问:
How do I make spirograh diagrams using software?
There are these "string math artworks" I believe they are called "spirographs" where people usually take a circle or other geometric shape and put in nails on the shape and then use yarn and weave the yarn in a particular way such as to make these artworks.

I am wanting to make similar diagrams using software instead of physically making them with nails and yarn. I found this one link where someone is using Tikz with Latex to generate a circle with nodes. This might work, but I don't know exactly how to use Latex and this Tikz (plug-in) software.

If someone can point me in the right direction that would be great. Thanks.

Here are some links:

https://tex.stackexchange.com/questions/415330/make-graph-where-nodes-and-connecting-edges-are-circles?newreg=03724c430452487a8f756a4c311bbdc3

https://www.youtube.com/watch?v=0K2YbYTyEqY

答:
Here's a recreation of the figure from that youtube video. It's written in asympote. If you have one of the popular LaTeX distributions like MacTex or tex-live, then you already have asymptote!

```<code>/*  Set up the format of the figure.                                          */
import settings;
settings.outformat = "pdf";

/* Size of the output figure.                                                 */
size(256);

/*  Number of points along the circle.                                        */
int N_Points = 1000;

/*  Number of points to skip for a and b.                                     */
int a_skip = 4;
int b_skip = 3;

/*  ints to correspond to the current points on the circle.                   */
int a, b;

/*  The starting values for a and b. Set them with angles 0 and pi.           */
a = 0;
b = (int)(N_Points / 2);

/*  Loop over a and b and stop once the two values correspond to the same     *
 *  point on the circle. This occurs when a-b = 0 mod N_Points.               */
while ((a-b) % N_Points != 0)
{
    draw(expi(2*pi*a/N_Points) -- expi(2*pi*b/N_Points));
    a += a_skip;
    b += b_skip;
}

/*  Draw the outline of the circle.                                           */
draw(circle((0, 0), 1));
</code>```

Saving this as spiral.asy, you can then run it with asy spiral.asy which will produce spiral.pdf.

Body:
Here's a recreation of the figure from that youtube video. It's written in asympote. If you have one of the popular LaTeX distributions like MacTex or tex-live, then you already have asymptote!

/*  Set up the format of the figure.                                          */
import settings;
settings.outformat = "pdf";

/* Size of the output figure.                                                 */
size(256);

/*  Number of points along the circle.                                        */
int N_Points = 1000;

/*  Number of points to skip for a and b.                                     */
int a_skip = 4;
int b_skip = 3;

/*  ints to correspond to the current points on the circle.                   */
int a, b;

/*  The starting values for a and b. Set them with angles 0 and pi.           */
a = 0;
b = (int)(N_Points / 2);

/*  Loop over a and b and stop once the two values correspond to the same     *
 *  point on the circle. This occurs when a-b = 0 mod N_Points.               */
while ((a-b) % N_Points != 0)
{
    draw(expi(2*pi*a/N_Points) -- expi(2*pi*b/N_Points));
    a += a_skip;
    b += b_skip;
}

/*  Draw the outline of the circle.                                           */
draw(circle((0, 0), 1));

Saving this as spiral.asy, you can then run it with asy spiral.asy which will produce spiral.pdf.


问:
Continuous injective curve has "direction of travel"
Let $\gamma,\hat{\gamma}:[0,1]\to\mathbb{R}^n$ be continous, injective curves such that $\gamma([0,1])=\hat{\gamma}([0,1])$. How can I show that if the two curves start in the same point and we have that $\gamma(t_1)=\hat{\gamma}(s_1)$, $\gamma(t_2)=\hat{\gamma}(s_2)$ and $t_1<t_2$ then it follows that $s_1<s_2$.

答:
Since $\gamma, \hat{\gamma}$ are bijections onto their (same) ranges, we see that $\pi = \hat{\gamma}^{-1} \circ \gamma$ is a bijection $\pi:[0,1]\to[0,1]$. In particular it must be monotonic, and since $\gamma(0)= \hat{\gamma}(0)$ it must be increasing.

Hence given the conditions in the question we see that $t_1 < t_2$ iff$s_1 < s_2$.

Addendum:

As @AnthonyCarapetis pointed out, I skipped over a detail.

I need to show that $\hat{\gamma}^{-1}$ is indeed continuous.

Suppose $\hat{\gamma}(t_n) \to \hat{\gamma}(t)$ but $t_n \not\to t$ ($t_n,t$ are well defined since $\hat{\gamma}$ is injective). Then there is some $\delta >0$ such that $|t_n-t| \ge \delta$ infinitely often and hence $t_n \overset{K}{\to} t^* \neq t$ for some subsequence $K$. Then $\gamma(t_n) \overset{K}{\to} \gamma(t^*) \neq \gamma(t)$ (again because of injectivity) which is a contradiction and so $t_n \to t$.

ContentLicense: CC BY-SA 4.0


问:
Proof that $\inf A = -\sup(-A)$

Let $A$ be a nonempty subset of real numbers which is bounded below. Let $-A$ be the set of of all numbers $-x$, where $x$ is in $A$. Prove that $\inf A = -\sup(-A)$

So far this is what I have

Let $\alpha=\inf(A)$, which allows us to say that $\alpha \leq x$ for all $x \in A$. Therefore, we know that $-\alpha \geq -x$ for all $x \in -A$. Therefore we know that $-\alpha$ is an upper bound of $-A$. $\ \ \ \ $

Now let $b$ be the upper bound of $-A$. There exists $b \geq-x \implies-b \leq x$ for all $x \in A$. Hence,
\begin{align}
-b & \leq \alpha\\
-\alpha & \leq b\\
-\alpha & = - \inf(A) = \sup (-A)
\end{align}

By multiplying $-1$ on both sides, we get that $\inf(A) = -\sup (-A)$

Is my proof correct?

答: Aside from dropping a $-$ sign in your last line, that's good work. (Also, I'd say "let $b$ be an upper bound," instead, but it's clear that's what is meant.)

ContentLicense: CC BY-SA 3.0

Title: Proof that $\inf A = -\sup(-A)$


问:
The functions $f,g$ are non constant, differentiable, real valued functions..

I have found that $\cos{x}$ and $\sin{x}$ satisfies the given condition. But how to prove in general. Please give some hint

答:
Letting $x=y=0$ gives us straight away that $f(0)=1$.

Differentiating equation (1) we obtain
$$f'(x+y)=f'(x)f(y)-g'(x)g(y)$$
Set $x=0$, using $f'(0)=0$:
$$f'(y)=-g'(0)g(y)$$
for every $y$. Similarly differentiating equation (2) gives
$$g'(x+y)=f'(x)g(y)+g'(x)f(y)$$
and setting $x=0$ gives
$$g'(y)=g'(0)f(y)$$
for every $y$.

These equations show that $f$ and $g$ are twice differentiable, moreover
$$f''(y)=-g'(0)g'(y)=-g'(0)^2f(y)$$
by combining them. Let $\omega=g'(0)$. This is a simple differential equation whose general solution is
$$f(y)=A\cos(\omega y)+B\sin(\omega y)$$
Our initial conditions $f(0)=1$ and $f'(0)=0$ show us that
$$f(y)=\cos(\omega y)$$
Finally letting $x=y$ in equation (1) gives:
$$\cos(2\omega y)=\cos^2(\omega y)-g(y)^2$$
But since $\cos(2\omega y)=2\cos^2(\omega y)-1$ it follows
$$g(y)^2=1-\cos^2(\omega y)=1-f(y)^2$$
hence
$$f(y)^2+g(y)^2=1$$


问:
solve the equation of the following
the real number $a, b, c, d, e$ have some equations as follows

\begin{array}{lcl} ac=1 \\ ad+bc=-1 \\ ae+bd=1 \\ be=-6 \end{array}

how can I find the value of $a+b+c+d+e$?

答:
(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

Hint: $x^3 - x^2 + x - 6 = (x-2) ( x^2 + x + 3)$.

Title: solve the equation of the following


问:
inverse image under the inclusion and extension by zero
Suppose $X$ consists of two points and has discrete topology, namely  $X=\{p_1\} \coprod \{p_2\}$. Let $\iota_i: \{p_i\} \to X$ be the inclusion map. Let $\underline{\mathbb{Z}} \in \mathcal{Sh}(X)$ be the constant sheaf. What is $\iota_1^{-1}\underline{\mathbb{Z}}$ the inverse image of $\underline{\mathbb{Z}}$.

First at the stalk level we have $\iota_1^{-1}\underline{\mathbb{Z}}_{p_1} \cong \underline{\mathbb{Z}}_{p_1} = \mathbb{Z}$, which implies the global $\Gamma(p_1, \iota_1^{-1}\underline{\mathbb{Z}})= \mathbb{Z}$. Or we can  compute by definition the colimit of
$\mathbb{Z} \oplus \mathbb{Z} \xrightarrow{(1,0)}\mathbb{Z}$ which also gives $\Gamma(p_1, \iota_1^{-1}\underline{\mathbb{Z}})= \mathbb{Z}$.

However if we consider the short exact sequence of sheaves
$$0\to j_!\iota_2^{-1}\underline{\mathbb{Z}} \to\underline{\mathbb{Z}} \to \iota_1^\ast\iota_1^{-1}\underline{\mathbb{Z}} \to 0 $$
where $ j_!\iota_1^{-1}\underline{\mathbb{Z}}$ is  extension by zero,  i.e the sheaf with $ j_!\iota_2^{-1}\underline{\mathbb{Z}}(U) = \iota_2^{-1}\underline{\mathbb{Z}}(U)$ if $U \subset p_1$, and $0$ otherwise.
Then if we take the global section, since $\Gamma(X, \iota_1^\ast\iota_1^{-1}\underline{\mathbb{Z}})= \Gamma(p_1, \iota_1^{-1}\underline{\mathbb{Z}}) = \mathbb{Z}$, we get
$0\to0\to\mathbb{Z} \oplus \mathbb{Z}\to \mathbb{Z}$ which is not exact.

I cannot see what went wrong.

答:
You said that $j_!\underline{\mathbb{Z}}(V)=0$ if $V\not\subset\{p_1\}$. But this is wrong. More generally, if $j:U\to X$ is the inclusion of an open subset, then the property $j_!\mathcal{F}(V)=0$ if $V\not\subset U$ does not hold in general.

In fact the association $V\mapsto \begin{cases}\mathcal{F}(U)&\text{if $V\subset U$}\\ 0&\text{if $V\not\subset U$}\end{cases}$ only define a presheaf and $j_!\mathcal{F}$ is the sheaf associated to it.

The two points set is a striking example. You get many others when you realize that $\mathcal{F}(U_1\sqcup U_2)=\mathcal{F}(U_1)\times\mathcal{F}(U_2)$, so if $U_1$ is a subset of $U$ and $U_2$ is not, $U_1\cup U_2\not\subset U$ but $\mathcal{F}(U_1\sqcup U_2)\neq 0$ in general. You can even have connected $V\not\subset U$ with the property that $j_!\mathcal{F}(V)\neq 0$.

As for the pullback of a constant sheaf, there is an easy answer : it is always the constant sheaf with the same stalks. In fact if $E$ is any set then the constant sheaf $\underline{E}$ on $X$ is the pullback $a_X^{-1}E$ where $a_X:X\to\{*\}$. It follows that for any map $f:Y\to X, f^{-1}\underline{E}=f^{-1}a_X^{-1}E=(a_Xf)^{-1}E=a_Y^{-1}E=\underline{E}$.


问:
Using the definition, prove that following limit exists

Suppose that $x_n$ is a sequence of real numbers that converges to $1$ as $n\to\infty$. Using the definition, prove that following limit exists.
$$\frac{x^2_n − e}{\sqrt{n}}→ 1 − e \quad\text{ as }\quad n \to\infty.$$

Here is my attempt, but I'm not sure. If it is wrong give me hint to prove that's enough.

My Attempt

By hypothesis, given $\epsilon>0$ there is an $N\in\mathbb{N}$ such that $n\geq N$ implies $|x_n-1|<\epsilon$.

Next, apply $\epsilon=1$ to choose $n\geq N_2$, which implies $|x_n-1|<1$ (i.e $x_n<2$).  So,

\begin{align}\left|\frac{x^2_n − e}{\sqrt{n}}- (1 − e)\right| &=\left|\frac{(x_n −1)(x_n +1)}{\sqrt{n}}+\frac{1-e}{\sqrt{n}}-1 +e\right|\\
&=\left|\frac{(x_n −1)(x_n +1)}{\sqrt{n}}+\frac{(1-e)(1-\sqrt{n})}{\sqrt{n}}\right|\\
&\leq \left|\frac{(x_n −1)(x_n +1)}{\sqrt{n}}\right|+\left|\frac{(1-e)(1-\sqrt{n})}{\sqrt{n}}\right|.
\end{align}

Since $|1-e|<2$ and given $ϵ>0$, there is an  $N\in\mathbb{N}$  such that $n≥N$ implies $|\frac{1}{\sqrt{n}}|\lt \epsilon$ so choose $n≥N_1 $ implies $\frac{1}{\sqrt{n}}\lt 1$ so
$$\left|\frac{x^2_n − e}{\sqrt{n}}- (1 − e)\right|<\left|\frac{(x_n −1)(x_n +1)}{\sqrt{n}}\right|+2$$ since $x_n+1<3$ set $N=\max(N_1,N_2)$

$$\left|\frac{x^2_n − e}{\sqrt{n}}- (1 − e)\right|<3|x_n-1|+2<3\epsilon+2$$

Thank you!!!

答: You cannot prove that, since it is not true. In fact, since the sequence $(x_n)_{n\in\Bbb N}$ is bounded (since it converges), then so is the sequence $(x_n^{\,2}-e)_{n\in\Bbb N}$ and therefore$$\lim_{n\to\infty}\frac{x_n^{\,2}-e}{\sqrt n}=0.$$Concerning your attempted proof, note that, at the end, you get$$\left|\frac{x_n^{\,2}-e}{\sqrt n}-(1-e)\right|<3\varepsilon+2.$$So what? You should have got$$\left|\frac{x_n^{\,2}-e}{\sqrt n}-(1-e)\right|<\varepsilon$$instead.


问:
How do I understand the $\|a\|$ used in this equation which describes the force by linear springs between two particles?
I have the following equation:

$ F_{i j}^{e}(x)=k_{i j}\left(\left\|\mathbf{x}_{i}-\mathbf{x}_{j}\right\|-l_{i j}\right) \frac{\mathbf{x}_{i}-\mathbf{x}_{j}}{\left\|\mathbf{x}_{i}-\mathbf{x}_{j}\right\|} $

from this paper http://www.cs.jhu.edu/~misha/ReadingSeminar/Papers/Magnenat-Thalmann04.pdf on cloth simulation, it occurs on page 19. I just wanted to clarify what the meaning behind this equation would be.

So I believe it is saying that the Force between particles i and j is equal to the k constant multiplied by $ \left\|\mathbf{x}_{i}-\mathbf{x}_{j}\right\|-l_{i j} $ , which I would assume to be the distance (magnitude) between two particles, i and j, take away the rest length. But then in the fraction I can see that there is also $ \mathbf{x}_{i}-\mathbf{x}_{j} $, so without the ||, and I can't tell the difference in meaning between $ \mathbf{x}_{i}-\mathbf{x}_{j} $ and $ ||\mathbf{x}_{i}-\mathbf{x}_{j} || $ . I've looked for the meaning of || || in mathematics, but it appears to have many meanings, including the norm but in this context I am unsure as I haven't cowered norms in the limited university maths modules (I had one) and I don't really know how it applies here after looking for the definition. Could someone clarify?

