问:
Generators for the sheaf of regular functions
Reading this post I found out that my claim below is wrong in general for abstract varieties (I guess it is true for affine varieties, though). Below you'll find also my wrong naive proof... could you help me to understand where am I using implicitly that $X$ is affine? Or, more generally, to spot the mistakes in  my proof?

Claim (wrong!)

Let $X$ be an algebraic variety. Then for $f_i \in \mathcal{O}_X$ we have
$$ X = \bigcup_{i=1}^n D_X(f_i) \iff \mathcal{O}_X = (f_1, \dots, f_n) $$

Proof (wrong!)

Unwinding the definition of $D_X(f_i)$ we find
$$ X = \bigcup_{i=1}^n D_X(f_i) = \bigcup_{i=1}^n X \setminus \mathcal{Z}(f_i) = X \setminus \left( \bigcap_{i=1}^n \mathcal{Z}(f_i) \right) $$
        which implies that
        $$ \emptyset = \bigcap_{i=1}^n \mathcal{Z}(f_i) = \mathcal{Z}( f_1, \dots, f_n ) \iff (f_1, \dots, f_n) = (1) = \mathcal{O}_X. $$

答:
Ok, as pointed out by Brandon and Jerry the very last step of my proof is wrong: for algebraic varieties it's not true in general that
$$
\emptyset = \mathcal{Z}( f_1, \dots, f_n ) \implies (f_1, \dots, f_n) = (1).
$$

Nevertheless the above implication holds for an affine variety. This can be proved using Hilbert's Nullstellensatz, from which it follows that for ideals $A$ and $B$ we have
$$ \mathcal{Z}(A) = \mathcal{Z}(B) \implies \sqrt{A}=\sqrt{B}. $$
Therefore in the above proof (for an affine variety) this specializes to
$$ \mathcal{Z}(f_i) = \mathcal{Z}(1) \implies \sqrt{(f_i)}=\sqrt{(1)} \implies (f_i) = (1). $$


问:
Divergence free vector fields in general measure space
I'm reading Ambrosio's Gradient Flows in Metric Spaces and the Space of Probability Measures textbook, and I'm stuck on a part of Lemma 8.4.2. Rather than give the whole Lemma, I'll state the specific claim I'm confused by, and I'll simplify it by taking $p = 2$ ($p = 2$ is the only case I'm interested in), $X = \mathbb{R}^n$ (rather than some other separable Hilbert space $X$).

Here $\mu$ is a finite Borel measure on $X$, and $v,w$ are $L^2(\mu)$ vector fields (the integral of $|v|^2$ w.r.t. $\mu$ is finite). The statement is as follows:
\begin{equation*}
\int_X v \cdot w \, d \mu(x) = 0 \, , \text{ for any $w \in L^2(\mu)$ s.t. $\nabla \cdot (w \mu) = 0$}
\end{equation*}

is equivalent to
\begin{equation*}
v \text{ belongs to the $L^2(\mu)$ closure of } \{\nabla \phi \, : \, \phi \in C_c^\infty(X)\} \, .
\end{equation*}

Some clarifications:

$\nabla \cdot (w \mu) = 0$ means that, for any test function $\phi \in C_b^1(X)$, $$\int_X \nabla \phi \cdot w \, d \mu = 0 \, . $$

The second statement means that there is some sequence $\{\phi_n\} \subseteq C_c^\infty(X)$ so that $$\lim_{n \to \infty} \int_X |\nabla \phi_n - v|^2 \, d \mu = 0 \, . $$

Why is the stated equivalence true??? (The textbook states it without justification.)

I think the backward implication is clear: if we can write $v = \nabla \phi$ for some $\phi$, then $\nabla \cdot (w\mu) = 0$ means precisely that $\int_X v \cdot w \, d \mu = 0$.

However, the forward implication is unclear to me. If $\mu$ were just the Lebesgue measure/some absolutely continuous measure w.r.t. Lebesgue then I would try integration by parts (looks like some Sobolev thing going on), but I don't believe this is available to us here? I can't find any integration by parts formulas for arbitrary measures in the textbook.

答:
With all definitions given, the statement is more about Hilbert spaces (in this case $L^2(\mu)$), than anything else.

Notice that the first condition is that $v$ is orthogonal to the set of divergence free  vector fields, while the second condition is that $v$ is orthogonal to the orthogonal complement of smooth gradient fields (recall double orthogonal complement gives the closure of a subspace).

By definition, if $u$ is orthogonal to smooth gradient fields then it's divergence free (this can be justified by approximating $C^1_b$ functions by $C^\infty_c$ ones in an appropriate way) and so by hypothesis $u\perp v$, which gives the implication.

ContentLicense: CC BY-SA 4.0


问:
Mathematic notation $a:I\to \Bbb R^3$
I need a little bit of help with the notation for parametric curve $a$ given below:
$$a:I\longrightarrow\mathbb{R}^3.$$

What does this notation means?

Best regards,
Sergey

答: $a:I\to\Bbb R^3$ means that $a$ is a function with domain $I$ and codomain $\Bbb R^3$. Usually, if we're speaking about curves, $I$ is an interval (perhaps $I=[0,1]$) and $a$ is continuous (perhaps $C^1$).

Title: Mathematic notation $a:I\to \Bbb R^3$


答: $a: I \to \mathbb R^3$ means that $a$ is a function with domain $I$ and codomain $ \mathbb R^3.$

Title: Mathematic notation $a:I\to \Bbb R^3$

Body:
I need a little bit of help with the notation for parametric curve $a$ given below:
$$a:I\longrightarrow\mathbb{R}^3.$$

What does this notation means?

Best regards,
Sergey


问:
Area of an Ellipse using U-Substitution
I am supposed to determine the area of an ellipse using $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1.

I start by setting up an integral for the first quadrant, so I solved the above equation for y and got the integral $\int b\sqrt{1-\frac{x^2}{a^2}} $

The next step is to do a change of variables using u = $\frac xa$, and this is where I get a little confused.  I then have to finish up by using geometry to evaluate the integral.  Could someone explain how to do this? Thanks for any help in advance!

答:
By substituting $u = \frac xa$ we get
$$ \int_0^a b\sqrt{1-\frac{x^2}{a^2}}\,dx = ab\int_0^1 \sqrt{1-u^2}\,du.$$

Now substitute $u = \sin t$ to get

$$ \int_0^1 \sqrt{1-u^2}\,du = \int_0^{\pi/2} \cos t\sqrt{1-\sin^2 t}\,dt = \int_0^{\pi/2} \cos^2 t\,dt = \int_0^{\pi/2} \frac{1+\cos 2t}2\,dt.$$

Can you finish from here?

Title: Area of an Ellipse using U-Substitution


问:
How do you solve a simple simultaneous equation to get unkowns for a sinusoid (with no phase shift) given a couple of points?
This is probably a very simple question, but I am struggling to find the answer.

Given:

\begin{equation}
y_1 = A \cos(\omega t_1)\\
y_2 = A \cos(\omega t_2)\\
y_3 = A \cos(\omega t_3)
\end{equation}

(I can go up to as much as $y_n = A \cos(\omega t_n)$. Using only $y_1$ and $y_2$ is also, probably, fine.)

So I know the $y's$ and $t's$. How do I find $A$ and $\omega$?

I, essentially, want to use that to compute the peak of a discrete signal I'm assuming is a cosine for a very short interval which includes that peak. So I thought if I can obtain the equation of the sinuoid then I can differentiate it, set it to $0$ and then solve for the $t$. That way I can obtain $t$ and $y$ at the maximum/peak.

答:
Considering that you have $n$ data points $(t_i,y_i)$, you want to adust $A$ and $\omega$ in order to minimize
$$\text{SSQ}=\frac12\sum_{i=1}^n \big[A \cos(\omega t_i)-y_i\big]^2$$ Computing the partial derivatives and setting them equal to $0$
$$\frac{\partial \text{ SSQ}}{\partial A}=\sum_{i=1}^n \cos(\omega t_i)\big[A \cos(\omega t_i)-y_i\big]=0\tag1$$
$$\frac{\partial \text{ SSQ}}{\partial \omega}=-A\sum_{i=1}^n t_i \sin (\omega  t_i)\big[A \cos(\omega t_i)-y_i\big]=0\tag2$$

From $(1)$
$$A=\frac {\sum_{i=1}^n y_i \cos(\omega t_i) } {\sum_{i=1}^n \cos^2(\omega t_i) }\tag 3$$ and plugging in $(2)$ you are left with an equation in $\omega$ which can write
$$\Big[\sum_{i=1}^n y_i \cos(\omega t_i)\Big]\Big[\sum_{i=1}^n t_i \sin(2\omega t_i)\Big]-2\Big[\sum_{i=1}^n \cos^2(\omega t_i)\Big] \Big[\sum_{i=1}^n t_iy_i \sin(\omega t_i)\Big]=0$$

Example

Let us use the following data
$$\left(
\begin{array}{cc}
t & y\\
 13 & 49 \\
 14 & 90 \\
 15 & 114 \\
 16 & 124 \\
 17 & 112 \\
 18 & 85 \\
 19 & 42
\end{array}
\right)$$

I "know" that $0.3 \leq \omega \leq 0.5$. Just plotting the equation fot this range, the solution is close to $0.4$.

Starting Newton iterations, the path to solution is
$$\left(
\begin{array}{cc}
 n & \omega_n \\
 0 & 0.4000000000 \\
 1 & 0.3946667642 \\
 2 & 0.3945546651 \\
 3 & 0.3945546065
\end{array}
\right)$$

Plug the solution in $(3)$ to get $A=123.17165$. The recomputed values of the $y$'s are

$$\{49.87,89.33,115.1,123.1,112.3,84.14,43.10\}$$  and the peak is at $t=15.9248$


问:
How to show that R^n is not totally bounded for n>1
I can show that R is not totally bounded, but how do I show that R^n is not totally bounded for n>1? Is this suppose to be a induction proof?</p>

答: Take $\varepsilon=1$ and suppose we can cover $\mathbb{R}^n$ with finitely many open balls of radius $\varepsilon$. Since there are finitely many, their centres are all distance at most $M<\infty$ from the origin (taking the maximum of each of the distances). Any point of distance larger than $M+1$ from the origin will hence not lie inside any of these balls.


问:
$au_x+bu_y=0$ type equations
I know how to solve an equation like this and even a bit more complex but only because I've done several exercises like these a while back. In a problem like this I typically just do $dy/dx= b/a$ and get the constant of integration by itself and write $u(x,y)=f(C)$ But I'm realizing that I'm not really understanding what is going on in the geometric (directional derivative solution) sense. I've looked at youtube but nothing satisfactory. I feel like it's a stupidly simple fix or push in the right direction I need. I'm missing something.

Anyways here is a picture:

Why do we have $f(bx-ay)$? I understand 2.3 but not 2.4. I don't understand the "line containing the point (x,y)..."

if $(x,y)=(a,b)$  (or any multiple of this vector) we get f(0) but what does that mean? at $(-b,a)$ which is perpendicular to $(a,b)$ we get $f(-b^2-a^2)$.

Suppose $a=2$ and  $b=3$ then $u(x,y)=f(3x-2y)$ . like what is f? I know it's an arbitrary function which needs to be specified initially but it's still a bit confusing. at (1,0) we get u(1,0)=f(3). at another point it changes. how can it be constant? Please provide clarity

答:
This equation is the transport equation. It might help you if you think of $y$ as time. The solution $u(x,t)$ is constant along the characteristic lines, that is, to find the solution $u(x,t)$ for $t>0$ you just have to travel back in time along the characteristic line passing by the point $(x,t)$. Since the equation is linear, the characteristic are parallel lines, so it is uniquely determined.

To get a well posed-problem (i.e giving a unique solution), you need some information on $u$ at some point $t_0$. Let's say you know $u(x,t_0)=f(x)$, $\forall x \in\mathbb R$. This additional assumption will fix your solution, in the sense that you won't have an infinite number of solutions anymore.

What the fact that $u$ is constant along the characteristics tells us is that to find the value of $u$ at the point $(x,t)$, we just need to travel down the characteristic passing by the point (dotted line in the drawing) until we reach a point $(\bar x, \bar t)$ at which we know the solution. In the pic, I travel down the dotted line until it intersects the $x$ axis because I assumed that we know $u(t=0)=f$. Hence the value of $u(x,t)=f(\xi)=f(ax-bt)$.

Obviously with this approach you can also take a point $(x,t)$ with $t<0$ and travel up the right characteristic to find the solution.

Title: $au_x+bu_y=0$ type equations


问:
Trigonometry limit:: $\lim _ { x \rightarrow \pi/2 } \frac { \ln \sin x } { \cos ^ { 2 } x }$
$$\lim _ { x \rightarrow \frac { \pi } { 2 } } \frac { \ln \sin x } { \cos ^ { 2 } x }$$

I need to find the limit without using L'Hopital's theorem

I tried to solve the limit using the first remarkable limit, but I couldn't find an equivalent function because $x$ tends to $\pi/2$ and not to zero. In general, I'm at a dead end

I would be grateful if you specify the true path

答: Hint: Since $\ln (1+s) \sim s$ as $s \to 0$ we get $\lim_{s\to 0} \frac {\ln (1+s)} {1-(1+s)^{2}}=-\frac 1 2$. Now put $s=\sin x-1$ in this.


问:
Rouché Theorem to calculate the number of zeros
How can I calculate the number of zeros of $\cos z+3z^3$ using the Rouché Theorem?</p>

答:
Let $f(z) = 3z^3$ and $g(z) = \cos z$. Then on $|z|=1$, we have $|f(z)| = 3$ and
$$
|g(z)| = \left| \frac{e^{iz}+e^{-iz}}{2} \right| \le \frac12 ( |e^{iz}| + |e^{-iz}| ) \le e < 3.
$$

Rouché implies that $f+g$ and $f$ have the same number of zeros inside the unit circle, i.e. $3$. (The estimate on $|\cos z|$ is very rough, but enough for this purpose.)

ContentLicense: CC BY-SA 3.0


问:
rigid measure preserving transformation
I want to proof Lemma 6.7.2 in C.E. Silva's Book "Invitation to Ergodic Theory" (the proof is left as an excercise).

A finite measure-preserving transformation $ T $ is said to be rigid if for all measurable sets $ A $ and for every $ \varepsilon > 0 $, there exists an integer $n>0$, such that $\mu(T^{-n}(A)\triangle A) <\varepsilon$.

