问:
How to show $P^{n}_{B} \simeq P^{n}_{A} \times_{\operatorname{Spec} A} \operatorname{Spec} B$ as in Hartshorne?
In order to define projective $n$-space, Hartshorne claimed that 'if $A \rightarrow B$ is a homomorphism of rings, and $\operatorname{Spec} B \rightarrow \operatorname{Spec} A$ is the corresponding morphism of affine schemes, then $P^{n}_{B} \simeq P^{n}_{A} \times_{\operatorname{Spec} A} \operatorname{Spec} B$', where $P^{n}_{A}$ means $\operatorname{Proj} A[x_0,...,x_n]$.

My questions are what is the morphism between $P^{n}_{A}$ and $\operatorname{Spec} A$ and how do I show that $P^{n}_{B} \simeq P^{n}_{A} \times_{\operatorname{Spec} A} \operatorname{Spec} B$?

I think, if $f:A \rightarrow A[x_0,...,x_n]$ is the embedding of rings, then the morphism between $P^{n}_{A}$ and $\operatorname{Spec} A$ just sends homogeneous prime ideals of $P^{n}_{A}$ to its preimage under $f$, however I am not entirely sure how to show this is indeed a morphism of scheme.

Also, since there is a homomorphism of rings $A \rightarrow B$, I think we have a graded homomorphism of graded rings $A[x_0,...,x_n] \rightarrow B[x_0,...,x_n]$, which by Hartshorne exercise 2.14(b), induces a natural morphism of schemes $P^{n}_{B} \rightarrow P^{n}_{A}$.

答:
There is an alternative approach using the "language of representable functors" that might be of interest. It uses the fact that the projective space bundle represents a certain functor. You avoid having to check that the isomorphisms (defined locally) above glue to global isomorphisms.

If $\phi: A \rightarrow B$
with $S:=Spec(B), T:=Spec(A)$ and where $V:=A\{e_0,..,e_n\}$ and $V^*:=A\{x_0,..,x_n\}$.
Let $E^*:=p^*V^*:=B\otimes_A V^*$ where

$p: S \rightarrow T$

is the canonical map. We may form the projective space bundle $\mathbb{P}(V^*):=Proj(Sym_A^*(V^*))$ and projection map

$\pi_T: \mathbb{P}(V^*) \rightarrow T$.

By functoriality of this construction it follows the fiber product $S\times_T \mathbb{P}(V^*) $ satisfies the formula

I0. $S\times_T \mathbb{P}(V^*) \cong \mathbb{P}(p^* V^*)\cong \mathbb{P}(E^*).$

Let $F_{V^*}: Sch(T)\rightarrow Sets$

be the "well known" functor

$F_{V^*}(X,f):=\{ f^*V^* \rightarrow L \rightarrow 0\}/\cong$

where $L \in Pic(T)$. There is an isomorphism of functors

$F_{V^*}(-) \cong h_{\mathbb{P}(V^*)}(-)$.

(I believe this follows from Proposition II.7.12 in Hartshorne.)

Let

$F_{p^*V^*}: Sch(S) \rightarrow Sets$

be the functor

$F_{p^*V^*}(Y,g)=\{ g^*p^*V^* \rightarrow L \rightarrow 0 \}/\cong$

where $L\in Pic(S)$. Since $F_{p^*V^*}(-)\cong h_{\mathbb{P}(p^*V^*)}(-)$
and since there is a canonical projection map

$\pi_S : \mathbb{P}(p^*V^*) \rightarrow S$, we get a canonical map

$j:\mathbb{P}(p^*V^*) \rightarrow S\times_T \mathbb{P}(V^*)$.

At the level of functors, it maps the element $g^*p^*V^*\rightarrow L \rightarrow 0$ in $F_{p^*V^*}(Y,g)$ to the element $(p\circ g)^*V^* \rightarrow L \rightarrow 0$ in  $F_{V^*}(Y, p \circ g)$.

It should follow from Proposition II.7.12 in Hartshorne that the above constructed map is an isomorphism, proving that I0 is an isomorphism. To write down all the details in this proof is an exercise in the "language representable functors".

Since $\mathbb{P}(V^*):=Proj(A[x_0,..,x_n]):=\mathbb{P}^n_{Spec(A)}$ and $\mathbb{P}(E^*):=Proj(B[x_0,..,x_n]):=\mathbb{P}^n_{Spec(B)}$

we get an isomorphism

I1. $Spec(B)\times_{Spec(A)} \mathbb{P}^n_{Spec(A)} \cong \mathbb{P}^n_{Spec(B)}$.

Example 1. Let $A:=\mathbb{Z}$ be the ring of integers. It follows for any commutative unital ring $B$ there is an isomorphism

$ Spec(B)\times_{Spec(\mathbb{Z})}\mathbb{P}^n_{Spec(\mathbb{Z})}\cong \mathbb{P}^n_{Spec(B)}$.

Example 2. Let $u:Spec(\kappa(x))\rightarrow T$ be a point in $T$. it follows there is an isomorphism

$\pi_T(x):=Spec(\kappa(x))\times_T \mathbb{P}(V^*)\cong \mathbb{P}(\kappa(x)\otimes_A V^*) \cong \mathbb{P}^n_{\kappa(x)}$.

Hence the fiber of $\pi_T$ at $x$ is projective $n$-space over the residue field $\kappa(x)$ of $x$. Hence for different points $x,y \in T$ it follows
the fibers of $\pi_T$ at $x,y$ will be projective spaces over "different fields". Their characteristic may differ.

Example 3. Rings of integers in number fields. Let $K$ be any number field and let $E$ be a rank $d+1$ projective $\mathcal{O}_K$-module. Let $T:=Spec(\mathcal{O}_K)$ and let $\pi: \mathbb{P}(E^*)\rightarrow T$ be the projective space bundle of $E$. For any closed point $x\in T$ it follows the fiber of $\pi$ at $x$ is $\mathbb{P}^d_{\kappa(x)}$ and the residue field $\kappa(x)$ is a finite field. Let $U:=Spec(A) \subseteq T$ be an open subscheme where $E$ trivialize: $E(U)\cong A\{e_0,..,e_d\}$. It follows

$\pi^{-1}(U) \cong U\times_T \mathbb{P}^d_{\mathbb{Z}}$.

This example illustrates the difference between this viewpoint and the viewpoint in complex analytic geometry and differential geometry. The fiber of a complex holomorphic vector bundle (or projective bundle) is affine space (or projective space) over the complex numbers. The fibers of the map $\pi$ constructed above are defined over fields where the characteristic may vary.


问:
What is answer of $\sum_{i=m}^{n} a^i=?$
How can i solve the $\sum_{i=m}^{n} a^i=?$, I thought this formula was like this $\sum_{i=m}^{n} a^i=a^m\frac{1-a^n}{1-a}$ but it's not true!

答:
You counted the number of term wrongly. There are a total of $n-m+1$ terms.

Hence the formula should be $$a^m \cdot \frac{1-a^{n-m+1}}{1-a}$$

if $a \ne 1$.

You should handle the case where $a=1$ separately.

ContentLicense: CC BY-SA 4.0


答:
You are almost there:
$$\sum_{i=m}^na^i=a^m\sum_{i=0}^{n-m}a^i=a^m\frac{1-a^{n-m+1}}{1-a}$$
with the sum at $a=1$ being $n-m+1$, of course.

Title: What is answer of $\sum_{i=m}^{n} a^i=?$

Body: How can i solve the $\sum_{i=m}^{n} a^i=?$, I thought this formula was like this $\sum_{i=m}^{n} a^i=a^m\frac{1-a^n}{1-a}$ but it's not true!


问:
Probability: best chance of picking a desired marble out of 10
following are the two questions I've made myself, but I need help in solving them.

1) Suppose there are 10 marbles in a box. One out of them is the desired marble, or you can say one is black others are white,etc etc. Case 1 : You pick 3 marbles altogether out of the box. What is the probability that you'll choose the black marble ? Case 2: You pick 3 marbles out, but one at a time. Now what is the probability of getting the black marble in those 3 attempts ?

2) Now suppose there are 10 marbles arranged linearly, each with position say 1,2....10. again, one is black & rest are white.You are to pick 3 marbles in order to have best chances of getting a black marble. Mathematically, is it better to pick 3 successive marbles,ie position 3,4,5 or would it be better to pick up randomly, like position 3,6,9 or it makes no difference whether you do it either way ??

Note: Although both questions can be put into one single question, it'd be better to solve it, & for me to understand it, if put up as two seperate questions.

答:
An intuitive explanation:

Picking "3 marbles altogether out of the box" is the same than randomly shuffling an ordered list of the ten marbles and keeping the three initial ones.

Now, if you shuffle such a list (with no bias of course), you expect all permutations of the list to share the very same probability. Then, the black marble can be at any position in the list with the same probability.

Thus probability for the black marble to be at the first position is 1/10, the same for all other positions. Which leads to the expected answer: the black marble is among the three initial ones with probability 3/10.

ContentLicense: CC BY-SA 3.0


问:
Series of $\int_0^z \zeta^{-1} \sin \zeta d \zeta$
This is a homeworkquestion so I would appreciate some good hints. I have $f(z) = \int_0^z \zeta^{-1} \sin \zeta d \zeta$. Can this be written as a power-series in $\mathbb C$ around $z = 0$?</p>

答: Hint: What is the power series for $\frac{\sin z}{z}$ around $z=0$?</p>


问:
How is this 20square root3 here?

The question that we are given a polygon on left on diagram with each side 20m. Then in the right side I made a triangle of ABC. I drew head and tail of the vector on my own. Angle I got between vector AB and BC is 60. Therefore I drew components of BC . Then the triangle (green colour) I considered as my triangle whose length AC I have to find.

My teacher got the answer 20$\sqrt{3}$. So what I did is that $20^2 + 10^2 + 10$$\sqrt{3}$ As sum of Base and perpendicular = $ H^2$

My sir answer is 20$\sqrt{3}$ but I am not getting the same answer.Where am I wrong ?

I have tagged vectors since I want to confirm whether the way I drew head and tail of vector and components is right or not.

答:
When working with trigonometry for vectors, only the magnitudes come into play.  As such, the direction of the vectors drawn do not make a difference in calculating magnitudes.

Now for your question, without drawing the outside right triangles, this problem can be solved by using the law of cosines $$c^2=a^2+b^2-2ab\cos\gamma,$$
where in your case, $a=b=AB=BC=20$, $c=AC$, and $\gamma=120^\circ$.  Just showing the math here:
\begin{align}
AC&=\sqrt{20^2+20^2-2(20)(20)\cos120^\circ}\\
&=\sqrt{800+400}\\
&=\sqrt{1200}\\
&=20\sqrt3.
\end{align}

Title: How is this 20square root3 here?


问:
When is $f(f^{-1}(B)) \neq B$?
$f: X \longrightarrow Y$ and $B \subseteq Y$

I am currentry trying to give an example where $f(f^{-1}(B)) \neq B$.

I know, that for a function to have an inverse, it must be bijective.

My thought process is, that the answer can only be equal to $B$
if the function $f$ is bijective, because only then can every element in $f^{-1}(B)$ be directly traced back to B.

Here's where I struggle: How can $f(f^{-1}(B)) \neq B$ if the function needs to be bijective in the first place? Wouldn't this be a contradiction? Or is that exactly what I need to disprove here?

答:
For the part where you struggled, it is true that for a function to have an inverse, it must be bijective. But $f^{-1}(B)$ is a subset of $X$ that is called the inverse image of B. It is defined for all function $f$ as
$$ f^{-1} (B) = \left\{ x \in X \ | \ f(x) \in B \right\}. $$

for $A \subseteq X$, $f(A)$ is also called the image of A and defined as
$$ f(A) = \left\{ f(a) \ | \ a \in A \right\}. $$

Back to the question. For every function, $f(f^{-1}(B)) \subseteq B$ is always true. You can prove this by using two definitions above.

However, $B \subseteq f(f^{-1}(B))$ is false if $f$ is not surjective. The reason is that not every element of $Y$ has an inverse-image. That particular element will be inside $B$, but not $f(f^{-1}(B))$.

For example, let $f(x) = x$, $X = [0,1]$, and $Y = [0,2]$. Let $B = Y$. We have $f^{-1}(B) = [0,1]$, so that $f(f^{-1}(B)) = f([0,1]) = [0,1]$. There, $B \nsubseteq f(f^{-1}(B))$.

Title: When is $f(f^{-1}(B)) \neq B$?


问:
Decompose an ideal into sum of a product with maximal ideal and a principal ideal.

Let $A$ be a domain and $J\subset A$ be an ideal. Let $m$ be a maximal
ideal in $A$, and $I=mJ$. Let $x\in J\setminus I$. I want to show that
$J=xA+I$.

As pointed out by Mindlack's answer, above statement is wrong. The full setup in the book is as the following:

Let $A$ be a Dedekind domain, then an ideal $I$ admits a unique factorizaiton by primes $I=p_1^{e_1}\dots p_r^{e_r}$. Let $J=p_1^{e_1-1}\dots p_r^{e_r}$. Let $x\in J\setminus I$. I want to show that $J=xA+I$.

The statement is used to prove that $A/I\cong  (S^{-1}A)/I (S^{-1}A)$ for a multiplicative set $S$ disjoints from each $p_i$ in decomposition of $I$. So could you please not use that conclusion, to avoid circular argument?

答:
Edit: I'm answering the modified question, my point about the original question is below. Note that by the Chinese remainder theorem $A/I \cong \prod_i{A/p_i^{e_i}}$, so $J/I \cong p_1^{e_1-1}A/p_1^{e_1} \cong p_1^{e_1-1}A_{p_1}/p_1^{e_1}A_{p_1}$ but $A_{p_1}$ is a DVR (thus a PID) and thus the above is a one-dimensional $A/p_1$-vector space, so is generated (over $A$) by the image of any nonzero element, for instance the image of $x$.

