答:
One trick is to introduce "ghost cells" like $x_{-1}$ with the definition per boundary condition
$$
\frac{x_1-x_{-1}}{2Δt}=\dot x(0)\implies x_{-1}=x_1-2Δt\,\dot x(0)
$$
Now you can either add this to the system, or directly eliminate $x_{-1}$ against the differential equation centered at $x_0$.

Another way to use the boundary condition with a sufficiently high order is to use a forward differentiation formula of that order, like in the second order
$$
\frac{-3x_0+4x_1-x_2}{2Δt}=\dot x(0).
$$

Title: Second Order Finite Difference Scheme to Solve Initial Value Problem

Body:
I would appreciate help in a problem that I cannot figure out where am I doing the mistake. The mistake could be something I am unaware of so please excuse my lack of knowledge.

I am trying to solve the equation of motion,
$$M \ddot{x} + K x = F,$$
where $M$, $K$, and $F$ are variables independent of $x$, and $\ddot{x}=\frac{d^2x}{dt^2}$. Using a central difference finite scheme, I rewrote the equation as,

$$M \frac{ x_{i+1} - 2x_i + x_{i-1} }{ {\Delta t} ^2 } + Kx = F.$$

Then, I am solving this equation for $x_{i+1}$ at every time step $i$ as,

$$ x_{i+1} = \frac{{\Delta t}^2}{M} \cdot \left( F - K x + \frac{M}{{\Delta t}^2} \cdot 2 x_i - \frac{M}{{\Delta t}^2} \cdot x_{i-1} \right).$$

Now to solve this problem, one needs two initial conditions, to my knowledge, namely, $x(0)$ and $\dot{x}(0)$. At the first time step, I am setting $x_i$ to the $x(0)$ initial condition. Now I am stuck with two problems:

What value should I set $x_{i-1}$ at first time step?

How can I apply the initial condition $\dot{x}(0)$?


问:
Union of closed points of $\mathbb{A}^2_k = \operatorname{Spec}(k[x,y])$
The question is as follows: For finitely many closed points $x_1,\dots,x_n \in \mathbb{A}^2_k$, for $k$ a field (not assumed algebraically closed) show that their union can be written as $V(f,g)$ for $f,g\in k[x,y]$.

This is part 2 of a problem, where in the first problem we classify closed points of $\mathbb{A}^2_k$. These correspond to maximal ideals in $k[x,y]$, and I can show that these are of the form $V(f,g)$ where $f\in k[x]$ and $g\in k[x,y]$ are irreducible.

One can try to prove the above by induction, so for two closed points $x_1= V(f_1,g_1)$ and $x_2=V(f_2,g_2)$, we have that $\{x_1,x_2\}=V(f_1,g_1)\cup V(f_2,g_2)=V((f_1,g_1)\cdot (f_2,g_2))$ where
$(f_1,g_1)\cdot (f_2,g_2)=(f_1f_2,f_1g_2,g_1f_2,g_1g_2)$ denotes the product of the ideals (also equal to the intersection by the Chinese Remainder Theorem). The problem is that is will have $4$ generators, and we need to cut that down to two and this is where I am stuck. The immediate try of $V(f_1f_2,g_1g_2)$ consists of four points rather than two.

Any help or hints are appreciated (the linked duplicates seem to deal with the case of an algebraically closed field. The closed points don't have coordinates in $k^2$)!

答:
I claim that we can always reduce to the case where all our points are $k$-rational points, and then the answer here shows that any finite collection of $k$-rational points in $\Bbb A^n_k$ can be defined by $n$ polynomials with coefficients in $k$.

To do this, we want to find a Galois extension $k\subset F$ so that all our closed points become rational points over $F$. To do this, we use the fact that any maximal ideal in $k[x_1,\cdots,x_n]$ contains a unique monic irreducible polynomial in $x_i$: simply take the minimal polynomial of the image of $x_i$ in $k[x_1,\cdots,x_n]/m$, a finite extension of $k$. Now compile all of these minimal polynomials in to a finite list, take the splitting field of the first polynomial, remove all the polynomials in the list that split over this extension, and repeat: because the splitting field of any polynomial is Galois and the composition of Galois extensions is Galois, we get a Galois extension $k\subset F$ so that after base change, our finite collection of points is $F$-rational.

From here, we may apply the linked answer to find $n$ polynomials which vanish exactly at the base change of our finite collection of points. By construction of these polynomials, we see that they are $Gal(F/k)$ invariant, and so they're all actually in $k[x_1,\cdots,x_n]$ and therefore their vanishing locus in $\Bbb A^n_k$ is exactly the collection of closed points we started with.

ContentLicense: CC BY-SA 4.0


问:
Difficulties to prove that a function is convex
Is the positive function $f(x) = \frac{{{x^D}}}{{x - A}}$ defined in $]A, + \infty[$ with $A \ge 1$ and $D > 1$ a convex function for all these $A$ and $D$ ?
I think so but I have a hard time proving it. Thanks.

答: It is not always convex, as demonstrated by this graph. If it were convex, then the purple dot should have been above the red one.</p>


问:
Optimum way to find the determinant of the following matrix.
Given a matrix like the following, is there a way, only by using elementary transformations between rows and columns, to find its determinant? I know it would be easy to compute it or to try and find a pattern, but is there a faster way using that method?

The matrix goes like:

1, if $i = j$.

$\lambda$, if $i = j > 1$.

k, if $i \neq j$.

Where $i$ and $j$ indicate the row and the column, respectively.

答:
$$M=\begin{bmatrix}1&k&k&k&\dots&k\\
                        k&\lambda&k&k&\dots&k\\
                        k&k&\lambda&k&\dots&k\\
                        \vdots&&\ddots&&\vdots\\
                        k&k&k&\dots&k&\lambda
                        \end{bmatrix}_{n\times n}$$

Subtract the first row from all the others, giving
$$M'=\begin{bmatrix}1&k&k&k&\dots&k\\
                        k-1&\lambda-k&0&0&\dots&0\\
                        k-1&0&\lambda-k&0&\dots&0\\
                        \vdots&&\ddots&&\vdots\\
                        k-1&0&0&\dots&&\lambda-k
                        \end{bmatrix}_{n\times n}$$

Subtract the last row from all other rows but the first, giving

$$M''=\begin{bmatrix}1&k&k&k&\dots&k&k\\
                        0&\lambda-k&0&0&\dots&0&k-\lambda\\
                        0&0&\lambda-k&0&\dots&0&k-\lambda\\
                        \vdots&&\ddots&&\vdots&&\vdots\\
                        k-1&0&0&\dots&&0&\lambda-k
                        \end{bmatrix}_{n\times n}$$

Of course $\det(M'')=\det(M)$

Expanding $\det(M'')$ by minors along the first column, $$
\det(M'')=(\lambda-k)^{n-1}-(-1)^n(k-1)\left\lvert\begin{matrix}k&k&k&\dots&k&k\\
                        \lambda-k&0&0&\dots&0&k-\lambda\\
                        0&\lambda-k&0&\dots&0&k-\lambda\\
                        \vdots&&\ddots&&\vdots&\vdots\\
                        0&0&\dots&&\lambda-k&k-\lambda\\
                        \end{matrix}\right\rvert_{n-1\times n-1}\\
                        =(\lambda-k)^{n-1}-(-1)^n(k-1)k(\lambda-k)^{n-2}
                        \left\lvert\begin{matrix}1&1&1&\dots&1&1\\
                        1&0&0&\dots&0&-1\\
                        0&1&0&\dots&0&-1\\
                        \vdots&&\ddots&&\vdots&\vdots\\
                        0&0&\dots&&1&-1\\
                        \end{matrix}\right\rvert_{n-1\times n-1}\\
                        $$
and the problem reduces to computing the last determinant.  I leave that to you, but I have a couple of remarks.  If you move the second row into the last spot, then the sign of the determinant changes, and we get a matrix like $M''$, upper triangular except fo a $1$ in the first column of the last row, so we can expand by minors along the first column easily.  A few experiments lead me to believe that the determinant of an $n\times n$ matrix of this form is $(-1)^{n+1}n$.  This looks easy to prove by induction, I think.

I hope I've typed all this correctly.  Check carefully.  I've made several mistakes today.


问:
Writing an expression using logic
Write an expression using letters $\land, \lor, and$ $\neg$ which has the following truth table:

$$\begin{array}{ccc|c}
P&Q&R&???\\ \hline
T&T&T&F\\
T&T&F&T\\
T&F&T&F\\
T&F&F&T\\
F&T&T&T\\
F&T&F&F\\
F&F&T&F\\
F&F&F&F
\end{array}
$$

How would you in general solve for ??? I was told there is a mechanical way of doing it but I don't see it.

答:
I’ll get you started. Look at the last column of the truth table: you want your statement to be true when

$$P\text{ is true and }Q\text{ is true and }R\text{ is false}\tag{1}$$

$$\mathbf{OR}$$

$$P\text{ is true and }Q\text{ is false and }R\text{ is false}\tag{2}$$

$$\mathbf{OR}$$

$$P\text{ is false and }Q\text{ is true and }R\text{ is true}\;,\tag{3}$$

and for no other combination of truth values of $P,Q$, and $R$.

Is there a simple statement that’s true exactly when $P$ and $Q$ are true and $R$ is false? Sure: $$P\land Q\land\neg R\;.$$

Now find a similar statement that is true exactly when $(2)$ is true, and another one that’s true exactly when $(3)$ is true. Once you have these three statements, how should you combine them to get what you want?

ContentLicense: CC BY-SA 3.0

Title: Writing an expression using logic


问:
Does $\prod_{n=1}^\infty\left(1+\frac{1}{p_n}\right)$ diverge?
$p_n$ is the $n^\text{th}$ prime number.

This looks like Euler's product for zeta function, but I don't know if they are related.

答:
Yes it does diverge.

$$\prod_{n=1}^\infty\left(1+\frac{1}{p_n}\right) \ge 1+\sum_{n=1}^{\infty} \frac{1}{p_n}$$

the RHS of which is well-known to diverge.

In fact, $\sum_{n=1}^{N} \frac{1}{p_n} = \Omega(\log \log N)$.


问:
Given $dx/dt=\sin(y)+0.1\sin(x),dy/dt=-\sin(x)+0.2\sin(y)$ and $F(x,y)=\cos(x)+\cos(y)$ Determine if $F$ is a lyapunov function near $(0,0)$

Given $dx/dt=\sin(y)+0.1\sin(x),dy/dt=-\sin(x)+0.2\sin(y)$ and $F(x,y)=\cos(x)+\cos(y)$ Determine if $F$ is a lyapunov function near $(0,0)$

I computed the jacobian matrix at $(0,0)$ as $\begin{pmatrix} 0.1&1\\-1&0.2\\\end{pmatrix}$

Which has eigenvalues $3/20 \pm i\sqrt{399}$

So both eigenvalues have positive real part, so this system should be asymptotically stable? Which I believe means it should also be lyapunov stable.

I have this lyapunov stability theorem,

Assume $U$ is an open neighbourhood of $x^*, F:U\to \mathbb{R}^n$ and $F(x^*)=0$. Further assume that there is a $C^1$-function $V:U\to\mathbb{R}$ satisfying $V(x^*)=0$, $V(x)>0$ for $x\neq x^*$. Then, V is called a lyapunov function for $x'=F(x)$ and:

a. If $V'(x)\leq 0$ for all $x\in U$, then $x^*$ is L-stable
b. If $V'(x)<0$ for all $x\in U\{x^*\}$, then $x^*$ is asymptotically stable.
c. If $V'(x)>0$ for all $x\in U\{x^*\}$ then $x^*$ is unstable.

I can show that $F'=-\sin(x)x'-\sin(y)y'=-\sin(x)\sin(y)-0.1\sin^2(x)+\sin(x)\sin(y)-0.2\sin^2(y)=0.1\sin^2(x)-0.2\sin^2(y)$

But I'm not sure how to relate that to the above theorem to say whether or not this is a lyapunov function.

答:
You missed to transfer a minus sign, it should be
$$
F'(x,y)=-0.1\sin^2x-0.2\sin^2y.
$$
As $F$ has a local maximum at $(0,0)$, it would be more in tune with the traditional definition to call $-F$ a Lyapunov function.

Per the sign of the derivative, the solutions close to $(0,0)$ move downwards in the levels of $F$, which means away from the origin.

That the eigenvalues have a positive real part means that the origin is unstable, you need negative real parts for a stable fixed point.


问:
Examples of inner products defined on a matrix vector space
Is $\langle A,B \rangle = \operatorname{tr}(A^TB)$ the only inner product defined for $M(2\times 2)$?
If there are others, what are some examples?

答:
Pick any basis $V_1,V_2,\ldots,V_{n^2}$ of $M_n(\mathbb R)$. Denote by $\langle X,Y\rangle_F$ the Frobenius inner product $\operatorname{tr}(X^TY)$. Then
$$
f:A\mapsto\left(\langle A,V_1\rangle_F,\,\langle A,V_2\rangle_F,\,\ldots,\langle A,V_{n^2}\rangle_F\right)
$$
is an isomorphism between $M_n(\mathbb R)$ and $\mathbb R^{n^2}$. Therefore we can define an inner product on $M_n(\mathbb R)$ by
$$
\langle A,B\rangle:=\langle f(A),f(B)\rangle_{\mathbb R^{n^2}}
=\sum_{j=1}^{n^2}\langle A,V_j\rangle_F\,\langle B,V_j\rangle_F,\tag{1}
$$
where $\langle \cdot,\cdot\rangle_{\mathbb R^{n^2}}$ is the standard inner product (i.e. dot product) on $\mathbb R^{n^2}$. Actually the converse is also true: every inner product on $M_n(\mathbb R)$ can be expressed in the form of $(1)$ for some basis $V_1,V_2,\ldots,V_{n^2}$ of $M_n(\mathbb R)$.

