问:
Evans, chapter 6 exercise 2
Consider the following question

I know I'm supposed to show boundedness and coercivity but haven't made much progress.
Do I need to integrate by parts? Also I'm really interested in seeing the coercivity step

The norm I am considering is the $H^1_{0}$ norm

答:
To prove that is bounded is straightforward as soon as the coefficients are bounded. For the coercivity, I guess you are assuming that the matrix $a$ of the coefficients is uniformly elliptic. Then you just have to use Poincaré inequality.

I just focus on coercivity. Call $A$ the uniformly elliptic matrix whose entries are the $a_{ij}$. The fact that $A$ is uniformly elliptic means that there exits $\lambda >0$ such that $A v\cdot v\geq \lambda |v|^2$ for every $v\in\mathbb{R}^n$. You have
$$
\int_\Omega A \nabla u\cdot \nabla u +c u^2 \geq \int_\Omega \lambda |\nabla u|^2 +c u^2 \geq \int_\Omega \frac{\lambda}{2} |\nabla u|^2 + \frac{\lambda}{2}|\nabla u|^2 +c u^2\;.
$$
Therefore, by the Poincaré inequality, i.e.,
$$
\int_\Omega |\nabla u|^2 \geq \frac{1}{c_P^2}\int_\Omega |u|^2 \, ,$$
applied to one of the $\lambda/2$ terms, you have
$$
\int_\Omega \frac{\lambda}{2} |\nabla u|^2 + \frac{\lambda}{2}|\nabla u|^2 +c u^2
\geq
\int_\Omega \frac{\lambda}{2} |\nabla u|^2 + (c+\frac{\lambda}{2 c_P^2})|u|^2\; .$$
Therefore it is sufficient to take the $\alpha$ you were looking for as
$$
\alpha:=\frac{\lambda}{2 c_P^2}
$$
because the $|u|^2$ term is nonegative as soon as $c+\frac{\lambda}{2 c_P^2}\geq 0$, i.e., as soon as $c\geq =-\frac{\lambda}{2 c_P^2}=:- \alpha$.

Title: Evans, chapter 6 exercise 2


问:
True or false: There is only one $n \times n$ matrix such that $(I_n-A)^{-1}=I_n-\frac{1}{n-1}A$
There is only one $n \times n$ matrix that suits:

$(I_n-A)^{-1}=I_n-\frac{1}{n-1}A$

We know that:$(I_n-A)^{-1}\neq I_n-A^{-1} $ but I dont know what to do next:

Thanks for any tip.

答: Since $I_n=(I_n-A)(I_n-\tfrac{1}{n-1}A)$ iff $(A-\tfrac12nI_n)^2=\tfrac14n^2I_n$, any square root $X$ of $I_n$ provides a solution $A=\tfrac12n(I_n+X)$.

ContentLicense: CC BY-SA 4.0


问:
On unions of independent events
If the events $\{ E_{\alpha}, \alpha\in A\}$ are independent, then so are the events $\{F_\alpha,\alpha\in A\}$, where each $F_\alpha$ may be $E_\alpha$ or $E_\alpha^c$; also if $\{A_\beta, \beta\in B \}$, where $B$ is an arbitrary index set, is a collection of disjoint countable subsets of $A$, then the events

$$\bigcup_{\alpha\in A_{\beta}}E_\alpha, \quad \beta\in B,$$

are independent.

Presumably if these were finite unions then independence should be easy to show I'm guessing. But here based on the notation these seem to be possibly uncountable unions.

答:
A set of events $\mathcal A$ is independent iff for any finite set $\{A_i\in\mathcal A:i=1\dots n\}$ we have $\mathbb P\left( \bigcap_{i=1}^n A_i\right) = \prod_{i=1}^n \mathbb P(A_i)$.

For unions We can say that if $\mathcal A$ are independent then $\mathbb P\left( \bigcup_{i=1}^n A_i\right) = 1-\prod_{i=1}^n(1- \mathbb P(A_i))$.

So we can assume $B$ is finite, which helps.

Set $$F_\beta = \bigcup_{\alpha\in A_\beta} E_\alpha$$

Any event with probability one or 0 is independent of everything
so we can assume without loss of generality that $\mathbb P\left(F_\beta\right)<1$.  By the second Borel-Cantelli lemma we must have $\sum_{\alpha\in A_\beta}\mathbb P (E_\alpha) <\infty$.

So choose some $\varepsilon>0$ and for every $\beta\in B$ let $A^\varepsilon_\beta\subset A_\beta$ be a finite set of events such that
$$\sum_{\alpha\in A_\beta}\mathbb P (E_\alpha) \leq\varepsilon + \sum_{\alpha\in A_\beta^\varepsilon}\mathbb P (E_\alpha).$$

Now the events $$F^\varepsilon_\beta = \bigcup_{\alpha\in A_\beta^\varepsilon} E_\alpha$$ are independent by messing about with finite products. We can arrange things so that $F^\varepsilon_\beta \subset F^\delta_\beta$ whenever $\delta >\varepsilon$ then we have

$$\begin{array}{rl}
\mathbb P\left(\bigcap_{\beta\in B} F_\beta\right) &= \lim_{\varepsilon\to 0}
\mathbb P\left(\bigcap_{\beta\in B} F_\beta^\varepsilon \right) \\
&= \lim_{\varepsilon\to 0} \prod_{\beta\in B} \mathbb P(F^\varepsilon_\beta) \\
& =  \prod_{\beta\in B} \mathbb P(F^\beta) \end{array}$$

Then as $B$ is an arbitrary finite subset we are done.

ContentLicense: CC BY-SA 3.0

Title: On unions of independent events


问:
Good book for analysis in particular sequences and series
I’m currently studying pure mathematics and I’m finding the analysis sections quite difficult to digest from the course text books. Could anybody recommend some good textbooks that explain sequence and series well in particular finding if they are convergent or divergent.

Thanks

答:
Problems in Mathematical Analysis I: Real Numbers, Sequences and Series.
It's a brilliant book. Basically focused on problem solving skills. If you read this book you will have good skills in problem solving in real analysis.</p>


问:
Volume of the body
Find the volume of body bounded with:$(z-2)^2\le x^2+y^2, z\ge 1,z-14+x^2+y^2\le 0$.
So we should have three volumes here. For first one(inside the paraboloid and on the cone) we have $V_1=\int\limits_{0}^{2\pi}d\phi\int\limits_{0}^{1}rdr\int\limits_{2-r}^{2+r}dz$. And I am not sure for the other two volumes, one should be from plane $z=1$ to paraboloid(with projection in OXY plane $9\le x^2+y^2\le 13$). But what is the third one(I mean I see one from the left, but it should be symmetric to this one).

答:
The integral can be split between $1 \leq z \leq 2$ and $2 \leq z \leq 5$

$\displaystyle \int_0^{2\pi} \int_1^2 \int_{2-z}^{\sqrt{14-z}} r \ dr \ dz \ d\theta \ + \int_0^{2\pi} \int_2^5 \int_{z-2}^{\sqrt{14-z}} r \ dr \ dz \ d\theta$

Title: Volume of the body


问:
Given a field $K$ of characteristic $p>0$ and $f \in K[x]$ an irreducible polynomial, find the separable closure of $K$ in $L=K[x]/(f).$
I have shown that we can find some irreducible, separable polynomial $g \in K[x]$ such that $f(x)=g(x^{p^n})$ for some $n \in \mathbb{Z}_{\geq 0}.$ What I have so far towards the problem in the title is the following:

Let $\alpha \in \overline{K}$ be a zero of $f \in K[x]$, so $L=K[x]/(f)=K(\alpha)$. By the first paragraph, $f(x)=g(x^{p^n})$ so $\alpha^{p^n}$ is a zero of the separable polynomial $g \in K[x]$. Therefore, $K(\alpha^{p^n})$ is a separable extension of $K,$ and $K(\alpha^{p^n}) \subset K(\alpha) = L$ so $K(\alpha^{p^n})$ is contained in $K_s,$ the separable closure of $K$ in $L$. I'm not sure what else I can say, or which directions to look in, so any hints would be useful.

答:
Hint: Working with what you have already shown, an important observation is the following: for any field, $K$, of characteristic $p>0$ and any element, $\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \al$, that is algebraic over $K$, we have $(K(\al)/K)_s=(K(\al^p)/K)_s$ (here $(E/K)_s$ means the separable closure of $K$ in $E$, i.e. the set/ field of all elements of $E$ that are separable over $K$). You should try proving this on your own, but an explanation is given below if needed.

Full explanation:

 The reason the above statement works is due to a criterion for separability: that an element, $\al$, algebraic over some field, $K$ with characteristic $p>0$, is separable over $K$ iff $K(\al)=K(\al^p)$ (this even holds if $\al$ is not assumed to be algebraic over $K$). On the one hand it is obvious that $(K(\al)/K)_s\geq (K(\al^p)/K)_s$  (as $K(\al)\geq K(\al^p)$).

 On the other hand, for any $\be\in K(\al)-K(\al^p)$, we have $\be^p\in K(\al^p)$ (you can see this by writing $\be=\sum_i k_i\al^i$ for some $k_i$s in $K$ and raising both sides to the $p$th power) and $\be\not\in K(\al^p)$, so $K(\be)\neq K(\be^p)$, i.e. $\be$ is inseparable over $K$. This means no element of $K(\al)-K(\al^p)$ is separable over $K$, so any element in $(K(\al)/K)_s$ must also lie in $(K(\al^p)/K)_s$, hence $(K(\al)/K)_s\leq (K(\al^p)/K)_s$. It follows that $(K(\al)/K)_s=(K(\al^p)/K)_s$.

 It now follows pretty easily that the separable closure you are looking for is just $K(\al^{p^n})$, as $\al^{p^n}$ is separable over $K$, although in the same vein you could also say that the separable closure is $K(\al^{p^{n+1}})$ or $K(\al^{p^{n+2}})$, etc.


问:
Disclude a number in a sequence so that the greatest divisor common of all the others at most. Find that GCD and the position of that discluded one.

Problem. Disclude a number in a sequence so that the greatest divisor common of all the others at most. By programming, find that GCD and the position of that discluded one.