答:
This looks like a force, and typically the force is not just a real number, it also usually has a direction.  That's what the $x_i-x_j$ is contributing. All the other quantities appearing combined to form a coefficient in front of the vector $x_i-x_j$.

I can't see anything to suggest that $\|\cdot\|$ means anything other than the Euclidean norm on real vectors.


答:
The curves generated by a spirograph are epi- and hypo-cycloids, or more generally epi- and hypo-trochoids. They are obtained by rolling a circle around another, with commensurate radii to obtain closed curve. The drawing pen need not be on the circumference.

https://en.wikipedia.org/wiki/Epicycloid
https://en.wikipedia.org/wiki/Epitrochoid

Title: How do I make spirograh diagrams using software?

Body:
There are these "string math artworks" I believe they are called "spirographs" where people usually take a circle or other geometric shape and put in nails on the shape and then use yarn and weave the yarn in a particular way such as to make these artworks.

I am wanting to make similar diagrams using software instead of physically making them with nails and yarn. I found this one link where someone is using Tikz with Latex to generate a circle with nodes. This might work, but I don't know exactly how to use Latex and this Tikz (plug-in) software.

If someone can point me in the right direction that would be great. Thanks.

Here are some links:

https://tex.stackexchange.com/questions/415330/make-graph-where-nodes-and-connecting-edges-are-circles?newreg=03724c430452487a8f756a4c311bbdc3

https://www.youtube.com/watch?v=0K2YbYTyEqY


问:
Economics-indifference curves and Utility Function
I will try to understand the theory about indifference curves correctly.
I have given a Utility function $u(x_1,x_2)=x_1x_2^2$. Then I have to illustrate the bundles that are indifference with the bundle $(x_1,x_2)=(2,4)$.

I think I the get the line $x_2(x_1)=(\frac{u}{x_1})^{1/2}$ for the indifference curve.
If we find $MRS(2,4)=-\frac{\frac{1}{2}x_1}{\frac{1}{2}x_2}=-\frac{1}{2}$.
I'm not quite sure how to plot that? I have not any values of u. Can someone help me?

答:
You have $u(x_1,x_2)=x_1x_2^2$. The indifference curves are the level sets of that function, that is, all values of $x_1,x_2$ such that $u(x_1,x_2)=K$ for some constant $K.$  In your particular problem, you want all $x_1,x_2$ such that $x_1x_2^2=2\times 4^2=32.$   don't know the name of that curve, but it should be possible to plot by hand or program.

Calculating the MRS is a good idea, because with the data from the problem you have both a point on the indifference curve and its slope at that point.  Of course in general the slope is $-\frac{1}{2}\frac{x_2}{x_1}$ so that can give you an idea of how the slope varies.

Also note that this is just plotting out indifference curves, so prices have not entered the questions yet, so we don't have to worry about them for this question, which is just about preferences.


问:
Flux through circle
Vector field

Question: The flux through a circle of r = 1 and center (0,0) (counterclockwise)?

答:
Flux is defined as

$$\int_{|r|=1} F\cdot d\vec S$$

Here, $d\vec S$ is an infinitesimal piece of the disc, pointing in the normal direction in counterclockwise orientation. Can you think of what $d\vec S$ would be? You may want to choose an appropriate coordinate system.

Title: Flux through circle


问:
proving with sequences that in a compact space $\bigcap_{n=1}^{\infty }F_{n}\neq \varnothing $
Let $(X,d)$ be a compact metric space then every descending sequence $F_{n}$ non-empty and closed sunsets of $X$ has $$\bigcap_{n=1}^{\infty }F_{n}\neq \varnothing. $$

my solution :

(i know there is a different one where you take $X=\bigcup_{n=1}^{\infty }F_{n}^{c}$ but my first thought was this)

since $F_{n}$ non-empty and closed and also $F_{n}\supseteq F_{n+1}$ $\Rightarrow $ $\bigcap_{n=1}^{m }F_{n}\neq \varnothing $

now  since $X$ is compact and $F_{n}$  closed we know that $F_{n}$ is also compact therefore exist $x_{(1)_{k_{n}}} \in F_{1}$

$x_{(1)_{k_{n}}}\rightarrow x_{1} \in F_{1}$

$x_{(2)_{k_{n}}}\rightarrow x_{2} \in F_{2}$  and $ x_{2} \in F_{1}$

$ \vdots $

If I continue like that  there will be at least one   element let say it $x$  in $\bigcap_{n=1}^{\infty }F_{n} $ and we showed that its non-empty.

Is this proof mathematically ok? Or this isn't correct(please explain).

Question: how many elements can be on the $\bigcap_{n=1}^{\infty }F_{n} $ infinite ,
finite or just one?

答: Hint : Take an element $x_k$ in each of the $F_k$. The sequence $(x_k)_{k \in \mathbb{N}^*}$ is a sequence in a compact, therefore there exists $\varphi : \mathbb{N} \rightarrow \mathbb{N}$ strictly increasing such that $(x_{\varphi(k)})_{k \in \mathbb{N}^*}$ converges to an element $x \in X$. Try to show that $$x \in \bigcap_{i=1}^{\infty} F_i$$


问:
Dini derivatives of Dirichlet function
I'm hoping someone can verify my answers. I have that for $x\in\mathbb{Q}$ we have $$D^+f(x)=\limsup_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\limsup_{h\to 0^+}\frac{f(x+h)-1}{h}$$ and since $\mathbb{Q}$ is dense in $\mathbb{R}$, $(x+h)$ is rational as $h\to 0^+$ and we have $$D^+f(x)=\limsup_{h\to 0^+}\frac{1-1}{h}=0$$  Similarly for $D^-f(x)$ we have $$D^-f(x)=\limsup_{h\to 0^+}\frac{f(x)-f(x+h)}{h}=\limsup_{h\to 0^+}\frac{1-1}{h}=0$$And for $D_+f(x)$ $$D_+f(x)=\liminf_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\liminf_{h\to 0^+}\frac{0-1}{h}=-\infty$$ Finally for $D_-f(x)$ we have $$D_-f(x)=\liminf_{h\to 0^+}\frac{f(x)-f(x+h)}{h}=\liminf_{h\to 0^+}\frac{1-0}{h}=\infty$$ Now for $x\in\mathbb{R}\setminus\mathbb{Q}$ we have that $f(x)=0$ and we have $$D^+f(x)=\limsup_{h\to 0^+}\frac{1-0}{h}=\infty$$ $$D^-f(x)=\limsup_{h\to 0^+}\frac{0-1}{h}=-\infty$$ $$D_+f(x)=\liminf_{h\to 0^+}\frac{0-0}{h}=0$$ $$D_-f(x)=\liminf_{h\to 0^+}\frac{0-0}{h}=0$$ I am having trouble convincing myself of the effect (and proper explanation) of $\liminf$ vs $\limsup$ and their impact on the derivates. Is this correct? Could someone help me better conceptually understand what is going on?

Thanks

答:
First comment

Your definition for the various $D^+, D_+, \dots$ are not the right ones. For example $D^-f(x)=\limsup_{h\to 0^+}\frac{f(x)-f(x-h)}{h}$ and not $D^-f(x)=\limsup_{h\to 0^+}\frac{f(x)-f(x+h)}{h}$. See Dini derivatives for the correct definitions.

Second comment

You got some right results in term of values. However, the way you did it is not the best one and can lead you to wrong results when applied to others functions.

Let's take the example of $D^+f(x)$ when $x \in \mathbb Q$ where you wrote $$D^+f(x)=\limsup_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\limsup_{h\to 0^+}\frac{1-1}{h}.$$

This is not correct as $f(x+h)$ depends on whether $x+h$ is rational or not. What you should say is that for any $\epsilon \gt 0$, $\{f(x+h) \mid 0 \lt h \lt \epsilon\} = \{0,1\}$ and therefore

$$\left\{\frac{f(x+h)-f(x)}{h} \mid 0 \lt h \lt \epsilon\right\} = \{0,-1/h\}.$$

As for $h \gt 0$ we have $\sup \{0,-1/h\} = 0$, we can conclude that

$$D^+f(x)=\limsup_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=0.$$

Title: Dini derivatives of Dirichlet function


问:
What is $(\operatorname{monad}(0), \leq)$ isomorphic to?
Suppose, given $\epsilon \in \operatorname{monad}(0)$ and $\epsilon \neq 0$, is it true for each $x \in \operatorname{monad}(0)$, $$x = \sum_{r_i \in s \subset \Bbb R, s \text{ is finite}}a_i \epsilon^{r_i}$$
$a_i \in \Bbb{R}$ is a constant coefficient.

Let $S$ be a subset of $\Bbb {R^{\Bbb R}}$ such that for all $f \in S$, $f(x)= 0$ for all but finite elements in $\Bbb R$

If so, it seems to me $(\operatorname{monad}(0), \leq)$ is isomorphic to $S$ with a lexicographic order.

答:
Suppose that $\epsilon$ is a positive infinitesimal. In the ultrapower construction let $\langle x_k:k\in\omega\rangle$ represent $\epsilon$; I’ll denote this relationship by $\epsilon=\langle x_k:k\in\omega\rangle_{\mathscr{U}}$. Without loss of generality we may assume that $x_k>0$ for all $k\in\omega$. Then $\epsilon^n=\langle x_k^n:k\in\omega\rangle_{\mathscr{U}}$ for each $n\in\omega$. For $n\in\omega$ let

$$y_n=\frac12\min\left\{x_k^n:k\le n\right\}\;;$$

then for each $n\in\omega$ we have $y_k<x_k^n$ for all $k\ge n$, so if $\delta=\langle y_k:k\in\omega\rangle_{\mathscr{U}}$, then $\delta<\epsilon^n$ for all $n\in\omega$. Thus, $\delta$ is smaller than any finite real combination of real powers of $\epsilon$: it’s infinitesimal compared with $\epsilon$.


问:
Question about $\delta$'s in the $\epsilon$-$\delta$ definition of continuity
If a function $f:M\to\mathbb{R}$ is continuous at point $x_0$ we know that for an arbitrary $\epsilon>0$ there exists a $\delta>0$ such that for all $x\in M$ and $|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$. Let's call those $\epsilon$ and $\delta(\epsilon)$ a pair, $(\epsilon,\delta)$.

What happens if I shrink the $\delta$? Does this imply that $|f(x)-f(x_0)|$ also shrinks?

My intuition says that we can't make any claim on the behaviour of $|f(x)-f(x_0)|$. Sure if $\delta$ attains a value which is very small then it will be smaller than another $\delta'$ which belongs to a pair $(\epsilon',\delta')$ where the $\epsilon'<\epsilon$. But if I shrink $\delta$ only a bit what happens then? How do I argue in a formal way?

答: If you've not yet played around with what pairs delta and epsilon work, you might want to try something like this: https://www.desmos.com/calculator/iejhw8zhqd which can give you a feel for what happens as you shrink delta.  The question you are asking is basically "if I narrow my x-window, does that necessarily shrink the y-window?" the answer could be yes or no, but depends on the behaviour of your function.  It might be instructive to try it for a function like f(x) = sin(1/x)  (with f(0) = 0) to test your intuition.</p>


问:
Proof of equality for triangle inequality
I am trying to prove the following statement here:

Let $(V,\left<\cdot,\cdot\right>)$ be an inner product space and let $\vec{u},\vec{v}$ be vectors in $V$. Under the assumption that $\vec{u}\neq0$ show that $||\vec{u}+\vec{v}||=||\vec{u}||+||\vec{v}||$ if an only if $\vec{v}=c\vec{u}$ for some non-negative scalar $c$.

Here's my attempt. I use the general proof of the triangle inequality here.

$||\vec{u}+\vec{v}||^2$

$=\left<\vec{u}+\vec{v},\vec{u}+\vec{v}\right>$

$=\left<\vec{u},\vec{u}\right>+2\left<\vec{u},\vec{v}\right>+\left<\vec{v},\vec{v}\right>\leq\left<\vec{u},\vec{u}\right>+2|\left<\vec{u},\vec{v}\right>|+\left<\vec{v},\vec{v}\right>$ (By Cauchy-Scharwz)

So,

$||\vec{u}+\vec{v}||^2\leq\left<\vec{u},\vec{u}\right>+2|\left<\vec{u},\vec{v}\right>|+\left<\vec{v},\vec{v}\right>$

$||\vec{u}+\vec{v}||^2\leq\left<\vec{u},\vec{u}\right>+2||\vec{u}||\hspace{1mm}||\vec{v}||+\left<\vec{v},\vec{v}\right>$

$||\vec{u}+\vec{v}||^2\leq(||\vec{u}||+||\vec{v}||)^2$

$||\vec{u}+\vec{v}||\leq(||\vec{u}||+||\vec{v}||)$

I have a few comments here though. From what I understand, if I can turn the inequality parts into an equality, I am essentially done. I'm not really sure how to do that though. If I let $\vec{v}=c\vec{u}$, then I get:

$=\left<\vec{u},\vec{u}\right>+2\left<\vec{u},c\vec{u}\right>+\left<c\vec{u},c\vec{u}\right>$

but I am stuck at this point.

I know that:

$\left<\vec{u},\vec{u}\right>+2\left<\vec{u},c\vec{u}\right>+\left<c\vec{u},c\vec{u}\right>\leq\left<\vec{u},\vec{u}\right>+2|\left<\vec{u},c\vec{u}\right>|+\left<c\vec{u},c\vec{u}\right>$

becomes:

$\left<\vec{u},\vec{u}\right>+2\left<\vec{u},c\vec{u}\right>+\left<c\vec{u},c\vec{u}\right>=\left<\vec{u},\vec{u}\right>+2|\left<\vec{u},c\vec{u}\right>|+\left<c\vec{u},c\vec{u}\right>$ if and only if $c\vec{u}$ and $\vec{v}$ are non negative real numbers.

So at this point, I've taken care of the Cauchy-Schwarz inequality portion since one vector is a scalar multiple of the other.

How can I show that $\vec{u}$ and $c$ are non-negative though? If $\vec{u}$ and $c$ are non-negative, that immediately implies $\vec{v}$ is non-negative.

Can someone guide me with this part here? Thank you!

答:
Suppose the equality is true for $u$, $v$. then, squaring both sides of the equation, using bilinearity of the inner product  and cancelling out terms gives $\left=|u||v|$. As $\left<x,y\right>=|x||y|cos\theta$ for all $x,y\in V$, we must have $\theta=0$, so there is $c\geq0$ such that $v=cu$. Conversely, if $v=cu$, with $c\geq 0$, we have $|u+cu|=(1+c)|u|=|u|+|cu|$ as we wanted.

Note that, in general, expressions such as "$cu$ and $v$  are non negative real numbers" make no sense, as $u$ and $v$ are vectors.

EDIT: if $\left=|u||v|$, then $\left<\frac{u}{|u|}-\frac{v}{|v|},\frac{u}{|u|}-\frac{v}{|v|}\right>=1+1-2=0$, so $\frac{u}{|u|}-\frac{v}{|v|}=0$, thus $v=cu$, where $c=\frac{|v|}{|u|}$.