The lemma I'm trying to proof states:

$ T $ is rigid $\iff$ there is a sequence $n_k\rightarrow\infty$ such that $ \lim\limits_{k\to\infty}\mu(T^{-n_k}A\triangle A)=0 $, for all measurable sets $A$.

So, I think $"\Leftarrow"$ is clear. The other implication seems to be a little more tricky to me.

I found in another paper on the topic, that it's sufficient to prove that for every measurable $A$ there is a rigidity sequence, i.e. a sequence $ n_k = n_k(A)$, such that $ \lim\limits_{k\to\infty}\mu(T^{-n_k}A\triangle A)=0 $.

At this point I got lost. If I take any measurable set $A$, I'm not sure how I can construct such a sequence, only knowing, that for every $\varepsilon>0$ there exists some $n$, such that $\mu(T^{-n}(A)\triangle A) <\varepsilon$.

Any help would be appreciated.

答:
You can construct $\{n_k\}$ recursively. For each $k\ge 1$, let
$$
n_k= \min\left\{n>n_{k-1}:\mu(T^{-n}(A)\Delta A)<2^{-k}\right\},
$$
with $n_0\equiv 0$. Then $n_k\to\infty$ and $\mu(T^{-n_k}(A)\Delta A)\to 0$ as $k\to\infty$.

It remains to show that $n_k$ is well defined for all $k$, i.e., for a given $n_{k-1}$, there exists $n>n_{k-1}$ s.t. $T^{-n}(A)$ approximates $A$. Suppose not. Then $\mu(T^{-n}(A)\Delta A)\ge 2^{-k}$ for all $n>n_{k-1}$. Therefore, by the rigidity of $T$, it must be true that $\mu(T^{-n_{k-1}}(A)\Delta A)=0$. Set $B:=T^{-n_{k-1}}(A)$. Since $T$ is rigid, there is $n>0$ s.t. $\mu(T^{-n}(B)\Delta A)=\mu(T^{-n}(B)\Delta B)<2^{-k}$, a contradiction.

Title: rigid measure preserving transformation


问:
$P(E_i) = 0, \forall i\in I \Leftrightarrow P(\bigcup_{i\in I} E_i) =0$ and $P(E_i) = 1, \forall i\in I \Leftrightarrow P(\bigcap_{i\in I} E_i) =1$

Prove that $P(E_i) = 0, \forall i\in I \Leftrightarrow P(\bigcup_{i\in I} E_i) =0$ and $P(E_i) = 1, \forall i\in I \Leftrightarrow P(\bigcap_{i\in I} E_i) =1$, where $I$ is a countable index set.

$P$ is a probability measure, and we have generalized Boole's inequality, to begin with:
$$0\le P\left(\bigcup_{i=1}^\infty E_i \right) \le \sum_{i=1}^\infty P(E_i)$$

Is the following proof okay?

$P(E_i) = 0$ for all $i\in I$ then $P(\cup_{i\in I}E_i)=0$ by Boole's inequality above. If $P(\cup_{i\in I}E_i)=0$ holds, then Boole's inequality is probably of no use. I tried a proof by contradiction. Let's suppose there exists $j\in I$ such that $P(E_j) \neq 0$. Then $P(\cup_{i\in I}E_i)=0$ is definitely absurd - but how do I put this in mathematical terms? $(\star \star \star)$

In the other case, we have $P(E_i^c) = 0$ for all $i\in I$, which tells us (with the help of the first part) that $$P\left(\bigcup_{i\in I}E_i^c\right)=0 \implies P\left(\bigcap_{i\in I}E_i\right)^c=0 \implies P\left(\bigcap_{i\in I}E_i\right) = 1$$ and the proof is complete.

If everything above sounds fine, I only need help with the line marked $(\star \star \star)$. Thanks!

答:
For (1), note that for any $i,j \ P(A_i \cap A_j)=P(A_i|A_j)P(A_j)=0$ and in general, using product rule, $P(A_1 \cap A_2 \cap \ldots \cap A_n)=P(A_1)P(A_2|A_1) \ldots P(A_n|A_{n-1} A_{n-1} \ldots A_1)=0$, therefore
$$
P(\cup_{i=1}^{n}A_j) = \sum_j P(A_j) -\sum_{j <i}P(A_j \cap A_i) +\ldots+(-1)^{n}P(A_1 \cap A_2 \cap \ldots \cap A_n)
$$
The first sum are $0$ by the definition, the second by the first argument above, the remaining terms by the second argument above, so the limit of the expression is $0$.

For (2), note that $A$ \ $B \cup B= A\cup B$. Since they are disjoint, $P(A$ \ $B)+P(B)=P(A \cup B)$, and, obviously $P(A \cup B) \geq P(B)=1 \text{ so } P(A \cup B)=1$ and $P(A$ \ $B=0$). Also, from $P(A)=P(A \backslash B)+P(A \cap B)$ it follows that $P(A|B)=1=P(A)$, so A and B are pairwise independent. Using the same logic, $P(A_1 \cup A_2 \cup A_3)=P(A_1 \cap A_2 \cap A_3)=1=P(A_1)P(A_2)P(A_3)$. This is the first step in the inductive proof.

Assume by induction, $P(\cup_{n}A_j)=P(\cap_n A_j)=1$. Next step,
$$
P(\cup_{n+1}A_j) = P(\cup_n A_j)+P(A_{n+1}) -P(\cap_{n+1}A_j)= 2 - P(\cap_{n+1}A_j)
$$
Hence $P(\cap_{n+1}A_j)=1$. Hence $A_j$ are independent, and P($\cap_n A_j)=1$.


Title: Trigonometry limit:: $\lim _ { x \rightarrow \pi/2 } \frac { \ln \sin x } { \cos ^ { 2 } x }$

Body:
$$\lim _ { x \rightarrow \frac { \pi } { 2 } } \frac { \ln \sin x } { \cos ^ { 2 } x }$$

I need to find the limit without using L'Hopital's theorem

I tried to solve the limit using the first remarkable limit, but I couldn't find an equivalent function because $x$ tends to $\pi/2$ and not to zero. In general, I'm at a dead end

I would be grateful if you specify the true path

Body:
$$\lim _ { x \to \frac { \pi } { 2 } } \frac { \ln \sin x } { \cos ^ { 2 } x }\\
=\lim _ { x \to \frac { \pi } { 2 } } \frac { \ln( \sin x -1 +1)} { 1-\sin ^ { 2 } x }\\
=\lim _ { x \to \frac { \pi } { 2 } } \frac { \ln( 1+ \sin x -1)} { -(\sin x-1)(1+\sin x) }\\
=\frac{-1}{2}$$
As $x \rightarrow \frac{\pi}{2}, \sin x \rightarrow 1$


答:
Let $t= \sin x$. Then we have to compute

$$\lim_{t \to 1} \frac{ \ln t}{1-t^2}.$$

We have

$$\frac{ \ln t}{1-t^2}=\frac{ 1}{1+t} \cdot \frac{ \ln t}{1-t}.$$

It is clear that $\frac{ 1}{1+t} \to 1/2$ as $t \to 1.$

Let $t=e^s$, then

$$\lim_{t \to 1} \frac{ \ln t}{1-t}= \lim_{s \to 0} \frac{s}{1-e^s}.$$

If $f(s):= e^s,$ then

$$ \frac{e^s-1}{s}= \frac{f(s)-f(0)}{s-0}=f'(0)=1.$$

Hence $\lim_{s \to 0} \frac{s}{1-e^s}=-1$

This gives

$$\lim _ { x \rightarrow \frac { \pi } { 2 } } \frac { \ln \sin x } { \cos ^ { 2 } x }=-1/2.$$


答:
I have a lenthier method, put sinx=$e^t$ As x→π/2,sinx→1, so the limit t tends to 0. ($e^0=1$)

we get

$lim_{t\rightarrow 0} \frac{t}{1-e^{2t}}$=

$lim_{t\rightarrow 0} \frac{t}{\left(1-e^t\right)\left(1+e^t\right)}$=

$lim_{t\rightarrow 0} \frac{t}{2\left(1-e^t\right)}$

Now using taylor series expansion for $e^x$ we easily get the limit as -1/2. Can you proceed or should I explain ahead?


问:
General solution of differential equation of order 3
Please ,how to find that the general solution of  $u'''(t)=e(t) , t\in [0,1]$ is given by
$u(t)=c_0+c_1t+c_2 t^2 +\frac12 \int_0^t (t-s)^2 e(s) ds$

$e:(0,1)\rightarrow \mathbb{R}$, and $e\in L(0,1)$.

i think that it is general homogeneous solution + particular solution ,

the general homogeneous is $c_0+c_1 t+ c_2 t^2$ , but i dont know how to finde that the paricular solution is $\frac12 \int_0^t (t-s)^2 e(s)ds $

Please;

Thank you

答:
You can find the particular solution by doing an integration by parts.

$\frac12 \int_0^t (t-s)^2 e(s)ds = [0-0] -2* \frac12 \int_0^t (t-s) E(s)ds $ where E(s) is a primitive of $e(s)$ such that $E(0)=0$

Similarly we have : $-\int_0^t (t-s) E(s)ds = - ( [0-0] - \int_0^t F(s)ds)$ where $F(s)$...

And u(t)=$\int_0^t F(s)ds$ is a solution of $u'''(t)=e(t)$

EDIT, to be clearer :

u(t)=$\int_0^t u'(s)ds + u(0)$

u(t)=$-\int_0^t (t-s) u''(s)ds + u'(0)t + u(0)$

$u(t)=\frac12 \int_0^t (t-s)^2 u'''(s) ds + u''(0)/2*t^2 + u'(0)t + u(0) = \frac12 \int_0^t (t-s)^2 e(s) ds + u''(0)/2*t^2 + u'(0)t + u(0) $


问:
Number of ways to choose $4$ items from $6$ the same items such that order doesn't matter using the number of ways when the order does matter
I have 2 problems and I was able to solve each.

Problem 1

How many ways you can draw 4 items from a box containing $6$ indistinguishable items such that the order is important?

My answer is $6^4$.

Problem 2

How many ways you can draw $4$ items from a box containing $6$ indistinguishable items such that the order is NOT important?

My answer is $84$ using the stars and bars concept.

My question is how can I use the answer in Problem 1 to solve Problem 2 and not using the stars and bars concept or any formula?

答:
The answer is no. Let us number the items from 1 to 6.

How many ways you can draw 4 items from a box containing 6 different but indistinguishable items (and put it back after drawn) such that the order is important?

This answer is $6^4$ as you written.

How many ways you can draw 4 items from a box containing 6 different but indistinguishable items (and put it back after drawn) such that the order is not important?

If you are trying to derive 2 from 1, you would need to basically cut out the dulplicates, so a case-by-case examine is required. For example, the 4 ways to draw in the first problem
$$
(1,2,2,2), (2,1,2,2), (2,2,1,2), (2,2,2,1)
$$
corresponds to only one way to draw $\{1,2,2,2\}$ in the second problem. And there are 24 ways to draw in the first problem that only correspond to only one ways to draw $\{1,2,3,4\}$. Till this point you can see how undesirable is to cut everying dulplicate from the $6^4$ down to 84.

And well, $6^4/84$ is not even an integer, you see.


问:
Is it possible to write this piecewise-defined function as a regular function?
I have this piecewise-defined function:
$$
g(a, b, e) = \left\{\begin{array}{ll}
1 - e & \text{if }a > b,\\
0.5 & \text{if }a = b,\\
e & \text{if }a < b
\end{array}\right.
$$
$$
e \in \mathopen(0,0.5\mathclose)
$$
I need it so that it can be written without the if else.
I already figured that it would be possible to have -1 or 1 depending on whether a is greater than b or vice versa:
$$
\text{sign} = \frac{a-b}{|a-b|}
$$
However, I am stuck here. Is this possible at all?

答:
Using the sign function you defined, you could take

$$g(x)=e+\frac{1-2e}{2}\left(1+\frac{a-b}{|a-b|}\right)=\frac{1}{2}+\frac{a-b}{|a-b|}\left(\frac{1}{2}-e\right)$$

This if $a>b$, we get $e+(1-2e)=1-e$ and if $a<b$ we get $e+0=e$. We still have zero-division problems when $a=b$, but I believe this will always be the case when trying to write functions that are discontinuous in their parameters as `regular' functions.

EDIT:
If you want to code this in Python, you can actually use the cases natively in your formula. If they are true, they will evaluate to $1$, else $0$. Hence, your function could look like

e*(a>b) + 0.5*(a==b) + (1-e)*(a<b)


问:
Check if a given set of vectors is the basis of a vector space
I'm given the following basis: $\beta = \{1 + X, 1 + X², X + X²\}$ and we are asked to verify that this is a basis for the vectorspace $(\mathbb{R}, \mathbb{R}[X]_{\leq2}, +)$. Now the definition of a basis says that $\beta$ is a basis for a vectorspace iff $\beta$ is linearly independent and $span<\beta>$ $= \mathbb{R}[X]_{\leq2}$.

I thus set out to prove the two properties. The first part of my question is more verification that the following method is allowed. I interpreted the basis to be of the form $\beta = \{(1, 1, 0), (1, 0, 1), (0, 1, 1)\}$ and solved the following linear systems of equations to check if the vectors of the basis are indeed linearly independant (the unknown variables are in the end represented as $\lambda_1, \lambda_2$ and $\lambda_3$):

$$ \begin{pmatrix}
1 & 1 & 0 & | & 0\\
1 & 0 & 1 & | & 0\\
0 & 1 & 1 & | & 0\\
\end{pmatrix} \begin{matrix} ~ \\ \xrightarrow{R_2 \rightarrow R_2 - R_1}\end{matrix} \begin{pmatrix}
1 & 1 & 0 & | & 0\\
0 & -1 & 1 & | & 0\\
0 & 1 & 1 & | & 0\\
\end{pmatrix} \begin{matrix} ~ \\ \xrightarrow{R_3 \rightarrow R_3 + R_2}\end{matrix} \begin{pmatrix}
1 & 1 & 0 & | & 0\\
0 & -1 & 1 & | & 0\\
0 & 0 & 2 & | & 0\\
\end{pmatrix} $$

From this matrix I could see that using backwards substitution, the values of $\lambda_3 = 0, \lambda_2 = 0$ and $\lambda_1 = 0$ and thus that the vectors are indeed linearly independent of each other.

The second part of the problem however I have no idea how to check. Is there a general method for checking if any basis spans the vectorspace? I have tried to look for this but cannot seem to find a satisfying answer to the question in proof-form. Any hint at how to start would already be much appreciated!

