问:
Formula for range of a projectile landing at some distance above the ground
What will be the formula for range of the projectile if it is launched from the ground level but lands at some height about the ground?
I have searched everywhere for the answer, but everyone has discussed the case of "launching from some height" but not "landing at some height".
The picture of the scenario is attached below.

答: As usual, $$ x(t) = u\cos \theta t \\ y(t) = u\sin\theta t - \frac 12 gt^2 $$ Say the projectile lands a distance $f$ above the ground. Now, $$y (t) = f \\ \iff gt^2 -2u\sin\theta t +2f =0 \\ $$ The larger root of this quadratic is $$\frac{2u\sin\theta + \sqrt{4u^2\sin^2 \theta -8gf}}{2g}=\tau $$ Then, the range required will be the value of $x$ at this time, i.e. $$x(\tau) = \frac{u^2 \sin2\theta + u\cos\theta\sqrt{4u^2\sin^2\theta -8gf}}{2g}$$


问:
How to prove $\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) \, dx=\ln 2$
How to prove the integral
$$\begin{align}&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\=&\int_0^\infty\left (\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right)dx\\=&\ln 2\end{align}$$

I try to use integration by parts
$$\begin{align}
&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\
=&\int_0^\infty \int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\\
=&x\left.\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right|_0^\infty+\int_0^\infty \cos x \int_x^\infty\frac{\sin t}{t}dx+\int_0^\infty \sin x  \int_x^\infty\frac{\cos t}{t}dx
\end{align}$$
But the last two integrals are divergent.

答:
Well, you can show that:

$$\int_0^\infty\text{Si}\left(x\right)\text{Ci}\left(x\right)\space\text{d}x=\frac{1}{4\pi}\int_0^\infty\left(\frac{\ln\left(1+\text{s}^2\right)}{\text{s}}\right)^2\space\text{ds}=\ln\left(2\right)\tag1$$

Using evaluating integrals over the positive real axis property of the Laplace transform.

ContentLicense: CC BY-SA 4.0


问:
Using induction to verify a statement
I have to prove that this statement is true.

For $n = 1, 2, 3, ...,$ we have $ 1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6$

Basically I thought I'd use induction to prove this. Setting $n = p+1$, I got this so far:

Left hand side: $1² + 2² + 3² +...+ p² + (p+1)² = p² + (p+1)² = 2p² + 2p + 1$

Right hand side: $(p + 1)(p + 1 + 1)(2 * (p + 1) +1)/6$

This is all I've got. I've never worked with induction so I don't even know if this correct to begin. Any insight or solution to my problem is appreciated.

答:
First of all, your statement
$1^2+2^2+⋯+p^2+(p+1)^2=p^2+(p+1)^2$
is false.  You have simply erased $1^2 + 2^2 + \cdots + (p-1)^2$ from the equation, which you really can't do.

It seems like you are trying to plug in $(p+1)$ to the LHS and the RHS of the equation.  That's fine, but now what you have to do is show that if the LHS and RHS side are equal when $n = p$, they are also equal when $n = p+1$.  So, you've calculated the new LHS and RHS for $n = p+1$ -- now show they're equal.

Inceptio's answer should be a sufficient hint at this point.

ContentLicense: CC BY-SA 3.0

Title: Using induction to verify a statement


问:
Continuity of Thomae's function

Here is my textbook proof of continuity of Thomae's function. I am unable to understand further from "it suffices to show $q_n \rightarrow \infty$ as $n \rightarrow \infty$ "

As I think if it happens then $x_n \rightarrow 0$ but we assumed $a$ is irrational I can't understand further please give some hint!

答: At this point, the author is assuming that $a\not\in\Bbb Q$ and that $(x_n)_{n\in\Bbb N}$ is a sequence of rational numbers which converges to $a$; furthermore, $x_n$, written in its reduced form, is $\frac{p_n}{q_n}$. The goal is to prove that$$\lim_{n\to\infty}f(x_n)=f(a)=0.$$Since $f(x_n)=\frac1{q_n}$, this is the same thing as asserting that $\lim_{n\to\infty}q_n=\infty$. Then, the author proves that, if this does not occur, then $(x_n)_{n\in\Bbb N}$ has a subsequence $(x_{n_k})_{k\in\Bbb N}$ with a rational limit, which is impossible, since$$\lim_{k\to\infty}x_{n_k}=\lim_{n\to\infty}x_n=a\notin\Bbb Q.$$

Title: Continuity of Thomae's function


问:
Finding a circumscribed circle around a plane compact set
This question is based on R. Osserman's proof of the four-vertex-theorem (see here: https://arxiv.org/pdf/math/0609268.pdf ).

Without explaining anything, neither in the original work of Mr. Ossermann, nor in the above cited paper is there a proof for the claim that there exists a minimal circle around a compact set. Of course, assuming that it exists it's easy to see why it is unique, but the existence itself is, according to my intuition, hard.

My own tries:

Consider the compact set $E$ and the set $\mathfrak{D}$ of all closed discs, such that $\forall D \in \mathfrak{D} : D \supseteq E$.

Let $(D_i)_{i \in I}$ be a decreasing family of closed discs in $\mathfrak{D}$. It's enough to show that $D:= \cap _{i\in I} D_i \in \mathfrak{D}$, because then it's shown that each monotonically decreasing sequence of discs has its minimum, therefore with Zorn's Lemma, there would exist a global minimum.

It follows immediately that $D$ is compact and $D \supseteq E$, as $\forall i \in I : D_i \supseteq E $. To show is, that $D$ is a disc.

Assuming $D$ wouldn't be a disc, then for any disc $D$', the set $\tilde{D} := D \Delta D' $ with the symmetric difference operator Delta wouldn't be empty.

Form here I have no idea how to get the contradiction.

Note: I didn't get any hints for how to do it, I'm just reading Ossermann's paper.

If there is a simpler or better proof, I'd like to get the source, as it's not obvious to me like Ossermann states it.

答:
HINT:

Let $r = \inf$ of radiuses of discs containing $A$. Consider a sequence of discs $D(c_n, r_n)$ containing $A$, with $r_n \to r$. Now, $c_n$ is a bounded sequence (why?), so it has a convergent subsequence. We may assume $c_n$ convergent to $c$. Recall that $r_n$ convergent to $r$, and $D(c_n, r_n)$ contains $A$. Now show that $D(c, r)$ contains $A$.


答:
I think you mean $$\begin{array}{rcl}1^2+2^2+\cdots+p^2+(p+1)^2 &=& \sum_{k=1}^p k^2+(p+1)^2\\ & = & \frac{p(p+1)(2p+1)}{6}+(p+1)^2\end{array}$$

In other words by writing just $1^2+\cdots +p^2+(p+1)^2=p^2+(p+1)^2$, the first $p-1$ terms from the left-hand side are missing, so it's not a valid equality.

At any rate, tidying up the displayed equation above, you should get what you need.

Title: Using induction to verify a statement

Body:
I have to prove that this statement is true.

For $n = 1, 2, 3, ...,$ we have $ 1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6$

Basically I thought I'd use induction to prove this. Setting $n = p+1$, I got this so far:

Left hand side: $1² + 2² + 3² +...+ p² + (p+1)² = p² + (p+1)² = 2p² + 2p + 1$

Right hand side: $(p + 1)(p + 1 + 1)(2 * (p + 1) +1)/6$

This is all I've got. I've never worked with induction so I don't even know if this correct to begin. Any insight or solution to my problem is appreciated.


问:
Counterexample to the product of subsets of a Lie Group
Let $G$ be a Lie group. Given $H,K \subset G$, let $HK= \{ hk \mid h\in H, k \in K\}$

Does the following hold?

If $H$ or $K$ is open then $HK$ is open. Also If $H$ and $K$ are compact then $HK$ is compact.

I tried finding counterexamples find I cant find any, clearly if we relaxed the condition that $G$ is not a Lie group then there are lots of counterexample. I also tried to prove it but I'm getting nowhere I've tried to used the smoothness but I'm still stuck.

Any hints and help would be much appreciated!

答:
If $H,K$ are compact, the product $H\times K$ with the product topology is compact. $f:H\times K\rightarrow G$ defined by $f(h,k)=hk$ is continuous, and its image $HK$ is compact since it is the image of a compact set by a continuous map.

Suppose that $H$ and $K$ are open, $H\times K$ are open $G\times G$ since the product $G\times G\rightarrow G$ is an open map, we deduce $HK$.


问:
Does $ \sum _{n=0}^{\infty}\frac{\sin(n)\cdot(n^2+3)}{2^n} $ converge or does it diverge?
Does this sum converge or diverge?

$$ \sum_{n=0}^{\infty}\frac{\sin(n)\cdot(n^2+3)}{2^n} $$

To solve this I would use $$ \sin(z) = \sum \limits_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1)!} $$

and make it to $$\sum \limits_{n=0}^{\infty}\sin(n)\cdot\frac{(n^2+3)}{2^n} = \sum \limits_{n=0}^{\infty}(-1)^n\frac{n^{2n+1}}{(2n+1)!} \cdot \sum \limits_{n=0}^{\infty}\frac{(n^2+3)}{2^n} $$

and since $$\sum \limits_{n=0}^{\infty}\frac{(n^2+3)}{2^n} \text{ and } \sum \limits_{n=0}^{\infty}(-1)^n\frac{n^{2n+1}}{(2n+1)!} $$

converges $$ \sum \limits_{n=0}^{\infty}\frac{\sin(n)\cdot(n^2+3)}{2^n} $$
would also converge.

Is my assumption true? I'm also a bit scared to use it since I've got the sin(z) equation from a source outside the stuff that my professor gave us

答:
By the same argument, since both series$$\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt n}\quad\text{and}\quad\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt n}$$are convergent (yes, they're equal), then the series$$\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt n}\times\frac{(-1)^n}{\sqrt n}\left(=\sum_{n=1}^\infty\frac1n\right)$$converges too. But it doesn't, right?

You can prove that your series converges using the comparison test or Dirichlet's test:

the series $\sum_{n=1}^\infty\sin(n)$ has bounded partial sums;

the sequence $\left(\frac{n^2+3}{2^n}\right)_{n\in\Bbb N}$ is monotonic;

$\lim_{n\to\infty}\frac{n^2+3}{2^n}=0$.


答:
Notice that for whatever $ p\geq 2 $, $ \lim\limits_{n\to +\infty}{\left|\frac{n^{p}\sin{n}\left(n^{2}+3\right)}{2^{n}}\right|}=0 $.

Thus $ \left|\frac{\sin{n}\left(n^{2}+3\right)}{2^{n}}\right|=\underset{\overset{n\to +\infty}{}}{\scriptsize\mathcal{O}}\left(\frac{1}{n^{p}}\right) $. Since $ \sum\limits_{n\geq 1}{\frac{1}{n^{p}}} $ converges for any $ p\geq 2 $, the series $ \sum\limits_{n\geq 1}{\left|\frac{\sin{n}\left(n^{2}+3\right)}{2^{n}}\right|} $ does also converge.

Our series does, hence, converge absolutely, which means it does converge.

Title: Does $ \sum _{n=0}^{\infty}\frac{\sin(n)\cdot(n^2+3)}{2^n} $ converge or does it diverge?

Body:
Does this sum converge or diverge?

$$ \sum_{n=0}^{\infty}\frac{\sin(n)\cdot(n^2+3)}{2^n} $$

To solve this I would use $$ \sin(z) = \sum \limits_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1)!} $$

and make it to $$\sum \limits_{n=0}^{\infty}\sin(n)\cdot\frac{(n^2+3)}{2^n} = \sum \limits_{n=0}^{\infty}(-1)^n\frac{n^{2n+1}}{(2n+1)!} \cdot \sum \limits_{n=0}^{\infty}\frac{(n^2+3)}{2^n} $$

and since $$\sum \limits_{n=0}^{\infty}\frac{(n^2+3)}{2^n} \text{ and } \sum \limits_{n=0}^{\infty}(-1)^n\frac{n^{2n+1}}{(2n+1)!} $$

converges $$ \sum \limits_{n=0}^{\infty}\frac{\sin(n)\cdot(n^2+3)}{2^n} $$
would also converge.

Is my assumption true? I'm also a bit scared to use it since I've got the sin(z) equation from a source outside the stuff that my professor gave us


问:
The numbers $\binom{n}{p}\binom{m}{q}$, $\binom{l}{p}\binom{j}{q}$ and $\binom{d}{p}\binom{r}{q}$ have a common divisor.
I am asking about the validity of the following claim

The numbers $\binom{n}{p}\binom{m}{q}$, $\binom{l}{p}\binom{j}{q}$ and $\binom{d}{p}\binom{r}{q}$ have a common divisor $>1$.

where the integers are all different.

答: It is false as stated.  You could have $n=p, m=q$ and the first one is $1$ which has no divisors.  Are there some conditions on the variables?


问:
Defining a $2-$category of "partial functors" to define a colimit functor even though all colimits don't exist.
Maybe a crazy idea but hear me out:

Not every category has all colimits/limits of a certain shape so it doesn't make sense to ask for a functor $\text{colim}: C^I \rightarrow C$. But what if we define a strict $2-$category $\text{PartialCat}$, which consists of categories and "partial functors" $F : A \rightarrow B$ defined only on a subcategory of $A$. So a partial functor between $A$ and $B$ is a diagram $A \leftarrow S \rightarrow B$ where $S$ is a subcategory of $A$.