For instance, for the Frobenius inner product on $M_2(\mathbb R)$, we can simply take $\{V_1,V_2,V_3,V_4\}$ as the standard basis
$$
V_1=\pmatrix{1&0\\ 0&0},
V_2=\pmatrix{0&0\\ 1&0},
V_3=\pmatrix{0&1\\ 0&0},
V_4=\pmatrix{0&0\\ 0&1}.
$$
One can directly verify that $\langle A,B\rangle_F$ is indeed equal to $\sum_{j=1}^4\langle A,V_j\rangle_F\,\langle B,V_j\rangle_F$.


问:
Monotone function with the properties below
Does there exist any strictly decreasing function $f: \Bbb R \to [0,+\infty)$ such that $f(x)-\sqrt{f(x)}$ is strictly increasing function on the whole $\Bbb R$?

Now, we had calculus at high school, but, at the moment, we haven't gone through it in analysis lectures, so, on the first exam we are supposed to use things mentioned in lectures only.
I tried to use the fact that if $f(x)$ is a monotonically decreasing function, then the composition $$f(f(x))$$ is monotonically increasing. I might have chosen the wrong path. I also considered inverses, but it didn't help me either.
Since I have a sum, I assumed, without deriving, I can't look at it as a real composition.
$$\operatorname{Ran}(f)=[0,+\infty)$$ implies $\sqrt{f(x)}$ would be defined well.
I constantly kept returning to some decreasing exponential function $g(x)=a^{x}, a\in(0,1)$ although a rational function with odd exponents seemed equally useful/useless...
I thought that the degree of $f(x)$ must be less than the degree of  $\sqrt{f(x)}$ in terms of that exponential function. I haven't proven it.
The role of the leading coefficient also seemed irrelevant in this case, it reminded me of polynomials, but they can't be injective and always positive at the same time globally, without restriction, am I right?
I "concluded" there is no such function that satisfies the constraints.

答:
Here is a bit different, maybe more elementary approach.

If $f:\Bbb R\to[0,+\infty)$ is strictly decreasing, then for any $x_1,x_2\in\Bbb R:$
$$x_1<x_2\implies f(x_1)<f(x_2)\implies\sqrt{f(x_1)}<\sqrt{f(x_2)}\space.$$

Let $g(x)=\sqrt{f(x)}$ and let $h(x)=f(x)-\sqrt{f(x)}=g(x)^2-g(x),$ so $$\begin{aligned}h(x_1)&=g(x_1)^2-g(x_1)\\h(x_2)&=g(x_2)^2-g(x_2)\space.\end{aligned}$$

We see that $g$ is decreasing and, by the assumption, $h$ is increasing. Therefore $$\begin{aligned}h(x_2)-h(x_1)&=g(x_2)^2-g(x_2)-g(x_1)^2+g(x_1)\\&=(g(x_2)-g(x_1))(g(x_2)+g(x_1)-1)\\&=\underbrace{(g(x_1)-g(x_2))}_{>0}(1-(g(x_1)+g(x_2)))>0\\\\\implies1-(g(x_1)+g(x_2))&>0\\\iff 1>g(x_1)+g(x_2)&>2g(x_2)\\\iff g(x_2)=\sqrt{f(x_2)}&<\frac12=\sqrt{\frac14}\\\iff f(x_2)<\frac14\end{aligned}$$
Since $x_1,x_2$ were arbitrary, $f(x)<\frac14\space\forall x\in\Bbb R.$


答:
There are exactly three rows which evaluate to "true = $T$".

We can take the first row, and see that $P \land Q \land \lnot R$ evaluates true.

Doing this with each row, we get that the truth table evaluates true when:

$$(P \land Q \land \lnot R) \lor (P \land \lnot Q \land \lnot R) \lor (\lnot P \land Q \land R)$$

This method gives you disjunctive normal form. And it is possible that it might be simplified. But this conveys precisely what the truth table conveys.

Title: Writing an expression using logic

Body:
Write an expression using letters $\land, \lor, and$ $\neg$ which has the following truth table:

$$\begin{array}{ccc|c}
P&Q&R&???\\ \hline
T&T&T&F\\
T&T&F&T\\
T&F&T&F\\
T&F&F&T\\
F&T&T&T\\
F&T&F&F\\
F&F&T&F\\
F&F&F&F
\end{array}
$$

How would you in general solve for ??? I was told there is a mechanical way of doing it but I don't see it.


问:
Finding a positive integer that can't be expressed in a certain form
I attended a math speech and the speaker left the following question as an exercise:

Which positive integer cannot be expressed in the form $$x^2+2y^2+5z^2+5w^2?$$

I've trying to solve it but I haven't accomplished it yet. Any help is welcome

答:
There are very few integers that are perfect squares, and none of them are negative. So we can order them starting from the smallest one, as $0$, $1$, $4$, $9$, $16$, $25$, etcetera. Now for any given positive integer $n$, there are only finitely many candidates for the squares $x^2$, $y^2$, $z^2$ and $w^2$ because all the coefficients are positive. Simply check them all to see whether $n$ can be expressed in this way.

Of course there is seemingly no guarantee that you will ever find a positive integer $n$ that is not of this form, even if it exists. But there is a wonderfully surprising theorem, the fifteen theorem, that states that if every positive integer up to $15$ can be expressed as such a sum of squares, then every positive integer can be expressed as such a sum of squares. So the approach described above only requires you to check up to $n=15$ to find a positive integer not of this form, if it exists.

A quick check shows that every integer $n<15$ is of this form, but that $n=15$ is not. A more precise version of the fifteen theorem then tells us that every integer $n>15$ is also of this form, thanks to lulu's comment with this link. So $n=15$ is the unique positive integer not of this form.


问:
Proving two sets E and G have no common elements
Say I have a set $E$ with an arbitrary number of elements which are of the form $f$ i.e. $E=\{f(x):x \in\Bbb N\}$ and a set $G$ with an arbitrary number of elements which are of the form $h$ i.e. $G=\{h(a):a\in\Bbb Z\}$.

What would be the general strategy of showing that these two sets are disjoint. I have a basic idea on how to start: Assume for the sake of contradiction that the two sets do have some common elements. How would I then show that this leads to a contradiction allowing me to conclude that $E$ intersection $G$ is the null set. Once again, I just want a general approach to solving this problem. Thanks.

答: Suppose there exists an element that is in both sets and reach a contradiction. Or that if $x$ is an element of one of the sets, then it can't be an element of the other set.


问:
Proof reviewer needed
Originally I wanted to elaborate one idea of mine in my answer to this one question:
How to prove that $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{|1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$

I'm trying to check whether: $$S(N) = \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|}\le \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$ for all natural numbers $N$ and all natural numbers $i<N+1$ is true.

The inequation $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|} + \frac{|y|}{1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$ is the special case for $i=2$ with $x=x_0, y=x_1, z=x_2$

For $N=0$ we have $\frac{|x_0|}{1+|x_0|} \le \frac{|x_0|}{1+ 0}$ for all $x_0$

Let's suppose for a certain $N$ that we have  $$S(N) = \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|} \le \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$ for all natural numbers $N$ and all natural numbers $i<N+1$.

and prove this inequality for $S(N+1)$ which means let's prove the following:

$$S(N+1) \le \sum_{i=0}^{N+1} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N+1} x_j|}$$  all natural numbers $i<N+2$.

We first will use $\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$ like proved in this link:Prove $\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$.

for $a = x_{N+1}$ and $ b= \sum_{i=0}^{N} x_i$

$$\frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|}$$

means that

$$\frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + S(N)$$

means that $$ \frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$

means that

$$S(N+1) \le \sum_{i=0}^{N+1} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N+1} x_j|}$$

Proved!

答:
You can show that

$$ \frac { | \sum_i x_i | } { 1 + |\sum_i x_i| } \leq \sum_i \frac{|x_i| } { 1 + \sum_j |x_j| } < \sum_i \frac{|x_i| } { 1 + \sum_{j\neq i} |x_j| }.$$

See my solution for the full details.

Title: Proof reviewer needed


问:
Trouble Understanding Proof of Lagrange's Identity Using Binomial Formula
I've been reading a book on elementary mathematics and am having trouble understanding a proof that they give for Lagrange's identity using the binomial formula.

Lagrange's identity is stated as :
\begin{equation}
\sum_{j=0}^{n} \binom{n}{j}^{2} = \binom{2n}{n}
\end{equation}
In the proof they say :

"Writing $(1+x)^{2n} = (1+x)^{n}(1+x)^{n}$ and applying Newton's binomial formula to both sides we get :
\begin{equation}
\binom{2n}{n} = \sum_{i+j=n} \binom{n}{i} \binom{n}{j} = \sum_{i=0}^{n} \binom{n}{i}\binom{n}{n-i} = \sum_{i=0}^{n} \binom{n}{i}^{2}
\end{equation}
$\square$."

A lot of detail seems to be missing here. Could someone provide the missing details here ? I cannot understand the proof due to the elision of details.

答:
They forgot to say they compute the coefficient of $x^n$ in the product and they apply the general formula for the product of two polynomials:
$$\sum_{i=0}^n a_ix^i\cdot \sum_{j=0}^p b_jx^j=\sum_{k=1}^{n+p}\Bigl(\sum_{i+j=k}a_ib_j \Bigr)x^k ?$$


问:
We have $2$ congruent, regular $n$-gons in the plane. Compute the size of the symmetry group of this configuration.

Suppose we have two congruent regular n-gons in the plane. The symmetry group of this configuration is the number of ways to pick up both polygons and put them down so that they cover the same points. Find the size of this group using the orbit-stabilizer theorem.

Conceptually, I believe that I understand what I am being asked:

Two $n$-gons are congruent if they are of the same shape and size but represented in terms of some reflections/rotations of each other. So, we have two copies of the same $n$-gon. The symmetry group of a regular $n$-gon is the dihedral group of order $2n$. If I need to place down two copies of the same $n$-gon such that they cover the same points in the $x$-$y$ plane, then I need to choose coordinates $(x,y)$ where $x, y \in D_{2n}$. In other words, I am being asked to compute the size of $D_{2n}\times D_{2n}$, which is simply $(2n)^2 = 4n^2$.

My issues lie with my understanding of group actions on sets and the orbit-stabilizer theorem. I have attempted the problem below. I am very confused about how to apply relevant definitions to the problem at hand (specifically, how to apply these definitions to both $n$-gons). Please let me know if there are any mistakes/misconceptions in my solution.

Defintions

We define $\pi: G \to \operatorname{Sym}(X)$ as the action of a group $G$ on some set $X$ under a homomorphism $\pi$. The group $G$ is represented in a fashion specified by the homomorphism $\pi$ as permutations of the set $X$.

The orbit of $x \in X$ under the group action $\pi$ is defined as $\operatorname{Orb}_{\pi}(x) = \{y: \pi(g)(x) = y$ for some $ g \in G\}$

The stabilizer of $x \in X$ in $G$ is $G_x = \{g \in G: \pi(g)(x)=x\}$. In other words, $G_x$ is the subset of $G$, given $\pi$, which maps $x$ to itself.

The orbit stabilizer theorem states that: $|G| = |\operatorname{Orb}_{\pi}(x)||G_x|$

Solution

Let $X$ be the Cartesian product of the two sets of vertices of both $n$-gons. Then, we defined the group action of $G$ under $\pi$ on $X$ as the set of permutations of $X$ such that the two sets of vertices cover the same points on the $x,y$-plane. For clarity, let $V_1$ and $V_2$ represent both sets of vertices. A vertex of $V_1$ can only correspond to one vertex of $V_2$ given that they are permutations of each other.

Any vertex of $V_1$ can be taken to any other vertex of $V_1$ via a rotation. The same is true for $V_2$. This implies that there are $n$ transformations we can apply to each vertex of an $n$-gon such that the vertex changes positions. So, there are $n(n) = n^2$ ways to combine all possible positions of both sets of vertices. This corresponds to the size of $\operatorname{Orb}_\pi(x)$. In other words, we should have that $|\operatorname{Orb}_\pi(x)| = n^2$

Now, if we consider what transformations fix each of the vertices of an $n$-gon, we arrive at the identiy rotation and a reflexion about a line between the origin and some vertex of the $n$-gon. So, there are $2$ transformations that we can apply which fix the vertex of an $n$-gon. Combining these transformations between the two $n$-gons yields 4 possible ways to fix $(v_1\in V_1,v_2\in V_2) \in X$. This corresponds to the size of $G_x$. In other words, we should have that $|G_x|=4$.

$\implies$ Applying the orbit-stabilizer theorem, we get that: $|G|=4n^2$.

答:
First a few remarks:

The group $D_{2n}$ describes the ways you can pick up a regular $n$-gon and place it back on the plane so that it covers the same points. Picking up two congruent regular $n$-gons and placing them back, there $2$ ways to choose where to place them, and then $|D_{2n}|=2n$ ways to orient them in that place. So you get more than $4n^2$ ways to place the two $n$-gons!

