问:
3D spline interpolation and shape manipulation
(I am new to this subject, so sorry if the question is too trivial).

I would like to find a method, possibly using polyharmonic splines, that would be a smooth mapping of  $\mathbb{R}^3 \mapsto \mathbb{R}^3$ and would also allow shape manipulation via control handles.

General polyharmonic spline interpolation consists of evaluating a function [1] of the nature of $$f(\mathbf{x}) = \sum_{i=1}^N w_i \phi(|\mathbf{x} - \mathbf{c}_i|) + \mathbf{v}^{\top}(1,\mathbf{x})$$

where $\mathbf{x} \in \mathbb{R}^d$ is the point to be interpolated, $\mathbf{c} \in \mathbb{R}^d$ denotes the control handles and the weights $\mathbf{w}$ and $\mathbf{v}$ are computed directly. But I am not sure if/how the mapping $f:\mathbb{R}^d \mapsto \mathbb{R}$ can be adapted to perform a $\mathbb{R}^3 \mapsto \mathbb{R}^3$ mapping, as well as being used for shape manipulation via control handles, without computing the weight matrices with every update of the control handles.

On a similar note, [2] offers a possible method for using a similar Thin Plate Spline (TPS) formulation for $\mathbb{R}^3 \mapsto \mathbb{R}^3$ mapping, but I am not sure how to manipulate the target/destination control handles without having to re-compute the weights in a closed form.

Any help/suggestion in this matter would be hugely appreciated.

[1] wikipedia.org/wiki/Polyharmonic_spline

[2] Hu, Y., Zhou, M. and Wu, Z., 2009. A dense point-to-point alignment method for realistic 3D face morphing and animation. International Journal of Computer Games Technology, 2009.

答:
The standard reference on interactive deformation of 3D space is a paper by Sederberg and Parry, SIGGRAPH 1986. There are many ways to modify and extend their method, but it will give you a place to start.

Actually, Bézier himself described this technique in his doctoral thesis, in 1977, but people almost always cite the Sederberg/Parry paper, instead.

More here:
https://en.wikipedia.org/wiki/Free-form_deformation


问:
Why this integral is $0$? $ \int_{0}^{2 \pi} \frac{2 \cos(2t) (\sqrt{2}ie^{it})}{4e^{i4t}-2\sqrt{2}e^{i2t}+1}dt$
I was unable to calculate this integral directly $$ \int_{0}^{2 \pi} \frac{2 \cos(2t) (\sqrt{2}ie^{it})}{4e^{i4t}-2\sqrt{2}e^{i2t}+1}dt$$

so I put it into wolfram and it says that it is equal to zero. Does anyone know why?

EDIT:

The original integral was $$\int_{|z| = \sqrt{2}} \frac{\operatorname{Re}(z^2)}{z^4-\sqrt{2}z^2+1} dz$$ and I did the substitution $z = \sqrt{2}e^{it}$.

答:
You have
$$\int_{|z| = \sqrt{2}} \frac{\operatorname{Re}(z^2)}{z^4-\sqrt{2}z^2+1} dz
=\int_{|z| = \sqrt{2}} \frac{z^2/2+2/z^2}{z^4-\sqrt{2}z^2+1} dz
=\int_{|z| = R} \frac{z^2/2+2/z^2}{z^4-\sqrt{2}z^2+1} dz
$$
for any $R>1$ (as the singularities are in the closed unit disk). Let $R\rightarrow +\infty$ and you obtain the wanted result.

ContentLicense: CC BY-SA 4.0


问:
Show that $\forall n\ge3,(2 n-1) \times(2 n-3) \times(2 n-5) \times \cdots \times 5 \times 3 \times 1\ge2 \times 7^{n-2}$
Here's my question: Let $f(n)=(2 n-1) \times(2 n-3) \times(2 n-5) \times \cdots \times 5 \times 3 \times 1$, and I need to prove that for all $n\ge3,f(n)\ge2\times7^{n-2}$.

I have figured out some interesting points, but still I have no clue how to make use of them.

$f(n+1)=2n\times(2n-2)\times\dots\times4\times2=2^n\times n!$

$f(n+1)\times f(n)=(2n)(2n-1)\dots(3)(2)(1)=(2n)!$

Then everything stops here. I'm trying to make sense with the $7^{n-2}$ with the statement, but failed. So is there any other hints for this question? Thanks a lot for your help!

答:
we put $\sigma_{2n-1}=(2n-1)!!$ for $n>3$ :
We have $\sigma_5=9×7×5×3×2×1>2(7^{5-2})$

and $\sigma_7=9×\sigma_5>7\sigma_5$

By recurence we obtain $\sigma_{2n-1}>7^{n-6}\sigma_5$


问:
Can we apply $2$ different quantifiers to the same variable
So I'm very new to FOL and I've been trying to wrap my head around quantifiers and the scope of them however I got a bit confused on this example

$\forall x\in D [p(x) \implies \exists x d(x)]$

Does this mean verbally for all $x \in D$ if p(x) is true then there exsists an $x \in D$ such that property $d(x)$ is true since the $\forall x$ scopes everything in the implication then everything in the implication must be an element of $D$ .
however I  know that $2$ different quantifiers cannot quantify the same variable so is it just then for all $x \in D$ if $p(x)$ is true then there exsists some x possibly not in $D$ such that $d(x)$ is true?.

Thanks in advance.

答:
The quantifiers $\forall x\in D~$ and $\exists x$ each bind all free occurrences of $x$ inside their scope. However, any such occurrence can be bound to only one quantifier, and its immediate container has precedence.

Since such occurrences of $x$ inside the scope of $\exists x$ are bound to that quantifier, therefore they are not considered free inside the scope of $\forall x\in D$.

$\forall x\in D~.(p(x)\to\exists x~d(x))$ is equivalent to $\forall x\in D~.(p(x)\to\exists y~d(y))$ (with the caveat that $y$ does not occur free in $d(x)$ and is thus substitutable).

"For anything in $D$, should it satisfy $p(~)$ it will imply that something$^\dagger$ will satisfy $d(~)$."

$\dagger$ not necessarily the same thing, nor does it even need be in $D$.


问:
A simple proof of the Cauchy–Schwarz inequality with vectors
I have seen many proofs online, but none of them is as simple as my potentially correct and simple proof. Please let me know if the following is true, and if not where my mistake is.

$$\begin{align*}
|\vec{x} \cdot \vec{y}| &= ||\vec{x}|| \cdot ||\vec{y}|| \cdot \cos(\theta)\\[5pt]
\underbrace{-1 \le \cos(\theta) \le 1}_\le\quad\underbrace{||\vec{x}||}_{\ge0} \cdot \underbrace{||\vec{y}||}_{\ge0} &=||\vec{x}|| \cdot ||\vec{y}||\\[5pt]
\end{align*}$$

答:
Your proof is true for some dot products. It depends on the way you define a dot product. You used the fact that
$$
x \cdot y = ||x|| \times || y|| \times \cos(\theta)
$$
where $\theta$ is the angle between $x$ and $y$,
which is true with the canonical dot product over $\Bbb R^n$. But for a random dot product (positive definite symmetric bilinear form) the existence of such $\theta$ isn't acquired yet. In fact, it's thanks to the Cauchy-Schwartz inequality that we get it.


问:
Simplifying $f(\sqrt{7})$, where $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$

If $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$ ; then $f(\sqrt {7})=\; ?$

I tried solving this equation through many methods, I tried rationalizing, squaring, etc. But after each of them, the method became really lengthy and ugly.

I also noted that once we substitute $\sqrt {7}$ the inner part of the radical becomes imaginary. How to proceed with this piece of information?

Please help me with this problem. Any more innovative methods would be appreciated.

Answer: $4$.

Edit $[7^{th}$ March, $2021]$:

I was told yesterday that this question was wrong and that we were supposed to find $f(2\sqrt5)$. Although I can solve with $2\sqrt5$ the same way as given below in the answers.... What I don't understand is that why $\sqrt7$ doesn't work. Can someone please help?

答:
Set $\sqrt{x-4}=u \to x=u^2+4\\$so
$$\sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}=\\ \sqrt{u^2+4-4u}+\sqrt{u^2+4+4u}=\\
|u-2|+|u+2|=\\
|\sqrt{x-4}-2|+|\sqrt{x-4}+2|$$ but when deals to imaginary numbers absolute sign is not necessary.


问:
Taking notes as a mathematics student
a couple of years back I was in class with a  student that was copying notes in class from the whiteboard and after class he would make his own notes out of the already written notes (he would somehow rewrite them). On theoretical tests (tests where they tested knowledge of theorems definitions and proofs of them) he always got 100 % marks. I have lost connection with this student, so I want to ask this community, those of you who make their own notes(after copying from the whiteboard) how do you make them(what information do you rewrite). I know that this is a personal preference, but I want to try this approach and see how it goes.</p>

答: Personally, I very much like to rewrite the teacher's notes using the following approach. I rewrite all the definitions and propositions on two different sheets of paper. This allows me to focus first on definitions (as in a play, the characters are introduced from the beginning). And when I master them all, I can start reading the propositions and theorems. What I find most important in a course is not the theorems and propositions, but the definitions, without them it's impossible to understand something in math.</p>

Title: Taking notes as a mathematics student


问:
Evaluate the integral $\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx$
I try to calculate the integral
\begin{align}
f(y,z):=\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx
\end{align}
on the set $y>0$ and $0<z<1$. I could show by some calculation that in the limit $z\to0$ holds
\begin{align}
f(y,0)=e^{y/2}K_0(y/2),
\end{align}
where $K_0$ denotes the modified Bessel function of the second kind and
\begin{align}
\lim_{z\to1}f(y,z)=f(y,1)=\infty,\\
\lim_{y\to0}f(y,z)=f(0,z)=\infty.
\end{align}
But I can't evaluate the integral in the interior of the defined set. Any help, ideas or hints how to solve this would be appreciated.

Here is a selection of what I've tried:

several substitutions (e.g. $\sinh$,$\dots$),

integration by parts,

tried to evaluate the integral $$\partial_zf(y,z)=\int_{\mathbb R} -yx\cdot \exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)\mathrm dx, $$

tried to evaluate the integrals $$\partial_yf(y,z)=\int_{\mathbb R} -(x^2+zx\sqrt{x^2+1})\cdot \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx $$ and
$$f(y,z)-\partial_yf(y,z)=\int_{\mathbb R} (\sqrt{x^2+1}+zx)\cdot \exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)\mathrm dx. $$

答:
\begin{align}
f(y,z)=&\int_{\mathbb R} \frac{\exp\left(-y(x^2+z\cdot x\sqrt{x^2+1} ) \right)}{\sqrt{x^2+1}}\mathrm dx\\
=&\frac{1}{2}\int_{\mathbb R} \exp\left(-y(\sinh^2\frac{u}{2}+z\cdot \sinh\frac{u}{2}\cosh\frac{u}{2} ) \right)\mathrm du\\
=&\frac{1}{2}\int_{\mathbb R} \exp\left(-\frac{y}{2}(\cosh u -1+z\cdot \sinh u ) \right)\mathrm du\\
=&\frac{1}{2}e^{y/2}\int_{\mathbb R} \exp\left(-\frac{y}{2}(\cosh u+z\cdot \sinh u ) \right)\mathrm du
\end{align}
Until this point I used the substitution $x=\sinh\frac{u}{2}$ and the equation $\sinh^2\frac{u}{2}=\frac{1}{2}(\cosh u -1)$, as well as $\sinh\frac{u}{2}\cosh\frac{u}{2}=\frac{1}{2}\sinh u$. From this I could see for the case $z=0$, that we have $f(y,0)=e^{y/2}K_0(y/2)$ using $$K_0(\phi)=\int_0^\infty\exp\left(-\phi\cosh x \right)\mathrm dx,\qquad \forall \phi>0.$$
To go on from this someone helped me with the hint $$ \cosh u+z\cdot \sinh u =\sqrt{1-z^2}\cosh(u+\tanh^{-1} z).$$
This brings us to
\begin{align}
f(y,z)=&\frac{1}{2}e^{y/2}\int_{\mathbb R} \exp\left(-\frac{y}{2}\sqrt{1-z^2}\cosh(u+\tanh^{-1} z) \right)\mathrm du\\
=&\frac{1}{2}e^{y/2}\int_{\mathbb R} \exp\left(-\frac{y}{2}\sqrt{1-z^2}\cosh u \right)\mathrm du\\
=&e^{y/2}\int_{0}^\infty \exp\left(-\frac{y}{2}\sqrt{1-z^2}\cosh u \right)\mathrm du\\
=&e^{y/2}K_0\left(\frac{y}{2}\sqrt{1-z^2}\right).
\end{align}

The idea behind the hint was to find values $\beta, v$ such that we can use the addition theorem
\begin{align}
\cosh u+z\cdot \sinh u=\frac{1}{\beta}(\beta\cosh u+z\beta\cdot \sinh u)\overset{!}{=}&\frac{1}{\beta}(\cosh v\cosh u+\sinh v\sinh u)\\
=&\frac{1}{\beta}\cosh(u+v),
\end{align}
as a shift $v$ of the argument doesn't affect the integral since we are integrating from $-\infty$ to $\infty$. To find these values we need to have $\beta=\cosh v$ and $z\beta=\sinh v$. This brings us to
$$v=\tanh^{-1}z$$
and $$\beta=\frac{1}{\sqrt{1-z^2}}.$$


问:
Prove that there is a unique $A\in\mathscr{P}(U)$ such that for every $B\in\mathscr{P}(U), A\cup B = B$
$U$ can be any set.

For the existence element of this proof, I have $A = \varnothing$

But it's for the uniqueness element of this proof where I am having trouble.  So far I have:

$\forall(C\in\mathscr{P}(U))(\forall(B\in\mathscr{P}(U))(C\cup B=B)\rightarrow C\in\mathscr{P}(U) = A\in\mathscr{P}(U))$

Assume $C\in\mathscr{P}(U)$ is arbitrary and assume $\forall(B\in\mathscr{P}(U))(C\cup B=B)$ then:

What I have left to prove is:
$C\in\mathscr{P}(U) = A\in\mathscr{P}(U) $

which is where I am having trouble.  I can say that $C = \varnothing$ which would then equal $A$ but it is not certain that $C = \varnothing$.

