答:
To find a basis of the image of a linear transformation $T(\vec{x})=A\vec{x}$, you first find a generating set, then remove elements until you have a basis.

A generic set is given by the columns of $A$. To remove redundant columns, you need to know which columns are linear combinations of other columns. This you can do by doing row operations on $A$ and reducing to reduced row echelon form $A_{red}$. The columns in the reduced matrix $A_{red}$ having a pivot element correspond to the columns in $A$ that are part of a basis.

Title: How to find basis for an image of a subset under a transformation?

Body:
I was watching the khan acadamey videos on linear algebra.

Sal shows that the transformation $T$ of a subspace $S$ is another subspace $T(S)$. Since $S$ and $T(S)$ can have different dimensions, how do I find the basis for $T(S)$ if know the matrix $\mathbf A$ such that $T(\mathbf x) = \mathbf A\mathbf x$ and I know the basis for $S$?

For example if $S$ is a line in $\mathbb R^2$, and $T:\mathbb R^2 \rightarrow \mathbb R^3$; does the transformation of the basis of $S$ form a basis of $T(S)$? Is it possible $T(S)$ be a plane in $\mathbb R^3$ or must it be a line or a point?


问:
What is $(A\Delta B) \cap (A\Delta C)$ ??
What is $$(A\Delta B) \cap (A\Delta C)$$
without using a Venn Diagram.

My try: we have
$$(A\Delta B) \cap (A\Delta C)$$
$$\implies ?? $$

please I need help tried (commutative, associative, distributive) rules for ∩ and ∪ and I know the result(I used the table of truth ) but I want the demonstration

答:
Notice that
$$x \in A \Delta B \cap A \Delta C \iff x \in [(A -B)\cup(B-A)] \cap [(A -C)\cup(C-A)].$$
But
$$
x \in [(A -B)\cup(B-A)] \cap [(A -C)\cup(C-A)] \iff x \in [(A-B) \cap (A-C)] \cup [(B-A) \cap (C-A)].
$$
Since
$$(A-B) \cap (A-C)=(A \cap B^c) \cap(A\cap C^c)=A\cap B^c \cap C^c$$
and
$$(B-A) \cap (C-A)= (B \cap A^c) \cap(C \cap A^c)=A^c \cap B\cap C$$
we have that
$$
x \in A \Delta B \cap A \Delta C \iff x \in [A\cap B^c \cap C^c] \cup [A^c \cap B\cap C].
$$

ContentLicense: CC BY-SA 4.0


问:
Selfadjoint bounded isometric operators and orthogonal projections
Let $H$ a Hilbert space, $ T = T^{\star} \in B(H)$ with $ || T || = 1$.
Prove that for any $v\in H$ $ \lim_{n \to \infty} T^nv= Pv$,
where $P\in B(H)$ is the orthogonal projection onto $\text{ker}(Id-T) $.

答:
This is false.  If $T=\pmatrix{1&0\cr 0&-1}$, then
$$
T^n\pmatrix{x\cr y}=\pmatrix{x\cr (-1)^ny},
$$
which does not converge if $y\neq0$.

EDIT: From now on let us assume that $T≥0$.

It is well known that  $σ(T)$ consists of a sequence of eigenvalues converging to zero in $[0, 1]$,
including 0 (which may or may not be an eigenvalue).  Moreover the spectral
radius of $T$ coincides with its norm,  so 1 is necessarily in the spectrum of $T$.

Denoting by  $H_λ$ the eigenspace for each $λ∈σ(T)$,  we have that
$$
  H=\bigoplus_{λ∈σ(T)}H_λ = H_1\oplus K,
  $$
where
$$
  K=\bigoplus_{λ∈σ(T)\setminus \{1\}}H_λ.
  $$

Letting $S=T|_K$,  observe that the spectrum of $S$ coincides with $σ(T)\setminus \{1\}$ and, since  1 is necessarily an isolated point of
$σ(T)$, we see that
$σ(S)⊆[0,a]$,
for some $a<1$.  In particular
$$
  \|S\|\leq a<1.
  $$

Observing that $T=I_{H_1}\oplus S$, we have that
$$
  T^n=I_{H_1}\oplus S^n \to I_{H_1}\oplus 0,
  $$
because $\|S^n\|≤\|S\|^n\to 0$.  Finally notice that $I_{H_1}\oplus 0$ is the orthogonal projection onto $H_1$, as desired.

This in fact proves a little more than required, namely that $T$ converges to $I_{H_1}$ in norm.


问:
What properties do the solutions of linear difference equations have?
If I remember correctly, a solution to a linear difference equation is another equation parameterized by some values. Therefore isn't it true that for a linear difference equation $g(p) = xg(p - 1) + yg(p + 1)$, where x and y are nonzero, any scalar multiple of its solution is also a solution? How about the product and/or sums of its solutions? Are they not also solutions?

答:
$$g(p)=xg(p - 1) + yg(p + 1)$$

For any $\lambda,\mu$ such that $\lambda^2+\mu^2\ne 0$, we trivially verify that $\lambda g(n)+\mu g(n)$ is also a solution. Indeed
$$\lambda g(n)+\mu g(n)=\lambda\left(xg(p - 1) + yg(p + 1)\right)+\mu\left(xg(p - 1) + yg(p + 1)\right)=\\=(\lambda+\mu)x g(p-1)+(\lambda+\mu)y g(p+1)=x'g(p-1)+y'g(p+1)$$

Hope this can help

Bonus: I calculated the general solution, but it was useless for the proof:
$$g(n)=a \left(\frac{1-\sqrt{1-4 x y}}{2 y}\right)^p+b \left(\frac{\sqrt{1-4 x y}+1}{2 y}\right)^p$$


问:
Proving a graph has two disjoint cycles
For a graph A (does not contain any loops), that has every vertice of degree 6 or more, how can we prove that there exist two cycles such that s1 and s2 have set of edges disjoint from one another?

I can explain it, but I'm not sure how to write the proof formally. I can say a cycle s1 is a path that contain no repeated edges and then we can trace s2 to have vertices more than degree 2 now, and a path exist. But I'm not sure how to formally write this as a proof

答:
Hint: Find one cycle, remove it's edges, prove you can find another.

For a full proof, click the spoiler:

 It suffices to show that a graph with minimum degree $2$ or more has a cycle. Start at any vertex, keep moving to adjacent vertices, until you get a repeat. Then you have yourself a cycle. So if $G$ is our original graph with minimum degree at least $4$, it has a cycle $S_1$. Every vertex of the cycle has $2$ edges in the cycle. So the graph $G-E(S_1)$ has minimum degree at least $2$. Thus there is a cycle $S_2$ in $G-E(S_1)$.

Title: Proving a graph has two disjoint cycles


问:
Combinatorial Proof for $\binom{3n}{n}$
I am trying to prove the following equality:

$\sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} \binom{n}{k+l} = \binom{3n}{n}$

This is what I have so far:

$\sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} \binom{n}{k+l} = \sum_{k=0}^{n}\binom{n}{k} \sum_{l=0}^{n}\binom{n}{l} \sum_{k=0}^{n} \sum_{l=0}^{n}\binom{n}{k+l}$

$\implies 2^{n} \cdot 2^{n} \sum_{k=0}^{n} \sum_{l=0}^{n}\binom{n}{k+l}$

$\implies 2^{2n} \cdot \sum_{k=0}^{n} \sum_{l=0}^{n}\binom{n}{k+l}$

I got stuck here, and I'm having trouble manipulating the second term. I would appreciate any help on this proof!

答:
Let us call the sum of the given series as $S$.
$$S = \sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} \binom{n}{k+l}$$
For the inner summation,$\binom{n}{k}$ is a constant and can be moved outside. We are left with:
$$S = \sum_{k=0}^{n} \binom{n}{k} \sum_{l=0}^{n}  \binom{n}{l} \binom{n}{k+l}$$

We first need to focus on how to find the value of the sum $\sum_{l=0}^{n}  \binom{n}{l} \binom{n}{k+l}$. Call it $S_1$ for now.

To do so, we can apply a bit of Binomial Theorem.
We know that:
$$(1+x)^n = 1 + \binom{n}{1}x + ... + \binom{n}{k}x^k + \binom{n}{k+1}x^{k+1} + ... + \binom{n}{n}x^n$$
We can also once again write the series in reverse:
$$(x+1)^n = x^n + \binom{n}{1}x^{n-1} + ... + \binom{n}{n-1}x + \binom{n}{n}$$

If you notice carefully, when we multiply out the above two Binomial sequences, all of the terms of the sequence $S_1$ appear as coefficients of $x^{n+k}$, i.e. the sequence
$$S_1 = \sum_{l=0}^{n}  \binom{n}{n-l} \binom{n}{k+l} = \binom{n}{k}\binom{n}{0} + \binom{n}{k+1}\binom{n}{1} + ... + \binom{n}{k+n}\binom{n}{n} $$
is nothing but the coefficient of $x^{n+k}$ in $(1+x)^{2n}$ = $\binom{2n}{n+k}$

In fact, we have discovered a useful result known as the Vandermonde Identity which states that in general:
$$\binom{m+n}{r}=\sum_{k=0}^r \binom{m}{k} \binom{n}{r-k}$$

Now that we have figured out $S_1$, $S$ reduces to:
$$S = \sum_{k=0}^{n} \binom{n}{k} \binom{2n}{n+k}$$

Rewriting $\binom{n}{k}$ as $\binom{n}{n-k}$ and applying Vandermonde again, we finally get:
$$S = \binom{3n}{2n} = \binom{3n}{n}$$

Title: Combinatorial Proof for $\binom{3n}{n}$


问:
Riemann-Stieltjes integral of simple functions
I quote Øksendal (2003).

Let us consider a probability space
$\left(\Omega,\mathbb{P},\mathcal{A},\right)$ and a class of functions
$f:\left[0,\infty\right]\times\Omega\mapsto\mathbb{R}$.
 For $0\le
S<T$, $\left(B(t)\right)_{t\ge0}$ a Brownian motion and $f(t,\omega)$
given, we want to define: $$\int_S^T f(t,\omega)dB(t)(\omega)$$
It is reasonable to start with a definition for a simple class of functions $f$ and then extend by some approximation procedure. First assume that $f$ has the form:
$$\phi(t,\omega)=\sum_{j\ge0}e_j(\omega)\cdot1_{[j\cdot2^{-n}, (j+1)2^{-n})}(t)$$
where $1$ denotes the indicator function and $n$ is a natural number.

For such functions, it is reasonable to define:
$$\int_S^T\phi(t,\omega)dB_t(\omega)=\sum_{j\ge0}e_j(\omega)\left[B_{t_{j+1}}-B_{t_j}\right](\omega)\tag{1}$$
where:
$$t_k=t_k^{(n)}=\begin{cases}k\cdot 2^{-n}\hspace{0.3cm}\text{if } S\le k\cdot 2^{-n}\le T\tag{2}\\ S\hspace{0.3cm}\text{if } k\cdot 2^{-n}<S\\
T\hspace{0.3cm}\text{if } k\cdot 2^{-n}>T
\end{cases}$$

My doubts concern the part in italics. Namely:

Questions

Why, according to $(1)$:
\begin{align}\int_S^T\phi(t,\omega)dB_t(\omega)&=\int_S^T\sum_{j\ge0}e_j(\omega)\cdot1_{[j\cdot2^{-n}, (j+1)2^{-n})}(t)dB_t(\omega)\\&=\sum_{j\ge0}e_j(\omega)\left[B_{t_{j+1}}-B_{t_j}\right](\omega)\end{align}?

 Is that due to the fact that $\sum_{j\ge0}e_j(\omega)\cdot1_{[j\cdot2^{-n}, (j+1)2^{-n})}(t)$ do not depend on the variable of integration $B_{t}(\omega)$, hence they go outside sign of integration and one has:
\begin{align}\int_S^T\phi(t,\omega)dB_t(\omega)&=\int_S^T\sum_{j\ge0}e_j(\omega)\cdot1_{[j\cdot2^{-n}, (j+1)2^{-n})}(t)dB_t(\omega)\\&=\sum_{j\ge0}e_j(\omega)\left[B_{t_{j+1}}-B_{t_j}\right](\omega)\end{align} with $t_k$ as specified in $(2)$ and $\sum_{j\ge0}\left[B_{t_{j+1}}-B_{t_j}\right]=B_{T}-B_{S}$?

Besides, could you please detail the reason why $(2)$ is defined that way? In particular, is there in place the choice of left-end point of every time interval? Why does the value $t_k$ depend on whether $k\cdot2^{-n}$ is positioned? What I would expect instead is something like:
$$t_k=t_k^{(n)}=\begin{cases}t_k\hspace{0.4cm}\text{if } k\cdot 2^{-n}\le t_k \le (k+1)\cdot2^{-n}\tag{2.bis}\\ 0\hspace{0.5cm}\text{otherwise}
\end{cases}$$

答:
(1) It is the definition of a stochastic integral for elementary functions w.r.t. a BM. (See the beginning of the next section.) Why is it reasonable? Consider a discrete-time analogue. Let $\{X_n\}$ be a martingale adapted to $\{\mathcal{F}_n\}$ and let $\{H_n\}$ be a bounded, previsible process, i.e., $H_n\in\mathcal{F}_{n-1}$. Then we define
$$
(H\cdot X)_n:=\sum_{i=1}^n H_i \Delta X_i,\quad (H\cdot X)_0=0
$$

as our discrete time stochastic integral (in fact, it is called the martingale transform of $X$). The standard example is that if you bet 1\$ each time (i.e., $H_n=1$), your total gain/loss at time $n$ is exactly $(H\cdot X)_n$. A nice property of this process is that it is a martingale (is it crucial that $H$ is predictable; take, for example, $H_n=\operatorname{sgn}(\Delta X_n)$). The corresponding processes in your case are $H_n=e_{n-1}$ and $X_n=B_{t_n}$ (setting $S=0$).

(2) The "logic" behind the definition of $t_k$ is related to the definition of elementary functions. For each $k$, such a function is constant on $[k 2^{-n},(k+1)2^{-n})$ and $(B_t)$ is "sampled" at the corresponding end-points.


问:
Can the group of rotation be considered as a cyclic group
I read from some physics books that the 2D rotation can be expressed in exponential form:

$$R(\theta) = e^{X(\theta)}, \theta \in R $$

In physics, X relates to the momentum of a particle, and is defined to be the generator of the rotation. Some authors also state that X is the generator of the rotation group .

Is that statement correct, and if it is, can we consider the rotation group ($N\ge2$) as a cyclic group. As if I understand correctly, only cyclic groups have a generator.

答:
In case of a rotation group, this is called (in mathematics) an infinitesimal generator and is not a generator in the sense you mention.

