问:
what is the difference between Strong Fréchet derivative and Fréchet derivative
I am confused about the relationship between Frechet differentiable and strong Fréchet differentiable.

Assume the function $f(x) \in \mathbb{R}, x\in \mathbb{R}^n$ is Fréchet differentiable at $x$.

Then we have $$\lim_{x \neq y^1 \to x} \frac{f(y^1) - f(x) - \nabla f(x)^T(y^1-x)}{\Vert y^1-x\Vert} = 0$$

$$\lim_{x \neq y^2 \to x} \frac{f(y^2) - f(x) - \nabla f(x)^T(y^2-x)}{\Vert y^2-x\Vert} = 0$$

Then use the first equation minus the second one and the inequality $\Vert y^1-x\Vert \leq \Vert y^1-y^2\Vert + \Vert y^2 - x\Vert$, then we get

$$\lim_{\substack{y^1 \neq y^2 \\ (y^1, y^2) \to (x, x)}} \frac{f(y^1) - f(y^2) - \nabla f(x)^T(y^1-y^2)}{\Vert y^1 - y^2 \Vert} = 0, $$
which is the definition of strong F-derivative.

It seems I did not use any other condition but F-derivative to get strong F-derivative. I am quite confused about their difference.

答:
The typical counterexample is $f \colon \mathbb R \to \mathbb R$,
$$f(x) = \begin{cases} 0 & \text{if } x = 0, \\ x^2 \sin(x^{-1}) & \text{if } x \ne 0. \end{cases}$$
This function is Fréchet differentiable on all of $\mathbb R$, but not strictly/strongly Fréchet differentiable at $x = 0$.

Since $f'(0) = 0$, it is also very instructive (and not too hard) to use this counterexample and see where your argument fails.

If I remember correctly, one even has the following: If the function $f$ is Fréchet differentiable everywhere, then the strict/strong differentiability is equivalent to the continuity of the derivative.


问:
maximum and minimum singular values
I was studying the Stephen Boyd's textbook on convex optimization and have a question. The book says the following:

The singular values of $A$, $\sigma_1 \geq \sigma_2 \geq ... \geq \sigma_n$ are the square roots of the eigenvalues of $\lambda_1 \geq \lambda_2 \geq ... \geq \lambda_n$ of $G$ ($G$ is the Gram Matrix, $A^T A$). Therefore $\sigma_1^2$ is a convex function of $G$, and $\sigma_n^2$ is a concave function of $G$.

Can anybody explain why the maximum and minimum eigenvalues are a convex and concave function of a Gram matrix?

Thanks.

答:
If $A$ is a real symmetric matrix, then its maximal eigenvalue is
$$\lambda_1(A)=\max_{\|v\|=1} \langle Av, v\rangle \tag1  $$
and its smallest eigenvalues is
$$\lambda_n(A)=\min_{\|v\|=1} \langle Av, v\rangle \tag2  $$
For each fixed vector $v$ the function $A\mapsto \langle Av, v\rangle$ is a linear function of $A$. The maximum of any family of linear functions is convex. The minimum of any family of linear functions is concave. </p>

ContentLicense: CC BY-SA 3.0

Title: maximum and minimum singular values


问:
Trivialising homeomorphisms for a principal $G$-bundle as $G$-space morphisms
Let $P$ by a principal $G$-fibre bundle over a locally-compact Hausdorff space $X$. Denote by
$$
h: U \times G \to P|_{U}
$$
a trivialising homeomorphism for a trivialising open set $U \subseteq X$. Now the domain and range of $h$ are both $G$-spaces in an obvious way. Is $h$ a morphism of $G$-spaces? If so why? If true and not evident, I would be grateful for a reference.

答: I am not sure which definitions you use, but this is more or less true by definition. I suppose you would start by requiring the transition functions between the trivializations to be given by left multiplications in the group. Then you define the principal right action on $P$ by requiring that in the trivialization it is given by multiplication from the right in $G$. But this exactly means that the map $h$ you consider is $G$-equvariant for $(x,g)\cdot h:=(x,gh)$ and the principal right action on $P|_U$.

ContentLicense: CC BY-SA 4.0


问:
Find the probability mass function of W.
Let X and Y be independent Geom(P) random variables.

Let

\begin{equation}
   W=
      \begin{cases}
         0, & \text{if X<Y}\\
         2, & \text{if X=Y}\\
         3, & \text{if X>Y}
      \end{cases}
\end{equation}

Find the probability mass function of W.

My attempt:

$W\quad\;\;$ | $\quad\quad\quad\quad\quad\quad0\quad\quad\quad\quad\quad\quad\;\;\;\quad\quad1\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad2$

$\mathbb{P}(W)\;\;$| $(1-p)^{k-1}p\left(1-\sum^{k-1}_{i=0}(1-p)^ip\right)\quad\left((1-p)^{k-1}p\right)^2\quad(1-p)^{k-1}p\left(1-\sum^{k-1}_{i=0}(1-p)^ip\right)$

(Sorry, I don't know how to draw tables on stack exchange.)

I've been looking at this problem and it seems really similar to this problem: Find the probability mass function of $V=\min(X,Y)$

I think the difference is that we do not sum up the values. My question then, is when do we know when to sum up the probabilities and when to leave them when calculating a pmf. I know this is a stupid question, but I've seen pmfs expressed both ways in my probability class.

答:
You need to actually perform the summation.  In particular, $$\begin{align}
\Pr[X = Y] &= \sum_{k=0}^\infty \Pr[X = k]\Pr[Y = k] \\
&= \sum_{k=0}^\infty \left( (1-p)^k p \right)^2 \\
&= p^2 \sum_{k=0}^\infty ((1-p)^2)^k \\
&= \frac{p^2}{1-(1-p)^2} \\
&= \frac{p}{2-p}. \end{align}$$

By symmetry, $\Pr[X < Y] = \Pr[X > Y]$ because $X$ and $Y$ are independent and identically distributed.  Therefore, $$2 \Pr[X < Y] + \Pr[X = Y] = 1,$$ or $$\Pr[X < Y] = \frac{1}{2}\left(1 - \frac{p}{2-p}\right).$$


问:
Find the locus of the point
Let P be a variable point on a given line and let A be any given fixed point. Find the locus of the point Q lying on AP or AP produced such that AP/AQ= constant.I think if A doesn't lie on the given line
then the locus of the line will be any line parallel to the given line . In order to  show this, let us assume that P and P' be any two points on the given line and let Q and Q' be two points on AP and AP'
such that AQ/AP=AQ'/AP' in triangle APP'.Consequently,QQ' will be parallel to the given line .Conversely,
one can show that for QQ' being any line parallel to the given line, AQ/AP=AQ'/AP'.However,if A lies on the given line what should be the locus of Q?Please help me verify this.</p>

答: The locus will the given line itself since every point on this line can be regarded as some $Q$ for a given constant.

Title: Find the locus of the point


问:
book for study finite element method
I want to study finite element method for study partial differential equations in particular for parabolic type equation. Can you recommend a book which includes espesially mathematical background?

Thanks in advance...

答:
I came up too late, but I recommend a series of books for finite element method with different approach for future reading.

For engineering approach:

Huebner - The Finite Element Method For Engineers.

For a introduction to the mathematical theory:

Sayas - A gentle Introduction to Finite Element Method  link.

Brenner & Scott - The Mathematical Theory of Finite Element Method.

Braess - Finite Elements Theory, Fast Solvers, and Applications in Elasticity Theory.

Solin et al. - Higher-Order Finite Element Methods. (Implementation aspects)

Introductory books for research purposes.

Demkowicz- Mathematical Theory of Finite Element Method link (For DPG method with optimal test functions).

Gatica - A Simple Introduction to the Mixed Finite Element Method.

Sayas - First steps in the boundary element method link

Girault & Raviart - Finite element methods for Navier-Stokes equations.

Monk - Finite Element Methods for Maxwell's Equations.

Online Courses:

Introduction to Finite Element Methods

High Performance Finite Element Modeling-i-and-ii-1.769003

Title: book for study finite element method


问:
Two exercises on mathematical induction
Studying for a test and can't work out how to do two questions on the sample test.

(1) Suppose a sequence of numbers $a_1$, $a_2$, $\dots$ is defined recursively by:
$$a_1 = 1\qquad\text{and}\qquad a_{n+1} = \frac{a_n}{a_n + 1}$$

Prove, using mathematical induction, that for each integer $n\ge 1$

$$a_n = \frac{1}{n}.$$

(2) Using mathematical induction, prove that the formula
$$ g^n = \begin{pmatrix}1 & 0 \\ -2n & 1\end{pmatrix}$$
where
$$ g = \begin{pmatrix}1 & 0 \\ -2 & 1\end{pmatrix}$$
is correct for each integer $n \geq 1$.

If someone could show me how there done would be a big help thanks.

答:
Hints:

$$\bullet\;\;\;\;\;\;\;\;\;\;\;\;\;a_{n+1}=\frac{a_n}{a_n+1}\stackrel{\text{Ind. Hypot.}}=\frac{\frac1n}{\frac1n+1}=\frac1{n+1}$$

Since I've no idea what $\,g\,$ is in the second question I can't help there.

Title: Two exercises on mathematical induction


问:
Prove by strong induction that $9^n$ ends in $1$ if $n$ is even and ends in $9$ if $n$ is odd.

Prove by strong induction that $9^n$ ends in $1$ if $n$ is even and ends in $9$ if $n$ is odd.

Would it be valid to work separately with even and odd exponents, making two separate proofs, each using weak induction?

How can I prove this using strong induction in one proof?

答:
Well you have three options.

You can prove if $9^{2k}$ ends with $1$, then $9^{2k+2}$ also ends with $1$; and then prove if $9^{2k-1}$ ends with $9$ then $9^{2k+1}$ also ends with $9$

or

You can prove if $9^n$ ends with $1$ then $9^{n+1}$ ends with $9$; and then prove if $9^{n}$ ends with $9$ then $9^{n+1}$ ends with $1$.

You can prove if $9^n$ ends with whatever, then $9^{n+2}$ also ends with whatever

Those are all very easy proves.

And 1) and 3) can be proven by noting $9^{n+2} = 9^n\cdot 9^2 = 9^n\cdot 81 = (10something + whatever)\times (80 + 1)= 800something + (80whatever + 10something) + whatever$.

And 2) can proven that $9^{n+1} = 9^n\cdot 9 = 9(10something + \begin{cases}1\\9\end{cases}) = 90something + \begin{cases}9\\81\end{cases})$.


问:
Divergence of the sum of convergent and divergent series (proof check)
I need to prove that if  $\sum_{n=1}^{\infty} a_{n}$ converges and $\sum_{n=1}^{\infty} b_{n}$ diverges then the sum $\sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right)$ diverges.

By defining $k_{n}=s_{n}+t_{n}$, where $s_{n}=\sum_{k=1}^{n} a_{k} \text { and } t_{n}=\sum_{k=1}^{n} b_{n}$ (the nth partial sums),  can I prove this simply by noting that

$$\lim_{n \to \infty}  k_{n}= \lim_{n \to \infty}  (s_{n} + t_{n}) = L \pm \infty = \pm \infty $$

$$?$$
The book I'm following uses a proof by contradiction but this approach seems more straightforward, although I might be missing something.

答: No, that is not correct. You seem to assume that the series $\sum_{n=1}^\infty b_n$ diverges if and only if $\lim_{n\to\infty}t_n=\pm\infty$. Actually, the series $\sum_{n=1}^\infty b_n$ diverges if and only if the sequence $(t_n)_{n\in\Bbb N}$ diverges. And, if the sequence $(s_n+t_n)_{n\in\Bbb N}$ was convergent with limit $L'$, then$$\lim_{n\to\infty}t_n=\lim_{n\to\infty}\bigl((s_n+t_n)-s_n\bigr)=L'-L.$$And this is a contradiction.


问:
Finding $\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)$
I'm trying to find out :

$$\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)$$

First I tried by inspection. I suspect that
$$\lim_{x\rightarrow \infty} \left( \frac{x+1}{x-1}\right)=1$$
So I expect $\ln (\cdots)\rightarrow 0$ as $x\rightarrow \infty$. While If I multiply this by $x$, One might expect this to go infinite.

Next what I have done is to use a substitution, I don't know if it's right to do or not.
$$x'=\frac{x+1}{x-1}\Rightarrow x=\frac{x'+1}{x'-1}$$

So that
$$\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)\Rightarrow \lim_{x'\rightarrow 1}\left(\frac{x'+1}{x'-1}\right)\ln x'$$
Now I can use L'Hospital rule
$$\lim_{x'\rightarrow \infty}\frac{\ln x'+\frac{x'+1}{x'}}{1}=2$$

Can I use such a substitution always to solve problems? What's wrong with the reasoning I did earlier?

答:
$f(x):=\log \left (\dfrac{ (x+1)^x}{(x-1)^x} \right)=$

$\log \left (\dfrac {(1+1/x)^x}{(1-1/x)^x} \right)=$

$\lim_{x \rightarrow \infty} f(x)= \log \dfrac {e^1}{e^{-1}}=2.$

Used:

$1)\log$ is a continuos function;

$2)\lim_{x \rightarrow \infty} (1+a/x)^x= e^a$, $a$ real.


问:
A problem on compact space
Given $X$  a compact space and $Y$ a metric space. $f$ and $g$ are continuous mappings from $X$ to $Y$. Does the set $A=\{(f(x),g(x))|x\in X\}$ need to be a compact set in $Y\times Y$?

答:
$\textit{(There is actually no issue, see edit below)}$

I think there is a little issue with working with the 'diagonal' $D = \{(x,x): x \in X$}, as is done in the other answer and the comments. For this argument to be sound I think you need the diagonal to be closed (hence compact as a subset of a compact space), but the diagonal is only closed in Hausdorff spaces, which is not given for $X$. But I think one can proceed as follows:

Let $h:X \to Y \times Y, x \mapsto (f(x),g(x))$. This function is continuous since $f$ and $g$ are continuous. Now we have

$$ h(X) = \{h(x) : x \in X\} = \{(f(x), g(x)): x \in X \} = A $$

Since $X$ is compact and $h$ is continuous, $A$ is also compact as the image of a compact set under a continuous function.

Edit: As Jihai Zhu pointed out in the comment, there is actually no problem with the approach using the diagonal, since the diagonal is compact when $X$ is. No need to argue with closed sets.

Title: A problem on compact space


问:
Show that $\lambda<\alpha<\lambda h(x)+\sum_{i=1}^n\lambda _if_i(x) \quad\forall x\in E$
Let $E$ a vector space and $h,f_1,\cdots,f_n : E \to \mathbb R$ linear such that $\displaystyle \bigcap_{i=1}^n \ker(f_i) \subset\ker h.$
Let $F : E\to \mathbb R^n : x\mapsto(h(x),f_1(x),\cdots,f_n(x))$.

Show that $(1,0,\cdots,0)\not \in Im(F)$.

