问:
Smooth convex functions
Let $E\subseteq\mathbb R^d$ be a convex set, $\beta\geq 0$ be a given real number and $f:E\to\mathbb R$ be a convex and differentiable function satisfying:
$$f(y)\leq f(x)+\nabla f(x)^\top (y-x) +\frac{\beta}{2}\|x-y\|_2^2, \quad \forall x,y\in E.$$
Show that $\nabla f$ is $\beta$-Lipschitz, i.e.,
$$\|\nabla f(x)-\nabla f(y)\|_2\leq\beta\|x-y\|_2, \quad \forall x,y\in E.$$

Edit: I am interested in the case when $E$ can be any convex subset of $\mathbb R^d$. When $E=\mathbb R^d$, the claim can be proven as proposed by the answer below. However, in many convex optimisation textbooks, the claim is stated for general domains $E$ without proof.

答:
Let $E$ be open, convex and assume that $f$ is convex, differentiable and satisfies your inequality.
It is easy to check that $\nabla f$ is continuous:

Let $(x_n) \subset E$ be a sequence with $x_n \to x \in E$. Since $\nabla f(x_n)$ is bounded ($f$ is locally Lipschitz), we have $\nabla f(x_{n_k}) \to g$ for a subsequence. A standard argument shows that $g \in \partial f(x) = \{ \nabla f(x)\}$. A subsequence-subsequence argument shows that the entire sequence $(\nabla f(x_k))$ converges towards $\nabla f(x)$.

The answer by daw shows that we get
$$
\| \nabla f(y) - \nabla f(x) \| \le \beta \, \| y - x \|
$$
whenever $x,y \in E$ satisfy
$$
x + \frac1\beta \, \Big(\nabla f(y) - \nabla f(x)\Big),\;
y + \frac1\beta \, \Big(\nabla f(x) - \nabla f(y)\Big) \in E.$$

Now, since $E$ is open, we can check that there is $\varepsilon_x > 0$,
such that all $y \in B_{\varepsilon_x}(x)$ satisfy this condition.

Finally, let $x,y \in E$ be arbitrary. Since the segment $[x,y]$ is compact,
we can find finitely many $z_k \in [x,y]$, such that the corresponding balls cover $[x,y]$.
A repeated application of the triangle inequality shows the desired
$$
\| \nabla f(y) - \nabla f(x) \| \le \beta \, \| y - x \|.
$$

Title: Smooth convex functions


问:
If $B$ commutes with a positive semidefinite matrix $A$, then $B$ commutes with $\sqrt A$
How to prove if $B$ commutes with a positive semidefinite matrix $A$, then $B$ commutes with $\sqrt A$.

The logic of notes I refer to is like this: first if A is positive semidefinite, then
$$\sqrt A=\text{polynomial of } A\\
=P\begin{pmatrix}\sqrt{\lambda_1}&&\\&\sqrt{\lambda_2}\\&&\ddots\end{pmatrix}P^T.$$
And by this theorem we can deduce that if $B$ commutes with a positive semidefinite matrix $A$, then $B$ commutes with $\sqrt A$. Using the fact that if $B$ commutes with  $A$, then B commutes with some polynomials of $A$. But actually I don't quite understand what is a polynomial of a matrix. And how can we show that if $B$ commutes with  $A$, then B commutes with some polynomials of $A$.

答:
A polynomial of a matrix means a "substitution" of the matrix for the indeterminate $x$ into a polynomial in $x$. E.g. if $f(x)=c_0+c_1x+c_2x^2+\cdots+c_mx^m$ is a polynomial with complex coefficients, then $f(A)$ means the sum $c_0I+c_1A+c_2A^2+\cdots+c_mA^m$.

Let $f$ be any polynomial such that $f(\lambda_i)=\sqrt{\lambda_i}$ for every $i$ (e.g. take $f$ as a Lagrange interpolation polynomial). Then $f(A)=f(P\Lambda P^{-1})=Pf(\Lambda)P^{-1}=P\sqrt{\Lambda}P^{-1}=\sqrt{A}$.

Now, if $B$ commutes with $A$, then $B$ also commutes with all nonnegative integer powers of $A$: $BA^k=(BA)A^{k-1}=ABA^{k-1}=A(BA)A^{k-2}=AABA^{k-2}=\cdots=A^kB$. Hence $B$ commutes with all linear combinations of nonnegative integer powers of $A$, i.e. $B$ commutes with all polynomials in $A$. Since $\sqrt{A}$ is a polynomial in $A$, the conclusion follows.

ContentLicense: CC BY-SA 4.0


问:
Inner product on Fock spaces
I'm trying to understand Fock spaces by myself and so I got some very basic doubts I'd like to clarify with you. Let $\mathcal{H}$ be a Hilbert space and $\mathcal{F}(\mathcal{H}):= \bigoplus_{n=0}^{\infty}\mathcal{H}^{\otimes n}$ its associated Fock space. Here, $\mathcal{H}^{\otimes n}$ is the $n$-fold tensor product $\mathcal{H}^{\otimes n} := \mathcal{H}\otimes \cdots \otimes \mathcal{H}$. Now, I'm most interested in bosonic/fermionic Fock spaces. Let $\wedge^{n}\mathcal{H}$ to be the subspace of all antisymmetric tensors of $\mathcal{H}^{\otimes n}$. Then, the fermionic Fock space is $\mathcal{F}_{f}(\mathcal{H}) := \bigoplus_{n=0}^{\infty}\wedge^{n}\mathcal{H}$. If all these being said, let me ask:

(1) As far as I understand, the direct sum $\bigoplus_{n=0}^{\infty}\mathcal{H}^{\otimes n}$ simply means the space of all sequences $(x_{0},x_{1},...)$ with all but finitely many nonzero entries, with $x_{n}\in \mathcal{H}^{\otimes n}$. Thus, it seems natural to define an inner product $\langle \cdot, \cdot \rangle$ on $\mathcal{F}(\mathcal{H})$ by setting:
$$\langle x, y \rangle := \sum_{n=0}^{\infty}\langle x_{n},y_{n}\rangle_{\mathcal{H}^{\otimes n}}$$
With this inner product, I believe $\mathcal{F}(\mathcal{H})$ becomes a Hilbert space. Is this reasoning correct?

(2) Related to the above question, some texts I know actually define the same notion of inner product I defined above but directly for fermionic and bosonic Fock spaces. This sounds okay to me as well, but it seems more natural to me to define the inner product on $\mathcal{F}_{f}(\mathcal{H})$ and then restric it to each subspace. Are these constructions equivalent?

(3) Some texts define the Fock space as the space of all sequences $x = (x_{0},x_{1},...)$ with $x_{n} \in \mathcal{H}^{\otimes n}$ satisfying $||x||^{2} := \sum_{n=0}^{\infty}||x_{n}||^{2}_{\mathcal{H}^{\otimes n}} < +\infty$. If the reasoning of my first question is correct, it seems that this definition is just a consequence of my inner product on $\mathcal{F}(\mathcal{H})$, right?

答:
As mentioned in one of the comments, you need a construction as in $(3)$ for your space to be complete whereas your construction in $(1)$ is not necessarily complete, so they are not equivalent. However, if you take the Hilbert space completion of the space you defined in (1), this space will be the same as the space defined in (3).

See the discussion here:
https://en.wikipedia.org/wiki/Direct_sum_of_modules#Direct_sum_of_Hilbert_spaces.
See also
https://en.wikipedia.org/wiki/Fock_space#Definition, the fourth paragraph.

Title: Inner product on Fock spaces


问:
Show $\log{a_n}\rightarrow\log{a}$ as ${n\rightarrow\infty}$
Let $(a_n)_{n\in\mathbb{N}}\in\mathbb{R}$ and assume $a_n>0$ and $a_n\rightarrow a>0$ as $n\rightarrow\infty$.

The sequence $a_n$ clearly converges and I am inclined to think that if $a_n$ is convergent with the above statement being true, then

$\lim_{n\rightarrow\infty}{\log(a_n)=\log(\lim_{n\rightarrow\infty}{a_n})}$ but I am not sure how to prove it.

答:
The result can come directly from the continuity of $\log$... But, just for using something different, Lagrange's theorem allows you to establish that
$$
|\log a_n - \log a| = \frac{1}{\xi_n}|a_n -a|, \quad \xi_n \in (a_n, a)
$$

From the previous equality, we get
$$
\lim |\log a_n -\log a| = \lim \frac{1}{\xi_n}|a_n-a| = \frac 1a \cdot 0=0,
$$

which in turn means that $\lim a_n =a$.


问:
Invertible elements of $\mathbb Z_9$
I have this problem: Prove that $x=[1]_{9}+[3]_{9}[b]_{9}$ is invertible in $\mathbb{Z}_{9}$, for all $[b]_{9}\in \mathbb{Z}_{9}$ and find the inverse.

In empirical way i found that $x\in\{[1]_{9}, [4]_{9}, [7]_{9}\}$,  so $x$ is invertible for all $[b]_{9}\in \mathbb{Z}_{9}$, since the inverses of $\mathbb{Z}_{9}$ are $\{[1]_{9},[2]_{9}, [4]_{9}, [5]_{9},[7]_{9}, [8]_{9},\}$, but is there a rigorous proof to prove that?

Thanks in advance. Best wishes.

答:
More generally, this holds in any ring:

If $u^n=0$ then $1+u$ is invertible and the inverse is $1-u+u^2-\cdots+(-1)^{n-1} u^{n-1}$.

In your case, $u=3b$ and $u^2=0$ and so the inverse is $1-u=1-3b$.

Title: Invertible elements of $\mathbb Z_9$


问:
Show $\lim_{n\rightarrow \infty}(1-x^{a \log(n)}) =0$ when $a <\frac{-1}{\log(x)}$
I'd like to show that
$$\lim_{n\rightarrow \infty}(1-x^{a \log(n)}) =1$$ when $a >\frac{-1}{\log(x)}$ and
$$\lim_{n\rightarrow \infty}(1-x^{a \log(n)}) =0$$ when $a <\frac{-1}{\log(x)}$

Where $x \in (0,1)$. I have managed to show the first limit using $a = \frac{-1}{\log(x)}+\epsilon$ where $\epsilon>0$. However this approach did not work for the second one (taking $a = \frac{-1}{\log(x)}-\epsilon$).

My idea was since we want the limit to be $0$, we would want $\lim_{n\rightarrow \infty} x ^{a\log(n)}=1$. This only happens when $\lim_{n\rightarrow \infty} a \log(n) = 0$. Substituting $a = \frac{-1}{\log(x)}-\epsilon$ does not help solving this limit though. I was wondering if someone maybe could give me a hint on how to approach this.

Thanks in advance

答:
When $0 < x < 1$ then
\begin{aligned}
\lim_{n\to\infty}\big( 1 - x^{a\log n} \big) = \left\{
\begin{array} & 0 & \text{if } a = 0 \\
1 & \text{if } a > 0 \\
-\infty & \text{if } a < 0
\end{array}
\right.
\end{aligned}


答:
If $\{v_1,\ldots,v_n\}$ is a basis of $\mathbb R^n$ with eigenvectors of $A$, i.e., $Av_k=\lambda_kv_k$, $\lambda_k\ge 0$, then
$$
ABv_k=BAv_k=B(\lambda_kv_k)=\lambda_k Bv_k, \quad k=1,\ldots,n.
$$
Let $V(\lambda_k)$ be the eigen-space of $\lambda_k$, i.e., $Aw=\lambda_kw$, for all $w\in V(\lambda_k)$.

Claim. $B[V(\lambda_k)]\subset V(\lambda_k).$

Proof of the Claim. If $\lambda_k=0$, then $Av_k=0$, and $BAv_k=0$, and hence $ABv_k=BAv_k=0$. So $Bv_k\in V(\lambda_k)$. If $\lambda_k>0$, then
$ABv_k=\lambda_kBv_k$, implies again that  $Bv_k\in V(\lambda_k)$.

Let now $Cv_k=\lambda_k^{1/2}v_k$, $k=1,\ldots$, and hence $C^2=A$.
By virtue of the Claim, $CBv_k=\lambda^{1/2}_kBv_k$.

If $v\in\mathbb R^n$, then $v=c_1v_1+\cdots+c_nv_n$, for some
$c_1,\ldots,c_n\in\mathbb R$, and
$$
BCv=\sum_{k=1}^nc_kBCv_k=\sum_{k=1}^nc_k\lambda^{1/2}_kBv_k=
\sum_{k=1}^n c_kCBv_k=CBv.
$$

Title: If $B$ commutes with a positive semidefinite matrix $A$, then $B$ commutes with $\sqrt A$

Body:
How to prove if $B$ commutes with a positive semidefinite matrix $A$, then $B$ commutes with $\sqrt A$.

The logic of notes I refer to is like this: first if A is positive semidefinite, then
$$\sqrt A=\text{polynomial of } A\\
=P\begin{pmatrix}\sqrt{\lambda_1}&&\\&\sqrt{\lambda_2}\\&&\ddots\end{pmatrix}P^T.$$
And by this theorem we can deduce that if $B$ commutes with a positive semidefinite matrix $A$, then $B$ commutes with $\sqrt A$. Using the fact that if $B$ commutes with  $A$, then B commutes with some polynomials of $A$. But actually I don't quite understand what is a polynomial of a matrix. And how can we show that if $B$ commutes with  $A$, then B commutes with some polynomials of $A$.


问:
pseudoconvex functions (proof)
We have the following theorem. Let $f:\mathbb{R}^{n}\to\mathbb{R}$ be differentiable convex function and $g:\mathbb{R}\to\mathbb{R}$ nondecreasing and pseudoconvex function. Prove that $g(f(x))$ is pseudoconvex function.

I used this idea: We have $x,y\in\mathbb{R}^{n}$ and we need to verify the definition of pseudoconvexity:
$$(\nabla {g(f(x))}^{T}(y-x)\ge 0\implies g(f(y))\ge g(f(x)).$$ From the assumption we get $$g{'}(f(x)) (\nabla f(x)(y-x))\ge 0  $$ The first term $g{'}(f(x))\ge 0$, because $g$ is nondecreasing function. So the second term $(\nabla f(x)(y-x))\ge 0$ and because $f$ is differentiable convex function we know that $$f(y)-f(x)\ge \nabla f(x)(y-x)$$ So we get $f(y)\ge f(x)$ and because $g$ is a nondecreasing function we have $g(f(y))\ge g(f(x))$, which we wanted to show.

My question is why do we need the assumption of pseudoconvex function $g$? I didn't use it at all in my idea, but I suppose it's important, otherwise it would not be there. What is wroing in my idea then? Thanks for advice.

