问:
Bound for order of a group depending on number of elements of maximal order
In a paper On the Number of Elements of maximal order in a Group it is proven that an arbitrary group $G$ with a finite number of elements of maximal order has bounded size. Namely: $$|G|\leq\frac{mk^2}{\varphi(m)},$$ where $m$ is the maximal order and $k$ the number of elements that have order $m$.

I wanted to characterize all groups $G$, where the limit is sharp, i.e. $|G|=\frac{mk^2}{\varphi(m)}$. Using GAP I found all groups with this property up to order 1023 and was able to state a conjecture. It is easy to see in the paper, that a group has this property only if all elements of maximal order are conjugated. So we need this as as a requirement.

I wanted to prove the following conjecture, but missing some tiny part. Maybe someone knows a way, I would be really happy.

Conjecture.

Let $G$ be a group with $k<\infty$ elements of maximal order $m$, in which all elements of maximal order are conjugated. Then the following are equivalent.

$i)$ $|G|=\frac{mk^2}{\varphi(m)}$

$ii)$ $k=\varphi(m)$

$iii)$ $G$ has a unique subgroup of order $m$

$iv)$ $C_m \cong C_G(x)=C_G(y)\trianglelefteq G$ for all $x,y\in G$ with maximal order

Proof.

$i) \implies ii)$
This is the part, I could not prove:

I only could prove, that all elements of order $m$ commute:

Let $C_G(x)$ be the stabilizer of an element of maximal order. Orbit-Stabilizer-Theorem tells us, that $|C_G(x)|=\frac{mk}{\varphi(m)}$. Assume there exists an element of order $m$, not contained in $C_G(x)$. $\langle x \rangle$ operates via left-multiplication on $C_G(x)$. $C_G(x)$ is partitioned into $\frac{|C_G(x)|}{m}$ orbits. According to Lemma 3 of  the paper linked above, in each orbit exist at least $\varphi(m)$ elements of order $m$, i.e. in $C_G(x)$ exist at least $\varphi(m)\frac{|C_G(x)|}{m}$ elements of order $m$. Our assumption tells us $\varphi(m)\frac{|C_G(x)|}{m} < k$, which leads to the contradiction $|C_G(x)| < \frac{mk}{\varphi(m)}$. It follows that all elements of order $m$ commute.

This is where I can't proceed further. Maybe someone has an idea?

$ii) \iff iii)$
If $k=\varphi(m)$, an element of order $m$ generates a cyclic subgroup which contains $\varphi(m)$ elements of order $m$, that all generate this subgroup. So there can't be other elements of order $m$ in different subgroups. Otherwise, if there is only one cyclic subgroup of order $m$, then it contains $\varphi(m)$ elements of order $m$, no additional elements of order $m$ can exist, as they would generate a second cyclic subgroup of order $m$.

$iii) \implies iv)$ Let $Z$ be the unique subgroup of order $m$ and $X=\{x_1,\dots,x_k\}$ the set of elements of order $m$. As all $x\in X$ generate $Z$, $Z$ must be contained in all centralizers of elements in $X$. Note that $G$ operates on itself via conjugation. Orbit-Stabilizer-Theorem tells us for $x \in X$: $$|G|=|^Gx||G_x|=k|G_x|=\frac{mk^2}{\varphi(m)}=mk$$ This follows as all elements of order $m$ are conjugated and $k=\varphi(m)$ holds. It follows, that $|G_x|=m$, which leads to $G_x=Z\cong C_m$ for all $x \in X$.

For the normal subgroup part, note that $\phi(x_i)=x_j$ for an inner automorphism $\phi$ and $i,j\in \{1,\dots k\}$. Let $y \in Z$ arbitrary, then $y=x_1^\alpha$ for $\alpha \in \mathbb{N}$. Let $\phi$ be an arbitrary inner automorphism. It follows that there is a $i \in \{1,\dots k\}$ with $$\phi(y)=\phi(x_1^\alpha)=\phi(x_1)^\alpha=x_i^\alpha \in Z$$ It follows that $Z$ is invariant under inner automorphisms, i.e. normal.

$iv) \implies i)$ Orbit-Stabilizer-Theorem tells us that $|G|=|^Gx||G_x|=mk$. As all stabilizers of elements of order $m$ are equal to the same cyclic group of order $m$, it follows, that there exist only one cyclic group of order $m$, it follows $k=\varphi(m)$ and $|G|=mk=\frac{mk^2}{\varphi(m)}$.

Thanks to anyone, who read till here ;)

Another property, which my GAP-study suggests to be equivalent is :

$v)$ $G'$ is cyclic

This proof has low priority, as I first want to have my circle-implications. I guess I can show, that $G'$ is contained in the unique cyclic group $Z$ of order $m$, by proving, that $G/Z$ is abelian. I did not succeed yet, though.

答:
This is not a complete answer, but provides a bit more information about a group $G$ with $|G| = mk^2/\phi(m)$.

Since you have proved that all elements of order $m$ commute, the elements of order $m$ generate an abelian normal subgroup $N$ of $G$ of exponent $m$.

Let $g \in G$ have order $m$. We claim that $C_G(g) = N$. To prove this, let $h \in C_G(g)$. We want to show that $h \in N$. This is clear if $h \in \langle g \rangle$. Otherwise, since $m$ is the largest order of any element in $G$, $\langle g,h \rangle$ is a $2$-generator abelian group of exponent $m$, and so it is equal to $\langle g \rangle \times \langle hg^i \rangle$ for some $i$ with $0 \le i < m$. But then $hg^{i+1}$ has order $m$ and hence lies in $N$, so $h \in N$, which establishes the claim.

So $[G:N] = [G:C_G(g)] = k$, and hence $|N| = mk/\phi(m)$.

I think that an abelian group $N$ of exponent $m$ with $k$ elements of order $m$ can only have order $mk/\phi(m)$ when $k=\phi(m)$ and $N$ is cyclic, and it should be possible to prove this, but I have done so yet. You can easily check for example that if $m=p$ is prime then it is only possible when $|N|=p$


问:
Conditional probability hard problem
Suppose that we have three coins. One has two heads, while the others are standard. Two coins are flipped. In each case, the result is heads.

If you pick one of these two coins randomly, what is the probability that you will get the two headed coin?

If you pick the unflipped coin, what is the probability that you will get the two headed coin?

My attempt at 1:

Let A be the event: "We obtain two heads" and let B be the event: "We got the two-headed coin." We are looking for $$P(B|A)=\frac{P(B\bigcap A)}{P(A)}$$
Now, from Baye's Theorem: $$P(B\bigcap A)=P(A|B)P(B)=(\frac{1}{2})(\frac{1}{3})=\frac{1}{6}$$
I still need to find $P(A)$, could anyone help?

答:
Let's conduct $12$ trials.  We'll pick each of the three possible pairs of coins in $4$ of those trials.  Then we'll flip both of the coins we picked.

We'll get a pair of heads once out of the four trials where we picked two fair coins.  The other two pairs of coins will each give us a pair of heads twice.

That means we end up with $5$ trials where we flipped two heads.  In $1$ of those $5$ trials, we picked two fair coins.  In the other $4$, one of our coins is the two-headed coin.  That means there is a $\frac 45$ probability that one of our coins is two-headed so our chance of picking the two-headed coin if we randomly select one of the two flipped coins is $\frac 25 (= \frac 45 \cdot \frac 12)$ and the chance that the unflipped coin is the two-headed coin is $\frac 15$.

ContentLicense: CC BY-SA 4.0

Title: Conditional probability hard problem


问:
Why is the Frobenius norm (energy) of a rectangular matrix $A$ equal to the trace of $A^T A$ or $AA^T$
Reading through a book, it is mentioned that $||{A}||^2_F=\text{Energy}(A)=\text{tr}(AA^T)=\text{tr}(A^TA)$. I understand that for square matrices the square Frobenius norm would be the the squared sum of all elements within the matrix, but I cannot intuitivley get why for rectangular matrices it would be the trace of the matrix multiplied by its transpose (or the other way around). For instance it would be that $\text{tr}(CD^T) = \text{tr}(DC^T) = \displaystyle\sum_{i=1}^n\sum_{j=1}^dc_{ij}d_{ij}$ for some matrices $D, C$. of size $n \times d$. Maybe some sort of proof would help?

SOURCE: Linear Algebra and Optimization for Machine Learning: A Textbook (page 20)

答:
In frobenius norm of matrix $A_{m\times n}$, we consider $A$ as an $m \times n$ vector that its entries  are written one after the other $($ because $ M_{m\times n}(\mathbb{R})\equiv \mathbb{R}^{m\times n})$ and then frobenius norm of $A$ is norm-$2$ of this vector.

For example if
$$
A=\begin{bmatrix}
    x_{11}       & x_{12} \\
    x_{21}       & x_{22}  \\
\end{bmatrix}
and \ v=\begin{bmatrix}
    x_{11}   \\     x_{12} \\
    x_{21}    \\    x_{22}  \\
\end{bmatrix}$$
then $||A||_F=||v||_2.$

Now
$$||A||_F=||v||_2=\sqrt{\sum_{i=1}^{m}\sum_{j=1}^{n}|a_{ij}|^2}=\sqrt{\sum_{i=1}^{m}\sum_{j=1}^{n}(a_{ij})(a_{ij})}=\sqrt{tr(AA^T)}$$


问:
Symmetric Matrices of $I_{2}$

Find $10$ symmetric matrices $ A = \begin{pmatrix}
a &b \\
 c&d
\end{pmatrix}$ such that $A^{2}=I_{2}$

(I'm going to call matrix A the "square root" of $A^{2}$. If this is the incorrect name for it, may someone please tell me what it is actually called?)

My professor posed this question in class and told us there was an infinite amount of square roots. (Assuming I understood him correctly). However I don't see how there would be many of these, as I was under the impression that a matrix only has one inverse, for $A  A^{-1}=I_{n}$. If someone could tell me if I either misunderstood the professor or if I'm thinking about something incorrectly, please correct me.

My other question is other than blatant guess and check, is there a method to think of these symmetric square roots?

Thanks in advance.

答:
You can do this by writing down equations, writing out the multiplication explicitly: You'll get four quadratic equations in four variables.