答:
For the uniqueness if $A$ and $A'$ are two sets verifying the hypothesis then we have
$$A\cup A'=A'=A'\cup A=A$$
so $A=A'$ and obviously this set is $\emptyset$</p>

ContentLicense: CC BY-SA 3.0


问:
Statement of the Perron-Frobenius Theorem
Basically, I'm having trouble proving the last part of statement 4 below. The problem comes down to showing that if $A\in\mathbb{R}^{n\times n}$ is an irreducible non-negative matrix, $x>0$ is an eigenvector of $A$ associated with the eigenvalue $\rho_A$ and there is a number $\mu$ such that $Ax\leq \mu x$, then $\rho_A=\mu$.

答:
By applying the first part of theorem 1 to $A^T$, we see that $A$ has a positive left eigenvector $v$ corresponding to the eigenvalue $\lambda_\max$. It follows that $\lambda_\max v^Ty=v^TAy\le \mu v^Ty$. Since $y$ is nonzero and nonnegative, $v^Ty$ is positive. Therefore $\lambda_\max\le\mu$.

When $\mu=\lambda_\max$, we have $\lambda_\max v^Ty=v^TAy\le \mu v^Ty=\lambda_\max v^Ty$. Therefore $v^T(\mu y-Ay)=0$. As $v>0$ and $\mu y-Ay\ge0$, we must have $\mu y-Ay=0$. Thus $y$ is also an eigenvector of $A$ corresponding to the eigenvalue $\mu=\lambda_\max$.

In that case, if $y$ is not a scalar multiple of $x$, then for some appropriate $t>0$, the vector $z=x-ty$ is nonnegative and it has both positive and zero entries. Let $\mathcal I\sqcup\mathcal J$ be a partition of $\{1,2,\ldots,n\}$ such that $z(\mathcal I)>0$ and $z(\mathcal J)=0$. From $Az=\lambda_\max z$, we obtain $A(\mathcal J,\mathcal I)z(\mathcal I)=0$ and in turn $A(\mathcal J,\mathcal I)=0$. This contradicts the assumption that $A$ is irreducible. Hence $y$ must be a scalar multiple of $x$.


问:
The set of all $x$ satisfying, $\sqrt{4x+1} + \sqrt{7-x} = 6 $, consists of:

The set of all $x$ satisfying, $\sqrt{4x+1} + \sqrt{7-x} = 6 $, consists of:

$A)$ Two rational numbers. $B)$ An irrational number. $C)$ Complex number. $D)$ None.

How to solve the above question using proper method?

I got the values by substitution. I just considered the values after evaluation of the radical to be $(5,1)$ ,$(3,3)$ and $(4,2)$.

This leads me to nice and whole values of $x = 6$. I can't find any other solutions.

I tried squaring, but the solution is getting really nasty at a point. I wonder if there could be any more methods to solve this problem. Any help or suggestion would be appreciated. Thanks!

答: Note that\begin{align}\sqrt{4x+1}+\sqrt{7-x}=6&\implies4x+1+7-x+2\sqrt{4x+1}\sqrt{7-x}=36\\&\iff2\sqrt{4x+1}\sqrt{7-x}=28-3x\\&\implies4(4x+1)(7-x)=(28-3x)^2\\&\iff25 x^2-276 x+756=0\\&\iff x=6\text{ or }x=\frac{126}{25}.\end{align}And you can easily check that $6$ and $\frac{126}{25}$ are actually roots of the original equation.


问:
Graph convergence on probability
Good afternoon.
In one book on probability theory I came across this problem which I found myself difficult to understand and prove. Would appreciate any help on that:

Let $X_n$ be the number of complete subgraphs on 4 vertices in $G(n,\frac{1}{2})$. Prove that $\frac{X_n-E(X_n)}{E(X_n)} \xrightarrow{P} 0$

答:
Let $Z_n=(X_n-\mathsf{E}X_n)/\mathsf{E}X_n$. Then for each $\epsilon>0$,
$$
\mathsf{P}(|Z_n|\ge \epsilon)\le \epsilon^{-2}\operatorname{Var}(Z_n)=\frac{\operatorname{Var}(X_n)}{\epsilon^2(\mathsf{E}X_n)^2}.
$$
It suffices to show that the last term converges to $0$ as $n\to\infty$.

First, since the probability that a given set of $r$ nodes is a complete subgraph is $p^{\binom{r}{2}}$ (where $p$ denotes the connectivity parameter; $p=1/2$ in your case),
$$
\mathsf{E}X_n=\binom{n}{r}p^{\binom{r}{2}}.
$$
Second, since one can choose $\binom{n}{r}\binom{r}{i}\binom{n-r}{r-i}$ ordered pairs of sets of $r$ nodes with $i$ nodes in common,
$$
\mathsf{E}X_n^2=\sum_{i=0}^r\binom{n}{r}\binom{r}{i}\binom{n-r}{r-i}p^{2\binom{r}{2}-\binom{i}{2}}
$$
because the probability of two complete subgraphs of $r$ nodes with $i$ nodes in common is  $p^{2\binom{r}{2}-\binom{i}{2}}$ (we assume that $n\ge 2r$).

Finally, it is easy to show that
$$
\frac{\operatorname{Var}(X_n)}{(\mathsf{E}X_n)^2}=\binom{n}{r}^{-1}\sum_{i=2}^r\binom{r}{i}\binom{n-r}{r-i}\left(p^{-\binom{i}{2}}-1\right).
$$

Plugging $r=4$ and $p=1/2$, we get
$$
\frac{\operatorname{Var}(X_n)}{(\mathsf{E}X_n)^2}=\left(6\binom{n-4}{2}+28n-49\right)\binom{n}{4}^{-1}\to 0
$$
as $n\to\infty$.

Title: Graph convergence on probability


问:
Burnside's Lemma Giving Incorrect Answer For Number of Non-Isomorphic Size 5 Tournaments?
The question I am looking to answer asks the following:

How many non-isomorphic size 5 tournaments exist?

The most straightforward way to approach this problem is Burnside's Lemma, which states that:

$$|X/G| = (\frac{1}{|G|})\sum_{∈G}|_|$$

where |X/G| is the number of orbits, |G| is the size of the group, and $|_|$ is the number of fixed points of set X under a given symmetry  ∈ G. In terms of a tournament, T,  of size 5, the size of the group is equal to the number of permutations of the vertices of T, which is simply 5! = 120.

Further, the number of fixed points varies with the cycle structure of T. There are seven cycle structures of T, namely:

(1, 1, 1, 1, 1): This is the identity; fixes every tournament. There are 5 choose 2 edges; each edge can go both ways. Results in $2^{\binom{5}{2}}=1024$.

(2, 1, 1, 1): If two vertices are swapped naturally the resulting graph isn't the original since the edge can't change direction.

(2, 2, 1): Same as above.

(3, 1, 1): Formula, (given by
this post gives the number of fixed points as 20.

(3, 2): See 2; the number of fixed points is 0.

(4, 1): The number of fixed points is 0 because the diagonals can't be preserved.

(5): The aforementioned formula gives the number of fixed points to be 24.

Applying Burnside's Lemma gives an answer of $\frac{1024+20+24}{120}$ which is not only incorrect but is not even an integer. It's well known that there are exactly 12 non-isomorphic tournaments of size 5, so what went wrong exactly in my reasoning? No matter how many times I check my work I can't figure out how the numerator can possibly ever equal 1440.

答:
You have already (correctly) shown that the only permutation types with fixed points are  the identity (one permutation) and those of types "3+1+1" (20 permutations) and 5 (24 permutations).

Now, as you have (also correctly) stated, the identity fixes all 1024 tournaments, but you haven't taken into account the numbers of fixed tournaments of the other two permutation types. These are

16 tournaments for a permutation of type "3+1+1" and

4 tournaments for a permutation of type "5".

Thus, the number of non-isomorphic tournaments is

$$\frac{1\cdot1024+20\cdot16+24\cdot4}{120}=\frac{1440}{120}=12.$$


问:
Quotient spaces in linear algebra: proof of linear independence
Let $V$ be a vector space and $F$ a vector subspace of $E$. Proof that, for $v_1, \dots, v_k \in E$, the classes $v_1 + F, \dots, v_k + F \in E/F$ are linearly independent if, and only if, dim(span($v_1, \dots,v_k$)) $=k$ and span$(v_1, \dots, v_k) \cap F = \{0\}$.

$(\implies)$ I wanted to use the fact that they are linearly independent, thus $\lambda_1 (v_1 + F) + \dots + \lambda_k (v_k + F) = \lambda_1 v_1 + \dots + \lambda_k v_k + F = F$. Therefore, $\lambda_1 v_1 + \dots + \lambda_k v_k \in F$. But now I don't know how to proceed

$(\Longleftarrow)$ I have no clue how to even start.

Can someone help me?

答:
$(\implies)$ First, we need to show that $\dim(\text{span}(v_1,...,v_k))=k$, i.e that the vectors are linearly independent. So suppose $\lambda_1 v_1+...+\lambda_k v_k=0$. Since $0\in F$ this tells that $\lambda_1 v_1+...+\lambda_k v_k\in F$, and so:

$F=\lambda_1 v_1+...+\lambda_k v_k+F=\lambda_1(v_1+F)+...+\lambda_k(v_k+F)$

Since $v_1+F,...,v_k+F$ are linearly independent it follows that $\lambda_1=...=\lambda_k=0$.

Note that during the proof we also showed that $\text{span}(v_1,...,v_k)\cap F=\{0\}$, as we have shown that $\lambda_1 v_1+...+\lambda_k v_k\in F$ implies $\lambda_1=...=\lambda_k=0$.

$(\Longleftarrow)$ Suppose $\lambda_1(v_1+F)+...+\lambda_k(v_k+F)=F$. This means that $\lambda_1 v_1+...+\lambda_k v_k\in F$. Since $\text{span}(v_1,...,v_k)\cap F=\{0\}$ it follows that $\lambda_1 v_1+...+\lambda_k v_k=0$. Since $v_1,...,v_k$ are linearly independent (as $\dim(\text{span}(v_1,...,v_k))=k$) it follows that $\lambda_1=...=\lambda_k=0$. So $v_1+F,...,v_k+F$ are linearly independent, as required.


问:
How do you prove that if $f(n)\in O(g(n))$ then $g(n)\in\Omega(f(n))$

Prove that if $f(n)\in O(g(n))$ then $g(n)\in\Omega(f(n))$

I saw a proof of this from the site from way back. I am not convinced this is true with the statement THEN between.  This is Transpose symmetry I think and the statement in between Prove should be Prove that if $f(n)\in O(g(n))$ IF AND ONLY IF  $g(n)\in\Omega(f(n))$.  Can anyone elaborate

答:
The if and only if is true.

If $f(n) \in O(g(n))$ then there is some $N$ and $k>0$ such that for $n \geq N$,
$$f(n) \leq k g(n).$$
This means that for $n \geq N$, $g(n) \geq \frac{1}{k} f(n)$,i.e. , $g(n) \in \Omega(f(n))$.

The other way around is very similar.


答: If $U$ is empty then this is trivial.  If $U$ is not empty then suppose some such $A\ne \varnothing$ exists.  Let $x\in A$.  There is at least one subset of $U$ not containing $x$, since $\varnothing$ is such a set.  For any subset $B$ of $U$ such that $x\not\in B$, we have $A\cup B\ne B$.</p>

Title: Prove that there is a unique $A\in\mathscr{P}(U)$ such that for every $B\in\mathscr{P}(U), A\cup B = B$

Body:
$U$ can be any set.

For the existence element of this proof, I have $A = \varnothing$

But it's for the uniqueness element of this proof where I am having trouble.  So far I have:

$\forall(C\in\mathscr{P}(U))(\forall(B\in\mathscr{P}(U))(C\cup B=B)\rightarrow C\in\mathscr{P}(U) = A\in\mathscr{P}(U))$

Assume $C\in\mathscr{P}(U)$ is arbitrary and assume $\forall(B\in\mathscr{P}(U))(C\cup B=B)$ then:

What I have left to prove is:
$C\in\mathscr{P}(U) = A\in\mathscr{P}(U) $

which is where I am having trouble.  I can say that $C = \varnothing$ which would then equal $A$ but it is not certain that $C = \varnothing$.


问:
Determining divergence and gradient using chain rule
Suppose $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$, and $r=|r|$. Also, let $f$ be a scalar function of $r$ and let $\vec{A}$ be a vector function of $r$. We wish to determine $\nabla f$ and $\nabla.\vec{A}$.

It seems to be obvious that the chain rule is the way to go here:

$$\nabla f= \dfrac{\partial f}{\partial x}\hat{i}+\dfrac{\partial f}{\partial y}\hat{j}+\dfrac{\partial f}{\partial z}\hat{k} $$
$$= \dfrac{\partial f}{\partial r}\dfrac{\partial r}{\partial x}\hat{i}+\dfrac{\partial f}{\partial r}\dfrac{\partial r}{\partial y}\hat{j}+\dfrac{\partial f}{\partial r}\dfrac{\partial r}{\partial z}\hat{k} $$

$$= \dfrac{\partial f}{\partial r}(\dfrac{\partial r}{\partial x}\hat{i}+ \dfrac{\partial r}{\partial y}\hat{j}+\dfrac{\partial r}{\partial z}\hat{k}) $$

$$= \dfrac{\partial f}{\partial r}\hat{r} $$

Which is correct. However,

$$\nabla.\vec{A}=\dfrac{\partial \vec{A}}{\partial x} .\hat{i}+\dfrac{\partial \vec{A}}{\partial y} .\hat{j}+\dfrac{\partial \vec{A}}{\partial z} .\hat{k}$$

$$=\dfrac{\partial \vec{A}}{\partial r}\dfrac{\partial r}{\partial x} .\hat{i} + \dfrac{\partial \vec{A}}{\partial r}\dfrac{\partial r}{\partial y} .\hat{j}  + \dfrac{\partial \vec{A}}{\partial r}\dfrac{\partial r}{\partial z} .\hat{k}$$

$$=\dfrac{\partial \vec{A}}{\partial r}.(\dfrac{\partial r}{\partial x} \hat{i} + \dfrac{\partial r}{\partial y} \hat{j}+ \dfrac{\partial r}{\partial z} \hat{k})$$

$$=\dfrac{\partial \vec{A}}{\partial r}.\hat{r}$$

This expressions turns out to be incorrect.