Body:
Suppose the equality is true for $u$, $v$. then, squaring both sides of the equation, using bilinearity of the inner product  and cancelling out terms gives $\left<u,v\right>=|u||v|$. As $\left<x,y\right>=|x||y|cos\theta$ for all $x,y\in V$, we must have $\theta=0$, so there is $c\geq0$ such that $v=cu$. Conversely, if $v=cu$, with $c\geq 0$, we have $|u+cu|=(1+c)|u|=|u|+|cu|$ as we wanted.

Note that, in general, expressions such as "$cu$ and $v$  are non negative real numbers" make no sense, as $u$ and $v$ are vectors.

EDIT: if $\left<u,v\right>=|u||v|$, then $\left<\frac{u}{|u|}-\frac{v}{|v|},\frac{u}{|u|}-\frac{v}{|v|}\right>=1+1-2=0$, so $\frac{u}{|u|}-\frac{v}{|v|}=0$, thus $v=cu$, where $c=\frac{|v|}{|u|}$.


问:
Demonstration using the Pigonhole principle
I was thinking about the following problem:

Let $n\in\mathbb N$ be odd. If I have a symmetric matrix in $M_n(\mathbb{N})$, i.e. a square symmetric matrix of size $n$, for which each column and each row consists of all numbers between $1$ and $n$, then the diagonal consists of all the numbers from $1$ to $n$.

I want to prove this using the pigeonhole principle. Thanks for helping me.

答:
I am not sure that I understood correctly, but wouldn't
$$
\left(\begin{array}{ccc}
1&2&3\\
3&1&2\\
2&3&1
\end{array}\right)
$$
be a counterexample? All the rows and columns have all the numbers from $1$ to $3$, yet the diagonal has only $1$s.

Adding a proof for the non-existence of such a matrix, when the requirement of symmetry is added.

Each number must occur an odd number of times. But because of the symmetry, all the numbers appear off the diagonal an even number of times. Therefore they all appear in the diagonal an odd number of times, i.e. at least once. Because all there are $n$ numbers appearing on the diagonal, they must all appear there exactly once.


问:
a (bigger than 1) is always bigger than its roots
how can I formally proof that for a > 1: $$ a> \sqrt a > \sqrt[3]a > \sqrt[4]a ...  $$?
Can someone help? ;)

答:
Hint: Use logarithms to show that
$$a^{\frac{1}{n}}>a^{\frac{1}{n+1}}$$


问:
Lattice points on a hyperbola.
A lattice point in the $xy$-plane is a point both of whose coordinates are integers (not necessarily positive). How many lattice points lie on the hyperbola $x^2-y^2=17$?

I think the answer should be $4$, because $x^2-y^2 = (x+y)(x-y) = 17$. $17$ has $4$ factors: $1, 17, -1,$ and $-17$. But I don't know if these numbers actually work.

答:
Yes, they work:

$x+y=17$ and $x-y=1\implies 2x=18\implies x=9\implies y=8$

$x+y=1$ and $x-y=17\implies 2x=18\implies x=9\implies y=-8$

$x+y=-17$ and $x-y=-1\implies 2x=-18\implies x=-9\implies y=-8$

$x+y=-1$ and $x-y=-17\implies 2x=-18\implies x=-9\implies y=8$

Title: Lattice points on a hyperbola.


答:
To show $\sqrt[n]a>\sqrt[n+1]a$, raise both sides to $n(n+1)$ to obtain
$$
a^{n+1}>a^n
$$
and prove the latter inequality from $a>1$.

Title: a (bigger than 1) is always bigger than its roots

Body:
how can I formally proof that for a > 1: $$ a> \sqrt a > \sqrt[3]a > \sqrt[4]a ...  $$?
Can someone help? ;)


答:
Note: This answer was for the original question, which inadvertently omitted the requirement that the matrix be symmetric.

It’s not true: consider, for instance, the matrix

$$\begin{bmatrix}1&2&3\\3&1&2\\2&3&1\end{bmatrix}\;.$$

You don’t even have to work that hard: with $n=2$ already you have

$$\begin{bmatrix}1&2\\2&1\end{bmatrix}\;.$$

Title: Demonstration using the Pigonhole principle

Body:
I was thinking about the following problem:

Let $n\in\mathbb N$ be odd. If I have a symmetric matrix in $M_n(\mathbb{N})$, i.e. a square symmetric matrix of size $n$, for which each column and each row consists of all numbers between $1$ and $n$, then the diagonal consists of all the numbers from $1$ to $n$.

I want to prove this using the pigeonhole principle. Thanks for helping me.


问:
Norm of $ T:(C([0,1]),||\cdot||_\infty)\rightarrow\mathbb{R}$ where $Tf=\sum_{k=1}^n a_kf(t_k)$?
I need to calculate the operator Norm of the linear operator defined as:
$$ T:(C([0,1]),||\cdot||_\infty)\rightarrow\mathbb{R} \text{ where } Tf=\sum_{k=1}^n a_kf(t_k)$$
for $0\leq t_1<t_2<...<t_n\leq 1$ and $a_1,...,a_n\in \mathbb{R}$.

I have been able to show that $||T||\geq\left|\sum_{k=1}^n a_k\right|$ and $||T||\leq \sum_{k=1}^n|a_k|$ but I don't seem to able to bound it further than that. Would appreciate any help. Thank you.

答:
It might help to separate $T$ into the composition of two simpler maps.

Note that the map $s:C[0,1] \to \mathbb{R}^n$ defined by
$s(f) = (f(t_1),...,f(t_n))$ has norm one (using the $l_\infty$ norm on $\mathbb{R}^n$) and is surjective.
Furthermore, for any $y$ with $\|y\|_\infty =1$ it is easy to construct (by interpolation & extrapolation) some $f \in C[0,1]$ with $\|f\| = 1$ such that $s(f) = y$.

Let the operator $\tau:\mathbb{R}^n \to \mathbb{R}$ be given by $\tau(y) = \sum_k a_k y_k$. It is straightforward to show that $\|\tau\| = \|a\|_1$ and there is some
$y \in \mathbb{R}^n$ such that $\|y\|_\infty = 1$ and $\tau(y) = \|a\|_1$.

Note that $T = \tau \circ s$.

Hence $\|T\| \le \|s\| \|\tau\| = \|a\|_1$.
Suppose $\tau(y) = \|a\|_1$ with $\|y\|_\infty = 1$ and $s(f) = y$ where $\|f\| = 1$, then
$Tf = \tau(s(f))  = \tau(y) = \|a\|_1$ and so $\|T\| = \|a\|_1$.


问:
Conservation Laws: Difference and Reasonability Weak solutions, Integral solutions, distributional solutions
Suppose we have for $T\in\mathbb{R}_{>0}$ a conservation law of the following general form
\begin{align}
\dot{u}(t,x)+(a(t,x,u(t,x)))_{x}&=0 && (t,x)\in(0,T)\times\mathbb{R}\\
u(0,x)&=u_{0}(x) &&x\in\mathbb{R}
\end{align}
with $u_{0}$ and $a$ given and $a$ sufficiently smooth. We call it initial value problem (IVP).

If one is interested in defining non-classical solutions, there are several different definitions of "weak" solutions:

$u\in C([0,T];L^{1}_{\text{loc}}(\mathbb{R}))$ is called weak solution of the IVP, iff
\begin{equation}
\int_{0}^{T}\int_{\mathbb{R}}u(t,x)\phi_{t}(t,x)+a(t,x,u(t,x))\phi_{x}(t,x)\ d x\ dt+\int_{\mathbb{R}}u_{0}(x)\phi(0,x)\ dx=0\quad \forall \phi\in C^{1}_{\text{c}}((-1,T)\times\mathbb{R}).
\end{equation}
In my understanding this is quite a nice function space, since we can indeed evaluate $u$ at every time $t$ (reasonable for a kind of time evolution and the semi group property) and in particular for $t=0$ as a $L^{1}_{\text{loc}}$ function.

$u$ is called distributional solution if $u$ is a measureable function on $(0,T)\times\mathbb{R}$ and satisfies the integral equality in item 1.

$u$ is called integral solution, if $u\in L^{\infty}((0,T)\times\mathbb{R})$ is satisfying the integral equality in item 1.

Now we are interested in existence, uniqueness, and the corresponding entropy conditions. As I have seen, in the literature there is not a very precise statement of which kind of weak solution (sometimes every of these 3 types of solutions are called weak solutions) is necessary to apply entropy conditions and to get kind of uniqueness and existence. Even more, there seems to be some mixing between them.

So, if anybody has a good and mathematical rigorous reference concerning the different types of weak solutions, it would be great.

In addition, when will these three types of "weak" solutions be identical?
Could it make sense to consider solutions which do not change continuously in time if one measures the space in the $L^{p}$ topology?

I am thankful for every comment or help and thank everyone for reading in advance!

Alex

答: Any weak solution is a distributional solution.. However when we are dealing with initial value problem, it makes more sense to talk about weak solution as one would expect in some sense solution takes the initial data ($u(\cdot,t) \rightarrow u_0(\cdot)$  as $t \rightarrow 0$ in $L^1_{loc}$). Since it is an evolution equation it is natural to seek for a solution which is continuous in time in some norm...


问:
Why is $x^4-3x^2+18$ irreducible over the $3$-adics?
Let $f = x^4 - 3x^2 + 18 \in \mathbb{Q}_3[x]$. Since there is an LMFDB page of an extension defined by this polynomial, I assume that $f$ is irreducible. Could you verify if this is true and also why that is (not) the case?

It is not an Eisenstein polynomial (as $3^2$ divides $18$, the constant coefficient of $f$). I also tried the substitution $y = x^2$ and asked myself if $g = y^2 - 3y + 18$ is irreducible. Its reduction mod $3$ is $y^2$ which is not irreducible, so I cannot say whether $g$ is irreducible or not.

答:
The polynomial $x^2 - 3 x + 18$ is irreducible over $\mathbb{Q}_3$.  Consider the quadratic extension  $F=\mathbb{Q}_3[\sqrt{-7}]$ over which it decomposes as $(x-x_1)(x-x_2)$.  So we have
$$x^4 - 3 x^2 + 18= (x^2-x_1)(x^2 - x_2)$$
Now it is enough to see that $(x^2 - x_1)$, $(x^2 -x_2)$ are irreducible over $F$, equivalently $x_i$ are not squares in $F$.

Consider the extension of the valuation $v_3$ to $F$. We have $v_3(x_i) = 1$. It is easy to see that there are no elements in $F$ of value $\frac{1}{2}$.

Note: the extension of the valuation $v_3$ to $F$ is $v(a + b \sqrt{-7}) = \frac{1}{2}v_3(a^2 + 7 b^2)$


问:
Find the Infimum and supremum of M set
$$M = \bigcap_{n\in N}{I_n},$$
$$I_n = \bigcup_{k = 1}^{3^n - 2} \left(\frac{k}{3^n},\frac{k+1}{3^n}\right)$$

答:
Hint

First prove that if $M=\bigcap_{n\in\Bbb N}I_n$, then
$$
\sup M=\inf_n\sup I_n\\
\inf M=\sup_n\inf I_n
$$

Title: Find the Infimum and supremum of M set


问:
Abstract Algebra - Dummit Exercise 3.1.1

This question in Abstract Algebra by Dummit seemed ill-posed to me:

Let $\varphi: G \rightarrow H$ be a homomorphism and let $E$ be a subgroup of $H .$ Provethat $\varphi^{-1}(E) \leq G$ (i.e., the preimage or pullback of a subgroup under a homomorphism is a subgroup). If $E \trianglelefteq H$ prove that $\varphi^{-1}(E) \trianglelefteq G .$ Deduce that ker $\varphi \trianglelefteq G$

In this problem, how do we know $\phi^{-1}(E)$ is well defined? I.e., how do we know there is no $e \in E, \; s.t. \; \phi(g)\neq e \;\; \forall g \in G$? Homomorphisms don't have to be surjective, so I'm thinking $E$ could inherit such and element from $H$.

Edit: Here is my attempted proof:

First we prove some things about $\phi^{-1}$. Let $\phi(x),\phi(y) \in E$:

\begin{align*}
    \phi^{-1}(\phi(x)\phi(y))= \phi^{-1}(\phi(xy)) = xy = \phi^{-1}(\phi(x))\phi^{-1}\phi((y))
\end{align*}

Thus $\phi^{-1}$ is also a homomorphism. Also:

\begin{align*}
    \phi^{-1}(x)^{-1} &= \phi^{-1}(1_G)\phi^{-1}(x)^{-1}\\
    &= \phi^{-1}(x^{-1}x)\phi^{-1}(x)^{-1} \\
    &= \phi^{-1}(x^{-1}) \phi^{-1}(x) \phi^{-1}(x)^{-1}\\
    &= \phi^{-1}(x^{-1})
\end{align*}

Armed with this knowledge, let $e, e' \in E$, $\phi^{-1}(e),\phi^{-1}(e') \in \phi^{-1}(E)$:

\begin{align*}
    \phi^{-1}(e)\phi^{-1}(e')^{-1} &= \phi^{-1}(e(e')^{-1})\\ &
    = \phi^{-1}(e^{*}) \in \phi^{-1}(E)
\end{align*}

Thus $\phi^{-1}(E) \leq G$ .

If $E \trianglelefteq H$, then we know $N_H(E) = H, \; hEh^{-1} \forall h \in H$. Let $e \in E$, $\phi^{-1}(e)\in \phi^{-1}(E)$:

\begin{align*}
    g\phi^{-1}(e)g^{-1} &= \phi^{-1}(h)\phi^{-1}(e)\phi(h)^{-1}\\
    &= \phi^{-1}(h e h^{-1})\\
    &= \phi^{-1}(e^*) \in \phi^{-1}(E)
\end{align*}

Since ${1} \trianglelefteq H$, and $\phi^{-1}(1) = \ker \phi$, $\ker\phi \trianglelefteq G$.

Is this correct?

答:
Given a set $A$, let $\mathcal{P}(A)$ denote the "power set" of $A$, which is the set of all subsets of $A$. So, $X\in\mathcal{P}(A)$ if and only if $X\subseteq A$. In particular, note that $\emptyset\in \mathcal{P}(A)$ and $A\in\mathcal{P}(A)$. There is also a canonical injection from $A$ to $\mathcal{P}(A)$ given by $a\mapsto\{a\}$ for any $a\in A$.

Power sets are one way of thinking about the notation you are confused by, based on the following: a function $f:A\rightarrow B$ between sets $A,B$ induces two maps $f^\to:\mathcal{P}(A)\rightleftarrows\mathcal{P}(B):f^\gets$. Can you think about what the action of these maps $f^\to$ and $f^\gets$ might be?

 For $X\in\mathcal{P}(A)$ – ie for $X\subseteq A$ – we define $f^\to(X)=\{f(x):x\in X\}$. Since the range of $f$ is $B$, we have $f^\to(X)\subseteq B$ and thus $f^\to(X)\in\mathcal{P}(B)$ as desired.

 Conversely, for $Y\in\mathcal{P}(B)$ – ie for $Y\subseteq B$ – we define $f^\gets(Y)=\{a\in A:f(a)\in Y\}$. Clearly this a subset of $A$, and so $f^\gets(Y)\in\mathcal{P}(A)$ as desired. Note that there can be elements in $Y$ that are not in the image of $f$! Indeed, even if every element of $Y$ is outside the image of $f$, this construction is still just fine, and we just get $f^\gets(Y)=\emptyset\in\mathcal{P}(A)$. So $f$ absolutely does not need to be surjective for this construction to work.