For a geometric solution, here's a hint: Reflection across any line through the origin would be such a linear map (matrix). For example, $\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$ and  $\begin{bmatrix} 0&1\\1&0\end{bmatrix}$ are, respectively, reflection across the $x$-axis and the line $y=x$.

ContentLicense: CC BY-SA 3.0

Title: Symmetric Matrices of $I_{2}$


问:
The special function $P(s)=\int^\infty_0 \frac{\ln(x)dx}{1+x^s}$

Evaluate the integral
$$P(s)=\int^\infty_0 \frac{\ln(x) \,dx}{1+x^s}$$

I tried integration by parts, using $u=\ln(x),du=\frac{x}{dx},v=\frac{1}{1+x^s}$, but I did not manage to get anything.

I thought about using a power series, but I did not find one that converges on the entire domain.

Using the substitution $u=\ln(x),du=\frac{dx}{x}$ we get $$\int^\infty_{-\infty} \frac{ue^{-u}}{1+e^{su}} ,$$
which doesn't seem useful either.

The previous similar question Integral of $\int^{\infty}_0 \frac{x^n}{x^s+1}dx$ gives
$$\int^\infty_0 \frac{x^n \,dx}{1+x^s}=\frac{\Gamma(\frac{1}{s})\Gamma\left(1-\frac{1}{s}\right)}{s(n+1)} .$$

Another possibly relevant question is Integral of $\int^{\infty}_0 \frac{e^{-x}}{x^s+1}\,dx$.

答:
$$I=\int_{0}^{\infty} \frac{\ln x}{1+x^s} dx.$$
Let $x=e^t$, then
$$I=\int_{-\infty}^{\infty} \frac{t e^t}{1+e^{st}}=\int_{-\infty}^{0} \frac{t e^t}{1+e^{st}} dt+\int_{0}^{\infty} \frac{t e^t}{1+e^{st}} dt$$
In the first one let $t=-z$, then
$$I=-\sum_{k=0}^{\infty}\int_{0}^{\infty} ze^{-(1+ks)z} dz+\sum_{0}^{\infty}
\int_{0}^{\infty} te^{-(s+ks-1)t} dt$$
$$I=-\sum_{k=0}^{\infty} \frac{1}{(1+ks)^2}+ \sum_{k=0}^{\infty} \frac{1}{[s(1+k)-1]^2}=\frac{\psi^{(1)}(1-1/s)-\psi^{(1)}(1/s)}{s^2}$$
Using the property of poly-Gamma function:
https://en.wikipedia.org/wiki/Polygamma_function

Hoping to come back.


问:
Computing a Killing vector field from flow
I am given the following manifold $N=\{(x,y)\in \mathbb{R}^2, y>0\}$ with metric:

$$ds^2=\frac{dx^2+dy^2}{y^2}$$

There is a suggestion to take $z=x+iy$ and consider the transformations:

$$z\to z+c\,, \quad z\to cz\,,\quad z \to \frac{z}{cz+1}\,,\quad c\in \mathbb{R}$$

I have to find three independent Killing vector fields.

I think the idea is to define a flow from those transformations and then compute the Killing vectors associated to that flow. Is there any systematic way to do this?

答:
You have the right idea. First, define
\begin{align}
\varphi_1(t,z) &= z + t &
\varphi_2(t,z) &= e^{2t}z &
\varphi_3(t,z) &= \dfrac{z}{tz +1}
\end{align}
They correspond to the matrices in $PSL_2(\mathbb{R})$, generating $1$-parameter subgroups
\begin{align}
A_1 &=\begin{pmatrix}
1 & t \\
0 & 1
\end{pmatrix}&
A_2 &=\begin{pmatrix}
e^t & 0 \\
0 & e^{-t}
\end{pmatrix}&
A_3 &=\begin{pmatrix}
1 & 0 \\
t & 1
\end{pmatrix}
\end{align}
Show they are flows acting by isometry on the hyperbolic plane. Then, compute $X_i(z) = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\varphi_i(t,z)$. They will be Killing vector fields.

For example, fix $z$ in the hyperbolic plane. Then the vector field $X_1(z)$ at $z$ is the tangent vector in $T_z\mathbb{H} \simeq \mathbb{C}$
$$
X_1(z) = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\varphi_1(t,z) = 1 = 1 + i\cdot 0 = \partial_x
$$


问:
How can I derive the leading term of asymptotic expansion from WKB approximation.
I would like to derive the the asymptotic expansion from the WKB approximation. I thought that is standard and could be found in textbooks, but I did not find any useful materials.

To make the question clear, let us start with an example, say, I have a Bessel's equation,
$$
z^2 y''+z y' +(z^2-\nu^2)y=0,
$$
whose solution is Bessel's functions $J_{\pm\nu}(z)$.
Everyone is familiar with the leading term of asymptotic expansion as $z\to\infty$, that is
$$
J_\nu(z)\sim \sqrt{\frac{2}{\pi z}}\cos\left(z-\frac{\nu \pi}{2}-\frac{\pi}{4}\right).
$$
My task is to derive this formula from Bessel's equation by WKB approximation.

To use the WKB method, I prefer to make a transformation first $y\to y/\sqrt{z}$, which gives
$$
y''+Q(z) y=0,\qquad Q(z)=1+\frac{1/4-\nu^2}{z^2}.
$$
The WKB phase can be calculated analytically
$$
I=\int^z dz \sqrt{Q(z)}
$$
and as $z\to \infty$, one will have
$$
I\sim z+\frac{ \pi}{4}   \sqrt{4 \nu ^2-1}
$$
but the additional constant here is not the phase shift in above cosine function.

Could someone give me a clue of solution, or full derivation, or any relevant references? Thanks in advance.

The similar question for first order differential equation can be found here.

答:
Your equation may be written in the form
$$
\frac{{d^2 y(z)}}{{dz^2 }} = ( - f(z) + g(z))y(z)
$$
with
$$
f(z) = 1,\quad g(z) = \frac{{\nu ^2  - 1/4}}{{z^2 }}.
$$
This equation has solutions $y_\pm(z)$ satisfying
$$
y_ \pm  (z) = \exp \left( { \pm i\int_{}^z {\sqrt {f(t)} dt} } \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{z}} \right)} \right) = C_ \pm  \exp ( \pm iz)\left( {1 + \mathcal{O}\!\left( {\frac{1}{z}} \right)} \right)
$$
as $z\to +\infty$, with arbitrary constants $C_\pm$. This can be established from Theorem 2.2 on p. 196 of F. W. J. Olver's Asymptotics and Special Functions. The Bessel function $\sqrt{z}J_\nu(z)$ will be an appropriate linear combination of these two solutions.


问:
How is the volume of a cross-polytope in $\mathbb{R}^n = \frac{2^n}{n!}$?
I'm fairly new to convex geometry, and I wish to prove that the volume of a cross-polytope in $\mathbb{R}^n$ is $\frac{2^n}{n!}$.

A cross-polytope is the convex-hull of $2n$ points in $\mathbb{R}^n$, namely, $(\pm 1,0,...,0), (0,\pm 1,0,...,0),...,(0,...,0,\pm 1)$. Since this is the same as the unit ball of $l_1$ norm in $\mathbb{R}^n$, we denote it by $B^n_1$.

$B^n_1$ is made up of $2^n$ pieces similar to the piece whose points are in the convex hull of $(\pm 1,0,...,0), (0,\pm 1,0,...,0),...,(0,...,0,\pm 1)$ and are all non-negative. So the total volume is $2^n$ times the volume of this piece (denoted by $P_n$ hereafter), which (from what I understand) has vertices $(1,0,...,0), (0, 1,0,...,0),...,(0,...,0, 1)$.

How do I find the volume of $P_n$? I have read an inductive proof that I don't quite understand. It says that this piece $P_n$ is a cone of height $1$ (how?) over a base, which is the analogous piece in $\mathbb{R}^{n-1}$, i.e. $P_{n-1}$ (how?).

If we take the above for granted and use that the volume of a cone in $\mathbb{R}^n$ is $\frac{Bh}{n}$ (and how do we prove this?), we'll get the desired result ($B$ = volume of base, $h$ = height of the cone) with the help of $$\operatorname{vol}(P_n) = \frac{\operatorname{vol}(P_{n-1})}{n}$$

A cone is defined as the convex hull of a single point and a convex body of dimension $\mathbb{R}^{n-1}$ in $\mathbb{R}^n$.

In general, I have found it difficult carrying forward results from $\mathbb{R}^3$ to $\mathbb{R}^n$ -  a lot of the stuff is not intuitive!

Edit 1:

Based on a comment by @Arthur, I have tried out integration to find the volume of $P_n$ as follows:
$$\int \int ... \int dx_1 dx_2 ... dx_n$$
We have $x_1 + x_2 + ... + x_n = 1$ and $x_i \ge 0 \forall i = 1,...,n$. So the limits for $x_1$ goes from $0$ to $1 - x_2 - x_3 - ... - x_n$
We get
$$\int \int ... \int (1 - x_2 - x_3 - ... - x_n) dx_2 ... dx_n$$

How do I go ahead from here? What are the limits for $x_2$? Is there a pattern I should see?

答:
These are fairly intuitive inductions if you go from $1$ to $2$ dimensions and from $2$ to $3$ dimensions.  Similar steps in higher dimensions are essentially the same.

For $n=1$ you are measuring the length from $-1$ to $+1$ which is $2=\frac{2^1}{1!}$

For the induction step you start at $n$ dimensions with a cross-polytope in black with hypervolume $V_{n} = \frac{2^n}{n!}$.  You then introduce an orthogonal  vertical line in red in the $n+1$th dimension with height $h$ from $-1$ to $+1$ and see that similar copies in grey of the cross-polytope have linear proportion $1-|h|$ and so $n$-dimensional hypervolume proportion $(1-|h|)^n$; it is easier just to consider the top half and double the result.  So integrating over the slices to find the next hypervolume: $$V_{n+1} = 2\int\limits_{h=0}^1  \frac{2^n}{n!}(1-h)^n\,dh = 2\left[\frac{2^n}{n!} \times \frac{-1}{n+1}(1-h)^{n+1}\right]^1_0=\frac{2^{n+1}}{(n+1)!}$$


答:
Your professor is right, there's an infinite number of square roots, kind of like how there's two square roots of $1$ (namely, $1$ and $-1$).

