问:
Is this correct?
How many distinct $(a,b,c)$ satisfy $a^2+bc=b^2+ac$ where $a,b,c$ are integers between $1$ and $5$ inclusive. $a(a-c)=b(b-c)$, $a/b=(b-c)/(a-c)=(a+b-c)/(a+b-c)=1$, so $a=b$, so the answer is 25 solutions. My only concern is that there seem to be no restrictions on the variable $c$. Is this ok?</p>

答:
Note that from you equation you get $a^2-b^2=ac-bc$ which is $$(a-b)(a+b)=(a-b)c$$

One option is thus $a=b$. Now assume $a\neq b$. Then you get $a+b=c$. And this has infinitely many solutions!

Title: Is this correct?


问:
Finding the constants so the integration formulas are accurate to an accuracy level of $3$

$$\int^1_{-1}f(x)dx=af(-1)+bf(1)+cf'(-1)+df'(1)$$

I need to find constants so that the integration formulas are accurate to an accuracy level of $3$ (i.e. will give results identical to the true value for a polynomial of no more than $3$). I started to study this subject alone last week, and I got lost.

Any hints?

答:
There are four degrees of freedom (corresponding to the four weights $a, b, c, d$). A cubic polynomial is determined by four pieces of information. Thus we should expect that doing this for a generic cubic polynomial will exactly uniquely determine the weights.

The rest of this answer will follow two steps:

There will be a unique cubic polynomial $F(x)$ having prescribed values at $-1$ and $1$, and with prescribed derivatives at $-1$ and $1$. We find this polynomial.

Integrate the above polynomial to find the weights.

(Bonus --- we'll check on an example)

Finding the cubic polynomial with prescribed values

There are many ways of doing this, but I'm lazy and don't want to think of a convenient interpolation basis. So instead I'll use linear algebra. Consider a cubic polynomial of the form $F(x) = Ax^3 + Bx^2 + Cx + D$. Form the matrix system whose rows are (in order) the equations $F(-1), F'(-1), F(1), F'(1)$. This system is then
$$
\begin{pmatrix}
-1 & 1 & -1 & 1 \\
3 & -2 & 1 & 0 \\
1 & 1 & 1 & 1 \\
3 & 2 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
A \\
B \\
C \\
D
\end{pmatrix} =
\begin{pmatrix}
F(-1) \\ F'(-1) \\ F(1) \\ F'(-1)
\end{pmatrix}.
$$
The solution to this system will be the unique polynomial $F$. Perhaps the easiest way to determine this is to note that the inverse of the matrix is
$$
\frac{1}{4} \begin{pmatrix}
1 & 1 & -1 & 1 \\
0 & -1 & 0 & 1 \\
-3 & -1 & 3 & -1 \\
2 & 1 & 2 & -1
\end{pmatrix},
$$
and thus the solution is
$$
\begin{pmatrix} A \\ B \\ C \\ D \end{pmatrix} =
\frac{1}{4}\begin{pmatrix}
F(-1) + F'(-1) - F(1) + F'(1) \\
-F'(-1) + F'(1) \\
-3F(-1) - F'(-1) + 3F(1) - F'(1) \\
2F(-1) + F'(-1) + 2F(1) - F'(1)
\end{pmatrix}. \tag{1}
$$

Let's make sure this is right by checking this on the arbitrary, made-up data $F(-1) = 1, F'(-1) = 2, F(1) = 3, F'(1) = 4$. Then this system indicates that the unique cubic polynomial satisfying these conditions is $F(x) = x^3 - x^2/2 + 3/2$, and a quick check confirms this.

Integrating

In terms of the generic cubic $F(x) = Ax^3 + Bx^2 + Cx + D$, we compute that
$$ \int_{-1}^1 F(x) dx = \frac{2B}{3} + 2D.$$
From $(1)$ above, we find that in terms of the values $F(-1), F'(-1), F(1), F'(1)$, this is
$$
F(-1) + \frac{1}{3}F'(-1) + F(1) - \frac{1}{3}F'(1). \tag{2}
$$

We claim that $(2)$ is perfect for cubic polynomials, and is our approximation for generic functions.

Checking

Let's check on a variety of polynomials.

The constant $c$ polynomial has $f(-1) = f(1) = c$ and $f'(-1) = f'(1) = 0$. The integral is $2c$, and formula $(2)$ gives that it's approximately $c + 0 + c + 0 = 2c$. Good.

The polynomial $ax + c$ has $f(-1) = (c - a)$, $f(1) = (a + c)$, $f'(-1) = f'(1) = a$. The integral is $2c$ still. Formula $(2)$ gives that it's approximately $(c-a)  + (a/3) + (a + c) - (a/3) = 2c$. Good.

The polynomial $x^2 + 1$ has $f(-1) = f(1) = 2$, $f'(-1) = -2$, and $f'(1) = 2$. The integral is $8/3$. By $(2)$, we expect this to be $2 - 2/3 + 2 - 2/3 = 8/3$. Good.

The cubic $5x^3 + 3x^2 - x + 1$ has integral $4$. Then $f(-1) = 0, f(1) = 8, f'(-1) = 8, f'(1) = 20$, and $(2)$ indicates that the integral should be $0 + 8/3 + 8 - 20/3 = 4$. Good.

So this works out on lots of examples. For fun, let's also apply this to one case where it won't work without error. Consider the function $f(x) = e^x \sin x$. Then the integral over $[-1, 1]$ is actually about $0.6634$. The approximation $(2)$ indicates that the integral should be about
$$
f(-1) + f'(-1)/3 + f(1) - f'(1)/3 \approx
-0.3095 - 0.0369 + 2.2873 - 1.2520
\approx 0.6888,
$$
which is pretty close to the correct value of $0.6634$.

Generalization and alternative method

I hope it's clear that this approach readily generalizes to general degree. There is a bit of linear algebra in the middle, but there's not much more to it.

However there is an alternative approach --- still linear algebra, but more direct. This argument should make it clear that the resulting approximation will be linear in the prescribed values. What this means is that if you choose almost any four random cubic polynomials (or polynomials of an appropriate power more generally), compute the relevant values of the polynomial and its derivative, set up a linear system from these random polynomials, and solve --- then you'll also find the weights (unless by some unlucky coincidence your random polynomials aren't independent --- then remove the dependent polynomial and replace it with a different random polynomial).

For example, let's use the four polynomials I used above as checkers, specialized to (say) $f_1(x) = 1, f_2(x) = 2x + 1, f_3(x) = x^2 + 1, f_4(x) = 5x^3 + 3x^2 - x + 1$. If we believe that the integral should be exactly given by a linear combination of the values of $f$ and $f'$ at $-1$ and $1$ (which I proved occurs from above), then we can try to find this linear combination from the linear system
$$
\begin{pmatrix}
f_1(-1) & f_1'(-1) & f_1(1) & f_1'(1) \\
f_2(-1) & f_2'(-1) & f_2(1) & f_2'(1) \\
f_3(-1) & f_3'(-1) & f_3(1) & f_3'(1) \\
f_4(-1) & f_4'(-1) & f_4(1) & f_4'(1)
\end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}
=
\begin{pmatrix}
\int_{-1}^1 f_1(x) dx \\
\int_{-1}^1 f_2(x) dx \\
\int_{-1}^1 f_3(x) dx \\
\int_{-1}^1 f_4(x) dx
\end{pmatrix},
$$
which specializes for my choices of polynomial above to
$$
\begin{pmatrix}
1 & 0 & 1 & 0 \\
-1 & 2 & 3 & 2 \\
2 & -2 & 2 & 2 \\
0 & 8 & 8 & 20
\end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}
= \begin{pmatrix} 2 \\ 2 \\ 8/3 \\ 4 \end{pmatrix}.
$$
Solving this system gives the solution $a = 1, b = 1/3, c = 1, d = -1/3$, which is exactly the answer $(2)$ above.

I rather like how the two solutions I describe are dual to each other, and how both actually end up being reduced to a game of linear algebra.

ContentLicense: CC BY-SA 4.0


问:
Close-form for integral $T(n)=\int_0^{\pi/2}\frac{1}{1+\sin^n(x)}dx$
Context:

I have recently become interested in integrals of the form $$T(n)=\int_0^{\pi/2}\frac{dx}{1+\sin(x)^n},$$
and I conjecture that $T(n)$ has a closed form evaluation for all $n\in\Bbb Z_{\ge0}$, or at least all non-negative even integers $n$. Trivially, one has $T(0)=\pi/4$. Less trivially, there is $T(1)=1$ (easy with $t=\tan(x/2)$), and $T(2)=\frac\pi{2\sqrt{2}}$ as seen here. The integral $J=T(3)$, however, is not so easy.
We see that this integral is $$3J=\int_0^{\pi/2}\frac{dx}{1+\sin x}-\int_0^{\pi/2}\frac{\sin (x)dx}{\sin(x)^2-\sin(x)+1}+2\int_0^{\pi/2}\frac{dx}{\sin(x)^2-\sin(x)+1}.$$
The first integral is easy, and we get
$$3J=1-J_1+J_2.$$
The next integral is
$$J_1=\int_0^{\pi/2}\frac{\sin (x)dx}{\sin(x)^2-\sin(x)+1},$$
which is, from $t=\tan(x/2)$,
$$J_1=2\int_0^1\frac{\frac{2t}{1+t^2}}{(\frac{2t}{1+t^2})^2-\frac{2t}{1+t^2}+1}\frac{dt}{1+t^2},$$
which is the awful
$$J_1=4\int_0^1\frac{tdt}{t^4-2t^3+6t^2-2t+1}.$$
I found that nothing but brute force could tackle this integral, so I used partial fractions and got
$$J_1=4\sum_{a\in A}\frac{1}{f'(a)}\left(1+a\log(1-a)-a\log(-a)\right)$$
where $$A=\left\{\frac12\left(1+p_1i\sqrt3+p_2\sqrt{-6+2p_3i\sqrt3}\right):p_1,p_2,p_3\in\{-1,1\}\right\}$$
is the set of roots of the polynomial $f(z)=z^4-2z^3+6z^2-2z+1,$
and $\log(z)$ is the complex logarithm.

However, the remaining integral is a little worse. We have from the substitution $t=\tan(x/2)$ the awful
$$J_2=4\int_0^1\frac{1+t^2}{1-2t+5t^2-2t^3}dt.$$
We can do the same sort of trick here as with the last integral and get
$$J_2=4\sum_{b\in B}\frac1{g'(b)}\int_0^1\frac{t^2+1}{t-b}dt$$
where $$B=\{z\in\Bbb C: 1-2z+5z^2-2z^3=0\}$$
is the set of roots of the polynomial $g(z)=1-2z+5z^2-2z^3$. These roots do indeed have explicit evaluations. The integral in the summation is easy enough to calculate, but I'm not going to, as we already see that the integral has a closed form.

Next up, Wolfram evaluates $$T(4)=\int_0^1\frac{dx}{1+\sin(x)^4}=\frac\pi4\sqrt{1+\sqrt2},$$
as well as $$T(6)=\int_0^{\pi/2}\frac{dx}{1+\sin(x)^6}=\frac{\pi}{12}(\sqrt{2}+2\sqrt{3}),$$
which is here.

In fact, we may evaluate $T(2n)$ in terms of hypergeometric functions, which may have a general closed form. We do so by noting that
$$\frac{1}{1+\sin(x)^{2n}}=\sum_{k\ge0}(-1)^k\sin(x)^{2nk}$$
so that
$$T(2n)=\sum_{k\ge0}(-1)^k\int_0^{\pi/2}\sin(x)^{2nk}dx=\frac{\sqrt\pi}{2}\sum_{k\ge0}(-1)^k\frac{\Gamma(nk+\tfrac12)}{(nk)!}.$$
This is
$$T(2n)=\frac\pi2\,_{n}F_{n-1}\left(1-\tfrac{1}{2n},A_n;B_n;-1\right)$$
where
$$\begin{align}
A_n&=\left\{\frac{2r+1}{2n}:0\le r\in\Bbb Z\le n-2\right\}\\
B_n&=\left\{\frac{r}{n}:1\le r\in\Bbb Z\le n-1\right\}.
\end{align}$$
Whether or not this hypergeometric has a closed form I am unsure, but it looks simple enough to be evaluated exactly.

Questions:

Can $T(n)$ be computed in closed form of all $n$? If not, when can it be computed in closed form?

And at the very least, what is $T(5)$? It seems to be very nasty.

Thanks!

答:
The close-form result, valid for both odd and even $n$, is
$$T(n) = \frac{1-(-1)^n}{2n}+ \frac1n \sum_{k=1}^{[\frac n2]}\beta_{k}^+(\frac\pi2+\tan^{-1}\beta_k^-)- \beta_k^- \coth^{-1} \beta_k^+
\tag1$$
with
$$\beta_k^{\pm }= \frac1{\sqrt2}\left(\sqrt{\cot\frac{(2k-1)\pi}{2n}}  \pm \sqrt{\tan\frac{(2k-1)\pi}{2n}} \right)
$$

Proof: Note

$$\frac1{1+y^n}=-\frac1n\sum_{k=1}^n \frac{e^{i\theta_k}}{y-e^{i\theta_k}}, \>\>\>\>\>
\theta_ k = \frac{(2k-1)\pi}n$$
and combine conjugate pairs

$$\frac1{1+y^n}=\frac{1-(-1)^n}{2n(1+y)} + \frac2n \sum_{k=1}^{[\frac n2]} \frac{1-y \cos\theta_k }{y^2-2y\cos\theta_k+1} $$
Set $y = \cos x$ to integrate

\begin{align}
T(n)&=\int_0^{\pi/2}\frac{dx}{1+\sin^n x}
= \int_0^{\pi/2}\frac{dx}{1+\cos^n x} \\
&= \frac1{2n}\int_0^{\frac\pi2}\frac{1-(-1)^n}{1+\cos x}dx
 + \frac2n \sum_{k=1}^{[\frac n2]} \int_0^{\frac\pi2} \frac{1-\cos x \cos\theta_k }{\cos^2x-2\cos x\cos\theta_k+1}dx\\
&= \frac{1-(-1)^n}{2n}
+ \frac2n \sum_{k=1}^{[\frac n2]} \int_0^{1} \frac{\tan^2\frac{\theta_k}2-t^2}{\tan^2\frac{\theta_k}2 +t^4}dt
\end{align}
where the half-angle substitution $t=\tan\frac x2$ is applied. Then, integrate over $t$ to arrive at the close-form expression (1). Note that the general result (1) simplifies considerably for even $n$ as the summation over the inverse trig functions vanishes. Explicitly, for $n =2m$
$$T(2m)= \frac\pi{4m} \sum_{k=1}^{m} \sqrt{\csc\theta_k+1}\tag2
$$
where $\beta_k^+ = \sqrt{\csc\theta_k+1}$ is recognized.