答: $\{1,X,X^{2}\}$ is a basis for your space. So the space is three dimensional. This implies that any three linearly independent vectors automatically span the space.


问:
I don't understand $\ln$ properties when it comes to absolute value.
Let's say we have a function with absolute value like:

$f(x) = \ln\vert x\vert$ where $x$ is any real number except 0

Now, when we get rid of the absolute value, we get this:

$f(x) = \ln(x)$ where $x$ is positive

$f(x) = \ln(-x)$ where $x$ is negative

But here is the thing, $\ln$ properties allow us to do something like

$\ln(xy) = \ln(x) + \ln(y)$

So we apply this to the second form of function

$f(x) = \ln(-x)$ means

$f(x) = \ln(-1) + \ln(x)$

But $\ln$ is undefined for negative real numbers so I don't get it, how does this work?

答: $\ln (xy)=\ln (x)+\ln (y)$ holds only for positive numbers $x$ and $y$. You cannot use negative numbers in this.


答:
The absolute value is present exactly to avoid this.

When you write $f(x) = ln|x|$ It means that a negative number cannot come inside the bracket.

so when you apply the second form of function, x is a negative number and negative and negative cancel out. What you have done is similar to

$ln(1)=ln(-1*-1)=ln(-1)+ln(-1)$, which cannot be done, as the property is defined only for positive real numbers :~)

Title: I don't understand $\ln$ properties when it comes to absolute value.

Body:
Let's say we have a function with absolute value like:

$f(x) = \ln\vert x\vert$ where $x$ is any real number except 0

Now, when we get rid of the absolute value, we get this:

$f(x) = \ln(x)$ where $x$ is positive

$f(x) = \ln(-x)$ where $x$ is negative

But here is the thing, $\ln$ properties allow us to do something like

$\ln(xy) = \ln(x) + \ln(y)$

So we apply this to the second form of function

$f(x) = \ln(-x)$ means

$f(x) = \ln(-1) + \ln(x)$

But $\ln$ is undefined for negative real numbers so I don't get it, how does this work?


问:
Equations Proof Solvers
Can Someone solve

$$x(x+1)(x+2)...(x+2020)=1$$

analitically rather than numerically and show me all steps. Also maybe with some preconditions, like only natural solutions, so that it‘s adaptable to diophantine equations, etc.

Thank you all in advance!!!

答:
Write the equation like $$(x+1) (x+2) (x+3) (x+4) (x+5)\ldots(x+k)=\frac{1}{x}$$ The intersection between the polynomial $y=(x+1) (x+2) (x+3) (x+4) (x+5)\ldots(x+k)$ and the hyperbola $y=1/x$ are exactly $k$ if $k>3$.

This happens because $P_k(x)=(x+1) (x+2) (x+3) (x+4) (x+5)\ldots(x+k)$ has $k$ zeros and
in any interval $(j-1,j);\;j\in\mathbb{Z},j<0$ there is a minimum $\xi_j$ and a maximum $\zeta_{j-1}$ for $j=\ldots,-3,-2,-1$.

TO DO

We must prove that the local minima $\xi_j$ have negative values $P(\xi_j)<\frac{1}{\xi_j}$ for $|j|>3$ therefore the curve $y=P_j(x)$ intersects the hyperbola in $|j|$ points for $|j|>3$.

$$...$$

Title: Equations Proof Solvers


答: The equality $\ln (xy) = \ln x + \ln y$ only holds if all components are well defined, namely if $x,y >0$. The same happens in situations that do not involve the absolute value... For instance $\ln (x^2)$ is well defined for any $x \ne 0$, but the property $\ln(x^2) = 2 \ln x$ only holds for $x>0$.


问:
For $x_{n+1}=x_n-x_n^3$, with $|x_1|>1$. What about the convergence?
I see Let $\{ x_n \}_{n=1}^{\infty}$ such that $x_{n+1}=x_n-x_n^3$ and $0<x_1<1$ that if $|x_1| \le 1$, then $x_n\to 0$. However, if $|x_1|>1$, what can we say about this sequence? It seems hard to find the tendency of it. For $x_1$ near $1$, it converges to $0$, but for $x_1$ large, it is not the case.

答:
Let us first state some useful facts.

If some $x_m = -1, 0, 1$, then all $x_n = 0$ for $n > m$ and thus $x_n \to 0$.

$x_{n+1} = x_n(1 - x_n^2) = x_n(1 - \lvert x_n \rvert^2)$.

If $\lvert x_n \rvert \in (0,1)$, then $\lvert x_{n+1} \rvert = \lvert x_n \rvert (1 - \lvert x_n \rvert^2) \in (0, \lvert x_n \rvert)$:

Obvious, for $\lvert x_n \rvert \in (0,1)$ we have $\lvert x_{n+1} \rvert = \lvert x_n \rvert (1 - \lvert x_n \rvert^2)$ and $0 <  \lvert x_n \rvert (1 - \lvert x_n \rvert^2)  < \lvert x_n \rvert$.

Hence, if some $\lvert x_m \rvert \in (0,1)$, then the sequence $(\lvert x_n \rvert)_{n \ge m}$ is strictly decreasing with values in $(0,1)$.

If some $\lvert x_m \rvert \in [0,1]$, then $x_n \to 0$:

If $\lvert x_m \rvert = 0,1$, then 1. applies. If $\lvert x_m \rvert \in (0,1)$, then by 4. the sequence $(\lvert x_n \rvert)$ converges to some $\xi \in [0,1)$. But then $\xi = \xi(1- \xi^2)$ by 3. which has the unique solution $\xi = 0$.

If all $\lvert x_n \rvert > 1$, then $x_{n+1} = -x_n(\lvert x_n \rvert^2 -1)$ for all $n$. Thus the sequence $(x_n)$ has alternating signs and thus does not converge.

Note that if $\lvert x_n \rvert > 1$, then $\lvert x_{n+1} \rvert= \lvert x_n \rvert(\lvert x_n \rvert^2 -1)$.

Let us prove the follwing result which gives a complete answer of the question:

All $\lvert x_n \rvert > 1$ if and only if $\lvert x_1 \rvert \ge \sqrt 2$.

If $\lvert x_1 \rvert = \sqrt 2$, then we see by induction that $\lvert x_n \rvert = \sqrt 2$ for all $n$.

If $\lvert x_1 \rvert > \sqrt 2$, then $(\lvert x_n \rvert)$ is strictly increasing:
By induction using 7. we get $\lvert x_{n+1} \rvert= \lvert x_n \rvert(\lvert x_n \rvert^2 -1) > \lvert x_n \rvert (\lvert x_1 \rvert^2 -1) > \lvert x_n \rvert$. One can moreover show that $\lvert x_n \rvert \to +\infty$, but that is irrelevant.

Let us finally look at $1 < \lvert x_1 \rvert < \sqrt 2$:

Assume that all $\lvert x_n \rvert > 1$. Then by 7. (using induction) we get $1 < \lvert x_{n+1} \rvert < \lvert x_n \rvert < \sqrt 2$ for all $n$. Therefore $(\lvert x_n \rvert)$ is strictly decreasing. Thus $(\lvert x_n \rvert)$ converges to some $\xi \in [1, \sqrt 2)$. But then $\xi = \xi(\xi^2 -1)$ which has solutions $\xi = 0, \pm \sqrt 2$. None of them lies in $[1, \sqrt 2)$, thus the above assumption was wrong and we get some $\lvert x_n \rvert \in [0,1]$.


问:
Singularities of $\operatorname{arctanh}(x)$
The function $\operatorname{arctanh}(x)$ is undefined for all $\vert x\vert\ge 1$.

Does this mean, because the graph shoots up to infinity at $1$ and $-1$, that these are the singularities of the function,

or would this not be true since it actually breaks down for every point after as well?

For example, $\tan(x)$ has singularities at $\pi\over 2$, but is also defined before and after this point, unlike $\operatorname{arctanh}$.

答:
$$\tanh x=\frac{e^{2 x}-1}{e^{2 x}+1}$$
The range of $\tanh x$ is $(-1,1)$ therefore the domain of $\text{arctanh }x$ is $(-1,1)$.

Solving $\frac{e^{2 x}-1}{e^{2 x}+1}=y$ we get $x=\frac{1}{2} \log \left(\frac{1+y}{1-y}\right);\;|y|<1$.

The graph of arctanh x is the symmetric of the graph of $\tanh x$ wrt the line $y=x$. Domain and range of the function and its inverse swap.

See the picture below

$$...$$


问:
Really easy algebra problem
Can somebody show me how to go about solving the following (really easy) equation for $\alpha$ please?
$$\displaystyle 0 = \sum_{n=1}^{N}\left( \frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)$$</p>

答:
Multiplying through by $\alpha(1-\alpha)$ you get
$$\begin{align*}
0 &= \sum_n\left(\frac{y_n}{\alpha} + \frac{1-y_n}{1-\alpha}\right)\\
0 &= \sum_{n}\left( (1-\alpha)y_n + \alpha(1-y_n)\right)\\
0 &= \sum_n \left( y_n + (1-2y_n)\alpha\right)\\
0&= \sum_n y_n + \sum_n \alpha(1-2y_n)\\
\sum_n \alpha(2y_n -1) &= \sum_n y_n\\
\alpha\sum_n(2y_n-1) &= \sum_n y_n\\
\alpha &= \frac{\displaystyle \sum_n y_n}{\displaystyle\sum_n (2y_n-1)}.
\end{align*}
$$</p>

Title: Really easy algebra problem


问:
A simple 2 grade equations system
If we have:
$$x^2 + xy + y^2 = 25 $$
$$x^2 + xz + z^2 = 49 $$
$$y^2 + yz + z^2 = 64 $$

How do we calculate $$x + y + z$$

答:
$$x^2 + xy + y^2 = 25 \dots (1)$$
$$x^2 + xz + z^2 = 49 \dots(2)$$
$$y^2 + yz + z^2 = 64 \dots(3)$$

$(2)-(1)$

$x(z-y)+(z+y)(z-y)=24$

$\Rightarrow (x+y+z)(z-y)=24$

Similarly we get ,

$(x+y+z)(y-x)=15$ by $(3)-(2)$

$(x+y+z)(z-x)=39$ by $(3)-(1)$

Clearly Let $1/\lambda=(x+y+z)\ne 0$

Then we have,

$(z-y)=24\lambda$

$(y-x)=15\lambda$

$\Rightarrow x+z-2y=9\lambda$

$\Rightarrow 3y=1/\lambda+9\lambda\Rightarrow y=1/3\lambda+3\lambda$

Now solve for x and put in the first equation to find the value of $\lambda$.

Title: A simple 2 grade equations system


答:
The hyperbolic tangent is a bijective (strictly increasing) function $\tanh:\Bbb{R}\to (-1,1)$. Note that here the domain is all of $\Bbb{R}$, and the image is $(-1,1)$. This is in contrast to the tangent function whose domain is not all of $\Bbb{R}$, and the image is all of $\Bbb{R}$, and the tangent function is only injective once you restrict the domain appropriately (for example to $(-\frac{\pi}{2},\frac{\pi}{2})$).

The reason for this is that $\tanh$ is defined as $\frac{\sinh}{\cosh}$, and $\cosh(x)\geq 1$ for all $x\in\Bbb{R}$. Hence, the quotient never goes off to $\pm\infty$, unlike for $\tan$. So, yes, the inverse $\text{argtanh}: (-1,1)\to\Bbb{R}$ is also a strictly increasing function with $\lim\limits_{x\to -1^-}\text{argtanh}(x)=-\infty$ and $\lim\limits_{x\to 1^+}\text{argtanh}(x)=\infty$. I suggest using wolfram alpha/desmos to sketch the graphs of $\tanh$ and $\text{argtanh}$. So, sure if you wish, you can call $-1$ and $1$ singularities of $\text{argtanh}$, because it "blows up" there, and hence there is no way of continuously extending the definition outside of $(-1,1)$.

Title: Singularities of $\operatorname{arctanh}(x)$

Body:
The function $\operatorname{arctanh}(x)$ is undefined for all $\vert x\vert\ge 1$.

Does this mean, because the graph shoots up to infinity at $1$ and $-1$, that these are the singularities of the function,

or would this not be true since it actually breaks down for every point after as well?

For example, $\tan(x)$ has singularities at $\pi\over 2$, but is also defined before and after this point, unlike $\operatorname{arctanh}$.


问:
Scaling a parallelogram to fit a triangle
How do I, with straightedge and compass, scale the red parallelogram to the green one, given the blue triangle so that both top corners of the green parallelogram touches the triangle?

答: Draw a line, parallel to the sides of the parallelogram, through the apex of the triangle. The point where this line intersects the base of the triangle is the centre of the scaling operation, so if you draw lines from there to the corners of the red parallelogram they will go through the corners of the green parallelogram as well exactly where they intersect the sides of the triangle.


问:
Example of global holomorphic form which is not closed.
Let $X$ be a compact complex manifold. Any global holomorphic $k$-form $\omega \in H^0(X, \Omega^k)$ is automatically harmonic with respect to any choice of hermitian structure on $X$. When $X$ is Kähler, by the Kähler identities, this implies that $\mathrm{d}{\omega} = 0$.

I am wondering if the same is true when $X$ is not Kähler. If not, what is an example of a nonclosed global holomorphic $k$-form.

答:
This happens on the Iwasawa manifold.

Let $G$ be the Heisenberg group of complex upper-triangular $3 \times 3$ matrices
$$
G = \begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}
$$
and let $\Gamma \subset G$ be the discrete subgroup generated by matrices whose non-diagonal entries are Gaussian integers. The Iwasawa manifold is the quotient $X := G / \Gamma$. It is a compact complex manifold.

There are three linearly independent holomorphic forms on $X$: $\alpha = dx$, $\beta = dy$ and $\gamma = dz - x \, dy$. They're defined by looking at the entries of $M^{-1} \, dM$ for $M(x,y,z) \in G$.

We see that $d \gamma = - \alpha \wedge \beta \not= 0$, so $\gamma$ is a non-closed holomorphic form. In particular, this proves $X$ is not Kahler.

The facts here are pretty much verbatim from Michel Schweitzer's manuscript Autour de la cohomologie de Bott-Chern.