To see how to get it in general, notice that, for a symmetric matrix, you have
$$
\begin{pmatrix}a&b\\b&c\end{pmatrix}^2 = \begin{pmatrix}a^2+b^2&b(a+c)\\b(a+c)&b^2+c^2\end{pmatrix}
$$
So, for the right side to be equal to the identity, you must have
$$
a^2+b^2=1\\
b^2+c^2=1\\
b(a+c) = 0
$$
What solutions does this system of equations admit?

To demonstrate that multiple solutions exist directly, consider that
$$
\begin{pmatrix}0&1\\1&0\end{pmatrix}^2 = \begin{pmatrix}1&0\\0&1\end{pmatrix}
$$
and so is a square root of the identity matrix.

Title: Symmetric Matrices of $I_{2}$

Body:
Find $10$ symmetric matrices $ A = \begin{pmatrix}
a &b \\
 c&d
\end{pmatrix}$ such that $A^{2}=I_{2}$

(I'm going to call matrix A the "square root" of $A^{2}$. If this is the incorrect name for it, may someone please tell me what it is actually called?)

My professor posed this question in class and told us there was an infinite amount of square roots. (Assuming I understood him correctly). However I don't see how there would be many of these, as I was under the impression that a matrix only has one inverse, for $A  A^{-1}=I_{n}$. If someone could tell me if I either misunderstood the professor or if I'm thinking about something incorrectly, please correct me.

My other question is other than blatant guess and check, is there a method to think of these symmetric square roots?

Thanks in advance.


问:
Evaluare $I(y)=\int_0^1\frac{1}{y+\cos(x)}dx$. Hence determine $J(y)=\int_0^1\frac{1}{(y+\cos(x))^2}dx$

Consider the following integral:
$$I(y)=\int_0^1\frac{1}{y+\cos(x)}dx $$
with Weierstrasse substitution I showed that

$$ I(y)=\frac{2}{\sqrt{y^2-1}}\arctan\left(\sqrt{\frac{{y-1}}{y+1}}\right). $$

It is this next part that I am not sure about: Hence detertemine
$$J(y)=\int_0^1\frac{1}{(y+\cos(x))^2}.$$

My attempt:

Since $I(y)\pm J(y)$ yields no simplication, the most obvious approach would be to rewrite $I(y)$ as
$$I(y)=\int_0^1\frac{(y+\cos(x))}{(y+\cos(x))^2}=yJ(y)+\underbrace{\int_0^1\frac{cos(x)}{(y+\cos(x))^2}dx}_{K(y)}$$
Applyling Weierstrasse substitituion for $K(y)$:
$$K(y)=2\int_0^1\frac{(1-x^2)}{((y-1)x^2+(y+1))^2}dx=-2\int_0^1\frac{\frac{1}{y-1}((y-1)x^2+(y+1))-\frac{y+1}{y-1}-1}{((y-1)x^2+(y+1))^2}$$
$$=-2\left(\underbrace{\frac{1}{y-1}\int_0^1\frac{1}{(y-1)x^2+(y+1)}dx}_{\text{elementary integral}\to \arctan()}-\underbrace{\frac{2}{y-1}\int_0^1\frac{1}{((y-1)x^2+(y+1))^2}dx}_{\text{evaluated with }
 \tan(u)=\sqrt{\frac{y-1}{y+1}}x}\right)$$
so what follows is quite elmenatary.

However it seems to me that $K(y)$ is by no means simpler than $J(y)$, (i.e. the same approach can be used for $J(y)$ but without having to use the result for $I(y)$) so this is clearly not the point of the question?

答:
I suppose that the trick is

$$\frac{1}{(y+\cos (x))^2}=\frac{y}{\left(y^2-1\right) (y+\cos (x))}-\frac{y \cos (x)+1}{\left(y^2-1\right) (y+\cos (x))^2}$$


问:
Complex polynomials and lines of constant argument
Here are three domain plots of a complex polynomial of degree $5$. The left picture is very zoomed out, and the right picture is more zoomed into the zeroes. (Pictures are taken from Elias Wegert's book "Visual Complex Functions".) The color indicates the argument of the function; the modulus is not featured.

Say we focus our attention on one color, like yellow. Then we can see that the yellow lines coming in from infinity seem to "end" at the zeroes of the polynomial. My confusion is: I cannot justify why this should be the case in general. Why is it true that every yellow line (that is, a line of constant argument) coming in from infinity should terminate at a zero of the polynomial?

It makes sense to me that around a zero of order $n$, the polynomial should look like $z^n$ -- that is, there should be $n$ yellow lines emanating from that point. What is not clear to me, though, is why every yellow line coming in from infinity should terminate at a zero in particular. Why can't it terminate at any other point? Any suggestions / hints would be greatly appreciated. Thanks!

答:
Around a nonzero point, the argument is virtually constant.

$$f(z+\epsilon e^{i\theta})\approx f(z)+\epsilon e^{i\theta} f'(z)$$

and when $\epsilon\ll |f(z)|$,

$$f(z+\epsilon e^{i\theta})\approx f(z).$$

The condition $\epsilon\ll |f(z)|$ cannot be met at a zero.


问:
question on summation?
Please, I need to know the proof that

$$\left(\sum_{k=0}^{\infty }\frac{n^{k+1}}{k+1}\frac{x^k}{k!}\right)\left(\sum_{\ell=0}^{\infty }B_\ell\frac{x^\ell}{\ell!}\right)=\sum_{k=0}^{\infty }\left(\sum_{i=0}^{k}\frac{1}{k+1-i}\binom{k}{i}B_in^{k+1-i}\right)\frac{x^k}{k!}$$

where $B_\ell$, $B_i$ are Bernoulli  numbers.

Maybe we should replace $k$ with $j$?

Anyway, I need to prove how to move from the left to right.

Thanks for all help.

Title: question on summation?

Body:
$$\left(\sum_{k=0}^{\infty} \dfrac{n^{k+1}}{k+1} \dfrac{x^k}{k!} \right) \left(\sum_{l=0}^{\infty} B_l \dfrac{x^l}{l!}\right) = \sum_{k,l} \dfrac{n^{k+1}}{k+1} \dfrac{B_l}{k! l!} x^{k+l}$$

$$\sum_{k,l} \dfrac{n^{k+1}}{k+1} \dfrac{B_l}{k! l!} x^{k+l} = \sum_{m=0}^{\infty} \sum_{l=0}^{m} \dfrac{n^{m-l+1}}{m-l+1} \dfrac{B_l}{(m-l)! l!} x^{m}$$
This gives us
$$\sum_{m=0}^{\infty} \sum_{l=0}^{m} \dfrac{B_l}{(m-l+1)! l!}n^{m-l+1} x^{m} = \sum_{m=0}^{\infty} \left(\sum_{l=0}^m \dfrac1{m-l+1} \dbinom{m}{l} B_l n^{m-l+1}\right)\dfrac{x^m}{m!}$$


问:
Difference of equal derivative functions value
$f_1(x) = \text{tan}^{-1}(x)$

$f_2(x) = \text{tan}^{-1}\big(\frac{x-1}{x+1}\big)$

Where

$f_1^\prime(x) = f_2^\prime(x)=\frac{1}{x^2+1}$

Therefore

$f_1(x) - f_2(x) = \theta$

How to find values of $\theta$ mathematically. I have solved it graphically and the answers are $\frac{\pi}{4}$ and $-\frac{3\pi}{4}$ in my opinion. Thanks in advance.

答:
Since $f_2(x)$ is nondifferentiable at $x=-1$, if you are integrating it, I think you technically need to integrate it piecewise, resulting in: $F'_2(x) = \arctan(x) + $C1 for $x > -1$ and $F_2'(x)= \arctan(x) + $C2 for $x < -1 $ with C1 and C2 constants.

From there, you can just find C1 = $f_1(x) - f_2(x)$ for some $x > -1$ and C2 = $f_1(x) - f_2(x)$ for some $x < -1$.


问:
Orthogonal polynomials with respect to $e^{-|x|} \mathrm{d} x$ on the entire real line?
The Laguerre polynomials
https://en.wikipedia.org/wiki/Laguerre_polynomials
form a system of orthogonal polynomials with respect to the measure $e^{ -x} \mathrm{d} x$ on $(0,\infty)$.

Is anything known about the orthogonal polynomials with respect to the measure $e^{ -|x|} \mathrm{d} x$ on $(-\infty,\infty)$?

Thank you!

答:
I'll quote from T. J. Sullivan's Introduction to Uncertainty Quantification (SpringerLink). Let $\mathfrak P$ be the set of polynomials (as a subset of $L^2(\mu)$, where $\mu$ is a measure). Then by Gram-Schmidt on monomials, we can get the existence of orthogonal polynomials for $\mu=\exp(-|x|)dx$-

Theorem 8.5. If the $L^{2}(\mu)$ inner product is positive definite on $\mathfrak{P},$ then there exists an infinite sequence of orthogonal polynomials for $\mu$.

The book also mentions that $\int \exp(a|x|)d\mu(x)<\infty$ for some $a>0$ implies the completeness of the above constructed set of orthogonal polynomials; thus this works for $\mu=\exp(-|x|)dx$.