Defining an action as a group homomorphism $\pi:\ G\ \longrightarrow\ \operatorname{Sym}(X)$, an element $g\in G$ acts on $x\in X$ by mapping $x$ to $\big(\pi(g)\big)(x)$. It is common to omit the homomorphism $\pi$ from the notation, as it is usually clear from the context, and to simply write $g\cdot x$ or even $gx$. I'll do that from here on.

A cleaner, more common way to define the orbit is
$$\operatorname{Orb}_{\pi}(x)=\{gx:\ g\in G\}.$$
The orbit is usually denoted $Gx$, with the homomorphism $\pi$ implicit from the context.

Now for your solution:

The question concerns two congruent regular $n$-gons in the plane, and the symmetries obtained by picking them up and putting them back to cover the same region. Together the two $n$-gons have $2n$ vertices, but of course not every permutation of the vertices corresponds to an allowed symmetry. This should also be clear because you have already computed the total number of symmetries; it isn't $(2n)!$.

 Now you start off by taking for $X$ the Cartesian product of the sets of vertices of the $n$-gons, and for $G$ the group of all symmetries of this set. This Cartesian product has $n\times n=n^2$ elements, and $(n^2)!=\text{huge}$ permutations. That can't possibly be right. Stick to the original geometric setup!

Reading further, I cannot make sense of any of your arguments, so I'll just give you a rough framework for how to approach the question:

Take for $X$ a finite set that captures everything that goes on in the geometric picture. The $n$-gons are fully determined by their vertices, for example. The group acting on this set is simply the group given in the question; the group of symmetries of the two $n$-gons. There is no need to describe this group in any alternative way.

Now to apply the orbit-stabilizer theorem, consider the orbit and stabilizer of a vertex. First the easy part: What is the size of its orbit? That is, how many vertices can this vertex be mapped to?

Then what is the stabilizer of a vertex: How many symmetries of the two $n$-gons are there that keep one given vertex in place? How many symmetries for the $n$-gon containing that vertex? How many for the other one?

Now you have the sizes of the orbit and the stabilizer of a vertex, so you can use the orbit-stabilizer theorem to determine the size of the group of symmetries.


问:
Dense set in $L^2$
Let $ \Omega\subset \mathbb{R}^n$ with $m(\Omega^c)=0 $. Then how can we show that $ \mathcal{F}(C_{0}^{\infty}(\Omega))$ (here $ \mathcal{F}$ denotes the fourier transform) is dense in $L^2$(or $L^p$)?

Besides,  I'm also interested to know if the condition that $m(\Omega^c)=0$ can be weakened to some more general set.

Thanks for your help.

答: To deal with the case $\Omega=\Bbb R^n$, take $f\in L^2$. Then by Plancherel's theorem, theorem 12 in these lecture notes, we can find $g\in L^2$ such that $f=\mathcal F g$. Now approximate $g$ by smooth functions with compact support and use isometry. </p>

Title: Dense set in $L^2$


答:
Let $|x_i|=a_i$ and $\sum\limits_{i=1}^{n}a_i=S.$

Thus, since by triangle inequality $$\frac{|\sum\limits_{i=1}^nx_i|}{1+|\sum\limits_{i=1}^nx_i|}=1+\frac{|\sum\limits_{i=1}^nx_i|}{1+|\sum\limits_{i=1}^nx_i|}-1=1-\frac{1}{1+|\sum\limits_{i=1}^nx_i|}\leq$$
$$\leq1-\frac{1}{1+\sum\limits_{i=1}^n|x_i|}=1-\frac{1}{1+S}=\frac{S}{1+S},$$
it's enough to prove that:
$$\sum_{i=1}^n\frac{a_i}{1+S-a_i}\geq\frac{S}{1+S}.$$
Indeed, by C_S
$$\sum_{i=1}^n\frac{a_i}{1+S-a_i}=\sum_{i=1}^n\frac{a_i^2}{a_i+Sa_i-a_i^2}\geq\frac{S^2}{S+S^2-\sum\limits_{i=1}^na_i^2}\geq\frac{S^2}{S+S^2}=\frac{S}{1+S}.$$

Title: Proof reviewer needed

Body:
Originally I wanted to elaborate one idea of mine in my answer to this one question:
How to prove that $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{|1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$

I'm trying to check whether: $$S(N) = \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|}\le \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$ for all natural numbers $N$ and all natural numbers $i<N+1$ is true.

The inequation $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|} + \frac{|y|}{1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$ is the special case for $i=2$ with $x=x_0, y=x_1, z=x_2$

For $N=0$ we have $\frac{|x_0|}{1+|x_0|} \le \frac{|x_0|}{1+ 0}$ for all $x_0$

Let's suppose for a certain $N$ that we have  $$S(N) = \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|} \le \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$ for all natural numbers $N$ and all natural numbers $i<N+1$.

and prove this inequality for $S(N+1)$ which means let's prove the following:

$$S(N+1) \le \sum_{i=0}^{N+1} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N+1} x_j|}$$  all natural numbers $i<N+2$.

We first will use $\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$ like proved in this link:Prove $\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$.

for $a = x_{N+1}$ and $ b= \sum_{i=0}^{N} x_i$

$$\frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|}$$

means that

$$\frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + S(N)$$

means that $$ \frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$

means that

$$S(N+1) \le \sum_{i=0}^{N+1} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N+1} x_j|}$$

Proved!


问:
Find $P(X_1 < X_2 < X_3)$
Suppose $X_1, X_2, X_3$ are independently and identically distributed random variables of an exponential distribution with rate $\beta$. What is the probability $P(X_1 < X_2 < X_3)$?

Attempt:

$$P(X_1<X_2) = P(X_2<X_3) = \frac{\beta}{2\beta}$$

Now, $P(X_1 < X_2 < X_3) = P(X_1<X_2, X_2<X_3) = P(X_1<X_2 | X_2<X_3)\times P(X_2<X_3)$

Am I moving in the right direction? How to calculate $P(X_1<X_2 | X_2<X_3)$? Is it $0.5$ due to the memoryless property?

答:
Hint

If $X_1,X_2$ and $X_3$ are iid with distribution $f$, then

$$\mathbb P\{X_1<X_2<X_3\}=\int_{0}^\infty \int_{x_1}^\infty \int_{x_2}^\infty f(x_1)f(x_2)f(x_3)\,\mathrm d x_3\,\mathrm d x_2\,\mathrm d x_1.$$

Title: Find $P(X_1 < X_2 < X_3)$


问:
Confusion about definition of an almost everywhere continuous function
Let $1_\mathbb{Q}$ be the indicator of the rational numbers. Is $1_\mathbb{Q}$ continuous almost everywhere?

From what I understand, this is not true. I would say that $1_\mathbb{Q}$ is continuous nowhere. But if I restrict the domain to $\mathbb{R}\setminus\mathbb{Q}$ (i.e. I remove a set of measure zero), then $1_\mathbb{Q} \equiv 0$ and is continuous. Some of the responses I've seen seem to consider this as being "continuous almost everywhere". For example: Measurability of almost everywhere continuous functions

Which definition is correct/standard?

答:
When you restrict to $\mathbb{R} \setminus \mathbb{Q},$ you're changing the topology. This is a subtle but important point.

The criterion you describe says that in the original topology, the function is continuous at almost every point. This is not the same as "when you remove a null set, the restricted function is continuous on that subspace topology."

A similar confusion arises in Lusin's theorem, which says that if $f : [0, 1] \rightarrow\mathbb{R}$ is measurable, then for any $\epsilon > 0$ there's $F$ so that $f$ restricted to $F$ is continuous and $m(F) > 1-\epsilon.$

Just because $f$ restricted to $F$ is continuous at a point $p \in F,$ doesn't mean that $f$ is continuous at $p$. $f$ and the restriction of $f$ onto $F$ are two different functions on two different topological spaces.


问:
Smoothness and composition
I have a following problem:

Suppose f is a k-smooth function ($ \in C^k$) but not $(k+1)$-smooth ($f \notin C^{k+1}$) and $g(x)=ax+b, a\neq 0$).

Is it true, and if it is than how to prove it, that $f \circ g $ is also a k-smooth function ($ \in C^k$) but is not $k+1$-smooth ($ f \circ g \notin C^{k+1}$)? \

I can't see anywhere such questions. All I see is that the composition doesn't reduce smoothness.

答:
Suppose $f$ is in $C^k$ but not in $C^{k+1}$.  We want to show that $f \circ g$ is also in $C^k$ but not in $C^{k+1}$.  Since $f \circ g(x) = f(ax+b)$, it is not difficult to see that $f \circ g$ belongs to $C^k$.  The problem is to show that $f \circ g$ does not belong to $C^{k+1}$.

Suppose by way of contradiction it does belong to $C^{k+1}$.  Let $h(x) = \frac{1}{a}x - \frac{b}{a}$.  It is not difficult to see that $g(h(x)) = x$.  Since $f \circ g$ belongs to $C^{k+1}$, so does $(f \circ g) \circ h$.  But

$$ (f \circ g) \circ h(x) = f(g(h(x)) = f(x)$$
In other words, $(f \circ g) \circ h = f$.  This is a contradiction, since we assumed that  $f$ did not belong to $C^{k+1}$.

Title: Smoothness and composition


问:
Intersection of a Hyperplane and a Subspace
Let $Y$ be a dense linear subspace of a normed space $X$, and let $M$ be a closed hyperplane in $X$. I'm trying to show that $M \cap Y$ is a hyperplane in $Y$ and dense in $M$.

I've been trying to show that the codimension of $M \cap Y$ in $Y$ is $1$, and relying on the identity:

$$\dim(A + B) + \dim(A \cap B) = \dim(A) + \dim(B)$$

and keeping in mind that $\dim(X) - \dim(M) = 1$, but this just keeps bringing me round in circles.

答:
Yes,  $Y\cap M$ is dense in $M$.  Here is why.

Write $Y=(Y\cap M) \oplus Y'$ (algebraic direct sum) and notice that the projection $\pi :X\to X/M$ is injective on $Y'$ because $Y'\cap M=\{0\}$.  Since
$\text{dim}(X/M)=1$ we deduce that $\text{dim}(Y')\leq 1$.

If $\text{dim}(Y')=0$ then $Y\subseteq M$, and hence $Y$ is obviously dense in $M$, so let us suppose that  $\text{dim}(Y')=1$.  We
then have that $X=M\oplus Y'$ as a topological direct sum (the projections are continuous).

Setting $M'=\overline {Y\cap M}$, suppose that $M'\neq M$.  Then $M'\oplus Y'$ is a proper closed  subspace but this is a
contradiction because $Y\subseteq M'\oplus Y'$.


问:
Solution verification of $\lim_{n\to\infty}\left(1+\left(\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)\right)^{-1}\right)^{n}$
$$\lim_{n\to\infty}\left(1+\left(\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)\right)^{-1}\right)^{n}$$

Source: Romanian Mathematical Magazine Spring edition 2022, problem UP. 354.

I work out in following manner.

Clearly, $\displaystyle \frac{k\pi}{n}>0$  for all $k\in[ 1, n-1]$ so due to    Taylor series around $x=0$ we have $\sin x \leq x$ for all $ x>0$. Choose $\displaystyle x=\frac{k\pi}{n}$  we have then $$\sin\left(\frac{k\pi }{n}\right)\leq \frac{k\pi}{n}\Rightarrow S(n)= \sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)\leq \pi\sum_{k=1}^{n-1}\frac{k^2}{n^2}=\frac{\pi}{6}\left[\frac{(n-1)(2n-1)}{n}\right]\cdots(1)$$ Since  the partial sum of the later sum diverges as $n\to \infty$ so $S(n)\to \infty$ (due to comparison test). Therefore,$$L=\lim_{n\to \infty} \left(1+\left(S(n)\right)^{-1}\right)^{n}=1^{\infty}$$ and hence we have $$L= \lim_{n\to\infty}\exp\left(\left(S(n)\right)^{-1}\right)=\exp\left(\lim_{n\to \infty}n\left(\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)\right)^{-1}\right)$$ since latter expression is Riemann integrable function on $[0,1]$ so the limit reduces to $$L= \exp\left(\int_0^1 x\sin(\pi x) dx\right)^{-1} \underbrace {=}_{IBP}e^{\left(\pi^{-1}\right)^{-1}}=e^{\pi}$$

The reason of $(1)$ in above is to show that for large value  of $n$ the original sum is divergent so that its reciprocal is $0$ giving us $1^{\infty}$ form.

Question:
Is my work and inequality $(1)$ supports the valid proof?

I manage  to find the limit of problem  in other way by showing that $$\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)=\frac{1}{2}\cot\left(\frac{\pi}{2n}\right)$$ giving us $e^{\pi}$.

答:
As an alternative by

$$\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right) =\Im\left(\frac1n \sum_{k=1}^{n-1} ke^{\frac{k\pi}{n}i}\right)$$

we can use that

$$\sum_{k=1}^{n-1} kx^k=\frac{(n-1)x^{n+1} - nx^n +x}{(x-1)^2}$$

and by $x=e^{\frac{\pi}{n}i}$ we obtain

$$\Im\left(\frac1n \sum_{k=1}^{n-1} ke^{\frac{k\pi}{n}i}\right)=\Im\left(\frac1{1-e^{\frac{\pi}{n}i}}\right)=\frac{\sin \left(\frac{\pi}{n}\right)}{2-2\cos \left(\frac{\pi}{n}\right)}=\frac{1}{2}\cot\left(\frac{\pi}{2n}\right)$$

and then

$$\lim_{n\to\infty}\left(1+\left(\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{k\pi}{n}\right)\right)^{-1}\right)^{n}
=\lim_{n\to\infty}\left(1+\frac {1}{\frac{1}{2}\cot\left(\frac{\pi}{2n}\right)}\right)^{n} =e^\pi$$


问:
Find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, $a > 0$ converges or not.
Find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, $a >0$ converges or not.