For a cyclic group, there is a generator $G$ such that the elements of the group are $G^k$ with $k \in \mathbb Z$.  For a rotation group, there is an "infinitesimal generator" $X$ such that the elements of the group are $e^{tX}$ with $t \in \mathbb R$.

====

If $R(t) = e^{tX}$ is the rotation group, then the derivative is
$$
R'(t) = \lim_{t \to 0}\frac{R(t) - I}{t} = X
$$
(the identity matrix $I = R(0)$) and that is the sense it is an "infinitesimal generator".  We have
$$
R(t) = I + tX + O(t^2)
$$
so if $t$ is very small, then rotation is approximately $I+tX$.  So $X$ is the "tangent direction" to the rotation.


问:
Two questions about integral scheme
Following are examples of 3.0.1 in Hartshorne's textbook

If $X=\operatorname{Spec}A $ is an affine scheme, then

i) $X$ is reduced if and only if $\operatorname{Nil}A=0$

$(\Rightarrow)$ for every open in $U$ in X, $\mathcal{O}_X(U) $ has no nilpotent.  Thus, take by $U=X$, $\mathcal{O}_X(X)=A$. Thus, $\operatorname{Nil}A=0$. The trouble is $(\Leftarrow)$. At the outset, I attampted to use contravariance propery of sheaf. since $U \subset X$, (for any open in $U$),  $\mathcal{O}_X(U) \supset \mathcal{O}_X(X)=A  $ , I expect to hold $\operatorname{Nil}\mathcal{O}_X(U) \subset \operatorname{Nil}\mathcal{O}_X(X)=\operatorname{Nil} A = 0  $...(★). However, (★) does not hold.

ii) $X$ is integral if and only if $A$ is an integral domain. : Also this has similar trouble .. $(\Rightarrow)$ is obvious by taking $U=X$, $\mathcal{O}_X(X)=A$ is an integral domain. But, $(\Leftarrow)$ is not easy for me.

答:
The first idea is to show that if $A$ is reduced (resp. integral), then all its localizations are reduced (resp. integral). This shows $\Leftarrow$ for affine open subsets.

In the reduced case, $\Leftarrow$ follows formally from the statement for affine open subsets.

In the integral case, if $A$ is integral, then all restriction maps $\mathcal{O}_X(U) \rightarrow \mathcal{O}_X(V)$ are injective – start when $U$ is affine and $V$ principal in $U$, then extend to the case $U$ affine and any $V \subset U$, and finally show it for general $U$ and any $V \subset U$. The conclusion (of $\Leftarrow$) follows since the ring of global sections over any principal open subset of $X$ is integral.

Title: Two questions about integral scheme


问:
Basic logic: $A\to C$ and $B\to C$ is equivalent to $A\lor B\to C$?
Consider the two results: "if A holds, then C", "if B holds, then C".
From these, is it correct to write a theorem that says "if A or B hold, then C"?</p>

答:
Maybe I’m wrong here but that proof tree looks massively convoluted, and the solution is straightforward. I’ll write my answer like pseudocode.

Suppose (A or B) is true.

$\quad$ If A is true then C is true.

$\quad$ Else B is true, so C is true.

In either case, C is true.


问:
Find the area of the parallelogram.
Hit this in my book. Can someone help me out?

Let
$$
\begin{align}
u &= i - j \\
v &= 2i - 3j + k.
\end{align}
$$

Find the area of the parallelogram defined by $u$ and $v$.

Am I right to take a 3rd point $v \times u$?

答:
The area of a parallelogram =$|\vec u \times\vec v  |$  where $\vec u$ and $\vec v$ are vectors of conjugate sides of parallelogram.
$\vec u=i-j$ and $\vec v=2i-3j+k$
$$\vec u \times \vec v=
\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\1 & -1 & 0 \\2 & -3 & 1 \end{vmatrix}$$
$$\vec u \times \vec v=\mathbf{i}(-1-0)-\mathbf{j}(1-0)+\mathbf{k}(-3+2)$$
$$\vec u \times \vec v=\mathbf{-i}-\mathbf{j}\mathbf{-k}$$
$$|\vec u \times\vec v  |=\sqrt3$$
area of parallelogram is $\sqrt 3$

here $\mathbf{i}$,$\mathbf{j}$,$\mathbf{k}$ are unit vectors.

ContentLicense: CC BY-SA 3.0

Title: Find the area of the parallelogram.


问:
Proving $\sup \left( {A + B} \right) = \sup A + \sup B$ using the usual definition.
Prove $\sup \left( {A + B} \right) = \sup A + \sup B$ using the definition given in this problem (link).
 I was able to prove using lemma of the given definition.
My attempt is here,

let $s = \sup A$, and $t = \sup B$

Now, to show $s + t$ is least upper bound for $A + B$ I need to show that for all  $\varepsilon  > 0$ there is $c \in A + B$  such that $s + t \le c$.
 Since,$s = \sup \left( A \right)$ , therefore, there is some  $a \in A$ such that $s - {\varepsilon  \over 2} \le a$ . Similarly, there is $b \in B$  such that$t - {\varepsilon  \over 2} \le b$ . Hence, for any $c \in A + B$,

  $s - {\varepsilon  \over 2} + t - {\varepsilon  \over 2} \le a + b = c$
which completes the proof.

My question is using the usual definition rather than lemma to prove that $\sup \left( {A + B} \right) = \sup A + \sup B$

I could show $s + t$ is upper bound for $A + B$. Then lets chose $x$ be an arbitrary upper bound for $A+B$ and temporary fix $a \in A$. How do I show this,

$t \le x- a$,

and then finally conclude that
$\sup \left( {A + B} \right) = \sup A + \sup B$.

Please, if possible, explain each step, and why that step is taken and how the definition is used in steps.

EDIT:

My question is specific to using the definition. Here is what I have tried, since $x$ is upper bound of $A+B$, therefor, I can write, $a+b \le x$, for all $a \in A$ and $b \in B$. For fixed $a \in A$ I can proceed further and can write, $b \le x - a$, giving $x-a$ as an upper bound of $B$. Now I know that $t$ is least upper bound of $B$, so $t \le x-a$ which proves part of my problem.  Going by similar argument it can be achieved that $s \le x-b$.
 Adding both the inequality gives,

$s+t \le 2x-(a+b)$

Since, it is already known that $a+b \le x$, therefore, the maximum value of $a+b$ will be $x$. Which gives,

$s+t \le 2x-(x)$

$s+t \le x $

Now, since $t+s \le x$, therefore $t+s$ is least upper bound of $A+B$.
 $\sup \left( {A + B} \right) = \sup A + \sup B$.
 Proved
Please check and suggest if there is something wrong with the solution or its complete.

答:
In order to show  $\sup \left( {A + B} \right) = \sup A + \sup B$ . We need to verify the following:

$(s+t)$ is an upper bound for the set $A+B$ , where s = $ \sup A$ and  t = $ \sup B$

if  $b$ is any upper bound for set $A+B$, then $(s+t)≤b $

Proof: (Direct)

Lets try to verify (1) first:

Since $s$ is the $\sup A$

So, $a ≤ s$, for all $a \in A$ .
We can write:

$a+b≤(s+b)$   for all $ a \in A$ and $b \in B$

Similarly, since $t$ is the $\sup B$ .

So , $b ≤ t$,    for all $b \in B$ . Hence

$(s+b)≤(s+t)  $  for all $b \in B$ . So,

$(a+b)≤ (s+b)≤(s+t) $

$(a+b)≤ (s+t) $ for all $ a \in A$ and $b \in B$

Thus, $(s+t)$ is the upper-bound of set $A+B$

So  (1) is verified.

To verify (2):

Assume $u$ be an arbitrary upper-bound for set $A+B$. Let's temporarily fix $a \in A$.

So, $(a+b)≤ u$ ,  for all $b \in B$ .

$b≤ (u-a)$  ,  for all $b \in B$ .

This means, $(u-a)$ is an upper-bound for set $B$.

Since  t = $ \sup B$. So, $t≤ (u-a)$.

Rearranging, $a≤ (u-t)$ for all $a \in A$ .

This means, $(u-t)$ is an upper-bound for set $A$.

Since  s = $ \sup A$. So, $s≤ (u-t)$.

Therefore, $ (s+t) ≤ u$

This means $(s+t) $ is smaller than an arbitrary upper-bound of set $A+B$

Hence (2) is verified.

In conclusion, we have shown :

$(s+t) $ is an upper-bound for set $A+B$

$(s+t)$ is smaller than (or equal to) to any arbitrary upper-bound for set $A+B$

So, $s+t= \sup(A+B)$

$\sup(A) + \sup(B)= \sup(A+B)  $


问:
Why is the condition $\|Z\| < 1$ equivalent to $I - ZZ^{\top} > 0$?
As the title says, for a matrix $Z \in \mathbb{R}^{p \times q}$, the condition $\begin{Vmatrix}Z\end{Vmatrix} < 1$ equivalent to $I - ZZ^{\top} > 0$. How can I show the equivalence?

Attempt:

$\begin{Vmatrix}Z\end{Vmatrix} = \sup_{|x| = 1} \begin{Vmatrix}Zx\end{Vmatrix}$
$\implies$ $\begin{Vmatrix}Zx\end{Vmatrix} < 1$ $\implies$ $(Zx)^{\top}(Zx) < 1$ $\implies$ $x^{\top}Z^{\top}Zx < 1$.

Multiplying by the identity matrix on both sides,

$\implies$ $(x^{\top}Z^{\top}Zx)I  0 $.

答:
Alternatively, thanks to a comment by @Rodrigo de Azevedo, from Maximize $\langle \mathrm A , \mathrm X \rangle$ subject to $\| \mathrm X \|_2 \leq 1$,

\begin{align}
\|Z\| < 1
&\Leftrightarrow \sigma_{\max}(Z) < 1 \\
&\Leftrightarrow \lambda_{\max}(Z^{\top}Z) < 1 \\
&\Leftrightarrow \lambda_{\max}(ZZ^{\top}) < 1 \quad (\text{SVD decomposition}) \\
&\Leftrightarrow 1 - \lambda_{\max}(ZZ^{\top}) > 0 \\
&\Leftrightarrow \lambda_{\min}(I - ZZ^{\top}) > 0 \\
&\Leftrightarrow (I-ZZ^{\top}) \succ 0_{p\times p}.
\end{align}


问:
A connection between the Gelfand Naimark Theorem and the GNS construction
Let $\mathcal{A}$ be an abelian $C^*$ algebra. By the Gelfand Naimark theorem, we know $\mathcal{A}$ is isometrically $*$ isomorphic to $\mathcal{C}_0(\Delta_{\mathcal{A}})$ where $\Delta_{\mathcal{A}}$ is the space of non zero characters on $\mathcal{A}$. We have that $\Delta_{\mathcal{A}}$ is a locally compact Hausdorff space.

Now by the GNS construction, we know every $C^*$ algebra $\mathcal{A}$ admits a faithful representation into some $\mathcal{B(H)}$ where $\mathcal{H}$ is a Hilbert space i.e. $\mathcal{A}$ is a closed subalgebra of a  $\mathcal{B(H)}$.

My question is whether for abelian $C^*$ algebras, can we say something more specific regarding the embedding $\mathcal{B(H)}$ using the Gelfand Naimark Theorem?

Note: I was thinking along the lines of getting the irreducible representations of $\mathcal{A}$ which are in direct correspondence with the pure states of $\mathcal{A}$ which are exactly the characters on $\mathcal{A}$ as it is abelian.

Thanks.

答:
The characters of $C_0(X)$ are the maps $\{ ev_x\mid x\in X\}$ where $ev_x: C_0(X)\to\Bbb C$ is given by $f\mapsto f(x)$. As such the associated semi-definite inner-product $(,)_{ev_x}$ is given by:
$$(f,g)_{ev_x}:= ev_x(f^* g) =\overline{f(x)}\cdot g(x)$$
And clearly $C_0(X) /N_{ev_x}\cong\Bbb C$ where $N_{ev_x}$ is the null space of $(,)_{ev_x}$. The isomorphism $C_0(X)/N_{ev_x}\to \Bbb C$ is given by $[g]\mapsto g(x)$ as you can check explicitly. Now the action of $C_0(X)$ on this Hilbert space is given by:

$$f\cdot [g] = [f\cdot g]$$
which, under the above isomorphism $C_0(X)/N_{ev_x}\cong \Bbb C$, corresponds to:
$$f\cdot z = f(x)\cdot z$$

Now if you put all this together what you get is that the GNS space is:
$$H= \bigoplus_{x\in X} \Bbb C = \ell^2(X)$$
and the representation is defined by:
$$(\pi(f) v)_{x} = (f(x)\cdot v_x)$$
For some $v = (v_x)_{x\in X}\in \ell^2(X)$.


问:
When a polynomial $f\left(x,d\right)$ of degree=$d$ with complex roots have only real outputs?
The question by itself is simple: i have a polynomial $f\left(x,d\right)=\prod_{n=1}^{d}\left(x-r_{n}\right)$ where $x$ is real and $r_n$ is complex and i want to know what are the rules to write $r_n$ such that $f\left(x,d\right)$ have a real codomain.

答:
Complex roots always occur in conjugate pairs in a polynomial with real coefficients.

The way you have constructed your polynomial $f(x,d)$, you can choose arbitrary complex roots $r_0, \dots, r_{d-1}$ such that they don't occur in conjugate pairs. This will cause the coefficients of $f(x,d)$ to be in $\mathbb{C}$ if you choose all complex roots with an imaginary part.

So, your rule is really choose $r_i$ such that they occur in conjugate pairs. BTW, $\mathbb{R} \subset \mathbb{C}$. So, if your degree is odd, you need to choose at least one root from $\mathbb{R}$ after choosing complex conjugate pairs of roots from $\mathbb{C}$.

The simplest example is a quadratic polynomial that doesn't touch the $X$-axis. It has complex roots and they occur in conjugate pairs.


答:
They are equivalent, as Player says.  But beware.  Some students are confused into thinking (falsely) that
$$
(\forall x)A(x) \rightarrow C \quad\wedge\quad (\forall x) B(x) \rightarrow C
$$
is equivalent to
$$
(\forall x)(A(x) \vee B(x)) \rightarrow C
$$

Title: Basic logic: $A\to C$ and $B\to C$ is equivalent to $A\lor B\to C$?

Body:
Consider the two results: "if A holds, then C", "if B holds, then C".
From these, is it correct to write a theorem that says "if A or B hold, then C"?</p>


问:
Deriving closed form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}\ln(n+1)}{n} $
I wanted to derive closed form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}\ln(n+1)}{n}$ which I converted into a integral,

$$\sum_{n=1}^\infty\frac{(-1)^{n+1}\ln(n+1)}{n}=\int_0^1\int_0^1\frac{u^v}{1+u^v}\text{d}u\text{d}v$$

after some more manipulations I got,

$$\int_0^1\int_0^1\frac{u^v}{1+u^v}\text{d}u\text{d}v=1-\int_0^1\frac{\psi^{(0)}\left(\frac{1}{v}\right)-\psi^{(0)}\left(\frac{1}{2v}\right)-\ln(2)}{v}\text{d}v$$
not sure how to proceed further...

