|^6ds < \infty,$ so the process $\{ \int_0^t X^3_s dX_s, t \geq 0 \}$ is a martingale and therefore, $E \int_0^t X^3_s dX_s = 0.$

Thus, taking expectations in (1)
$$E|X_t|^4 \leq 6 E[\sup_{0 \leq s \leq t} |X_s|^2 \langle X \rangle_t]$$
and by the Holder inequality
$$E|X_t|^4 \leq 6 \left(E\sup_{0 \leq s \leq t} |X_s|^4\right)^{\frac{1}{2}} \left(E\langle X \rangle^2_t \right)^{\frac{1}{2}}. \tag{2}$$
Now (apart from the above) by Doob's martingale inequality
$$E \sup_{0 \leq s \leq t}|X_s|^4 \leq \left(\frac{4}{3}\right)^4 E|X_t|^4. \tag{3}$$
Substituting (2) in (3)
$$E \sup_{0 \leq s \leq t}|X_s|^4 \leq \left(\frac{4}{3}\right)^4 6 \left(E\sup_{0 \leq s \leq t} |X_s|^4\right)^{\frac{1}{2}} \left(E\langle X \rangle^2_t \right)^{\frac{1}{2}}.$$
Thus
$$E \left( \sup_{0 \leq s \leq t}X_s \right)^4 \leq E \sup_{0 \leq s \leq t}|X_s|^4 \leq \left(\frac{4}{3}\right)^8 6^2  E\langle X \rangle^2_t.$$


答:
There is no analytical function that will give you x = f(c). I can recommend Wolfram Alpha to get a quick insight into the problem: (https://www.wolframalpha.com/input/?i=%28x+-+sinx%29+%2Fx%5E0.4+%3D+10+solve+for+x).

You get immediately a plot of your function. Hope that helps a bit


问:
Measure theoretic reasoning that $\mathbb E(Y\mid Z)=\mathbb E(Y)$ where $Y$ and $Z$ are independent

Show that $\mathbb E(Y\mid Z)=\mathbb E(Y)$ where $Y$ and $Z$ are
independent.

I know how to show this without measure theory:

$$\mathbb E(Y\mid Z)=\int_{\mathcal Y} y f_{Y\mid Z}(y\mid z)\overset{\text{ind}}{=}\int_{\mathcal Y} y f_{Y}(y)=\mathbb E(Y)$$

where $\mathcal Y$ is the support of $Y$. However, I'm trying to give a measure theoretic justification of this. I'm new to conditional expectation as it relates to measure theory so please bear with me. I'm given the hint to integrate over the set $\Lambda\in\mathcal G$. Here I suppose that the hint is referring to $\Lambda$ as being an atom of a $\sigma$-algebra $\mathcal G$?

My initial thoughts were to consider the probability triple $(\Omega,\mathcal F, P)$. We have that $\mathbb E(Y\mid Z)=\mathbb E(Y\mid\sigma(Z))$. From here should I consider $\sigma(Z)$ to be $\mathcal G$? I'm just not sure how $\Lambda$ and $\mathcal G$ come into play here. I assume at some point I need to use $\mathbb P(Y\mid Z)=\mathbb P(Y)$ but I'm not sure how to use it.

I assume the proof is a one-liner but I would appreciate someone explaining it to me like I'm $5$.

答:
We can come back to the definition of a conditional expectation.

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and $\mathcal{G} \subset \mathcal{F}$. We define a r.v. $X$  such that $\mathbb{E}[|X|] < \infty$.

The conditional expectation of $X$ knowing $\mathcal{G}$ is defined as any random variable $Y$ verifying the following two conditions:

$Y$ is $\mathcal{G}$-measurable

For all $A \in \mathcal{G}$, $\int_A Xd\mathbb{P} =\int_A Yd\mathbb{P} $ (i.e. $\mathbb{E}[X\mathbf{1}_A] = \mathbb{E}[Y\mathbf{1}_A]$  a.s.)

In fact, we define $Y = \mathbb{E}[X|\mathcal{G}]$ a.s.

Now using this definition, if $X$ is independent of $\mathcal{G}$ (i.e knowing $\mathcal{G}$ does not give any information on $X$) then we can show that $\mathbb{E}[X|\mathcal{G}] = \mathbb{E}[X]$. Let's verify the two points:

$\mathbb{E}[X]$ is a degenerated r.v. integrable and $\mathcal{G}$-mesurable

Let $A \in \mathcal{G}$, we have $\mathbb{E}[Y\mathbf{1}_A] = \mathbb{E}[X\mathbf{1}_A] = \mathbb{E}[\mathbf{1}_A]\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X]\mathbf{1}_A]$

The second equality comes from the independence.


问:
Tangent line to a surface
I have to find the tangent line to the surface
$$z=\frac{x^{2}}{4}+\frac{y^{2}}{9}$$
at
$$P(2,-3,2)$$
such that this line pass through the $x$-axis.

I evaluated the directional derivative of $z$ in the direction of $\vec{u}=(2,-3)$ that is equal to
$$\frac{4}{\sqrt{13}}$$
and so, the line equation is
$$y+3=\frac{4}{\sqrt{13}}(x-2)$$
But I don't think this is the right way.

答: Hint. Find the plane tangent to the surface at the given point $P$. Such plane intersect the $x$-axis at $Q$. The line through $P$ and $Q$ is what you are looking for.

Title: Tangent line to a surface


问:
For any $11$-vertex graph $G$, show that $G$ and $\overline{G}$ cannot both be planar
Let $G$ be a graph with 11 vertices. Prove that $G$ or $\overline{G}$ must be nonplanar.

This question was given as extra study material but a little stuck. Any intuitive explanation would be great!

答: It seems the following. Euler formula implies that $E\le 3V-6$ for each planar graph. If both $G$ and $\bar G$ are planar, then $55=|E(K_{11})|\le 6|V(K_{11})|-12=54$, a contradiction.</p>


答:
Hint:
Suppose $f_0$ has a graph that looks like this (with $y=x$ in blue):

Let $f_t$ be obtained by moving this graph slightly up on the left and slightly down on the right.

Title: Convergence of fixed points

Body:
Problem: Let $f_t$ and $f_0$ be continuous functions where $f_t,f_0:[0,\infty) \to [0,\infty)$ and $t \in [0,1]$. Suppose that $f_0$ has a unique fixed point $x_0$ and $f_t$ has a unique fixed point $x_t$. If $f_t \to f_0$ uniformly on $[0,\infty)$ as $t \to 0$, is it true that $x_t \to x_0$?

My attempt: Define $f_t = tx$ then each $f_t$ has a unique fixed point at $0$. However, $f_1$ has infinitely many fixed points. This is where I'm having troubles with. Is there any hint or suggestion?


问:
Limit and existence of this function
I have the following function f(x) = $\lim_{m \to \infty} \lim_{n \to \infty} (\cos(\pi m!x))^{2n}$

I have to prove that f(x) exists  and calculate its value. I was wondering if using $e^{i\pi x}$ to direct calculate it was enough or I need the definition with $\epsilon$.

After that, I need, as well, to  find {x$\in\mathbb{R}$| f is continuous at x}.

I decided to consider the série $e^u = \sum_{k=0}^{n}\frac{u^k}{k!}$. Here, u = $i\pi x m!$
I choose r > 0 and x$\in$[-r,r], so we can $i\pi x m! \leq i \pi r m!$ and using the Abel's theorem (don't know how it's called in english) then the serie is normally convergent.

Thus, f is continuous when x $\in \mathbb{R}$

答:
Define $f_m(x) =  \lim_{n\to \infty} (\cos(\pi m! x))^{2n}$.

If $x$ is rational, then $x=p/q$ for some integers $p,q$. If $2q$ divides $m!$, then $(\cos(\pi m! x))^2 = 1$, which will be the case for $m$ sufficiently large. Hence $f_m(x)=1$ eventually for all $m$ after some point.

If $x$ is irrational, then $0<(\cos(\pi m! x))^2<1$, so $f_m(x)=0$ for all $m$.

Therefore $f(x) = 0$ for irrational $x$ and $f(x)=1$ for rational $x$. This function is not continuous anywhere.

Title: Limit and existence of this function


问:
I'm looking for references for generalized confluent hypergeometric differential equation
According to wolfram, A generalization of the confluent hypergeometric differential equation is given by;

$$y''+\left(\frac{2R}{x}+2F'+p\frac{H'}{H}-H'-\frac{H''}{H'}\right) y'+\left[\left(p\frac{H'}{H}-H'-\frac{H''}{H'}\right)\left(\frac{R}{x}+F'\right)+\frac{R(R-1)}{x^2}+\frac{2R}{x}F'+F''+(F')^2-\frac{q}{H}(H')^2 \right]y=0$$

Which has the solutions $y_1=x^{-R} e^{-F} M(q,p,H)$ and $y_2=x^{-R} e^{-F} O(q,p,H)$, where $M(q,p,H)$  is the confluent hypergeometric function of the first kind and $O(q,p,H)$ is the confluent hypergeometric function of the second kind. Meanwhile, $R,F$ and $H$ are fucntions of $x$.

I tried to look on google for more details about this equation but i didn't find anything, can anyone here please give me more references about this particular equation? Like how it was deriven, the relation between the parameters $p$ and $q$..etc.

答:
Theory of hypergeometric functions:
The book presents a geometric theory of complex analytic integrals representing hypergeometric functions of several variables. Starting from an integrand which is a product of powers of polynomials, integrals are explained, in an open affine space, as a pair of twisted de Rham cohomology and its dual over the coefficients of local system. It is shown that hypergeometric integrals generally satisfy a holonomic system of linear differential equations with respect to the coefficients of polynomials and also satisfy a holonomic system of linear difference equations with respect to the exponents. These are deduced from Grothendieck-Deligne’s rational de Rham cohomology on the one hand, and by multidimensional extension of Birkhoff’s classical theory on analytic difference equations on the other.

Hypergeometric summation. An algorithmic approach to summation and special function identities:
Modern algorithmic techniques for summation most of which have been introduced within the last decade are developed and carefully implemented via computer algebra system software (which can be downloaded from the Web). The algorithms of Gosper, Zeilberger, and Petkovsek on hypergeometric summation and recurrence equations and their $q$-analogues are covered, and similar algorithms on differential equations are considered. An equivalent theory of hyperexponential integration due to Almkvist and Zeilberger completes the volume. The combination of all results considered gives work with orthogonal polynomials and (hypergeometric type) special functions a solid algorithmic foundation. Hence, many examples from this very active field are given. The book is designed for use as framework for a seminar on the topic, but is also suitable for use in an advanced lecture course.


问:
area of square between tangent(externally) circles
Two externally tangent circles,have a square between them,standing on the same base as the two circles.The circles have a radius of 1 unit each.The top two vertices of the square are touching one circle each.How do I find the area of the square?</p>

答:
A picture would help. I hope it will be clear how to draw it.

Let the two circles have equations $(x+1)^2+(y-1)^2=1$ and $(x-1)^2+(y-1)^2=1$.
So they are tangent to the $x$-axis, and symmetrical about the $y$-axis.  The centre of the bottom side of the square is then at $(0,0)$.

Let the square have side $2t$. Then the point $(t,2t)$ is on the right-hand circle.

It follows that $(t-1)^2+(2t-1)^2=1$. This simplifies to $5t^2-6t+1=0$. The relevant root is $t=\frac{1}{5}$. So the square has area $\frac{4}{25}$.

Remark: We could strip away the coordinatization, and write the solution using only the Pythagorean Theorem. But coordinatization is a (provably) powerful tool in geometry, so one might as well use it.