Composition of $A \leftarrow S \xrightarrow F B$ and $B \leftarrow S' \xrightarrow G C$ is defined by the diagram $A \leftarrow F^{-1}(S') \rightarrow S' \xrightarrow G C$, giving us the diagram $A \leftarrow F^{-1}(S') \rightarrow  C$ i.e a partial functor $A \rightarrow C$.

If we find a suitable notion of $2-$morphisms between partial functors we can define $\text{colim}$ as the left adjoint of the diagonal functor $C \xleftarrow{id} C \xrightarrow \Delta C^I$ in the $2-$category $\text{PartialCat}$.

$\text{colim}$ would then ideally be defined on the full subcategory of $C^I$ consisting of the diagrams $I \rightarrow C$ which have a colimit in $C$ and take those diagrams to their colimit.

This would allow us to bypass a lot of the diagram manipulation used to prove simple results such as "Left adjoint functors preserve colimits" for categories where all colimits don't exist and instead use $2-$categorical machinery. We generally prove results about colimits by using their universal property but we could bypass this and just use facts about adjunctions in $2-$categories basically.

We could also generalize this construction to define colimits of diagrams in $(\infty,1)-$categories easier since these are generally harder to understand just in terms of their universal property. Proving that $(\infty, 1)-$left adjoints preserve $(\infty, 1)-$colimits is not an easy result to prove analytically but becomes easy if we assume that there is a $(\infty,1)-$functor $\text{colim}: C^I \rightarrow C$ for quasicategories $C$ and $I$. This is just a formal consequence of the theory of adjunctions in a $2-$category. Specifically the homotopy $2-$category of quasicategories.

As for what the $2-$morphisms in $\text{PartialCat}$ should be I have no idea, maybe if $A \leftarrow S \rightarrow B$ and $A \leftarrow S' \rightarrow B$ are two partial functors called $F$ and $G$ respectively we can define $\text{Hom}(F,G)$ as the set of natural transformations between the two functors but restricted to $S \cap S' \rightarrow B$ so they have a common domain.

Explicitly: $\text{Hom}(F,G) = \text{Nat}(\left.F\right|_{S \cap S'} , \ \left.G\right|_{S \cap S'})$

I don't know how these natural transformations are supposed to compose though so it's probably wrong. Can't quite figure it out.

Thoughts about all this?

答:
I suggest to use profunctors induced by (or simply instead of) partial functors.

First of all, a profunctor $H:A^{op}\times B\to {\rm Set}$ is said to be functorial if $H(a,-):B\to {\rm Set}$ is a representable functor for each object $a\in A$. This is equivalent to saying that $H\simeq \hom_B(F-,\, -)\,=:F_*$ for a functor $F:A\to B$.

And actually the profunctor of the colimit functor can be defined for any category, even if it lacks some colimits of the given shape: namely, consider
$$H:(C^I)^{op}\times C\to {\rm Set}\quad (\underset{I\to C}D,\,a)\mapsto
\{D\to a\text{ cocones in }C\}\,,$$
where a $D\to a$ cocone can be thought of as a natural transformation $D\to a$ to the constant functor.

Note that a diagram $D\in C^I$ has a colimit, by the very definition, if and only if $H(D,-)$ is representable.

Secondly, not only partial functors but also all categorical relations (that is, subcategories of $A\times B$), moreover any span of categories $A\overset{P}\leftarrow R\overset{Q}\to B$ do determine a profunctor in a natural way, namely,
$$P^*\otimes Q_*:A\not\to R\not\to B$$
where $P^*=\hom_A(-,\,P-);\ \ Q_*=\hom_B(Q-,\,-)$ and $\otimes$ means composition (tensor product) of profunctors, which is a colimit/coend construction that yields a quotient of $\displaystyle\bigsqcup_{r\in R}P^*(a,r)\times Q_*(r,b)$.

I guess, the $2$-category you are defining would naturally embed into the bicategory of profunctors this way.


答: Both claims are true in any topological group. For the first claim, assume $H$ is open. Then if $x \in HK$, then $x = hk$ for some $h \in H$ and $k \in K$, but then $Hk$ is an open neighbourhood of $x$ that is contained in $HK$, so $HK$ contains an open neighbourhood of any of its points and hence is open. Similarly if $K$ is open. For the second claim, $HK$ is a continuous image of the compact set $H\times K$.

Title: Counterexample to the product of subsets of a Lie Group

Body:
Let $G$ be a Lie group. Given $H,K \subset G$, let $HK= \{ hk \mid h\in H, k \in K\}$

Does the following hold?

If $H$ or $K$ is open then $HK$ is open. Also If $H$ and $K$ are compact then $HK$ is compact.

I tried finding counterexamples find I cant find any, clearly if we relaxed the condition that $G$ is not a Lie group then there are lots of counterexample. I also tried to prove it but I'm getting nowhere I've tried to used the smoothness but I'm still stuck.

Any hints and help would be much appreciated!


问:
A generalization of contraction mapping principle
Let $f_1,\dots,f_n:R^d\rightarrow R^d$ be diffeomorphisms.
$\exists 0<c<1$ s.t. $\parallel Df_i(x)\parallel\leq c,\forall x\in R^d,\forall i$. Where the norm is the operation norm in $L(R^d)$.

How to proof that there exists a unique non-empty compact subset $K\subset R^d$ s.t. $K=\cup_{i=1}^nf_i(K)$

答:
Let me flesh out the suggestion made in the comments to use Hausdorff distances. Let $C(\mathbb{R}^d)$ be the set of compact subsets of $\mathbb{R}^d$. Define $F: C(\mathbb{R}^d) \to C(\mathbb{R}^d)$ by $F(x) = \bigcup_{i} f_{i}(x)$.

Let's show that $F$ is a contraction on $C(\mathbb{R}^d)$. Let $X$ and $Y$ be two compact subsets of $\mathbb{R}^d$ with $d_{H}(X,Y) = \epsilon$, and let $M \subset X \times Y$ be a matching realizing this distance, i.e. $\forall (x,y) \in M$, $\|x - y\| \leq \epsilon$. Note that the Hausdorff distance, which is usually a infimum, can be taken to be a minimum for compact sets, although this isn't strictly necessary to make the proof work, it just makes it a little cleaner. Consider now $F(X)$ and $F(Y)$. Taking $x' \in F(X)$, we know $x' = f_{i}(x)$ for some $x \in X$ and some $i$. Let $y$ be paired with $x$ in $M \subset X \times Y$. Then $y' = f_{i}(y)$ is in $F(Y)$, and $\|x' - y'\| \leq c \|x - y\|$. A symmetric argument for an arbitrary element $y' \in F(Y)$ shows how to build a matching $M' \subset F(X) \times F(Y)$ realizing a Hausdorff distance of at most $c\epsilon$.

Applying the contraction mapping theorem shows that there is a unique fixed point for $F$, which is the set $K$ in question.


问:
What is $-i$ exactly?
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?

答:
You need to impose an orientation on the plane. Since we mathematicans like counterclockwise to be the positive direction, we think of $i$ as being "one-up" from zero.  You could equally well work with the opposite.  It's a "tom(ay)to vs. tom(ah)to" situation.  I refer all readers to this (humorously)....

http://www.youtube.com/watch?v=zZ3fjQa5Hls

Title: What is $-i$ exactly?


问:
Continuity and Intermediate Value Theorem.
$$Duplicate$$
Please disregard this post.

答:
If $ f(1) \neq 0$ and $f$ is supposed to be continuous at $1$ then

$$\lim\limits_{x \to 1} f(x) f\left(\frac{1}{x}\right) = \left(\lim\limits_{x \to 1} f(x)\right)\left(\lim\limits_{x \to 1} f\left(\frac{1}{x}\right)\right) =f^2(1) \gt 0.$$

In contradiction with the given hypothesis $f(x) f\left(\frac{1}{x}\right) \lt 0$ which implies $ \lim\limits_{x \to 1}f(x) f\left(\frac{1}{x}\right) \le 0$. Therefore $f(1) =0$.


问:
how to prove $a+b< (a^{\alpha}+b^{\alpha})^{\frac{1}{\alpha}}$
how to prove $a+b< (a^{\alpha}+b^{\alpha})^{\frac{1}{\alpha}}$ when $0<\alpha<1$.(Im not suure if is for every a,b) I am really dont try anything, but i think that maybe is using integration in some region, i will appreciate if you give me one hint, thank you so much.

答:
You can prove it directly just using strict monotonicity of an appropriate function.

First note that you are dealing with an inequality among real numbers and there  expressions like $a^{\alpha}, b^{\alpha}$ with real $0<\alpha<1$ are defined for $a,b\geq 0$.

Your inequality is wrong if $a=0$ or $b=0$. So, you may want to prove it for $a,b>0$.

Without loss of generality we can assume that $0<b\leq a$ and set $x=\frac ba$:
$$a+b < \left(a^{\alpha}+ b^{\alpha}\right)^{\frac 1{\alpha}}\Leftrightarrow (a+b)^{\alpha} < a^{\alpha}+ b^{\alpha}$$
$$\stackrel{x=\frac ba}{\Leftrightarrow}(1+x)^{\alpha} < 1+x^{\alpha} \text{ for } 0<x\leq 1$$

The function $f(x) = 1+x^{\alpha} - (1+x)^{\alpha}$ is continuous on $[0,1]$ and differentiable on $(0,1)$. Furthermore you have $f(0) = 0$. So, it is enough to show that $f'(x) >0$ on $(0,1)$:

$$f'(x) = \alpha \left(\underbrace{\frac 1{x^{1-\alpha}}}_{>1 \text{ on } (0,1)}  - \underbrace{\frac 1{(1+x)^{1-\alpha}}}_{<1 \text{ on } (0,1)}\right) > 0$$


问:
Dimension of normalizer of closed connected subgroup

Let $G$ be unipotent and let $H$ be proper closed, connected subgroup. Show that $\operatorname{dim}(N_G(H)) > \operatorname{dim} H$.

We know that $H \triangleleft Z_G(H) \triangleleft N_G(H)$, therefore I want to prove that one of the inclusion is proper.

I think the latter inclusion is proper, but I have no idea how to prove.

Any help will be appreciate..

答:
We should assume that $G$ is connected, otherwise we can let $H = G^{\circ} \Rightarrow \operatorname{dim}G \geq \operatorname{dim}N_GH > \operatorname{dim}H = \operatorname{dim}G$, contradiction.

Following is my proof, may have some mistakes.

Let $Z = C(G)^{\circ}$, i.e., the identity component of $C(G)$, then $Z$ is closed in $C(G)$, so $Z$ is closed in $G$ since $C(G)$ is closed in $G$.

Since $G$ is unipotent algebraic group, i.e. $G = G_u$, $G$ is solvable group. And by [Lemma 6.3.4,
, T. A. Springer, Linear algebraic groups.], $\operatorname{dim}C(G)\geq1$.

If $Z \nsubseteq H$, then $\operatorname{dim}Z \geq 1$ and $|ZH/H| \geq 2$. But $ZH/H$ is connected since $ZH$ is connected, so $\operatorname{dim}(ZH) - \operatorname{dim}(H) = \operatorname{dim}(ZH/H) > 0$. Note that $ZH \subset N_G(H)$, so in this case we have $\operatorname{dim}N_G(H) \geq \operatorname{dim}ZH > \operatorname{dim}H \Rightarrow \operatorname{dim}N_G(H) > \operatorname{dim}H$.

Now we prove this by induction on $\operatorname{dim}G$. If $\operatorname{dim}G = 1$, then $G$ is commutative by [Prop 3.1.3, , T. A. Springer, Linear algebraic groups.], so $Z = G^{\circ} = G$, thus $Z \nsubseteq H$. By the above proof, we konw $\operatorname{dim}(N_G(H)) > \operatorname{dim}H$.

Suppose this is right for all $G$ satisfies $\operatorname{dim}G \leq n-1$. Consider this case $\operatorname{dim}G = n$. By the above proof, we may assume $Z \subseteq H$, then $G/Z$ is a unipotent connected group ($G$ is unipotent) and
$$
     \operatorname{dim} (G/Z) = \operatorname{dim}G - \operatorname{dim}Z = \operatorname{dim}G - \operatorname{dim}C(G) < \operatorname{dim}G.
  $$
Also, $H/Z$ is a proper connected closed group of $G/Z$. By induction, we have
$$
    \operatorname{dim}N_{G/Z}(H/Z) > \operatorname{dim}H/Z,
  $$
and
$$
    \operatorname{dim}N_GH - \operatorname{dim}Z = \operatorname{dim}N_GH/Z = \operatorname{dim}N_{G/Z}(H/Z) > \operatorname{dim}H/Z = \operatorname{dim} H- \operatorname{dim}Z.
  $$
Therefore, $\operatorname{dim}N_GH> \operatorname{dim} H$.


问:
How to derive rotation matrix for quaternion
I've been following the Wikipedia article on quaternions and spatial rotations and I've come across something I don't understand:

.

Everything up to this point is clear yet I can't see how the first equation leads to the second. If someone could give a more clear explanation that would be greatly appreciated.