Listed below are some actual results per the close-form (1) or (2).

Even $n$:
\begin{align}
T(2) &= \frac\pi4 \sqrt{\csc\frac\pi2+1}=\frac{\pi}{2\sqrt2}\\
T(4) &= \frac\pi8 \left(\sqrt{\csc\frac\pi4+1}
 + \sqrt{\csc\frac{3\pi}4+1} \right)=\frac{\pi}{4}\sqrt{\sqrt2+1}\\
T(6) &= \frac\pi{12} \left(\sqrt2+\sqrt{\csc\frac\pi6+1}
 + \sqrt{\csc\frac{5\pi}6+1} \right)=\frac{\pi}{12}(\sqrt2+2\sqrt3)\\
T(8) &= \frac\pi8 \left(\sqrt{\csc\frac\pi8+1}
 + \sqrt{\csc\frac{3\pi}8+1} \right)\\
&=\frac{\pi}{8}\left(\sqrt{\sqrt{2(2+\sqrt2)}+1}
 + \sqrt{\sqrt{2(2-\sqrt2)}+1}\right)\\
T(10) &= \frac\pi{10} \left(\frac1{\sqrt2}+\sqrt{\csc\frac\pi{10}+1}
 + \sqrt{\csc\frac{3\pi}{10}+1} \right)\\
&=\frac{\pi}{10}\left(\frac1{\sqrt2}+ \sqrt[4]5+\sqrt{2+\sqrt5}\right)\\
T(12) &= \frac{\pi}{12} \sqrt{1+\sqrt2}\left(1+ \sqrt{2+2\sqrt3}\right)
\end{align}

Odd $n$:
\begin{align}
T(3) &=\frac13 +\frac{\sqrt[4]3+\frac1{\sqrt[4]3}}{3\sqrt2} \left(\frac\pi2+ \tan^{-1} \frac{\sqrt[4]3-\frac1{\sqrt[4]3}}{\sqrt2} \right)
- \frac{\sqrt[4]3-\frac1{\sqrt[4]3}}{3\sqrt2} \coth^{-1} \frac{\sqrt[4]3+\frac1{\sqrt[4]3}}{\sqrt2}
\\
 T(5) &=\frac15 +\frac15\sqrt{ \sqrt{2+\frac2{\sqrt5}}+1}
\left(\frac\pi2+ \tan^{-1} \sqrt{ \sqrt{2+\frac2{\sqrt5}}-1}\right)
\\
 &\hspace{10mm} +\frac15\sqrt{ \sqrt{2-\frac2{\sqrt5}}+1}
\left(\frac\pi2- \tan^{-1} \sqrt{ \sqrt{2-\frac2{\sqrt5}}-1}\right)
\\
 &\hspace{10mm} -\frac15\sqrt{ \sqrt{2+\frac2{\sqrt5}}-1}
\coth^{-1} \sqrt{ \sqrt{2+\frac2{\sqrt5}}+1}
\\
 &\hspace{10mm} +\frac15\sqrt{ \sqrt{2-\frac2{\sqrt5}}-1}
\coth^{-1} \sqrt{ \sqrt{2-\frac2{\sqrt5}}+1}
\\
 T(7) &=\frac17 +\frac17\sqrt{ \csc\frac\pi7+1}
\left(\frac\pi2+ \tan^{-1} \sqrt{ \csc\frac\pi7-1}\right)
\\
 &\hspace{10mm} +\frac17\sqrt{ \csc\frac{3\pi}7+1}
\left(\frac\pi2+ \tan^{-1} \sqrt{ \csc\frac{3\pi}7-1}\right)
\\
 &\hspace{10mm} +\frac17\sqrt{ \csc\frac{5\pi}7+1}
\left(\frac\pi2- \tan^{-1} \sqrt{ \csc\frac{5\pi}7-1}\right)
\\
 &\hspace{10mm} -\frac17\sqrt{ \csc\frac{\pi}7-1}
\coth^{-1} \sqrt{ \csc\frac{\pi}7+1}
\\
 &\hspace{10mm} -\frac17\sqrt{ \csc\frac{3\pi}7-1}
\coth^{-1} \sqrt{ \csc\frac{3\pi}7+1}
\\
 &\hspace{10mm} +\frac17\sqrt{ \csc\frac{5\pi}7-1}
\coth^{-1} \sqrt{ \csc\frac{5\pi}7+1}
\\
\end{align}


问:
If F is a commutator subgroup to G then prove that G/F is abelian.
I am trying to understand some basic Algebraic topology for Physics.

The relation between the fundamental group and the homology group is given by an isomorphism $\phi:\pi_1(|K|)/F\to H_1(K)$. Where F is the commutator subgroup of $\pi_1(|K|)$ and K is a connected simplicial complex.

As I understand from the internet the map $\psi:\pi_1(|K|)\to\pi_1(|K|)/F$ is called abelianization of the fundamental group. It makes sense intuitively that if you quotient out all the non-commutating commutators you get a new relation i.e $(x_i;r_m)\to(x_i;r_m,[x_i,x_j])$. But how do you prove that $\pi_1(|K|)/F$ is abelian?

答:
This answer presumes you know the basics of group theory:

With your notation, observe that $\;G/F\;$ is abelian iff

$$\forall\,x,\,y\in G\;,\;\;xFyF=yFxF\iff xyF=yxF\iff (yx)^{-1}(xy)=x^{-1}y^{-1}xy=:[x,y]\in F$$

But $\;F=G'=\;$ the commutator subgroup of G is generated precisely by all the commutators $\;x,y\;$ , as $\,x,y\;$ run over all the possible choices of elements in G...


问:
If $A^3B-B$ is invertible then $BA-B$ is invertible.
I assume this is a simple proof but i'm stuck here.

I need to prove that if $A^3B-B$ is invertible then $BA-B$ is invertible.

So $A^3B-B=(A^3-I)B$  and then both $(A^3-I)$ and $B$ are invertible.

So I need to show that $BA-B=B(A-I)$ are both invertible.

I thought maybe I could somehow show that if $(A^3-I)$ is invertible then $(A-I)$ is invertible but I guess there's a simpler way to get there. What am I missing?

Thanks!

答:
Note that

$A^3B - B = (A^3 - I)B; \tag 1$

thus the invertibility of $A^3B - B$ implies both $A^3 - I$ and $B$ are invertible.  Since

$A^3 - I = (A - I)(A^2 + A + I), \tag 2$

the invertibility of $A^3 - I$ yields the same for $A - I$ and
$A^2 + A + I$.  Since now $A - I$ and $B$ are invertible, so is their product

$B(A - I) = BA - B. \tag 3$


问:
Convexity of expected value
I am trying to understand if the expected value of a variable is convex in that variable or not.
I know that expectation is a linear operator, so must be convex.
But I do not see why it does not depend on what characteristics the probability vector $p(x)$ has.
In other words, why $E(x)= x^Tp(x)$ is convex for any $p(.)$ such that $p\ge0, p(x)^T1=1$? Is $p$ constant wrt $x$?
Confused.</p>

答: The expected value does not depend on $x$ as we integrate out $x$. So, your question is not meaningful.</p>

ContentLicense: CC BY-SA 3.0

Title: Convexity of expected value


问:
Immersion of a map and regular submanifold
I know $f:\mathbb{R}\rightarrow S^1\times S^1$ with $f(t)=(e^{iat}, e^{ibt})$ with fixed $a$ and $b$ is an immersion. What kind of constraint can we impose on $a$ and $b$ so that the image of  $f$ is a regular sub-manifold of $S^1\times S^1$?

答: The case where both $a$ and $b$ are equal to zero then $f$ is a constant function. In such a case you have that the image is a point. A point is a $0$ manifold and it's inclusion is an injective immersion which is homeomorphic to it's image so we can consider the case where $b\neq 0$. $f$ will be an immersion in this case. But You have that if $a/b \notin \mathbb{Q}$ then the image will be dense. In such case it can't be a submanifold since given any chart $\varphi:U\rightarrow \mathbb{R}^2$ will send $f(\mathbb{R})\cap U$ to a dense subset of $\mathbb{R}^2$ Which can't be a subspace since it's all of $\mathbb{R}^2$. So one must have $a/b\in\mathbb{Q}$ If $a/b\in\mathbb{Q}$ then the image is homeomorphic to $S^1$. Identifying the torus as being $\mathbb{R}^2/\mathbb{Z}^2$ you have that $f$ can be lifted to a function $\tilde{f}$ to $\mathbb{R}^2$ which will have as image a line with slope $a/b$ passing through the origin. Since it's rational we have $\tilde{f}$ will pass through a point with integer coordinates at $n\cdot b$ where $n\in \mathbb{Z}$. This means that $f$ will have period $b$. By identifying $\mathbb{R}/b\mathbb{Z}\equiv S^1$ one can define a homeomorphism from $S^1$ to the image of $f$ through the quotient map. This map will be an embedding that has the same image of $f$. So the only condition is that either $a/b\in \mathbb{Q}$ or $b/a\in\mathbb{Q}$ or $a=0$ and $b=0$. (There's probably a neater way to write this)


问:
Find the area of an equilateral triangle inscribed in a unit square
If inside the square $ABCD$ there is an equilateral triangle $CMN$ inscribed, as shown in the figure. If the area of the square is 1, find the area of the triangle.

I attempted to say, I state that $x=MD$

then $MC^2=1+x^2$. $AM=1-x$ and I thought that something might come from Pythagoras, but it didn't. $\triangle AMN$ looks isosceles, but I can't prove that it is, although if it is proved to be isosceles, then the question can easily be solved by use of the pythagorean theorem. We also have that $\angle CNB+\angle CMD=150^o$. I got stuck here, could you please explain to me how to solve the question?

答:
Given square ABCD with AB=BC=CD=1, MNC an equilateral triangle MC=BC=MN=a.

For S$_{\triangle MNC}$, if we know MN=a, then we can directly apply S$_{\triangle MNC} =\frac{3}{4}a^2$.

Set NB = x, NC=MC=MN=a= $\sqrt{x^2+1}$, MD=$\sqrt{MC^2-1}$=x=NB, AM=AN=1-x.

Apply Pythagorean theroem to $\triangle ABN$

\begin{align} 2(1-x)^2=x^2+1
, x^2-4x+1=0, x=2\pm \sqrt{3} \end{align}

Since x <1, keep only  x=$2-\sqrt{3}$ $\to$ MN$^2$=a$^2$= 1+x$^2$=8-4$\sqrt 3$ \begin{align}
S_{\triangle MNC}=\frac{\sqrt{3}}{4}a^2=2\sqrt{3}-3 \end{align}


问:
Question on domain and ranges.
Suppose that $f(x) = 13x-4$ and $g(x) = \frac{1}{6x-18}$. Then clearly \begin{align*}
\text{Dom}f &=(-\infty, \infty)\\ \text{Ran}f &= (-\infty
, \infty )\\ \text{Dom}g &= (-\infty, 3)\cup (3, \infty)\\ \text{Ran}g &= (-\infty, 0)\cup (0, \infty )
\end{align*}
Now considering $(f\circ g)(x)$ we have $$(f\circ g)(x) = 13\left( \frac{1}{6x-18} \right) -4 = \frac{13}{6x-18}-4\frac{6x-18}{6x-18} = \frac{13-24x+72}{6x-18}=\frac{85-24x}{6x-18}$$ so that \begin{align*}
\text{Dom}\left( f\circ g \right) &= (-\infty, 3)\cup (3, \infty)\\ \text{Ran}\left( f\circ g \right) &=(-\infty, -4)\cup (-4,0)\cup (0, \infty)
\end{align*} My question is regarding the range. I am aware that if the composition of the functions punctures a new hole in the domain then I have to exclude it from the domain of the composition. But my question is: does the same follow for the range? That is, should the range of the composition above really be $(-\infty, -4) \cup (-4, \infty )$ or is it correct as is? If so why?

答: The range of $f\circ g$ is $\Bbb R\setminus\{-4\}$. The number $0$ belongs to the range, since$$f\left(g\left(\frac{85}{24}\right)\right)=0.$$

Title: Question on domain and ranges.


问:
weak convergence to a constant means convergence in probability
Suppose $Z_k$ converges weakly to a constant $c\in \mathbb R$. Show that $Z_k$ converges in probability to $c$ by only using one of the statements of the Portmanteau theorem. I think part $(ii)$ of the statement is given? And we need to somehow apply part $(iii)$ of the statement since $\{c\}$ is a closed set? Thanks for the help. I am not understanding what are $Z_k$ to be used here?

答:
Let $\mu_k$ be the measure induced by $Z_k$, i.e. $\mu_k(E) := P(Z_k \in E)$ for any measurable set $E \subseteq \mathbb{R}$. Similarly let $\mu$ be the probability measure that puts all mass on $c$. We are given that $\mu_k \Rightarrow \mu$.