If we really want to use Pythagorean Theorem only, let $O$ be the centre of the right-hand circle, and let $P$ be the point where the top right corner of the square meets the circle. Drop a perpendicular from $O$, towards the $X$-axis, and let $Q$ be the point where this perpendicular meets the top edge of the square, extended.

Then $PQ=1-t$ and $OQ=1-2t$. By the Pythagorean Theorem, $(PQ)^2+(OQ)^2=(OP)^2$, and therefore $(1-t)^2+(2-t)^2=1^2$. Expand, and continue as in the main solution.


问:
Codomain as part of the definition of a function
For most practical purposes, the codomain is part of the definition of a function. The codomain is specified at the moment we write $f:A \to B$, and is exactly what makes the word "surjective" meaningful.

EDIT: by "part of the definition", I mean "as an item of the ordered pair below".

However, in set theory, the definition of function as a binary relation (i.e. a subset $R$ of $A \times B$) does not include its codomain - if we define the function as a set $R$ satisfying certain properties, then we cannot really decide what the codomain is from $R$. (The domain and image can, however,  be extracted from the set $R$ easily.)

Of course, to define the relation $R$ using a formula $\phi$, we also need to specify a set of objects from which we are choosing to form $R$. See Axiom schema of specification in ZFC. For example, we may define $R=\{(x,y)\in \mathbb Z^2, x=y\}$, but the set $\mathbb Z^2$ (the "domain" and "codomain") is not part of the structure of $R$.

And in fact, sometimes, we need to change the codomain of a function, just like we often need to restrict or extend the domain. For example, when we use the word "embedding", we usually mean "isomorphism onto the image", so we are implicitly and temporarily "seeing" the image as the "codomain".

So, when we write done a function $f$, do we actually mean the pair $(R, B)$, the graph of $f$ and the codomain? Does this depend on the writing style and the convenience of notation for the problem we are solving?

答:
TL;DR I don't think you need one to describe a function, but I think you need a codomain to define a function.

(The original iteration)

However, in set theory, the definition of function as a binary relation (i.e. a subset  of $×$) does not include its codomain

Sure it does, you just did it: it's $B$.  You can't specify ordered pairs without mentioning $B$. What does $(a,b)$ even mean if you don't know what set $b$ is from? (Edit: Of course you can build an ordered pair with two arbitrary sets. What I wanted to express, but failed to make clear, is that to use comprehension to select ordered pairs it would be necessary to mention $B$. The discussion is in the commentary to Arthur's comment below.)

It's not entirely unnatural to discern based on codomains

Also, $(,)$ does mean something if, for example, we change $$ to a larger set, and this does not affect $$.

This is part of the "is the codomain a part of the function?" question.  I would say it is not so clear that it "does not affect $R$."

In Mac Lane's formulation of category theory, for example, each arrow has two associated objects of categories: the domain, and the codomain. This definition allows two technically different arrows which otherwise might be "the same function" to have different codomains, one properly containing the other.

Now look what happens when we define extensions of functions: say that we have a function $R\subseteq A\times B$ and a function $S\subseteq C\times D$ where $A\subseteq C$ and $B\subseteq D$.  We say that $S$ extends $R$ if $R(a)=S(a)$ for every $a\in A$.

Of course, if $A=C$ the "bigger codomain" is just a special case of extension.  From this viewpoint one can avoid confusing $R$ with its extension $S$, although for all intents and purposes we usually think of them as being the same thing "with different codomain."

While Mac Lane's formulation of category theory is largely independent of set theory, I think the mindset above is a thing in set theory too.

Regarding Arthur's comment:

A set $$ of ordered pairs where each element of  appears exactly once as the first element of a pair" avoids the mention of anything like a codomain.

I will concede that this is a description of a function, but I would argue it does not allow one to define a function.  (I'm operating under ZF only.)

Think about what it would take to try to define a function using this description. You have in hand, we suppose, the sets that will be elements of the domain and the sets that will be their images. For each assignment $a\mapsto b$, you can form $\{\{a\},\{a,b\}\}$ using the axiom of pairing, no problem.

But now you need something to pull all of these into a single set that is your set of ordered pairs. But how does one do that? I do not believe the axiom of comprehension can handle this$^\ast$.  As the wiki notes, "the axiom schema of specification can only construct subsets".  I think attempting to pull all the pairs into a single set would amount to using unrestricted comprehension, which is of course not safe.

A thought experiment

What I'm trying to get at is that to realize functions as sets, picking a codomain seems as necessary as specifying a domain.

How about we double-down on the logic that the codomain can be done away with completely with this thought experiment: if one believes that a function doesn't need a codomain, should we also believe that a partial function don't need a domain specified either? It's just "a set of ordered pairs, period." Is the domain not meaningful because we can extend it at will?

$^\ast$ (With a finite domain, you might be able to combine pairing, powerset and union to assemble them into a set, but of course I want to discuss infinite domains too.)


问:
Show that if $X$ is uncountable and $(X,F)$ is the co-countable topology on $X$, is not compact.

Show that if $X$ is uncountable, and $F$ is the collection of sets $E$ such that $X\setminus E$ is empty or at most countable. Then $(X,F)$ the co-countable topology on $X$, is not compact.

I know I've got to make a collection of sets in $F$ which don't have a finite sub cover. I want to say I can get a collection $\bigcup_{n=1}^{\infty}V_n=X$ such that each $V_{1}^c\supset V_{2}^c\supset ...$ are countable. Since if I then take only a finite collection of these, I'd have $(\bigcup_{i=1}^n V_i)^c$ isn't empty, as one of them has the smallest complement, which is non empty. So they don't cover $X$. But I'm not sure how to make such a set.

Maybe I can just take any countable subset of $X$, call it $U$ and define $V_1$ as $X\setminus U$, then define $V_2$ as $X\setminus (U\setminus \{u\})$ for some $u\in U$, and continue this by saying $V_i=X\setminus(V_{i-1}\setminus \{u\})$ for some $u\in U$. Which I believe will cover $X$ eventually because any $x\in X$ is either in $V_1$ at the start or is in $V_i=X\setminus (V_{i-1}\setminus \{x\})$ but I can't cover $X$ with only finitely many since I'll always miss some subset of $U$.

I'm not sure if I can actually construct such a set though.

答:
Your idea will work if you execute it carefully. You need to start with a countably infinite set $U=\{u_n:n\in\Bbb Z^+\}$. (Countable normally means finite or countably infinite, and you need to exclude the finite case.) Now define your sets $V_n$ recursively: $V_1=X\setminus U$, and $V_{n+1}=V_n\cup\{u_n\}$ for each $n\in\Bbb Z^+$.

Note that $V_{n+1}$ isn’t $X\setminus(V_n\setminus\{u_n\})$, as you tried to define it: we need to remover $u_n$ from the complement of $V_n$, not from the open set $V_n$ itself. If we set $U_n=X\setminus V_n$ for each $n\in\Bbb Z^+$, $U_1=U$, and $U_{n+1}=U_n\setminus\{u_n\}$ for each $n\in\Bbb Z^+$. We’re removing one point at a time from the countable complements and thereby adding it to the open sets themselves.

Now let $x\in X$. If $x\notin U$, then $x\in V_1$. If $x\in U$, then $x=u_n$ for some $n\in\Bbb Z^+$, so $x\in V_{n+1}$. Thus, $\mathscr{V}=\{V_n:n\in\Bbb Z^+\}$ is an open cover of $X$. If $\mathscr{V}_0$ is any finite subset of $\mathscr{V}$, there is a largest $n$ such that $V_n\in\mathscr{V}_0$, and it’s easy to see that $\bigcup\mathscr{V}_0=V_n$. The construction ensures that $u_n\notin V_n$, so $\mathscr{V}_0$ does not cover $X$, and $\mathscr{V}$ is an open cover of $X$ with no finite subcover.


问:
Show that $(a − b)(b − c)(a − c)$ is always an even integer if $a, b,$ and $c$ are integers.
Show that $(a − b)(a − c)(b − c)$ is always even if $a, b, c$ are integers. It should be solved with the pigeonhole principle but I could not find a solution.

答:
If a, b and c are odd or even it is obvius that difference between each pair is even number.

If there are two odd numbers and an even number, the difference between the odd numbers is even.

If there are two even numbers and an odd number, the difference between the even numbers is even.


答:
Hint: Remember that when adding/multiplying integers,

$\text{even} \pm \text{even} = \text{even}$

$\text{odd} \pm \text{odd} = \text{even}$

$\text{even} \pm \text{odd} = \text{odd}$

$\text{even} \times \text{even} = \text{even}$

$\text{even} \times \text{odd} = \text{even}$

$\text{odd} \times \text{odd} = \text{odd}$

Title: Show that $(a − b)(b − c)(a − c)$ is always an even integer if $a, b,$ and $c$ are integers.

Body: Show that $(a − b)(a − c)(b − c)$ is always even if $a, b, c$ are integers. It should be solved with the pigeonhole principle but I could not find a solution.


问:
Find supremum and infimum of a set given by: for $n \in N , k \in N,$. Got supremum, need infimum.
As in the title, I have to find supremum and infimum of the set $$S=\left\\mid n \in \mathbb N,\, k \in \mathbb N, 2|k\right\}$$

I know that $k^6$ should "grow faster" than $n^2$ if both are natural numbers. Then I can transform that expression into: $$$$ For $n \implies \infty$ I get: $$\frac{n^2}{8n^2}=\frac{1}{8}$$ That would be my supremum. BUT there is a question of infimum and the information that $2|k$. I don't know how to use that information in the process of finding the infimum. Any help would be much appreciated.

Important note: in that case, we take $\mathbb N as \{1, 2, 3, 4, 5, 6, 7...\}$ (without $0$).

答:
Denote for $n,k \in \mathbb N$

$$s(n,k)= \frac{n^2k^6}{n^2+8n^2k^6+9}.$$

All the elements of $S$ are less than $1/8$. Now taking $n=1$ and $k=2l$ we have

$$\lim\limits_{l \to \infty} s(1,2l) \to 1/8$$

and $s(1,2l) \in S$ for all $l \in \mathbb N$, proving that $\sup S=1/8$.

Regarding the $\inf$, consider the map

$f(x) = \frac{n^2 x}{n^2 + 8n^2x + 9}$ defined for $x \in [2^6, \infty)$. You have
$$f^\prime(x) = \frac{n^2(n^2 + 8n^2x + 9) - 8n^4x}{(n^2 + 8n^2x + 9)^2}= n^2\frac{n^2+9}{(n^2 + 8n^2x + 9)^2}$$ which is positive, proving that f achieve its minimum when $x=2^6$, i.e. $k=2$. We get

$$s(n, k) \ge \frac{64n^2}{513n^2+9}$$ for $n,k \in \mathbb N$ when $k$ is even. Analyzing the map $g(x) = \frac{64x}{513x+9}$ for $x \in [1,\infty)$, you'll find that it is increasing on that interval. Therefore its minimim is achieved for $x=1$ and the minimum of $S$ is $s(1,2)=\frac{32}{261}$.


问:
How to show that $A(A+B)^{-1}B = B(A+B)^{-1}A$

I am trying this problem. Could not proceed . Please give a hint.

The exercise, if the above image is not showing clearly, is the following: Let $A$ and $B$ be two square $m \times m$ matrices for some integer $m \ge 2$ such that $A+B$ is invertible. Then show that
the equation $A(A+B)^{-1}B = B(A+B)^{-1}A$. [Note that either $A$ or $B$ or each of both may not be invertible, and that $A$ and $B$ may not commute.]

答:
Hint: Write
$$
A(A + B)^{-1}B =
[(A + B) - B](A + B)^{-1}B
$$
and distribute.