Thanks for any help

答:
It comes down to straightforward manipulations, but a lot of them. First, simplify the scalar term and expand the cross product:
\begin{align}
s(-\mathbf v \cdot \mathbf p q_r + q_r \mathbf v\cdot \mathbf p&,
\mathbf v(\mathbf v \cdot \mathbf p) + q_r^2 \mathbf p + q_r \mathbf v \times \mathbf p + \mathbf v \times (q_r \mathbf p + \mathbf v \times \mathbf p))
\\ & =
s(0,
\mathbf v(\mathbf v \cdot \mathbf p) + q_r^2 \mathbf p + q_r \mathbf v \times \mathbf p + \mathbf v \times (q_r \mathbf p + \mathbf v \times \mathbf p))
\\ & =
s(0,
\mathbf v(\mathbf v \cdot \mathbf p) + q_r^2 \mathbf p + q_r \mathbf v \times \mathbf p + q_r\mathbf v \times \mathbf p +
\mathbf v \times (\mathbf v \times \mathbf p))
\\ & =
s(0,
\mathbf v(\mathbf v \cdot \mathbf p) + q_r^2 \mathbf p + 2q_r \mathbf v \times \mathbf p+
\mathbf v \times (\mathbf v \times \mathbf p)).
\end{align}
Now, rewrite each term as a matrix multiplied by $\mathbf p$:

$\mathbf v(\mathbf v \cdot \mathbf p) = (\mathbf{v \otimes v})\mathbf p$

$q_r^2\mathbf p = q_r^2 \mathbf I \mathbf p$

$2q_r \mathbf v \times \mathbf p = 2q_r[\mathbf v]_\times \mathbf p$

$\mathbf v \times (\mathbf v \times \mathbf p) = [\mathbf v]_\times^2 \mathbf p$.

From there, we simply factor the product.

$$
(\mathbf{v \otimes v})\mathbf p + q_r^2 \mathbf I \mathbf p + 2q_r[\mathbf v]_\times \mathbf p + [\mathbf v]_\times^2 \mathbf p =
(\mathbf{v \otimes v} + q_r^2 \mathbf I + 2q_r[\mathbf v]_\times + [\mathbf v]_\times^2)\mathbf p.
$$


问:
Nontrivial subspace of a vector space has a unique complement?

Can it happen that a non-trivial subspace of a vector space $V$ (i.e.,
a subspace different from both $0$ and $V$) has a unique complement?

I have trouble understanding why this question is worth arguing, and feel like I am missing something. If I have a complement $U$ of $V$ in $X$, then by definition $U\cap V$ is trivial and $U\cup V$ is $X$, so wouldn't $U$ be $(X-V)\cup\{0\}$?

答:
The condition of complementarity is not that $U \cup V = X$, but that $U \oplus V = X$. In other words, the vectors in $X \setminus U$ don't need to be in $V$, but they need to be written as sums $u + v$ for $u \in U$ and $v \in V$. This gives some freedom in choosing $V$. For example, if $X = \mathbb{R}^2$, and $U$ is the $x$-axis, then $V$ can be any other line passing through the origin.

Put another way, the bases for $U$ and $V$ need to be linearly independent, but together span all of $X$.


答:
Since you don't know how the $sin(x)$ as $x\to \infty$ behaves, you can use the fact that, if a series converges absolutely, then the original series converges. So you can see that

$\sum^{\infty}_{n=0} \frac{sin(n)(n^2+3)}{2^n} \leqslant \sum^{\infty}_{n=0} \frac{n^2+3}{2^n}$

and I'll leave the rest to you... I just wanted to give some mental momentum.


问:
Estimating the sum $\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}$
By integral test, it is easy to see that
$$\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}$$
converges. [Here $\ln(x)$ denotes the natural logarithm, and $\ln^2(x)$ stands for $(\ln(x))^2$]

I am interested in proving the following inequality (preferrably using
  integral calculus) $$\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}>2$$

By wolfram alpha, the actual value of the sum is about 2.10974. Since $\frac{1}{k \ln^2(k)}$ is decreasing, we have

$$
\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}\ge \int_2^{\infty} \frac{1}{x \ln^2(x)}
dx=\frac{1}{\ln(2)}\approx 1.4427$$
So this lower bounded is weaker than desired.

My motivation for asking this question is that by being able to estimate this particular sum will hopefully teach me a general technique which I may try applying to sums of the form $\sum_{k=1}^{\infty} f(k)$.

Thanks!

答:
\begin{align*}
\sum_{k=2}^{\infty} \frac1{k\ln^2k} &= \frac1{2\ln^22} + \frac1{3\ln^23} + \sum_{k=4}^{\infty} \frac1{k\ln^2k} \\
&\ge \frac1{2\ln^22} + \frac1{3\ln^23} + \int_4^{\infty} \frac{dx}{x\ln^2x} \\
&= \frac1{2\ln^22} + \frac1{3\ln^23} + \frac{1}{\ln4} > 2.038.
\end{align*}
(I see Andrew just wrote this in a comment.)</p>


答:
Clearly,

$$\left|\sum_{n=0}^\infty\frac{\sin(n)(n^2+3)}{2^n}\right|<\sum_{n=0}^\infty\frac{(n^2+3)}{2^n}.$$

Then by the ratio test, for $n\ge2$,

$$\frac12\frac{(n+1)^2+3}{n^2+3}\le\frac67$$ and the series converges.


问:
Ramanujan Tau function and Modular discriminant
The following is a question from my assignment in number theory:

$$ \sum_{n=1}^\infty \tau(n)e^{2\pi in\tau} =
(2\pi)^{-12}\Delta(\tau) = e^{2\pi i\tau}
\prod_{m=1}^\infty (1-e^{2\pi im\tau})^{24}.$$

3. Let $p$ be a prime and let $k$ be an
integer, $1\le k\le p-1.$ Show that there exists
integer $h$ such that
$$ \tau^{12}\Delta\Big(\frac{\tau+h}p\Big) =
\Delta\Big(\frac{k\tau-1}{p\tau}\Big) $$
and that $h$ runs through a reduced residue
system mod $p$ with $k$.

I am unable to get an idea which result should I use for proving the existence of h. I used 1 st formula in image but  unable to prove it.

So, please guide.

Original image: https://i.stack.imgur.com/4nxwY.jpg

答:
There is some $\gamma\in SL_2(Z)$ such that $\gamma(k/p)= i\infty$

$$\gamma(\tau) = \frac{a\tau+b}{p\tau-k},\qquad -ak-bp=1$$

Then $$\Delta(\frac{k\tau-1}{p\tau})=(p\frac{k\tau-1}{p\tau}-k)^{-12}\Delta(\gamma(\frac{k\tau-1}{p\tau}))=\tau^{-12} \Delta(\frac{\tau+r}{p})=\tau^{-12} \Delta(\frac{\tau+r}{p}-\lfloor r/p\rfloor)$$


问:
Why is it important to have a discrepancy between image and codomain?
A function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $f(x)=x^2$ has $\mathbb{R}_{\geq0}$ as its image and $\mathbb{R}$ as its codomain.

What's the need for this discrepancy? Why don't we just write $f:\mathbb{R}\rightarrow \mathbb{R}_{\geq0}$?

I believe the reason for this is that there may be some functions in which stating the image this way may be impractical - if this hypothesis is true, what is this class of functions? If yes or no, is there any other reason for this?

答:
What is a function? Informally, it is a process, or an assignment, from an input set to an output set. It is not just the process or assignment that forms a function, but specifying the input and output is part of what it is. Formally, a function is a triple $(A,B,f)$ where $A,B$ are sets and $f\subseteq A\times B$ is a subset of the cartesian product, i.e., a relation from $A$ $to$ $B$. If instead one wanted to view the function $(A,B,f)$ as equivalent to $(A,Im(f),f)$ then one needs to add a condition to an otherwise very clean definition. So, to force the codomain to be the range actually burdens the definition rather than simplify things. That is one reason not to force such things.

Another reason is so that one has no problems talking about composition when it's obvious one should be able to talk about composition (so actually this is a category theoretic reasoning). Suppose that $f:A\to B$ and $g:B\to C$. If I want to know if the composition $g\circ f$ exists I don't care what the actual range of the functions is. Just a glance as the domain of $g$ and the (ta ta ta taaaaaa) codomain of $f$. In other words, if the input type of $g$ matches the output type of $f$, then the composition is defined.

If we insisted that codomain=range, then the condition above will have to be replaced by "the range of $f$ is contained in the domain of $g$". And now I'll get into a bit of technical lingo from category theory. The resulting category from forcing codomain=range will be the category $Set_{Surj}$ of sets and surjections. It's a perfectly legit category, but it has far fewer nice properties when compared to $Set$, the category of sets and all functions. For instance, the empty set is characterized in $Set$ as an initial object (i.e., there is precisely one function from it to any other given set) and is dual to singletons, which are terminal. This is no longer the case in $Set_{Surj}$. The disjoint union of two (or more) sets is an example of a categorical product in $Set$, and is dual to the cartesian product. Disjoint unions are no longer coproducts in $Set_{Surj}$. Many other of the nice properties of $Set$ are lost if passing to $Set_{Surj}$.

In particular, many very convenient injections will no longer be allowed if we require all functions to be surjective. For instance, it is very convenient to be able to speak of the functions $f_y:\mathbb R \to \mathbb R^2$, given by $f(x)=(x,y)$, or various curves in the plane being parametrized by some function $\gamma:[a,b]\to \mathbb R^2$.


Title: The numbers $\binom{n}{p}\binom{m}{q}$, $\binom{l}{p}\binom{j}{q}$ and $\binom{d}{p}\binom{r}{q}$ have a common divisor.

Body:
I am asking about the validity of the following claim

The numbers $\binom{n}{p}\binom{m}{q}$, $\binom{l}{p}\binom{j}{q}$ and $\binom{d}{p}\binom{r}{q}$ have a common divisor $>1$.

where the integers are all different.

Body:
Its false, take
$$
\binom{3}{2}\binom{9}{1}, \binom{4}{1}\binom{5}{2}, \binom{6}{1}\binom{7}{2}
$$


问:
$cos(ax)=a\cdot cos(x)$ Which are all the values of $a$?
How do I know for which values of $a$ is true the following?$\space$

$cos(ax)=a\cdot cos(x)$ for all $x\in \mathbb{R}$

It is trivial that it is true for $a=1$. But are the more values of $a$ for which it is true?

答:
If $|a|>1$, then $|a\cos(x)|=|a|>1$ for some values of $x$, but you never have $|\cos(ax)|>1$. So, it's not true then that$$(\forall x\in\Bbb R):\cos(ax)=a\cos(x).\tag1$$And, if $|a|<1$, you never have $a\cos(x)=1$, but $\cos(ax)=1$ for some value of $x$. So, gain, it's not true that you have $(1)$.

On the other hand, if $a=-1$, it's clear that you also don't have $(1)$. So, the only solution is $a=1$.


答: Replacing with $x=0$ gives $1 = a$, so $a=1$ is the only value.

Title: $cos(ax)=a\cdot cos(x)$ Which are all the values of $a$?

Body:
How do I know for which values of $a$ is true the following?$\space$

$cos(ax)=a\cdot cos(x)$ for all $x\in \mathbb{R}$

It is trivial that it is true for $a=1$. But are the more values of $a$ for which it is true?


问:
Given inner parabola curve $y=0.1x^2$ and thickness $d$, find outer curve
As illustrated in the picture, the thickness $d$ is known and the parabolic formula of the inner (red) curve is given, say $y = 0.1x^2$. Then, how to obtain the formula for the outer (green) curve?

I think the formula should be in this form: $y_2 = kx^2 - d$ but I have no idea how to derive the value of $k$.

Can anyone give me some hint?

Thank you!

答:
For any point it $(t,at^2)$ on the parabola $y=ax^2$, the corresponding point on the outer curve is

$$x=t+d\sin \theta,\>\>\>\>\> y = at^2 -  d\cos \theta$$
where $\theta $ is the tangent angle at $(t,at^2)$, given by
$\tan \theta = y’ =2ax$. Thus, the outer curve is

$$x=t+\frac {2adt}{\sqrt{1+4a^2t^2}}, \>\>\>\>\>
 y= at^2- \frac d{\sqrt{1+4a^2t^2}}
$$

which is in a parametrized form with $t$ as the parameter. Note that it is no longer a parabola.


问:
Locating proof of if $L/K$ is abelian,then $N_{L/K}C_L$ is an open subgroup of $C_K$, of finite index.
In Cassels and Frohlich, the chapter Global Class Field Theory by Tate, section 12 Proof of the Existence Theorem, on page number 201, it says

We have already shown that if $L/K$ is abelian,then $N_{L/K}C_L$ is an open subgroup of $C_K$, of finite index.

Although I could not find this proof in the book. Can anyone tell me where the proof of the above is located in the book (or perhaps somewhere else)?

Finite index follows as part of second inequality. But I need proof of openness.

答:
Here's a simple enough argument.

It's enough to show that it contains an open neighborhood of the unit element. So, if we consider the "full" idele group $I_K$ (without quotienting by the field elements -- sorry, I didn't check the most standard notations), we want to show that $\nu=N_{L/K}(I_L)$ contains a neighborhood of $1 \in I_K$.

We'll show that $\nu$ contains a product $\prod_v{T_v}$ where $T_v=\mathbb{R}^{+*}$ for complex or real places, $T_v=\mathcal{O}_v^{\times}$ for all but finitely many $v$, and for all finite $v$, $T_v$ is an open subgroup of $\mathcal{O}^{\times}_v$.

We consider instead the subgroup $N_{L/K}(I'_K) \subset \nu$ (where $I'_K \subset I_L$ is the subgroup of elements where all the coordinates are in the appropriate completion of $K$ inside the given completion of $L$) : it contains indeed the product of the $\mathbb{R}^{+*}$ (at the real/complex places), and the $\mathcal{O}_v^{\times,[L:K]}$ at the finite places.