答:
We cannot get $\left(\nabla f(x)\right)^T(y-x)\geq 0$ from $g'\left(f(x)\right)\left(\nabla f(x)\right)^T(y-x)\geq 0$ and $g'\left(f(x)\right)\geq 0$ because $\left(\nabla f(x)\right)^T(y-x)$ could be negative if $g'\left(f(x)\right)=0$.

Now, $g$ is a non-decreasing and pseudoconvex function.

Case 1: $g'(f(x))>0$, then $\left(\nabla f(x)\right)^T(y-x)\geq 0$. You have proved it.

Case 2: $g'(f(x))=0$, then $g'(f(x))(f(y)-f(x))=0$ satisfies the condition $g'(f(x))(f(y)-f(x))\geq0$, so, $f(y)\geq f(x)$ due to the pseudoconvexity of $g$.

Title: pseudoconvex functions (proof)


问:
How to show that $f(x) = o{(x^2)}$ as $x \to \infty$. when $f$ is differentiable on $(0, \infty)$ and $f'(x) = o(x)$ as $x\ \to \infty$
Suppose $f$ is differentiable on $(0, \infty)$ and $f'(x) = o(x)$ as
$x \to \infty$.

How to show  that $f(x) = o{(x^2)}$ as $x  \to \infty$?

答:
Let $\epsilon>0$. Fix $x_0>0$, such that $|f'(x)|<\epsilon x$ for $x>x_0$. It holds
$$ |f(x)| = \left|f(x_0) + \int_{x_0}^x f'(\xi)d\xi\right|
\le|f(x_0)|+\epsilon\left|\int_{x_0}^x\xi\,d\xi\right|
=|f(x_0)|+\frac{\epsilon}{2}\left|x^2-x_0^2\right|\,.$$
Now fix $x_1>x_0$, such that $|f(x_0)|<\frac\epsilon2x^2$ for $x>x_1$. Then for all $x>x_1$ we have
$$ |f(x)| < \frac\epsilon2x^2+\frac\epsilon2\left|x^2-x_0^2\right| \le \epsilon x^2\,, $$
qed.</p>

ContentLicense: CC BY-SA 3.0


问:
Finding the power of a complex valued matrix with only the characteristic polynomial
Hi I have this question here.

Suppose that $A \in M_4(\mathbb{C})$ has the characteristic polynomial $$p_A(x)=(x^2+3)(x^2+\sqrt{3}x+3)$$ Find $A^{12}$.

I get my eigenvalues are $\frac{-\sqrt{3}+3i}{2},\frac{-\sqrt{3}-3i}{2},\sqrt{3}i,-\sqrt{3}i$. Since the eigenvalues are distinct I know that $A$ is diagonalizable over the complex numbers if $A$ is of size $4 \times 4$ (Which it is).

I also know that $A^{12}=PD^{12}P^{-1}$. How do I figure out $P$ (And subsequently $P^{-1}$ ) though? I don't even know if that's possible...

答: If you expand $D^{12}$, you get a constant multiple of $I_{4}$, so $A^{12}=D^{12}$.


问:
Derivative as first order variation in output?
In reference to method shown at 4:00 of this video,

Consider,

$$f(x) = x^x$$

Then,

$$ f(x+h) = (x+h)^{x+h} = x^{x+h} ( 1 + \frac{h}{x})^{x+h}$$

Now, consider right most term in brackets,

$$ ( 1 + \frac{h}{x})^{x+h} = 1+ (x+h) \frac{h}{x} + \frac{ (x+h)(x+h-1)}{2!} ( \frac{h}{x})^2...= 1+h+ stuff$$

This suggests derivative of $x^x$ is $x^x$ which is wrong! So, how would I prove the derivative using this 'perturbation method'

答:
May be, logarithmic differentiation would me life easier
$$f(x)=x^x \implies \log(f(x))=x \log(x)\implies \frac {f'(x)}{f(x)}=\big[x \log(x) \big]'$$ Now, consider
$$g(x)=x  \log(x)$$
$$g(x+h)=(x+h)\log(x+h)=(x+h)\big[\log(x)+\log \left(1+\frac{h}{x}\right) \big]$$ Since $h$ is small, using equivalents (or Taylor series)
$$\log \left(1+\frac{h}{x}\right)\sim \frac{h}{x}$$ Replacing
$$g(x+h)\sim (x+h)\big[\log(x)+\frac{h}{x} \big]$$  Expanding
$$g(x+h)\sim x \log(x) + x\frac{h}{x}+h \log(x) + h\frac{h}{x}=x \log(x)+h+h \log(x) + \frac{h^2}{x}$$

$$\frac{g(x+h)-g(x)}h\sim\frac{h+h \log(x) + \frac{h^2}{x} }h=1+\log(x) + \frac{h}{x}  $$
$$\lim_{h\to 0} \, \frac{g(h+x)-g(x)}{h}=1+\log(x)$$
$$\frac {f'(x)}{f(x)}=1+\log(x)\implies f'(x)=f(x)\times\frac {f'(x)}{f(x)}= \big[1+\log(x)\big]\,x^x$$


问:
Two random variable density functions.
I'm new to this site and was wondering if anyone could help me on a math question!

I am currently studying probability and statistics and have come across a challenging question for me. See link below.

So previous I have worked with joint density functions in order to determine probabilities but I'm afraid I'm totally stuck when dealing with independent density functions.

Could anyone help me on this matter please. Kind regards.

Suppose that the random variable $X$ has density
$$
f_X(x) =
\begin{cases}
\frac{x}{2}, & 0<x<2 \\
0, & \text{otherwise}
\end{cases}
$$
and suppose that the random variable $Y$ has density
$$
f_Y(y) =
\begin{cases}
4y^3, & 0<y<1 \\
0, & \text{otherwise}
\end{cases}
$$
also suppose that the variables $X$ and $Y$ are independent.

Determine $P(Y>X)$ to three decimal places.

question link

答:
Easy answer:

$$\mathbb{P}[Y>X]=\int_{0}^{1}f_Y(y)\Bigg[\int_0^y f_X(x)dx\Bigg]dy$$

Title: Two random variable density functions.


问:
Proving $x|x|$ is differentiable for all $x$
$f:\mathbb{R} \to \mathbb{R}$ be defined by $ f(x)= x|x| $ for all $x\in \mathbb{R}$.

Prove that the function $f$ is differentiable, and find a formula for its derivative  $f':\mathbb{R} \to \mathbb{R}$.

solution:

I have attempted using  $\lim_{x\to a} = \frac{f(x)-f(a)}{x-a}$ = $\frac{x|x|-a|a|}{x-a}$

I don't know how to simplify this or go any further?

答:
Your definition is not quite correct.  The derivative should be $$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a},$$ that is to say, the derivative of $f$ evaluated at $a$ is the limit as $x$ approaches $a$, not $0$.

If you use the correct definition, then you have three cases:  first, if $a > 0$, then $|a| = a$, and for a sufficiently small neighborhood around $a$, we would also have $|x| = x$ as $x \to a$, thus the quotient becomes $$\frac{x^2 - a^2}{x-a} = x + a$$ and the limit is $2a$.

Second, if $a < 0$, then $|a| = -a$ and again, in a sufficiently small neighborhood around $a$, we may assume $|x| = -x$, and the rest of the computation is similar.

The third case is if $a = 0$.  Here, we need to evaluate $$\lim_{x \to 0} \frac{x|x|}{x} = \lim_{x \to 0} |x| = 0.$$  Therefore, the derivative exists for all $x \in \mathbb R$.


答:
Hints:

If $a>0$, you can work in a neighborhood of $a$ such that $x>0$, and

$$\frac{x|x|-a|a|}{x-a}=\frac{x^2-a^2}{x-a}.$$

If $a<0$, you can work in a neighborhood of $a$ such that $x<0$, and

$$\frac{x|x|-a|a|}{x-a}=-\frac{x^2-a^2}{x-a}.$$

If $a=0$,

$$\frac{x|x|-0}{x-0}$$ simplifies to $|x|$ for $x\ne0$.

Alternatively, the derivative of $|x|$ is known to be $\text{sgn}(x)$ for $x\ne0$ and be undefined for $x=0$.

Hence for $x\ne0$

$$(x|x|)'=|x|+x\text{ sgn}(x)=2|x|.$$

For $x=0$, we can use the left and right derivatives, which are both $0$.

Title: Proving $x|x|$ is differentiable for all $x$

Body:
$f:\mathbb{R} \to \mathbb{R}$ be defined by $ f(x)= x|x| $ for all $x\in \mathbb{R}$.

Prove that the function $f$ is differentiable, and find a formula for its derivative  $f':\mathbb{R} \to \mathbb{R}$.

solution:

I have attempted using  $\lim_{x\to a} = \frac{f(x)-f(a)}{x-a}$ = $\frac{x|x|-a|a|}{x-a}$

I don't know how to simplify this or go any further?


问:
$\lim_{n\rightarrow\infty} ||T^n||$ where $T$ is left-shift operator on $l^2$
Let $T$ be the left-shift operator on $l^2$. I want to calculate the two limits $\lim_{n\rightarrow\infty} \|T^nx\|_2\;\text{and}\; \lim_{n\rightarrow\infty} \|T^n\|$

I have no problems with the first one $$ \lim_{n\rightarrow\infty} \|T^nx\|_2 = \lim_{n\rightarrow\infty}  \left(\sum_{i=n}^\infty|x_i|^2 \right)^\frac{1}{2} = 0.$$

But I'm not sure about the second one. I tried to write it out as $$ \|T^n\|= \inf\{c>0: \forall x_i \; \|T^nx\|_2 \leq c\|x\|_2\} $$

but I can't find a proper $c$ as $n\rightarrow \infty$.

Body: $\|T^{n}x\|\leq \|x\|$, so  $\|T^{n}\| \leq 1$. Also $T^{n}(e_{n+1})=e_{1}$ Hence $\|T^{n}\| \geq 1$. So $\|T^{n}\| = 1$ for all $n$.


问:
Bound for Inner Product Between Matrix Vector and Vector i.e $Ax$ and $x$
Assume $A\in \mathbb{R}^{n\times n}$ is invertible, what kind of assumptions on $A$ do I need to have a bound of the form

$$c_1 \|x\|^2\leq \langle Ax,x\rangle \leq c_2 \|x\|^2,~~\text{for any} ~x\in \mathbb{R}^n$$

for some $c_1,c_2$ constants.

答:
This is true for arbitrary $A \in \mathbb{R}^{n \times n}$. You don't even need $A$ to be invertible. Any inner product $\langle x , y \rangle$ on $\mathbb{R}^n$ can be expressed as $y^T S x$ for a symmetric, positive-definite matrix $S \in \mathbb{R}^{n \times n}$. So
$$
|\langle Ax , x \rangle| = |x^T S (Ax)| \leq \|x\|\|(SA)x\|
$$ But linear maps like $x \mapsto (SA)x$, composed of matrix-vector products, are bounded maps i.e. $\|(SA)x\| \leq C\|x\|$ for some $C \geq 0$. Hence,
$$
|\langle Ax , x \rangle| \leq \|x\|(C\|x\|) = C \|x\|^2
$$


问:
cardinality of the union of 2 sets with at least two elements and cardinality of cartesian product
Suppose X,Y are sets with at least 2 elements. Show that
$X\cup Y\le X\times Y$

So my first thought was that cardinality $|X|\ge 2$ and the same for $|Y|\ge 2$ but by the inclusion-exclusion principle we have $|X\cup Y|=|X|+|Y|-|X\cap Y|$ but the problem does not say if they are disjoint or not.
If we assume that they are disjoint then $|X\cup Y|\ge 4$ right?
But from the definition of the cardinality for the cartesian product we have similar
$|X\times Y|\ge 4$. But I have to find an injective function (including the number of elements) s.t $f:X\cup Y\to X\times Y$ and thus conclude the result.
How should I do it?

答: HINT: max$(|X\cup Y|)$=$|X|+|Y| \leq |X|.|Y|$ f0r the given conditions.</p>


问:
Struggling to find solution: question 5 (ODE) (Separable Equations) (Polynomials)
Could someone help me with this problem?
$$y'=\frac{x^3+x^2y+3y^3}{x^3+3xy^2}$$

I have tried $y=ux$ since it kind of resembles $f(\frac{y}{x})$ structure, and tried to gather terms in parentheses to resolve it but couldn't.

答: TRICK: Divide both numerator and denominator by $x^3$, so your EDO would be $$y'=\frac{1+(y/x)+3(y/x)^3}{1+2(y/x)^2}$$ and with your substitution $$u'x+u=\frac{1+u+3u^3}{1+3u^2}$$


问:
Continuity of the function $f(x,y)=\frac{\cos(x+y)+\cos(x-y) -2}{x^2 +y^2 }$

Find the natural domain of the function

$$f(x,y)=\frac{\cos(x+y)+\cos(x-y) -2}{x^2 +y^2 }$$

find the points where it's continuous and for each accumulation point of it's natural domain, determine whether or not the function can be expanded to a continuous one.

My attempt:

I think the natural domain is $\mathscr D_f=\Bbb R^2\setminus\{0,0\}$ and that the function is continuous at every $x\in\mathscr D_f$. The denominator is always positive.

I also think $f(x,y)$ does have a limit as $(x,y)\to (0,0)$.

I was looking at the Taylor expansion of cosine:
$\cos x = 1 -\frac12x^2 +\frac{1}{24} x^4 + \cdots$
so
$\begin{aligned}\cos(x+y)&=1-\frac12(x+y)^2+\frac1{24}(x+y)^4 + \cdots\\\cos(x-y)&=1-\frac12(x-y)^2+\frac1{24}(x-y)^4 + \cdots\end{aligned}$

The point is that $(x+y)^2 +(x-y)^2 =2x^2 +2y^2,$ so
$\cos(x+y)+\cos(x-y)-2=-(x^2 +y^2) +\frac1{12} (x^4 +6x^2 y^2 +y^4 )+\cdots$

which shows that
$\lim\limits_{(x,y)\to(0,0)} f(x,y) = -1$
so there is then a removable discontinuity at $(0,0)$ and therefore, we can extend $f(x,y)$ there.

Is this correct? If so, is there anything I could improve or any easier approach?

Thank you very much!

答:
Corrrect. Another way: $\cos(x\pm y)=\cos x \cos y \mp \sin x \sin y$.

So $\cos(x+y)+\cos(x-y)=2\cos x \cos y.$


答:
Whenever $|X| \ge 2$ and $|Y| \ge 2$ you have
$$|X| + |Y| - |X \cap Y| \le |X| + |Y| \le |X| \cdot |Y|$$
so it doesn't matter whether or not they're disjoint.

When $|X|=1$ or $|Y|=1$ the result isn't true in general, e.g. $|\{0\} \cup \{1,2\}| = 3$ but $|\{0\} \times \{1,2\}| = 2$.