My code in the environment C++ is:

```<code>#include <iostream>
using namespace std;
int gcd(int x, int y) {
    while (y) {
        int r=x%y;
        x=y;
        y=r;
    }
    return x;
}
int pcd(int arr[], int u, int w) {
    int temp=arr[u];
    arr[u]=0;
    int result=0;
    for (int i=1; i<=w; i++) {
        result=gcd(result, arr[i]);
    }
    arr[u]=temp;
    return result;
}
int main() {
    int n;
    cin>>n;
    int a[100000];
    for (int i=1; i<=n; i++) {
        cin>>a[i];
    }
    int t=0;
    int v;
    for (int j=1; j<=n; j++) {
        if (pcd(a, j, n)>t) {
            t=pcd(a, j, n);
            v=j;
        }
    }
    cout<<v<<" "<<pcd(a, v, n);
}
</code>```

This is buffalo-way, I substituted the discluded one by $0,$ then replace $0$ by that discluded number and compare all the GCDs we have, that consumed time. How can we speed up the process ? I need to the help, thanks a real lot !

答:
Think of the FFT.

The numbers are $a_k$

$b_k=gcd(a_{2k-1},a_{2k})$.

Solve the problem for the $n/2$ numbers $b_k$, get $c_k$.

Do gcd $(c_k,a_{2k_-1})$ and gcd $(c_k,a_{2k})$.


问:
Find $a,b,c \in \mathbb {Q}$
Find $a,b,c \in \mathbb {Q}$ such that:
$\left\{\begin{array}{rl} x^3&\in \mathbb Q \\ x&\notin \mathbb{Q}\\ ax^2+bx+c &=0\end{array}\right.$

I tried Vieta's formulas, but seem like it didn't help.

I think $a=b=c=0$ is only solution.

答:
$$a^2x^3=-ax(bx+c)=-b(ax^2)-cax=b(bx+c)-cax=x(b^2-ac)+bc$$

$$x^3=\frac{x(b^2-ac)+bc}{a^2}$$ will be $\in Q$ iff $x$ $\in Q$ unless $b^2-ac=0$

If $b^2=ac, \frac cb=\frac ba=r$(say) $\implies b=ar,c=br=ar^2$

$\implies  ax^2+bx+c=a(x^2+rx+r^2)=0\implies x^2+rx+r^2=0$ as $a\ne0$

$\implies x=r\frac{-1\pm\sqrt3i}2$ which is not $\in Q$ for $r\in Q$

But $x^3=r^3$ which  $\in Q$ for $r\in Q$

Title: Find $a,b,c \in \mathbb {Q}$


问:
Finding the integral of $1/(x^2 +2x +5)$
I am working through a Pure Maths book as a hobby. I am struggling with the second part of this problem:

Differentiate $\arctan(x + a)$ with respect to x and use your result to find

a) $\displaystyle\int \frac{dx}{x^2 + 4x +5}$

b) $\displaystyle\int \frac{dx}{x^2 + 2x +5}$

I differentiated $\arctan(x + a)$ with respect to x:

$$\frac {dy}{dx} \arctan(x + a) = \frac{1}{(x + a)^2 +1}$$

From this I was able to say:

$$\int \frac{dx}{x^2 + 4x +5} = \int \frac{dx}{(x + 2)^2 +1}
\implies \int \frac{dx}{x^2 + 4x +5} = \arctan(x + 2) + c$$

But I cannot apply this technique to b) because $x^2 + 2x + 5$ won't factorise.

答: You can do it as follows: $\displaystyle x^2+2x+5=(x+1)^2+4=4\left(\left(\frac{x+1}2\right)^2+1\right)$, and therefore\begin{align}\int\frac{\mathrm dx}{x^2+2x+5}&=\frac14\int\frac{\mathrm dx}{\left(\frac{x+1}2\right)^2+1}\\&=\frac14\times2\arctan\left(\frac{x+1}2\right)+C\\&=\frac12\arctan\left(\frac{x+1}2\right)+C.\end{align}


问:
Simple question about Power/Voronoi diagrams on the sphere
I am reading "Laguerre Voronoi Diagram on the Sphere" by Kokichi Sugihara (I have provided a link to the paper at the bottom of this post). In this paper the author defines a notion of Laguerre Voronoi diagram on the sphere (also called a power diagram). One can define a power diagram in $\mathbb{R}^d$ by starting with a set of $n$ points $p_i$ equipped with weights $w_i$, and decomposing $\mathbb{R}^d$ into cells $$W_i=\{x\in \mathbb{R}^d | d(x,p_i)-w_i < d(x,p_j)-w_j, \forall j\neq i\}$$ (for those familiar with power diagrams there is another way to phrase the weights in terms of radii of circles centered at points $p_i$, where $w_i=r_i^2$). The intuition is that (like Voronoi diagrams) the cell $W_i$ corresponds to all $x\in \mathbb{R}^d$ such that $x$ is closer to $p_i$ than any other $p_j$ in our set of points, where closeness is defined with respect to $d(x,p_i)-w_i$. In particular notice that this is a generalization of Voronoi diagrams, where if all $w_i=0$ we recover the Voronoi decomposition of $\mathbb{R}^d$.

Sugihara defines a version of this construction on the unit 2-sphere, where now proximity between a point $p_i$ in our defining set $\{p_i\}_{i=1}^n$ and an arbitrary point $p\in S^2$ is given by $d_L(p_i,p)$, which is defined to be proportional to $\cos(d(p_i,p))$ where $d$ is the geodesic distance on the surface of the sphere. I am suppressing the dependence of $d_L$ on a set of weights, but for the purposes of my question, we can assume all weights are equal (say to 0), in which case one would think we would recover the Voronoi diagram. Once again, these diagrams are defined by sets $$W_i=\{p\in S^2|d_L(p_i,p)<d_L(p_j,p)\}.$$

Here is my question: Since $d_L(p_i,p)$ is proportional with $\cos(d(p_i,p))$, it is inversely proportional with $d(p_i,p)$, since the maximum distance between two points on the unit sphere is $\pi$, and cosine is monotonically decreasing in $[0,\pi]$. But then since $d_L$ decreases as distance increases, and the sets $W_i$ are defined by minimizing $d_L$, $W_i$ should consist of all $p\in S^2$ that are further away from $p_i$ than all other $p_j$ in terms of the geodesic distance on the sphere. This seems to be a very unintuitive definition, and at odds with Euclidean Voronoi/Power diagrams, where $W_i$ consists of points which are close to $p_i$.

Am I missing something? Why would one want to define a version of Voronoi diagrams by points that are further away, rather than closer? Or have I made an error in reasoning somewhere?

Link to the paper in question: https://www.heldermann-verlag.de/jgg/jgg06/jgg0606.pdf

答:
It appears there is an error in the sign of the inequality in equation (14). Sugihara notes that the "Laguerre Proximity" (this expression involving the cosine) is inversely related to distance; it decreases as we move away from point / circle.

In each of the results(e.g., the definition of the Laguerre bisector in (8), Theorem 1), the Laguerre proximity is appears in equality expressions. Sugihara is analyzing the boundaries between Laguerre Voronoi cells. Everything is ok except the sign of the inequality in the definition of the Laguerre Voronoi definition is backwards: to be "nearer" a circle  means to that a larger (not smaller) Laguerre proximity.


问:
No of possible binary sequences with at most 2 consecutive zeroes
The problem:

How many such sequences of ten $1's$ and ten $0's$ are possible such that they don't contain three or more consecutive zeroes? (Eg. $00101001001101110011$)

I know there are $20!/10!10!$ total sequences. How do I filter out the ones which have more than two consecutive zeroes somewhere?

答:
The answer can be written as

$${11\choose0}{11\choose10}+{11\choose1}{10\choose8}+{11\choose2}{9\choose6}+{11\choose3}{8\choose4}+{11\choose4}{7\choose2}+{11\choose5}{6\choose0}$$

That is, imagine you've got a string of $10$ ones and you now need to insert $10$ zeroes in doublets and singlets. There are $11$ places where the doublets and singlets can go: $9$ places between ones and $2$ places at the far left and far right.  If you first insert $k$ doublets, you'll be left with $10-2k$ singlets to insert, with $11-k$ places to insert them, for ${11\choose k}{11-k\choose10-2k}$ choices in all. The number of doublets can range from $0$ to $5$, hence the sum.

Remark: The expression calculates out to $24{,}068$, which is is divisible by $547$, as the OP says (in a comment below true blue anil's answer) must be the case.


答:
But I cannot apply this technique to b) because $x^2+2x+5$ won't
factorise.

I think you mean, ...because $x^2+2x+5$ doesn't have remainder $1$ when you've completed the square. Which is true. However, you can get $$\frac{1}{x^2+2x+5}$$ into the form

$$ \frac{1}{(x + a)^2 +1}$$

by some basic manipulation:

\begin{align} \frac{1}{(x+1)^2 + 4} \\
\\
= \frac{1}{4\left(\frac{1}{4}(x+1)^2 + 1\right)} \\
\\
= \frac{1}{4} \cdot\frac{1}{\frac{1}{4}(x+1)^2 + 1} \\
\\
= \frac{1}{4} \cdot\frac{1}{\left(\frac{1}{2}\right)^2(x+1)^2 + 1} \\
\\
= \frac{1}{4} \cdot\frac{1}{\left(\frac{1}{2}\left(x+1\right)\right)^2 + 1} \\
\\
= \frac{1}{4} \cdot\frac{1}{\left(\frac{1}{2}x+\frac{1}{2}\right)^2 + 1}\\
\\
\end{align}

Now substitute $u = \frac{1}{2}x\ $ and apply the reverse chain rule (i.e. $ dx = 2du$), and then you can apply your formula.

Title: Finding the integral of $1/(x^2 +2x +5)$

Body:
I am working through a Pure Maths book as a hobby. I am struggling with the second part of this problem:

Differentiate $\arctan(x + a)$ with respect to x and use your result to find

a) $\displaystyle\int \frac{dx}{x^2 + 4x +5}$

b) $\displaystyle\int \frac{dx}{x^2 + 2x +5}$

I differentiated $\arctan(x + a)$ with respect to x:

$$\frac {dy}{dx} \arctan(x + a) = \frac{1}{(x + a)^2 +1}$$

From this I was able to say:

$$\int \frac{dx}{x^2 + 4x +5} = \int \frac{dx}{(x + 2)^2 +1}
\implies \int \frac{dx}{x^2 + 4x +5} = \arctan(x + 2) + c$$

But I cannot apply this technique to b) because $x^2 + 2x + 5$ won't factorise.


答: Hint: A nice, simple rational number to start your investigations with is $1$. Maybe for simplicity, try assuming that $x^3=1$. Maybe one of $a$, $b$, or $c$ could be $1$. Experiment and see what happens!

Title: Find $a,b,c \in \mathbb {Q}$

Body:
Find $a,b,c \in \mathbb {Q}$ such that:
$\left\{\begin{array}{rl} x^3&\in \mathbb Q \\ x&\notin \mathbb{Q}\\ ax^2+bx+c &=0\end{array}\right.$

I tried Vieta's formulas, but seem like it didn't help.

I think $a=b=c=0$ is only solution.


答:
HINT:

Recall that
$$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}$$

If you need any more help please don't hesitate to ask.


问:
If $f:(0,1)\times (0,1)\to \mathbb R$ is continuous and integrable, then so is $F(x)=\int_0^1 f(x,t)dt$
Let $f:(0,1)\times (0,1)\to \mathbb R$ continuous, and $f(x,\cdot )\in L^1$ for all $x$. Prove or disprove that $$F(x):=\int_0^1 f(x,t)dt$$
is continuous on $(0,1)$.

So, if $f$ is extendable by continuity on $[0,1]\times [0,1]$, then $f$ is uniformly continuous, and I could solve the problem. But How can I do if $f$ is not extendable by continuity ?

So $$|F(x+h)-F(x)|\leq \int_0^1 |f(x+h,t)-f(t)|dt.$$
Let $\varepsilon >0$. Thereis $\delta >0$ s.t. $|(x,t)-(x+h,t)|<\delta \implies |f(x+h,t)-f(x,t)|$, but here, $\delta $ depend on $t$. Is there a way to avoid this problem ?

答:
The answer is negative.

Consider $\frac{2t+2x-1}{(2x-1)^2} I_{x+ t < \frac12}.$


答:
Complete the square in the denominator:
$$\int \frac{\mathrm dx}{x^2 + 4x +5}=\int \frac{\mathrm d(x+1)}{(x+1)^2 + 4}$$
and use the basic formula
$$\int\!\frac{\mathrm du}{u^2+a^2}=\frac1a\,\arctan \frac ua.$$


问:
Approximating a continuous function uniformly when f is continuous on $[0,\infty)$ and $f(x) \to a$ as $x\to \infty$
This question was asked in my assignment of real analysis and I am having a trouble attempting it.

If  f is continuous on $[0,\infty)$ and if $f(x)\to a$ as $x \to \infty$ then f can be uniformly approximated on $[0,\infty)$ by a function g of the form g(x)=$p(e^{-x})$ , where p is a polynomial.

Wierestrauss approximation theorem comes to mind but $[0,\infty) $ is not compact. and I am at loss of ideas on how this question should be approached. I can say that range of  f is connected but applying Weirestrauss theorem seems impossible. Also, Joining of intervals $[0,x]\cup[x,y]$... seems not correct approach as I will still not get $[0,\infty)$.

It is my humble request to you to some time on this?

答: Let $g(x)=f(- \ln x)$ if $x>0$ and $a$ if $x=0$. Check that $g$ is continuous on $[0,1]$ and apply Weierstrass Approximation Theorem to $g$.


问:
Any measurable function $f:X \to \mathbb R, \Sigma=\{X,\emptyset\}$(where $X$ is a non empty set) must be constant.
Any measurable function $f:X \to \mathbb R, \Sigma=\{X,\emptyset\}$(where $X$ is a non empty set, $\Sigma$ is a sigma algebra on $X$) must be constant.

My attempt:-
We know that $f:X \to \mathbb R$ is measurable.  Assume on contrary that $f$ is not constant. I could write the proof till here. I don't know how to proceed? Please help me.

答: Assume that $f$ takes at least two different values $a<b$, $f(x)=a,f(y)=b$. Then $A=f^{-1}((a,+\infty))$ must be in the $\sigma$-algebra $\Sigma$ because $f$ is measurable. But this means that $A=∅$ or $A=X$. We definitely cannot have $A=∅$ because $y \in A$. Thus, $A= X$. But this would imply $f(x) = a \in (a,+\infty)$, which is a contradiction.


问:
How to calculate $\lim_{t\to 0^+} \frac{1}{\sqrt{t}}\int_{0}^{+\infty}\exp\frac{x+1-e^x}{t}dx$
How to calculate $$\lim_{t\to 0^+} \frac{1}{\sqrt{t}}\int_{0}^{+\infty}\exp\frac{x+1-e^x}{t}dx$$

I have known that it has connection with $$\int_{0}^{+\infty}e^{-x^2}dx=\frac{\sqrt{\pi}}{2}$$
and I try to let $u=\frac{1}{\sqrt{t}}$ but then I can't find a way to continue...


答: Take a primitive cube root of $1$. Then $x^3 = 1$, $x\notin\mathbb{R}$ and $x^2+x+1 = 0$.</p>

Title: Find $a,b,c \in \mathbb {Q}$


问:
Proving $||v + u|| = ||v||+||u|| \iff \exists \alpha > 0: v = \alpha u$ on a real vector space