问:
How to solve $at + b = 0 \pmod {(a-t)}$?
Is there any way (except trying for $t=0,1,2,\ldots,a-1)$ to solve the following equation for $t$ when $a$ and $b$ are known?

$$ at +b = 0 \pmod{(a-t)} \text{ with } a,b,t \in N $$

Body: $at +b = 0 \mod\ (a-t)$ then $at+b=-a^2+at \mod\ (a-t)$ we have $$b+a^2=0 \mod\ (a-t)$$ and $$b+a^2=k(t-a)$$ $$t=\frac{b}{k}+\frac{a^2}{k}+a: k\in \mathbb Zk|b ,k|a^2$$


问:
Interchanging lim and log
Suppose I have a n-dependent function $f_n(X,Y)$, and I have the following result

$$
\lim_{n \rightarrow \infty} \frac{1}{n} \ln\left( f_n(X,Y)\right) = g(X,Y)
$$

From this, can we deduce that $f_n(X,Y) = \exp( n g(X,Y))$ as $n \rightarrow \infty$?

答:
Assuming we are talking about pointwise limits, we have that
$$
\lim \frac 1n \ln(f_n(x,y)) = \lim \ln (f_n(x,y)^{1/n}) = \ln \left( \lim (f_n(x,y)^{1/n})\right)
$$

So, what you can say is

$$
\lim f_n(x,y)^{1/n} = e^{g(x,y)}.
$$

Note: the interchange between $\ln$ and $\lim$ is justified by the continuity of the logarithm function.

Title: Interchanging lim and log


问:
Permutation-invariant embedding preserving as much information as possible
Suppose there are 2 discrete random variables $x$ and $y$ that each can take $n$ possible values, and $n$ is not too large, at most 10. Then their joint PMF $p \in \mathbb{R}^{n^2}$ is a histogram of $n^2$ bins, which are all non-negative and sum up to 1. I am interested in finding a map $f: \mathbb{R}^{n^2} \rightarrow \mathbb{R}^m$ which satisfies the following:

It is symmetric under exchange of the random variables
$$f(p[x, y]) = f(p[y, x])$$

It preserves as much information about its argument as possible. That is, given $f(p)$ is should be possible to reconstruct as much of $p$ as possible. I'm not quite sure how exactly to formalize this statement, I welcome suggestions.

The dimension $m$ is a free parameter and can be chosen sufficiently large to satisfy the above constraints. Smaller $m$ are preferable if constraints are satisfied. The goal is to find an algorithm that implements any solution to this problem and demonstrate that it satisfies the conditions. In case optimal solution is intractable, an approximate solution may also be of interest.

My attempts:

$f(p)$ = const. For sure satisfies first condition but not the second.

$m = n^2$, two indices go over the values of $x$ and $y$ respectively and

$$
f_{ij} = \begin{cases}
  \min(p_{ij}, p_{ji}) & i < j \\
  \max(p_{ij}, p_{ji}) & i \geq j
\end{cases}$$

This looks like a good candidate solution, but I don't know how to check if it is optimal or if one can do better. I strongly suspect it is not optimal, because, for example, given $p$ I can test if $x$ and $y$ are independent, but given this version of $f$ I can't, and this information should not depend on their order.

Edit: Let me make the independence example more clear. If $x$ and $y$ are statistically-independent, then $p[x,y] = p[x]p[y]$, where $p[x] = \sum_y p[x,y]$. The test for statistical independence is $\Delta = p[x,y] - p[x]p[y] = 0$. $\Delta$ is invariant of the order of $x$ and $y$, and thus an optimal $f(p)$ should preserve that information. It is easy to verify that my attempt 2. does not preserve this information, so it is not optimal

Edit 2: There was a request to know the purpose of this exercise. It is as follows. I have a recording of many variables (e.g. 50) over many trials (e.g. 1000). I want to explore what sort of pairwise and triple-wise statistical relations are likely among these variables. For this purpose, I can compute all pairwise and triple-wise empirical PMF's. Now I would like to perform some classification/clustering on those PMF's. A standard tool to use a low-dimensional embedding, e.g. tSNE. However, to do the embedding, the dimensions of the PMFs have to have the same meaning for all PMFs. This is where the problem comes in - if I can succeed to filter out the permutation-asymmetric part from the PMFs, the resulting vectors will all have the same meaning, and can be clustered. For this reason, there are other properties of $f$ that are desirable for good result. Namely, it would be good if $f$ does not experience discontinuous jumps due to small perturbations. Generally, it would be good if $f$ does not amplify small differences and does not dampen large differences.

答:
This is not a complete answer to the intended question -- but it aims to show points where the question might need additional clarification. I will also borrow the title of a movie: The Good, the Bad and the Ugly.

Contrary to the usual order, I'll start with the (obvious) Bad news: The first condition forces us to drop at least some of the information from the original PMF; since $f(p[x,y])=f(p[y,x])$, knowing only the value of $f$ does not allow us to determine, for example, the probability $P(x>y)$. In general, any question whose answer is not invariant under swapping $x$ and $y$ is going to become unanswerable.

Now, the Good news is that aside from this inevitable loss of information, we don't need to drop anything else as long as $m\geq(n^2-1)$. Choose one fixed reading order $o(p[])$ of the $n\times n$ table $p[]$ (e.g. reading it row-by-row, or column-by-column, or any other ordering which lists every item exactly once) which results in a list of $n^2$ numbers.

Since lists of numbers of equal length can be compared lexicographically (i.e. comparing them element-by-element and once we encounter differing elements, the list with the greater element is declared to be greater of the two) and $p[y,x]$ can be obtained from $p[x,y]$ without any additional information, we can define $$f(p[x,y])=\max\left(o(p[x,y]),o(p[y,x])\right)$$
Since $o(p[x,y])$ determines $p[x,y]$ completely, $f(p[x,y])$ determines the pair $\{p[x,y], p[y,x]\}$ completely too. As such, $f$ allows us to distinguish all the values we are allowed to distinguish and is thus optimal in this sense. As each list sums to $1$, we can actually omit the last element in $o(p[])$, reducing the dimension to $m=(n^2-1)$.

Having said that, this map $f$ has one bad property: Small change in one element can change the value of $f$ considerably. For example, consider the following table $p[x,y]$ (all numbers are understood to be divided by 10 to avoid messy decimals) and the row-by-row reading order:

$$p[x,y]=\left(\begin{array}{l}
0 & 2 & 0 \\
2 & 0 & 2 \\
0 & 3 & 1 \\
\end{array}\right)$$

We have
$$\begin{array}{lcl}
o(p[x,y]) & = & \left(0,2,0,2,0,2,0,3,1\right)\\
o(p[y,x]) & = & \left(0,2,0,2,0,3,0,2,1\right)\\
\end{array}$$

and thus $f(p[x,y])=\left(0,2,0,2,0,3,0,2\right)$ (the last element removed). However, a small change in the number $2$ in upper row yields a different picture:
$$p'[x,y]=\left(\begin{array}{l}
0 & 2+\varepsilon & 0 \\
2 & 0 & 2 \\
0 & 3 & 1-\varepsilon \\
\end{array}\right)$$

We have
$$\begin{array}{lcl}
o(p'[x,y]) & = & \left(0,2+\varepsilon,0,2,0,2,0,3,1-\varepsilon\right)\\
o(p'[y,x]) & = & \left(0,2,0,2+\varepsilon,0,3,0,2,1-\varepsilon\right)\\
\end{array}$$
so $f(p'[x,y])=o(p'[x,y])=\left(0,2+\varepsilon,0,2,0,2,0,3\right)$, a big difference from f(p[x,y]).

If this kind of "discontinuous" behaviour is acceptable for the desired mapping, just wait what the Ugly brings: We can actually have $m=1$ and lose no information (other than the inevitable loss) at all! It is not actually all that surprising; since $|\mathbb{R}^k|=\mathbb{R}|$, there is a bijection $h:\mathbb{R}^{n^2}\to R$ so we can define $g:\mathbb{R^{n^2}}\to R$ as $$g(p[x,y])=h(f(p[x,y]))$$ and stuff everything into a single real number. There is no need to worry about computability either; as long as the numbers in $p[x,y]$ are computable, so is the combined value.

Of course, most of this breaks down if we impose some additional restrictions on the map $f$; continuity being the most obvious candidate. What is the actual goal you are trying to achieve by having a mapping with those two properties?


问:
How can I factorise this union of intersections of sets?
How can I rewrite
$$
F = F_1 \cup \bigcup_{i=2}^{n \geq2} \left(
F_i \cap \bigcap_{j=1}^{i-1} S_j
\right)
$$
without explicitly writing $F_1$?

Assume
\begin{align}
F_i \cap S_i &= \emptyset \\
\bigcup_{i=1}^n F_i \cup S_i &\subset \mathbb{R}^m.
\end{align}

For example, if $n=3$, then
$$
F = (F_1) \cup (F_2 \cap S_1) \cup (F_3 \cap S_1 \cap S_2).
$$

答:
If you are working in a whole space called $X$, let's say, then define
$$
S_0 = X
$$
then clearly
$$
F_1 = F_1 \bigcap S_0
$$
and
$$
F_i \cap \bigcap_{j=1}^{i-1} S_j =F_i \cap \bigcap_{j=1}^{i-1} S_j \bigcap S_0 =  F_i \cap \bigcap_{j=0}^{i-1} S_j
$$
so we can rewrite it as
$$F = \bigcup_{i=1}^{n \geq2} \left(
F_i \cap \bigcap_{j=0}^{i-1} S_j
\right)$$
well it's a little bit cheating by introducing new variable but it works.


问:
Why is there no constant introduced when computing the arc length?
I've recently started learning multivariable calculus. When reading about how to compute the length of a curve there is this formula:

$\int_{a}^{b} |\vec{r'}(t)| \,dt $

From my past understanding of integration, one should always introduce a arbitrary constant $C$, since the derivative of such a function that arbitrary constant is lost. Why is this not the case when integrating over the length of a curve?

An example from the litterature I'm reading:

"Find the length of the curve $\vec{r} = t^2\vec{i}+t^2\vec{j}+t^3\vec{k}$ from $t=0$ to $t=1$.

As a first step i'll take the derivative of $\vec{r}$.

$\vec{v}=\vec{r'}=2t\vec{i}+2t\vec{j}+3t^2\vec{k}$

Next I'll use the formula of finding the arc length.

$\int_{0}^{1} |\vec{r'}(t)| \,dt = $
$ \int_{0}^{1} \sqrt{8t^2+9t^4} \,dt = $
$ \int_{0}^{1} \sqrt{t^2(8+9t^2)} \,dt = $
$\int_{0}^{1} t\sqrt{8+9t^2} \,dt =\frac{1}{18}\frac{2}{3}*(8+9t^2)^{3/2}\bigg\rvert_{0}^{1}=1.7579$ (about this, more decimals of course).

This is correct with the answer in the book, but why do I not introduce an arbitrary constant when integrating? Is it simply because if we'd introduce such a constant, we'd not be able to figure out it's lenght? Given that we'd then have too many variables? Perhaps I have missed some key insight, or my mind is playing tricks..!

答:
Let $f:[a,b] \to \mathbb R$ be a Riemann integrable function an $F$ an antiderivative of $f$. If $c$ is a constant, then $G:=F+c$ is also an antiderivative of $f$.

We get

$$\int_a^b f(x) dx= G(b)-G(a)= F(b)+c-(F(a)+c)=F(b)-F(a).$$

Do you see what happens wit the contant $c$ ?


问:
Book in which Calculus is explained in the form of a Teacher-Student Conversation
I remember reading the preface of a Physics textbook in which the author mentions a book on Introductory Calculus in which the matter is explained in the form of an innovative conversation between a teacher and his student. I am eager to lay my hands on a copy of book as I think it is a very effective way of learning. Does anyone know what it's called?</p>

答: Concepts, Problems and Solutions in School Calculus : A Dialogue Approach by Lev Tarasov and Chandra Shekhar Kumar ( ISBN-13 : 978-9354372926 )


问:
Solutions of the following Ode
I am a master student in mathematical physics. I study soliton and traveling wave solutions of the differential equations.

Let's consider the following ODE:
$$Q^{\prime}(\xi)=ln(A)(\alpha+\beta Q(\xi)+\sigma Q^2(\xi))$$
where
$A \neq 0,1.$

In a book, the solutions of the ODE are given as follows: (But, I don't understand how to derive it.)

There are twelve solution cases w.r.t coefficients of ODE.
Any help would be appreciated.

CASE I) When $\beta^{2}-4 \alpha \sigma<0$ and $\sigma \neq 0$, then
$$
\begin{array}{l}
Q_{1}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma} \tan _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2} \xi\right) \\
Q_{2}(\xi)=-\frac{\beta}{2 \sigma}-\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma} \cot _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2} \xi\right) \\
Q_{3}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma}\left(\tan _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right) \pm \sqrt{p q} \sec _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right) \xi}\right)\right) \\
Q_{4}(\xi)=-\frac{\beta}{2 \sigma}-\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma}\left(\cot _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right) \pm \sqrt{p q} \csc _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right)\right) \\
Q_{5}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{4 \sigma}\left(\tan _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{4} \xi\right)-\cot _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha\right) \sigma}}{4} \xi\right)\right)
\end{array}
$$

CASE II:

$\vdots$

CASE XII:
When $\beta=\lambda, \sigma=m \lambda(m \neq 0)$ and $\alpha=0,$ then
$$
Q_{37}(\xi)=\frac{p A^{\lambda \xi}}{q-m p A^{\lambda \xi}}
$$
where triangular functions are defined as
\begin{array}{l}
\sin _{A}(\xi)=\frac{p A^{i \xi}-q A^{-i \xi}}{2 i}, \quad \cos _{A}(\xi)=\frac{p A^{i \xi}+q A^{-i \xi}}{2} \\
\tan _{A}(\xi)=-i \frac{p A^{i \xi}-q A^{-i \xi}}{p A^{i \xi}+q A^{-i \xi}}, \quad \cot _{A}(\xi)=i \frac{p A^{i \xi}+q A^{-i \xi}}{p A^{i \xi}-q A^{-i \xi}} \\
\sec _{A}(\xi)=\frac{2}{p A^{i \xi}+q A^{-i \xi}}, \quad \csc _{A}(\xi)=\frac{2 i}{p A^{i \xi}-q A^{-i \xi}}
\end{array}
where $\xi$ is an independent variable, $p$ and $q$ are constants greater than zero and called deformation parameters.