Show the exitence of $(\lambda,\lambda_1,\cdots,\lambda_n)\in \mathbb R^{n+1}, \alpha\in \mathbb R$ :
$\lambda<\alpha<\lambda h(x)+\sum_{i=1}^n\lambda _if_i(x) \quad\forall x\in E$

The first question is easy to prove bythe absurd but I got stuck in the second one. I know the factorisation of a map by other maps  as mentioned here but this exercice is a proof of it .
Any help is appreciated.

答:
Note that $F(E) \subset \mathbb{R}^{n+1}$ is a closed subspace and $e_1 = (1,0,...,0) \notin F(E)$.

Hence the Hahn Banach theorem gives the existence of some linear $\phi$ and some $\alpha$ such that $\phi(e_1) < \alpha < \phi(F(x))$ for all $x$.

Since $\phi$ is linear, it has the form
$\phi((h,f_1,...,f_n)) = \lambda h + \sum_k \lambda_k f_k$ from which we get the desired result.


问:
A triangle; given side and two heights
A triangle $ABC$ is given $BC=8$ and $AE=4$ and $BK=6,4$ are heights. Find $AC$ and $AB$.

So we can find the area of $ABC$: $$S=\dfrac{BC\cdot AE}{2}=16.$$
So $AC=\dfrac{2S}{BK}=\dfrac{32}{6,4}=5.$
How can I find $AB$ from here?

答:
Just use Pythagorean theorem. To be specific:

we have $AC=5, AE=4$, use Pythagorean theorem for triangle AEC,so $EC=3\rightarrow BE=BC-EC=5$

use Pythagorean theorem for triangle AEB, we have $AB=\sqrt{4^{2}+5^{2}}=\sqrt{41}$

Title: A triangle; given side and two heights


问:
Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x^2 + y + f(y)) = 2y + (f(x))^2$

Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $$f(x^2 + y + f(y)) = 2y + (f(x))^2$$

I tried something like this but that inner $f(y)$ makes things worse to my opinion. Any help is appreciated.

答:
Here's my solution:

We claim that the only solution to the functional equation is $f(x)=x$, indeed it is easy to see that it is a solution. Now, we prove that it is the only one.

Let $P(x,y)$ denote the given assertion, by comparing $P(x,y)$ and $P(-x,y)$ we see that $f(x)^2=f(-x)^2$.

Then, we claim that $f$ is surjective, indeed, $$P(0,x): f(x+f(x))=2x+f(f(0))^2$$ and take any $k \in \mathbb{R}$, let $x=\frac{k-f(f(0))^2}{2}$, we get $f\left(\frac{k-f(f(0))^2}{2}+f\left(\frac{k-f(f(0))^2}{2}\right)\right)=k$.

Let $a\in \mathbb{R}$ such that $f(a)=a$, from $f(x)^2=f(-x)^2$, by letting $x=a$, we see that $f(-a)=0$. Moreover, we have $$P(0,a): 0=f(a+f(a))=2a+f(f(0))^2$$ $$P(0,-a): 0=f(-a+f(-a))=-2a+f(f(0))^2$$ subtracting both equation yields $a=0$. Therefore, we also have $f(x+f(x))=2x$ and $P(x,0) : f(x^2)=f(x)^2\ge 0$.

We now prove that $f$ is odd, notice that since $$P\left(x,-\frac{f(x)^2}{2}\right): f\left(x^2-\frac{f(x)^2}{2}+f\left(-\frac{f(x)^2}{2}\right)\right)=0$$ by the fact above that if $f(a)=0$, then $a=0$, we have $$-\frac{f(x)^2}{2}+f\left(-\frac{f(x)^2}{2}\right)=-x^2 \implies f\left(-\frac{f(x)^2}{2}+f\left(-\frac{f(x)^2}{2}\right)\right)=f(-x^2).$$ On the other hand, we have $$P\left(0,-\frac{f(x)^2}{2}\right): f\left(-\frac{f(x)^2}{2}+f\left(-\frac{f(x)^2}{2}\right)\right)=-f(x)^2$$ which means that $f(-x^2)=-f(x^2)=-f(x)^2 \le 0$. Therefore, we can see that $f(x)=-f(-x)$.

Now, there are two ways to proceed:

1st way: We show that $f$ is injective, denote $Q(x,y)$ by the new assertion $$P(\sqrt{x},y): f(x+y+f(y))=2y+f(\sqrt{x})^2=2y+f(x) \ \forall x\ge 0$$ if $x\le 0$, then use $P(\sqrt{-x},y)$. (Or just use the fact that $f$ is odd.) Then, by comparing $Q(x,y)$ and $Q(y,x)$, if $f(s)=f(t)$, then $s=t$. So, $f$ is bijective.

Next, since $f$ is bijective, so as $x+f(x)$. Thus, $f$ is additive! As $$f(x^2+y+f(y))=2y+f(x)^2=f(x^2)+f(y+f(y)).$$ And so, since $f$ is also positive on the positive reals, it is linear, by checking, $f\equiv x$.

2nd way: As we have shown that $f(x)=-f(-x)$, $$P(x,-x^2): -f(f(x)^2)=f(f(-x^2))=-2x^2+f(x)^2$$ also since $x^2-\frac{f(x)^2}{2}+f\left(-\frac{f(x)^2}{2}\right)=0 \implies -2x^2+f(x)^2=2f\left(-\frac{f(x)^2}{2}\right)=-2f\left(\frac{f(x)^2}{2}\right)$, we have $$f(f(x)^2)=2f\left(\frac{f(x)^2}{2}\right)\implies f(x)=2f\left(\frac{x}{2}\right) \ \forall x\ge 0$$ by surjectivity.

Next, as $f(x+f(x))=2x$, by replacing $x$ with $f(x)$ gives $$f(f(x)+f(f(x)))=2f(x)=f(2x) \implies f(x)+f(f(x))=2x \ \forall x\ge 0.$$ Since $f$ is positive on the positive reals, we use recurrence relation to solve for $f$, let $x_n=f^n(x)$, where $f$ is iterated $n$ times and define $x_0=x$ for a arbitrary real positive number, then this yields the following recurrence relation $$x_{n+1}+x_n-2x_{n-1}=0$$ which $x_n=\lambda_1\cdot (1)^n+\lambda_2 \cdot (-2)^n$ and $x_0=\lambda_1-\lambda_2$, $x_1=\lambda_1-2\lambda_2$ since $a_n\ge 0$, we must have $\lambda_2 =0$ otherwise $a_n$ would be negative for some arbitrary large, so we have $f(x_0)=x_1=\lambda_1=x_0$. Hence, as $x_0$ is arbitrary, we have $f(x)=x$ for all positive reals and since $f$ is odd $f\equiv x$.


问:
Maximal Tori of $U(n)$ and $Sp(n)$
I am reading Section 3.6 of the Naive Lie Theory by Stillwell. The author claims that we can argue that $T^n$ is the maximal torus of $Sp(n)$ exactly as in the case of $U(n)$. The proof of the maximal torus of $U(n)$ makes use of a matrix A in the maximal torus commuting with all elements R in the torus $T^{n}$ of $U(n)$, where R has the form
$$R=\begin{pmatrix}
e^{i\theta_1} & & & \\
 & e^{i\theta_2} & & \\
 & & \ddots & \\
 & & & e^{i\theta_n} \\
\end{pmatrix}.
$$
If A also has the form of R, then $T^{n}$ is the maximal torus of $U(n)$. After several steps, we arrive at the condition that $A\mathbf{e}_k=c_k\mathbf{e}_k$. Therefore $c_k=e^{i\phi_k}$ and we arrive at the conclusion.

However, in the case of $Sp(n)$, after carrying out the same steps, I arrived at the condition that $c_k=e^{\mathbf{u}_k\phi_k}$, where $\mathbf{u}_k$ is an arbitrary unit quaternion, and $A$ does not necessarily follow the form of $R$. I wonder how to argue that $\mathbf{u}_k=\mathbf{i}$ for all $k$'s.

答:
Let $D$ be diagonal in $\mathrm{Sp}(n)$ with entries $\lambda_1,\cdots,\lambda_n$ which we may freely choose from $S^1$.

Suppose $S$ is an $n\times n$ quaternionic matrix which commutes with all such $D$. That is, $DS=SD$. This means that $\lambda_is_{ij}=s_{ij}\lambda_j$. If $s_{ij}\ne0$ for distinct $i,j$ then we could solve for $\lambda_j$ in terms of $\lambda_i$, which is impossible since we can choose the $\lambda$s independently of each other. Thus, $S$ is a diagonal matrix. Moreover, if we set $\lambda_i=\mathbf{i}$, then the equation $\mathbf{i}s_{ii}=s_{ii}\mathbf{i}$ implies $s_{ii}$ has zero $\mathbf{j}$ and $\mathbf{k}$ components, i.e. $S$ is a complex matrix. Then to finish, notice $s_{ii}\in S^1$ follows if $S\in\mathrm{Sp}(n)$.

Title: Maximal Tori of $U(n)$ and $Sp(n)$


答:
Your answers are right except you should take the sum from k=1 to infinity for all three cases. It's possible to do the summation to find $P(X<Y)$:

$$P(X=1)P(Y>1)+P(X=2)P(Y>2)+...=$$

$$\sum_{i=0}^\infty P(X=i+1)\underbrace{\sum_{j=i+1}^\infty P(Y=j+1)}$$

The underlined portion is

$$1-\sum_{j=0}^i P(Y=j+1)\\
=1-p\sum_{j=0}^i (1-p)^j\\
=1-p\frac{1-(1-p)^{i+1}}{p}\\
=(1-p)^{i+1}$$

Now the whole thing is

$$p\sum_{i=0}^\infty (1-p)^{2i+1}\\
=p(1-p)\sum_{i=0}^\infty \left[(1-p)^2\right]^i\\
=p(1-p)\frac 1 {1-(1-p)^2}\\
=\frac{1-p}{2-p}$$

Now of course $P(Y<X)=\frac{1-p}{2-p}$ and $P(X=Y)=1-2\frac{1-p}{2-p}=\frac{p}{2-p}$

Body:
Let X and Y be independent Geom(P) random variables.

Let

\begin{equation}
   W=
      \begin{cases}
         0, & \text{if X<Y}\\
         2, & \text{if X=Y}\\
         3, & \text{if X>Y}
      \end{cases}
\end{equation}

Find the probability mass function of W.

My attempt:

$W\quad\;\;$ | $\quad\quad\quad\quad\quad\quad0\quad\quad\quad\quad\quad\quad\;\;\;\quad\quad1\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad2$

$\mathbb{P}(W)\;\;$| $(1-p)^{k-1}p\left(1-\sum^{k-1}_{i=0}(1-p)^ip\right)\quad\left((1-p)^{k-1}p\right)^2\quad(1-p)^{k-1}p\left(1-\sum^{k-1}_{i=0}(1-p)^ip\right)$

(Sorry, I don't know how to draw tables on stack exchange.)

I've been looking at this problem and it seems really similar to this problem: Find the probability mass function of $V=\min(X,Y)$

I think the difference is that we do not sum up the values. My question then, is when do we know when to sum up the probabilities and when to leave them when calculating a pmf. I know this is a stupid question, but I've seen pmfs expressed both ways in my probability class.


答:
Well, to prove inductively that a statement holds for all $n \geq 0$, you need to show that

The statement holds for $n_0$

Whenever the statement holds for $n \geq n_0$, it also holds for $n+1 \geq n_0$.

In your case, you thus need to show that

$a_1 = \frac{1}{1}$

If $a_n = \frac{1}{n}$ ($n \geq 1$) then $a_{n+1} = \frac{a_n}{a_n + 1} \overset{\text{?}}= \frac{1}{n+1}$.

Title: Two exercises on mathematical induction

Body:
Studying for a test and can't work out how to do two questions on the sample test.

(1) Suppose a sequence of numbers $a_1$, $a_2$, $\dots$ is defined recursively by:
$$a_1 = 1\qquad\text{and}\qquad a_{n+1} = \frac{a_n}{a_n + 1}$$

Prove, using mathematical induction, that for each integer $n\ge 1$

$$a_n = \frac{1}{n}.$$

(2) Using mathematical induction, prove that the formula
$$ g^n = \begin{pmatrix}1 & 0 \\ -2n & 1\end{pmatrix}$$
where
$$ g = \begin{pmatrix}1 & 0 \\ -2 & 1\end{pmatrix}$$
is correct for each integer $n \geq 1$.

If someone could show me how there done would be a big help thanks.


问:
Truth table logic (includes negation and conjunction) (Haskell)
¬(P ∧ ¬Q)

Complete truth table for the following sentence.

EDIT: The image below shows where I am at in the process. With all of the negation symbols, I'm getting a little confused on how to fill out the connectives.

$$\def\true{\mathsf T}\def\false{\mathsf F}
\begin{array}{|l:l|l:l:l:l:l|}\hline
\rm P &\rm Q &\lnot &\rm(P &\land &\lnot &\rm Q)\\\hline
\true &\true &      &\true &      &      &\true \\\hdashline
\true &\false&      &\true &      &      &\false\\\hdashline
\false&\true &      &\false&      &      &\true \\\hdashline
\false&\false&      &\false&      &      &\false\\\hline
\end{array}$$

答:
The rightmost $\lnot$ column contains the negation of the $Q$ column. Obviously this is $\lnot Q$

The $\land$ column contains the conjunction of the $P$ and rightmost $\lnot$ column. This is $P\land\lnot Q$.

The leftmost $\lnot$ column contains negation of the $\land$ column.  This is $\lnot(P\land\lnot Q)$.

So, here is the first row:

$$\def\true{\mathsf T}\def\false{\mathsf F}
\def\trueblue{\color{blue}{\mathsf T}}\def\falseblue{\color{blue}{\mathsf F}}
\begin{array}{|c:c|c:c:c:c:c|}\hline
      &      &\tiny\rm\lnot(P\land\lnot Q)      &      &\tiny\rm P\land\lnot Q      &\tiny\lnot\rm Q      &      \\\hdashline
\rm P &\rm Q &\lnot &\rm(P &\land &\lnot &\rm Q)\\\hline
\true &\true &\trueblue      &\true &\falseblue      &\falseblue      &\true \\\hdashline
\true &\false&      &\true &      &      &\false\\\hdashline
\false&\true &      &\false&      &      &\true \\\hdashline
\false&\false&      &\false&      &      &\false\\\hline
\end{array}$$


问:
If $H\le G$, there is a homomorphism $\phi:G\to A(H\backslash G)$ and $\ker\phi$ is the largest normal subgroup of $G$ contained in $H$

Let $G$ be a group. If $H\le G$ (notation for $H$ is a subgroup of $G$), then show that there is a homomorphism $\phi: G\to A(H\backslash G)$ and that $\ker\phi$ is the largest normal subgroup of $G$ contained in $H$. Here, $A(H\backslash G)$ denotes the set of all permutations on the set $H\backslash G$, of all right cosets of $H$ in $G$.