I used d'Alembert's criterion and I found $\lim_{n\to \infty} \frac{x_n}{x_{n+1}}=1$

Moving to Raabe–Duhamel's test, I found $$\lim_{n\to \infty} n(\frac{x_n}{x_{n+1}} -1) =  \lim_{n\to \infty} n(\frac{1}{\sqrt[n+1]{a}} - 1) $$

What can I do from there? Or is there a better way to find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, $a >0$ converges or not?

答:
hint

You can use the fact that the sequence
$$\sum_{k=1}^n\frac 1k-\ln(n)$$

converges to Euler's constant $\gamma$.


答:
First owing to the concavity of the logarithm function, we have

$$\int_1^n \frac1x\,dx\le \sum_{k=1}^n\frac1k \le 1+\int_1^n \frac1x\,dx$$

Therefore, for $1>a>0$ we have

$$a^{\log(n)+1}\le a^{\sum_{k=1}^n \frac1k}\le a^{\log(n)}$$

while for $a\ge 1$ we have

$$a^{\log(n)}\le a^{\sum_{k=1}^n \frac1k}\le a^{\log(n)+1}$$

Thus, using $a^{\log(n)}=n^{\log(a)}$ we find from by "the p-test" that the series converges whenever $0<a<1/e$ and diverges for $a\ge 1/e$.

Title: Find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, $a > 0$ converges or not.

Body:
Find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, $a >0$ converges or not.

I used d'Alembert's criterion and I found $\lim_{n\to \infty} \frac{x_n}{x_{n+1}}=1$

Moving to Raabe–Duhamel's test, I found $$\lim_{n\to \infty} n(\frac{x_n}{x_{n+1}} -1) =  \lim_{n\to \infty} n(\frac{1}{\sqrt[n+1]{a}} - 1) $$

What can I do from there? Or is there a better way to find if $\sum\limits_{n=1}^{\infty} a^{1+\frac1{2}+\frac1{3}+\dots+\frac1{n}}$, $a >0$ converges or not?


答:
There is no need to work so hard for this question.

Think of it this way.  There are only six possible outcomes (except for those outcomes for which at least two of the three random variables are exactly equal, but these occur with probability zero):

$$X_1 < X_2 < X_3 \\ X_1 < X_3 < X_2 \\ X_2 < X_1 < X_3 \\ X_2 < X_3 < X_1 \\ X_3 < X_1 < X_2 \\ X_3 < X_2 < X_1$$  Because each $X_i$ is independent and identically distributed, the probability of any one of these is equal to the other.  The underlying distribution makes no difference.  Therefore, the answer is $1/6$.  Had these not been IID then you'd need to do more work and the answer would depend on how they are distributed, but in this case, the variables are exchangeable.

Title: Find $P(X_1 < X_2 < X_3)$

Body:
Suppose $X_1, X_2, X_3$ are independently and identically distributed random variables of an exponential distribution with rate $\beta$. What is the probability $P(X_1 < X_2 < X_3)$?

Attempt:

$$P(X_1<X_2) = P(X_2<X_3) = \frac{\beta}{2\beta}$$

Now, $P(X_1 < X_2 < X_3) = P(X_1<X_2, X_2<X_3) = P(X_1<X_2 | X_2<X_3)\times P(X_2<X_3)$

Am I moving in the right direction? How to calculate $P(X_1<X_2 | X_2<X_3)$? Is it $0.5$ due to the memoryless property?


答:
The binomial coefficient $\binom{2n}{n}$ is the coefficient of $x^n$ in the left-hand side.

In the right hand side you get $x^n$ from terms $x^i$ and $x^j$ where $i+j=n$. The term $x^i$ has coefficient $\binom{n}{i}$ and similarly for $x^j$, so you have
$$
\binom{2n}{n}=\sum_{i+j=n}\binom{n}{i}\binom{n}{j}
$$
Now observe that if $i+j=n$, then $j=n-i$ and that
$$
\binom{n}{n-i}=\binom{n}{i}
$$
and that $i$ can run from $0$ to $n$, so we finally get
$$
\binom{2n}{n}=\sum_{i=0}^n\binom{n}{i}^{\!2}
$$

As an example with $n=3$: in the right hand side we have
$$
\binom{3}{0}+\binom{3}{1}x+\binom{3}{2}x^2+\binom{3}{3}x^3
$$
multiplied by itself, but we just need to consider the terms with $x^3$, so
$$
\binom{3}{0}\cdot\binom{3}{3}x^3+\binom{3}{1}x\cdot\binom{3}{2}x^2
+\binom{3}{2}x^2\cdot\binom{3}{1}x+\binom{3}{3}x^3\cdot\binom{3}{0}
$$

Title: Trouble Understanding Proof of Lagrange's Identity Using Binomial Formula

Body:
I've been reading a book on elementary mathematics and am having trouble understanding a proof that they give for Lagrange's identity using the binomial formula.

Lagrange's identity is stated as :
\begin{equation}
\sum_{j=0}^{n} \binom{n}{j}^{2} = \binom{2n}{n}
\end{equation}
In the proof they say :

"Writing $(1+x)^{2n} = (1+x)^{n}(1+x)^{n}$ and applying Newton's binomial formula to both sides we get :
\begin{equation}
\binom{2n}{n} = \sum_{i+j=n} \binom{n}{i} \binom{n}{j} = \sum_{i=0}^{n} \binom{n}{i}\binom{n}{n-i} = \sum_{i=0}^{n} \binom{n}{i}^{2}
\end{equation}
$\square$."

A lot of detail seems to be missing here. Could someone provide the missing details here ? I cannot understand the proof due to the elision of details.


问:
How to find a circle that passes through $2$ points and its diameter is on a given line.
Suppose I have $2$ points $A,B$, and a line $\ell$. Something like this:

I want to find a circle that passes through both points $A$ and $B$ and whose diameter also passes through the given line $\ell$. Playing around with Geogebra, I use the "Circle through $3$ points" option trough $A$ and $B$ and some other point $C$, and from what I can tell, it seems to me that I can always find a third point on $\ell$ to satisfy the desired requirements:

However, I only seem to be able to do this by trial and error, and moving the third point $C$ by hand until it seems to fit.

I know that, analytically, if I say that
\begin{align*}
A &= (x_A,y_A) \\
B &= (x_B,y_B) \\
\ell &: \{y = mx +b\}
\end{align*}
I can find the coordinates for the center of the circle and it's radius by solving the system of equations
$$
 \begin{cases}
  (x_A-h)^2 +(y_A-k)^2 = r^2\\
    (x_B-h)^2 +(y_B-k)^2 = r^2\\
    k = mh+b
   \end{cases}
$$
for $h,k$ and $r$. However, WolframAlpha gives some pretty ugly solutions to the system (as seen here). To avoid this, I was wondering if there was a constructive way of finding this circle, maybe based on drawing lines and circles akin to how Euclid would do it. I gave it a shot but couldn't seem to find any simple way of doing this. Does anyone know if there's a simple way to construct this circle? Thank you!

答: As Christian said, the center has equal distance to $A$ and $B$ and therefore lies on the perpendicular bisector of $AB$. This center is therefore at the intersection of $AB$ and $ℓ$, and the radius is simply the distance of this center to $A$ or $B$.


问:
Can someone give a clear cut understanding of the usage of pi in trigonometric equations?
Through out my studies, I had been through a lot of trigonometry and differential equations with trigonometry.
Whenever 'pi' appears inside the sin/cos/tan..etc, I assumed it to be 180, and when it appeared outside due to integration or something else, I assumed it to be the irrational number-3.14, sometimes, both at a time.
I read a lot about radians and degrees. But Most math problems don't specify if it's radians or degrees.
Is it implied that whenever pi appears inside the trigonometric functions it is in degrees and outside in radians? or the other way round? Sorry I'm totally confused. I just wanna have some clarity.
Also, calculators seem to take the same 'pi' value inside and outside which is causing ridiculous errors. like sin 180 coming out as sin 22/7*.
Please help.

Please give an example: say this:
So, if there is an integral which, say x*sin(x), integrated over 0 and pi/5., After the integration is over, how do you apply the limits? Do you use pi=180 inside the -cos and the irrational just number on the plain old x?

答:
I think your confusion originates from misunderstanding radians and degrees.

$\pi$ is a number. It happens that the circumference of a circle is always $2\pi$ times the radius of the circle. So, in radians, an angle of $\pi$ does correspond to an angle of $180$ degrees.

However, $\pi \neq 180.$ When you see $\pi$ appearing in an integral, it's probably appearing as an area--$\pi$ is the area of the unit circle.

We may define $1^{\circ} = \pi/180$ to define degrees. But that just means 1 degree equals $\pi/180,$ not that $1 = \pi/180.$

Just like $100\text{cm} = 1\text{m}$ doesn't mean $100=1.$ There are different units.


问:
To show that there are $3$ numbers between $n^2$ and $(n+1) ^2$ such that they satisfy a certain property.
Here is the problem

Show that for any natural number $n$, one can find three distinct
natural numbers $a, b, c$ between $n^2$ and $(n + 1)^2$ such that $a^2 + b^2$ is divisible
by $c$.

It seems like here are several ways to express $a^2 + b^2$.
\begin{array}
 aa^2 + b^2 &= (a - b) ^2 +  2ab \\
 &= (a + b) ^2 - 2ab \\
 &= \frac { (a + b) ^2 + (a - b)^2 }{2} \\
 &= 2b^2 + (a + b)(a - b)
\end{array}
What the problem is literally saying: Find algebraic expressions $X,Y$ and $Z$ such that $n^2 \lt X,Y,Z \lt(n + 1)^2$ and $Y^2 + Z^2 = AX$, for some expression $A$. (Hence, $X|Y^2 + Z^2$)

In the original problem, $X$ is $c$, $Y$ and $Z$ are $b$ and $a$.

People with enough experience might know the answer at once, but is it possible to actually find the expressions from identities/relations? For example the relations that I have found above?

From my trying it is very difficult because one needs to search for expression that are restricted between $n^2$ and $(n + 1)^2$, and then the factoring. What I tried was just plugging random expressions to $Y$ and $Z$ and trying to find a suitable $X$ that suffices the factoring, but to no avail.

Please help, and thanks for that!

答:
There is in fact another way to do this:

Thm 1: Let $N$ be a sufficiently large integer. Then between $N$ and
$N +\lceil 2\sqrt{N} \rceil$ there are 3 integers $A$, $B$, $C$ such that
$C|(A^2+B^2)$.

To prove Thm 1, we make the following claim:

Claim 2: For each $M$ and $k$, the equation $(M-k)^2 \equiv_k (M+k)^2 \equiv_M k^2$.

We now finish the proof of Thm 1: So let $k_1,k_2$ be distinct positive integers both no larger than $\sqrt{N}$ such that $N+\sqrt{N} > k_1^2+k^2_2 > N$, picking
$k_1=\lfloor \sqrt{N} \rfloor$ and $k_2 = O(N^{1/4})$ should work.

Then let $C=k^2_1+k^2_2$, and then let $A=C+k_1$ and $B=C+k_2$. Then $A,B,C$ satisfy the conditions of Thm 1. $\surd$


答:
Since $\ell$ is a diameter, reflecting either $A$ or $B$ across $\ell$ will give a third point through which the circle passes.  You can then use this as the construction.

Alternatively, you may simply construct the perpendicular bisector of $AB$, which intersects $\ell$ at the center of the circle.

Title: How to find a circle that passes through $2$ points and its diameter is on a given line.

Body:
Suppose I have $2$ points $A,B$, and a line $\ell$. Something like this:

I want to find a circle that passes through both points $A$ and $B$ and whose diameter also passes through the given line $\ell$. Playing around with Geogebra, I use the "Circle through $3$ points" option trough $A$ and $B$ and some other point $C$, and from what I can tell, it seems to me that I can always find a third point on $\ell$ to satisfy the desired requirements:

However, I only seem to be able to do this by trial and error, and moving the third point $C$ by hand until it seems to fit.

I know that, analytically, if I say that
\begin{align*}
A &= (x_A,y_A) \\
B &= (x_B,y_B) \\
\ell &: \{y = mx +b\}
\end{align*}
I can find the coordinates for the center of the circle and it's radius by solving the system of equations
$$
 \begin{cases}
  (x_A-h)^2 +(y_A-k)^2 = r^2\\
    (x_B-h)^2 +(y_B-k)^2 = r^2\\
    k = mh+b
   \end{cases}
$$
for $h,k$ and $r$. However, WolframAlpha gives some pretty ugly solutions to the system (as seen here). To avoid this, I was wondering if there was a constructive way of finding this circle, maybe based on drawing lines and circles akin to how Euclid would do it. I gave it a shot but couldn't seem to find any simple way of doing this. Does anyone know if there's a simple way to construct this circle? Thank you!


答:
Saying that the diameter lies on a given line segment means that the center is on that line.  If A and B are points on the circle then the line segment between them is a chord and its perpendicular bisector also passes through the center of the circle.