Then, to find the polynomials, you can use the three-term recurrence-

Theorem
8.9. Let $\mathcal{Q}=\left\{q_{n} \mid n \in \mathcal{N}\right\}$ be the monic orthogonal polynomials for a measure $\mu .$ Then
$$
\begin{aligned}
q_{n+1}(x) &=\left(x-\alpha_{n}\right) q_{n}(x)-\beta_{n} q_{n-1}(x) \\
q_{0}(x) &=1 \\
q_{-1}(x) &=0
\end{aligned}
$$
where
$$
\begin{aligned}
\alpha_{n} &:=\frac{\left\langle x q_{n}, q_{n}\right\rangle_{L^{2}(\mu)}}{\left\langle q_{n}, q_{n}\right\rangle_{L^{2}(\mu)}}, & \text { for } n \geq 0 \\
\beta_{n} &:=\frac{\left\langle q_{n}, q_{n}\right\rangle_{L^{2}(\mu)}}{\left\langle q_{n-1}, q_{n-1}\right\rangle_{L^{2}(\mu)}}, & \text { for } n \geq 1, \\
\beta_{0} &:=\left\langle q_{0}, q_{0}\right\rangle_{L^{2}(\mu)} \equiv \int_{\mathbb{R}} \mathrm{d} \mu.
\end{aligned}
$$
Hence, the orthonormal polynomials $\left\{p_{n} \mid n \in \mathcal{N}\right\}$ for $\mu$ satisfy
$$
\sqrt{\beta_{n+1}} p_{n+1}(x)=\left(x-\alpha_{n}\right) p_{n}(x)-\sqrt{\beta_{n}} p_{n-1}(x)
$$
$$
\begin{aligned}
p_{0}(x) &=\beta_{0}^{-1 / 2} \\
p_{-1}(x) &=0.
\end{aligned}
$$

For instance, $q_{0,-1}$ are given, and to find $q_1$, we need to compute $\alpha_0$ and $\beta_0$, which are
$$\beta_0 = \int_{\mathbb R} d\mu = \int_{-\infty}^\infty \exp(-|x|)dx = 2,$$
and
$$ \alpha_{0} =\frac{\left\langle x q_{0}, q_{0}\right\rangle_{L^{2}(\mu)}}{\left\langle q_{0}, q_{0}\right\rangle_{L^{2}(\mu)}} = \frac12\int x\exp(-|x|)dx = 0 $$
so the first two in the list are
$$q_0=1,\quad q_1=x.$$
Then you can continue inductively. $\langle xq_1,q_1\rangle_{L^2(\mu)}$ is  zero (hence so is $\alpha_1$). In fact, a quick induction argument (once you have the three-term recurrence, and since $\mu$ is even; see comments) shows that $q_{2k}$ is an even polynomial, and $q_{2k+1}$ is an odd one. This also implies $\alpha_n\equiv 0$. A quick calculation shows that
$\beta_1 = \frac12\int x^2 d\mu = 2,$
and so
$$q_2=x q_1 -2 q_0 = x^2-2.$$
A less easy calculation gives
$ \quad \beta_2 = 10,$
which means that
$$ q_3 = xq_2 -20 q_1 = x^3 - 12x.$$
And it continues. Maybe there's a nice closed form for the recurrence, but I haven't tried.

Verifications: this link shows that $\langle q_0,q_2\rangle_{L^2(\mu)} = 0$. $\langle q_1,q_2\rangle_{L^2(\mu)} = 0 = \langle q_0,q_3\rangle_{L^2(\mu)}$  and also $\langle q_2 ,q_3\rangle_{L^2(\mu)} = 0$ by the parity of the polynomials. This link shows that $\langle q_1 ,q_3\rangle_{L^2(\mu)} = 0$.


问:
Proving that the limit of the Mandelbrot polynomial's co-efficients are the Catalan numbers.
According to Donald D. Cross, the characteristic polynomial of the $n^{\mathrm{th}}$ iteration of the Mandelbrot function, $z_{n+1} = z_n^2 + c$ where $z_0 = 0$, gives the Catalan numbers as the polynomial coefficients as $n \to \infty$.

For example, expanding $z_3$ gives you the coefficients $\{1, 1, 2, 1\}$, or $1c^1 + 1c^2 + 2c^3 + 1c^4$. With each iteration, the detail of the Mandelbrot boundary contour gets more detailed. Images below describe this, with the contour plot $|f(z)| =2$:

This is the plot for iteration 4, $\{1, 1, 2, 5, 6, 6, 4, 1\}$:

This is the plot for iteration 6, (see Cross' webpage for co-efficients):

An issue arrives when you attempt to find the summation equivalent to having Catalan co-efficients:$$\sum_{n=0}^{\infty} C_n x^{n+1} = x \sum_{n=0}^{\infty} C_n x^n.$$

If you use the generating function for the Catalan co-efficents we get the function:
$\frac{1-\sqrt{1-4x}}2$. Which very obviously is not a detailed representation of the Mandelbrot boundary contour. Image below:

This is obviously not the curve I am looking for, so what am I missing?

答:
You can use a Laurent series to approximate the equipotentials (level sets) of the Mandelbrot set [1]:

$$ \psi(w) = w + \sum_{n=0}^{\infty} \frac{b_n}{w^n} $$

This is an explicit conformal map from the exterior of the unit disk to the exterior of the Mandelbrot set, with a recursive formula for the coefficients $b_n$.  See also OEIS 3, 4.

Setting $w = r e^{2 \pi i t}$ and letting $r \to 1$ from above, gives a parameterized family of curves $w_r(t)$ that might converge to the boundary of the Mandelbrot set (some things might obstruct this, I think convergence might depend on whether "all (including irrational) external rays land" or "Mandelbrot set is locally connected" or some such, I'm not sure..).

But, practically, it is very slow to converge, so other methods may be better.

[1]: http://www.mrob.com/pub/muency/laurentseries.html , which cites [2]

[2]: John H. Ewing and Glenn Schober, The area of the Mandelbrot set, Numer. Math., vol. 61 (1992), pages 59-72.


答:
Hint:

Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

$$\arctan x+\arctan(-1)=\begin{cases} \arctan\frac{x-1}{1-(-1)x} &\mbox{if } x\cdot(-1)<1\\ \pi+\arctan\frac{x-1}{1-(-1)x} & \mbox{if } x\cdot(-1)>1\\\text{sign}(x)\cdot\dfrac\pi2 & \mbox{if } x\cdot(-1)=1\end{cases} $$

Title: Difference of equal derivative functions value

Body:
$f_1(x) = \text{tan}^{-1}(x)$

$f_2(x) = \text{tan}^{-1}\big(\frac{x-1}{x+1}\big)$

Where

$f_1^\prime(x) = f_2^\prime(x)=\frac{1}{x^2+1}$

Therefore

$f_1(x) - f_2(x) = \theta$

How to find values of $\theta$ mathematically. I have solved it graphically and the answers are $\frac{\pi}{4}$ and $-\frac{3\pi}{4}$ in my opinion. Thanks in advance.


Body:
In $(-1,\infty)$ put $x =0$ to get $\theta= f_1(x)-f_2(x) =0-\tan^{-1} (-1)=0-(-\frac {\pi} 4)=\frac {\pi} 4$ .

In $(-\infty, -1) $ note that $\theta =\lim_{x \to -\infty} [f_1(x)-f_2(x)]=-\frac {\pi} 2- (\frac {\pi} 4)=-\frac {3\pi} 4$.


问:
How do you solve $x^2−4\equiv 0 \mod 7$?
How do you solve $x^2−4\equiv 0 \mod 7$?

答: The most efficient method here is that $x^2-4=(x-2)(x+2)$, thus it implies that $7$ divides $(x-2)(x+2)$ but because $7$ is a prime, it must divide either $x+2$ or $x-2$, therefore the solutions are $x=±2$ (mod $7$)


答:
As you've described, it's enough to find the volume of the convex hull $P_n$ of the points $(0, \ldots, 0), (1, 0, \ldots, 0), (0, 1, 0, \ldots, 0), (0, \ldots, 0, 1)$, which is sometimes called the standard $n$-simplex.

It says that this piece $P_n$ is a cone of height $1$ (how?) over a base, which is the analogous piece in $\Bbb R^{n−1}$, i.e. $P_{n−1}$ (how?).

If we think of $\Bbb R^{n - 1}$ embedded in $\Bbb R^n$ in the usual way (as the $(n - 1)$-plane $\{x_n = 0\}$, via the usual inclusion $$\iota : \Bbb R^{n - 1} \hookrightarrow \Bbb R^n, \qquad (x_1, \ldots, x_{n - 1}) \mapsto (x_1, \ldots, x_{n - 1}, 0) ,$$
then we can see that $P_n$ is the cone over the $(n - 1)$ simplex $\iota(P_{n - 1}) \subset \iota(\Bbb R^{n - 1})$ with vertex $(0, \ldots, 0, 1)$, i.e., it is a cone (in fact, pyramid) of height $1$ over $\iota(P_n)$.

If we take the above for granted and use that the volume of a cone in $\Bbb R^n$ is $\frac{B h}{n}$ (and how do we prove this?)

To prove the cone formula, suppose a cone has height $h$ and base $S$ of $(n - 1)$-dimensional volume $B$, and consider the cross-section of the cone at height $x_n$. That cross section is similar to $S$, and by linearity any of its length dimensions is $(1 - \frac{x_n}{h})$ that of the corresponding length of the base. So, the $(n - 1)$-dimensional volume of the cross-section is $\left(1 - \frac{x_n}{h}\right)^{n - 1} B$. Integrating over the $x_n$-coordinate gives that the volume of the cone is
$$\int_0^h \left(1 - \frac{x_n}{h}\right)^{n - 1} B \,dx_n = \frac{B h}{n}.$$

Title: How is the volume of a cross-polytope in $\mathbb{R}^n = \frac{2^n}{n!}$?

Body:
I'm fairly new to convex geometry, and I wish to prove that the volume of a cross-polytope in $\mathbb{R}^n$ is $\frac{2^n}{n!}$.

A cross-polytope is the convex-hull of $2n$ points in $\mathbb{R}^n$, namely, $(\pm 1,0,...,0), (0,\pm 1,0,...,0),...,(0,...,0,\pm 1)$. Since this is the same as the unit ball of $l_1$ norm in $\mathbb{R}^n$, we denote it by $B^n_1$.

$B^n_1$ is made up of $2^n$ pieces similar to the piece whose points are in the convex hull of $(\pm 1,0,...,0), (0,\pm 1,0,...,0),...,(0,...,0,\pm 1)$ and are all non-negative. So the total volume is $2^n$ times the volume of this piece (denoted by $P_n$ hereafter), which (from what I understand) has vertices $(1,0,...,0), (0, 1,0,...,0),...,(0,...,0, 1)$.

How do I find the volume of $P_n$? I have read an inductive proof that I don't quite understand. It says that this piece $P_n$ is a cone of height $1$ (how?) over a base, which is the analogous piece in $\mathbb{R}^{n-1}$, i.e. $P_{n-1}$ (how?).

If we take the above for granted and use that the volume of a cone in $\mathbb{R}^n$ is $\frac{Bh}{n}$ (and how do we prove this?), we'll get the desired result ($B$ = volume of base, $h$ = height of the cone) with the help of $$\operatorname{vol}(P_n) = \frac{\operatorname{vol}(P_{n-1})}{n}$$

A cone is defined as the convex hull of a single point and a convex body of dimension $\mathbb{R}^{n-1}$ in $\mathbb{R}^n$.