For simplicity, let $S := H\backslash G$ hereafter. We know that $S = \{Hg: g\in G\}$ where $Hg:= \{hg:h\in H\}$. Consider the following mapping which I claim is a homomorphism:
$$\phi: G\to A(S), \phi: g\mapsto f_g$$ where $$f_g:S\to S, f_g: Hx\mapsto Hxg$$
$f_g$ is a bijection (permutation on $S$) because:

Injective. $f_g(Hx) = f_g(Hy) \implies Hxg = Hyg$. This happens iff $xg(yg)^{-1}\in H$ which gives $xy^{-1}\in H$ and thus $Hx=Hy$.

Surjective. Consider a right coset $Hx$. Clearly, $f_g(Hxg^{-1}) = Hxgg^{-1} = Hx$ showing surjectivity. Note that $xg^{-1}\in G$ since $G$ is a group, so $Hxg^{-1}$ is a valid right coset to choose.

I'm stuck while showing $\phi$ is a homomorphism because:

Take $g,h\in G$. We want $\phi(gh) = f_{gh} = f_g\circ f_h = \phi(g)\phi(h)$, but: consider arbitrary $Hx\in G/H$. $f_{gh}(Hx) = Hxgh = (Hxg)h = f_h(Hxg) = f_h(f_g(Hx))$ which is $f_h\circ f_g$ and not what we need. Is this not the right homomorphism? What do I do?

Certainly $\phi(g)$ is inside $A(S)$, so that is not a problem.

Lastly, I don't know what $\phi$ is yet (though intuition says it's closely related to the above stuff), leave alone $\ker\phi$. In the second part of the theorem, we want to show that $\ker\phi$ is the largest normal subgroup, i.e. any other normal subgroup of $G$ in $H$ is surely contained in $\ker\phi$. I hope this is easy to show once $\phi$ is known!

I'd appreciate any help, thank you!

答:
Let $g \in G$, and consider the mapping $\phi_g : S \to S$ given by $Hx \mapsto Hxg^{-1}$.

$\phi_g$ is injective, for if $Hxg^{-1} = Hyg^{-1}$, then$^\color{blue}1$ $xg^{-1}(yg^{-1})^{-1} = xy^{-1}$ is in $H$, but this means precisely that $Hx = Hy$.

$\phi_g$ is surjective, since $Hx = \phi_g(Hxg)$ for all $x \in G$.

Thus $\phi_g \in A(S)$ for each $g \in G$, and then the map $\phi : G \to A(S)$ given by $g \mapsto \phi_g$ is well-defined. Finally, $\phi$ is a group homomorphism: since
$$(\forall x \in G) \quad \phi_{gh}(Hx) = Hx(gh)^{-1} = Hxh^{-1}g^{-1} = \phi_g(Hxh^{-1}) = \phi_g(\phi_h(Hx))$$ we have $\phi_{gh} = \phi_g \circ \phi_h$.

$^\color{blue}1$ Recall that $Na = Nb$ iff $ab^{-1} \in N$.


问:
Integration of dirichlet kernel
How do I find $\lim_{n \to +\infty}$ $\int_{-1/10}^{1/10} |D_n(t)|dt$?

答:
For $a \in (0, \pi)$ you have

$$\begin{aligned}
\int_{-a}^a \vert D_n(t) \vert \ dt &\ge \int_0^a \frac{\vert \sin \left((2n+1)t\right) \vert}{t} \ dt\\
&= \int_0^{(2n+1)a} \frac{\vert \sin t \vert}{t} \ dt
\end{aligned}$$

as $\int_0^{\infty} \frac{\vert \sin t \vert}{t} \ dt$ diverges $\lim\limits_{n \to \infty} \int_{-a}^a \vert D_n(t) \vert \ dt = \infty$.

Title: Integration of dirichlet kernel


问:
Is Independence of two events generalised by conditional independence of those two events?
If $\mathbb{P}(A\mid C)* \mathbb{P}(B\mid C) = \mathbb{P}(A \cap B\mid C)$ i.e A and B are independent given C has happened, then can we generalise and say that A and B are independent irrespective of C?

My explanation is that as A and B are independent given C that means given C and A happened , our beliefs about B does not change, i.e we gain no new knowledge about B's happening .

Then removing our old knowledge (i.e conditional C is removed), we have no more knowledge of C's happening . With lesser information , we gain no new insight about B , thus with A happening our beliefs or chance of B happening doesn't change implying A and B are independent in generalised way?

答:
You cannot conclude that $A$ and $B$ are independent. A simple counter-example: Let $A$ be any event with $0 <P(A)<1$ and take $A=B=C$.

$\mathbb{P}(A\mid C)=1, \mathbb{P}(B\mid C) =1$ and  $\mathbb{P}(A \cap B\mid C)=1$ in this case.


问:
Reformulation of LMI
In a paper I have read, the authors reformulated the following LMI

$$X_{t} \succeq \left[\begin{array}{cc}\alpha_{i}\left(B_{t} U+C_{t}\right)^{\top} E_{i}\left(B_{t} U+C_{t}\right) & \left( B_{t} U+C_{t}\right)^{\top} e_{i} \alpha_{i} \\\\ \alpha_{i} e_{i}^{\top}\left( B_{t} U+C_{t}\right) & e_{i}^{0} \alpha_{i}-\beta_{t}\end{array}\right]$$

into two LMIs

$$X_{t} - \left[\begin{array}{cc}\alpha_{i}P_{t}^{i} & \left( B_{t} U+C_{t}\right)^{\top} e_{i} \alpha_{i} \\\\ \alpha_{i} e_{i}^{\top}\left( B_{t} U+C_{t}\right) & e_{i}^{0} \alpha_{i}-\beta_{t}\end{array}\right] \succeq 0$$

and

$$\left[\begin{array}{cc}P_{t}^{i} & \left(B_{t} U+C_{t}\right)^{\top} E_{i}^{1 / 2} \\\\ E_{i}^{1 / 2}\left(B_{t} U+C_{t}\right) & \mathbb{I}_{n}\end{array}\right] \succeq 0 $$

where $\alpha_i$ and $e_i^0$ are scalar and $e_i$ is vector. The decision variables are $X_{t} \in S_{+}^{N d+2}$ and $P_{t}^{i} \in S_{+}^{N d+1}$.

I know the second LMI could be reformulated into $$P_{t}^{i} - \left(B_{t} U + C_{t}\right)^{\top} E_{i} \left(B_{t} U + C_{t}\right) \succeq 0$$ according to Schur complements. Just wondering why the two LMIs are equivalent to the first LMI and why the reformulation is necessary.

Any hint would be appreciated!

答: The first semidefinite constraint is not an LMI as they have quadratic term in the top-left block (assuming $U$, $B_t$ or $C_t$ is a decision variable). However, it is convex in the standard form $A-B^TCB$ (generic variable names here) so they introduce an intermediate variable to upper bound the nonlinear part, and then apply a Schur complement to arrive at two LMIs. An alternative could be to apply a Schur complement directly on the first semidefinite constraint arriving at one larger LMI but without any extra variable.

Title: Reformulation of LMI


答:
If you are an engineer, you may find the following intuition helpful.

Consider $Z_1, Z_2$ to be independent noise random variables. Let $S$ be a random signal.

Let $Y_1 = S + Z_1$ and $Y_2 = S + Z_2$ be two noisy observations of the signal $S$. These observations can be at different times or using different instruments hence we use different, independent random variables $Z_1, Z_2$ to model the noise in each observation.

Conditioned on $S = s_0$, we have $Y_1 = s_0 + Z_1$ and $Y_2 = s_0 + Z_2$. Hence, $Y_1$ and $Y_2$ are conditionally independent given $S = s_0$.

But it is clear that $Y_1$ and $Y_2$ are not independent.

Title: Is Independence of two events generalised by conditional independence of those two events?

Body:
If $\mathbb{P}(A\mid C)* \mathbb{P}(B\mid C) = \mathbb{P}(A \cap B\mid C)$ i.e A and B are independent given C has happened, then can we generalise and say that A and B are independent irrespective of C?

My explanation is that as A and B are independent given C that means given C and A happened , our beliefs about B does not change, i.e we gain no new knowledge about B's happening .

Then removing our old knowledge (i.e conditional C is removed), we have no more knowledge of C's happening . With lesser information , we gain no new insight about B , thus with A happening our beliefs or chance of B happening doesn't change implying A and B are independent in generalised way?


问:
Area of a triangle inside of other triangle
$BK$ and $CM$ are angle bisectors of angles $B$ and $C$. Let $L$ be the intersection point. If $AB=39,BC=42$ and $CA=45,$ find the area of $\triangle CKL$.

a) $756$

b) $252$

c) $126$

d) $140$

We can find the area of triangle $ABC$ by Heron's formula: $$S=\sqrt{63\cdot 21\cdot 18\cdot 24}=756.$$ I am stuck here.

答:
Use angle bisector theorem to find ratio between area of ABC ($S$) and area of CLK ($S_1$). To be specific:

BK is angle bisector of angles B, so $\frac{AK}{KC}=\frac{AB}{BC}=\frac{13}{14}$

Because of $AK+KC=AC=45 \rightarrow KC=\frac{14}{27}.AC=\frac{70}{3}$

CL is angle bisector of angles C, so $\frac{LK}{LB}=\frac{CK}{BC}=\frac{5}{9}$

$\frac{S_1}{S}=\frac{KC}{AC}.\frac{KL}{BK}=\frac{14}{27}.\frac{5}{14}=\frac{5}{27}$

Therefore, $S_1=140$


问:
Find the $o\left ( 1 \right )$ so that $2b_{n+ 1}\sim b_{n}^{2}+ o\left ( 1 \right )$

Original problem. Find the $\mathcal{o}\left(n\right)$ so that $a_n\sim\frac{\mathcal{o}\left(n\right)}{n}+\mathcal{O}\left(\frac{1}{n}\right){\rm as}\lim a_n=\infty$

Given a recursion $a_{n+ 1}= \dfrac{1}{2}\left ( a_{n}+ \dfrac{1}{a_{n}} \right )$ with $a_{1}> 0.$ First, we have $\lim a_{n}= 1,$ let $a_{n}= b_{n}+ 1,$ therefore
$$\ln\frac{1}{b_{n+ 1}}- \ln\frac{1}{b_{n}}= \ln b_{n}- \ln\left ( \frac{1}{2}\left ( b_{n}+ 1+ \frac{1}{b_{n}+ 1} \right )- 1 \right )\sim\ln\frac{2}{b_{n}}+ b_{n}+ \mathcal{O}\left ( b_{n}^{2} \right )\,{\rm as}\,n\rightarrow\infty$$
New problem. Find the $o\left ( 1 \right )$ so that $2b_{n+ 1}\sim b_{n}^{2}+ o\left ( 1 \right ).$ I still can't find such a way to approach this. To have the above formula I used the Moubinool Omarjee's technique for differential equation ${y}'= -y,$ if I chose ${y}'= -y+ \frac{y^{2}}{2}\Rightarrow x= \ln\left ( \frac{2}{y}- 1 \right ),$ that's not simple, so I need to a new way of thinking which is much more effective. Thank you so much.

Body:
This is based on the answer to the previous question. Let
$$
c: = \frac{{a_1  - 1}}{{a_1  + 1}},\quad \left| c \right| < 1.
$$
Then
$$
b_n  = a_n  - 1 = \frac{{1 - c^{2^{n - 1} } }}{{1 + c^{2^{n - 1} } }} - 1 =  - 2\frac{{c^{2^{n - 1} } }}{{1 + c^{2^{n - 1} } }}.
$$
Thus,
$$
2b_{n + 1}  - b_n^2  =  - 4\frac{{c^{2^n } }}{{1 + c^{2^n } }} - 4\frac{{c^{2^n } }}{{1 + 2c^{2^{n - 1} }  + c^{2^n } }} =  - 4c^{2^n } \left( {\frac{1}{{1 + c^{2^n } }} + \frac{1}{{1 + 2c^{2^{n - 1} }  + c^{2^n } }}} \right).
$$
By the geometric series,
$$
\frac{1}{{1 + c^{2^n } }} + \frac{1}{{1 + 2c^{2^{n - 1} }  + c^{2^n } }} = 2 + \mathcal{O}(c^{2^{n - 1} } )
$$
Therefore
$$
2b_{n + 1}  - b_n^2  \sim  - 8c^{2^n }
$$
as $n\to +\infty$. Note that by the Taylor series
$$
\frac{{1 - x}}{{1 + x}} = 1 + 2\sum\limits_{k = 1}^\infty  {( - 1)^k x^k } ,\quad \left| x \right| < 1,
$$
it holds that
$$
a_n  = 1 + 2\sum\limits_{k = 1}^\infty  {( - 1)^k c^{k2^{n - 1} } } .
$$


问:
Compute the probability that you will have to wait between $2$ to $4$ hours before you see $4$ leopards.
Suppose you are watching wild life and you expect to see a leopard once every $\frac{1}{2}$
hours. Compute the probability that you will have to wait between $2$ to $4$ hours before
you see $4$ leopards.

$\underline{Attempt}$

Exponential distribution with $\lambda =2$, so $f(x)=2e^{-2x}$

then I found $P(2<x<4)= 0.018$

Then the probability we want is $(0.018)^4$ Is it correct?

答: I concede that it is actually Gamma/Erlang. $X\sim Erlang(4, \frac 1 2)$. $P(2<X<4)\approx .12389$


答: No, $P(A|C)P(B|C)=P(A\cap B|C)$ does not imply $P(A)P(B)=P(A\cap B)$. Suppose $P(A|C)=P(B|C)=P(A\cap B|C)=1$. But suppose $P(A)=P(B)=\frac 2 3$ and $P(A\cap B)=\frac 1 3$.


问:
Vector Spaces and Projections
Q4. Suppose that $P$ is an $n \times n$ matrix such that $P^{2} = P$. Show that $\mathbb{R}^{n}$ is the direct sum of the range $R(P)$ and the nullspace $N(P)$ of $P$.  Show also that $P$ represents the projection from $\mathbb{R}^{n}$ onto $R(P)$.

A4. Again – not sure how to start. Its idemptotent... so???

Q5. Prove that for any real matrix $A$, $N(A) = \text{orthogonal complement of } \ R(A^{T})$. Prove that for any real matrix, $N(A^{T}) = \text{orthogonal complement of} \ R(A)$.

A5. I get the first bit of this. It’s the second part I’m not sure about. Let $A^{\ast} = A^{T}$. Let $R(A)^{\ast}$ be the orthogonal complement of $R(A)$. What I said was let $x \in N(A)$. Then $Ax = 0$. So $A^{\ast}(Ax)=A^{\ast}0=0$. So $A^{\ast}(Ax)=0$. So $A(A^{\ast}x)=0$ so $A^{\ast}x=0$. So x is in the nullspace of $A^{\ast}$. I’m not sure about show to show the row space bit.