答:
As you did
$$\int_0^1\frac{u^v}{u^v+1}\,du=\frac{H_{\frac{1}{2 v}}-H_{\frac{1}{2 v}-\frac12}}{2 v}$$

At this point, I am stuck with formal but we can notice that the rhs (our new integrand)
$$f(v)=\frac{H_{\frac{1}{2 v}}-H_{\frac{1}{2 v}-\frac12}}{2 v}\sim \frac{1}{2}- \left(\log (2)-\frac{1}{2}\right)v^{4/5}$$ This approximation matches the function value at the end points and corresponds to the minimum of the infinite norm.

This makes
$$\int_0^1f(v)\,dv \sim
\frac{1}{9} (7-5 \log (2))\approx 0.392696$$ while the exact value is $0.392259$.

For improvement, I think that we need an approximation at least able to reproduce the following values
$$\left(
\begin{array}{ccc}
v & f(v) & f'(v) \\
 0 & \frac{1}{2} & -\frac{1}{4} \\
 \frac{1}{2} & -1+2\log (2) & \frac{2}{3} \left(\pi ^2-6 (1+\log (2))\right) \\
 1 & 1-\log (2) & -\frac{\pi ^2}{12}+\log (2)
\end{array}
\right)$$

Being lazy, for the time being, I used a quintic polynomial matching all the above conditions and obtained as an approximation
$$\frac{1}{720} \left(\pi ^2+588 \log (2)-135\right)\approx 0.392278$$ which is better but not enough.

What is amazing is
$$\frac{1}{10} \left(\psi \left(\frac{4}{9}\right)+\psi
   \left(\frac{3}{5}\right)-\psi \left(\frac{1}{4}\right)-\psi
  \left(\frac{3}{10}\right)\right)\approx 0.392259435$$ while the exact value is                          $0.392259418$


问:
Math question partial derivatives
I have to find the $\partial^2 z/\partial x \partial y$ of $z=e^{xy}$. I know how to find  $\partial^2 z/\partial x^2$ which by the way is $y^2 e^{xy}$ but not this one...can you give me a little hint?
and please tell me how to find this type of "power" partial derivatives,like $\partial^3z/\partial x^2 \partial y$..</p>

答:
Hint : $$\frac{\partial^2z}{\partial x \partial y} = \frac{\partial}{\partial x} \left(\frac{\partial z}{\partial y}\right)$$

For all you others "types of power partial derivatives" as you say, simply split them.

Example : $$\frac{\partial^n z}{\partial x_1 \partial x_2 \partial x_3 \dots \partial x_n} = \frac{\partial}{\partial x_1}\left(\frac{\partial}{\partial x_2}\left(\frac{\partial}{\partial x_3}\left(\dots\left(\frac{\partial z}{\partial x_n}\right)\right)\right)\right) $$

Title: Math question partial derivatives


问:
Why is $1 - \lambda_{\max} (\mathrm A^\top \mathrm A) = \lambda_{\min} (\mathrm I - \mathrm A^\top \mathrm A)$
Let A be a non-square matrix. Why is the following equality true?

$$1 - \lambda_{\max} (\mathrm A^\top \mathrm A) = \lambda_{\min} (\mathrm I - \mathrm A^\top \mathrm A)$$

Attempt:

Since the eigenvalue of a square matrix $X$, is the reciprocal of the eigenvalue of $X^{-1}$, we have:

$$\lambda_{\max}(\mathrm A^\top \mathrm A) = \frac{1}{\lambda_{\min}((\mathrm A^\top \mathrm A)^{-1})}$$

I have the following:

$$\lambda(I - \mathrm A^\top \mathrm A) = 1 - \lambda(\mathrm A^\top \mathrm A)$$

$$\lambda_{\min}(I - \mathrm A^\top \mathrm A) = \frac{1}{\lambda_{\max}((I - \mathrm A^\top \mathrm A)^{-1})}$$

答: What you want to prove is pretty simple and is true for all square matrices, not just those of the form $A^TA$. $\nu$ is an eigenvalue of $P$ iff $1-\nu$ is an eigenvalue of $I-P$. Thus, any eigenvalue of $I-P$ is of the form $\nu=1-\lambda$ where $\lambda$ is an eigenvalue of $P$. Taking the minimum, $\nu_\min=(1-\lambda)_\min=1-\lambda_\max$ since $1-\lambda$ is minimized when $\lambda$ is maximum.


问:
Negation rules in natural deduction
There are various treatments of using negation in natural deduction for classical logic.

Let me quote bits of it:

$$\frac{}{\top}\top I$$

$$\frac{\bot}{A}\bot E\quad ex\ falso\ quodlibet$$

$$\frac{A\rightarrow\bot}{\neg A}\neg I$$

$$\frac{}{A\vee \neg A}tertium\ non\ datur$$

$$\frac{A\quad \neg A}{\bot}law\ of\ contradiction$$

$$\frac{A}{\neg\neg A}double\ negation\ 1$$

$$\frac{\neg\neg A}{A}double\ negation\ 2$$

$$\frac{\begin{equation}\begin{array}{c}{[\neg A]\\ \vdots\\ \bot}\end{array}\end{equation}}{A}indirect\ proof$$

What is a standard way of building this, what are the axioms and how are the other rules are proved?

You might answer this by referring to some reference.

答:
The only true "negation rule" needed is reductio ad absurdum aka proof by contradiction:

$$\dfrac{\begin{equation}\begin{array}{c}{[\neg A]\\ \vdots\\ \bot}\end{array}\end{equation}}{A}\bot$$

It is important to note that RAA is not an instance of negation introduction (see  below) but actually a separate rule.

$\neg A$ can be defined as syntactic sugar for what is actually $A \to \bot$, and $\top$ as $\neg \bot$ and hence $\bot \to \bot$. There is no deduction rule needed for switching between the two, they are treated as literally same formula that just looks different:

$$\neg A  \quad = \quad A \to \bot$$
$$\top \quad =  \quad\neg \bot \quad = \quad \bot \to \bot$$

Negation introduction and elimination is then a special case of implication introduction and elimination (negation elimination is what you call law of contradiction, and implication elimination is also known as modus ponens):

$$\frac{\begin{equation}\begin{array}{c}{[A]\\ \vdots\\ \bot}\end{array}\end{equation}}{\neg A}\neg I
\quad = \quad
\frac{\begin{equation}\begin{array}{c}{[A]\\ \vdots\\ \bot}\end{array}\end{equation}}{A \to \bot}\to I
$$

$$$$

$$\frac{A\quad \neg A}{\bot}\neg E
\quad = \quad
\frac{A\quad A \to \bot}{\bot}\to E
$$
EFQL is just RAA with no assumption discharged (see the comments for a discussion on the difference between the two):

$$\frac{\bot}{A}\text{EFQL}
\quad \rightsquigarrow \quad
\frac{\bot}{A}\bot
$$

The other rules can then be derived from these primitive rules:

For double negation,

$$\dfrac{A}{\neg\neg A}\neg \neg I
\quad \rightsquigarrow \quad
\dfrac{\dfrac{[\neg A]^1 \quad A}{\bot}\neg E}{\neg \neg A}\neg I^1
\quad = \quad
\dfrac{\dfrac{[A \to \bot]^1 \quad A}{\bot}\to E}{(A \to \bot) \to \bot}\to I^1
$$

and

$$\dfrac{\neg \neg A}{A}\neg \neg E
\quad \rightsquigarrow \quad
\dfrac{\dfrac{\neg \neg A \quad [\neg A]^1}{\bot}\neg E}{A}\bot^1
\quad = \quad
\dfrac{\dfrac{(A \to \bot) \to \bot \quad [A \to \bot]^1}{\bot}\to E}{A}\bot^1
$$

For the axioms,

$$\dfrac{}{\top}\top I
\quad = \quad
\dfrac{}{\neg \bot}\top I
\quad \rightsquigarrow \quad
\dfrac{[\bot]^1}{\neg \bot}\neg I
\quad = \quad
\dfrac{[\bot]^1}{\bot \to \bot}\to I^1
$$

and finally, $$
\dfrac{}{A\vee \neg A}\text{TND}
\quad \rightsquigarrow \quad
$$

see here; their $(* B)$ is our $* E$  (eliminiation) and $(* E)$ is $* I$ (introduction).

So natural deduction doesn't need axiom as primitives: Everything can be derived from the basic set of inference rules $\{\land I, \land E, \lor I, \lor E, \to I, \to E, \bot\}$.

Title: Negation rules in natural deduction


问:
Intuition behind multiplication in joint probability

In this example, If temp = high then there is chance for weather being sunny is 60%. And P(temp = high) = 30%.
The highlighted box represents days with high temperature and weather being sunny. In other words, P(temp=high ∩ weather = sunny) = 60% * 30% = 18%
As far as I know we multiply to get joint probability. So here I am trying to find intersection which is joint probability and I am doing multiplication operation.

Now, My question is, Why we need to do multiplication here? From kindergarten multiplication has been known as A*B. where A is added B times to itself. Which clearly does not signify the relationship of intersection. What I am missing here?

答:
Well, it is just by the definition of conditional probability that:

$$\mathsf P(\text{high_temp}\cap\text{sunny})=\mathsf P(\text{high_temp})\cdot\mathsf P(\text{sunny}\mid\text{high_temp})$$

Or in terms of generic events $A,B$:$$\mathsf P(A\cap B)=\mathsf P(A)\cdot\mathsf P(B\mid A)$$

Or with a little rearrangement: $\mathsf P(B\mid A)=\dfrac{\mathsf P(A\cap B)}{\mathsf P(A)}$

That is, the conditional probability of $B$ when given $A$ is just the measure of the probability for the intersection ($A\cap B$) relative to that for the given event ($A$).


问:
Find major and minor axis of given conic.
Find the equation of major and minor axis of the given conic.Also find length of major and minor axis.

$$5x^2 + 5y^2 +6xy + 22x - 26y + 29$$ =0

I tried finding the nature of conic and centre of conic by partially differentiating it but could not think ahead.

答:
Find the center of the ellipse then shoft your origin to it then you will get an equation like this
$$ax^2+2hxy+by^2=1$$
Now consider a circle $x^2+y^2=r^2$ a circle intersect ellipse in 4 points but if $r$ is semi major axis(or semi minor) the circle touches the ellipse in two points.

Now homogenize the equation using the circle
$$ax^2+2hxy+by^2=(x^2+y^2)/r^2$$
Now write a quadratic and set the discriminant to 0.

You will get two values of $r$ giving you the length of semi major and semi minor axes.


问:
Commutative ring: sum of nilpotent elements
I know that in a commutative ring, if $a$ and $b$ are nilpotent, then $a + b$ is nilpotent, and this can be proved using the binomial theorem. I cannot figure out, however, where the assumption of commutativity is used, because once this is lifted, the result is no longer necessarily true. Is commutativity required to invoke the binomial theorem? Is there another step where it's used?

答: The binomial theorem requires commutativity indeed (or, at least, that $a$ and $b$ commute). For instance, in a commutative ring you have$$(a+b)^2=a^2+2ab+b^2,$$but, in general, all that you can say is that$$(a+b)^2=a^2+ab+ba+b^2.$$Now, suppose that $a^2=b^2=0$. The fact that that $(a+b)^3=0$ (in a commutative ring) follows from the fact that$$(a+b)^3=a^3+3a^2b+3ab^2+b^3.$$But you would not be able to deduce that $(a+b)^3=0$ from$$(a+b)^3=a^3+a^2b+aba+ab^2+ba^2+bab+b^2a+b^3.$$


问:
Percentage change in volume of sphere calculation when there is an error in calculating diameter.
I am studying maths as a hobby and am doing a chapter on calculus and small changes and errors. I am trying to understand the following problem. I cannot get the answer in the text book, which is 6%. I have not tried to calculate the area change as I obviously am making a fundamental error.

If a 2% error is made in measuring the diameter of a sphere, find approximately the resulting percentage errors in the volume and surface area.

I have said:

Let V = volume and D = diameter

$V = \frac{4  \pi r^3}{3} = \frac{\pi D^3}{6}$

Now if $\delta V, \delta D$ represent small changes in V and D respectively:

$\frac{\delta V}{\delta D} \approx \frac{dV}{dD}$ and $\frac{dV}{dD} = \frac{\pi D^2}{2}$ so

$\delta V = \frac{\delta D.dV}{dD} = \frac{\delta D\pi D^2}{2} = \frac{2}{100}.\frac{\pi D^2}{2} = \frac{\pi D^2}{100}$

Percentage error = $\frac{\delta V}{V} = \frac{\pi D^2}{100} \div \frac{\pi D^3}{6} = \frac{\pi D^2}{100}.\frac{6}{\pi D^3} = \frac{6}{100D}$ but the answer is actually 6%.

Where have I gone wrong.

答:
$\delta V = \frac{\delta D.dV}{dD} = \frac{\delta D\pi D^2}{2} = \frac{2}{100} \times D \times \frac{\pi D^2}{2} = \frac{\pi D^3}{100}$

Please note $\delta D = \frac{2}{100} \times D $ (New $D$ minus old $D$).

$\frac{\delta V}{V} = \frac{\pi D^3}{100} \div \frac{\pi D^3}{6} = \frac{\pi D^3}{100}.\frac{6}{\pi D^3} = \frac{6}{100}$


问:
How to prove that $\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx$ converges
I have been trying really hard to prove that this integral $$\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx$$

converges. However, I did not succeed. For now, I proved that
$$\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx =  2  \int_1^\infty \frac{\ln(x)}{x^2-1}\, dx,$$
and that $f(x) = \frac{\ln(x)}{x^2-1}$ is continuous on $[1,+\infty)$.

If you can give me a hint, that would be nice.

答:
As you noticed, we have that

$$\int_0^\infty \frac{\ln x}{x^2-1} dx=\int_0^1\frac{\ln x}{x^2-1} dx+\int_1^\infty \frac{\ln x}{x^2-1} dx$$

and by $x=\frac1y$

$$\int_0^1\frac{\ln x}{x^2-1} dx=\int_{\infty}^1\frac{\ln\left(\frac1y\right)}{\frac1{y^2}-1} \left(-\frac1{y^2}\right)dy=\int_1^\infty \frac{\ln y}{y^2-1} dy$$

then

$$\int_1^\infty \frac{\ln x}{x^2-1} dx=\int_1^2 \frac{\ln x}{x^2-1} dx+\int_2^\infty \frac{\ln x}{x^2-1} dx$$

and the second integral converges bt limit comparison test with $\int \frac{dx}{x^{1.5}}$ while for the first integral by $t=x-1$ we have

$$\int_1^2 \frac{\ln x}{x^2-1} dx=\int_0^1 \frac{\ln (1+t)}{t(t+2)} dt$$

and as $t \to 0^+$

$$\frac{\ln (1+t)}{t(t+2)} \sim \frac{1}{t+2}$$


问:
How to prove or disprove that there are infinitely many different sequences that start with any finite set of initial terms)
Consider that we know some finite initial terms of a sequence.