These induced functions will always exists, without any conditions required on $f$. Now, the issue for you in this question is that it is common practice in mathematics to abuse notation slightly and refer to what we are calling $f^\to$ and $f^\gets$ as $f$ and $f^{-1}$, respectively. So, in particular, given a subset $X\subseteq A$ we denote $f(X)=\{f(x):x\in X\}\subseteq B$, and given a subset $Y\subseteq B$ we denote $f^{-1}(Y)=\{a\in A:f(a)\in Y\}\subseteq A$.

These maps do not act on $A$ and $B$, but on $\mathcal{P}(A)$ and $\mathcal{P}(B)$. It is unfortunate notation, for a number of reasons. For instance – and this is the source of your confusion – $f^\gets$ exists even if $f$ is not invertible. So the fact that $f^\gets$ exists says nothing about whether $f$ is invertible (or even surjective) or not. However, this abuse of notation is completely standard in much of mathematics, so you'll have to get comfortable with it.

(Another alternative to the clash of notation is as follows. For a subset $X\subseteq A$, we define $f[X]=f^\to(X)$ and, for a subset $Y\subseteq B$, we define $f^{-1}[Y]=f^\gets(Y)$. Using these square brackets when we are referring to subsets mitigates the clash of notation in question. This convention is fairly common, so you might want to adopt it for future use.)

Finally we can return to your question. What D&F call $\varphi^{-1}(E)$ is just the set $\varphi^\gets(E)=\{g\in G:\varphi(g)\in E\}$. There absolutely may be elements of $E$ not in the image of $\varphi$; this doesn't change anything. As an illustrative example, using this notation we have that $\ker(\varphi)=\varphi^{-1}(\{1\})$, where $1$ denotes the identity element of $H$; does this make sense?

So, armed with this correct definition in mind, do you think you can now solve the problem? (Note that your current proposed solution is incorrect, because it is assuming that $\varphi$ is bijective onto $E$. What D&F are calling $\varphi^{-1}$ is really $\varphi^\gets$; it acts on subsets of $H$ and is guaranteed to exist. On the other hand, what you denote by $\varphi^{-1}$ in your post acts on elements of $H$, and is not guaranteed to exist by any means.)

This might seem like a long-winded digression, but this is a very common convention in mathematics, and you may as well start getting used to it now!


问:
In a PID $R$, given $a,b\in R$, what are the ideals $\langle a\rangle,\langle b\rangle$ also generated by?
Let $R$ be a principal ideal domain and let $a$ and $b$ be two nonunit elements in $R$ then ideal generated by $a$ and $b$ is also generated by

$a+b$

$ab$

$\gcd(a,b)$

$\text{lcm}(a,b)$

I am not sure but I think that answer is 3.

Let $I=\langle a,b \rangle$ $=\{ax+by:x,y\in R\}$.  Let $d=gcd(a,b)$. Since $R$ is a PID we get $d=ax+by$ for some $x,y\in R$. So $\langle d \rangle \subseteq I$ as $d$ is a divisor of $a$ and $b$ clearly $a,b \in \langle d \rangle $ So $I \subseteq \langle d \rangle$. Hence $I=\langle d \rangle$. Is it correct?

答: Yes, it is correct. In fact, some authors go a step further and define the gcd of $a$ and $b$ of a ring to be a generator of the smallest principal ideal containing both $a$ and $b$.


问:
Closed form for an integral involving the incomplete Gamma function?
Let $\alpha>1$, $K>1$ and $n \in \mathbb{N}^+$. What is a closed form solution to a tough integral? $$I(\alpha,K,n)=-\int_{-\infty }^{\infty } \frac{2 i e^{-i K u} (K u-i) \Bigl[\alpha \, (-i u)^{\alpha } \,\Gamma (-\alpha ,-i u)\Bigr]^n}{u^2} \, du,$$
where $\Gamma(.,.)$ is the incomplete Gamma function:
$\Gamma (a,z)=\int _z^{\infty }d t\, t^{a-1} e^{-t}$.

I tried all manner of substitutions and various combinations of integration by parts.

答:
This is a partial answer, linked with the calculations via multiple convolution.

$\color{brown}{\textbf{Representation of the integral.}}$

Is known that

$$s^p\Gamma(-p,s) = E_{p+1}(s),$$

$$\dfrac1{2\pi i} \int\limits_{-i\infty}^{+i\infty}F(s)\,e^{ts}\,\text ds
= \mathcal L^{-1}_{s\mapsto t}(F(s)),$$

$$\mathcal L^{-1}_{s\mapsto t}(e^{s} E_{p+1}(s)) = \dfrac1{(t+1)^{p+1}},$$

where $\;E_{p+1}(s)\;$ is the exponential integral $\;E\;$ and $\;\mathcal L^{-1}_{s\mapsto t}\;$ is the inverse Laplace transform from $\;s\;$ to $\;t.$

Let $\;-iu =s,\quad K=n+t,\;$ then
$$I(\alpha,n+t,n) = \dfrac{4\pi}{2\pi i}\int\limits_{-i\infty}^{+i\infty}\dfrac{(n+t)s-1}{s^2}\big(\alpha s^\alpha\Gamma(-\alpha, s)\big)^ne^{(n+t)s}\text ds$$
$$ = \dfrac{4\pi\alpha^n}{2\pi i}\int\limits_{-i\infty}^{+i\infty}\dfrac{(n+t)s-1}{s^2}\big(e^s\,E_{\alpha+1}(s)\big)^n\,e^{ts}\,\text ds
=4\pi\alpha^n\mathcal L^{-1}_{s\,\mapsto t}\left(\dfrac{(n+t)s-1}{s^2}\big(e^s\,E_{\alpha+1}(s)\big)^n\right),$$
$$=4\pi\alpha^n\left((n+t)\int\limits_0^t G_n(\alpha,\lambda)\,\text d\lambda
-\int\limits_0^t\int\limits_0^\lambda G_n(\alpha,\mu)\,\text d\mu\,\text d\lambda \right)$$
$$=4\pi\alpha^n\left((n+t)\int\limits_0^t G_n(\alpha,\lambda)\,\text d\lambda
-\int\limits_0^t\int\limits_0^\lambda G_n(\alpha,\mu)\,\text d\mu\,\text d(n+\lambda) \right)$$
$$\;\overset{\text{IBP}}{=\!=}\; 4\pi\alpha^n \int\limits_0^t(n+\lambda)G_n(\alpha,\lambda)\,\text d\lambda,$$
$$I(\alpha,n+t,n)=4\pi\alpha^n\int\limits_0^t(n+\lambda)
G_n(\alpha,\lambda)\,\text d\lambda,\tag1$$
where
$$G_n(\alpha,t) = \underbrace{\dfrac1{(t+1)^{\alpha+1}}*\dfrac1{(t+1)^{\alpha+1}}*\dots*\dfrac1{(t+1)^{\alpha+1}}}_n\tag2$$
is the multiple convolution.

$\color{brown}{\mathbf{Case\; n = 1.}}$

\begin{cases}
G_1(\alpha,t)=\dfrac1{(t+1)^{\alpha+1}}\\
\big(G_1(\alpha,t)\big)'_t = -\dfrac1{(\alpha+1)(t+1)^{\alpha+2}}\\
\int\limits_0^t G_1(\alpha,t)\,\text dt
=\dfrac1\alpha \left(1 - \dfrac1{(t+1)^{\alpha}}\right).\tag3
\end{cases}
$$I(\alpha,t+1,1) = 4\pi\alpha\int\limits_0^t \dfrac{\text d\lambda} {(\lambda+1)^{\alpha}}
= \dfrac{4\pi\alpha}{\alpha-1}\left(1-\dfrac1{(t+1)^{\alpha-1}}\right),$$
$$I(\alpha,K,1) = \dfrac{4\pi\alpha}{\alpha-1}\left(1-\dfrac1{K^{\alpha-1}}\right).\tag4$$

$\color{brown}{\mathbf{Case\; n = 2.}}$

$$G_2(\alpha,t) = G_1(\alpha,t) * \dfrac1{(t+1)^{\alpha+1}}
=\int\limits_0^t G_1(\alpha,u)\,\dfrac{\text du}{(t-u+1)^{\alpha+1}},\tag5$$

$$I(\alpha,t+2,2) = 4\pi\alpha^2\int\limits_0^t \int\limits_0^\lambda \dfrac{(\lambda+2) G_1(\alpha,\mu)}{(\lambda-\mu+1)^{\alpha+1}}\,\text d\mu\,\text d\lambda$$
$$= 4\pi\alpha^2\int\limits_0^t \int\limits_0^\mu \dfrac{\big((\lambda-\mu+1) + (\mu+1)\big) G_1(\alpha,\mu)}{(\lambda-\mu+1)^{\alpha+1}}\,\text d\lambda\,\text d\mu$$
$$= 4\pi\alpha^2\int\limits_0^t \int\limits_0^\mu \left(
\dfrac{G_1(\alpha,\mu)}{(\lambda-\mu+1)^{\alpha}}
+\dfrac{G_1(\alpha-1,\mu)}{(\lambda-\mu+1)^{\alpha+1}}
\right)\,\text d\lambda\,\text d\mu$$
$$= -4\pi\alpha^2\int\limits_0^t \left(
\dfrac{G_1(\alpha,\mu)}{(\alpha-1)(\lambda-\mu+1)^{\alpha-1}}
+\dfrac{G_1(\alpha-1,\mu)}{\alpha(\lambda-\mu+1)^{\alpha}}
\right)\bigg|_0^\mu\,\text d\mu$$
$$= \dfrac{4\pi\alpha}{\alpha-1}\int\limits_0^t
\big(\alpha G_1(\alpha,\mu)(1-G_1(\alpha-2,-\mu))
+(\alpha-1)G_1(\alpha-1,\mu)(1-G_1(\alpha-1,-\mu))\big)
\,\text d\mu$$
$$= \dfrac{4\pi\alpha}{\alpha-1}\int\limits_0^t
(1-G_1(\alpha-2,-\mu))\,\text d G_1(\alpha-1,\mu)$$
$$- 4\pi\alpha \int\limits_0^t G_1(\alpha-1,\mu)(1-G_1(\alpha-1,-\mu)))
\,\text d\mu$$
$$= \dfrac{4\pi\alpha}{\alpha-1}
(1-G_1(\alpha-2,-\mu))\,G_1(\alpha-1,\mu)\bigg|_0^t
- 4\pi\alpha\int\limits_0^t G_1(\alpha-1,\mu) G_1(\alpha-1,-\mu) \,\text d\mu$$
$$- 4\pi\alpha (1-G_1(\alpha-2,-t)))
+ 4\pi\alpha \int\limits_0^t G_1(\alpha-1,\mu)G_1(\alpha-1,-\mu)\,\text d\mu$$
$$= \dfrac{4\pi}{\alpha-1}(1-G_1(\alpha-2,-t))\big(\alpha G_1(\alpha-1,t)
- \alpha+1)\big),$$

$$I(\alpha,K,2) = \dfrac{4\pi}{\alpha-1}(1-G_1(\alpha-2,K-2))\big(\alpha G_1(\alpha-1,K-2)
- \alpha+1)\big),\tag6$$

$$G_2(\alpha,t)=\int\limits_0^t\dfrac{\text du}{(u+1)^{\alpha+1}(t-u+1)^{\alpha+1}}
=\int\limits_1^{t+1}\dfrac{\text du}{u^{\alpha+1}(t+2-u)^{\alpha+1}}\\[4pt]
=\dfrac1{(t+2)^{2\alpha+1}}\int\limits_1^{t+1}\dfrac{\text d\left(\frac u{t+2}\right)}{\left(\frac u{t+2}\right)^{\alpha+1}\left(1-\frac u{t+2}\right)^{\alpha+1}}
=\dfrac1{(t+2)^{2\alpha+1}}\int\limits_{\large\frac1{t+2}}^{\large\frac{t+1}{t+2}} s^{-\alpha-1}(1-s)^{-\alpha-1}\text ds,$$
$$G_2(\alpha,t) = \dfrac{\text B_{\large\frac{t+1}{t+2}}(-\alpha,-\alpha)
-\text B_{\large\frac1{t+2}}(-\alpha,-\alpha)}{(t+2)^{2\alpha+1}}.\tag7$$

Obtained results show that algebraic closed form of the given integral can exist even if the convolution has not such form.

However, I have not any closed forms for $n\ge3.$


问:
Ackermann Function: Proof that n < A(m, n) for all m, n in N
I've been stuck at this problem for a while now and can't get it solved. The prof wants me to proof n < A(m,n) for all m and n being positive Integers.

More details on the Ackermann Function:
$$(1) A(0,n) = n + 1$$
$$(2) A(m + 1, 0) = A(m, 1)$$
$$(3) A(m + 1, n + 1) = A(m, A(m + 1, n))$$

My approach was the following:

Inductionstart_m: $m=0 ;  A(0, n) = n+1$

A(0, n) = n+1 is defined and can be assumed to be true

Inductionassumption_m: $n < A(m, n)$ for one m and all n

Inductionproof_m (What we want to proof): $n < A(m+1, n)$ for all n

InductionEnd_m -->

Inductionstart_n: $n=0 ; A(m+1, 0) = A(m, 1)$ --> $1 < A(m, 1)$

Inductionassumption_n: $n < A(m+1, n)$ for one n and all m

Inductionproof_n: $n+1 < A(m+1, n+1)$ for all m

InductionEnd_n: $n+1 < A(m+1, n+1) = A(m, A(m+1, n))$
we already proofed: $n < A(m+1, n)$
Now Im stuck. I dont know how to proof $n+1 < A(m+1, n)$

If that can be proven, then the proof would be done since we know that:
$A(m+1,n) < A(m, A(m+1,n)) $
is true from the Inductionassumption_m because it has the form: $n < A(m, n)$ for one m and all n

答:
This is a case in which it seems to be easier to carry out the induction with a stronger induction hypothesis. I used a three-part hypothesis:

Induction hypothesis:

$$\begin{align*}
&(1_m)\;A(k,n)>n\text{ for }0\le k\le m\text{ and }n\ge 0\\
&(2_m)\;A(k,\ell+1)>A(k,\ell)\text{ for }0\le k\le m\text{ and }\ell\ge 0\\
&(3_m)\;A(k+1,\ell)\ge A(k,\ell+1)\text{ for }0\le k<m\text{ and }\ell\ge 0
\end{align*}$$

$(1_m)$ is what we want to prove for all $m\ge 0$; the other two clauses just make that easier. $(2)$ says that $A$ is strictly increasing in its second argument, and $(3)$ is a stronger form of the assertion that it is strictly increasing in its first argument as well.

For the induction step we want to prove $(1)_{m+1},(2)_{m+1}$, and $(3)_{m+1}$. Specifically, we want to prove that $A(m+1,n)>n$ for $n\ge 0$, that $$A(m+1,n+1)>A(m+1,n)$$ for $n\ge 0$, and that $$A(m+1,n)\ge A(m,n+1)$$ for $n\ge 0$. We do this by induction on $n$.

To get the induction on $n$ started, we have $A(m+1,0)=A(m,1)>1$, which establishes $(1)_{m+1}$ and $(3_{m+1})$ for $n=0$. Assume that $(1)_{m+1}$ and $(3_{m+1})$ hold for $n$. Then

$$\begin{align*}
A(m+1,n+1)&=A\big(m,A(m+1,n)\big)\\
&\overset{(1)}\ge A\big(m,A(m,n+1)\big)\\
&\overset{(2)}\ge A(m,n+2)\\
&\overset{(3)}>n+2\,.