Q6. Let $A$ be a real matrix and let $R(A)$ denote its range. Show that the projection of a vector $b$ onto $R(A)$ parallel to the orthogonal complement of $R(A) – R(A)^{\ast}$ - is the vector in $R(A)$ closest to b.

Let A be a real matrix and let R(A) and R(A)* denote the range of A and the orthogonal complements of R(A) respectively. Show that the projection of a vector s onto R(A) parallel to R(A)* is the vector in R(A) closest to s.

A6. I think both of these questions are the same right? No idea where to start!!! :(

Im guessing these are all similar proofs - hence its on one thread.

答:
Hint: For the first question, Observe that if $v\in R(P)$ then $Pv=v$ (why?). Deduce from here that $N(P)\cap R(P)=\{{0\}}$. Now use the theorem of dimensions for $N(P),R(P)$ and for sums and intersections.

Added after OP's comment: Did you prove the hint? If yes then take any $v\in N(P)\cap R(P)$. Then you have $0=P(v)=v$ from the hint. Hence $N(P)\cap R(P)=\{{0\}}$. Since $\dim N(P)+\dim R(P)=\dim \mathbb{R}^n=n$ and $$\dim(N(P) + R(P))=\dim N(P)+\dim R(P) - \dim(N(P) \cap R(P))=n$$ we have $N(P) + R(P)=\mathbb{R}^n$

Title: Vector Spaces and Projections


问:
Isometry for polynomials of degree at most $n$

Let $P_n$ be the set of all polynomials of at most $n$ degrees such that $p(x)=a_n x^n +\cdots + a_1x+a_0$ and $q(x)=b_n x^n + \cdots + b_1x + b_0.$ Let $\langle p, q\rangle = a_n b_n + \cdots + a_1 b_1 + a_0 b_0$. For every $q \in P_2$ define the mapping $F_q : P_3 \to P_5$ such that $F_q(p) = qp.$ For what polynomials $q$ the mapping $F_q$ is an isometry from $(P_3, \langle \cdot, \cdot\rangle)$ to $(P_5, \langle \cdot, \cdot\rangle)$?

Every $q$ has the following form $q(x)= b_2x^2+b_1x+b_0$ so I have that $F_q(p) =(b_2x^2+b_1x+b_0)(a_3x^3+a_2x^2+a_1x+a_0)$ which indeed results in a polynomial of degree $5$, but I'm not sure how to find what polynomials makes this map isometric. I would need to have that $$\langle F_q(p), F_q(p')\rangle_{P_{5}} = \langle p, p'\rangle_{P_3}$$ where $p,p' \in P_3$? Any hints would be appreciated.

答:
Hint

Write the matrix $M_q$ of $F_q$ in the orthonormal basis $\{1, x, \dots, x^3\}$ of $P_3$ and  $\{1, x, \dots, x^5\}$ of $P_5$  and use the condition $M_q^T M_q = I_4$ that is equivalent to say that $M_q$ is the matrix of a linear isometry for the inner product $\langle \cdot, \cdot \rangle$.

Note: the matrix of $M_q$ is simple to find.

$$M_q=\begin{pmatrix}
b_0 & 0 & 0 & 0\\
b_1 & b_0 & 0 & 0\\
b_2 & b_1 & b_0 & 0\\
0 & b_2 & b_1 & b_0\\
0 & 0 & b_2 & b_1\\
0 & 0 & 0 & b_2
\end{pmatrix}$$


答:
I do not agree with the solution $X\sim Exp\left(\frac{1}{2}\right)$ is the distribution to see 1 leopard. The distribution to see 4 leopards is a Sum of 4 independent exponentials, thus it is

$$Y\sim Gamma\left(4;\frac{1}{2}\right)=\chi_{(8)}^2$$

Thus

$$\mathbb{P}[2<Y<4]=0.9810-0.8751=0.1239$$

Body:
Suppose you are watching wild life and you expect to see a leopard once every $\frac{1}{2}$
hours. Compute the probability that you will have to wait between $2$ to $4$ hours before
you see $4$ leopards.

$\underline{Attempt}$

Exponential distribution with $\lambda =2$, so $f(x)=2e^{-2x}$

then I found $P(2<x<4)= 0.018$

Then the probability we want is $(0.018)^4$ Is it correct?


问:
Affine-local Computation of Scheme-Theoretic Image for Reduced Schemes
The following is part of an argument in Vakil's Rising Sea, p. 238.

Let $\pi : X \to \operatorname{Spec}{B}$ be a morphism of schemes with $X$ reduced and $g \in B$. Consider the morphism $$\pi^{\#}_{B_g} : B_g \to \Gamma(\operatorname{Spec}{B_g}, \pi_*\mathcal{O}_X).$$ If $\pi^{\#}_{B_g}(r/g^n) = 0$, then $\pi^{\#}_B(rg) = 0$.

Vakil sketches the following: Because $r$ vanishes on $\operatorname{Spec}{B_g}$ and $g$ vanishes on $V(g)$, we get that $rg$ vanishes on $\operatorname{Spec}{B} = \operatorname{Spec}{B_g} \cup V(g)$.

What does this mean formally? The set $V(g)$ is not open, so we cannot just apriori restrict the action of the sheaf morphism to $V(g)$. So possibly $\pi^{\#}(g_{\mathfrak{p}}) = 0$ for $\mathfrak{p} \in V(g)$ but I'm not sure why or if this is true.

答:
In this case, Vakil is using "vanishing" to mean evaluating to $0$ in the residue field at every point of $V(g)$. Since $V(g)\subset\operatorname{Spec}B$ as a set consists of the prime ideals containing $g$, the evaluation map $B\to B_p/pB_p$ sends $g$ to zero.

(Let me point out that the reason this suffices for his purposes in this proof is that on a reduced scheme, any function which evaluates to zero in every residue field is actually zero.)


问:
Solution for nonhomogenous linear matrix ODE
For matrices $X, A, B\in \mathbb{R}^{n\times n}$, $X = X^\top, B=B^\top$, I am looking to find an explicit solution to the following linear first-order matrix ODE where $A$ and $B$ are constant;

$
\dot{X} = AX + XA^\top + B
$

I know that the solution for the homogenous part, $\dot{X} = AX + XA^\top $, is

$X(t) = \exp(tA)X(0)\exp(tA^\top)$

But i don't know how to find the solution for the non-homogenous form. Any help is appreciated.

答:
Hint.

Apply Lagrange's variation of constants. Make a particular solution $X_p = e^{At}C(t)(e^{At})'$ and after substitution into the full ODE we have

$$
\dot C(t) = e^{-At}B( e^{-At})'
$$

etc.


问:
Why do we need to invoke the cancellation laws to prove uniqueness of solution of ax = b in group theory?
Quoting Introduction to Analytic Number Theory by Tom M. Apostol:

Theorem 6.1 Cancellationi laws. If elements $a$, $b$, $c$ in $G$ satisfy
$$ ac = bc $$ or $$ ca = cb $$,
then $ a = b $.

Theorem 6.2 (d) For all $a$ and $b$ in $G$ thee equation $ax = b$ has the unique solution $ x = a^{-1}b$; the equation $ya = b$ has the unique solution $y = ba^{-1}$.

Quoting the proof of Theorem 6.2 (d) now:

Again by associativity we have

$$ a(a^{-1}b) = (aa^{-1})b = b\hspace{1.5em} \text{and}\hspace{1.5em} (ba^{-1})a = b(a^{-1}a) = b. $$
The solultions are unique because of the cancellation laws.

Why do we have to invoke the cancellation laws the prove that the solutions are unique?

In other words, which step of the proof will not work if we do not invoke the cancellation laws?

答:
In fact we don't need to use any additional argument using cancellation to prove uniqueness. Rather it follows immediately from the obvious existence proof, i.e. $\,ax = b\!\iff\! x = a^{-1}b,\,$ via scaling by $\,a^{\pm1}.\,$ Thus if $x_1,x_2\,$ are both roots then $\,x_1 = a^{-1}b = x_2,\,$ so $\,x_1 = x_2\,$

Remark $ $ This point is often misunderstood.  See here and here for much further discussion.


问:
On a sum related to alternating sign matrices
I'm trying to prove that
$$A_{n,k} = \binom{n+k-2}{k-1}\frac{(2n-k-1)!}{(n-k)!}\prod_{j=0}^{n-2}\frac{(3j+1)!}{(n+j)!}$$
implies
$$A_n = \sum_{k=1}^nA_{n,k}=\prod_{j=0}^{n-1}\frac{(3j+1)!}{(n+j)!}.$$
In other words, I want to show that
$$\sum_{k=1}^n\binom{n+k-2}{k-1}\frac{(2n-k-1)!}{(n-k)!}=\frac{(3n-2)!}{(2n-1)!}.$$
My first thought was that this would be easy, but now I'm not so sure. We can write $\frac{(2n-k-1)!}{(n-k)!}$ as $\binom{2n-k-1}{n-1}(n-1)!$ but I don't know what good that will do. Any help is greatly appreciated!</p>

答:
Let's, indeed, rewrite $\frac{(2n-k-1)!}{(n-k)!}$ as $\binom{2n-k-1}{n-1}(n-1)!$ and divide both sides by $(n-1)!$. Now we want to prove that
$$
\sum_{k=1}^n\binom{n+k-2}{k-1}\binom{2n-k-1}{n-1}=\binom{3n-2}{2n-1}
$$
or, equivalently, that
$$
\sum_{s=0}^{n-1}\binom{n-1+s}{n-1}\binom{2n-2-s}{n-1}=\binom{3n-2}{2n-1}
$$
— which is just a (Chu-)Vandermonde identity ($\sum\binom{a+s}c\binom{b-s}d=\binom{a+b+1}{c+d+1}$ — see e.g. identity (5.26) in «Concrete Mathematics»).</p>


问:
Want to find $A\subseteq M$ such that $(\partial A)^\circ = M$
$(M,d)$ is a metric space. I have been asked to produce $A\subseteq M$, such that $$(\partial A)^\circ  = M$$

Could someone help me find such a set? Here are my thoughts so far:

We want to find $A\subseteq M$ such that $(\partial A)^\circ = M$. For all $x\in A$ gives $x\in M$ and hence $x\in (\partial A)^\circ$. $x\in (\partial A)^\circ$ means that $x\in \partial A$ and there exists $r_x > 0$ such that $B(x,r_x) \subseteq \partial A$. $x\in\partial A$ additionally gives $B(x,r_x)\cap A\neq\emptyset$ and $B(x,r_x)\cap A^c\neq\emptyset$. We get $\partial A\cap A \neq \emptyset$ and $\partial A\cap A^c\neq \emptyset$. All this is good - but how do I find some $A$ that satisfies these properties?

Thanks a lot!

答: You cannot find such $A$ for all metric spaces, as $(\partial A)^\circ = M$ iff $\partial A = M$ iff $\overline{A}=M$ (so $A$ is dense) and $A^\circ = \emptyset$. So if $M$ has an isolated point, this point is in all dense sets and in their interior too, so no such $A$ can exist. E.g. in a set with the discrete metric all subsets $A$ have $\partial A = \emptyset$. It is well-known that if $M$ has no isolated points, such an $A$ does always exist; then we can split $M$ into two disjoint dense subsets and each of them well do the job.


问:
A exercise in basic algebra Cohn
This question appears in Basic Algebra written by Cohn and really stuck on me:

Let $G$ be a group and $K, L, N$ be subgroups such that $N$ is a normal subgroup of $K$, $K$ is a normal subgroup of $G$, $[L, K] = K$, $KL = G$, and $[L, N] = 1$. Show that $N$ lies in the center of $K$.

Note: If $A$ and $B$ are subgroups of a group $G$, then $[A, B]$ denote the subgroup of $G$ generated by all commutators $[a, b]$, where $a ∈ A$, $b ∈ B$.

Can anyone solve it out?

答:
As $K=[K,L]$ each element $k$ of $K$ is a product of commutators like $x:=k_1^{-1}(l^{-1}k_1l)$.

For each $n\in N$ we will have that $n^{x}=n^{k_1^{-1}l^{-1}k_1l}$.

Now use the facts  that $N$ is normalised by $K$ and that $L$ commutes with $N$ to simplify this,
$$
n^{x}
=n^{k_1^{-1}l^{-1}k_1l}
=n^{k_1^{-1}k_1l}
=n^{l}
=n.
$$

Hence we will have $n^{k}=n$ for all $k\in K$ and $n\in N$.

Title: A exercise in basic algebra Cohn


问:
How do I show that $e^{x+1} > 2x + 2$ for all $x>0$?
Hello I would need help to solve the equation above.
So far I have tried letting $$f(x) = e^{x+1}- 2x -2$$

$$f'(x)= e^{x+1} -2$$

$$e^ {x+1} > 2$$

$$x> \ln 2-1$$

but I do not know how to continue.

答:
Hint: at $x=0$,
\begin{align}
e^{1+x}&=e \text{ and } 2x+2=2 \, .
\end{align}
Then argue that the gradient of $e^{1+x}$ is greater than that of $2x+2$ for $x\geq0$.


问:
Example of a weak inverse that is not an inverse

In the theory of semigroups, a weak inverse of an element $x$ in a semigroup $(S, \cdot)$ is an element $y$ such that $y \cdot x \cdot y = y$.

What is an example of this that is not also an inverse?

答:
Consider a set $X$ and call $S$ the monoid of the partial functions $g:X\dashrightarrow X$, i.e. $S$ is the set of functions with codomain $X$ and domain a subset of $X$, endowed with restricted composition $f\circ g:=f\circ \left.g\right\rvert_{g^{-1}[\operatorname{dom} f]}$. Then, the empty function $\varnothing:\varnothing \to X$ is a weak inverse of all elements (even of those which do have an inverse).

If $X\ne\varnothing$, then the empty function is not an actual inverse of any function.

More generally, an absorbing element in any semigroup has the same property.


答:
Let $f(t) = e^{t}- 2t$

You have to show that $f(t) > 0$ for $t>1.$

We have

$$f(1)=e-2>0$$

and

$$f'(t)=e^t-2 >e-2>0$$

for $t>1.$

Hence $f$ is strictly increasing on $[1,\infty).$ Thus $f(t)>f(1)>0$ for $t>1.$

Title: How do I show that $e^{x+1} > 2x + 2$ for all $x>0$?

Body:
Hello I would need help to solve the equation above.
So far I have tried letting $$f(x) = e^{x+1}- 2x -2$$

$$f'(x)= e^{x+1} -2$$

$$e^ {x+1} > 2$$

$$x> \ln 2-1$$

but I do not know how to continue.