In general, I have found it difficult carrying forward results from $\mathbb{R}^3$ to $\mathbb{R}^n$ -  a lot of the stuff is not intuitive!

Edit 1:

Based on a comment by @Arthur, I have tried out integration to find the volume of $P_n$ as follows:
$$\int \int ... \int dx_1 dx_2 ... dx_n$$
We have $x_1 + x_2 + ... + x_n = 1$ and $x_i \ge 0 \forall i = 1,...,n$. So the limits for $x_1$ goes from $0$ to $1 - x_2 - x_3 - ... - x_n$
We get
$$\int \int ... \int (1 - x_2 - x_3 - ... - x_n) dx_2 ... dx_n$$

How do I go ahead from here? What are the limits for $x_2$? Is there a pattern I should see?


问:
Relationship between Partial Harmonic Sum and Logarithm.
Let $H_n$ denote the partial harmonic sum

$$H_n = \sum_{j=1}^n \frac{1}{j} \ \ .$$

I saw in some lecture notes that $H_n \approx \ln(n)$ without any explanation given. I tried to understand this approximation using Taylor series. Recall that the Taylor expansion of $\ln(x)$ for $x>0$ is

$$\ln(x) = 2 \sum_{j=1}^{\infty} \frac{ \bigg[ \displaystyle\frac{(x-1)}{(x+1)} \bigg]^{(2j-1)} }{2j-1} \ \ .$$

I tried to use this to obtain the connection, however I was unable to see the relationship between $H_n$ and $\ln(n)$.

I performed some crude numerical calculations and I'm not sure whether the approximation is correct because $H_n$ does not appear to be close to $\ln(n)$ for the values of $n$ that I have tried.

Can someone please help me understand this approximation or confirm my suspicions that there is an error in the notes.

If there is an error in the notes, does a relationship exist between the partial harmonic sum and the logarithm however?

答:
You may also use creative telescoping by noticing that $\frac{1}{k}$ is pretty close to $\log(k+1)-\log(k)=\log\left(1+\frac{1}{k}\right)$ for any large $k$. Indeed

$$ H_n -\log(n+1) = \sum_{k=1}^{n}\underbrace{\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right)}_{\Theta(1/k^2)}=O(1) $$
and the approximation can be further improved by noticing that $\frac{1}{k}-\log\left(1+\frac{1}{k}\right)$ is pretty close to $\log\left(\frac{12k+1}{12k-5}\right)-\log\left(\frac{12k+13}{12k+7}\right)$:

$$ H_n = \log(n+1)+\gamma-\sum_{k>n}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right) $$

$$ H_n = \log(n+1)+\gamma-\log\left(\frac{12n+13}{12n+7}\right)+\sum_{k>n}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)-\log\left(\frac{12k+1}{12k-5}\right)-\log\left(\frac{12k+13}{12k+7}\right)\right) $$

$$\boxed{H_n=\log(n+1)+\gamma+\log\left(\frac{12n+7}{12n+13}\right)+O\left(\frac{1}{n^3}\right)}$$

where $\gamma$ is the Euler-Mascheroni constant, corresponding to $\sum_{k\geq 1}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right)$.


问:
How to dynamically visualize $3$-spheres?
I am no mathematician, but I am really curious.

In my limited knowledge (for the sake of analogy, the following paragraph may sound strange), I know:

A point is $0$-sphere, it is the surface of $1$-ball (a segment); A circle is $1$-sphere, it is the surface of $2$-ball (a disk); A sphere is $2$-sphere, it is the surface of $3$-ball (a ball); A $3$-phere is the surface of $4$-ball. An $n$-sphere is a set of points with distance from a point equal to a length in $n+1$ dimensions, it has finite $n$-area and borderless in $n$ dimensions, infinity to the power of $n$ (I know it's still the same...thing) axes of symmetry, and you can travel in any direction tangent to it in $n$ dimensions without changing direction forever. $n$-sphere is the intersection of $(n+1)$-sphere and $(n+1)$-plane, as the $(n+1)$-sphere moves through the $(n+1)$-plane, $n$-sphere's radius starts at $0$, increases until reaching radius of $(n+1)$-sphere and decreases until reaching $0$ again(and after that becomes non-existent).

And that's all I know. I had found this image on Wikipedia, and I really can't quite understand it:

My brain really isn't evolved to deal with higher dimensions, will somebody please make some gif files to dynamically visualize how $3$-spheres rotate, and explain it like I am five? And please use perspective projection.

答:
You can look at this nice video of the Hopf fibration to visualize a sphere in 4 dimension.

https://www.youtube.com/watch?v=AKotMPGFJYk

Of course to understand it you need some knowledge of what a bundle is, but see it in this way: in the video you see different parts of the 4-sphere.


问:
Integration by parts disconnect
I'm trying to integrate $\displaystyle E(Y^2) = \int^\infty_0 y^2\lambda e^{-\lambda y} dy$
doing it by parts this is my logic.

$\displaystyle E(Y^2) = \int^\infty_0 y^2\lambda e^{-\lambda y} dy$ where $u=y^2$, $du=2y\,dy$, $dv=\lambda e^{-\lambda y}\, dy$ and $v = -e^{-\lambda y}$ solving by parts we get $\displaystyle (y^2)(-e^{-\lambda y})-\int^\infty_0 -e^{-\lambda y} 2y \,dy$ which then simplifies to $\displaystyle 0 + 2\int^\infty_0y e^{-\lambda y} dy$ which is the same as $0 + 2 E(Y) = \dfrac{2}{\lambda}$ where $E(Y)=\dfrac{1}{\lambda}$

however the solutions shows a different answer of $\dfrac{2}{\lambda^2}$ can someone explain why this is?
http://oi40.tinypic.com/b5fwxs.jpg

答:
Integrating by parts, take $u= y^2 \Rightarrow du= 2y \cdot dy$ and $dv=e^{-\lambda y} \Rightarrow v = \dfrac{e^{-\lambda x}}{-\lambda}.$

$$\begin{align}
\int y^2\lambda e^{-\lambda y} dy & = \lambda \int y^2 e^{-\lambda y} dy \\
& = \lambda \left [\frac{y^2e^{-\lambda y}}{-\lambda} - \int \frac{2y e^{-\lambda y}}{(-\lambda)} dy\right] \\
& = \lambda \left [\frac{-y^2e^{-\lambda y}}{\lambda} + \color{red}{\frac{2}{\lambda}\int ye^{-\lambda y} dy}\right]. \\
\end{align}$$

Integrating the red colored by parts again, let $u= y \Rightarrow du= dy$ and $dv=e^{-\lambda y} \Rightarrow v = \dfrac{e^{-\lambda y}}{-\lambda}.$
$$\begin{align}
\color{red}{\frac{2}{\lambda}\int ye^{-\lambda y} dy}
& = \frac{2}{\lambda} \left[\frac{ye^{-\lambda y}}{-\lambda} - \int \dfrac{e^{-\lambda y}}{(-\lambda)} dy \right] \\
& = \frac{2}{\lambda} \left[\frac{-ye^{-\lambda y}}{\lambda} + \int \dfrac{e^{-\lambda y}}{\lambda} dy \right] \\
& = \frac{2}{\lambda} \left[\frac{-ye^{-\lambda y}}{\lambda} + \dfrac{e^{-\lambda y}}{(-\lambda)\lambda} dy \right] \\
& = \color{blue}{\frac{2}{\lambda} \left[\frac{-ye^{-\lambda y}}{\lambda} - \dfrac{e^{-\lambda y}}{\lambda^2} \right]}.
\end{align}$$

Put the blue part back, we have
$$\begin{align}
\int y^2\lambda e^{-\lambda y} dy
& = \lambda \left [\frac{-y^2e^{-\lambda y}}{\lambda} + \color{red}{\frac{2}{\lambda}\int ye^{-\lambda y} dy}\right] \\
& = \lambda \left [\frac{-y^2e^{-\lambda y}}{\lambda} + \color{blue}{\frac{2}{\lambda} \left[\frac{-ye^{-\lambda y}}{\lambda} - \dfrac{e^{-\lambda y}}{\lambda^2} \right]}\right] \\
& = e^{-\lambda y} \left(-y^2 - \frac{2y}{\lambda} - \frac{2}{\lambda^2} \right).
\end{align}$$

Therefore
$$\begin{align}
\int_0^\infty y^2\lambda e^{-\lambda y} dy
& = \left[e^{-\lambda y} \left(-y^2 -\frac{2y}{\lambda} - \frac{2}{\lambda^2} \right) \right]_0^\infty \\
& = 0 - \left(- \dfrac{2}{\lambda^2}\right) \\
& = \dfrac{2}{\lambda^2}.
\end{align}$$

Title: Integration by parts disconnect


问:
If $A$ is bounded then the outer measure of $A$ is finite
I got entangled with the following question:

I need to prove or disprove:

Let $A \subseteq \mathbb R$

If $A$ is bounded then $m^*(A) \lt {\infty }$

If $m^*(A) \lt {\infty }$ then $A$ is bounded

答:
The first statement is true (I will leave the writing of the proof to you), assuming $m^*$ means the Lebesgue outer measure on $\mathbb{R}$. Note that:

$A$ is bounded (i.e. we have $A\subseteq [-M,M]$ for some $M>0$).

For closed intervals $I=[a,b]$ (which are measurable) we have $m^*(I)=\ell(I)=b-a$.

The outer measure is monotone.

Conversely, the second statement is not true. For instance, let $A=\mathbb{Q}$ (again, fill in the gaps).


问:
How to check symmetric space is locally symmetric space?
Definition Riemannian manifold $\mathcal M$ is symmetric if $\forall x\in \mathcal M$, there exist an isometric $\varphi: \mathcal M \to \mathcal M$, such that for any $\gamma$  geodesic satisfying $\gamma (0)=x$, $\forall t \in \text{Dom}(\gamma) $, we have $\varphi(\gamma(t))= \gamma(-t)$.

Definition Let $R$ be the curvatural tensor of Riemannian manifold $\mathcal{M}$, we say $\mathcal M$ is loally symmetric if $DR=0$.