Alternatively, we can reduce our consideration to the case that $A$ is invertible and then deduce the general result using the continuity of matrix operations. In particular, if $A$ is invertible we can write
$$
A(A + B)^{-1}B = A([I + BA^{-1}]A)^{-1}B \\
= A(A^{-1} [I + (BA^{-1})]^{-1})B \\
= [I + (BA^{-1})]B\\
= [I + (BA^{-1})](BA^{-1})A\\
= (BA^{-1})[I + (BA^{-1})]A.
$$


问:
Does $\forall x (\phi(x) \vee \neg \phi(x))$ imply (2) $\forall x \phi (x) \vee \neg \forall x \phi(x)$ in intuitionistic logic?
Does (1) $\forall x (\phi(x) \vee \neg \phi(x))$ imply (2) $\forall x \phi (x) \vee \neg \forall x \phi(x)$ in intuitionistic logic?

It seems to me that it does not, and my heuristic is this: (1) "says" that we have an algorithm that shows, for every object, that it is $\phi$ or not $\phi$. (2) "says" that either we have an algorithm that shows that every object is $\phi$, or we don't have it. (1) doesn't give us such an algorithm, nor does it show that there is no such thing, so it does not entail (2).

But, unless am I simply mistaken, the question is: how to show this more formally?

答:
Usually, the way to show that an implication does not hold is to give a countermodel.

This requires some notion of semantics. There are a few ways to give semantics to intuitionistic first-order logic, but the simplest is probably Kripke semantics.

It will suffice to find a countermodel to $\forall x\,(P(x)\lor \lnot P(x))\models \forall x\,P(x)\lor \lnot \forall x\, P(x)$, where $P$ is a unary relation symbol. That is, we want to find a Kripke model $M$ and a world $w$ in $M$ such that $w\models \forall x\,(P(x)\lor \lnot P(x))$ but $w\not\models \forall x\,P(x)\lor \lnot \forall x\, P(x)$.

The meaning of the first sentence is: for all worlds $w'\geq w$ and elements $x\in D(w')$, we have that $P$ holds of $x$, or for all worlds $w''\geq w'$, $P$ does not hold of $x$. Part of the definition of Kripke model is that if $P$ holds of $x$ in $w'$, then for all worlds $w''\geq w'$, $P$ holds of $x$. So $P$ cannot switch from "on" to "off", and this sentence says additionally that $P$ cannot switch from "off" to "on".  So intuitively, this says that the truth value of $P$ never switches.

The meaning of the second sentence is: Either it's true that for all $w'\geq w$ and  elements $x\in D(w')$, $P$ holds of $x$, or it's true that for all $w'\geq w$, there exists $w''\geq w'$ and an element $x\in D(w'')$ such that $P$ does not hold of $x$. Intuitively, either $P$ holds of everything in sight, or counterexamples to $P$ are "dense".

To get the second sentence to fail, we need:
(1) There exists $w'\geq w$ and $x\in D(w')$ such that $P$ does not hold of $x$, and  (2) there exists $w'\geq w$ such that for all $w''\geq w$ and all $x\in D(w'')$, $P$ holds of $x$.

Ok, now we can construct our countermodel. Let $W = \{w,w_1,w_2\}$ with $w\leq w_1$, $w\leq w_2$, and $w_1,w_2$ incomparable. Let $D(w) = D(w_2) = \{a\}$ and $D(w_1) = \{a,b\}$, with $P = \{a\}$ in all three worlds.

The truth value of $P$ never switches on an element, so $w\models \forall x\,(P(x)\lor \lnot P(x))$. The world $w_1\geq w$ witnesses condition (1), since $P$ does not hold of $b\in D(w_1)$, and the world $w_2\geq w$ witnesses condition (2), since $P$ holds of all elements of all worlds $\geq w_2$. So $w\not\models \forall x\,P(x)\lor \lnot \forall x\, P(x)$.


问:
How can I prove $\sum_{i=0}^{n-2} B_iB_{n-i-2} = \sum_{i=1}^{n-1} M_iM_{n-i}$?
I have a recursion $B$ below:

$$\begin{align}
B_{n-1} &= \sum_{i=0}^{n-2} B_iB_{n-i-2} \\
B_1 &= 1 \\
B_0 &= 1
\end{align}$$

And another $M$ below:

$$\begin{align}
M_n &= \sum_{i=1}^{n-1} M_iM_{n-i} \\
M_2 &= 1 \\
M_1 &= 1
\end{align}$$

How can I show that these two recursions are equal??

答:
It’s just an induction on $n$. Suppose that $M_k=B_{k-1}$ for $k=1,\ldots,n-1$. Then

$$\begin{align*}
M_n&=\sum_{k=1}^{n-1}M_kM_{n-k}\\
&=\sum_{k=1}^{n-1}B_{k-1}B_{n-1-k}\\
&=\;...\;.
\end{align*}$$

Now just shift the indices. I’ve finished it off in the spoiler-protected block below.

 $\sum_\limits{k=0}^{n-2}B_kB_{n-2-k}=B_{n-1}$


问:
A smooth (n-1)-surface $S \subset \mathbb {R}^n$ is orientable $ \implies S $ has a continuous field of normal vectors to it.
I would like to be able to show that: if a (n-1)-surface $S \subset \mathbb {R}^n$ of class $ C ^ 1 $ is orientable $ \implies S $ has a continuous field of normal vectors to it.
Below I show where I have arrived and where I stop.

$ S $ is a surface of class $ C ^ 1 $, so for each of its points $ x \in S $ there exists the tangent space to $S$ at that point, $ TS_x $. A basis for each of these spaces can be found by taking the columns of the Jacobian (evaluated in $ x $) of any chart that contains $ x $ in its domain of action. Given the basis, therefore, we can find the two normals to $ S $ in $ x $, that is the only two vectors with unit norm that are orthogonal to all vectors of the first basis found. Between these two normals, one can be chosen using the following criterion: the basis of $ \mathbb {R} ^ n $ formed by the (n-1) vectors of the basis of $ TS_x $ previously found plus the normal, must be oriented as the canonical basis of $ \mathbb {R} ^ n $. At this point it remains only to show that the normal thus selected, varies continuously on $S$. I have to show that:

$ \lim_ {S \ni x \to x_0 \in S} \mathbf {n} (x) = \mathbf {n} (x_0) $

Thanks to the fact that we are interested in the limit value, we can forget that $ S $ is described by several charts and consider any chart $ \phi \in \mathcal {A} $, fixed, which contains $ x_0 $ in its domain of action.
I feel that somehow I have to tie in the fact that $ J _ {\phi} (x) $ and $ J _ {\phi} (x_0) $ are $ \delta $-close (in any matrix norm) in order to conclude that the normal $ \mathbf {n} (x) $ and $ \mathbf {n} (x_0) $ are $\epsilon $-close, but I can't formalize this argument.

Can anyone help me, please?

EDIT: Proof attempt

$S$ is a surface of class $C^1$, therefore for each of its points $x\in S$ there exists the space tangent to it in that point, $TS_x$, of dimension $n-1$. Since the orthogonal complement of $TS_x$ in $\mathbb{R}^n$ has dimension 1, we can find the two normals to $TS_x$, ie the only two vectors with unit norm that are orthogonal to all vectors of a basis of $TS_x$. Between these two normals one can choose one using the following criterion: the basis of $\mathbb{R}^ n$ formed by the (n-1) vectors of the basis of $TS_x$ previously found and by the normal, must be oriented as the canonical basis of $\mathbb{R}^ n$. Such normal can be found practically as:

$$\mathbf{n}(x)=\frac{\det\begin{pmatrix}
\mathbf{e}_1 & |  & | & | \\
... & \mathbf{v}_1 & ... & \mathbf{v}_{n-1}  \\
\mathbf{e}_n & | & | & |
\end{pmatrix}}{\left | \det\begin{pmatrix}
\mathbf{e}_1 & |  & | & | \\
... & \mathbf{v}_1 & ... & \mathbf{v}_{n-1}  \\
\mathbf{e}_n & | & | & |
\end{pmatrix} \right |_{\mathbb{R}^n}}$$

where $\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x)\}$ are the columns of the Jacobian in $x$ of any chart that contains $x$ in its domain of action (they are a basis of $TS_x$).
We must therefore show that $\mathbf{n}(x)$ thus defined is such that:

$\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x),\mathbf{n}(x)\}$ is oriented like $\{\mathbf{e}_1 (x),…, \mathbf{e}_{n} (x)\}$;

The above formula for $\mathbf{n}(x)$ always gives the same vector if you use another map of the orienting atlas of $S$ (covering the point $x$), i.e. a different set $\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x)\}$;

$\mathbf{n}(x)$ is continuous $\forall x\in S$.

As for the first point, it follows from:

$$\det\begin{pmatrix}
|  & | & | & |\\
 \mathbf{v}_1 & ... & \mathbf{v}_{n-1} &\mathbf{n}  \\
 | & | & |&|
\end{pmatrix} = C_1^2+...+C_n^2>0$$

where the determinant has been developed along the last column (the normal) and the various $C_i$ are the cofactors of the Laplace development (which are also the components of $\mathbf{n}$, as the normal was first defined).
Since the possible normals for each $TS_x$ are only two, even changing the map used to define $\mathbf{n} (x)$, by point 1 one would always obtain the one capable of orienting $\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x),\mathbf{n}(x)\}$ as $\{\mathbf{e}_1 (x),…, \mathbf{e}_{n} (x)\}$, so point 2 is proved.
Finally, the continuity $\forall x \in S$ can be inferred by taking any card covering x (thanks to point 2 and thanks to the fact that continuity is a local property) and noting that the single components of the various $\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x)$  are continuous, being the chart of class $C^1$, and that $\det$ and $|\cdot |_(\mathbb{R}^n)$ are also continuous functions.

However, it seems that this 'proof' still does not go well, because if I am not mistaken it can still be applied to each step to the mobius strip. Still, what @neal told me to do I think I managed to prove it in point 2 above.

答:
Your method of selecting a normal vector field is correct. Continuity will follow from continuity of the determinant and, as you intuit, can be handled in each individual coordinate chart, as it is a local argument.

What's missing is actually the crux of the argument, which is that the normal field thus selected is consistent between coordinate charts. You need to start with the fact that $S$ has an orienting atlas, and then you can use its properties to show the normal field is invariant under coordinate change.


问:
Showing $f_{n}$ is a normal family, where $f_{n}$ is the $n^{th}$ iterate of $f$
Assume that $f$ is an entire function with $f(0)=0$. Consider the family of functions {$f_{n}$} where $f_{n}$ is the $n^{th}$ iterate of $f$, i.e. $f_{n}=f \circ f \circ \ldots \circ f$ ($n$ times).

I'm trying to show that if $|f^{'}(0)|<1$, then there is an open set $U$ containing zero so that the family {$f_{n}$} is normal on $U$. I have been able to show $|f_{n}^{'}(0)|<1$ and not much more.

Lastly if we take $|f^{'}(0)|>1$ then there does not exist an open set $U$ containing zero such that {$f_{n}$} is normal, but I'm not sure how to show this either.
I'm pretty sure we can do this problem without relying on Fundamental Normality Test. Thank you for the help and guidance.

答:
If $|f'(0)|<1$, then there is a constant $m$, $0<m<1$, and a disk $D$ centered at $0$ such that $|f'(z)|\leq m$ for all $z\in D$.  Then for all $z\in D$, $|f(z)|=|\int_{[0,z]}f'(w)dw|\leq m|z|$.  Inductively you get that $f_n(D)\subseteq m^n D$, so $(f_n)_n$ converges uniformly to $0$ on $D$.