Fix $\epsilon > 0$. Consider the closed set $A:=\{x \in \mathbb{R} : |x-c| \ge \epsilon/2\}$. Part (iv) implies
\begin{align}
\limsup_{k \to \infty} P(|Z_k - c| > \epsilon)
&\le \limsup_{k \to \infty} P(|Z_k - c| \ge \epsilon/2)
\\
&= \limsup_{k \to \infty} \mu_k(A)
\\
&\le \mu(A)
\\
&= 0.
\end{align}


问:
Are we using Abel's theorem here?
We can show by using the geometric series that $$\arctan x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}, \forall x \in (-1, 1).$$
Now, we know that the series in the RHS converges when $x=\pm 1$ (by the Leibniz test for instance), so we may write that $$\arctan x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}, \forall x \in [-1, 1].$$
My qustion is whether this last equality is due to Abel's theorem. I think that it is, but I just want to make sure because the statement of Abel's theorem puzzles me sometimes.

答:
In general, suppose $f(x) = \sum_{n=0}^\infty a_n x^n$ converges for all $x \in (-1,1)$. If by some other means you show that  the series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty a_n (-1)^n$ converge, then by defining $f$ appropriately you can always write that $f(x) = \sum_{n=0}^\infty a_n x^n$ for all $x \in [-1,1]$.

What Abel's theorem really provides is that $f$ defined in this way is continuous at the endpoints, that is

$$\lim_{x \to 1-} f(x) = \sum_{n=0}^\infty a_n, \quad \lim_{x \to -1+} f(x) = \sum_{n=0}^\infty a_n(-1)^n$$

In this case, $f(x) = \arctan x$.

Title: Are we using Abel's theorem here?


问:
Newton Binomial to find $(x_1+x_2+x_3)(x_4+x_5+x_6+x_7)=77$
What is the number of solutions to $(x_1+x_2+x_3)(x_4+x_5+x_6+x_7)=77$, where $x_1,x_2,\dots,x_6,x_7$ are non negative integers?
Obviously the $2$ multipliers are $1 \times 77$ and $7 \times 11$, so we have $x_1+x_2+x_3 = 1,7,11,77$ and $x_4+x_5+x_6+x_7 = 1,7,11,77$.

答:
Recall the method of "stars and bars," which tells us that the number of nonnegative integer solutions to the equation $$ x_1 + x_2 + \cdots + x_k = n $$ is $$ \binom{n+k-1}{n}. $$

There are four cases, as you pointed out. The tuple $(x_1 + x_2 + x_3, x_4 + x_5 + x_6 + x_7)$ can only equal $(1, 77)$, $(7, 11)$, $(11, 7)$, or $(77, 1)$. In the first case, where $x_1 + x_2 + x_3 = 1$ and $x_4+x_5+x_6+x_7=77$, the number of solutions is $$ \binom{3}{1}\binom{80}{77}. $$

Applying this reasoning for the other three cases, we get that the total number of solutions is $$ \binom{3}{1}\binom{80}{77} + \binom{9}{7}\binom{14}{11} + \binom{13}{11}\binom{10}{7} + \binom{79}{77}\binom{4}{1}. $$


问:
A question regarding least norm element of a closed, convex subset of a Hilbert space

Suppose $(V, (\cdot, \cdot))$ is a Hilbert space, $U$ is a nonempty closed convex subset of $V$, and $g \in U$ is the unique element of $U$ with smallest norm. Prove that $\Re(g, h)\ge \|g\|^2 \ \forall h\in U$.

I was thinking about considering $f=\frac{1}{2}(g+h)\in U$. So $\|f\|^2 \ge \|g\|^2$. Unable to get the desired inequality. Any help?

答:
As you noticed, for $t \in [0,1]$, using the convexity of $U$,
$$
\|tg + (1 - t)h\|^2 \geq \|g\|^2
$$
Which, when expanding the inner product and dividing by $(1 - t)$ gives
$$
2t\Re(g, h) \geq (1 +t)\|g\|^2 - (1 - t)\|h\|^2
$$
Letting $t = 1$ gives the result.

Edit

As pointed out in the comments, 'letting $t = 1$' is imprecise and should be replaced with 'let $t \uparrow 1$'.


问:
Power series radius of convergence question
Find the radius of convergence of the following series: $\sum_{n=1}^\infty \frac {(x-1)^{2n}}{n9^n} $

What I did was:

I used the ratio test and got $\lim \limits_{n \to \infty }|\frac {(x-1)^{2n+2}}{(n+1)9^{n+1}}\cdot \frac {n9^{n}}{(x-1)^{2n}} |$ I can remove the absolute value since all $n$ is positive and $(x-1)^2$ is positive, $(x-1)^2$ $\lim \limits_{n \to \infty } \frac {n}{9n+9}$ and lastly I got that the $limit$ is $\frac {1}{9}$ so the radius is $9$

then I checked what happens in the series when I put $-9$ and $9$
And I got that it diverges on both of them so my answer is $(-9,9)$ but according to online calculators it is not right.. what did I miss or do wrong?
thanks for any help and tips!

答: Actually, the limit is $\frac{(x-1)^2}9$. So, the series converges absolutely if $\frac{(x-1)^2}9<1(\iff|x-1|<3)$ and diverges if $\frac{(x-1)^2}9>1(\iff|x-1|>3)$. So, the radius of convergence is $3$.


问:
Nets of Geodesic spheres
I would realize the papercraft of a geodesic sphere like this:

It is the dual of the one discussed in THIS OTHER QUESTION .

Where can I find the printable nets, or the online resources to create them?

In the other discussion I learned that there are 3 classes of possible tessellations.
One of these classes needs less pentagons to tessellate the sphere?

I don't exactly know how much faces I need, indicatively between 100 and 200.
Which chord factors I should consider?

Thanks to all

答:
Where can I find the printable nets, or the online resources to create them?

Sorry, don't know that. But once you know how to compute the relevant 3D coordinates, it should be possible to construct them yourself.

In the other discussion I learned that there are 3 classes of possible tessellations. One of these classes needs less pentagons to tessellate the sphere?

No: all of them are based on the symmetry of the icosahedron, so they all require 12 pentagons, matching the 12 vertices of the icosahedron.

I don't exactly know how much faces I need, indicatively between 100 and 200. Which chord factors I should consider?

For 100 to 200 faces in total, you'd want as many vertices in the dual. This means 196 to 396 triangles in that dual, computed using $\frac{6a-12}3$. The idea is that you count vertex-edge combinations, which are the same in both primal and dual. You can count these by multiplying the number of faces by the number of corners they each have, taking care of the 12 pentagons, and then dividing by the number of corners in a triangle. The easiest nets are Class I, I think, even though the Image you included in your question is Class III. For Class I, the number of triangles per icosahedron face has to be a square number. The only square number in the desired range is $4^2=16$ triangles per icosahedron face, leading to 320 triangles in total, which translates into 162 faces for your surface: 150 hexagons and 12 pentagons.

So here is what I'd do:

Start with the 3D corners of the icosahedron. Place these corners on the unit sphere.

Divide each icosahedron edge into 4 segments of equal length.

Compute the inner points by joining matching points along the edges, and intersecting such lines inside the icosahedral face.

Project all these inner points out onto the sphere.

Compute the barycenter of each small triangle.

Project these again out onto the sphere.

Join the resulting points to form your hexagons and pentagons.

Compute edge lengths from 3D coordinates.

Place them one next to the other in the plane to get a printable grid.

Notice that 3. and 4. can be done together using homogenous coordinates. If you want to compute the intersection between some line $AB$ and another line $CD$, and you have 3D coordinates for both of them, simply compute $(A\times B)\times(C\times D)$ and normalize the result to length $1$. The first cross products $A\times B$ will give you a vector orthogonal to both $A$ and $B$. It is the normal vector of the plane spanned by these two points and the origin. The connection has to lie within that plane. The outer cross product gives you a vector orthogonal to two such normal vectors, i.e. a vector along the line where two planes intersect. By normalizing the result, you choose the point on that line which also lies on the sphere. Some care might be neccessary to get the sign right; you don't want the point on the opposite side of the sphere.

Steps 5. and 6. can be combined into summing three vectors and normalizing the result. 5., 7. and 9. are the steps where you have to get your combinatorics right, which will probably take the most time, even though there is nothing deep going on there.

Title: Nets of Geodesic spheres


问:
Holomorphic automorphisms on $\mathbb C$
I'm reading Gong Sheng's Concise Complex Analysis, where it introduced the concept of automorphism first in Ch2.5:

Let $U$ be a region contained in $\mathbb C$. The group of holomorphic
automorphisms on $U$ is defined as follows:

Suppose $f(z)$ is a holomorphic function defined on $U$. If $f(z)$
maps $U$  one-to-one and holomorphically to itself, then $f(z)$ is
called a holomorphic  automorphism on $U$. All of the holomorphic
automorphisms on $U$ form a  group, called the group of holomorphic
automorphisms of $U$ and is denoted  by $Aut( U)$.

So apparently here automorphism is only a bijection, not preserving any operation on $\mathbb C$, i.e. if $a(z) \in Aut(U)$ and $z_1, z_2 \in U$, it's not required to have $a(z_1+z_2) = a(z_1) + a(z_2)$ etc.

Then, in Ch3.4, to find the group of holomorphic automorphisms on $\mathbb C$, denoted as $Aut(\mathbb C)$, it says:

Let $a(z) \in Aut(\mathbb C)$. Then $a(z)$ maps $\infty$ to $\infty$. Since the map is an automorphism, it is one-to-one.
Thus, $z = \infty$ is a simple pole of $a(z)$.  According to the preceding result, $a(z)$ must be
a polynomial of degree  one, $a(z) = az + b$,
where $a,b \in \mathbb C$ and $a\ne 0$. Conversely, it is easy to  see
that $az + b \in Aut(\mathbb C)$ if $a, b \in \mathbb C$ and
$a \ne 0$. So $Aut(\mathbb C)$ consists of all  linear transformations $az+b$
with $a, b \in \mathbb C$ and $a\ne 0$. In other words, every

element of $Aut(\mathbb C)$ is the composition of a translation
$\alpha(z) = z + b$ and a  dilation $\alpha ( z) = az$.

I'm confused in the bold line

$z = \infty$ is a simple pole of $a(z)$

This actually means

$$a(z) = \sum_{n=0}^\infty \frac{c_{-n}}{z^n} + c_1 z$$

But why? could $a(z)$ near $\infty$ has power higher than 1, or even $\infty$?

答:
I'm not sure how much machinery you've already seen, but here is an approach (eg this is how it's treated in Henri Cartan's text). A-priori, since $a$ is holomorphic, there are only two possibilities for the behavior of the function $a$ near $\infty$:

Either $\infty$ is an isolated essential singularity of $a$.

Or $\infty$ is not an isolated essential singularity of $a$ (and hence $a$ is a polynomial).

We can now use Weierstrass's theorem which says that the image under a holomorphic map of any neighborhood of an isolated essential singularity is dense in the complex plane. We use this theorem as follows: let $D$ be the open unit disc. Then, the image $a(D)$ is a non-empty open set in $\Bbb{C}$. The set $U=\Bbb{C}\setminus\overline{D}=\{z\in\Bbb{C}:\,\, |z|>1\}$ is an open neighborhood of the point at infinity, so $a(U)$ is dense in the complex plane by the theorem. However, we have $a(D)\cap a(U)=\emptyset$, contradicting the fact $a(U)$ is dense in $\Bbb{C}$. Therefore, option (1) is not possible.


问:
linear program formulation discontinuous function
I have the following discontinuous function:

$f(x) = 10 $ if $x = 0$ or $f(x) = 5 + 8x $ if $x > 0$ where $ x \in [0,30]$

I'd like to formulate as a bilinear program using binary variables but I am unclear as to how to capture the following relationship between binary variable $y = [0 \ or \  1]$ and x for the following constraints/equations:

$x = 0 \implies y = 0$

$x > 0 \implies y = 1$

If I can represent the above then f(x) becomes a single function of x and y such that:

$f(x,y) = 10(1-y) + (5+8x)y$

答:
Specify a small constant tolerance $\epsilon>0$ and impose linear "big-M" constraints:
\begin{align}
\epsilon y \le x &\le 30 y \\
(5-10) y \le f - 10 &\le (245-10) y \\
f - (5+8x) &= (10-5) (1-y) \\
\end{align}

If $y=0$, then $x = 0$ and $f=10$.

If $y=1$, then $\epsilon \le x \le 30$, $5 \le f \le 245$, and $f = 5+8x$.


问:
Linear operators $T: X \rightarrow Y$ for which $\dim (\operatorname{Im}(T)) \succ \dim(X)$.

Theorem: Let $X, Y$ be $\mathbb{k}$-vector spaces, and
$T: X \rightarrow Y$ be a linear operator. Then, if $\dim(X)$ is finite,
$\dim(\operatorname{Im}(T)) \preceq \dim(X)$.

I'm looking for examples where $\dim(X)= \infty$, and $\dim (\operatorname{Im}(T)) \succ \dim(X)$.

Edit: Is this result true for infinite-dimensional vector spaces?

答:
The result is always true, even in the infinite-dimensional setting.

Indeed, if $\{e_i: i \in I\}$ is a basis of $X$, then $\operatorname{Im}(T) = \operatorname{span}\{Te_i: i \in I\}$ which shows that $\dim \operatorname{Im(T)}\le |I| = \dim(X)$.

Alternatively, by the first isomorphism theorem
$$\operatorname{Im}(T) \cong X/\ker T$$
and thus $$\dim \operatorname{Im}(T) = \dim X/\ker T \leq \dim X.$$
This inequality can be strict: take $X$ your favorite vector space of dimension $n \geq 2$ (infinite dimension is also allowed) and consider any linear map
$$T: X \to \mathbb{k}.$$


问:
Multiplication/addition of Intervals
Let

$M=[0,2) \cup \{3\}$, $N=[4,9]$

Determine 3M, M+M, 2N, M+N

So I have no clue how to determine these, would $2N$ be $[8,18]$? Could someone help me? I can not find similar examples.