The original question is wrong in general. Indeed, assume the result holds with $J=m$ and $A$ Noetherian. Then $m/m^2$ as a vector space over $A/m$ is generated by $[x]$ so has dimension one. As $A$ is an integral domain, $A_m$ (thus $A$) is a field or $A_m$ is a regular local ring of dimension $1$ hence a DVR.

If you want an explicit example, take $A=\mathbb{C}[X,Y]$, $m=J=(X,Y)$. Then $Y \in m \backslash (XA+m^2)$.


问:
Repeated roots in exponential ansatz method of solving second order differential equations
Consider the differential equation,

$$ y'' - 10y' + 25y=0$$

Suppose $ y = e^{rx}$, we will find $r=5$ as solution. THis suggests:

$$ y = A e^{5x}$$

But it can also be found that the family of functions of form $Bxe^{5x}$ also solves this differential equation. What exactly is the intuition behind this happening?

i.e: repeated root equations having $xe^{rx}$ also as solution?

答:
By variation of the constant,

$$A''(x)e^{5x}+10A'(x)e^{5x}+25A(x)e^{5x}-10A'(x)e^{5x}-50A(x)e^{5x}+25A(x)e^{5x}=0$$

or

$$A''(x)=0.$$

Then

$$A(x)=Cx+D.$$

An empirical explanation is that when the two roots are very close, say $r$ and $r+\delta$ with $\delta\ll$, the solution is

$$Ce^{rx}+D e^{(r+\delta)x}=(C+D e^{\delta x})e^{rx}.$$

Then plugging some initial conditions, you get

$$C+D=y_0,\\\delta D+r(C+D)=y'_0.$$

Then

$$C+De^{\delta x}=C+D+D\delta x+D\delta^2\frac{x^2}2+\cdots=y_0+(y'_0-ry_0)x+O(\delta)$$ which tends to $$y_0+(y'_0-ry_0)x.$$


问:
A Combinatorial/Algebraic Solution to Combinatorics Identities Exercise
The exercise:

$$\sum_{k=r}^n {n \choose k} {k \choose r} 2^k = {n \choose r} 2^r 3^{n-r}$$

I have tried everything. Combinatorial proof, differentiating a function to reach both sides, finding patterns of combinatorial identities, binomial coefficient identity and mixtures of all the above. Nothing. One idea which I thought was promising is summing both sides with $\sum_{r=0}^n$ so the right-hand side becomes $5^{n}$ by binomial identity but (a) I'm not sure that is even valid and (b) I couldn't make anything of the left-hand side.

Thanks.

答:
Here is a combinatorial proof. Suppose there are $n$ people, and I want to give them jobs. There are $3$ jobs, let's give them numbers $1,2,3$. Jobs number $1$ and $2$ can be done either at day time or night time, while job number $3$ can only be done at night. I don't care how many people will be at each job, but I want exactly $r$ people to work at day time. So how many options do I have to make such choice? First, I need to choose $r$ people who will work at day. Then for each one of them I need to give either job $1$ or job $2$, so it is $2$ options for each. Finally, for each of the $n-r$ people who will work at night I have three options: to give him either job $1$, job $2$ or job $3$. So the number of options is $\binom{n}{r}2^r3^{n-r}$.

Now let's compute the number of options in a different way. Suppose I want $k$ to be the number of people who will work at either job $1$ or job $2$. Note that $k\geq r$, because all $r$ people who will work at day are counted here. So I have to choose $k$ people out of $n$, the ones who will work either at job $1$ or $2$. Then from these $k$ people I have to choose the $r$ who will work at day time, the others will work at night. Finally, for each of these $k$ people I have two options: to give him job $1$ or to give him job $2$. Note that I don't have to choose anything else-the remaining $n-k$ people are exactly the ones who will do job $3$, and they will obviously do it at night time. And now we sum over the possible values of $k$, and get $\sum_{k=r}^n\binom{n}{k}\binom{k}{r}2^k$.


答:
Here is a way to derive it without any confusion.

Let $S = a^m + a^{m+1} + \cdots + a^n$. Then we see that
$$
S = a^m + a(a^m + a^{m+1}+ \cdots + a^{n-1}) = a^m + a(S-a^n)
$$
Then by solving for $S$ you have a formula for $S$. Hope you can finish that.


答:
Let $ n\in\mathbb{N} $, if $r\leq n $, then what is the $ r^{\textrm{th}}$ derivative of $ x\overset{f}{\mapsto}\sum\limits_{k=0}^{n}{\binom{n}{k}x^{k}} $ ?

Well $ \left(\forall x\in\mathbb{R}\right) $ : $$ f^{\left(r\right)}\left(x\right)=\sum_{k=r}^{n}{\binom{n}{k}\frac{k!}{\left(k-r\right)!}x^{k-r}} $$

On the other hand, since $ \left(\forall x\in\mathbb{R}\right),\ f\left(x\right)=\left(1+x\right)^{n} $, then : $$ f^{\left(r\right)}\left(x\right)=\frac{n!}{\left(n-r\right)!}\left(1+x\right)^{n-r} $$

Thus, for all $ x\in\mathbb{R} $, we have : $$ \fbox{$\begin{array}{rcl}\displaystyle\sum_{k=r}^{n}{\binom{n}{k}\binom{k}{r}x^{k}}=\binom{n}{r}x^{r}\left(1+x\right)^{n-r} \end{array}$}$$

Title: A Combinatorial/Algebraic Solution to Combinatorics Identities Exercise

Body:
The exercise:

$$\sum_{k=r}^n {n \choose k} {k \choose r} 2^k = {n \choose r} 2^r 3^{n-r}$$

I have tried everything. Combinatorial proof, differentiating a function to reach both sides, finding patterns of combinatorial identities, binomial coefficient identity and mixtures of all the above. Nothing. One idea which I thought was promising is summing both sides with $\sum_{r=0}^n$ so the right-hand side becomes $5^{n}$ by binomial identity but (a) I'm not sure that is even valid and (b) I couldn't make anything of the left-hand side.

Thanks.


答:
You have for $z\in\mathbb{C}$,
$${\sin(z)\over z} = \sum_{n=0}^\infty {(-1)^n z^{2n}\over (2n + 1)!}$$
from the Taylor series expansion. This is an entire function.</p>

Title: Series of $\int_0^z \zeta^{-1} \sin \zeta d \zeta$

Body: This is a homeworkquestion so I would appreciate some good hints. I have $f(z) = \int_0^z \zeta^{-1} \sin \zeta d \zeta$. Can this be written as a power-series in $\mathbb C$ around $z = 0$?</p>


问:
Doesn't the position of unit circle affect the answer in this question?
This is a KVPY-SA 2017 Question:

Section 5 - PartB-Mathematics

61)Let S be the circle in $xy$-plane which touches the x-axis at point A, the y-axis at point B and the unit circle $x^2 + y^2 = 1$ at point C externally. If O denotes the origin, then the angle OCA equals

A) $5π/8$

B) $π/2$

C) $3π/4$

D) $3π/5$

So, I'm preparing for KVPY this year and that's how I came upon this question. There's nothing wrong with the question. According to Wikipedia, It's normally placed in the origin. So I placed it and got the answer as $5π/8$. When I checked the KVPY answer key, I found my answer's right.

On googling the question (to see whether my procedure was right,)  I found a lot of solutions to this question like this one by toppr. Here's the accompanying figure:

(Please note that the axes in their image are interchanged.)

I know there are a lot of ways to find the answer but it's not the method in which they found the answer that makes me mad. In their solution, they didn't even keep the unit circle on the origin and yet they found the answer!!!!

So doesn't the position of the unit circle affect the answer.

Can anyone please explain the logic behind this?

Please forgive me if this is a dumb question. I can't just get the idea.

答:
The figure given in the toppr answer is wrong.
As a consequence, that answer also gives the wrong value of the radius of the circle.

But you don't actually need to know the radius of the circle.
All you really need to know is that the centers of both circles are on
the line $y = x$ and therefore so is the point of tangency of the circles, $C.$

So with the incorrect small circle removed from the figure from toppr, and with the $x$ and $y$ axes in the conventional orientation
($x$ axis horizontal), we have the following.
Note that the point of tangency $C$ is exactly where the toppr figure said it would be; they just used an incorrect circle to get the correct tangency point.
Using the correct circle would tell us that the distance $OC$ is $1,$
but since that information is not used in answering the question,
I have omitted the unit circle from the figure.

Now observe that $\triangle APC$ is isosceles. You know the angle at the apex $P$, so you can work out the two base angles (which must be equal).
And now that you know $\angle APC$ and $\angle CAP$
you can add them to get the exterior angle $\angle OCA$,
or you can subtract $\angle ACP$ from $\pi$ as the toppr answer does;
either way gets the correct answer.


问:
Why does the Minus-1 trick to get particular and general solutions work?
Minus-1 trick: A practical trick where the reduced row-echelon form of a system of equations is augmented with -1 rows. It is used to get particular and general solutions.

While it is a handy trick, I was wondering why does it work.

The following is an example of the minus-1 trick from the book Mathematics for Machine Learning Book

答:
It can easily be verified that, if:

$$\tilde{A}=\begin{bmatrix}1&3&0&0&3\\0&-1&0&0&0\\0&0&1&0&9\\0&0&0&1&-4\\0&0&0&0&-1\end{bmatrix}$$

then $\tilde{A}$ is an involutory matrix, i.e.:

$$\tilde{A}^{-1}=\tilde{A}$$

We will prove that later. However, let's see how this helps. The equation $Ax=0$ now becomes equivalent to $A\tilde{A}\tilde{A}x=0$, which is equivalent to:

$$\begin{bmatrix}1&0&0&0&0\\0&0&1&0&0\\0&0&0&1&0\end{bmatrix}\tilde{A}x=0$$

(As the rows of $A$ are some of the rows of $\tilde{A}$, then the rows of $A\tilde{A}$ are the corresponding rows of $\tilde{A}\tilde{A}=I$.)

Set $y=\tilde{A}x$, and solve the system:

$$\begin{bmatrix}1&0&0&0&0\\0&0&1&0&0\\0&0&0&1&0\end{bmatrix}y=0$$

It has obvious solutions:

$$y=\lambda_2\begin{bmatrix}0\\1\\0\\0\\0\end{bmatrix}+\lambda_5\begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}$$

However, then:

$$x=\tilde{A}^{-1}y=\tilde{A}y=\lambda_2\tilde{A}\begin{bmatrix}0\\1\\0\\0\\0\end{bmatrix}+\lambda_5\tilde{A}\begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}$$

and one can easily see that the two columns $\tilde{A}\begin{bmatrix}0\\1\\0\\0\\0\end{bmatrix}$ and $\tilde{A}\begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}$ multiplied by $\lambda_2$ and $\lambda_5$ are indeed the $2$nd and $5$th column of $\tilde{A}$.$\quad\blacksquare$

Now, why is $\tilde{A}^{-1}=\tilde{A}$? One can recall the algorithm for finding the inverse matrix for a given matrix, say $\tilde{A}$, via Gaussian elimination:

$$\left[\begin{array}{ccccc|ccccc}1&3&0&0&3&1&0&0&0&0\\0&-1&0&0&0&0&1&0&0&0\\0&0&1&0&9&0&0&1&0&0\\0&0&0&1&-4&0&0&0&1&0\\0&0&0&0&-1&0&0&0&0&1\end{array}\right]$$

Once you apply the following elementary transformations to bring the left-side matrix to $I$:

Add the second row multiplied by $3$ to the first

Add the fifth row multiplied by $3,0,9,-4$ to $1$st,$2$nd,$3$rd,$4$th row, respectively

Multiply the second and fifth row by $-1$

you will indeed get the matrix $I$ on the left side, but notice that you will actually create the very same matrix $\tilde{A}$ on the right side. Thus, $\tilde{A}^{-1}=\tilde{A}$.$\quad\blacksquare$


问:
Fundamental group of the projective plane minus two points
I want to compute the fundamental group of $\mathbb R \mathbb P_2 \backslash \{A,B\}$ (where $A$,$B$ are points) but I am stuck. I tried to use van Kampen's theorem bit I can't find two good subset (I tried $A=\mathbb R \mathbb P_2 \backslash \{A,B,p\}$ (where $p$ is anorher point) and $B=(\mathbb R \mathbb P_2 \backslash \{A,B\})^\circ$, but I can't manage to compute their fundamental groups).

Any hints? Am I on the right way with using Van Kampen, or should I try something else?

答:
Well, you can think of $\mathbb{R}\text{P}_{2}$ as the upper hemisphere of a $2$-sphere with antipodal maps on the boundary identified.  (See here: Real Projective Plane is Same as Identifying Antipodal Boundary Points of The $2$-Disc.).
Then, choose any two points on the upper hemisphere, and get rid of those. You can continually deform the resulting space into the boundary of $\mathbb{R}\text{P}_{2}$ along with an arc joining any two points on the boundary. This space is the same the wedge sum of two copies of $S_{1}$, and by Van - Kampen's theorem, the fundamental group of this space is the free product of two copies of $\mathbb{Z}$, i.e. $\mathbb{Z} * \mathbb{Z}$.


问:
Preimage of semi-infinite interval — terminology
For a function $f:\mathcal X \to \mathbb R$ and $L \in \mathbb R$, does the the preimage of the semi-infinite interval

$$ f^{-1}(\infty, L] = \big\{x \in \mathcal X\,:\,f(x) \leq L\big\}$$

have a general name? I.e., like "level set" is used for for $f^{-1}\{L\}$.