If $|f'(0)|>1$, then $f_n'(0)\to\infty$ by the chain rule, while $f_n(0)=0$ for all $n$, which implies that local uniform convergence of a subsequence is impossible (even if you allow $\infty$).  If the sequence were normal on a neighborhood of $0$, then some subsequence $(f_{n_k})_k$ would converge uniformly on a neighborhood of $0$ to a holomorphic function $g$ (with $g(0)=0$).  But this would imply that $f_{n_k}'(0)\to g'(0)\neq\infty$ as a consequence of Cauchy's integral formula.


问:
Find intersection of all plan $n^{2}x+(2n-1)y+nz=3$
Problem :
Let $(P_{n})_{n}$ the families of plan equation :

$$P_{n}~~:~~n^{2}x+(2n-1)y+nz=3$$
And note that $E$ are the intersection of all this plan e.o :

$$E=\{~M(x,y,z)~/~\forall n\in\mathbb{N}~,M\in P_{n} \}$$

Choose the right option :

$E=?$

$•~\varnothing$

$•~E~\text{is the plan}~ x+y+z=3$

$•~E~\text{is the droit}~\begin{cases}x+y+z=3,\\y=-3\end{cases}$

$•~E~\text{is the point} (0,-3,6)$

I don't know how but for $n=1$ and $n=0$ the intersection is the suggestion 3
But for all I don't know
Can you assist!

Thanks!

答:
If you solve the system you get for $n\in\{1,2,3\}$:
\begin{eqnarray}x+y+z&=&3\\
4x+3y+2z&=&3\\
9x+5y+3z&=&3\\
\end{eqnarray}
which is a single point $(0,-3,6)$. Now check if it lies in every plane:
$$ n^2\cdot 0+(2n-1)\cdot (-3) +n\cdot 6 = 3$$ which is true for all $n$,
so you chose last answer.


Title: How to show that $A(A+B)^{-1}B = B(A+B)^{-1}A$

Body:
I am trying this problem. Could not proceed . Please give a hint.

The exercise, if the above image is not showing clearly, is the following: Let $A$ and $B$ be two square $m \times m$ matrices for some integer $m \ge 2$ such that $A+B$ is invertible. Then show that
the equation $A(A+B)^{-1}B = B(A+B)^{-1}A$. [Note that either $A$ or $B$ or each of both may not be invertible, and that $A$ and $B$ may not commute.]

Body:
You can try prove:

$$
A(A+B)^{-1} B * (B(A+B)^{-1}A)^{-1}=I
$$


问:
Traveling distance for a flea flying from one dog's nose to another repeatedly until collision
The question is:

Two dogs, each traveling 10 ft/sec, run toward each other from 500 feet apart. As they run, a flea flies from the nose of one dog to the nose of the other at 25 ft/sec. The flea flies between the dogs in this manner until it is crushed when the dogs collide. How far did the flea fly?

I know the dogs will collide after 25 seconds I used some cumbersome process to find the flea travels 610 ft. I've seen another solution indicate 25*25 = 625 ft (The rate of fly * the seconds before collison) which doesn't feel right to me.

What's the answer?

答:
This isn't linear algebra....

Regardless, this is a variation off the fameous "problem that foold Von Neuman."

The obvious solution is to treat this a geometric series and sum the series.

The "clever solution" is to say, the dogs will meet in 25 seconds.  The flea travels at 25 ft/sec for 25 sec = 625 ft.

When this question of was of Von Neuman, he answered almost immediately.  His interlocutor said, "you must have spotted the trick, most people try to sum the series."

Von Neuman responded that he did sum the series.

Summing the series is not that hard.

$(500)(\frac {25}{25+10})\sum_\limits{n=0}^{\infty} (1-\frac {4}{7})^n\\
(500)(\frac {5}{7})(\frac {7}{4}) = 625$


问:
If $T$ is $\kappa$-stable, then $|T|\leqslant\kappa$
I have heard that if $T$ is a $\kappa$-stable theory (say over a language $\mathcal{L}$), then there is an $\mathcal{L}$-theory $T'$ such that $|T'|\leqslant\kappa$ and $T'\models T$. I think I have a proof of this as follows:

Denote the constant symbols of $\mathcal{L}$ by $C$. First note that we can assume distinct symbols of $C$ are distinct modulo $T$. (Indeed, given any $c,d\in C$ distinct constant symbols, if $T\models c=d$ then we can replace every instance of $d$ in a sentence of $T$ by $c$ and so have that $T\upharpoonright_{\mathcal{L}\setminus\{d\}}\models T$.)

Now note that we may suppose $|C|\leqslant\kappa$. Indeed, for any $c\neq d\in C$, by the above we have that $\text{tp}(c)\neq\text{tp}(d)\in \text{S}_1(T)$, and so $|C|>\kappa$ would contradict $\kappa$-stability of $T$.

Indeed, because $T$ is $\kappa$-stable, we have that $|\text{S}_1(T)|\leqslant\kappa$. For any $p\neq q\in\text{S}_1(T)$, choose $\varphi_{p,q}(v)$ an $\mathcal{L}$-formula such that $\varphi_{p,q}\in p$ and $\neg\varphi_{p,q}\in q$. We claim that any $\mathcal{L}$-formula $\psi(v)$ consistent with $T$ is equivalent to a boolean combination of finitely many of the $\varphi_{p,q}(v)$.

Indeed, for any $r(v)\in\text{S}_1(T)$ such that $\psi\in r$, let $\Sigma_r(v)=\{\varphi_{r,q}\}_{q\neq r\in\text{S}_1(T)}$. We claim $\Sigma_r(v)\models\psi(v)$. If not, then $\Sigma_r(v)\cup\{\neg\psi(v)\}$ is consistent and so is contained in some complete type $s\in\text{S}_1(T)$. Note that $r\neq s$ as the former contains $\psi$ and the latter contains $\neg\psi$, and so we have $\varphi_{r,s}\in\Sigma_r(v)$. However, by construction $\Sigma_r(v)\subseteq s$, so this would mean $\varphi_{r,s}\in s$, a contradiction. Thus indeed $\Sigma_r(v)\models\psi(v)$ and so by compactness there is a finite subset $\Delta_r(v)\subseteq\Sigma_r(v)$ such that $\Delta_r(v)\models\psi(v)$. For each $r\ni\psi$, let $\theta_r(v)=\bigwedge\Delta_r$, so that $\theta_r\in r$, $\theta_r\models\psi$, and $\theta_r$ is a conjunction of (finitely many of) the $\varphi_{r,q}$.

Now we claim that $\{\neg\theta_r\}_{r\ni\psi}\models\neg\psi$. Indeed, if not, then $\{\neg\theta_r\}_{r\ni\psi}\cup\{\psi\}$ would be consistent and hence contained in a complete type $t\in\text{S}_1(T)$. But then $t\ni\psi$, so we would have $\theta_t\in t$ and $\neg\theta_t\in t$, a contradiction. Thus indeed $\{\neg\theta_r\}_{r\ni\psi}\models\neg\psi$, and so again by compactness there are $r_1,\dots,r_k\in \text{S}_1(T)$ such that $\{\neg\theta_{r_i}\}_{i=1}^k\models\neg\psi$. Let $\theta=\bigvee_{i=1}^k\theta_{r_i}$. Then $\neg\theta\models\neg\psi$, so by contraposition $\psi\models\theta$, and also each $\theta_{r_i}\models\psi$, so $\theta\models\psi$, and so $\psi$ is logically equivalent to $\theta$. In addition, each $\theta_{r_i}$ is a conjunction of finitely many $\varphi_{r,q}$, and so $\theta$ is a boolean combination of (finitely many of) the $\varphi_{p,q}$, as desired.

Now, $T\subset\{\psi(c):c\in C,\psi\text{ an }\mathcal{L}\text{-formula}\}$. Hence, letting $T'=\{\varphi_{p,q}(c):c\in C,p\neq q\in\text{S}_1(T), T\models\varphi_{p,q}(c)\}$, the above argument shows that $T'\models T$. Also, $|C|\leqslant\kappa$, and – since $|\text{S}_1(T)|\leqslant\kappa$ – also $|\{\varphi_{p,q}\}_{p\neq q\in\text{S}_1(T)}|\leqslant\kappa$, so $|T'|\leqslant\kappa$ as desired.

So, I have two questions – first, is this proof correct? Second, is there a quicker way of seeing this result? The condition that $|\text{S}_1(T)|\leqslant\kappa$ seems to me to be much weaker that $\kappa$-stability, so I feel like there should be a more immediate way of seeing this result that doesn't involve fiddling around with constant symbols, etc. Can we use the strength of full $\kappa$-stability somehow?

Edit: I think I have found a hole in my argument; in paragraph 5, I think the fact that $\Sigma_r(v)\cup\{\neg\psi(v)\}$ is consistent does not necessarily mean it is contained in an element of $\text{S}_1(T)$, for which we need also the condition that $\Sigma(v)\cup\{\neg\psi(v)\}$ is consistent with $T$. (For instance, if $\psi(v)$ is actually a sentence (so that the variable $v$ does not appear (free) in it) then $\neg\psi$ will never be consistent with $T$ by the assumption that $T$ is complete and that $\psi$ is consistent with $T$.) Is there a way to fix this hole? Note that the cases where $\psi(v)$ is a sentence are actually very important, as they are the only way to account for those sentences of $T$ that contain no constant symbols.

答:
Unfortunately, the thing you're trying to prove is not true (althought the slogan "If $T$ is $\kappa$-stable, then $|T|\leq \kappa$" is essentially correct - see below the horizontal line for a way to make this precise).

For a counterexample, consider a language with $\aleph_1$-many constant symbols $\{c_\alpha\mid \alpha<\aleph_1\}$ and the theory which says there are at least two elements in the domain and all the constants are equal: $T = \{\exists x\exists y\, x\neq y\}\cup \{c_\alpha = c_{\beta}\mid \alpha,\beta<\aleph_1\}$. Then $T$ is $\aleph_0$-stable, but $T$ is not axiomatizable by a countable set of sentences. If $|T'| = \aleph_0$, then $T'$ only mentions countably many constant symbols. Picking two constant symbols $c_\alpha$ and $c_\beta$ which are not mentioned in $T'$, we can find a model of $T'$ in which $c_\alpha\neq c_\beta$ by taking an arbitrary model of $T'$ and changing it to interpret $c_\alpha$ and $c_\beta$ to be two distinct elements of the domain. So  $T'\not\models T$.

This issue crops up in the first paragraph of your proposed proof: You say that if $c$ and $d$ are constants such that $T\models c = d$, then letting $T'$ be the theory obtained by replacing $d$ by $c$ everywhere, we have $T'\restriction_{\mathcal{L}\setminus \{d\}}\models T$. But that's not true, because in a model of $T'\restriction_{\mathcal{L}\setminus \{d\}}$, we can interpret the constant $d$ as any element, not necessarily $c$, so  $T'\restriction_{\mathcal{L}\setminus \{c\}}\not\models c = d$.

Also, as you note in the edit, what you've really proved in the fourth and fifth paragraphs of the proof is that there is a finite subset $\Delta(v)\subseteq  \Sigma(v)$ such that  $T\cup \Delta(v)\models \psi(v)$. What happens in the type space $S_1(T)$ is all happening modulo $T$, so the idea of focusing on the type space in one variable and then thinking of sentences in $T$ as formulas in one free variable isn't going to get you anywhere - any sentence in $T$ will automatically be in every type in $S_1(T)$, otherwise the type wouldn't be consistent with $T$! But actually, your idea of separating types and seeing that a small set of formulas suffice to define arbitrary formulas is on the right track... once you clarify what it is you should be trying to prove.