How can we prove or disprove that there can be infinitely many sequences which have these terms as there initial terms?

答:
There are infinitely many sequences which have given finitely many initial terms.

For example, after the initial terms we could have $1,0,0,0,...; 0,1,0,0,...; 0,0,1,0,...; $ etc.

(So if you are given the first few terms of a sequence (e.g., $1,2, 4, 8, 16, 32$),

you can't say what the rest of the terms of the sequence are, without further information.)


答:
To answer the question you can either use calculus via Math Lover's method, or use knowledge of scale factors. Diameter is a length, so the length scale factor is either $0.98$ or $1.02$.

So the area scale factor is $0.98^2$ or $1.02^2$ which is $ \sim 0.96$ and $1.04 $ respectively: either way an error of 4%.

The volume scale factor is $0.98^3$ or $1.02^3$ which is $ \sim 0.94$ and $1.06 $ respectively: either way an error of $ \sim 6$%.

Here I am using the well-known result:

For small positive $x, \quad (1+x)^n \sim 1+nx$, which comes from the Binomial expansion of $(1+x)^n$.

You could argue that this answer is unsatisfactory, because proving that area scale factor = (length scale factor)$^2$ and volume scale factor = (length scale factor)$^3$ probably requires calculus...

Title: Percentage change in volume of sphere calculation when there is an error in calculating diameter.

Body:
I am studying maths as a hobby and am doing a chapter on calculus and small changes and errors. I am trying to understand the following problem. I cannot get the answer in the text book, which is 6%. I have not tried to calculate the area change as I obviously am making a fundamental error.

If a 2% error is made in measuring the diameter of a sphere, find approximately the resulting percentage errors in the volume and surface area.

I have said:

Let V = volume and D = diameter

$V = \frac{4  \pi r^3}{3} = \frac{\pi D^3}{6}$

Now if $\delta V, \delta D$ represent small changes in V and D respectively:

$\frac{\delta V}{\delta D} \approx \frac{dV}{dD}$ and $\frac{dV}{dD} = \frac{\pi D^2}{2}$ so

$\delta V = \frac{\delta D.dV}{dD} = \frac{\delta D\pi D^2}{2} = \frac{2}{100}.\frac{\pi D^2}{2} = \frac{\pi D^2}{100}$

Percentage error = $\frac{\delta V}{V} = \frac{\pi D^2}{100} \div \frac{\pi D^3}{6} = \frac{\pi D^2}{100}.\frac{6}{\pi D^3} = \frac{6}{100D}$ but the answer is actually 6%.

Where have I gone wrong.


问:
Eigenvalue bound for product of matrices
For two square matrices $A$ and $B$ that are diagonalizable and have equal dimension, let $\lambda_i(\cdot)$ be the $i^{th}$ eigenvalue and $\lambda_{\max}(\cdot)$ be the max eigenvalue.

Is it true that for all $i$, $\lambda_i(AB)\leq \lambda_i(A)\lambda_{\max}(B)$ and if so, how can one show this?

答: No. Look at $A=\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}$ and $B=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}$.


答:
Let

$I=\int_0^\infty \frac{\ln(x)}{(x^2-1)} dx$.

We can rewrite this as

$-I=\int_0^\infty \frac{\ln(x)}{(1-x^2)} dx$.

Now the problem arises because a singularity point exists in this. You can try to resolve that in any one of the two ways. You can take the function as complex logarithm and integrate by finding the correct contour. Another way is to repeat your steps and get the equality

$\int_0^\infty \frac{\ln(x)}{(x^2-1)} dx=2\int_0^1 \frac{\ln(x)}{(x^2-1)} dx$.

Now in $(0, 1)$ we have taylor series

$\frac{1}{(1-x^2)}=\Sigma_{k=0}^\infty x^{2k}$

Now using dominated convergence theorem you can convert above integration into a summation of integrals and get the value for original integral.

Title: How to prove that $\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx$ converges

Body:
I have been trying really hard to prove that this integral $$\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx$$

converges. However, I did not succeed. For now, I proved that
$$\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx =  2  \int_1^\infty \frac{\ln(x)}{x^2-1}\, dx,$$
and that $f(x) = \frac{\ln(x)}{x^2-1}$ is continuous on $[1,+\infty)$.

If you can give me a hint, that would be nice.


问:
Proving line is tangent to circle.
Let $W_1$ be a circumcircle of triangle $ABC$. $D$ be any point on segment $AC$. And $W_2$ be a circle which is tangent to $BD$, $AD$ and circle $W_1$. $M$ be a tangent point on $AD$. Then prove that the line parallel to $BD$ that passes through the point $M$ is tangent to incircle of $ABC$.

My try: If we take the point which is the intersect of 2 circles as point $E$. With homotethy we achieve that $F$ is the midpoint of the arc $AC$. (Whereas $F$ is intersection of $W_1$ and $EM$). So $BF$ is angle bisector of angle $ABC$. And if we take the line that is parallel to $BD$ as $l$. Intersection of $l$ and $AB$ is $K$. Since angle $DNM$ $DMN$ and $KMN$ are equal. $MN$ is angle bisector of $KMD$. (WHEREAS $N$ is tangent point on $BD$).Now if we can prove angle bisector of $BCA$ or $BAC$ passes through the point where $BF$ and $MN$ intersected we will achieve that quadrilateral $BKMC$ is tangential one.

答: It's another way to phrase Sawayama's Lemma (statement and hints here). It states that if $N$ is the tangency point of the yellow circle to $BD$ then $I$, $M$ and $N$ are collinear. But $\triangle MDN$ is isosceles so $\angle DMI = \frac{1}{2} \angle CDB$. You already know that $MD$ is one tangent from $M$ to the incircle, so the other one must be symmetrical to it about $MI$, hence form the angle $2 \angle DMI = \angle CDB$ with $AC$, implying that it is indeed parallel to $BD$.

Title: Proving line is tangent to circle.


问:
Finite quotients of ring of integers of local field
Let $K$ be a non-Archimedean local field, so either a finite extension of $\mathbb{Q}_p$ or a finite extension of $\mathbb{F}_q((t))$. Let $\mathcal{O}$ denote its ring of integers and $\pi$ a uniformizer.

Is there a simple description of the finite ring $\mathcal{O}/ \pi^k \mathcal{O}$, for example in terms of the degree of the finite extension?

The case of $\mathbb{Q}_p$ giving $\mathbb{Z}/p^k\mathbb{Z}$ is what I would call very simple, and the case of $\mathbb{F}_q((t))$ giving $\mathbb{F}_q[t] / t^n$ is what I would call simple. But I have no intuition about how the case of finite extensions of these two fields could look like.

答:
$F=Q_p$ or $F_p((t))$, $$O_K/(\pi_K^k) = \{ \sum_{m=0}^{k-1} \pi_K^m s_m, s_m \in S\}$$ where $S$ are representatives of the residue field, from which the $O_F/(\pi_F^{\lceil  k/e \rceil})$-module structure follows easily (free module if $e | k$, non-free otherwise).

The ring structure is less obvious, it depends on the minimal polynomial of $\pi_K$.
$$O_K = O_F[\zeta_{q-1}][\pi_K]=O_F[x]/\psi(x)[y]/(f(x,y))$$ For $k$ large enough then $$O_K/(\pi_K^k)=O_F/(\pi_F^{\lceil  k/e \rceil})[x]/\psi(x)[y]/(f(x,y))$$

When $p\nmid e$ (tamely ramified) then $f(x,y) = y^e-\pi_F y^r$ for some $r$.

(If $e\nmid k$ then quotient by $y^k$)


问:
What is the geometric meaning of this vector equality? $\vec{BC}\cdot\vec{AD}+\vec{CA}\cdot\vec{BD}+\vec{AB}\cdot\vec{CD}=0$
I was doing some exercises for linear algebra. One of them was to prove that for any four points $A, B, C, D \in \mathbb{R}^3$ the following equality holds:
$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0$$
The proof is easy; you just make three vectors starting in $A$ and then see that all the terms cancel out.

My question is: what is the geometric interpretation of this equality? How can I visualize it or understand its deeper meaning? Does this equality have a name or where can I read more about it?

I'm asking this because it turns out that it is not just a random equality and is rather useful. For example, if we want to prove the existence of orthocenter, we can do it surprisingly easily and quickly using this equality.

答:
This equation is true for any $4$ points in $\mathbb{R}^n$, for $n\ge1$. Since any $4$ points in $\mathbb{R}^n$, for $n\ge3$, live in a $3$-dimensional hyper-plane, we get full generality from $\mathbb{R}^3$. However, the result is just as easy to prove in $\mathbb{R}^n$, so we will.

This equation is true in each coordinate; the dot product then simply sums the zeros in the coordinates. The equation in each coordinate is simply a statement about vanishing triple products in $\mathbb{R}^3$:
$$
\begin{align}
&\color{#090}{(C-B)}\cdot\color{#00F}{(D-A)}+\color{#090}{(A-C)}\cdot\color{#00F}{(D-B)}+\color{#090}{(B-A)}\cdot\color{#00F}{(D-C)}\\[3pt]
%&=\sum_{k=1}^n\begin{bmatrix}a_k&b_k&c_k\end{bmatrix}\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}\begin{bmatrix}d_k-a_k\\d_k-b_k\\d_k-c_k\end{bmatrix}\\
&=\sum_{k=1}^n[\color{#090}{(c_k-b_k)}\color{#00F}{(d_k-a_k)}+\color{#090}{(a_k-c_k)}\color{#00F}{(d_k-b_k)}+\color{#090}{(b_k-a_k)}\color{#00F}{(d_k-c_k)}]\tag1\\
&=\sum_{k=1}^n\color{#090}{\begin{bmatrix}1\\1\\1\end{bmatrix}\times\begin{bmatrix}a_k\\b_k\\c_k\end{bmatrix}}\cdot\color{#00F}{\begin{bmatrix}d_k-a_k\\d_k-b_k\\d_k-c_k\end{bmatrix}}\tag2\\
&=\sum_{k=1}^n\color{#090}{\vec u_k\times\vec v_k}\cdot\color{#00F}{(d_k\vec u_k-\vec v_k)}\tag3\\[9pt]
&=0\tag4
\end{align}
$$
Each triple product $\vec u_k\times\vec v_k\cdot(d_k\vec u_k-\vec v_k)=0$ because it represents the volume of the parallelepiped generated by $\vec u_k$, $\vec v_k$, and $d_k\vec u_k-\vec v_k$. Since these three vectors lie in the plane generated by $\vec u_k$ and $\vec v_k$, the parallelepiped is degenerate and has a volume of $0$.


问:
Proving that a function is nowhere differentiable so it cannot be analytically continued.
Here is a problem from Stein and Shakarchi

Where the definiton of analytic continuity is given here

This is what I have so far. At the unit circle, $g(z)=\sum_{n=0}^{\infty} 2^{-n \alpha} e^{i 2^n}$.
If g(z) is analytic, then $\forall z \in D$ the coefficients are given by $2^{-n \alpha} = \frac{g^(n)(z_0)}{n!}$.Can I somehow show that these $g^(n)$ don't exist.

答:
Another way of proving the result is as follows: suppose that $z_0\in C$ is a regular point. Then there's an open neighbourhood of $z_0$ such that $f$ has an analytic extension there.  Thanks to Abel's theorem, it is easy to see that such an extension must satisfy, on the intersection of the neighbourhood with the unit circle $$f(e^{i\vartheta})=\sum 2^{-n\alpha}e^{i2^n\vartheta}=\sum\frac{\cos(2^n\vartheta)}{2^{n\alpha}}+i\sum\frac{\sin(2^n\vartheta)}{2^{n\alpha}}$$

Both the real and the immaginary part of $f$ are Weierstrass' functions and so are continuous and non-differentiable. Thus $f$ cannot be analytically continued on any point of the unit circle (since $f$ has to be smooth at regular points).


问:
Two questions about integral scheme
Following are examples of 3.0.1 in Hartshorne's textbook

If $X=\operatorname{Spec}A $ is an affine scheme, then

i) $X$ is reduced if and only if $\operatorname{Nil}A=0$

$(\Rightarrow)$ for every open in $U$ in X, $\mathcal{O}_X(U) $ has no nilpotent.  Thus, take by $U=X$, $\mathcal{O}_X(X)=A$. Thus, $\operatorname{Nil}A=0$. The trouble is $(\Leftarrow)$. At the outset, I attampted to use contravariance propery of sheaf. since $U \subset X$, (for any open in $U$),  $\mathcal{O}_X(U) \supset \mathcal{O}_X(X)=A  $ , I expect to hold $\operatorname{Nil}\mathcal{O}_X(U) \subset \operatorname{Nil}\mathcal{O}_X(X)=\operatorname{Nil} A = 0  $...(★). However, (★) does not hold.

ii) $X$ is integral if and only if $A$ is an integral domain. : Also this has similar trouble .. $(\Rightarrow)$ is obvious by taking $U=X$, $\mathcal{O}_X(X)=A$ is an integral domain. But, $(\Leftarrow)$ is not easy for me.

答:
It's obviously reduced. And it has an unique minimal prime ideal $0$. So it's irreducible. Note that every $\{0\} \in V(\alpha) \Rightarrow \alpha = 0 \Rightarrow V(\alpha)=X$.

Now by the theorem following the example, we've done. You can also avoid using the theorem by translating it.