The only scope for error seems to be in the usage of the chain rule, however if it(my usage of the chain rule) worked for the gradient, why did it fail here ?

答:
Ok, so here's the problem, you didn't account for the unit vectors changing.

$$ \nabla \cdot \vec{A} = \left[\hat{x}\cdot \frac{\partial}{\partial x}  + \hat{y} \cdot \frac{\partial}{\partial y} + \hat{z} \cdot\frac{\partial}{\partial z} \right]  \vec{A}$$

Now, let's consider the dot product with $\hat{x}$:

$$ \hat{x} \cdot \frac{\partial}{\partial x} (\vec{A}) \tag{1}$$

We can write the differential operator as:

$$ \frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial }{\partial r}  + \frac{\partial \theta}{\partial x} \frac{\partial }{\partial \theta}$$

Hence from (1), we get:

$$ \hat{x} \cdot ( \frac{\partial \vec{A}}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial \vec{A} }{\partial \theta})$$

Particularly speaking, the term of interest is $ \frac{\partial \vec{A} }{\partial \theta}$, this we can write as(*);

$$ \frac{\partial \vec{A} }{\partial \theta} = \frac{\partial |A(r)| \hat{r} }{\partial \theta} = |A| \frac{\partial \hat{r}}{\partial \theta} = |A|\hat{\theta}$$

Even though we the function is not directly dependent on $\theta$ or $\phi$ the unit vectors are and hence contribute to derivative. Hope this helped!

For a reference on the derivative of unit vector with coordinates see wiki

*: I assumed $\vec{A} = |\vec{A}| \hat{r}$


问:
Center of Mass Problem
Let $E=[0,3] \times  [-2,2] \times  [0,2]$ in Rectangular Coordinates with $0\le x \le 3, -2 \le y \le 2, 0 \le z \le 2$. Now let the density at each point E be equal to the distance from the z-axis. This means $(x,y,z)$ will be the center of mass E.

Now how can I find the two coordinates for the center of mass via symmetry and the third with an integral?

答:
Distance of any point from $z-$axis is given by $\sqrt{x^2+y^2}$, so density $\rho = \sqrt{x^2+y^2}$.

Note that the density $\rho$ is not a function of $z$ and hence the  $z$ coordinate of center of mass will simply be $z = 1$.

You can also use the symmetry about $x-$axis to conclude that the $y$ coordinate of the center of mass will be $0$.

So you are left with finding $x$ coordinate for the center of mass. For that, consider mass of unit height (unit $z$) in $XY$ plane above $x-$axis (as there is symmetry above and below $x-$axis).

Mass $m = \displaystyle \int_{0}^{2} \int_0^3  \sqrt{x^2+y^2} \ dx \ dy$

And, $\overline{x} = \displaystyle \frac{1}{m} \int_{0}^{2} \int_0^3 x \sqrt{x^2+y^2} \ dx \ dy$

Just as a side note, the second integral can be evaluated over $dx$ using $u = y^2+x^2$ substitution. Then hyperbolic substitution works over $dy$.

To make it easier to evaluate the first integral - if you are familiar with line integral of vector fields and Green's theorem, please note line integral of the vector field $\frac{1}{3}(-y \sqrt{x^2+y^2}, x \sqrt{x^2+y^2})$ over the closed curve which is perimeter of the rectangle with vertices $A(0,0), B(3, 0), C(3,2), D(0,2)$, is same as the first integral to find mass. Line integral over $AB$ and $DA$ is zero. So after parametrization, this translates to line integral over $BC$ and $CD$ as below.

$\displaystyle \int_0^2 \sqrt{t^2+9} \ dt + \frac{2}{3}\int_0^3 \sqrt{t^2+4} \ dt$

Title: Center of Mass Problem


答:
Here's is a variation of Jose's answer, which appeared as I was typing this up:

$$\sqrt{4x+1} + \sqrt{7-x} = 6 \\
\iff \sqrt{4x+1} = 6 - \sqrt{7-x} $$
$$\implies 4x+1 = 36+(7-x) - 12\sqrt{7-x}\tag{1} \\
\iff 5x-42=12\sqrt{7-x} $$
$$\implies 25x^2+1764-420x=144(7-x) \tag{2}\\
\iff 25x^2-276x+756=0 \\
\iff x=6\text{ or }\frac{126}{25}$$

The steps labelled (1) & (2) aren't invertible and might have created extraneous solutions; a quick check verifies that both $6$ and $\frac{126}{25}$ indeed satisfy the original equation.

Body:
The set of all $x$ satisfying, $\sqrt{4x+1} + \sqrt{7-x} = 6 $, consists of:

$A)$ Two rational numbers. $B)$ An irrational number. $C)$ Complex number. $D)$ None.

How to solve the above question using proper method?

I got the values by substitution. I just considered the values after evaluation of the radical to be $(5,1)$ ,$(3,3)$ and $(4,2)$.

This leads me to nice and whole values of $x = 6$. I can't find any other solutions.

I tried squaring, but the solution is getting really nasty at a point. I wonder if there could be any more methods to solve this problem. Any help or suggestion would be appreciated. Thanks!


问:
Let $X$ be a Banach space and let $Y\subseteq X$ be a closed subspace such that $\operatorname{dim}(X/Y)$ is finite. Show that Y is complemented
$X$ is a Banach space and $Y$ is a closed subspace of $X$ such that $\operatorname{dim}(X/Y)$ is finite. Then show that $Y$ is complemented. I have two ideas.

First one-

Claim- $Z= \operatorname{span} \{x_1, x_2, x_3, \ldots, x_n\}$ is the complement of $Y$, where $[x_i]$, $i= 1,\ldots, n$ are equivalence classes:then, since $\operatorname{dim}(X/Y)$ is finite, the equivalence classes $[x_n]$ are finite.

I want to prove that $Z$ is closed and $Z$ intersect with $Y$ is empty. Can we do this by a contradiction? Then I need to show $X= Z+Y$.

Second one-

Or else I thought of finding a projection $P$ from $X$ to $X$ such that $\operatorname{range}(P)=Y$. I could not come up with such a projection.

Which idea is good for a proof?

答:
Your first idea is correct but you have not written the proof correctly. Let $[x_1],[x_2],..,[x_n]$ be  basis for $X/Y$. Define $Z$ to be the span of $x_1,x_2,...,x_n$. [$x_1,x_2,...,x_n$ are not unique but that doesn't matter. Just pick some $x_i$'s in the equivalence classes].

If $x \in X$ then $[x]=\sum c_i[x_i]$ for some $c_i$'s. This means $x-\sum c_ix_i \in Y$. Hence, $x=(x-\sum c_ix_i)+\sum c_ix_i \in Y+Z$.

It remains to check that $Z\cap Y=\{0\}$. If $x \in Z \cap Y$ then (by definition of $Z$) $x =\sum a_ix_i$ for some $a_i$'s. But $x \in Y$ so $\sum a_ix_i \in Y$ implying that $[\sum a_ix_i]=0$, so $\sum a_i[x_i]=0$. By independence of $[x_i]$'s this implies $a_i=0$ for all $i$, so $x=0$.


问:
What makes $5$ and $6$ so special that taking powers doesn't change the last digit?
Why are $5$ and $6$ (and numbers ending with these respective last digits) the only (nonzero, non-one) numbers such that no matter what (integer) power you raise them to, the last digit stays the same? (by the way please avoid modular arithmetic) Thanks!</p>

答:
It holds for any base (radix) of the form $\rm\,2p^j,\,$ for $\rm\ p\ $ an odd prime, since
$$\begin{eqnarray}\rm  n^2 \equiv n\ \ (mod\,\ 2p^j)\! &\iff&\rm 2p^j\mid n(n\!-\!1) \\ &\iff&\rm \ \ p^j\mid n(n\!-\!1),\qquad\ \ \ by\ \ \ 2\mid n\ \ or\ \ n\!-\!1,\ \ \ (2,p)=1\\ &\iff&\rm n \equiv 0,1\ \ (mod\ p^j),\ \ \ by\ \ \ p\ \ prime,\ \ n,n\!-\!1\ \ coprime \\ &\iff&\rm n\equiv 0,1,p^j,p^j\!+\!1\ \ \, (mod\,\ 2p^j)\end{eqnarray}$$</p>


问:
Invertible matrix product of non-square matrices
From a singular value decomposition I obtain $\mathbf{F}\in\mathbb{R}^{2m\times 2n}$ with $\mathbf{F}^T\mathbf{F}=\mathbf{I}$ and $m\geq n$. I have to find a matrix $\mathbf{E}\in\mathbb{R}^{m\times n}$ such that the product $\mathbf{M} = \mathbf{F}^T\begin{bmatrix}\mathbf{E} & \mathbf{0} \\ \mathbf{0} & \mathbf{E}\end{bmatrix}\in\mathbb{R}^{2n\times2n}$ is invertible. In case $m=n$ a possible choice for matrix $\mathbf{E}$ is the identity matrix, but in the general case $m>n$ it is not clear for me how to choose $\mathbf{E}$. Is there a guarantee that it will always work out? Any idea or references on this topic are highly appreciated. Thanks a lot in advance!

答: This is not always possible. E.g. when $F=\pmatrix{I_2\\ 0}\in\mathbb R^{4\times2}$, the matrix product $F^T\pmatrix{E&0\\ 0&E}=\pmatrix{E&0}$ is singular because it has some zero columns.


问:
Fermat's Little Theorem: Induction Proof
I was verifying various proofs of Fermat's Little Theorem lately and stumbled upon a proof by induction, which I think, uses some kind invalid circular argument. Correct me if I am wrong please here.

The steps are as follows:

Assume: $n^p \equiv n \pmod p$

Work out lemma: $(n+m)^p \equiv n^p + m^p \mod p$ using Binomial expansion

Define base case: $n = 1$ (which is true) and prove that it is true for $n < m < p -1$

$(n + 1)^p \equiv n^p + 1 \mod p$ by lemma

use the assumption from point 1. to prove the assumption from point 1?:):

$n^p + 1 \equiv n + 1 \mod p$

My problem is obviously with step 5. where you are using what you want to proof in the actual proof. (or you have to know that $n^p \equiv n \pmod p$ is somehow true by using other proof, thus rendering this inductive proof useless).

Sources:

https://proofwiki.org/wiki/Fermat%27s_Little_Theorem#Proof_4

https://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem#Multinomial_proofs

答:
Hint: it is the special case $\,f_n:= n^p\,$ below, working $\bmod p$

Lemma $ $ If $\,f_1\equiv 1\,$ and $\,\forall n\ge 1\!:\ \color{#0a0}{f_{n+1}}\equiv f_n+1\,$ then $\,\forall n\ge 1\!:\ f_n\equiv n$

Proof $ $ We induct on $\,n.\,$ Our hypothesis $\,f_1\equiv 1\,$ is the base case, and the inductive step is just the following easy inference $\,\color{#c00}{f_n\equiv n}\Rightarrow \color{#0a0}{f_{n+1}}\equiv \color{#c00}{f_n}+1\equiv \color{#c00}n+1$


问:
Find number of terms and coefficient of $x^5$ in $(1+x+x^2)^7$
I tried using the multinomial theorem to find general term ie.

$$T_n =\frac{7!}{a!b!c!} 1^a \cdot x^b \cdot x^{2c} $$

Now $b+2c=5$ and $a+b+c=7$

The second equation can be interpreted as distributing $7$ things in $3$ groups, ie $\binom {9}{2}$ but I don’t know what to do with the second condition, and how there solutions will overlap

How should I proceed?

Body:
with respect to $$(a+b+c)^5=\sum_{x+y+z=5} (5!)\frac{a^xb^yc^z}{x!y!z!}$$
$ax^5$ made from
$$2+1+2=5\to 1^2.x^1.(x^2)^2\to \frac{5!}{2!.1!.2!}\\
1+3+1=5 \to 1^1.(x^3).(x^2)^1\to \frac{5!}{1!.3!.1!}\\0+5+0=5 \to 1^0.(x^5).(x^2)^0 \to \frac{5!}{0!.5!.0!}$$so $$a=\frac{5!}{2!.1!.2!}+\frac{5!}{1!.3!.1!}+\frac{5!}{0!.5!.0!}$$


问:
Show $\lim\limits_{x\to0}x\ln(x)=0$ by using Taylor series
Is it possible to show the limit $\lim\limits_{x\to0}x\ln(x)=0$ by expanding the corresponding Taylor series?

My approach:

I know that the Taylor series of $\ln(x)$ converges uniformly on a closed interval $(1-\delta,1+\delta)$ where $\delta>0$ is sufficiently small.

So let be $$S_n(x):=\sum\limits_{k=1}^n(-1)^{k+1}(x-1)^k\frac{1}{k}\\
\lim\limits_{n\to\infty}S_n(x)=\sum\limits_{k=1}^{\infty}(-1)^{k+1}(x-1)^k\frac{1}{k}=\ln(x)\\
T_n(x):=(x-1)+\sum\limits_{k=2}^n(-1)^{k}(x-1)^k\frac{1}{(k-1)k}\\
\lim\limits_{n\to\infty}T_n(x)=(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}(x-1)^k\frac{1}{(k-1)k}\overset{?}{=}x\ln(x),
$$
where $\lim\limits_{n\to\infty}S_n(x)$ denotes the Taylor series of $\ln(x)$ at $x=1$ and $\lim\limits_{n\to\infty}T_n(x)$ the Taylor series of $x\ln(x)$ at $x=1$.

What I can observe so far is that $T_n(x)$ converges uniformly on $(0,1+\delta)$ (the point $x=0$ doesn't cause problems anymore) and this allows us to swap limit taking:
$$
\lim\limits_{x\to0}\lim\limits_{n\to\infty}T_n(x)=\lim\limits_{n\to\infty}\lim\limits_{x\to0}T_n(x)=0.
$$

However, I don't know what the limit function is or in other words how to show that $\lim\limits_{n\to\infty}T_n(x)=x\ln(x)$.

Any suggestions how to complete the proof?

答:
I'm surprised this question still didn't produce the answer the OP is missing (badly). First off: the Taylor series of $x\,\ln x$ around $x=1$ is indeed (correcting a typo from above) $$x-1+\sum^\infty_{k=2}\frac{(-1)^k}{k\,(k-1)}\,(x-1)^k.\tag{1}$$ That can be done (as the OP did) by multiplying the known series for $\ln x$ by $x=1+(x-1)$, or directly by calculating the derivatives at $x=1$, observing $$\frac{d}{dx}\,x\,\ln x=\ln x+1.$$ Whether (1) really converges to $x\,\ln x$, is another question, but if we assume (as the OP obviously did) that the series for $\ln x$ converges for $x\in(0,2)$, their derivation shows that (1) converges in that interval, too. Otherwise, we might just estimate the remainder term to come to the same conclusion.