答:
This is my approach

The mapping $F_q$ is an isometry from $(P_3, \langle \cdot, \cdot\rangle)$ to $(P_5, \langle \cdot, \cdot\rangle)$ mean $\forall p,p' \in P_3$ $$\langle F_q(p), F_q(p')\rangle_{P_{5}} = \langle p, p'\rangle_{P_3}$$

So, I think, we can make it easy by choose particular values of $p,p' \in P_3$ and make
equations of $b_0,b_1,b_2$

With $p=1,p'=1 \rightarrow 1=b_0^{2}+b_1^{2}+b_2^{2}$

With $p=1,p'=x+1 \rightarrow 1=b_0^{2}+b_1^{2}+b_2^{2}+2b_1b_2$

With $p=1,p'=x^{2}+1 \rightarrow 1=b_0^{2}+b_1^{2}+b_2^{2}+b_0b_2$

With $p=x,p'=x+1 \rightarrow 1=b_0^{2}+b_1^{2}+b_2^{2}+b_0b_1+b_1b_2$

Therefore, $b_0b_1=b_1b_2=b_0b_2=0$, $1=b_0^{2}+b_1^{2}+b_2^{2}$
$\rightarrow (b_0,b_1,b_2)$ must have 2 values = 0, and the other equal 1 or -1

So, q can be $1,-1,x,-x,-x^{2},x^{2}$ ( Test back and it's true )

Title: Isometry for polynomials of degree at most $n$

Body:
Let $P_n$ be the set of all polynomials of at most $n$ degrees such that $p(x)=a_n x^n +\cdots + a_1x+a_0$ and $q(x)=b_n x^n + \cdots + b_1x + b_0.$ Let $\langle p, q\rangle = a_n b_n + \cdots + a_1 b_1 + a_0 b_0$. For every $q \in P_2$ define the mapping $F_q : P_3 \to P_5$ such that $F_q(p) = qp.$ For what polynomials $q$ the mapping $F_q$ is an isometry from $(P_3, \langle \cdot, \cdot\rangle)$ to $(P_5, \langle \cdot, \cdot\rangle)$?

Every $q$ has the following form $q(x)= b_2x^2+b_1x+b_0$ so I have that $F_q(p) =(b_2x^2+b_1x+b_0)(a_3x^3+a_2x^2+a_1x+a_0)$ which indeed results in a polynomial of degree $5$, but I'm not sure how to find what polynomials makes this map isometric. I would need to have that $$\langle F_q(p), F_q(p')\rangle_{P_{5}} = \langle p, p'\rangle_{P_3}$$ where $p,p' \in P_3$? Any hints would be appreciated.


问:
Strategy while solving quadratics by splitting the middle term
While solving a quadratic equation in one variable by "splitting the middle term", sometimes I just can't arrive at the two numbers I need and it takes a lot of time.

For eg: $x^2 -124x+1344 = 0$

For solving the above equation by splitting the middle term, I first multiplied the coefficient of $x$ which is $1$ in this case by $1340$ (the constant term). I prime factorised the product, and now I have to pick a group of two numbers such that their sum is $-124$ (coefficient of $x$).

Do we always have to pick random numbers? If only hit and trial works here, what must be our thinking process for picking the other pair of number if the former pair of number doesn't work?

Eg: $x^2 -124x+1344 = 0$

Two numbers from the prime factorization of ($1 \cdot 1344$) is $-64$ and $-21$ but they do not add up to $-124$. What should be my strategy to pick a different pair of numbers?

(I know about the quadratic formula, but it becomes too lengthy sometimes)

答:
You need to find two numbers $a$ and $b$ such that
\begin{align}
a+b &= -124 \\
a \times b &= 1344 \, .
\end{align}
If you want a way of doing this that doesn't require trial and error, then let $a=-62+k$ and $b=-62-k$ for some constant $k$ (this step involves halving the number that $a$ and $b$ need to add up to, in this case $-124$). Notice that
$$
a+b=(-62+k)+(-62-k)=-124 \, .
$$
Furthermore,
$$
a \times b= (-62+k)(-62-k)=1344 \, ,
$$
and so
\begin{align}
62^2-k^2 &= 1344 \\
3844-k^2 &= 1344 \\
k^2 &= 2500 \\
k &= 50 \, .
\end{align}
Hence, $a=-62+50=-12$ and $b=-62-50=-112$. We find that
$$
x^2-124x+1344  = (x+a)(x+b)=(x-12)(x-112)
$$
and so $x=12$ or $x=112$.


答:
You can use the known inequality $e^x > x+1$.
Multiplying by $e$:  $e^{x+1}> ex+e$.
Now, I would like to use that $e>2$ to conclude, but maybe $x+1<0$. However, in this case, since $exp$ is a positive function, the inequality holds anyways, so I can conclude in the following way:

If $x+1 < 0$, then since $e^{x+1}>0$, we have $e^{x+1}>2x+2$.

In the other case, $2x+2<ex+e<e^{x+1}$.


问:
Question about non-separable ODE
I have a problem with my assignment. The problem is as follows:

$(x^2+y^2e^\frac{x}{y})\frac{dx}{dy}=xy$

I don't think this is an exact ODE, and I have tried to solve it using integrating-factor method but I cannot find the result. I also tried to multiply each side by $y^{-2}$ and set u = $\frac{x}{y}$ but unable to solve it properly.

I would really appreciate if anybody could help me with this problem. It would be a very big help for me. Thank you very much.

答:
We have $$(x^2+y^2e^\frac{x}{y})\frac{dx}{dy}=xy$$
Dividing through by $y^2$ as you suggested yields
$$\left(\frac{x^2}{y^2}+e^\frac{x}{y}\right)\frac{dx}{dy}=\frac{x}{y}$$
Again, we will use your suggestion: let $$u=\frac{x}{y}\iff x=uy$$
We can differentiate both sides implicitly with respect to $y$:
$$\frac{dx}{dy}=u+y\frac{du}{dy}$$
But we know that $$\frac{dy}{dx}=1\Big/\frac{dx}{dy}$$
Hence, substituting back into our original equation, we have
$$\begin{align}
\frac{u^2+e^u}{u+y\frac{du}{dy}}=u&\iff u+y\frac{du}{dy}=\frac{u^2+e^u}{u}=u+u^{-1}e^u\\
&\iff y\frac{du}{dy}=u^{-1}e^u=\frac{e^u}{u}\\
&\iff \frac{du}{dy}=\frac{e^u}{uy}\\
\end{align}$$
We are now in a position to separate the variables! We have:
$$\int ue^{-u}~du=\int \frac{1}{y}~dy$$
Integrating by parts on the left we eventually come to
$$-ue^{-u}-e^{-u}=\ln\lvert y\rvert +C=\ln\lvert y\rvert +\ln k$$
So finally
$$-\frac{x}{y}e^{-\frac{x}{y}}-e^{-\frac{x}{y}}=\ln \lvert ky\rvert$$

Title: Question about non-separable ODE


问:
Find the radius of convergence of two series
I have to tell the convergence-radius of the following series:

$\sum_{n=0}^\infty n^5x^n$

Here I thought about using Euler and I get $\lim \frac{1}{(1+\frac{1}{n})^5} = 1$ So the convergence-radius is $1$?

Instead I could have used the Cauchy-Hadamard and I would get:
$r = \frac{1}{\limsup (\sqrt[n](|n^5|)}$ - what now?

$\sum_{n=1}^\infty (x+1)^n\frac{(-1)^n}{2^nn^2}$

What exactly should I do here?

Body:
Hints:

$$\bullet\;\;\;\;\sqrt[n]{n^5}=n^{5/n}=\left(\sqrt[n]n\right)^5\xrightarrow[n\to\infty]{}1^5=1$$

$$\bullet\;\;\;\left|\frac{a_{n+1}}{a_n}\right|=\left|\;\frac{(x+1)^{n+1}}{2^{n+1}(n+1)^2}\cdot\frac{2^nn^2}{(x+1)^n}\;\right|=\frac12\,|x+1|\frac{n^2}{(n+1)^2}\xrightarrow[n\to\infty]{}\frac12|x+1|<1\iff$$

$$\iff |x+1|<2\iff -3<x<1$$


问:
Integral form of balance law
Let $\rho: \mathbb{R} \times [0, \infty) \to \mathbb{R}$ be the density, $f: \mathbb{R} \to \mathbb{R}$ the flux of the density and $g: \mathbb{R} \to \mathbb{R}$ the source (or loss) term.
The equation

$$
\partial_t\rho(x,t) + \partial_x f(\rho(x,t)) + g(\rho(x,t)) = 0
$$

is called one-dimensional balance law.

For $g=0$ the balance law reduces to a conservation law, for which the integral-form is given by

$$
\int_{x_1}^{x_2} \rho(x,t_2) \,\mathrm{d}x
-
\int_{x_1}^{x_2} \rho(x,t_1) \,\mathrm{d}x
=
\int_{t_1}^{t_2} f(\rho(x_1,t)) \,\mathrm{d}t
-
\int_{t_1}^{t_2} f(\rho(x_2,t)) \,\mathrm{d}t.
$$

which implies that the change of mass inside an interval $[x_1,x_2]$ during a time interval $[t_1,t_2]$ is defined by the difference of the inflow and the outflow at $x_1$ and $x_2$ during this time interval.

Is there any literature for a derivation of the integral form for the balance law with $g \neq 0$ inclusive physical explanation? Any hints or suggestions are highly appreciated.

答:
I would say that you just need to proceed the same way to derive the balance law. For interpretations, assume that $f_\rho$, $g$ are both positive. If we integrate the local form of the balance law $$\rho_t + f(\rho)_x = -g(\rho)$$ over the domain $x\in [x_1, x_2]$ and times $t \in [t_1, t_2]$, we find
$$
\left[\int_{x_1}^{x_2} \rho\, \text d x \right]^{t_2}_{t_1} = \left[\int_{t_1}^{t_2} f(\rho)\, \text d t \right]^{x_1}_{x_2} - \underset{[x_1,x_2] \times [t_1,t_2]}{\iint} g(\rho)\, \text{d}x \text d t \, .
$$
The left-hand side is the variation of mass $\int \rho \, \text d x$ between the times $t_1$ and $t_2$. The right hand side includes a first term which represents the variation of mass flux between both boundaries of the spatial domain (the mass entering $[x_1,x_2]$ at $x=x_1$ minus the mass which has left at $x=x_2$). The last term represents the loss of mass over the domain considered here, or if you prefer, $\iint -g \, \text d x \text d t$ represents the mass production.

Title: Integral form of balance law


问:
A concrete meaning of quasi-compact and quasi-separated schemes
This is exercise 5.1.H in Vakil's note 'FOAG':

Show that a scheme $X$ is quasi-compact and quasi-separated iff $X$ can be covered by a finite number of affine open subsets, any two of which have intersection also covered by a finite number of affine open subsets.

My idea is as follows:

If $X$ is quasi-compact and quasi-separated, then $X$ can be covered by a finite number of affine open subsets, any two of which have intersection also covered by a finite number of affine open subsets by the previous exercise 5.1.F:

A scheme is quasi-separated iff the intersection of any two affine open subsets is a finite union of affine open subsets.

This is easy and I have solved it. My question is below:

Conversely, we set the scheme $X$ can be covered by a finite number of affine open subsets, any two of which have intersection also covered by a finite number of affine open subsets. Let $X=\bigcup_{i=1}^nU_i$ where $U_i$ are affine open subsets such that any two of which have intersection also covered by a finite number of affine open subsets. So $X$ is obviouly quasi-compact. We just need to prove that the intersection of any two affine open subsets is a finite union of affine open subsets by the previous exercise 5.1.F as above.

(*) Take any affine open subset $Y$ and if we can prove that $Y\cap U_i$ can be written as a finite union of affine open subsets, then the question is finished:

Take $A,B$ are two affine open subsets of $X$,then $A\cap B=\bigcup_{i=1}^n((A\cap U_i)\cap(B\cap U_i))$. Use (*) we know that $A\cap U_i=\bigcup_{\alpha}C_{\alpha},B\cap U_i=\bigcup_{\beta}D_{\beta}$ where $\alpha,\beta$ are finite and $C_i,D_i$ are affine open subsets. Then we find that
$$(A\cap U_i)\cap(B\cap U_i)=\left(\bigcup_{\alpha}C_{\alpha}\right)\cap\left(\bigcup_{\beta}D_{\beta}\right)\subset U_i.$$
Since $U_i$ is affine, then it is quasi-separated, then $$\left(\bigcup_{\alpha}C_{\alpha}\right)\cap\left(\bigcup_{\beta}D_{\beta}\right)=\bigcup_{\alpha,\beta}(C_{\alpha}\cap D_{\beta})$$ is a finite union of affine open subsets, then so is $A\cap B$.

So the question is: how to prove (*)? or how to prove the exercise without using this?

答: I’m sorry, this is a trivial problem. Just need to prove that $Y\cap U_i$ is quasi-compact. This is very easy.


答:
As previously mentioned, you need to find two integers $a,b$ such that $a+b=124, ab = 1344$ due to Vieta's formulas. Now $124$ is divisible by $4$ and $1344$ is divisible by $4^2 = 16$, so we can write $a/4 + b/4 = 124/4, \frac{a}{4} \frac{b}{4} = 1344/4$, and changing variables $m = a/4, n = b/4$ gives $m + n = 31, mn=84$.

If you check the factor pairs of $84$ systematically, for example $(m,n) = (84, 1) = (42, 2)$ and compute their sum, you will find that $m = 28, n = 3$ or vice versa. Thus convert back to the roots by multiplying by $4$ to get $x = 112, 12$.

Title: Strategy while solving quadratics by splitting the middle term

Body:
While solving a quadratic equation in one variable by "splitting the middle term", sometimes I just can't arrive at the two numbers I need and it takes a lot of time.

For eg: $x^2 -124x+1344 = 0$

For solving the above equation by splitting the middle term, I first multiplied the coefficient of $x$ which is $1$ in this case by $1340$ (the constant term). I prime factorised the product, and now I have to pick a group of two numbers such that their sum is $-124$ (coefficient of $x$).

Do we always have to pick random numbers? If only hit and trial works here, what must be our thinking process for picking the other pair of number if the former pair of number doesn't work?

Eg: $x^2 -124x+1344 = 0$

Two numbers from the prime factorization of ($1 \cdot 1344$) is $-64$ and $-21$ but they do not add up to $-124$. What should be my strategy to pick a different pair of numbers?

(I know about the quadratic formula, but it becomes too lengthy sometimes)


问:
Drawing from an urn - probability of drawing maximum 2 of each color.
I have an urn with 100 balls; 10 colors with each 10 balls.

Now I draw 10 balls from the urn without replacement - I want to calculate the probability of getting 3 or more of any color.

How do I calculate this? My problem is that for example the event "more than 3 greens" and "more than 3 yellows" are not exclusive, there are possible draws which fit into both categories. So I cant the hypergeometric distribution and multiply it by 10.

I feel like this question is school level.... I got a math degree, so the answer doesnt need to be super basic, I just suck at stochastics. Help greatly appreciated.

答:
Observe that the event you are looking for is complementary to the event "there are no more than two balls of the same color". In terms of the generating functions the number of combinations corresponding to the latter event is:
$$
N_2=[x^{10}]\left(1+\binom{10}1x+\binom{10}2x^2\right)^{10},
$$
where the function $[x^n]$ extracts the coefficient at $x^n$ in the following expression.