Title: Multiplication/addition of Intervals

Body:
Yeah, it's as easy as you think it is.

$2N = \{2n| n \in N\} = \{k=2n| n\in [4,9]\}=$

$\{k=2n| 4 \le n \le 9\} = \{k= 2n| 8 \le 2n \le 18\}$

$= \{k| 8 \le k\le 18\} = [ 8, 18]$.

...

$M+N$ con be tricky though.

$M + N = \{m + n|m \in M; n \in N\} = $

$\{m+n| 0\le m \le 2$ or $m=3$ and $4\le n\le 9\}=$

$\{k| k = m+n; 0\le m \le 2; 4\le n \le 9\} \cup \{k|k = m+n; m = 3; 4\le n\le 9\}=$

$\{k| k=m+n; 4 \le m+n \le 11\}\cup \{k|k = n+3; 4\le n \le 9\}=$

$\{k| 4\le k \le 11\} \cup \{k| 7 \le k \le 12\} =$

$[4,11]\cup [7\cup 12] = [4, 12]$.


问:
Determine invariant subspaces
imagine that a matrix of an endomorphism has the characteristic polynomial
$(\lambda-2)^2(\lambda-3)$
now i was wondering whether all invariant subspaces can be determined by $0,V$ and $\ker(A-2)^2, \ker(A-2), \ker(A-3)$? or how do I find them? </p>

答:
No, you miss a few invariant subspaces. One in the non-diagonalizable case. Infinitely many (assuming the field is infinite) in the diagonalizable case.

So $A$ is a $3\times 3$ matrix over a field $K$. By Cayley-Hamilton, $(A-2I)^2(A-3I)=0$ and $K^3=\ker (A-2I)^2\oplus \ker (A-3I)$. Note that $\{0\}\subsetneq \ker (A-2I)\subseteq \ker (A-2I)^2$, with $\dim \ker(A-2I)^2=2$ and $\dim\ker(A-3I)=1$.

Let $T$ be the linear operator whose matrix is $A$ in the canonical basis. Of course it has the same chracteristic polynomial and eigenvalues/spaces as $A$.

The invariant subspaces of dimension $0$ and $3$ are $\{0\}$ and $K^3$. It remains to determine those of dimension $1$ and $2$.
Let $F$ be such an invariant subspace. In a basis starting with a basis of $F$, the matrix of $T$ becomes block upper triangular. In particular, the upper left block is the matrix of $T_{|F}$ whose characteristic polynomial divides that of $T$.

Case 1: $A$ is diagonalizable, i.e. $\dim \ker (A-2I)=2$ and $\ker (A-2I)=\ker (A-2I)^2$.

If $\dim F=1$, then $F$ is the span of an eigenvector of $T$. So it is either $\ker (A-3I)$, or any one-dimensional susbpace of $\ker (A-2I)$.

If $\dim F=2$, then the characteristic polynomial of $T_{|F}$ is either $(\lambda-2)^2$ or $(\lambda-2)(\lambda-3)$. In the first case, $F=\ker (A-2I)^2=\ker (A-2I)$. In the second case, $F=Kv\oplus \ker(A-3I)$ for any eigenvector $v\in\ker (A-2I)$.

Case 2: $A$ is not diagonalizable, i.e. $\dim\ker(A-2I)=1$.

If $\dim F=1$, now $F=\ker(A-3I)$ or $\ker(A-2I)$.

If $\dim F=2$ and the characteristic polynomial of $T_{|F}$ is $(\lambda-2)(\lambda-3)$, then $F=\ker(A-2I)\oplus\ker(A-3I)$. And if it is $(\lambda-2)^2$, then $F=\dim\ker (A-2I)^2$.

Remark: an easy case, in any dimension, is when $A$ is diagonalizable with pairwise distinct eigenvalues. Then the invariant subspaces are the $2^n$ direct sums built with the eigenspaces.

Title: Determine invariant subspaces


问:
Integral being zero implies measure of set is zero.

Let $E$ be a Lebesgue measurable set. How does
$$\int_{E}e^{-x^2/2}dx=0.\tag{1}$$
imply that the $m(E)=0$, where $m$ is the Lebesgue measure over $\mathbb{R}$.

This is a smaller part of a larger question, but I can't seem to see how it holds. Any help is appreciated.

I know that $(1)$ implies $f(x):=e^{-x^2/2}=0$ Lebesgue almost everywhere on $E$, since $f$ is a non-negative measurable function and $(1)$ holds.

答:
It is sufficient to show that $m(E \cap [a, b]) = 0$ for every $a$ and $b$ with $a < b$. Let $t = \inf_{x \in [a, b]}{e^{-x^2/2}}$. Then $t > 0$ and we have
$$0 = \int_{E}e^{-x^2/2}dx \ge \int_{E\cap [a, b]}e^{-x^2/2}dx \ge \int_{E\cap [a, b]}tdx = m(E \cap [a, b])t$$ and as $t > 0$, this implies $m(E \cap [a, b]) = 0$.


答:
You can also do it with Cauchy-Hadamard Theorem. First, observe that we have the power series

$$\sum_{n=0}^\infty\frac{(x-1)^{2n}}{n\,9^n}=\sum_{n=0}^\infty a_n(x-1)^n$$

with

$$a_n=\begin{cases}0\;\;\;\;\;\;\;\;\;,&n\;\text{is odd}\\{}\\
\cfrac1{\frac n2\,9^{n/2}}\;,&n\;\text{ is even}\end{cases}$$

Thus, by the CHT, we get that

$$\frac1R=\limsup_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}\frac1{\sqrt[n]\frac n2\,9^{1/2}}=\frac1{\sqrt9}=\frac13\implies R=3$$

Title: Power series radius of convergence question

Body:
Find the radius of convergence of the following series: $\sum_{n=1}^\infty \frac {(x-1)^{2n}}{n9^n} $

What I did was:

I used the ratio test and got $\lim \limits_{n \to \infty }|\frac {(x-1)^{2n+2}}{(n+1)9^{n+1}}\cdot \frac {n9^{n}}{(x-1)^{2n}} |$ I can remove the absolute value since all $n$ is positive and $(x-1)^2$ is positive, $(x-1)^2$ $\lim \limits_{n \to \infty } \frac {n}{9n+9}$ and lastly I got that the $limit$ is $\frac {1}{9}$ so the radius is $9$

then I checked what happens in the series when I put $-9$ and $9$
And I got that it diverges on both of them so my answer is $(-9,9)$ but according to online calculators it is not right.. what did I miss or do wrong?
thanks for any help and tips!


问:
Euclidean algorithm and Multiplicative inverse in RSA Cryptosystem
I understand RSA Cryptosystem, the Euclidean algorithm, and mod, however, I can't seem to understand how to solve the following problem.

-Use Euclidean algorithm to compute the multiplicative inverse $s = r^{-1}\bmod (p − 1)(q − 1)$, where $r = 89$ and $p = 173$, $q = 103$.

So, I've found the equation for the multiplicative inverse ($89x + 17544y = 1$) and some GCD:

$\gcd(17544, 89)= 11 (197×89 +11) $

$11=17544-197×89$

$\gcd(89, 11)= 1 (8×11+1)$

$1= 89 -8×11$

$1=89-8×11 \implies 1= ?$
That is the part I'm confused about. I'm not sure with what I should substitute it with.

答:
Use the extended Euclidean algorithm to obtain a direct answer, using the recursive relation $\:u_{n+1}=u_{n-1}-q_nr_n$:

\begin{array}{rrrr}
r_n &u_n&v_n &q_n \\\hline
17544&0&1 \\
89 &1&0&197 \\\hline
11&-197 &1 & 8 \\
1 & \color{red}{1577} &-8  \\
\hline
\end{array}


问:
Moment of sum of independent random variables
I'm stuck at the following problem. Let $(\Omega,\mathcal{F}, \mathbb{P})$ be a probability space with $X$ and $Y$ independent random variables defined over it. Then, if $\mathbb{E}[\mid X+Y \mid ^p]<+\infty$ for some $p>0$, we have that $\mathbb{E}[\mid X\mid^p]<\infty$ and $\mathbb{E}[\mid Y\mid^p]<\infty$.

I really have no intuition about this result. It looks like those results derived from convexity type arguments (for the case $p>1$), which I never really understood all that well. I also tried to break up the first integral into integrals over sets I could manage but also got nowhere. I see that independence is the key here since this is obviously not true in general without this requirement: if $X$ is any non integrable random variable, taking $Y = -X$ would suffice (but of course, they are not independent in this case..).

答: This is an exercise on Fubini's Theorem. $\int\int |x+y|^{p} dF_X(x)dF_Y(y) <\infty$ and this implies that $\int |x+y|^{p} dF_X(x)<\infty$ for almost all $y$. We can fix a $y$ for which this holds. Now use the inequality $|x|^{p} \leq 2^{p} ({|x+y|^{p}+|y|^{p}})$ to conclude that $\int |x|^{p} dF_X(x) <\infty$, so $E|X|^{p} <\infty$. Similarly,  $E|Y|^{p} <\infty$.


问:
Alternative definition of product measure.
While reading some notes the concept of product measures showed up and it was defined by $\int \phi(x,y)d\mu\otimes\nu(x,y)=\int\phi(x,y)d\mu(x)d\nu(y)$ (I think there should have some conditions on $\phi$ here). I was surprised because I thought that to define a measure one need to define how it acts on measurable set (I thought that the product measure was the only one such that $\mu\otimes\nu(A\times B)= \mu(A)\nu(B)$ for all $A$ $\mu$-measurable and $B$ $\nu$-measurable). But is this the same?

答:
The first equation holds for all non-negative measuarble functions $\phi$ iff the second equation holds for all measurable sets $A$ and $B$.

If the first one holds just take $\phi (x,y)=1$ for $x \in A, y \in B$ and $0$ for all other $(x,y)$ to get the second equation.

Suppose the second equation holds. Consider the collection of all measuarble sets $E$ in the product for which the first equation holds with $\phi (x,y)=1$ for $(x,y) \in E$ and $0$ for all other $(x,y)$. This is a sigma algebra which contains all measurable rectangles $A \times B$. Hence, it contains all measurable sets $E$ in the product. Now use a straightforward simple function approximation to finish the proof.


问:
When can we quit a game of War?
Consider the game of War. (The rules are below.) It would be nice to be able to end the game early. Suppose, for example, one player has 50 of the 52 cards. It is very likely that he's going to win. Without knowing what's in his hand,[1] can we say how likely it is that he's going to win? More generally, what is the likelihood of a win for a player once he has $n$ of the 52 cards?