I was trying to prove the following:

Let $V$ be an inner-product space over $\mathbb R$, and let $||v||=\sqrt{\left<v,v \right>}$ be a norm on $V$. Prove that: $||v + u|| = ||v||+||u|| \iff \exists \alpha >  0: v = \alpha u$

I was easily able to show that $\exists \alpha >  0: v = \alpha u \Rightarrow ||v + u|| = ||v||+||u||$, but I'm having some trouble with proving the opposite, how can I do this? Can someone give me a hint?

P.s: I'm aware the there is another post concerning this topic, but that other post assumes that $V$ is a complex vector space. This post is only concerning real vector spaces.

edit:

I did the following:
$||u+v||=||u||+||v|| \iff \sqrt{+2+<v,v>}=||u||+||v|| \iff ||u||^2 + 2 + ||v||^2 = ||u||^2+||v||^2+2||u||||v|| \iff =||u||||v||$ which means $ \ge 0$. However, I'm not sure what to do now to prove that this implies that $u=\alpha v$, for $\alpha>0$.

edit:
I was able to prove it "going back" on the proof for Cauchy Schwarz inequality. Thank you.

答: The claim is false in, for example, $L^{1}$. That is if u and v are non-negative and disjoint support.


问:
Proof of $rms(x)^2=avg(x)^2+std(x)^2$
I am trying to derive the proof of $rms(x)^2=avg(x)^2+std(x)^2$ but I am stuck at one line.

$std(x)^2= (1/n)||x-(1^Tx/n)1||^2=(1/n)[x^Tx-2x^T(1^Tx/n)1+n(1^Tx/n)^2]$

So from this, how can I proceed? I know that:

$std(x)^2=rms(x)^2+(1/n)[-2x^T(1^Tx/n)1+n(1^Tx/n)^2]$ is left for me. The book that I use used some sort of property I guess to get the equality:

$-2x^T(1^Tx/n)1=(2/n)(1^Tx)^2$

I know that we can move around $n$ because it's a constant, but I don't understand how $x^T1^Tx1=(1^Tx)^2$.

答:
Okay, I solved it. We want to prove that $x^T1^Tx1=(1^Tx)^2$

1:

Let $x=\begin{bmatrix}
           x_{1} \\
           x_{2} \\
           \vdots \\
           x_{n}
         \end{bmatrix}
 $  and we know that $1^Tx= \sum x_i$.

Then $1^Tx1=\begin{bmatrix}
           \sum x_i \\
           \sum x_i \\
           \vdots \\
           \sum x_i
         \end{bmatrix}$.

Then $x^T1^Tx1= \begin{bmatrix}
           x_1, x_2
           \dots
           x_n
         \end{bmatrix} \begin{bmatrix}
           \sum x_i \\
           \sum x_i \\
           \vdots \\
           \sum x_i
         \end{bmatrix}=$

$x_1\sum x_i+x_2\sum x_i\dots+x_n\sum x_i=\sum x_i\sum x_i=(\sum x_i)^2=(1^Tx)^2$

If there's a more cleaner and more intuitive method I'd glad to hear it!

Title: Proof of $rms(x)^2=avg(x)^2+std(x)^2$


问:
Side limits of ${\displaystyle \lim_{x\to0}\frac{|\sin\left(\frac{1}{x}\right)|}{1 + 2^{1/x}}}$
I need to find the side limits of this function if they exist. I've tried dividing both denominator and numerator with $\frac1x$, but that didn't help either. According to the book, the left-side limit doesn't exist, and the right-side limit is $1$. How could I solve this problem?

$${\displaystyle \lim_{x\to0}\frac{|\sin\left(\frac{1}{x}\right)|}{1 + 2^{1/x}}}$$

答:
Hint:

in both cases, set $u=\dfrac1x$.

For the right-hand side, use  that $|\sin u|\le 1$.

For the left-hand side, use that if there were a limit, every subsequence would have the same limit and use that the denominator tends to $1$. Can you prove that $\sin\frac1x$ has no limit at $0$?


问:
Finding probability $P(X<Y)$
How can I find this probability $P(X<Y)$ ? knowing that X and Y are independent random variables.</p>

答:
Two of the previous answers are specific for continuous random variables. For discrete independent random variables $X$ and $Y$ the probability of $X > Y$ can be written as
$$\mathbb{P}(X < Y) = \sum_{x \in R_{X}} \, \sum_{y \in R_{Y}\, | \, y > x} \mathbb{P}(x,y)$$
or can be written as
$$\mathbb{P}(X < Y) = \sum_{y \in R_{Y}} \, \sum_{x \in R_{X}\, | \, x < y} \mathbb{P}(x,y)$$
where in each case the subscript on the second is only for $x < y$ and where

$R_{X}$ is the range of random variable $X$

$R_{Y}$ is the range of random variable $Y$

$\mathbb{P}(x,y)$ is the joint probability of $X$ and $Y$

Since $X$ and $Y$ are independent random variables
$$\mathbb{P}(x,y) = \mathbb{P}_{X}(x) \cdot \mathbb{P}_{Y}(y)$$
where $\mathbb{P}_{X}$ is the marginal probability of $X$ and $\mathbb{P}_{Y}$ is the marginal probability of $Y$.

Thus, the summations are over each allowed pair of $X$ and $Y$

$$\mathbb{P} = \sum_{x \in R_{X}} \, \sum_{y \in R_{Y}\, | \, y > x} \mathbb{P}_{X}(x) \cdot \mathbb{P}_{Y}(y)$$
or
$$\mathbb{P} = \sum_{y \in R_{Y}} \, \sum_{x \in R_{X}\, | \, x < y} \mathbb{P}_{X}(x) \cdot \mathbb{P}_{Y}(y)$$
In either case, the summation is over all pairs of $X$ and $Y$ such that $X < Y$.

Title: Finding probability $P(X<Y)$


问:
What is the domain of this function in 3 variables?
Consider the function
$$f(x, y, z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+xyz.$$
Which is the domain of this function?

Certainly, we have to ask $(x, y, z)\neq(0, 0, 0)$ but also the points of the type
$$ (0, y, z), \ (x, 0, z), \ (x, y, 0), \ (0, 0, z), \ (x, 0, 0), \ (0, y,0)$$
are not allowed. How to write in an "easy" way?

I would say
$$D(f)=\{(x, y, z)\in\mathbb{R}^3 : (x, y, z)\neq (0, 0, 0), ..\}$$
but I don't know how to continue. Could anyone please help me?

Finally, all these points to be excluded, consitute a straight line, a plane or something else?

Thank you in advance!

答:
For your first question consider labeling the domain:

$D(f)=\{(x, y, z)\in\mathbb{R}^3 : x,y,z \neq 0\}$.

The following is a response to your second question from the bottom of your post:

As $\mathbb{R^{3}}$ represents all of $3$-dimensional space, $D(f)$ is $\mathbb{R^{3}}$ without the points previously mentioned. The origin $(0,0,0)$ is removed, WLOG points of the form $(0,0,z)$ lie on one of the $3$ axes, so each axis is removed. WLOG points of the form $(0,y,z)$ represent an arbitrary point on the $xy$ plane. Altogether, we end up with $\mathbb{R^{3}}$ without the $xy$,$xz$,and $yz$ planes.


问:
what will be the value of $x$ and $y$ and why?
Compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies above the $xy-plane$ and is cut off by the sphere $x^2+y^2+z^2=2ax$.

My attempt : According to given question , we can take

$$r(u, \theta) = u\cos \theta \hat{i} + u\sin\theta \hat{j} + u\hat{k}$$

$$r_u = \cos\theta\hat{i} + \sin\theta\hat{j} + \hat{k}$$
$$r_\theta = -u\sin\theta\hat{i} + u\cos\theta\hat{j}$$

Therefore $$\|r_u\times r_\theta\| = \sqrt2u$$

It is given in question that  area cutoff by the sphere, so we have
$$x^2 + y^2 + z^2 \leq 2ax$$
$$\implies 2z^2 \leq 2ax $$
$$\implies u \leq a\cos\theta$$

$$Area = \int \int_{S} dS=\int_{x}^{y} \int_{0}^{acos \theta } \sqrt 2 u du d \theta$$

My confusion : How to find   the  $x$ and $y$ in Area.

Im finding difficulty  in choosing the  $x$ and $ y$ .  what will be the value of  $x$ and $y$ and why ?

答:
The equation of the sphere can be re-written as $(x-a)^2 + y^2 + z^2 = a^2$.

It is a sphere centered at $(a, 0, 0)$, with radius $a$.

In $XY$ plane (at $z = 0$), you get the circle with the equation,

$x^2 + y^2 = 2 a x \implies r = 2a \cos \theta$. Hence, $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

So your integral to find surface area should be,

$\displaystyle \int_{-\pi/2}^{\pi/2} \int_{0}^{a \cos \theta } \sqrt2 \ u \ du \ d\theta$

Edit: Added a diagram as requested that explains limits of $\theta$ for the circle, with center on $x$-axis and touching the origin.


问:
Non-negative random variables (ask about the definition)
As I understand, a non negative random varialbe which takes non-negative values. But I've seen in that in many books the definition for non-negative random varialbles is a random variable $ X $ such that

$ \mathbb{P}\left(X<0\right)=0 $.

How come those two definitions equivalent? $ \mathbb{P}\left(X<0\right)=0 $ does not imply that $ X(\omega) \geq 0 $ for any $ \omega \in \varOmega $.

What is the acceptable definition between the two that  I mentioned?

答:
You probably have heard about Murphy's law. Aside all the rhetoric and myths around it, the Murphy's law actually is quite important. An event can be possible or impossible. Probability is only defined over possible event. A possible event can be improbable, i.e., its probability can be zero, but it doesn't mean that it cannot occur.

But as you mentioned, it is customary to assign zero probability to impossible events. Even though it is ultimately a bad practice, it usually works. The good practice however, is to always make a clear distinction between impossible and improbable events.


答:
Hints:

Do you know about cube roots of unity? There are two other cube roots of $1$ other than $1$ which are imaginary. Those are $e^{i2\pi/3}$ and $e^{i4 \pi/3}$. You have sufficient conditions now. Good luck.

Title: Find $a,b,c \in \mathbb {Q}$


问:
How do you prove$ \frac{1-\cos x}{\sin x}=\tan\left(\frac{x}{2}\right)$?
Now I've seen this before but I forgot how to do it. (by the way, this is without the use of t-formula). It gets difficult when the angles are not the same but I split the fraction on the LHS up and got and stuff in terms of $\csc{x}$ and $\cot{x}$. I then used the identity $1 + cot^2{x} = csc^2{x}$ but am still stuck on reaching something in terms of $\frac{x}{2}$.Is there any neat efficient solution?

答:
You can use the Weierstraß substitution: setting $t=\tan\frac x2$, you have
$$\cos x=\frac{1-t^2}{1+t^2},\qquad \sin x =\frac{2t}{1+t^2}.$$


问:
Obtain the conditional distribution of $X$ given $X^2=t$
Consider the model $X \sim N(\theta, 1)$ for $\theta \in \mathbb R$. Obtain the conditional distribution of $X$ given $X^2=t$ for $t\ge 0$.

Hint: This is a discrete distribution.

Current work:
$f(X=x|X^2=t)=\displaystyle \frac{f(X=x, X^2=t)}{f(X^2=t)}=\frac{f(X=x, X^2=t)}{\frac{1}{2\sqrt{2\pi t}}(e^{(\sqrt t-\theta)^2/2}+e^{(\sqrt t + \theta)^2/2})}=\frac{f(X=\sqrt t)}{\frac{1}{2\sqrt{2\pi t}}(e^{(\sqrt t-\theta)^2/2}+e^{(\sqrt t + \theta)^2/2})}=\frac{\frac{1}{\sqrt {2\pi}}e^{-(\sqrt t-\theta)^2/2}}{\frac{1}{2\sqrt{2\pi t}}(e^{(\sqrt t-\theta)^2/2}+e^{(\sqrt t + \theta)^2/2})}$? But this can't be right because there's no $x$ in the answer. What does one do here?

Body:
MAJOR EDIT

Better check @angryavian's answer.

$\Pr\{X=x|X^2=t\}$

$\;\;\;=\Pr\{X=x|X>0, X^2=t\}\Pr\{X>0|X^2=t\}$

$\;\;\;\;\;\; +\Pr\{X=x|X<0, X^2=t\}\Pr\{X<0|X^2=t\}$

Now:

$\Pr\{X=x|X>0, X^2=t\}=1$, if $x=\sqrt{t}$, and zero otherwise,

also $\Pr\{X=x|X<0, X^2=t\}=1$, if $x=-\sqrt{t}$, and zero otherwise.

And:

$\Pr\{X>0|X^2=t\} = \dfrac{\Pr\{X^2\approx t,X>0\}}{\Pr\{X^2\approx t\}}$

$\;\;\;\;
=\dfrac{f_X(\sqrt{t})\left|d(x^2)|_{x=\sqrt{t}}\right|}{f_X(\sqrt{t})\left|d(x^2)|_{x=\sqrt{t}}\right|+f_X(-\sqrt{t})\left|d((-x)^2)|_{x=-\sqrt{t}}\right|}$

$\;\;\;\;= \dfrac{\frac{1}{\sqrt{2\pi}}\exp\{-0.5(\sqrt{t}-\theta)^2\}}{\frac{1}{\sqrt{2\pi}}\exp\{-0.5(\sqrt{t}-\theta)^2\}+\frac{1}{\sqrt{2\pi}}\exp\{-0.5(-\sqrt{t}-\theta)^2\}}
= \dfrac{e^{\theta\sqrt{t}}}{e^{\theta\sqrt{t}}+e^{-\theta\sqrt{t}}}$

In a similar way:

$\Pr\{X<0|X^2=t\} = \dfrac{e^{-\theta\sqrt{t}}}{e^{\theta\sqrt{t}}+e^{-\theta\sqrt{t}}}$

So,

$\Pr\{X=x|X^2=t\}
=\dfrac{e^{-\theta\sqrt{t}}}{e^{\theta\sqrt{t}}+e^{-\theta\sqrt{t}}}\delta(x+\sqrt{t})+\dfrac{e^{\theta\sqrt{t}}}{e^{\theta\sqrt{t}}+e^{-\theta\sqrt{t}}}\delta(x-\sqrt{t})
$


问:
Integral of a measurable function over N
I want to calculate the integral of the function $a:\mathbb{N}\to\mathbb{R}$, $a(n)= \frac{(-1)^n}{2^n}$ over the set $\mathbb{N}$, with the cardinal measurement.

I've seen how to calculate this type of integrals on simple functions, but this is not a simple function, and I can't find any examples on how to calculate this type of integrals.

Any kind of help will be much appreciated, thank you.

答:
Note that the integration with respect to the counting measure is just the (infinite) sum, see for example here for a proof. So the the desired integral is $$\int_{\Bbb N}a~\mathrm{d}\mu=\sum_{n=1}^\infty \frac{(-1)^n}{2^n}=\cdots$$

Edit: the proof in the link only shows the result for non-negative bounded functions. But of course our case follows from that by splitting it up into a negative and positive part.


答: $${1-\cos x\over \sin x}={2\sin^2 \frac x2 \over 2\sin \frac x2 \cos \frac x2}=\tan \frac x2$$

Title: How do you prove$ \frac{1-\cos x}{\sin x}=\tan\left(\frac{x}{2}\right)$?

Body: Now I've seen this before but I forgot how to do it. (by the way, this is without the use of t-formula). It gets difficult when the angles are not the same but I split the fraction on the LHS up and got and stuff in terms of $\csc{x}$ and $\cot{x}$. I then used the identity $1 + cot^2{x} = csc^2{x}$ but am still stuck on reaching something in terms of $\frac{x}{2}$.Is there any neat efficient solution?


问:
What is a proof?
I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).

Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.

Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.

Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".