Hence the probability in question
is
$$
1-\frac{N_2}{\binom{100}{10}}\approx0.53.
$$

EDIT:

In general case we can assume that the balls are of $m$ colors, and there are $N_i$ balls of $i$-th color.
The question is: how many combinations of $n$ drawn balls contain no more than $k_i$ balls of the $i$-th color ($k_i$ are given for each $i$, they must not be the same). The answer to the question is:
$$
[x^n] \prod_{i=1}^m \sum_{k=0}^{k_i}\binom{N_i}k x^k.
$$


问:
Prove that $ABCD$ is a trapezoid
$ABCD$ is a cyclic quadrilateral inscribed of circle $O$. The tangent line of circle $O$ at $B$ intercepts $CD$ at $K$. The tangent line of circle $O$ at $C$ intercepts $AB$ at $M$.

if $AB = BM$  and $CD=CK$, prove that $ABCD$ is a trapezoid.

I was going down the route of:

$\frac{BM}{\sin \angle BCM} = \frac{BC}{\sin \angle BMC}$, so $BM = \frac{BC \sin \angle BCM}{\sin \angle BMC}$

$\frac{AB}{\sin \angle ACB} = \frac{BC}{\sin \angle BAC}$, so $AB = \frac{BC \sin \angle ACB }{\sin \angle BAC}$

and $\angle BCM = \angle BAC$

$\frac{BM}{AB} = \frac{\sin \angle BCM \sin \angle BAC}{\sin \angle BMC \sin \angle ACB} = 1$

but the relationship is awkward here.

答:
It's enough to prove that $BD=AC$.

Observe that, $\triangle BCK\sim \triangle DBK$ and $\triangle CBM\sim \triangle ACM$.

Hence, $\frac{BK}{DK}=\frac{CK}{BK}$

$\Rightarrow  BK=CK\sqrt{2}$

Thereafter, $\frac{BD}{BC}=\frac{BK}{CK}=\sqrt{2}$

Similarly yield $\frac{AC}{BC}=\sqrt{2}$ and thus $BD=BC\sqrt{2}=AC$.

Title: Prove that $ABCD$ is a trapezoid


问:
Subgroups of product of two finite groups with coprime order
I want to explain the following proposition (if true).

Let $G_1$ and $G_2$ be finite groups with coprime orders and let $H$ be a subgroup of $G_1 \times G_2$. Then there exists subgroups $H_1$ and $H_2$ of $G_1$ and $G_2$, respectively, so that $H = H_1 \times H_2$.

Of course, if familiar, it can be handled by the the Gourat's lemma. Note that it suffices to prove that 'if the images of $H$ under the natural projections are $G_1$ and $G_2$, then $H=G_1 \times G_2$'. By the lemma the quotients of $G_j$ by the kernels are isomorphic to each other and hence, should be trivial because of the coprime orders. This implies that $H = G_1 \times G_2$.

I want to explain the proposition to the students of undergraduate sophomore level. Can you help to develop a suitable explanation or proof of the proposition? (By the way, they are familiar with Sylow blah, I guess)

PS: Sadly(?) I just found the following question. This post may be treated as duplicate.

$ |G_1 |$ and $|G_2 | $ are coprime. Show that $K = H_1 \times H_2$

答:
Here is a completely elementary proof, which does not use nine-lemma, Goursat's lemma or Sylow theorems.

Let $H_1$ the set of elements $h_1\in G_1$ such that $H$ contains an element of the form $(h_1,g_2)$ for some $g_2\in G_2$, and similarly, let $H_2$ the set of elements $h_2\in G_2$ such that $H$ contains an element of the form $(g_1,h_2)$ for some $g_1\in G_1$.

Clearly, $H_i$ is a subgroup of $G_i$ and $H\subset H_1\times H_2$.

Let us prove the reverse inclusion. Let $(h_1,h_2)\in H_1\times H_2$.

By definition, $(h_1,g_2)\in H$ and $(g_1,h_2)\in H$ for some $g_i\in G_i$.

Let $m_i$ the order of $g_i$. Since $H$ is a subgroup of $G_1\times G_2$, we get that $(h_1^{am_2},1), (1,h_2^{bm_1})\in H$ for all $a,b\in\mathbb{Z}$.

Since $m_2$ divides the order of $G_2$, it is coprime to the order of $G_1$, hence to the order of $h_1$. Hence $uo(h_1)+vm_2=1$ for some $u,v\in\mathbb{Z}$.

Thus $h_1^{vm_2}=h_1$ , so taking $a=v$ above yields $(h_1,1)\in H$. Similarly, $(1,h_2)\in H$, and finally $(h_1,h_2)\in H$, that is $H_1\times H_2\subset H$. a

Consequently, $H=H_1\times H_2$.


问:
Maximum and minimum of the function $xy+z^2$
Find the maximum and minimum values of the function $f(x,y,z)=xy+z^2$ in the circumference obtained by intersections between the sphere $x^2+y^2+z^2=4$ and the plane $y-x=0$.

I did Lagrange and found the points $(0,0,\pm2)$ I would call them maximum values and that the function have no minimum values. Not sure if that's correct... Thanks.

Edit: Lagrange calculations:

$\begin{cases}f_x\rightarrow y=4x\gamma\\
f_y\rightarrow x=0\\
f_z\rightarrow 2z=2z\gamma\\
2x^2+z^2=4
\end{cases}$

答:
The Lagrange calculation as you've written it is mixing equations from both the one-constraint and two-constraint versions.

If you choose to combine the information from the spherical constraint ($ \ x^2 \ + \ y^2 \ + \ z^2 \ = \ 4 \ $ ) and the planar constraint ( $ \ y \ - \ x \ = \ 0 \ \ \Rightarrow \ \ x \ = \ y \ $ ) , the function to be extremized becomes $$ \ f(x,y,z) \ = \ xy \ + \ z^2 \ \ \rightarrow  \ \ \phi (x,z) \ = \ x^2 \ + \ z^2 \ \ , $$ subject to the single constraint $$ \ g(x,y,z) \ = \ x^2 \ + \ y^2 \ + \ z^2 \ = \ 4 \ \ \rightarrow \ \ \gamma (x,z) \ = \ 2x^2 \ + \  z^2 \ = \ 4 \ \ . $$

The Lagrange equations $ \ \nabla \phi \ \ = \ \ \lambda \ \nabla \gamma \ $ are then

$$ \ \phi_x \ = \ \lambda \ \gamma_x \ \ \Rightarrow \ \ 2x \ = \ \lambda \ · \ 4x \ \ \ , \ \ \ \phi_z \ = \ \lambda \ \gamma_z \ \ \Rightarrow \ \ 2z \ = \ \lambda \ · \ 2z \ \ .  $$

It is generally not a good idea to solve such a system by simply cancelling factors across each equation; one should instead bring all terms to one side of each equation and consider the products of factors, thus:

$ ( \ 2 \ - \ 4 \lambda \ ) \ x \  =  \ 0 \ \ \Rightarrow \ \ \lambda \ = \ \frac{1}{2} \ \ $ or $ \ \ x \  =  \ 0 \ \ . $  In the latter case, we have

$$ x \  =  \ 0 \ \ \Rightarrow \ \ y \ = \ x \ = \ 0 \ \ , \ \ z^2 \ = \ 4 \ \ \Rightarrow \ \ z \ = \ \pm 2 \ \ , $$

while in the former case,

$$ \lambda \ = \ \frac{1}{2} \ \ \Rightarrow \ \  2z \ = \ \frac{1}{2}  ·  2z \ = \ z \ \ \Rightarrow \ \ z \ = \ 0 \ \ \Rightarrow \ \ 2x^2 \ = \ 4 \ \ \Rightarrow \ \ x \ = \ \pm \sqrt{2} \ = \ y \ \ .  $$

Hence, there are four critical points, $  \ ( \ 0 \ , \ 0 \ , \ \pm 2 \ ) \ \ ,  $ at which  $ \ f \ = \ 2^2 \ = \ 4 \ \ , \ $  and  $ \ ( \ \pm \sqrt{2} \ , \ \pm \sqrt{2} \ , \ 0 \ ) \ \ , $  where $ \ f \ = \ (\pm \sqrt{2}) · (\pm \sqrt{2}) \ = \ 2 \ \ . $

$ $

Applying the two constraints individually, which then uses all three coordinate variables, we define the second constraint function $ \ h(x,y,z) \ = \ y \ - \ x \ = \ 0 \ \ . $  The corresponding Lagrange equations $ \ \nabla f \ \ = \ \ \lambda \ \nabla g  \ + \ \mu \ \nabla h \ $ are now

$$  f_x \ = \ y \ = \ \lambda \ · \ 2x \ + \ \mu \ · \ (-1) \ \ \ , \ \ \ f_y \ = \ x \ = \ \lambda \ · \ 2y \ + \ \mu \ · \ 1 \ \ \ , \ \ \  f_z \ = \ 2z \ = \ \lambda \ · \ 2z \ + \ \mu \ · \ 0  \ \ .  $$

This time, we'll start with the third equation, $ ( \ 2 \ - \ 2 \lambda \ ) \ z \  =  \ 0 \ \ \Rightarrow \ \ \lambda \ = \ 1 \ \ $ or $ \ \ z \  =  \ 0 \ \ . $  The latter case will bring us directly to the critical points $ \ ( \ \pm \sqrt{2} \ , \ \pm \sqrt{2} \ , \ 0 \ ) \ \ . $  The case for $ \ \lambda \ = \ 1 \   $ calls for a bit of "playing around" to find a satisfactory approach (as with other sorts of systems of equations, methods of solution are not always obvious).  The third equation, $ \ 2z \ = \ 1 · 2z \  $ leaves $ \ z \ $ unspecified for the moment.  The other equations become $ \ y \ = \ 1 · 2x \ - \ \mu \ $ and $ \ x \ = \ 1 · 2y \ + \ \mu \ \ . $  Solving for $ \ \mu \ $ gives us

$$ \mu \ \ = \ \ 2x \ - \ y \ \ = \ \ x \ - \ 2y \ \ \Rightarrow \ \ x \  = \  -y \ \ . $$

But our planar constraint requires that $ \ x \ = \ y \ $ , so it must be that $ \ \ x \ = \ y \ = \ 0 \ \ .  $  (This incidentally tells us that $ \ \mu \ = \ 0 \ $ , but we don't actually care about that any longer.)  The spherical constraint then produces the remaining critical points $  \ ( \ 0 \ , \ 0 \ , \ \pm 2 \ ) \ \ .  $

$ $

For this particular problem, a method along the lines of what user252450 is much easier.  We could simply note that after "substituting" $ \ y \ = \ x \ $ , we are looking for the extrema of the "distance-squared" from the origin, $ \ x^2 \ + \ z^2 \ $ , of the points on the ellipse $ \ 2x^2 \ + \ z^2 \ = \ 4 \ \ \rightarrow \ \ \frac{x^2}{( \sqrt{2})^2} \ \ + \ \ \frac{z^2}{2^2} \ \ = \ \ 1 \ \ . $  The largest distance-squared of  $ \ 4 \ $ is reached for the endpoints $  \ ( \ 0  \ , \ \pm 2 \ ) \   $ of its major axis, with the smaller distance-squared of  $ \ 2 \ $ attained at the endpoints of its minor axis, $ \ (  \ \pm \sqrt{2} \ , \ 0 \ ) \ \ . $


答:
The ratio test is sufficient, in these cases.

(1) Set $a_n=n^5x^n$. Then

$$
 \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
 =\lim_n\to\infty \frac{(n+1)^5\,|x|}{n^5}=|x|
$$

so the series converges for $|x|<1$ and diverges for $|x|>1$.

(2) Set $y=x+1$ and $a_n=\frac{(-1)^ny^n}{2^nn^2}$; then
$$
\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
=\lim_{n\to\infty}\frac{2^{n}n^2}{2^{n+1} (n+1)^2}|y|=\frac{|y|}{2}.
$$
Therefore the series converges for $|y|<2$ and diverges for $|y|>2$.

Title: Find the radius of convergence of two series

Body:
I have to tell the convergence-radius of the following series:

$\sum_{n=0}^\infty n^5x^n$

Here I thought about using Euler and I get $\lim \frac{1}{(1+\frac{1}{n})^5} = 1$ So the convergence-radius is $1$?

Instead I could have used the Cauchy-Hadamard and I would get:
$r = \frac{1}{\limsup (\sqrt[n](|n^5|)}$ - what now?

$\sum_{n=1}^\infty (x+1)^n\frac{(-1)^n}{2^nn^2}$

What exactly should I do here?


问:
The FONC and the optimal step size $\alpha_k$. Help me with the intuition please
I was hoping someone could clarify something for me. For a one-dimensional line search problem

$x^{k+1} = x^{k} + \alpha d_{k}$ where $d_k$ is the descent direction and $\alpha$ the step size, how do we find the optimal step size, $\alpha_k$ at the point $x_k$?

I am confused about a question from my tutorial set and the solution provided by the model answers does not convince me. I argue that just because we find the optimal $\alpha_k$ that does not mean that $x_k+ \alpha_k d_k$ is the point at which we have the global minima so why does the First Order Necessary Condition (FONC) hold? Each step k in the algorithm will have its own optimal $\alpha_k$ and only at the actual optima $x^*$ will the FONC hold

The later steps about orthogonality I follow but I struggle with the intermediate steps

EDIT: Clarified meaning of FONC as the First Order Necessary Condition which means that the directional derivative is zero for a local minima

答:
If I'm not mistaken:

consider the functions $\phi_k: \mathbb{R} \rightarrow \mathbb{R}$ given by $\phi_k(\alpha)=f(x^k+\alpha d^k)$. Note that we have a different function for every index $k$

since $\alpha_k$ is a global minimum of the differentiable function $\phi_k$, then $\frac{d}{d\alpha}\phi_k(\alpha_k)=0$. So, we are applying the "FONC" at each step to the composed function$\phi_k$, and that doesn't mean that $x_k+\alpha_k d^k$ is a critical point of $f$ (by coincidence, it would be if your method converged at the current iterate to a local solution of $f$, for instance, but this is generally not the case)


问:
Area of a triangle inside of other triangle
$BK$ and $CM$ are angle bisectors of angles $B$ and $C$. Let $L$ be the intersection point. If $AB=39,BC=42$ and $CA=45,$ find the area of $\triangle CKL$.

a) $756$

b) $252$

c) $126$

d) $140$

We can find the area of triangle $ABC$ by Heron's formula: $$S=\sqrt{63\cdot 21\cdot 18\cdot 24}=756.$$ I am stuck here.

答:
Here are two observations that can get you to the answer quickly -

i) As $L$ is the incenter of the triangle, altitude from $L$ to all sides of the triangle is the same and that is inradius of the triangle $( = r)$.

ii) $A = r \cdot s$ where $A$ is the area of the triangle, $r$ is the inradius of the triangle and $s$ is sub-perimeter.