I imagine if it exists it may naturally arise in convexity or measure theory.

答: Sublevel set is a reasonable name for those.</p>


问:
How to modified the Gauss–Laguerre quadrature in the case of $\int\limits_{x=0}^{+\infty}{{e^{-ax}}f(x)}dx$?
1/ How to modified the integral  Gauss–Laguerre quadrature rule so that we could approximate the following integral:

$I = \int\limits_{x = 0}^{ + \infty } {{e^{ - ax}}f\left( x \right)} dx$

The things that make me feel uneasy is that although the exponential term is scaled by a positive constant $a$.

In that case what should we do to modified the original Gauss–Laguerre quadrature that is

$I = \int\limits_{x = 0}^{ + \infty } {{e^{ - x}}f\left( x \right)} dx \approx \sum\limits_{i = 1}^{i = n} {{w_i}f\left( {{x_i}} \right)}$

Where ${{x_i}}$ is the i-th root of the  Laguerre polynomial ${L_n}\left( x \right)$ and the weight is calculated by

${w_i} = \frac{{{x_i}}}{{{{\left( {n + 1} \right)}^2} \times {{\left[ {{L_{n + 1}}\left( {{x_i}} \right)} \right]}^2}}}$

2/ What does it really mean by saying the "the i-th root of the  Laguerre polynomial". Suppose I was looking for the root of
${L_3}\left( x \right) = \frac{1}{6}\left( { - {x^3} + 9{x^2} - 18x + 6} \right)$ then between 3 of these root

$\begin{gathered}
  x \approx 0.416 \hfill \\
  x \approx 2.294 \hfill \\
  x \approx 6.290 \hfill \\
\end{gathered} $

Which root would be consider as the ${1^{st}},{2^{nd}},{3^{rd}}$. In my opinion, the order seem to be quite irrelevant in this case

Please help me with this, thank you very much !

答:
The Gauss-Laguerre formula is for integrals of the form $\int_0^{+\infty}e^{-x} f(x) dx$. In your case, you just need to perform a change of variable $y = ax$
$$
\int_0^{+\infty} e^{-ax} f(x) dx = \frac 1a \int_0^{+\infty} e^{-y} \underbrace{f(y/a)}_{= \tilde f(y)} \,dy,
$$
and apply the usual rule for $\tilde f$, i.e.
$$
\int_0^{+\infty}e^{-ax} f(x) dx \approx \frac 1a \sum_{i=1}^n w_i f\left(\frac{x_i}{a}\right).
$$

The order of the points is irrelevant... as long as you assign the correct weights.


问:
How to systematically simplify deduction in formal first order logic into daily English
Suppose the function
\begin{equation}
h(x,y) = \frac{\sin\sqrt{1-x^2-2y^2}}{\sqrt{1-x^2-2y^2}}
\end{equation}
is defined on $\left\{(x,y)\vert x^2 + 2y^2 < 1\right\}$. We need to prove that
\begin{equation}
\forall x_0 \in \mathbb{R}, \forall y_0 \in \mathbb{R}, {x_0}^2 + 2{y_{0}}^{2} = 1 \longrightarrow \lim_{(x,y)\to(x_0,y_0)}h(x,y) = 1
\end{equation}

In my proof, I assumed the existence of the following conclusions:
\begin{equation}
\lim_{u\to 0}\frac{\sin u}{u} = 1,
\end{equation}
\begin{equation}
\forall u, \forall v, u^2 + 2v^2 = 1 \longrightarrow \lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0.
\end{equation}

To strictly follow the rules of first order logic, I applied the fitch style. The proof is divided into two parts. In the first part, I deduced the following proposition:
\begin{equation}
\begin{aligned}
&\left(\forall u, \forall v, u^2 + 2v^2 = 1 \longrightarrow \lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0\right) \longrightarrow\\
&\left(\forall u, \forall v, u^2 + 2v^2 = 1 \longrightarrow
\forall \varepsilon > 0, \exists \delta > 0, \forall x, \forall y, \left(0 < \sqrt{(x-u)^2+(y-v)^2} < \delta \wedge x^2+2y^2 < 1\right) \longrightarrow \sqrt{1-x^2-2y^2} < \varepsilon\right)
\end{aligned}
\end{equation}
The formal proof is as follows:

In the second proof, the final conclusion is deduced. The formal proof is given below:

The formal proof of this seemingly simple conclusion is very tedious. However, without proper use of first order logic, there is no systematic way to examine a deduction. Recalling my first classes in undergraduate, I feel very upset about the type of proof I learned at that time. Every deduction was based on intuitive use of math rules, especially for quantification. I am wondering how to train myself to prove in English while have a picture of first order logic in mind.

答:
Actually the proof you wrote is structurally very close to what a mathematician would write, the trick is to make the right omissions, and to be conservative when expanding definitions.

For example, by making the right omissions, translating a hypothesis by "suppose", universal elimination by "let", a conclusion by "then", and existential elimination by "there exists ... such that", then your first proof almost literally translates to:

Suppose $$\forall u. \forall v. (u^2 + 2v^2 = 1 \longrightarrow\lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0),$$ let $u_0,v_0$ be arbitrary, then suppose  $$u_0^2 + 2v_0^2 = 1,$$ then $$\lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0,$$ let $\epsilon_0$ be arbitrary, suppose $\epsilon_0
> 0$, then there exists $\delta_0$ such that $$\delta_0>0,$$ $$\forall x.\forall y. 0 < \sqrt{(x - u_0)^2 + (y - v_0)^2} < \delta_0\longrightarrow|1x^2-2y^2|<\epsilon_0^2,$$

let $x_0,y_0$ be arbitrary, then suppose $$0 <\sqrt{(x_0-u_0)^2+(y_0-v_0)^2}<\delta_0,$$ $$x_0^2 + 2y_0^2 < 1,$$ then $$1 - x_0^2 - 2y_0^2 > 0,$$ $$|1 - x_0^2 - 2y_0^2| < \epsilon_0^2,$$ $$\sqrt{1 - x_0^2 - 2y_0^2} < \epsilon_0,$$ thus $$\forall \varepsilon > 0. \exists \delta > 0. \forall x. \forall y. \left(0 <\sqrt{(x-u)^2+(y-v)^2} < \delta \wedge x^2+2y^2 < 1 \\\longrightarrow\sqrt{1-x^2-2y^2} < \varepsilon\right).$$

You could write this in even more natural language as

Suppose that $$\forall u,v. (u^2 + 2v^2 = 1 \longrightarrow \lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0).$$ Now let $u_0,v_0$ be arbitrary, and suppose
that $u_0^2 + 2v_0^2 = 1,$ then $\lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0.$
So for $\epsilon_0 > 0$,
there exists $\delta_0>0$ such that for all $x$ and $y$,
$0 < \sqrt{(x - u_0)^2 + (y - v_0)^2} < \delta_0$ implies $|1x^2-2y^2|<\epsilon_0^2.$

Now let $x_0,y_0$ be arbitrary such that
$0 <\sqrt{(x_0-u_0)^2+(y_0-v_0)^2}<\delta_0$ and
$x_0^2 + 2y_0^2 < 1,$
then since $1 - x_0^2 - 2y_0^2 > 0,$
we know
$$1 - x_0^2 - 2y_0^2 = |1 - x_0^2 - 2y_0^2| < \epsilon_0^2,$$
which implies that $\sqrt{1 - x_0^2 - 2y_0^2} < \epsilon_0$, concluding the proof.

So even though the natural language proof looks informal and contains omissions, the idea is that it should contain enough information to be able to recover the fully formal proof you have written. However different people write proofs with varying degree of details, depending on how much they expect the reader to be familiar with the topic.


问:
Why are the order-of-operations conventions good?
Children are sometimes taught silly mnemonics like "PEMDAS" to remember conventions on order of operations.  (I never heard of "PEMDAS" until long after graduating from college, as far as I can recall.  I think it means (1) parentheses, (2) exponentiation, (3) multiplication and division, and (4) addition and subtraction.)

I think it would be better to help them understand why those particular conventions, rather than some others, are a good thing.  Maybe even demonstrably optimal by some precisely definable desiderata?

How would one make the case for the usual conventions against possible alternatives?

I think I might say that if one operation distributes over another, it should come first, and we go from left to right because we normally read that way.  Maybe I'll post my own answer to this if I'm so inspired at some point.

答: I think that the conventions now in use are not necessarily better than any other possible convention; that they are what they are simply is a codification of historical usage; and the way to “make the case for the usual conventions against possible alternatives” is to observe that to change them would cause an untold amount of unnecessary difficulty, frustration, and anger.</p>


问:
Why aren't the values true and false part of logic, which is concerned only with the preservation of truth in arguments?
I quoted from the author's "Introduction for Instructors" on p. 15 where he's trying to distinguish his book from standard logic books that "did not address
significant problems and confusions—those that were common among the
students in those classes and those that appeared among students in other
philosophy courses, even when such students had taken more than one class
in logic and critical thinking."

I'm trying to decide if this book will introduce me comfortably to logic. But I don't understand the emboldened sentences. Can someone explain them so that a 16 year old high schooler can understand?

How can logic be "concerned only with the preservation of truth in arguments", if the values true and false aren't part of logic? Then again, I don't understand what the authors means by "minimalist account of the values true and false" or "preservation of truth".

      Critical thinking, as understood in this book, includes logic—formal and
informal—but in addition all rigorous methods for reliably arriving at true
propositions. Hence the book contains a minimalist account of the values true
and false. Such an account is not part of logic, which is concerned only with the
preservation of truth in arguments. The account of truth and falsity given here
is minimal in the sense that it includes all schemas of the form "It is true that P
if and only if P," where that P is a proposition which may be expressed in some
language and P picks out the state of affairs which the proposition purports
to describe. This account is stated less formally in the text by distinguishing
the state of affairs picked out by the proposition "P" from the proposition
"P." While this account is close to deflationary accounts such as those given
by Horwich$^1$, it is not meant to be a philosophically adequate account of truth
and falsity, but rather, in keeping with the pedagogical purposes of the book,
is meant to help students avoid common confusions by distinguishing the
language with which we talk about the world from the world about which
we talk. Moreover, this account of truth and falsity is not intended to rise to
the level of a correspondence theory of truth, since it proposes no detailed
articulation of states of affairs in accord with the articulation of propositions.

The Elements of Arguments: An Introduction to Critical Thinking and Logic by Philip Turetzky. I screen shot the whole p. 16 if you like more context.

答:
What is meant is that logic isn't concerned with the actual truth value of statements in the real world, or what exactly it means for a statement to be true. Logic is only concerned with how the truth value of a complex expression is determined by the assumed truth values of its component expressions, and how truth is preserved when reasoning from premises to a conclusion. Logic is about the structure of sentences and arguments, whereas assumptions about which primitive statements are true to begin with have to be made from the outside.

Logic will tell you that if you accept the premises that all humans are mortal and that Sokrataes is a human, then you have to accept the conclusion that Sokrates is mortal.

Logic will not tell you whether to accept the premise that all humans are mortal (and in consequence, whether you have to accept that Sokrates is mortal). Whether these statements are actually true is a matter of biology, not logic.


问:
Using a smooth function to replace a coordinate on a manifold
I don't know very much about differential geometry, so this may be trivial but I will try to make this question as clear as I can.

Suppose I have a $d$ dimensional smooth manifold $\mathcal{M}^d$, with a smooth maximal atlas $\mathcal{A}$. In the neighborhood of a point $m$ I have a chart $(U,\phi) \in \mathcal{A}$ that gives some local coordinates for my manifold, say $\phi(m) = (x^1(m), ..., x^d(m))$ and $\phi : U \subseteq \mathcal{M}^d \rightarrow \mathbb{R}^d$ is a homeomorphism. Suppose also that I have a smooth function defined on my manifold, $f: \mathcal{M}\rightarrow \mathbb{R}$.

My question is the following: Are there any conditions that guarantee that the chart $(V, \psi)$, with $m \in U \cap V$, $\psi = (x^1, ..., f)$ is also in $\mathcal{A}$?

In other words, how can I be sure that I can replace one of my coordinates with the smooth function $f$?

答:
Take a look at Tu's Introduction to manifolds, Section 6.3 on the Inverse Function Theorem.

The way the IFT is written there is that some local functions $(f_1,\dots,f_n):U\to\mathbb R^n$ define a coordinate chart around $p\in U$ iff there's a chart $(x_1,\dots,x_n)$ around $p$ such that
$$
\det\begin{bmatrix}\dfrac{\partial f_i}{\partial x^j}\end{bmatrix}\neq 0
$$

So we get a sufficient a condition:
$(x_1,\dots,x_{n-1},f)$ defines a chart if
$$
\det
\begin{bmatrix}
\dfrac{\partial x_1}{\partial x^1} &\cdots &\dfrac{\partial x_1}{\partial x^{n-1}}&\dfrac{\partial x_1}{\partial x^n}\\
\vdots & \ddots &\vdots & \vdots \\
\dfrac{\partial x_{n-1}}{\partial x^1} &\cdots &\dfrac{\partial x_{n-1}}{\partial x^{n-1}}&\dfrac{\partial x_{n-1}}{\partial x^n}\\
\dfrac{\partial f}{\partial x^1}&\cdots&\dfrac{\partial f}{\partial x^{n-1}}&\dfrac{\partial f}{\partial x^n}
\end{bmatrix}
=
\det
\begin{bmatrix}
1 & \cdots &0&0\\
\vdots & \ddots &\vdots & \vdots \\
0 &\cdots &1&0\\
\dfrac{\partial f}{\partial x^1}&\cdots&\cdots&\dfrac{\partial f}{\partial x^n}
\end{bmatrix}
=
\dfrac{\partial f}{\partial x^n}
\neq 0.
$$


问:
Weak convergence in Sobolev Space: does this integral converge to $0$?
Let $\Omega$ be an open bounded domain in $\mathbb{R^N}$ and $A(x)\in L^{\infty}(\Omega)$. Let $s\geq 1$ and $(u_n)_n$ be a sequence such that
$$|u_n|^s u_n\rightharpoonup|u|^s u\quad\mbox{ in } W_0^{1,p}(\Omega).$$
If we suppose $u=0$, it is true that
$$\int_{\Omega} A(x) \nabla(|u_n|^s u_n) dx\to 0?$$

About me the answer is yes, I reasoned in this way. Since $|u_n|^s u_n\rightharpoonup|u|^s u$ in $W_0^{1, p}(\Omega)$ and $u=0$, thus $|u_n|^s u_n\rightharpoonup 0$ in $L^p(\Omega)$ and $\nabla(|u_n|^s u_n)\rightharpoonup 0$ in $L^p(\Omega)$. Moreover, since $1\in L^p(\Omega)$ and since $A(x)\in L^{\infty}(\Omega)$, a constant $c\geq 0$ exists such that
$$\int_{\Omega} |A(x)| \nabla(|u_n|^s u_n) dx\leq c \int_{\Omega} \nabla(|u_n|^s u_n) dx\to 0.$$
It is true or am I missing something? Could anyone please help?