So construct the perpendicular bisector of AB.  The point where it crosses the given line is the center of the circle.  Set the leg of compasses at that point and draw the circle through A and B.


问:
How an interval [a, b] in a Boolean algebra B can be made into a Boolean algebra?
How an interval $[a, b]$ in a Boolean algebra $B$ can be made into a Boolean algebra?

答:
Assuming $a\le b$ (that is, $a\wedge b=a$ or, equivalently, $a\vee b=b$), the operations are exactly the same as in $B$, except for complementation; the minimum is $a$, the maximum is $b$ and the complement is
$$
x^*=(x'\wedge b)\vee a=(x'\vee a)\wedge b
$$
where $x'$ denotes the complement in the algebra $B$. We're “cutting $x'$ above by $b$ and below by $a$”.

Check the axioms.


问:
Justification of the tangent name for the goniometric operator $\tan$
When we give a proof that the tangent is the sine to cosine ratio of an oriented angle,

$$\bbox[5px,border:2px solid #C0A000]{\tan \alpha=\frac{\sin\alpha}{\cos \alpha}}$$
with $\cos \alpha \neq 0$, we take the tangent $t$ in $A(1,0)\equiv S$ to the circle of center in $O(0,0)$ ad radius $r=1$. See the image

The name tangent has been given because we consider the tangent to the circle of radius $1$ at point $A\equiv S$ or for another reason?

答:
This is a good reason and it is consistent also with the other equivalent definition for the tangent

(credit)

Refer also to The Etymology of Trig Functions.


问:
How do I find the bounds for an integral between a plane and a sphere?

So this is what I got so far:

$$\int _0^{2\pi }\:\int _0^{\sqrt{5}}\:\int _3^{\sqrt{16-r^2}}\:\left(r\right)dzdrd\theta $$

And the answer is shown above in the screenshot. I'm not exactly sure which bounds I'm messing up.

I know it's a sphere, so it's gonna be $0$ to $2\pi$ for theta. The middle one I plug in $z=3$ and subtract and replace the squared $x$ and $y$ with $r$ squared. And then the furthest right one I do the same thing pretty much there.

What's my problem? Thank you.

答:
Perhaps it would be easier to start by finding the volume of a small slice of the sphere created by leaving only the portion of the sphere that lies in the range $z=a\dots b,\,b>a$.

This section would essentially be a flat pancake with its symmetry axis along z-axis. What is the radius of that pancake? Call it $\rho$. From basic Pythagoras theorem $\rho=\sqrt{r^2-z^2}$, where $r$ is the radius of the sphere. Clearly this only makes sense for $z^2\le r^2$.

So now you can find the volume of this pancake (note how positioning of $\int$ and $d\dots$ can be used to clarify the bounds in the integrals):

$$
V_{pancake}\left(a,b,r\right)=\int_a^b dz\int_0^\sqrt{r^2-z^2} \rho d\rho\int_0^{2\pi} d\phi=2\pi\int_a^b dz \frac{\left(\sqrt{r^2-z^2}\right)^2}{2}=\dots
$$

From here it is easy to get the result for any portion of the sphere intersected by the plane


问:
Are all numbers greater than the product of their prime divisors?
Saw a lemma in a lecture
for a number $M$, all prime divisors of $M\geq 2$so if $M$ has $t$ distinct prime divisors then $|M| > 2^t$
This only seems to follow from the title statement, is there proof for this?

答:
This should be obvious.

Every positive integer has a prime factorization.  So it may be written as $n = p_1^{a_1}p_2^{a_2}......p_t^{a_t}$ where $p_i$ are the distinct prime factors and $a_i$ are the power of the prime factors.

Now $p_i^{a_i} \ge p_i^1=p_i$ with equality holding if and only if $a_i = 1$.  So $n = p_1^{a_1}p_2^{a_2}......p_t^{a_t} \ge p_1p_2......p_t \ge 2\cdot 2\cdot ..... 2=2^t$.

So $n \ge p_1p_2......p_t=$ the product of it's prime factors (and is equal if and only if $n$ has no prime factors to more than $1$ power; that is if and only if $n$ is "squarefree" and has no square factors).

And $p_1p_2......p_t \ge 2^t$ with equality holding if and only if $t=1$ and $p_1 = 2$.


问:
Writing an expression using logic
Write an expression using letters $\land, \lor, and$ $\neg$ which has the following truth table:

$$\begin{array}{ccc|c}
P&Q&R&???\\ \hline
T&T&T&F\\
T&T&F&T\\
T&F&T&F\\
T&F&F&T\\
F&T&T&T\\
F&T&F&F\\
F&F&T&F\\
F&F&F&F
\end{array}
$$

How would you in general solve for ??? I was told there is a mechanical way of doing it but I don't see it.

答:
There are several methods available to solve this.  One method that's fairly easy for smaller circuits is to use a Karnaugh Map.  It requires a bit of study in order to understand what functions will look like on the map, but it makes the task fairly easy after Karnaugh maps are understood.

A second way is using the following trick:
We know that $x \lor y \lor z$, read "x or y or z", will have only one false value, and the rest will be true.  So you can substitute $P$, $Q$, $R$ or their negations, $\overline{P}$, $\overline{Q}$,or $\overline{R}$ into this to get one false value and seven true values.  Then you can essentially "paste" functions like this together to get the result.

For example, "(not $p$) or (not $q$) or (not $r$)" gives the truth table:

T
T
T
T
T
T
T
F

Then, "(not $p$) or (not $q$) or ($r$)" gives the truth table:

T
T
T
T
T
T
F
T

...and combining them gives : "((not $p$) or (not $q$) or (not $r$)) and ((not $p$) or (not $q$) or ($r$))":

T
T
T
T
T
T
F
F

...We can proceed in this fashion mechanically to slowly add one false value until we reach the desired function.

Title: Writing an expression using logic


问:
Baby Rudin 11.11 remarks (b)
I am reading Baby Rudin's Chapter 11 (The Lebesgue Theory) and find some gaps hard to fills in the 11.11 (b) remark. (pp309)

If $A \in \mathscr{M}(\mu)$ and $\varepsilon >0$, there exists sets $F$ and $G$ such that
  $$
F \subset A \subset G,
$$
  $F$ is closed, $G$ is open, and
  $$\mu(G-A)<\varepsilon, \mu(A-F) < \varepsilon.
$$
  The first inequality holds since $\mu*$ (outer measure) was defined by means of coverings by open elementary sets.

I find it hard to relate the hint "since $\mu*$ (outer measure) was defined by means of coverings by open elementary sets." to the proof of the first inequality above.

答:
$A \in \mathscr{M}(\mu) \Rightarrow A = \bigcup\limits_{k=1}^{\infty}A_i$ for some collection of $\{A_i\} \subset \mathscr{M}_F (\mu)$.

For each $n$, since $A_n \in \mathscr{M}_F (\mu) $, we can choose a countable covering $\{G_{ni}\}$ of elementary open sets with $\bigcup\limits_{i=1}^{\infty} G_{ni} \supset A_n$, $((\bigcup\limits_i G_{ni}) \in \mathscr{M}_F (\mu)$ (by 11.11(a)),  and $\mu((\bigcup\limits_i G_{ni}) -A_n)<\epsilon/2^n$.

Let $G=\bigcup\limits_{k}\bigcup\limits_{i}G_{ki}$. Then $G$ is open (countable union of open sets), and $$\mu(A -G)\leq \sum_k\mu(A_k-(\bigcup\limits_{i}G_{ki}))\leq \sum_k\epsilon/2^k=\epsilon$$

Title: Baby Rudin 11.11 remarks (b)


问:
Why substitution in irrational equation doesn't give equivalent equation?
I have two examples of irrational equations:

The first example:  $\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$

In solution, they take cube of both sides and do following:

\begin{eqnarray*}
&\sqrt[3]{3-x} &+ \sqrt[3]{6+x}=3\\
&\iff&
3-x+ 3\sqrt[3]{(3-x)(6+x)}(\sqrt[3]{3-x} + \sqrt[3]{6+x}) +6+x=27\\
&\iff&
3-x+ 9\sqrt[3]{(3-x)(6+x)} +6+x=27   \iff \sqrt[3]{(3-x)(6+x)} = 2\\
&\iff&
x^2+3x-10=0\\
&\iff&   x=2\quad  \text{ or }\quad x= -5
\end{eqnarray*}
They conclude that both values are solutions, they satisfy the original equation.

The second example:  $\sqrt[3]{x+1} +  \sqrt[3]{3x+1} = \sqrt[3]{x-1}$

Here they do the following:
\begin{eqnarray*}
&\sqrt[3]{x+1}& +  \sqrt[3]{3x+1} = \sqrt[3]{x-1}\\
&\iff&
x+1 + 3 \sqrt[3]{(x+1)(3x+1)} (\sqrt[3]{x+1} +  \sqrt[3]{3x+1} ) + 3x+1 = x-1\\
&\implies&
3 \sqrt[3]{(x+1)(3x+1)(x-1)} = -3x-3\\
&\iff&
(x^2-1)(3x+1)= -(x+1)^3\\
&\iff&
x= 0 \quad\text{ or }\quad x = -1
\end{eqnarray*}

but the only solutions is $x= -1$, because $0$ doesn't satisfy the equation.

What is difference between these examples, why do we have all equivalence signs in the first, and implication in the secondnd example?

The explanation in solution is that we have substituted  $\sqrt[3]{3-x} + \sqrt[3]{6+x}$ by a number in the 1st example, and in the 2nd example we substituted by another expression which depends on x. I don't understand that, when does substitution  give equivalent equation? Do we always have to verify if solutions satisfy the original equation in the end (is that the answer)?

Please help with this explanation, I need to understand better the irrational equations. Thanks a lot in advance.

答:
Very good question! The first example is misleading in writing equivalences everywhere; the second equivalence
$$3-x+ 9\sqrt[3]{(3-x)(6+x)}(\sqrt[3]{3-x} + \sqrt[3]{6+x})+6+x$$
$$\iff$$
$$3-x+ 9\sqrt[3]{(3-x)(6+x)} +6+x=27$$
should (at least conceptually) be an implication $(\implies)$. It is true that it is in fact an equivalence, but this is not yet clear at this point. Let me explain:

The argument starts from the assumption that if $x$ satisfies
$$\sqrt[3]{3-x} + \sqrt[3]{6+x}=3,$$
then it also satisfies the expressions that follow. In both examples the first equivalence stems from the simple fact that
$$x^3=y^3\quad\iff\quad x=y,$$
but the second equivalence uses a substitution that need not be reversible; a number $x$ can satisfy
$$3 \sqrt[3]{(x+1)(3x+1)(x-1)} = -3x-3,$$
but this does not imply that it should satisfy
$$\sqrt[3]{x+1} +  \sqrt[3]{3x+1} = \sqrt[3]{x-1}.$$
This is illustrated by the solution $x=0$. In the first example it just so happens that we do not get any extra solutions, and hence, with hindsight, this implication turns out to be an equivalence. But that is not at all clear before checking whether all solutions to the last equation are also solutions to the original equation.

So to answer your question; yes, you should check whether all solutions to the last equation are also solutions to the original equation. If you are certain that every step along the way is unambiguously reversible, then you do not need to check. But if it is not clear whether a step is reversible (as in both examples here), you should check.


问:
Equivalent measures and their topological supports
The topological support $\sigma(\mu)$ of a Borel measure $\mu$ can be defined as the smallest closed Borel set $X$ that supports $\mu$, in the sense that $\mu(\mathbb{R}\setminus X)=0$. One can show that the topological support exists and is given by the set of its growth points, i.e.
\begin{equation*}
\sigma(\mu)=\left\{x\in\mathbb{R}:\forall\delta>0,\,\mu(I_\delta(x))>0\right\},
\end{equation*}
with $I_\delta(x)=(x-\delta,x+\delta)$.

Let $\mu$, $\nu$ be two Borel measures and suppose that $\mu\ll\nu$ and $\nu\ll\mu$, i.e. each of them is absolutely continuous with respect to the other one; in this sense, I will say that $\mu$ and $\nu$ are equivalent.

I would like to show $\sigma(\mu)=\sigma(\nu)$. If I am not wrong, this equality can be shown by noticing that, for a given $\mu$, $\sigma(\mu)$ coincides with the spectrum of the self-adjoint multiplication operator $(A_\mu f)(x)=xf(x)$ on the Hilbert space $L^2_\mu(\mathbb{R})$ (when defined on the proper domain); if $\mu$ and $\nu$ are equivalent, the operators $A_\mu$ and $A_\nu$ can be readily shown to be unitarily equivalent and thus isospectral. However, I expect a more immediate argument to show this equality to exist.

答:
That was easy.
Let $x\in\sigma(\mu)$ and suppose $x\notin\sigma(\nu)$. Then there is $\delta_0>0$ such that $\nu(I_{\delta_0}(x))=0$. But, since $\mu\ll\nu$, this implies $\mu(I_{\delta_0}(x))=0$, contradicting the initial assumption. Hence $\mu\ll\nu$ implies $\sigma(\mu)\subset\sigma(\nu)$. Repeating the argument exchanging $\mu$ and $\nu$ shows the claim.


答:
Yes, "tangent" takes its name from its association with a tangent line of the unit circle. Likewise, "secant" is named for a secant line. And "sine" for (a mistranslation of) semi-chord.