Let $v$ be a finite place of $K$ for which every prime dividing $[L:K]$ is invertible. Then it follows from Hensel's lemma that $\mathcal{O}_v^{\times,[L:K]}=\mathcal{O}_v^{\times}$.

For the other finite places (there are finitely many of them), we show that $\mathcal{O}_v^{\times,[L:K]}$ contains a $1+\mathfrak{m}_v^k\mathcal{O}_v$ for some $k \geq 1$ and is thus an open subgroup of $\mathcal{O}_v^{\times}$. Let $p$ be the prime above which $v$ is, with $p|[L:K]$ with valuation $\alpha$ and $e$ be the ramification index of $v$ above $\mathbb{Q}$. Let $x=1+y$ with $v(y) \geq e+1+e\alpha$.

Let $z=\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}y^k}{k}}$, then $v(z) \geq e+1+e\alpha$ by ultrametric inequality, and we can define $z'=\frac{z}{[L:K]} \in \mathfrak{m}_v^{e+1}\mathcal{O}_v$ and $x'=\sum_{k=0}^{\infty}{\frac{z'^k}{k!}}$, and it's easy to check that $x'^{[L:K]}=x$, and this concludes.


问:
If we know the eigenvalues of a matrix $A$, and the minimal polynom $m_t(a)$, how do we find the Jordan form of $A$?
We have just learned the Jordan Form of a matrix, and I have to admit that I did not understand the algorithm.

Given $A = \begin{pmatrix} 1 & 1 & 1 & -1 \\ 0 & 2 & 1 & -1 \\ 1 & -1 & 2 & -1 \\ 1 & -1 & 0 & 1 \end{pmatrix} $, find the Jordan Form $J(A)$ of the matrix.

So what I did so far:

(I) Calculate the polynomial: $P_A(\lambda) = (\lambda - 1)^2(\lambda -2)^2$.

(II) Calculate the minimum polynomial: $m_A(\lambda) = P_A(\lambda) =(\lambda - 1)^2(\lambda -2)^2 $

But I am stuck now, how do we exactly calculate the Jordan Form of $A$?

And an extra question that has been confusing me. In this case, does $A$ have $4$ eigenvalues or $2$ eigenvalues?

答:
Since the minimal polynomial has degree $4$ which is the same order of the matrix, you know that $A$'s smith normal form is
$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & m_A(\lambda )\end{pmatrix}$.

Therefore the elementary divisors (I'm not sure this is the correct term in english) are $(\lambda -1)^2$ and $(\lambda -2)^2$.

Theory tells you that one jordan block is $\color{grey}{(\lambda -1)^2\to} \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}$ and the other is $\color{grey}{(\lambda -2)^2\to} \begin{pmatrix} 2 &1\\ 0 & 2\end{pmatrix}$.

Therefore a possible JNF for $A$ is $\begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 2 &1\\ 0 & 0 & 0 & 2\end{pmatrix}$.

Regarding the extra question, it seems to be asking about the geometric multiplicity. In this case it has two eingenvalues and not four.


问:
Find the first four non-zero terms of the Maclaurin series of $f(z)=e^\frac{1}{1-z}$
Consider the following complex function:

$$f(z)=e^\frac{1}{1-z}$$

Find the first four non-zero terms of the Maclaurin series for $f(z)$, so around $z_0=0$.

I've asked my lecturer what's the best way to solve this and she simply said "just differentiate the function 4 times and find its value at $z=0$". I'm refusing to accept this and I'm sure there's a smarter way to approach this.

Unfortunately I'm clueless about this one and have no idea how to even start. I tried thinking of a function $g(z)$, which I know its Maclaurin series already, such that $g(z)f(z)=1$ and then comparing coefficients but no luck of finding one.

Any help would be grealy appreciated!

答:
From the Maclaurin series of $$\dfrac{1}{1-z}=1+z+z^{2}+z^{3}+O\left( z^{4}\right) $$ and $$\exp(z)=  1+z+\dfrac{1}{2!}z^{2}+\dfrac{1}{3!}z^{3}+O(
z^{4}) ,$$  we can obtain the expansion of $\exp \left( \frac{1}{1-z}\right)$ as explained in the following steps:

\begin{eqnarray*}
\exp \left( \frac{1}{1-z}\right)  &=&\exp \left( 1+z+z^{2}+z^{3}+O\left(
z^{4}\right) \right)\\&=& \exp \left( 1 \right)\exp \left( z \right)\exp \left( z^2 \right)\exp \left( z \right)\exp \left( z^3 \right) \exp \left( O(z^4) \right)  \\
&=&\underset{\exp \left( 1\right) }{\underbrace{e}}\,\underset{\exp \left(
z\right) }{\underbrace{\left( 1+z+\frac{1}{2}z^{2}+\frac{1}{6}z^{3}+O(
z^{4})\right)  }}\times  \\
&&\qquad\underset{\exp \left( z^{2}\right) }{\times \underbrace{\left(
1+z^{2}+O( z^{4}) \right) }}\, \underset{\exp \left( z^{3}\right) }{
\underbrace{\left( 1+z^{3}+O( z^{4}) \right ) }} \\
&=&e\left( 1+z+\frac{1}{2}z^{2}+\frac{1}{6}z^{3}+O( z^{4})
\right) \underset{\exp \left( z^{2}\right) \exp \left( z^{3}\right) }{
\underbrace{\left( 1+z^{2}+z^{3}+O( z^{4}) \right ) }} \\
&=&e\left( 1+z+( 1+\frac{1}{2}) z^{2}+( 1+1+\frac{1}{6})
z^{3}+O( z^{4}) \right) \\
&=&e+ez+\frac{3}{2}ez^{2}+\frac{13}{6}ez^{3}+O( z^{4}) .
\end{eqnarray*}


问:
Dimension of the space of symmetric traceless matrices
A set of symmetric $n \times n$ matrices have $ \frac{1}{2}n^{2} - \frac{1}{2}n$ independent elements.  But how do you get to this result?  I understand that a general $n \times n$ matrix would have $n^{2}$  independent elements.  How exactly does the constraint of symmetry lead to the piece above?  Furthermore, how would I determine the number of elements for a symmetric traceless matrix?</p>

答:
An $n\times n$ matrix, $\mathbf T$, has, as you said, $n^2$ independent components, namely $[\mathbf{T}]_{ij}$, where each index can run from $1$ to $n$.
Symmetry, in this context, means that the matrix is equal to its transpose, $\mathbf{T}=\mathbf{T}^T$. What does this mean about the components? Let's see...

$[\mathbf{T}]_{ij}=[\mathbf{T}^T]_{ij}\\
   \hspace{0.85cm}=[\mathbf{T}]_{ji}$

That was simple enough! Symmetry implies that the $(i,j)$ component of the matrix equals the $(j,i)$ component of the matrix. What does this look like visually?

Neat - when you write out a symmetric matrix you'll notice that almost every component has a twin on the other side of the matrix. A helpful way to visualize a transpose is to flip the matrix across its diagonal and realize that symmetry means doing so does not change the matrix whatsoever. Long story short, when you get to counting independent components, be careful to not count both twins as if they were independent - only count pairs of twins.

Now, which elements DON'T have a twin? If you stare at the matrix, you'll soon figure out that it's the elements on the diagonal. That's 'cause $[\mathbf{T}]_{ii}=[\mathbf{T}]_{ii}$ trivially - symmetry did not impose any additional pairing up for these elements. Going back to our visual representation, if you flip a matrix across its diagonal, the diagonal elements stay put.

Phew, that was a lot, but now we're ready to count the number of independent components of a symmetric matrix. First the diagonal - there are $n$ independent components there. Now the pairs of twins. Removing the diagonal elements, we now have $n²-n$ components. If we twin them up, then we'll have $\frac{n²-n}{2}$ independent pairs. Then we just have to add the independent diagonal components back to get the total number of independent components:

$\frac{n²-n}{2}+n=\frac{n²-n+2n}{2}=\frac{n(n+1)}{2}$

Voila!

If you didn't like how we counted, then we can totally count them another way. Let's look at each row of the matrix and count how many independent components there are. Remember we're only counting the diagonal components and those on one side of the diagonal. The first row has $n$ independent components. The second row has $n-1$ independent components. We continue in this fashion till we get to the $n^{th}$ row which has but one independent component. Now we just have to add all these up:

$1+2+3+\cdots+(n-1)+n = \frac{n(n+1)}{2}$

If you're interested in how this sum was done, check out the really cool triangle numbers.

Now, about those traceless matrices. The trace is equal to the sum of all the diagonal components of the matrix. For a matrix to be traceless means that its trace is equal to zero. But that means that you can rewrite one of the diagonal components in terms of the others, so we have lost an independent component! Hence the number of independent components of a traceless symmetric matrix will be one less than the number of independent components of a symmetric matrix, namely

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

Hope that helps!


问:
Prove that $\sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}=2^{2n}$

Prove that $$\sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}=2^{2n}$$

I was having trouble with this one because all four of the indexes are varying here. So I tried to think of a combinatiorial argument, but wasn't quite able to tie everything together. I was thinking along the lines of imagining $2n$ elements which were separated into $n$ pairs, and then partitioned into two subsets (total $2^n$ subset combinations possible). How should I proceed? (PS no proofs using induction please)

答:
Let $ x\in\mathbb{R} $, such that $ \left|x\right|<\frac{1}{4} $. Using cauchy product theorem, we have :

\begin{aligned}\sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}\right)x^{n}}&=\left(\sum_{n=0}^{+\infty}{\binom{2n}{n}x^{n}}\right)^{2}\\ &=\left(\sum_{n=0}^{+\infty}{\binom{-\frac{1}{2}}{n}\left(-4x\right)^{n}}\right)^{2}\\ &=\left(\frac{1}{\sqrt{1-4x}}\right)^{2}\\ &=\frac{1}{1-4x}\\ \sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}\right)x^{n}}&=\sum_{n=0}^{+\infty}{4^{n}x^{n}}\end{aligned}

Thus for any $ n\in\mathbb{N} $, we have : $$ \sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}=4^{n} $$


答:
By no means a real proof, but if you series-expand teach trig term and re-combine powers in $x$, it's pretty clear:

\begin{align*}\cos\left(ax\right) &= a \cos x \\ 1 - \frac{a^2x^2}{2!} + \frac{a^4x^4}{4!} - \cdots &= a - a\frac{x^2}{2!} + a\frac{x^4}{4!} - \cdots \\ \left(1-a\right) - \frac{x^2}{2!}\left(a^2-a\right) + \frac{x^4}{4!}\left(a^4 - a\right) - \cdots&= 0\end{align*}

For the above to hold for any $x$, the terms in parentheses must each be zero. This only works with $a=1$.


问:
$\gcd(a+b, p^4)$ if $\gcd(a,p^2) = p$ and $\gcd(b,p^3)=p^2$
I'm having some trouble solving the following exercice:

Let $a,b \in \mathbb Z$ and $p$ be a prime number. If $\gcd(a,p^2) = p$ and $\gcd(b,p^3)=p^2$ determine:

$\gcd(ab,p^4)$

$\gcd(a+b, p^4)$

I was able to determine the first one and got $\gcd(ab,p^4) = p^3$ (The part I'm having trouble with is the second one, but I would appreciate If someone could also confirm this first answer).

I'm not being able to determine the value of $\gcd(a+b, p^4)$. How can I do this?

答:
Hint: From the given information, we can be sure that there are integers $x$ and $y$ for which $a = px$, $b = p^2 y$, and $\gcd(x,p) = \gcd(y,p) = 1$ (i.e. $x,y$ are not divisible by $p$). In terms of $x$ and $y$, we have
$$
ab = p^3 xy, \quad a + b = px + p^2y = p(x + py).
$$
How do we know that $x + py$ is not divisible by $p$?


问:
Prove that $\triangle AMC$ is an isosceles right triangle

In $\triangle ABC$ construct squares $ABDE$ and $BCFG$. Let $M$ be the midpoint of $EF$ then, prove that $\triangle AMC$ is an isosceles right triangle.

I was able to prove that $\triangle AMC$ is an isosceles only and here is my solution:

The area of $\triangle BAF$ is equal to the area of $\triangle EBC$ so we get
\begin{equation}
AE×AF\sin \angle BAF=CF×CE\sin \angle ECB
\end{equation}
And using law of cosines on $EF$ we get
\begin{equation}
AE^2+AF^2=CF^2+CE^2
\end{equation}
By the length formula of median $AM$ and $CM$ we get that $AM=CM$. I know this solution is kinda rush and not give much detail but I'm not really good at Latex, please don't mind me. Please help prove that it is the right triangle. Thanks!

BTW, $N$ in the photo is the point where $AE$ meet $CF$

答:
This is a stronger statement of Bottema's Theorem, which merely indicates that (in the notation here) $M$ is independent of $B$.

The recent Math.SE questions "A generalization of Bottema's theorem" and "A new generalization of Bottema's theorem" abandon the squares, introducing an arbitrary scale factor (here, $1$) and rotation angle (here, $90^\circ$) for determining the positions of $E$ and $F$ relative to points $A$ and $C$. Solutions given to the first generalization (using vectors, complex numbers, or in my case, a diagram) can be adapted for the second, and also specialized for the current question.

My diagrammatic proof of the generalization is a bit of overkill for this case, so here's a simpler alternative:


问:
computing the speed of decreasing of the temperature of a function that depends on the position on the space
Let's suppose that the temperature of a particle at a given position $(x,y,z)$ is given by $ T(x,y,z)=e^{-(x^2+2y^2+3z^2)}$ where $x,y,z$ let's consider that $x,y,z$ are given in meters=m$

If the particle moves at a speed of $ 8 m$ per second, compute the velocity of decreasing of the temperature.