I'm looking for a formula. However, I ran a simulation, and will share the results in case it helps someone find a formula (and for general interest). I used war.pl (from version 1.45 of the Games::Cards Perl module) to play 1,000,000 games of war.[2] For each ($i$th) game, I noted the greatest number $m_i$ of cards held by the loser at any time during the game. Then, for each integer $n\in[27,52]$, I found the number $w$ of games for which $m_i<n$.[3] Those results are:

```<code>cards n   games w
     27    117488
     28    202416
     29    280208
     30    332080
     31    396849
     32    439204
     33    502378
     34    536453
     35    594499
     36    621622
     37    675878
     38    697617
     39    747468
     40    764932
     41    811182
     42    825481
     43    867215
     44    878880
     45    917070
     46    925920
     47    959246
     48    965522
     49    987942
     50    989722
     51    999968
     52   1000000
</code>```

Rules of War:

You start with a standard 52-card deck, shuffled, and two players.

Distribute 26 cards to each player.

Start of a round: Each player displays his top card; the higher rank wins the round, and proceed to the "Disposition of a round" step. If there's a tie, proceed to the next step:

Each player displays four more cards off the top. Among the respective last cards displayed by each player, the higher rank wins the round, and proceed to the "Disposition of a round" step. If there's a tie, repeat this step.

If a player has run out of cards, his last card is always the one to compare to the other player's, no matter how many times the other player displays four more cards.

If both players have run out of cards and there's a tie, the game is over: it's a draw.[4]

Disposition of a round: The winner of the round takes all the cards that have played this round and puts them at the bottom of his set of cards. Note the ambiguity in this last step — I haven't said what order the player should take them in. Indeed, in my experience playing War, there is no rule, and each player can do what he wants, including taking them in differing orders during different rounds. For the purposes of this question, I'd be happy with an answer that answers the question according to any rule of your choosing.

I don't know what rules war.pl uses (I didn't trace through its dependencies), and you should feel free to use those rules instead of my own.

[1] This is unlikely, but let's assume it anyway.

[2] I edited the script to not give up after any finite number of turns, but every game ended anyway. I also edited it so it would give me the information noted in the text above. (See the next footnote.) But I didn't change the rules it used for the game.

[3] In case anyone is wondering (or finds a bug), here's the script I used. To war.pl I added the line map {$size[$_] = $Hands[$_]->size if $size[$_] < $Hands[$_]->size} keys @Hands to the start of each turn and print $size[$other[$winner]] to sub win. Then in a separate script I collected the $m_i$s by for (1..1_000_000) {my $score = `perl war.pl`; $scores{$score}++;} and displayed the $w$s by $\=$/;for my $score (sort{$a<=>$b} keys %scores) {print (($score+1) . ' ' . ((sum0 map { ($_<=$score) * $scores{$_} } keys %scores) / sum0 values %scores))}.

[4] I made that rule up for the purposes of this question, so the game is well defined. I don't think it has ever actually come up in my years of playing War, and I don't know what I would do if it did. If you prefer to use a different rule in answering this question, by all means do so.

答:
According to the rules I used, when there is a tie, each player displays only 1 cards. If one of the two cards is higher than the other one, then the player that showed it wins all four cards (the two cards of the tie and the new two cards); if there is another tie, the players keep showing one card at a time, until there is no tie or one of the two players ends his deck.

It seems to me that with these rules I proposed, what really matters is the number of Aces (the highest card) in each deck. When a player has all 4 Aces in his deck, we can say that he will eventually win.

A rule of thumb about when to end the game (which me and my friends actually used when we played this version of "War") is that when we become aware that one of the two players has all 4 Aces, then the game can be ended.

A lower bound of the winning probability having $n$ cards can thus be obtained computing the probability of having all 4 Aces in $n$ cards.

To have a better estimate, one could consider a Markov chain where the state is given by the couple $(n,k)$, where k is the number of Aces in a player's deck. For instance, one could manage to estimate what is the probability of eventually catching the fourth Ace, when having already three of them (and the complementary probability of losing one).

I hope these remarks are helpful.

Title: When can we quit a game of War?


问:
Non-linear bijections preserving the relation $fg = f^2$
Let $X$ and $Y$ be compact spaces and let $C(X)$ and $C(Y)$ be the corresponding spaces of continuous, real-valued functions. Suppose $T\colon C(X)\to C(Y)$ is a non-linear bijection such that

$$fg = f^2 \iff (Tf)(Tg) = (Tf)^2\qquad (f,g\in C(X))$$

Once can deduce (see Lemma 3.1 here) that

(#) if $fg = 0$, then $(Tf)(Tg) = 0$ and in this case $T(f+g) = Tf + Tg$.

Addendum: One can prove that the condition concerning preservation of the relation $fg = f^2$ in both ways is equivalent to $T$ and $T^{-1}$ satisfying (#), which may make the picture cleaner.

Suppose that $(f-h)(g-h) = 0$ constantly. Do we have $(Tf-Th)(Tg-Th) = 0$?

The motivation comes from the fact that I'd like to simplify/amend considerations in the linked preprint. Maps $T$ with this property need not be continuous, but I am happy to assume continuity, or something else, if that helps.

答:
The condition you wrote is not valid.

Let $\preceq$ be the compatibility ordering in your paper.

Let $X=Y=[-2,2]$. Let $u(x)=\max(0,1-|x|)$ and $v=-u$. (The graph of $u$ is looks like a triangle above $0$.)

Define $T\colon C(X)\to C(Y)$ as the $\preceq$-isomorphism which "interchanges $u$ and $-u$".

Formally,

If $u\preceq f$ or $-u\preceq f$, $T(f)(x)=\begin{cases}f(x)&\text{ if }|x|>1\\-f(x)&\text{ if }|x|\leq 1\end{cases}$.

Otherwise, $T(f)=f$.

Let $f(x)=\max(1-x,0)$, $g(x)=\max(1+x,0)$. Then $u$ is always equal to either $f$ or $g$, which means that $(f-u)(g-u)=0$. However, $(T(f)-T(u))(T(g)-T(u))=(f+u)(g+u)$ is non zero (at $x=0$ its value is $4$).

However, $T$ is not continuous wrt to any of the usual topologies on $C(X)$.


答:
It's important to be neat and methodical when performing the extended Euclidean algorithm, so you don't get tangled up. ;)

Your goal is to find an $x,y$ pair that satisfy
$$89x + 17544y = 1$$

You have calculated all the bits you need. You just need to replace the $11$ in
$$1=89-8×11$$
with
$$11=17544-197×89$$
Thus,
$$1=89-8×(17544-197×89)$$
which can easily be simplified.

Title: Euclidean algorithm and Multiplicative inverse in RSA Cryptosystem

Body:
I understand RSA Cryptosystem, the Euclidean algorithm, and mod, however, I can't seem to understand how to solve the following problem.

-Use Euclidean algorithm to compute the multiplicative inverse $s = r^{-1}\bmod (p − 1)(q − 1)$, where $r = 89$ and $p = 173$, $q = 103$.

So, I've found the equation for the multiplicative inverse ($89x + 17544y = 1$) and some GCD:

$\gcd(17544, 89)= 11 (197×89 +11) $

$11=17544-197×89$

$\gcd(89, 11)= 1 (8×11+1)$

$1= 89 -8×11$

$1=89-8×11 \implies 1= ?$
That is the part I'm confused about. I'm not sure with what I should substitute it with.


问:
Vector bundles with inner products induced from the metric
If $E \to M$ is a tensor bundle over a Riemannian manifold (M, g) then the metric $g$ induces an inner product on the fibers of $E$ in the same way that one can give tensor products of inner product spaces an inner product.

We can also consider direct sums of tensor bundles which will also have an inner product induced from the metric in the same way that the direct sum of some inner product spaces is an inner product space with the inner product taken to be the sum of the inner products on the projections.

In this way one can build a collection of vector bundles over $M$ which have some sort of natural inner product structure induced from the metric $g$. Does this collection of vector bundles have a name and is this talked about in any texts?

答: I do not know if this is what you are looking for, but this recall me the notion of "bundle metric". This is treated in Jost's Riemannian Geometry and Geometric Analysis, chapter 2. Another book (but very difficult to find unless you live in Brazil) is "Tópicos de Geometria Diferencial" written by Antonio Caminha and Muniz Neto. In this last reference, they call the bundles with metric something like "Riemannian bundles".</p>


问:
Calculate the next Inverse laplace transform
This question may be very basic, but I dont know how to get the next inverse laplace's transform:

$${\scr L}^{-1}\left\{\frac{1}{(s-1)^2}\right\}$$

I can only use these two formulas:
${\scr L}\{e^{at}\}=\frac{1}{(s-a)}$ and
${\scr L}\{t^n\}=\frac{n!}{s^{n+1}}$

I can also use all the laplace's transform properties but the definition of the laplace transform definition

答:
Using the property

$$\mathcal{L}[tf(t)] = -F'(s), $$

where $F(s)$ is the Laplace transform of $f(t)$ and the fact

$$ \mathcal{L}[e^{t}] = \frac{1}{(s-1)}  $$

the result follows.


问:
How many $8$ letter words can be formed from the letters $PAAAARRTTM$?
How many $8$ letter words can be formed from the letters $PAAAARRTTM$?

This is what I thought:

Having $8$ letter words is basically just counting the ways of arranging the letters, after we choose to remove $2$ of the letters from the $10$ given.

Thus, I will count the ways of arranging the letters, after I remove some two letters.

The letters I will remove come in $13$ cases and are:

$$\{P,A\},\{P,R\},\{P,T\},\{P,M\},\{A,R\},\{A,T\},\{A,M\},\{A,A\},\{R,T\},\{R,M\},\{R,R\},\{T,M\},\{T,T\}.$$

Removing $P$ and some other letter after:

$$= \frac{8!}{3!2!2!} + \frac{8!}{4!2!} + \frac{8!}{4!2!} + \frac{8!}{4!2!2!} = 3780.$$
Similarly for $A$:

$$= \frac{8!}{3!1!2!} + \frac{8!}{3!2!1!} + \frac{8!}{3!2!2!}+ \frac{8!}{2!2!2!} = 13440$$
Similarly for $R$:

$$=\frac{8!}{4!2!} + \frac{8!}{4!2!2!} + \frac{8!}{4!2!} = 2100$$
Similarly for $T$:

$$=\frac{8!}{4!2!2!} + \frac{8!}{4!2!}= 1260$$

So total number of ways is $20580$.

However, the answer is $22260$, with a difference of $1680 = \frac{8!}{4!}$...

Did I go wrong somewhere?

Also, for these type of questions, (given $n$ letters where some repeat and we have to form $k < n$ words from them), is this the systematic way to approach it?

答:
$\newcommand{gtxt}[1]{\bbox[lightgray,4px]{\text{#1}}}$
You can condense this process by considering the possible multiplicity patterns of the 8 selected letters. This avoids the confusion of working through individual combinations.

Here we have the multiset $\{A^4,R^2,T^2,M,P\}$ to choose $8$ items from.

The multiplicity patterns can be written using tokens as $\gtxt{ccccddee}$, $\gtxt{ccccddef}$, $\gtxt{ccccdefg}$, $\gtxt{cccddeef}$, $\gtxt{cccddefg}$, $\gtxt{ccddeefg}$. These are constrained by both the availability of different letters and their multiplicity in the source set.

This pattern also determines how many arrangements there are of each, through the corresponding multinomial, and we also need to calculate how many way there are to fill each pattern. In a pattern with several groups of two, for example, we would pick possible choices to fill those in one step. Then we multiply these two factors to give options for each pattern, and add across all patterns.

$$\begin{array}{|c|c|} \hline
\text{For this pattern:} & \text{options to fill} & \text{arrangements} &  \text{total options} \\[1ex] \hline
 \gtxt{ccccddee} & \binom 11 \binom 22  = 1  & \binom{8}{4,2,2} = 420  &  420  \\[1ex]
 \gtxt{ccccddef} & \binom 11 \binom 21 \binom 32 = 6  & \binom{8}{4,2,1,1} = 840  & 5040 \\[1ex]
 \gtxt{ccccdefg} & \binom 11 \binom 44 = 1  & \binom{8}{4,1,1,1,1} = 1680 & 1680 \\[1ex]
 \gtxt{cccddeef} & \binom 11\binom 22\binom 21 = 2  & \binom{8}{3,2,2,1} = 1680 & 3360 \\[1ex]
 \gtxt{cccddefg} & \binom 11\binom 21 \binom 33= 2  & \binom{8}{3,2,1,1,1} = 3360  & 6720 \\[1ex]
 \gtxt{ccddeefg} & \binom 33\binom 22 = 1  & \binom{8}{2,2,2,1,1} = 5040 & 5040 \\[1ex]\hline
 &   & \text{total}&  22260 \\[1ex]\hline
\hline
\end{array}$$

as given.


问:
Why $X\times\emptyset=\emptyset?$
If is true that $X\times\emptyset=\emptyset$, so take $A\times B$ with $U,V\subset A$ and $R,S\subset B$. We know that $(U\cup V)\times(R\cup S)=(U\times R)\cup (V\times S)$. So, take $X$ with $|X|\ge2$, then $X\times X=(X\setminus\{x\}\cup\{x\})\times(X\cup\emptyset)=(X\setminus\{x\}\times X)\cup(\{x\}\times\emptyset)=X\setminus\{x\}\times X?$. It's a paradox or I miss anything?