Concerning the limit as $x\to0$, we have Abel's theorem (https://en.wikipedia.org/wiki/Abel%27s_theorem), saying that $$\lim_{x\to0} x\,\ln x=-1+\sum^\infty_{k=2}\frac1{k\,(k-1)}=-1+\sum^\infty_{k=2}\left(\frac1{k-1}-\frac1k\right)=-1+1=0,$$ since the series on the RHS converges.


问:
Simplifying $f(\sqrt{7})$, where $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$

If $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$ ; then $f(\sqrt {7})=\; ?$

I tried solving this equation through many methods, I tried rationalizing, squaring, etc. But after each of them, the method became really lengthy and ugly.

I also noted that once we substitute $\sqrt {7}$ the inner part of the radical becomes imaginary. How to proceed with this piece of information?

Please help me with this problem. Any more innovative methods would be appreciated.

Answer: $4$.

Edit $[7^{th}$ March, $2021]$:

I was told yesterday that this question was wrong and that we were supposed to find $f(2\sqrt5)$. Although I can solve with $2\sqrt5$ the same way as given below in the answers.... What I don't understand is that why $\sqrt7$ doesn't work. Can someone please help?

答:
Let $f(x) = 4 g(t)$, $x = 4(t^2+1)$. Then, provided that $\Re t \notin\{ 1, -1\}$,

$$\begin{split}
  g(t) &= \tfrac{1}{2}\left(
    \sqrt{(1-t)^2}+\sqrt{(1+t)^2}\right) \\
    &= \begin{cases}
    -t & \Re t < -1 \\
    1 & -1 < \Re t < 1 \\
    t & \Re t > 1
    \end{cases}\text{.}
\end{split}$$

That expression is found by the complex-number identity

$$\sqrt{a^2}=\begin{cases}
  a & \Re a > 0 \\
  -a & \Re a < 0
\end{cases}\text{,}$$
and the constraint on the domain comes from ignoring the branch cut. Then when $x=\sqrt{7}$, $t=\pm \mathrm{i}\sqrt{1-\tfrac{\sqrt{7}}{4}}$, so $g(t)=1$ and $f(x)=4$.


问:
Are there practical applications to the new prime pair proof?
I've recently heard that its been proven that the set of prime pairs that are separated by no more than 70,000,000 is infinite.

Does this have any impact on cryptography or another practical application?

答: What you say you recently heard is false. [I see you've edited the question in light of my answer, so now it says something that is actually true.] There are indeed pairs of primes separated by more than 70 million composite numbers.  What was proved is that there are infinitely many pairs of primes separated by no more than 70 million composite numbers.  The fact that it's true in infinitely many cases does not mean it's true in all cases.  For example, it is conjectured that there are infinitely many twin primes, i.e. primes separated by exactly $2$, so just one composite number is between them.  But if so, that obviously doesn't mean that EVERY pair of consecutive primes is separated by only two.</p>


问:
What closed subspace definition is used here?
Lemma 2.2 If $L$ is a subcomplex of $K$, then the polytope $\lvert L \rvert$ is a closed subspace of the polytope of $K$, denoted $\lvert K \rvert$. In particular, if $\sigma \in K$, then $\sigma$ is a closed subspace of $\lvert K \rvert$.

Proof: Suppose $A$ is closed in $\lvert L \rvert$. If $\sigma$ is a simplex of $K$, then $\sigma \cap \lvert L \rvert$ is the union of those faces $s_i$ of $\sigma$ that belong to $L$. Since $A$ is closed in $\lvert L \rvert$, the set $A \cap s_i$ is closed in $s_i$, and hence closed in $\sigma$. Since $A \cap \sigma$ is the finite union of the sets $A \cap s_i$, it is closed in $\sigma$. We conclude that $A$ is closed in $\lvert K \rvert$. Conversely, if $B$ is closed in $\lvert K \rvert$, then $B \cap \sigma$ for each $\sigma \in K$, and in particular for each $\sigma \in L$. Hence $B \cap \lvert L \rvert$ is closed in $\lvert L \rvert$.

I am using Elements of Algebraic Topology by Munkres and it does not give the definition of a closed subspace. It also does not appear to be using the definition here: What is a "closed subspace" of a topological space?.

答:
To sum up what we have said in the comments. Munkres is indeed using the same definition of a closed subspace, in this case, with respect to the weak topology: a subspace $C$ of $|K|$ is closed if and only if $C\cap \sigma$ is closed in $\sigma$ for every simplex $\sigma$ in $|K|$. So he shows that this is true for $|L|$ by showing it for every closed subspace $A$ of $|L|$ (in particlar, for $|L|$ itself).

In addition, in the last line he shows that the subspace topology of $|L|$ in $|K|$ is the same as the weak topology of $|L|$, so $|L|$ is naturally a subspace of $|K|$.

Finally, to also add the answer to your follow up question, $s_i$ stands for a (not necessarily top-dimensional) face of $\sigma$. Since a simplex has finitely many faces, then we can guarantee that there are finitely many $A\cap s_i$.

The reason why an $n$-simplex has finitely many faces is a consequence of the definition of a face. A face of an  $n$-simplex consists of those points $(t_0,\dots, t_n)$ such that at least one $t_i=0$. There are only finitely many possibilities for that.


问:
Find limit (without using L' Hospital Rule) I can find this limit using L' Hospital Rule, I do not know how to do it without that
Find limit (don't use Lophital rule)
$$\lim _{x\to 0}\left(\frac{\sqrt{1+x}\:-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}\:}\right)$$

I can find this limit using L' Hospital Rule, I do not know how to do it without that.

Body:
Hint: put $\;a=\sqrt[3]{1+x}\;,\;\;b=\sqrt[3]{1-x}\;$ , then your eexpression is is

$$\frac{a^{3/2}-b^{3/2}}{a-b}=\frac{(a^{1/2}-b^{1/2})(a+a^{1/2}b^{1/2}+b)}{a-b}=\frac{a+a^{1/2}b^{1/2}+b}{a^{1/2}+b^{1/2}}$$


问:
The set of all $x$ satisfying, $\sqrt{4x+1} + \sqrt{7-x} = 6 $, consists of:

The set of all $x$ satisfying, $\sqrt{4x+1} + \sqrt{7-x} = 6 $, consists of:

$A)$ Two rational numbers. $B)$ An irrational number. $C)$ Complex number. $D)$ None.

How to solve the above question using proper method?

I got the values by substitution. I just considered the values after evaluation of the radical to be $(5,1)$ ,$(3,3)$ and $(4,2)$.

This leads me to nice and whole values of $x = 6$. I can't find any other solutions.

I tried squaring, but the solution is getting really nasty at a point. I wonder if there could be any more methods to solve this problem. Any help or suggestion would be appreciated. Thanks!

Body:
WLOG $\sqrt{4x+1}=6\cos^2t$  and $\sqrt{7-x}=6\sin^2t$

$\implies4x+1=36\cos^4t,7-x=36\sin^4t$

$\implies36\cos^4t+144\sin^4t=\cdots=29$

$\implies36\cos^4t+144(1-\cos^2t)^2=\cdots=29$

$\iff180\cos^4t-288\cos^2t+115=0$

$\implies\cos^2t=\dfrac56,\dfrac{23}{30}$

$\implies x=\dfrac{36(\cos^2t)^2-1}4=?$


问:
Prove that $f_n$ converges uniformly on $[0,1]$ iff $\phi(1) = 0$
Let $\phi: [0,1] \to \mathbb{R}$ be continuous. Let $f_n :[0,1] \to \mathbb{R}$ be defined by $f_n(x) = x^n \phi(x)$. Prove that $f_n$ converges uniformly on $[0,1]$ iff $\phi(1) = 0$

The forward implication part was easy but I'm having trouble in proving the converse part.

Attempt: Conversely, suppose $\phi(1)=0$ then $f_n(x)= x^n\phi(x) \to 0$ for each $x\in [0,1]$. Let $g:[0,1] \to \mathbb{R}$ be defined by $g(x) = 0, \ \forall x \in [0,1]$, Then $f_n \to g$ pointwise on $[0,1]$

And $|f_n(x)- g(x)| = |f_n(x)|=|x^n \phi(x)|$

(I'm stuck here, I tried estimating the inequality by using the maximum of $\phi$ but didn't get anything.  Give me hints so that I can proceed, Thanks)

答: Hint: if $M_n = sup |f_n(x) - g(x)|$, then $f_n$ is uniformly convergent to $g$ iff $M_n$ goes to $0$ as $n \to \infty$


答:
Let $\varepsilon > 0$. Since $\phi$ is continuous and $\phi(1) = 0$, there exists $\delta > 0$ such that $$x \in [1-\delta,1] \implies |\phi(x)| \le \varepsilon.$$

Now, denoting $\|\phi\|_\infty = \sup_{x\in[0,1]}|f(x)|$, pick $n \in \Bbb{N}$ large enough such that $(1-\delta)^n\|\phi\|_\infty \le \varepsilon$.

Then for $x \in [0,1-\delta]$ we have
$$|x^n\phi(x)| \le (1-\delta)^n\|\phi\|_\infty \le \varepsilon$$
while for $x \in [1-\delta,1]$ we have
$$|x^n\phi(x)| \le \varepsilon.$$
Therefore for all $x \in [0,1]$ we have $|x^n\phi(x)| \le \varepsilon$ which proves uniform convergence.

Body:
Let $\phi: [0,1] \to \mathbb{R}$ be continuous. Let $f_n :[0,1] \to \mathbb{R}$ be defined by $f_n(x) = x^n \phi(x)$. Prove that $f_n$ converges uniformly on $[0,1]$ iff $\phi(1) = 0$

The forward implication part was easy but I'm having trouble in proving the converse part.

Attempt: Conversely, suppose $\phi(1)=0$ then $f_n(x)= x^n\phi(x) \to 0$ for each $x\in [0,1]$. Let $g:[0,1] \to \mathbb{R}$ be defined by $g(x) = 0, \ \forall x \in [0,1]$, Then $f_n \to g$ pointwise on $[0,1]$

And $|f_n(x)- g(x)| = |f_n(x)|=|x^n \phi(x)|$

(I'm stuck here, I tried estimating the inequality by using the maximum of $\phi$ but didn't get anything.  Give me hints so that I can proceed, Thanks)


问:
Vectors as differential operators act on functions.
Within my studies of differential geometry the concept of tangent space has just arisen. Now, a tangent vector $v_p$ at point $p \in M$ on a manifold $M$ is defined to be the differential operator $v_p^i \frac{\partial}{\partial x^i}$. Given an arbitrary smooth function $f$, I understands how this vector corresponds to a directional derivative of $f$.

However, what I don't understand -- even after consulting numerous google results and similar questions posted on this forum -- is why I need this function for my vector to be defined? To be more specific, what has this function (a specific function? a general function? a class of functions?) to do with my vector.

I suppose, from an naive point of view, when dealing with vectors I want to e.g. describe mechanical situations, maybe a collision. Perhaps I want to describe the electrical field of an electron. But what do I have to think of when talking about an action of my vector field (electric field) on some function? What physical relevance has this function. Why would I want my definition of a vector to depend on an additional function? Is there a correspondence to a more intuitive but less abstract object (e.g. in 3-d euclidean space) or maybe a low-level but didactically promising example? This is all so confusing.

$${}$$

EDIT:
To formalise my concerns:

The "usual" interpretation of a vector field as I am used to it is a function $V: \mathbb{R}^3 \longrightarrow \mathbb{R}^3$.
The "new" interpretation of a vector field as a differential operator is a function
$V: C^{\infty}(M) \longrightarrow C^{\infty}(\mathbb{R})$.

But this is not the object I want to have, is it? In fact this is a scalar-function, where I would want to have a vector.

答:
Suppose you want to explain the definition of a quadratic function to somebody and you give an example, $f(x)=x^2$. Then they ask "Why do you need this $x$ for the function to be defined? What does this $x$ have to do with my function?" What would be your answer? Mine would be that without this $x$ there is no function since the whole point of functions is to act on elements of the domain and the letter $x$ is a placeholder for a generic element of the domain of $f$.

The same about vectors as derivations. Yes, this definition is unintuitive, but it is technically convenient. If you prefer a more intuitive definition, think about manifolds embedded in some $R^N$, then vectors can be identified with directed segments, geometrically tangent to the submanifold. This is useful. But this intuitive picture has various drawback such as dependence on an embedding. Also, the picture of a tangent bundle becomes a mess since you have to worry about intersections between different tangent spaces.

Another intuitive picture is to draw vectors geometrically in domains of charts on your manifold. The problem then is that you always have to keep track of transition maps and how things depend on the choice of a chart.

How do these intuitive definitions correspond to derivations? By associating with each "geometric" vector $v$ the directional derivative $D_v$, as it is done in vector calculus classes.

If the abstract definition bothers you, when reading the textbook  and working through proofs and problems, keep both formal and geometric definitions at your disposal, like training wheels on a bicycle when learning how to ride.

Edit. Once you have the standard definition of vector fields on $M$ as certain maps $C^\infty(M)\to C^\infty(M)$, you can define their action on $C^\infty(M, {\mathbb R}^k)$ for any $k$, simply by
$$
X(f_1,...,f_k)= (X(f_1),...,X(f_k)).
$$
On the other hand it would not make sense to define vector fields by their action on ${\mathfrak X}(M)$, where the latter is the space of vector fields on $M$, since that would be circular. However, again, once you defined ${\mathfrak X}(M)$ you may want to define a notion of "differentiation" ${\mathfrak X}(M)\to {\mathfrak X}(M)$. This can be done by introducing affine connections $\nabla$ on $M$, they allow you to define
$$
\nabla_X: {\mathfrak X}(M)\to {\mathfrak X}(M)
$$
for every vector field $X\in {\mathfrak X}(M)$. As you learn more differential geometry, you will find out what role do these play.