And locally symmetric space has an equivalent description as follows: for any piecewisely smooth curve (needn't geodesic) $\gamma:[0,1] \to \mathcal M$, and any unit orthogonal tangent vectors $e_1, e_2 \in T_{\gamma(0)} \mathcal M$, the sectional curvature $K(e_1,e_2) = K(P_\gamma (e_1), P_\gamma (e_2))$, where $P_\gamma $ is the parallel movement along $\gamma$.

Then how can we check that if $\mathcal M$ is symmetric space, then it is locally symmetric space?

答:
Suppose $\mathcal{M}$ is symmetric and $x \in \mathcal{M}$. Consider $s_x$ the symmetry at $x$, that is
$$
s_x(\exp_x(tv)) = \exp_x(-tv)
$$
It is an isometry by definition of symmetric space. Thus, $R$ is invariant under the action of $s_x$,. Now, one can show this general lemma: every $(r,s)$ tensor with $r+s$ odd that is invariant under $s_x$, is zero at $x$.

Apply this lemma to $\nabla R$, this will show that $(\nabla R)_x = 0$.


问:
Prove the edges of a multigraph may be oriented such that the net-degree of any vertex is $\leq 1$.
The net-degree of a vertex $v$, denoted $\text{netdeg}(v)$, in a digraph $G$ is defined by

$$ \text{netdeg}(v)=| ~ \text{outdeg}(v) - \text{indeg}(v) ~| $$

where $\text{outdeg}(v)$ and $\text{indeg}(v)$ are the out-degree and in-degree of $v$.

Show that the edges of any multigraph may be oriented so that $\text{netdeg}(v) \leq 1$ for all its vertices $v$.

My first thought was to order the vertices and alternate edges in and out beginning at the first vertex then move on to the next vertex and orient the remaining edges depending on the net-degree of the remaining vertices, but this approach does not seem to work.

答: For any graph $G$, perhaps you could add an auxiliary vertex $x$ and make it adjacent to all vertices of odd degree. Call this new graph $G'$. Since there must be an even number of such vertices (so as to keep the degree sum even), $G'$ has an Eulerian circuit. Assign the orientation to $G'$ realizing the circuit starting from $x$. Removing $x$ yields an orientation for $G$ where for each vertex $v\in V(G)$, $netdeg(v)$ is 0 if $deg(v)$ is even and $netdeg(v)$ is 1 if $deg(v)$ is odd.</p>


问:
Calculate integral of exponential to the x suared (Simple Gaussian Integral)
For: $\int_{-1}^{1}{\frac{1}{\sqrt{2\pi}} e^{\frac{-x^{2}}{2}}}dx $

A method I saw (but did not get the right result) was to set:

$I = \int_{-1}^{1}{\frac{1}{\sqrt{2\pi}} e^{\frac{-x^{2}}{2}}}dx $

and

$I = \int_{-1}^{1}{\frac{1}{\sqrt{2\pi}} e^{\frac{-y^{2}}{2}}}dx $

then by calculating  $I^2 $ I could make use of polar coordinates since $x^2+y^2=r^2$ and $dxdy=rdxd\theta$

Any other process I may use instead? I just need to be pointed in the right direction

答: The polar coordinates are useful only to calculate the integral: $$\int_{-\infty}^{\infty}{\frac{1}{\sqrt{2\pi}} e^{\frac{-x^{2}}{2}}}dx $$ You cannot calculate the integral $I$ exactly,  because the primitive of the gaussian can't be expressed in terms of elementary functions. You can read here:The easiest way to evaluate Gaussian integral


问:
What is the greatest number of points in the plane such that the distance between any two of them is an odd integer?
Suppose the origin is one of the points, we can take $(3,0)$ as the second point. If $(x,y)$ is some other point in the set I think we can use the fact that $x^2+y^2$, for $x$ and $y$ odd, is never an integer. There is also a lot of lines that the points certainly cannot belong to. But I could not go beyond.

答:
The answer is $3$.

Assume the contrary, let's say we can find four points $O,A,B,C$ such that following six distances are all odd integers:
$$(a,b,c,a_1,b_1,c_1) = (BC,CA,AB,OA,OB,OC)$$

Since $O, A, B, C$ lies in the plane, they form a degenerate tetrahedron with volume $V = 0$. Express $V$ in terms of corresponding Cayley Menger determinant,
the distances satisfy

$$\left|\begin{matrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & a_1^2 & b_1^2 & c_1^2\\
1 & a_1^2 & 0 & c^2 & b^2 \\
1 & b_1^2 & c^2 & 0 & a^2\\
1 & c_1^2 & b^2 & a^2 & 0 \\
\end{matrix}\right| = 288V^2 = 0\tag{*1}$$

Recall if $u$ is an odd integer, then $u^2 \equiv 1 \pmod 8$.

Taking modulo $8$ on both sides of $(*1)$, we arrive at a contradiction.

$$4 = \left|\begin{matrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & 1 & 1 & 1\\
1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 & 1\\
1 & 1 & 1 & 1 & 0 \\
\end{matrix}\right| \equiv 0 \pmod 8$$

This means it is impossible to find four points in the plane whose pairwise distances are all odd integers.

Since it is trivial to find three points at odd integral distances from each other (eg. the vertices of an equilateral triangle of side $1$), the greatest number we seek is $3$.


问:
How to make my algorithm work for case: irrational number $\sqrt{2}$ so that $\left | \frac{p}{q}- \sqrt{2} \right |< 0.001$
On one day, I read a magazine then I'm so interested in the algorithms, one of my favorites is RATCONVERT, i.e., if you have $\dfrac{142}{727}\cong 0.195323246..$ then how do you find $0.195323246.. \cong\dfrac{142}{727},$ the generalization of this problem here:

Given two natural numbers $K, M.$ Find two numbers $u, v$ so that $0\leq u, v< N, \left | \dfrac{u}{v}- \dfrac{K}{M} \right |\leq \dfrac{1}{M}$ with a natural numbers given $N.$

In 1981 Paul Wang had researched this problem in case $N= \sqrt{\dfrac{M}{2}}.$ With that we can prove that there only one solution exists, that's true, assume that there are two solutions $\left ( u_{1}, v_{1} \right ), \left ( u_{2}, v_{2} \right )$ being right by the above conditions, we observe that:
$$\left | \frac{u_{1}}{v_{1}}- \frac{u_{2}}{v_{2}} \right |\leq \left | \frac{u_{1}}{v_{1}}- \frac{K}{M} \right |+ \left | \frac{u_{2}}{v_{2}}- \frac{K}{M} \right |\leq \frac{2}{M}$$
$$\Rightarrow \left | u_{1}v_{2}- u_{2}v_{1} \right |\leq \frac{2v_{1}v_{2}}{M}< \frac{2N^{2}}{M}\leq 1\Rightarrow \frac{u_{1}}{v_{1}}= \frac{u_{2}}{v_{2}}$$
On the other hand,
$$\left | \frac{u}{v}- \frac{K}{M} \right |\leq \frac{1}{M}\Rightarrow \left | Mu- Kv \right |\leq v< N$$
then if $\dfrac{u}{v}\cong \dfrac{K}{M},$ there exists an integer $r$ so that $\left | r \right |< N, r\equiv Kv\mod M.$ I generalized it and wrote it as an algorithm so that there must exist two conditions $0\leq \left | r \right |< \sqrt{M/2}> v> 0$ and $r\equiv Kv\mod M$ to find $\left ( v, r \right ),$ that's easier. So how to make it work for finding an approximately fraction $\dfrac{p}{q}$ of irrational number $\sqrt{2}$ so that

$$\left | \frac{p}{q}- \sqrt{2} \right |< 0.001$$

I remember Lagrange states that any positive quadratic surd $\sqrt{a}$ has a regular continued fraction which is periodic after some point. There is a few of relationship between two problems, so I need to your help to complete this. Thanks a real lot !

答:
Your algorithm is applicable to approximate rationals by rationals, but it doesn’t applicable directly to approximate reals by rationals. On the other hand, best rational approximations of a given a real number are provided by its simple continued fraction. Moreover, for rational approximants, listed here, we see that $\left | \tfrac {41}{29}
- \sqrt{2} \right |< 4.21\cdot 10^{-4}$.


问:
Automorphism of $\mathbb Q (\sqrt[5]{2}, \zeta_5)$

Show that if $\sigma \in \mathrm{Aut}(\mathbb Q (\sqrt[5]{2}, \zeta_5)/\mathbb Q)$, then $\sigma (\zeta_5) = \zeta^i_5$ for some $i=1,2,3,4$ and $\sigma (\sqrt[5]{2}) = \zeta^j_5 \sqrt[5]{2}$ for some $j=1,2,3,4,5$.

Help please.

答:
Hint:

$(\zeta_5)^4+(\zeta_5)^3+(\zeta_5)^2+(\zeta_5)+1=0$

$(\sqrt[5]{2})^5-2=0$

For any $\sigma$, we have $\sigma(a)=a$ for all $a\in\mathbb{Q}$ (by definition, elements of $\mathrm{Aut}(L/K)$ are automorphisms of $L$ that fix $K$).


问:
If $a^2 + b^2 +c^2 =4$ and $a^3 + b^3 + c^3 = 8$ then find $a^4+b^4+c^4$
This is a popular question, but I can find $a^4+b^4+c^4$ if I know $a+b+c, a^2+b^2+c^2, a^3 +b^3 +c^3$ or in special case $a+b+c=0$ we only need a more condition like $a+b+c$. I have tried in many times and can't find out the way to archive it.

答:
A somewhat simpler solution is given by Lagrange multipliers. The stationary points of $f(a,b,c)=a^3+b^3+c^3$ over $a^2+b^2+c^2=4$ occur where $3(a^2,b^2,c^2)=2\lambda(a,b,c)$, i.e. at $\pm\left(\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}\right)$ and at the permutations of $\pm(0,0,2)$ and $\pm(0,\sqrt{2},\sqrt{2})$. In particular
$$ \max_{a^2+b^2+c^2}a^3+b^3+c^3 = 8 $$
and in a point of absolute maximum we have $a^4+b^4+c^4=16$.