答:
$(U\cup V) \times (R\cup S) \ne (U\times R) \cup (V\times S)$.  It just doesn't.

Consider a $u\in U$ but $u \not \in V$ and $s \in S$ but $s \not \in R$.

then $u \in U\cup V$ and $s \in R\cup S$ so $(u,s )\in (U\cup V)\times (R\cup S)$

but $s\not \in R$ so $(u,s) \not \in U\times R$.  And $u \not \in V$ so $(u,s) \not \in V\times S$.  So $(u,s) \not \in (U\times R) \cup (V\times S)$ so $(U\cup V) \times (R\cup S) \ne (U\times R) \cup (V\times S)$.

.....

Any how $X \times \emptyset = \{(x,y)| x \in X; y \in \emptyset\}$.  But there are no $y \in \emptyset$ so there are no $(x,y) \in X\times \emptyset$.

So $X \times \emptyset$ is empty.

Title: Why $X\times\emptyset=\emptyset?$


问:
If F is a commutator subgroup to G then prove that G/F is abelian.
I am trying to understand some basic Algebraic topology for Physics.

The relation between the fundamental group and the homology group is given by an isomorphism $\phi:\pi_1(|K|)/F\to H_1(K)$. Where F is the commutator subgroup of $\pi_1(|K|)$ and K is a connected simplicial complex.

As I understand from the internet the map $\psi:\pi_1(|K|)\to\pi_1(|K|)/F$ is called abelianization of the fundamental group. It makes sense intuitively that if you quotient out all the non-commutating commutators you get a new relation i.e $(x_i;r_m)\to(x_i;r_m,[x_i,x_j])$. But how do you prove that $\pi_1(|K|)/F$ is abelian?

答:
Suppress the quotient map $G\to G^{ab}$ in the notation. You can calculate

\begin{align} gh= gh1 = gh(h^{-1}g^{-1}hg) = hg \end{align}

in $G^{ab}$.


问:
How to parametrize a complex curve
I just started Complex Analysis course in my Uni! We are at the part of Complex Line Integrals! There is something there saying something about the Parametrization of line integrals and i wanted to know how you can do that! I read the description in my slides but i can't figure it out how to do it!
Can someone help me understand how to do them! Ideally by providing some examples! If you can't can you help me solve this?

C segment from 1 + i to 3

f(z)=iz

答:
To parameterize the line segment, do it just like you would do it in $\mathbb{R}^2$: $$z(t) = (1-t)(1+i)+3t,\ 0\leq t\leq 1$$

By definition, $\mathrm{d}z=z'(t)\mathrm{d}t= (-1-i+3)\mathrm{d} t$, so the integral is $$\int_0^1 i((1-t)(1+i)+3t))(2-i)\mathrm{d}t$$

Title: How to parametrize a complex curve


答:
To calculate an integral over a complex line froma definition (and not using theorems like the Cauchy's theorem) you need a parametrization of the curve.

A paramterization of a straight line from $z_1$ to $z_2$ is
$$ z(t) = z_1 + t (z_2-z_1), \qquad t\in[0,1]$$
Another useful curve (not in your specific problem, just in general) is an arc of a circle. It can be parametrized as
$$ z(t) = z_0 + R e^{it}$$
when going counterclockwise or
$$ z(t) = z_0 + R e^{-it}$$
when going clockwise, where $z_0$ is the center of the circle, $R$ is the radius, and you need to choose the range of $t$ to fit the initial and final points of the arc.

One you have the parametrization, the integral is defined as
$$ \int_\gamma f(z) dz : = \int_{t_1}^{t_2} f(z(t)) \frac{dz}{dt} dt = \int_{t_1}^{t_2} {\rm Re}\big( f(z(t)) \frac{dz}{dt} \big)  dt + i \int_{t_1}^{t_2} {\rm Im}\big( f(z(t)) \frac{dz}{dt} \big) dt$$

If a curve is more complex, but it can be decomposed into straight lines and arcs, you write the integral as sum of the integrals over each segment and calculate them separately. If it involves a more difficult curve, you need a parametrization of this curve or use some theorems about integrals over complex curves.

Taking your specific example, you need a straight line from $1+i$ to $3$. It can be parametrized as $$ z(t) = 1+i+ t (2-i), \qquad t\in[0,1]$$
Therefore we have
$$ \int_\gamma f(z)dz = \int_0^1 iz(t) \frac{dz}{dt} dt = \int_0^1 i\big(1+i+ t (2-i) \big)(2-i) dt$$

Title: How to parametrize a complex curve

Body:
I just started Complex Analysis course in my Uni! We are at the part of Complex Line Integrals! There is something there saying something about the Parametrization of line integrals and i wanted to know how you can do that! I read the description in my slides but i can't figure it out how to do it!
Can someone help me understand how to do them! Ideally by providing some examples! If you can't can you help me solve this?

C segment from 1 + i to 3

f(z)=iz


问:
The antipodal map $\mathbb{C}P^1\to\mathbb{C}P^1$ in coordinates
I know that on $S^2$, we have the antipodal map $a:S^2\to S^2, (x,y,z)\mapsto(-x,-y,-z)$. But I know that we can identify $S^2$ with $\mathbb{CP}^1$, using the diffeomorphism $$
\Psi:\mathbb{CP}^1\to S^2,
\Psi([w_0:w_1])=\frac{1}{|w_0|^2+|w_1|^2}\left(2\text{Re}(w_1\overline{w_0}), 2\text{Im}(w_1\overline{w_0}),|w_0|^2-|w_1|^2\right)
$$
But now, I want to see how the antipodal map looks like in coordinates on $\mathbb{CP}^1$. I know it should be something like $[z_0:z_1]\mapsto[\overline{z_1}:\overline{z_0}]$, but I have been trying to compute it all day now and did not really progress. Could anyone help please?

答:
$\mathbb{CP}^1$ is defined as the quotient space
$$(\mathbb C^2 \setminus \{0\})/\sim$$
where $(w_0,w_1) \sim (w'_0,w'_1)$ if there exists $\lambda \in \mathbb C$ such that $\lambda (w_0,w_1)  = (\lambda w_0,\lambda w_1) = (w'_0,w'_1)$. We have
$$\Psi([w_0 :w_1]) = \frac{1}{|w_0|^2+|w_1|^2}\left(2w_1\overline{w_0},|w_0|^2-|w_1|^2\right) \in \mathbb C \times \mathbb R = \mathbb R^3 .$$
You want to see what $\Psi^{-1}(-\Psi([w_0:w_1]))$ looks like. Define
$$A : \mathbb C^2 \setminus \{0\} \to \mathbb C^2 \setminus \{0\}, A(w_0,w_1) = (-\overline{w_1},\overline{w_0}) .$$
Clearly $A$ is a diffeomorphism.
We have
$$A(\lambda w_0,\lambda w_1) = (-\overline\lambda \overline{w_1},\overline\lambda \overline{w_0}) = \overline\lambda (-\overline{w_1},\overline{w_0}) = \overline\lambda A(w_0,w_1)$$
which means that $(w_0,w_1) \sim (w'_0,w'_1)$ implies $A(w_0,w_1) \sim A(w'_0,w'_1)$. Therefore $A$ induces a diffeomorphism $\bar A : \mathbb{CP}^1 \to \mathbb{CP}^1$ given by $\bar A ([w_0:w_1]) = [-\overline{w_1}: \overline{w_0}]$. We get
$$\Psi(\bar A ([w_0:w_1])) = \Psi([-\overline{w_1}: \overline{w_0}]) = \frac{1}{|-\overline{w_1}|^2+|\overline{w_0}|^2}\left(2\overline{w_0}(-w_1),|-\overline{w_1}|^2-|\overline{w_0}|^2\right) \\ = \frac{1}{|w_0|^2+|w_1|^2}\left(-2w_1\overline{w_0},|w_1|^2-|w_0|^2\right) = - \Psi([w_0:w_1]) .$$

This shows that $\Psi^{-1}(-\Psi([w_0:w_1])) = [-\overline{w_1}: \overline{w_0}]$.


问:
How do I reexpress the equation $\nabla \times (\nabla \times gs)) \times (\nabla \times \nabla(f\nabla \cdot t))$?
How do I go about reexpressing

$\nabla \times (\nabla \times bs)) \times (\nabla \times \nabla(c\nabla \cdot t))$

where s and t are vector properties and b and c are scalar.

I don't know where to even start on this one, I'm just asking if anyone can give me some hints/point me in the right direction.
I've looked at a lot of grad/div/curl combination rules but it's quite overwhelming for a beginner

Thank you

答:
You just need to know that for any scalar function $f$:
$$ \nabla \times \nabla f = {\rm curl}\, {\rm grad}\, f = 0$$
That means that
$$ \nabla \times \nabla(c \nabla\cdot t) = 0$$
$$ \big((\nabla \times (\nabla \times bs)\big) \times \big(\nabla \times \nabla(c \nabla\cdot t)\big) = \big((\nabla \times (\nabla \times bs)\big)  \times 0 = 0$$


问:
On the vector spaces of Taylor Series and Fourier Series
Taylor series expansion of function, $f$, is a vector in the vector space with basis: $\{(x-a)^0, (x-a)^1, (x-a)^3, \ldots, (x-a)^n, \ldots\}$. This vector space has a countably infinite dimension. When $f$ is expressed as linear combination of the basis vector the scalar multiple for the $n$-th basis vector is $\operatorname{Diff}_n{f}(a)/n!$

Fourier series expansion of function, $f$, is a vector in the vector space with basis: $\{\sin(1x), \cos(1x), \sin(2x), \cos(2x), \ldots, \sin(nx), \cos(nx), \ldots\}$. This vector space has a countably infinite dimension. When $f$ is expressed as linear combination of the basis vector the scalar multiple for the $n$-th basis vectors are $\operatorname{Int}\{f\cdot\sin(nx)\}$ and $\operatorname{Int}\{f\cdot\cos(nx)\}$.

Questions:

The vector space for the Fourier series has an inner product, $\operatorname{Int}\{f\cdot g\}$, and it's this inner product that provides the above expressions like $\operatorname{Int}\{f\cdot\sin(nx)\}$ and $\operatorname{Int}\{f\cdot\cos(nx)\}$. Is there a similar inner product based derivation of the scalar multiples for the vector space of spanned by the polynomial basis in Taylor series?

What is the relationship, if any, between the vector space produced by Taylor Series and that of Fourier Series? E.g. is one a subspace of the other?

When Fourier series is taught, why isn't Taylor Series re-explained in the vector space framework used for Fourier series? And would this approach not lead the discussion of the implication of the choice of basis (and perhaps the choice of inner product) for function spaces?

Just as Fourier series get generalized to Fourier Transform (the summation of the series becomes an integral), is there something equivalent to Taylor series?

Are there any recommended resources (books, courses, etc.) available which can help clarify my thinking regarding these issues?

答:
I would add a comment, not an answer, but I don't have any reputation whatsoever. I also know it's an old thread, but it may help someone nowadays.

Regarding the 4th point, I would recommend looking for the Mellin tranform. It's widely used in analytic number theory. Just as the Fourier tranform is closely related to the coeficients of a fourier series, the Mellin transform is related to the coefficient of a Taylor series, through Ramanujan's interpolation theorem.


问:
How can a parallel edge be identified?
An edge will have the same vertices as another edge that it is parallel to, so how can it be uniquely described? </p>

答:
For an undirected multigraph, we simply have a multiset of edges, not a set. Thus, if $V=\{a,b,c\}$, we might have
$$E=\{[(a,b),2],[(a,c),3]\},$$
which specifies that there are two edges connecting vertices $a$ and $b$, and three edges connecting vertices $a$ and $c$. We still can't technically distinguish "individual edges" between the same vertices; all we know is how many edges there are between any two vertices. However, for many purposes, that's all that's really necessary.

If you want to point to individual edges, you could take the approach typically used for directed multigraphs: we take a quadruple $(V,E,s,t)$, where $V$ is the set of vertices, $E$ is the set of edges, $s:E\to V$ is the function taking an edge to the vertex it starts at (its "source") and $t:E\to V$ is the function taking an edge to the vertex it ends at (its "target"). Then edges $e_1,e_2\in E$ are parallel with the same direction if $s(e_1)=s(e_2)$ and $t(e_1)=t(e_2)$, and parallel with opposite directions if $s(e_1)=t(e_2)$ and $t(e_1)=s(e_2)$.

Title: How can a parallel edge be identified?


问:
Finding the orthogonal trajectories of the family of cissoids
Find the orthogonal trajectories of the cissoids, $y^2=\frac{x^3}{a-x}$.

My solution:

$\left( a-x\right)y^2=x^3$

$\left( a-x\right)2y dy-(2x^2+y^2)dx=0$

$\left( a-x\right)2ydx + (2x^2+y^2)dy=0$

$a=\frac{x^3}{y^2}+x$

$2y \left( \frac{x^3}{y^2}\right)dx+(2x^2+y^2)dy=0$

$2x^3dx+(2x^2+y^2)dy=0$

I do not know how to go further. What method could be used to solve this differential equation?

答:
Taking a derivative of your equation
\begin{align}
y&=\sqrt{\frac{x^3}{a-x}},\\
\frac{\mathrm dy}{\mathrm dx}&=\frac{3a-2x}{2}\sqrt{\frac{x}{(a-x)^3}},
\end{align}
and substituting $a=x^3/y^2+x$ gives
\begin{align}
\frac{\mathrm dy}{\mathrm dx}&=\frac{3x^3/y^2+x}{2}\sqrt{\frac{x}{(x^3/y^2)^3}}.
\end{align}
The orthogonal curves have slope that's the negative of the inverse, i.e.,
\begin{align}
\frac{\mathrm dy_{\bot}}{\mathrm dx}=\frac{-2}{3x^3/y_{\bot}^2+x}\sqrt{\frac{(x^3/y_{\bot}^2)^3}{x}}.
\end{align}
Substituting $y_{\bot}=xu$ gives that
\begin{align}
x\frac{\mathrm du}{\mathrm dx}+\frac{2+u(3+u^2)}{3+u^2}=0.
\end{align}
This is separable,
\begin{align}
\int\left( \frac{3u+u^3}{2+u(3u+u^3)}\frac{\mathrm du}{\mathrm dx}+\frac{1}{x}\right)\mathrm dx&=\int 0\mathrm dx\\
\ln(u^2+1)-\frac{1}{2}\ln(u^2+2)+\ln x&=k\\
\frac{u^2+1}{\sqrt{u^2+2}}&=\frac{k}{x}.
\end{align}

Solving for $y_{\bot}=u/x$,
\begin{align}
y_{\bot}^2=k\pm\sqrt{k(2x^2+k)}-x^2.
\end{align}

Fun problem!


答:
In general, $X \times Y$ is defined as the set that contains all pairs $(x,y)$ such that $x \in X$ and $y \in Y$. But, if $Y=\emptyset$, there are no such pairs! That is true since there is no $y$ such that $y \in \emptyset$. Therefore, according to the general definition, $X \times \emptyset = \emptyset$.