Curiously, according to the Earliest Known Uses of Some of the Words in Mathematics site (which has entries for all of the terms), "Vieta did not approve of the term tangent because it could be confused with the term in geometry." To me, the connection with geometry is a feature of the terminology.

FYI: The geometry is a focus of my note "(Almost) Everything You Need to Remember about Trigonometry, in One Simple Diagram (PDF link via tricochet.com)".

Title: Justification of the tangent name for the goniometric operator $\tan$

Body:
When we give a proof that the tangent is the sine to cosine ratio of an oriented angle,

$$\bbox[5px,border:2px solid #C0A000]{\tan \alpha=\frac{\sin\alpha}{\cos \alpha}}$$
with $\cos \alpha \neq 0$, we take the tangent $t$ in $A(1,0)\equiv S$ to the circle of center in $O(0,0)$ ad radius $r=1$. See the image

The name tangent has been given because we consider the tangent to the circle of radius $1$ at point $A\equiv S$ or for another reason?


问:
Why do I get a negative value for this integral?
I am trying to compute the integral: $$\int_{4}^{5} \frac{dx}{\sqrt{x^{2}-16}}$$
The question is related to hyperbolic functions, so I let $x = 4\cosh(u)$ therefore the integral becomes: $$-\int_{0}^{\ln(2)}\frac{4\sinh(u)}{\sqrt{16-16\cosh^{2}(u)}}du = -\int_{0}^{\ln(2)}1du = -\ln(2)$$

The answer is $\ln(2)$ so if someone could point out where I went wrong that would be great, thanks

答:
Setting $x = 4 \cosh(u)$, gives us $dx = 4 \sinh(u)du$. Hence the integral becomes
$$\int_0^{\ln(2)} \dfrac{4 \sinh(u) du}{\sqrt{16\cosh^2(u) - 16}} = \int_0^{\ln(2)} \dfrac{4 \sinh(u) du}{4 \sinh(u)} = \ln(2)$$</p>


问:
Given $R$, a symmetrical relation over $A$, if $(R \circ R)$ is a function, then $R$ is a function
Given a non-empty set $A$ and the symmetrical relation $R \subset A\times A$ such that $(R\circ R)$ is a function. The exercise asks if $R$ is a function.

I'm not sure if my reasoning is correct here. Here's what I came up with so far.

If $R$ is symmetrical, then $\forall x,y \in (R\circ R)$, $x=y$.

So every element of $R$ is of the form $(x,x)$, which is the identity function. Is that right?

答:
$\newcommand{\dom}{\operatorname{dom}}$It’s true that $R\circ R$ is the identity function on $\dom R$, but $R$ itself need not be. For instance, let $A=\{0,1\}$ and $R=\{\langle 0,1\rangle,\langle 1,0\rangle\}$; then $R$ is symmetric, $R\circ R$ is the identity function on $A$, and $R$ is a function, but not the identity function on $A$.

Suppose that $x,y,z\in A$ and $\langle x,y\rangle,\langle x,z\rangle\in R$; we want to show that $y=z$. By symmetry we must have $\langle y,x\rangle\in R$, so $\langle y,y\rangle,\langle y,z\rangle\in R\circ R$. And $R\circ R$ is a function, so $y=z$, as desired, and $R$ is indeed a function.


问:
Can I solve this multi-variable limit with polar coordinates: $\lim_{ (x,y)\to(0,0)} \frac{y^2\sin^2(x)}{x^4+y^4}$?
$$\lim_{ (x,y)\to(0,0)} \frac{y^2\sin^2(x)}{x^4+y^4}$$

I know that I could let $x=0$ and solve the limit, and then the same for $y=x$.
But I want to know if I could solve it only with polar coordinates.

答:
If we want proceed by polar coordinates as a first step we can use

$$\frac{y^2\sin^2(x)}{x^4+y^4}=\frac{\sin^2(x)}{x^2}\frac{y^2x^2}{x^4+y^4}$$

with $\frac{\sin^2(x)}{x^2} \to 1$ and then use polar coordinates for the second term.


问:
Homeomorphism between topological spaces
I need to check the following exercise:

Let $X=[0,1] \times [0,1]$ and let $\sim$ be a relation defined by $(x,0) \sim (x,1)$. Show that $X/ \sim$ is homeomorphic to the cylinder $S^1 \times [0,1]$.

Solution:

Consider the function $f: X \rightarrow [0,1] \times S^1$ defined as $f(\theta, t)=(t(t-1), \cos(2 \pi \theta), \sin (2 \pi \theta))$. Since $f$ is conituous, bijective and $f(0,0)=(0,1,0)=f(0,1)$ we have that $[f]:X/ \sim \longrightarrow [0,1] \times S^1$ is continuos too.

Moreover, since $f^{-1}:[0,1] \times S^1 \rightarrow X$ defined by $f^{-1}(x,y,z)=\left ( \sqrt{x-1}, \frac{\arccos y}{2 \pi}, \frac{\arcsin z}{2 \pi},  \right )$ is continuos on its and bijective on its domain, we have that $[f]^{-1}$ is continuous, hence homeomorphism.

Does it work?

答:
Your $f$ is not injective. For instance

$$
f(0, 1/4) = (1, 0, 3/16)
$$
and
$$
f(0, 3/4) = (1, 0, 3/16).
$$

I think you might want to look at the order of the two factors.

NB

The question has been changed (to yet another non-injective function) after I provided this answer. The function, when I answered, was
$$
f: X \rightarrow S^1 \times [0,1]: (\theta, t) \mapsto (\cos(2 \pi \theta), \sin (2 \pi \theta), t(t-1)).
$$

To critique this new answer, which says (I quote this because I anticipate it being changed again and again and again)

$f: X \rightarrow [0,1] \times S^1$ defined as $f(\theta, t)=(t(t-1),
\cos(2 \pi \theta), \sin (2 \pi \theta))$. Since $f$ is conituous,
bijective and $f(0,0)=(0,1,0)=f(0,1)$ we have that $[f]:X/ \sim
\longrightarrow [0,1] \times S^1$

It's worth nothing that when $t = 0.5$, the first output component, $t(t-1) = 0.5 (-0.5) = -0.25$ is not actually in the interval $[0, 1]. So this answer's still wrong.


问:
Proof that if the measure of the complement of A is 0, then A is dense in $\mathbb{R}^n$.
If $m(A^c) = 0$, then $A$ is dense in $\mathbb{R}^n$. $A \subset \mathbb{R}^n$.

答:
We have to use that a set $B$ with zero outer measure is Lebesgue measurable, which implies that $B^\mathsf c$ is Lebesgue measurable too;  we'll also use the monotonicity of the Lebesgue measure, i.e., if there are two measurable sets $B\subset C$ then $m(B)\le m(C)$; and that a nonempty open ball $B$ has positive measure (and it's Lebesgue measurable of course), since it contains a non-degenerated interval, which has positive measure.

If $U$ is open and nonempty and $A\cap U=\emptyset$ then $U\subset A^\mathsf c$, so $m(U)\le m(A^\mathsf c)=0$, and thus $m(A)=0$.

However, since $U$ is open and nonempty, for $x\in U$ there is some $r$ such that $B(x,r)\subset U$, so $0<m(B(x,r))\le m(U)$, which is clearly a contradiction.

Therefore $A\cap U\neq\emptyset$, and $A$ must be dense.


问:
Matrix $A$ can be written as a linear combination of $n \times n$ matrices with all the entries zero except one of them which is one.
Let $\mathbb{K}$ be a field with characteristic zero, $n \in \mathbb{N}$ and $A \in \cal{M}_{n}\mathbb{(K)}.$ I want to prove that $A$ can be written as a linear combination of $n \times n$ matrices with all the entries zero except one of them which is one.

For example, $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = 1 \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + 2 \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} + 3 \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + 4 \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.$

For that I defined the matrix $M_{ab} = (m_{ij})_{n}$ where ${m_{ab}}_{ij} = \begin{cases} 1, & \text{if } i = a, j = b \\ 0, & \text{otherwise} \end{cases}$ (with $a,b \in \mathbb{N}$ such that $1 \leq a,b \leq n$).

Then we that

$$\begin{align}
A & = (a_{ij})_{n} \\
& = (a_{ij} \cdot 1)_{n} & \text{($1$ is the identity for multiplication in $\mathbb{K}$)}\\
& = (a_{ij} \cdot {m_{ij}}_{ij})_{n} & \text{(by definition of $M_{ab}$)}\\
& = \sum_{1 \leq i,j \leq n} {a_{ij}M_{ij}}
\end{align}$$

I just don’t know how I can justify the last step. Do you have any ideas? Or any other approaches to prove this?

Thank you so much!

答: You may consider the $ij^\text{th}$ entry of $B=A-\sum_{x,y}a_{xy}M_{xy}$.$$b_{ij}=a_{ij}-\sum_{x,y}a_{xy}[M_{xy}]_{ij}=a_{ij}-a_{ij}\times1=0$$since $[M_{xy}]_{ij}=\begin{cases}1,&x=i,y=j\\0,&\text{otherwise}\end{cases}$.


问:
Simplify this summation
I have $\frac{(m-1)!}{(m+n)!} \sum_{i=0}^{m-1} \frac{(m+n-i-1)!}{(m-1-i)!}$. And i'm fairly certain that this equals 1/(n+1) but i'm not sure how I would simplify the big summation term to show that.

答: It helps to write your sum as $\frac{(m-1)!n!}{(m+n)!}\sum_{j=0}^{m-1}\binom{n+j}{j}$, where we take $j=m-1-i$. With the famous result$$\sum_{j=0}^{m-1}\binom{n+j}{j}=\binom{m+n}{m-1},$$we get the desired$$\frac{(m-1)!n!(m+n)!}{(m+n)!(m-1)!(n+1)!}=\frac{n!}{(n+1)!}=\frac{1}{n+1}.$$

Title: Simplify this summation


问:
Numerical Approximation of the derivative
In high school students are typical presented with the one sided definition of the derivative. Let $\epsilon > 0 $ and $f(x)$ be a mapping between two vector spaces then by definition
$$f'(x) = \lim_{\epsilon \to 0} \frac{f(x + \epsilon) - f(x)}{\epsilon}$$

Another definition is given as
$$f'(x) = \lim_{\epsilon \to 0} \frac{f(x + \epsilon) - f(x - \epsilon)}{2\epsilon}$$
From a theoretical point of view these two are the same since $\epsilon$ truly approaches zero. However, from a practical point of view in the world of numerical computation we can only approximate the limit. It turns out that the second definition yiels an error of $O(\epsilon^2)$ whereas the first one yields an error of $O(\epsilon)$. Why is this the case? Is there a known proof of this?

答:
Recall Taylor expansion: for $f \in C^3$, you have:

$f(x+h) = f(x) + f'(x)h + \dfrac{1}{2} f''(x) h^2 + O(h^3)$

$f(x-h) = f(x) - f'(x)h + \dfrac{1}{2} f''(x) h^2 + O(h^3)$

Then with appropriate differences, you have:

$\dfrac{f(x+h) - f(x)}{h} - f'(x) = \dfrac{1}{2} f''(x) h + O(h^2) = O(h)$

$\dfrac{f(x+h) - f(x-h)}{2h} - f'(x) = O(h^2)$


问:
Prove $\int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx} = 2\pi\log^2{2} + \frac{\pi^3}{6}$
How to prove

$$\int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx} = 2\pi\log^2{2} + \frac{\pi^3}{6}$$

I got this result using Fourier representation of
$$|\log(\cos(\frac{x}{2}))|$$
and Parseval's identity. But I am wondering if there is a direct way to calculate this integral.

答:
In my favorite CAS is has two identic forms integrate in the sense of an unbounded general integral. I can only do screenshots because the MathML is ill-posed. This can be looked up in books with integreal relationsships and representations like Abramovitch, Stegun.

First a longer one:

Now are either integration done automatically or looked up in formula collections.