Thank you in advance!

答:
Your idea is right and your proof is almost correct. However your last estimate
$$\int_{\Omega} |A(x)| \nabla(|u_n|^s u_n) \,\mathrm{d}x\leq c \int_{\Omega} \nabla(|u_n|^s u_n) \,\mathrm{d}x\to 0$$
does not imply that $\int_{\Omega} A(x) \nabla(|u_n|^s u_n) \,\mathrm{d}x$ vanishes in the limit, as these quantities can be signed (note $\nabla(|u_n|^s u_n)$ is not necessarily non-negative!).

Instead observe that since $\Omega$ is bounded and $A \in L^{\infty}(\Omega),$ we have $A \in L^{p'}(\Omega) \cong L^p(\Omega)^*.$ Hence as $\nabla(|u_n|^su_n) \rightharpoonup 0$ in $L^p(\Omega),$ we have

$$ \lim_{n \to \infty} \int_{\Omega} A(x) \nabla(|u_n|^su_n)\,\mathrm{d} x = \int_{\Omega} A(x) \cdot 0 \,\mathrm{d} x = 0 $$
as required.


问:
What is the relationship between rank and solutions of a linear system with multiple columns?
We can determine whether or not a linear system $Ax = b$ has a unique solution or a solution at all based on the rank of the augmented matrix rank$(A|b)$ and the rank of the matrix rank$(A)$. (e.g.: Linear system solutions)

What happens in a more complex linear system where $b$ is replaced by $B$ which has multiple columns? For example:

Be $A \in K^{i,j}$ and $B \in K^{i,k}$ with $i,j,k \in \mathbb{N}$. When does the linear system $AX = B$ have at least one solution $\hat{X} \in K^{j,k}$?

答: Let $X = (x_1,\ldots,x_k)$ and $B = (b_1,\ldots,b_k)$. Then $AX = B$ is equivalent to $Ax_n = b_n$ for all $n \in \{1,\ldots,k\}$. Now you can apply your theorem recursively to obtain that $AX = B$ has a solution if and only if $\mathrm{rank}(A|B) = \mathrm{rank}(A)$.


问:
Maximum value(LPP)
Let $\alpha \in R $ . If $(3,0,0,\beta)$ is an optimal solution  of LPP,

Minimize $x+y+z-\alpha t$

s.t

$ 2x-y+z=6$

$-x+y+t=3$

$x,y,z,t \ge 0 $

The the maximum value of $\beta - \alpha$ is ?

My attempt: I tried by simply plugging in the value of optimal solution in constraint equation(2nd one) and got the value of $\beta = 6$.
Now to find the value of $\alpha$, I put the Min= - Max
That is the problem is reduced to

-Maximize $-x-y-z+ \alpha t$

Also, it can be observed that z and t seems to be slack variables for this problem. Hence, as the problem is not Maximum type but of -Maximum type, therefore slack variables must be subtracted from the -maximization equation instead of adding.(Is this correct?)

Therefore to make the above equation as
-Maximize $-x-y-z-t$
we have to put $\alpha = -1$
So, the maximum value must be 6-(-1)=7

答:
You have arrived at the right numerical answer, but, to be honest, I don't follow your logic in the last two paragraphs.

Also, it can be observed that z and t seems to be slack variables for this problem. Hence, as the problem is not Maximum type but of -Maximum type, therefore slack variables must be subtracted from the -maximization equation instead of adding.(Is this correct?)

Rewriting the objective function doesn't change the constraint.

It's possible that a max problem contains a '≥' constraint, even though it's "bizarre" in the sense of the "SOB" method.

There's no such thing callled "maximization equation".  The function to maximize is called the "objective function, while the constraints in the given LPP are all equations.

Therefore to make the above equation as -Maximize $-x-y-z-t$ we have to put $\alpha=-1$.  So, the maximum value must be $6-(-1)=7$.

I don't follow your idea as I'm already lost in the previous paragraph.

I agree with your observation that $z$ and $t$ "seem" to be a slack variable.  To put this more concretely, transform the given LPP into a new LPP having only two decision variables $x$ and $y$.  To make things simple, I keep the LPP type (minimization) unchanged throughout my answer.

Make each of $z$ and $t$ the subject in the corresponding equality constraint, then substitute them into the objective function in the given LPP, so that it becomes $\text{constant} -(\alpha+1)x - (\alpha-2)y$.  The constant term can be neglected.

Use the "slackness" $z \ge 0$ and $t \ge 0$ to get two inequality constraints $2x-y \le 6$ and $-x+y \le 3$.

(Optional) Plot the feasible region of the transformed LPP.

Invoke the Fundamental Theorem of LP to conclude that an optimal solution is attained at one of the extreme points of the feasible region:

coordinates
objective function value
optimal solution?

$(0,0)$
$0$
maybe

$(0,3)$
$-3\alpha+6$
no

$(3,0)$
$-3\alpha-3$
sure

$(9,12)$
$-21\alpha+15$
maybe

Deduce from the given optimal solution that $\alpha \ge -1$ and $\alpha \le 1$.

The previous step justifies the choice of $\alpha = -1$ for calculation the maximum possible value of $\beta - \alpha$.

Title: Maximum value(LPP)


问:
Determining the position of a vertex of a triangle

A triangle has a vertex $A$ at $(0,3)$, vertex $B$ at $(4,0)$, and vertex $C$ at $(x,5)$. If the area of the triangle is $8$, what is the value of $x$?

I did this problem out by using the formula of the given coordinates of the vertices. My answer was 2.66.

答:
The area of the triangle with vertices $(x_{i},y_{i})$, $i=1,2,3$ is given by $$\Delta = \frac{1}{2} \cdot \left| \begin{array}{cc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\\ x_{3} & y_{3} & 1 \end{array}\right| = \frac{1}{2}\left|\begin{array}{cc} 0 & 3 & 1 \\  4 & 0 & 1 \\ x & 5 & 1 \end{array}\right|$$

Now substitute $\Delta=8$, and the corresponding values for $(x_{i},y_{i})$ and solve for $x$. Solving you get $16 = -3(4-x) + 1 \times 20 = -12 + 3x +20$. Therefore you have $3x=8$ which says $x=\frac{\pm{8}}{3}$. You get the minus sign if you interchange the vertices. (Thanks to Arturo for pointing out the error.)


问:
Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.

Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.

I tried many method, such as AM-GM, but all of them failed.

Thank you.

答: Geometrically the given equation represents a sphere and $x+y+z=t$ represents a plane which can be shifting around. So the min will be achieved on $x,y$ or $z$-axes intersecting with the sphere. Hence the possible points are $(\frac{\sqrt{14}-1}{2},0,0),(0,\frac{\sqrt{17}}{2}-1,0),(0,0,\frac{\sqrt{22}-3}{2})$, and the min is $\frac{\sqrt{22}-3}{2}$.</p>


问:
Localization of Dedekind domain at a prime ideal is a P.I.D
Let $A$ be a Dedekind domain and $\mathfrak{p}\subset A$ be a prime ideal. Then the localization $A_\mathfrak{p}$ is also a Dedekind domain. I can show it has a unique maximal ideal $\mathfrak{p}':=\mathfrak{p}A_\mathfrak{p}$. Thus any ideal in $A_\mathfrak{p}$ can be expressed as $\mathfrak{p}'^n$. I want to show that for any $x\in\mathfrak{p}'\setminus \mathfrak{p}'^2$, we have $\mathfrak{p}'^n=x^nA_\mathfrak{p}$. This in particular implies $A_\mathfrak{p}$ is a P.I.D.

Any help or hint is much appreciated.

答:
Given the comments, the question is actually as follows: why is a local Dedekind domain a PID?

Let $A$ be a local Dedekind domain, and let $m$ be its maximal ideal. For each $x \in A \backslash \{0\}$, define $v(x) \in \mathbb{N}$ as the nonnegative integer such that $m^{v(x)}=xA$.

Let $x \in m$ be such that $v(x)$ is minimal: then, for each $y$, $yA=m^{v(y)} \leq m^{v(x)}=xA$ so that $y \in xA$. Therefore $m \subset xA$ and $m$ is a principal ideal, thus so are its powers, and therefore so is every ideal of $A$.


问:
What is the limit as $x\to\infty$ of the incomplete elliptic integral of the third kind with imaginary argument $\phi=-ix$?
I am trying to evaluate the limit as $x\to\infty$ of the incomplete integral of the third kind,

$$\Pi(n;\phi|m), \quad \mathrm{(using~the~Wolfram~language~convention)}$$

with $\phi=-ix$.

In particular I need to consider

$$\lim\limits_{x\to\infty}\Pi(a+ib;-ix|(-1)^{1/3}).$$

I've been unable to find anything in Abramovitz and Stegun, but experimenting on Wolfram Alpha (WA) shows that the limit can be calculated for a given $a+ib$. With $a+ib=1+i$, WA calculates the variation of the argument of the limit with $x$ as

<img src="https://i.stack.imgur.com/mznVn.png" alt="Plot of $\Pi(1+i;-ix|(-1)^{1/3})$ against $x$" />,

showing that, for $a+ib=1+i$ at least, the limit is approached quickly. WA calculates the limit to be $-0.24-0.69i$ to 2s.f. in this case.

Is there a closed form for the limit in terms of $a$ and $b$? I would like to compute the limit over a range of $a$ and $b$, this would be much simpler if I didn't have to take the limit each time. Thanks in advance for your help.

答:
Have a look at my Elliptic Integrals and Functions monograph. The relevant part is in section 2.3:

Let $m^*=1-m$ and $\tan\psi=-i\sin\varphi$, then (BF 161.01, .02, DLMF 19.7.7, 22.6(iv))
$$\Pi(n,\varphi,m)=\frac i{1-n}(F(\psi,m^*)-n\Pi(1-n,\psi,m^*))$$

If $\varphi=-ix$ with $x\ge0$, $\psi=-\tan^{-1}\sinh x$; as $x\to\infty$, $\psi\to-\frac\pi2$. Thus the integrals become complete in the limit:
$$\lim_{x\to\infty}\Pi(n,-ix,m)=\frac i{1-n}(F(-\pi/2,1-m)-n\Pi(1-n,-\pi/2,1-m))$$
$$=\frac i{1-n}(n\Pi(1-n,1-m)-K(1-m))$$


问:
difference of recursive equations
Lets have two recursive equations:

\begin{align}
f(0) &= 2  \\
f(n+1) &= 3 \cdot f(n) +  8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot  g(n) + 12
\end{align}

We want a explicit equation for f(x) - g (x).

I firstly tried to do in manually for first $n$ numbers

\begin{array}{|c|c|c|c|}
\hline
n & f(n) & g(n) & f(n) - g(n) \\ \hline
0 & 2 & -2 &    4 \\ \hline
1 & 6  & 6 &    0 \\ \hline
2 & 26  & 30 &  -4 \\ \hline
3 & 94 & 102 &  -8 \\ \hline
4 & 306  & 318 & -12 \\ \hline
\end{array}

We can deduce that $f(n) - g(n) = 4 - 4n$

But now we have to prove it.

Lets extend recursive equation $f(n)$:
$$f(n) = 3^n \cdot f(0) + 8 \cdot (3^{n+1} \cdot 0 + \dots + 3^{0}(n-1))$$
for $g$ we get $$g(n) = 3^n \cdot g(0) + 12 \cdot (3^{n-1} + \dots + 3^0)$$

We can simply check this by induction but I will skip it, so the question won't be so long.

Now lets  put it together:

$$
f(n) - g(n) = 2 \cdot 3^n + 8 * 3^{n-1} \cdot 0 + ... + 3^0 \cdot (n-1) + 2 \cdot 3^n - 4 \cdot 3 \cdot ( 3^{n-1}+ ... + 3^0)= \\
= 4 \cdot 3^n  + 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^n + 3^{n-1} + ... + 3^1) = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 \cdot 3^0 = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4
$$

As we can see, we already got the $4$, so to get $-4n + 4$, the rest of the equation must equal $-4n$. But this is where I don't know how to continue.

How to prove that:

$$8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = -4n$$

All I could do is this:

\begin{align}
&8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = \\
&= 4 \cdot (\frac{0}{2}3^{n-1} + \dots + \frac{1}{2} \cdot (n-1) - 4 * (\frac{2}{2}3^{n-1} + \dots + \frac{2}{2}3^0) = \\
&= 4 \cdot (3^{n-1} \cdot (\frac{0- 2}{2}) + \dots + 3^0 \cdot \frac{(n-1)-1}{2}) = \\
&= 4 \cdot (-\frac{2}{2}3^{n-1} + \dots + \frac{n-2}{2})
\end{align}

And I made sum function out of it:
$\sum^{n-1}_{i=0}{\frac{i - 2}{2}\cdot 3^{n-1-i}}$

What to do next? Did I go the wrong direction anywhere?