EDIT : How is the term "wrong" defined in mathematics then? You can say that proving Fermat's Last Theorem using the number theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms.

Is mathematics as rigorous and as thought-through as it is believed and expected to be? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals.

答:
Starting from the end, if you take Pythagoras' Theorem as an axiom, then proving it is very easy. A proof just consists of a single line, stating the axiom itself. The modern way of looking at axioms is not as things that can't be proven, but rather as those things that we explicitly state as things that hold.

Now, exactly what a proof is depends on what you choose as the rules of inference in your logic. It is important to understand that a proof is a typographical entity. It is a list of symbols. There are certain rules of how to combine certain lists of symbols to extend an existing proof by one more line. These rules are called inference rules.

Now, remembering that all of this happens just on a piece of paper - the proof consist just of marks on paper, where what you accept as valid proof is anything that is obtained from the axioms by following the inference rules - we would somehow like to relate this to properties of actual mathematical objects. To understand that, another technicality is required. If we are to write a proof as symbols on a piece of paper we had better have something telling us which symbols are we allowed to use, and how to combine them to obtain what are called terms. This is provided by the formal concept of a language. Now, to relate symbols on a piece of paper to mathematical objects we turn to semantics. First the language needs to be interpreted (another technical thing). Once the language is interpreted each statement (a statement is a bunch of terms put together in a certain way that is trying to convey a property of the objects we are interested in) becomes either true or false.

This is important: Before an interpretation was made, we could still prove things. A statement was either provable or not. Now, with an interpretation at hand, each statement is also either true or false (in that particular interpretation). So, now comes the question whether or not the rules of inference are sound. That is to say, whether those things that are provable from the axioms are actually true in each and every interpretation where these axioms hold. Of course we absolutely must choose the inference rules so that they are sound.

Another question is whether we have completeness. That is, if a statement is true under each and every interpretation where the axioms hold, does it follow that a proof exists? This is a very subtle question since it relates semantics (a concept that is quite illusive) to provability (a concept that is very trivial and completely mechanical). Typically, proving that a logical system is complete is quite hard.

I hope this satisfies your curiosity, and thumbs up for your interest in these issues!

Title: What is a proof?


问:
The normal order of $\nu(n)$ and its normal distribution
I understand that for $\nu(n)$, the number of divisors of $n$, let me denote $\log_2 x=\log\log x$. Then $$\sum_{n\le x}\Big(\nu(n)-\log_2x\Big)^2=O(x\log_2x)$$

The book told me about the theorem (due to Erdos and Kac), for $S(x;\alpha,\beta)=\#\left\{ n\le x : \alpha\le\dfrac{\nu(n)-\log_2n}{\sqrt{\log_2(n)}}\le\beta\right\}:=\#S$, then $$\lim_{x\rightarrow\infty}\dfrac{S(x;\alpha,\beta)}{x}=\dfrac{1}{\sqrt{2\pi}}\int_{\alpha}^\beta e^{-t^2/2} dt$$

I want to prove that $\displaystyle\sum_{n\le x}\Big(\nu(n)-\log_2n\Big)^2\ll x\log_2 x$ by using the theorem stated above.

What I am confused is that if $n\in S$ then $\displaystyle \delta(n):=\left|\dfrac{\nu(n)-\log_2 n}{\sqrt{\log_2 (n)}}\right|\le \max(|\alpha|,|\beta|)$ and if we choose $|\alpha|,|\beta|\le \sqrt x$ then the statement I want follows for $n\in S$, but what about $n\not\in S, n\le x$? I confuse about the phrase "normally distributed". What does it mean exactly in this sense?

As I guess, if we choose $\alpha,\beta$ that $\beta-\alpha$ is wide enough then $\delta(n)$ that satisfies the condition in $S$ would be wide enough, too. Therefore, the number of terms left from the distribution (the term on the left and right part of normal distribution) would be small.

How can I add the left terms that I mentioned to cover all $n\le x$?

答:
I proved it by using the method that resembled Turan that
\begin{equation}
\begin{split}
\sum_{1<n\le x}\nu(n) & =x\log_2x+O(x) \\
\sum_{n\le x} \nu(n)^2&= x(\log_2x)^2+O(x\log_2x)
\end{split}
\end{equation}
and by partial summation
\begin{equation}
\begin{split}
\sum_{1<n\le x}(\log_2 n)^2 & =\sum_{1<n\le x}1\cdot(\log_2n)^2 = (\left[\,x\right]-1)\log^2_2x-\int_2^x (\left[\,t\right]-1)\cdot 2\log_2 t\cdot\left(\dfrac{1}{t\log t}\right) dt\\
&= x(\log_2 x)^2+O(\log^2_2 x)+O(x\log_2 x)= x(\log_2 x)^2+O(x\log_2 x)\\
\sum_{1<n\le x} \nu(n)\log_2 n & = \sum_{n\le x} \nu(n)\cdot \log_2 n=\Big(x\log_2 x+O(x)\Big)(\log_2 x)-\int_2^x (t\log_2 t+O(t))\left(\dfrac{1}{t\log t}\right) dt \\
& = x(\log_2 x)^2+O(x\log_2 x)
\end{split}
\end{equation}

Hence, we obtain
\begin{equation}
\begin{split}
\sum_{1<n\le x}(\nu(n)-\log_2 n)^2 & = \sum \nu(n)^2-2\sum\nu(n)\log_2n+\sum(\log_2 n)^2 = O(x\log_2 x)
\end{split}
\end{equation}
as desired. Therefore, almost all numbers $n$ have $\log_2 n$ numbers of prime factors as Hardy and Ramanujan stated (I find their theorem indeed elegant).


问:
How to prove $\lim_{n\to \infty} \int_{0}^{+\infty} \frac{dx}{e^x+x^n}=1-\frac{1}{e}$?
How to prove $$\lim_{n\to \infty} \int_{0}^{+\infty} \frac{1}{e^x+x^n}\ \mathrm{d}x=1-\frac{1}{e}$$?

答:
Let $ n\geq 2 $, we have : \begin{aligned}\int_{0}^{+\infty}{\frac{\mathrm{d}x}{\mathrm{e}^{x}+x^{n}}}&=\int_{0}^{1}{\frac{\mathrm{d}x}{\mathrm{e}^{x}+x^{n}}}+\int_{1}^{+\infty}{\frac{\mathrm{d}x}{\mathrm{e}^{x}+x^{n}}}\\ &=1-\frac{1}{\mathrm{e}}-\int_{0}^{1}{\frac{x^{n}}{\mathrm{e}^{x}\left(\mathrm{e}^{x}+x^{n}\right)}\,\mathrm{d}x}+\int_{1}^{+\infty}{\frac{\mathrm{d}x}{\mathrm{e}^{x}+x^{n}}}\end{aligned}

Since : $$ \left|\int_{0}^{1}{\frac{x^{n}}{\mathrm{e}^{x}\left(\mathrm{e}^{x}+x^{n}\right)}\,\mathrm{d}x}\right|\leq\int_{0}^{1}{x^{n}}=\frac{1}{n+1}\underset{n\to +\infty}{\longrightarrow}0 $$

And : $$ \left|\int_{1}^{+\infty}{\frac{\mathrm{d}x}{\mathrm{e}^{x}+x^{n}}}\right|\leq\int_{1}^{+\infty}{\frac{\mathrm{d}x}{x^{n}}}=\frac{1}{n-1}\underset{n\to +\infty}{\longrightarrow}0 $$

Then $ \int_{0}^{1}{\frac{x^{n}}{\mathrm{e}^{x}\left(\mathrm{e}^{x}+x^{n}\right)}\,\mathrm{d}x} $, and $ \int_{1}^{+\infty}{\frac{\mathrm{d}x}{\mathrm{e}^{x}+x^{n}}} $ both approaches $ 0 $ as $ n $ goes to $ +\infty $, Thus : \begin{aligned}\lim_{n\to +\infty}{\int_{0}^{+\infty}{\frac{\mathrm{d}x}{\mathrm{e}^{x}+x^{n}}}}=1-\frac{1}{\mathrm{e}}\end{aligned}


问:
Split-complixificatinon of Lie algebras and Lie groups
The complexification of the Lie algebras and groups over $\mathbb{C}$ is a well-studied topic. However, I can not find any reference to the similar process for "split-complexification" over $\mathbb{D}$ (say over ${}^2\mathbb{R}$). Obviously, the results from both processes shall be different. (Please do not point me to Clifford algebras, as my interest is precisely in their specific representation via Lie groups).

The general problem:

Let $\mathfrak{h}(\mathbb{R})$ be n-dimensional Lie algebra over the reals with the basis $E$ as $$e_1,e_2,..,e_n$$ with given commutator relations. Now consider the "split-complixification" of the algebra $\mathfrak{h}(\mathbb{D})$. Thus, the basis is doubled by adding $E'$: $$e_{n+1},e_{n+1},..,e_{2n}$$
such as $\quad e_{n+k}=j \:e_{k}\quad,k=0,1...n \quad (j^2=1, j \neq 1)$. The commutator relations for $E'$ appear to be the same as for $E$. Thus, algebra $\mathfrak{h}(\mathbb{D})$ is a new 2n-dimensional algebra. Hence, the new basis can generate a new split-complexified Lie group.

$E'$ with $E$ give $[e_{n+1},e_{1}]\neq 0$,$[e_{n+2},e_{2}]\neq 0 $ etc. Though, in case of "two copies" they would commute.

As an example, the particular problem is Split-biquaternion $\mathbb{H}'(\mathbb{D})$ representation in terms of the Lie groups.

Split-quaternion $\mathbb{H}'(\mathbb{R})$ isomorphic to $Mat(2,\mathbb{R})$ or
$SL(2,\mathbb{R}) \cong SU(1,1)$. Therefore, split-biquaternion is split-complixification of $SL(2,\mathbb{R})$ i.e. matrices with entries of split-complex numbers $SL(2,\mathbb{D})$.
The split-biquaternions is also $\mathbb{H}\otimes\mathbb{D}=\mathbb{H}\oplus\mathbb{H}$, hence the resulting group is also two copies of $SU(2)$.

Then, looking at the properties of $SL(2,\mathbb{D})$ my guess is that there is a homomorphism $\phi$: $SL(2,\mathbb{D}) \to SO(2,2)$.

As always, thanks for any clue for further study.

答:
Let $\mathbb{D}=\mathbb{R}[j]/(j^2-1)$ be the split complex numbers. It is well-known that $\mathbb{D}\cong\mathbb{R}\oplus\mathbb{R}$ as rings, with the isomorphism given by linearly extending $(1,0)\leftrightarrow\frac{1}{2}(1+j)$ and $(0,1)\leftrightarrow\frac{1}{2}(1-j)$ (these are the orthogonal primitive idempotents for its Peirce decomposition).

Thus $\mathfrak{g}\otimes_{\mathbb{R}}\mathbb{D}\cong\mathfrak{g}\otimes(\mathbb{R}\oplus\mathbb{R})\cong(\mathfrak{g}\otimes\mathbb{R})\oplus(\mathfrak{g}\otimes\mathbb{R})\cong\mathfrak{g}\oplus\mathfrak{g}$ as real lie algerbas. The isomorphism is actually the same as above: $$(X,Y)\mapsto \frac{1}{2}(1+j)X+\frac{1}{2}(1-j)Y=\frac{1}{2}(X+Y)+\frac{1}{2}(X-Y)j.$$

Let's also examine this claim:

Split-quaternion $\mathbb{H}'(\mathbb{R})$ isomorphic to $Mat(2,\mathbb{R})$ or
$SL(2,\mathbb{R}) \cong SU(1,1)$. Therefore, split-biquaternion is split-complixification of $SL(2,\mathbb{R})$ i.e. matrices with entries of split-complex numbers $SL(2,\mathbb{D})$.

First off, let's talk about language. There's a difference between a "normed" algebra and its group of elements of norm $1$. You can say $\mathbb{H}'$ is isomorphic to $M_2(\mathbb{R})$ but not to $SL(2,\mathbb{R})$, the latter isn't even an algebra! So here's the language-fixed version of your statement:

Split-quaternions $\mathbb{H}'(\mathbb{R})$ are isomorphic to $M_2(\mathbb{R})$. Therefore, the split-biquaternions $\mathbb{H}\otimes\mathbb{D}$ are the split-complixification of the split quaternions $\mathbb{H}'$.

But this is not true: the split biquaternions are the split-complexification of the normal quaternions, not of the split quaternions! The split biquaternions are isomorphic to $\mathbb{H}^2$ whereas the split-complexification of the split quaternions are isomorphic to $M_2(\mathbb{R})^2$, which is inequivalent.

That said, there is a $2$-to-$1$ surjective homomorphism $SL_2(\mathbb{D})\to SO(2,2)$. Note that $\mathbb{D}\cong\mathbb{R}^2$ so $SL_2(\mathbb{D})\cong SL_2(\mathbb{R})^2$ which acts on $M_2(\mathbb{R})$ via $(A,B)X:=AXB^{-1}$. This preserves the determinant, which is a quadratic form on $M_2(\mathbb{R})$ of signature $(2,2)$.


答:
$\newcommand{\R}{\mathbb{R}}$
You've already found an answer by yourself. You can just set the domain of your function $f:X \rightarrow Y$ to
$$
\begin{align}
X
&= \{(x,y,z) \in \R^3 ~ | ~ x \neq 0 \land y \neq 0 \land z \neq 0 \}\\
&= \R^3 \setminus \{(x,y,z) \in \R^3 ~ | ~ x = 0 \lor y = 0 \lor z = 0 \}\\
&= \R^3 \setminus \left[\left\{(x,y,z) \in \R^3 ~ | ~ x = 0\} \cup \{(x,y,z) \in \R^3 ~ | ~ y = 0\} \cup \{(x,y,z) \in \R^3 ~ | ~ z = 0\right\}\right]
\end{align}
$$
It constitutes an entire three-dimensional Euclidean space with the $x-y$, $y-z$, and $z-x$ planes being removed.

Meanwhile, the codomain should be $Y = \R$.

Title: What is the domain of this function in 3 variables?

Body:
Consider the function
$$f(x, y, z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+xyz.$$
Which is the domain of this function?

Certainly, we have to ask $(x, y, z)\neq(0, 0, 0)$ but also the points of the type
$$ (0, y, z), \ (x, 0, z), \ (x, y, 0), \ (0, 0, z), \ (x, 0, 0), \ (0, y,0)$$
are not allowed. How to write in an "easy" way?

I would say
$$D(f)=\{(x, y, z)\in\mathbb{R}^3 : (x, y, z)\neq (0, 0, 0), ..\}$$
but I don't know how to continue. Could anyone please help me?

Finally, all these points to be excluded, consitute a straight line, a plane or something else?

Thank you in advance!


问:
When can you define addition on a set of equivalence classes?
Suppose I am given an equivalence relation on set A which is endowed with addition & closure under addition.

Then, to have addition on equivalence classes of A to be well-defined, it is to show:

$[a]+[b] = [a+b]$

Suppose I have shown that if $\forall a,b \in A, a \sim a', b\sim b'$ then $a+b \sim a'+b'$, then to my understanding, I proved that there exist an equivalence class $[a+b]$ so that $[a]+[b] \subset [a+b]$.