Edit: So your real task it so prove that $m+n \le mn$ whenever $m,n \ge 2$. Suppose wlog that $2 \le m \le n$, then
$$m+n \le n+n = \cdots$$
You finish it off.

Title: cardinality of the union of 2 sets with at least two elements and cardinality of cartesian product

Body:
Suppose X,Y are sets with at least 2 elements. Show that
$X\cup Y\le X\times Y$

So my first thought was that cardinality $|X|\ge 2$ and the same for $|Y|\ge 2$ but by the inclusion-exclusion principle we have $|X\cup Y|=|X|+|Y|-|X\cap Y|$ but the problem does not say if they are disjoint or not.
If we assume that they are disjoint then $|X\cup Y|\ge 4$ right?
But from the definition of the cardinality for the cartesian product we have similar
$|X\times Y|\ge 4$. But I have to find an injective function (including the number of elements) s.t $f:X\cup Y\to X\times Y$ and thus conclude the result.
How should I do it?


问:
Right adjoint to forgetful functor from $G$-Sets to $H$-Sets?
Let $H$ be a subgroup of group $G$.

Let $F_1$ be a forgetful functor from $G$-Sets (sets with action of group $G)$ to $H$-Sets.

Is it true that right-adjoint functor to $F_1$ exists if and only if $H$ is a normal subgroup of $G$ ?

答:
No, $F_1$ has adjoints on both sides, whether $H$ is normal or not.

Indeed, the category of $G$-sets is isomorphic to the category of functors $[G,\mathbf{Set}]$ (where $G$ is seen as a one-object category), and the functor $F_1$ coincides with the precomposition with the inclusion functor/homomorphism $i:H\to G$. Thus $F_1$ has left/right adjoints, which are constructed as left/right Kan extensions along $i$; these Kan extensions exist because $\mathbf{Set}$ is complete and cocomplete.

In fact this is true even for an arbitrary morphism $\varphi:H\to G$, or if $G$ and $H$ are monoids, or if one replaces sets with vector spaces, modules, topological spaces, or any (co)complete category (although it could make the constructions more complex).


问:
What is the Real Prime?
There seems to be an importance to the ring of adeles for the rational numbers (discussed here), with valuations for every $\mathbb{Q}_p$, but also one "infinite" valuation "$\mathbb{Q}_∞$", seemingly equal to $\mathbb{R}$.

Why would something like $\mathbb{Q}_∞$ be used in the first place, and how is that equal to the reals? Is there something like a $∞$-adic metric that works like the usual one?

Moreover, it seems to suggest that $∞$ here is a sort of an infinite prime number, i.e. the real prime, having some occult-sounding books written about it. So, does it exist as some sort of a describable object here, or is it just notation?

答:
It is a formal notation. We treat $|\cdot|$ as an absolute
value $|\cdot|_{\infty}$ coming from an “infinite prime”, so that we obtain, among other things, a product formula
$$
\prod_{p\le \infty} |\alpha|_p=1
$$
for every $\alpha\in \Bbb Q^{\times}$. Of course, $p=\infty$ is not really a prime. So $\Bbb Q_{\infty}:=\Bbb R$ is just a notation. We can summarize all completions of $\Bbb Q$ by

$$
\Bbb Q_2,\Bbb Q_3,\Bbb Q_5,\cdots ,\Bbb Q_{\infty}
$$

Title: What is the Real Prime?


问:
Shilov or Axler for Linear Algebra?
I'm looking for a book on linear algebra. I found that Axler's and Shilov's books have a good reputation. Which of them is better? Which is more complete and suitable for theoretical study?</p>

答:
I see that the author of the accepted answer only owns Axler's book.
For me the opposite is true - I only own Shilov's book.

If Axler's text delays the concept of determinants, Shilov does the opposite - determinants are the first concept examined in his book.

I can only talk about this part, since I got stuck:
It is rigorous, and rather formal - it features a definition of determinants based on permutations - very helpful if you were previously only confronted with things like the Leibniz-Formula.
It then continues with cofactors and minors, with the usual "Lemma, Definition, Theorem" - Style. I think this is great if you want a treatment without any surplus to slow you down - personally I am looking for a book with a little more commentary to accompany Shilov's text for my studies.

Title: Shilov or Axler for Linear Algebra?


问:
Coinflip statistics
There is a betting strategy for coinflips, called Martingale, in which the player should pick heads or tails and keep betting on it (doubling their bet on every loss) to eventually make a win when it appears. In theory this is watertight, but with one caveat: there can be very long strings of losses, and the player runs out of money, and looses it all.

Now, some people wait until they have seen a very long string of outcomes for one side, and start betting at the other side, with the idea that has a larger chance now. I know that mathematicly this is nonsense, both sides still have 50% chance no matter what happend before.

Although I can understand a dice/coin/ball has no notion of past/future, so will always remain 50% chance, I never quite understanded the following. You can proof there is only a very slight possibility of me ever witnessing 50 heads in a row during my lifetime. So after 49 spins, I will either observe something extraordinary (50 heads) or something less special (first tail), how can that not effect chance?

Maybe the answer is obvious, that for math/statistics it doesnt matter whether someone is observing the event or not, but please enlighten me.

答:
I would say that if it's the case that you have already seen 49 heads in a row (history), you have already witnessed something extremely unlikely.

What you presently see on the $50$th flip, heads or tails, is a "toss-up" (excuse the pun!): the outcome of which is hardly extraordinary: tails or heads.

You're right, Joshua. The probability of getting $49$ heads followed by a head $\left(\dfrac 1{2}\right)^{50}$ is equal to that of getting $49$ heads, followed by a tails: $\left(\dfrac 1{2}\right)^{49} \cdot \dfrac 12.\;$ Both extremely unlikely events, taken as a whole.

Title: Coinflip statistics


问:
Hypothesis testing Z-test:Where does the normal curve come from?
When performing a hypothesis testing z-test we can in the end determine the p-value using a standard normal curve after determine the z-score. Is the standard normal curve derived from my sample, that is using central limit theorem and then standardize the curve?</p>

答: The standard normal curve is independent of your sample. Given a certain p-value, we can calculate the confidence interval for your estimation. You can look up central limit theorem MIT on youtube for 2 good examples (50 min video).</p>


问:
Is $\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}{2})}{\Gamma(\tfrac{n}2+1)}=\frac{2}{\sqrt{\pi}}$ true?

Prove/disprove $$\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}=\frac{2}{\sqrt{\pi}}.\tag 1$$

As far as I can tell, this is true, although it seems to converge very slowly.

I came up with a proof but I don't know if it's valid.

Let $$J=\int_0^\pi \frac{xdx}{1+\sin x}.$$
On one hand, we have
$$\frac1{1+\sin x}=\sum_{n\ge0}(-1)^n\sin(x)^n,$$
so that
$$J=\sum_{n\ge0}(-1)^n p_n,\tag 2$$
where
$$
\begin{align}
p_n&=\int_0^\pi x\sin(x)^ndx\\
&=\int_\pi^0 -(\pi-x)\sin(\pi-x)^ndx\\
&=\pi\int_0^\pi\sin(x)^ndx-p_n\\
\Rightarrow p_n&=\frac\pi2\int_0^\pi\sin(x)^ndx.
\end{align}
$$
And since $\sin(x)=\sin(\pi-x)$, $$p_n=\pi\int_0^{\pi/2}\sin(x)^ndx=\frac{\pi^{3/2}}{2}\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}.\tag 3$$

On the other hand, we have $1+\sin x=2\sin^2(\tfrac{x}2-\tfrac\pi4)$, so that
$$\begin{align}
J&=\frac12\int_0^\pi\frac{xdx}{\sin^2(\tfrac{x}2-\tfrac\pi4)}\\
&=2\int_{\pi/4}^{3\pi/4}\frac{tdt}{\sin^2t}-\frac\pi2\int_{\pi/4}^{3\pi/4}\frac{dt}{\sin^2 t}\\
&=2\left(\ln\sin x-x\cot x\right)\bigg|_{\pi/4}^{3\pi/4}-\frac\pi2\left(-\cot x\right)\bigg|_{\pi/4}^{3\pi/4}\\
&=2\pi-\frac\pi2\cdot2=\pi.
\end{align}$$
Then from $(2)$ and $(3)$, we have
$$\frac{\pi^{3/2}}{2}\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}=\pi,$$
which is equivalent to $(1)$. $\square$

Can you come up with any other proofs to $(1)$? Thanks!

Edit (11/12/2020):

Here is a proof that the interchange of the sum and integral in $(2)$ is valid.

The partial sums $$S_M(x)=\sum_{n=0}^M(-1)^n\sin(x)^n$$
form a uniformly convergent sequence of functions for $x$ in $[0,\pi/2)$ or $(\pi/2,\pi]$, and they converge to the limit $$\lim_{M\to\infty}S_M(x)=\frac1{1+\sin x},\qquad x\in[0,\pi]\setminus\{\pi/2\}.$$
Choose $\varepsilon>0$ and notice that
$$J=\int_{0}^{\pi}\frac{xdx}{1+\sin x}=\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}+\int_0^{\pi/2-\varepsilon}\frac{xdx}{1+\sin x}+\int_{\pi/2+\varepsilon}^\pi\frac{xdx}{1+\sin x}.$$
The sums $S_M(x)$ converge uniformly as $M\to\infty$ when $x\in[0,\pi/2-\varepsilon]\cup[\pi/2+\varepsilon,\pi]$, so we can interchange the sum and integral to get
$$J=\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}+\sum_{n\ge0}(-1)^n(a_n(\pi/2-\varepsilon)+b_n(\pi/2+\varepsilon)),$$
where
$$\begin{align}
a_n(t)&=\int_0^t x\sin(x)^ndx\\
b_n(t)&=\int_t^\pi x\sin(x)^ndx.
\end{align}$$
We have $a_n(t)+b_n(t)=p_n$ for all $t\in[0,\pi]$. As $\varepsilon$ approaches $0$, $\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}$ approaches $0$.

And since $a_n(t), b_n(t)$ are continuous, $a_n(\pi/2-\varepsilon)+b_n(\pi/2+\varepsilon)$ approaches $a_n(\pi/2)+b_n(\pi/2)=p_n$, so that
$$J=\sum_{n\ge0}(-1)^np_n$$
as desired. $\square$

答:
Under the Legendre duplication formula,
$$\Gamma(z)\Gamma(z+\frac12)=2^{1-2z}\sqrt{\pi}\Gamma(2z)$$
making:
$$\Gamma(z+\frac12)=2^{1-2z}\sqrt{\pi}\frac{\Gamma(2z)}{\Gamma(z)}$$
and so:
$$\frac{\Gamma(z+\frac12)}{\Gamma(z+1)}=2^{1-2z}\sqrt{\pi}\frac{\Gamma(2z)}{\Gamma(z)\Gamma(z+1)}=2^{1-2z}\sqrt{\pi}\frac{\Gamma(2z)}{z\Gamma^2(z)}$$
could you use this?


答:
$f$ is certainly differerentiable in all points $a \ne 0$ because
$$f(x) = \begin{cases} x^2 & x \ge 0 \\ -x^2 & x \le 0 \end{cases}$$
Certainly you know that a function $\phi$ is differentiable at $a$ if and only if both one-sided limits $\phi'_+(a) = \lim_{x \to a^+} \dfrac{\phi(x)- \phi(a)}{x-a}$ and $\phi'_-(a) = \lim_{x \to a^-} \dfrac{\phi(x)- \phi(a)}{x-a}$ exist and are equal.

In your case we get $f'_+(0) = f'_-(0) = 0$.


答:
Only $x=0$ the suspect point for differentiability.
$f(x)=x|x|=x^2, x\ge 0, x|x|=-x^2$ the left derivative is $Lf'(x)=-2x$ and right derivative is $Rf'(x)=-2x$. So at $x=0$, we have $Lf'(0)=0=Rf'(0)$, so $f(x)$ is differentiable even at $x=0$. Hence, $f(x)$ is differentiable everywhere.


问:
Algorithm to find roots of a scalar field
Let $f : D\subset \mathbb{R}^2 \longrightarrow \mathbb{R}$ be smooth (e.g. Lipschitz continuous). I'm trying to develop a numerical algorithm which gives a large number of roots of $f$ with a minimum amount of function evaluations. This can also be understood as getting the contour 'line' $$\{x \in D \,|\, f(x)=0\}.$$

My attempt: Fix $\epsilon,\delta>0$ and perform the following recursion to get a sequence of roots $r_i$:

Sample a point $p_0 \in D$ uniformly at random such that $|f(p_0)| > \epsilon$.

Sample a point $p_1 \in D$ uniformly at random such that $|f(p_1)| > \epsilon$.

If $f(p_0)$ and $f(p_1)$ have opposite signs apply 'secant method' to get root $r_1$ with $|f(r_1)| < \epsilon$. Otherwise go to 2.

Calculate the tangent to $r_1$ and get $p_2$ on it a distance $\delta$ appart from $r_1$.

Do 3. with $p_2$ and either $p_0$ or $p_1$ (such that they have opposite signs) to get $r_2$.

Do 4. with $r_2$ and so on to trace the contour.

Restart at 1. to find other unconnected parts of the contour.

How could this algorithm be improved? Or do you know a better method? Is it better than 'marching squares'? Would it even work to get complicated contours like that:

答:
Let's try to improve the 'marching squares' algorithm. Consider only Lipschitz continuous functions.
W.l.o.g assume that $D$ is a square, $f \geq 0$ and $\forall x,y\in D : |f(x)-f(y)| \leq \|x-y\|$.

Algorithm:
Fix $\epsilon >0$ and cover $D$ by a regular $m \times m$ grid of square cells, where $x_{i,j}$ is the center of cell $(i,j)$. Let $A$ be a $m \times m$ zero matrix.

set $x=x_{1,1}$

set $A$ to $1$ at the index of $x$ and

if $f(x) \geq \epsilon$ then $\forall (k,l) : ||x-x_{k,l}|| < f(x)$ set $A_{k,l}=1$

else mark $x$ as a root

go serpentine-like through $A$ to find the next zero entry $x$ and go back to 2.

Example: Let $D=[0,8]\times[-4,4]$ and
$$f(x,y)=\frac{1}{2} \left|\sin(x)\sin(x-y) - \sin(y) \sin(x+y) -\frac{1}{2}\right|, \ (x,y)\in D.$$

Plot of $f$ with contour (red).

The algorithm uses 90.2% less evaluations than 'marching squares' to achieve the same accuracy.

Combining this algorithm and contour tracing should reduce the evaluations even further.