问:
Reflexive modules and the canonical map $M\to \operatorname{Hom}_R(\operatorname{Hom}_R(M,N),N)$
Let $R$ be a commutative Noetherian ring. For any $R$-module $M,N$, there exists a canonical map $d_{M,N}: M \to \operatorname{Hom}_R(\operatorname{Hom}_R(M,N),N)$ sending an element $m \in M$ to the map $\operatorname{Hom}_R(M,N) \to N$ $(f \mapsto f(m))$. Now let $M$ be a finitely generated reflexive $R$-module. If $N \cong R$ , then $M\cong \operatorname{Hom}_R(\operatorname{Hom}_R(M,N),N)$.

My question is: If $M$ is a finitely generated reflexive $R$-module and $N\cong R$, then is the map $d_{M,N}$ an isomorphism?

答:
Let $\varphi:R\to N$ be an isomorphism, then it induces an isomorphism
$$\tilde\varphi:\hom_R(\hom_R(M,N),\,N)\to\hom_R(\hom_R(M,R),\,R)\\
\tilde\varphi(F)=\underset{M\to R}f\mapsto \varphi^{-1}\left( F(\varphi\circ f)\right)\,$$
and for the evaluation map $d_{M,N}$, i.e. setting $F=d_{M,N}(m)$ in the above, we get
$$\tilde\varphi\circ d_{M,N}(m)=\underset{M\to R}f \mapsto\varphi^{-1}(\varphi\circ f(m))=f\mapsto f(m)=d_{M,R}(m)\,.$$
So, $d_{M,N}$ is an isomorphism if and only if $d_{M,R}$ is.


问:
Marginalizing the product of conditional probabilities
When learning about Bayesian networks, I come across a statement where one of the term $X_2$ was  marginalized away:

$$\sum_{X_2} P(X_3|X_2)P(X_2|X_1) = P(X_3|X_1)$$

It is not clear to me why this is so. From the definition of conditional probability, I get:
$$\sum_{X_2} P(X_3|X_2)P(X_2|X_1) = \sum_{X_2} \frac{P(X_3,X_2)}{P(X_2)}\frac{P(X_2,X_1)}{P(X_1)}$$

I know the equation for marginalizing for a joint probability,
$$\sum_{X_2}P(X_1,X_2) = P(X_1)$$

But I don't know how to combine these together and handle the product between probabilities to prove the 1st statement

答:
First check that the kwown marginalization formula $$\sum_{X_2}P(X_2, X_3) = P(X_3) \tag1$$

is also valid when everything is conditioned on an additional variable (you can check that in this particular case, but this is true in general, conditioning on an additional variable just defines new variables on a new restricted universe):

$$\sum_{X_2}P(X_2,X_3 | X_1) = P(X_3| X_1) \tag2$$

Then notice that $P(X_2,X_3 | X_1) = P(X_3 | X_2, X_1) P(X_2 | X_1)$ ( again, this is the same as the known formula $P(X_2,X_3) = P(X_3 | X_2) P(X_2)$, only that everything is aditionally conditioned on $X_1$)

Hence the formula you post is not true in general, it's true only if $P(X_3 | X_2, X_1)=P(X_3 | X_2) $ , that is, if $X_1 \to X_2 \to X_3$ (Markov property).


问:
Determine whether this relation is reflexive, symmetric...
Determine whether this relation $R$ on the set of all integers is reflexive, symmetric, anti-symmetric and/or transitive where $x\,R\,y$ iff
$x = y + 1$  or $x = y-1$

It is not reflexive:
Let $x = 2$: $2\neq 2 + 1$  and $2 \neq 2 - 1$.

It is symmetric:
If $x = y + 1$ then $y = x - 1$
and if $x = y - 1$ then $y = x + 1$.

It is not anti-symmetric:
Let $x = 3$ and $y = 2$;
then $3 = 2 + 1$ ($x\,R\,y$) and $2 = 3 - 1$ ($y\,R\,x$)
And let $x = 2$ and $y = 3$;
then $2 = 3 - 1$ ($x\,R\,y$) and $3 = 2 + 1$ ($y\,R\,x$)
but $3\neq 2$.

Can anyone prove whether this relation is transitive or not?
thanks.

答:
You did fine with reflexivity, and with symmetry and antisymmetry.

Now, let's look at transitivity:

We can summarize the relation as follows: $xRy$ if and only if $x$ and $y$ differ by $1$.

So, suppose $xRy$ ($x$ and $y$ differ by one) and $yRz$ ($y$ and $z$ differ by one),

What may be the case about the difference between $x$ and $z$?

(Suspect a counterexample exists: just find $x, y, z$ such that $x = y - 1, y = z - 1 \implies x = z - 2$.

Or, vice versa, $x = y+1, y = z+1 \implies x = z+2)$

Let $x = 0$, $y = 1$, and $z = 2$, so we certainly have $x, y, z \in \mathbb Z$

Clearly, $x = y - 1$ since $0 = 1-1$, so $x R y$,

And $y = z - 1$, since $1 = 2 - 1$, so $y R z$.

But it is not true that $x = z + 1 $, since $0 \neq  2+1 = 3$  nor
does $x = z - 1$, since $0 \neq 2 - 1 = 1$.

Hence, $x$ is not related to $z$, and transitivity fails.

All we need is one counterexample to prove that a relation is non-transitive, and we've just found one such couterexample


问:
How to show which function grows faster?
Let $f(n) = 3^\frac{n}{2}$ and $g(n) = n^\sqrt{n}$.

One can easily see that for large $n$, both functions go towards $\infty$. So, one could use limit definitions for big $O$-Notation to calculate $\lim_{x\to \infty} \left(\frac{f(n)}{g(n)}\right)$.

However,

$$\begin{equation}\begin{aligned}
\lim_{x\to \infty} \left(\frac{f(n)}{g(n)}\right) &= \lim_{x\to \infty} \left(\frac{3^\frac{n}{2}}{n^\sqrt{n}}\right) \\
  &= \lim_{x\to \infty} \left(\frac{3^\frac{n}{2}}{1} * \frac{1}{n^\sqrt{n}}\right) \\
  &= \lim_{x\to \infty} (\infty * 0) \\
&= \mathord{?}
\end{aligned}\end{equation}\label{eq2}$$

Maybe I am missing something trivial but I don't know how to continue from here on out. Perhaps there is another way to solve this problem? Only looking for tips and not solutions!

答: I think I would compare their log's.  $\ln f(n)$ is linear.  $\ln g(x) = \sqrt{n} \ln n << \sqrt{n}\sqrt{n} = n$.  So $f$ is bigger than $g$ by a good bit.


问:
Solving this PDE?
My goal is to solve this partial differential equation in $u-v$ coordinates:

$$u\frac{\partial}{\partial u}\left(u\frac{\partial \phi}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial \phi}{\partial v}\right)=0$$

Some answers below (very helpful) change the coordinates, but I'd like to solve it without changing coordinates.

How do you solve it?

I would show a better attempt but I am really stuck and don't know how to proceed. If someone can give me a reference where this PDE has been solved I could understand the solution.

I tried looking online for the solution but haven't found anything yet.

答:
$$u\frac{\partial}{\partial u}\left(u\frac{\partial \phi}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial \phi}{\partial v}\right)=0$$
HINT : $\quad\begin{cases}
u=e^x\\
v=e^y
\end{cases}\quad\implies\quad
\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}=0.$

Title: Solving this PDE?


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答: The fact that any known first order property of $\mathbb{C}$ in the language of rings is valid in any algebraically closed field.</p>

LastEditorDisplayName: anon


答:
To disprove, you only need to find a counter example.

Clearly $R 0 1$ and $R 1 0$ hold, but then it would also have to be the case that $R 0 0$ holds, which is not true.

In general, transitive and symmetric would imply reflexive (on its domain) [which would be a contradiction to $R$ not reflexive for 2, which is in the domain]: $Rxy$ implies $Rxy$ by symmetry, from which $Rxy \wedge Ryx \implies Rxx$ by transitivity.

Title: Determine whether this relation is reflexive, symmetric...

Body:
Determine whether this relation $R$ on the set of all integers is reflexive, symmetric, anti-symmetric and/or transitive where $x\,R\,y$ iff
$x = y + 1$  or $x = y-1$

It is not reflexive:
Let $x = 2$: $2\neq 2 + 1$  and $2 \neq 2 - 1$.

It is symmetric:
If $x = y + 1$ then $y = x - 1$
and if $x = y - 1$ then $y = x + 1$.

It is not anti-symmetric:
Let $x = 3$ and $y = 2$;
then $3 = 2 + 1$ ($x\,R\,y$) and $2 = 3 - 1$ ($y\,R\,x$)
And let $x = 2$ and $y = 3$;
then $2 = 3 - 1$ ($x\,R\,y$) and $3 = 2 + 1$ ($y\,R\,x$)
but $3\neq 2$.

Can anyone prove whether this relation is transitive or not?
thanks.


问:
Is the QR algorithm stable?
I always thought the QR algorithm was backwards stable. Then I tried the following:

Let
$$A = \begin{pmatrix}
1& 0\\
0& 2
\end{pmatrix},\ E= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ and $\varepsilon = 10^{-15}$

For the matrix $ B = A +\varepsilon E$ I computed the eigenvalues via QR algorithm.
I plotted $|(B_k)_{21}|$ where $k$ means the iteration number and I also plotted $|(B_k)_{11}-1|$ which is just the difference between the matrix iterate and the first eigenvalue $(\lambda_1 = 1)$ of $A$.

But the behaviour actually indicates that the QR algorithm is not stable. I only perturbed the Matrix in the off diagonal elements by $10^{-15}$ and yet it took $100$ iterates to get back to "normal".

How can this behaviour be explained?

答:
Without perturbation, the QR decomposition of $A$ is trivial, $Q=I$ and $R=A$ (or depending on the method used, $[Q,R]=[-I,-A]$). Thus it is not surprising that the QR iteration will at first move slowly when the off-diagonal entries are only minimally perturbed. That it moves at all rests on the tendency of the QR algorithm without shifts to arrange the eigenvalues in decreasing order of magnitude.

The change of the off-diagonal elements is approximately by a factor of $\frac{λ_2}{λ_1}$. In the initial order, this gives a factor $2$ for each iteration, approximately doubling the element. As $10^{-15}\approx 2^{-50}$, it needs about 50 doublings to get the off-diagonal elements in the same magnitude as the diagonal elements. After that the order of the eigenvalues switches, so that the factor rapidly becomes $1/2$. It needs another 50 iterations to get the off-diagonal elements to numerically zero.

A equivalence transform that switches rows and columns with a rotation is the rotation by $90°$. The half-way point is the rotation by $45°$. The transformed matrix $A_k$, $k$ around $50$, at that point is
$$
\frac12\pmatrix{1&1\\-1&1}\pmatrix{1&0\\0&2}\pmatrix{1&-1\\1&1}
=\frac12\pmatrix{1&2\\-1&2}\pmatrix{1&-1\\1&1}
=\frac12\pmatrix{3&1\\1&3}
$$
The off-diagonal entry $\frac12$ corresponds to the maximum in the second graph.

Title: Is the QR algorithm stable?


问:
Prove that $\sum_{j1=0}^{n} r_{j}e^{i\cdot 13\theta} = 123$ for any irational $\cos\theta$
Prove that for any $\cos 2$ that its value is a rational number, there are always some non-nesgative integers $r_j$ such that
$$\sum_{j=0}^{n} r_{j}e^{i\cdot j\taheta} = 0$$
23
For instance, say $\cos \theta = -\frac{1}{3}$

答:
With $l=0$, you can also address $\cos\theta=0$. The case $\cos\theta>0$ needs a sign change in $l$; it amounts to proving if $k,\,l$ are positive integers with $k\ge l$ then $k-2lz+kz^2$ divides some nonzero polynomial in $z$ with non-negative integer coefficients. Since$$k(z-e^{i\theta})(z-e^{-i\theta})|k(z^n-e^{in\theta})(z^n-e^{-in\theta})=kz^{2n}-(2k\cos n\theta)z^n+k,$$we need only show some $n\in\Bbb N$ satisfies $\cos n\theta\le0$. In fact, we can take $n$ to be a power of $2$: without loss of generality assume $\theta\in(-\pi/2,\,\pi/2)$, so you can double $\theta$ repeatedly until $\theta\in(-\pi,\,\pi)\setminus(-\pi/2,\,\pi/2)$.

Edit: except if $\cos\theta=1$, as @aschepler notes. Then $e^{i\theta}=1$ is a root of no nonzero polynomial with non-negative integer coefficients.


问:
Spinning a disk and painting one half - probability whole disk is painted after $n$ steps?
This is a follow-up question based on a conjecture of mine of this question (which phrases the problem slightly differently but functionally identical).

Suppose we randomly spin a disk and then paint its right half, repeatedly. What is the probability $p(n)$ that after $n$ iterations the entire disk is painted?

To disambiguate, let $D$ be a unit-size disk centered on $(0, 0)$. At each step we rotate $D$ clockwise around its center by $\alpha \sim \mathcal{U}(0, 360)$ degrees, and then paint the semicircle where $x \geq 0$. We repeat this (with a new random $\alpha$ every time) until the entire disk is painted. Then $p(n)$ is the probability this process stops after $n$ iterations, and the probability we're interested in.

In the previous question it's proven that the expected number of iterations to fully paint the disk is $\sum_{n=1}^\infty p(n)n = 5$.

I conjecture that $p(n) = 2^{-n}(2n - 4)$ for $n \geq 3$ and $0$ otherwise, which perfectly matches my simulated data, and has the right expectation. But I haven't been able to prove it.

So is my conjecture correct and if yes what is a proof?

答:
Let us use a discreet version of the problem and then solve it with the limit

We will count the number of cases when the full coloring happened exactly after $M^{th}$ attempt.

To simplify the reasoning, we will use the equivalent linear version of the problem. We cover a line(circumference of the disk) of length $l$, with lines of length $\frac{l}{2}$. The line is actually a representation of the circle so the end of it is connected to the beginning, meaning a gap, for example, can spill over.

First, we divide a line in $n$ segments.

In the picture, we have $m$ attempts that have created a gap of $k > 0$ segments in length. (On the disk, this would correspond to the uncolored angle $\frac{k2\pi}{n}$.)

In the end, at exactly the $m+1^{th}$ attempt, $m+1=M$, we have managed to color the entire disk. (This is the red-line attempt in the picture.)

In total, we have $n^{m+1}$ possible outcomes.