But then, now I need to show the other direction. I found it is easiest to define $c \in A$ so $b+c = 0$. So if $x \in [a+b]$ then $x \sim a+b$, and using what I proved, $x+c \sim a+(b+c) = a$. Letting $r = x-b$ and $s = b$ shows the other direction.

My question now is, to do this, (A,+) seems to need to have the property that all elements must have an identity, an inverse, and is associative, so what exactly must (A,+) be for all this to work, a group? abelian group? and why?

Note: Also I think I'm failing to consider a lot of things when thinking about this, so it will be very helpful if anyone could point many of those things out.

答:
$(A,+)$ does not need to be a group. For example, we may take $A=\Bbb N$ (with usual addition) and define $a\sim b$ as "$a$ and $b$ end in the same decimal digit". Then $(A/{\sim},+)$ is a group (cyclic of order $10$).

Similary, $(A/{\sim},+)$ may be an abelian group even if $(A,+)$ is non-abelian. (But not vice versa).


答:
You do not want any of $x,y$ or $z$ to vanish. Since $xyz$ vanishes if any one of the factors vanishes, then for no factor to vanish the product too must not vanish. (Another way to see this is to add the fractions, whose denominator will be $xyz.$)

Hence, the maximal domain is the set of all points in space except for those on the surfaces $xyz=0,$ namely the coordinate planes.


问:
GCSE: The volume of frustum
Looking please for any help! The density of marble used for the pedestal shown in the figure below is $2750\text{ kg/m}^3$
. A crane can lift a maximum of 2 tons. Can the crane raise this pedestal?

My attempt was the following: To find the volume of the frustum: I use the formula of the trapezoid surface:  $$\frac{(1+0.4)\cdot 1.5}{2}=1.05\text{ m}^2$$
And then $1.05\times 1.5=1.575\text{ m}^3\longrightarrow$ the mass is $1.575\times 2750=4331\text{ kg}=4.3\text{ tons}$

And my answer was the crane can’t raise the pedestal.

I don’t know if this is the right answer! All help/solutions appreciated. Thanks

答: There's no information in the question as to what the length l could be. The length l could be anything, so the volume could be anything and the answer is indeterminate.

Title: GCSE: The volume of frustum


问:
Product of permutation matrices
I want to prove that the product of two permutation matrices is itself a permutation matrix. But I don't know how. Please help!</p>

答:
Method 1 - Using Matrices:

Suppose you have two permutation matrices $X$ and $Y$. So by definition both $X$ and $Y$ are $n\times n$ matrices having precisely one non-zero entry in each row and column, and these entries all take the value 1.

If the non-zero entry in the first row of $X$ is in the $i$-th column, then what will the first row of the matrix $X\cdot Y$ be? By definition of matrix multiplication it will be

$
\left(\sum_{j=1}^{n}X_{1,j}Y_{j,1}, \sum_{j=1}^{n}X_{1,j}Y_{j,2},\ldots,\sum_{j=1}^{n}X_{1,j}Y_{j,n}
\right)
$

However, we know that $X_{1,j}=\delta_{i,j}$ (where $\delta_{i,j}$ is the Kronecker delta taking the value $1$ when $i=j$ and $0$ otherwise). Hence the first row of $X\cdot Y$ will be

$
\left(X_{1,i}Y_{i,1}, X_{1,i}Y_{i,2},\ldots,X_{1,i}Y_{i,n}
\right)
$

which since $X_{1,i}=1$ will be

$
\left(Y_{i,1}, Y_{i,2},\ldots,Y_{i,n}
\right).
$

So the first row of $X\cdot Y$ corresponds to the $i$-th row of $Y$. Continuing in this way, you see that the $m$-th row of $X\cdot Y$ will be the $l$-th row of $Y$, where the $1$ in the $m$-th row of $X$ appears in the $l$-th position (I hope that makes sense!)

Thus when you multiply two permutation matrices together, you are just rearranging the rows of one of the matrices.

Method 2 - Using Permutations:

A slightly easier approach is to use permutations. Suppose $X$ and $Y$ are $n\times n$ permutation matrices as above. Then define $\sigma_{X}:\{1,\ldots, n\}\rightarrow\{1,\ldots,n\}$ as follows:

$\sigma_{X}(i)=j$ where the $1$ in the $i$-th column of $X$ occurs in the $j$-th position. Similarly you may define $\sigma_{Y}$.

Then you can check that $\sigma:X\mapsto\sigma_{X}$ respects multiplication (in a similar way as above).

Method 3 - Using Vectors:

You can also show this using vectors (see joriki's comment above for more details).

Note: All of these methods are essentially the same, but are just different ways of viewing the problem.

Title: Product of permutation matrices


问:
How to find area between curves? ($\frac{1}{x}$ problem)
I need to find the area between those curves:
$f(x)=\frac{1}{x}$, $f(x)=6e^x$, $f(x)=1$, $f(x)=6$

I calculated the limits of integration, it is $\ln(\frac{1}{6})$ to $1$.
But we cannot integrate through this interval, since the sign changes and $\frac{1}{x}$ is not defined for zero. How to find area in this case?

Desmos(I added inequality for colored the target area)

答:
The easiest way to do it is to consider $x$ as a function of $y$ in each case.

But to proceed your way, note that you have to split the integral into three parts, to get $$\int_{-\log 6}^0 (6e^x -1)dx+\int_0^{1/6} (6-1)dx+\int_{1/6}^1 \left(\frac{1}{x}-1\right) dx.$$


问:
Can $S_n=a_1a_2+a_2a_3+\cdots+a_na_1=0$ if $a_i=±1$ for $n=28$ and $n=30$?

Let $$S_n=a_1a_2+a_2a_3+\cdots+a_na_1$$
If $a_i=\pm 1$, can $S_{28}=S_{30}=0$?

My approach: Start from small $n$ if we can see a pattern

For $n=1$:

$S_1=a_1=±1 \neq0$

For $n=2$:

$S_2=a_1a_2+a_2a_1=2a_1a_2=±2 \neq0$

For $n=3$:

$S_3=a_1a_2+a_2a_3+a_3a_1$

Because $a_1a_2$, $a_2a_3$ and $a_3a_1$ are odd numbers and $0$ is an even number, $S_3$ cannot be $0$. (this can be applied to other $S_n$ if $n$ is an odd number)

For $n=4$:

$S_4=a_1a_2+a_2a_3+a_3a_4+a_4a_1=(a_1+a_3)(a_2+a_4)$

However, I can't do the same with $n=6$ and above...

Body:
Write $$S_{30}=S_{28} -a_{28}a_1+a_{28}a_{29}+a_{29}a_{30}+a_{30}a_{1}$$

If $S_{30}=0=S_{28}$,

$$0=-a_{28}a_1+a_{28}a_{29}+a_{29}a_{30}+a_{30}a_{1}$$

$$a_{28}(a_{1}-a_{29})=(a_{29}+a_{1})a_{30}$$

But one of $(a_{1}-a_{29})$, $(a_{29}+a_{1})$ is zero, while other is not.

Hence impossible.


问:
"Proof" that every ultrafilter has a least element
Ultrafilters come in the principal and the free variant. Elsewhere it is said that principal is equivalent to the ultrafilter having a least element, i.e. one that is contained in every other element of the ultrafilter. Obviously this then should not be true for free ultrafilters.

Here is my "proof" that all ultrafilters have a least element:

Let $\mathcal F$ be an ultrafilter and $A\in\mathcal F$.

If $A\cap B = A$ for all $B\in \mathcal F$, then $A$ is minimal.

Otherwise $\exists B\in\mathcal F$ such that $C=A\cap B \subsetneq A$. Since $\mathcal F$ is a filter, $C\in \mathcal F$. Now re-apply (inductively so to say) (1) to $C$.

Ultimately the above process will single out the least element.

Since free ultrafilters do not have a least element, something is wrong with this proof. What is it? Does it have to do with the kind of induction which may involve an uncountable number of steps?

答:
All you proved is that there is a sequence $A_n\in\cal F$ such that $A_{n+1}\subseteq A_n$. You did not prove that this sequence has an intersection inside $\cal F$, which may or may not be the case. And even then, you did not prove that the intersection is minimal, etc. So this is certainly not a transfinite recursion, since those require you to deal with the limit steps as well.

But even if you did deal with the limit step somehow, all you can conclude is that there is a limit ordinal $\delta$, and a decreasing sequence of sets $A_i\in\cal F$ for $i<\delta$ such that $\bigcap A_i\notin\cal F$.


答:
Hint: Show that if $ S_n = 0$, then $4 \mid n$.

Your work supports this hypothesis, so prove it.

Corollary: $S_{30}$ is never 0.

 Clearly we need $ 2 \mid n$. (Your work hints at this strongly.)

 Suppose that there are $a$ terms of the form $+1$ and $b$ terms of the form $-1$.

 Then, $ n = a + b$, and $ 0 = a - b$.

 What is $ 1^a (-1)^b$ in 2 different ways?

 Thus, show that $b$ is even, so $ 4 \mid n$.

Title: Can $S_n=a_1a_2+a_2a_3+\cdots+a_na_1=0$ if $a_i=±1$ for $n=28$ and $n=30$?

Body:
Let $$S_n=a_1a_2+a_2a_3+\cdots+a_na_1$$
If $a_i=\pm 1$, can $S_{28}=S_{30}=0$?

My approach: Start from small $n$ if we can see a pattern

For $n=1$:

$S_1=a_1=±1 \neq0$

For $n=2$:

$S_2=a_1a_2+a_2a_1=2a_1a_2=±2 \neq0$

For $n=3$:

$S_3=a_1a_2+a_2a_3+a_3a_1$

Because $a_1a_2$, $a_2a_3$ and $a_3a_1$ are odd numbers and $0$ is an even number, $S_3$ cannot be $0$. (this can be applied to other $S_n$ if $n$ is an odd number)

For $n=4$:

$S_4=a_1a_2+a_2a_3+a_3a_4+a_4a_1=(a_1+a_3)(a_2+a_4)$

However, I can't do the same with $n=6$ and above...


问:
Base of Kernel and pre-image of base of Image are linear independent and span V.
I have the following two tasks before me:

Let $\psi:V\to W$ be a linear Transformation and $V, W$ are finite-dimensional over $K$.

$u_1,\dots,u_m$ is a basis of ker $\psi$ and $w_1,\dots,w_n$ is a basis of Im $\psi$.

The Vectors $v_1,\dots,v_n \in V$ are choosen such that: $\psi(v_j)=w_j$ for $ j=1,\dots,n$

Show that $u_1,\dots,u_m,v_1,\dots,v_n$ are linear independent in $V$

Show that $u_1,\dots,u_m,v_1,\dots,v_n$ is a generating system of $V$.

What I have so far:

Since {$w_1,\dots,w_n$} is a basis of Im $\psi$ and thus in particular linear independent in $W$$\implies$ $v_1,\dots,v_n$ are l.i. in $V$.

Also, $u_1,\dots,u_m$ are linear independent in $V$, since ker $\psi$ is a subspace of $V$.
We now need to show that $u_1,\dots,u_m,v_1,\dots,v_n$ is still linear independent: Assume $u_1,\dots,u_m,v_1,\dots,v_n$ was linear dependent, it then follows that $\psi(u_1),\dots,\psi(u_m),\psi(v_1),\dots,\psi(v_n)$ was also linear dependent. That however, can't be true since $\psi(u_1),\dots,\psi(u_m),\psi(v_1),\dots,\psi(v_n) \leftrightarrow 0,\dots,0,w_1,\dots,w_n \leftrightarrow w_1,\dots,w_n$ (Here I am not so sure if I can just remove all the zero vectors like I did.)

Again we can use the fact that ker $\psi$ is a subspace of $V$ and expand its base into a base for $V$: {$u_1,\dots,u_m,v_1,\dots,v_k$} If we manage to show that $k=n$ we have proven that $u_1,\dots,u_m,v_1,\dots,v_n$ is indeed a generating system. We can achieve this by observing that $\psi(u_1),\dots,\psi(u_m),\psi(v_1),\dots,\psi(v_k) \leftrightarrow w_1,\dots,w_k$ Since ${w_1,\dots,w_k}$ must be a generating system of Im $\psi$ we simply need to prove linear independence to conclude that $k=n$.

The more I think about it the more confused I am. As pointed out I am especially confused about just removing the zero vectors to show linear independence. Also if ${w_1,\dots,w_k}$ is indeed linear independent, that means every base of $V$ gets mapped to a base of Im $\psi$?

答:
Your problem is that there is no prompt to say that the dimension of $V$and $W$ is same.So you can not think that if $e$ is the basis of $V$ so the $\psi(e)$ is also the basis of $W$.

I will try to give a detailed explanation:

1.$u_1,u_2,\dots,u_m$ is the bais of $\ker{\psi}$, so you can find $\forall k_1,K_2,\dots,k_m \in K$, that $\psi(k_1u_1+k_2u_2+\dots+k_mu_m)=0$.But we know that in $V$,only $0=0$,but through $\psi$, many $u=0$ in the $V$ will become $\psi(u)=0$,if as you think, only $\psi(0)=0$ in the W, obviously it's wrong.

2.The similar as 1....

Ok,I know my explanation is terrible...

The key point is that if $e$ is the basis of $V$, but $\psi(e)$ is not the basis of $W$,cause you don't know that they have a same dimension, so your inference in 1. 2. is not true fundamentally...

We can have a extreme example,we let $V$to be $K=\mathbb{R}^2$ and $W$ to be $\{0\}$, so the $\psi$,$\forall x \in K, \psi(x)=0$

We can figure that the on of the basis of $V$ is $((0,1),(1,0))$,but the basis of $W$...

That's why you fell confused


答:
For part I start like this,

Suppose $\{u_1, \cdots, u_m, v_1,\cdots, v_n\}$ is linearly dependent.

Then $a_1u_1 + \cdots +a_m u_m + b_1v_1 + \cdots + b_n v_n = \bf{0}$ and at least one
of these scalars is nonzero.

Now take $T$ of the statement above and you'll get $b_1 w_1 + \cdots + b_n w_n =
T(\bf{0})$

What can you now conclude?

For part II, first employ rank - nullity, i.e dim$(V)$ = rk$(A)$ + nullity$(A) = n+m$. Now use the fact that the dimension of $V$ is the same as the number of vectors in the lin indep. set - we have a result that tells you automatically the lin indep. set is also spans, hence it's a basis.

Comment : I know the second part is not by brute force, but we can pair two theorems and get it done pretty quickly.

Title: Base of Kernel and pre-image of base of Image are linear independent and span V.

Body:
I have the following two tasks before me:

Let $\psi:V\to W$ be a linear Transformation and $V, W$ are finite-dimensional over $K$.

$u_1,\dots,u_m$ is a basis of ker $\psi$ and $w_1,\dots,w_n$ is a basis of Im $\psi$.

The Vectors $v_1,\dots,v_n \in V$ are choosen such that: $\psi(v_j)=w_j$ for $ j=1,\dots,n$

Show that $u_1,\dots,u_m,v_1,\dots,v_n$ are linear independent in $V$

Show that $u_1,\dots,u_m,v_1,\dots,v_n$ is a generating system of $V$.