问:
Polynomial in $\mathbb{Z}[x]$ which is irreducible but not prime.
I know that in an integral domain, prime implies irreducible. Moreover, in a principal integral domain, these notions are equivalent. The ring of polynomials $\mathbb{Z}[x]$ is not a principal integral domain (if it were, an irreducible ideal $I$ would be maximal, and the quotient $\mathbb{Z}[x]/I$ would be a field, but the quotient by the irreducible ideal $(x)$ is $\mathbb{Z}$ which is not a field). Is there a polynomial in $\mathbb{Z}[x]$ which is irreducible but not prime?

答: As said in comment, you cannot find such an example in $\mathbb{Z}[X]$. In other rings, this could be possible : for example, $2 \in \mathbb{Z}[2i]$ is irreducible but not prime.


问:
Zero derivative for a map between connected manifolds (Unnecessary hypothesis?)
I have a quick question. I am asked to prove that if $F : X \to Y$  is a smooth map between connected manifolds with zero jacobian, then it is constant.

I found a proof which I am quite satisfied with. Basically, we restrict our study to charts around $x$ and $F(x)$, and define the map induced by $F$ between these two charts (so from $\mathbb{R}^n$ to $\mathbb{R}^m$ for some $n, m$). It is straightforward to check that this induced map has zero derivative, because it takes every vector in $T_xX$ to the zero vector in $T_{F(x)}Y$. So then, we see that $F$ is constant on this given chart for $X$, and from the connected hypothesis of $X$, $F$ is always constant.

My question is... why do we need connectedness of $Y$? If, in any case, $F$ is going to be constant on any chart of $X$, I don't understand why it would matter that $Y$ is connected.

If any one has any insight on something I may have missed, I'd be grateful!

Thanks :D

答: It is irrelevant whether $Y$ is connected. Since $F$ is locally constant on $X$, it is constant on every connected component of $X$. Thus we only need $X$ connected. Another way to see it is to observe that $F(X)$ is connected, thus contained in a component $Y_0$ of $Y$. But $Y_0$ is a connected manifold.


问:
$\frac{z_1+z_2}{1+z_1z_2} \in \mathbb{R}$. Is my solution correct? Why?
I have to show that if numbers:

$z_1$ and $z_2 \in \mathbb{C} $

$ |z_1| = |z_2| = 1 $

$z_1*z_2 \neq -1$

then:
$$\frac{z_1+z_2}{1+z_1z_2} \in \mathbb{R}$$

So I thought I can transform that expression like that:
\begin{align*}
\frac{z_1+z_2}{1+z_1z_2}
&= \frac{(z_1+z_2)(1-z_1z_2)}{(1+z_1z_2)(1-z_1z_2)}
= \frac{z_1-z_1*z_1z_2+z_2-z_2*z_1z_2}{|1+z_1z_2|^2}
\\
&= \frac{z_1-|z_1|^2*z_2+z_2-|z_2|^2*z_1}{|1+z_1z_2|^2}
\\
&= \frac{z_1-z_2+z_2-z_1}{|1+z_1z_2|^2} = \frac{0}{|1+z_1z_2|^2}
\\
&= 0
\end{align*}

Is that correct? Could you tell me exactly why $(1+z_1z_2)(1-z_1z_2) = |1+z_1z_2|^2$ ?

I know that $z * \bar z = |z|^2$. But that equality gets me confused (if it is even right).

答:
I would like to add a geometric interpretation explaining this property:

if $z_1 \ne z_2$, $z=z_3=\frac{z_1+z_2}{1+z_1z_2}$ is the intersection of line $[z_1,z_2]$ with the real axis.

Indeed, line $[z_1,z_2]$ has equation:

$$\begin{vmatrix}z_1&\overline{z_1}&1\\
z_2&\overline{z_2}&1\\
z&\overline{z}&1
\end{vmatrix}=0 \tag{1}$$

(see here).

As we are looking for a real $z$, we have $\overline{z}=z$.

Taking into account the fact that $\overline{z_1}=\tfrac{1}{z_1}$ and $\overline{z_2}=\tfrac{1}{z_2}$, it is not difficult, by multiplying by $z_1z_2$  to obtain by Laplace expansion of (1) followed by a factorization by $(z_2-z_1$):

$$(z_2-z_1)(z(1+z_1z_2)-(z_1+z_2))=0$$

$$\iff z_1=z_2 \ \ \text{or} \ \ z=\frac{z_1+z_2}{1+z_1z_2}$$

as desired.

Remark: if $z_1=z_2$, by continuity, point $z$ is the intersection point of the tangent to the unit circle with the real axis.


问:
Calculating expected value of 3 dice
I'm having a little trouble with calculating the expected value. So in my question, I can roll $3$ dice and I make $\$100$ for every value.
E.g. If I role $(1,2,3) = 1 + 2 + 3= 6$, so I'd make $\$600$.
Not sure how to calculate the expected value?

答:
Since $E(D_6) = 3.5$ for a single dice roll, by linearity,

$$E(D_6 + D_6 + D_6) = 3 \cdot E(D_6) = 10.5 \implies \boxed{\$1050} $$

Title: Calculating expected value of 3 dice


问:
If $|f(x) − f(y)| ≤ |\sin x − \sin y|,\text{for all} ~~x, y \in\mathbb{R}$ then $\text{Show that} ~f ~~\text{is}~~ 2\pi−\text{periodic}~?$
Problem :

Let : $f : \mathbb{R} ~
\to ~\mathbb{R}~~\text{be such that}~: $

$$\color{orange}{|f(x) − f(y)| ≤ |\sin x − \sin y|,\text{for all} ~~x, y \in\mathbb{R}}$$

$$\color{red}{•~\text{Show that} ~~f ~~\text{is}~~ 2\pi−\text{periodic}~?}$$

$$\color{red}{•\text{Show that} ~f~ \text{is differentiable at }~\frac{\pi}{2}~\text{then compute}~ f'\left(\frac{\pi}{2}\right)~?}$$

I know that :

function $f$ is $2\pi−$periodic if $∀x\in\mathbb{R}~, f(x + 2\pi) = f(x)$ but how I applied here? Also second question how ? Can some one give me a hints or ideas?

Edit

Thanks to Fred's for given a good solution but not complete now we need prove that there no $k≤2π$ such satisfy the equation $f(x+k)=f(x)$.

Body:
$|f(x+ 2 \pi)-f(x)|  \le | \sin(x+2 \pi)- \sin x|=0.$  Conclusion ?

$|\frac{f(x)-f( \pi/2)}{x- \pi/2}| \le |\frac{\sin(x)-\sin ( \pi/2)}{x- \pi/2}| $

What is $ \lim_{x \to \pi/2}\frac{\sin(x)-\sin ( \pi/2)}{x- \pi/2}$ ?


问:
$\int_{-1}^1{x^4}dx$ by substitution
My student asked me: While solving $\int_{-1}^1{x^4}dx$, why can't we substitute as $t=x^2$
$$x^2 =t, 2xdx = dt, x=\pm \sqrt t, \int_{-1}^{1}(x^2 \cdot x \cdot x) dx = \int_1^1{\pm t
\sqrt{t}dt} =0 \ne \frac{1}{5}$$
I think the problem comes from $x = \pm \sqrt t$, but I can't properly explain why this happens. How can I prove that this substitution is wrong?

答:
$$I=\int_{-1}^{1} x^4 dx=2\int_{0}^{1} x^4=2\int_{0}^{1} u^2 \frac{du}{2\sqrt{u}}=\int_{0}^{1} u^{3/2} du=\frac{2}{5}$$

Other method: $$I=\int_{-1}^{0} x^4 dx+ \int_{0}^{1} x^4 dx$$
$x^2=u \implies x=\mp \sqrt{u}$, for the first part $x=-\sqrt{u}$ and for the second one let $x=+\sqrt{u}$, then
$$I=\int_{1}^{0} u^2 \left(-\frac{du}{2\sqrt{u}}\right)+\int_{0}^{1} u^2 \left(\frac{du}{2\sqrt{u}}\right)=\frac{2}{5}.$$

Note that the substitution needs to be increasing or decreasing in the domain of integration. Here $=x^2$ has a minimum so break the integral in two domains $(-1,0)$ and $(0,1)$, because there is a minimum at $x=0$. For the first domain use $x=-\sqrt{u}$ and for the other use $x=\sqrt{u}$ as done in above.

Title: $\int_{-1}^1{x^4}dx$ by substitution


问:
Why over $\mathbb{Z}/n\mathbb{Z}$ projectivity, injectivity and flatness coincide for cyclic modules?
Assume $R=\mathbb{Z}/n\mathbb{Z}$ ($n\neq0$) and let $M$ be a cyclic $R$-module. Could you tell me how to prove that $M$ is projective if and only if it is injective if and only if it is flat? And also why $M$ is free if and only if it is cofree?

By cardinality reasons I think I proved that if $M$ is free then it is equal to $R$, but I don't get the rest.

答:
Let me expand on the comment I made (the other answer is great, by the way. But, I did have fun working it all out by hand).

First, since $M$ is cyclic it is of the form $\mathbb{Z}/m\mathbb{Z}$ for some $m \mid n$ and $m > 1$. I claim that $M$ is projective (resp. flat, injective) if and only if $(m,n/m) = 1$.

Suppose that $m\mid n$ and $m'\mid n$. The key is the knowledge of $(*)$:
\begin{equation*}
\operatorname{Hom}_{\mathbb{Z}/n\mathbb{Z}}(\mathbb{Z}/m'\mathbb{Z}, \mathbb{Z}/m\mathbb{Z}) \simeq \mathbb{Z}/\operatorname{gcd}(m,m')\mathbb{Z} \simeq \mathbb{Z}/m\mathbb{Z} \otimes_{\mathbb{Z}/n\mathbb{Z}} \mathbb{Z}/m'\mathbb{Z}.
\end{equation*}
Neither of these are so hard, so I'll leave it to you.

Now, consider the short exact sequence $(**)$:
\begin{equation}
0 \rightarrow \mathbb{Z}/(n/m)\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z} \rightarrow 0.
\end{equation}
Let $d = \gcd(n/m,m)$. Here are some of the proofs.

Projective, resp. injective, resp. flat implies $d=1$: Take $\operatorname{Hom}_{\mathbb{Z}/n\mathbb{Z}}(\mathbb{Z}/m\mathbb{Z},-)$, resp. $\operatorname{Hom}_{\mathbb{Z}/n\mathbb{Z}}(-,\mathbb{Z}/m\mathbb{Z})$, resp. $-\otimes_{\mathbb{Z}/n\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}$ of the sequence (**) and using the "key fact'' $(*)$ I gave.

$d = 1$ implies projective: $\mathbb{Z}/m\mathbb{Z}$ is a direct summand of $\mathbb{Z}/n\mathbb{Z}$ by the Chinese remainder theorem: $$\mathbb{Z}/n\mathbb{Z} \simeq \mathbb{Z}/m\mathbb{Z} \oplus \mathbb{Z}/(n/m)\mathbb{Z}.$$

$d=1$ implies flat: projective things are flat and $d = 1$ implies projective.

$d = 1$ implies injective: This one was the trickiest one (for me). By Baer's criterion it is enough to show that the natural map
\begin{equation*}
\mathbb{Z}/m\mathbb{Z} \simeq \operatorname{Hom}_{\mathbb{Z}/n\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) \rightarrow \operatorname{Hom}_{\mathbb{Z}/n\mathbb{Z}}(\mathbb{Z}/k\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) \simeq \mathbb{Z}/\gcd(k,m)\mathbb{Z}
\end{equation*}
is surjective for every $k\mid n$ (the groups $\mathbb{Z}/k\mathbb{Z}$ are all the possible ideals inside $\mathbb{Z}/n\mathbb{Z}$). Note that the inclusion $\mathbb{Z}/k\mathbb{Z} \hookrightarrow \mathbb{Z}/n\mathbb{Z}$ is given by $x \mapsto (n/k)x$. Thus, if you stare at this long enough you will see that the composition $(***)$
\begin{equation*}
\mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/\gcd(k,m)\mathbb{Z}
\end{equation*}
is given by $x \mapsto {n \over mk'}x$ where $k = \gcd(k,m)k'$. Since $k'$ is invertible modulo $m$ (by definition), the map $(***)$ is surjective if and only if $n/m$ is invertible module $\gcd(k,m)$. Since $\gcd(n/m,m) = 1$, this is clearly true. Indeed, $\gcd(k,m)\mid m$ and so $\gcd(n/m,\gcd(k,m)) = 1$.


问:
Optimizing a function with no local maximum?
suppose we have a function f(x)=$\sqrt{\left(\ x\ \right)}\ +\ \left(\frac{C-x}{4}\right)$ and we want to maximize the function when C=<4 and x<C. How do we procede with the problem?

Note that I've never studied optimization, I'm just trying to solve a budget constraint problem in microeconomics.

答:
Solving $f'(x)=0$ gives
$$
\frac1{2\sqrt x} -\frac14=0,
$$
where the solution is $x=4$. Since $f'(x)>0$ for $0<x<4$, the function is strictly monotonically increasing on $[0,C]$, hence the maximum is at the right boundary of the intervall, hence at $x=C$.


答:
Calling

$$
f(x,C) = \sqrt{x}+\frac{C-x}{4}
$$

In the following plot in blue we have the feasible region $(0 < x < C, C\le 4)$ to the maximization problem, and in black the level curves to the surface $f(x,C)$

As we can easily verify, the maximum is located at $x = C = 4$

Title: Optimizing a function with no local maximum?

Body:
suppose we have a function f(x)=$\sqrt{\left(\ x\ \right)}\ +\ \left(\frac{C-x}{4}\right)$ and we want to maximize the function when C=<4 and x<C. How do we procede with the problem?

Note that I've never studied optimization, I'm just trying to solve a budget constraint problem in microeconomics.


问:
Is the following set relatively compact in $C[0,1]$
Let $C_1 > 0$ and $C_2 > 0$.
We define the following set
$$
M = \{f\in C^1[0,1]: |f(0)| \leq C_1, \ \int_{0}^1|f'(t)|^2dt \leq C_2 \} \subset C[0,1]
$$
I want to know if $M$ a relatively compact set in $C[0,1]$.

My attempt. If we define replace the condition $\int_{0}^1|f'(t)|^2dt \leq C_2$ to condition
$\int_{0}^1|f'(t)|dt \leq C_2$, then we can show that this set is not relatively compact by Arzelà–Ascoli theorem. Since the subset $\{C_2x^n, \ n\in \mathbb{N}\}$ is not uniformly equicontinuous.  But I can not to give a similar example in case $\int_{0}^1|f'(t)|^2dt \leq C_2$. Also I can not show
uniformly equicontinuous for $M$.

答: For $x <y$ we have $|f(x)-f(y)| =|\int_x^{y} f'(t)dt| \leq  \sqrt {\int_x^{y}|f'(t)|^{2}dt} \sqrt {|x-y|} \leq \sqrt {C_2} \sqrt {|x-y|}$. Hence the family is equi-continuous. Together with the first condition this makes the family also uniformly bounded. Hence it is relatively compact in $C[0,1]$.