Thank you for your fast responses.

答:
I think you can just use an induction proof, since you already have an intuition of the result, it is much easier to check :

$$\text{Let : } H_n :"w_n=f(n)-g(n)=4-4n" $$

First, for $n=0$ :
$$w_0=4=4-4\times 0 $$
Hence, $H_0$ is true.

Let $n\in\mathbb{N} $, such that $H_n$ is true, let us show $H_{n+1}$
$$w_{n+1}=3w_n+8n-12=3(4-4n)+8n-12\\
= 12-12n+8n-12=-4n=4-4(n+1) $$

Hence, $H_{n+1} $ is true, so we can conclude that :

$$\forall n \in \mathbb{N}, f(n)-g(n)=4-4n $$

Title: difference of recursive equations


问:
Trouble finding multivariable limit of a piecewise function
We are required to find the limit of $f(x,y)$, as $(x,y)$ approaches $(1,1)$, where

$f(x,y)=1$,      if $x+y \geq2$

$f(x,y)=-1$,      if $x+y <2$

Approach If we approach the point $(1,1)$ through 2 different paths, viz: $x+y \geq2$ and $x+y <2$, we get the limits as $1$ and $-1$, i.e different. Therefore, the limit does not exist

This is indeed the correct answer, (i.e the limit does not exist) however, is this the correct solution?  Am I missing something in my idea of "approach"?

If its not the correct approach, then I cant think of any other method to solve the question: so any alternate method will be welcomed.

答: I don't see why is it that you call $x+y\geqslant2$ a path. A path would be the set $\{(t,1)\mid t\geqslant1\}$. That is, $\lim_{t\to1^+}f(t,1)=1$. Furthermore, $\lim_{t\to1^-}f(t,1)=-1$. So, the limit $\lim_{(x,y)\to(1,1)}f(x,y)$ doesn't exist.


问:
What's an $\mathcal O_X$-algebra when $X= \operatorname{Spec} R$?
Take $X= \operatorname{Spec}R$, $R$ a commutative ring with unit. What is an $\mathcal O_X$-algebra in that case? Is there more than just ordinary $R$-algebras?

Thank you in advance.

答:
In the following thread

Noetherian $R$-algebra corresponds to a coherent sheaf of rings on $\operatorname{Spec}(R)$

you find the following: Let $X:=Spec(A)$ with $A$ a commutative unital ring. I believe the following "Theorem" holds:

"Theorem". There is an equivalence of categories between the category of sheaves of quasi coherent commutative $\mathcal{O}_X$-algebras and maps of $\mathcal{O}_X$-algebras and the category of commutative $A$-algebras and maps of $A$-algebras.

Proof. Given a commutative $A$-algebra $R$ and an open set $U \subseteq X$. Define

C1. $\mathcal{R}(U):=\mathcal{O}_X(U)\otimes_A R.$

It follows C1 defines $\mathcal{R}$ as a quasi coherent sheaf of commutative $\mathcal{O}_X$-algebras on $X$. Conversely given a quasi coherent sheaf of commutative $\mathcal{O}_X$-algebras $\mathcal{R}$, it follows $R:=\Gamma(X, \mathcal{R})$ is a commutative $A$-algebra. You may check that this defines an equivalence of categories.

In the non-commutative setting the situation becomes complicated due to the fact that you cannot localize non-commutative rings. There is a class of non-commutative rings that localize well - rings of differential operators.


问:
How to prove that this graph is non-planar?

I'm having trouble proving that the above shown graph is non planar, I think that I have to look for a subdivision of the graph $K_{3,3}$ using Kuratowski's theorem, can someone give me any tips to find this?

答:
A good way to visualize $K_{3,3}$ for the purposes of planarity is as a hexagon together with the diagonals connecting opposite vertices.

To find a subdivision of such a thing in a graph, look for a long cycle - here, we might take the cycle $ABCDEHGFA$, for example, which includes all the vertices. Then, look for chords of that cycle that make the $K_{3,3}$ pattern: essentially, you want three edges such that if you draw them all inside the cycle, each one crosses the other two. (One choice, here, is the edges $AD, BG, CF$.) Then if you just take the cycle and those edges, you have a subdivision of $K_{3,3}$.

If the cycle we found did not contain all the vertices, we'd have a tougher time of it, because the other pieces of $K_{3,3}$ don't necessarily have to be single edges, but we could do it.

Alternatively, you could notice that the graph you have is bipartite, but all planar bipartite graphs on $n \ge 3$ vertices have $m \le 2n-4$ edges; here, we have $8$ vertices and $13$ edges, and $13 > 2\cdot 8 - 4$.


答: Consider the cycle $ADEHCF$. The edges between them form $K_{3,3}$ minus the edge $AH$; the path $ABGH$ links $A$ and $H$ without using vertices in the chosen cycle. This is a $K_{3,3}$ minor in the graph, so it is non-planar.

Title: How to prove that this graph is non-planar?

Body: I'm having trouble proving that the above shown graph is non planar, I think that I have to look for a subdivision of the graph $K_{3,3}$ using Kuratowski's theorem, can someone give me any tips to find this?


答:
So, denoting $w_n = f_n -g _n$, you can write down the equation
\begin{align}
w_0 = f_0-g_0 = 4, \qquad &w_{n+1} = f_{n+1}-g_{n+1} = 3(f_n-g_n)+8n -12, \quad\text{i.e.}\\
w_0 = 4, \qquad &w_{n+1} = 3 w_n +8n -12.
\end{align}

This a liner difference equation for which you can have an explicit solution, in the same way that you could have computed explicitly $f_n$ and $g_n$ right from the beginning.

The general solution of the homogeneous equation $w_{n+1} = 3 w_n$ is $w_n^h = c 3^n$, and if you search for a particular solution of the form $w_n^*  =  a n + b$, you'll get $w_n^* = 4 - 4n$. So the general solution is
$$
w_n = c 3^n +4 -4n.
$$

Finally, using the initial condition $w_0=4$, we get $c=0$ and the solution is $w_n = 4 - 4n$.

Title: difference of recursive equations

Body:
Lets have two recursive equations:

\begin{align}
f(0) &= 2  \\
f(n+1) &= 3 \cdot f(n) +  8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot  g(n) + 12
\end{align}

We want a explicit equation for f(x) - g (x).

I firstly tried to do in manually for first $n$ numbers

\begin{array}{|c|c|c|c|}
\hline
n & f(n) & g(n) & f(n) - g(n) \\ \hline
0 & 2 & -2 &    4 \\ \hline
1 & 6  & 6 &    0 \\ \hline
2 & 26  & 30 &  -4 \\ \hline
3 & 94 & 102 &  -8 \\ \hline
4 & 306  & 318 & -12 \\ \hline
\end{array}

We can deduce that $f(n) - g(n) = 4 - 4n$

But now we have to prove it.

Lets extend recursive equation $f(n)$:
$$f(n) = 3^n \cdot f(0) + 8 \cdot (3^{n+1} \cdot 0 + \dots + 3^{0}(n-1))$$
for $g$ we get $$g(n) = 3^n \cdot g(0) + 12 \cdot (3^{n-1} + \dots + 3^0)$$

We can simply check this by induction but I will skip it, so the question won't be so long.

Now lets  put it together:

$$
f(n) - g(n) = 2 \cdot 3^n + 8 * 3^{n-1} \cdot 0 + ... + 3^0 \cdot (n-1) + 2 \cdot 3^n - 4 \cdot 3 \cdot ( 3^{n-1}+ ... + 3^0)= \\
= 4 \cdot 3^n  + 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^n + 3^{n-1} + ... + 3^1) = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 \cdot 3^0 = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4
$$

As we can see, we already got the $4$, so to get $-4n + 4$, the rest of the equation must equal $-4n$. But this is where I don't know how to continue.

How to prove that:

$$8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = -4n$$

All I could do is this:

\begin{align}
&8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = \\
&= 4 \cdot (\frac{0}{2}3^{n-1} + \dots + \frac{1}{2} \cdot (n-1) - 4 * (\frac{2}{2}3^{n-1} + \dots + \frac{2}{2}3^0) = \\
&= 4 \cdot (3^{n-1} \cdot (\frac{0- 2}{2}) + \dots + 3^0 \cdot \frac{(n-1)-1}{2}) = \\
&= 4 \cdot (-\frac{2}{2}3^{n-1} + \dots + \frac{n-2}{2})
\end{align}

And I made sum function out of it:
$\sum^{n-1}_{i=0}{\frac{i - 2}{2}\cdot 3^{n-1-i}}$

What to do next? Did I go the wrong direction anywhere?

Thank you for your fast responses.


问:
How easily to check: $f$ is a closed map?
Assume that $f:\mathbb R\to\mathbb R$ is a function where $\mathbb R$ is the real number and the usual topology is defined on $\mathbb R$.

I have two questions:
   1. Let $C$ be closed. Then what is the proper form of $C$? For example, without the loss of generality, open set is replaced by the open interval.
   2. To check that $f$ is a closed map, we need some information: $f(C)$ is closed. How to easily check this?

答:
As I said in the comments, it will depend on the map $f$, in general.

In the case of $f(x) = x^2$ (also see comments) we can write $\mathbb{R}$ as $A \cup B$, where $A = \{x: x \le 0 \}, B = \{ x: x \ge 0\}$, so $A$ and $B$ are closed and $f | A$ and $f|B$ are homeomorphisms onto $[0, \infty)$. If then $C$ is closed, then $f[A \cap C]$ is closed in $[0,\infty)$, as homeomorphisms are closed maps and $A \cap C$ is closed in $A$, and likewise for $f[B \cap C]$. Hence their union which equals $f[C]$ is closed in $[0, \infty)$ as well and thus closed in $\mathbb{R}$ (closed in a closed sets is closed in the whole space). So $f$ is a closed map.


问:
Basic Question: Possible Jordan Canonical forms of a given $5 \times 5$ matrix
I have this exercise question from the book "A second course in linear algebra" and I am struggling to get a sense of this. This is probably embarrassingly simple and I do apologize in advance.

If $A \in M_5$, $(A−2I)^3 = 0$, and $(A−2I)^2 \neq 0$, what are the possible Jordan canonical forms for $A$?

答:
Hints: [Assuming that we're working over $\Bbb C$ (or some other algebraically closed field).]

Do you see that all the eigenvalues of $A$ must be $2$? (Why? Suppose $\lambda \neq 2$ is an eigenvalue and $v \neq 0$ is a corresponding eigenvector, what happens when you compute $(A - 2 I)v$? $(A - 2I)^2v$?...)

Suppose that all the Jordan blocks are of size at most $2$. What happens to $(A - 2I)^2$ then? Conclude that there is at least one Jordan block of size at least $3$.

What happens if there's a Jordan block of size at least $4$? Compute $(A - 2I)^3$ and conclude that that's not possible.

Conclude that there is one Jordan block of size $3$. What choices are left for the other block(s)? Show that both are possible.

In the above, you may assume that $A$ itself is in its Jordan form. (Why?)

It may be useful to recall how block matrices are multiplied.


答:
$$d(n+1)=3d(n)+8n-12,\\d(0)=4.$$

The homogeneous solution is

$$d_h(n)=3^nd_h(0).$$

Plugging the initial condition,

$$3^0d_h(0)=3\cdot4+8\cdot0-12=0.$$

Then with the ansatz $d_a(n)=an+b,$

$$an+a+b=3an+3b+8n-12$$ or by identification,$$d_a(n)=4-4n=d(n).$$

Title: difference of recursive equations


问:
Problems with understanding rotor's conception
By definition rotor is:
$$ rot \vec A =
    \begin{bmatrix}
    (\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z})  \\
    (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x})  \\
    (\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y})  \\
    \end{bmatrix}
$$

And I'm trying to visualize it. So if I have a vector field, take one vector, and find a projection, and I don't understand how we can take derivative on, for example for $\hat i$, $A_z$ if there is no y axis anymore (see pic.1) and for $A_y$ there is no z axis anymore

答:
As an aside, while it is "ротор" in Russian (basically "rotor") and it's sometimes written $\mathop{\rm rot}\vec A$, this quantity would usually be called "curl" in English (and perhaps very occasionally "rotation" or "rotational", but probably never "rotor").

Your question was initially about curl, but it seems the main thing you are having trouble grasping is the meaning of a partial derivative of "the wrong" component, like $\dfrac{\partial A_x}{\partial y}$, where $\vec A=\left\langle A_x,A_y,A_z\right\rangle$.

It's important to keep in mind that for a particular piece of a vector field $\vec A$, there are not just three numbers to worry about ($A_x, A_y,A_z$), but actually $6$: the three inputs are important too. And things like partial derivatives and curl measure sensitivity to changes in input.

So we have something more like $\vec A(a,b,c)=\left\langle A_x,A_y,A_z\right\rangle=\left\langle A_x(a,b,c),A_y(a,b,c),A_z(a,b,c)\right\rangle$. $\dfrac{\partial A_x}{\partial y}$ then measures how much the $x$-component of the output changes with a small change to $b$, the $y$-component of the input. The $A_z$ displayed in the picture in the OP only shows the value of $A_z$ for a particular input $(a,b,c)$. And to approximate the derivative $\dfrac{\partial A_x}{\partial y}$, we need to look at both $(a,b,c)$ and $(a,b+h,c)$  for a small value of $h$.