问:
Direct sums of co-hopfian modules
I recently saw this question If $G \bigoplus H$ is isomorphic to a proper subgroup of itself, then must the same be true of one of $G$ and $H$?, which turned out to apparently be an open problem. (There s a reference given in the comments.) We can ask a similar question about $R$-modules for $R$ not equal to $\mathbb{Z}$... a $R$-module is called "co-hopfian" if it is not isomorphic to a proper submodule of itself, so the question is whether a direct sum of two co-hopfian modules can be non-co-hopfian. Are there rings $\neq\mathbb{Z}$ over which the answer to this question turns out to be easier?

答:
I won't give a ring that answers the question, but I'll show that there is a ring with two co-Hopfian modules whose direct sum is not co-Hopfian. It wouldn't surprise me if somebody came up with an explicit example that makes this answer look a bit silly.

For the dual problem, whether the direct sum of two Hopfian abelian groups can be non-Hopfian, Corner produced an example in 1965:

Corner, A. L. S., Three examples on Hopficity in torsion-free Abelian groups, Acta Math. Acad. Sci. Hung. 16, 303-310 (1965). ZBL0145.03302.

Let $G$ and $H$ be the groups in Corner's example. There is a small abelian subcategory $\mathcal{C}$ of the category of abelian groups that contains $G$ and $H$.

By the Freyd-Mitchell embedding theorem, there is a full exact embedding $F:\mathcal{C}^{\text{op}}\to R\text{-Mod}$ of the opposite category of $\mathcal{C}$ into the module category of some ring $R$.

Then $F(G)$ and $F(H)$ are co-Hopfian $R$-modules, but $F(G\oplus H)$ is not co-Hopfian.

Title: Direct sums of co-hopfian modules


问:
How to show that $\frac{a + b}{\gcd(a,b)^2}$ is a Fibonacci number when $ \frac{a+1}{b}+\frac{b+1}{a}$ is an integer?

Let $a, b$ be positive integers such that the number $ \dfrac{a + 1}{b} + \dfrac{b + 1}{a}$ is also integer. Then, show that $\dfrac{a + b}{\gcd{(a, b)^{2}}}$ is a Fibonacci number.

I prove that : $ \displaystyle{\forall a, b \in \mathbb{N}} $ such that $ \displaystyle{ k = \frac{a + 1}{b} + \frac{b + 1}{a} \in \mathbb{N} \Rightarrow k = 3 }$ or $ \displaystyle{ k = 4 } $. Also, we have that (I use the method Vieta Jumping) :

$ \displaystyle{ \mathbb{S} = \Big \{ \big( a, b \big) \in \mathbb{N} \times \mathbb{N} : \frac{a + 1}{b} + \frac{b + 1}{a} = k \in \mathbb{N} \Big \} = \Big \{ \big( a, b \big) \in \mathbb{N} \times \mathbb{N} : \frac{a + 1}{b} + \frac{b + 1}{a} = 3 \text{  } or \text{  } \frac{a + 1}{b} + \frac{b + 1}{a} = 4 \Big \} } $

Fibonacci sequence : $\displaystyle{f_{0} = 0, f_{1} = 1}$ and $\displaystyle{f_{n+2} = f_{n+1} + f_{n}, \forall n \in \mathbb{N} \cup \{ 0 \} = \{ 0, 1, 2, 3, . . . \}}$

$ \displaystyle{ k = 3 } $ : If $ \displaystyle{ \big( a, b \big) \in \mathbb{S} } $ with $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 3 } $, then $ \displaystyle{ \big( 3 \cdot a - 1 - b, a \big) \in \mathbb{S} } $ so we have $ \displaystyle{ \big( 2, 2 \big), \big( 3 = f_{3} + 1, 2 \big), \big( 6 = f_{5} + 1, 3 \big), \big( 14 = f_{7} + 1, 6 \big), \big( 35 = f_{9} + 1, 14 \big), . . . \in \mathbb{S} } $ (I checked some couples and I found that the request is valid) but I can not porve that $ \displaystyle{ \frac{a + b}{\gcd{\big(a, b \big)^{2}}} } $ is a Fibonacci number.

$ \displaystyle{ k = 4 } $ : If $ \displaystyle{ \big( a, b \big) \in \mathbb{S} } $ with $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 4 } $, then $ \displaystyle{ \big( 4 \cdot a - 1 - b, a \big) \in \mathbb{S} } $ so we have $ \displaystyle{ \big( 1, 1 \big), \big( 2, 1 \big), \big( 6, 2 \big), \big( 21, 6 \big), \big( 77, 21 \big), . . . \in \mathbb{S} } $ (I checked some couples and I found that the request is valid) but also I can not porve that $ \displaystyle{ \frac{a + b}{\gcd{\big(a, b \big)^{2}}} } $ is a Fibonacci number.

I need some help, because I can not solve the problem. So my questions are :

if $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 3 } $ where $ \displaystyle{ a = f_{2 \cdot n - 1} + 1, b = f_{2 \cdot n + 1} + 1 \in \mathbb{N}, \forall n \geqslant 2, \forall n \in \mathbb{N} } $, how I can show that $ \displaystyle{ \frac{a + b}{\gcd{\big( a, b \big)^{2}}} } $ is a Fibonacci number ?

and

if $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 4} $ where $ \displaystyle{ a, b \in \mathbb{N} } $, how I can show that $ \displaystyle{ \frac{a + b}{\gcd{\big( a, b \big)^{2}}} } $ is a Fibonacci number ?

答:
First of all we assume that $gcd(a,b)=d $ so we may assume that there exist coprime integers $x,y$ such that $a=dx , b=dy$ then from the hypothesis of the problem: $ab|a^2+b^2+a+b
\rightarrow ab|(a+b)(a+b+1)
\rightarrow dxy|(x+y)(dx+dy+1)$
because $d$ is coprime to $dx+dy+1$ , $d|x+y$ (first result) and because $x,y$ are coprime it is easy to conclude that $xy$ is coprime to $x+y$ and therefore $xy|dx+dy+1$ (second result) now instead of the first result, assume that there is a positive integer $t$ such that $y=td-x$ then the second result will be $x(td-x)|td^2+1 $.
Now assume a positive integer $k$ such that $td^2+1=ktdx-kx^2$ which is the equation $kx^2-ktdx+(td^2+1)=0$ and by multiplying both sides by $4k$, $4k^2x^2-4k^2tdx+(4ktd^2+4k)=0$ and this is actually $(2kx-ktd)^2=(k^2t^2d^2-4k-4ktd^2)        $.
From this result, the existance of such $x$ is clearly equivalent to these two:
i) $k^2t^2d^2-4k-4ktd^2$(we'll call this $r$) is a perfect square
ii) $2k|\sqrt{r}+ktd$. Now just let's say that we choose $k$ to be odd. Then this condition will be changed to the following two conditions:

$2|\sqrt{r}+ktd$ which is clear because $\sqrt{r}$ has the same parity with $r$ and $r$ has the same parity with $k^2t^2d^2$ so actually $\sqrt{r}$ has the same parity with $k^2t^2d^2$ and thus, $\sqrt{r}+ktd$ is even because $ktd+k^2t^2d^2$ is even, so the first condition is satisfied.

$k|\sqrt{r}+ktd$ iff $k|\sqrt{r}$ iff $k^2|r$ iff $k^2|4k+4ktd^2$ iff $k|4+4td^2$ but because we've chosen $k$ to be odd, that's equivalent to $k|td^2+1$
So, summing up; if we choose $s$ to be a number that satisfies $td^2+1=ks$, it's enough to prove that $r=4k^2s-k^2t^2d^2$ is a perfect square, which is again equivalent to $4s-t^2d^2$ being a perfect square.
So, for avoiding any confusion here I restate the new problem:
We have positive integers that satisfy $td^2+1=ks$ and $4s-t^2d^2$ is a perfect square and want to prove that $t$ is a fibbonachi number.
Now $k$ doesn't have a role anymore so instead of $td^2+1=ks$, we say that $s|td^2+1$.
Now again assume $u$ to be a number that $4s-t^2d^2=u^2$.
Mixing the two hypotheses, we'll have $u^2+t^2d^2=4s|4td^2+4$.Now because RHS shall be bigger than or equal to the LHS, $4td^2+4\leqslant u^2+t^2d^2$.
Now considering $t$ as our variable,
$t^2d^2-4td^2+(u^2-4)$ is less than or equal to zero.
On the other hand that is actually $4+4d^2-u^2\leqslant(t-2)^2d^2$.
Now two important notes here:

$d\not=1$ because then the problem's original statement leads to a contradiction so I assume it meant $d\not=1$.

$u\leqslant2$:
If $u=0$ , then back to the $u^2+t^2d^2|4td^2+4$, $t|t^2d^2|4td^2+4$ and thus $t|4$ but if $t=4$ then $16d^2|16d^2+4$ a contradiction. Otherwise $t=1or2$ which are both fibbonachi numbers.
If $u=1$, back to the definition of $u$, $4s=t^2d^2+1$ which is a contradiction $mod4$. So $u\leqslant2$.
Now back to our inequality, $4+4d^2-u^2\leqslant(t-2)^2d^2$. This is equivalent to $sqrt{\frac{4+4d^2-u^2}{d^2}}+2\leqslant t$
Now by the second note we just mentioned, the LHS is maximized when u=2 , and so clearly $2\leqslant t$ and so $t$ is a fibbonachi number.
In the same way you can consider that $k$ is even and get to the numbers {3,5} and so all of the possible values of $t$ are {1,2,3,5} which are all fibbonachi numbers.


问:
Why does the following limit exist?
I was doing a problem regarding infinite sums of functions, but I don't understand one thing. If we define the following functions:
$$ f_n(x)=\sin^n(x)$$
And the following inifnite series:
$$ \sum_{n=0}^\infty f_n(x)=\sum_{n=0}^\infty \sin^n(x)=\frac{1}{1-\sin x}=f(x) $$
The question asks to say whether the following limit exists or not:

$$ \lim_{x\to\frac\pi2} f(x)= \lim_{x\to\frac\pi2} \frac{1}{1-\sin x} $$

To me it would seem that the limit shouldn't exist as the function goes to inifinity and/or the series diverges and so doesn't converge to a function, so its limit shouldn't exist, however the answer says it does. How can this be?