Another way to see that you want $X \times \emptyset$ to be the empty set is that you want $|X \times Y| = |X| \cdot |Y|$ to hold, where $|X|$ denotes the number of elements in set $X$. Let's focus on finite sets for now, although this is true for infinite sets as well. That is, in order to construct a pair $(x,y) \in X\times Y$, you have $|X|$ choices for $x$, and for every such choice, you have $|Y|$ choices for $y$. For $Y=\emptyset$, that tells you that you'd like $|X\times\emptyset|$ to be $0$, i.e. $X\times\emptyset = \emptyset$, since the empty set is the only set with $0$ elements.

Regarding the rest of your question, $(U\cup V)\times(R\cup S)=(U\times R)\cup (V\times S)$ isn't true. Consider any example with finite sets where $U \cap V = R \cap S = \emptyset$, and then simply count the size of each side. For the LHS we have:
$$|(U \cup V)\times(R \cup S)|= |U \cup V| \cdot |R \cup S| = (|U| + |V|) \cdot (|R| + |S|).$$
For the RHS:
$$|(U \times R) \cup (V \times S)| = |U\times R| + |V \times S| = |U|\cdot |R| + |V|\cdot |S|.$$
So, as long as none of $U,V,R,S$ is empty, the LHS can not be equal to the RHS.

Here is a concrete counterexample:
$$(\{0\}\cup\{1\})\times(\{a\}\cup\{b\})=\{0,1\}\times\{a,b\}=\{(0,a),(0,b),(1,a),(1,b)\}$$
whereas
$$\{0\}\times\{a\}\cup\{1\}\times\{b\}=\{(0,a),(1,b)\}.$$

Title: Why $X\times\emptyset=\emptyset?$

Body: If is true that $X\times\emptyset=\emptyset$, so take $A\times B$ with $U,V\subset A$ and $R,S\subset B$. We know that $(U\cup V)\times(R\cup S)=(U\times R)\cup (V\times S)$. So, take $X$ with $|X|\ge2$, then $X\times X=(X\setminus\{x\}\cup\{x\})\times(X\cup\emptyset)=(X\setminus\{x\}\times X)\cup(\{x\}\times\emptyset)=X\setminus\{x\}\times X?$. It's a paradox or I miss anything?


问:
Analytic semigroups of linear operators
can someone explain the physical meaning of an analytic semigroup of linear bounded operators? Why are they important with respect to the strongly continuous case?</p>

答: Analytic semigroups are often related to parabolic equations. The prototype example is the heat equation (more generally diffusion equations). The analyticity of the semigroup means in particular that the solution is smooth (analytic) for positive times. This fact is physically interpreted by a smoothing effect  of the heat far away from zero even for non-smooth initial conditions (initial temperatures).</p>

Title: Analytic semigroups of linear operators


问:
Existence for the vectors with the same length and isotropic position.
Suppose that $v_1,\ldots,v_m\in\mathbb{R}^n$ satisfying $$\sum_{i=1}^m v_iv_i^T=I_n$$ and $$\|v_i\|^2=\frac{n}{m}$$ for each $i$.

Do the vectors $\{v_i\}_{1\leq i\leq m}$ always exist? If not, what's the condition for the existence?

Note that $[v_1,\ldots,v_m]$ can be viewed as a submatrix extracted $n$ rows from an orthogonal matrix, and each columns have the same length.

答:
We have $n = \mathrm{rank}\left(I_n\right) = \mathrm{rank}\left(\sum_{i=1}^m v_iv_i^T\right)\leq m.$ Therefore, we know that $n\leq m.$

We now show that the vectors $v_1,\ldots,v_m$ can be constructed for all $m$ and all $n\leq m.$ If $n$ divides $m,$ there is a trivial way of doing it. So let us assume that $m$ is not a multiple of $n.$

We use induction to show that the $v_i$ can be constructed. Assume that a construction exists for all $m' < m$ and all $n' \leq m'.$ The base case of the induction is trivial. (Set $m=1$ and show that the statement holds.) From this, we construct $V=[v_1,\ldots,v_m]$ as follows:

Let $a=\left\lfloor\frac{m-n}{n}\right\rfloor$ and $b=\left\lceil\frac{m-n}{n}\right\rceil.$ Then $b-a=1,$ because $n$ does not divide $m.$ Let $c=bn-(m-n)$ and $d=(m-n)-an.$ Then $c+d=n$ and $ac+bd=m-n.$

Let $\lambda = \sqrt{\frac{n}{m}}.$ Let $e_1,\ldots,e_n$ be the canonical basis of $\mathbb{R}^n.$

Now we set $v_i = \lambda\,e_1$ for the first $a$ columns of $V,$ $v_i = \lambda\,e_2$ for the next $a$ columns and so on until we have $c$ groups of $a$ identical vectors. Then we set $v_i = \lambda\,e_{c+1}$ for the following $b$ columns of $V,$ $v_i = \lambda\,e_{c+2}$ for the next $b$ columns and so on until we have $d$ more groups of $b$ identical vectors. In the end, we have filled $V$ except for the last $n$ columns. Example: $m=11,\;n=4$. Then $a=1,\,b=2,\,c=1,\,d=3$ and
$$
V=\begin{pmatrix}
\lambda & & & & & &          & \vdots & \vdots & \vdots & \vdots \\
 & \lambda & \lambda & & & & &   v_8  &   v_9  & v_{10} & v_{11} \\
 & & & \lambda & \lambda & & & \vdots & \vdots & \vdots & \vdots \\
 & & & & & \lambda & \lambda & \vdots & \vdots & \vdots & \vdots
\end{pmatrix}
$$
Due to our induction assumption, we know that we can construct a $c\times n$ matrix $V_1$ with $V_1V_1^T=I_c$ and all columns of $V_1$ have length $\sqrt{\frac{c}{n}}.$ We extend this matrix to an orthonormal $n\times n$ matrix. Let $V_2$ be the $d\times n$ matrix that forms this extension. We know that $V_2V_2^T=I_d$ and all columns of $V_2$ have length $\sqrt{\frac{d}{n}}.$

Now we set $\mu_1 = \sqrt{\frac{n+d}{m}}$ and $\mu_2 = \sqrt{\frac{n-c}{m}}$ and
$$
\left[v_{m-n+1},\ldots,v_m\right]
=\begin{pmatrix} & \mu_1 V_1 & \\
& \mu_2 V_2 &
\end{pmatrix}
$$
It can easily be verified that the resulting matrix $V$ has all required properties.


问:
Prove this Cauchy integral $\oint_C \frac{f'(z)}{z-z_0}dz = \oint_C \frac{f(z)}{(z-z_0)^2}dz$
How can I prove that?

I know if I do $\frac{d}{dz}(f(z)/(z-z_0)) = \frac{f'(z)}{z-z_0} - \frac{f(z)}{(z-z_0)^2}$, but if I put this derivative as zero, I will have these two terms on the right being equal.

Can I then just put that term in the Cauchy integral and say that they are the same?

The question: http://www.math.ubc.ca/~cautis/calcIV02/cpx3.pdf -> page 12, ex 9b

答:
You need to use the fact that C is a closed curve (which is marked by the symbol $\oint$.

For an open curve from $z_1$ to $z_2$ we'd have
$$ \int_C \frac{d}{dz} \big(\frac{f(z)}{z-z_0}\big) dz = \frac{f(z_2)}{z_2-z_0} -\frac{f(z_1)}{z_1-z_0}$$
but for a closed curve we have $z_1=z_2$, so
$$ \oint_C \frac{d}{dz} \big(\frac{f(z)}{z-z_0}\big) dz = 0$$
Calculating the derivative you get the equality you want.

Another way: If $f(z)$ is holomorphic in the whole region bounded by $C$ (which we didn't have to assume in the previous method) we can use the Cauchy formula
$$ \oint_C \frac{g(z)}{z-z_0}dz = g(z_0)$$
with $g(z)=f'(z)$; then we get
$$ \oint_C \frac{f'(z)}{z-z_0}dz = f'(z_0)$$
On the other hand
$$ f'(z_0) = \frac{d}{dz_0} f(z_0) = \frac{d}{dz_0} \oint_C \frac{f(z)}{z-z_0}dz =  \oint_C \frac{\partial}{\partial z_0}\frac{f(z)}{z-z_0} dz=  \oint_C \frac{f(z)}{(z-z_0)^2} dz$$


问:
The number of odd-degree vertices is even in a finite graph?
if you add up the degrees of all the vertices of a finite graph then the result will be twice the number of edges, provided that the vertices are odd-degree. What is the name of this theorem? why does it require that the graph be undirected?</p>

答: It is known as the handshaking lemma because if a bunch of people shake hands and you add up the number of hands each person shook, you get an even number.  Each handshake adds two to the total.  It does not require that each vertex be of odd degree, but it shows there are an even number of vertices of odd degree.  For a directed graph, this is still true if you add all the indegrees and outdegrees for the same reason, but the sum of indegrees need not be even.  Think of a directed graph with two vertices and one directed edge connecting them.</p>


问:
How to find the rotation of $A$, $B$ and $C$ about a fixed point like $(x_1,y_1)$?
The problem is as follows:

Let the points:

$A=(-6,-1)$

$B=(-5,1)$

$C=(-2,-2)$

$D=(-3,2)$

These represent a quadrilateral on the $x-y$ coordinate system. Find the sum of the coordinates from all the vertex of such quadrilateral after the figure has been moved $6$ units to the right and rotated $90^{\circ}$ degrees about point $E=(5,-4)$ in the counterclockwise direction.

The choices given by my workbook are as follows:

$\begin{array}{ll}
1.&-25\\
2.&-22\\
3.&-23\\
4.&-24\\
\end{array}$

I'm not very sure how to solve this problem. But I can tell that when you first move six units to the right the new coordinates become as follows:

$A_2=(0,-1)$

$B_2=(1,1)$

$C_2=(4,-2)$

$D_2=(3,2)$

Then when it indicates that it has been rotated $\frac{\pi}{2}$ to the counterclockwise direction about point $(5,-4)$.

$A_3=(2,-9)$

$B_3=(0,-8)$

$C_3=(3,-5)$

$D_3=(-1,-6)$

In short I ommited the steps on how I did find these, but to my intuition I believe that if you make a right triangle you can then use this as an aid to rotate such right triangle to get the new coordinate.

For example what I did for point $A_3$.

With respect of point $E=(5,-4)$, point $A_2=(0,-1)$ means that $A_2$ is five units to the left of point $E$ on the $\textrm{x-direction}$, therefore when it is rotated $\frac{\pi}{2}$ to the counterclockwise direction it will became the new $\textrm{y-direction}$ thus this will be added to the previous $y$ coordinate on point $E$ moving it further downwards.

$y=-4-5=-9$

Conversely this point $A_2$ is located three units above point $E$ thus when rotated $\frac{\pi}{2}$ to the counterclockwise direction it will become the new $\textrm{x-direction}$. But because it is to the left of point $E$, it will be moved three units to the left of that point.

$x=5-3=2$

Therefore $A_3=(2,-9)$.

Thus adding all of these results into:

$2-9+0-8+3-5-1-6=-24$

Thus I believe the answer is$ -24$ or choice 4.

But I cannot be sure if this is the correct answer. Since I'm not good with spatial abilities. I really beg someone could instruct me on how to see these in the cartesian plane?. Can someone put some drawing for these rotations?. By the way does it exist a formula for these?

I can't see it clearly so I need help on this one.

答:
You have calculated correctly.

As for visualization, the rotation through $90º$ is one of the easiest angles to work with.  For example, to move $B_2$ to $B_3$, I noted that $B_2$ was $4$ units left and $5$ units up from the rotation center, so I started from the rotation center and moved $4$ units down and $5$ units left.  Up $\rightarrow$ Left $\rightarrow$ Down $\rightarrow$ Right $\rightarrow$ Up...and so on.


问:
Computing center manifolds
Consider the following system $x' = y, y'=-y+3x+xy$.
find the center manifold and by converting to coordinates. Is the origin stable, semi-stable, or unstable?

I am confused if I am doing it correctly or if I answered the question.

For linearization, $\frac{\partial x'}{\partial x} = 0$, $\frac{\partial x'}{\partial y} = 1$

$\frac{\partial y'}{\partial x} = 6x + y$, $\frac{\partial y'}{\partial y} = -1 + x$.

So, $J(x, y)= \begin{bmatrix} 0 & 1\\ 6x+y & -1+x \end{bmatrix}$

$J(0, 0) = \begin{bmatrix} 0 & 1\\ 0 & -1 \end{bmatrix}$

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

$x' = y$ and $y'=-y$

Eigenvalues of matrix are $\lambda (\lambda + 1) = 0 \Longrightarrow \lambda = 0, 1$.

答:
Assuming that $\dot y = -y + 3x^2 + xy$, the work is correct, but the 0 eigenvalue begs for a center manifold reduction. I will leave some of the details for you to fill in.

First, we can show that the eigenvector of $\lambda = 0$ is $(1,0)$, which implies that we should seek an approximation of the center manifold given by $y = y(x)$. Differentiation gives $\dot y = y'(x) \dot x $ and so
$$ -y + 3 x^2 + x y = y'(x)y. $$
Let's look for a power series representation $y = \sum_{n=2}^\infty a_n x^n$ (why can I skip the first two terms?). Plugging this into the above expression leads to
$$ x \sum_{n=2}^\infty a_n x^n + 3 x^2 - \sum_{n=2}^\infty a_n x^n = \left(\sum_{n=2}^\infty a_n n x^{n-1} \right) \left(\sum_{m=2}^\infty a_m x^m \right). $$
We balance terms to find the first few coefficients:
\begin{align*}
& \mathcal O(x^2): \quad 3 - a_2 =0 \\
& \mathcal O(x^3): \quad a_2 - a_3= 2a_2^2
\end{align*}
This implies that $a_2 = 3$ and $a_3 = -15$ and so the center manifold is approximated by $y(x) = 3x^2 - 15 x^3$ for $x \ll 1$. Plugging this into the equation for $\dot x$ we see that
$$\dot x = 3x^2 + \mathcal O(x^3)$$
thus the point is actually a saddle.

Title: Computing center manifolds


问:
Largest possible number of multiplicity in a regular polygon
Is there a conjecture that states the biggest possible number diagonals of a regular polygon intersecting at one single point in the regular polygon outside of the intersection point of all the diameters?</p>

答:
This is known — see this amazing paper by Poonen and Rubinstein!

The maximum number of diagonals incident to a non-center point in the interior of a regular $n$-gon (for $n > 4$) is

$$\begin{cases}
2 &: n~\text{is odd} \\
3 &: n~\text{is even and not divisible by}~6 \\