Title: Two questions about integral scheme


问:
vector matrices
Given that $u=(1, 1, 1)$, $v = (-2, 1, 0)$ and $w = (0, 3, 0)$, calculate the following (where possible):

```<code>(a) u . v
(b) u x v
(c) (u x v) x w
(d) (u . v) x w
(e) u . (v x w)
</code>```

Is $u \cdot v$ and $u \times v$ not the same thing?

答: Hint: Look at the definitions of the dot product ($\cdot$) and the cross product ($\times$).

Title: vector matrices


问:
Translating sentence into symbolic logic containing 'only'
I am very new to formal logic and have not encountered 'only' in a statement yet. How can we translate this into formal logic:

"Only yellow bananas are in the purple basket. Domain is set of all bananas."

I need to use symbols Y and P for this translation. Thanks for any help to write this in formal logic.

答:
Saying that only yellow bananas are in the basket is equivalent to saying that anything that is not a yellow banana is not in the basket. Hence:

$\forall x (\neg Y(x) \to \neg P(x))$

By contraposition, this is equivalent to:

$\forall x (P(x) \to Y(x))$


问:
Volume of cylinder between planes
Calculate the volume inside the cylinder $x^2+4y^2=4$, and between the two planes $z=12-3x-4y$ and $z=1$.

Converting to cylindrical coordinates gives $$r^2\cos^2\theta +4r^2\sin^2\theta=4\\ z=12-3r\cos\theta-4r\sin\theta\\ z=1$$.

$r$ goes from $0$ to $1$ and $2$. Can i use $0\le r\le 1.5$?

$0\le \theta \le 2\pi$

$1\le z \le 12-3r\cos\theta-4r\sin\theta$

$$\int_{0}^{2\pi}\int_{0}^{1.5}\int_{1}^{12-3r\cos\theta-4r\sin\theta} rdzdrd\theta $$

What have i done wrong?

答:
$z=12-3x-4y\implies z=12-3\color{red}{r\cos\theta}-4\color{blue}{r\sin\theta}$.

Note that the area over which you perform the integration is not a circle (it is an ellipse) so you will not be able to keep constant bounds for $r$. The upper bound of $r$ depends on $\theta$: for $\theta=0,r$ ranges till $2$ while for $\theta=\pi/2,r$ ranges till $1$. The upper bound on $r$ is given by:$$\frac{x^2}4+y^2=1\implies r^2(\cos^2(\theta)/4+\sin^2\theta)=1$$which gives $r_\max=(\cos^2(\theta)/4+\sin^2\theta)^{-1/2}$.

These are the two errors in your integral besides the typo on the bounds of $\theta$ (it should be from $0\to2\pi$ as you wrote earlier).

As an alternative, you could use the substitution $x=2r\cos\theta,y=r\sin\theta$ instead of the regular polar substitution. The upper bound of $r$ will become $1$, independent of $\theta$. This should be relatively easier to integrate.

Title: Volume of cylinder between planes


问:
$F$ is in the neighorhood of $L$ , $\lim_{z\to L} F(z) = F(L),$ and $\lim_{x\to x_0} f(x) = L.$ Show $\lim_{x\to x_0}F(f(x))=F(L)$
How do I prove this using $\delta$ and $\epsilon$?

Would the $\delta$ and $\epsilon$ be defined with regard to the limit of $f$ or the limit of $F$?

Or do each of them need their own $\delta$ and $\epsilon$ defined separately?

答:
$\lim_{z\to L} F(z) = F(L)$ means $$\forall \epsilon_1>0\exists\delta_1>0: d(z,L)<\delta_1\implies d(F(z),F(L))<\epsilon_1$$
and $\lim_{x\to x_0} f(x) = L$ means $$\forall \epsilon_2>0\exists\delta_2>0: d(x,x_0)<\delta_2\implies d(f(x),L)<\epsilon_2.$$
You want to show $\lim_{x\to x_0} F(f(x)) = F(L)$, which means  $$\forall \epsilon>0\exists\delta>0: d(x,x_0)<\delta\implies d(F(f(x)),F(L))<\epsilon.$$
Take any $\epsilon>0$. Now you have to show that such a $\delta>0$ exists.

Choose $\epsilon_1=\epsilon$, so you find a $\delta_1>0$ according the first equation with $d(F(z),F(L))<\epsilon$ for all $z$ with $d(z,L)<\delta_1$. Now choose $\epsilon_2=\delta_1$, so you find a $\delta_2>0$ according the second equation with $d(f(x),L)<\epsilon_2=\delta_1$ for all $x$ with $d(x,x_0)<\delta_2$. Alltogether you found a $\delta_2>0$ with $d(F(f(x)),F(L))<\epsilon$ for all $d(x,x_0)<\delta_2$.


答:
Well, this is just using the definitions of dot product and cross product.

Namely, if $u=(u_1,u_2,u_3)$ and $v=(v_1,v_2,v_3)$, then $$u \cdot v = u_1v_1+u_2v_2+u_3v_3.$$

So the dot product of two vectors is a real number. However, the cross product is a new vector. Here is a formula:

$$ u \times v = (u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-v_1u_2).$$

You should be able to do it from here.

Title: vector matrices

Body:
Given that $u=(1, 1, 1)$, $v = (-2, 1, 0)$ and $w = (0, 3, 0)$, calculate the following (where possible):

```<code>(a) u . v
(b) u x v
(c) (u x v) x w
(d) (u . v) x w
(e) u . (v x w)
</code>```

Is $u \cdot v$ and $u \times v$ not the same thing?


问:
How to solve the indefinite integral $\int \frac{x^{1 / 7}+x^{1 / 2}}{x^{8 / 7}+x^{1 / 14}} \ \mathrm{d} x$?
I need to calculate a complicated indefinite integral below:
$$\int \frac{x^{1 / 7}+x^{1 / 2}}{x^{8 / 7}+x^{1 / 14}} \ \mathrm{d} x$$
Can you help me to find out how it solved?I think my answer by using the exchange methods three times:
$$
70\ln \left|4x^{5/7}-4x^{5/14}+4\right|-\frac{1}{\sqrt{3}}\arctan \left(\frac{2x^{5/14}-1}{\sqrt{3}}\right) + x^{5/14} + C
$$
is wrong.

Any help would be greatly appreciated!

答:
You should apply the substitution: $x = t^{14}$

This will get you a rational function of $t$.

And then... as we know all rational functions can be integrated,

there is a well-known procedure for that.


问:
Count a subset of a countably infinite set
This is related to subset of finite set

A countably infinite set S is a set that has a bijection  M to Z+ ={1,2,3...}. The problem is to show that a subset of such a set is countable, that is that it can be bijectively mapped to Z+ or to [n]={1,2,3... n}.

It's not clear to me what can be taken as given axioms for this problem.

It is possible to describe an algorithm that constructs such a mapping: The superset  S maps to Z+, so start with the element that maps to 1 under M. For the map we are constructing, map it to one. Iterate ( infinitely) over the elements of S. For the k-th element, map it to the successor (i.e. add one) of the image of the k-1 th element.

This sounds like induction.

Is it permissible to describe an algorithm and call that a proof? Does it matter that the algorithm might have an infinite number of steps?

答:
Let $M:S \to \mathbb{Z}_+$ be a bijection. Let $U \subseteq S$ and notice that $M:U \to M(U)$ is a bijection. So, proving that $U$ is countable is equivalent to prove that $M(U)$ is countable.

So we just need to prove that "Every subset $T \subseteq \mathbb{Z}_+$ is countable".

If $T$ is finite then the result follows. So suppose it is not the case and let's enumerate $T$. Choose $x_1 \in T$ to be the  least element of $T$ (using the well-ordering principle). Now, suppose that $x_1 < x_2 < \cdots < x_n$ are defined and write $T_n=\{x_1, \cdots, x_n\}$. Since $T_n \subseteq T$ and $T$ is not finite we have that $R_n=T \setminus T_n \neq \emptyset$. So we define $x_{n+1}$ to be the  least element of $R_n$(again using the well-ordering principle). Then if we show that $T=\{x_1, \cdots, x_n, \cdots \}$ we are done. Suppose there is $x \in T$ such that $x \notin \{x_1, \cdots, x_n, \cdots \}$. Then $x \neq x_n$ for all $n$. Hence $x \in R_n$ for all $n$ and $x > x_n$ for all $n$. So $x$ is a positive integer greater then all elements of $\{x_1, \cdots, x_n, \cdots \} \subseteq \mathbb{Z}_+$ which implies that $\{x_1, \cdots, x_n, \cdots \}$ is limited. This is a contradiction because $X \subseteq \mathbb{Z}_+$ is finite $\iff X$ is limited (You can check this). Hence $T=\{x_1, \cdots, x_n, \cdots \}$ as we wnated.


Title: How to solve the indefinite integral $\int \frac{x^{1 / 7}+x^{1 / 2}}{x^{8 / 7}+x^{1 / 14}} \ \mathrm{d} x$?

Body:
I need to calculate a complicated indefinite integral below:
$$\int \frac{x^{1 / 7}+x^{1 / 2}}{x^{8 / 7}+x^{1 / 14}} \ \mathrm{d} x$$
Can you help me to find out how it solved?I think my answer by using the exchange methods three times:
$$
70\ln \left|4x^{5/7}-4x^{5/14}+4\right|-\frac{1}{\sqrt{3}}\arctan \left(\frac{2x^{5/14}-1}{\sqrt{3}}\right) + x^{5/14} + C
$$
is wrong.

Any help would be greatly appreciated!

Body:
Hint:

As $[2,7,14]=14,$

Let $x^{1/14}=y, x=y^{14}, dx=14y^{13}dy$

$$\int\dfrac{y^2+y^7}{y^{16}+y}\cdot14y^{13}dy$$

$$=14\int\dfrac{y^{15}(1+y^5)}{y(1+y^{15})}dy$$

$$=14\int\dfrac{y^{10}}{1-y^5+y^{10}}\cdot y^4dy$$

Set $y^5=z$

The rest is left as an exercise!


答:
Combine
\begin{align}
\int_0^1 \frac{\ln x}{1-x^2}dx
= \int_0^1 \frac{\ln x}{1-x}dx - \int_0^1 \overset{x^2\to x}{\frac{x\ln x}{1-x^2} }dx = \frac34\int_0^1\frac{\ln x}{1-x}dx
\end{align}

$$2\int_0^1  \frac{\ln x }{1-x^2}dx
= \int_0^1\frac{\ln x}{1-x}dx+ \int_0^1\frac{\ln x}{1+x}dx
$$
to establish
$$ \int_0^1  \frac{\ln x }{x^2-1}dx=- \frac32 \int_0^1\frac{\ln x}{1+x}dx
$$
Then
$$ \int_0^\infty \frac{\ln x }{x^2-1}dx = 2\int_0^1  \frac{\ln x }{x^2-1}dx=- 3\int_0^1\frac{\ln x}{1+x}dx <-3\int_0^1{\ln x}dx=3
$$

hence, converging.


问:
Finding anti difference for a given expression
I am given:

$$ a_n = \frac{1}{n^2 +3n+2}$$

And am asked to find the anti difference (i.e: quantity which when differences over gives the expression)

So, I started by partial fractions

$$ a_n = \frac{1}{n+1} + \frac{-1}{n+2} $$

Now the answer to this is:

$$ -\Delta \frac{1}{n+1}$$

But its wrong for some reason ( doesn't match with given answers)..

Options:

答:
If you denote $$b_n = -\frac{1}{n+1}$$

then $$b_{n+1}-b_n = a_n = \frac{1}{n^2 +3n+2}$$

So $b_n$ is the desired antidifference of $a_n$.


问:
No of solutions for $x_1x_2x_3x_4 = 770$
Question:
Let $N$ be the set of all integral solutions of the equation $x_1x_2x_3x_4 = 770$. Find $N$

So, for this question I used cases. Using prime factorization we know that the factors are $7 , 2 ,5 ,11$.

Case 1:When $x_1,x_2,x_3,x_4$ are $7 , 11 , 5 ,2$ in all permutations then number of ways of arranging :$4!$

Case 2:When two of the numbers are already multiplied. Ex. $7,11,10,1$ and here first I have to choose two numbers and then arrange so:${4 \choose 2} \cdot 4!$

Case 3: When a pair of two numbers are multiplied. Ex. $77 , 10 , 1 , 1$ and here I have to choose two and then arrange:${4 \choose 2} \cdot \frac{4!}{2!}$

Case 4: When three numbers are multiplied. Ex.$7 , 110 , 1,1$ and here I have to choose three numbers then arrange:${4 \choose 3} \cdot \frac{4!}{2!}$

Case 5: When one number is $770$ and the others are $1$ then ways of arranging is $4$

Thus total no of ways ($N$) $= 292$. However the answer is given as $256$. Which case have I missed and is there a better method to approach this question. Any help will be appreciated!

答:
There is a simpler way to solve the problem without making multiple cases.

The number $770$, as you rightly said, has $2,5,7$ and $11$ as it's prime factors. The positive integers $x_1$,$x_2$,$x_3$,$x_4$ must exist as some combination of these factors.

Let $a_1$, $a_2$,$a_3$, $a_4$ represent the exponents of 2 in each of these numbers. Similarly, assume $b_n$, $c_n$ and $d_n$ as the exponents of 5, 7 and 11 respectively. Consider the exponents of 2. These will add up as we multiply each of the $x_n$s and the sum of the exponents of 2 from each $x$ must equal the total exponent of 2 in the number 770. Hence we get the equation:

$$a_1 + a_2 + a_3 + a_4 = 1$$

Similarly for other exponents,

$$b_1 + b_2 + b_3 + b_4 = 1$$
$$c_1 + c_2 + c_3 + c_4 = 1$$
$$d_1 + d_2 + d_3 + d_4 = 1$$

The formula for the number of solutions to such an equation is quite well known, and is equal to $\binom{4+1-1}{1} = 4$
Since these are $4$ such simultaneous equations, we can easily see that the total number of solutions must be $4^4 = 256$.

In fact, you can extend this method to solve for any equation of the form $x_1.x_2.x_3....x_n = k$ as long as you know the prime factors of the number $k$.


问:
Example where an inverse function does not equal the elements
I'm trying to find an example of a function $f: A \to B$ and $X \subset A$ so that $f^{-1}(f(X)) \ne X$, and similarly where $Y \subset B$ so that $f(f^{-1}(Y)) \ne Y$.

I thought to have $f = x^2$, which has no inverse, thus making it vacuously true that $f^{-1}(f(X)) \ne X$ and $f(f^{-1}(Y)) \ne Y$, but that seems like a copout. Any help is greatly appreciated.

答: Consider $f : \mathbb{R} \rightarrow \mathbb{R}$ defined for all $x \in \mathbb{R}$ by $f(x)=0$, and $X = \lbrace 0 \rbrace$ and $Y = \mathbb{R}$.


问:
How to model the arc tangent like function in a high school way, given the points on its graph?
I'm trying to come up with a regression function for a student of mine that best fits the the attached graph, but using a method for high school students, who're only familiar with trigonometry and calculus of one variable.

To be precise, they're given $n=30$ pair of points $(x_i,y_i)$ (which I'm not write here, but they do have them!) and they're also given the attached graph that the function generates. They're asked to model the function.

If it was not high school, then I'd have modeled it like:

$$y= f(x):= a \hspace{1mm} arctan (bx + c) + d,  $$
and I'd have tried the minimize the sum of squared errors $\sum_{i=1}^{n} (y_i - f(x_i))^2$, and would've minimized it w.r.t. $a,b,c,d$, form the normal equations and either analytically or numerically solve them.

But for high school, I can't do all that - they've barely seen regression and they don't know partial derivatives, so I'm not completely sure how to proceed here. Any help appreciated!