What I have so far:

Since {$w_1,\dots,w_n$} is a basis of Im $\psi$ and thus in particular linear independent in $W$$\implies$ $v_1,\dots,v_n$ are l.i. in $V$.

Also, $u_1,\dots,u_m$ are linear independent in $V$, since ker $\psi$ is a subspace of $V$.
We now need to show that $u_1,\dots,u_m,v_1,\dots,v_n$ is still linear independent: Assume $u_1,\dots,u_m,v_1,\dots,v_n$ was linear dependent, it then follows that $\psi(u_1),\dots,\psi(u_m),\psi(v_1),\dots,\psi(v_n)$ was also linear dependent. That however, can't be true since $\psi(u_1),\dots,\psi(u_m),\psi(v_1),\dots,\psi(v_n) \leftrightarrow 0,\dots,0,w_1,\dots,w_n \leftrightarrow w_1,\dots,w_n$ (Here I am not so sure if I can just remove all the zero vectors like I did.)

Again we can use the fact that ker $\psi$ is a subspace of $V$ and expand its base into a base for $V$: {$u_1,\dots,u_m,v_1,\dots,v_k$} If we manage to show that $k=n$ we have proven that $u_1,\dots,u_m,v_1,\dots,v_n$ is indeed a generating system. We can achieve this by observing that $\psi(u_1),\dots,\psi(u_m),\psi(v_1),\dots,\psi(v_k) \leftrightarrow w_1,\dots,w_k$ Since ${w_1,\dots,w_k}$ must be a generating system of Im $\psi$ we simply need to prove linear independence to conclude that $k=n$.

The more I think about it the more confused I am. As pointed out I am especially confused about just removing the zero vectors to show linear independence. Also if ${w_1,\dots,w_k}$ is indeed linear independent, that means every base of $V$ gets mapped to a base of Im $\psi$?


问:
Proving a limit using squeeze rule
I'm trying to figure out how to calculate the limit of

$f(x)= \begin{cases}
           \frac{3}{2}\cdot x^{\frac{1}{2}}\cdot\sin\left(\frac{1}{x}\right)-\frac{1}{\sqrt{x}}\cdot\cos\left(\frac{1}{x}\right) & x  > 0 \\
            \\
           0    & x≤0 \\ \end{cases}$

as $x$ tends to $0^{+}$ I know it doesn't exist and tends to $-∞$, I've been trying forever to prove there exist two functions that are bigger/smaller than $f$ in order to use the squeeze rule but I'm begging to think there is some other easier way..

答:
$\displaystyle \lim_{x\to 0^-} f(x)=0.$
But $\displaystyle \lim_{x\to 0^+} f(x)$ does not exist.

Use sequential criterion to prove  that $\displaystyle \lim_{x\to 0} f(x)$ does not exist.
Consider two sequences : $x_n={2\over (4n+1)\pi}$ and $x'_n={1\over 2n\pi}.$

See that: $\displaystyle \lim_{n\to \infty} f(x_n)=\lim_{n\to \infty}\textstyle \frac 32\cdot \frac{\sqrt 2}{\sqrt {(4n+1)\cdot \pi}}=0.$ But $\displaystyle \lim_{n\to \infty} f(x'_n)=\lim_{n\to \infty} -\sqrt{2n\pi}\textstyle \to -\infty$.
Thus $\displaystyle \lim_{n\to \infty} x_n=\lim_{n\to \infty} x'_n=0$ but $\displaystyle \lim_{n\to \infty} f(x_n)\neq \lim_{n\to \infty} f(x'_n)$.

$\therefore \displaystyle \lim_{x\to 0} f(x)$ does not exist.

Title: Proving a limit using squeeze rule


问:
I need help with simple algebra with logarithms
Can anyone help me with this?
$$xe^{ht} = y $$
How do I eliminate "t" from the LHS of the equation? What I want to end up with is the LHS being a function of "h".

Body:
$$xe^{ht}=y$$
$$ln(xe^{ht})=ln(y)$$
$$ln(x)+ht=ln(y)$$
$$ht=ln(y)-ln(x)=ln(\frac{y}{x})$$
$$h=\frac{ln(\frac{y}{x})}{t}$$


问:
Incomplete space
I must show that $C_0(\mathbb{R})$ space of all continuous real functions $f:\mathbb{R} \longrightarrow \mathbb{R}$ with compact support is not a complete space endowed by norm
$\|f\|=\sup\limits_{t \in \mathbb{R}}|f(t)|$.

Thank for any help.

答:
You could also use Baire's theorem to obtain $n\in \mathbb N$ such that every element of $C_0(\mathbb R)$
would have support in $[-n,n]$. </p>

Title: Incomplete space


问:
Difficult Cauchy Problem
Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality:
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$

I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried using knowing what is $a^2 + b^2 + c^2$ in terms of $a$, $b$ and $c$ to try and get $a+b_c$ in terms of them using the sum of squares where $(a+b)^2 - 2ab = a^2 + b^2$ but I reached a dead end where I could not simplify any further.

I think assuming $a=b=c$ did get me somewhere but since it is a proof problem I cannot assume but rather have to prove that $a=b=c$. I cannot figure that bit out.

I know Cauchy but very little bit, so I do not now if it can be applied too well here but I think it is a right approach.

答:
By Cauchy-Schwarz,
\begin{align}
 \left(\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \right)(a+b+c) & \geq \left(\dfrac{a}{bc} + \dfrac{b}{ca} + \dfrac{c}{ab}\right) ^2 \\
& =  \left (\dfrac{a^2+b^2+c^2}{abc} \right)^2 \\
& = \left (\dfrac{3abc}{abc} \right)^2 \\
& = 9. \\
\end{align}
Thus, dividing both sides of the preceding inequality by $a+b+c$, we have that
$\left(\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \right) \geq \dfrac{9}{a+b+c}$, and we are done.

Title: Difficult Cauchy Problem


答: You have that $$e^{ht}=\frac yx,$$ or that $$ht=\log\left(\frac yx\right).$$ Can you continue from here?

Title: I need help with simple algebra with logarithms

Body:
Can anyone help me with this?
$$xe^{ht} = y $$
How do I eliminate "t" from the LHS of the equation? What I want to end up with is the LHS being a function of "h".


问:
Suggestion for mistake on computation of a pull-back $2$-form on the sphere
Let $f : S^3 \to S^2$ be the function given by
$$
f(x , y , z , t) = (p = (2 (x z + y t) , q = 2 (- x t + y z) , r = - x^2 - y^2 + z^2 + t^2).
$$
Let $\omega = (p \, dq \wedge dr + q \, dr \wedge dp + r \, dp \wedge dq)$ be a $2$-form in $S^2$. How can I show that $f^*(\omega) = 4(dx \wedge dy + dz \wedge dt) \in {\Omega}^2(S^3)$? My attempt is the next:
$$
f^*(\omega) = p f^*(dq) \wedge f^*(dr) + q f^*(dr) \wedge f^*(dp) + r f^*(dp) \wedge f^*(dq).
$$
For instance,
$$
f^*(dq) = \frac{\partial q}{\partial x} \, dx + \frac{\partial q}{\partial y} \, dy + \frac{\partial q}{\partial z} \, dz + \frac{\partial q}{\partial t} \, dt = 2 (- t \, dx + z \, dy + x \, dz + y \, dt).
$$
If we continue with this process, do we reach that $f^*(\omega) = 4(dx \wedge dy + dz \wedge dt)$? I guess no, because
$$
p f^*(dq) \wedge f^*(dr) = 2 (x z + y t) 2 (- t \, dx + z \, dy + x \, dz + y \, dt) \wedge 2 (- x \, dx - y \, dy + z \, dz + t \, dt) =
$$
$$
= \ldots = 8 (x z + y t) (t y + x z) \, dx \wedge dy + \ldots
$$
(the coefficient of $dx \wedge dy$ in the expression of $p f^*(dq) \wedge f^*(dr)$ is $8 (x z + y t) (t y + x z)$. If we take the coefficient of $dx \wedge dy$ from $q f^*(dr) \wedge f^*(dp)$ and $r f^*(dp) \wedge f^*(dq)$, we obtain
$$
8 (- x t + y z) (- x t + y z) \quad \text{and} \quad 4 (- x^2 - y^2 + z^2 + t^2) (z^2 + t^2).
$$
Then using that $x^2 + y^2 + z^2 + t^2 = 1$ in $S^3$, we have that
$$
f^*(\omega) = (8 (x z + y t) (t y + x z) + 8 (- x t + y z) (- x t + y z) + 4 (- x^2 - y^2 + z^2 + t^2) (z^2 + t^2)) \, dx \wedge dy + \ldots = \ldots = 4 (z^2 + t^2) \, dx \wedge dy + \ldots
$$
if I am not wrong, and it is not compatible with $f^*(\omega) = 4(dx \wedge dy + dz \wedge dt)$. Where am I getting mistake? Also I have to keep in mind that $x \, dx + y \, dy + z \, dz + t \, dt = 0$ but it is because $x^2 + y^2 + z^2 + t^2 = 1$.

答:
Thanks to Eliot Yu, I could build the answer for the question. The whole expression for $p f^*(dq) \wedge f^*(dr)$ is
$$
p f^*(dq) \wedge f^*(dr) = 2 (x z + y t) 2 (- t \, dx + z \, dy + x \, dz + y \, dt) \wedge 2 (- x \, dx - y \, dy + z \, dz + t \, dt) =
$$
$$
= 8 (x z + y t) ((t y + z x) \, dx \wedge dy + (- t z + x y) \, dx \wedge dz + (- t^2 - x^2) \, dx \wedge dt +
$$
$$
+ (z^2 + y^2) \, dy \wedge dz + (z t - x y) \, dy \wedge dt + (y t + x z) \, dz \wedge dt).
$$
We display $(- t z + x y) \, dx \wedge dz$ using $x \, dx + y \, dy + z \, dz + t \, dt = 0$ and the multilineality from $\wedge$ as he said to get:
$$
(- t z + x y) \, dx \wedge dz = - t \, dx \wedge (z dz) + y \, (x \, dx) \wedge dz =
$$
$$
= - t \, dx \wedge (- x \, dx - y \, dy - t \, dt) + y (- y \, dy - z \, dz - t \, dt) \wedge dz =
$$
$$
= y t \, dx \wedge dy + t^2 \, dx \wedge dy - y^2 \, dy \wedge dz + y t \, dz \wedge dy.
$$
Similarly,
$$
(z t - x y) \, dy \wedge dt = x z \, dx \wedge dy - z^2 \, dy \wedge dz + x^2 \, dx \wedge dt + x z \, dz \wedge dt.
$$
If we replace these two values in the expression obtained for $p f^*(dq) \wedge f^*(dr)$ and computing we get
\begin{equation}\label{1}\tag{1}
p f^*(dq) \wedge f^*(dr) = 16 {(x z + y t)}^2 \, (dx \wedge dy + dz \wedge dz).
\end{equation}
Similarly we can obtain easier expressions for $q f^*(dr) \wedge f^*(dp)$,
\begin{equation}\label{2}\tag{2}
q f^*(dr) \wedge f^*(dp) = 16 {(- x t + y z)}^2 \, (dx \wedge dy + dz \wedge dz)
\end{equation}
and for $r f^*(dp) \wedge f^*(dq)$,
\begin{equation}\label{3}\tag{3}
r f^*(dp) \wedge f^*(dq) = 4 {(- x^2 - y^2 + z^2 + t^2)}^2 \, (dx \wedge dy + dz \wedge dz).
\end{equation}
Then using \eqref{1}, \eqref{2} and \eqref{3},
$$
f^*(\omega) = 4 \left({(x z + y t)}^2 + {(- x t + y z)}^2 + {(- x^2 - y^2 + z^2 + t^2)}^2\right) \, (dx \wedge dy + dz \wedge dz) =
$$
$$
= 4 (p^2 + q^2 + r^2) \, (dx \wedge dy + dz \wedge dz) = 4 \, (dx \wedge dy + dz \wedge dz)
$$
because $p^2 + q^2 + r^2 = 1$, as $f(S^3) \subset S^2$ (is well defined).


问:
Variation of Bin Packing/Knapsack problem

Statement of Problem

We are given four different collections of objects (apples, bananas, oranges and melons), and must store them in N bins. Bins can only store a single type of object, but the amount of each which can be stored may vary; for a given bin n, it may store 3 apples, 2 bananas, 2 oranges, and no melons. A different bin m may store 2 apples, 2 bananas, 2 oranges, and 2 melons.

Given some collection of objects, and some collection of bins, is it possible to allocate bins to store some type of object, such that all of the items may be stored?

Example - Successfully storing all objects

Objects: 4 apples, 3 bananas, 2 oranges, no melons.
Bins:

Bin 1; Space for 2 of any kind.

Bin 2; 3 Apples, 2 Bananas, 2 Oranges, 2 Melons

Bin 3; Space for 3 of any kind.

Bin 4; 1 orange.

Bin 5; 1 orange.

In this case, the objects may be stored by the following distribution: store apples in bins 1 and 2, store bananas in bin 3, store oranges in bins 4 and 5.

Current work

In my current implementation, I employ a brute force method.

(This problem is the problem of "how many farm animals may be stored in a cave," from the board game Caverna)

答:
Suppose we have $n$ bins and fruits $f_1,\ldots,f_m$. Let the upper bound on the number of fruits of type $f_k$ in bin $s$ be $b(f_k,s)$. A rephrase of the problem is as follows:

Given any term in the product below, can you find which terms of the product multiply to form the term in the product?

$$\prod\limits_{i=1}^n\left(f_i^{b(f_1,i)}+f_2^{b(f_2,i)}+\ldots+f_m^{b(f_m,i)}\right)$$

The reason why this works is that each multiplicand in the product represents a bucket, and we pick one fruit in each bucket.

You can use generating functionology from here.


问:
Group of order $pqrt$
Let $G$ be a group of order $pqrt$, where $p>qrt$ and $p,q,r,t$ are prime ($\underline{\text{not necessarily distinct}}$).

I have to show that $\exists$ $M,N,L \le G$ of order $pq,pr$ and $pt$.

The condition that I underlined is my doubt in this exercise because if I 'd start considering all the possible cases it would become really long and tedious.

I wanted to share on MSE my ideas about this exercise because I found several observations which I think are plausible.

The first way we can follow to show the property is using the fact that every group of a prime order is cyclic.

In $G$, the product of  two cyclic subgroups of order $k\in\{p,q,r,t\}$ will be a cyclic group whose order is the product of the two prime numbers that we chose.

The second method is the following:

I noticed that, since $|G|=pqrt$ with $p>qrt$, $G$ must have only one $p$-Sylow group. All $p$-Sylow  are conjugated, so we have that $\exists! P\trianglelefteq G$ of order $p$.

For the condition $P\trianglelefteq G$, I considered the quotient group $G/P$, whose order is $qrt$.

Let $n_q$ be the number of $q$-Sylow in $G/P$ $\implies n_q\equiv 1\text{ (mod q)}$ and $n_q\in\{1,\not r,\not t,rt\}$. The possibilities are $n_q=1,rt$.

If $n_q=rt\implies\exists Q^{(1)},\dots,Q^{(rt)}\le G/P$ of order $q$. Since their order is a prime, they are all cyclic groups, so they are abelian and $\forall i,j=1,\dots,rt$ ($i\ne j$) $Q^{(i)}Q^{(j)}=Q^{(j)}Q^{(i)}\implies Q^{(i)}Q^{(j)}\le G$. The order of the product of the $q$-Sylow is
$$|Q^{(i)}Q^{(j)}|=\dfrac{|Q^{(i)}||Q^{(j)}|}{|Q^{(i)}\cap Q^{(j)}|}\Big|qrt\text{ and } Q^{(i)}\cap Q^{(j)}\le Q^{(i)},Q^{(j)}\implies |Q^{(i)}\cap Q^{(j)}|\in\{1,q\}.$$
If $|Q^{(i)}\cap Q^{(j)}|=1\implies|Q^{(i)}Q^{(j)}|=q^2\Big|qrt\implies q\Big|rt$ but $q$ is a prime, hence $q\Big|r$ or $q\Big|t$, which is a contradiction.