5 &: n~\text{is divisible by}~6~\text{and not divisible by 30} \\
7 &: n~\text{is divisible by}~30
\end{cases}$$


问:
If $\triangle ABC$ is isosceles and right-angle, with $AB=AC=x$ and $M,N$ are the midpoints of $AC$ and $AB$ respectively, find the area of $MON$
If $\triangle ABC$ is isosceles and right-angle, with $AB=AC=x$ and $M,N$ are the midpoints of $AC$ and $AB$ respectively, find the area of $\triangle MON$

My solution goes as follows, If $D$ is a point on $CB$ such that $MD\perp CB$, we have that the area of $\triangle CMB$ is $\frac{x^2}{4}$. Hence $\frac{MD*x\sqrt{2}}{2}=\frac{x^2}{4}$ we then from this find the height of $\triangle COB$ and $\triangle AMN$, and since we also have the height of $\triangle CAB$ from $CB$, so from this we get the height of $\triangle MON$ and hence we have that its area is $\frac{x^2}{24}$. My solution is very long and tedious, is there a faster and easier approach to this problem?

答:
Note that the triangles $OMN$ and $OBC$ are similar with $\frac{ON}{OC}=\frac12$. Then
$$[OMN]= \frac{ON^2}{OC^2}[OBC]=\frac14\cdot \frac23 [NBC]= \frac16\cdot\frac12[ABC]=\frac1{24}x^2
$$


答:
I'm going to continue from here.

$$A_2=(0,−1)$$

$$B_2=(1,1)$$

$$C_2=(4,−2)$$

$$D_2=(3,2)$$

To rotate these points 90º degrees about point $E=(5,−4)$ in the counterclockwise direction you could use a rotation matrix, or even better, treat vectors as complex numbers. That's what I'm going to do. That means $\vec{EA_3}=i\vec{EA_2}$ and so forth.

We have
\begin{align*}
\vec{EA_2}&=(0,−1)-(5,-4)=-5+3i &\Rightarrow& \vec{EA_3}=-3-5i\\
\vec{EB_2}&=-4+5i &\Rightarrow& \vec{EB_3}=-5-4i\\
\vec{EC_2}&=-1+2i &\Rightarrow& \vec{EC_3}=-2-i\\
\vec{ED_2}&=-2+6i &\Rightarrow& \vec{ED_3}=-6-2i
\end{align*}

Therefore, the sums of the coordinates of $A_3,B_3,C_3$ and $D_3$ is $-3-5-5-4-2-1-6-2+\underbrace{4(5-4)}_{\text{4 times the sum of the coordinates of $E$.}} = -24$, as you've found.

Title: How to find the rotation of $A$, $B$ and $C$ about a fixed point like $(x_1,y_1)$?

Body:
The problem is as follows:

Let the points:

$A=(-6,-1)$

$B=(-5,1)$

$C=(-2,-2)$

$D=(-3,2)$

These represent a quadrilateral on the $x-y$ coordinate system. Find the sum of the coordinates from all the vertex of such quadrilateral after the figure has been moved $6$ units to the right and rotated $90^{\circ}$ degrees about point $E=(5,-4)$ in the counterclockwise direction.

The choices given by my workbook are as follows:

$\begin{array}{ll}
1.&-25\\
2.&-22\\
3.&-23\\
4.&-24\\
\end{array}$

I'm not very sure how to solve this problem. But I can tell that when you first move six units to the right the new coordinates become as follows:

$A_2=(0,-1)$

$B_2=(1,1)$

$C_2=(4,-2)$

$D_2=(3,2)$

Then when it indicates that it has been rotated $\frac{\pi}{2}$ to the counterclockwise direction about point $(5,-4)$.

$A_3=(2,-9)$

$B_3=(0,-8)$

$C_3=(3,-5)$

$D_3=(-1,-6)$

In short I ommited the steps on how I did find these, but to my intuition I believe that if you make a right triangle you can then use this as an aid to rotate such right triangle to get the new coordinate.

For example what I did for point $A_3$.

With respect of point $E=(5,-4)$, point $A_2=(0,-1)$ means that $A_2$ is five units to the left of point $E$ on the $\textrm{x-direction}$, therefore when it is rotated $\frac{\pi}{2}$ to the counterclockwise direction it will became the new $\textrm{y-direction}$ thus this will be added to the previous $y$ coordinate on point $E$ moving it further downwards.

$y=-4-5=-9$

Conversely this point $A_2$ is located three units above point $E$ thus when rotated $\frac{\pi}{2}$ to the counterclockwise direction it will become the new $\textrm{x-direction}$. But because it is to the left of point $E$, it will be moved three units to the left of that point.

$x=5-3=2$

Therefore $A_3=(2,-9)$.

Thus adding all of these results into:

$2-9+0-8+3-5-1-6=-24$

Thus I believe the answer is$ -24$ or choice 4.

But I cannot be sure if this is the correct answer. Since I'm not good with spatial abilities. I really beg someone could instruct me on how to see these in the cartesian plane?. Can someone put some drawing for these rotations?. By the way does it exist a formula for these?

I can't see it clearly so I need help on this one.


问:
Given $z, w\in\mathbb{C}$ prove that $|z| = 1$ implies $|(z + w)/(1 + \overline{z}w)| = 1$
I know I am supposed to write my process, but the thing is I have no clue.

I tried to rewrite $z$ as $a + bi$ and $w$ as $x + yi$ and try to rearrange the left side in a way I can get to the other, but no success.

Then I tried to solve like $|(z + w)/(1 +  \overline{z}w)| = 1$ iff $|z| = 1$ because i thought it would be easier to make $w$ "disapear" and end up with $|z| = 1$, but no success as well.

(i am not putting what i did, because it's all wrong and useless)

I hope you know and help me :(

答:
Hint: if $|z|=1$, then $\bar z=1/z$. See if you can use this to rearrange the expression
$$\frac{z+w}{1+\bar zw}$$
into something more manageable.


问:
Proving the convergence of a complex power series
I have been studying series' and sequences' recently and I have a hard time really internalizing some of the concepts, so I tackled some problems and I came across the following one, where I am not quite sure my proof works.

Let $z \in \mathbb{C}$. Prove that the following series converges.

$$\sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)!}$$

I tried applying the ratio test, which gave me this:

$$\lim_{n \to \infty} \sup |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} \sup |\frac{\frac{z^{2(n+1)+1}}{(2(n+1)+1)!}}{\frac{z^{2n+1}}{(2n+1)!}}|$$

$$= \lim_{n \to \infty} \sup |\frac{\frac{z^{2n+3}}{(2n+3)!}}{\frac{z^{2n+1}}{(2n+1)!}}| = \lim_{n \to \infty} \sup |\frac{z^{2n+3} * (2n+1)!}{z^{2n+1} * (2n+3)!}|$$

$$= \lim_{n \to \infty} \sup |\frac{z^2}{(2n+2)(2n+3)}| = \lim_{n \to \infty} \sup \frac{|z|^2}{(2n+2)(2n+3)}$$

Since $|z|^2$ is just some constant real number the limit of the above will be $0$, which is of course less than $1$ and therefore the series converges. Is this a correct use of the ratio test and would this count as a proof or is there something left to show?

答:
Yep -- this is a great proof!

(Answering under Community wiki so that the question can be closed as answered.)


问:
Prove that function of two variables is continuous
I was reading my class examples and I got the function
$$f(x,y)=
\begin{cases}
      \frac{1}{2}x^2 + y^2-1& \quad\text{if } x^2+y^2>1,\\
      -\frac{1}{2}x^2 & \quad\text{if } x^2+y^2\leq1.\\
   \end{cases}
$$
Is it continuous over $\mathbb{R}^2$?

Basically the notes explain that making $\lvert f(x,y) - f(x_{0},y_{0}) \lvert$  (if $(x_{0},y_{0})=1$) you end up with $\lvert f(x,y) - (-\frac{1}{2}x_{0}^2) \rvert \leq \lvert \frac{1}{2}x^2 + y^2 - 1 + \frac{1}{2}x_{0}^2\rvert + \lvert -\frac{1}{2}x^2+\frac{1}{2}x_{0}^2 \rvert$.
So, just taking the limit as $(x,y) \rightarrow (x_{0},y_{0})$ (using that $||(x_{0},y_{0})||=1$) you have that the limit is zero.

If $||(x_{0},y_{0})|| \neq 1$ we have that $f$ is continuous because it is the sum of continuous functions and $f$ is continuous over $\mathbb{R}^2$.

I'd like to know how to derive the first part when proving that $f$ is continuous, because I didn't got it 100% clear, please! D:

答:
You need to show that, given any $(x_0,y_0)$ with $x_0^2+y_0^2=1$ and any $\epsilon>0$, there exists some $\delta>0$ so that
$$\left|(x-x_0,y-y_0)\right|\leq\delta\implies |f(x,y)-f(x_0,y_0)|\leq \epsilon.$$
For this, you can use the expression you've written; since $f(x,y)$ is one of $\frac12x^2+y^2-1$ or $-\frac12x^2$, you know
$$\left|f(x,y)-f(x_0,y_0)\right|\in\left\{\left|\frac12x^2+y^2-1+\frac12x_0^2\right|,\left|-\frac12x^2+\frac12x_0^2\right|\right\},$$
so you can definitely get
$$\left|f(x,y)-f(x_0,y_0)\right|\leq\left|\frac12x^2+y^2-1+\frac12x_0^2\right|+\left|-\frac12x^2+\frac12x_0^2\right|.$$
You can then write
$$\frac12x^2+y^2-1+\frac12x_0^2=\frac12x^2+y^2-x_0^2-y_0^2+\frac12x_0^2=\frac{x^2-x_0^2}2+(y^2-y_0^2),$$
so
$$\left|f(x,y)-f(x_0,y_0)\right|\leq |x^2-x_0^2|+|y^2-y_0^2|.$$
Can you use this to finish the bounding?


问:
Variance Definition: question on expansion
Wiki defines variance as follows.  I have a question on the expansion which I will point out.

$$
Var(X) = E[(X - E[X])^2] = E[X^2 - 2XE[X] + E[X]^2]
$$

I'm assuming $E[X - Y] = E[X] - E[Y]$ applies next.  Is this true?  Also, does anyone know how $-2E[X]E[X]$ is derived in the next term?

$$
Var(X) = E[X^2] - 2E[X]E[X] + E[X]^2 = E[X^2] - E[X]^2
$$

答: Hint : with the equality you pointed out you can 'split' the expected value. For the derivation instead, remember that $E[X]$ is a number; what is then $E[2XE[X]]$?


答:
You are wrong on that line. It should read: $|f(x,y) - f(x_0,y_0)|=\left|\frac{x^2}{2}+y^2-1+\frac{x_0^2}{2}\right|= \left|\frac{x^2}{2}+y^2-(x_0^2+y_0^2)+\frac{x_0^2}{2}\right|= \left|\frac{x^2-x_0^2}{2}+y^2-y_0^2\right|\le \frac{1}{2}|x^2-x_0^2|+|y^2-y_0^2|= \frac{1}{2}|x-x_0||x+x_0|+|y-y_0||y+y_0|< \frac{\delta}{2}|x+x_0|+\delta|y+y_0|\le \frac{\delta}{2}(|x-x_0|+|2x_0|)+\delta(|y-y_0|+|2y_0|)< \frac{\delta}{2}(\delta+|2x_0|)+\delta(\delta+|2y_0|)< \frac{\delta}{2}(1+|2x_0|)+\delta(1+|2y_0|)< \epsilon $ if you take $\delta = \text{min}\left(1,\frac{2\epsilon}{3+2|x_0|+4|y_0|}\right)$. Thus $f$ is continuous at $(x_0,y_0)$ on the unit circle, and therefore on $\mathbb{R^2}$ too.
Ah, I found out I missed the part that $f$ could be taking the "other value". But it's a small technical detail that can be added to it. The main thought is still there....

Title: Prove that function of two variables is continuous

Body:
I was reading my class examples and I got the function
$$f(x,y)=
\begin{cases}
      \frac{1}{2}x^2 + y^2-1& \quad\text{if } x^2+y^2>1,\\
      -\frac{1}{2}x^2 & \quad\text{if } x^2+y^2\leq1.\\
   \end{cases}
$$
Is it continuous over $\mathbb{R}^2$?

Basically the notes explain that making $\lvert f(x,y) - f(x_{0},y_{0}) \lvert$  (if $(x_{0},y_{0})=1$) you end up with $\lvert f(x,y) - (-\frac{1}{2}x_{0}^2) \rvert \leq \lvert \frac{1}{2}x^2 + y^2 - 1 + \frac{1}{2}x_{0}^2\rvert + \lvert -\frac{1}{2}x^2+\frac{1}{2}x_{0}^2 \rvert$.
So, just taking the limit as $(x,y) \rightarrow (x_{0},y_{0})$ (using that $||(x_{0},y_{0})||=1$) you have that the limit is zero.

If $||(x_{0},y_{0})|| \neq 1$ we have that $f$ is continuous because it is the sum of continuous functions and $f$ is continuous over $\mathbb{R}^2$.

I'd like to know how to derive the first part when proving that $f$ is continuous, because I didn't got it 100% clear, please! D:


问:
is this prime probability function that generated from Eratosthenes Sieve can predict numbers of prime in closed interval?
I have two question regarding this prime probability $P(p)$ for $p$ that exists for $[p_{k-1}^2 , x]$

$P(p)=\prod^k_{i=1}\big{(} 1 -\frac{1}{p_i}\big{)}$

Where $x<p_k^2$ and $k$ was index such that $p_1=2$, $p_2=3$ and so on.

Is it trivial probability function of prime for the interval? Or it's not working because prime wasn't equally distributed enough even in $[p_{k-1}^2, p_k^2]$ or there was another reason so it doesn't work?

Secondly, if this works and i want to use $P(p)$ into my proof works, did i need to proof the probability first?

I'm trying looking into this but didn't find the answer. I hope my question was clear enough.

答:
This probability $P(p)$ is a reasonable heuristic for the probability that a random number in $[p_{k-1}^2,x]$ is prime, but it doesn't actually equal the probability.

This number was presumably derived using the fact that the probability that $p_i$ divides a "randomly chosen number" is $1/p_i$, and then multiplying together the probability that $p_i$ doesn't divide a randomly chosen number across all $i$. The first (less pressing) issue is that the probability isn't exactly $1/p_i$ if you're choosing the number from an interval of a length that isn't a multiple of $p_i$. This isn't a terrible issue, but it is definitely important; we can't have all of our primes dividing the length of $[p_{k-1}^2,x]$, since $p_1\cdots p_{k-1}$ is very large.

The second issue is that the events "divisible by $p_i$" and "divisible by $p_j$" aren't independent. They are likely roughly independent (and are independent if $p_ip_j$ divides the length of your interval), but for a smaller interval it's possible to get unlucky and have one number that has a lot of prime factors, or for your prime factors to be more spread out than normal. These events are both hard to estimate and quantify.

In sum, $P(p)$ doesn't actually equal the probability that a number in your interval is prime -- for it to equal the probability that no small prime divides a random number, your interval would need to be much, much larger, and numbers in such an interval will have too many possible prime factors.