You have already found that $s = 63$ and you used that to find $A = 756$.

So area of $\triangle CLK = \frac{1}{2}r \cdot CK = \displaystyle \frac{756}{2 \cdot 63} \times \frac{42 \cdot 45}{42+39} = 140$


问:
Kerodon exercise 1.1.2.8
I'm doing the exercise 1.1.2.8. from Lurie's Kerodon: https://kerodon.net/tag/000T

The link is straight to the exercise does not contain anything else but the exercise itself, so I see no reason to retype it here. However, if someone thinks this should be done, I will do it.

Let $\Sigma = \{ (\sigma_0,\dots, \sigma_n) \in S^n_{n-1} \mid S_{\partial^{n-1}_j}(\sigma_k) = S_{\partial^{n-1}_{k-1}}(\sigma_j) \text{ for } 0 \leq j < k \leq n \}$. We have an injection $\Phi\colon \mathsf{Hom}_{\mathsf{Set}_{\Delta}}(\partial \Delta^n, S) \to \Sigma$ given by $f \mapsto (f\circ \Delta^{\partial^n_j})_{0 \leq j \leq n}$. It is asked to prove that $\mathrm{im}(\Phi) = \Sigma$. The implication $\mathrm{im}(\Phi) \subseteq \Sigma$ is straightforward. The implication $\Sigma \subseteq \mathrm{im}(\Phi)$ is doesn't seem hard as well, but I can't complete a proof due to my lack of experience in combinatorial matters, probably.

We clearly have $\partial \Delta^n = \bigcup_{0 \leq j \leq n} \Delta^{[n]\setminus\{j\}}$. There are unique order-preserving bijections $a_j\colon [n]\setminus\{j\} \to [n-1]$. Let $(\sigma_0,\dots, \sigma_n) \in \Sigma$. Via the Yoneda lemma, we can idenity $\sigma_i$ with a map $\tilde{\sigma}_i\colon \Delta^{n-1}\to S$. Explicitly, $(\tilde{\sigma}_i)_m(u) = S_u(\sigma_i)$ for any $[m] \in \Delta$ and any morphism $u\colon [m]\to [n-1]$. Thus, we can take the union $\bigcup \tilde{\sigma}_i$ to obtain a map $f\colon \partial \Delta^n\to X$ with sends $u\colon [m]\to [n]$ such that $\mathrm{im}(f) \subseteq [n]\setminus\{j\}$ to $(\tilde{\sigma}_i)(a_j\circ u)$. My idea (of which I'm almost sure) is that the condition $S_{\partial^{n-1}_j}(\sigma_k) = S_{\partial^{n-1}_{k-1}}(\sigma_j)$ for $j < k$ should guarantee that the map $f$ is well-defined, that is, that it agrees on $\Delta^{[n]/\{j\}}\cap \Delta^{[n]\setminus\{k\}}$. Of course, $u$ factors through $[n]\setminus\{j,k\}$ by assumption.

What can be done next? I've spent some time playing with simplicial operators, but it led me nowhere.

答:
I think I've got it.

We have to note that ${{(a_j)}_{\restriction}}_{[n]\setminus\{j,k\}} = \partial^{n-1}_{k-1}\circ a_{jk}$ and ${{(a_k)}_{\restriction}}_{[n]\setminus\{j,k\}} = \partial^{n-1}_{j}\circ a_{jk}$ where $a_{j,k}$ is the unique order-preserving bijection $[n]\setminus\{j,k\} \to [n-2]$.

I hope this settles this. Any comments are welcome.

Title: Kerodon exercise 1.1.2.8


问:
Find the asymptotic analysis of $\frac{a_{n+ 1}^{2}}{\prod_{0}^{n+ 1}a_{i}}- \sqrt{5}$

Given a recursion $a_{n+ 1}= a_{n}^{2}- 2$ with $a_{0}= 3.$ Find the asymptotic analysis of
$$\frac{a_{n+ 1}^{2}}{\prod_{0}^{n+ 1}a_{i}}- \sqrt{5}$$
Maybe this helps - https://artofproblemsolving.com/community/c4h2308107p18323223 -

And this is my way of thinking, I use two inequalities
$$\sqrt{5}\leq \frac{a_{n+ 1}^{2}}{\prod_{0}^{n+ 1}a_{i}}\leq\sqrt{5}+ \frac{2}{3}\cdot\left ( \varphi- 1 \right)^{-2^{n+ 1}}$$
with $\varphi= \frac{3+ \sqrt{5}}{2}.$ From the recurrence sequence $a_{n+ 1}= a_{n}^{2}- 2\Rightarrow a_{n+ 1}= \varphi^{2^{n}}+ \varphi^{-2^{n}},$ we have
$$\frac{a_{n+ 1}^{2}}{\prod_{i= 0}^{n+ 1}a_{i}}= \frac{\varphi^{2^{n+ 1}}+ \varphi^{-2^{n+ 1}}}{\prod_{i= 1}^{n}\left ( \varphi^{2^{i}}+ \frac{1}{\varphi^{2^{i}}} \right )}= (\varphi- \frac{1}{\varphi})\left ( 1+ \frac{2\varphi^{-2^{n+ 2}}}{1- \varphi^{-2^{n+ 2}}} \right )> \varphi- \frac{1}{\varphi}= \sqrt{5}$$
Because I'm busy now.. OK, I'll get it done by tonight, I haven't posted the solution for the rightside inequality, and how can we use it to find the $o\left ( 1 \right ).$ I need to the help, thanks a real lot !

Edit. The rightside inequality is just single variable of $f\left ( n \right )\leq g\left ( n \right )$ with
$$f\left ( n \right )= \sqrt{5}+ \frac{2}{3}\cdot\varphi^{-2^{n+ 1}}, g\left ( n \right )= \sqrt{5}\left ( 1+ \frac{2\varphi^{-2^{n+ 2}}}{1- \varphi^{-2^{n+ 2}}} \right ), 0< \varphi^{-2^{n+ 1}}\leq\frac{7-3\sqrt{5}}{2}$$

答:
First of all I think it's strange your product goes up to $i=n+1$, as that factor cancels with the numerator.

Assuming that is what you mean, notice that

$$\prod_{i=0}^{n+1} a_i = F_{2^{n+3}},$$

where $F_k$ is the $k$th Fibonacci number. That alongside with the formula

$$ a_n = \left(\frac{2}{\sqrt{5}+3}\right)^{2^n}+\left(\frac{1}{2} \left(\sqrt{5}+3\right)\right)^{2^n} $$

should do the trick.


问:
Calculus2 integration by parts
$$\int e^{3x}(e^{4x}+7)~dx$$
(integration by parts)
I was solving this problem with
\begin{align*}
u & =e^{3x} & dv & =e^{4x}+7\\
du & =3e^{3x}dx & v & = \frac{1}{4}e^{4x}+7x
\end{align*}
and I got
$$e^{3x}\left(\frac{1}{4}e^{4x}+7\right) - \int \left(\frac{1}{4}e^{4x}+7x\right)3e^{3x}~dx$$
This equation doesn't seem simpler than the previous equation.  What should I do next? or should I change $u$ to a different function?

答: You can define the variable $u=e^x$, which implies that $dx=\frac{du}{u}$. Then your integral is very easy to calculate. Of course, this does not require any integration by parts, as some people in the comments already pointed out.

Title: Calculus2 integration by parts


问:
Extending a question on a subset of transpose matrices and linear independence
A previous question, which appears in the text: Linear Algebra by Friedberg, Insel, and Spence, asks ``prove that if $\{A_{1},A_{2},\dots,A_{k}\}$ is a linearly independent subset of $M_{n\times n}(F),$ then $\{A_{1}^{t},A_{2}^{t},\dots,A_{k}^{t}\}$ is also linearly independent.''

Can the statement be rephrased as an if and only if? Namely, prove that $\{A_{1},A_{2},\dots,A_{k}\}$ is a linearly independent subset of $M_{n\times n}(F)$ if and only if $\{A_{1}^{t},A_{2}^{t},\dots,A_{k}^{t}\}$ is a linearly independent subset of $M_{n\times n}(F).$

I believe this is the case, as the previous solution shows one direction. Then, the other direction follows, if we suppose $\{A_{1}^{t},A_{2}^{t},\dots,A_{k}^{t}\}$ is a linearly independent subset of $M_{n\times n}(F).$ Namely, suppose $0 = \sum_{i=1}^{k}a_{i}A_{i},$ where $a_{i} \in F$ for $i=1,\dots,k.$ Then, $0 = (\sum_{i=1}^{k}a_{i}A_{i})^{t} = \sum_{i=1}^{k}a_{i}A_{i}^{t}$ implies $a_{1}=\cdots=a_{k}=0.$ Hence, $\{A_{1},A_{2},\dots,A_{k}\}$ is a linearly independent subset of $M_{n\times n}(F).$

答:
The answer goes in both directions because $(A^t)^t=A$. More precisely, if $\{A_1^t,\cdots,A_k^t\}$ are linearly independent then we use the fact already shown in the first direction to show that the set transpose matrices $\{(A_1^t)^t,\cdots,(A_k^t)^t\}$
is lienarly independent.

Now remember that $\{(A_1^t)^t,\cdots,(A_k^t)^t\}=\{A_1,\cdots,A_k\}$.


答:
Hint:
$$
e^{3x}(e^{4x}+7)=e^{7x}+7e^{3x} \, .
$$
Don't use integration by parts if you can avoid it.

Title: Calculus2 integration by parts

Body:
$$\int e^{3x}(e^{4x}+7)~dx$$
(integration by parts)
I was solving this problem with
\begin{align*}
u & =e^{3x} & dv & =e^{4x}+7\\
du & =3e^{3x}dx & v & = \frac{1}{4}e^{4x}+7x
\end{align*}
and I got
$$e^{3x}\left(\frac{1}{4}e^{4x}+7\right) - \int \left(\frac{1}{4}e^{4x}+7x\right)3e^{3x}~dx$$
This equation doesn't seem simpler than the previous equation.  What should I do next? or should I change $u$ to a different function?


问:
Senators in a senate using pigeonhole principle. (NOTE: not duplicate, only clarification)
There are $30$ senators in a senate. The senate needs to be divided into $n$ committees such that each senator is on exactly one committee. Each senator hates exactly $6$ other senators and is hated by exactly $6$ senators. (If senator A hates senator B, then senator B does not necessarily hate senator A). Find the smallest value of $n$ such that it is always possible to arrange the committees in a way that no senator hates another senator on his or her committee.

Yes, I know, there is already a question asked like this, but what I REALLY want is to clarify the answer of the other similar question on the site: Pigeonhole: Committee of senators who hate one another

Do you always start with $7$ when guessing the number of possible committees? Or is there a formula, or rule that states what number to start on when doing this guessing? The answer refers to something called the "induction hypothesis", but I'm not familiar about it, so it could really help if anyone explain these questions.

This question is not about finding the answer to the question in the first paragraph, but to go into detail about the method, and solution to solving this question. Any help would be wonderfully appreciated.

:)

答:
Induction is the idea that for natural numbers $\Bbb N$ (so $1,2,3 \cdots$), if we can prove a statement is true for any natural number and also for the generic number $n + 1$, then it is true for all natural numbers. The induction hypothesis, specifically, means that we prove for some minimum case (in this problem, $7$) then assume (aka, hypothesise) that it is true for some generic natural number $n$. If we can then prove that if it's true for $n$, it is always true for $n + 1$, then it is true for all natural numbers (because we proved it is true for $7$, so if we let $n = 7$ then we've proven it's also true for $8$, so for $9$, and so on.)

Your other question, on why 7, is discussed in the comments of the answer for your linked question - the answer uses 7 because the solution given in the question says 7, and then there is further discussion into how one would find that it is 7 without already knowing that - sTertooy's answer gives a very intuitive approach of why 6 or fewer would not work.


问:
Prove that to fully connect $n$ nodes we need at least $n-1$ pairwise links
Suppose we have $n$ points/nodes, and we are given $m$ pairwise links/edges that we can use to connect arbitrary pairs of nodes.

Our goal is to arrange the edges in order to fully connect the nodes, that is, to have a configuration such that one can go from any node to any other node passing through the edges.

For example, for $n=4$ and $m=3$, denoting with $(ij)$ an edge between $i$-th and $j$-th node, such fully connected networks could be $(12)(23)(34)$, or $(12)(13)(14)$.

By trying this in a few cases, it seems clear that the smallest $m$ for which this is possible is $m=n-1$. What's a good way to prove that this is indeed the case?

答: First create a single link between two nodes. As all the nodes are to be connected, the network of links must be connected. Adding a new link to the existing network further connects at most $1$ new node, and we have $n-2$ nodes to add, hence $n-1$ links.


答:
Hi you can easily prove it by induction.

Base case: $n=2$. You need $n-1=1$ edges. Graph connected.

Assume it is true that for $n$ nodes you need $n-1$ edges.

Now add a node, hence $n+1$ nodes. Just connect the last node to another node, and the number of edges become $n$.

Title: Prove that to fully connect $n$ nodes we need at least $n-1$ pairwise links

Body:
Suppose we have $n$ points/nodes, and we are given $m$ pairwise links/edges that we can use to connect arbitrary pairs of nodes.

Our goal is to arrange the edges in order to fully connect the nodes, that is, to have a configuration such that one can go from any node to any other node passing through the edges.

For example, for $n=4$ and $m=3$, denoting with $(ij)$ an edge between $i$-th and $j$-th node, such fully connected networks could be $(12)(23)(34)$, or $(12)(13)(14)$.

By trying this in a few cases, it seems clear that the smallest $m$ for which this is possible is $m=n-1$. What's a good way to prove that this is indeed the case?


问:
If a sequence $a_n$ has exactly three partial limits, and a sequence $b_n$ has exactly two partial limits. could the sequence $c_n=a_n+b_n$ converge?
If a sequence $a_n$ has exactly three partial limits, and a sequence $b_n$ has exactly two partial limits. is it possible for the sequence $c_n=a_n+b_n$ to converge?

I think this question doesn't demand a proof for every two sequences that have the same features, i think there is some examples for sequences that work and some that don't. for now I could think of examples for sequences that only disprove this allotment.

For example $a_n= 1,2,3,1,2,3,1,2,3,...$
and $b_n=-1,1,-1,1,-1,1,-1,..$

then $c_n=0,3,2,4,1,4,0,3,2...$

do you gues have an example for sequences that might work?

答:
Suppose $(a_n+b_n)$ converges to $L$. Suppose further that $(a_n)$ has a subsequence that converges to $M$. Then, since any subsequence of $(a_n+b_n)$ converges to $L$, it follows that $(b_n)$ has a subsequence (corresponding to the subsequence of $(a_n)$ mentioned above) converging to $L-M$.

The same holds with the roles of $(a_n)$ and $(b_n)$ interchanged. From this, it follows that the set of limit points of $(a_n)$ and the set of limit points of $(b_n)$ have the same cardinality.