I don't understand, since the problem talk about time, then the position is a function of the form $ r(t)= (x(t),y(t),z(t)) $ and the temperature also depends on time. I thought that I would need to use the gradient, but if I derivate with respect to $t$ I can not find the derivative explicitly.

答:
The chain rule says that
$$\frac d{dt} T(r(t)) = \frac d{dt}T(x(t),y(t),z(t)) = \nabla T(r(t))\cdot \frac{dr}{dt} =
\frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt} + \frac{\partial T}{\partial z}\frac{dz}{dt}\,.$$
To complete the problem, we need to know more than the speed (which is the magnitude of the velocity vector). We need to know both the position and the velocity vector at the desired instant $t_0$. Then we can evaluate the gradient vector at the desired point ($r(t_0)$) and dot it with the velocity vector.


问:
Non-existence of this stochastic process?
I am intrigued by the following statement in Oksendal , Stochastic differential equations, Chapter 3:

(A)

There does not exist a 'reasonable' stochastic process satisfying the following conditions:

(i) $t_1 \ne t_2 \implies$ $W_{t_1}$ and $W_{t_2}$ are independent

(ii) $W_t$ is stationary ie the joint distribution of $\{W_{{t_1}+t}...W_{{t_k}+t}\}$ does not depend on t

(B)

Moreover if we require $E[W_t^2]=1$ then the function $(t,\omega)\rightarrow W_t(\omega) $ cannot be measurable with respect to the $\sigma$-algebra $B \text{ x } F$ where $B$ is the Borel $\sigma$-algebra on $[0,\infty[ $ (See Kallianpur 1980 p10)

I do not have access to that reference. Is there another one that someone could suggest to explain these statements?

答:
The variance computation shows that $\int_I X_t\,dt=0$ a.s.; the null set $N_I$ just acknowledges the fact that this "almost surely" depends on the interval $I$. The grand null set $N$ then has the property that
$$
\omega\notin N \Longrightarrow \int_IX_t(\omega)\,dt=0,\quad\hbox{for all } I\hbox{ with rational endpoints}.
$$
To be extra careful, notice that the $E\left[\int_0^1 |X_t|\,dt\right]<\infty$, so there is one more null set $N'$ such that
$$
\omega\notin N'\Longrightarrow t\mapsto X_t(\omega)\in L^1([0,1]).
$$
It $\omega\notin N'$ then $\int_a^b X_t(\omega)\,dt$ is a continuous function of $a$ and $b$; it now follows that in $\omega\notin N\cup N'$ then $\int_a^b X_t(\omega)\,dt = 0$ for all real $a,b$ with $0\le a\le b\le 1$. You can now apply the Lebesgue differentiation theorm to conclude that if $\omega\notin N\cup N'$ then $X_t(\omega)= 0$ for (Lebesgue) a.e. $t\in[0,1]$. This leads, via Fubini, to a contradiction  of the assumption that $E[X_t^2]=1$ for each $t\in[0,1]$.


问:
Non trivial homomorphism into injective envelope of a simple composition factor
Let $A$ be an artin algebra and let mod$(A)$ be the category of finitely generated $A$-modules.

Let $S$ be a simple module and let $M \in$ mod$(A)$. We further assume that $S$ is a composition factor of $M$. Then there exists a non trivial homomorphism $M \rightarrow I(S)$, where $I(S)$ is the injective envelope of $S$.

What I have so far:
We just need to show that there is a non trivial momomorphism $S \rightarrow M$. If we have that, then the existence of a non trivial homomorphism $M \rightarrow I(S)$ follows from the infectivity of $I(S)$.

So let $S$ be a composition factor of $M$. Then there exist submodules $L \subseteq K$ of $M$, such that $S \cong K/L$. Now one needs to construct a non trivial monomorphism $g: K/L \rightarrow M$. Thats the point where I have trouble continuing the proof. Intuitively I would say we construct $g$ by $k+L \mapsto k$. But I don't think it is a well defined.

答:
There is not necessarily a nontrivial monomorphism $S\to M$.

But there is a nontrivial homomorphism $K\to I(S)$: namely, the composition of the obvious maps $K\to K/L\cong S\to I(S)$. And then, by injectivity of $I(S)$, this homomorphism extends to a homomorphism $M\to I(S)$.


问:
Proof $\overline X_n \rightarrow \mu$ a.s. given $\sqrt{n}(\overline X_n - \mu) \rightarrow Z$ in distribution
Given $X_1...X_n...$ i.i.d. random variables with $E|X_1|<\infty$ and $\sqrt{n}(\overline X_n - \mu) \rightarrow Z$ in law for a certain random variable Z and a certain real number $\mu$ (not necessarily equal to $EX_1$), I need to proof that $\overline X_n \rightarrow \mu$ a.s.

It is very similar to the central limit theorem but the problem is I don’t know if the variance is finite, so I cannot use it. Any hint on how can I approach this problem?

答:
As $\sqrt{n}(\overline X_n - \mu) \rightarrow Z$ in distribution and $\frac{1}{\sqrt{n}} \to 0$ hence by Slutsky's theorem we have $\frac{1}{\sqrt{n}}\sqrt{n}(\overline X_n - \mu) \rightarrow Z \cdot 0$ and $\overline X_n - \mu \to 0$ in distribution. As it is  convergence in distribution to $const$ then  $\overline X_n - \mu \to 0$ in probability. Hence $\overline X_n \to \mu  $ in probability.

From S.L.L.N. we have $\overline X_n \to EX_1$ a.s. Thus  $\overline X_n \to EX_1$ in probability.

As $\overline X_n \to \mu  $ in probability and $\overline X_n \to EX_1$ in probability we get that $\mu = EX_1$.

As $\overline X_n \to EX_1$ a.s. and $\mu = EX_1$. we get that $\overline X_n \to \mu$  a.s.


问:
What method or subject deals with the following?
Can you point me to areas and methods that deals with modeling output from input data?  Say we think our output depends on certain parameters, and say we have samples of this output versus time and samples of the parameters versus time, but we don't know how the two relate.  More specifically, consider the following:

$ f \in \mathbb{R}^m $ and $A \in \mathbb{R}^{m \times n} $

Using least squares, we may find $ x \in \mathbb{R}^n $ such that $ \| f - Ax\|_2 $ is minimized.  But what other ways are there to model this relationship, if, for example, we don't believe it is linear?

Thanks.

答:
When you do OLS regression, you are actually assuming that your data comes from a Gaussian distribution whose mean is a linear function of covariates. Then, to estimate the coefficients of such a function, you use maximum likelihood; there are other methods of predicting the coefficients but ML is the simplest.

For data that is not (or can not be) Gaussian distributed, there are two straightforward ways to proceed.

We could just apply a transformation to the data poinys to make them have a close enough disyribution to use OLS. This isn't usually accurate.

Rather, we estimate the mean as a composition of a lunk function and a linear combination of covariates. This is called a Generalized Linear Model.

Past that, we can use techniques building on the GLM but I recommend first looking up all of the above. In my opinion, concise explanations of the theory are hard to come by, so I'll be more than happy to provide an in depth explanation when at a computer.


问:
Showing that there exists a certain function
We have the following functions:

$f(x) = x-\sin(x); \ \ \ \ g(x) = 1-\cos(x)$

I have to prove that there exists a function h such that h(f(x)) = g(x). But I don't exactly how to show that there exists such a function

It looks like I don't need to find the exact function $h$, only to show the existence of one. I already proved that $f(x)$ is bijective. But what about that function $h$ which should exists? What can I do? Can I use the fact that $f$ is bijective?

I also trying to find the derivative of $f(x)$. Yes, I am able to do that with my own knowledge of derivatives (and it is equal to $g(x)$), but I am also trying with the definition of the derivative. I tried the following:
$f(x+h) = x+h - \sin(x+h)$, but then I have write is as $f(x) + Df_x + r(h)$ where $r(h) \to 0 \text{ as } h \to 0$. But how can I rewrite $f(x+h)$ to get in into those terms?

答:
Since $f$ is bijective just take $h(x) = g(f^{-1}(x))$. Then
$$h(f(x)) = g(f^{-1}(f(x))) = g(x) $$
as required.


问:
Prove that any non-negative real number $a$ has $n$th root.
Prove that any non-negative real number $a$ has $n$th root.

By definition an $n$th root of a number $x$ is a number $r$ such that $r^n=x$.

Consider the set $\mathcal B=\left\{x:  0\le x \le 1+a\right\}$,then $(\mathcal B, \le)$ is a partially ordered set containing minimum and maximum element,define a function $f: \mathcal B \to \mathcal B$ with $$ x \mapsto x+ \frac{a-x^n}{n(1+a)^{n-1}}$$

Such function is order-preserving since for $0 \le x \le y \le 1+a$ we have that:

$$f(x)-f(y)=(x-y)(1-\frac{\sum_{k=0}^{n-1}x^ky^{n-1-k}}{n(1+a)^{n-1}})$$

Besides we know that $0 \le c \le d \implies c^n \le d^n$ for $n \in \mathbb N$,using this we have that $$0 \le \sum_{k=0}^{n-1}x^ky^{n-1-k} \le n(1+a)^{n-1}$$

Which implies that $f(x) \le f(y)$,using Knaster–Tarski theorem follows that exists $r$ in $\mathcal B$ such that : $$r+ \frac{a-r^n}{n(1+a)^{n-1}}=f(r)=r$$

And from the definition of $n$th root the desired result does hold.

I like to know if my proof is right and if it's not true then I want to know the errors.

答:
Your proof is correct!
Perhaps, a less-involved approach (or another alternative) is as follows; first $0^n=0$, $1^n=1$, and $a^{1}=a$, so we may assume that $a\notin\{0,1\}$ and $n\ge 2$.  Now, consider the continuous function
$$f\colon [0,a]\to\mathbb{R}\,,~\,~\,~\,~ x\mapsto
\left\lbrace
\begin{array}{ll}
x^n-a &\mbox{if $a>1$}\\
a-x^n &\mbox{if $a<1$}
\end{array}\,.
\right.
$$
It follows that $f(0)$ and $f(a)$ are non-zero and of opposite signs; hence, by the Intermediate Value Theorem, there exists $r\in(0,a)$ such that $f(r)=0$——that is, $a=r^n$.


问:
Determine if edge is a bridge in a graph
I would like to implement Fleury’s algorithm to find Eulerian trails in a graph. This algorithm requires me to tell if a given edge is a cut edge (bridge).

Is there a more effective way of doing this besides the one described here?

答:
The algorithm you link to checks if an edge $uv$ is a bridge in the following way:

Do a depth-first search starting from $u$, and count the number of vertices visited.

Remove the edge $uv$ and do another depth-first search; again, count the number of vertices visited.

Edge $uv$ is a bridge if and only if these counts are different.

Asymptotically, this is probably the best we can do, because depth-first searches are the easiest way to check for bridges. But we can reduce the number of depth-first searches that have to be done.

First, there is absolutely no reason to do two depth-first searches per edge we check. The most straightforward way to check if edge $uv$ is a bridge in $G$ is to do a single depth-first search in $G-uv$, starting at $u$. Rather than count the vertices visited, just check if $v$ is one of them! (We can even stop the search as soon as we find $v$.) This is the biggest improvement; if the DFS part of the algorithm is the majority of the running time, we've just sped things up by a factor of $2$.

A second big shortcut is specific to Fleury's algorithm. Suppose we're currently at vertex $u$ and we've discovered that edge $uv$ is a bridge. That means that (assuming a tour is possible) edge $uv$ must be the last edge we use out of $u$. In particular, assuming a tour is possible, there can be no other bridges out of $u$ - otherwise multiple edges would have to be the last edge we take, which is a contradiction. So we can simplify the iterative step at each vertex to the following:

Let $v_1, v_2, \dots, v_k$ be the neighbors of $u$.

If $k=1$, use $uv_1$, the only possible edge.

Check if $uv_1$ is a bridge. If it's not, use it.

If $uv_1$ is a bridge, use the edge $uv_2$ instead. We don't have to do another check on this edge: we know it's not a bridge.

Finally, we can adapt Fleury's algorithm somewhat to a recursive procedure that passes to a subgraph whenever it encounters a bridge. Suppose we've done a depth-first search from vertex $u$ and discovered that $uv$ is a bridge. In that case, we know that the Eulerian trail we want will begin by exploring $u$'s component of $G-uv$, then take edge $uv$, then continue with the rest of the graph.

So when this happens, we can actually remove edge $uv$ from the graph for the moment; recursively call the Eulerian trail procedure to find a tour (of the current component) from $u$; then, print edge $uv$ and continue finding an Eulerian trail from $v$.


问:
Why doesn't dividing identical cases not matter anymore in combinatorics when related with probability?
Consider six people are divided into two groups (with at least one person)

Now if we find the ways that there will be a group with two and four, we have $\frac{6C4\cdot4C4}{2}$ because picking the four first is the same as picking the two first.

However, when we find the probability that there will be a group with two and four in the other, it becomes:
$|S|=2^6 - 2$

$|E| = 6C4\cdot4C4$, so $P = \frac{6C4\cdot4C4}{2^6 - 2}$

Why was dividing by 2 not necessary?

答:
I think you're a bit confused about the basic concepts. So I'll try my best to explain the whole problem simpler so that you could understand where you're going wrong.