答:
The model being
$$y(x)=a \tan ^{-1}(b x+c)+d$$

Looking at the graph, a few approximations can be visually done

$$y(0)=a \tan ^{-1}(c)+d \approx 0$$

$$y(\infty)=a \frac \pi 2 +d \approx 4000$$

Around $x=6$, there is an inflection point; so $c+6b \approx 0$

Now, let them play with $c$ by trial and error.


答:
Hint: If $b_{n+1}-b_n = a_n$ then $b_n = b_0 + a_0 + \cdots + a_{n-1}$.
Since $a_n = \frac{1}{n+1} + \frac{-1}{n+2}$, the sum telescopes.

Title: Finding anti difference for a given expression

Body:
I am given:

$$ a_n = \frac{1}{n^2 +3n+2}$$

And am asked to find the anti difference (i.e: quantity which when differences over gives the expression)

So, I started by partial fractions

$$ a_n = \frac{1}{n+1} + \frac{-1}{n+2} $$

Now the answer to this is:

$$ -\Delta \frac{1}{n+1}$$

But its wrong for some reason ( doesn't match with given answers)..

Options:


答:
Here $f^{-1}$ does not denote inverse function of $f$ but inverse image of a set, which is defined even when $f$ is not invertible. In fact your search for an invertible example will be in vain as for invertible $f,f(f^{-1}(Y))=Y$ and $f^{-1}(f(X))=X$ for all $X\subseteq A,Y\subseteq B$.

$y(x):\Bbb R\to\Bbb R,y=x^2$ is a valid example. Take $X=\{1\}\implies f^{-1}(f(X))=f^{-1}(\{1\})=\{\pm1\}$.

Take $Y=\{0,-1\}$. Then $f(f^{-1}(Y))=f(\{0\})=\{0\}$.

Title: Example where an inverse function does not equal the elements

Body:
I'm trying to find an example of a function $f: A \to B$ and $X \subset A$ so that $f^{-1}(f(X)) \ne X$, and similarly where $Y \subset B$ so that $f(f^{-1}(Y)) \ne Y$.

I thought to have $f = x^2$, which has no inverse, thus making it vacuously true that $f^{-1}(f(X)) \ne X$ and $f(f^{-1}(Y)) \ne Y$, but that seems like a copout. Any help is greatly appreciated.


问:
Example on coherent scheme but not noetherian, $\mathrm{Spec}\underset{n \in \mathbb{N}}{\cup}k[[t^{\frac{1}{n}}]]$.
In class, my teacher gave an example of coherent scheme that is not noetherian, namely $\mathrm{Spec}\underset{n \in \mathbb{N}}{\cup}k[[t^{\frac{1}{n}}]]$.

The definition, of a coherent sheaf of module over a scheme $(X,\mathcal{O}_X)$, is a sheaf of $\mathcal{O}_X$-module locally (on $\mathrm{Spec}{A} \subset X$) being $\tilde{M}$ with $M$ a finitely generated $A$-module, and every kernel of arbitrary $A^{\oplus n} \rightarrow M$ is finitely generated.

Going back to the example. $k[[t^{\frac{1}{n}}]]:=A$ is obviously not noetherian. But I don't know how to show that kernel of arbitrary $A^{\oplus n} \rightarrow A$ is finitely generated.

答:
Define $A_n = k[[t^{\frac{1}{n}}]]$ for every $n \in \mathbb{N}$. Note that if $n$ divides $m$, then $A_m$ is a flat $A_n$-module. Indeed, we have
$$A_{m} = \bigoplus_{j=1}^{\frac{m}{n}}A_n\cdot t^{\frac{j}{m}}$$

So it is even a free $A_n$-module.

Consider a morphism $\phi:A^{\oplus s} \rightarrow A$. We denote its kernel by $K$. We have $\phi(e_i) = f_i$ for $1\leq i \leq s$ (here $\{e_i\}_{1\leq i\leq s}$ is the standard basis of a free module $A^{\oplus s}$). There exists $n_0$ such that $f_1,...,f_s\in k[[t^\frac{1}{n_0}]] = A_{n_0}$. Define
$$\mathcal{F} = \{n\in \mathbb{N}\,|\,n\geq n_0\mbox{ and }n_0\mbox{ divides }n\}$$
For $n\in \mathcal{F}$ define $\phi_n:A_n^{\oplus s}\rightarrow A_n$ by $\phi(e_i) = f_i$ for every $1\leq i\leq s$ (this time $\{e_i\}_{1\leq i\leq s}$ is the standard basis of a free module $A_n^{\oplus s}$). We also denote by $K_n$ the kernel of $\phi_n$.
Note that
$$1_{A_n}\otimes_{A_{n_0}}\phi_{n_0} = \phi_n$$
for $n\in \mathcal{F}$. Since for $n\in \mathcal{F}$ algebra $A_n$ is flat over $A_{n_0}$, we derive that the tensor product $A_n\otimes_{A_{n_0}}(-)$ preserves kernels and hence
$$K_n = A_n\otimes_{A_{n_0}}K_{n_0}$$
Thus
$$K = \bigcup_{n\in \mathcal{F}}K_n = \mathrm{colim}_{n\in \mathcal{F}}K_n = \mathrm{colim}_{n\in \mathcal{F}}\left(A_n\otimes_{A_{n_0}}K_{n_0}\right) =$$
$$= \left(\mathrm{colim}_{n\in \mathcal{F}}A_n\right)\otimes_{A_{n_0}}K_{n_0} = A\otimes_{A_{n_0}}K_{n_0}$$
Due to the fact that $K_{n_0}$ is finitely generated over $A_{n_0}$ ($A_{n_0}$ is noetherian), we derive that $K$ is finitely generated over $A$.

I think that more generally this proves that a filtered colimit of a diagram of flat ring extensions in which every ring is coherent is coherent itself.


问:
Second order inhomogeneous equation: $y''-2xy'-11y=e^{-ax}$
My question relates to an a second order inhomogeneous equation:

$$y''-2xy'-11y=e^{-ax}$$

First I need to investigate the homogeneous equation:

$$y''-2xy'-11y=0$$

$$y''-2xy'=11y$$

Forms Hermite's Equation where $\lambda = 11$

So I need a general solution of the homogeneous equation
$Ly(x) = 0$.

To do this I need to linearly independent solutions, $y_1(x)$ and $y_2(x)$ say, and
then the general solution of $Ly(x) = 0$.

becomes:

$$Ay_1(x) + By_2(x)$$

where $A$ and $B$ are arbitrary constants.

I am struggling to find two independent solutions of the hermite equation above:

My attempt

If I take a solution to be if the form $$y=\sum_{n=0}^{\infty}a_{n}x^n$$
Then putting this into the ODE I get the following:

$$\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=1}^{\infty}na_{n}x^{n}-11\sum_{n=0}^{\infty}a_{n}x^{n}=0$$

Which I can reduce to:

$$\sum_{n=0}^{\infty}[(n+2)(n+1)a_{n+2}-2(n+\frac{11}{2})a_{n}]x^{n}=0$$

And I have found that rearranging gives:

$$a_{n+2} = \frac{-2(n+\frac{11}{2})a_{n}}{(n+2)(n+1)}$$

This can be used to develop a recurrence relation...

How can I use this to find my two independent solutions $y_1(x)$ and $y_2(x)$ which I need in order to calculate the inhomogeneous solution?

Maybe there are more efficient ways of calculating this ODE that I am unaware of.

Edit

I displayed the equation incorrectly and have since edited it.

Should my solution be in the form of an infinite sum rather than a finite polynomial because I believe that the solution could only be constructed as a series solution which terminates if and only if $\lambda = −2n$ where $n \in \mathbb N$.

In my case $\lambda$ isn't of the above form hence an infinte series is required.

答:
You have correctly found the recurrent relation which gives $a_{n+2}$ as function of $a_{n}$
Hence, if you start with $a_0$ you get all the even coefficients and if you start with $a_1$ you get all the odd coefficients. Hence we get
$$y_1(x)=\sum_{k=0}^\infty a_{2k}x^{2k}\quad \text{and}\quad y_2(x)=\sum_{k=0}^\infty a_{2k+1}x^{2k+1}.$$
This solutions are defined up to a constant ($a_0$ and $a_1$ respectevely) that you can arbitrarly choose (but not null, otherwise we get the trivial solution).

Using this ansatz we are implicitly supposing that the solution is analytic in some interval. Hence you can check that $y_1-c y_2 \equiv 0$ means that each coefficient of the series is $0$ and hence both $y_1$ and $y_2$ should be zero. Hence if we choose $a_0,a_1 \neq 0$ we get independent solutions.

We could use the 2 homogeneus solutions with the method of variation of parameters to get the inhomogeneous solution.

Otherwise we could expand $e^{-a x}=\sum_{k=0}^\infty \frac{(-a)^k x^k}{k!}$ and we find the coefficients $a_i$ terms by terms.


问:
Solutions to equations involving hyperoperations
Are there any texts on solutions to equations involving hyperoperations?
Let's define $H_n(x,y) : \mathbb{N}_0 \rightarrow \mathbb{N}_0 $ to  be the n-th hyperoperation, in particular:

$H_n(x,y)=\begin{cases}
y+1 & n=0 \\
x & n=1,y=0 \\
0 & n=2,y=0 \\
1 & n\geq3,y=0 \\
H_{n-1}(x,H_{n}(x,y-1)) & \text{otherwise} \\
\end{cases}$

This comes from Hyperopration article on wikipedia.
The particular problem I am looking at is for fixed $m, n.$ What inverses or other operations are required to give a closed form solution of $H_m(x,x)=H_n(x,x)$ for $x?$ I would expect these to be some generalizations of Lambert W function, but I cant find what are they called.

答: The special case of $H_m(x,x)=H_n(x,x)$ is remarkably simple to solve. Assuming $m<n$, one can easily verify that $x\ge3$ has no solutions since $H_m(x,x)$ will be significantly smaller than $H_n(x,x)$. One can then verify that all of $x=0,1,2$ are always solutions except for a small handful of cases when $m<2$.


问:
Using calculus of residues to evaluate a trig integral

This is my partial attempt at the solution. I am unsure how to proceed further.

答:
Hint: You have a pole at $z=0$ of order $5001$ and you need the formula to calculate the residue.

Added: Another approach for finding the residue is to compute the Laurent series of the integrand. Rearranging the integral as

$$\frac{2\pi}{2^{5000}} \frac{1}{2\pi i}\int_{|z|=1}\frac{(1+z^2)^{5000}}{z^{5001}}dz.$$

Then computing the Laurent series using the binomial theorem gives

$$ \frac{(1+z^2)^{5000}}{z^{5001}} = \sum_{m=0}^{5000} {5000 \choose m} z^{2m-5001}.  $$

Now, the residue is the coefficient of $z^{-1}$ which corresponds to $m=2500$ in the above series and equals

$$ r = \frac{2\pi}{2^{5000}}{5000 \choose 2500} . $$


问:
Calculate $ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{\sin\left(x^{3}+y^{3}\right)}{\sin\left(x^{2}+y^{2}\right)} $
I have to calculate $ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{\sin\left(x^{3}+y^{3}\right)}{\sin\left(x^{2}+y^{2}\right)} $

From wolfram calculator I know the limit is $ 0 $. The onl way I cant think of proving it is switch to polar, and to show that $ \lim_{r\to0}\frac{\sin\left(r^{3}\left(\cos^{3}\theta+\sin^{3}\theta\right)\right)}{\sin\left(r^{2}\right)} $ is $ 0$.

If I'll treat $ \theta $ as a constant and I'll get that the limit is zero, is that mean that from any direction that the function getting closer to zero, the limit is zero?

If so, I could show it using l'Hospital's rule and I guess it would be easy, but I'm not sure its legit.

Thanks in advance

答:
By your way the result is not immediately clear, as previously suggested by Mark Viola, we can use that

$$\frac{\sin\left(x^{3}+y^{3}\right)}{\sin\left(x^{2}+y^{2}\right)}=
\frac{\sin\left(x^{3}+y^{3}\right)}{x^3+y^3}
\frac{x^2+y^2}{\sin\left(x^{2}+y^{2}\right)}\frac{x^{3}+y^{3}}{x^{2}+y^{2}}$$

and using standard limits we reduce to evaluate the simpler

$$\lim_{\left(x,y\right)\to\left(0,0\right)}\frac{x^{3}+y^{3}}{x^{2}+y^{2}}$$

Note that assuming $\theta$ constant corresponds to take the limit by linear path (i.e. $y=mx$) and we can't conclude that the limit is zero in this way.

More in general we can't prove the limit existence by this way, as noticed by Mark Viola in the comments, "we can show a limit fails to exist by showing that the value of the limit has different results along different paths. But we don't show the existence by looking at the limit along paths".

We need to use some bounding and squeeze theorem to conclude, that is in this case

$$\frac{x^{3}+y^{3}}{x^{2}+y^{2}}=r (\cos^3 \theta + \sin^3 \theta) \to 0$$

since by squeeze theorem

$$\left|r (\cos^3 \theta + \sin^3 \theta)\right|=r\left|\cos^3 \theta + \sin^3 \theta\right| \le 2r \to 0$$


答:
You can use inequality $\frac 12|u|\le |\sin(u)|\le |u|$ for small $|u|<1$ (it is true on a bit larger interval, but this is not very important).

$$0\le \dfrac{|\sin(x^3+y^3)|}{|\sin(x^2+y^2)|}\le 2\dfrac{|x^3+y^3|}{|x^2+y^2|}\le 2\dfrac{|x|^3+|y|^3}{x^2+y^2}\le 2\,\underbrace{\max(|x|,|y|)}_{\to 0}\ \underbrace{\dfrac{x^2+y^2}{x^2+y^2}}_1\to 0$$

Title: Calculate $ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{\sin\left(x^{3}+y^{3}\right)}{\sin\left(x^{2}+y^{2}\right)} $

Body:
I have to calculate $ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{\sin\left(x^{3}+y^{3}\right)}{\sin\left(x^{2}+y^{2}\right)} $

From wolfram calculator I know the limit is $ 0 $. The onl way I cant think of proving it is switch to polar, and to show that $ \lim_{r\to0}\frac{\sin\left(r^{3}\left(\cos^{3}\theta+\sin^{3}\theta\right)\right)}{\sin\left(r^{2}\right)} $ is $ 0$.

If I'll treat $ \theta $ as a constant and I'll get that the limit is zero, is that mean that from any direction that the function getting closer to zero, the limit is zero?

If so, I could show it using l'Hospital's rule and I guess it would be easy, but I'm not sure its legit.

Thanks in advance


问:
Is there an object such that its unique existence follows from choice axiom, while its existence cannot be proven without choice axiom?
Let $\varphi$ be a formula, and suppose $\text{ZFC}\vdash \exists x\varphi(x)\wedge\forall x\forall y(\varphi(x)\wedge\varphi(y)\to x=y)$

Is $\text{ZF}\vdash \exists x\varphi(x)$ true?