\end{align*}$$

Here $(1)$ uses $(2)_m$ and the assumption that $(3)_{m+1}$ holds for $n$; $(2)$ uses $(1)_m$ to conclude that $A(m,n+1)\ge n+2$ and then $(2)_m$, and $(3)$ again uses $(1)_m$. This establishes $(1)_{m+1}$ and $(3_{m+1})$ for $n+1$. Finally,

$$\begin{align*}
A(m+1,n+1)&=A\big(m,A(m+1,n)\big)\\
&>A(m+1,n)
\end{align*}$$

by $(1)_m$, which establishes $(2)_{m+1}$. Thus, the main induction goes through to establish $(1)_m,(2)_m$, and $(3)_m$ for all $m\ge 0$.


问:
How to solve this limit of piecewise function
Basically, I am trying to solve this problem here, but I am stuck as to what to do first. My only idea is to substitute the x-values, but I do not know what to do after. What happens when they are true and false and how can I solve a? Can someone please help me. Thanks. Here is the problem</p>

答:
You appear to be saying that you do not know what a "limit" is.
$\lim_{x\to a} f(x)= \lim_{x\to a+} f(x)$
and
$\lim_{x\to a} f(x)= \lim_{x\to a-} f(x)$
where "$x\to a+$ means "using only numbers above a"
and "$x\to a-$ means "using only numbers below a".

If those two "one-sided" limits are not the same then the limit does not exist.

Here, for x<-1, f(x)= 1.  What is the limit of that as x goes to -1?  For x> -1 (but close to -1) f(x)= -x.  What is the limit of that as x goes to -1?  Are those the same?

For x< 0, but close to 0, f(x)= -x.  What is the limit of that as x goes to 0?  For x> 0, but close to 0, f(x)= -x.  What is the limit of that as x goes to 0?  Are those the same?

For x< 1, but close to 1, f(x)= -x.  What is the limit of that as x goes to 1?  For x> 1, but close to 1, f(x)= 1.  What is the limit of that as x goes to 1?  Are those the same?


问:
Norm of a functional on $C[0,1]$
Calculate the norm of the functional
$$
\phi (f) = \displaystyle\int^1_0 f(t) a(t) dt, \hspace{1cm} a \in C[0,1],
$$
on $C[0,1]$

I would like to know if my proof is correct. I consider
$$
\begin{align*}
|\phi (f)| \le \max_{f\in C[0,1]} \left|\displaystyle\int^1_0 f(t) a(t) dt\right|& \\ \le \max_{f\in C[0,1]} \displaystyle\int^1_0 \left| f(t)| |a(t) \right|dt \\ \le \left| |a| \right|_{\infty} \max_{f\in C[0,1]} \displaystyle\int^1_0 \left| f(t)\right| dt \\ = \left| |a| \right|_{\infty} \left| |f| \right|_{\infty}
\end{align*}
$$
Then, $\left| |\phi| \right|_{op} \le \left| |a| \right|_{\infty}$.

For equality, suppose $\epsilon > 0$, there are $[a,b] \subset [0,1]$ such that $\left| |a| \right|_{\infty} - \epsilon < |a(t)|$ for $a<t<b$. Then
$$
\begin{aligned}
\left| |\phi| \right|_{op}\ge \frac{\left| |\phi (\chi_{[a,b]})| \right|_{\infty}}{\left||\chi_{ [a,b]}|\right|_{\infty}} = \frac{1}{(b-a)} \max_{f\in C[0,1]}\displaystyle\int^1_0 \left| \chi_{[a,b]}(t)| |a(t) \right|dt &\\ = \frac{1}{(b-a)} \max_{f\in C[0,1]}\displaystyle\int^b_a \left| a(t) \right|dt > \frac{\left| |a| \right|_{\infty} - \epsilon}{(b-a)}(b-a)
\end{aligned}
$$
Finally $\left| |\phi| \right|_{op} \ge \left| |a| \right|_{\infty}$.

I appreciate the comments and corrections

答:
Based on the comments above, it's easy to see that
$$
||\phi||_{op} \le \int^1_0 |a(t)|dt
$$
and for equality you can consider the sign function $f = sgn(a)$, then for $a(t) \neq 0$
$$
||\phi||_{op} \ge |\phi(f)|= \left|\int^1_0 sign(a) \cdot a(t)dt\right|
$$
but the product, $sign(a) \cdot a(t) = |a(t)|$. Finally
$$
||\phi||_{op} \ge \int^1_0 |a(t)|dt
$$
For the case $a(t)=0$, equality is trivial

Title: Norm of a functional on $C[0,1]$


问:
Show there exists a sequence of positive real numbers s.t. ...
Let $f_n$  be a sequence of measurable functions on $[0,1]$ with $|f_n(x)|\lt\infty$  a.e. Show there exists a sequence $c_n$  of positive real numbers s.t. $f_n(x)/c_n\to0$  for almost every $x$ in $[0,1]$. Hint: use the borel-cantelli lemma- pick the sequence $c_n$  so that $$m[x{\rm\ in\ }[0,1]:|f_n(x)|/c_n\gt1/n]\lt2^{-n}$$ where $m$ is the measure.</p>

答:
Fact 1: If $f:[0,1]\to\mathbb R$ is  measurable and $|f(x)|$ is finite for almost every $x$, then for any $\epsilon>0$ there is a number $M$ such that the set $\{x:|f(x)|>M\}$ has measure at most $\epsilon$.

Reason: Since the intersection $\bigcap_{k=1}^\infty \{x:|f(x)|>k\}$ has measure $0$ and the sets in this intersection are nested, it follows that $m[\{x:|f(x)|>k\}]\to 0$ as $k\to\infty$.

Choose   $c_n$ on the basis of Fact 1, so that   $m[\{x:|f(x)|/c_n>1/n\}]<2^{-n}$.

The rest is an application of Borel-Cantelli: the set of number $x$ such that $|f(x)|/c_n>1/n$ infinitely often has measure zero. Rephrase this as: for almost every $x$ we have $|f(x)|/c_n\le 1/n$ for all sufficiently large $n$.


问:
Exponential distribution and floor(X)/n
Suppose $X$ is an exponential distribution given by $0$ for $x<0$ and $ae^{-ax}$ for $x\geq 0$

Let $X$ have the exponential distribution. For $n = 1,2, \ldots$ , let $Y_n$ equal $\lfloor X \rfloor/n$. Prove that $Y_n$ is a random variable, and calculate and identify its distribution function.

答:
Ignore $/n$ for now. The r.v. $Z=\lfloor X\rfloor$ has the following pmf (we say pmf because both $Z$ and $Y$ are discrete random variables)
$$
p(z){
=\Pr\{Z=z\}\\=\Pr\{\lfloor X\rfloor=z\}
\\=\Pr\{z\le X<z+1\}
}
$$
I hope the rest of the procedure is straightforward.

Remark

The pdf of $X$ must be $-ae^{ax}$ for $x>0$ since $a<0$.

Title: Exponential distribution and floor(X)/n


问:
Find $\int_{0}^{1}f(x)\ dx$ given $\int_{0}^{e^2} \frac{f(ln(\sqrt{x}))}{x}dx = 2e$.
Find $\int_{0}^{1}f(x)\ dx$ given $\int_{0}^{e^2} \frac{f(ln(\sqrt{x}))}{x}dx = 2e$.

My attempt

Make the substitution $u=\ln(\sqrt{x})$, then $du = \frac{1}{2x}dx$. Hence:

$$\int_{0}^{e^2} \frac{f(ln(\sqrt{x}))}{x}dx = 2\lim\limits_{b \to 0} \int_{\ln{\sqrt{b}}}^{1} f(u)du=2e$$

We note that $\int_{0}^{1}f(x)\ dx= \lim\limits_{b \to 0}\int_{b}^{1} f(x)dx = 4e$.

Is this solution valid?

答:
There's a typo in the problem.

There's no way to calculate the value of $\int_0^{e^2} \frac{f(\ln(\sqrt x))}{x}\,dx$ knowing the value of $\int_0^1 f(x) \,dx$.

To prove that observe that $f(\ln(\sqrt x))$ evaluates $f$ at negative values since $\ln(\sqrt x)$ is negative for $x \in (0,1)$. That means that $\int_0^{e^2} \frac{f(\ln(\sqrt x))}{x}\,dx$ depends on the value of $f$ at negative numbers, but $\int_0^1 f(x) \,dx$ doesn't.

The correct problem should say:
Find $\int_0^1 f(x) \,dx$ given $\int_1^{e^2} \frac{f(\ln(\sqrt x))}{x}\,dx = 2e$

With that correction the substitution you tried works.


问:
I have a recursive sequence and need to answer with a function
$a_n=a_{n-1}+4$, $a_1=-3$

What function can I use to determine the nth term?

Body:
$a_n-a_{n-1}=4$ $\Rightarrow$ $a_{x+n}=a_x+4((n+x)-x)$

$a_{x+n}=a_x+4n$ $\Rightarrow$ $a_n=a_0+4n$


问:
Demonstration using the Pigonhole principle
I was thinking about the following problem:

Let $n\in\mathbb N$ be odd. If I have a symmetric matrix in $M_n(\mathbb{N})$, i.e. a square symmetric matrix of size $n$, for which each column and each row consists of all numbers between $1$ and $n$, then the diagonal consists of all the numbers from $1$ to $n$.

I want to prove this using the pigeonhole principle. Thanks for helping me.

答:
You don't use the pigeonhole principle, but the double counting principle.

Note that since the matrix is symmetric, if an entry appears in the strictly upper right triangle, it must appear again in the strictly lower left triangle, hence appear an even number of times.

The only way for a number to appear an odd number of times, is for it to appear an odd number of times on the diagonal. Hence, each number must appear at least once on the diagonal.


问:
Function f that has infinitely many superstable points
Suppose $f: R \rightarrow R$. Point $x^*$ is called superstable when it's fixed $(f(x^*)=x^*)$ and $f'(x)\vert_x* = 0$

Is this possible to find $f$ that has an infinite number of supestable points?

Definitely, it's possible to find some functions, that have infinitely many fixed points (e.g. $x$ itself, $x\sin x, x\cos x$...). But I can't figure anything for supestability case.

答: Of course. There are basically two constraints on $f$. The first constraint require that the graph of $f$ intersect with the graph of $x$. Then at those intersections, let the slop of tangent line on $f$ be 0.


答:
From a geometric point of view, you are asking for a function whose graph intersects (infinitely many times) the graph of $g(x)=x$ and the tangent in the intersection points has slope 0: it's not difficult to figure out a lot of examples.

There are also function with infinite "$\infty$-superstable" points: $f(x)=\lfloor x\rfloor+\frac{1}{2}$ has infinite stable points in which all the derivatives are $0$.

Title: Function f that has infinitely many superstable points

Body:
Suppose $f: R \rightarrow R$. Point $x^*$ is called superstable when it's fixed $(f(x^*)=x^*)$ and $f'(x)\vert_x* = 0$

Is this possible to find $f$ that has an infinite number of supestable points?

Definitely, it's possible to find some functions, that have infinitely many fixed points (e.g. $x$ itself, $x\sin x, x\cos x$...). But I can't figure anything for supestability case.


问:
i need to proove that An∪B→A∪B
If $A_n\rightarrow A$, show that $(A_n \cup B) \rightarrow (A \cup B)$.

My attempted solution used $\lim \inf(A_n \cup B)$ and lim sup but I'm not sure how to proceed.

答:
Hint

Define

$$
C_n=(A_n\cup B)-(A\cup B)\\
D_n=(A\cup B)-(A_n\cup B)
$$
and prove that $C_n\to \emptyset$ and $D_n\to \emptyset$.

Title: i need to proove that An∪B→A∪B


问:
Showing that a subspace equals the null space or the whole space.
Let $\varphi$ be a linear functional on the vector space $V$ and suppose $U$ is a subspace of $V$ so that null $\varphi \subseteq U$. Show that either $U=$ null $\varphi$ or $U=V$.

To prove this I thought that the rank nullity theorem as well as properties of the kernel and image would come in handy. I have the following:

dim(null) $\leq$ dim($V$) ; dim(null) $\leq$ dim ($U$); dim(Image) $\leq$ dim(F) ; dim $(U)\leq $ dim ($V$)

With various rearranging of the above inequalities, I haven't found a contradiction. Is there another way I should be thinking about this problem? Thank you.

答:
If $\varphi$ is the null function, then $U=V=\ker\varphi$. Otherwise, if $\ker\varphi\varsubsetneq U$, take $u\in U\setminus\ker\varphi$. Then $V=\ker\varphi\bigoplus\langle u\rangle\subset U$, and so $V=U$.

Note that the equality $V=\ker\varphi\bigoplus\langle u\rangle$ comes from the fact that, if $v\in V$, then $v$ is the sum of $v-\frac{\varphi(v)}{\varphi(u)}u\in\ker\varphi$ with $\frac{\varphi(v)}{\varphi(u)}u\in\langle u\rangle$.


问:
Two inequalities involving real positive numbers
Let $a, b, c, d$ be some real positive numbers. I would like to understand if these inequalities are true:

$1) \max(a, b)\leq\max(a+c, b+d);$

$2) \max(a, b) +\max(c, d)\geq\max(a+c, b+d).$

About me the first inequality is trivially true because we deal with real positive numbers. Moreover, also the second inequality is true (about me). I proceed in this way:

if $\max(a,b)=a$ and $\max(c, d)=c$, thus $a+c=a+c$;

if $\max(a,b)=a$ and $\max(c, d)=d$, thus $a+d> a+c$ and $a+d> b+d$ and in particular it is greater than the maximum;

if $\max(a,b)=b$ and $\max(c, d)=c$, thus $b+c>a+c$ and $b+c> b+d$ and in particular it is greater than the maximum;

if $\max(a,b)=b$ and $\max(c, d)=d$, thus $b+d = b+d$.

Could anyone please tell me if my reasonings are valid and if the two inequalities are true?

Thank you in advance!

答:
Yes, the inequalities are true and your reasonings are correct. From the very casework that you've done, you can see that
$$\max(a, b) + \max(c, d) = \max(a + c, a + d, b + c, b + d)$$
and hopefully from here it should be clear why the second inequality holds.

Note that the second inequality holds for any real numbers $a, b, c, d$.


问:
Prove $rs=sr^{-1}$ in ${\rm Dih}(2n)$

Let $r$ and $s$ be the rotation and reflection symmetries respectively in ${\rm Dih}(2n)$, the dihedral group of order $2n$. Show that $rs=sr^{-1}$.

I also need to show by induction that $r^js=sr^{-j}$ but I think that will follow easily.

答: Hint:  First you want to show that if $r$ is a rotation and $s$ is a reflection, then $rs$ must be a reflection.  What is the square of any reflection?  What is the inverse of any reflection?</p>

Title: Prove $rs=sr^{-1}$ in ${\rm Dih}(2n)$


问:
parameter estimation by the method of moments
Let $X_1,X_2,\dots X_n$ be a random sample from the density $$f(x;\theta)=\theta x^{-2}, \quad 0<\theta \le  x<\infty$$
Find the method of moments estimator of $\theta$.