Imagine there are 6 people A,B,C,D,E,F

From these 6 people, if we take two people (say A and B) and put them into a separate team. The rest four people (C,D,E,F)  will simultaneously become the other team. Thus our purpose of splitting them into two teams is fulfilled!

Wait! we are not finished yet. Putting A and B in a separate team is only one possibility out of many.

Out of 6 people, 2 people can be selected in $^{6}C_{2}$ ways and 4 people from the remaining 4 people is $^{4}C_{4}$ ways which is equal to 1 (obviously! as the other 4 are automatically sorted, as mentioned before)

So the total number of ways teams of 2 people and 4 people can be formed = $^{6}C_{2}\times^{4}C_{4}$ (OR) $^{6}C_{2}$ ways  (as $^{4}C_{4}=1$)

Now we need to find the total number of ways in which 6 people can be split into two teams of different compositions (ie) Teams with 1 and 5 or teams with 2 and 4 and so on...

Which can be calculated as: $(^{6}C_{1}\times\ ^{5}C_{5}) + (^{6}C_{2}\times\ ^{4}C_{4}) + (^{6}C_{3}\times\ ^{3}C_{3}) + (^{6}C_{4}\times\ ^{2}C_{2}) + (^{6}C_{5}\times\ ^{1}C_{1})$

=$\ ^{6}C_{1} + ^{6}C_{2} + ^{6}C_{3} + ^{6}C_{4} + ^{6}C_{5}$

NOTE 1: We are adding each term because only one can occur at a time (ie) we cannot choose two compositions of teams simultaneuosly (I cannot split them into teams of 4 and 2 and 5 and 1 simultaneously. I can only do one of them at a time.). We multiply if two events were to occur simultaneously.

NOTE 2:  Team 1 having 5 people and Team 2 having 1 people is the same as Team 1 having 1 people and Team 2 having 5 people  (Secret: You can just rename the team! lol. The essence of a team is their constituting members). Thus we divide the total possibilities by 2! (As the total number of ways we can arrange the two teams is 2! . Similarly, if we were to split 6 people into 3 teams, the teams can be arranged in 3! ways) In other words, order doesn't matter

(ie) total possibilities = $\frac{^{6}C_{1} + ^{6}C_{2} + ^{6}C_{3} + ^{6}C_{4} + ^{6}C_{5}}{2!}=\frac{62}{2\times1}=31$

P(splitting into teams of 2 and 4 members)= $\frac{^6C_{2}\times^{4}C_{4}}{31}=\frac{^{6}C_{2}}{31}=\frac{15}{31}$

Now, coming to your query on why don't we divide by 2, I've already posted it in the comment section of your question. I hope that this helps you.


问:
Local obstruction to having a complete extension
This question arose while I was reading Helgason's book on symmetric spaces. In chapter IV section 5, one can read the following:

Let $M$ be a Riemannian manifold, $p$ a point in $M$. In general it is impossible to find any neighborhood $N$ of $p$ which can be extended to a complete Riemannian manifold $\tilde{M}$.

To me this is very surprising. This means that there are local "obstructions" to the existence of a complete extension, so the completeness of a manifold $M$ impacts its local geometry.

My first thought was that locally, say in a chart $B(0,\varepsilon)\subset \mathbf{R}^n\to  M$, the metric doesn't vary to much from its value at $p$. So maybe we can extend it to a metric on $\mathbf{R}^n$ using partitions of unity and we should be able to ask that the metric is the standard metric outside a ball $B(0,\varepsilon+r)$. If this is possible I guess it would give a complete metric on $\mathbf{R}^n$ (here I'm probably wrong).

But apparently this is not going to work. Hence my question is:

What are some examples of manifolds $M$ and point $p$ of $M$ such that no neighborhood of $p$ can extend to a complete manifold ?

From Helgason's book I know that locally symmetric spaces have these neighborhoods that extend to a complete manifold.

Any kind of help will be greatly appreciated.

EDIT : In fact the argument of partition of unity works in the $C^\infty$ category. However it doesn't work if everything is assumed to be real-analytic, which seems to be the what Helgason does implicitly. So I am still interested in a detailed example in the case where everything is real analytic.

答:
One can take any real-analytic Riemannian manifold $(M,g)$ such that for each point $p\in M$ there is a geodesic path $c: [0, 1)\to M$ (of finite length),   satisfying

$c(0)=p$.

The absolute value of scalar curvature $|S(c(t))|$ of $(M,g)$ diverges to   infinity as $t\to 1-$.

For instance, one can take the surface of revolution in ${\mathbb R}^3$ obtained by rotating the graph
$$
y= x^{1/2}, 0<x<\infty
$$
around the $y$-axis. Then for a point $p=(x,y,z)$ in this surface $S$ the curve $c$ is obtained by intersecting $S$ with a half-plane passing through $p$ and the $y$-axis; the curve $c$ is chosen so that $\lim_{t\to 1} c(t)=(0,0,0)$. (One uses, of course a suitable constant speed parameterization to ensure that the curve is indeed geodesic.)

To check both (1) and (2) it suffices to consider points $p$ in the $xy$-plane. The finiteness of the length of $c$ is, hopefully clear.

Then the lines of curvature of $S$ through the point $c(t)$ are:

(a) the curve $y= x^{1/2}$ itself.

(b) the circle obtained by rotating $c(t)$ around the $y$-axis.

The curvature of the circle, of course, diverges to $\infty$ as $x\to 0$, while the curvature of the curve $x=y^2$ converges (as $x\to 0$) to the curvature of this parabola at $(0,0)$, which equals $2$.

Hence, the product of principal curvatures, which is the value of the Gaussian curvature of $S$, diverges (in absolute value) to infinity as $x\to 0$.

I claim that properties (1) and (2) imply that $(M,g)$ does not locally isometrically embed in a complete real-analytic Riemannian manifold $(N,h)$ of the same dimension.

Indeed, suppose there is such an isometric embedding $\iota: U\to N$, $U$ is an open disk in $M$ centered at a point $q$. Let $p=\iota(q)$, $c$ is a geodesic as above through $q$, $c(0)=q$, $c([0,T])\subset U$, $T>0$, $b=\iota\circ c$. By the completeness of $(N,h)$, the geodesic $b$ extends to $[0,\infty)$.

Since $g, h$ are real-analytic, the scalar curvatures of $g, h$ along $c, b$ are real-analytic functions of the parameter $t$. They agree on the interval $[0,T)$, hence, agree on the interval $[0,1)$. Thus, the scalar curvature of $(N,h)$ blows-up at $b(t)$ as $t\to 1$. A contradiction.

Edit. Geodesics in real-analytic Riemannian manifolds are real-analytic curves. Indeed, the Levi-Civita connection $\nabla$ of a real-analytic Riemannian metric is also real analytic (since Christoffel symbols are expressed in terms of partial derivatives of the tensor metric). Geodesics  themselves are then solutions of the ODE $\nabla_{c'}c'=0$ with real-analytic coefficients. Therefore, by the Cauchy–Kovalevskaya theorem, geodesics are also real-analytic.

Lastly, the example given by Ryan does not work since it embeds isometrically in the real line (just the isometric embedding is not given by the identity map). It is the fact that we do not know what an isometric embedding might look like forces one to consider the behavior of some intrinsic invariants such as the scalar curvature.


问:
Prove that $\mathcal F_\tau$ is a $\sigma -$algebra
Let $(\Omega ,\mathcal F,\{\mathcal F_t\},\mathbb P)$ a probability filtred space and $\tau$ a stopping time. I'm trying to prove that $$\mathcal F_\tau=\{A\in \mathcal F\mid A\cap \{\tau\leq t\}\in \mathcal F_t\}.$$

I could prove that $\Omega ,\varnothing \in \mathcal F_\tau$, and that $\mathcal F_\tau$ is stable by countable union. But I have difficulties to prove that it's stable by complementary. Let $A\in \mathcal F_\tau$. So, $A\cap\{\tau\leq t\}\in \mathcal F_t$. How can I prove that $A^c\cap \{\tau\leq t\}$ ?

答:
$A^c\cap \{\tau\leq t\} = \{\tau\leq t\} \backslash  (A\cap \{\tau\leq t\} $).

But $\{\tau\leq t\} \in \mathcal{F}_t$  as $\tau$ is Markov moment,

$ (A\cap \{\tau\leq t\} ) \in \mathcal{F}_t $ as $A \in \mathcal{F}_{\tau}$

and finally $\mathcal{F}_t$ is sigma-algebra.

Thus $\{\tau\leq t\} \backslash  (A\cap \{\tau\leq t\} ) \in \mathcal{F}_t $  and it follows that $A^c\cap \{\tau\leq t\} \in \mathcal{F}_t $


问:
Information propagation rate in 2nd-order PDEs
(Apologies if some details are wrong, this is my first time passing through this stuff, and would appreciate any feedback on correcting the problem statement as well as getting the solution.)

Three common characteristics of 2nd order PDEs are elliptic, hyperbolic, and parabolic. For systems with many variables, I am going to describe them as
$$
0 = \underbrace{\left(\frac{1}{2}x^TH x+x^Tg + c\right)}_{F} u
$$
where $x = (x_1,...,x_n)$ and $x_i$ stands in place for $\partial f/\partial u_i$, $H$  is a symmetric matrix, and $g$ is a vector. The elements $H_{ij}$ and $g_i$ may be constants or functions of $u$. Let us assume they are constants for now. I am writing $F$ as an operator on $u$.

Then, according to Strauss 1.6, the system is

elliptic if $H$ is nonsingular and all the eigenvalues have the same sign

hyperbolic if $H$ is nonsingular, with some eigenvalues positive and some negative

parabolic if $H$ is singular, with exactly one 0 eigenvalue.

(My guess is that singular $H$ with more than 1 eigenvalue that is 0 can be reduced to parabolic with changes of variables.)

First question: is that interpretation correct?

I am trying to stare at these equations and use them to make sense of the following statement: "hyperbolic systems have finite information propagation, and parabolic/elliptic systems have infinite information propagation rate". I wish to make this concept precise.

What exactly is meant by "information propagation"? Without necessarily including a time variable, can I assume that this is some bound on $\max_i|\partial u_i/\partial u_j|$?

Why would some systems result in finite information propagation rate and others necessarily have infinite information propagation rate?

I understand how wave propagation works in the specific wave equation case, but 1) I'm not super clear on why diffusion works out to be infinite propagation rate and 2) I'm not sure how this relates to the known properties of the different PDE classes.

Any clarification or discussion would be appreciated!

答:
By your notation I will assume you are looking at operators of the form
$$ F(D)u = \frac12\sum_{i,j=1}^n H_{ij}\partial_{x_ix_j}u + \sum_{i=1}^n g_i\partial_{x_i}u + cu = 0. $$

First question: is that interpretation correct?

The classification given in Strauss is for PDEs in two dimensions only; in higher dimensions you still have a classification, but it's no longer exhaustive. Also you implicitly assume $H \neq 0$ so the system is genuinely of second order, which is why the parabolic case assumes exactly one eigenvalue is zero.

In general dimensions this is somewhat convention-dependent, but you have the following classification:

The equation is elliptic if $H$ is non-singular and all eigenvalues have the same sign.

The equation is hyperbolic if $H$ is non-singular and all but one eigenvalue has the same sign.

The equation is parabolic if $H$ has one zero eigenvalue, and all other eigenvalues have the same sign. This is somewhat convention-dependent however, and one usually assumes extra structure in this case.

Note this is not a complete classification; if $H$ is non-singular but neither elliptic nor hyperbolic, then it is sometimes referred to as ultrahyperbolic - however this is limited literature on equations of these type.

Generally you should think of the model equations: Laplace's equation, the heat equation and the wave equation. The classification also extends more generally to nonlinear equations, etc, and while the convention varies this is generally based on how solutions to these equations behave.

For the parabolic and hyperbolic case, if $H$ is constant we see we have a distinguished direction $v$ - in the parabolic case this corresponds to the zero eigenvector, while in the hyperbolic case this corresponds to the eigenvector whose eigenvalue has different sign. By a change of variables we can assume $v = e_n,$ and that writing $x_n=t$ the equation takes the form
$$ g_n \partial_{t}u + \frac12\sum_{i,j=1}^{n-1} H_{ij}\partial_{x_ix_j}u + \sum_{i=1}^{n-1} g_i \partial_{x_i} u + cu = 0 $$
in the parabolic case, and
$$ \frac12H_{nn}\partial_{t}^2u + \frac12\sum_{i,j=1}^{n-1} H_{ij}\partial_{x_ix_j}u + \sum_{i=1}^{n-1} g_i \partial_{x_i} u +g_n\partial_tu+ cu = 0 $$
in the hyperbolic case (I am omitting the details here, but some linear algebra is involved). Therefore we see we naturally obtain a distinguished direction, which we suggestively denote as $t$ - which was completely determined by the equation! Note in the parabolic case we usually assume $g_n \neq 0,$ as otherwise the equation reduces to an elliptic equation in the first $(n-1)$ variables, and in the hyperbolic case $H_{nn} < 0$ necessarily because $H$ is non-degenerate (assuming the other eigenvalues are positive).

Generally when working with parabolic and hyperbolic equations we assume there is a distinguished direction $\partial_t,$ even for more general variable coefficient and nonlinear equations${}^\ast$ - this is because in the applications we are interested in there is a natural time direction, and it is more convenient for purposes of analysis.