答:
$$Q^{\prime}(\xi)=ln(A)(\alpha+\beta Q(\xi)+\sigma Q^2(\xi))$$
In order to simplify the typing, change of symbols :
$\begin{cases}
Q=y \\
\xi=x \\
\ln(A)\alpha=a \\
\ln(A)\beta=b \\
\ln(A)\sigma=c
\end{cases}$
$$y'(x)=a+by(x)+c\big(y(x)\big)^2$$
This is a Riccati ODE. The usual change of function to solve it is :
$$y(x)=\frac{-1}{c}\frac{u'(x)}{u(x)}$$
which leads to :
$$u''(x)-b\:u'(x)+a\:c\:u(x)=0$$
This is a second order linear ODE with constant coefficients. I suppose that you can continue.

Title: Solutions of the following Ode


问:
Find the derivative of the given function.
How to find the value of $$f_x \ \text{and} \ f_y$$

If:
$$f(x,y) = \int_y^x e^{t^2} \, dt$$

答:
Have you studied the fundamental theorem of calculus? It shows how to differentiate indefinite integrals. In this case,

$$
f_x = e^{x^2}, \qquad f_y = -e^{y^2}.
$$


答:
The derivative of $\int_0^xf(t)dt$ is $f(x)$.

In this case, $f(x,y)=\int_0^x{t^2}dt-\int_0^ye^{t^2}dt$
We deduce that $f_x=e^{x^2}$ and $f_y=-e^{y^2}$.

Title: Find the derivative of the given function.

Body:
How to find the value of $$f_x \ \text{and} \ f_y$$

If:
$$f(x,y) = \int_y^x e^{t^2} \, dt$$


问:
Are singletons limit points?
So I just started learning topology and I came across this problem.

Consider the topological space $(X,\tau)$where the set $X=\{a,b,c,d,e\}$ and $\tau=\{X,\emptyset,\{a\},\{c,d\},\{a,c,d\},\{b,c,d,e\}\}$

Determine the limit points of

i) $\{a\}$

Is $a$ a limit point for itself since the singleton a is an open set in the topology.
From the definition of limit point of a set  every neighbourhood of the point must contain a point of the set that id  different from a however when considering the singleton there are no different points of a so what is the correct answer im very confused.

Thanks in advance

答:
The limit points of a set $S$ are not an intrinsic property of $S$. They depend on how the set $S$ is immersed into an ambient topological space $X$.

To answer your question, you have to check if there are points $x\in X$ such that every punctured neighborhood of $x$ (i.e., the result of removing from an open neighborhood of $x$ the point $x$) passes through $a$.

Now the punctured neighborhoods of ${a}$ cannot contain $a$.

So let's examine the punctured neighborhoods of ${b}$. These are
$\{c,d,e\}$ and $X-\{b\}$. At least one does not contain $a$. So $b$
is not a limit point of $\{a\}$.

The punctured neighborhoods of ${c}$ are $\{d\}$ and others. But you
already found one that does not contain $a$ so $c$ cannot be a limit
point of $a$.

And so on...

If you want some intuition about the concept of a limit point, then you can think that an element $x\in X$ is a limit point of a subset $S\subseteq X$ if $x$ can be well approximated by $S-\{x\}$.

Title: Are singletons limit points?


问:
Determining the significance level for a Neyman-Pearson hypothesis test when $X_1,...,X_n \sim \text{Uniform}(0,\theta)$?
We want to test $H_0: \theta=\theta_0$ against $H_1: \theta=\theta_1$ where $0<\theta_1<\theta_0$ and where $X_1,...,X_n\sim$ Uniform$(0,\theta)$ are IID. We have critical value $c$.

My problem is: I want to justify why the significance level, $\alpha$, must be such that $\alpha\geq\frac{\theta_1^n}{\theta_0^n}$ in the case where $c\leq\frac{\theta_0^n}{\theta_1^n}$

I've computed the Neyman-Pearson test statistic to be
$$ T(\textbf{x})=\frac{f(\textbf{x};\theta_1)}{f(\textbf{x};\theta_0)}=\frac{\frac{1}{\theta_1^n}\mathbb{1}\{x_1,...,x_n\in[0,\theta_1]\}}{\frac{1}{\theta_0^n}\mathbb{1}\{x_1,...,x_n\in[0,\theta_0]\}}= \frac{\theta_0^n \mathbb{1}\{x_1,...,x_n\in[0,\theta_1]\}}{\theta_1^n\mathbb{1}\{x_1,...,x_n\in[0,\theta_0]\}}  $$

This implies that the power function is $$\mathbb{P}\bigg( \frac{\theta_0^n \mathbb{1}\{x_1,...,x_n\in[0,\theta_1]\}}{\theta_1^n\mathbb{1}\{x_1,...,x_n\in[0,\theta_0]\}}\geq c ;\theta\bigg) $$

Now in the case where $x_1,...,x_n\in[0,\theta_1]$, if $\frac{\theta_0^n}{\theta_1^n}\geq c$ then clearly the power function is $1$, and if $\frac{\theta_0^n}{\theta_1^n}< c$ then clearly it is $0$.

Now in the case where there exists $i$ such that $x_i\in(\theta_1,\theta_0]$ then the power function becomes $\mathbb{P}(0\geq c)=0$.

So going back to the problem: when $c\leq\frac{\theta_0^n}{\theta_1^n}$, how are we able to deduce from what I've done that $\alpha=\mathbb{P}(T(\textbf{X})\geq c;\theta_0)\geq \frac{\theta_1^n}{\theta_0^n} $. Perhaps I've done something wrong because all I am getting is that the power function is either $0$ or $1$ and hence $\alpha$ is either $0$ or $1$?

答:
Your likelihood ratio is

$$\frac{L(\theta_1|\mathbf{x})}{L(\theta_0|\mathbf{x})}=\dots=\begin{cases}
\Big(\frac{\theta_0}{\theta_1}\Big)^n,  & \text{if $0<x_{(n)}\leq \theta_1$ } \\
0, & \text{if $\theta_1<x_{(n)}\leq \theta_0$}
\end{cases}$$

thus the likelihood ratio is monotonic and we can apply a well known theorem and the critical region is

$$C=\{\mathbf{x};x_{(n)}<k\}$$

Thus

$$\alpha=\int_0^c \frac{n y^{n-1}}{\theta_0^n}dy=\frac{c^n}{\theta_0^n}$$

Now I think it is self evident that you cannot justify your statement.

Counterexample:

Set $\theta_0=3$,$\theta_1=2$, $n=2$

If $c<\Big(\frac{3}{2}\Big)^2=\frac{9}{4}$, say $c=\frac{6}{4}$, $\alpha=\frac{1.5^2}{9}=\frac{1}{4}<\frac{4}{9}$

This is a graphical explanation: The two drawing represent the one sided test with your uniform. I type error (significance level) is the purple area. As you can see,

in the left drawing, that is the case when $\theta_1<\theta_0$ as $c$ decreases, so do $\alpha$.

in the right drawing, that is the case when $\theta_1>\theta_0$ as $c$ decreases,  $\alpha$ increases.

Between $c$ and $alpha$ there is an easy relation:

$$\alpha=\frac{c^n}{\theta_0^n}$$

$$\alpha=1-\frac{c^n}{\theta_0^n}$$

respectively, in the two cases.


答:
Let  $y$ be constant , then we get, by the FTC:

$$f_x(x,y)= e^{x^2}.$$

Let  $x$ be constant , then $f(x,y)=- \int_x^y e^{t^2}$ and we get, by the FTC:

$$f_y(x,y)= -e^{y^2}.$$


问:
Symbolic solutions to $\log(x)=\frac{1}{2}(x-1)$
Wolfram Alpha gives a numerical solution to this equality, but is it not possible to find a symbolic answer? Why not? What types of functions with log in it can and can't you solve and why?

Thanks!

答:
It is surprising (I repeated the calculation with Wolfram Alpha), that you did not receive the sybolic answer since
$$\log(x)=a x+b \implies x=-\frac 1a W\left(-a e^b\right)$$ where $W(t)$ is Lambert function.

Applied to your case, this gives, beside the trivial $x=1$
$$x=-2 W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right)\sim 3.51286$$


问:
Degree of the map $\mathbb{C} \times \mathbb{R} \longmapsto \mathbb{C} \times \mathbb{R}$ $: X(z,t) = (z^2,t)$
Consider the vector field $X : \mathbb{C} \times \mathbb{R} \longmapsto \mathbb{C} \times \mathbb{R}$ such that $X(z,t) = (z^2,t)$.

I'd like to prove that $X$ has only one zero of index $2$. This seems reasonable since $\mathbb{S}^{1} \longmapsto \mathbb{S}^{1}$ the power to the square has degree $2$ and basically I'm doing nothing on the second component. But I'm having trouble dealing with computation.

It's clear that $(0,0)$ is the only zero here, hence isolated. But I have to compute $\text{ind}_{X}((0,0))$ this is by definition the degree of the map from a ball centered in the zero, with any radius in this case to $\mathbb{S}^{2}$ in this case since we are in $\mathbb{R}^{3}$ (But it's not clear to me how I'm seeing $\mathbb{S}^{2}$ here as subset of $\mathbb{C} \times \mathbb{R}$ ) which is $X(z,t) \longmapsto \frac{X(z,t)}{\lvert\lvert X(z,t)\rvert\rvert}$.

How to compute the degree of $X(z,t) \longmapsto \frac{X(z,t)}{\lvert\lvert X(z,t)\rvert\rvert}$ ? It's not even clear to me wether I should think this map as a composition or not, and try to guess who the regular values are.

The definition of degree I'm using is : Let $M$ be compact $N$ connected with dim$M=$dim$N$, both without boundary, orientable and oriented. Take $f : M \longmapsto N$ of class $C^{\infty}$ then deg($f$) := $\sum\limits_{x \in f^{-1}(y)} \text{sgn}df_x$ for $y \in \text{RegVal}(f)$.

The index of an isolated zero $p$ is defined as the pullback through a parametrization of the manifold of the index on an open set of $\mathbb{R}^{n}$ so I'm going to give this definition : $\text{ind}_X(p) := \text{deg}(f_{r})$, where $f_r : \partial B(p,r) \longmapsto \mathbb{S}^{n-1}$ with $f_r(q) := \frac{X(q)}{\lvert \lvert X(q) \rvert \rvert}$ for every $r \in (0,\epsilon)$ given $\epsilon$ the radius of the ball where there are no other zeros.

Any help will be appreciated, thanks in advance.

答:
We can take $r=1$. Then your definition of the index is
$$ind_X(0)=deg(f_1)$$
where $f_1: S^2\to S^2$ is given by
$$f_1\left(se^{i\theta}, t\right) = \frac{\left(s^2e^{2i\theta}, t\right)}{\sqrt{s^4+t^2}},$$
where $s$ and $t$ are real positive numbers so that $s^2+t^2=1$. Now you must find a regular value to compute the degree of $f_1$. I suggest you use $(1,0)$. The inverse image are the two points $(-1,0)$ and $(1,0)$.

Is this clear so far? Can you take it from here?


问:
About Cycles And Permutations
Comsider a $m$ cycle $$\phi =(1, 2, ........., m)$$. Let $k < m$ be any positive integer. Prove that ${\phi}^k$ is again a $m$ cycle if and only if $$\gcd(m, k)=1$$.
$$$$Let us assume that $1 \neq d=(m, k)$ and ${\phi}^k$ is a $m$ cycle. Now after applying $\phi$, $k$ times, we get $${\phi}^k=(1, k+1, r_1, r_2, .........., r_{m-2}),$$  where $r_1$ is such that $$(k+1)+k \equiv r_1 \mod m$$, and for all $2 \leq i \leq m-2$, $r_i$ is defined such that $$r_{i-1}+k \equiv r_i \mod m$$, also each $r_i\leq m $. Now as $d$ divides both $m$ and $k$ so we get $$r_i \equiv 1  \mod d$$ for every $1 \leq i \leq m-2$. Now as the cycle $$(1, k+1, r_1, ........, r_{m-2})$$ is a $m$ cycle hence all of $r_is$ are different and also different from $k+1$. Hence $k$ and all the $r_i-1$ for $1 \leq i \leq m-2$ are $m-1$ distinct multiples of $d$ and all are less than $m$. But as $d\mid m$ and $d>1$ hence we cannot have $m-1$ distinct multiples of $d$ all of which are less than $m$, hence a contradiction.
$$$$Now suppose that $(m, k)=1$ and there is a cycle $$(1, k+1, r_1, r_2, ...... r_{n-2})$$ where $r_is$ are defined in similar manner and $n<m$. Now we see that $$(k+1)+k \equiv r_1 \mod m$$, $$r_1 +k \equiv r_2 \mod m , \\\vdots\\\vdots$$ $$r_{n-3}+k \equiv r_{n-2}  \mod  m,$$ $$ r_{n-2} +k \equiv 1  \mod m.$$ Adding all these we get $$m|2k + k(n-2)$$. Now as $(m, k)=1$ hence we get $m\mid n$, but as $n<m$, it is a contradiction. Hence we get the required result.
$$$$Is My Proof Correct??

答:
Well done! Your proof is correct although there are other much shorter ways of completing the proof. For example:

Let ${\phi}^k=\psi $ and note that  ${\phi}^m=I$, the identity.

If  $ d=(m, k)\ne 1$, then $ {\psi}^{m/d}=({\phi}^{mk/d})=I^{k/d}=I$ and so $\psi$ is not an $m$-cycle.

If interested, you might like to work out a similar proof for the case when $(m, k)=1$.

Title: About Cycles And Permutations


问:
Given the minimal polynomial of a matrix $A^2$, what could the minimal polynomial of $A$ be?
It is given that the minimal polynomial of $A^2$ is $\phi_{A^2}(x) = (x-1)^2$, where $A$ is a complex $4\times4$ matrix. The question is, what are the possible minimal polynomials for $A$?

From the given that $\phi_{A^2}(x) = (x-1)^2$, I can derive several things. First off, I know that the characteristic polynomial $f_{A^2}$ of $A^2$ must have the same irreducible factors, which implies that the only eigenvalue of $A^2$ is $1$. Secondly, it is easily seen that $(A^2 -I)^2 = O$ and $A^2 - I \neq O$. Therefore, the matrix $A^2 - I$ is nilpotent with index 2. This implies that the invariant system is either $\{2,1,1\}$ or $\{2,2\}$. Putting this together, there are two possible Jordan forms of $A^2$:
$$ J_{A^2,1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix} \quad\text{or}\quad J_{A^2,2} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix} $$
Great, all these things I know about $A^2$... but how does any of it lead to information about the minimal polynomial of $A$? Or just any information about $A$ at all? I'm kind of stuck on this.