问:
Proving $x|x|$ is differentiable for all $x$
$f:\mathbb{R} \to \mathbb{R}$ be defined by $ f(x)= x|x| $ for all $x\in \mathbb{R}$.

Prove that the function $f$ is differentiable, and find a formula for its derivative  $f':\mathbb{R} \to \mathbb{R}$.

solution:

I have attempted using  $\lim_{x\to a} = \frac{f(x)-f(a)}{x-a}$ = $\frac{x|x|-a|a|}{x-a}$

I don't know how to simplify this or go any further?

答:
This is mostly to show that you can do things with only two cases ($a=0$ and $a\not=0$) instead of three ($a=0$, $a\gt0$, and $a\lt0$), using the fact that $|u|^2=u^2$.

If $a=0$ we have

$${x|x|-a|a|\over x-a}=|x|\to0$$

For $a\not=0$, we have

$$\begin{align}
{x|x|-a|a|\over x-a}
&={x^2|x|^2-a^2|a|^2\over(x-a)(x|x|+a|a|)}\\
&={x^4-a^4\over(x-a)(x|x|+a|a|)}\\
&={x^3+ax^2+a^2x+a^3\over x|x|+a|a|}\\
&\to{4a^3\over2a|a|}=2a|a|
\end{align}$$


问:
Is the function injective if the Jacobian has full column rank?
Let $f:\mathbb{R}^m \to \mathbb{R}^n: x \to f(x)$ be a continuous and differentiable function with $m < n$. If the Jacobian $J_f$ has full column rank (i.e., rank=$m$) $\forall x \in \mathbb{R}^m$, does this imply that $f$ is an injective function? If yes, can I get a reference for this result?

答: No, take $f(t) =\pmatrix{ \sin t\\ \cos t}$.


问:
Questions related to group theory
If $H$ and $K$ are two normal subgroups of a group $G$, such that $H \subseteq K$, then show that $K/H$ is a normal subgroup of $G/H$.

Here how do I prove that $hk=kh$ because in this question as we know $H$ is normal to $G$ so $hg=gh$ and $K$ is also normal to $G$ so $kg=gk$. And also $H$ is contained in $K$. Which means all $h$ belong to $K$.

答:
By definition:

\begin{alignat}{2}
K/H \unlhd G/H &\iff &&\forall g\in G,\forall k\in K, \space \space gH(kH)(gH)^{-1}\in K/H \\
&\stackrel{H\unlhd G}{\iff} &&\forall g\in G,\forall k\in K, \space \space gH(kH)(g^{-1}H)\in K/H \\
&\stackrel{H\unlhd K}{\iff} &&\forall g\in G,\forall k\in K, \space \space gkH(g^{-1}H)\in K/H \\
&\stackrel{H\unlhd G}{\iff} &&\forall g\in G,\forall k\in K, \space \space gkg^{-1}H\in K/H \\
\end{alignat}

and the latter holds because $K\unlhd G$.

Title: Questions related to group theory


问:
quotient of consecutive primes, an inequality
A friend of mine asked me, if the following statement is true or not:

$\frac{p_{n+1}}{p_n}>1+\frac{1}{n}$ for sufficiently large $n$.

Here is my solution:

Assume that the statement is true. It is equivalent to $p_{n+1}-p_n>\frac{p_n}{n}$. Using this and Zhang-polymath8 result (https://en.wikipedia.org/wiki/Yitang_Zhang), we would have $246>\frac{p_n}{n}$ for infinitely many $n$. Contradiction, because $p_n>n\ln n$ for all $n$.

Is there a simpler (with the use of less powerful theorem) solution?

答:
It seems that the question relates to the prime gap

Results of Goldston, Pintz and Yldirim (2005) :

$$\lim \inf_{n\to \infty} \frac{p_{n+1} - p_n}{\log n} = 0$$

is enough to show the result. In fact, even the weaker inequality $< 1$ would do.

However, note other older results that
$$\lim \sup _{n\to \infty}\frac{p_{n+1} - p_n}{\log n} = \infty$$


问:
Given $\angle A = 60°, \angle B = 80°, \angle C = 40°$, find $\alpha$

I tried making an equilateral triangle with side $b$ but it didn't seem to be of any use. Please help.

答:
Construct $BC=AC=a+b+c$

Mark $EF=ED=c$. So $FB=b$

$\triangle DEF$ is $100^{\circ}-40^{\circ}-40^{\circ}$ triangle, so is $\triangle CDF$

$CD=DF=b=FB$.

$\angle BFD=140^{\circ} \Rightarrow \angle FBD = \angle FDB = 20^{\circ}$

In $\triangle ACD$, $$ \dfrac{b}{\sin \alpha} = \dfrac{a+b+c}{\sin 60^{\circ}}$$
In $\triangle BCD$, $$ \dfrac{b}{\sin 20^{\circ}} = \dfrac{a+b+c}{\sin 120^{\circ}}$$
We conclude $$ \boxed{\alpha = 20^{\circ}}$$

Thanks to @Nanayajitzuki, we can also finish without resorting to sine law.

Construct equilateral $\triangle GCD$ with $G$ on $AD$.

In $\triangle CGA$ and $\triangle CDB$, $\angle CGA = 120^{\circ} = \angle CDB$ . $CD=b=CG$. $CA=CB$.

So $\triangle CGA \cong \triangle CDB$ by SSA congruence condition** $$\boxed{\therefore \alpha = 20^{\circ}}$$

Note**

The ambiguity in SSA congruence condition arises only when involved angle is acute (it leads to precisely two possible triangles as shown below $\triangle ABD$ and  $\triangle ABE$). If obtuse angle is involved, there is no ambiguity. So its use here is justified.


问:
Applications of design theory
I have recently started reading up on design theory, with the ultimate purpose of doing some original research in that area. I understand the mathematics fairly well, but am not understanding the application. Can someone here explain that? Alternatively are there some articles, web-sites, papers explaining the application of design theory in agriculture, biological experiments etc.

For example after establishing the existence of/constructing a certain type of design mathematically, how do we use that knowledge in practice?

答:
I'll try to give a basic idea of how these are used in exploratory medical research.

Traditional medical experiments:  Patients are given a single candidate treatment and are observed for a given response (i.e. dose-response).  If the response is statistically significant, then it is considered for further studies, and possibly may end up on the market.  Observations may necessarily take years to conduct, and suitable patients may be hard to find.

This slow process is undesirable if there is only a small chance of the candidate treatment working.  Instead we can use a scheme along the lines:

We have a number of patients (or test animals) with a disease (or diseases).

We have a large number of candidate treatments.

We treat the patients with some subset of the possible treatments.

We observe for a large number of responses.

Combinatorial designs are used to determine which patients receive which treatments in such a way that if a given response is observed, then the structure of the design would indicate the treatment that caused it.  We also want to minimise the cost of the experiment (so having fewer patients is better).  We don't need to deduce for sure that treatment X causes response Y, we just need to reduce the possibilities significantly; more in-depth and specialised studies can be conducted for plausible candidates.

Essentially, it's faster and cheaper to conduct exploratory research using a scheme based on a combinatorial design.  Moreover, inadvertent discoveries, such as the highly profitable Viagra, may also be made this way.

This is an example in a medical setting, but hopefully it's clear how the general idea extends to experiments throughout science.  There are other ways, such as factorial experiments, in which combinatorial designs are used.

For example after establishing the existence of/constructing a certain type of design mathematically, how do we use that knowledge in practice?

It usually doesn't work that way.  Most often, we start off with a hypothesis (or a family of hypotheses) and design an experiment around it/them.  Real-world constraints often result in mathematically inelegant questions.  Futher, understanding why a certain design is or isn't suitable may require domain-specific knowledge.

Charlie Colbourn once told an interesting anecdote at a summer school I was attending; some chemists asked him to design an experiment according to some criteria.  He gave them a design that met their criteria, but for some reason the chemists ignored his reply.  He asked some other chemists about what happened, who explained that if they mixed the chemicals according to his design, they would blow up the lab.  (Of course, this is second-hand information, but hopefully we can appreciate how domain-specific knowledge is necessary.)

Title: Applications of design theory


问:
Find general solution of linear congruence equation
Congruences are beyond my understanding, I do not understand at all, if you could explain it to me as simply as possible on this example, I would be very grateful:

Find a general solution of linear congurence $2x\equiv 5 \pmod{13}$

答:
The important properties of congruences are

if $a\equiv b\pmod{n}$ and $c\equiv d\pmod{n}$ then $a+c\equiv b+d\pmod{n}$;

if $a\equiv b\pmod{n}$ and $c\equiv d\pmod{n}$ then $ac\equiv bd\pmod{n}$.

If $x$ is a solution to $2x\equiv5\pmod{13}$, then also

$$
2kx\equiv 5k\pmod{13}
$$
for every integer $k$, because $k\equiv k\pmod{13}$ and you can apply property 2.

How does this simplify the situation? Well, if you choose $k=7$, you get
$$
14x\equiv 35\pmod{13}
$$
and therefore
$$
x\equiv9\pmod{13}
$$
So if $x$ is a solution, then $x\equiv 9\pmod{13}$. But also the converse is true, because from $x\equiv 9\pmod{13}$ we obtain $2x\equiv18\equiv5\pmod{13}$.

In this case it is quite the same as solving a degree one equation: $2x=5$ becomes $x=5/2$ after multiplying both sides by $1/2$.

In the case of congruences we cannot “multiply by $1/2$”; but we can see that $\gcd(2,13)=1$, so by the general theory, we know there exists $k$ such that $2k\equiv1\pmod{13}$. It's just a matter of finding it.

Trial will work for small numbers, the extended Euclidean algorithm will do for bigger numbers.

When we have this $k$, then the congruence becomes
$$
2kx\equiv 5k\pmod{13}
$$
and so $x\equiv5k$. The steps can be done backwards, because upon multiplying this by $2$, the right-hand side becomes $5\cdot2k\equiv5\cdot1\equiv5\pmod{13}$.


问:
Index of regularity equals regularity of homogeneous ideal?
Notation:

Let $HF_{M}(d)$ and $HP_M(d)$ denote the Hilbert-function resp. the
Hilbert-polynomial.
$\newcommand{\reg}{\operatorname{reg}}\newcommand{\depth}{\operatorname{depth}}$

Let $i_{\reg}(I)$ denote the index of
regularity of $I$, i.e. the smallest integer s.t. $HF_{R/I}(d) =
   HP_{R/I}(d)$ for all $d \geq i_{\reg}(I)$.

Let $\reg(I)$ denote the Castelnuovo-Mumford regularity of $I$.

I'm trying to prove the following statement:

If $I\subset R = k[x_1,\ldots,x_n]$ is a homogeneous ideal and $I_d = R_d$ for $d \gg 0$, then $i_{\reg}(I) = \reg(I)$.

Apparently this should follow directly from the following Corollary from Eisenbud's  Book:

Corollary 4.15: Let $M$ be a finitely generated graded Cohen-Macaulay $R$-module. If $s$ is the smallest number such that $HF_M(d) = HP_{M}(d)$ for all $d \geq s$,
then $s = 1 − \depth(M) + \reg(M)$.

What I did was this:

From $I_d = R_d$ follows that $HF_{R/I}(d) = 0$ for $d \gg 0$, and therefore the Hilbert-polynomial is zero. As far as I've understood, this implies that $R/I$ has krull-dimension zero.

This then implies that $\depth(R/I) = {\rm kdim}(R/I) = 0$, i.e $R/I$ is Cohen-Macaulay.
Now Corollary 4.15 implies that $i_{\reg}(I) = 1-0+\reg(R/I) \neq \reg(I)$.

What have I missed?

I think $\reg(R/I) = \reg(I) + 1$, so the result would be $2$ apart from what I'm trying to achieve.

答:
Let
$$0\to F_k\to F_{k-1}\to\cdots\to F_0\to I\to 0$$
be the minimal free resolution of $I$, with $F_i=\oplus_jR(-a_{i,j})$. Then
$$0\to F_k\to F_{k-1}\to\cdots\to F_0\to S\to S/I\to 0$$
is the minimal free resolution of $S/I$.

By definition, ${\rm reg}(I)=\max_{i,j}a_{i,j}-i$. In the minimal free resolution of $S/I$, the free module $F_i$ is in the $(i+1)$-th position and the $0$-th term $S$ doesn't affect the regularity. Thus
$${\rm reg}(R/I)=\max_{i,j}a_{i,j}-(i+1)=\max_{i,j}a_{i,j}-i-1={\rm reg}(I)-1.$$


问:
30 coin flips, what is $\mathbb{P}(\text{at least } k \text{ are }H \text{, and at least } k \text{ are T})$?
This is my attempt at starting, am I going in the right direction? -

We know at least $k$ are H, so the probability of $k$ being H is

$\begin{pmatrix} 30\\k \end{pmatrix}\frac{1}{2^k} \frac{1}{2^{30-k}}= \begin{pmatrix} 30\\k \end{pmatrix}\frac{1}{2^{30}}$

Now $k$ of the 30 flips have been taken up, so out of the remaining $30-k$ flips, we need $k$ to be T, and the probability of this is:

$\begin{pmatrix} 30-k\\k \end{pmatrix}\frac{1}{2^{30}}$

So multiplying should give the result?

Have I thought about this correctly?

答:
The probability of exactly $j$ heads in $30$ flips is
$$
\binom{30}{j}\Bigl(\frac{1}{2}\Bigr)^{30}
$$
hence for any fixed $k\in\{0,...,15\}$, the probability that in $30$ flips you get at least $k$ heads and at least $k$ tails is
$$
\sum_{j=k}^{30-k} \binom{30}{j}\Bigl(\frac{1}{2}\Bigr)^{30}
=
\Bigl(\frac{1}{2}\Bigr)^{30}\left(\sum_{j=k}^{30-k}\binom{30}{j}\right)
$$
where the lower bound $j=k\;$is to guarantee at least $k$ heads, and the upper bound $j=30-k\;$is to guarantee at least $k$ tails.

By symmetry, the result can be simplified to
$$
\Bigl(\frac{1}{2}\Bigr)^{30}
\left(\left(2\sum_{j=k}^{14} \binom{30}{j}\right)
+
\binom{30}{15}\right)
$$
Equivalently, by complementary counting, the result can be rewritten as
$$
1-\Bigl(\frac{1}{2}\Bigr)^{29}\left(\sum_{j=0}^{k-1} \binom{30}{j}\right)
$$
which requires less work if $k$ is small.


答:
Since for $r>0$, $\dfrac{\partial x^r}{\partial x}=r\times x^{r-1}$ and if $r-1+r=4$ then $r=\dfrac{5}{2}$.