The correct statement is this: If an $L$-theory $T$ is $\kappa$-stable (in fact, $|S_n(T)|\leq \kappa$ for all $n$ is enough), then $T$ is a definitional expansion of an $L'$-theory $T'$ with $|L'|\leq \kappa$ (and this implies $|T'|\leq \kappa$). We can't hope to have $T'\models T$ in general, but still $T'$ is essentially the same as $T$.

Lemma: Suppose $\Delta$ is a set of $L$-formulas in $n$ free variables such that for any types $p\neq q$ in $S_n(T)$, there exists $\varphi\in \Delta$ such that $\varphi\in p$ and $\lnot \varphi\in q$. Then every $L$-formula in $n$ free variables is equivalent modulo $T$ to a finite Boolean combination of formulas in $\Delta$.

You can prove the lemma with two applications of the compactness theorem, along the lines of your attempted proof. But, for fun, here's a fancier topological proof. The formulas in $\Delta$ generate a Hausdorff topology on $S_n(T)$ which is coarser than the usual topology. If a Hausdorff topology on a set $X$ is coarser than a compact topology on $X$, then these topologies are equal. So $\Delta$ in fact generates the standard topology on $S_n(T)$. But in a Stone space (compact Hausdorff and totally disconnected) with a fixed clopen basis, every clopen set is a finite Boolean combination of the elements of the basis.

Ok, now let's fix $n$. For each pair $p\neq q$ in $S_n(T)$, let $\varphi_{p,q}$ be a formula distinguishing these types, and let $\Delta_n = \{\varphi_{p,q}\mid p\neq q\}$. By the Lemma, every formula in $n$ free variables is equivalent modulo $T$ to a finite Boolean combination of the formulas in $\Delta_n$.

Let $\Delta = \bigcup_{n\in \omega} \Delta_n$, and let $L'$ consist of all symbols in $L$ which appear in some formula in $\Delta$. Since $|S_n(T)| \leq \kappa$,  $|\Delta_n| \leq \kappa$, for all $n$, so $|\Delta|\leq \kappa$, and hence $|L'|\leq \kappa$.

For each $n$-ary relation symbol $R$ in $L\setminus L'$,  the formula $R(x_1,\dots,x_n)$ is equivalent modulo $T$ to a finite Boolean combination of formulas in $\Delta_n$, so it is definable by an $L'$-formula $\psi_R(x_1,\dots,x_n)$. Similarly, for every $n$-ary function symbol $f$, $f(x_1,\dots,x_n) = x_{n+1}$ is definable by an $L'$-formula $\psi_f(x_1,\dots,x_{n+1})$, and for every constant symbol $c$, $c = x_1$ is definable by an $L'$-formula $\psi_c(x_1)$.

Let $T'$ by the $L'$-formula obtained by replacing all symbols in $L\setminus L'$ by their definitions in the sentences of $T$. Then $T$ is a definitional expansion of $T'$, i.e. if we extend $T'$ by the definitions of all the symbols in $L\setminus L'$, the resulting theory is equivalent to $T$.


问:
How to formally show that $f(z)$ is analytic at $z=0$?
Let $z$ be a complex number. Let $$f(z)=\dfrac{1}{\frac{1}{z}+\ln(\frac{1}{z})}.$$ How to formally show that $f(z)$ is analytic at $z=0$?
I know that for small $z$ we have $$\left|\tfrac{1}{z}\right|>\left|\ln(\tfrac{1}{z})\right|$$ and that implies $|f(0)|=0.$
Are there multiple ways to handle this ?</p>

答:
$$f(z)=\frac1{\frac1z+\text{Log}\frac1z}=\frac z{1+z\,\text{Log}\frac1z}$$

Now,

$$\text{Log}\frac1z:=\log\frac1{|z|}+i\arg\frac1z\implies z\,\text{Log}\frac1z=z\log\frac1{|z|}+iz\arg\frac1z$$

If you now choose a branch cut for the complex logarithm (and you better do if you have any hope to make any sense in this particular stuff), say the usual half negative real axis and zero ($\;\Bbb R_-\;$), you get a nice analytic function in the rest (and we don't care what happens with $\,z\to\infty\,$ since we're far away from there) , so

$$z\,\text{Log}\frac1z\xrightarrow[z\to 0\,,\,z\notin\Bbb R_-]{}0$$ and you can talk of a removable singularity of your function at $\,z=0\,$...although this would be a little absurd since you already removed this point by choosing the above branch cut for the logarithm, and this will be the situation with any branch cut you choose as any of them must contain the zero point...

In short, definitely not analytic there, though...


问:
Numerical method to find critical points (Roots of Jacobian)
Let $f : D\subseteq\mathbb{R}^2 \longrightarrow \mathbb{R}^2$ be differentiable. Denote by
$$\mathcal{C}=\{x\in D \,|\, \det \mathfrak{J}_f(x)=0 \}$$ the set of all critical points, where the Jacobian matrix of $f$ is singular.

Assume that $f$ is known only numerically. Is there a nice numerical method to get those points where $\det \mathfrak{J}_f \approx 0$, in particular, when the calculation of $\mathfrak{J}_f$ is elaborating?

Thanks in advance for some tips, literature, etc.!

Idea 1: Let $p \in D$. From $$f(x) -  f(p) = \mathfrak{J}_f(p)(x- p)^t + o(\|x -p\|) \quad (\text{as } x \to p)$$ one can derive for $\epsilon >0$ that
$$\mathfrak{J}_f(p)_{i,j} \approx\frac{f_i(p+\epsilon e_j)-f_i(p)}{\epsilon}, \ i,j\in \{1,2\}.$$
This allows to get $\det \mathfrak{J}_f$ approximately. The 'Marching squares' algorithm can then be used to find the contour where it vanishes. Unfortunately, this idea is brute-force and might require a fairly dense grid.

Idea 2: Sample points $x_1, \dots, x_n$ uniformly at random in $D$. The regions where the pattern $$\{ r(x_i) \,|\, i=1, \dots,n \}$$ is dense are regions where $|\det \mathfrak{J}_f(x_i)|$ is low due to 'Change of Variables Theorem'. Unfortunately, this might be costly as well.

答:
Consider a point $x\in D$ and an infinitesimal rectangular triangle $S \subset D$ such that $x\in S$. The triangle is transformed by $f$ to some curved region $f(S)$. In contrast, the best linear approximation $\mathfrak{J}_f(x)$ of $f$ around $x$ maps $S$ again to a triangle $P$ with $$\text{signarea}(P) = \det \mathfrak{J}_f(x) \cdot \text{signarea}(S),$$
where $\text{signarea}$ denotes the signed area. Instead of using $P$ one might use the triangle $Q$ given by the $f$-transformed vertices of $S$. It is $$\det \mathfrak{J}_f(x) = \frac{\text{signarea}(P)}{\text{signarea}(S)}\approx \frac{\text{signarea}(Q)}{\text{signarea}(S)}.$$
If $S=\Delta(x_1,x_2,x_3)$ the signed area can be calculated using the 'Shoelace formula' by
$$\text{signarea}(Q)=\frac{1}{2}\left(f_1(x_1) f_2(x_2) + f_1(x_2) f_2(x_3) + f_1(x_3) f_2(x_1) - f_1(x_2) f_2(x_1) - f_1(x_3) f_2(x_2) - f_1(x_1) f_2(x_3) \right).$$
For instance using $x_1=x+\epsilon e_1, x_2=x+\epsilon e_2, x_3=x$ leads to
$$\det \mathfrak{J}_f(x) \approx F_\epsilon(x)=\frac{1}{\epsilon^2}\left(f_1(x+\epsilon e_1) f_2(x+\epsilon e_2) + f_1(x+\epsilon e_2) f_2(x) + f_1(x) f_2(x+\epsilon e_1) - f_1(x+\epsilon e_2) f_2(x+\epsilon e_1) - f_1(x) f_2(x+\epsilon e_2) - f_1(x+\epsilon e_1) f_2(x) \right).\tag{1}$$

Now a multivariate root finding algorithm can be applied to $F_\epsilon$.


问:
Solving for (or approximating) $x$ in $\frac{x-\sin x}{x^{0.4}} = c$
I'm an engineer and I'm deducting a formula, and I found a BIG problem: I can't isolate $x$.

The formula is:
$$\frac{x-\sin x}{x^{0.4}} = c$$
where $c$ is a constant.

How to find the value for $x$, or at least a good approximation?

答:
Here is the graph of the inverse of
$$f(x)=\frac{x-\sin x}{x^{0.4}}$$
as you can see it is not a function. For some value of $c$ there are more than one $x$, even three in some intervals.

For $c=3$ we have $x_1\approx 4.5;\;x_2\approx 6.35;\;x_3\approx 7.83$.

For $c=3.3$ we have $x\approx 8.55$.

I used GeoGebra, plotted $f(x)$ and to get the inverse I took its symmetric with respect to the bisector of the first quadrant. This gives fine approximations for any $c$.

$$
$$


问:
Quotient space. Understanding and proving things
Let $E$ be a vector space and $F$ a subspace of $E$. The quotient space $E/F$ is $\{x+F:x\in E\}.$ Prove that if $E/F$ has dimension $n$, then there exists a linearly independent set $\{l_1,\dots, l_n\}$ in the algebraic dual space $E^{\#}$ such that $F = \bigcap_{i=1}^{n} \ker(l_i)$.

The only thing I can think of is, if $a \in E/F$ then $a = x+f$ with $f \in F$ and $x \in E$. For example, if $a \in F$ then $x=0$.

Since $E/F$ has dimension $n$, there is a Hamel base $\{h_1,\dots,h_n\} \subset E/F$ such that $$a = c_1h_1+\dots+c_nh_n.$$ Also, $\ker(l_i) = \{x \in E: l_i(x) = 0\}$, so if $x \in \bigcap_{i=1}^{n} \ker(l_i)$ the $l_i(x) = 0$ for all $i=1,\dots,n$.

But I don't see how can I relate all of this to get the result. Any hints?

We don't know any big result on this topic. The only thing we have is the very definition of the quotient space.

答:
It's helpful to think of the elements in $\{x + F: x \in E\}$ as equivalence classes mod F, i.e. for $y,y' \in E / F$ we say $ y =_{mod ~ F} y'$ if $y - y' \in F$.

In this context, we can define the application $\pi: E \to E / F$ given by $\pi x = \bar{x}$ where $\bar{x}$ is the set of all $x' \in E$ such that $x - x' \in F$. Observe that $\pi$ is linear, surjective and $\ker(\pi) = F$. This last sentence means that all $y \in F$ are in the zero class: $y \in \bar{0} \in E / F$. On the other hand if $x \in \bar{0}$ then $x = 0  + f$ for some $f \in F$, so the zero class is precisely $F$.

If $E / F$ has dimension $n$, then we can represent any class $\bar{x}$ as a linear combination of $n$ independent classes:
\begin{equation}
\bar{x} = \lambda_1 \bar{e}_1 + ... + \lambda_n \bar{e}_n
\end{equation}

Since the components for each $\bar{x}$ in this basis are unique, we can define the linear functionals $\bar{l}_i : E / F \to \mathbb{K}$, $\bar{l}_i(\bar{x}) = \lambda_i$. It's clear that $\bar{l}_i(\bar{x}) = 0 $ if the i-th component of $\bar{x}$ is zero. Over this construction, we define $l_i : E \to \mathbb{K}$ by $l_i(x) = \bar{l}_i(\pi(x)) = \lambda_i$. In this case we have, $\cap_{i =1,..,n} \ker(l_i) = \{x \in E : l_i(x) = 0 ~~\forall i = 1,...,n\} = \{x \in E: \pi(x) = \bar{0} \} = \{ x \in E : x \in F\} = F$.