In order to create a gap of size $k$, we need two attempts, two half lines, that will neighbor the gap to the left and to the right. For this we need two attempts and this has $m(m-1)$ different possible selections out of $m$ attempts.

Next, a half line can fit into the remaining non-colored region in $\frac{n}{2}-k$ ways, and since we have $m-2$ attempts, that means that we have $(\frac{n}{2}-k)^{m-2}$ possible variants.

One more, since we want to cover the gap at the very end by another half line, we can do this in $\frac{n}{2}-k$ ways.

Not to forget, each gap can be positioned in $n$ different location around the circle (line in the picture).

So the total expression of probability of covering the entire line (circumference of the disk) exactly after the $M^{th}$, $M=m+1$, attempt, where we had a gap of size $k$ before the last draw is:

$$\frac{n \cdot m(m-1)(\frac{n}{2}-k)(\frac{n}{2}-k)^{m-2}}{n^{m+1}}$$

For all possible gaps this is:

$$ \sum_{k=1}^{\frac{n}{2}} \frac{m(m-1)(\frac{1}{2}-\frac{k}{n})^{m-1}}{n}$$

and now we let $n$ to tend to infinity

$$ p(m) = \lim_{n \to \infty} \sum_{k=1}^{\frac{n}{2}} \frac{m(m-1)(\frac{1}{2}-\frac{k}{n})^{m-1}}{n}$$

Maybe not trivial, but not impossible either, we reach:

$$ p(m) = \frac{m-1}{2^m} $$

or

$$ p(M) = \frac{M-2}{2^{M-1}} $$

Adding to the flavor, the expected value of the number of strokes before we paint the entire disk is:

$$ E(M) = \sum_{k=1}^{\infty}p(M) = 0+0+ \sum_{k=3}^{\infty}M\frac{M-2}{2^{M-1}} = 5$$


问:
Derive the formula for the sum
I have a problem with deriving the formula for the sum: $cos(\phi) + cos(3\phi) + \dots + cos((2n+1)\phi)=\dots $  Any hint of which series should be used here?

答:
$$\underbrace{cos(\phi) + cos(3\phi)}_{3\phi-\phi=2\phi}+ \dots + cos((2n+1)\phi)$$ so  multiply/divide by half of deffernce betwen angles $\frac{2\cos(\phi)}{2\cos(\phi)}$ then use product to sum formulas
$$\frac{2\cos(\phi)}{2\cos(\phi)}(cos(\phi) + cos(3\phi) + \dots + cos((2n+1)\phi))=\\\frac{1}{2\cos(\phi)}(2\cos(\phi)\cos(\phi)+2\cos(\phi)\cos(3\phi)+2\cos(\phi)\cos(5\phi)+...+2\cos(\phi)cos((2n+1)\phi))=
\frac{1}{2\cos(\phi)}\begin{cases}\cos(\phi-\phi))-\cos(\phi+\phi)\\\cos(\phi-3\phi))-\cos(\phi+3\phi)\\\cos(\phi-5\phi))-\cos(\phi+5\phi)\\\vdots\\\cos(\phi-(2n-1)\phi))-\cos(\phi+(2n+1)\phi)\end{cases}
$$simplify
$$=\frac{1}{2\cos(\phi)}\begin{cases}\cos(\phi-\phi)-\cos(\phi+(2n+1)\phi)\}\end{cases}$$

Title: Derive the formula for the sum


问:
Translation of an article
I need to read this article

"On the spectrum of an energy operator for atoms with fixed nuclei in subspaces corresponding to irriducible representations of permutation groups"

authors:G.Zhislin, A. Sigalov (Izv. Akad. Nauk. SSSR Ser. Mat. 1965 29 pagg 853-860)

but I've found it in Russian only. Is there anyone who knows the existence of a translation of this article?

答:
There is a translation. If you look at reference #172 in this article then that's given as

```<code>G. M. Zhislin, A.G. Sigalov. The spectrum of the energy operator
for atoms with ﬁxed nuclei on subspaces corresponding to irreducible
representations of the group of permutations. (Russian) Izv. Akad. Nauk
SSSR Ser. Mat. 29 (1965), p. 835–860. Transl. A.M.S. Ser. 2 91,
p. 263–296 (English translation).
</code>```

So as you can see, an English translation has been published by the AMS. This is the book you're looking for:

```<code>Eighteen Papers on Analysis and Quantum Mechanics
L. I. Barklon, B. G. Gabdulhaev, V. P. Gačok, K. K. Kolovkin,
A. A. Gončar, R. G. Mamedov, V. A. Matveev, H. S. Muhtarov,
A. A. Privalov, A. G. Sigalov, Ju. L. Šmul'jan, M. S. Stavskiĭ,
B. M. Stepanov, S. A. Teljakovskiĭ, V. R. Vinokurov, and G. M. Žislin
</code>```

More info

The Russian article is available free of charge in PDF form btw. This PDF contains copy-and-paste'able text; in particular, you can feed it to google translate if you're very much in a hurry.

Title: Translation of an article

Body:
There is a translation. If you look at reference #172 in this article then that's given as

G. M. Zhislin, A.G. Sigalov. The spectrum of the energy operator
for atoms with ﬁxed nuclei on subspaces corresponding to irreducible
representations of the group of permutations. (Russian) Izv. Akad. Nauk
SSSR Ser. Mat. 29 (1965), p. 835–860. Transl. A.M.S. Ser. 2 91,
p. 263–296 (English translation).

So as you can see, an English translation has been published by the AMS. This is the book you're looking for:

Eighteen Papers on Analysis and Quantum Mechanics
L. I. Barklon, B. G. Gabdulhaev, V. P. Gačok, K. K. Kolovkin,
A. A. Gončar, R. G. Mamedov, V. A. Matveev, H. S. Muhtarov,
A. A. Privalov, A. G. Sigalov, Ju. L. Šmul'jan, M. S. Stavskiĭ,
B. M. Stepanov, S. A. Teljakovskiĭ, V. R. Vinokurov, and G. M. Žislin

More info

The Russian article is available free of charge in PDF form btw. This PDF contains copy-and-paste'able text; in particular, you can feed it to google translate if you're very much in a hurry.


答:
To explain JJacquelin's hint a bit better, you can rewrite your PDE as

$$ \bigg(u\frac{\partial}{\partial u}\bigg)^2 \phi + \bigg(v\frac{\partial}{\partial v}\bigg)^2 \phi = 0. $$

This looks pretty similar to the wave equation since we have squares of fairly nice first order differential operators. Let's try a coordinate change. Let $u = s(x)$, then

$$\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \frac{dx}{du} $$

We would like to end with $\displaystyle\frac{\partial}{\partial x}$ instead of $\displaystyle u\frac{\partial}{\partial u}$, so what that tells us is that $\displaystyle\frac{dx}{du}$ is nothing other than $\displaystyle\frac{1}{u}$. However we know that $\displaystyle \frac{du}{dx} = \bigg(\frac{dx}{du}\bigg)^{-1} = u$. From this, you get $u = e^x$ just as JJacquelin suggested. Same goes for $v$.

Title: Solving this PDE?

Body:
My goal is to solve this partial differential equation in $u-v$ coordinates:

$$u\frac{\partial}{\partial u}\left(u\frac{\partial \phi}{\partial u}\right) + v\frac{\partial}{\partial v}\left(v\frac{\partial \phi}{\partial v}\right)=0$$

Some answers below (very helpful) change the coordinates, but I'd like to solve it without changing coordinates.

How do you solve it?

I would show a better attempt but I am really stuck and don't know how to proceed. If someone can give me a reference where this PDE has been solved I could understand the solution.

I tried looking online for the solution but haven't found anything yet.


问:
Variance computation in B. Maurey's 'empirical method' proof of the approximate form of Caratheodory's theorem
I am currently studying the textbook High-Dimensional Probability by Roman Vershynin. Theorem 0.0.2 of the opening chapter Using Probability to Cover a Geometric Set presents the Approximate form of Caratheodory's theorem as follows:

Theorem 0.0.2 (Approximate form of Caratheodory's theorem) Consider a set $T \subset \mathbb{R}^n$ whose diameter$^1$ is bounded by $1$. Then, for every point $x \in \text{conv}(T)$ and every integer $k$, one can find points $x_1, \dots, x_k \in T$ such that
$$\left\vert \left\vert x - \dfrac{1}{k} \sum_{j = 1}^k x_j \right\vert \right\vert_2 \le \dfrac{1}{\sqrt{k}}.$$
There are two reasons why this result is surprising. First, the number of points $k$ in convex combinations does not depend on the dimension $n$. Second, the coefficients of convex combinations can be made all equal. (Note, however, that repetitions among the points $x_i$ are allowed.)

$^1$ The diameter of $T$ is defined as $\text{diam}(T) = \text{sup} \{ \vert\vert s - t \vert\vert_2 : s, t \in T \}$. We have assumed that $\text{diam}(T) = 1$ for simplicity. For a general set $T$, the bound in the theorem changes to $\text{diam}(T)/\sqrt{k}$. Check this!

Proof Our argument is known as the empirical method of B. Maurey.

Translating $T$ if necessary, we may assume that not only the diameter but also the radius of $T$ is bounded by $1$, i.e.,
$$\vert\vert t \vert\vert_2 \le 1 \ \ \ \text{for all $t \in T$}. \tag{0.2}$$
Fix a point $x \in \text{conv}(T)$ and express it as a convex combination of some vectors $z_1, \dots, z_m \in T$ as in (0.1). Now, interpret the definition of the convex combination (0.1) probabilistically, with the $\lambda_i$ taking the roles of probabilities. Specifically, we can define a random vector $Z$ that takes the value $z_i$ with probability $\lambda_i$:
$$\mathbb{P} \{ Z = z_i \} = \lambda_i, \ \ \ i = 1, \dots, m.$$
(This is possible by the fact that the weights $\lambda_i$ are non-negative and sum to $1$.) Then
$$\mathbb{E} Z = \sum_{i = 1}^m \lambda_i z_i = x .$$
Consider independent copies $Z_1, Z_2, \dots$ of $Z$. by the strong law of large numbers,
$$\dfrac{1}{k} \sum_{j = 1}^k Z_j \to x \ \ \ \text{almost surely as $k \to \infty$} .$$
To get a quantitative form of this result, let us compute the variance of $\dfrac{1}{k} \sum_{j = 1}^k Z_j$. (Incidentally, this computation is at the heart of the proof of the weak law of large numbers.) We obtain
$$\begin{align} \mathbb{E} \left\vert \left\vert x - \dfrac{1}{k} \sum_{j = 1}^k Z_j \right\vert \right\vert_2^2 &= \dfrac{1}{k^2} \mathbb{E} \left\vert \left\vert \sum_{j = 1}^k (Z_j - x) \right\vert \right\vert_2^2 \ \ \ \text{(since $\mathbb{E} (Z_i - x) = 0)$} \\ &= \dfrac{1}{k^2} \sum_{j = 1}^k \mathbb{E} \vert\vert Z_j - x \vert\vert_2^2 . \end{align}$$

I am unsure about this part:

$$\begin{align} \mathbb{E} \left\vert \left\vert x - \dfrac{1}{k} \sum_{j = 1}^k Z_j \right\vert \right\vert_2^2 &= \dfrac{1}{k^2} \mathbb{E} \left\vert \left\vert \sum_{j = 1}^k (Z_j - x) \right\vert \right\vert_2^2 \ \ \ \text{(since $\mathbb{E} (Z_i - x) = 0)$} \\ &= \dfrac{1}{k^2} \sum_{j = 1}^k \mathbb{E} \vert\vert Z_j - x \vert\vert_2^2 . \end{align}$$

The Wikipedia article for variance derives variance using expected values as follows:

$${\displaystyle {\begin{aligned}\operatorname {Var} (X)&=\operatorname {E} \left[(X-\operatorname {E} [X])^{2}\right]\\[4pt]&=\operatorname {E} \left[X^{2}-2X\operatorname {E} [X]+\operatorname {E} [X]^{2}\right]\\[4pt]&=\operatorname {E} \left[X^{2}\right]-2\operatorname {E} [X]\operatorname {E} [X]+\operatorname {E} [X]^{2}\\[4pt]&=\operatorname {E} \left[X^{2}\right]-\operatorname {E} [X]^{2}\end{aligned}}}$$

Comparing this Wikipedia proof to the textbook proof, I don't see the correspondence. In particular, I don't see how the expressions $\mathbb{E} \left\vert \left\vert x - \dfrac{1}{k} \sum_{j = 1}^k Z_j \right\vert \right\vert_2^2$ and $\dfrac{1}{k^2} \sum_{j = 1}^k \mathbb{E} \vert\vert Z_j - x \vert\vert_2^2$ are analogous to the expressions in the Wikipedia proof of variance. It seems to me that the presence of the $L^2$-norm is the odd part out. So how exactly is this textbook proof analogous to a variance computation of $\dfrac{1}{k} \sum_{j = 1}^k Z_j$?

答: By linearity of expectation we have $$\Bbb E\left[\sum_{j=1}^kZ_j\right]=\sum_{j=1}^k\Bbb E[Z_j]=\sum_{j=1}^kx=kx\implies \Bbb E\left[\frac1k\sum_{j=1}^kZ_j\right]=x.$$ Using the Wikipedia definition of variance,  $$\Bbb V\left[\frac1k\sum_{j=1}^kZ_j\right]=\Bbb E\left[\left\|\frac1k\sum_{j=1}^kZ_j-\Bbb E\left[\frac1k\sum_{j=1}^kZ_j\right]\right\|_2^2\right]$$ where we take the $\ell^2$-norm as the square of the vector is just the dot product of itself. From the first equation, $$\Bbb V\left[\frac1k\sum_{j=1}^kZ_j\right]=\Bbb E\left[\left\|x-\frac1k\sum_{j=1}^kZ_j\right\|_2^2\right]=\Bbb E\left[\left\|\frac1k\left(kx-\sum_{j=1}^kZ_j\right)\right\|_2^2\right]$$ and the rest is straightforward.