答:
Thankyou for your comment @Mauro ALLEGRANZA

I totally got your point.

Let $\varphi$ be a formula such that $\text{ZF}\vdash \exists_{unique} x\varphi(x)$

Then $\text{ZF}\not\vdash\exists x(\varphi(x)\wedge\text{AC})$ while $\text{ZFC}\vdash \exists_{unique} x(\varphi(x)\wedge \text{AC})$


问:
How to prove that $\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx$ converges
I have been trying really hard to prove that this integral $$\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx$$

converges. However, I did not succeed. For now, I proved that
$$\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx =  2  \int_1^\infty \frac{\ln(x)}{x^2-1}\, dx,$$
and that $f(x) = \frac{\ln(x)}{x^2-1}$ is continuous on $[1,+\infty)$.

If you can give me a hint, that would be nice.

答:
Near $1$,
$$\lim_{x \rightarrow 1 \\ x >1}\frac{\ln(x)}{x^2-1}  = \frac{1}{2}$$ so the function  is integrable near $1$, and near $+\infty$,
$$\frac{\ln(x)}{x^2-1} = o \left( \frac{1}{x^{3/2}}\right)$$ which is integrable near $+\infty$. Because the function is continuous over $(1,+\infty)$, then it is integrable over $(1,+\infty)$.


问:
Recall max$\{x_1,x_2\}=\frac{x_1+x_2}{2}+\frac{|x_1-x_2|}{2}$. Using this identity, derive a formula for max$\{x_1,x_2,...,x_n\}$
I understand the identity when it's dealing with two numbers

Again, recall that max$\{x_1,x_2\}=\frac{x_1+x_2}{2}+\frac{|x_1-x_2|}{2}$

I get stuck when dealing with $n$ numbers, but I can get the first part down

Here's what I have: max$\{x_1,x_2,\ldots,x_n\}=\frac{x_1+x_2+\cdots+x_n}{2}+{}$ "stuff"

I can't figure out how to use absolute value here.

Would the "stuff" be something like $\frac{|x_1-x_2-\cdots-x_n|}{2}$ or would it be

$$\frac{|x_1-x_2|+|x_2-x_3|+\cdots+|x_{n-1}-x_n|}{2} \text{ ?}$$

Or is it neither of those?

Body:
I suspect you may be expected to use things like $\max \{x_1,x_2, x_3\} = \max \{\max \{x_1,x_2\} , x_3\}$

That would then be
$$\frac{\frac{x_1+x_2}{2}+\frac{|x_1-x_2|}{2}+x_3}{2}+\frac{\big|\frac{x_1+x_2}{2}+\frac{|x_1-x_2|}{2}-x_3\big|}{2} \\= \frac{x_1+x_2+|x_1-x_2|+2x_3+\big|x_1+x_2+|x_1-x_2|-2x_3\big|}{4} $$

etc.


问:
Optimal Control - difference between indirect/direct approaches
In an indirect method, my understanding is we convert the Continuous Optimal Control problem to a 2-point boundary problem by using initial conditions on the states and terminal conditions on the costates.

Then there are the direct methods based on input and/or control discretization. In direct collocation, both states and controls are discretized and a single NonLinear Program is solved.

3 questions then:

Is the control trajectory found through indirect/direct approaches is a locally optimal one? (I think it is using the maximum principle so I don't see why this can be a global optimal)

How exactly is direct collocation is different then discretizing states/controls/time?

Lasty, why is indirect apprach is also called "first optimize, then discretize", where is the discretization part?

答:
The solution of the Euler-Lagrange equations (2-point BVP) which comes from Pontryagin's Maximum Principle (PMP) might not even be a local minimum since it constitutes a necessary condition for the input $\boldsymbol u$ to be a minimizer of $J(\boldsymbol u)$. You could also have calculated a local maximum or even a saddle point of $J(\boldsymbol u)$. In order to check if it is, in fact, a local minimizer you would have to solve the Hamilton-Jacobi-Bellman (HMB) equations which constitute a necessary and sufficient condition.

Collocation is just the general idea to discretize a continuous ODE
in time. More specifically, in optimization, collocation means that
you include each discretization step of the "collocation" scheme
that you choose, e.g., Euler, Simpson, Trapezoidal, ..., into the
the optimization problem as a equality constraints. This means you
implicitly integrate the ODE with the chosen scheme by means of
these equality constraints. This also means that it is just a
synonymous way of saying you discretize the states and inputs. In my
opinion this always comes along with including the discretization to
your problem. This, however, is not always necessary! You can
exclude the solution of the ODE from the optimization problem and
only discretize the control inputs in time which yields a less
complex (w.r.t. problem size) optimization problem. This approach is
known as control vector parameterization (and some other names). So
collocation => ODE part of problem; discretization => ODE not
necessarily part of problem

The term direct approach comes from the fact that you directly work with the continuous optimal control problem. You discretize the integral and all time-dependent functions (ODE, constraints, etc.) in order to obtain a static nonlinear program. For this nonlinear program you can construct optimality conditions given by the KKT equations. So first you discretize the OCP, then you optimize the resulting nonlinear static program to obtain the (approximate) optimal solution of the optimal control problem.
In the indirect approach, you first look at the optimality conditions of the continuous-time optimal control problem which leads you to the 2-point BVP, the Euler Lagrange equations (which, again, only constitute a necessary condition for a minimizer). This 2-point BVP is still continuous. So in order to solve it, you discretize it in time to obtain an approximate numerical solution.


答:
You can just note that
$$
\lim_{(x,y)\to (0,0)}\dfrac{\sin(x^3+y^3)}{\sin(x^2+y^2)} = \lim_{(x,y)\to (0,0)}\frac{x^3+y^3}{x^2+y^2}
$$

and, regarding this last limit, since

$$
\left|\frac{x^3+y^3}{x^2+y^2} \right|\leq \frac{|x|^3+|y|^3}{x^2+y^2}\leq \frac{2(x^2+y^2)^{3/2}}{x^2+y^2}=\sqrt{x^2+y^2} \to 0
$$

we conclude that the original limit exists and is zero.

note: I'm using the fact that $|x|, |y| \leq \sqrt{x^2+y^2}$.


问:
Line equation - parametric and canonical
Let's say I have a line in R3:

$$
l:\begin{cases}
x-3y+3z=0\\
x+2y-2z=2
\end{cases}
$$
How to change it to canonical and parametric equation?

答:
To get a parametric equation, add an equation:
$$ax+by+cz=t$$
Where $t$ is your parameter and $a$, $b$, $c$ are such that the system of three equations is regular. Here it's enough to let $t=y$.
Then solve for $x$, $y$, $z$ and you get them as functions of $t$.

Notice that with $t=x$, the system is not regular, since after removing $x$ from the first two equations, you get a system in $y$ and $z$ with null dterminant.

What do you call a canonical equation of the line ?


答:
Let $\varphi(x)$ say: (AC and $x=\varnothing$) or ($\lnot$AC and $x\neq x$).

ZFC proves the existence of a unique $x$ satisfying $\varphi(x)$, namely $\varnothing$.

ZF does not prove the existence of an $x$ satifying $\varphi(x)$. If it did, it would prove (AC and $x=\varnothing$) and hence prove AC, but AC is independent of ZF.

Edit: Oops, the second disjunct is not necessary here. (AC and $x=\varnothing$) works fine, as shown in Yuz's answer.

Title: Is there an object such that its unique existence follows from choice axiom, while its existence cannot be proven without choice axiom?

Body:
Let $\varphi$ be a formula, and suppose $\text{ZFC}\vdash \exists x\varphi(x)\wedge\forall x\forall y(\varphi(x)\wedge\varphi(y)\to x=y)$

Is $\text{ZF}\vdash \exists x\varphi(x)$ true?


答:
In your solution, you take $\theta$ constant with respect to $r$ and use l'Hospital. This proves that the limit is $0$ along straight line paths $y=x\tan\theta$ through the origin. But this doesn't prove that the limit exists in general. While here the limit exists and is equal to $0$, here is a question of mine posted about an year ago where I found that the limit may not exist despite having the same value along all straight line paths.

You can solve it like this: if $x^3+y^3=0$ i.e. along $y=-x$, the numerator is $0$ so the limit is $0$.

If $x^3+y^3\ne0$, then$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^3+y^3)}{x^3+y^3}\times\frac{x^2+y^2}{\sin(x^2+y^2)}\times\frac{x^3+y^3}{x^2+y^2}$$The first two terms are standard limits equal to $1$. You can write $\frac{x^3+y^3}{x^2+y^2}$ in polar coordinates as $r(\cos^3\theta+\sin^3\theta)$, so the limit is$$\lim_{r\to0}r(\cos^3\theta+\sin^3\theta)=0$$


问:
A proof problem in mathematical statistics

Let $X_{1}, \ldots, X_{n}$ be independent and identically distributed random variables, with uniform distribution in $(0,1)$. Consider $Y_{1}, \ldots, Y_{n}$ the associated order statistics and define $$V_{i} = \frac{Y_{i}}{ Y_{i + 1}}, i = 1, ..., n -1 \quad \text{and} \quad V_{n} = Y_{n}$$Show that $V_{1}, \ldots, V_{n}$ are independent and $V_{i}\sim  \textbf{Beta}(i, 1), i = 1, \ldots, n$.

How could you prove this result?

My approach: We need two prove two parts,  for to prove that $V_{i}$ are independent, we need to show something like $$f_{V_{1},V_{2},\ldots,V_{n}}(v_{1},v_{2},\ldots,v_{n})=f(v_{1})f(v_{2})\cdots f(v_{n})$$
and for the second part we need to show that $V_{i}\sim \textbf{Beta}(i,1)$, so we need to show that something like $$f(v_{i})=\frac{v_{i}^{i-1}(1-v_{i})^{1-1}}{B(i,1)}$$ is PDF of beta distribution.

Also, I think that I can prove that the joint distribution of order statistics for uniform distribution is also a uniform distribution and maybe use that result in the solution of problem.

I think I can use fundamental transformation theorem, but I don't sure how can I approach this problem.

答:
Here is the idea: You can condition on the ordering of the random variables! Due to symmetry and the fact that $X_1,\dots,X_n$ are continuous iid. RVs, it follows that $$\textbf{P}(X_1\leq X_2\leq\dots\leq X_n)=\frac{1}{n!}.$$ Therefore for any $(x_1,\dots,x_n)\in (0,1)^n$ it follows that \begin{align*}\textbf{P}(V_1\leq x_1,\dots,V_n\leq x_n)&=\textbf{P}(Y_1\leq x_1Y_2,Y_2\leq x_2Y_3,\dots,Y_n\leq x_n)\\&=n!\textbf{P}(Y_1\leq x_1 Y_2,\dots,Y_n\leq x_n|X_1\leq\dots\leq X_n)\textbf{P}(X_1\leq\dots\leq X_n)\\&=n!\textbf{P}(X_1\leq x_1 X_2,\dots,X_n\leq x_n|X_1\leq\dots\leq X_n)\textbf{P}(X_1\leq\dots\leq X_n)\\&=n!\textbf{P}(X_1\leq x_1 X_2,\dots,X_n\leq x_n)\\&=n!\frac{1}{n!}x_1x_2^2\cdots x_n^n.\end{align*}

In the first equality we have used the fact that $X_1,\dots,X_n$ are positive with probability 1. The second follows from symmetry and the law of total probability. The third uses the fact that $Y_i=X_i$ for this specific ordering and the fourth is really interesting: Since $X_1\leq x_1X_2$ is known, $X_1\leq X_2$ is redundant information! The same follows for the rest. Thus, you have to do some integration to get to the final result, which is exactly the formula you were looking for, since $F_{Y_i}(x)=\int^x_0 \frac{t^{i-1}}{1/i}\;dt=x^i$ for any $1\leq i\leq n$. You can get the marginal CDFs with the same trick. Hope this helped!


问:
Summing all the values of a function
The following function
$$f(x)=\left\{\begin{array}{ll} 50, &0\leq x\leq10 \\ 10+6x-0.2x^2, & 10<x\leq30\end{array}\right.$$
gives me the number of products that are sold in the $x$-th day of a month.

How am I supposed to find the approximated total number of sold products in a month? Is there a easier way than not doing
$$f(0)+f(1)+f(2)+\ldots+f(29)+f(30)?$$

Title: Summing all the values of a function

Body:
Yes.

$$\sum_{k=0}^{30} f(k)= \sum_{k=0}^{10} f(k) + \sum_{k=11}^{30} f(k) = \sum_{k=0}^{10} 50 + \sum_{k=11}^{30}(10 + 6k -0.2k^2)$$
$$=50 \times 11 + 10 \times 20 + 6 \left( \frac{30\times 31}{2} -\frac{10\times 11}{2} \right) - 0.2 \left(  \frac{30\times 31 \times 61}{6} -\frac{10\times 11 \times 21}{6} \right) = 1396$$


问:
Is disjoint union of discrete spaces is discrete?
I mean internal, literal and external as in here: Is there a notion that $0 + \mathbb Ri$ or $\mathbb R + 0i$ has an 'internal complexification' equal to 'external complexification' of $\mathbb R$?

According to Wiki, 'every disjoint union of discrete spaces is discrete'. This refers to 'external' disjoint unions like the disjoint union of topological spaces without (necessarily) some common ambient topological space.

2 questions:

For 'internal' but not 'literal' disjoint unions:

Let $\{A_i\}_{i \in I}$ now be topological spaces that are topological subspaces of a common ambient topological space $S$. Then disjoint union is the subset of $S \times I$ given by $A := \bigcup_{i \in I} A_i \times \{i\}$. I've seen this definition in geometry/topology texts where $I$ is actually a manifold such that $I$ has a topological structure.

For all $A_i$ discrete, what conditions (eg on cardinality or topology of $I$, or topology of $S$) make $A$ discrete?

For 'internal' and 'literal' disjoint unions:

Consider $S$ as the 'literal' and 'internal' union of $S = \bigcup_{i \in I} \{A_i\}$ of disjoint subspaces. (Disjoint is the original definition that each pair is disjoint: $A_i \cap A_j = \emptyset$.)

For all $A_i$ discrete, what conditions (eg on cardinality or topology of $I$, or topology of $S$) make $S$ discrete?

Context for Question 2: I notice that for non-constant holomorphic map of connected Riemann surfaces $F: X \to Y$ ($X$ and $Y$ need not be compact, and $F$ need not be surjective), that $F$ is 'discrete' meaning that each fibre of $F$ is discrete. In this case,

$$X = \bigcup_{y \in Y} F^{-1}(y),$$

such that $X$ is a 'disjoint union' of discrete subspaces. However, this is not quite the same 'disjoint union' as talked about in wiki.