We can look at this situation pictorially in this example:

There are two vectors from a vector field $\vec A(x,y,z)$ floating off in space (not shown) positioned with tails at $(a,b,c)=(?,-1.1,?)$ and $(a,b+h,c)=(?,-1.1+0.5,?)=(?,-0.6,?)$. You can see the $y$-coordinates of the tails/inputs marked in purple on the $y$-axis.

Those vectors have been translated to have tails at the origin as orange and yellow vectors (with yellow corresponding to $\vec A(a,b+0.5,c)$). The $x$-coordinates of these outputs are displayed as blue points on the $x$-axis.

The difference quotient $\dfrac{\vec A(a,b+0.5,c)-\vec A(a,b,c)}{0.5}$ is displayed as a gray vector with its tail at the head of the orange vector (which corresponds to $\vec A(a,b,c)$). This grey vector is also translated to have its tail at the origin, representing an approximation to $\dfrac{\partial \vec A(x,y,z)}{\partial y}$ at $(a,b,c)$.

Finally, we look at the projection of this gray vector onto the $x$-axis to see the $x$ component: an approximation to $\dfrac{\partial \vec A_x(x,y,z)}{\partial y}$ at $(a,b,c)$.

You can vary the perspective, values, and function in this math3d interactive.


问:
What will be the function for this to be an homeomorphism
Consider this problem: Define an equivalence relation ~ on $X =\mathbb{R}^2$ by $(x_1,y_1)~ (x_2,y_2)$ iff $y_1 =y_2$ . Let (X/ ~ ,U) be identification space and let T denote the usual topology on $\mathbb{R}$. Prove that (X/~ ,U) is homeomorphic to $(\mathbb{R},T)$.

The problem is that I am not able to define the homeomorphism . Can you please define the map and give a reasoning why you chose that particular  map?

I am a beginner in this course and classes are online. So, I am a bit weak in problem solving and Trying to learn.

答:
HINT:

The first step for you is to define a function which takes an equivalence class and maps it to an element of $\mathbb{R}$. Once you've done this, the rest shouldn't be too tricky - you need to check that the function is continuous, and define a continuous, two-sided inverse. I'll give some hints about this first step and leave the rest to you.

We'll go step by step. You need a function from $(X/ \sim, U)$ to $\mathbb{R}$. Let's first inspect this space $(X/ \sim, U)$. Elements of $(X/ \sim, U)$ are equivalence classes; as you say in the comments, an equivalence class here consists of all elements of $\mathbb{R}^2$ with the same $y$ coordinate. To be specific, an element of $(X/ \sim, U)$ therefore corresponds precisely to one particular $y$ coordinate - that is, a particular element of $\mathbb{R}$. Can you see how to define a function of the sort we need, using this information?


问:
How do I do this counting problem?
There are 17 mathematicians and 3 official languages. Every pair of mathematicians communicate in one of the official languages. Prove that there are 3 mathematicians communicating in the same language, pairwise.

I guess pigeonhole principle must be applied but I cannot see how to move forward with it.
So, let us consider the three languages (say, A, B and C) as pigeonholes. By pigeonhole principle, at least one of A, B and C must contain at least ceil(17/3) = 6 pigeons. So, 6 mathematicians must be there and all of them know the same language.
I can only see this.

答:
Let's look at it from the perspective of Mathematician A.

By PHP, there will be at least $6$ mathematicians such that 'A' talks with all the mathematicians in the same language(Language L).

If you find any 2 mathematicians among those 6 mathematicians talking in language L we are done because then we found 3 people(A and the other 2 mathematicians) who speak in the same language pairwise

So basically 6 people have to talk in the other two languages

Hence the question simplifies to proving that 3 people talk in the same language pairwise with 6 mathematicians and 2 languages. Using the same argument as above, you can prove that.

Title: How do I do this counting problem?


问:
About this simple linear system $Ax=0$ under the condition $Dim(Ker(A))>0$
Let $k>n>0$ be two integers. Let $A$ be a real $k \times k$ matrix. We suppose that $Dim(Ker(A))>0$. We look at the system

$$Ax=0$$

We know that there is infinitely many solutions to this system because of $Dim(Ker(A))>0$. But can we always find $k-n$ different solutions $x^{(1)}=(x_1^{(1)}, \ldots, x_{k}^{(1)}),\ldots, x^{(k-n)}=(x_1^{(k-n)}, \ldots, x_{k}^{(k-n)})$ satisfying $x_i^{(i)} \neq 0$ for $i=1,\ldots, k-n$? If not, what extra conditions should we add on $A$ to get this property ?

答:
No. Suppose for example that $A = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$. Then the kernel of $A$ is the linear span of $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$. In particular, every $x \in \ker(A)$ satisfies $x_2 = 0$.

A sufficient (but of course not necessary) condition to achieve what you want would be that every combination of $k-1$ columns is linearly independent. In that case your eigenvectors don't have a zero entry and you get what you want trivially.

Also note that your property is not basis-independent. So almost every change of basis would achieve this as well.


问:
General rule on partial fraction expansion?
I have fractions of the form

$$F(p) = \prod_{i=1}^{n} \frac{1}{1+\alpha_i p} $$

and it appears that the partial fraction expansion is

$$F(p) = \sum_{i=1}^{n} \frac{1}{(1+\alpha_i p)} \prod_{j=1,j\neq i }^{n} \frac{\alpha_i}{\alpha_i-\alpha_j}$$

at least for $n\in{1,2,3,4}.$ Is that a general relationship? Or did I just get lucky?

答:
When you are looking at the partial fraction expansion, you want to solve this problem :

$$F(p)=\sum_{i=0}^{n}\frac{A_i}{1+\alpha_ip} $$ And solve for the $A_i$. First, you can note that $-\frac{1}{\alpha_i}$ is a root of $F(p)$, but it is not a root of $(1+\alpha_ip)F(p) $
Hence, let : $G(p)=(1+\alpha_ip)F(p)  $

$$\forall i \leq n, G\left(-\frac{1}{\alpha_i}\right)=\underbrace{A_i}_{\text{from the sum}}=\underbrace{\prod_{j=0,i\neq j}^n\frac{1}{1-\alpha_j/\alpha_i}}_{\text{from the product}}\\
=\prod_{j=0,i\neq j}^n\frac{\alpha_i}{\alpha_i-\alpha_j} $$

Hence, you get the result :

$$ F(p)=\sum_{i=0}^n\frac{1}{1+\alpha_ip} \prod_{j=0,i\neq j}^n\frac{\alpha_i}{\alpha_i-\alpha_j}$$


问:
Minimizing the variance of my MLE

A random sample of size $n_1$ is to be drawn from a normal population
with mean $\mu_1$ and variance $\sigma^2_1$.

A second random sample of size $n_2$ is to be drawn from a normal
population with mean $\mu_2$ and variance $\sigma^2_2$. The two samples are independent.

What is the maximum likelihood estimator of $\alpha =\mu_1-\mu_2$?

Assuming that the total sample size $n=n_1-n_2$ is fixed, how should
the  observations be divided between the two populations in order to
minimise the variance of α̂ ?

I already came up with my MLE of $\alpha =\mu_1-\mu_2$. Now I need to divide my n observations, provided that $n=n_1-n_2$, in such a way that I can minimize the variance of α. Any leads on how I could go about this?

答:
Hints:

Find the maximum likelihood estimator of $\mu_1-\mu_2$ given $n_1$ and $n_2$; here this is is also the natural estimator, and is the same as the MLE of $\mu_1$ minus the MLE of $\mu_2$

Find the variance of the MLE of $\mu_1-\mu_2$  given $n_1$ and $n_2$; it will be a function of $n_1, n_2, \sigma^2_1, \sigma^2_2$ and you can use the independence of the to parts of the MLE

Minimise that variance of the MLE subject to $n_1+n_2=n$; probably easiest using calculus to find $n_1$ as a function of $\sigma^2_1, \sigma^2_2$ though note that $n_1$ should be an integer

Title: Minimizing the variance of my MLE


答:
This is an elegant problem. You can use induction on the difference and skip the brute force computation.

Proof by induction:
Let S(n) be the claim that $f(n)-g(n)=4-4n$.

Base case: trivial.

Inductive case: Suppose S(n-1) is true. That is, suppose $f(n-1)-g(n-1)=4-4(n-1)$. Then $f(n)-g(n)=[3f(n-1)+8(n-1)]-[3g(n-1)+12]=3[f(n-1)-g(n-1)]+8n-20.
$

By induction, that equals:$3[4-4(n-1)]+8n-20=12-12(n-1)+8n-20=12-12n+12+8n-20=24-4n-20=4-4n$

and you're done.

Title: difference of recursive equations


问:
Definition of Differentiability (equivalent form)

Suppose $E$ is an open set in $\mathbb R^n$, $f$ maps $E$  into $\mathbb R^m$, and $x\in E.$   If there exists a linear transformation $A$ of  $\mathbb R^n$ into $\mathbb R^m$ such that
$$ \lim_{h\to 0} \frac{|f(x+h)-f(x)-Ah|}{|h|}=0$$
then we say that $f$ is differentiable at $x$, and we write
$$f'(x)=A$$

Question: Can we rewrite the above definition to the following form:
$$ f(x+h)-f(x)=f'(x)h + r(h)$$
where the remainder $r(h)$ satisfies
$$ \lim_{h\to 0} \frac{|r(h)|}{|h|}=0. $$ If so, how to justify this.

答:
We can. And in fact, it's close to the version I prefer, personally: $f$ is differentiable at $x\in E$ if there exist a linear map $f'(x):\mathbb R^n\to\mathbb R^m$ and a map $r:E-x\to\mathbb R^n$ such that for all $h\in\mathbb R^n$ for which $x+h\in E$ we have that

$$f(x+h)=f(x)+f'(x)h+r(h),$$

where $\lim_{h\to0}\frac{\vert r(h)\vert}{\vert h\vert}=0$ (that is, $r\in o(\vert h\vert)$). By $E-x$ I mean $E$ translated by $-x$.

This is Taylor's theorem for once differentiable functions, which is why I like this version: It says that a function is differentiable if there is a first degree Taylor "polynomial" with which  it can be approximated.

Justifying this version (or yours, which is really the same via a simple rearrangement of the equation) is as simple as solving for $r(h)=f(x+h)-f(x)-f'(x)h$ and inserting it into the limit condition placed on $r$. We obtain the usual definition of differentiability. And the other way around we just choose $r(h)=f(x+h)-f(x)-f'(x)h$ and quickly see that $f(x+h)=f(x)+f'(x)h+r(h)$, and that $r$ satisfies the limit condition by definition.


问:
Classification of compact smooth manifolds of dimension 3.
I know classification of compact dimension 2 smooth manifolds. They are diffeomorphic to a sphere with n "ears" (connected sum of n tori) or a sphere with m mobius strips (connected sum of m real projective planes).
I know only that geometric hypothesis proved by Perelman says something about 3 manifolds, but I can't find similar as above precise classification for compact smooth manifolds of dimension 2. Is there a similarsimple classification? IF Yes, could You leave a link to it or write it in comment?</p>

答:
A manifold is called prime if whenever it is homeomorphic to a connected sum, one of the two summands is homeomorphic to a sphere.

In dimension two, the closed prime manifolds are $S^2$, $\mathbb{RP}^2$, and $S^1\times S^1$. By the classification of surfaces, every closed two-dimensional manifold is homeomorphic to a connected sum of prime manifolds. In the orientable case, the connected summands are unique up to $S^2$ summands (you can always connected sum with $S^2$ without changing anything). In the non-orientable case, we no longer have uniqueness as $(S^1\times S^1)\#\mathbb{RP}^2$ is homeomorphic to $\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2$. However, one can recover uniqueness (up to spherical summands) if one prohibits the use of $S^1\times S^1$ summands.

There is a similar story for closed three-manifolds. The prime decomposition theorem for three-manifolds states that every closed three-manifold is homeomorphic to a connected sum of prime manifolds. If the orientable case, the connected summands are unique up to $S^3$ summands. If $M$ is non-orientable, then uniqueness no longer holds, however one can recover uniqueness by prohibiting the use of $S^2\times S^1$ as one of the connected summands.

The key difference between dimensions two and three is that there are infinitely many prime three-manifolds.  In the orientable case, they fit into three categories:

those manifolds covered by $S^3$,

the manifold $S^2\times S^1$, and

orientable aspherical manifolds.

These categories can also be characterised via the fundamental group: namely finite, infinite cyclic, and infinite non-cyclic respectively.

In the non-orientable case however, there are too many prime manifolds to admit a classification; see the answer to this question of mine.

In dimension four, we no longer have uniqueness, even in the orientable case. For example, $(S^2\times S^2)\#\overline{\mathbb{CP}^2}$ is homeomorphic to $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}\#\overline{\mathbb{CP}^2}$. Note the similarity to the fact that $(S^1\times S^1)\#\mathbb{RP}^2$ is homeomorphic to $\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2$.


问:
What is $E[E[Y|X]|X]$?
What is $E[E[Y|X]|X]$ ?

Can I use the law of total expectation and say that it's equal to $E[Y|X]$ ?

If it's correct, could someone explain why?

答: The law of total expectation would $E[Y]=E[E[Y|X]]$. In this expression $E[Y|X]$ is a random variable that is a function of $X.$ $E[Y|X]$ changes with $X,$ assuming different realizations depending on different outcomes of $X=x,$ and by the law of total expectation, its expected value, across the support of $X$ is just $E[Y].$ Conditioning on $X$ the expression $E[Y|X]|X$ is redundant $E[Y|X]|X=E[Y|X],$ and in this sense it amounts to applying the same law of total expectation $E[E[Y|X]]=E[E[Y|X]|X]=E[Y].$ $\;$

Title: What is $E[E[Y|X]|X]$?