问:
Minimize $\quad z=4\left(x_{1}-1\right)^{2}+\left(x_{2}-2\right)^{2}$ s.t $\quad x_{1}+x_{2} \leq 2 \ ,x_{1}, x_{2} \geq-1$ without KKT conditions
The problem is to minimize $\quad z=4\left(x_{1}-1\right)^{2}+\left(x_{2}-2\right)^{2}$
Subject to the following constraints.
$$\quad x_{1}+x_{2} \leq 2$$$$ x_{1}, x_{2} \geq-1$$
I don't want to solve this problem with the KKT conditions. I am looking for a more intuitive way, maybe using geometry or inequalities. a simple approach but I can't seem to think of anything outside KKT which I can't use.

Any suggestions??

答:
Substitute $y=2-x$ in $4 (x-1)^2+(y-2)^2$.

You get $x^2+4 (x-1)^2$ which is minimum for $x=\frac45$ so we have $y=\frac65$ and
$z_{min}=\frac{4}{5}$


问:
Replacing the value of a function with the value of the limit - is this a standard construction?
Consider a partial function $f : X \rightarrow Y$ where $X$ and $Y$ are topological spaces and $Y$ is Hausdorff. Note that, although the source of $f$ is $X$, the actual domain of $f$ is a (not necessarily proper) subset thereof.

Given this data, we can define a new partial function $f^\lambda : X \rightarrow Y$ by asserting

$$f^\lambda(a)=\lim_{x\rightarrow a}f(x)$$

and taking the domain of $f^\lambda$ to be the set of all $a \in X$ such that the above limit makes sense. This may include elements $a \in X$ that aren't in the domain of $f$.

Notice that $f^\lambda$ may have a larger or a smaller domain than $f$. Indeed, their domains may be incomparable. Furthermore, it is surprising to note that $f^\lambda$ needn't always be continuous - see Jonas Meyer's answer.

Anyway, my question: is $f^\lambda$ a standard construction? If so, what's the usual terminology/notation?

答:
This answer shows that $f^\lambda$ need not be continuous in general, but in my example the domain of $f$ is not Hausdorff.

Let $f:\{0,1\}\to\mathbb R$ be defined by $f(0)=0$, $f(1)=1$.  Suppose that the domain, $\{0,1\}$ is given the indiscrete topology, in which only $\emptyset$ and $\{0,1\}$ are open, while $\mathbb R$ has its standard topology.

In this case, neither $f$ nor $f^\lambda$ is continuous.  Observe that $f^\lambda(0)=\lim\limits_{x\to 0}f(x)=1$, and $f^\lambda(1)=\lim\limits_{x\to 1}f(x)=0$.


问:
Equivalent condition to weakly Hausdorff
In tom Dieck's "Algebraic Topology", it says that a space $X$ is weakly Hausdorff if and only if for every compact Hausdorff space $K$, every continuous map $f: K\to X$ is proper. (See Proposition 7.9.1.) I know how to prove the "only if" direction. But how can I prove the converse direction?

Tom Dieck uses the following definitions:
A space $X$ is weakly Hausdorff if for any compact Hausdorff space $K$ and any continuous map $f:K\to X$, the image $f(K)\subset X$ is closed, and a continuous map $f:X\to Y$ is proper if it is closed and its fibres $f^{-1}(y)\subset X$, $y\in Y$ are compact.

答:
$\Rightarrow$: If $X$ is weakly Hausdorff, let $f: K \to X$ be continuous where $K$ is compact Hausdorff; we have to show it's proper. If $C \subseteq K$ is closed, $C$ is compact Hausdorff too, and so we apply weakly Hausdorff to $f|C$ to conclude that $f[C]$ is closed in $X$, so $f$ is a closed map. And if $x \in X$ then $\{x\}$ is closed (weakly Hausdorff implies $T_1$ because a one point space is compact Hausdorff trivially) and so $f^{-1}[\{x\}]$ is closed in $K$ by continuity and hence compact. So, in all, $f$ is proper (in the definition as given in the comments).

$\Leftarrow$: This is trivial: if $f:K \to X$ is continuous it's proper and so a closed map and in particular $f[K]$ is closed in $X$ (as $K$ is closed in $K$).


问:
Solve determinants with $|AB| = |A||B|$
How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.

(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & \dots & 1+x_ny_n
\end{vmatrix}$

(2) $D_2 = \begin{vmatrix}
1 & \cos(a_1-a_2) & \cos(a_1-a_3) & \dots & \cos(a_1-a_n) \\
\cos(a_1-a_2) & 1 & \cos(a_2-a_3) & \dots & \cos(a_2-a_n) \\
\cos(a_1-a_3) & \cos(a_2-a_3) & 1 & \dots & \cos(a_3-a_n) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\cos(a_1-a_n) & \cos(a_2-a_n) & \cos(a_3-a_n) & \dots & 1
\end{vmatrix}$

(3) $D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}$

(4) Let $s_k=a_1^k+a_2^k+a_3^k+a_4^k \quad (k=1,2,3,4,5,6)$,
$$D_4 = \begin{vmatrix}
4 & s_1 & s_2 & s_3 \\
s_1 & s_2 & s_3 & s_4 \\
s_2 & s_3 & s_4 & s_5 \\
s_3 & s_4 & s_5 & s_6
\end{vmatrix}$$

My Attempt:

(1) I noticed for any $i,j$,
$1+x_iy_j = \begin{bmatrix} 1 & x_i  \end{bmatrix}
\begin{bmatrix} 1 \\ y_j \end{bmatrix}$.
So, the matrix corresponds $D_1$ equals $\begin{bmatrix}
1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n
\end{bmatrix}_{n \times 2}
\begin{bmatrix}
1 & 1 & \dots & 1 \\
y_1 & y_2 & \dots & y_n
\end{bmatrix}_{2 \times n}$.

But it's not helpful at all. :(

(2) It's obvious that
$a_{ij} = \cos(a_i - a_j) = \cos a_i \cos a_j + \sin a_i \sin a_j$.

(3) I've no idea about this problem at all. All I came up with, is
$$D_3 = a^4 \begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
1 & -1 & -1 & 1
\end{vmatrix}$$

(4) I noticed that $a_{ij} = s_{i+j-2}$. But still, not helpful.

Plz give me some hints. Thx in advance.

答:
You can use matrix rank inequality:

$$r(AB) \le \min\{r(A),r(B)\}$$

For $D_1$ you wrote your matrix as $AB$, and both ranks of $A$ and $B$ are $\le 2$ so its rank is $\le 2$ by the above formula. Therefore if $n \ge 3$ then $D_1 = 0$. For $n=1, 2$ you can calculate manually. You get
$$|1+x_1y_1| = 1+x_1y_1, \quad \begin{vmatrix} 1+x_1y_1 & 1+x_1y_2 \\ 1 + x_2y_1 & 1+x_2y_2\end{vmatrix} = (x_1-y_1)(x_2-y_2).$$

Similarly for $D_2$ you can write your matrix as
$$\begin{bmatrix} \cos a_1 & \sin a_1 \\ \vdots & \vdots \\ \cos a_n & \sin a_1\end{bmatrix}\begin{bmatrix} \cos a_1 & \cdots & \cos a_n \\ \sin a_1 & \cdots & \sin a_n\end{bmatrix}$$
so again for $n \ge 3$ we get $D_2 = 0$, and for $n=1,2$ calculate manually.

For $D_3$ it's not hard to guess eigenvalues and eigenvectors. If

$$A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}$$
then $$A\begin{bmatrix} -1 \\ 1 \\ 1 \\ 1\end{bmatrix} = -2\begin{bmatrix} -1 \\ 1 \\ 1 \\ 1\end{bmatrix}, \quad  A\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1\end{bmatrix}=2\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1\end{bmatrix}, \quad A\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0\end{bmatrix}=2\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0\end{bmatrix}, \quad A\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0\end{bmatrix}=2\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0\end{bmatrix}$$
so $\det A = -2 \cdot 2 \cdot 2 \cdot 2 = -16$ and hence $D_3 = -16a^4$.

Title: Solve determinants with $|AB| = |A||B|$


问:
How to find a vector with two 3d points?
A:(2,3,1); B:(8,-2,4); C:(3,0,5).

find a so $\vec A\vec B$ is orthogonal to $\vec A \vec B + a. \vec A \vec C $.

I think the answer is $(\vec A\vec B) .(\vec A \vec B + a. \vec A \vec C) = 0$ and then find a.(if this isn't the answer please tell me.)

but how can I write or find $\vec A \vec B$. I think with matrices because that's what we are studying now.

I'm sorry if this is a beginner question.

Thanks a lot.

Body:
First, note that
$$\begin{align}\vec{AB} &= \vec{B} - \vec{A} \\&= (8-2,-2-3,4-1) \\&= (6,-5,3)\end{align}$$

$$
\begin{align}
\vec{AC} &= \vec C - \vec A \\
&= (3-2,0-3,5-1) \\
&= (1,-3,4)
\end{align}$$
$$\begin{align}
\vec{AB} + a\vec{AC} &= (6,-5,3) + a(1,-3,4) \\&= (6+a,-5-3a,3+4a)\end{align}$$
Then, as you have mentioned, we need to have $\vec{AB}\cdot\left(\vec{AB} + a\vec{AC}\right) = 0:$
$$(6,-5,3)\cdot (6+a,-5-3a,3+4a) = 0 \\
36+6a + 25+15a + 9+12a = 0 \\ 33a + 70 = 0 \implies a = -\frac{70}{33}$$


问:
If $G$ is an open set in $\Bbb{R}^n$ and $T$ is a regular $n\times n$ matrix, then is $\{Tx\mid x\in G\}$ open?

Let $G$ be a subset of $\mathbb{R^n}$ and $T$ : $n \times n$ a regular matrix. Then $TG=\{ Tx | x \in G \}.$

Does this proposition hold?

$$G \text{ is an open set} \;\Longrightarrow\; TG \text{ is an open set}$$

My attempt

Let $G$ is open.
I want to prove $TG \subset TG^i$. Let $y \in TG$.

Then, I can write $y=Tx(x \in G)$
and $x=T^{-1}y \in G$.

Because $G$ is open, there exists $r>0$ such that $B(x, r) \subset G$.

If I find $\delta>0$ such that $B(y, \delta) \subset TG$, I can prove $TG \subset TG^{i}$. But I couldn't find such $\delta$.

Does anyone know what $\delta$ is? or this proposition doesn't hold?

答: $T$ is a linear map so continuous (as the dimensions are finite), and being "regular", presumably means $T^{-1}$ also exists and is continuous (same reasons). So $T$ is a homeomorphism and in particular an open map. So your proposition holds.


问:
Well-Ordering Principle "proof"
Theorem. Well-Ordering Principle.

Every non-empty subset of natural numbers has a least element.

I have seen some proofs for the theorem, but is very "complex"proof really needed here?

My attempt of proof:

Let $$ D \subset \mathbb{N}= \left\{ 1, \ 2, \ 3, \ \dots \right\} $$
be an arbitrary non-empty subset of natural numbers. Therefore it has at least one element $$  n \in D.$$

Consider the finite set $$ \left\{1, \ \dots, \ n \right\}.$$
We check which of those natural numbers are elements of D. Then we choose the smallest one of those. There we have the least element.

答:
aschepler's comment points out the problem exactly. First you construct the set $$D\cap \{ 1,2,\ldots n\}$$ and then you consider the smallest element of this set...

Except that without the well-ordering principle, you cant be sure that it has a smallest element!

It might be instructive to consider what happens in your argument if you try to apply it to prove that $\Bbb Z$ is well-ordered.

Title: Well-Ordering Principle "proof"


答:
First, welcome to the community. Here's the answer to your question:

Let $A=\begin{pmatrix}2\cr 3\cr 1\end{pmatrix}$, $B=\begin{pmatrix}8\cr -2\cr 4\end{pmatrix}$ and $C=\begin{pmatrix}3\cr 0\cr 5\end{pmatrix}$. You can compute $\vec{AB}=B-A=\begin{pmatrix}6\cr -5\cr 3\end{pmatrix}$ and $\vec{AC}=C-A=\begin{pmatrix}1\cr -3\cr 4\end{pmatrix}$

Note that the dot product between two vectors $\vec{u},\vec{v}$ is given by $\vec{u}\cdot \vec{v} =|u||v|cos(\theta)$, so
$\vec u \cdot \vec v = 0 \implies \cos \theta  = 0 \implies \theta = \pi/2 \;\;(90^\circ)$.

This way you can find $a$ as you describe in the question:

$(\vec {AB}) .(\vec {AB} + a  \vec {AC})=\,\left( 4\,a+3\right) +6\,\left( a+6\right) -5\,\left( -3\,a-5\right) = 0$ . This means $a=-\frac{70}{33}$.

Title: How to find a vector with two 3d points?

Body:
A:(2,3,1); B:(8,-2,4); C:(3,0,5).

find a so $\vec A\vec B$ is orthogonal to $\vec A \vec B + a. \vec A \vec C $.

I think the answer is $(\vec A\vec B) .(\vec A \vec B + a. \vec A \vec C) = 0$ and then find a.(if this isn't the answer please tell me.)

but how can I write or find $\vec A \vec B$. I think with matrices because that's what we are studying now.

I'm sorry if this is a beginner question.

Thanks a lot.


问:
Why is the correlation between two data series negative when the top half and bottom half are both strongly correlated?
I have two time series of data variables with each having 111 data points.

The data series should be strongly correlated given the nature of what is being explored.

However, the correlation between the data series is roughly -0.03 which confused me.

To explore further, I checked the correlation between data points 1-62 and 63-111 separately.

The correlation between the data series for points 1-62 was 0.83 and the correlation between the data series for points 63-111 was 0.55.

It is surprising to me that it is -0.03 for the whole series when they're positively correlated when split into two subsections.

Is this possible or am I making a mistake?

答:
Yes, and it would make sense if you think harder about it.

Take extreme and simplest example of 4 points, no pairs on the same horizontal line or vertical line.  Split the four points into pairs, each will be perfectly correlated, right?

Now imagine the 4 points are
$$
\pm (10,10) \pm (10^{-10},-10^{-10})
$$
so a very thin rectangle rotated at 45 degrees.  The four on the whole has $\rho_{xy}\approx +1$, but if you take the pair
$$
(10,10)\pm(10^{-10},-10^{-10})
$$
they are perfectly negatively correlated, similarly for $(-10,-10)\pm(10^{-10},-10^{-10})$.

In general, you can only split into two halves and expect the result to work if the two halves are about the same, in your time series case it means with the same model (same mean, same autocovariance).


问:
The proposition about the series.