Thus,

\begin{align}\int_{-1}^1 x^4\,dx&\overset{x\rightarrow x^4 \text{is even function}}=2\int_0^1 x^4 dx\\
&\overset{y=x^{\frac{5}{2}}}=2\times \frac{2}{5}\int_0^1 ydy\\
&=\frac{4}{5}\int_0^1 y dy\\
&=\frac{4}{5}\left[\frac{y^2}{2}\right]_0^1\\
&=\frac{4}{5}\times \frac{1}{2}\\
&=\boxed{\frac{2}{5}}
\end{align}

Title: $\int_{-1}^1{x^4}dx$ by substitution

Body:
My student asked me: While solving $\int_{-1}^1{x^4}dx$, why can't we substitute as $t=x^2$
$$x^2 =t, 2xdx = dt, x=\pm \sqrt t, \int_{-1}^{1}(x^2 \cdot x \cdot x) dx = \int_1^1{\pm t
\sqrt{t}dt} =0 \ne \frac{1}{5}$$
I think the problem comes from $x = \pm \sqrt t$, but I can't properly explain why this happens. How can I prove that this substitution is wrong?


问:
if $\sin24^\circ = p$ what is $\cos24^\circ$?
Let $p=\sin 24^\circ$

Then what would $\cos (24^\circ)$ be in terms of $p$?

What would $\sin (168^\circ) \cdot  \sin(-78^\circ)$ be in terms of $p$?

I'm not sure how to approach these as we have only started this section.

Thanks.

答:
Part 1

Use the identity
\begin{equation*}
\sin^2A+\cos^2A=1\\
\cos^2 A={1-\sin^2 A}
\end{equation*}
so
\begin{align*}
\cos^2 24^\circ&={1-\sin^2( 24^\circ) }\\
\\
\\cos^2 24^\circ&={1-p^2 }
\\
\\
\cos24^\circ&=\sqrt{1-p^2 }
\end{align*}
Here we only take positive value of square root because $24^\circ$ is in I quaderent and cos ratio in quaderent I is positive.

Part 2

\begin{align*}
\sin 168^\circ \cdot \sin(-78^\circ)
&=-\frac{2\sin 168^\circ\cdot\sin78^\circ}{2}
\end{align*}
Using the formula ${2\sin A\cdot\sin B}=\cos (A-B)-\cos(A+B)$
\begin{align*}
\sin(168^\circ)\cdot \sin(-78^\circ)&=-\frac{\cos 90^\circ-\cos 246^\circ}{2} \\
&=\frac{\cos 246^\circ}{2}\\
&=\frac{\cos(270^\circ-24^\circ)}{2}\\
&=-\frac{\sin 24^\circ}{2}\\
&=-\frac{p}{2}
\end{align*}


问:
In $\Delta ABC, AC > AB.$ The internal angle bisector of $\angle A$ meets $BC$ at $D,$ and $E$ is the foot of the perpendicular from $B$ onto $AD$.

In $\Delta ABC, AC > AB.$ The internal angle bisector of $\angle A$ meets $BC$ at $D,$ and $E$ is the foot of the perpendicular from $B$ onto $AD$. Suppose $AB=5,BE=4$ and $AE=3$ . Find $\Big(\frac{AC + AB}{AC - AB}\Big)ED$ .

What I Tried: Here is a picture :-

The picture only shows it. I have no other idea to try it, first that angle-chasing has no use here. Second is that I only have the required information of one right triangle, so no Pythagoras Theorem here.
I got no similar triangles, or even if I got one pair, I cannot show they are similar and so on, and in the end I am weak at Trigonometry.

Is there any way to solve this using basic techniques in Geometry? (Like angle-chasing , similarity , area , pythagorean theorem and so on) . Thank You.

Edit:- After @Michal Adamaszek's hint, I was able to do a bit of progress. (everything is there in the picture). Some congruent and similar triangles came up, but now I don't know what to do next. I need to find the value of $AC$ and $DE$ , is there any way to do it?

答:
I use a bit of trigonometry.(I would love to see something without this)

let $BC=a,CA=b,AB=c$.

By Napiers Analogy:$$\tan {\frac{B-C}{2}}=\frac{b-c}{b+c}\cot (A/2) \tag 1$$
Notice that $$\sin (A/2)=\frac{4}{5}$$
$$\cot (A/2)=? \tag 2$$
Also $\angle EDB=90+\frac{C-B}{2}$
$$\tan{\angle EDB}=\tan (90+\frac{C-B}{2})=\cot {\frac{B-C}{2}}=\frac{4}{ED} \tag 3$$
By $(1),(2),(3)$ you should be able to finish.


答:
You write in a comment that the formula
$$\int_{\varphi (a)}^{\varphi (b)}{g(x)}dx = \int_{a}^{b} {g(\varphi (x))\varphi '(x)dx}$$
doesn't require $\varphi$ to be injective. This is true. But if you try and use this formula to evaluate your integral with $\varphi(x)=x^2$, you need to choose $g$ so that
$$g(\varphi(x))=\frac12 x^3$$
on $[a,b]=[-1,1]$. But this would require that $g(\varphi(-1))=-\frac12$ and $g(\varphi(1))=\frac12$, which is impossible, because $\varphi(-1)=\varphi(1)$.

So $\varphi$ has to be invertible on $[a,b]$. Otherwise no suitable $g$ can exist.

Taken together with the requirement that $\varphi$ be continuously differentiable on $[a,b]$, this implies that $\varphi$ is $monotonic$ on $[a,b]$. And this is where your substitution breaks down. As others have explained, you can get round this by integrating separately over $[-1,0]$ and $[0,1]$, because $\varphi$ is monotonic on each of these intervals.

This monotonicity requirement seems to be ignored by the various web pages which state the appropriate theorem. True, it is a logical consequence of the theorem, but not, I would say, an obvious consequence.

Title: $\int_{-1}^1{x^4}dx$ by substitution


问:
cardinality of the union of 2 sets with at least two elements and cardinality of cartesian product
Suppose X,Y are sets with at least 2 elements. Show that
$X\cup Y\le X\times Y$

So my first thought was that cardinality $|X|\ge 2$ and the same for $|Y|\ge 2$ but by the inclusion-exclusion principle we have $|X\cup Y|=|X|+|Y|-|X\cap Y|$ but the problem does not say if they are disjoint or not.
If we assume that they are disjoint then $|X\cup Y|\ge 4$ right?
But from the definition of the cardinality for the cartesian product we have similar
$|X\times Y|\ge 4$. But I have to find an injective function (including the number of elements) s.t $f:X\cup Y\to X\times Y$ and thus conclude the result.
How should I do it?

答: We can do far better: suppose $(\kappa_{i})_{i \in I},(\lambda_{i})_{i \in I}$ are two families of cardinals with $ \kappa_{i} <\lambda_{i} \forall i \in I$, then $ \coprod_{i\in I} k_i < \prod_{i\in I}\lambda_i$ (the equality case is easily treated separatedly).</p>


问:
Condition for a submanifold to have a trivial normal bundle
Is the following assertion true?

Suppose $M$ is a smooth manifold of dimension $n$, and $S$ an embedded submanifold of dimension $k$. If there is an embedding $S\times \Bbb R^{n-k}\to M$ (with $S\times 0$ corresponding to $S$ in the obvious way), then the normal bundle of $S$ in $M$ is trivial.

I think this can be proved as follows, but I'm not sure about my argument: If $f$ is such an embedding, then the image of $f$ should be an open subset of $M$, so the normal bundle of $S$ in $M$ and the normal bundle of $S$ in $f(S\times \Bbb R^{n-k})$ are the same, but the latter is a trivial bundle.

答:
The sub-assertion

... the normal bundle of $S$ in $M$ and the normal bundle of $S$ in $f(S \times \mathbb R^{n-k})$ are the same...

is false as stated literally. They aren't the "same". But it contains a grain of truth which you can then use to turn into a proof.

The key thing to keep in mind is that whenever you are tempted to use the "s" word, you should instead ask yourself: What kind of isomorphism are we talking about here?

The true statement which you should have in place of the one above is

... the derivative $Df : T(S \times \mathbb R^{n-k}) \to TM$ restricts to an isomorphism between the normal bundle of $S \times 0$ in $S \times \mathbb R^{n-k}$ and the normal bundle of $f(S \times 0)$ in $M$...

And now you can go on to prove it.


问:
Is there a “Cheat Sheet” for the different types of convergence?
When studying different topics in mathematics, such as Probability and Analysis, often new types of convergence show up. For example, Probability talks about convergence in distribution, also calling it sometimes weak convergence, which also appears in Functional Analysis, but there is also weak$*$ convergence, epi-convergence ... Anyways, sometimes things get confusing.

Hence, I was wondering if anyone knows of a “Cheat Sheet” or something similar, which presents and “organizes” all this different types of convergence. When searching on Google, I mostly find things regarding convergence of series, instead of what I’m looking for.

Just to clarify. I’m looking for a reference where the many concepts of convergence (as many as possible) are defined and possibly connected, as the case with weak convergence and convergence in distribution.

答:
Generally speaking, the best cheat sheet for you is the one you write yourself. I can't imagine a better way of reaching an understanding of all the different notions of convergence than to write such a sheet.

However, I will say that the one unifying concept here is that of topology. Each different type of these convergences is really convergence of sequences (or of nets) taking place in some appropriate metric space (or topological space). In probability and analysis, the spaces in question are usually normed vector spaces (or other non-normed topological vector spaces). So if you want to look for a unifying book on the topic, I would look for a book on topological vector spaces.


问:
Cokernel of group homomorphism
We know $H=\{ (1),(12) \}$ is subgroup of $S_3$. Consider inclusion '$\varphi: H \hookrightarrow S_3$', this is clearly group homomorphism. Prove that coker$\varphi$ is trivial. I can't understand how to apply universal property of cokernel to this homomorphism $\varphi$.

答: The cokernel of $\varphi$ is, by definition, initial among all morphism $\psi: S_3\to G$ such that $\psi\circ \varphi$ is the trivial morphism. So you need to prove that if $\psi\circ \varphi$ is trivial then $\psi$ is trivial. Note that $\psi\circ \varphi$ is trivial if and only if $H\leq \ker(\psi)$, so it suffices to prove that any normal subgroup of $S_3$ containing $H$ must be equal $S_3$ itself.

Title: Cokernel of group homomorphism


答:
For any sets $A,B$ (either disjoint or not) the cardinality of the Cartesian product is the product of their cardinality, while the cardinality of their union is less or equal the the sum of their cardinality. Let $a:=|A|$ and $b:=|B|$, then
$$
|A\times B| = ab \quad\text{and}\quad |A\cup B|=|A|+|B|-|A\cup B|\leq a+b
$$
It is sufficient to show that if $a,b\geq 2$ then $a+b\leq ab$. This is trivial since if $b\geq 2$ then
$$
a=a\cdot1\leq a(b-1) \quad\Rightarrow\quad a+b\leq ab
$$
Note that it is not necessary that both sets have more than 2 elements and that the inequality trivially holds even considering infinite cardinality sets.</p>


答:
If I extend the line $BE$ to meet $AC$ at $M$, $\triangle AEB \sim \triangle AEM$ (A-A-S).

So $AM = AB = 5, EM = EB = 4$

$CM = AC - AB$

In $\triangle BMC$,

$\frac{sin \angle C}{8} = \frac{sin \angle DBE}{AC-AB}$ ...(i)

Also in $\triangle BDE$,

$\frac{sin (90^0 - \angle DBE)}{4} = \frac{\sin \angle DBE}{ED}$

$ED = 4 \tan \angle DBE$ ...(ii)

$\angle C = 90^0 - \angle CAE - \angle DBE$

$\sin C = \cos (\angle CAE + \angle DBE) = \frac{3}{5} \cos \angle DBE - \frac{4}{5} \sin \angle DBE \, \,$ ($\angle CAE = \angle BAE$) ...(iii)

$AC - AB = AC - 5 =  \frac{8 \sin \angle DBE}{\sin C}$

$AC + AB = AC - AB + 10 = \frac{8 \sin \angle DBE}{\sin C} + 10 = \frac{6 \cos \angle DBE}{\sin C}$

$\frac{AC + AB}{AC-AB}ED = 3$

EDIT:

Your modified diagram is probably the easiest solution.

$\frac{AC+AB}{AC-AB}ED = \frac{5x + 5 + 5}{5x} \times 3x \times \frac{4}{4x + 8} = 3$

Body:
In $\Delta ABC, AC > AB.$ The internal angle bisector of $\angle A$ meets $BC$ at $D,$ and $E$ is the foot of the perpendicular from $B$ onto $AD$. Suppose $AB=5,BE=4$ and $AE=3$ . Find $\Big(\frac{AC + AB}{AC - AB}\Big)ED$ .

What I Tried: Here is a picture :-

The picture only shows it. I have no other idea to try it, first that angle-chasing has no use here. Second is that I only have the required information of one right triangle, so no Pythagoras Theorem here.
I got no similar triangles, or even if I got one pair, I cannot show they are similar and so on, and in the end I am weak at Trigonometry.

Is there any way to solve this using basic techniques in Geometry? (Like angle-chasing , similarity , area , pythagorean theorem and so on) . Thank You.

Edit:- After @Michal Adamaszek's hint, I was able to do a bit of progress. (everything is there in the picture). Some congruent and similar triangles came up, but now I don't know what to do next. I need to find the value of $AC$ and $DE$ , is there any way to do it?


问:
Can every element of Hilbert space be represented by a linear combination?
Let $H$ be an infinite dimensional Hilbert space and $f$ be an element of $H$. Let $T:H \rightarrow H$ be a bounded linear operator. Then, can $Tf$ be represented by
$$ Tf = \sum_{k=1}^\infty f_k\langle f,g_k\rangle$$
where $f_k,g_k \in H?$

If the dimension of the image of $T$ is finite, it would be. But if not?

答:
If $H$ is separable you may always find a countable orthonormal basis $\{e_n\}_n$, so each $f$ in $H$ may be expressed as
$$
f = \sum_n \langle f, e_n \rangle e_n
$$
whence
$$
T(f) = \sum_n \langle f, e_n \rangle T( e_n).
$$


答:
This proof can probably be simplified or made more natural, but here is my take:

Construct $CG$ parallel to $FB$ as shown above. We have:

$$\triangle AFE \cong \triangle ABE\ (ASA), \quad \triangle AFE \sim \triangle ACG, \triangle BED \sim \triangle CGD\ (AAA)$$

So we have the following ratios:

$$AF = AB, \ \frac {AE}{EG} = \frac {AF}{FC},\ \frac {ED}{DG} = \frac {DB}{DC} = \frac {AB}{AC}$$

The last ratio is by Angle Bisector Theorem.