${}^\ast$Note the classification is also used to distinguish nonlinear PDEs, but here there is no standard way of doing this. Typically we require a suitable linearised equation to be elliptic/parabolic/hyperbolic, but what this means depends on the particular application. Also in this case we don't always have a distinguished time direction, which notably arises in the context of general relativity.

Now that I've explained how a time direction naturally arises in the parabolic and hyperbolic cases, I can turn to your question about propagation of information.

What exactly is meant by "information propagation"? Without necessarily including a time variable, can I assume that this is some bound on $\max_i|\partial u_i/\partial u_j|$?

Propagation of information generally refers to how local changes in the initial/boundary data is reflected in the corresponding solution. For this however, note that the type of boundary conditions varies depending on the equations we consider.

For elliptic equations we consider boundary value problems, where for a bounded domain $\Omega \subset \Bbb R^n$ we seek solutions $u$ subject to conditions on the boundary; e.g. $u = g$ on $\partial\Omega$ (Dirichlet) or $\partial_{\nu}u = h$ on $\partial\Omega$ where $\partial_{\nu}$ is the normal derivative (Neumann).

For parabolic and hyperbolic problems we consider initial boundary value problems, typically on domains of the form $\Omega' \times (0,T)$ where $\Omega' \subset \Bbb R^{n-1}$ is a bounded domain. In both cases we prescribe $u$ on $\partial\Omega' \times (0,T)$ (the boundary part) and in the parabolic case $u$ on $\Omega' \times \{0\}$ also (the initial part). For hyperbolic problems we need to prescribe initial values for both $u$ and $\partial_tu$ at $t=0.$

The reason we impose these conditions is because it turns out this is necessary to ensure the equation is well-posed; if you prescribe suitably regular initial/boundary data, then you can show a unique solution exists. There are many examples that show well-posedness fails if you don't impose the correct conditions, and for instance this explains why there is no time direction if your equation is elliptic.

In the elliptic and parabolic cases, local changes to initial/boundary data can change the solution at every point, which is what we mean by infinite speed of propagation. For wave equations this is not the case however, and changes to initial data is propagates along a 'wave cone.'

The details of this are lengthy, so I refer you to Chapter 2 of Strauss' text which discusses these phenomena in the context of the wave and diffusion (heat) equations.

Why would some systems result in finite information propagation rate and others necessarily have infinite information propagation rate?

This is a rather deep question, but as I've alluded to in the above discussion the behaviour of solutions to PDEs greatly varies depending on their classification. This is why for instance there is no unified theory of PDEs - small changes to the equation can result is very different behavior, which is to be expected because different equations model different physical phenomena.


问:
Root-finding algorithms for discrete function
Assume that we have function f(x)

f(x) has 3 possible constant outputs

f(x)= a if x>b

f(x)= 0   if x =b

f(x)= -a if x<b

Is there an iteration method that can be used to generate initial guess then iterates to find b?

答:
For initial guess

Take $u_1 = 0$ and $\epsilon = 1$ if $f(u_0) = -a$ / $\epsilon = -1$ if $f(u_0) = a$. Compute $u_n = u_0 + n \epsilon$ until $f(u_0)f(u_n) \lt 0$.

Then use Bisection method algorithm.


答:
If you have $l$ and $r$ such that $f(l) = -a$ and $f(r) = a$ then the bisection method should converge to $x$.

That is, set $l_1 = l$ and $r_1 = r$. Then if $f\left(\frac{l_n + r_n}{2}\right) = 0$ we are done. Otherwise if it is $-a$, set $l_{n+1} = l_n$ and $r_{n+1} = \frac{l_n + r_n}{2}$ and finally if is $a$, set $l_{n+1} = \frac{l_n + r_n}{2}$ and $r_{n+1} = r_n$. $l_n$ and $r_n$ should both converge to $x$.

Now to find initial guesses, try $l = -2^n$ and $r = 2^n$ which will find initial guesses in the asymptotic order of binary digits of $b$.

Title: Root-finding algorithms for discrete function

Body:
Assume that we have function f(x)

f(x) has 3 possible constant outputs

f(x)= a if x>b

f(x)= 0   if x =b

f(x)= -a if x<b

Is there an iteration method that can be used to generate initial guess then iterates to find b?


问:
Find the general term.
I tried to solve the wave equation:

After some calculation I reached at this step where I have to find the constant alpha.

For even value of n alpha is 0

For n =  1, 3, 9, 11, 17, 19 . . . . . . alpha is $\sqrt{2}$

For n =  5, 7, 13, 15, 21, 23 . . . . . . alpha is $-\sqrt{2}$

Can anyone help me to find the general term for 2 and 3?
I tried my best but failed. Please help.
I want something like $\frac{{(-1)^{something}}8\sqrt{2}}{{\pi^2}{n^2}}$

答:
I used Mathematica powerful function

```<code>FindSequenceFunction[{1, -1, 1, -1, -1, -1, -1, -1, 1, -1,
   1, -1, -1, -1, -1, -1, 1, -1, 1, -1}, x] // FullSimplify
</code>```

$$\small f(x)=\frac{1}{4} \left(\sqrt{2} \sin \left(\frac{\pi  x}{4}\right)+\sqrt{2} \sin \left(\frac{3 \pi  x}{4}\right)-\sqrt{2} \sin \left(\frac{5 \pi  x}{4}\right)-\sqrt{2} \sin \left(\frac{7 \pi  x}{4}\right)-2 \cos (\pi  x)-2\right)$$

Title: Find the general term.

Body:
I used Mathematica powerful function

FindSequenceFunction[{1, -1, 1, -1, -1, -1, -1, -1, 1, -1,
   1, -1, -1, -1, -1, -1, 1, -1, 1, -1}, x] // FullSimplify

$$\small f(x)=\frac{1}{4} \left(\sqrt{2} \sin \left(\frac{\pi  x}{4}\right)+\sqrt{2} \sin \left(\frac{3 \pi  x}{4}\right)-\sqrt{2} \sin \left(\frac{5 \pi  x}{4}\right)-\sqrt{2} \sin \left(\frac{7 \pi  x}{4}\right)-2 \cos (\pi  x)-2\right)$$


问:
Find CDF of $Y=-X+2$
i have a task to find CDF of $Y=-X+2$, while:

$\begin{align}F_X (x) & = \begin{cases} 0 & :x \leq 0 \\ \tfrac 12x & : 0\lt x \leq 2 \\ 1 & : x\gt 2\end{cases}\end{align}$

i did something like this:

$P(Y<x) = P(-X+2<x)=P(X>-x+2)=1-P(X<-x+2)=1-F_X(-x+2)$

and i have no idea what to do in the next step. I've heard i need to define specific ranges for the arguments in $F_Y$ but i dont know how to do that. I have tried a lot of things but i still haven't got the right answear.

The answear i have to get is:

$\begin{align}F_Y (x) & = \begin{cases} 0 & :x \leq 1 \\ \tfrac 12x & : 1\lt x \leq 2 \\ 1 & : x\gt 2\end{cases}\end{align}$

答:
You are almost there.

Since $P(Y<x)=1-F_X(-x+2)=\begin{cases}1=1-0& -x+2\le 0\\1-\frac 12(-x+2)&0< -x+2\le 2\\ 1-1&-x+2>2\end{cases}$
as you noticed, we have $$F_Y(x)=\begin{cases}0&x<0\\\frac 12 x&0\le x< 2\\ 1& x\ge 2\end{cases}.$$

Title: Find CDF of $Y=-X+2$


答:
General term is:

$$f(x) = {-1}^{\lfloor2\{x / 8 \} \rfloor +1} $$

Here is the proof.

As mentioned by me in the comments the first row contains numbers of the form $8k+1,8k+3$ and the second row contains numbers of the form $8k+5,8k+7$

Case 1: First Row

8k+1 or 8k+3

{(8k+1)/8} or {(8k+3)/8} = 0.125 or 0.375

2{(8k+1)/8} or 2{(8k+3)/8} = 0.25 or 0.75

When you take the floor function it becomes 1 because floor(0.25 or 0.75) = 1

Therefore ${-1}^2  = 1$

Case 2: 2nd Row

{(8k+5)/8} or {(8k+7)/8} = 0.625 or 0.875

2{(8k+1)/8} or 2{(8k+3)/8} = 1.25 or 1.75

When you take the floor function it becomes 1 because floor(1.25 or 1.75) = 2

Therefore ${-1}^3 = -1$

Title: Find the general term.

Body:
I tried to solve the wave equation:

After some calculation I reached at this step where I have to find the constant alpha.

For even value of n alpha is 0

For n =  1, 3, 9, 11, 17, 19 . . . . . . alpha is $\sqrt{2}$

For n =  5, 7, 13, 15, 21, 23 . . . . . . alpha is $-\sqrt{2}$

Can anyone help me to find the general term for 2 and 3?
I tried my best but failed. Please help.
I want something like $\frac{{(-1)^{something}}8\sqrt{2}}{{\pi^2}{n^2}}$


答:
If the domain were a compact interval such as $[a,b]$, you'd simply take the boundary points as initial guesses and proceed by binary bisection. Here, we can use a suitable function to map $\Bbb R$ to a bounded interval and extend it to its closure.

For example, $\arctan$ maps $\Bbb R$ to $(-\frac\pi2,\frac\pi2)$, so we might do the following

Set $L\leftarrow-\frac \pi2$ and $R\leftarrow \frac\pi2$.

If $L>-\frac\pi2$ and $R<\frac\pi2$, go to step 7

Set $M\leftarrow \frac{L+R}2$ and compute $Y=f(\tan M)$

If $Y=0$, terminate with result $\tan M$

If $Y=a$, set $R\leftarrow M$ and go back to step 2

Otherwise, set $L\leftarrow M$ and go back to step 2

Set $L\leftarrow \tan L$, $R\leftarrow\tan R$

Set $M\leftarrow \frac{L+R}2$ and compute $Y=f(M)$

If $Y=0$, terminate with result $M$

If $Y=a$, set $R\leftarrow M$ and go back to step 8

Otherwise, set $L\leftarrow M$ and go back to step 8


问:
Linear Regression: Expectation Proof
I found the following proof in my notes:

$E(Y_i) = E[\beta_0 + \beta X_i + \varepsilon_i] =\cdots= \beta_0 + \beta X_i$.  This does not seem right to me, however.  Why would $E(\beta_1 X_i) = \beta_1 X_i$?  I wonder if i might have written it down incorrectly, with the actual proof meaning to be for the estimated value Yi hat (I don't know how to code this unfortunately).  Does anyone recall this property of linear regression?

答: In this sort of regression problem, $X_i$ may be random in the sense that if you take another sample, all the $X_i$ values change, but one behaves as if one seeks the conditional expected value of $Y_i$ given $X_i$, so that in effect $X_i$ is treated as if it were constant rather than random.  And the $\beta$s are also being treated as constant.

Title: Linear Regression: Expectation Proof


问:
Sturm separation and comparison theorem result:
Let $u"+p_1(t)u=0$ and $v"+p_2(t)v=0$ with $p_2(t)>p_1(t)$ in $(a,b)$. Suppose that $u(a)=v(a)=0$ and $u'(a)=v'(a)=x>0$. Show that there exists $\epsilon>0$
such that $v(t)>u(t)$ in $(a,a+\epsilon)$.

My attempt:

First with both the equations I write $$u"-v"+(p_1-p_2)(t)(u-v)=0$$ and
$$(u-v)(a)=0 , (u-v)'(a)=0$$
So, from an application of sturm theorem that if suppose $p(t)<0$ then the solution of equation $y"+p(t)y=0$ has no more than one zero.
I am trying to use this result here.
But how to show $(u-v)(t)$ has exactly one zero in $(a,b)$.
Please suggest a solution to this!!

答:
Since $u(a)=v(a)=0,u'(a)=v'(a)=x>0$, there is $\epsilon>0$ such that $v(t)>0$ for $t\in(a,a+\epsilon)$. Note
$$ \bigg(\frac uv \bigg)'(t)=\frac{u'(t)v(t)-u(t)v(t)'}{v^2(t)}. \tag1$$
Multiplying the 1st equation by $v$ and by integration by parts, one has
$$ u'(t)v(t)=\int_a^tu'(s)v'(s)ds-\int_a^tp_1(s)u(s)v(s)ds. \tag2 $$
Similarly, from the 2nd equation, one has
$$ u(t)v'(t)=\int_a^tu'(s)v'(s)ds-\int_a^tp_2(s)u(s)v(s)ds. \tag2 $$
From (1), (2) and (3), one has
$$ \bigg(\frac uv \bigg)'(t)=\frac{u'(t)v(t)-u(t)v(t)'}{v^2(t)}=-\frac1{v^2(t)}\int_a^t(p_1(s)-p_2(s))u(s)v(s)ds<0 $$
which means that $\bigg(\frac uv \bigg)(t)$ is decreasing in $(a,a+\epsilon)$. So for $s,t\in(a,a+\epsilon)$ ($s<t$), one has
$$ \bigg(\frac uv \bigg)(s)>\bigg(\frac uv \bigg)(t). $$
Letting $s\to a$ gives
$$ \bigg(\frac uv \bigg)(t)<1 \text{ or }u(t)<v(t). $$


问:
Line of Intersection of 2 Planes
I'm trying to understand how to find line of intersection between 2 planes. In almost every explanation I've seen, it involves removing one of the unknowns by substituting it with 0.

For example in this video https://www.youtube.com/watch?v=q2FO675dmyM

```<code>For the planes:
x + 2y - 4z = 16
2x - y + 3z = 6

Substituting x = 0,
We get the equations
2y - 4z = 16
2y + 6z = 12
</code>```

And eventually we can solve the equation.