答:
HINT

You know a polynomial satisfied by $A$, namely $(A^2-1)^2$. So the minimal polynomial has to be a factor of $(x-1)^2(x+1)^2$.

Which of the possible factors could give the correct minimal polynomial for $A^2$?

If neither power of $(x-1)^i(x+1)^j$ is $2$ then we would have one of $A-I=0,A+I=0,(A+I)(A-I)=0$. In all these cases $A^2=I$, a contradiction. This cuts your work considerably.

Finally, to check the remaining possibilities easily, use simple 2x2 Jordan blocks e.g.

$A=\begin{pmatrix}1&1&0&0 \\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}$ has minimal polynomial $(x-1)^2$. What is the minimal polynomial for $A^2$?


问:
Calculus Books, preferably Soviet.
I would like some book recommendations on calculus books for novices. I love old Russian texts and if you you suggest such books(no issue if they are rigorous), I would be very grateful. If the text is quite challenging, I have no problem. But it should be interesting with proper exposition of topics.

The book should be good for self-study, preferably a textbook(again, for beginners). And it's not necessary for it to be Soviet. Just anything good. I merely like Soviet/Slavic books and so, the title.

Thanks to you all.

Edit again:Please try to abstain from suggesting analysis books. All I want is a nice, rigorous Book on Integral and Differential Calculus. Doesn't matter which origin(at this point), but should be translated into English. Thanks.

答: The book Concepts, Problems and Solutions in School Calculus : A Dialogue Approach by Lev Tarasov and Chandra Shekhar Kumar ( ISBN-13 : 978-9354372926 ) can be a good starting point for beginners because it doesn't assume any previous knowledge on calculus.

Title: Calculus Books, preferably Soviet.


问:
Why does the solution for the derivative of the quadratic function on x value, equals the first coefficient on the linear function?
I have the following function
$$() = 2² − 3 − 2$$
and I have to find the angle made by the tangent to this function at
$$  =(1, −3)$$ with the  −axis.

I know the linear function
$$() =  + n$$ and
$$  =(1, −3)$$
so $$(1) =  +  = −3$$
and then to find m there is the derivative of the quadratic $$′() = 4 − 3$$
$$′(1) = 4 − 3 = 1 ⟹ m$$

I assumed that because the derivative and the linear function belong to the same quadratic, they will have the same properties, and that's why we use the derivative.

But why m equals 1?
Is there some flaw in my reasoning?

答: The slope of the line $L(a)$ that is tangent to the graph of $f$ at $(a, f(a))$ is $f'(a).$ The slope of the line whose equation is $y=mx+n$ is $m,$ so if this line is $L(a)$ then $m=f'(a).$ In particular if $a=1$ and $f(a)=-3$ and $(a,f(a))=(1,-3)=A$ then  $m=f'(a)=f'(1)=1.$ Your reasoning is correct.


问:
Inclined parabola
Please help me to solve the inclined parabola.
Is it that we have to convert this entire equation in the form of axis  and tangent at vertex. That would be tiresome. Any other method?
For this  parabola,$x^2 +y^2 +2xy -6x -2y + 3=0$,How can we find it's focus?</p>

答:
Let There be a parabola $(ax+by)^2+2gx+2fy+c=0 $

This can be written as $(ax+by)^2=-2gx-2fy-c$
Now we will add arbitrary constant in square term,
$$(ax+by+\lambda)^2=xf_1(\lambda)x+f_2(\lambda)y+f_3(\lambda)-(1)$$

Now simply choose $\lambda$ such that the lines,
$ax+by+\lambda$ and $xf_1(\lambda)x+f_2(\lambda)y+f_3(\lambda)$
are perpendicular.

now $xf_1(\lambda)x+f_2(\lambda)y+f_3(\lambda)=bx-ay+\mu$

we can write (1) as,
$$(\frac{ax+by+\lambda}{\sqrt{a^2+b^2}})^2=4\rho(\frac{bx-ay+\mu}{\sqrt{a^2+b^2}})$$
which is of the form $Y^2=4\rho X$

Now we can do our stuff on this reduced form,

1)LATUSRECTUM:- $4\rho$=$\frac{1}{\sqrt{a^2+b^2}}$

2)AXIS:- Y=0 or ax+by+$\lambda$=0

3)EQUATION OF TANGENT AT VERTEX:- X=0 or bx-ay+$\mu$=0

4)VERTEX:- Intersection of Y=0 and X=0

5)EQUATION OF DIRECTRIX:- X+$\rho$=0

6)EQUATION OF LATUS RECTUM:- X-$\rho$=0

7)FOCUS:- By solving X-$\rho$=0 and X=0

well now we can apply all the basic parabola properties in short {^w^} .

Title: Inclined parabola


问:
Choice of numbers in calculating $\lim_{x\rightarrow 0}\frac{e^{2x}-1}{\sin 3x}$
I'm studying the limits to $\lim_{x\rightarrow 0}\frac{e^{2x}-1}{\sin 3x}$. According to my text book, I should calculate it like this:

$$\frac{e^{2x}-1}{\sin 3x} = \frac{e^{2x}-1}{2x} \times \frac{3x}{\sin 3x} \times \frac{2}{3} \rightarrow 1 \times 1 \times \frac{2}{3}$$

While it makes sense, I don't understand the use of 2 and 3. Wouldn't any other numbers make as much sense? Why are these chosen and how?

答: The fractions involving $e$ and $\sin$ turn out to be the very expressions (for $\sin$: reciprocal) that you use to calculate the derivatives of $\exp$ and $\sin$ at $0$. (This amounts to what lab said in a comment, but maybe show sth emotivation behind it)</p>


问:
The nature of variables within proofs
I am confused about how best to view certain kinds of variables that show up in the course of an elementary proof. For example, to prove that the square of every odd integer is odd, I would write:

Let $m \in \mathbf{Z}$ be arbitrary. Suppose $m$ is odd. Choose $k \in \mathbf{Z}$ such that $m = 2k + 1$. Then $m^2 = (2k + 1)^2 = 2(2k^2 + 2k) + 1$, so since $2k^2 + 2k \in \mathbf{Z}$, it follows that $m^2$ is odd.

The symbol $m$ is introduced as part of the technique of universal generalization; Velleman [1] describes this step as "introducing a new variable [$m$] ... to stand for an arbitrary object." The symbol $k$ is introduced as part of the technique of existential instantiation; [1] describes this as "introducing a new variable [$k$] to stand for an object for which [the predicate $m = 2k + 1$] is true."

In what sense are $m$ and $k$ variables? They do not seem to be bound variables, as they are in the scope of no quantifier. They also can't be free variables, because if they were, then, for example, the sentence being proved true after the introduction of arbitrary integer $m$ -- namely that if $m$ is odd then $m^2$ is odd -- would be a predicate and not even have a truth value; I am also obviously not free to substitute values into $m$ and $k$, if for no other reason than $m$ and $k$ are dependent.

Stoll [2] says a variable is free if it is not bound, so there is no third option.

Is it more correct to think of the phrases "let $m \in \mathbf{Z}$ be arbitrary" and "choose $k \in \mathbf{Z}$ such that ..." as introducing new constants rather than variables, as their values are fixed despite being unspecified ($m$) or unknown ($k$)? More generally, the idea of a symbol that stands for a particular but arbitrary element of some set seems incompatible with the typical description of a variable as a placeholder as seen in, e.g., Epp [3].

[1] Velleman, D. (2006). How to Prove It: A Structured Approach (2nd ed.). Cambridge: Cambridge University Press.

[2] Stoll, Robert Roth (1963). Set Theory and Logic. W.H. Freeman.

[3] Epp, Susanna S. Variables in Mathematics Education. https://condor.depaul.edu/~sepp/VariablesInMathEd.pdf

答:
During a proof, there are two things present at any point in time:

the thing(s) you want to proof right now (which I will call the goal(s) )

and the things you are assuming right now, the things you can use to continue in your proof (referred to as the context).

Let's go through the proof you want to do, keeping track of both the context on the left and the goal on the right.
At the start we have
$$
\boxed{\begin{array}{c|c}
\text{Context} & \text{Goal} \\
\hline
 & \hspace{1em} \forall m \, (m \text{ is odd } \Rightarrow m^2 \text{ is odd})  \hspace{1em}
\end{array}}
$$
So our goal is something of the form $\forall x \, \varphi(x)$. How do we show this in general?

We can conclude $\forall x \, \varphi(x)$ from the context $\Gamma$ if we can show $\varphi(y)$ for a variable $y$ which does not appear as a free variable in either the context $\Gamma$ or $\varphi(x)$.

In our case $y$ would actually satisfy this and thus we get a new goal
$$
\boxed{\begin{array}{c|c}
\text{Context} & \text{Goal} \\
\hline
 &  \hspace{1em} y \text{ is odd } \Rightarrow y^2 \text{ is odd}  \hspace{1em}
\end{array}}
$$
Why is it fine to simply get rid of the quantification and replace $m$ with $y$? Well, we did not simply get rid of the quantification. We had to make sure that $y$ is not free in the context and the right side! This condition assures that the context can not have any information about the variable $y$ for us. So we are not assuming anything about $y$, we know nothing about it and thus $y$ can truly be considered to be anything/general. And so if we show the right side for a variable we can consider general, the argument is that we have shown $\forall m [\dots]$.

Now let's continue with the proof. To show the implication you assume whats on the left and try to show whats on the right.
$$
\boxed{\begin{array}{c|c}
\text{Context} & \text{Goal} \\
\hline
\hspace{1em}  y \text{ is odd }  \hspace{1em} &
\hspace{1em}  y^2 \text{ is odd} \hspace{1em}
\end{array}}
$$
In order to continue we need to write out the definition of odd
$$
\boxed{\begin{array}{c|c}
\text{Context} & \text{Goal} \\
\hline
\hspace{1em} \exists k \, (y = 2k + 1) \hspace{1em} &
\hspace{1em} \exists k \, (y^2 = 2k + 1)  \hspace{1em}
\end{array}}
$$
Now at this point you can apply existential instantiation on the statement $\exists k \, (y = 2k + 1)$ in your context. What happens here is that you are allowed to replace $\exists k \, (y = 2k + 1)$ by $(y = 2z + 1)$ for any variable $z$ that does not appear free in your context and goal. Similarly to before, this now makes sure that the only thing you know about $z$ is that it satisfies $y = 2z + 1$. So we get
$$
\boxed{\begin{array}{c|c}
\text{Context} & \text{Goal} \\
\hline
\hspace{1em}  y = 2z + 1 \hspace{1em} &
\hspace{1em} \exists k \, (y^2 = 2k + 1) \hspace{1em}
\end{array}}
$$

This explanation is basically a verbalization of some of the rules for natural deduction, which is a formal way of describing proofs. So if you haven't yet and really want to learn more about proofs in detail, this would be a topic to get into; although it's a deep dive.

Title: The nature of variables within proofs


问:
Rouché Theorem to calculate the number of zeros
How can I calculate the number of zeros of $\cos z+3z^3$ using the Rouché Theorem?</p>

答: Try $g(z) = \cos z +3 z^3$, $f(x) = -3 z^3$. Then $|g(z)+f(z)| = |\cos z|$. If $|z| = 1$, then $|\cos z| = \cosh (\operatorname{Im} z) \le e < 3$. Since $|f(z)| = 3$ when $|z|=1$, we have $|g(z)+f(z)| < |f(z)|$ for $z$ on the unit circle. Both $f,g$ are entire, hence $f$ and $g$ have the same number of zeros (counted by multiplicity) inside the unit circle. Hence the answer is three.</p>


问:
Total Probability and Bayes Rule

The probability that a person has 0, 1, 2, 3, 4, or 5 or more siblings is 0.15, 0.49, 0.27, 0.06, 0.02, 0.01, respectively.

a) Three friends who are not siblings are gathered. What is the probability that they combined have three siblings?

b) Emma and Jacob are not siblings, but combined they have a total of 3 siblings. What is the probability that Emma has no siblings?

So for a) I have calculated that there are 7 possible ways this can occur: 1 situation where all three friends have 1 sibling, 3 cases where two friends have zero siblings and the other one has 3 siblings, and 3 cases where one person has 1 sibling, one person has 2 siblings and the last person has 0 siblings. So calculating this I obtain:

Using the notation that $P(X = N)$ means the probability that one person has $N$ siblings.

$$P(X = 1)^3 + 3*P(X = 0)^2 * P(X = 3) + 3*P(X = 2)*P(X = 1) * P(X = 0) \approx 0.181$$ Is this correct?

For part b) I used the notation $P(E = N)$ to mean the probability that Emma has N siblings, and similarly $P(J = N)$ to mean the probability that Jacob has $N$ siblings.

I interpreted the problem as asking us to calculate $$P(E = 0 | E + J = 3) = \frac{P (E = 0 \cap E + J = 3)}{P(E + J = 3)} = \frac{P(E+J = 3| E = 0) * P(E = 0)}{P(E+J = 3)} = \frac{P(J = 3)*P(E = 0)}{2*P(E = 0)*P(J = 3) + 2 * P(E = 1)*P(J = 2)} \approx 0.4416$$ where I have used Baye's rule to transform the expression.

Are my calculations correct?

答:
So for a) I have calculated that there are 7 possible ways this can occur

I calculated 10 possible ways thus I did not check your calculations

$0-0-3$

$0-1-2$

$0-2-1$

$0-3-0$

$1-0-2$

$1-1-1$

$1-2-0$

$2-0-1$

$2-1-0$

$3-0-0$

Summing the single probabilities I get $0.240769\approx 24\%$

For (b) I get

$$\frac{0.15\times0.06}{0.15\times0.06+0.49\times0.27+0.27\times0.49+0.06\times0.15}=0.031847\approx 3\%$$

Title: Total Probability and Bayes Rule


答:
In first part of the question, when you are counting cases for $2, 1 $ and $0$ siblings, there are $3! = 6$ ways, instead of $3$. They are $\{0, 1, 2\}, \{0, 2, 1\}, \{1, 0, 2\}, \{1, 2, 0\}, \{2, 0, 1\}, \{2, 1, 0\}$. So the probability should be

$P(X = 1)^3 + 3  \ P(X = 0)^2 \  P(X = 3) + 6 \ P(X = 2) \ P(X = 1) \ P(X = 0) \approx 0.2408$

For the second one, you have written the formula correctly but the answer seems wrong.