答:
If we have $$\lim_{h\to 0} \frac{|f(x+h)-f(x)-Ah|}{|h|}=0$$ then our left hand side will be equal to some function $|g(h)|$ some function $r: U\to \mathbb{R}^m$ with $U$ being some small neighborhood near $0$. in otherwords, $$\frac{|f(x+h)-f(x)-Ah|}{|h|}=|g(h)|$$
Then we can multiply the $|h|$ away to get $$|f(x+h)-f(x)-Ah|=|h||g(h)|$$
If the previous equation is true then there ought to be a function $g(h)$ such that we have $$f(x+h)-f(x)-Ah=g(h)|h|$$ or $$f(x+h)-f(x)=Ah+g(h)|h|$$ By our previous supposition, we must have $\lim_{h\to 0} g(h)=0$ if we let $r(h)=g(h)|h|$ then we have $$\lim_{h\to 0} \frac{r(h)}{|h|}=\lim_{h\to 0} g(h)=0$$ meaning that if our function $f$ satisfies the first definition, then it satisfies our new one.

Title: Definition of Differentiability (equivalent form)

Body:
Suppose $E$ is an open set in $\mathbb R^n$, $f$ maps $E$  into $\mathbb R^m$, and $x\in E.$   If there exists a linear transformation $A$ of  $\mathbb R^n$ into $\mathbb R^m$ such that
$$ \lim_{h\to 0} \frac{|f(x+h)-f(x)-Ah|}{|h|}=0$$
then we say that $f$ is differentiable at $x$, and we write
$$f'(x)=A$$

Question: Can we rewrite the above definition to the following form:
$$ f(x+h)-f(x)=f'(x)h + r(h)$$
where the remainder $r(h)$ satisfies
$$ \lim_{h\to 0} \frac{|r(h)|}{|h|}=0. $$ If so, how to justify this.


问:
An integral identity, general multidimensional case
A follow-up to this  question. Which of the following equations is true? Can someone give a proof to the correct equation. Thanks.

Please note the prefactors $N$ and $N!$

$$\int_0^A dx_1 \ldots  \int_0^A dx_N \quad f(x_1) \ldots f(x_N) = N
\int_0^A dx_1 \ldots\int_0^{x_{N-1}}dx_N \quad f(x_1) \ldots f(x_N)
\qquad \mathrm{Eq. 1}$$

$$\int_0^A dx_1 \ldots  \int_0^A dx_N \quad f(x_1) \ldots f(x_N) = N!
\int_0^A dx_1 \ldots\int_0^{x_{N-1}}dx_N \quad f(x_1) \ldots f(x_N)
\qquad \mathrm{Eq. 2}$$

答: The second is correct. Consider the unit cube and try to see that one sixth of it satisfies $x\le y\le z$. More generally, there are $N!$ permutations of any given $N$-tuple of values, and exactly one of them satisfies $x_1\le\dotso\le x_N$.</p>


问:
Minimizing the number of slopes when given $x_1$ and $x_2$ number of points
Let's say that we have the total of $x_1 \in \mathbb{N}$ points lying on the negative side of the $x$-axis (let us call this set $S_1$) and $x_2 \in \mathbb{N}$ points lying on the positive side of the $y$-axis (let us call this set $S_2$). We want to determine the minimum number of slopes these points create by connecting a each point from $S_1$ to each point in $S_2$. By 'minimizing the number of slopes' I am referring to an optimal way of placing the points in $S_1$ and $S_2$ so that the number of slopes determined in the described way is as small as possible.

My idea was the following: the points on both, the $x$-axis (set $S_1$) and the $y$-axis (set $S_2$) would have to be evenly spaced from one another. This makes sense to me, I am just not entirely sure how to prove it. Going by contradiction seems that it could potentially work, but this is where I got stuck. I know that trivially the upper bound for the slopes would have to be $x_1 \cdot x_2$. The lower bound is much more difficult to think of. Would the crossings of the slopes have to be minimized as well when wanting to minimize the number of slopes?

Also, I noticed that when we want to minimize the slopes, we would have to maximize the number of slopes that appear multiple times. Such examples would be $2$, $\frac{4}{2}$, $\frac{8}{4}$ and so on. We can call these 'duplicates'.

答:
I assume that the points on each semi-axis must be distinct.

Statement: There are at least $x_1+x_2-1$ distinct slopes.

Proof, by descent on $x_1$:

Consider removing the left-most point of the $x_1$ points on the negative x-axis. It has $x_2$ lines connecting it to the $x_2$ points on the positive y-axis. The line connecting it to the lowest point on the y-axis has a slope that is unique, i.e. none of the other lines can have that same slope because the two sets of points lie on different sides of that line. Therefore, when you remove the left-most point and the lines through it, the number of unique slopes decreases by at least $1$.

If you repeat the above $x_1-1$ times, you get down to the case with $x_1=1$ which has $x_2$ lines that must all have unique slopes since the y-axis points are distinct.

Therefore the starting arrangement must have had at least $x_2+(x_1-1)$ unique slopes to begin with.

Statement: The lower bound of $x_1+x_2-1$ distinct slopes is achievable.

Proof: Consider the arrangement of points given by the coordinates $(-2^1,0)...(-2^{x_1},0)$ and $(0,2^1)...(0,2^{x_2})$. The slopes are all powers of two, ranging from $2^{-x_1+1}$ to $2^{x_2-1}$, and these form exactly $x_1+x_2-1$ distinct values.


答:
Let $\mathscr F$ be the $\sigma$-algebra generated by $X.$ Then
$E[Y|X]=E[Y|\mathscr F]$ is an $\mathscr F$-measurable random variable and we know that if $Z$ is a $\mathscr F$-measurable random variable, we have $E[Z|\mathscr F]=Z.$

Title: What is $E[E[Y|X]|X]$?

Body:
What is $E[E[Y|X]|X]$ ?

Can I use the law of total expectation and say that it's equal to $E[Y|X]$ ?

If it's correct, could someone explain why?


答:
I do not think this is the low of total expectation, at least not in its standard form.

Recall instead that:

$E[A|B]=A$ if $A$ is $B-$measurable.

$E[Y|X]$ is $X-$measurable.

So now put in the first property $A=E[Y|X]$ and $B=X$, that is your result.

Title: What is $E[E[Y|X]|X]$?


问:
Prove that there exists a number α > 0 such that f(x) > α for all $x \in I$.
I = [a, b] and let $f : I \to \mathbb{R}$ be a continuous function such that f(x) > 0 for each
$x \in I$. Prove that there exists a number α > 0 such that f(x) > α  for all $x \in I$.

答: Let $L = inf_{x \in I} f(x)$. Since $f$ is a continuous function,  $\exists x_0 \in I $ such that $0<f(x_0)=L$. Take $a = \frac L2$. Then $f(x)>a \forall x \in I$.


问:
Rings with zero divisors
Is there a ring  $~R~$ with non-trivial multiplication (i.e. $~\exists a,b\in R ~~~ ab\neq 0$) such that each non-zero element of $~R~$ is a zero-divisor? </p>

答: To get a general class of examples, you can take the nilradical of any commutative ring (ie, the set of nilpotent elements). This forms an ideal consisting solely of nilpotent elements (and hence of zero-divisors), but will usually have non-trivial multiplication.</p>

Title: Rings with zero divisors


问:
What's the $\lim _{x \to \frac{π}{4}} \frac{\ln (\cot x)}{{1-\tan x}}$?

$$
\lim_{x \to \pi/4} \frac{\ln (\cot x)}{{1-\tan x}}
$$

I'm trying to solve it without using L'Hopital or Taylor series.

答:
Set $1 + y = \cot x$. Then $1 - \tan x = \frac y{1+y}$, and hence

$$\lim_{x \to \pi/4} \frac {\ln \cot x}{1 - \tan x} = \lim_{y \to 0} \frac {\ln (1+y)}{y}(1+y) = 1\cdot 1 = 1$$


答:
\begin{gather*}
\varliminf _{x\rightarrow \frac{\pi }{4}}\frac{\ln( \cot x)}{1-\tan x} =\varliminf _{h\rightarrow 0}\frac{\ln\left( \cot\left(\frac{\pi }{4} -h\right)\right)}{1-\tan\left(\frac{\pi }{4} -h\right)}\\
=\varliminf _{h\rightarrow 0}\frac{\ln\left(\frac{1+\tan( h)}{1-\tan( h)}\right)}{1-\frac{1-\tan( h)}{1+\tan( h)}} =\varliminf _{h\rightarrow 0}\frac{\ln\left( 1+\frac{2\tan( h)}{1-\tan( h)}\right)}{\frac{2\tan( h)}{1+\tan( h)}}\\
\end{gather*}
Can you take it from here?

Title: What's the $\lim _{x \to \frac{π}{4}} \frac{\ln (\cot x)}{{1-\tan x}}$?

Body:
$$
\lim_{x \to \pi/4} \frac{\ln (\cot x)}{{1-\tan x}}
$$

I'm trying to solve it without using L'Hopital or Taylor series.


问:
Eigenvector Perturbation with a symmetric matrix
Consider the matrix $ \mathbf{M} = \lambda_1 x x^T + \frac{\lambda_2}{2}(\mathbf{A}+\mathbf{A}^T)$, where $x \in \mathbb{R}^n$ is a unit norm vector, and $\mathbf{A}$ is a full rank matrix in $\mathbb{R}^{n \times n}$ with $\|\mathbf{A}\|_F = 1$. Let $u$ be the top eigenvector of $\mathbf{M}$, and suppose that $\lambda_1 > \lambda_2$, is there any theorem lower bounding $\langle x,u \rangle$?

答:
I assume that we have $\lambda_1 > \lambda_2 \geq 0$.

By the Rayleigh-Ritz theorem, we note that
$$
\lambda_{\max}(M) = \max_{\|y\|=1} y^TMy = \max_{\|y\|=1} \lambda_1 |x^Ty|^2 +
\lambda_2\,y^TAy,
$$
and the vector $y$ for which this maximum is attained is an eigenvector. Thus, we see that
$$
\lambda_{\max}(M) \geq x^TMx = \lambda_1 + \lambda_2\,x^TAx.
$$
Because $y = u$ maximizes $y^TMy$ over the unit vectors $y$, we have
\begin{align}
u^TMu &\geq x^TMx \implies\\
\lambda_1|u^Tx|^2 + \lambda_2 u^TAu &\geq \lambda_1 + \lambda_2 x^TAx \implies\\
|u^Tx|^2 & \geq \frac{\lambda_1 + \lambda_2(x^TAx - u^TAu)}{\lambda_1}.
\end{align}
Now, because $\|A\|_F = 1$, we have $-1 \leq y^TAy \leq 1$ for all unit-vectors $y$. Thus, we can relax the above bound to get
$$
|u^Tx|^2 \geq \frac{\lambda_1 - 2\lambda_2}{\lambda_1}.
$$

Another approach: with the assumption with $u^Tx = 0$, we want a lower-bound to $\lambda_2$. Let $\lambda = \lambda_{\max}(M)$. Define $S = \frac {1}2(A + A^T)$. We know that
$$
Mu = \lambda u \implies \lambda_2 Su = \lambda u.
$$
That is, $\lambda$ is an eigenvalue of $\lambda_2 S$. However, $\|S\|_F \leq 1$ implies that $|\lambda| \leq \lambda_2$. Moreover, we note that
$$
|x^T(\lambda_2 S)x| \leq \sqrt{\lambda_2^2 - \lambda^2}.
$$
Now, we have
\begin{align}
u^TMu &\geq x^TMx \implies\\
\lambda &\geq \lambda_1 + \lambda_2 x^TSx \implies\\
\lambda & \geq \lambda_1 - \sqrt{\lambda_2^2 - \lambda^2}\implies\\
\lambda + \sqrt{\lambda_2^2 - \lambda^2}
& \geq \lambda_1.
\end{align}
For any $0\leq \lambda_2< \lambda_1$, consider the function
$$
f(x) = x + \sqrt{\lambda_2^2 - x^2}.
$$
We note that $f(0) = \lambda_2 < \lambda_1$, but $f(\lambda)\geq \lambda_1$. So,
$\lambda_2$ must be such that there is a solution to $f(x) = \lambda_1$ over the domain $0 \leq x \leq \lambda_2$. We rewrite
$$
x + \sqrt{\lambda_2^2 - x^2} = \lambda_1 \implies\\
\sqrt{\lambda_2^2 - x^2} = \lambda_1 - x \implies\\
\lambda_2^2 - x^2 = x^2 - 2 \lambda_1x + \lambda_1^2 \implies\\
2x^2 - 2\lambda_1 x + (\lambda_1^2 - \lambda_2^2) = 0.
$$
First of all, in order for a solution to this equation to exist, the discriminant  of the equation must be non-negative. We find
$$
(2\lambda_1)^2 - 8(\lambda_1^2 - \lambda_2^2) \geq 0 \iff 8\lambda_2^2 - 4\lambda_1^2 \geq 0,\\
$$
which is to say that we have $\lambda_2 \geq \lambda_1/\sqrt{2}$

so a solution $x \in \Bbb R$ to the above equation exists. This equation has two positive roots, and the smaller of these two roots must satisfy $x \leq \lambda_2$. In other words, $\lambda_2$ must be such that
$$
\frac{4\lambda_1^2 - \sqrt{8\lambda_2^2 - 4\lambda_1^2}}{4} \leq \lambda_2.
$$


问:
A wrong configuration in Geometry?
The mid-point of the hypotenuse of a right angled triangle ABC, right angled at B is M. A line is drawn perpendicular to the hypotenuse through M, in such a way, that the portion of it lying inside the triangle is 3 cm long and outside the triangle, up to the other side is 9 cm. Find the length of the hypotenuse.