Evaluating this at the limits of the given integral is:

```<code>+∞
</code>```

doubly singular. The signs do not change for -/+ . So the limits have to be taken. These are not too easy and need complex theory and residue methodology.

The problem gets simpler with the substitution t=x/2 and the knowledge that the function is even.

For t=0 and     t=/2 the term is positive ∞.

This is the graph of the function of the question:

With the substitution     z=Cos[t] from @z-ahmed we get something even easier:

The process path can be gone in this direction and reverse. Mind the minus sign.

Our limit for     z->0 is now 0.

The limit for     z->1 is     1/24  (^2 + 3 Log5^2)

We remember our factor 4.

So the result is

.

This can be simplified with 4 Log[2]^2 - Log[4]^2==0.

Body:
In my favorite CAS is has two identic forms integrate in the sense of an unbounded general integral. I can only do screenshots because the MathML is ill-posed. This can be looked up in books with integreal relationsships and representations like Abramovitch, Stegun.

First a longer one:

Now are either integration done automatically or looked up in formula collections.

Evaluating this at the limits of the given integral is:

+∞

doubly singular. The signs do not change for -/+ . So the limits have to be taken. These are not too easy and need complex theory and residue methodology.

The problem gets simpler with the substitution t=x/2 and the knowledge that the function is even.

For t=0 and     t=/2 the term is positive ∞.

This is the graph of the function of the question:

With the substitution     z=Cos[t] from @z-ahmed we get something even easier:

The process path can be gone in this direction and reverse. Mind the minus sign.

Our limit for     z->0 is now 0.

The limit for     z->1 is     1/24  (^2 + 3 Log5^2)

We remember our factor 4.

So the result is

.

This can be simplified with 4 Log[2]^2 - Log[4]^2==0.


答: Using user's answer $\frac{x^2y^2}{x^4+y^4}= \frac{r^4 \sin^2a\cos^2a}{r^4(\sin^4a+\cos^4a)}= \frac{\sin^2a\cos^2a}{\sin^4a+\cos^4a}$, where angle $a$ can assume any value as $r\to 0$.  Therefore the expression has no limit as both $x,y\to 0$.

Title: Can I solve this multi-variable limit with polar coordinates: $\lim_{ (x,y)\to(0,0)} \frac{y^2\sin^2(x)}{x^4+y^4}$?

Body:
$$\lim_{ (x,y)\to(0,0)} \frac{y^2\sin^2(x)}{x^4+y^4}$$

I know that I could let $x=0$ and solve the limit, and then the same for $y=x$.
But I want to know if I could solve it only with polar coordinates.


问:
$u_n\to f\in L^1([0,1])$ but $u_n'\to g\notin L^1([0,1])$?
Is it possible to have a sequence of functions $(u_n)_{n\ge 1} \subset C^1([0,1])$ such that $u_n\to f\in L^1([0,1])$ but the sequence of derivatives are such that $u_n'\to g\notin L^1([0,1])$?

Body: Yes, for example $u_n(x)=(1+\frac{1}{n})e^{-1/x}$ if $0<x$ and $u_n(0)=0$.


问:
Zeno-like riddle with additional complication: Runner with dog running back and forth at different speeds
I came across this little riddle:

An amateur runner trains on a 1 250 m long track at a speed of $v_L = 2 m/s$. His dog runs ahead to the target, turns around and then runs back to him, then he runs back to the  target, back, etc. His speed is when running forth $v_H = 5 m/s$ and when running back (because of the headwind) $v_Z = 4 m/s$. Calculate the distance traveled by the dog.

It is little bit like Zeno's paradoxes but with the additional complication that here the dog is running back and forth at different speeds.

Could anyone please give some hints how to elegantly tackle the problem or references where a solution can be found (Perhaps this riddle even has a common name under which it can be found in the literature)? Thank you!

答:
A variation on the classic elegant approach works.  The runner will take $625$ seconds to complete the course.  The dog will take $250$ seconds for the first leg.  After that, the dog averages $\frac{2}{\frac{1}{5}+\frac{1}{4}}=\frac{40}{9} \frac{m}{sec}$ and you can apply the usual approach.  The main points are 1)use the harmonic mean to find average speed, as you are averaging over common distances 2)after the first leg, the dog makes round trips so you can average the speed.  As the dog is making round trips for $375$ seconds, it covers $375*\frac{40}{9}=1666\frac{2}{3}$ meters in that time, which when added to the $1250$ it already did makes $2916\frac{2}{3}$ meters total.

Added after the comment to Tim van Beek's answer:  the classic version has the dog at a single speed, say $5$ m/sec.  Then the simple answer comes from the runner taking $625$ seconds, so the dog travels $625*5=3125$ meters.  The complex answer is to sum the infinite series of legs traveled by the dog.  The first leg is $1250$ meters, after which the runner has covered $250$ meters.  So the dog runs back $\frac{5}{7}$ of $1000$ meters and returns to the end.  Then you figure out the next leg of the dog.  It is a geometric series which you can sum.


问:
How can i resolve this point (a,b) combinations?
In a given programming language, an identifier is a sequence of a certain number of characters in which the first character must be a letter of the English alphabet and the rest can be a letter or a digit.
a How many identifiers are there of length 5?

b. In particular, some implementations of Pascal and identifier is a sequence of 1 to 8
characters with the above description. How many Pascal identifiers are there?
I tried to do this: there are five spaces, the first is 26 and the others could be 36 because 10+26, but I don’t know how to resolve this.
I don’t know how to resolve b.

答:
first character must be a letter of the English alphabet and the rest can be a letter or a digit. How many identifiers are there of length 5?

Well, there are $26$ possibilities for the first character, and $26+10$ possibilities for the remaining 4 spots.
That's: $(26)(36^4)=43,670,016$ possible combinations

As for $n$ characters, there will be $(26)(36^{n-1})$ possible combinations


问:
Show, using the $N$, $\epsilon$ definition of convergence, that the sequence $\frac{2}{\sqrt{n+2}}$ converges to $0$.
I have tried to do the following: let $\epsilon >0$, and we want to find a positive integer $N$ such that for all $n>N$, we have that $|2/(\sqrt{n+2})-0|$ and this should be less than $\epsilon$, but I'm not sure where to go from there.

答: From the Archimedian property, you can always find $n_0\in\Bbb N$ such that $1/n_0<\epsilon/2$. Take $n_0>1$.  Then $\forall m\ge n_0^2-2,\sqrt{m+2}\ge n_0\implies \frac1{\sqrt{m+2}}\le\frac1{n_0}<\epsilon/2$.


问:
Properties of the "distinguished logarithm"
It's been a long time since I was working in complex analysis. I remember that we can (uniquely?) extend the exponential function to a continuous function $\mathbb C\to\mathbb R$, while the natural logarithm can only be extended to $\mathbb C$ in the sense of a multivalued function.

While I don't want to dive to deep into this topic, there is one particular result I need to work with and which I'm currently using in a black-box fashion: If $\varphi\in C^0(\mathbb R^d,\mathbb C\setminus\{0\})$ with $\varphi(0)=1$, there is a unique $\psi\in C^0(\mathbb R^d,\mathbb C)$ with $\psi(0)=0$ and $\varphi=e^\psi$ and, abusing notation, we define $\ln\varphi=\psi$ (which is not to be confused with the composition of a function called "$\ln$" with $\varphi$). $\ln\varphi$ is called the distinguished logarithm of $\varphi$.

On the one hand, the notation introduced above seems plausible, since $\psi$ is uniquely determined by $\varphi$. On the other hand, I wonder whether this notation is compatible with the "usual properties" of the logarithm on $\mathbb R$ which we are used to work with. For example,

if $\varphi_1,\varphi_2$ are two instances of $\varphi$, can we verify that $\ln(\varphi_1\varphi_2)=\ln(\varphi_1)+\ln(\varphi_2)$?

if $d=1$ and $\varphi(x)=e^{{\rm i}x}$, can we verify that $\ln\varphi(x)={\rm i}x$?

In (2.), if I remember correctly, the "usual" multivalued logarithm $\log$ on $\mathbb C$, would yield a set $\log(e^{{\rm i}x}$. Does the distinguished logarithm somehow select the single element ${\rm i}x$ of this set?

答:
The result you state holds because $\mathbb{R}^d$ is simply connected.  And (1) and (2) are both true, because you can verify in each case that the right hand side satisfies the properties of the distinguished logarithm on the left hand side, which is uniquely determined by those properties.

Nevertheless, this notation has a potentially confusing issue which you have already pointed out, namely:

$\ln \varphi$ is not to be confused with the composition of a function called "ln" with $\varphi$.

So there is no function $\ln$, only a function $\ln \varphi$. This means that in (2), when you write $\ln \varphi(x) = ix$, the left side must be read as $(\ln \varphi)(x)$ and not as $\ln (\varphi(x))$.  If you read it as the latter, then from $\ln(\varphi(x)) = ix$, you would get an apparent contradiction by substituting $x=0$ and then $x = 2 \pi$, since $\varphi(0) = \varphi(2 \pi)$ but $0 \ne 2 \pi i $.

For your last question, the distinguished logarithm does select $ix$ in (2), but it is not as magical as it might initially seem. From the definition of the distinguished logarithm, it is a function $\psi(x)$ satisfying $e^{\psi(x)} = \varphi(x) = e^{ix}$.  Of course $\psi(x) = ix$ satisfies that.


问:
Proof that $\frac{d}{dt} \log |A(t)| = \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A \right]$
For a symmetric $n\times n$  matrix $A(t)$ that depends on the variable $t$, I want to prove that
$$
\frac{d}{dt} \log |A(t)| = \text{Tr}\left[A(t)^{-1} \frac{d}{dt} A(t) \right].
$$

Here, $|A|$ denotes the determinant of $A$ and $\text{Tr}[X]$ denotes the trace of $X$. How can I prove this equality?

The reason I ask this question: The equality is presented in equation C.22 of Pattern Recognition and Machine Learning of C. Bishop and the proof is left "as an exercise to the reader".

My attempt so far (to not clutter the notation, I leave out the dependency on $t$):

We know that $|A|=\prod_{i=1}^n \lambda_i$, where $\lambda_i$ denotes the $i$-th eigenvalue.
Hence, we can write
$$
\frac{d}{dt} \log|A|
=\frac{d}{dt}\log\prod_{i=1}^n\lambda_i
=\frac{d}{dt}\sum_{i=1}^n\log(\lambda_i)
=\sum_{i=1}^n\frac{d}{dt}\log(\lambda_i)
=\sum_{i=1}^n\frac{1}{\lambda_i} \frac{d}{dt}\lambda_i.
$$

The textbook further suggests to make use of the equalities
$$
A = \sum_{i=1}^n \lambda_i u_i u_i^T
\quad \text{and} \quad
A^{-1} = \sum_{i=1}^n \frac{1}{\lambda_i} u_i u_i^T,
$$
where $u_i$ denotes the $i$-th eigenvector.

I do not know how to proceed from here. I also do not know if this is going into the right direction.

答:
Provided the problem is well-defined, i.e. matrix is diagonalizable, non-singular and determinant is positive, one should be able to use the orthonormality of eigenvectors, i.e.

$$
\mathbf{u}^T_i.\mathbf{u}_j=\sum_s u^\left(s\right)_i u^{\left(s\right)}_j=
\begin{cases}
1,\quad i=j \\
0,\quad otherwise
\end{cases}
$$

Where sum over $\dots^{\left(s\right)}$ is over individual components of the eigen-vectors.

Now:

$$
\begin{align}
\mbox{Tr}\left(\mathbf{A^{-1}}.\frac{d}{dt}\mathbf{A}\right)=&\mbox{Tr}\left(\left[\sum_i \mathbf{u}_i.\frac{1}{\lambda_i}.\mathbf{u}_i^T\right]\:\cdot\:\frac{d}{dt}\left[\sum_j \mathbf{u}_j .\lambda_j.\mathbf{u}_j^T\right]\right) \\
=&\sum_{sp}\sum_{ij}u^{(s)}_i\lambda_i^{-1}u^{(p)}_i\cdot\left(\frac{du^{(p)}_j}{dt}\lambda_ju^{(s)}_j+u^{(p)}_j\frac{d\lambda_j}{dt} u^{(s)}_j+u^{(p)}_j \lambda_j \frac{du^{(s)}_j}{dt}\right)\\
=&\sum_i \frac{1}{\lambda_i}\cdot\frac{d\lambda_i}{dt}+\sum_p\sum_i u^{(p)}_i\frac{du^{(p)}_i}{dt}+\sum_s\sum_i u^{(s)}_i\frac{du^{(s)}_i}{dt}\\
=&\sum_i \frac{1}{\lambda_i}\cdot\frac{d\lambda_i}{dt}+2\sum_s\sum_i u^{(s)}_i\frac{du^{(s)}_i}{dt}\\
=&\sum_i \frac{1}{\lambda_i}\cdot\frac{d\lambda_i}{dt}+\frac{d}{dt}\sum_s\sum_i u^{(s)}_i u^{(s)}_i \\
=&\sum_i \frac{1}{\lambda_i}\cdot\frac{d\lambda_i}{dt}
\end{align}
$$

Since

$$
\sum_i \sum_s u^{(s)}_i u^{(s)}_i = \sum_i 1=\mbox{number of components}=\mbox{const}
$$

Which, I think, was what you were after


问:
Is the action functional "locally well defined up to an additive constant"?
I'm looking at an arbitrary symplectic manifold $(P, \omega)$, and the action functional on the set of contractible loops $\Lambda P:$ $$A: \Lambda P \to \mathbb{R}, \hspace{5pt} A(z) = \int_{D^2} \bar{z}^*\omega,$$where $\bar{z}$ is some extension of $z: S^1 \to P$ to $D^2$, which exists by contractibility.