It should be $ = \frac{0.06 \times 0.15}{2 \times 0.06 \times 0.15 + 2 \times 0.49 \times 0.27} \approx 0.0318$

Title: Total Probability and Bayes Rule

Body:
The probability that a person has 0, 1, 2, 3, 4, or 5 or more siblings is 0.15, 0.49, 0.27, 0.06, 0.02, 0.01, respectively.

a) Three friends who are not siblings are gathered. What is the probability that they combined have three siblings?

b) Emma and Jacob are not siblings, but combined they have a total of 3 siblings. What is the probability that Emma has no siblings?

So for a) I have calculated that there are 7 possible ways this can occur: 1 situation where all three friends have 1 sibling, 3 cases where two friends have zero siblings and the other one has 3 siblings, and 3 cases where one person has 1 sibling, one person has 2 siblings and the last person has 0 siblings. So calculating this I obtain:

Using the notation that $P(X = N)$ means the probability that one person has $N$ siblings.

$$P(X = 1)^3 + 3*P(X = 0)^2 * P(X = 3) + 3*P(X = 2)*P(X = 1) * P(X = 0) \approx 0.181$$ Is this correct?

For part b) I used the notation $P(E = N)$ to mean the probability that Emma has N siblings, and similarly $P(J = N)$ to mean the probability that Jacob has $N$ siblings.

I interpreted the problem as asking us to calculate $$P(E = 0 | E + J = 3) = \frac{P (E = 0 \cap E + J = 3)}{P(E + J = 3)} = \frac{P(E+J = 3| E = 0) * P(E = 0)}{P(E+J = 3)} = \frac{P(J = 3)*P(E = 0)}{2*P(E = 0)*P(J = 3) + 2 * P(E = 1)*P(J = 2)} \approx 0.4416$$ where I have used Baye's rule to transform the expression.

Are my calculations correct?


问:
Lagrangian mechanics problem
i have managed to find the kinetic energy but it cant be right because i didnt do anything to do with the rotational inertia because i have no idea how to do part (a)

for part 3 the lagrangian is just $L=T-U$ where $T$ is the kinetic energy and $U$ is the potential energy?

part 4 seems straight forward but i need to calculate L first, but i'm struggling to find the potential energy and the correct kinetic energy, i've only been doing examples with one mass this time there is multiple so its confusing me.

答:
Calling $I_O$ the hoop inertia moment regarding $O$ and

$p_M = a(\sin\phi,-\cos\phi),\ \ p_m = p_M + a(\sin\theta,-\cos\theta)$ we have

$$
T = \frac 12\left(I_O\dot\phi^2+m \|\dot p_m\|^2\right)
$$

and

$$
V = g\left(M p_M + m p_m\right)\cdot(0,1)
$$

then $L = T - V$ and the movement equations are

$$
\left\{
\begin{array}{rcl}
\phi'' & = &  -\frac{a \left(2 a m \sin (\phi-\theta ) \left(\phi '^2 \cos (\phi-\theta)+\theta '^2\right)+g (m+2 M) \sin (\phi)+g m \sin (\phi
  -2 \theta)\right)}{2 \left(a^2 (m+M)+a^2 (-m) \cos ^2(\phi -\theta )+I_O\right)} \\
\theta'' & = &  -\frac{a^3 m \theta '^2 \sin (2 (\phi-\theta ))-g \sin (\theta) \left(a^2 (m+M)+2 I_O\right)+a^2 g (m+M) \sin (2 \phi-\theta)+2
   a \phi '^2 \left(a^2 (m+M)+I_O\right) \sin (\phi-\theta )}{2 \left(a^3 m \cos ^2(\phi-\theta)-a \left(a^2
   (m+M)+I_O\right)\right)} \\
\end{array}
\right.
$$

Considering now small variations $(|\alpha| << 1, \sin\alpha = \alpha,\ \cos\alpha = 1,\ \alpha^2=0)$ we have

$$
\left(
\begin{array}{c}
\phi''\\
\theta''
\end{array}
\right)=
\left(
\begin{array}{cc}
 -\frac{a g (m+M)}{I_O} & \frac{a g m}{I_O} \\
 \frac{a g (m+M)}{I_O} & -\frac{g \left(a^2 m+I_O\right)}{a I_O} \\
\end{array}
\right)\left(
\begin{array}{c}
\phi\\
\theta
\end{array}
\right)
$$

Title: Lagrangian mechanics problem


问:
Equation relating Chern-Simons and Bott-Chern secondary characteristic forms
Let $\overline{\mathcal E}: 0\to \overline S\to \overline E\to \overline Q\to 0$ be a short exact sequence of Hermitian vector bundles over a complex manifold $X$. Here $\overline E=(E,h^E)$ is a holomorphic vector bundle $E$ together with a Hermitian metric on $E$, and similarly for $S$ and $Q$. We do not require $h^S$ and $h^Q$ to be induced from $h^E$. Then each bundle has a Chern connection, they will be denoted $\nabla^S,\ \nabla^E$ and $\nabla^Q$. We have the Bott-Chern form $\widetilde{ch}_{BC}(\overline{\mathcal E})$ which satisfies a double transgression formula
$$
d d^c\overline{ch}_{BC}(\overline{\mathcal E}) = ch(\nabla^E)-ch(\nabla^S)-ch(\nabla^Q)
$$
where $ch(\nabla)$ is the Chern-Weil form representing the Chern-character computed from the connection $\nabla$.
Smoothly $\overline{\mathcal E}$ is split in a canonical way, namely the orthogonal projection $E\to S$. Therefore $E$ carries, in addition to the Chern connection $\nabla^E$, also the connection $\nabla^S\oplus\nabla^Q$. My question is:

How does the Chern-Simons transgression form $\widetilde{ch}_{cs}(\nabla^E, \nabla^S\oplus\nabla^Q)$ relate to the Bott-Chern double transgression form $\widetilde{ch}_{BC}(\overline{\mathcal E})$?

My conjecture is that $\widetilde{ch}_{cs}(\nabla^E, \nabla^S\oplus\nabla^Q) =d^c \widetilde{ch}_{BC}(\overline{\mathcal E})$. My only evidence, however, is that both these forms are solutions to the equation $d\omega = ch(\nabla^E)-ch(\nabla^S)-ch(\nabla^Q)$. If my conjecture is a well known formula, I would greatly appreciate a reference!

答: The formula I conjectured is almost true. The correct formula is $ch_{CS}(\overline{\mathcal E})=\overline\partial ch_{BC}$. See the last line of the proof of proposition 3.15 of "Hermitian vector bundles and the equidistribution of the zeroes of holomorphic sections", in which Bott and Chern introduce the Bott-Chern form.


问:
Intersection of all subgroups with a finite index of an infinite group is a normal subgroup
Let $A(G)$ be the intersection of all subgroups of finite index of an infinite group $G,*$. I have to show that $A(G)$ is a normal subgroup: $A(G) \lhd G$.
Unfinished proof:

Because $A(G)$ is an intersection of subgroups, $A(G)$ itself will be a subgroup of $G$. That means that I can use the following equivalence: \begin{align*}
A(G) \lhd G  \iff \forall g \in G , \forall h \in A(G):g \ * h \ *g^{-1} \in A(G).
\end{align*}
So let $g \in G$ and $h \in A(G)$. To prove that $g \ * h \ *g^{-1} \in A(G)$, I have to show that $g \ * h \ *g^{-1} \in D$ for every subgroup $D$ with finite index. So let $D$ be a subgroup of $G$ with finite index (this is possible because $G$ itself is a subgroup with finite index). However, now I am stuck. I cannot seem to prove that $g \ * h \ *g^{-1} \in D$. I know that $h \in D$ because $h \in A(G)$, but appart from that I have used everything I know (I think). I would like to know the next step in my proof (except if there is an error in my reasoning, but I don't know anything about actions).

答: Hint: If $D$ is any subgroup with finite index then $g^{-1}Dg$ is also a subgroup with finite index, and so $h$ belongs to it.


答:
Every subgroup of finite index contains a normal subgroup of finite index: For $G$ group and $H$ subgroup of finite index, prove that $N \subset H$ normal subgroup of $G$ of finite index exists

And so the intersection of all subgroups of finite index is the same as the intersection of all normal subgroups of finite index.

And the intersection of arbitrary family of normal subgroups is normal, as you can read here: https://proofwiki.org/wiki/Intersection_of_Normal_Subgroups_is_Normal

Title: Intersection of all subgroups with a finite index of an infinite group is a normal subgroup

Body:
Let $A(G)$ be the intersection of all subgroups of finite index of an infinite group $G,*$. I have to show that $A(G)$ is a normal subgroup: $A(G) \lhd G$.
Unfinished proof:

Because $A(G)$ is an intersection of subgroups, $A(G)$ itself will be a subgroup of $G$. That means that I can use the following equivalence: \begin{align*}
A(G) \lhd G  \iff \forall g \in G , \forall h \in A(G):g \ * h \ *g^{-1} \in A(G).
\end{align*}
So let $g \in G$ and $h \in A(G)$. To prove that $g \ * h \ *g^{-1} \in A(G)$, I have to show that $g \ * h \ *g^{-1} \in D$ for every subgroup $D$ with finite index. So let $D$ be a subgroup of $G$ with finite index (this is possible because $G$ itself is a subgroup with finite index). However, now I am stuck. I cannot seem to prove that $g \ * h \ *g^{-1} \in D$. I know that $h \in D$ because $h \in A(G)$, but appart from that I have used everything I know (I think). I would like to know the next step in my proof (except if there is an error in my reasoning, but I don't know anything about actions).


问:
Is the set of surjective functions from $\mathbb{N}$ to $\mathbb{N}$ uncountable?
I want to use Cantor's diagonalisation argument to prove that the set S of surjective functions of the form $\Bbb{N} \to \Bbb{N}$ is uncountable. The normal procedure is creating a matrix and filling it with elements of S and introducing a new surjective function $p$.The problem that I'm having is that I don't know in which way to change the diagonal elements so that $p$ should, but cannot be in the matrix.

Can someone help me, or at least give me some tips?

答:
$\begin{array}{ccc}
f_1(1) && f_1(2) && f_1(3)...\\
f_2(1) && f_2(2) && f_2(3)...\\
f_3(1) && f_3(2) && f_3(3)...\\
\vdots
\end{array}
$

Form a function $g$ with $g(1)$ the least natural number other than $f_1(1)$, $g(2)$ the least natural number other than $f_2(2)$ which is not already used for $g(1)$ and so on.

If $g$ is a surjection, done. If $g$ is not a surjection, then there exist constants $k,c$ such that for all $n \geq k$, $f_n(n)=c$. All functions with exactly one member that maps to $c$ are in the sequence, and for any $n>k$ there are infinitely many such functions with $f(n)=c$, but only one can go in the $n$th place. Contradiction.


答:
Alternatively you could use your $m$-cycle to define a cyclic group of order $m$:

$$\langle (1,2,3,...,m) \rangle = \langle \phi\rangle$$

It turns out that $\langle \phi^k \rangle = \langle \phi^{(k,m)} \rangle$ and the generators of the cyclic group are precisely when $(k,m)=1$ i.e. $\phi^k$ is an $m$-cycle when $(k,m)=1$.

Proof:

Since $(k,m)|k$, $\phi^k \in \langle \phi^{(k,m)}\rangle$ so $\langle \phi^k \rangle \subseteq \langle \phi^{(k,m)} \rangle$.

Similarly $(k,m)= ky+mx$ for some $x,y \in \mathbb{Z}$, $\phi^{(k,m)}= \phi^{ky+mx}=\phi^{ky}\phi^{mx}=\phi^{ky}$ since $\phi^{m}=(1)$. Thus $\phi^{(k,m)} \in \langle \phi^k \rangle$. $\langle \phi^{(k,m)} \rangle \subseteq \langle \phi^k \rangle$.

Title: About Cycles And Permutations

Body:
Comsider a $m$ cycle $$\phi =(1, 2, ........., m)$$. Let $k < m$ be any positive integer. Prove that ${\phi}^k$ is again a $m$ cycle if and only if $$\gcd(m, k)=1$$.
$$$$Let us assume that $1 \neq d=(m, k)$ and ${\phi}^k$ is a $m$ cycle. Now after applying $\phi$, $k$ times, we get $${\phi}^k=(1, k+1, r_1, r_2, .........., r_{m-2}),$$  where $r_1$ is such that $$(k+1)+k \equiv r_1 \mod m$$, and for all $2 \leq i \leq m-2$, $r_i$ is defined such that $$r_{i-1}+k \equiv r_i \mod m$$, also each $r_i\leq m $. Now as $d$ divides both $m$ and $k$ so we get $$r_i \equiv 1  \mod d$$ for every $1 \leq i \leq m-2$. Now as the cycle $$(1, k+1, r_1, ........, r_{m-2})$$ is a $m$ cycle hence all of $r_is$ are different and also different from $k+1$. Hence $k$ and all the $r_i-1$ for $1 \leq i \leq m-2$ are $m-1$ distinct multiples of $d$ and all are less than $m$. But as $d\mid m$ and $d>1$ hence we cannot have $m-1$ distinct multiples of $d$ all of which are less than $m$, hence a contradiction.
$$$$Now suppose that $(m, k)=1$ and there is a cycle $$(1, k+1, r_1, r_2, ...... r_{n-2})$$ where $r_is$ are defined in similar manner and $n<m$. Now we see that $$(k+1)+k \equiv r_1 \mod m$$, $$r_1 +k \equiv r_2 \mod m , \\\vdots\\\vdots$$ $$r_{n-3}+k \equiv r_{n-2}  \mod  m,$$ $$ r_{n-2} +k \equiv 1  \mod m.$$ Adding all these we get $$m|2k + k(n-2)$$. Now as $(m, k)=1$ hence we get $m\mid n$, but as $n<m$, it is a contradiction. Hence we get the required result.
$$$$Is My Proof Correct??


问:
Finding Natural Domain of functions and expressing answer in terms of intervals
Really have no clue on how to do this. Some of the questions I have are

(a) $f(x)=2x^2$

(b) $f(x)=x^8$

(c) $f(x)= \frac{1}{x}-7$

My attempt was for the answer for the first one was(−∞,∞) because it stood for all the real numbers.
Would this be the same for $x^8$? And how do I go about the third question? Thank you

答:
The natural domain of a function is the set of all inputs it can accept. It does not include any inputs that'll make the particular function undefined.