答:
If by disjoint union of topological spaces, you are talking about the union of disjoint subsets of a topological space, the answer is no. Here is a counter example.

In $\mathbb{R}$, with the usual topology, every singleton $\{x\}$ is discrete, but $\mathbb{R} = \bigcup_{x\in \mathbb{R}}\{x\}$ is not discrete.


问:
Calculate the integral $\int \frac{\exp(kx)}{x}dx$
How can you calculate the integral $$\int_a^b\frac{\exp(kx)}{x}\,\mathrm dx$$for any $k>0$?</p>

答: The integral cannot be expressed by elementary functions, instead what one obtain is $$\int_{a}^{b}\frac{e^{kx}}{x}dx=Ei(bk)-Ei(ak)$$ that is, when the integral is defined. So, if I want to calculate the integral $\int_{1}^{2}\frac{e^{2x}}{x}dx$, I use the this result to obtain $$\int_{1}^{2}\frac{e^{2x}}{x}dx=Ei(4)-Ei(2) \approx 14.6766$$ Now, if you would want to approximate the integral then you could try to express the function as a serie. In our case that would be $$\int_{1}^{2}\sum_{n=-1}^{∞}\frac{2^{n+1}x^{n}}{(1+n)!}dx $$</p>


问:
Average density of a continuous random variable

Suppose, $X$ is a continuous random variable on $[0; 1]$, and $p_X$ is its PDF. Suppose $E[X] = \lambda$. What is the smallest possible value of $E[p_X(X)]$?

I managed to prove the following bound:

$$E[p_X(X)] \geq 3 \lambda^2$$

Indeed,

$$\lambda = E[X] = \int_0^1 t p_X(t) dt \leq \sqrt{(\int_0^1 p_X(t)^2 dt)(\int_0^1 t^2 dt)} = \sqrt{\frac{E[p_X(X)]}{3}}$$

by Hoelder inequality.

However, I do not know, whether this bound is tight…

答:
For $\ 0<\lambda<\frac{1}{3}\ $, $\ E\big[p_X(X)\big]\ge\frac{4}{9\lambda}\ $, and
$\ E\big[p_X(X)\big]=\frac{4}{9\lambda}\ $ for
$$
p_X(t)=\cases{\frac{2}{3\lambda}-\frac{2t}{9\lambda^2}&if $\ 0\le t\le3\lambda$\\
0&if $\ 3\lambda< t\le1\ $.}
$$

For $\ \frac{1}{3}\le\lambda\le\frac{2}{3}\ $, $\ E\big[p_X(X)\big]\ge4-12\lambda+12\lambda^2\ $, and $\ E\big[p_X(X)\big]=$$4-12\lambda+12\lambda^2\ $ for
$$
p_X(t)=4-6\lambda+(12\lambda-6)t\ .
$$

For $\ \frac{2}{3} <\lambda<1\ $, $\ E\big[p_X(X)\big]\ge\frac{4}{9(1-\lambda)}\ $, and
$\ E\big[p_X(X)\big]=\frac{4}{9(1-\lambda)}\ $ for
$$
p_X(t)=\cases{0&if $\ 0\le t<3\lambda-2$\\
\frac{4-6\lambda}{9(1-\lambda)^2}+\frac{2t}{9(1-\lambda)^2}&if $\ 3\lambda-2\le t\le1$.}
$$

$\ 0<\lambda<\frac{1}{3}\ $

It is straightforward to check that for
$$
 p_X(t)=\cases{\frac{2}{3\lambda}-  \frac{2t}{9\lambda^2}&if $\ 0\le t\le3\lambda$\\
0&if $\ 3\lambda< t\le1\ $}
$$
we have
\begin{align}
p_X(t)&\ge0\ \ \text{ for }\ 0\le t\le1\\
\int_0^1p_X(t)\,dt&=1\ ,\\
\int_0^1tp_X(t)\,dt&=\lambda\ \text{, and}\\
\int_0^1p_X(t)^2\,dt&=\frac{4}{9\lambda}\ .
\end{align}
Moreover, if $\ p\ $ is any probability density satisfying $\ \displaystyle\int_0^1p(t)\,dt=1\ $ and $\ \displaystyle\int_0^1tp(t)\,dt=\lambda\ $, then
\begin{align}
\int_0^1p(t)p_X(t)\,dt&=\int_0^{3\lambda}p(t)\left(\frac{2}{3\lambda}-  \frac{2t}{9\lambda^2}\right)\,dt\\
&\ge\int_0^1p(t)\left(\frac{2}{3\lambda}-  \frac{2t}{9\lambda^2}\right)\,dt\\
&=\frac{4}{9\lambda}
\end{align}
Therefore, for any such probability density,
\begin{align}
\int_0^1p(t)^2\,dt&=\int_0^1\big(p(t)-p_X(t)+p_X(t)\big)^2\,dt\\
&=\int_0^1\big(p(t)-p_X(t)\big)^2\,dt\\
&\ +2\int_0^1\big(p(t)-p_X(t)\big)p_X(t)\,dt\\
&\ \ +\int_0^1p_X(t)^2\,dt\\
&\ge\int_0^1\big(p(t)-p_X(t)\big)^2\,dt+ \frac{4}{9\lambda}\\
&\ge \frac{4}{9\lambda}\ .
\end{align}

$\ \frac{1}{3}\le\lambda\le\frac{2}{3}\ $

Again, it is straightforward to check that for $\
   p_X(t)=$$4-6\lambda+(12\lambda-6)t\ $ we have

\begin{align}
p_X(t)&\ge0\ \ \text{ for }\ 0\le t\le1\\
\int_0^1p_X(t)\,dt&=1\ ,\\
\int_0^1tp_X(t)\,dt&=\lambda\ \text{, and}\\
\int_0^1p_X(t)^2\,dt&=4-12\lambda+12\lambda^2\ .
\end{align}
Moreover, if $\ p\ $ is any probability density satisfying $\ \displaystyle\int_0^1p(t)\,dt=1\ $ and $\ \displaystyle\int_0^1tp(t)\,dt=\lambda\ $, then
\begin{align}
\int_0^1p(t)p_X(t)\,dt&=4-6\lambda+\lambda(12\lambda-6)\\
&=4-12\lambda+12\lambda^2\ .
\end{align}
Therefore, for any such probability density,
\begin{align}
\int_0^1p(t)^2\,dt&=\int_0^1\big(p(t)-p_X(t)+p_X(t)\big)^2\,dt\\
&=\int_0^1\big(p(t)-p_X(t)\big)^2\,dt\\
&\ +2\int_0^1\big(p(t)-p_X(t)\big)p_X(t)\,dt\\
&\ \ +\int_0^1p_X(t)^2\,dt\\
&=\int_0^1\big(p(t)-p_X(t)\big)^2\,dt+4-12\lambda+12\lambda^2\\
&\ge 4-12\lambda+12\lambda^2\ .
\end{align}

$\ \frac{2}{3} <\lambda<1\ $

If $\ \int_0^1p(t)\,dt=\lambda\ $, then $\ \int_0^1p(1-t)\,dt=1-\lambda\ $, so the result for this case follows from application of that of the the first case to $\ 1-X\ $.


问:
Count a subset of a finite set
Finite sets are defined as sets that can be bijectively mapped to [n]={1,2,...n}. Prove that a subset T of a finite set S is finite. That's the stated problem.

It is not clear to me what can be taken as given axioms.

The resolution is clearly to show the existence of a bijection from a subset of [n] to [m] where m<n.

It's trivial to describe an algorithm that does this: start with the smallest element and map it to 1. Proceed through S, mapping each element to an element of [m]. For example if T ={2,5,6...} map it to {1,2,3...}. For the k+1 th element of S, map it to the successor of the image of the k-th element of S. (By successor, I mean add one)

This sounds like induction.

Question: is it permissible to use this description of an algorithm and call it a proof?

答:
First you should try to prove the particular case "if $a \in X$, $X$ is finite then $X \setminus \{a\}$ is finite". This follows from the following lemma which has a simple proof

"If $f:X \to Y$ is a bijection, $x \in X$, $y \in Y$ then there is a bijection $g:X \to Y$ such that $g(x)=y$."

Then let $f:\{1, \cdots, n\} \to X$ be a bijection and suppose wlg that $f(n)=a$. Hence if $n=1$, $X \setminus \{a\}=\emptyset$ and if $n >1$, the restriction $f |_{\{1, \cdots, n-1\}}:\{1, \cdots, n-1\} \to X \setminus \{a\}$ is a bijection.

For the general case suppose $X$ is finite and let $S \subseteq X$. Take a bijection $f:\{1, \cdots, n\} \to X$ and let's prove the result by induction. For $n=1$, $S=X$ or $S = \emptyset$ and the result follows. Now the induction hypothesis is "If $X$ is finite, $f:\{1, \cdots, n\} \to X$ is a bijection and $S \subseteq X$ then $S$ is finite".  Let $f:\{1, \cdots, n+1\} \to X$ be a bijection. If $S=X$ there is nothing to prove so suppose $S \neq X$. Choose $a \in X$ such that $x \notin S$. Hence $S \subseteq X \setminus \{a\}$. From the particular case there is a bijection between $\{1, \cdots, n\}$ and $X \setminus \{a\}$. Then $S$ is finite by the induction hypothesis.

Title: Count a subset of a finite set


问:
To prove change of variables

If G is an antiderivative of f,G exists since f is continous.
$G.(\phi.\gamma))'(t)= G'(\phi .\gamma(t)) . (\phi.\gamma)'(t)$.

How do I proceed after this?

答:
Let $\gamma : [a, b] \to U$ be a path and $g: U \to \mathbb{C}$ continuous. The complex line integral of $g$ along $\gamma$ is defined as follow:
$$ \int_{\gamma} g(z) dz = \int_a^b g(\gamma(t)) \gamma'(t) dt. $$
Since $\phi$ is holomorphic, then $(\phi \circ \gamma)' = (\phi' \circ \gamma) \gamma'$.
Using the above definition twice, you get
\begin{align}
 \int_{\phi \circ \gamma} g(z) dz &= \int_a^b g((\phi \circ \gamma)(t)) (\phi \circ \gamma)'(t) dt = \int_a^b g((\phi \circ \gamma)(t)) \phi'(\gamma(t)) \gamma'(t) dt \\ &= \int_{\gamma} g(\phi(z)) \phi'(z) dz.
\end{align}

Title: To prove change of variables


问:
Is that directional derivative OK?
Recently I solved a directional derivative and I think it’s okay but I’m not sure.
The question of the exercise is: find the directional derivative at P=(1,1) of f(x,y)=|x+y-2|
I substituted and did the math, and it gave me cos(θ)+sin(θ).
Thanks in advance.</p>

答:
The directional derivative along $v=(\cos \theta, \sin \theta)$ would be given by
\begin{align*}
f'_v(1,1)= &\lim_{t \to 0}\dfrac{f(1+t \cos \theta, 1+ t \sin \theta)-f(1,1)}{t}=\lim_{t\to 0} \dfrac{|1+t\cos \theta+1+t\sin \theta-2|}{t}\\
 =& |\cos \theta+\sin \theta| \lim_{t\to 0}\frac{|t|}{t}.
\end{align*}

But, since this limit does not exist, the directional derivative is not defined at the point.

Title: Is that directional derivative OK?


问:
An analytic function having maximum modulus inside a domain
A function is given to be analytic in the closed disk:$|z|\leq 4$ with $Min|f(z)|=3$ on the circle $|z|=4$ and with $f(1)=2i.$

Can there be an example of such a function? I am having a hard time in coming up with such a function, as it violates the maximum modulus principle. Does such a function exist? Thanks beforehand.

答:
Pick a unit disc automorphism st $g(1/4)=2i/3$ (a composition of $\frac{4z-1}{4-z}$ with $\frac{3z+2i}{3-2iz}$ will do as $1/4 \to 0 \to 2i/3$) and let $f(z)=3g(z/4)$. Clearly $|f(z)|=3, |z|=4$ and $f(1)=2i$

No violation of maximum modulus since $1/f$ is not analytic in the given disc as it has a zero somewhere.


答:
If $0$ is in interval $(a,b)$ then integral does not converge.

In other cases this integral has no elementary derivation. The result in this case is

$$\mathrm{Ei}(k\cdot b)-\mathrm{Ei}(k\cdot a)$$

where $\mathrm{Ei}(x)$ is function given by
$$\mathrm{Ei}(x)=\int_{-\infty} ^{x} \dfrac{e^t}{t} \mathrm{d}t$$

Title: Calculate the integral $\int \frac{\exp(kx)}{x}dx$

Body: How can you calculate the integral $$\int_a^b\frac{\exp(kx)}{x}\,\mathrm dx$$for any $k>0$?</p>


Title: Summing all the values of a function

Body:
The following function
$$f(x)=\left\{\begin{array}{ll} 50, &0\leq x\leq10 \\ 10+6x-0.2x^2, & 10<x\leq30\end{array}\right.$$
gives me the number of products that are sold in the $x$-th day of a month.

How am I supposed to find the approximated total number of sold products in a month? Is there a easier way than not doing
$$f(0)+f(1)+f(2)+\ldots+f(29)+f(30)?$$

Body: An approximation might be $$\int_{-0.5}^{10.5} 50 \,dx +  \int_{10.5}^{30.5} (10+6x-0.2x^2)\, dx$$ which would suggest $\frac{4187}{3} = 1396-\frac13$, so not far away


问:
diffusion equation with time and space derivatives in boundary conditions
I think the system I'm working with can be modeled with the following initial boundary value problem based on the 1D diffusion equation.

partial differential equation

$u_t - Du_{xx} = 0$

boundary conditions

$u_t(0,t) = \beta D u_x(0,t)$

$u(L,t) = 0$

initial condition

$u(x,0) = \phi(x)$

Where $u$ is the dependent variable, $0<x<L$ is location, $t$ is time, $D$ is the diffusion coefficient, $\beta$ is a constant, and $\phi$ is some arbitrary function of $x$ specifying the initial value of $u$ over the domain.

What solution methods exist for this sort of IBVP? Would the BC at $x=0$ be considered a Neumann boundary condition? The methods I've read about so far don't deal with any scenarios where both time and space derivatives are present in the BC.

答:
Starting from @Leon Avery's answer , you have to solve
$$\cos (\omega )-\frac{\omega }{\beta }\, \sin (\omega )=0$$ Admitting that you are concerned by the first solution in $\omega$, rewrite the equation as
$$\omega \tan(\omega)=\beta$$ and have a look at this question of mine where is proposed, as a quite good approximation

$$\frac{\pi^2 \omega^2\left(\left(\frac{\pi ^2}{3}-\frac{17}{4}\right) \omega^2 +\pi^2\right) }{\omega^4-\frac{17 \pi ^2 }{4}\omega^2+\pi ^4 }=\beta$$ which reduces the problem to a quadratic equation in $\omega^2$.