问:
Number of homomorphisms from $D_6$ to $A_4$
Clarification for notation: $|D_6|=6$. Also $|A_4|=12$

So I know that there are $2$ elements of order $3$ and $3$ of order $2$ in $D_6$ (and $1$ of order $1$) and in $A_4$ there are $3$ elements of order $2$ and $8$ elements of order $3$.

So my answer is that I must send the $3$ elements of order $2$ in $D_6$ to the $3$ elements of order $2$ in $A_4$ and then I have $\frac{8!}{2!6!}=28$ possible homomorphisms (choosing $3$ elements of order $3$ from the $8$ available in $A_4$).

I feel $28$ is too much compared to other examples so I don't think the answer is right. I have also tried to build an homomorphism to $S_4$ since every group is isomorphic to a subgroup of $S_n$.

答:
There's no surjection.  Since the image of $h$ is a subgroup, it either has order $6,4,3$ or $2$.  But $4\nmid6$.

Now, if the image has order $6$, then $h$ is an isomorphism. But that's out:  $A_4$ doesn't have a copy of $S_3$ as a subgroup.  In fact it has no subgroup of order six.

So the image has order $3$ or $2$.  If the image has order $3$, the kernel has order two.  There are $3$ subgroups of order $2$, but they are not normal.  So the image can't have order $3$.

If the image has order $2$, the kernel has order $3$.  There is one subgroup of order $3$ in $D_6$.

There are $3$ copies of $C_2$ in $A_4$.

Thus we get $4$, when we throw in the trivial one.


问:
Conditions for the function to produce Positive Values
I have the following function

```<code>2* Pi *r^2*(3*a + 8*b^3 + 6*b*r^2)
</code>```

r is the radius of a small circle, hence the value of r is extremely small, where a,b are the coordinates in an x-y axis, hence a,b can be either positive or negative values .
To find whether the above function produces output > 0 , the condition 3 a + 8 b^3 > 0  must be met, since r^2 will be positive, albeit a small positive value. Hence for the above function to be greater than 0, 3 a + 8 b^3 > 0 must be met.

Now take the example of the below function