But what if we cannot substitute x = 0?

```<code>Consider the planes:
x + y = 8
x = 5
</code>```

In this case, the line of intersection will never touch x = 0 or y = 0.
I know that with this equation, we can already easily solve this question as there is only 2 unknowns (technically 1 unknown since x = 5).

But I'm just curious if it is possible for such a case to occur with 3 unknowns.
Where we have 3 unknowns but the line of intersection will never pass through one of the axis.

答:
Rather than assign the value $0$ to $x$, for the reasons you mention, (or $y$ or $z$, for that matter), it is better to eliminate one of the letters from the two equations first.

So, for example, taking the equations you mention:$$x+2y-4z+16$$ and $$2x-y+3z=6$$

If we decide to eliminate $y$ for example we get $$5x+2z=28$$
Whereupon, $$5x=28-2z=\lambda$$

We can then get each of $x$ and $y$ and $z$ in terms of the parameter $\lambda$, which gives us an equation of the line.

Title: Line of Intersection of 2 Planes


问:
Series convergence and euler number
a belong to real set.
Find the range of x such that the series converge:
$\sum_{0}^{\infty}(\frac{a(a-1)...(a-k+1)}{k!})x^{k}$

Okay, i found that for |x| < 1 the series converge, by the ratio rest. Otherwise, and for |x| different of 1, the series diverges. But i think i am seeing a contradiction here:

See that the series is: $1 + ax + a(a-1)x²/2 + ...$.

I notice that $1 + ax + a(a-1)x²/2 + ... < 1 + ax + (a²)x²/2 + (a³)x³/3!...$

So that we can call the right side of the inequality as $e^{ax}$
And this is just a number! So that, technically, the series converge for any x. What am i missing here?

Title: Series convergence and euler number

Body:
$$\sum_{k=0}^{\infty}\left(\frac{a(a-1)...(a-k+1)}{k!}\right)x^{k}=\sum _{k=0}^{\infty } \frac{a! x^k}{k! (a-k)!}=\sum _{k=0}^{\infty } x^k \binom{a}{k}$$
if $|x|<1$ the series converges to $(1+x)^a$


问:
Given vectors spaces $V,W$ with finite dimension,is that true to claim that if $W \subseteq V$ and $\text{dim}(V)=\text{dim}(W)$ implies that $W=V$?
Given vectors spaces $V,W$ with finite dimension,is that true to claim that if $W \subseteq V$ and $\text{dim}(V)=\text{dim}(W)$ implies that $W=V$?

For finite sets $A$ and $B$ with $\left|A\right|=\left|B\right|$ I know that if $A \subseteq B$  then $A=B$,but I'm not sure if the same does hold for vector spaces with finite dimension.

If it does hold then please give me a proof,if it's not true please provide a counterexample.

答:
Yes.

Pick a basis of $W$. That's an independent family of vectors in $V$ and hence can be extended to a basis of $V$, but without adding any vectors.


问:
Using generating functions, Find a closed formula to next expression: $\sum_{k=0}^m{k(k+2)}$
Using generating functions, Find a closed formula to next expression:

$\sum_{k=0}^m{k(k+2)}$

If i use calculus power series rules, The question is fairly simple. But how can i find the proper relation with generating functions?

答:
First note that

$$\sum_{k=0}^nk(k+2)x^k=\sum_{k=0}^n(k+1)^2x^k-\sum_{k=0}^nx^k\;.$$

The last summation is a geometric series, so you know a closed form for it. Next, check that

$$\sum_{k=0}^n(k+1)^2x^k=\sum_{k=1}^{n+1}k^2x^{k-1}=\left(\sum_{k=1}^{n+1}kx^k\right)'=\left(\sum_{k=0}^{n+1}kx^k\right)'\;.$$

If you can find a closed form for $\displaystyle\sum_{k=0}^{n+1}kx^k$, you can differentiate it to get one for $\displaystyle\sum_{k=0}^n(k+1)^2x^k$. And

$$\sum_{k=0}^{n+1}kx^k=\sum_{k=1}^{n+1}kx^k=x\sum_{k=1}^{n+1}kx^{k-1}=x\left(\sum_{k=1}^{n+1}x^k\right)'=x\left(\sum_{k=0}^{n+1}x^k\right)'\;,$$

for which you can easily find a closed form.


问:
Asymmetry between definitions of curvature in 2-space and 3-space
Edit

I am coming back to this question because although the answer I received cleared the matter a little bit I am still confused by the difference between the 2D and 3D case.

So in 2D case one can define abstractly the normal versor N(s) as the unique vector such that the ordered set $\{T(s), N(s)\}$ forms an orthonormal, positively oriented basis of $\mathbb{R}^2.$

Then one sees that since $\dot{T}(s)=\frac{d}{ds}T(s)$ is orthogonal to $T(s),$ one may take $N(s)$ to be equal to $\frac{\dot{T}(s)}{\|\dot{T}(s)\|}$ for all $s$ such that $\text{det}(T(s),\dot{T}(s))=1$ and equal to $- \frac{\dot{T}(s)}{\|\dot{T}(s)\|}$ for all $s$ such that the determinant is equal to $-1.$ This determines the sign of the curvature at each point $s.$

In 3D case instead, one starts with $T(s)$ and $N(s)=\frac{\dot{T}(s)}{\|\dot{T}(s)\|},$ which amounts to choosing by definition that the curvature must be positive. Then since $B= T \land N$ must be orthogonal to both $T$ and $N,$ those three vectors are linearly independent and thus form an orthonormal basis of $\mathbb{R}^3$.

Question

Is $B$ the unique vector such that $\{ T, N, B \}$ forms a positively oriented basis of $\mathbb{R}^3$?

How can I have an interpretation of the sign of the torsion in a way similar to the one I gave above for curvature?

答: You haven't been explicit in your definitions. In the plane, you can require that $T,N$ form a right-handed orthonormal basis, and then you find that $dT/ds$ is possibly a negative multiple of $N$ (what is going on here is convexity/concavity). However, in three dimensions, there is no way to define $N$ other than by saying it's the unit vector in the direction of $dT/ds$ (assuming that vector is non-zero, of course). The right-handed orthonormal basis comes in when you add in the binormal vector $B$. It then makes sense to ask if the curve is twisting toward $B$ (positive torsion, by most people's definition) or away from $B$ (negative torsion); here is the sign you were attaching to curvature in the case of a plane curve.


答:
If we note $k=\dim W$, we can write $W= \text{Vect} (e_1,...,e_k)$ where $(e_1,...e_k)$ is a basis of $W$.

Then $\text{Vect} (e_1,...,e_k) \subset V$, so $(e_1,...e_k)$ is a family of $k$ elements of $V$ that is linearly-independent, and as $\dim V = k$, it's a basis of $V$.

So $W=V=\text{Vect} (e_1,...,e_k)$.

Body:
Given vectors spaces $V,W$ with finite dimension,is that true to claim that if $W \subseteq V$ and $\text{dim}(V)=\text{dim}(W)$ implies that $W=V$?

For finite sets $A$ and $B$ with $\left|A\right|=\left|B\right|$ I know that if $A \subseteq B$  then $A=B$,but I'm not sure if the same does hold for vector spaces with finite dimension.

If it does hold then please give me a proof,if it's not true please provide a counterexample.


问:
Dividing a polynomial with integer coefficients by a quadratic
In the following question, one of the coefficients of the divisor quadratic polynomial is unknown; so, it cant be factored for performing synthetic division. I tried to perform long division hoping to get some clues but ran into a conflict.

Appreciate your help. Thanks!!

Here is the question:
$x^2 -x +a$  evenly divides $x^8 + 5x^6 + 13x^4 + 20x^2 +36$. Determine positive integer $a$.

A few things I noted are:

$a$ must be a factor of 36 so, I am thinking $a$ would be from $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$

All the coefficients of the dividend polynomial are integers and degrees of all the terms are even. Not sure how to use this info.

$P(x) = Q(x)(x^2 -x +a) + mx + n$.  Since the remainder is zero, $m$ and $n$ are zero. When I tried long division, I got two equations; but the value of $a$ would come as imaginary which conflicts with the first observation and the question asks for real integer $a$.

答:
Avoiding the long division with a polynomial of degree $8$, you may note that
$$\begin{align}
\color{blue}{x^8} + \color{red}{5x^6} + \color{blue}{13x^4} + \color{red}{20x^2} +\color{blue}{36}
&=(x^8 + 13x^4 +36)+ (5x^6+ 20x^2)\\
&=(x^4+9)(x^4+4)+5x^2(x^4+4)\\
&=(x^4+5x^2+9)(x^4+4).
\end{align}$$
Now long division is easier to perform. Can you take it from here?


问:
A question about intuitionistic logic
I find it hard to understand the following formula in intuitionistic logic, where $P\neq\perp$ and $P\not\leftrightarrow(\neg\neg\exists xQ(x)\rightarrow\exists xQ(x))$:
$$P\rightarrow(\neg\neg\exists xQ(x)\rightarrow\exists xQ(x))\quad\quad(*)$$
Intuitively, it says that from $P$, infer $\neg\neg\exists xQ(x)\rightarrow\exists Q(x)$. Clearly, the consequent $\neg\neg\exists xQ(x)\rightarrow\exists xQ(x)$ does not hold in intuitionistic logic. So, my question is under what conditions, this formula is true. The only way I can come up with is that $(*)$ holds if $P\rightarrow\exists xQ(x)$. But is that the only way to derive $(*)$? If that's the only way, can we infer $P\rightarrow\exists xQ(x)$ from $(*)$? This looks quite odd. So, is there any other way to derive $(*)$ and what can we infer from $(*)$? Thanks in advance!

答:
We definitely do not need $P \to \exists x Q(x)$ for this. For example, we could have that $P$ is $R \wedge (\neg \neg \exists x Q(x) \to \exists x Q(x))$, where $R$ is some new propositional symbol. So $(*)$ trivially holds. Then it is easy to see that $P \to \exists x Q(x)$ cannot be derived from $(*)$. Just consider a model where $R$ is true and $\forall x \neg Q(x)$ holds.

I do not think we can say much more in general than that. Because in the above example, once we would also add the axiom that $R$ is always true, we just get that $P$ is equivalent to $\neg \neg \exists x Q(x) \to \exists x Q(x)$ and $R$ has no real function. So no information is gained from $(*)$.

Title: A question about intuitionistic logic


问:
Count the numbers in which the product of the digits is 48 (1 not allowed)

Count the numbers in which the product of the digits is 48 (1 not allowed as digit).

Proposed solution: I addressed the problem by brute force, first noticing that
$$48=2^4\times 3.$$

Then I found the possible groups of digits to be permuted for each number: (1) 2,2,2,2,3; (2) 2,2,2,6; (3) 4,2,2,3; (4) 4,2,6; (5) 4,4,3; (6) 8,2,3; (7) 8,6.

By adding the permutations with repetition for each case, I've found 38 possible numbers. Which agrees with the answer provided in the source.

My problem: I found this solution is not elegant and I'm wondering if there is a better way to address problems like these. By brute force it is easy to undercount terms (as I did as first solving the problem). Any help?

答:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $1\le\text{n}\le10^7$.

I wrote and ran some Mathematica-code:

```<code>In[1]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[Product[
       Part[IntegerDigits[n], k], {k, 1, 1 + Floor[Log10[n]]}] == 48 &&
      MemberQ[IntegerDigits[n], 1] == False], n, Nothing], {n, 1,
   10^7}] //. {} -> Nothing
</code>```

Running the code gives:

```<code>Out[1]={68, 86, 238, 246, 264, 283, 328, 344, 382, 426, 434, 443, 462, 624,
642, 823, 832, 2226, 2234, 2243, 2262, 2324, 2342, 2423, 2432, 2622,
3224, 3242, 3422, 4223, 4232, 4322, 6222, 22223, 22232, 22322, 23222,
32222}
</code>```

So, the sum of the numbers is given by $179141$. I found that using:

```<code>In[2]:=Clear["Global`*"];
Total[ParallelTable[
   If[TrueQ[
     Product[Part[IntegerDigits[n], k], {k, 1,
         1 + Floor[Log10[n]]}] == 48 &&
      MemberQ[IntegerDigits[n], 1] == False], n, Nothing], {n, 1,
    10^7}] //. {} -> Nothing]

Out[2]=179141
</code>```

Body:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $1\le\text{n}\le10^7$.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[Product[
       Part[IntegerDigits[n], k], {k, 1, 1 + Floor[Log10[n]]}] == 48 &&
      MemberQ[IntegerDigits[n], 1] == False], n, Nothing], {n, 1,
   10^7}] //. {} -> Nothing

Running the code gives:

Out[1]={68, 86, 238, 246, 264, 283, 328, 344, 382, 426, 434, 443, 462, 624,
642, 823, 832, 2226, 2234, 2243, 2262, 2324, 2342, 2423, 2432, 2622,
3224, 3242, 3422, 4223, 4232, 4322, 6222, 22223, 22232, 22322, 23222,
32222}

So, the sum of the numbers is given by $179141$. I found that using:

In[2]:=Clear["Global`*"];
Total[ParallelTable[
   If[TrueQ[
     Product[Part[IntegerDigits[n], k], {k, 1,
         1 + Floor[Log10[n]]}] == 48 &&
      MemberQ[IntegerDigits[n], 1] == False], n, Nothing], {n, 1,
    10^7}] //. {} -> Nothing]

Out[2]=179141