Hence:

$$\frac {AC+AB}{AC-AB} = \frac {AC}{CF} + \frac {AB}{CF} = \frac {AB}{CF}\left(1+\frac {AC}{AB}\right)= \frac{AF}{CF}\left(1+\frac {DG}{ED}\right)=\frac{AE}{EG}\frac {EG}{ED}=\frac{AE}{ED}$$

Giving the result:

$$\left(\frac{AC+AB}{AC-AB}\right)ED = AE$$

Title: In $\Delta ABC, AC > AB.$ The internal angle bisector of $\angle A$ meets $BC$ at $D,$ and $E$ is the foot of the perpendicular from $B$ onto $AD$.


问:
When a rock hits the ground, the velocity is zero. But, what is acceleration?
Disclaimer: I am not a student trying to get free internet homework help. I am an adult who is learning Calculus from a textbook. I am deeply grateful to the members of this community for their time.

When graphing acceleration or v'(t), should you put a "hole" in the endpoints of the graph?

Let's look at the typical graph of a rock being thrown, where s(t) = inverted parabola.

v(t) is a diagonal line going from positive to negative (eg: y=-32t+4)

a(t) = v'(t) = constant (negative)    (eg: y=-32)

Let's say the rock hits the ground at t=10.  What should acceleration be EXACTLY at t=10?   Is it still -6, like the times before it?  Or does it not exist since at t=10, the rock has stopped.  Can you have acceleration when velocity = 0?   I guess you can b/c acceleration at the peak position (vertex of parabola) is still negative -6, even though the rock is not moving in that peak's instant.  Can you take the derivative of the point at t=10 on the velocity graph?

I saw the sample problem does NOT create a hole at the very end of the A(t) graph when the rock hits the ground.  So, the graph is saying the Acc. is still the same as when it was falling a moment before.  It also fills in the A(t) at t=0, even though the rock is not yet moving and has velocity = 0 at t=0

etc etc etc....

答:
The answer given several years before this one completely answers the mathematical question. I will give some thoughts here on how this relates to the physical question.

The short answer is:

It is common in textbooks to gloss over what happens at the endpoints of actions like the thrown rock and to plot the graphs as if neither the thrower nor the ground exists.
The rock is just magically already rising at vertical velocity $v'(0)$ at time $t=0$ and we just stop looking at it when $t=10.$
We never actually model the rock hitting the ground, so we have nothing to say about what the acceleration might be when that happens.

If the displacement graph actually plots the rock as stationary before $t=0$ and after $t=10,$ that is, if it is a plot of the function

$$
y(t) =
\begin{cases}
0 & t < 0, \\
-32t^2 + 320 t & 0 \leq t \leq 10, \\
0 & t > 10, \\
\end{cases} \tag1
$$

then the mathematical answer is that the velocity and acceleration are both undefined at $t=0$ and at $t = 10.$
The velocity and acceleration in this model are the first and second derivatives of the function in Equation $(1),$

$$
y'(t) =
\begin{cases}
0 & t < 0, \\
160 - 32t & 0 < t < 10, \\
0 & t > 10, \\
\end{cases}
$$
$$
y''(t) =
\begin{cases}
0 & t < 0, \\
-32 & 0 < t < 10, \\
0 & t > 10, \\
\end{cases}
$$

Both of these functions have "holes" at $t=0$ and at $t=10.$
The holes are the result of an incomplete physical model:
the function in Equation $(1)$ ignores the thrower's motion that accelerates the rock from zero velocity to vertical velocity $160$ upward,
and it ignores the interaction of the rock with the ground that decelerates the rock (usually in a very short period of time) from vertical velocity $-160$ to zero.

It is much simpler not to think about details like this, so in simple textbook exercises we don't.


问:
Evaluate $P(x)=5+3x+5x^2+4x^4$ using Horner's method
I want to evaluate $P(x)=5+3x+5x^2+4x^4$ using Horner's method

So I rewrite $P(x)$ as :

$$P(x)=5+3x+5x^2+0x^3+4x^4\tag{1}$$

and applying Horner's method to get:
$$P(x)=5+x(3+x(5+x(0+4x)))\tag{2}$$
Is it done? or I should rewrite it as :
$$P(x)=5+x(3+x(5+x\times4x)))\tag{3}$$
What about this:
$$P(x)=5+x(3+x(5+4x^2)))\tag{4}$$Which one of these three form is ok for Horner's method?

答:
Horner's method is understood as a systematic algorithm, not as a formula.

$$p\leftarrow4,\\p\leftarrow p\cdot x+0,\\p\leftarrow p\cdot x+5,\\p\leftarrow p\cdot x+3,\\p\leftarrow p\cdot x+5.$$

Note that this is not truly equivalent to the expression

$$\left(\left(\left(\left(4\right)\cdot x+0\right)\cdot x+5\right)\cdot x+3\right)+5$$

and even less

$$5+x\cdot\left(3+x\cdot\left(5+x\cdot\left(0+x\cdot\left(4\right)\right)\right)\right)$$

as executed by a calculator or computer, which would use a stack to handle the nested parenthesis and possibly end up in a stack overflow (low-end calculator or high degree polynomial). Horner's scheme only requires constant additional space.


问:
Right adjoint to forgetful functor from $G$-Sets to $H$-Sets?
Let $H$ be a subgroup of group $G$.

Let $F_1$ be a forgetful functor from $G$-Sets (sets with action of group $G)$ to $H$-Sets.

Is it true that right-adjoint functor to $F_1$ exists if and only if $H$ is a normal subgroup of $G$ ?

答:
This is a complement to Arnaud D's excellent answer (+1). In this situation, it is actually easy to compute explicitly the right adjoint $R$.

For any $G$-set $X$ and any $H$-set $Y$, you want a natural isomorphism :
$$\operatorname{Hom}_G(X,R(Y))=\operatorname{Hom}_H(F_1(X),Y)$$
If this holds for any $X$, in particular, it should hold for $X=G$ where the $G$-action is given by left multiplication. Thus :
$$\operatorname{Hom}_G(G,R(Y))=\operatorname{Hom}_H(F_1(G),Y)$$
But $\operatorname{Hom}_G(G,R(Y))=R(Y)$ as sets by the bijection $\varphi\mapsto \varphi(e)$ whose inverse is $y\mapsto (g\mapsto gy)$. It follows that, as sets $R(Y)=\operatorname{Hom}_H(F_1(G),Y)=\operatorname{Hom}_H(G,Y)$ (where $G$ is a $H$-set via left multiplication). It remains to find the correct $G$-action tu put on $R(Y)$ for the adjunction to hold. Well, there is nothing fancy here, the only reasonable action is the following : for any $\psi\in\operatorname{Hom}_H(G,Y)$, the action of $g$ is $g\psi=(g'\mapsto \psi(g'g))$ (check that this is indeed a left action).

So you have it : $R(Y)=\operatorname{Hom}_H(G,Y)$ with the action $g\psi=(g'\mapsto\psi(g'g))$. It remains to check the adjunction property. I leave it to you.


问:
Translating plain english to math-ish using quantifiers and modulus

There is a rational number and an irrational number which are 2/3 apart from each other.
So far I have :

$$\forall \delta >0.\exists x\in\mathbb{Q}. \exists y\in\left(\mathbb{R}/\mathbb{Q}. | x-y| > \frac{2}{3}\delta\right)$$

I'm not too sure on the $2/3\ \delta$ part, any help is appreciated.

Thanks in advance!

答:
You don't need that initial universal quantifier. What you write says that for any $\delta$ there is a rational and an irrational more that $2/3\delta$ apart -- a much stronger claim!

∃x ε ℚ, ∃y ε (ℝ \ ℚ) s.t. |x - y| > 2/3

is what you want to render the claim that there is a rational and an irrational whose difference is more than 2/3. Or at least, that works in the sort of syntax you are using.


答:
$\frac{AB}{AC}=\frac{DB}{DC}$ (This is a well-known property of the angle bisector, which certainly has some non-trigonometric proof)

$\frac{ED}{CM}=\frac{BD}{BC}$ (similarity)

$CM=\frac{3}{5}CF=\frac{3}{5}(AC-AF)=\frac{3}{5}(AC-AB)$

From there you can put it together.


问:
Dual of a finitely generated projective module is again finitely generated and projective.
Let $M$ be a module over the commutative ring $k$. Put $M^* = \operatorname{Hom}_k(M,k)$.

If $M$ is finitely generated and projective, is the same true for $M^*$?

I tried using a finite dual base to construct a finite dual base for $M^*$ but it does not work out. How can I show this?

答: Suppose $M \oplus N \cong k^n$, then $M^* \oplus N^* \cong \mathrm{Hom}(M \oplus N,k) \cong \mathrm{Hom}(k^n,k)=k^n$. Thus $M^*$ is projective. As we have a surjection $k^n \to M^*$, $M^*$ is finitely generated.


答:
You know that $\sin^2+\cos^2=1$. As $\sin(24)=p$ you have $p^2 + \cos^2(24)=1$. Solve this equation for $\cos(24)$.
For the second part you can use some trigonometric theorems.
For example
$$ \sin(\alpha) \cdot \sin(\beta) = -\frac{1}{2} (\cos(\alpha+\beta) -\cos(\alpha - \beta))$$

This gives you
$$\sin(168^\circ) \cdot \sin(-78^\circ)=-\frac{1}{2} ( \cos(90^\circ)- \cos(246^\circ))$$
We know that $\cos(90^\circ)=0$ and that $\cos(x+180^\circ)=-\cos(x)$ hence we have
$$-\frac{1}{2} \cos(66^\circ)=-\frac{1}{2}\cos(90^\circ-24^\circ)=-\frac{1}{2} \sin(24^\circ)$$

Title: if $\sin24^\circ = p$ what is $\cos24^\circ$?

Body:
Let $p=\sin 24^\circ$

Then what would $\cos (24^\circ)$ be in terms of $p$?

What would $\sin (168^\circ) \cdot  \sin(-78^\circ)$ be in terms of $p$?

I'm not sure how to approach these as we have only started this section.

Thanks.


问:
An example about a non-commutative division ring with finite characteristic
After reading the proof of the theorem

“For every central division $F$-algebra $D$ with $D$ $\neq$ $F$, $D$ contains a separable extension $K \supsetneqq F$“,

I have a question: dose there exist a non-commutative division ring $D$ with finite characteristic, which contains an $x\in D$ such that $x^n \not\in Z(D)$ for all positive integers n?

First I tried quaternion algebra over  field $F_p(X)$.

Let $i^2= x$, $j^2= y$, $ij=k$, $ji=-k$ where $x, y$ are any non-zero elements in $F_p(X)$, the quaternion algebra $H(F_p(X))$ is a vector space over $F_p(X)$ with a basis $\{1,i ,j, k\}$.

Let $α =a+ib+jc+kd$, $\bar{α}=a-ib-jc-kd$, $α\bar{α}= \bar{α}α= a^2-xb^2-yc^2+xyd^2$, then
$H(F_p(X))$ is a division ring if and only if $α\bar{α}= 0$ implies $α= 0$.
After trying some examples, I find that the consequent processes are hard. Does there exist some good example that satisfies the property in the question?

答:
All you need to do is to produce an Ore domain which isn't commutative and has finite characteristic.

In fact there's an easy way to produce left principal ideal domains (which are necessarily left Ore).  The construction is that of the skew-polynomial ring $k[X;\sigma]$, where $\sigma$ is an automorphism of $k$ that isn't the identity.  Addition is plain addition of polynomials, but multiplication is governed by the rule $Xa=\sigma(a)X$.  When the automorphism isn't the identity, the ring is not commutative.

To achieve this we'll select $k$ to be $F_4$, the field of four elements, and use the Frobenius automorphism $a\mapsto a^2$ (which isn't the identity on $F_4$.)

So now you have $F_4\langle X;\sigma\rangle$, a not-commutative left Ore domain. By Ore's theorem, the classical left ring of fractions exists, and is a division ring that is not commutative, and it's easy to see that it has characteristic $2$.

With regards to the way you were going: there are quaternion-like constructions that do produce division rings from finite fields. You have elements $i,j,k$ such that $i^2=a$, $j^2=b$ and $ij=-ji=k$.

It is possible for this to be a division ring. When the characteristic of $F$ is not $2$, and $-1$ is not the sum of two squares in $F$, then the quaternion construction I described is a division ring (See Corollary 4.23 of K. Conrad's notes.

But also note Corollary 4.24: every quaternion algebra over $F_p$ is isomorphic to $M_2(F_p)$ (hence not a division ring.) So the only hope would be to use more complex fields of finite characteristic.


问:
Meaning of $B^2$ for $B\in \Omega^*(M,\operatorname{End}(E))$
Let $E\rightarrow M$ a vector bundle over a smooth manifold and let $B\in  \Omega^*(M,,\operatorname{End}(E))$ be an $,\operatorname{End}$-valued form. Is there a natural meaning to $B^2$?

I guess that if there is we would like to have $B^2\in  \Omega^*(M,,\operatorname{End}(E))$ as well so if we write $B$ as a sum of $\omega\otimes f$ with $\omega\in\Omega^*(M)$ and $f\in\Gamma(,\operatorname{End}(E))$ we would want something like a sum of
$$\omega\wedge \eta\otimes f\circ g$$
Is there something I am missing?

答:
You are not missing anything, except this operation is usually denoted by $B \wedge B$ and not $B \circ B$ or $B^2$ (so that one won't be confused with the pointwise composition) although this notation also has its drawbacks. This is an instance of the following general construction:

Let's assume that $V,W$ are vector spaces over $k$ and consider the space $\Lambda(V) \otimes W$. You can always define the product of two elements in $\Lambda(V) \otimes W$ to be an element of $\Lambda(V) \otimes W \otimes W$ by the formula
$$ (\alpha \otimes w) \wedge (\beta \otimes u) := (\alpha \wedge \beta) \otimes (w \otimes u). $$
If, in addition, $W$ has the structure of a $k$-algebra so that you have a map $m : W \otimes W \rightarrow W$, you can apply it to the $W \otimes W$ factor and get a multiplication map $\wedge_m \colon (\Lambda(V) \otimes W) \times (\Lambda(V) \otimes W) \rightarrow \Lambda(V) \otimes W$ defined by:
$$ (\alpha \otimes w) \wedge_m (\beta \otimes u) := (\alpha \wedge \beta) \otimes m(w,u). $$
This resulting algebra $\Lambda(V) \otimes W$ is precisely the tensor product of the algebras $\Lambda(V)$ and $W$ as algebras.

The two most important cases that appear in Differential Geometry are where:

$W = \operatorname{End}(E)$ with the multiplication being composition. In this case, the corresponding product on $\Omega^{*}(M;\operatorname{End}(E))$ is usually denoted by $\wedge$ for the lack of better notation.