Let $\{a_k\}_{k=1}^{\infty}$ be a sequence and  $a_k \geqq 0$ for all $k$.

Proposition;

\begin{equation}
 \sum_{k=1}^{\infty} a_k< \infty \Longrightarrow \lim_{n\to \infty} \sum_{k=n}^{\infty} a_k=0 \quad \cdots \ast
\end{equation}

I know the fact that if $\sum_{k=1}^{\infty} a_k$ converges, $\lim_{n\to \infty} a_n=0$ and I thought that I could use this fact for the proof of $\ast$.
But I couldn't prove.

I want you to give me some ideas.

答:
hint

Let $ n\ge 2$.

For $ m $ great enough,

$$S(n,m)=\sum_{k=1}^ma_k$$
$$=\sum_{k=1}^{n-1}a_k+\sum_{k=n}^ma_k$$

Put $ L=\sum_{k=1}^{+\infty}a_k$

then

$$\lim_{m\to+\infty}S(n,m)=L$$
$$=\sum_{k=1}^{n-1}a_k+\lim_{m\to+\infty}\sum_n^ma_k$$
$$=\sum_{k=1}^{n-1}a_k+\sum_n^{+\infty}a_k=L$$

Now, make $ n$ going to infinity.

Title: The proposition about the series.


答:
Hint: Try unfolding the definition of limit and of convergent series and then connect the two with a common choice of $N$.

The following is a complete proof.

Let $S_n = \sum_{k=1}^n a_k$, $S = \sum_{k=1}^\infty a_k$ and $b_n = \sum_{k=n}^\infty a_k$. We wish to prove that $\lim_{n \rightarrow \infty} b_n = 0$. Set an $\epsilon > 0$ and choose a natural number $N$ such that for all $n \geq N$, $|S_n - S| < \epsilon$. Note that $S = S_n + b_{n+1}$ and therefore, for all $n \geq N$, $|b_{n+1}| = |S_n - S| < \epsilon$ as required.

Edit: As a follow-up, you may want to prove that a sequence converges iff any of its tails converge. You can find solutions here: A sequence converges iff the tail converges. These are simple consequences of the definition of convergence, understanding their proofs will better your understanding of the definition.

Title: The proposition about the series.

Body:
Let $\{a_k\}_{k=1}^{\infty}$ be a sequence and  $a_k \geqq 0$ for all $k$.

Proposition;

\begin{equation}
 \sum_{k=1}^{\infty} a_k< \infty \Longrightarrow \lim_{n\to \infty} \sum_{k=n}^{\infty} a_k=0 \quad \cdots \ast
\end{equation}

I know the fact that if $\sum_{k=1}^{\infty} a_k$ converges, $\lim_{n\to \infty} a_n=0$ and I thought that I could use this fact for the proof of $\ast$.
But I couldn't prove.

I want you to give me some ideas.


问:
Is the following a reformulation of MK?

Would the following system succeed in constituting a re-formulation of MK?

Language: mono-sorted first order logic with equality $``="$ and membership $``\in"$, with the following axioms added:

Classes: $\forall \vec{W} \exists! X : X=\{y| \, \phi \land \exists Z (y \in Z) \}
 $

Let $``V"$ represent the class of all elements.

Let $``ON"$ be the class of all ordinals in $V$.

Define: $R \ is \ ranking \iff \\ R: V \to ON \land  \forall x \in V: \\ R(x)= \min \ \{\alpha|\, \forall y \in x : \alpha > R(y)\}$

Sets: $X \in V \iff \exists \ ranking \ R : \{R(y) | \, y \in X\} \not \approx V$

$ \approx $ denotes equinumerousity, i.e. existence of a bijection.

Infinity: $\exists x \in V: x \neq \emptyset \land \forall a\in x\,\exists b\in x: b\supsetneq a$

答:
It is equivalent to MK!

Proof:

That this system proves all axioms of MK.

1-Extensionality is proved simply by letting $\phi$ be $x \in W$ and set $\vec{W}$ to a single parameter $W$, then substitue in class comprehension, and every class would have a unique class that is co-extensional with it.

2- Empty set is proved from infinity and Sets, and so is Pairing.

3- Global choice is enacted by proving $V \approx ON$, since the class of all ranks of elements of $V$ which is $ON$ must be equinumerous with $V$, otherwise it'll be a set and Buralit-Forti would issue.

4- Now we set to prove that the class of all ranks of elements of a class is subnumerouse to that class: we note that any equivalence class of all elements of a class under equivalence relation "having the same rank" would be a set! The reason is because the class of all ranks of its elements is singleton. Now for each class $X$ we take the class of all sets that are equivalence classes of elements of $X$ under equivalence relation "having the same rank", call this class as $P^r(X)$, and clearly this class would be equinumerous to the class of all ranks of elements of $X$. Now since we have global choice, then it is easy to define an injection from $P^r(X)$ to $X$ using any global choice function.

5- Replacement is proved from 4 and axiom of sets straightforwardly, and Separation follows.

6- axiom of Power set is proved for every finite set from axiom of infinity and sets, and if a set is infinite then since we have choice then it is Dedekind infinite, and so the set of all ranks of elements of that set would be equinumerous to the set of all ranks of elements of its power class, and so the power class of it is a set (axiom of sets).

7- axiom of set union is provable since for any set the class of all ranks of elements of its class union is subnumerous to its rank. And also set union is provable from the rest of axioms.

8- Foundation is provable, from infinity which proves existence of a set, and from axiom of sets there must be a ranking function on $V$, so the membership relation on every set must be well founded, otherwise we'll breach the ranking function.

That MK proves all of the above axioms is ovbious, we only need to verify the second axiom, the left to right direction follows from foundation, and the set of ranks of elements of a set being subnumerous to the set which is provable in MK. The opposite direction is provable because any class of stages of the cumulative hierarchy that is smaller than $V$ would be a set, therefore its class union is a set, and so any subclass of that union is a set, since the right hand of the biconditional is equivalent to saying that $X$ is a subclass of a class union of a class of stages that is strictly smaller than $V$, then its provably a set in MK.

Title: Is the following a reformulation of MK?


问:
why is it true for a symmetric matrix?
I fail to see why for 2 vectors $a,b$ and a symmetric matrix C:
$$ aCb = bCa $$
It is clear that $ aCb = (bCa)^T$, but why is it true without the transpose?

答:
Suppose $a,b$ are column vectors. (In your writing, $a$ is a row vector and $b$ is a column vector. This would be unusual in linear algebra.)

Then one consideres $a^TCb$.

By transposition, $(a^TCb)^T  = b^T C^T a = b^T C a$, when $C$ is symmetric.

Since $a^TCb = (a^TCb)^T$ as the result is a scalar, we obtain

$a^TCb = b^T C a$.

Title: why is it true for a symmetric matrix?


问:
Find supremum of a function
Let $x,y\in \mathbb{R}^n$. Find $$\sup_x (<x,y> - e^{<x,1>})$$ or we can rewrite this as $$\sup_x (\sum_i^nx_iy_i - e^{\sum_i^nx_i})$$ We can find the derivative, it gives $y = e^{<x,1>}$, but it looks like it can not possible to get an explicit formula if we put this into the initial one. Moreover, it looks like this function does not have the supremum at all (i.e it's $\infty$). Is this true? How to properly show it then?

P.s if we consider $<x,1> = 0$ then the sup is just $0$, but what if $<x,1> \not =$0 ?

答:
This depends on $y$.

If there is $y_i$<0, then take $x_n=(0,\cdots,0,x_{i,n},0,\cdots,0)$, where $x_{i,n}=-n$. Then
$$\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}n|y_i|-e^{-n}=\infty$$
So $\operatorname{sup}_xf(x)=\infty$.

Assume $y_i, \forall i$ positive, but there exist $y_j\neq y_k$. Assume w.l.o.g. $y_j>y_k$, take $x_n=(0,\cdots,0,x_{j,n},0,\cdots,0,x_{k,n},0,\cdots,0)$ where $x_{j,n}=n$ and $x_{k,n}=-n$. Then
$$\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}n(y_j-y_k)-1=\infty$$
So $\operatorname{sup}_xf(x)=\infty$.

At last assume all $y_i$s are the same and positive, denote $y_i=a>0$. Then the function takes the form
$$f(x)=a<x,1>-e^{<x,1>}$$
Lets look at the function $g(z)=e^z-az$, $z\in\mathbb{R}$.
$$\lim_{z\to-\infty}g(z)=\infty$$
Obviously. Also
$$\lim_{z\to\infty}\frac{az}{e^z}=0,\ \lim_{z\to\infty}\frac{z}{e^z-az}=0$$
So for all $z$ big enough we have
$$z<e^z-az$$
And this means that
$$\lim_{z\to\infty}g(z)=\infty$$
So the function $g(z)$ is coercive, and this implies it has global minimum. Taking the derivative and equating to $0$ we find $e^z-a=0$, so $z^*=\ln a$. This means that
$$a(1-\ln a)\leq e^z-az,\ \forall z\in\mathbb{R}$$
or
$$a(\ln a-1)\geq az-e^z,\ \forall z\in\mathbb{R}$$
This still holds for $<x,1>,\ x\in\mathbb{R^n}$
$$a(\ln a-1)\geq a<x,1>-e^{<x,1>}$$
So we can take $x_1=z^*,\ x_i=0,\ i>1$ and state that $\operatorname{sup}_xf(x)=a(\ln a-1)$.

Title: Find supremum of a function


问:
Why cannot $\varepsilon$ be negative?
I understand why $\varepsilon$ can not be equal to zero, but why cannot it be negative? One possible explanation might be $|f(x)-L|<\varepsilon$. But why can't we remove the modulus (even though distance cannot be negative)?

答: Going further from Peter's comment,$$ -\epsilon<f(x)-L<\epsilon$$ $\varepsilon$ must be   greater than $0$, for otherwise, $$\varepsilon<f(x)-L<-\varepsilon$$ which does not make quite sense.

Title: Why cannot $\varepsilon$ be negative?


问:
Proving $\sum_{i=0}^{\phi(n)-1} a^i\equiv 0\pmod n$, where $\gcd(a,n)=\gcd(a-1,n)=1 $, from Euler's theorem

If $\gcd(a,n)=\gcd(a-1,n)=1 $ Then:
$$\sum_{i=0}^{\phi(n)-1} a^i\equiv 0\pmod n$$

I don’t know how to prove this result, i’ve just made some stuffs :
$$\sum_{i=0}^{\phi(n)}(a^i)-a^{\phi(n)}= \sum_{i=0}^{\phi(n)-1}a^i$$
But i don’t where should i use this.

答:
Note that $$(a-1)\sum_{i=0}^{\phi(n)-1}a^i=a^{\phi(n)}-1$$
Since $\gcd(a,n)=1$ we get $a^{\phi(n)}-1\equiv 0\pmod n$. As $\gcd(a-1,n)=1$ we may cancel $a-1$ from both sides: $$(a-1)\sum_{i=0}^{\phi(n)-1}a^i\equiv 0\pmod n\implies\sum_{i=0}^{\phi(n)-1}a^i\equiv0\pmod n$$


问:
Prove that if $X$ has $n$ points then $C(X,Y)$ is homeomorphic to $Y^n$
Let $X$ be a finite set with $|X|=n$ equipped with the dicrete topology.Show that $C(X,Y)$ is homeomorphic to $Y^n$ where $Y^n=\overbrace{Y\times Y \times Y \cdots \times Y}^{n-times}$.

In the book there are a theorem that can help me

Theorem:Let Let $X, Y, Z$ be spaces, and assume all function
spaces have the compact-open topology. There exists a natural one-to-one map
$\gamma:C((X \times Y),Z) \longrightarrow C(X,C(Y \times Z))$

Futhermore:
If $Y$ is locally compact and Hausdorff $\gamma$ is bijective.

If $X$ is locally compact and Hausdorff $\gamma$ is one to one.

If $X,Y$ are both locally compact and Hausdorff $\gamma$ is homeomorphism.

The problem here is that I need have the $CO$(Compact-open) topology and I don´t have it, thus I don´t know how to apply this result to the demostration of our result.

I don´t have a idea about how to I should attack this problem.If someone may provide me a idea about how to attack the problem in this case, was useful.Thanks in advice

答:
Compact-open on $C(A,B)$ is the usual product $B^A$ (*) if $A$ is discrete, since all maps are continuous (so they are the same underlying set) and all compacts in $A$ are finite so the subbase $[K,U]=\{f\in C(X,Y)\mid f(K)\subseteq U\}$ (over all compact $K\subseteq A$ and open $U\subseteq Y$) of compact-open can be thinned down to $[\{a\},U]$ (over all $a\in A$ and open $U\subseteq Y$), hence is the same as the subbase $\bigcap_{a\in K}\pi_a^{-1}U$ of the product topology.

(*): cheating a little in the notation here.  We are taking $B^A$ in Sets, and equip it with the product topology.


答:
This is a proof by induction. The theorem states that every non-empty subset of $\mathbb{N}$ has a least element.

Let $A\subseteq \mathbb{N}$ be a set with no least element. We want to prove that $A=\varnothing $, that is $\forall n$ $\in \mathbb{N}$, $n\notin A$.
For induction on $n$ we have that:

$0\notin A$, in fact if $0 \in A$ we would have that: $0=$min$\mathbb{N}=$min$A$, but $A$ has no least element.

Suppose now that $\begin{Bmatrix}
0,1,...,n
\end{Bmatrix}\cap A=\varnothing $. What we want to prove is that: $n+1\notin A$.
If $n+1\in A$, infact, $n+1\neq$ min$A$ , since $A$ has no minimum for hypotesis. So $\exists m\in A$, $m<n+1$, which is a contradiction with the inductive hypothesis. So $n+1 \notin A$.

Hence $A=\varnothing$, as we wanted.

Title: Well-Ordering Principle "proof"

Body:
Theorem. Well-Ordering Principle.

Every non-empty subset of natural numbers has a least element.

I have seen some proofs for the theorem, but is very "complex"proof really needed here?

My attempt of proof:

Let $$ D \subset \mathbb{N}= \left\{ 1, \ 2, \ 3, \ \dots \right\} $$
be an arbitrary non-empty subset of natural numbers. Therefore it has at least one element $$  n \in D.$$

Consider the finite set $$ \left\{1, \ \dots, \ n \right\}.$$
We check which of those natural numbers are elements of D. Then we choose the smallest one of those. There we have the least element.