```<code> (Pi *r^2*(8*a + 3*(4*b^2 + r^2)))/4
</code>```

For the above function to be positive what are the conditions for a & b.
The above function when expanded yields:

```<code>(Pi *r^2*(8*a + 12*b^2 + 3r^2)))/4
</code>```

12*b^2 will always be positive, hence for the above function to produce values greater than 0, shouldn't the condition required only be 8a > 0? Or am i missing something here?

答:
Your discussion of the first expression is incomplete.

When $a = 25/6$, $b = -1$, and $r = 1$, we compute $3a + 8b^3 = 9/2 > 0$ and $3a + 8b^3 + 6 b r^2 = -3/2 < 0$, so the original expression can be negative when the condition you write is met.

When $a = -19/6$, $b = 1$, and $r = 1$, we compute $3a + 8b^3 = -3/2 < 0$ and $3a + 8b^3 + 6 b r^2 = 9/2 > 0$, so the original expression can be positive when the condition you write fails.

Here is a correct method to understand the positivity of the first expression...
$$ 2 \pi r^2 (3a + 8b^3 + 6 b r^2)  $$
is the product of a positive constant, $2\pi$, a nonnegative constant, $r^2$, and the expression $3a + 8b^3 + 6 b r^2$.  We are told that $r$ is the radius of a circle, not a point, so $r > 0$ and the expression $r^2$ is positive.  The expression is linear in $a$, so is positive when $a > \frac{-8b^3 - 6 br^2}{3}$.

Then you study
$$  \frac{\pi r^2 (8a + 3(4b^2 + r^2))}{4}  \text{.}  $$
The denominator is positive, so this is positive when its numerator is.  The numerator is the product of a positive constant, $\pi$, a nonnegative number, $r^2$, and the expression $8a + 3(4b^2 + r^2)$.  Again, since $r$ is the radius of a circle, $r > 0$, so $r^2$ is positive.  Again, this is linear in $a$, so we obtain $a > \frac{-12b^2 - 3r^2}{8}$.  Alternatively, this is quadratic in $b$, so we can write an equivalent expression (using the quadratic formula to partition the $ab$-plane and testing on which side(s) the inequality is satisfied):
$$  a > \frac{-3r^2}{8} \text{ or } \frac{1}{2}\sqrt{\frac{-8a}{3} - r^2} < b  \text{ or } b < \frac{-1}{2}\sqrt{\frac{-8a}{3} - r^2}  \text{.}  $$
(For an example where this $a$ condition is not met and the $b$ condition is, consider $a = -9/8$, $b = -1$, and $r = 1$.)


问:
How to find the complex number based on its argument?
Suppose I've the following:

$$\arg(2+i5-z) = 0.9$$

How can I find $z$? Wolfram says it's $-1.96$, but I didn't understand how to get this value.

EDIT: I wrote originally that the answer was 2, but it's in fact 0.90.

答:
hint

Put
$$z=a+ib$$

then

$$Z=2+5i-z=(2-a)+i(5-b)$$

with

$$\frac{5-b}{2-a}=\;\tan(0.9)$$

There are a lot of solutions.

If you look for real solution, with $ b=0$, you will get
$$\frac{5}{2-a}=\tan(0.9)=1.26$$
So
$$a=z=-1.96$$


问:
Definite Integral of square root of polynomial
I need to learn how to find the definite integral of  the square root of a polynomial such as:
    $$\sqrt{36x + 1}$$ or $$\sqrt{2x^2 + 3x + 7} $$

EDIT: It's not guaranteed to be of the same form. It could be any polynomial that can't be easily factored into squares.

This isn't homework, I'm studying for a final. And for context, I'm finding the arc length of a function.

答:
Recall that for $\alpha \neq -1$, we have
$$\int (ax+b)^{\alpha}dx = \dfrac1a \cdot \dfrac{(ax+b)^{\alpha+1}}{\alpha+1} + \text{ constant}$$
A way to see the above is as follows. Let $y = ax+b$. We then have $dy = adx$. Hence for $\alpha \neq -1$,
$$\int (ax+b)^{\alpha}dx = \int y^{\alpha} \dfrac{dy}a = \dfrac1a \dfrac{y^{\alpha+1}}{\alpha+1} + \text{ constant} = \dfrac1a \dfrac{(ax+b)^{\alpha+1}}{\alpha+1} + \text{ constant}$$
If $\alpha = -1$, we then have
$$\int \dfrac{dx}{ax+b} = \dfrac{\log(ax+b)}a + \text{ constant}$$

In general, there is no easy way to get $$\int \sqrt{P(x)} dx,$$ if degree of $P(x)$ is greater than $2$.

If $P(x)$ is linear, i.e., has degree $1$, I have mentioned above how to proceed.

Below we will see how to proceed if $P(x)$ is quadratic, i.e., $$P(x) = ax^2 + bx + c = a ((x+b_1)^2 + c_1).$$
$$b_1=\frac{a}{2b}$$ $$c_1=\frac{c}{a}-b_1^2$$

We now have the following
$$\sqrt{P(x)} = \sqrt{a} \sqrt{(x+b_1)^2 \pm c_1}$$
which gives us that
$$\int \sqrt{P(x)} dx = \sqrt{a} \int \sqrt{(x+b_1)^2 + c_1} dx$$
and
$$\int \sqrt{(x+b_1)^2 + c_1} dx = \dfrac{(b_1+x)\sqrt{P(x)} + c_1 \log \left(b_1 + x + \sqrt{P(x)}\right)}2 + \text{constant}$$


问:
How to interpret this minimum expression?
I am trying to understand the below conditional equation - it comes from Example 1 (Page 9) of A Tutorial on Variable Neighborhood Search (Hansen & Mladenovic, 2003). If more context is needed, find the paper here: https://www.researchgate.net/publication/2906122_A_Tutorial_on_Variable_Neighborhood_Search

$$\begin{equation}
  x_{ij}=\left\{
  \begin{array}{@{}ll@{}}
    1, & \text{if}\ c_{ij}=\min_{\ell|y_\ell=1} c_{\ell j} \text{(with minimum index}\ \ell\ \text{in case of ties)} \\
    0, & \text{otherwise}
  \end{array}\right.
\end{equation}$$

$\min_{\ell|y_\ell=1} c_{\ell j} \text{(with minimum index}\ \ell\ \text{in case of ties)}$

Could anyone please explain how to interpret this?

答: You didn't copy it correctly.  It says $$c_{i,j}=\min_{\ell|y_\ell=1} c_{\ell,j},$$ and it means that $i$ is the cheapest cost open facility that can serve customer $j$.


问:
Only one-way implication
This is a basic question about implication. In constructive logic, we have $\Rightarrow$ for implication and $\Leftrightarrow$ for bi-implication. If we want to express only one-way implication, we can perhaps have the following:
$$(A\Rightarrow B)\wedge(B\not\Rightarrow A).$$
That is, $A$ implies $B$ but not vice versa. But as far as I know, $\not\Rightarrow$ is not a connective in the object language but is used at the metal-level. Is there any way to express $B\not\Rightarrow A$ in the object language? Thanks in advance!

答:
$$\lnot (B \implies A)$$
equivalently,
$$\lnot (\lnot B \lor A)$$
equivalently,
$$B \land \lnot A$$

Title: Only one-way implication


答:
You can use material implication as

$$(P\implies Q)\iff (\lnot P \vee Q)$$

So, you can replace $ B$ does not imply $ A $ by
$$\lnot( \lnot B \vee A) $$
or
$$B \wedge \lnot A$$

Title: Only one-way implication

Body:
This is a basic question about implication. In constructive logic, we have $\Rightarrow$ for implication and $\Leftrightarrow$ for bi-implication. If we want to express only one-way implication, we can perhaps have the following:
$$(A\Rightarrow B)\wedge(B\not\Rightarrow A).$$
That is, $A$ implies $B$ but not vice versa. But as far as I know, $\not\Rightarrow$ is not a connective in the object language but is used at the metal-level. Is there any way to express $B\not\Rightarrow A$ in the object language? Thanks in advance!


问:
Proof of the Stirling Number Inequality, $S(m, n-2) \leq \binom{n}{3}S(m,n)$
I came across the following textbook problem: Prove that

$S(m, n-2) \leq \binom{n}{3}S(m,n)$ for $n \geq 3, m \geq 2n - 3$.

I tried to prove the inequality by induction on $n$ using the fact that, by definition, the Stirling number $S(m, j)$ is given by

$S(m, j) = \frac{1}{j!} \sum_{i=0}^{j-1} (-1)^j \binom{j}{i} (j - i)^m$

Nonetheless, I am unable to complete the inductive step. Is there any other way to prove this inequality? I would appreciate any help on this!

答:
Here is a combinatorial proof: Let $\mathbb{S}(m,n)$ denote the set of partitions of $[m]=\{1,2,\cdots , m\}$ into $n$ non empty disjoint blocks. Denote, also, $\binom{[n]}{k}$ the set of subsets of $[n]$ of size $k.$ Consider the function $\varphi : \mathbb{S}(m,n)\times \binom{[n]}{3}\longrightarrow \mathbb{S}(m,n-2)$ defined as follows:

let $\pi =\{B_1,\cdots ,B_n\}$ be a partition of $[m]$ into $n$ blocks and let $X=\{x_1,x_2,x_3\}$
$$\varphi ((\pi,X)) =\pi \setminus \{B_{x_1},B_{x_2},B_{x_3}\}\cup \{B_{x_1}\cup B_{x_2}\cup B_{x_3}\},$$
so essentially you are replacing $3$ blocks (indexed with elements in $X$) with their union. Notice that this functions is well-defined and try to show that this function is onto (use pidgeon-hole to see that because $m\geq 2n-3=2(n-2)+1$ then there is, necessarily, at least one block with $3$ elements, so you can split it).

 This implies your inequality because if a function is onto then the codomain has no more elements than the domain.


Title: Definite Integral of square root of polynomial

Body:
I need to learn how to find the definite integral of  the square root of a polynomial such as:
    $$\sqrt{36x + 1}$$ or $$\sqrt{2x^2 + 3x + 7} $$

EDIT: It's not guaranteed to be of the same form. It could be any polynomial that can't be easily factored into squares.

This isn't homework, I'm studying for a final. And for context, I'm finding the arc length of a function.

Body:
Using the $u$-substitution $u = 36x + 1$, we have $du = 36dx$. Then, it follows that:

$$\int_{a}^{b} \sqrt{36x+1}dx = \frac{1}{36}\int_{36a+1}^{36b+1}u^{\frac{1}{2}}du = \frac{1}{36}*\frac{2}{3}u^{\frac{3}{2}}|^{36b+1}_{36a+1} = \frac{1}{54}(36x+1)^{\frac{3}{2}}|^{b}_{a}$$


问:
Minimal cost flow problem - How to balance supply and demand by adding a node and edges?
I have the following graph for a minimal cost flow problem. Usually in this type of problem the demand = supply. However here we have 30 of supply and only 16 of demand.

I'm tasked with adding a node to the graph to make sure that the surplus of the supply is absorbed by this node. However, I'm not sure how to do that. It must not change an optimal solution to the problem. My instinct is to just add a node linked to all three suppliers. Each edge will have a maximum capacity of 6 like all the other edges in the graph.

答: Yes, add a dummy sink node with demand 14.  (Your figure shows supply 40 and demand 10, though.)  Yes, add links from suppliers to the dummy sink node, but you do not need any capacities or costs for the dummy links.</p>


问:
How can I solve this by induction?
The statement: $ \root{n} \of{n} + \frac{1}{n} < 2$ where $n>2$

My partial solution(?):

$ \root{n+1} \of{n+1} + \frac{1}{n+1} < 2$

$ \root{n+1} \of{n+1}(n+1) + 1 < 2(n+1)$

$ \root{n+1} \of{(n+1)^{n+2}} + 1 < 2n+2$

$ \root{n+1} \of{(n+1)^{n+2}} < 2n+1$

$ (n+1)^{n+2} < (2n+1)^{n+1}$

And I'm stuck. Would appreciate some help.

Thanks in advance.

答:
You don't really need induction, you just need the inequality $(1+x)^n\gt1+nx$ for $n\gt1$ (which, if you like, can be proved by induction), from which it follows that

$$\left(2-{1\over n}\right)^n=\left(1+{n-1\over n}\right)^n\gt1+n\cdot{n-1\over n}=n$$

which implies $2-{1\over n}\gt\sqrt[n]n$.

Title: How can I solve this by induction?


答:
Let $P$ be a partition of $[m]$ into $n-2$ parts. If $$m\ge 2n-3=2(n-2)+1\,,$$ there must be a part with at least $3$ members. Among all such parts let $p$ be the one with the smallest minimum element. Let $j,k$, and $\ell$ be the three smallest elements of $p$, and replace $p$ by $p\setminus\{j,k\}$, $\{j\}$, and $\{k\}$ to get a partition $p'$ of $[m]$ and a $3$-element subset $\big\{p\setminus\{j,k\},\{j\},\{k\}\big\}$ of $[n]$. The map

$$p\mapsto\left\langle\big\{p\setminus\{j,k\},\{j\},\{k\}\big\},p'\right\rangle$$

is an injection from the set $\mathscr{P}$ of partitions of $[m]$ into $n-2$ parts to the set $[n]^3\times\mathscr{P}'$, where $[n]^3$ is the set of $3$-element subset of $[n]$, and $\mathscr{P}'$ is the set of partitions of $[m]$ into $n$ parts. Thus,

$${m\brace{n-2}}=|\mathscr{P}|\le\left|[n]^3\times\mathscr{P}'\right|=\binom{n}3{m\brace n}\,.$$

Title: Proof of the Stirling Number Inequality, $S(m, n-2) \leq \binom{n}{3}S(m,n)$

Body:
I came across the following textbook problem: Prove that

$S(m, n-2) \leq \binom{n}{3}S(m,n)$ for $n \geq 3, m \geq 2n - 3$.

I tried to prove the inequality by induction on $n$ using the fact that, by definition, the Stirling number $S(m, j)$ is given by

$S(m, j) = \frac{1}{j!} \sum_{i=0}^{j-1} (-1)^j \binom{j}{i} (j - i)^m$

Nonetheless, I am unable to complete the inductive step. Is there any other way to prove this inequality? I would appreciate any help on this!


问:
How to show that $A(A+B)^{-1}B = B(A+B)^{-1}A$

I am trying this problem. Could not proceed . Please give a hint.

The exercise, if the above image is not showing clearly, is the following: Let $A$ and $B$ be two square $m \times m$ matrices for some integer $m \ge 2$ such that $A+B$ is invertible. Then show that
the equation $A(A+B)^{-1}B = B(A+B)^{-1}A$. [Note that either $A$ or $B$ or each of both may not be invertible, and that $A$ and $B$ may not commute.]

答:
First note the following equation:

$$A(A+B)^{-1}(A+B) = (A+B)(A+B)^{-1}A.$$

[Indeed, both sides of this equation just above are $A$.] That hint may or may not be enough. If you want more help, see below:

 Distributing on each side gives.....

 $$A(A+B)^{-1}A +A(A+B)^{-1}B = A(A+B)^{-1}A + B(A+B)^{-1}A.$$

 But then subtracting both sides by $A(A+B)^{-1}A$ gives

 $$A(A+B)^{-1}B = B(A+B)^{-1}A,$$

 which is what you want.


问:
If $f(x)$ is a Borel function, is $f(x+a)$ also a Borel function?
I've been recently covering Borel functions and wondering are they preserved by the following property.

If $f(x)$ is a Borel function then $f(x+a)$ is also a Borel function where $a$ is a real number.

It makes sense to me as if $f(x)$ is a Borel set $(-\infty ,b)$ in a set in the Borel algebra so $f(x+a)$ would have a similar set $(-\infty, b-a)$ in the Borel $\sigma$-algebra and therefore is a Borel function.

Any help would be great thanks

答: A composition of Borel functions is Borel, so $f$ composed with $g(x)=x+a$ is Borel, because both $f$ and $g$ are Borel.


问:
stability and asymptotic stability: unstable but asymptotically convergent solution of nonlinear system
Consider nonlinear systems of the form $X(t)'=F(X(t))$, where $F$ is smooth (assume $C^\infty$). Is it possible to construct such a system (preferably planar system) so that $X_0$ is an unstable equilibrium, but all nearby solution curves tend to $X_0$ as $t \to \infty$? If so, how? A conceptual construction is enough. What will the phase portrait look like?</p>

答:
Definition (Verhulst 1996)

The equilibrium solution $X_c$ is called asymptotically stable if there exists a $\delta(t_0)$ such that
\begin{equation}
||X_0 - X_c|| \le \delta(t_0) \implies \lim_{t\rightarrow\infty} ||X(t;t_0,X_0) - X_c|| =0
\end{equation}

So no, you're essentially asking can a system be both asymptotically stable and unstable which is a contradiction.

EDIT:

Glendinning's text refers to this definition as quasi-asymptotic stability to differentiate it from 'direct' asymptotic stability (solutions are both quasi-asymptotically stable and Lyupanov stable).

As an example he presents this equation:
\begin{equation}
\dot{r} = r(1-r^2) \qquad \dot{\theta} = 2\sin^2\left(\frac{1}{2}\theta\right),
\end{equation}

which has an unstable critical point at $(0,0)$ a quasi-asymptotically stable critical point at $(1,0)$.

Points are attracted to the invariant circle at $r=1$. On this curve, the flow is semi-stable. If $\pi>\theta >0$ then the flow wraps around until it passes $\theta = \pi$. If $ -\pi> \theta>0$ the trajectory falls into the critical point at $(1,0)$. This is an example of a homoclinic orbit.

In the example provided, the critical point at $(1,0)$ is a saddle. Which, if all we considered was the linearised system, is unstable. Indeed it is Lyapunov unstable as there is a trajectory that leaves any $\epsilon$-ball about $(1,0)$.
However, in nonlinear systems the linearisation is often not sufficient,
especially with saddles, which are often the cause of a whole bunch of interesting behavior that aren't represented in the linearisation, like homoclinic and heteroclinic connections.

I think the confusion comes from how you/they define instability.
The texts I refer to define a critical point as unstable if it is not Lyapunov stable or 'Quasi-Asympotically' stable. However if all you go off is the linearisation then i'd imagine the example is what you're after.


问:
Showing that an integral is convergent
I was doing a problem where I had to show that the integral is convergent to prove the existence of the expected value of a probability density function. But I don't fully understand the solution:

$ \frac{|x|e^x}{(e^x + 1)^2} \sim_{-\infty} |x|e^{-x} $ and $ \int_{-\infty}^{0} |x|e^x $ converges by comparing to $\frac{1}{x^2}$

How do we get the idea of using $\frac{1}{x^2}$ ? Is it just being used to solve these kinds of problems or is there a specific technique ?

答: Because for $x<0$ we have $xe^x<\frac{1}{x^2}$ and $1/x^2$ converges on $(-\infty,-\varepsilon]$ for positive $\varepsilon$.

Title: Showing that an integral is convergent


问:
Find at least one root of $\sin(t^2-1)=\sin(t)$
I have solved one exercise then I have to solve the trigonometric equation at last to finish the exercise.
The equation $\sin(t^2-1)=\sin(t)$

So I start to solve the following. Please give me a recommend or something to recover my solution:

$\begin{cases}
t^2-1=t+2k\pi\\
t^2-1=\pi-t+2k\pi
\end{cases} , k\in\mathbb{Z}$

Then I started to solve the quadratic equation using Discriminant.

Did I do this in correct way ? Or something give me a hint about this.

答:
Note that :

$sin(x) =sin(y) $$\Rightarrow$ $x=y+2k\pi $ or $x=\pi-x+2k\pi ( k \in \mathbb{Z} $)

So

$sin(t^2 - 1)=sin(t)$$\Rightarrow $$ t^2 - 1=t+2k\pi $
or $t^2 - 1=\pi-t+2k\pi $

if :

$t^2 - 1=t+2k\pi $

$\Rightarrow $ $t^2 - t-(1+2k\pi)=0$

$\triangle =5+8k\pi≠0$

If $\triangle <0 $ this equation have not a solution

If $\triangle >0 $the equation have two solutions

$t_1=\frac{1-\sqrt{5+8k\pi}} {2}$ and $t_2=\frac{1+\sqrt{5+8k\pi}}{2} $

Suppose :

$t^2 - 1=\pi-t+2k\pi $

$\Rightarrow $$ t^2 +t-(1+\pi+2k\pi)=0 $

$\triangle =1+4(\pi+1+2k\pi)≠0$

If$\triangle <0$ this equation have not a solution

If $\triangle >0$ this equation have two solutions

$t_1=\frac{-1-\sqrt{5+4(2k\pi+\pi)}}{2}$
And $t_2=\frac{-1+\sqrt{5+4(2k\pi+\pi)}}{2}$


问:
Showing that a certain signal to noise ratio is bounded by one (matrix inequality)
Let $A, \, B \in \mathbb{R}^{n \times n}$ be $n \times n$, symmetric, positive definite matrices.

I would like to show that for any column vector $x \in \mathbb{R}^{n \times 1}$, it holds that
$$ 0 \le \frac{(x^TA^{-1}x)^2}{(x^TB^{-1}x)(x^TA^{-1} B A^{-1}x) }\le 1.$$

In signal processing literature, if $A$ and $B$ are certain covariance matrices, then the fractional quantity I write above is sometimes referred to as the normalized signal to interference plus noise ratio. Showing this quantity is bounded below by zero is trivial, because of the positive definiteness of all the matrices involved. On the other hand, I probably need to use the Cauchy-Schwarz inequality in a clever way to show it bounded from above by one. This is where I am stuck. Hints or solutions are greatly appreciated.

答:
Apply the Cauchy-Schwarz inequality, noticing that

$$x^T A^{-1} x = x^T B^{-1/2} B^{1/2} A^{-1}x = (B^{-1/2}x)^T B^{1/2} A^{-1}x.$$


问:
How can I prove that the solutions of this differential equation is monotone?
I'm trying to proof if $x:I\to \mathbb R$ a maximal regular solution of $x'=f(x)$, such that the image $x(I)\subset \mathbb R$ is bounded and $f:\mathbb R\to \mathbb R$ is $C^1$, then $x$ is strictly monotone. How can I prove that $f$ doesn't change sign?

I need help

Thanks a lot

答:
If $f(r) = 0$, $x=r$ is a constant solution of $x' = f(x)$, and it certainly isn't strictly monotone.  Moreover, by the Existence and Uniqueness Theorem every solution on an interval $I$ that takes the value $r$ at some point of $I$ is constant.  On the other hand, if $x: I \to \mathbb R$ is a solution and $f \ne 0$ at all points of the interval $x(I)$, by the Intermediate Value Theorem either $f > 0$ on $x(I)$ or $f < 0$ on $x(I)$.  In the first case,
$x' > 0$ so $x$ is strictly increasing; in the second case $x' < 0$ so $x$ is strictly decreasing. </p>


问:
Proof Verification of $|\cos(a) - \cos(b)| \leq |b-a|$
Prove of $|\cos(a) - \cos(b)| \leq |b-a|$

My Attempt:

Let $g(x) =1$ and $f(x) = \sin(x)$. We know that $|f(x)| \leq|g(x)|$ for all $x \in R$.
Integration theorems tell us that:

$$\int_{a}^{b}|f(x)|dx \leq \int_{a}^{b} |g(x)|dx$$

$$\left|\int_{a}^{b}\sin(x)dx\right| \leq \left|\int_{a}^{b}1\, dx \right|$$

$$|\cos(a)-\cos(b)| \leq |b-a|$$

Does this work?

Update

By theorem: $|\int_{a}^{b} f(x)dx| \leq\int_{a}^{b}|f(x)|dx$ the inequality holds from line one to line 2. Also $|g(x)| = g(x)$.

答:
No it doesn't. In general, it is not true that $$|f(x) |\leq |g(x) | \Rightarrow  \left| \int_a^b f(x) \, dx \right| \leq \left| \int_a^b g(x)\, dx \right|.$$

E.g. $f(x) = 1, g(x) = \begin{cases} -1 &  x \leq 0 \\ 1 & x \geq 0. \end{cases}$
Then, integration from $ - 1 $ to $1$ gives us $ 2 \leq 0$.

What is true (and you might have to prove it to use it) is that

$$ |f(x) | \leq g(x) \Rightarrow \left|\int_a^b f(x) \, dx \right| \leq \int_a^b g(x) \, dx . $$

This still holds in your case since $g(x) = 1$.

Of course, there could be other integration theorems to use.


问:
kernel and image subspaces proof
Let $f:U \to V$ be a linear map between two vector spaces $U$ and $V$ over a field $IF$. The kernel is $\ker(f)=\{u \in U \mid f(u)=0\}$ and image $\operatorname{im}(f)=\{v \in V \mid \exists u \in U s.t. f(u)=v\}$

i)  Prove that $\ker(f)$ is a subspace of $U$

ii) Prove that $\operatorname{im}(f)$ is a subspace of $V$

For i) I got using the subspace criterion 2: clearly $\ker(f)$ is a subset of U. and clearly $0$ is contained in $\ker(f)$. And let u and v be in $\ker(f)$ and a be in $IF$, so
$f(u)=f(v)=0$. Then $au+v=af(u)+f(v)=a0+0=0$ which is in $\ker(f)$. I am not sure if this is correct because it seems a bit abstract for me. and do i need more justification that $0$ is in $\ker(f)$?

Also for the image, i have no idea...

答:
To prove that image of a linear transformation is a subspace of domain vectorial space you need use definitions.

Let $T$ a linear transformation such that $T:U\to V$ and the $Im(T)$ is defined as:

$Im(T)=\{v \in V: \exists u\in U, T(u)=v\}$

So, you should always get a $u \in U$ that holds the definition.

So, to prove that $Im(T)$ is a subspace of $V$ we need to prove that is closed under the  $(+, \cdot)$ operators and have the $0$ vector of $V$.

Have zero vector

Since $T(0_U)=0_V \implies \exists0_U\in U: T(0_U)= 0_V\in V$ so $0 _V \in Im(T)$

Closed under $(+)$

Let $v_1, v_2 \in Im(K)$, this implies that $T(u_1)=v_1$ for some $u_1 \in U$ and same for $T(u_2)=v_2$ for some $u_2 \in U$.

And $v_2+v_1=T(u_2)+T(u_1)=T(u_1+u_2)$ and here you can see that exist a $(u_1+u_2)$ such that $T(u_1+u_2) = v_1 + v_2$ so is closed under sum.

And to prove $(\cdot)$ is the same process.

Title: kernel and image subspaces proof


问:
Is a $R \otimes S$-module of the form $V \otimes W$ with $V$ simple $R$-module and $W$ simple $S$-module a simple module?

Let $R$ and $S$ be $\mathbb{C}$-algebras. Show that $X$ is a simple $R \otimes S$-module if and only if it is isomorphic to a $R \otimes S$-module of the form $V \otimes W$ with $V$ simple $R$-module and $W$ simple $S$-module.

Starting with the backwards direction, I know that if we assume the semisimplicity of $R$ and $S$, I can show that the endomorphism ring of $V \otimes W$ is a division ring (it is isomorphic to $\mathbb{C}$), since we can also show $R \otimes S$ is finite dimensional and semisimple, that would imply that $V \otimes W$ is simple.

But now I would like to show both directions $without$ assuming the semisimplicty of $R$ and $S$, and I'm lost.

答:
(I assume all tensor products are taken over $\mathbb{C}$ here. Also, this is not a complete answer.)

This is false if $R$ and $S$ can be infinite-dimensional. Take $R = S = \mathbb{C}(t)$; then the only tensor product of simple modules available is the free module $R \otimes S$ of rank $1$, but $R \otimes S$ is not a field (e.g. it admits a natural surjective map to $\mathbb{C}(t)$ given by multiplication which has nontrivial kernel, so it has a nonzero proper ideal) so it is not simple as a module over itself.

For a more complicated counterexample involving the Weyl algebra see this math.SE answer. According to that question it should be true if we assume that $X$ is finite-dimensional but I'm not sure how to prove that.

If $R$ and $S$ are finite-dimensional, then a simple $R \otimes S$-module is a module over the quotient $(R \otimes S)/J(R \otimes S)$ by the Jacobson radical, and the tensor product $V \otimes W$ of a simple $R$-module and a simple $S$-module is a module over the tensor product $R/J(R) \otimes S/J(S)$. There's a natural map

$$R/J(R) \otimes S/J(S) \to (R \otimes S)/J(R \otimes S)$$

and it would suffice to prove that this is an isomorphism, because then we'd be reduced to the semisimple case. I think this is true (we will need to use the fact that $\mathbb{C}$ is algebraically closed, or at least that it's perfect; this is false over a non-perfect field) but I also don't know how to prove this off the top of my head.


答:
The key inequality you really want to invoke is

$$\left|\int_a^bf(x)\,dx\right|\le\int_a^b|f(x)|\,dx$$

if $a\le b$.  It follows from this that, if $a\le b$, then

$$|\cos a-\cos b|=\left|\int_a^b\sin(x)\,dx\right|\le\int_a^b|\sin(x)|\,dx\le\int_a^b1\,dx=b-a=|b-a|$$

The case $b\le a$ amounts to interchanging $a$ and $b$.

Remark: In a way, you shouldn't need any calculus at all. If you draw a picture of the unit circle with angles $a$ and $b$, then you can see that $|b-a|$ is the length of the arc of the unit circle lying above a horizontal line segment of length $|\cos a-\cos b|$. The trick is, I'm describing a picture with $0\le a,b\le\pi$, so you need to do a bit of finagling to show it holds for all $a$ and $b$.

Title: Proof Verification of $|\cos(a) - \cos(b)| \leq |b-a|$

Body:
Prove of $|\cos(a) - \cos(b)| \leq |b-a|$

My Attempt:

Let $g(x) =1$ and $f(x) = \sin(x)$. We know that $|f(x)| \leq|g(x)|$ for all $x \in R$.
Integration theorems tell us that:

$$\int_{a}^{b}|f(x)|dx \leq \int_{a}^{b} |g(x)|dx$$

$$\left|\int_{a}^{b}\sin(x)dx\right| \leq \left|\int_{a}^{b}1\, dx \right|$$

$$|\cos(a)-\cos(b)| \leq |b-a|$$

Does this work?

Update

By theorem: $|\int_{a}^{b} f(x)dx| \leq\int_{a}^{b}|f(x)|dx$ the inequality holds from line one to line 2. Also $|g(x)| = g(x)$.


答:
You can't just move the absolute value bars as you did from the first line to the second. If you're allowed to use the integral inequality $\left|\int_{a}^{b}f(x)dx\right|\leq \int_{a}^{b}|f(x)|dx$, then you can rigorously deduce that

$$\left|\int_{a}^{b}-\sin(x)\text{ }dx\right|\leq \int_{a}^{b}|-\sin(x)|\text{ }dx\leq \int_{a}^{b}1\text{ }dx$$

and from there conclude that $|\cos(b)-\cos(a)|\leq |b-a|$.

That being said, I feel like this approach buries a bit of the "circularity" at play here. You see, in order to use the Fundamental Theorem of Calculus like you did, you need to prove that $\cos(x)$ is an antiderivative of $-\sin(x)$, which in turn needs you to prove that the derivative of $\cos(x)$ is $-\sin(x)$. By proving the inequality like you did, you're implicitly assuming all of this, so it's better to simply leverage the derivative and any theorems related to it. To be specific, if you have proven that $\cos(x)$ is differentiable everywhere and its derivative is $-\sin(x)$, you can apply the mean value theorem to the function $f(x)=\cos(x)$ to deduce that for any pair of numbers $(a,b)$ with $a<b$, there exists a number $c$ such that $a<c<b$ and

$$\left|\frac{\cos(b)-\cos(a)}{b-a}\right|=|-\sin(c)|\leq 1$$

and thus arrive at your result: $|\cos(b)-\cos(a)|\leq |b-a|$.

Of course, you could instead opt to prove that $\int_{a}^{b}-\sin(x)dx=\cos(b)-\cos(a)$ directly from the definition of the integral and any of the equivalent definitions of the trigonometric functions, but this is usually extremely difficult to do.


答:
Integrals: who needs 'em? Without loss of generality, assume $a<b$. By the Mean Value Theorem for derivatives, for some $c\in (a,b)$ we have
$$
\left|\frac{\cos(b)-\cos(a)}{b-a}\right| = |-\sin(c)|\le 1
$$