In particular, the situation you describe can not occur.


问:
If $Y$ is an open subspace of $X$ then the boundary in $Y$ of an open set $A$ of $Y$ is the intersection of the boundary of $A$ in $X$ and $Y$
Statement

If $Y$ is an open subspace of $X$ then the boundary $\partial_Y$ in $Y$  of an open set $A$ of $Y$ is the intersection of the boundary $\partial_X$ of $A$ in $X$ and $Y$, that is
$$
\partial_Y A=\partial_X A\cap Y
$$

Proof. So if $x\in\partial_Y A$ then $x\in Y$ and for any open set $U$ of $X$ containing $x$ it must be $(U\cap Y)\cap A\neq\emptyset$ and $(U\cap Y)\cap(Y\setminus A)\neq\emptyset$ but $(U\cap Y)\subseteq U$ and $Y\setminus A\subseteq X\setminus A$ so that $x\in\partial_X A$ and thus $\partial_Y A\subseteq(\partial_X A\cap Y)$. Conversely if $x\in(\partial_X A\cap Y)$ then $x\in Y$ and if $U\cap Y\setminus A=\emptyset$ for some neighborhood of $x$ then $\overset{\circ}{\overbrace{U\cap Y}}=(U\cap Y)\subseteq A$ and thus $x\in\overset{\,\,\circ}A=A$ that would implies $\partial_X A\cap A\neq\emptyset$ that is not possible.

So is the statement true and in that case are my arguments correct?

答:
Your proof looks good to me.

Alternatively, it might be a bit easier to use that $\partial A=\overline{A}\setminus A$ for open sets $A$.

Let $\overline{A}$ be the closure of $A$ in $X$. Then $\overline{A}\cap Y$ is the closure of $A$ in $Y$, hence
$$ \partial_Y A = (\overline{A}\cap Y)\setminus A = (\overline{A}\setminus A)\cap Y = \partial_X A \cap Y$$


问:
exponential distribution - get some meaning
I read that a continuous random variable having an exponential distribution can be used to model inter-event arrival times in a Poisson process. Examples included the times when asteroids hit the earth, when earthquakes happen, when calls are received at a call center, etc. In all these examples, the expected value of the number of events per unit of time, lambda, is known and is constant over time. Moreover, each event's occurrence is independent of previous events' occurrences. And the exponential variable that models inter-event arrivals has the same lambda parameter as the Poisson variable that models the number of events.

And now, my problem. It don't get the connection between the intuition behind the exp distrib and its pdf. It seems obvious that the more time it passes by without an earthquake happening, the more likely it is than an earthquake will happen. Assume my understanding of lambda is correct, i.e., lambda = the rate at which the event happens, e.g., 5 earthquakes per minute on some remote, angry planet. On the pdf graphic, the prob of generating a value between 0 and 1 is greater than the prob of choosing a value between 4 and 5 for instance. How is this graphic related to the fact that on average we need to have 5 earthquakes per minute?

答:
The link between the Poisson point process and the exponential is probably the key to understanding this.

Let's say that asteroids hit your planet as a Poisson point process rate $\lambda$ per minute.  This means that if $X_i$ is the number of asteroids that arrive in the $i$th minute then the sequence $X_i$ are independent Poisson distributed random variables with mean $\lambda$.  Furthermore the exact arrival times of the $X_i$ asteroids in minute $i$ are distributed uniformly in the interval $(i-1,i)$.

The "magic" of the Poisson distribution then means that in any period of $T$ minutes, the number of asteroids that arrive will be a Poisson distribution mean $\lambda T$. Even if $T$ is not an integer.

So, what is the probability that the first asteroid lands  in the $4$th minute?
That's two events, firstly at least one asteroid lands in the $4$th minute, and secondly, that no asteroid arrives in the first three minutes.

That's reasonably easy to work out. The probability that at least one asteroid lands in the $4$th minute is $1-e^{-\lambda}$ and the probabilty that no asteroid lands in the first three minutes. is $e^{-3\lambda}$.  So the probability that the first asteroid lands in the $4th$ minute is

$$(1-e^{-\lambda})e^{-3\lambda} = e^{-3\lambda}-e^{-4\lambda} = \frac 1\lambda\int_3^4e^{-\lambda t}dt.$$

The maths works out the same for any interval.  But the intuition is hopefully a bit clearer as to why the pdf is decreasing.  If the first asteroid arives at time $t$ it's because no asteroids have arrived in the first $t$ minutes.

As the (infinitesimal) probability that an asteroid arrives at time $t$ doesn't depend on $t$, the probability density that the first asteroid arrives at time $t$ is directly proportional to the  probabilty that no asteroids arrive in the interval $(0,t)$ which must be decreasing.


答:
There are simpler way to calculate this integral, but if you want to use the integration by parts, do it the other way around, that is
\begin{align}u &= e^{4x} + 7 & dv = e^{3x}dx \\
du &= 4e^{4x}dx & v= \frac13 e^{3x}\end{align}
Then you'll get
$$ \frac13 e^{3x}(e^{4x}+7) - \frac43 \int e^{3x}e^{4x} dx$$
which is simpler.

Integration by parts relies on dividing the integrated functions into two factors, one that has a simple derivative and the other that has a simple antiderivative. The trick is not not just about finding a way to split the function into two factors, but also about choosing which is which.

Title: Calculus2 integration by parts


答: Assume $n \ge 2$. Let $(i, j)$ be an edge, with $i\not = j$. Simplify the graph by removing that edge and identifying the vertices $i$ and $j$. The new graph has one less egde and one less vertex. Hence the difference between the number of edges and the number of vertices remains the same. By induction, this difference must be $\ge -1$.


问:
Finding the order of the ideal class group
This is an understanding question : to search for the order $n$ of the ideal class group, why do we only need to try to split the $(p)$ where p is a prime $\leq M$ ($M$ being the Minkowski bound) ? I think it's because every ideal class contains an ideal of norm $\leq M$ and that every ideal can be split in a product of primes, and for every ideal $I$, $I^n$ is principal but i can't put it together properly.

Thank you !

答:
Any ideal class may be represented by a ideal $I$ of norm $\leq M$.

In particular,  any prime ideal $P$ dividing $I$ has norm $\leq M$.
Hence the ideal $I$ is the product of prime ideals with norm $\leq M$.

Note that any prime ideal $P$ is lying above a prime number $p$.  The norm of $P$ is a non trivial power of $p$, so $p\leq M$

So the classes of prime ideals with lying above a prime $p\leq M$ generate the class group.

Now if $p$ is inert, then $P=(p)$ is principal, and its class is trivial in the class group. SO, you just need to consider the non inert $p$.

This is all you can say if you work with an arbitrary number field.

If you are only interested in quadratic fields, it is true that you can stick to split  or totally ramified primes, because in this case $p$ is inert, split or totally ramified.

However, there are cases where the unique prime ideal lying above a ramified prime is not principal (so a priori you need to take it into account in the class group)

For example, for $\mathbb{Q}(\sqrt{-30})$, $M\approx 6.97$, so you can only consider $p=2,3,5$. All of them are totally ramified, and you can show that the class group is the Klein group, and generated by the classes of $\mathfrak{p}_2$ and $\mathfrak{p}_3$ (see https://kconrad.math.uconn.edu/blurbs/gradnumthy/classgpex.pdf, Example 4).

So you cannot only stick to split $p\leq M$.


问:
Fixed field of $G(K/\Bbb Q)\cap V$
I've been trying to solve the following question, but I wasn't able to.

Given $p\in\Bbb Q[x]$ an irreducible quartic polynomial, and $a,b,c,d$ its roots, consider $K=\Bbb Q(a,b,c,d)$ its splitting field, and $G=G(K/\Bbb Q)\subseteq S_4$ its Galois group. Prove that $G\cap V$, with $V=\{\textit{Id},(12)(34),(13)(24),(14)(23)\}$ the Klein group, has $L=\Bbb Q(ab+cd,ac+bd,ad+bc)$ as fixed field.

I see that each element of $V$ fixes one of the three generators of $L$, but that's not enough, since every generator must be fixed. I also know that we can get rid of one of the generators using Vieta's formulas, but there are still two and it's not clear $V$ fixes both of them. I've read something related to this problem about a cubic resolvent, but I've never used that before, so it's unlikely it's necessary for the solution.

Any advice? Thanks in advance.

答:
It follows immediately from the definition of $V$ that it fixes $\alpha= ab+cd, \beta= ac+bd, \gamma= ad+bc$. Conversely, it is easily seen (by tedious checking)that any permutation of the roots $a,b,c,d$ fixing $\alpha, \beta, \gamma$ belongs to $V$. Hence $L=\mathbf Q(\alpha, \beta, \gamma)$ is the fixed field of $G\cap V$, and $G/G\cap V$ is the Galois group of $L/\mathbf Q$.

NB. 1) The "resolvent cubic" of your irreducible quartic $P(X)$ is $(X-\alpha)(X-\beta)(X-\gamma)$. The name comes from the fact that the degree $n$ of the splitting field of the resolvent determines completely $G$, unless when $n=2$, in which case a supplementary computation is necessary.
2) The base field need not be $\mathbf Q$, it could be any field of characterisic $\neq 2$.

Title: Fixed field of $G(K/\Bbb Q)\cap V$


问:
Does every set of generators of a finite group contain a minimal set of generators?
Suppose that $G$ is a finite group that can be generated from $n$ elements $g_1,...,g_n\in G$.

Now, if $S\subseteq G$ is another set of generators of $G$, can I always find a subset $S'\subseteq S$ of size $\le n$ that still generates $G$?

答:
No.

For example, $S_4$ is generated by $(1,2)$ and $(1,2,3,4)$, so it is $2$-generated. However, no subset of size $2$ in $\{(1,2),(2,3),(3,4)\}$ generates $S_4$.


问:
How do I compute this integral with a Dirac's delta?
While studying probability I encountered this integral
$$I=\int_{\mathbb{R}^2}\exp\left({-\frac{x_1^2+x_2^2}{2}}\right)\delta\left(r-\sqrt{x_1^2+x_2^2}\right)\,dx_1\,dx_2$$
If I compute this in polar coordinates i get
$$I=\int_0^{2\pi}\,d\theta \int_0^{+\infty}\exp\left(-\dfrac{\rho^2}{2} \right)\rho\delta(r-\rho)\,d\rho=2\pi r\exp\left(-\dfrac{r^2}{2}\right)$$
but in cartesian coordinates I only get
$$I=\exp\left(-\frac{r^2}{2}\right)$$
I don't understand why. I just thougth that I was using the Dirac's delta's properties in both cases. I think the first result is the correct one and there is something I don't know about Dirac's delta with more than one variable.

Which result is correct and why?

答:
You may also make direct calculations in Cartesian coordinates - integrating, for instance, over $x_1$ first and then over $x_2$.
We can multiply and divide the argument of $\delta$-function ($r-\sqrt{x_1^2+x_2^2}$) by  ($r+\sqrt{x_1^2+x_2^2}$) (because ($r+\sqrt{x_1^2+x_2^2}$) is always positive). We can also replace the power of exponent by $-\frac{r^2}{2}$ - due to the condition imposed by $\delta$-function

$I(r)=\int_{\mathbb{R}^2}\exp\left({-\frac{x_1^2+x_2^2}{2}}\right)\delta\left(r-\sqrt{x_1^2+x_2^2}\right)\,dx_1\,dx_2=\int_{\mathbb{R}^2}\exp\left({-\frac{r^2}{2}}\right)\delta\left(\frac{r^2-(x_1^2+x_2^2)}{r+\sqrt{x_1^2+x_2^2}}\right)\,dx_1\,dx_2$

We see that $x_2$ contributes if only $x_2\in[-r,r]$, otherwise $\delta()=0$

$I(r)=\int_{-r}^rdx_2\int_{-\infty}^{\infty}dx_1\exp\left({-\frac{r^2}{2}}\right)\delta\left(\frac{(\sqrt{r^2-x_2^2}-x_1)(\sqrt{r^2-x_2^2}+x_1)}{r+\sqrt{x_1^2+x_2^2}}\right)$

But $\delta(\frac{(a-x_1)(x_1+b)}{A})$$=|\frac{A}{a-x_1}|\delta(x_1+b)+|\frac{A}{x_1+b}|\delta(a-x_1)=|\frac{A}{a-x_1}|\delta(x_1+b)+|\frac{A}{x_1+b}|\delta(x_1-a)$

We get

$I(r)=\int_{-r}^rdx_2\int_{-\infty}^{\infty}dx_1\exp\left({-\frac{r^2}{2}}\right)\left(r+\sqrt{x_1^2+x_2^2}\right)$$\left(\frac{1}{\sqrt{r^2-x_2^2}-x_1}\delta(\sqrt{r^2-x_2^2}+x_1)+\frac{1}{\sqrt{r^2-x_2^2}+x_1}\delta(\sqrt{r^2-x_2^2}-x_1)\right)=$
$\int_{-r}^rdx_2\int_{-\infty}^{\infty}dx_1\exp\left({-\frac{r^2}{2}}\right)2r$$\left(\frac{1}{2\sqrt{r^2-x_2^2}}\delta(\sqrt{r^2-x_2^2}+x_1)+\frac{1}{2\sqrt{r^2-x_2^2}}\delta(x_1-\sqrt{r^2-x_2^2})\right)=\int_{-r}^r\exp\left({-\frac{r^2}{2}}\right)\frac{2r}{\sqrt{r^2-x_2^2}}dx_2$
$$I(r)=\int_{-1}^1\exp\left({-\frac{r^2}{2}}\right)\frac{2r}{\sqrt{1-t^2}}dt=2\pi{r}e^{-\frac{r^2}{2}}$$


问:
An integral problem why:$\int_a^xf(x)dt = (x-a)f(x) $
Let $f(x)$ be continuous on $[a,b]$ and $F(x)=\frac{1}{x-a}\int_a^xf(t)dt$

Proof: The functions $F(x)$ and $f(x)$ have the same monotonicity on $(a, b]$.

The answer is here,but I don't know why $\int_a^xf(x)\,dt = (x-a)f(x) $

答: The integrand $f(x)$ is a constant (we're integrating with respect to $t$). And the integral of a constant is equal to the product of that constant with the width of the interval you're integrating over, which is $x-a$.


答: If $x$ is fixed, then in $\int_a^xf(x)dt$ you integrate over the constant $f(x)$ with respect to $t$. Thus $\int_a^xf(x)dt =f(x) \int_a^x 1 dt=(x-a)f(x).$

Title: An integral problem why:$\int_a^xf(x)dt = (x-a)f(x) $

Body:
Let $f(x)$ be continuous on $[a,b]$ and $F(x)=\frac{1}{x-a}\int_a^xf(t)dt$

Proof: The functions $F(x)$ and $f(x)$ have the same monotonicity on $(a, b]$.

The answer is here,but I don't know why $\int_a^xf(x)\,dt = (x-a)f(x) $