Of course, this function is not well defined, because the value of $A$ will almost certainly depend on the choice of extension $\bar{z}$ (unless for example $\pi_2(P) = 0$). However, according to Floer in his paper "The unregularized gradient flow of the symplectic action," this function is locally well-defined, up to an additive constant, so we can talk about its derivative $dA$.

For a fixed $z \in \Lambda P$, how does one choose the neighborhood $z \in U \subset \Lambda P$ on which $A$ can be well-defined, with a suitable, coherent choice of extensions of loops in $U$?

For what it's worth, here's my attempt, even though I don't think it needs to be that complicated.

Choose $z$ and a "small," contractible neighborhood $U$ of $z$ (small enough so that all loops in $U$ are homotopic to $z$). Choose also an arbitrary extension $\bar{z}$ of $z$ to $D^2$. For any other $z' \in U$, choose a homotopy $h: z \simeq z'$. Since $D^2 \times \{0\} \cup S^1 \times I$ is a retract of $D^2 \times I$, there exists a map $\tilde{h}: D^2 \times I \to P$ with $\tilde{h}(x, 0) = \bar{z}(x)$ and $\tilde{h}|_{S^1 \times I} = h$. Thus I can define a lift of  $\bar{z}' := \tilde{h}|_{D^2 \times \{1\}}$, and set $$A(z') := \int_{D^2} \bar{z}'^* \omega.$$However, this is no better than choosing an arbitrary extension, because the homotopy $h$ was arbitrary. So instead of looking at the extension problem

I should be looking at all of the homotopies simultaneously, something like

where ideally I'd be able to choose the homotopies to depend continuously (even smoothly) on $z' \in U$. Since $U \times D^2 \times I$ is contractible, there's definitely an extension up to homotopy, but that's not what I need. If $U \times D^2 \times I$ had a CW-complex structure, I could find a solution, but it doesn't in general because $U$ lives in an infinite-dimensional manifold, and they don't admit CW-structures, according to https://mathoverflow.net/questions/357477/cw-structure-on-infinite-dimensional-manifolds/357522#357522.

P.S: I know that people usually work with some quotient of the universal cover of $\Lambda P$ instead etc, but I'm just interested to see if this approach can work too.

答:
I think I figured it out: I was wrong to look at extensions $\bar{z}$ themselves rather than the homotopy classes. I will assume that $P$ is path connected, so that $\Lambda P$ is also.

Denote by $\widetilde{\Lambda P}$ the universal covering of $\Lambda P$. This is, as a set, for some choice of $p \in P$, $$\widetilde{\Lambda P} = \{ \overline{z}: I \to \Lambda P \hspace{3pt} | \hspace{3pt} \overline{z}(0) = c_p\}/ \simeq (\textrm{rel } \{0,1\}),$$and this can be reinterpreted as $$\widetilde{\Lambda P} = \{\overline{z}:D^2 \to P \hspace{3pt} | \hspace{3pt} \overline{z}(0) = p\}/\simeq (\textrm{rel } S^1 \cup \{0\}),$$with the covering map $p: \widetilde{\Lambda P} \to \Lambda P$ being nothing but the restriction to $S^1$. Then we can define $$\widetilde{A}: \widetilde{\Lambda P} \to \mathbb{R}, \hspace{5pt} \widetilde{A}[\overline{z}] := \int_{D^2} \overline{z}^* \omega.$$If $\overline{z}, \overline{z}'$ belong to the same equivalence class, then, if $h: \overline{z} \simeq \overline{z}'$, we have $$\int_{D^2} \overline{z}^* \omega - \int_{D^2} \overline{z}'^* \omega = \int_{\partial(D^2 \times I)} (h^* \omega)|_{\partial(D^2 \times I)} = \int_{D^2 \times I} d h^* \omega = 0,$$so $\widetilde{A}$ is well-defined.

Now, take $z \in \Lambda P$, and a neighborhood $z \in U \subset \Lambda P$ such that $$ p^{-1}(U) = \bigsqcup_{[\overline{z}] \in p^{-1}(z)} U_{[\overline{z}]},$$ and $p: U_{[\overline{z}]} \to U$ is a smooth diffeomorphism. Choosing some such $[\overline{z}]$ (this is the additive constant), we can define $$A: U_z \to \mathbb{R}, \hspace{5pt}, A(z') := \widetilde{A}[\overline{z}'],$$where $[\overline{z}']$ is the unique element of $U_{[\overline{z}]} \cap p^{-1}(z').$

EDIT: Actually, I still don't know how to prove the "up to an additive constant" part. Suppose I have two homotopy classes of extensions of $z$, $[\overline{z}]$ and $[\widetilde{z}]$. If I take $z' \in U_z$ and $[\overline{z'}]$ and $[\widetilde{z'}]$, two extensions of $z'$ in $U_{[\overline{z}]}$ and $U_{[\widetilde{z}]}$, respectively, then I need to show that $$\int_{D^2} \overline{z'}^* \omega - \int_{D^2} \widetilde{z'}^* \omega = \int_{D^2} \overline{z}^* \omega - \int_{D^2} \widetilde{z}^* \omega.$$


答:
Hint: write down the inequation, and solve it

Step by step:

$\left\vert \dfrac{2}{\sqrt{n+2}} \right\vert < \varepsilon$

$\dfrac{2}{\sqrt{n+2}} < \varepsilon$

$\sqrt{n+2} > \dfrac{2}{\varepsilon}$

$n > \left( \dfrac{2}{\varepsilon} \right)^2 - 2$

So take $N = \left\lceil \left( \dfrac{2}{\varepsilon} \right)^2 - 2 \right\rceil$, ensuring your above inequality for all $n \geq N$.

Title: Show, using the $N$, $\epsilon$ definition of convergence, that the sequence $\frac{2}{\sqrt{n+2}}$ converges to $0$.

Body: I have tried to do the following: let $\epsilon >0$, and we want to find a positive integer $N$ such that for all $n>N$, we have that $|2/(\sqrt{n+2})-0|$ and this should be less than $\epsilon$, but I'm not sure where to go from there.


问:
List of interesting math videos / documentaries
This is an offshoot of the question on Fun math outreach/social activities. I have listed a few videos/documentaries I have seen. I would appreciate if people could add on to this list.

$1.$ Story of maths Part1 Part2 Part3 Part4

$2.$ Dangerous Knowledge Part1 Part2

$3.$ Fermat's Last Theorem

$4.$ The Importance of Mathematics

$5.$ To Infinity and Beyond

答: One interesting documentary that I haven't seen mentioned is "the code" by Prof Marcus du Sautoy (available on netflix-https://www.netflix.com/gb/title/80063658)</p>


问:
An integer $n \geq 2$ is called square-positive- proof?
An integer $n \geq 2$ is called square-positive if there are $n$ consecutive positive integers whose sum is a square. Determine the first four square-positive integers.

So I have found the first four square-positive numbers, but I need to prove that why it $4$ is not a square-positive number and I also need to write a general formula for determining whether a number is square-positive or not. I have tried to write the sum of consecutive positive integers like this $a + a +1 + a + 2 + a+3 \dots a - 1$ and I wrote it like this for all numbers, and part of the proof for why $4$ isn't a square-positive number is that $4a + 6$ is not divisible with $4$. But I haven't got so far.

Here is my answer:

2 : 4 + 5 = 9  which is 3^2

3 : 2 + 3 + 4 = 9 which is 3^2

5: 18 + 19 + 20 + 21 + 22 = 100  which is 10^2

6: 35 + 36 + 37 + 38 + 39 + 40 = 225  which is 15^2

Interesting fact is that for all odd numbers and some even numbers like 6 and 10, you can find out which number is the first (the one you start with and then go forward here like 3, 2, 18 and 35) using this formula :

(I show it in an example because I still can't write it algebraically):

For example: the sum of 95 subsequent numbers is 5n + 10

(10^2 - 10) /5 = 18

So your first number is 18

And if you keep adding, 18 + 19 + 20 + 21 + 22 you get 100 which is 10^2, the same number you squared.

答:
Let $b-2, b-1, b, b+1$ be four consecutive numbers. The sum is $4b -2$.  Let $m^2 = 4b-2$  Then $\frac {m^2}2 = 2b-1$.  And $2|m^2$.  So $2|m$.  Let $m=2k$ then $4k^2 = 4b -1$ and $2k^2 = 2b-1$.  That's impossible as the LHS is even and the RHS is odd.  So $4$ is not square positive.

=====

In general:

........

If the first integer is $a$ and there are $n$ consecutive integers the numbers are $a, a+ 1,......, a+(n-1)$ and the sum of the the integers is $a + (a+ 1) + ..... + (a + (n-1)) = n\cdot a + (1+ 2 + ...... + (n-1)) = n\cdot a + \frac {(n-1)n}2$.

If $n\cdot a + \frac {(n-1)n}2 = M^2$ then by completing the square

$\frac 12n^2 - (a-\frac 12)n -M^2 = 0$ and .....

well, we can keep going but.... let's try to be clever....

If $n=2k+1$ is an odd number and the middle number is $b$ then the first number is $b-k$ and the last number is $b+k$ and the sum of the number is $$(b-k) + (b-k+1) + ..... + (b-1) + b+ (b+1) + ..... + (b+k) =\\ nb + (-k+(-k+1)+...... + (-1) + 0 + 1 + 2 + ..... k) =\\nb + 0= nb$$.  ANd $nb$ can be square positive whenever $b= n$.  (and other times).  For intsance.  $1=1^2$ and $2+3+4=3^2$ and $3+4+5+6+7 = 5^2$ and $(n-\frac {n-1}2) + ...... + (n + \frac {n-1}2) = n^2$.

So all odd numbers are square positive.

And if $n= 2k$ is even the there is no middle number but if the "middlish" number (off by one half) is $b$:Then we can notice that first $k$ of these numbers are $b-k ,b-(k-1) , b-(k-2),........, b-1$; and the second $k$ of these numbers are $b, b+1, b+2, ......, b+(k-1)$; and the sum of these numbers are $$nb + (-k,-(k-1),-(k-2),......., (k-2),(k-1)) =\\ nb - k =\\ 2kb -k = k(2b-1)$$.  And that can be a square positive every time we have $2b-1 = k=\frac n2$.

But this is possible only if  $\frac n2$ is odd.

But if $k=2m$ is even we must have $2m(2b-1)$ is a perfect square which requires $m$ to be even.

By induction will prove $n$ is square positive if and only if $n = 2^{odd}\cdot odd$.

So the list of all square positive numbers are $2,3,5,6,7,8,9,10,11,13,14,15,17,18,19,21,22,32,24,......$ etc.


答:
Usually in problems like these it helps to first 'work backwards' And only then write out the formal proof. For your problem, the 'backwards work' looks like

$$\epsilon>\left|\frac{2}{\sqrt{n+2}}-0\right|=\frac{2}{\sqrt{n+2}}$$

$$\sqrt{n+2}>\frac{2}{\epsilon}$$

$$n+2>\frac{4}{\epsilon^2}$$

$$n>\frac{4}{\epsilon^2}-2$$

Thus, choose $N=\max\left\{\frac{4}{\epsilon^2},1\right\}$. First, if $\epsilon\geq 2$, then $N=1$. This implies $n\geq N=1$ and therefore

$$n\geq 1$$

$$n+2\geq 3$$

$$\sqrt{n+2}\geq \sqrt{3}>1$$

$$\left|\frac{2}{\sqrt{n+2}}-0\right|=\frac{2}{\sqrt{n+2}}< 2\leq \epsilon$$

and we are done. Now, if $\epsilon<1$ then $N=\frac{4}{\epsilon^2}$. Thus $n\geq N=\frac{4}{\epsilon^2}$ implies

$$n\geq N=\frac{4}{\epsilon^2}>\frac{4}{\epsilon^2}-2$$

$$n+2>\frac{4}{\epsilon^2}$$

$$\epsilon^2>\frac{4}{n+2}$$

$$\epsilon>\frac{2}{\sqrt{n+2}}=\left|\frac{2}{\sqrt{n+2}}-0\right|$$

We conclude that

$$\lim_{n\to\infty}\frac{2}{\sqrt{n+2}}=0$$


问:
Particular example of a non-vanishing analytic function f in unit disc |z|<1.
This question is from Ponnusamy and Silvermann complex analysis section : Cauchy inequality.

Also please note that they are not homework problems. I am trying myself.

Question: Give an example of a non -vanishing analytic function f in unit disc |z| <1 with infinitely many zeroes.

Keeping uniqueness theorem in mind I need to find a sequence in |z|<1 with no limit point.

Now I choose $ a_1 =0.1 , a_2=0.11, a_3 =0.111, a_4 =0.1111$ and so on.

I think this will not have any limit point but that's due to the reason that I am unable to find limit point of this sequence (I think it doesn't exists).

But can you please prove rigorously why limit point doesn't exists?

If I am wrong kindly tell an example that will work.

Many thanks!!

答:
You are searching for $$\prod_{n=1}^\infty (1-\frac{10^{-n}}{1-z})$$
Does it converge, is it analytic, where are its zeros


问:
Probability we have to draw x times (without replacement) before we get what we want
There was an example "fun" problem during one of my lectures that I am a bit confused about (prof went too fast).

The problem is:
Suppose we have cards in a bag. The cards are of 2 types. There are m monster cards and n spell cards. Suppose we pick a random card from the bag without replacement (one at a time) each uniformly at random until we get a spell card. If X is the number of cards we have to pick, what is P(X=k) if $k \in \{1,...,m\}$

The way it was done is that each monster card was marked with a number 1,...,m and we created Bernoulli Random Vars $X_i$ for each marked monster card. So $X_i=1$ means we pick monster card marked i before any spell card is picked. and $X_i=0$ otherwise. Then they said that $P(X_i=1) = 1/(1+n)$. This part I don't get. How did they get this? I get $X=X_1 +...X_m $ but using this fact how would you find what $P(X=k)$ is?

答:
In this answer I assume $X$ counts only the monster cards picked, i.e. $X$ does not include the spell card itself.

Calculating $P(X=k)$ explicitly is not hard, because it just means the first card must be a monster, then the second must be a monster, ..., then the $k$th card must be a monster, then the $(k+1)$th card must be a spell.  Define $t = m+n$ for convenience.  We have:

$$P(X=k) = {m \over t} { m-1 \over t-1 } \cdots {m-(k-1) \over t-(k-1)} {n \over t-k}$$

Incidentally, if you want $E[X]$, the above is not the easy way to go. Instead, the professor's approach is excellent.  We have $E[X_i] = P(X_i = 1) = 1/(n+1)$ so by linearity we have $E[X] = m/(n+1)$, nice and simple.