My attempt

$ E(X) = \int_{\theta}^{\infty}\theta x^{-2} = 1 $

I think that the integral bounds aren't correct.

答:
The integral bounds are correct since you are finding the expectation of $X$.

Another approach

What you need to maximize is
$$
f(x_1,\cdots,x_n|\theta)=\prod_{i=1}^n f(x_i|\theta)=\theta^n(x_1\cdots x_n)^{-2}
$$
where $\theta\le \min_i x_i$.


问:
Prove that $-\psi(y) \leq c(x,y) + c(x,\bar y) - \psi(\bar y)$
In the book Optimal Transport for Applied Mathematicians, Theorem 1.37, the authors states an inequality, but I'm having trouble seeing how one shows that the inequality is indeed true.

Definition: Given a function $c:X\times Y \to \mathbb R$, we say that $\Gamma \subset X \times Y$ is $c$-cyclically monotone ($c$-CM) if for every $k \in \mathbb N$, every permutation $\sigma$ and every finite family of points $(x_1,y_1),...,(x_k,y_k) \in \Gamma$ we have
$$
\sum^k_{i=1} c(x_i,y_i) \leq
\sum^k_{i=1} c(x_i,y_{\sigma(i)})
$$

Now, assume that $\Gamma$ is $c$-CM. Then, fix $(x_0, y_0) \in \Gamma$, and define:
$$
-\psi(y) = \inf\{
-c(x_n,y) +c(x_n,y_{n-1})-c(x_{n-1},y_{n-1})+...+ c(x_1,y_0) - c(x_0,y_0) : n \in \mathbb N,
(x_i,y_i) \in \Gamma \quad  \forall i=1,...,n
\}
$$

How does one then shows that fo $(x,y) \in \Gamma$ and $\bar y \in \text{Proj}_y \circ\Gamma$, we can then write:
$$
-\psi(y) \leq - c(x,y) + c(x,\bar y) - \psi(\bar y)
$$

In the proof, the author claims that this inequality follows from the very definition of the function $\psi$. What I found odd was that since $y$ and $\bar y$ are arbitrary, then I could just swap them, obtaining the opposite inequality, and hence, I actually have an equality. Which would be kind of odd. Anyways, how can you prove this inequality? And is it actually an equality?

答:
Ok, so after some struggle and the help of a friend I figured what I was misunderstanding. Here is the proof:

First, note that since $\bar y \in \pi_y\circ\Gamma$, this means that $\exists \quad  \bar x : (\bar x,\bar y) \in \Gamma$. Also, note that
$\{(x,y),(\bar x, \bar y)\} \subset \Gamma$, henece
$$
-\psi(y) \leq -c(x,y) + c(x,\bar y) - c(\bar x, \bar y)
$$

Now, fix the first two terms in the inequality. Note that for any $n \in \mathbb N$ and $(x_i,y_i) \in \Gamma$ for $i \in \{0,...,n\}$, we have:
$$
-\psi(y) \leq -c(x,y) + c(x,\bar y) - c(\bar x_n, \bar y) + c(\bar x_n,\bar y)-...-c(\bar x_0,\bar y_0)
$$

We can then conclude that
$$
-\psi(y) \leq -c(x,y) + c(x,\bar y) +
\inf \{-c(\bar x_n, \bar y) +...-c(\bar x_0,\bar y_0): n\in \mathbb N
, (\bar x_i,\bar y_i) \in \Gamma \quad \forall i \in \{0,...,n\} \} \\
-\psi(y) \leq -c(x,y) + c(x,\bar y) -\psi(\bar y)
$$


问:
How to explicitly describe the generator of the Galois group of the extension defined by $x^4-3x^2+18$?
Let $K=\mathbb{Q}_3$ and $L=K(\alpha)$ where the minimal polynomial of $\alpha$ over $\mathbb{Q}_3$ is $f = x^4-3x^2+18$. The LMFDB says that the Galois group is cyclic of degree $4$.

Suppose $\sigma$ is a generator of the Galois group. Since $L=K(\alpha)$, the generator $\sigma$ is uniquely determined by $\sigma(\alpha)$. Also, as $L$ is a $K$-vector space, the result of $\sigma(\alpha)$ can be written as a linear combination of powers of $\alpha$.

Question: How can I describe $\sigma(\alpha)$ explicitly as described above?

I know that there is a totally ramified subextension $L'=K(\alpha^2)$ which has minimal polynomial $x^2-3x+18$. Say that these roots in $L'$ are called $\beta$ and $\beta'$, and suppose $\alpha = \beta^2$. Then we may be able to describe $f$ in terms of these elements (maybe with $\beta+\beta' = 3$ or $\beta\beta' = 18$?) but I'm not sure if that really works.

答:
The Galois group of $x^4 - 3x^2 + 18$ over $\mathbf Q$ is dihedral of order 8, not cyclic of order $4$, so whatever argument is used has to involve something that is different between $\mathbf Q$ and $\mathbf Q_3$.

You can work out formulas for the roots of $f(x)$ by using the quadratic formula and square roots: $x^2 = 3(1 \pm \sqrt{-7})/2$, so
$$
x = \pm\sqrt{3\frac{1\pm\sqrt{-7}}{2}}.
$$
The number $-7$ is not a square in $\mathbf Q_3$ and $\beta := 3(1+\sqrt{-7})/2$ generates a quadratic extension $\mathbf Q_3(\beta) = \mathbf Q_3(\sqrt{-7})$, which as Lubin points out in a comment is unramified over $\mathbf Q_3$. Do you see why $\beta$ is a prime element of $\mathbf Q_3(\sqrt{-7})$? Letting $\alpha^2 = \beta$, the field $\mathbf Q_3(\alpha)$ has degree $4$ over $\mathbf Q_3$.

Let $\overline{\beta} = 3(1 - \sqrt{-7})/2$, so $\beta\overline{\beta} = 9(8)/4 = 9 \cdot 2$.  Since the ratio $2/(-7)$ is in $1 + 3\mathbf Z_3$, $2 = -7u^2$ for some $u \in \mathbf Q_3^\times$ (more precisely, $u \in 1 + 3\mathbf Z_3$).  In contrast, $2/(-7)$ is not a square in $\mathbf Q$ and that makes this a step that distinguishes studying the Galois closure of $x^3 - 3x^2 + 18$ over $\mathbf Q_3$ from studying it over $\mathbf Q$. Letting $\alpha'$ satisfy $\alpha'^2 = \overline{\beta}$, which only determines $\alpha'$ up to a sign, we have
$$
(\alpha\alpha')^2 = \beta\overline{\beta} = 9\cdot 2 = 9(-7u^2) = (3u)^2(-7),
$$
so
$$
\alpha\alpha' = \pm 3u\sqrt{-7} = \pm (2\beta - 3)u = \pm (2\alpha^2 - 3)u.
$$
Therefore we can pin down the sign on $\alpha'$ by requiring $\alpha\alpha' = (2\alpha^2-3)u$, which gives us the following formula for $\alpha'$ in terms of $\alpha$.
$$
\alpha'  = \frac{(2\alpha^2-3)u}{\alpha} \in \mathbf Q_3(\alpha).
$$
(It is not true that $\alpha' \in \mathbf Q(\alpha)$, since $u = \sqrt{2/(-7)}$ is not rational.)

The four roots of $x^4 - 3x^2 + 18$ are $\pm \alpha$ and $\pm \alpha'$, which are all in $\mathbf Q_3(\alpha)$. At this point you should work out why this means the Galois group over $\mathbf Q_3$ is cyclic of order $4$ and not the other group of order $4$.


答:
Both inequalities are true, but your reasoning for the second one is not completely correct. For example:

if $\max(a,b)=a$ and $\max(c, d)=d$, thus $a+d> a+c$ and $a+d> b+d$ and in particular it is greater than the maximum;

You can only conclude that $a+d \ge a+c$ and $a+d \ge b+d$ in that case, however that is sufficient to get the desired estimate. The same applies to the case

if $\max(a,b)=b$ and $\max(c, d)=c$, thus $b+c>a+c$ and $b+c> b+d$ and in particular it is greater than the maximum;

A perhaps simpler argument is that
$$
 a + c \le \max(a, b) +\max(c, d)
$$
and
$$
 b + d \le \max(a, b) +\max(c, d)
$$
and therefore
$$
\max(a+c, b+d) \le \max(a, b) +\max(c, d) \, .
$$

Title: Two inequalities involving real positive numbers

Body:
Let $a, b, c, d$ be some real positive numbers. I would like to understand if these inequalities are true:

$1) \max(a, b)\leq\max(a+c, b+d);$

$2) \max(a, b) +\max(c, d)\geq\max(a+c, b+d).$

About me the first inequality is trivially true because we deal with real positive numbers. Moreover, also the second inequality is true (about me). I proceed in this way:

if $\max(a,b)=a$ and $\max(c, d)=c$, thus $a+c=a+c$;

if $\max(a,b)=a$ and $\max(c, d)=d$, thus $a+d> a+c$ and $a+d> b+d$ and in particular it is greater than the maximum;

if $\max(a,b)=b$ and $\max(c, d)=c$, thus $b+c>a+c$ and $b+c> b+d$ and in particular it is greater than the maximum;

if $\max(a,b)=b$ and $\max(c, d)=d$, thus $b+d = b+d$.

Could anyone please tell me if my reasonings are valid and if the two inequalities are true?

Thank you in advance!


问:
Prove theorem similar to Egorov theorem
I am trying to prove the following theorem, which reminds me of Egorov's theorem.

Let $f$ be a bounded function on $[ -\pi , \pi]$ and suppose $\epsilon >0$. Then there is a step function $g$ and a set $E$ such that $|f-g|<\epsilon$ on the complement of E and $\mu(E)<\epsilon$

答:
I assume your function $f$ is supposed to be Lebesgue measurable, then I think I know how you can tackle the problem in 2 steps.

You can prove that if $f:[a,b]\longrightarrow \mathbb{R}$ is a measurable function then you can express it as the pointwise limit of a sequence of simple functions $(\psi_n)_n:\mathbb{R}\longrightarrow\mathbb{R}$ (I call a simple function one of the form $\psi=\sum\alpha_i\chi_{E_i}$ where $E_i$ are Lebesgue Measurable sets).

You can use that any Lebesgue Measurable set $E$ with $\mu(E)<\infty$ is "close to a finite union of intervals"
(or more precisely: $\forall \epsilon>0,\exists I_1,\ldots,I_n$ intervals such that $\mu(E\Delta\bigcup_{i=1}^{n}I_i)<\epsilon$)
to prove that you can also find a sequence $(\phi_n)_n$ of step functions that converges pointwisely to $f$ $\mu$-almost everywhere. Then using Egorov's theorem on this sequence gives you what you want.

There may be an easier way to do it, but these are the properties I'm familiar with right now.

Title: Prove theorem similar to Egorov theorem


问:
Find the value of summation
If
$$xy = 3 , yz = 6 , xz = 2$$
Find the value of $x+y+z$
My attempt:

Let $$k=x+y+z$$

$$k^2 = x^2 + y^2 + z^2 +2(xy+xz + yz)$$

$$k^2 = x^2 +y^2 + z^2 +22$$

But i stopped here i need a hint

答:
Hint: $x^2y^2z^2 = 36 \implies xyz = \pm6$

You already have value of $xy, yz, zx$.

Title: Find the value of summation


答:
hint

Observe that $$xyz\ne 0$$

$$\frac{yz}{xy}=\frac{6}{3}=2=xz$$

$$\implies x^2=1$$

Other approach :
The product gives
$$xyyzzx=(xyz)^2=36$$

$$\frac{x+y+z}{xyz}=\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}$$

The result you should find is $   \; \color{red}{\pm}6$.

Title: Find the value of summation

Body:
If
$$xy = 3 , yz = 6 , xz = 2$$
Find the value of $x+y+z$
My attempt:

Let $$k=x+y+z$$

$$k^2 = x^2 + y^2 + z^2 +2(xy+xz + yz)$$

$$k^2 = x^2 +y^2 + z^2 +22$$

But i stopped here i need a hint


问:
Elementary set theory problem: proof of equivalence
Let $A, B, C$ be arbitrary sets. Prove that:
$$(A\backslash B)\backslash C = A\backslash(B\backslash C) \Longleftrightarrow A\cap C = \emptyset$$

The left hand side of the equivalence:

$(A\backslash B)\backslash C = A\backslash(B\backslash C)$

is defined by the following logical expression

$\begin{align}
x\in (A\backslash B)\land x\notin C
&\Leftrightarrow x\in A \land \lnot(x\in B \land x\notin C)\\
&\Leftrightarrow x\in A\land (x\notin B \lor x\in C)\\
&\Leftrightarrow (x\in A \land x\notin B)\lor(x\in A \land x\in  C)\\
&\Leftrightarrow x\in(A\backslash B)\lor x\in (A\cap C)
\end{align}$

So I could rewrite the LHS as the following equivalence:

$x\in (A\backslash B)\land x\notin C \Leftrightarrow x\in(A\backslash B)\lor x\in (A\cap C)$

The right hand side of the equivalence:

$A\cap C = \emptyset$

is analogically defined by the expression:

$x\in A \Rightarrow x\notin C$

The whole problem, rewritten:
So the whole problem narrows down to proving:
$$\Big(x\in (A\backslash B)\land x\notin C \Leftrightarrow x\in(A\backslash B)\lor x\in (A\cap C) \Big)\stackrel{?}{\equiv} \Big(x\in A \Rightarrow x\notin C \Big)$$

Best further approach:
My question is generally: What would be the best and most comprehensive further approach as of this point?

rigorous proof based on boolean algebra with the given result and equivalent transformations;

splitting the proof into two directions: "$\Longrightarrow$" and "$\Longleftarrow$"

proof by contradiction (using terms like "suppose", "for an arbitrary" etc.)

other options?

I know these are all the same, but I'd love to keep going "the rigorous" way with equivalent transformations in the logical expressions. What would your advice be? Thanks in advance!

P.S. I'm also open for corrections of my notation (especially the $\equiv$ symbol, am I using that right?)

答:
I would not approach the problem that way at all: that kind of formal logical computation is in general unnecessarily hard to read and tends to obscure what is going on. It has an important place in formal logic, but it is generally a poor choice for communicating an argument in everyday mathematics, and it really isn’t any more rigorous than a careful verbal argument. See this answer for a little more on the subject.

Here is a straightforward argument written in a reasonably friendly style.

Suppose first that $A\cap C\ne\varnothing$, and let $x\in A\cap C$. Then $x\in C$ implies that $x\notin(A\setminus B)\setminus C$, but $x\in C$ also implies that $x\notin B\setminus C$, and $x\in A$ then implies that $x\in A\setminus(B\setminus C)$. Thus, $A\cap C\ne\varnothing$ implies that

$$(A\setminus B)\setminus C\ne A\setminus(B\setminus C)\,.$$

If, on the other hand, $A\cap C=\varnothing$, then $(A\setminus B)\cap C=\varnothing$, and therefore $(A\setminus B)\setminus C=A\setminus B$. Thus, we want to show that $A\setminus(B\setminus C)=A\setminus B$ as well. Clearly $B\setminus C\subseteq B$, so $A\setminus(B\setminus C)\supseteq A\setminus B$. Suppose that $x\in A\setminus(B\setminus C)$. Then $x\in A$, and $x\notin B\setminus C$, i.e., either $x\notin B$, or $x\in B\cap C$. Now $x$ cannot be in $B\cap C$, because then $x$ would be in $A\cap C$, which is empty. Thus, $x\notin B$, so $x\in A\setminus B$, and we’ve shown that $A\setminus(B\setminus C)\subseteq A\setminus B$. It follows that $A\cap C=\varnothing$ implies that

$$(A\setminus B)\setminus C=A\setminus(B\setminus C)\,.$$


问:
Every ridge in the convex polytope is an intersection of exactly two facets
This statement is used without proof in various sources (incl. Ziegler's "Lectures on polytopes" and Gruber's "Convex and discrete geometry"), yet I don't have any idea how to prove it, or have intuition about it in $n>3$ dimensions. Someone suggested me to use a concept of dual polytope to prove it, I have not explored it thoroughly and maybe someone can help with understanding how duality is convenient here.