So we state that $|Q^{(i)}\cap Q^{(j)}|=q$, then $\exists!Q\trianglelefteq G/P$.
For the correspondence theorem $\exists\mathcal Q\trianglelefteq G$ such that $|\mathcal Q|=pq$.

We can apply this idea for the other subgroups in $G/P$ and then use the correspondence theorem.

What abou the underlined Hypothesis??

答: Hint: by Cauchy's Theorem, $G$ possesses an element $g$ of order $q$. Now $\langle g \rangle P$ is a subgroup since $P$ is normal! And it has order $pq$. Same goes for the other primes.

Title: Group of order $pqrt$


问:
$X_n$ ($n\ge1)$ is sequence of real numbers such that $\lim_{n \to \infty}{\frac{x_n}{n}}=0.001$ then
if $X_n$ ($n\ge1)$ is sequence of real numbers such that $\lim_{n \to \infty}{\frac{x_n}{n}}=0.001$ then,

$(A)\ \ X_n$ is a bounded sequence

$(B)\ \ X_n$ is a unbounded sequence

$(C)\ \ X_n$ is a Convvergent sequence

$(D)\ \ X_n$ is a monotonically decreasing sequence

I can only think of this particular case.

$\text{convergent sequence}\cdot \frac{1}{n}=\text{convergent sequence}\cdot \text{convergent sequence} =$finite limit

Therefore it is convergent but I think its not a correct way to solve this problem.

How do I figure out what is our sequence behavior just on the basis of limits provided?

答:
In the comments you considered the example $x_n = 0.001 n$ and that correctly eliminated choices A, C, D.  So B is the only remaining choice.

It is useful to know why B holds true, that is, why any sequence $\{x_n\}$ that satisfies $x_n/n\rightarrow 0.001$ must be unbounded. Knowing that it is true for a particular example is not good enough.  One way to explore this is through "proof by contradiction": Assume $\{x_n\}_{n=1}^{\infty}$ is bounded, so that there is a finite number $M$ such that
$$ -M \leq x_n \leq M \quad \forall n \in \{1, 2, 3, ...\}$$
Now reach a contradiction with the assumption $x_n/n\rightarrow 0.001$.


问:
Differentiation of norm
How do I differentiate the "norm" of  $(x-μ)$, with respect to $μ$, where both $x$ and $μ$ are vectors ?

How will I start and proceed ?

Thank you in advance.

答:
Let $\;\phi=\|x-\mu\|\;\;{\rm and}\;\;w = (\mu-x)$

Use a dot product to rewrite the norm to obtain
$$\eqalign{
\phi^2 &= \|w\|^2 \;=\; w\cdot w \\
2\phi\,d\phi &= 2\,w\cdot dw \;=\; 2\,w\cdot d\mu \\
\frac{d\phi}{d\mu} &= \frac{w}{\phi} \;=\; \frac{\mu-x}{\|x-\mu\|} \\
}$$

Title: Differentiation of norm


答:
$(A)$ $\lim_{n \to \infty}{\frac{x_n}{n}} \leq \lim_{n \to \infty}{\frac{M}{n}} = 0 < 0.001$

$(C)$ $\lim_{n \to \infty}{\frac{x_n}{n}}=0.001 \implies 0=\lim_{n \to \infty}{\frac{L}{n}}=0.001$.

$(D)$ If $(x_{n})$ is eventually constant, we have the same case as $(C)$.

Otherwise, $(x_{n})$ is strictly decreasing and $(x_{n})$ is eventually always negative which implies that if $\lim_{n \to \infty}\frac{x_{n}}{n}$ exists, then $\lim_{n \to \infty}{\frac{x_n}{n}} < 0 < 0.001 $.

Then the answer must be $(B)$.

Finding an example (that satisfies the hypothesis but just so happens to not satisfy any of the other properties...) may also work as you have shown.

Body:
if $X_n$ ($n\ge1)$ is sequence of real numbers such that $\lim_{n \to \infty}{\frac{x_n}{n}}=0.001$ then,

$(A)\ \ X_n$ is a bounded sequence

$(B)\ \ X_n$ is a unbounded sequence

$(C)\ \ X_n$ is a Convvergent sequence

$(D)\ \ X_n$ is a monotonically decreasing sequence

I can only think of this particular case.

$\text{convergent sequence}\cdot \frac{1}{n}=\text{convergent sequence}\cdot \text{convergent sequence} =$finite limit

Therefore it is convergent but I think its not a correct way to solve this problem.

How do I figure out what is our sequence behavior just on the basis of limits provided?


问:
Which of the following statement is true and how?
An engineer measures THREE quantities $X, Y$ and $Z$ in an experiment. She finds that they follow a relationship that is represented in the figure below:

(the product of $X$ and $Y$ linearly varies with $Z$).

Then, which of the following statements is FALSE?

For fixed $Y$; $X$ is proportional to $Z$

For fixed $Z$; $X$ is proportional to $Y$

$XY/ Z$ is constant

For fixed $X$ ; $Z$ is proportional to $Y$.

My attempt: We see that $XY=\lambda Z\implies XY/Z=\lambda=\text{constant}$. Is it correct?

答:
Number 2 is wrong. If we fix $Z$ we have

$$ XY = C $$

So $X$ isn't directly proportional to $Y$, as in $ X = YC $.

Also: your attempt is correct.


问:
Validity of Inserting a Levi-Civita Identity
I want to calculate $ A_{fgh} e^f \wedge e^g \wedge e^h$ where $A_{fgh}$ is some complicated term depending on 3 of the sub-indices. I am using the convention that 1 upper and 1 lower index gets summed over. I have the identities $\epsilon_{b}\mathstrut^{fgh} \epsilon^{b}\mathstrut_{fgh}=4!$ and $\star e^b = \frac{1}{3!}\epsilon^{b}\mathstrut_{fgh}e^f\wedge e^g \wedge e^h$. I want to know if it is valid to do the following.

$$ A_{fgh} \frac{1}{4!} \epsilon_{b}\mathstrut^{fgh} \epsilon^{b}\mathstrut_{fgh} e^f \wedge e^g \wedge e^h = \frac{1}{4} A_{fgh}\epsilon_{b}\mathstrut^{fgh}\star e^b.$$

What confuses me is using indices that are summed over. Is everything okay? Thanks in advance.

Explicity,

$$A^{a}_{fgh} := \epsilon^{ij}\mathstrut_{gh}F_{af}F_{ij} - \epsilon^{ij}\mathstrut_{ah}F_{fg}F_{ij}$$

where $F$ can be assumed to be antisymmetric if necessary and $a$ is independent. I essentially want to be able to write $A^{a}_{fgh,b}\star e^b.$

答: You can use an index twice, but not four times. It should be$$A_{fgh}e^f\wedge e^g\wedge e^h=A_{fgh}\tfrac{1}{4!}e_b^{\;ijk}e^b_{\;ijk}e^f\wedge e^g\wedge e^h.$$What may be more useful for you is to average resullts such as$$A_{fgh}e^f\wedge e^g\wedge e^h=A_{gfh}e^g\wedge e^f\wedge e^h=-A_{gfh}e^f\wedge e^g\wedge e^h$$to get$$A_{fgh}e^f\wedge e^g\wedge e^h=A_{[fgh]}e^f\wedge e^g\wedge e^h.$$


问:
Fundamental group of a topological group: Inversion
Let $G$ be a topological group with neutral element $e$ and let $I:G\to G$ denote the inversion. Is it true that if $[\gamma]\in\pi_{1}(G,e)$, then $[I\circ\gamma]=[\overline{\gamma}]$, where $\overline{\gamma}$ denotes the inverse path, i.e. $\overline{\gamma}(t):=\gamma (1-t)$?

I know that there is  a similar statement for the multiplication $\mu:G\times G\to G$, namely that $\mu_{\ast}([\gamma_{1}],[\gamma_{2}])=[\gamma_{1}\cdot\gamma_{2}]$, where $\cdot$ denotes the concatenation.

答:
Let $(G,\bullet)$ be a topological group with identity element $e$. For any two continuous functions $\alpha,\beta:[0,1]\to G$, define $\alpha\bullet\beta:[0,1]\to G$ as $\alpha\bullet\beta(t)=\alpha(t)\bullet\beta(t)$ for all $t\in [0,1]$. Similarly define, $\alpha^{-1}:[0,1]\to G$ as $\alpha^{-1}(t):=\big(\alpha(t)\big)^{-1}$ for all $t\in [0,1]$. Let $C_e$ be the constant the loop in $G$ based at $e$. Also, $*$ stands for concatenation of two paths.

For two loops $\gamma,\delta$ in $G$ based at $e$ we have $\gamma*C_e\simeq_{\text{rel }e}\gamma\simeq_{\text{rel }e} C_e*\gamma$ as $C_e$ represents the identity element of $\pi_1(G,e)$, so that  $$\gamma*\delta=(\gamma*C_e)\bullet (C_e*\delta)\simeq_{\text{rel }e}\gamma\bullet\delta\simeq_{\text{rel }e}(C_e*\gamma)\bullet (\delta*C_e)=\delta*\gamma$$$$\text{ and putting }\delta=\gamma^{-1}\text{ we have, }$$$$\gamma*\gamma^{-1}\simeq_{\text{rel }e}\gamma\bullet \gamma^{-1}= C_e=\gamma^{-1}\bullet \gamma\simeq_{\text{rel }e}\gamma^{-1}*\gamma$$$$\implies \gamma^{-1}\simeq_{\text{rel }e}\overline \gamma$$$$\text{ as }\overline\gamma \text{ represents the inverse of  }[\gamma]\in \pi_1(G,e) .$$

In the above, to check the "$=$" just plugin an arbitrary $t\in [0,1]$ both left and right-hand sides.

Extra: From all the above facts we have $$\sigma*\overline \tau\simeq_{\text{rel }e}\sigma\bullet \tau^{-1}\simeq_{\text{rel
 }e}\tau^{-1}*\sigma\simeq_{\text{rel }e}\overline\tau*\sigma,$$ for
any two loops $\sigma,\tau$ in $G$ based at $e$. In particular,
$\pi_1(G,e)$ is abelian.


问:
Identity for a weighted sum of sines / sines with different amplitudes
I'm trying to simplify the following sum of sines with different amplitudes
$$
a \sin(\theta) + b \sin(\phi) = ???  \,\,\,\,\, (1)
$$

I know that
$$
a \sin(\theta) + a \sin(\phi) = a\cdot2\sin\left(\frac{\theta+\phi}{2}\right)\cos\left(\frac{\theta-\phi}{2}\right)
$$

But how do I do the same thing in the case where both sines have different amplitudes, as in (1).

Thanks!

答:
Answer is simple: you can't do that. This pattern is too complex to make some reasonable simplification.
Of course you can for example compare $a$ and $b$. Let's assume that $b>a$. Then $b=a+t$ for some positive $t$ and you can get
$$a\sin\theta+b\sin\phi=a\sin\theta+a\sin\phi+t\sin\phi=2a\sin\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}+t\sin\phi,$$
but I don't think that we could call that a simplification:)</p>


问:
When can a double integral be split into product of two integrals?
When can you do the following? I tried to understand from these lecture notes but I don't get whether it should be a "product function" or not.

答:
Notice that your formula is wrong. Nevertheless, if $g$ and $h$ are integrable on $[c,d]$ and $[a,b]$ respectively, it's always true that $$\int_a^b\int_c^d g(x)h(y)\,\mathrm d x\,\mathrm d y=  \left(\int_a^b h(y)\,\mathrm d y\right)\cdot \left(\int_c^d g(x)\,\mathrm d x\right).$$  Indeed, since $h(y)$ doesn't depend on $x$, $$\int_c^d g(x)h(y)\,\mathrm dx=h(y)\underbrace{\int_c^dg(x)\,\mathrm d x}_{=:J},$$
for all $y\in [a,b]$.
Finally, since $J$ doesn't depend on $y$,
\begin{align*}
\int_a^b\int_c^d g(x)h(y)\,\mathrm d x\,\mathrm d y&=\int_a^b h(y)J\,\mathrm d y\\
&=J\int_a^b h(y)\,\mathrm d y\\
&=\left(\int_c^d g(x)\,\mathrm d x\right)\cdot \left(\int_a^b h(y)\,\mathrm d y\right),
\end{align*}
as wished.


问:
Prove regular expressions are equal: $0^*1(1 + 00^*1)^*$ is $(0 + 1)^*1$
I tried:
$0^*1(1 + 00^*1)^* = 0^*1(1^* (00^*1)^*)^*$

but after that I'm not sure how to get further. I can definitely see that this expression always ends with 1 and that's what $(0 + 1)^*1$ but I don't see a rigorous chain of arguments to prove it.

I also tried constructing a DFA which is definitely useful in seeing the RE more clearly but still... no proof.

答:
I believe the DFA for the former regular expression would be too complicated to comprehend. Instead, as you observed that the latter regular expression gives all strings ending with $1$, we show that the regular expressions are equal by showing:

$0^*1(1+00^*1)^*$ can generate any string ending with $1$;

All strings generated by $0^*1(1+00^*1)^*$ ends with $1$.

I will show 1. and leave 2. as exercise. Pick any string ending with $1$, and we use the following algorithm to generate this string from $0^*1(1+00^*1)^*$:

If there are any leading $0$s, use $0^*$ for them.

A $1$ must come after the leading $0$s, and we have a $1$ next in line in the regex.

If there are any further $1$s, they either come with some leading $0$s (counting from the last $1$) or is directly after the previous $1$.

For the first case, we can handle it with $(00^*1)$.

For the second case, we can handle it with $1$.

For example, the string $00010010011$ can be handled as $0^*1(00^*1)(00^*1)(1)$ which is valid in $0^*1(1+00^*1)^*$. Now it remains to be shown that $0^*1(1+00^*1)^*$ always end in $1$.


问:
Quadratic equation with complex variable
I need to solve the following equation with $\omega \in \mathbb{C}$ and $\gamma, \omega_0 \in \mathbb{R}$ for its complex roots $\omega_1, \omega_2$
$$
  \omega^2 + \gamma \omega + \omega_0^2 = 0
$$

I've tried writing $\omega = x+iy$ but it seems to only make the equation more complicated. Because the variable $\omega$ is complex, I assume I can't use the standard quadratic formula. Should I just complete the square or even use $\omega$'s polar representation?

This seems like a very easy problem but complex equations don't click in my head yet and resources I have found on the internet just deal with real-valued equations that have complex solutions.

Thanks!

答:
The quadratic formula applies also to cases where the roots are complex. $$\omega = \frac{-\gamma\pm \sqrt{\gamma^2 -4\omega_0^2}}{2} $$

Depending on the sign of the $\gamma^2 -4\omega_0^2$, $\omega$ may be imaginary/real.


问:
Proving that $f'$ is not continuous
Let $f: \mathbb{R} \to \mathbb{R}$ be:

$$f(x)=\left\{\begin{matrix}
x^\frac{2}{3} \sin(\frac{1}{x}) & x>0\\
0 & x \le0
\end{matrix}\right.$$

I have calculated $f'$ here:

$$f'(x)=\left\{\begin{matrix}
\frac{3}{2}x^\frac{1}{2} \sin(\frac{1}{x})-\frac{1}{\sqrt x} \cos \frac{1}{x} & x>0\\
0 & x \le0
\end{matrix}\right.$$

So, I want to prove that $f'$ is not continuous (at least this is what I think).

Now I have 2 options: proving that $\lim_{x \to x_{0}}f'(x) \ne f'(x_{0})$ or checking if $f'$ is differentiable, meaning trying to prove that $f''$ does not exists.

I have tried both options with no luck...

Any hints will be appreciated! :)

Thanks a lot!

答:
If we assume that the function $f'$ is continuous, then for any sequence $x_n \rightarrow 0$ such that $x_n \geq 0$ for all $n$, the following must be true: $f'(x_n) = \frac{3}{2} x_{n}^{\frac{1}{2}} \sin (\frac{1}{x_n}) - \frac{1}{\sqrt{x_n}} \cos(\frac{1}{x_n}) \rightarrow 0$.

To prove that $f'$ is discontinuous at $0$, you need to find a sequence $x_n$ as mentioned above, but with $f'(x_n)$ being divergent.

Such a counter-example could be $x_n = \frac{1}{2\pi n}$.

It is clear that $x_n \rightarrow 0$ and that $x_n \geq 0$ for all $n$. However,

$f'(x_n) = \frac{3}{2} (\frac{1}{2\pi n})^{\frac{1}{2}} \sin (2\pi n) - \sqrt{2\pi n} \cos(2\pi n) = -\sqrt{2\pi n} \rightarrow - \infty$ as $n\rightarrow \infty$

In other words, $f'(x_n)$ is divergent, and thus, $f'(x)$ has a discontinuity at 0.

Title: Proving that $f'$ is not continuous


问:
Solving Modular Equations With Identities
$4+2x≡7 \pmod 8$
Find all possible solutions and note any identities.
Identify how you found the solutions.
Explain what identities are.</p>

答:
$\implies 2x\equiv3\pmod 8=3+8a$ for some integer $a$

So, as $3+8a$ is odd and $2x$ is even for integer $x,$ there is no solution

More generally, using Linear congruence theorem (Proof) $ax\equiv b\pmod n$ is solvable iff $(a,n)$ divides $b$


问:
Proving a limit using squeeze rule
I'm trying to figure out how to calculate the limit of

$f(x)= \begin{cases}
           \frac{3}{2}\cdot x^{\frac{1}{2}}\cdot\sin\left(\frac{1}{x}\right)-\frac{1}{\sqrt{x}}\cdot\cos\left(\frac{1}{x}\right) & x  > 0 \\
            \\
           0    & x≤0 \\ \end{cases}$

as $x$ tends to $0^{+}$ I know it doesn't exist and tends to $-∞$, I've been trying forever to prove there exist two functions that are bigger/smaller than $f$ in order to use the squeeze rule but I'm begging to think there is some other easier way..

答:
It is better to use a well known fact that as $x\to 0$ the functions $\sin(1/x),\cos (1/x)$ oscillate finitely between $1$ and $-1$. Most introductory calculus textbooks will present this example of oscillating function while dealing with the concept of limits.

Now as $x\to 0^+$ we can see that the first term of $f(x) $ namely $(3/2)x^{1/2}\sin(1/x)$ lies between $-(3/2)x^{1/2}$ and $(3/2)x^{1/2}$. Thus by Squeeze Theorem this term tends to $0$.

The next term $(1/\sqrt{x})\cos(1/x)$ consists of a factor which tends to $\infty $ and the other factor oscillates between $1$ and $-1$ and hence the term oscillates infinitely between $\infty$ and $-\infty $.

It follows that $f(x) $ is made of two terms one of which has limit $0$ and the other one oscillates infinitely. Hence $f(x) $ oscillates infinitely as $x\to 0^+$ and the limit of $f(x) $ as $x\to 0^+$ does not exist.

Title: Proving a limit using squeeze rule


问:
Difference between lim and lim sup of a function with a so called "blow up"?
Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a function with a "blow up" in finite time i.e. $$\limsup\limits_{t\uparrow T_{max}}|f(t)|=\infty.$$
I don't unterstand the difference between lim and lim sup in this case. How would the definition look like written in quantifiers?

答:
Here's the general definition:

Let $X$ be a topological space, $E\subset X$, $a\in\overline{E}$ (the topological closure) and $f:E\to [-\infty,\infty]$ a function. We define the limit superior as
\begin{align}
\limsup\limits_{x\to a}f(x):= \inf\limits_{\text{$a\in U$ open}} \sup\limits_{x\in (E\cap U)\setminus\{a\}}f(x).
\end{align}

The thing to notice here is that this ALWAYS exists as an element of $[-\infty,\infty]$. Contrast this with the ordinary limit which need not always exist. So, the point is that the limit superior captures the information about how large a function eventually gets as you get closer to the point $a$. Note that we can define $\liminf$ similarly in such generality (just interchange $\inf$ and $\sup$ above), and then one can show that $\lim$ exists if and only if $\limsup$ and $\liminf$ are equal, in which case $\lim=\liminf=\limsup$.

Here's a typical case of interest: $X=E=\Bbb{R}$ and $a\in\Bbb{R}$. In this case, the definition above reduces to the following:
\begin{align}
\limsup_{x\to a}f(x) = \inf_{\text{$a\in U$ open}} \sup_{x\in U\setminus \{a\}}f(x)
\end{align}
Now, notice that in $\Bbb{R}$, if $U$ is open and contains $a$ then there is an $\epsilon>0$ such that the $\epsilon$-ball is contained in $U$, i.e  $B_{a,\epsilon}:= \{x\in\Bbb{R}\,:\, |x-a|<\epsilon\}\subset U$. Conversely, for any $\epsilon>0$, there is an open set $U$ containing $a$ and contained in $B_{a,\epsilon}$ (this is trivial since $B_{a,\epsilon}$ works). Therefore, we can simplify the definition further:
\begin{align}
\limsup_{x\to a}f(x)&= \inf_{\epsilon>0} \sup_{0<|x-a|<\epsilon}f(x).
\end{align}
Finally, observe that the function $\epsilon\mapsto \sup\limits_{0<|x-a|<\epsilon}f(x)$ is weakly decreasing, because "taking the supremum over a smaller set gives a smaller value". Hence, we can replace $\inf\limits_{\epsilon>0}$ with $\lim\limits_{\epsilon\to 0^+}$ to get
\begin{align}
\limsup_{x\to a}f(x)&=\lim_{\epsilon\to 0^+}\sup_{0<|x-a|<\epsilon}f(x). \tag{$*$}
\end{align}

As you can see once again from this equality, this just reiterates what I said above: $\limsup$ captures the information about "how large" (this is the supremum part) the function becomes as you get closer and closer to $a$ (this is the $\lim\limits_{\epsilon\to 0^+}$ part). By the way, if you haven't taken some introductory topology, the previous stuff may be too abstract, so you can just take $(*)$ as your starting definition. Once again, the point here is that $\limsup\limits_{x\to a}f(x)$ is a quantity which for ANY function $f$, ALWAYS exists in $[-\infty,\infty]$.

At this point, we can also try to define things like $\limsup\limits_{x\to a, x\in S}$, where $S$ is a certain subset of the domain of your function; this will mean we take the $\limsup$ of the function while restricting $x$ to only come from the set $S$. So, in the case of the real line, we can define $\limsup\limits_{x\uparrow a}$ accordingly:
\begin{align}
\limsup_{x\uparrow a}f(x)&:= \lim_{\epsilon\to 0^+}\sup_{a-\epsilon<x<a}f(x).
\end{align}

With this, we can answer your question of what
\begin{align}
\limsup\limits_{t\uparrow T}|f(t)|:= \lim_{\epsilon\to 0^+}\sup_{T-\epsilon<t<T}|f(t)|=\infty
\end{align}
means in terms of quantifiers:

Explicitly, this means for every $R>0$, there is a $\delta>0$ such that for all $\epsilon>0$, if $0<\epsilon<\delta$, then $\sup\limits_{T-\epsilon<t<T}|f(t)|>R$.

As I've mentioned above, the limit exists if and only if $\limsup$ and $\liminf$ are equal, in which case all three are equal. So, the thing is that in your case, $\lim\limits_{t\uparrow T}|f(t)|$ need not exist, though the $\limsup$ always exists. But, in the case that the $\lim$ exists, then it will have to also equal the $\limsup$ which is $\infty$, which means:

For every $R>0$ there is a $\delta>0$ such that for all $t\in\Bbb{R}$, if $T-\delta<t<T$ then $|f(t)|>R$.


问:
Showing that $d\theta(\partial\Psi/\partial\theta)=1$
In Bachman's A Geometric Approach to Differential Forms, second edition, the following is exercise 8.4.

Often we parametrize surfaces in $\mathbb{R}^3$ by starting with cylindrical or spherical coordinates and expressing one of the parameters in terms of the other two. So, for example, by starting with spherical cooridnates and expressing $\rho$ as a function of $\theta$ and $\phi$, we may end up with a parametrisation $\Psi(\theta,\phi)=(x,y,z)$. However, we may also view $\theta(x,y,z)$ as a function on $\mathbb R^3$ that gives the $\theta$ coordinate of the point $(x,y,z)$, when it is expressed in spherical coordinates. Hence, it is often the case that we have parametrisations of the form $\Psi(\theta,\phi)$ such that $$\theta(\Psi(\theta,\phi))=\theta\text{ and }\phi(\Psi(\theta,\phi))=\phi.$$When we think of $\theta(x,y,z)$ as a 0-form on $\mathbb R^3$, we can differentiate to get a 1-form $d\theta$. Show that $$d\theta\left(\frac{\partial\Psi}{\partial\theta}\right)=d\phi\left(\frac{\partial\Psi}{\partial\phi}\right)=1$$and$$d\theta\left(\frac{\partial\Psi}{\partial\phi}\right)=d\phi\left(\frac{\partial\Psi}{\partial\theta}\right)=0.$$

I think I am mainly confused by how $d\theta$ acts on $\partial\Psi/\partial\theta$ (and similarly with $\phi$ replacing $\theta$). If given an explicit 1-form on some tangent space $T_pX$, say $$\omega_p=a_1dx+a_2dy+a_3dz,$$ and explicitly knowing some tangent vector $v=\langle b_1,b_2,b_3\rangle_p$ in that tangent space, I know how to compute $$\omega_p(v)=a_1b_1+a_2b_2+a_3b_3.$$
However here I am not sure how to geometrically visualise $\partial\Psi/\partial\theta$ as a tangent vector to $\operatorname{im}\Psi$ (the surface being paramatrised by $\Psi$), how to compute it, and hence how to find $d\theta(\partial\Psi/\partial\theta)$. For example since $\theta=\arctan(y/x)$ then I find that $$d\theta=-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$ and if we let $\Psi=(\Psi_1,\Psi_2,\Psi_3)$ I can find that $$\frac{\partial\Psi}{\partial\theta}=\left\langle\frac{\partial\Psi_1}{\partial\theta},\frac{\partial\Psi_2}{\partial\theta},\frac{\partial\Psi_3}{\partial\theta}\right\rangle.$$
Now, I evaluate that
$$d\theta\left(\frac{\partial\Psi}{\partial\theta}\right)=-\frac{y}{x^2+y^2}\frac{\partial\Psi_1}{\partial\theta}+\frac{x}{x^2+y^2}\frac{\partial\Psi_2}{\partial\theta}$$
which is clearly not $1$. Thus I get a feeling that there is some huge conceptual gap which is preventing me from solving the problem, one which I cannot seem to find myself.

Thanks in advance for any help---just pointing out some concept or idea that I'm missing, even if not a full solution, would already be of great assistance!

答:
Given $f : \mathbb{R}^2 \to \mathbb{R}$, we have $\frac{\partial f}{\partial x}(x,y) = d_{(x,y)} f \begin{pmatrix} 1 \\ 0\end{pmatrix}$. Therefore we have the following,

$$ \frac{\partial}{\partial \theta} \left( \theta \circ \Psi(\theta, \phi) \right) = d(\theta \circ \Psi)(\theta, \phi) \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

Now use chain rule on the RHS,

$$\begin{align*} d(\theta \circ \Psi)(\theta, \phi) \begin{pmatrix} 1 \\ 0 \end{pmatrix} &= d_{\Psi(\theta, \phi)} \theta \circ d_{(\theta, \phi)} \Psi \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ & = d\theta \begin{pmatrix} \frac{\partial \Psi}{\partial \theta} & \frac{\partial \Psi}{\partial \phi} \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} & \\ = d \theta \left(\frac{\partial \Psi}{\partial \theta}\right) \end{align*}$$

But isn't this $1$ since $\frac{\partial}{\partial \theta} \left( \theta \circ \Psi(\theta, \phi) \right) = \frac{\partial}{\partial \theta} (\theta)$ ? Take the other partial to get the zero result. You can then repeat this with $\phi$.


答:
Well: $$(x+iy)^2+\gamma(x+iy)+z^2=0\implies x^2-y^2+2xyi+\gamma x + i\gamma y +z^2=0$$

$$\implies x^2-y^2+x+z^2=0; iy(2x+\gamma)=0$$
by separating the real and imaginary parts.

Two cases are seen here: either $y=0$ (so $\omega$ is not complex) or $x=-\frac \gamma 2$, from which we solve the quadratic in $y$.

Title: Quadratic equation with complex variable

Body:
I need to solve the following equation with $\omega \in \mathbb{C}$ and $\gamma, \omega_0 \in \mathbb{R}$ for its complex roots $\omega_1, \omega_2$
$$
  \omega^2 + \gamma \omega + \omega_0^2 = 0
$$

I've tried writing $\omega = x+iy$ but it seems to only make the equation more complicated. Because the variable $\omega$ is complex, I assume I can't use the standard quadratic formula. Should I just complete the square or even use $\omega$'s polar representation?

This seems like a very easy problem but complex equations don't click in my head yet and resources I have found on the internet just deal with real-valued equations that have complex solutions.

Thanks!


答:
The sequence may have more limit points. In that case, the limit doesn't exist and limes superior is the supremum of those limit points. For example take sequence:$$
0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,8,\ldots
$$
This sequence has two limit points: $0$ and $\infty$. So the limit doesn't exist but limes superior does and it's $\infty$.

In your case, you are talking about a limit of a function but I gave you an example of limit of a sequence. Well, you can create a function which is analogous to this sequence, for example $f(x)=\frac{x}2\left(\sin x + 1\right)$. In this case, the limit points are all points in $[0,\infty]$ and limes superior is $\infty$.

Title: Difference between lim and lim sup of a function with a so called "blow up"?

Body:
Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a function with a "blow up" in finite time i.e. $$\limsup\limits_{t\uparrow T_{max}}|f(t)|=\infty.$$
I don't unterstand the difference between lim and lim sup in this case. How would the definition look like written in quantifiers?