I do not understand what is meant by part lying inside the triangle and outside, won't part outside be infinite?

Could someone please explain what this question wants and give a hint please?

答:
If my interpretation is correct, here is a diagram:

To solve for $x$, simply note that $\triangle CME \sim \triangle DMA$.

Title: A wrong configuration in Geometry?


答:
I think the question means that $DE = 3, EF = 9$.

Hint: you have two pairs of similar triangles that lead to the answer.

Title: A wrong configuration in Geometry?

Body:
The mid-point of the hypotenuse of a right angled triangle ABC, right angled at B is M. A line is drawn perpendicular to the hypotenuse through M, in such a way, that the portion of it lying inside the triangle is 3 cm long and outside the triangle, up to the other side is 9 cm. Find the length of the hypotenuse.

I do not understand what is meant by part lying inside the triangle and outside, won't part outside be infinite?

Could someone please explain what this question wants and give a hint please?


问:
Prove that a cylinder have a infinite number of planes of symmetry.
My definition of cylinder is:

A cylinder is the surface formed by parallel lines, where each line contains a point of a curve called guideline. Each lines is called a generatrix of that cylinder.

Intuitively I see that each plane orthogonal to a generatrix is a plane of symmetry, but I can't see an algebraic argument.

答:
Yes, I believe you're correct. A priori, the guideline curve may not be planar, but by shifting continuously along the generatrices, you may assume the guideline curve lies in a plane orthogonal to the generatrices. Then reflection across that plane or any plane parallel to it will preserve the cylinder.

Algebraically, parametrize the surface by $x(s,t) = \alpha(s) + tv$, where $v$ is a fixed unit vector. Using this formulation, you can replace $\alpha$ by $\tilde\alpha$ with $\tilde\alpha\cdot v = 0$, as I remarked earlier. Now check that sending $t$ to $c-t$ for any $c\in\mathbb R$ gives the reflection across the plane $v\cdot x = c/2$.


问:
Difference between independence of events and independence of random variables
I was wondering if someone could shed some light on the difference between independence of random variables and independence of events. The definitions look similar to me, but I am wondering if there are any conceptual differences between the two?</p>

答:
Events $A_1, \ldots, A_n$ are mutually independent if
$$P(\bigcap_{i \in S} A_i) = \prod_{i \in S} P(A_i)$$
for every subset $S \subseteq \{1,\ldots,n\}$. (In the case $n=2$, this is simply $P(A_1 \cap A_2) = P(A_1) P(A_2)$.)

The definition of independence of random variables relies on the definition of independence of events, which is probably why the definitions look similar to you. Random variables $X_1, \ldots, X_n$ are mutually independent if the events
$$\{X_1 \in E_1\}, \ldots, \{X_n \in E_n\}$$
are mutually independent for any measurable subsets $E_1, \ldots, E_n \subseteq \mathbb{R}$. (Equivalently, it suffices to check that $\{X_1 \le x_1\},\ldots,\{X_n \le x_n\}$ are mutually independent events for any real numbers $x_1, \ldots, x_n$.)


问:
Prove that $\mu = 0.$

Let $X$ be an infinite dimensional NLS and $\mu$ be a translation invariant measure on $\mathcal B(X)$ such that $\mu (F) \lt \infty$ for all $F \in \mathcal B (X)$ with finite diameter. Then show that $\mu = 0.$

How do I proceed? Any help will be appreciated.

Thanks for reading.

答:
Note that $X=\bigcup_{n\in \Bbb N} B_n(0)$ where $B_n(0)$ is the ball of radius $n$ around. Hence if $\mu(X)\neq0$ at least one of these balls has to have non-zero mass. Let $B_n(0)$ denote such a ball.

Now for any finite dimensional sub-space $V$ you can find an element $x\notin V$ of norm $≤4 n$ and so that $\|y-x\|≥2n$ for all $y\in V$. Use this to iteratively find a sequence $x_i$ so that $\|x_i\|≤4n$ and $\|x_i-x_j\|≥2n$ for all $i\neq j$. This means that the balls $B_n(x_i)$ are all disjoint and are further contained in the ball $B_{5n}(0)$.

Hence:
$$\infty > \mu(B_{5n}(0) ) ≥ \mu\left(\bigcup_{i\in\Bbb N} B_n(x_i)\right) =\sum_{i\in\Bbb N} \mu\left( x_i + B_n(0)\right) = \sum_{i\in\Bbb N} \mu \left ( B_n(0) \right ) = \infty$$
by translation invariance, a contradiction.

Title: Prove that $\mu = 0.$


答:
The relation is this: let $X,Y:\Omega \to \mathbb{R}$ two random variables. Then $X$ and $Y$ are independent if and only if their pre-image $\sigma $-algebras are independent.

In short: every two events of the form $X^{-1}(A)$ and $Y^{-1}(B)$, for arbitrary measurable $A,B\subset  \mathbb{R}$, are independent.

Title: Difference between independence of events and independence of random variables

Body: I was wondering if someone could shed some light on the difference between independence of random variables and independence of events. The definitions look similar to me, but I am wondering if there are any conceptual differences between the two?</p>


问:
Combinatorics summation problem - $\sum_{k=0}^r k.{n \choose k}. {m \choose {c-k}} = ?$
Is there any way to calculate this - $$\sum_{k=0}^r k.{n \choose k}. {m \choose {c-k}}$$ where $r, n, m$ and $c$ are constants and $k \leq c$.

I have no idea how to approach this.

Thanks!

EDIT - To be more specific I actually need to calculate this - $$\sum_{k=0}^{150} k.{1000 \choose k}. {14000 \choose {150-k}}$$

Body:
Note that the pmf of a hypergeometric distribution with parameters $N=15000,K=1000,n=150$ is given by $$p(k)=\frac{\binom{1000}{k}\binom{14000}{150-k}}{\binom{15000}{150}}$$
and the expected value is $$\sum _{k=0}^{150}k\cdot \frac{\binom{1000}{k}\binom{14000}{150-k}}{\binom{15000}{150}}=n\cdot \frac{K}{N}=150\cdot \frac{1000}{15000}=10$$
Multiplying both sides by $\binom{15000}{150}$, $$\sum_{k=0}^{150}k\cdot \binom{1000}{k}\binom{14000}{150-k}=10\binom{15000}{150}$$


问:
Prove that the general following family of vectors are a basis of $\mathbb{R}^{n+1}$
Given $a_0, a_1,...,a_n\in\mathbb{R}$ all different numbers.

Demonstrate that the following vectors are a basis of $\mathbb{R}^{n+1}$

$v_0=(1,a_0,a_0^2,...,a_0^n)$

$v_1=(1,a_1,a_1^2,...,a_1^n)$

$. . . $

$v_n=(1,a_n,a_n^2,...,a_n^n)$

$(1)$ Dimonstrate that is enough by showing that are linearly independent.

$(2)$ Generalize for every n by using Gauss elimination.

I recall the definition of linear independence as follows to use for $(1)$

Given $\left(v_i\right)_{i=1,..,k}$ as a family of $k$ vectors of $\mathbb{R}^n$, it is linearly independent if and only if:
$$\forall\lambda_1,\lambda_2,..\lambda_p\in\mathbb{R},\hspace{12pt}\left(\sum\limits_{i=1}^{k}\lambda_iv_i=0_{\mathbb{R}^n}\right)\hspace{-3pt}\Longrightarrow\hspace{-3pt}\left(\lambda_1=\lambda_2=..=\lambda_k=0\right)$$

Isn't my intention to ask for an exhaustive solution but for help given that I don't even know how to start with the point $(1)$ neither $(2)$. So please, I kindly ask for some orientation?

答: A linear system with such vectors as lines arises naturally when dealing with polynomial interpolation (The rhs is assembled with the values of the function to be interpolated at $x = a_0, a_1, \cdots, a_n$). As it was mentioned in a comment, this matrix is usually denoted as the VanderMonde matrix, which is invertible as long as the $a_i$'s are distinct. You can easily compute the determinant and see it is nonzero. This guarantees that all the lines are linearly independent. I'm not sure what 2. means... There is nothing to be generalised.


问:
Series of $\int_0^z \zeta^{-1} \sin \zeta d \zeta$
This is a homeworkquestion so I would appreciate some good hints. I have $f(z) = \int_0^z \zeta^{-1} \sin \zeta d \zeta$. Can this be written as a power-series in $\mathbb C$ around $z = 0$?</p>

答:
$sin(\xi)$ can be expressed in the form of Taylor expansion as $\sum\limits_{n=0}^{\infty}\frac{(-1)^n\xi^{2n+1}}{(2n+1)!}=\xi-\frac{\xi^3}{3!}+\frac{\xi^5}{5!}+...+$

$$\int_0^z \frac{sin\xi}{\xi}dz=\int_0^z\sum\limits_{n=0}^{\infty}\frac{(-1)^n\xi^{2n}}{(2n+1)!}dz=\sum\limits_0^{\infty}\int_0^z \frac{(-1)^n\xi^{2n}}{(2n+1)!}dz=\sum\limits_0^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)(2n+1)!}$$

The question is That I can't make sure the order of sum and integration can exchange.


问:
Limit of $\sum_{k=1}^\infty \frac{f(k)}{k!}$
Does this limit always exist when $f(k)$ is a polynomial?
$$
\sum_{k=1}^\infty \frac{f(k)}{k!}
$$
(My guess is yes, based on the root test, but I'm not entirely sure I applied it correctly.)

答: It suffices to show that $\sum_{k=1}^\infty \frac{k^m}{k!}$ converges for each $m \ge 0$. You can use the ratio test to show this series converges. Then note that your original series is a finite sum of series of this form.


答:
For sure as

$$F(f) =
\sum_{k=1}^\infty \frac{f(k)}{k!}
$$ is linear and defined for all monomial $f_n(x)=x^n$ as can be verified with the ratio test. Therefore it is defined for all polynomials by addition of a finite number of monomials.

Title: Limit of $\sum_{k=1}^\infty \frac{f(k)}{k!}$

Body:
Does this limit always exist when $f(k)$ is a polynomial?
$$
\sum_{k=1}^\infty \frac{f(k)}{k!}
$$
(My guess is yes, based on the root test, but I'm not entirely sure I applied it correctly.)


问:
Where am I wrong when calculating $\sum_{k=2}^{\infty}5k \frac{1}{6^{k-1}}$
I'm trying to calculate $\sum_{k=2}^{\infty}5k \frac{1}{6^{k-1}}$

so what I did is $\sum_{k=2}^{\infty}5k \frac{1}{6^{k-1}} = 5\sum_{k=2}^{\infty}k (\frac{1}{6})^{k-1} =  5\sum_{k=2}^{\infty} ((\frac{1}{6})^{k})' = 5 (\sum_{k=2}^{\infty} (\frac{1}{6})^{k})' = 5* \frac{\frac{1}{6^2}}{1-\frac{1}{6}} = \frac{1}{6}$

But the answer should be $\frac{11}{5}$.

I guess I should change the $k$ i'm summing for when taking the derivative maybe?

But I'm not sure why.

Can anyone pinpoint my mistake and explain why it is wrong?

答:
Can you differentiate a constant? Yes, but the result is zero. Sorry, but your method is completely wrong.

You can leave out the constant factor $5$ for the moment and consider
$$
f(t)=\sum_{k=2}^\infty kt^{k-1}
$$
If we integrate from $0$ to $x$, we get, for $|x|<1$,
$$
\int_0^x f(t)\,dt=\sum_{k=2}^\infty x^k=\frac{1}{1-x}-1-x
$$
so if we differentiate
$$
f(x)=\frac{1}{(1-x)^2}-1
$$
With $x=1/6$ and deploying the factor $5$ we obtain
$$
\sum_{k=2}^\infty 5k\frac{1}{6^{k-1}}=\frac{5}{(1-1/6)^2}-5=5\cdot\frac{36}{25}-5=\frac{11}{5}
$$


问:
Is the determinant of a matrix different depending on the field over which it is viewed
General question:

Suppose we have a square matrix over R.
Will its determinant change if we view the same matrix over a finite field (i.e. Z\5Z) or not?

答:
I am not sure how about general entries from $R$. Surely if the other field is containing $R$ as a subfield then nothing changes. If you are considering different fields you need entries in matrix $A$ to be in this different field (ie. in the intersection of $R$ and this field). I guess that to make it workable you need both of these fields to be contained in a bigger field.

In a less general setting, this is indeed true. Take for example matrix with only zeroes and ones (e.g. incidence matrix of some graph or set system).  Then by viewing it in a different field (real numbers, $F_2$, $F_p$ for $p$ prime) you can get different determinants:

A determinant is some sum over all permutations. These summands can be either zero or one since $A$ is 0-1 matrix (or -1 if the sign of the permutation is negative and fieůd is different from $F_2$).

For example, if determinant results to be $2$ for a matrix understood as a real matrix, over field $F_2$ you get determinant $0$ and  over field $F_3$ you still get $Det(A)=2$.

When applying methods of Linear Algebra in solving problems from combinatorics it can be indeed useful to understand incidence matrices over a different field (especially when asking whether it is a regular matrix).


问:
How to find a point on the border of a rectangle
I was wondering if it's possible to calculate points: A, B, C and D in the image I made below? And if it's possible, how?

Given are coords of points A2, B2, C2, D2. The distance between 'a1' - 'w', 'c' - 'v', 'b1' - 'b' and 'g' - 'p' is 7.071067811865475, in other words $5\sqrt 2$.

答: At each corner, you can draw a little 45-45-90 isosceles triangle. The sides of that triangle are each $5\sqrt2$, so the hypotenuse is $10$. The hypotenuse is parallel to a coordinate axis, so the inner vertices you want are 10 units away from the outer ones in just one coordinate.