答:
Your quest on proving that statement is just like proving that there is just a single parallel line to a given line. The latter seems obviously true within euclidean spaces, so one might be tempted to prove it, but it is neither true in spherical (less than 1) nor in hyperbolical ones (more than 1). In fact, in history right that trial to prove it by means of a contradictional attempt directly lead to the discovery of those other space types.

And in fact quite similarily, there are polytopes, which have 2 facets per ridge (the usual real space polytopes), but there are realizations of according abstract polytopal structures with 3 or more facets per ridge. This then is the realm of so called complex polytopes.

Cf. to https://en.wikipedia.org/wiki/Complex_polytope


答:
Obviously, $x,y,z\neq 0$. Thus, $x=\frac 3y$, $z\frac3y=2$, i.e. $3z=2y$. Now your second equation gives $\frac23y^2=6$ which gives $y=\pm 3$. Thus, $x=\pm 1$ and $z=\pm 2$.

The value of $x+y+z$ thus is either $6$ or $-6$.

Title: Find the value of summation


问:
Prove that there exist a tensor whose components relative to the product basis of $\pmb{any}$ basis are the values of the generalized kronecker delta
ACHTUNG!!!

To follow I adopt the Einstein notation!!!

Definition

The dual space $V^*$ of a vector space $V$ over the field $\Bbb F$ is the vector space of the linear functions from $V$ to $\Bbb F$.

Definition

Give a basis $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ of a vector space $V$ the dual set of $\mathcal E$ is the set $\mathcal E^*:=\{\vec e^{\,1},...,\vec e^{\,n}\}$ of $V^*$ whose elements are defined through the equality
$$
\vec e^{\,i}(\vec e_j)=\delta^i_j
$$
for each $i,j=1,..,n$.

Theorem

If $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ is a basis for a vector space $V$ then the dual set $\mathcal E^*$ of $\mathcal E$ as above defined is a basis of the dual space $V^*$ of $V$.

Theorem

Let be $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ and $\mathcal B:=\{b_1,..,b_n\}$ two basis of a vector space $V$ so that
$$
\vec b_k=B^s_k\vec e_s\,\,\,\text{and}\,\,\,\vec e_q=E^k_q\vec b_k
$$
for each $k,q=1,...,n$. So if this holds then
$$
\vec b^{\,\,q}=E^q_k\vec e^{\,\,k}\,\,\,\text{and}\,\,\,\vec e^{\,\,q}=B^q_k\vec e^{\,\,k}
$$
for each $k,q=1,...,n$ and moreover $B^q_kE^k_s=\delta^q_s$ for each $q,s=1,...,n$

Definition

If $V_1,...,V_s$ is a collection of vector spaces over the same field $\Bbb F$ then a $s$-linear function is a function $f$ form $V_1\times...\times V_s$ in $\Bbb F$ that is linear in each of its variables while the other variables are held costant. In particular if $V_i=V^*$ for some $r<s$ and $V_i=V$ otherwise we say that the function $f$ is a tensor of contravariant order $p:=r$ and covariant order $q:=(s-r)$.

Theorem

If $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ is a basis of $V$ then a tensor $\tau$ of contravariant order $p$ and covariant order $q$ is completely defined by the equality
$$
T^{i_1,...,i_p}_{j_1,..,j_q}:=\tau\big(\vec e^{\,i_1},...,\vec e^{\,i_p},\vec e_{j_1},...\vec e_{j_q}\big)
$$
for $i_1,...,i_p,j_1,...,j_q=1,...,n$. Furthermore if $\mathcal B:=\{\vec b_1,..,\vec b_n\}$ is another basis of $V$ then
$$
\tau(\vec b^{\,i_1},..,\vec b^{\,i_p},\vec b_{j_1},...,\vec b_{j_q})=E^{i_1}_{h_1}...E^{i_p}_{h_p}B^{k_1}_{j_1}...B^{k_q}_{j_q}T^{h_1,...,h_p}_{k_1,....,k_q}
$$
for each $i_1,...,i_p,j_1,...,j_q=1,...,n$.

Now let be $\mathcal E:=\{\vec e_1,...,\vec e_n\}$ a basis of a vector space $V$ so that we can define a tensor $\tau$ of contravariant and covariant order p letting that
$$
\tau(\vec e^{\,i_1},...,\vec e^{\,i_p},\vec e_{j_1},..,\vec e_{j_p})=\delta^{i_1,...,i_p}_{j_1,..,j_p}
$$
for each values of the indices $i_1,...,i_p,j_1,...,j_p$ from $1$ to $n$.
So I ask to prove that for any vector space and for any natural number $p$ there exist a tensor $\tau$ of order $p$ whose components relative to the product basis of any basis are the values of the generalized kronecker delta as here defined, that is more explicitly I ask to prove that if $\tau$ is the tensor above defined and if $\mathcal B:=\{\vec b_1,..,\vec b_n\}$ is another basis for the space $V$ above defined then
$$
\tau(\vec b^{\,i_1},...,\vec b^{\,i_p},\vec b_{j_1},..,\vec b_{j_p})=E^{i_1}_{h_1}...E^{i_p}_{h_p}B^{k_1}_{j_1}...B^{k_p}_{j_p}\delta^{h_1,...,h_p}_{k_1,...,k_p}
$$
for each values of the indices $h_1,...,h_p,k_1,...,k_p$.

So could someone help me, please?

答:
First of all we remember that if $\epsilon^{i_1,...,i_k}$ is the Levi-Civita symbol and if $A$ is a square matrix of order $n$ then
$$
\det A=\epsilon^{i_1,...,i_n}A^{i_1}_1....A^{i_n}_n=\epsilon_{i_1,...,i_n}A^1_{i_1}....A^n_{i_n}\,\,\,\text{and}\,\,\,\epsilon^{i_1,...,i_k}\epsilon_{j_1,...,j_k}=\delta^{i_1,...,i_k}_{j_1,...,j_k}
$$
and moreover
as showed here it holds the equality
$$
\delta^{i_1,...,i_k}_{j_1,...,j_k}=\text{det}\begin{pmatrix}\delta^{i_1}_{j_1}&\cdots&\delta^{i_1}_{j_k}\\
\vdots&\ddots&\vdots\\
\delta^{i_k}_{j_k}&\cdots&\delta^{i_k}_{j_k}\end{pmatrix}
$$
for each $k\in\Bbb N$ and for each value of the indices $i_1,...,i_k$ and $j_1,...,j_k$. So remembering this we observe that
$$
E^{i_1}_{h_1}...E^{i_p}_{h_p}B^{k_1}_{j_1}...B^{k_p}_{j_p}\delta^{h_1,...,h_p}_{k_1,...,k_p}=\epsilon^{h_1,...,h_p}E^{i_1}_{h_1}...E^{i_p}_{h_p}\epsilon_{k_1,...,k_p}B^{k_1}_{j_1}...B^{k_p}_{j_p}=\\
\det\begin{pmatrix}E^{i_1}_1&\cdots& E^{i_1}_p\\
\vdots&\ddots&\vdots\\
E^{i_p}_1&\cdots&E^{i_p}_p\end{pmatrix}\det\begin{pmatrix}B^{1}_{j_1}&\cdots& B^1_{j_p}\\
\vdots&\ddots&\vdots\\
B^{p}_{j_1}&\cdots& B^p_{j_p}\end{pmatrix}=\\
\det\begin{pmatrix}\begin{pmatrix}E^{i_1}_1&\cdots& E^{i_1}_p\\
\vdots&\ddots&\vdots\\
E^{i_p}_1&\cdots&E^{i_p}_p\end{pmatrix}&\begin{pmatrix}B^{1}_{j_1}&\cdots& B^1_{j_p}\\
\vdots&\ddots&\vdots\\
B^{p}_{j_1}&\cdots& B^p_{j_p}\end{pmatrix}\end{pmatrix}=\\
\det\begin{pmatrix}E^{i_1}_{s^1_1}B^{s^1_1}_{j_1}&\cdots&E^{i_1}_{s^1_p}B^{s^1_p}_{j_p}\\
\vdots&\ddots&\vdots\\
E^{i_p}_{s^p_1}B^{s^p_1}_{j_1}&\cdots&E^{i_p}_{s^p_p}B^{s^p_p}_{j_p}\end{pmatrix}=\det\begin{pmatrix}\delta^{i_1}_{j_1}&\cdots&\delta^{i_1}_{j_p}\\
\vdots&\ddots&\vdots\\
\delta^{i_p}_{j_1}&\cdots&\delta^{i_p}_{j_p}\end{pmatrix}=\delta^{i_1,...,i_p}_{j_1,...,j_p}
$$
for each values of the indices $i_1,...,i_p$ and $j_1,..,j_p$.


问:
Finding constants for second order diff. equation with repeated real root
http://personal.psu.edu/sxt104/class/Math251/Notes-2nd%20order%20ODE%20pt1.pdf

On Page 28 of the above link it shows example of how to find general form of second order diff. equation with repeated root.

The equation is $y″ − 4y′ + 4y = 0$ with condition $y(0) = 4$, $y′(0) = 5$.

But how they find the constants using initial conditions?

Body:
$$y'' − 4y′ + 4y = 0$$
Is equivalent to:
$$(ye^{-2x})''=0$$
Integrate twice.
$$y=e^{2x}(c_1x+c_2)$$
$$y'=e^{2x}(2c_1x+c_1+2c_2)$$
Apply intial condtions:
$$y(0)=4 \implies 4=c_2$$
$$y'(0)=5 \implies 5=c_1+2c_2 \implies c_1=-3$$


问:
How to simplify out of the gamma function $\Gamma(z)$?
$$\mathrm{d}t(x_i; \nu) = \frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2}) \sqrt{\pi\nu}} \Bigg( 1+\frac{x_i^2}{\nu} \Bigg)^{-\frac{\nu+1}{2} } \enspace, i=1,2$$

How to derive a closed-form solution of the above by simplifying out of the gamma function?

Is the gamma function that is being used above (for $z>0$) the same as
$$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\mathrm{d}x ?$$

答:
The symbol $\Gamma$ in the factor  $\frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2}) \sqrt{\pi\nu}} $ does denote the gamma function.

The expression for the pdf of the univariate Student's t-distributution $dt(x_i; \nu)$ is already in closed-form in the sense that the gamma function with an arbitrary real argument cannot be expressed in terms of more elementary functions other than in integral, infinite series, or infinite product representations (many of which are shown in the link). I don't believe that would qualify as "simplifying out of the gamma function".

If the argument is a positive integer, then we have the immense simplification $\Gamma(n) = (n-1)!$. However, for a degree-of-freedom parameter $\nu \not \in \mathbb{Z}$, both arguments $\nu/2$ and $(\nu+1)/2$ are not integers. Even if $\nu$ is an integer one or the other of the arguments is not an integer.

An alternative representation that "appears" less complicated uses the beta function which is related to the gamma function through

$$\int_0^1 t^{x-1}(1-t)^{y-1} \, dt := B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

Hence,

$$\frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2})\sqrt{\pi}}= \frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2})\Gamma (\frac{1}{2})}= \frac{1}{B \left(\frac{1}{2}, \frac{\nu}{2} \right)},$$

and $\displaystyle dt(x_i;\nu) = \frac{1}{\sqrt{\nu}B \left(\frac{1}{2}, \frac{\nu}{2} \right)} \Bigg( 1+\frac{x_i^2}{\nu} \Bigg)^{-\frac{\nu+1}{2} }.$

One may ask why the gamma function appears in this pdf to begin with. A random variable $X$ that is Student's t-distributed with $\nu$ degrees of freedom is obtained as $X = Z/S$ where $Z \sim \mathcal{N}(0,1)$ is a standard normal random variable and $S = \sqrt{Y/\nu}$ where $Y \sim \chi^2(\nu)$ has a chi-square distribution.

The cdf for $Y$ is

$$F_Y(y) = C \int_0^y u^{(\nu-2)/2}e^{-u/2} \, du,$$

where  the normalizing factor $C$ is chosen so that $F_Y(y) \to 1$ as $y \to \infty$, that is

$$\frac{1}{C}= \int_0^\infty u^{(\nu-2)/2}e^{-u/2} \, du = \int_0^\infty 2^{\nu/2 -1}t^{\nu/2 -1 }e^{-t} \, 2dt = 2^{\nu/2} \Gamma(\nu/2).$$

After many more steps of transformation the pdf $dt(x;\nu)$ is obtained, retaining the factor $\Gamma(\nu/2)$ and introducing $\Gamma((\nu+1)/2)$.


问:
Terminology for functions with preimage of finite cardinality?
I'm dealing with a class of continuous functions $f:[0, 1] \to \mathbb{R}$ such that the preimage of any $y \in f([0, 1])$ is of finite cardinality. I wonder whether there is a common terminology of such functions. Thanks!

答: I’ve seen the term “finite-to-one” being used in topology. No reference that I can recall.</p>


问:
Integrating $\int{\frac{1}{1+e^{x}}}dx$, Partial Fractions(?)
I need help with this integral:

$$H(x) = \int{\frac{1}{1+e^{x}}}dx$$

It should be easy, but I'm stuck. I thought about using a u-substitution but I didn't get any further. Am I meant to use partial fractions? I'm not yet very comfortable with partial fractions. I'd be thankful for someone's explanation!

答:
Let $e^x = t$. We then have
$$\int \dfrac{dx}{1+e^x} = \int \dfrac{e^xdx}{e^x+e^{2x}} = \int \dfrac{dt}{t+t^2}$$

Equivalently, let $e^{-x} = t$, we then have
$$\int \dfrac{dx}{1+e^x} = \int \dfrac{e^{-x}dx}{1+e^{-x}} = \int \dfrac{-dt}{1+t}$$

I trust you can take it forward using both/either of the above substitutions.


问:
Abstract Algebra - Dummit Exercise 3.1.1

This question in Abstract Algebra by Dummit seemed ill-posed to me:

Let $\varphi: G \rightarrow H$ be a homomorphism and let $E$ be a subgroup of $H .$ Provethat $\varphi^{-1}(E) \leq G$ (i.e., the preimage or pullback of a subgroup under a homomorphism is a subgroup). If $E \trianglelefteq H$ prove that $\varphi^{-1}(E) \trianglelefteq G .$ Deduce that ker $\varphi \trianglelefteq G$

In this problem, how do we know $\phi^{-1}(E)$ is well defined? I.e., how do we know there is no $e \in E, \; s.t. \; \phi(g)\neq e \;\; \forall g \in G$? Homomorphisms don't have to be surjective, so I'm thinking $E$ could inherit such and element from $H$.