The domain of $x^2$ is $\mathbb{R}$, or $(-\infty,+\infty)$ as you rightly mentioned. $x^2$ is a function which
can accept anything, but returns only real numbers, starting from $0$.

The domain of $x^8$ is $\mathbb{R}$ or $(-\infty,+\infty)$.What makes you think it'll be
anything else?

$\frac{1}{x}-7$ would consist of all real numbers, excluding zero. As you may know , $\frac{1}{0} $ is not defined, hence zero will not belong to the domain. So, we can write the domain as $(-\infty,+\infty)$ - {$0$}


问:
If $\phi: X\to Y$ is a bijection, is $\bar\phi : S^Y\to S^X$ an isomorphism?

Consider sets $X,Y \neq \emptyset$. $(S,\cdot)$ is a binary structure, i.e. $\cdot$ is the binary operation such that $\forall s_1,s_2\in S$, $s_1\cdot s_2 \in S$. $S^Y$ is the set of all functions from $Y$ to $S$. Similarly for $S^X$. Consider $$\bar\phi:S^Y \to S^X$$ such that for all $f:Y\to S$, $$\bar\phi(f)(x) = f(\phi(x)), \forall x\in X$$
Note that $S^X$ is also a binary structure (similarly for $S^Y$) with the operator $\otimes$, such that $$(f\otimes g)(x) = f(x)\cdot g(x), \forall x\in X$$

Prove that $\bar\phi$ is a homomorphism, i.e.
$$\bar\phi(f\otimes g)(x) = (\bar\phi(f)\otimes \bar\phi(g))(x), \forall x\in X$$

$$\bar\phi(f\otimes g)(x)  = (f\otimes g)(\phi(x)) = f(\phi(x))\cdot g(\phi(x)) = \bar\phi(f(x)) \cdot \bar\phi(g(x)) = (\bar\phi(f)\otimes\bar\phi(g))(x), \forall x\in X$$
I think this shows that this is a homomorphism. Is my proof fine?

Question: If $\phi: X\to Y$ is a bijection, is $\bar\phi : S^Y\to S^X$ an isomorphism? If yes, is this condition necessary for $\bar\phi$ to be a bijection, or will something weaker (such as injectivity of $\phi$) suffice?

Alright, let's say that $\phi$ is a bijection. Now I'll check if $\bar\phi$ is one-one. Assume $\bar\phi(f(x)) = \bar\phi(g(x)), \forall x\in X$ for some $f,g \in S^Y$. So, $f(\phi(x)) = g(\phi(x)), \forall x\in X$. I'm stuck here, what can I do next?

答:
Your proof that $\bar{\phi}$ is a homomorphism is fine.

For the proof of $\bar{\phi}$ being one-to-one note that since $\phi$ is surjective, the fact that $f(\phi(x))=g(\phi(x))$ for all $x\in X$ implies that $f(s)=g(s)$ for all $s\in S$, so necessarily $f=g$ as maps $X\to S$.

For surjectivity of $\bar{\phi}$, let $g\in S^X$ be a map $X\to S$. Define $f\in S^Y$ by $g(\phi^{-1}(y))$. This is well defined because $\phi$ is a bijection. Now, $\bar{\phi}(f)(x)=f(\phi(x))=g(\phi^{-1}(\phi(x)))=g(x)$, so $\bar{\phi}(f)=g$.

So in conclusion, $\bar{\phi}$ is an isomorphism.


问:
Adjoint of multiplication operator on Sobolev space
This is sort of an idle question, and I'll admit I didn't think very hard about it.

Let $H^1 = H^1(\mathbb{R}^n)$ be the Sobolev space with norm $||f||_{H^1}^2 = ||f||_{L^2(\mathbb{R}^n)}^2 + ||\nabla f||_{L^2(\mathbb{R}^n)}^2$.  Suppose $\psi \in C^1(\mathbb{R}^n)$ is a bounded function with bounded gradient.  Then the multiplication operator $Af = \psi f$ is a bounded operator on $H^1$ (with norm at most $\sqrt{||\psi||_\infty^2 + ||\nabla \psi ||_\infty^2}$ ).

What is the adjoint of $A$?

That is, given $g \in H^1$ we wish to find $h$ such that
$$(\psi f, g) + (\nabla(\psi f),\nabla  g) = (f,h) + (\nabla f,\nabla h)$$
for all $f \in H^1$.  I tried some simple manipulations on this expression but didn't get anywhere.

答:
First assume that $g\in H^2(\mathbb{R}^n)$ and $f\in C_c^\infty(\mathbb{R}^n)$. The equation for the adjoint implies
$$
\int f\psi (g-\Delta g)\,dx=\int(f-\Delta f)h\,dx.
$$
Thus $\Delta h\in L^2$ and $(1-\Delta)h=\psi(1-\Delta)g$, i.e., $h=(1-\Delta)^{-1}(\psi(g-\Delta g))$.

To see that this extends continuously to $H^1$, first note that $(1-\Delta)^{-1}$ is continuous from $H^{-1}$ to $H^1$. Moreover,
\begin{align*}
\left|\int\psi(g-\Delta g)\phi\,dx\right|&\leq\left|\int g\psi \phi\,dx\right|+\left|\int\nabla g\cdot\nabla (\phi\psi)\right|\\
&\leq \|\psi\|_\infty\|g\|_2\|\phi\|_2+\|\nabla g\|_2(\|\psi\|_\infty\|\nabla \phi\|_2+\|\nabla \psi\|_\infty\|\phi\|_2)\\
&\leq\max\{\|\psi\|_\infty,\|\nabla \psi\|_\infty\}\|g\|_{H^1}\|\phi\|_{H^1}
\end{align*}
for $\phi\in H^1$, hence $\|\psi(g-\Delta g)\|_{H^{-1}}\leq \max\{\|\psi\|_\infty,\|\nabla \psi\|_\infty\}\|g\|_{H^1}$.

Therefore the map $g\mapsto (1-\Delta)^{-1}(\psi(g-\Delta g))$ can be extended to a bounded linear operator on $H^1$.


问:
If $2x^4\dot{x}\ddot{x}+4x^3\dot{x}^3=-x\dot{x}$, show that $x^4\dot{x}^2=\frac{1}{2}(x_0^2-x^2)$, where $x(0)=x_0$ and $\dot{x}(0)=0$
I know that $\dot{x}\ddot{x}=\frac{1}{2}\frac{d(\dot{x}^2)}{dt}$, $x\dot{x}=\frac{1}{2}\frac{d(x^2)}{dt}$ and $x^3\dot{x}^3=\big(\frac{1}{2}\frac{d(x^2)}{dt}\big)^3$.

So this gives $x^4\frac{d(\dot{x}^2)}{dt}+\frac{1}{2}\big(\frac{d(x^2)}{dt}\big)^3=-\frac{1}{2}\frac{d(x^2)}{dt}$.

I feel like I need to integrate w.r.t $t$, but I'm not sure how to deal with the 2nd term on the LHS?

答:
$$2x^4\dot{x}\ddot{x}+4x^3\dot{x}^3=-x\dot{x} \\ \implies x^4 (2\dot x \ddot x) + (4x^3 \dot x)(\dot x^2) = -\frac 12 \cdot 2x\dot x \\ \implies \frac{d}{dt} (x^4 \dot x^2)= -\frac 12\frac{d}{dt} x^2$$

Integrating,
$$x^4 \dot x^2 =-\frac 12 x^2+ C $$

Plug in $t=0,x=x_0, \dot x=0$ to get $C=\frac 12 x_0^2 $, and your goal.


答:
$\bar\phi$ is one-to-one if $\phi$ is onto

To prove that $\bar\phi$ is one-to-one, we have to prove that $\bar\phi(f) = \bar\phi(g)$ implies $f=g$. And for this it is enough to have $f(y)=g(y)$ for all $y \in Y$.

$h$ is onto. Therefore for all $y \in Y$ it exists $x \in X$ such that $y = \phi(x)$ and

$$f(y) = f(\phi(x)) = \bar\phi(f(x)) = \bar\phi(f)(x)=\bar\phi(g(x))=g(\phi(x))=g(y)$$

proving that $\bar\phi$ is one-to-one as desired.

Body:
Consider sets $X,Y \neq \emptyset$. $(S,\cdot)$ is a binary structure, i.e. $\cdot$ is the binary operation such that $\forall s_1,s_2\in S$, $s_1\cdot s_2 \in S$. $S^Y$ is the set of all functions from $Y$ to $S$. Similarly for $S^X$. Consider $$\bar\phi:S^Y \to S^X$$ such that for all $f:Y\to S$, $$\bar\phi(f)(x) = f(\phi(x)), \forall x\in X$$
Note that $S^X$ is also a binary structure (similarly for $S^Y$) with the operator $\otimes$, such that $$(f\otimes g)(x) = f(x)\cdot g(x), \forall x\in X$$

Prove that $\bar\phi$ is a homomorphism, i.e.
$$\bar\phi(f\otimes g)(x) = (\bar\phi(f)\otimes \bar\phi(g))(x), \forall x\in X$$

$$\bar\phi(f\otimes g)(x)  = (f\otimes g)(\phi(x)) = f(\phi(x))\cdot g(\phi(x)) = \bar\phi(f(x)) \cdot \bar\phi(g(x)) = (\bar\phi(f)\otimes\bar\phi(g))(x), \forall x\in X$$
I think this shows that this is a homomorphism. Is my proof fine?

Question: If $\phi: X\to Y$ is a bijection, is $\bar\phi : S^Y\to S^X$ an isomorphism? If yes, is this condition necessary for $\bar\phi$ to be a bijection, or will something weaker (such as injectivity of $\phi$) suffice?

Alright, let's say that $\phi$ is a bijection. Now I'll check if $\bar\phi$ is one-one. Assume $\bar\phi(f(x)) = \bar\phi(g(x)), \forall x\in X$ for some $f,g \in S^Y$. So, $f(\phi(x)) = g(\phi(x)), \forall x\in X$. I'm stuck here, what can I do next?


问:
Change of variables during differentiation
If we have a function: $\psi(x)$ and $x=\rho\gamma$ where $\rho$ is a constant

Then is the following statement correct:

$1)~~$ $\psi(x) =\psi(\rho \gamma)=\rho \psi(\rho)$

What is the equation for : $\frac{d\psi(\gamma)}{d\gamma}$ and $\frac{d^2\psi(\gamma)}{d\gamma^2}$

Below is my attempt at finding it:

This question is the context of findings the solution of simple harmonic oscillator

I am having trouble understanding how $9.19$ is derived from $9.18$ .

答:
Question:  I am having trouble understanding how $9.19$ is derived from $9.18$ .

Answer: There are some mistakes (may be printing mistake) in their  calculation. You can see those in my answer which I have already fixed (in red color). Here we go:

$$-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi(x)}{dx^2}+\dfrac{mw^2x^2}{2}\psi(x)=E~\psi(x)\tag{9.18}$$Putting $~x=\rho\gamma\implies \psi(x) =\psi(\rho \gamma)=\rho \psi(\color{red}\gamma)~$ and the partial derivatives became $$\dfrac{d\psi(x)}{dx}=\dfrac{d\psi(\rho\gamma)}{d\gamma}\cdot\dfrac{d\gamma}{dx}=\dfrac{1}{\rho}\cdot\rho\dfrac{d\psi(\gamma)}{d\gamma}=\dfrac{d\psi(\gamma)}{d\gamma}~~~~~~~~\text{and}$$
$$\frac{d^2\psi(x)}{dx^2}=\dfrac{d}{dx}\left(\dfrac{d\psi(x)}{dx}\right)=\dfrac{d}{d\gamma}\left(\dfrac{d\psi(\gamma)}{d\gamma}\right)\cdot\dfrac{d\gamma}{dx}=\frac{d^2\psi(\gamma)}{d\gamma^2}\cdot\dfrac{d\gamma}{dx}=\color{red}{\dfrac 1\rho}\cdot\frac{d^2\psi(\gamma)}{d\gamma^2}~.$$

Now putting the above values in equation $(9.18)$ we have,
$$-\dfrac{\hbar^2}{2m}\dfrac 1\rho\dfrac{d^2\psi(\gamma)}{d\gamma^2}+\dfrac{mw^2\rho^2 \gamma^2}{2}\rho\psi(\gamma)=E~\rho\psi(\gamma)$$
$$\implies \dfrac{d^2\psi(\gamma)}{d\gamma^2}-\dfrac{2m\rho}{\hbar^2}\dfrac{mw^2\rho^2 \gamma^2}{2}\rho\psi(\gamma)=-\dfrac{2m\rho}{\hbar^2}E~\rho\psi(\gamma)$$
$$\implies \dfrac{d^2\psi(\gamma)}{d\gamma^2}-\dfrac{ m^2w^2\rho^4}{\hbar^2}\gamma^2\psi(\gamma)=-\dfrac{2mE\rho^2}{\hbar^2}\psi(\gamma)\tag{9.19}$$


问:
Combinatoric proof of an algebric expression
How can we prove that $\left(\frac{n}{k}\right)^k\le \binom{n}{k}$

I try with induction but I don't get the result I want.

答:
First multiply both sides by $k^k$.
$$n^k \le {n \choose k}k^k$$
Then the new $RHS$ gives you the number of ways in which you select $k$ out of $n$ people and hand them $k$ balls one by one, with no restriction to the number one man can receive. $n^k$ gives you the number of ways to hand $k$ balls to $n$ people without the restrictions.

But note that whenever you included person $A$ on the $RHS$, you counted the situation of him receiving all the $k$ balls all those times, in every combination of $k$ people in which he was included. On the $LHS$ you counted this situation only once. Similarly for any number of balls. Thus the $RHS$ is always greater than $LHS$, unless $k=1$ when you have equality, and you get the reason why.


答:
I'm going to assume that $k \neq 0$, thus $k\geq 1$

$\left(\frac{n}{k}\right)^k = \underbrace{\frac{n}{k}\frac{n}{k}\cdots \frac{n}{k}}_{\text{k-terms}}$

$\binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{k(k-1)\cdots 1}$ from which $\frac{n-i}{k-i} \geq \frac{n}{k}$ for $0 \leq i \leq k-1$.

It's unclear whether you need a combinatorial proof or not, induction doesn't constitute a combinatorial proof, and combinatorial proofs tend to verify that two different ways of counting things are equal, but you have an inequality. In any case here is a direct proof.

Title: Combinatoric proof of an algebric expression

Body:
How can we prove that $\left(\frac{n}{k}\right)^k\le \binom{n}{k}$

I try with induction but I don't get the result I want.