$W = \mathfrak{g}$ is a Lie algebra with the "multiplication" being the Lie bracket. In this case, the corresponding product on $\Omega^{*}(M;\mathfrak{g})$ is denoted by $\mu \wedge \nu$ or $[\mu \wedge \nu]$ (the latter is used to remind us that the product involves both $\wedge$ and $[,]$).

An example of a formula which uses this notation is the following "structure equation". If $\nabla$ is a connection on $E$ and you choose a local trivializing frame of $E$ over $U$, then $\nabla$ can be written on $U$ as $d + \omega$ where $d$ is the standard flat connection induced from the trivialization and $\omega \in \Omega^1(U;\operatorname{End}(E))$. Then
$$ d\omega + \omega \wedge \omega = d \omega + \frac{1}{2} [\omega \wedge \omega] = R_{\nabla} $$
where $R_{\nabla} \in \Omega^2(U;\operatorname{End}(E))$ is the curvature and $\omega \wedge \omega$ is defined using the wedge product and the composition while $[\omega \wedge \omega]$ is defined using the wedge product and the Lie algebra structure on $\operatorname{End}(E)$.

One can also write this formula globally if you choose some connection $\nabla_0$ and write $\nabla = \nabla_0 + \omega$. Then
$$  d_{\nabla_0}(\omega) + \omega \wedge \omega = R_{\nabla}. $$
The previous instance is obtained if you apply it locally with $\nabla_0 = d$ the standard flat connection associated to a specific trivialization.


答:
Given a group homomorphism $f : A \to B$, it is easy to see that $\operatorname{coker}(f) = B / [f(A)]$, where $[ - ]$ denotes the normal closure of a subgroup. In the language of category theory, more formally, $\operatorname{coker}(f)$ is the qoutient homomorphism  $p : B \to B / [f(A)]$.

To see this, let $q : B \to C$ be any homomorphism such that $q \circ f = 0$. This means $f(A) \subset \ker(q)$, thus $[f(A)] \subset \ker(q)$ because kernels are always normal subgroups. Therefore $q$ can be written as $q = \bar q \circ p$ with a unique homomorphism $\bar q :  B / [f(A)] \to C$.

Now let us look at $[H]$ in $S_3$. The only non-trivial normal subgroup of $S_3$ is the alternating group $A_3$, but certainly $H \not\subset A_3$. Thus $[H] = S_3$ which proves your claim.

Title: Cokernel of group homomorphism

Body: We know $H=\{ (1),(12) \}$ is subgroup of $S_3$. Consider inclusion '$\varphi: H \hookrightarrow S_3$', this is clearly group homomorphism. Prove that coker$\varphi$ is trivial. I can't understand how to apply universal property of cokernel to this homomorphism $\varphi$.


问:
Lindelof's Theorem And Second Countable Topological Space
Let $X$ be a second countable topological space and let a non empty open set $G$ can be written as a union of a class of $G_i$ of open sets, then $G$ can be written as a countable union of $G_i$'s.

It is given that
$$G = \bigcup G_i.$$
Now for any $x \in G$ we have $x \in G_{i_0}$. Now as $X$ is a second countable topological space and hence it has a countable base say
$$B_1, B_2, .........$$
Now as $x \in G_{i_0}$, so there is some $B_{n_0}$ such that
$$x \in B_{n_0} \subset G_{i_0}.$$
And as $x$ is an arbitrary element of $G$ so we have
$$G \subset \bigcup_{k=1}^{\infty} B_{n_k}.$$
Also for each $B_{n_k}$ there exists a $G_{i_k}$ such that
$$B_{n_k} \subset G_{i_k}$$
with the above property and hence we have
$$G \subset \bigcup_{k=1}^{\infty}B_{n_k} \subset \bigcup^{\infty}_{k=1}G_{i_k} \subset \bigcup G_i = G.$$
And hence we get
$$G = \bigcup^{\infty}_{k=1}G_{i_k}.$$
And hence $G$ can be written as a countable union of $G_i$'s.

Is My Proof Correct??

答:
Yes, the proof is correct. Maybe you could be more clear about what $(n_k)$ is: for every $x ∈ G$ you have $n_0$ depending on $x$. You collect all these $n_0$'s and enumerate them by $(n_k)$. So it would be easier to pick for every $x ∈ G$ a basic set $B_x$ and then just say that the collection $\{B_x: x ∈ G\}$ is countable, and possibly enumerate this collection.

Another pedantic note: $⋃G_i$ can mean $⋃_{i ∈ I} G_i$ as well as $⋃_{A ∈ G_i} A$, so I'd prefer to write at least $⋃_i G_i$.


问:
Struggling to find solution: question 4 (Ordinary Differential equation) (Reducible exact equation)
Could someone please help with this problem?
$$(2y+\frac{y^2}{x})dx+(2y+x)dy=0$$
I have tried and found the integrating factor $F(x)= 1/x$ and then re-arranged the equation to be:
$$(2\frac{y}{x}+\frac{y^2}{x^2})dx+(2\frac{y}{x}+1)dy=0$$
After this point it seemed so similar to a reducible seperable equation so I did this:
$$y=ux$$ $$y'=u'x+u$$
With this transformation
$$\frac{du}{dx}x+u=-\frac{2u+u^2}{2u+1}$$
$$\frac{du}{dx}x=-u-(u+1-\frac{1}{2u+1})$$

Body:
$$(2y+\frac{y^2}{x})dx+(2y+x)dy=0$$
$$(2xy+{y^2})dx+(2xy+x^2)dy=0$$
$$ydx^2+y^2dx+xdy^2+x^2dy=0$$
$$(ydx^2+x^2dy)+(y^2dx+xdy^2)=0$$
$$d(yx^2)+d(y^2x)=0$$
Integrate:
$$yx^2+y^2x=C$$
$$xy(x+y)=C$$


问:
Contractible manifolds of dimension $n\le 2$
I have to give a talk about the Whitehead manifold. I would like to stress the fact that it is the first given example of a contractible open (i.e. non compact) manifold which is not homeomorphic to $\mathbb{R}^n$ (when $n=3$). This happens because the Whitehead manifold is not simply-connected at infinity and as proved by Stallings this is equivalent to not be homeomorphic to $\mathbb{R}^n$. The first example of this kind occurs in dimension $3$ beacause as I read in many articles the only contractible manifold (up to homeomorphism) of dimension $n\le 2$ are: \begin{equation*}\{\text{pt.}\} \quad \mathbb{R} \quad \mathbb{R}^2
\end{equation*}
I'm thinking about a proof of this classification and I'm having problems with dimension $2$. So I have the following questions:

Question 1

Can you suggest me any reference for the classification in dimension $2$ with a detailed proof?

Question 2

Is the Stalling's result mentioned above valid for a general $n$-manifold or do we have to require that it is contractible?

答:
Forget the one-point compactification, it's a dead end. By the time you prove that a neighborhood of infinity looks like $[0,\infty) \times S^1$ (which is what you'd need to see the 1-pt compactification is a manifold) you're more or less already done.

Theorem: If $\Sigma$ is an open surface with $H_1(\Sigma; \Bbb F_2) = 0$, then $\Sigma$ is homeomorphic to $\Bbb R^2$.

Sketch of proof.

(1) First prove that $\Sigma$ has exactly one end; if $K \subset \Sigma$ is a compact subset, then $\Sigma \setminus \text{int}(K)$ may have many connected components, but only one of them is noncompact. The proof will be by contrapositive: if $\Sigma$ has two ends, show that $H_1(\Sigma; \Bbb F_2) \neq 0$. You will need to know that the inclusion $\partial \Sigma \to \Sigma$ is nonzero on first homology whenever $\Sigma$ is a noncompact surface with boundary.

(2) Use that $\Sigma$ has a compact exhaustion --- $\Sigma = \bigcup \Sigma_n$, where $\Sigma_n$ is a compact surface and $\Sigma_n \subset \text{int}(\Sigma_{n+1})$, so that $\Sigma_{n+1} = \Sigma_n \cup S_n$, where $S_n$ is also a compact surface. This can be justified using the fact that $\Sigma$ has a proper smooth function to $\Bbb R$, together with Sard's theorem.

(3) Using (1) and (2) together, observe that $\Sigma \setminus \text{int}(\Sigma_n)$ only has one noncompact piece. Modify your compact exhaustion so that $\Sigma \setminus \Sigma_n$ is connected and so that $S_n$ is connected.

(4) Prove that $\partial \Sigma_n$ is a single circle, as otherwise $\Sigma$ has positive genus (you'll glue on a pair of pants as $S_n$, because $S_n$ is connected), which would imply $H_1(\Sigma;\Bbb F_2) \neq 0$; again this involves some Mayer-Vietoris work.

(5) Prove that each $\Sigma_1$ is a disc and each $S_n$ is a cylinder. From here it follows that $\Sigma \cong \Bbb R^2$.

The details are not completely trivial, in particular on (3). I will not see or respond to comments, so please feel free to edit this answer as you desire. This gives a rough strategy for the general classification of noncompact surfaces without boundary, the ideas simplify when $\Sigma$ is contractible like this; similarly you can show that if $H_1(\Sigma)$ is finite, then $\Sigma$ is obtained by deleting finitely many points from a closed surface.


问:
Optimizing a function with no local maximum?
suppose we have a function f(x)=$\sqrt{\left(\ x\ \right)}\ +\ \left(\frac{C-x}{4}\right)$ and we want to maximize the function when C=<4 and x<C. How do we procede with the problem?

Note that I've never studied optimization, I'm just trying to solve a budget constraint problem in microeconomics.

答:
A little bit more formal derivation (although I think @Cesareo already has a nice, clear illustration).

This is an optimizitaion problem
$$\begin{aligned}
\text{max}_{x,C}\ &f(x,C)\\
\text{s.t.}\\
{h}_1(x,C)=x-c&\leq{0},\\
h_2(x,C)=C-4&\leq 0
\end{aligned}$$
in two variables, x and C,where $\vec{h}=[h_1, h_2]^T=[(x-C),\ (C-4) ]^T$. For a maximum (or minimum), the KKT equations
$$\begin{aligned}
\nabla f(x,C) + \nabla \vec{h}(x,C)\vec{\mu} &= \vec{0}\\
\vec{\mu}^T\vec{h}(x,C)&=0\\
\vec{\mu} &\geq \vec{0}
\end{aligned}$$
are a necessary condition, where $\nabla(\cdot)$ signifies the gradient of the respective quantity (which is the transpose of the Jacobian). The latter two equations are the so-called complementary slackness conditions, and $\vec{\mu}=[\mu_1,\mu_2]^T$ is a vector of Lagrange multipliers.

An inequality $i$ at the solution is said to be active if $h_i(x,C)=0$ and inactive if $h_i(x,C)<0$. The complementary slackness conditions say that if an inequality constraint is active, the corresponding Lagrange multiplier is $\mu_i>0$. If an inequality is inactive this implies that the corresponding $\mu_i=0$. This is why this condition is called complementary. Either the inequality is zero or its corresponding Lagrange multiplier is.

Applying this to your problem yields
$$\begin{align}
\frac{1}{2\sqrt{x}}-\frac{1}{4} + \mu_1&=0\\
 \frac{1}{4} -\mu_1 + \mu_2&=0\\
\mu_1(x-C)+\mu_2(C-4)&=0\\
\vec{\mu}&\geq \vec{0}.
\end{align}$$
Now, we have to distinguish between four cases:

both $h_1(x,C)$ and $h_2(x,C)$ are inactive at the solution which implies $\Rightarrow$ $\mu_1=\mu_2=0$. This, however, is not valid since this would result in $\frac{1}{4}=0$ (from the second equation above) which is not true.

$h_1(x,C)$ is active and $h_2(x,C)$ is inactive at the solution which implies $\Rightarrow$ $\mu_1=0$. From the second equation we now get $\mu_2=-\frac{1}{4}$ and from the first equation we get the candidate $x=1$. From the third equation, however, we see that $C=-16$ which violates $h_1(x,C)$ ($x$ must be smaller or equal to $C$).

$h_1(x,C)$ is inactive and $h_2(x,C)$ is active at the solution which implies $\Rightarrow$ $\mu_2=0$. From the second equation we now get $\mu_1=\frac{1}{4}$ and from the first equation we get $\frac{1}{2\sqrt{x}}=0$ which is impossible.

This only leaves us with the last possibility that both inequalities must be active at the solution so that $x^\ast=C^\ast$ and $x^\ast=4$ and $f(x^\ast,C^\ast)=2$.


问:
Identify quotient group on $\mathbb{Z} \times \mathbb{Z}_n$
Let $F: \mathbb{Z} \longrightarrow \mathbb{Z} \times \mathbb{Z}_n$ be the homomorphism of groups given by
$$F(a)=(ma,-[a]_n)$$
with fixed $m \in \mathbb{Z}$

What is, up to isomorphism, the quotient group $\mathbb{Z} \times \mathbb{Z}_n/\mathrm{Im}(F)$

答:
Consider the homomorphism $G : \mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/mn\mathbb{Z}$ given by:

$$
(x,y+n\mathbb{Z})\mapsto x+my+mn\mathbb{Z}
$$
Left to the reader: Show that this is well defined, that it is a group homomorphism, and that $\mathrm{Im}(F)\subseteq\mathrm{Ker}(G)$.

To show the other inclusion:

Suppose $G(x,y+n\mathbb{Z})=mn\mathbb{Z}$, then $x+my\in mn\mathbb{Z}$ so there is some k such that $x+my=mnk$, so $x=mnk-my=m(nk-y)$, and $y\equiv -(nk-y) \mod n$, that is $$(x,y+n\mathbb{Z})=(m(nk-y),-(nk-y)+n\mathbb{Z})=F(nk-y).$$

Now we have $\mathrm{Im}(F)=\mathrm{Ker}(G)$, so by the first isomorphism theorem we get

$$(\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z})/\mathrm{Im}(F)\cong \mathrm{Im}(G)$$

Now $\mathrm{Im}(G)=\mathbb{Z}/mn\mathbb{Z}$ because for all $x\in\mathbb{Z}$, $G(x,n\mathbb{Z})=x+mn\mathbb{Z}$

Note:

To get to the homomorphism above, I checked what $(x_1,y_1+n\mathbb{Z})+\mathrm{Im}(F)=(x_2,y_2+n\mathbb{Z})+\mathrm{Im}(F)$ in the quotient group means, which turned out to be (after some working out)

$$
\left\{\begin{aligned}
&x_1\equiv x_2&&\mod m\\
&x_1+my_1\equiv x_2+my_2&&\mod mn
\end{aligned}\right.
$$

And it turned out that the first congruence follows from the second, and the second congruence gives the homomorphism defined above.