Edit: Here is my attempted proof:

First we prove some things about $\phi^{-1}$. Let $\phi(x),\phi(y) \in E$:

\begin{align*}
    \phi^{-1}(\phi(x)\phi(y))= \phi^{-1}(\phi(xy)) = xy = \phi^{-1}(\phi(x))\phi^{-1}\phi((y))
\end{align*}

Thus $\phi^{-1}$ is also a homomorphism. Also:

\begin{align*}
    \phi^{-1}(x)^{-1} &= \phi^{-1}(1_G)\phi^{-1}(x)^{-1}\\
    &= \phi^{-1}(x^{-1}x)\phi^{-1}(x)^{-1} \\
    &= \phi^{-1}(x^{-1}) \phi^{-1}(x) \phi^{-1}(x)^{-1}\\
    &= \phi^{-1}(x^{-1})
\end{align*}

Armed with this knowledge, let $e, e' \in E$, $\phi^{-1}(e),\phi^{-1}(e') \in \phi^{-1}(E)$:

\begin{align*}
    \phi^{-1}(e)\phi^{-1}(e')^{-1} &= \phi^{-1}(e(e')^{-1})\\ &
    = \phi^{-1}(e^{*}) \in \phi^{-1}(E)
\end{align*}

Thus $\phi^{-1}(E) \leq G$ .

If $E \trianglelefteq H$, then we know $N_H(E) = H, \; hEh^{-1} \forall h \in H$. Let $e \in E$, $\phi^{-1}(e)\in \phi^{-1}(E)$:

\begin{align*}
    g\phi^{-1}(e)g^{-1} &= \phi^{-1}(h)\phi^{-1}(e)\phi(h)^{-1}\\
    &= \phi^{-1}(h e h^{-1})\\
    &= \phi^{-1}(e^*) \in \phi^{-1}(E)
\end{align*}

Since ${1} \trianglelefteq H$, and $\phi^{-1}(1) = \ker \phi$, $\ker\phi \trianglelefteq G$.

Is this correct?

答:
First and foremost, personally I'd avoid using the symbol "$\varphi^{-1}(E)$", as this misleads to thinking (as seemingly you did) that an inverse map exists, which in general is not the case. For this reason, I'll use in place of that a more neat (to me) "$\varphi^{\leftarrow}(E)$".

So, by definition, $\varphi^\leftarrow(E):=\{g\in G\mid \varphi(g)\in E\}$; since $E$ is a group, $\varphi^\leftarrow(E)\ne\emptyset$ as $e_G\in \varphi^\leftarrow(E)$. Now, let $g,g'\in \varphi^\leftarrow(E)$; then $\varphi(gg')=\varphi(g)\varphi(g')\in E$, because $E$ is a group, and hence  $gg'\in \varphi^\leftarrow(E)$. Likewise, let $g\in \varphi^\leftarrow(E)$; then, $\varphi(g^{-1})=\varphi(g)^{-1}\in E$, again because $E$ is a group; therefore, $g^{-1}\in \varphi^\leftarrow(E)$. By the subgroup criteria (closure and closure by inverses), indeed $\varphi^\leftarrow(E)\le G$.

For the second part, $\forall g \in G$:

\begin{alignat}{1}
g\varphi^\leftarrow(E)g^{-1} &= \{gkg^{-1}\in G\mid \varphi(k)\in E\} \\
&= \{k'\in G\mid \varphi(g^{-1}k'g)\in E\} \\
&= \{k'\in G\mid \varphi(g)^{-1}\varphi(k')\varphi(g)\in E\} \\
&= \{k'\in \varphi^{\leftarrow}(E)\mid \varphi(g)^{-1}\varphi(k')\varphi(g)\in E\}\cup \{k'\in G\setminus\varphi^{\leftarrow}(E)\mid \varphi(g)^{-1}\varphi(k')\varphi(g)\in E\} \\
&\stackrel{(E\unlhd H)}{=} \varphi^{\leftarrow}(E)\cup\emptyset \\
&= \varphi^{\leftarrow}(E)
\end{alignat}

and hence $\varphi^{\leftarrow}(E)\unlhd G$.

For the final corollary, note that $\operatorname{ker}(\varphi)=\varphi^{\leftarrow}(\{e_H\})$ and $\{e_H\}\unlhd H$.


问:
Showing the equivalence of two statements regarding the one-sided limits
I have the following definition for the right limit

If $a$ is a right accumulation point for $D$, we say that $L$ is the limit of $f$ from
right at $a$, denoted by $\lim\limits_{x\to a+} f(x) = L$,if for any neighbourhood $U \in V(L)$ there exists $r > 0$
such that $f((a, a + r) \cap D) \subseteq U$.

and I need to show the equivalence of the following two statements

$(i)\lim\limits_{x\to a+} f(x) = L$

$(ii)$ For any sequence $(a_n)_n$ with $a_n \in D$ and $a_n > a$ for all $n \in N$ such that
$\lim\limits_{x\to\infty} a_n = a$ we have $\lim\limits_{x\to\infty} f(a_n) = L.$

In the question, $L$ was an extended real number, but currently I am trying to show this only for $L\in\mathbb R$. So far, I was able to show that $(i)$ implies $(ii)$ but I am stuck with the other implication.

My proof so far for $(ii) \implies (i)$:

To show that $(i)$ holds, I am using the above definition of right limit. Take any neighbourhood $U \in V(L)$, then $(L-r,L+r) \subseteq U$ for some $r>0$. I want to show that $f((a,a+r)\cap D) \subseteq U$ holds exactly for that $r$ that we got from $U$. It is sufficient to show $f((a,a+r)\cap D) \subseteq (L-r, L+r)$.

At that point I am trying to make a transition to sequences to use $(ii)$ but I have no idea how to. If I were to have a sequence $(a_n)_n$ with $\lim\limits_{x\to\infty} a_n = a$ and with elements $a_n \in ((a,a+r)\cap D)$, it would mean that $a_n>a$ and $a_n \in D$ so it would be perfect for using $(ii)$. But I don't know if I have such a sequence or not.

Can you help please give me ideas on how to proceed on my proof or give me another starting point?

答:
hint

You want to prove that
$$(\forall U\in V(L))\; (\exists r>0) \;:$$
$$\;f((a,a+r)\cap D)\subset U$$

assume the negation,

$$(\exists U\in V(L))\; : \;(\forall r>0)$$$$\; \exists x\in(a,a+r)\cap D \;: f(x)\notin U$$

Take $ r=\frac 1n $  and $ x $ will be $ x_n$ satisfying $$a<x_n<a+\frac 1n$$


答: First of all notice that if $(a,b,c)$ is a solution to the system then so does $(-a,-b,-c).$ Therefore we need to have at least two solutions for $k.$ Multiplying given equations gives $$x^2y^2z^2=2.3.6=36$$ and therefore $xyz=\pm 6.$ Hence $x=\pm 1, y=\pm3, z=\pm2$ and finally $$x+y+z=\pm6.$$

Title: Find the value of summation


问:
Evaluating $\oint_C (\arctan{y} - 5y)dx + \left(\frac{x}{1+y^2} + 2x\right)dy$
Evaluate integral of

$$\oint_C (\arctan{y} - 5y)dx + \left(\frac{x}{1+y^2} + 2x\right)dy$$

where $C$ is the ellipse $x^2/16 + y^2/25 = 1$.

So I used the Green's Theorem and got double integral of $D$ of $7 \; dy\; dx$. What do I do now? How do I get the $x$ and $y$ bounds? Please show steps to how to solve this using Green's Theorem. Please don't use some crazy methods to solve it. I just want to keep it simple. Thank you

答: Now that you are at: $$\iint_{D}7dydx=7\iint_{D}dydx$$you actually don't need to worry about the integration limits. It turns out that: $$\iint_{D}1dxdy=(\mbox{Area of }D). $$ So you just need to get the area of that ellipse.</p>


问:
If $f(s)T(s)=a$ prove $\gamma (s) $ is a straight line
Let $\gamma:I\rightarrow \mathbb{R}^{3}$ be a curve with parameterized of arc length, and we suppose that for the tangent vector $T(S)$ of $\gamma(s)$ there exist a function $f:I\rightarrow \mathbb{R}$ such that $f(s)T(s)=a$ for some non-zero vector $a$ , prove the curve is a straight line.

I tried to take the derivative and show $k=0$ (where $k$ is the curvature of a curve )

using Frenet's formulas we have that ${f}(s)'T(s)+f(s)kN(s)=0$ but i am stuck how to show that $k=0$

any hints.

答:
By definition, $T(s)$ and $N(s)$ are perpendicular, hence linearly independent. Thus
$$
f'(s) T(s) + f(s) k N(s) = 0
$$
implies
$$
f'(s) = 0 \, ;  \quad f(s) k = 0
$$
Since $f'(s) = 0$ for all $s$, we may set $f(s) = C$. Thus $Ck = 0$. Since $a \neq 0$, we may not have $C = 0$. Therefore, $k = 0$.


问:
Random walk on a sphere: statistics of the solid angle?
Consider a random path on the surface of a 2-sphere, made of N discrete points (each picked with a uniform distribution across the surface). The path connects two successive points by the shortest line between them, and we connect the last point to the first one to make a closed path.

My question is, what is the statistics of the solid angle subtended by this path? I expect the mean to be zero, but I have no idea for the standard deviation.

答:
Here are important tracks. But it's not a complete solution.

As any solid angle is measured by the corresponding area, we need to use the formula for the area of a triangular triangle. Afterwards, there will be a summation process. Let $A,B,C$ be the angles of the triangle, we have

$$area=A+B+C-\pi$$

(angular excess: see Girard's theorem here).

But we must be able to compute these angles.

This is done by using this formula:

$$\cos c=\cos a\cos b+\sin a\sin b\cos C \ \ \iff \ \ \cos C=\dfrac{\cos c-\cos a\cos b}{\sin a\sin b}$$

and the 2 other formulas obtained by circular permutation:

$$\cos A=\dfrac{\cos a-\cos b\cos c}{\sin b\sin a} \ \ \text{and} \ \ \cos B=\dfrac{\cos b-\cos c\cos a}{\sin c\sin b}$$

When the path $M_1M_2\cdots M_nM_1$ is completed, one computes the sum of areas:

$$\Delta=s_1 \times area(M_1M_2M_3)+ s_2 \times area(M_1M_3M_4)+ \cdots + s_{n-2} \times  area(M_1M_{n-1}M_{n})\tag{1}$$

where the sign $s_k$ is $+$ if the triangle is positively oriented, $-$ otherwise.

(1) defines a random variable.

The mathematical expectation of $\Delta$ is the sum of the mathematical expectations of the triangular areas, which are clearly $0$ because there is a balance between the positive and negative areas.

The variance of $\Delta$, assuming mutual independence between the different areas, is the sum of variances of the triangle's areas:

$$E(area(M_1M_kM_{k+1})^{\color{red}{2}})$$

... which have to be computed by taking into account the variances and expectancies of sides lengths $\cos a, \cos b, \cos c$, themselves functions of the number $N$ of points "dropped" on the sphere, which doesn'tt look very easy to do, because we have to take as well into account the shortest path limitation. But I think rather good approximations can be obtained by taking a uniform enough distribution of points on the sphere.

Remark: Given a number $N$ of points on the sphere, set apart exceptional values of $N$, it is not possible to have an even distribution of points on the sphere, for example with $N=5$, but we can get rather close to it.


问:
Limit of $\frac{1}{\sqrt[n] n}$
I would like to show that the limit of the sequence $(\frac{1}{\sqrt[n] n})_{n=1}^{\infty}$ is equal to 1 using the definition of the limit of a sequence. (Another way to say this is that it converges to 1).

Definition: A sequence $(a_n)_{n=1}^{\infty}$ converges to a real number A iff for each $\epsilon > 0$ there is a positive integer N such that for all $n \geq N$ we have $|a_n - A| < \epsilon$.

Edit: Here is my work.

Let $\epsilon > 0$. There exists $N$ such that for $n \geq N$,

$$|\frac{1}{\sqrt[n] n}-1| = |\frac{n^{\frac{n-1}{n}}}{n} - 1| = |\frac{n^{\frac{n-1}{n}} - n}{n}| = \frac{n-n^{\frac{n-1}{n}}}{n}$$

Here is where I am stuck.

答:
Let's do this just using straightforward algebra and some crude (but clever) bounds.

For $n\ge1$ we have, using the algebraic identity $(1-a^n)=(1-a)(1+a+a^2+\cdots+a^{n-1})$ and the inequality $\sqrt[n]n\ge1$,

$$\begin{align}
0\le1-{1\over\sqrt[n]n}&=1-n^{-1/n}\\
&=(1-n^{-1/n}){1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}\over1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}}\\
&={1-n^{-1}\over1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}}\\
&\lt{1\over(n^{-1/2}+n^{-1/2}+\cdots+n^{-1/2})+(0+0+\cdots+0)}\qquad\text{(see note below)}\\
&\lt{1\over(n/3)n^{-1/2}}\\
&={3\over\sqrt n}\\
&\lt\epsilon\qquad\text{if $n\gt{9\over\epsilon^2}$}
\end{align}$$

The key step is the replacement of the (roughly) first third of the terms in $1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}$ with the smaller value $n^{-1/2}$ and the rest with $0$; one can, of course, go up close to the first half and replace the $3$ here with something close to $2$, but there's not much to be gained in doing so.

Title: Limit of $\frac{1}{\sqrt[n] n}$


问:
Find all real values of $x$ which satisfy $\sqrt{x^2 + 1} + x^2 + 1 = 90$.
Find all real values of $x$ which satisfy $\sqrt{x^2 + 1} + x^2 + 1 = 90$.

\begin{align*}
\sqrt{x^2+1} + x^2 &=  89 \\
{(\sqrt{x^2+1} + x^2)}^2 &=  89^2 \\
x^2+2x^2\sqrt{x^2+1}+x^4 &=89^2
\end{align*}

Letting $y = x^2$, so $y^2 = x^4$. This gives us
\begin{equation}
y^2+2y\sqrt{y+1}+y=7920
\end{equation}

I not sure how to proceed to eliminate the radical. I shouldn't square quantities that aren't both positive or else I'll lose the equivalency in the new equation right?

答:
Hint

Begin by putting
$$y=\sqrt{x^2+1}$$

with

$$y+y^2=90$$
or
$$y^2+y-90=0$$
$$y=\frac{-1\pm 19}{2}$$
and $ y>0$.

You should find that $$x=\pm \sqrt{80}$$


答:
Hint to solve

$x^2+1=(89-x^2)^2$

Second hint

Turn the $x^4$ to a variable like $z=x^2$ so you can solve the exercise using discriminant.

Body:
Find all real values of $x$ which satisfy $\sqrt{x^2 + 1} + x^2 + 1 = 90$.

\begin{align*}
\sqrt{x^2+1} + x^2 &=  89 \\
{(\sqrt{x^2+1} + x^2)}^2 &=  89^2 \\
x^2+2x^2\sqrt{x^2+1}+x^4 &=89^2
\end{align*}

Letting $y = x^2$, so $y^2 = x^4$. This gives us
\begin{equation}
y^2+2y\sqrt{y+1}+y=7920
\end{equation}

I not sure how to proceed to eliminate the radical. I shouldn't square quantities that aren't both positive or else I'll lose the equivalency in the new equation right?