答:
Whenever $|\sin(x)| < 1$, the geometric series $\sum_{n=0}^\infty (\sin(x))^n$ will always be converging, and its limit will indeed be $\frac{1}{1-\sin(x)}$. However, if $\sin(x)=1$ then the geometric series diverges (it is just an infinite sum of ones!).

This, in particular, means that $f(x)$ exists and is continuous in $[0,1)$, so you can do the limit... which is, of course, $+\infty$ (this can be proved explicitly with an $\epsilon$-$\delta$ argument, or perhaps with a clever inequality as well)

Title: Why does the following limit exist?


问:
Minimize Boolean function
I have got some silly task, but I am quite confused.

Need to minimize function.

$$f(x_1,x_2,x_3,x_4)=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4.$$

Thanks.

Sorry for my English. Minimize Boolean function

答: Type $$\rm minimize\ boolean$$ into a search engine and you'll be directed to lots of webpages that may help. Another keyphrase is $$\rm Karnaugh\ map$$ This will lead you to http://en.wikipedia.org/wiki/Karnaugh_map among other places. </p>

Title: Minimize Boolean function


问:
Does $\,x>0\,$ hint that $\,x\in\mathbb R\,$?

Does $x>0$ suggest that $x\in\mathbb R$?

For numbers not in $\,\mathbb R\,$ (e.g. in $\mathbb C\setminus \mathbb R$), their sizes can't be compared.

So can I omit "$\,x\in\mathbb R\,$" and just write $\,x>0\,$?

Thank you.

答: One might be able to decode that notation, but why force this puzzle on the reader?  </p>


问:
Remainder term for geometric series
Consider the Taylor series (about $0$) for the function $\dfrac{1}{1+t^2}$. We first consider the function:

$$f(x)=\dfrac{1}{1+x}=(1+x)^{-1}$$

We find that  $f^n(x)=(-1)^nn!(1+x)^{-(1+n)}$. The nth order Taylor polynomial $P_n(x)$is thus $\sum_{k=0}^n(-x)^k$ , and the remainder term $R_n(x)$  is  $ \dfrac{f^{n+1}(c)x^{n+1}}{n+1!}$= $\dfrac{(-x)^{n+1}}{(1+c)^{n+2}}$, for some $c$ in $(0,x).$

We therefore have, by Taylor's theorem:

$$f(x)=\dfrac{1}{1+x}=\sum_{k=0}^n(-x)^k+\dfrac{(-x)^{n+1}}{(1+c)^{n+2}}$$
The substitution $x=t^2$ results in:
$$\dfrac{1}{1+t^2}=\sum_{k=0}^n(-t^2)^k+\dfrac{(-t^2)^{n+1}}{(1+c)^{n+2}}$$
However, the top answer of: Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?, writes :
$$\dfrac{1}{1+t^2}=\sum_{k=0}^n(-t^2)^k+\dfrac{(-t^2)^{n+1}}{1+t^2}$$.

I am struggling to see how can we use the fact that $1<1+c<1+t^2$ ,(the only information we have about c), to reach this result, starting from $\dfrac{(-t^2)^{n+1}}{(1+c)^{n+2}}$.

答:
This is because it comes from a high-school identity, independent from Taylor's formula: from the factorisation formula $\;1-x^{n+1}=(1-x)(1+x+x^2+\dots +x^n)$, you can deduce instantly that
$$\frac1{1-x}=1+x+x^2+\dots +x^n+\frac{x^{n+1}}{1-x},$$
whence, substituting $x$ with $-t^2$,
$$\frac1{1+t^2}=1-t^2+t^4+\dots +(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}.$$

Title: Remainder term for geometric series


问:
The derivative of $\sin(\alpha x)/ \sin(x)$ is continuous at $0$?
Let $\alpha$ be a real number. Define $f = \sin(\alpha x)/ \sin(x)$ if $x \neq 0$ and $\alpha$ at $x = 0.$ How can I prove the derivative of $\sin(\alpha x)/ \sin(x)$ is continuous at $0$? I took the derivative and tried to apply L'Hopital's rule but I kept getting $0/0$ and could not make the argument work... Any explanation is appreciated.

答: The derivative at $0$ is$$\lim_{h\to0}\frac{\frac{\sin(\alpha h)}{\sin h}-\alpha}{h}=\alpha\lim_{h\to0}\frac{\frac{1-\tfrac16\alpha^2h^2+o(h^2)}{1-\tfrac16h^2+o(h^2)}-1}{h}=\alpha\lim_{h\to0}(\tfrac16(1-\alpha^2)h+o(h))=0.$$The derivative at some small but nonzero $x$ is$$\begin{align}&\frac{\alpha\cos\alpha x\sin x-\sin\alpha x\cos x}{\sin^2x}\\&\in\frac{\alpha(1-\tfrac12\alpha^2x^2+o(x^2))(x-\tfrac16x^3+o(x^3))-(\alpha x-\tfrac16\alpha^3x^3+o(x^3))(1-\tfrac12x^2+o(x^2))}{x^2+o(x^3)}\\&\in\alpha(1-\alpha^2)(\tfrac13x+o(x)),\end{align}$$which $\to0$ as $x\to0$. Therefore, the derivative is continuous at $x=0$.


答:
It really depends on context. But be safe; just say $x > 0, x\in \mathbb R$.

Omitting the clarification can lead to misunderstanding it. Including the clarification takes up less than a centimeter of space. Benefits of clarifying the domain greatly outweigh the consequences of omitting the clarification.

Besides one might want to know about rationals greater than $0$, or integers greater than $0$, and we would like to use $x \gt 0$ in those contexts, as well.

ADDED NOTE: That doesn't mean that after having clarified the context, and/or defined the domain, you should still use the qualification "$x\in \mathbb R$" every time you subsequently write $x \gt 0$, in a proof, for example. But if there's any question in your mind about whether or not to include it, error on the side of inclusion.

Title: Does $\,x>0\,$ hint that $\,x\in\mathbb R\,$?

Body:
Does $x>0$ suggest that $x\in\mathbb R$?

For numbers not in $\,\mathbb R\,$ (e.g. in $\mathbb C\setminus \mathbb R$), their sizes can't be compared.

So can I omit "$\,x\in\mathbb R\,$" and just write $\,x>0\,$?

Thank you.


问:
$P(\text{duck})$ given $P(\text{look})$ and $P(\text{swim})$ and $P(\text{quack})$

"If it looks like a duck, swims like a duck, and quacks like a duck,
then it probably is a duck"

We have the 3 probabilities:

  $P(\text{look}) = 80\%$

  $P(\text{swim}) = 70\%$

  $P(\text{quack}) = 90\%$

I would assume then that
$P(\text{duck}) = P(\text{look})·P(\text{swim})·P(\text{quack})$ $= 0.8 · 0.7 · 0.9$ $= 0.504$

This is obviously dead wrong. Common sense expects the resulted probability close to $1$. What would be right operation among the individual probabilities to represent the original saying correctly?

Update:

Giving some context: We have a problem in an industrial environment, where we have to validate if our calculated values are true or not. We can conduct a few independent measurements, each addresses a different aspect of the same setup and each measurment returns a yes or no with a probability. Then we have to summarize them and compare to the theoretical calculations. What we do is duck typing.

Update 2:

While reading up the answers did hit me: the more independent measurement events we do on properties of the duck, should prove or disprove the "duckness" of the object.
Should we have 6 independent measurements (look, swim, quack, fly, eat, walk), each around probability of 0.1, our calculation should conclude that this is certainly not a duck. Should we get confirming measurements (higher than 0.5), I expect higher "duckness" probability.

答:
The updates to the question imply that what you really want to do is classify an object based on independent measurements of the object.  The solution to that problem is known as the Naive Bayes classifier.  The math is similar to heropup's answer but without the constraint $P(L ~|~ D) = 1$.  Instead we start by writing Bayes' theorem in a convenient way:

$$
\frac{P(D ~|~ L)}{P(\neg D ~|~ L)} = \frac{P(D)}{P(\neg D)} \frac{P(L ~|~ D)}{P(L ~|~ \neg D)}
$$

This extends easily to multiple independent measurements:

$$
\frac{P(D ~|~ L,S,Q)}{P(\neg D ~|~ L,S,Q)} = \frac{P(D)}{P(\neg D)} \frac{P(L ~|~ D)}{P(L ~|~ \neg D)} \frac{P(S ~|~ D)}{P(S ~|~ \neg D)} \frac{P(Q ~|~ D)}{P(Q ~|~ \neg D)}
$$

Now use the first formula to simplify the second:

$$
\frac{P(D ~|~ L,S,Q)}{P(\neg D ~|~ L,S,Q)} = \left(\frac{P(D)}{P(\neg D)}\right)^{-2} \frac{P(D ~|~ L)}{P(\neg D ~|~ L)} \frac{P(D ~|~ S)}{P(\neg D ~|~ S)} \frac{P(D ~|~ Q)}{P(\neg D ~|~ Q)}
$$

Plugging in $P(D ~|~ L) = 0.8$ and so on gives:

$$
\frac{P(D ~|~ L,S,Q)}{P(\neg D ~|~ L,S,Q)} = \left(\frac{P(D)}{P(\neg D)}\right)^{-2}
\frac{0.8}{1-0.8} \frac{0.7}{1-0.7} \frac{0.9}{1-0.9}
$$

Even with the independence assumption, we cannot answer the question without $P(D)$.  For simplicity, I will take $P(D) = 0.5$, giving

$$
\frac{P(D ~|~ L,S,Q)}{P(\neg D ~|~ L,S,Q)} = 84
\\
P(D ~|~ L,S,Q) = \frac{84}{84 + 1} = 0.9882
$$

This approach has the intuitive quality that measurements with probability >0.5 will increase your confidence in $D$, measurements with probability <0.5 will decrease your confidence in $D$, and measurements with probability exactly 0.5 will have no effect (since this is the same as the prior probability $P(D)$ that I assumed).


答:
Both $\operatorname{sinc}(x)$ and its reciprocal are analytic, non-vanishing functions in a neigbourhood of zero, so it is
$$ \alpha\operatorname{sinc}(\alpha x)\frac{1}{\operatorname{sinc}(x)}. $$

Title: The derivative of $\sin(\alpha x)/ \sin(x)$ is continuous at $0$?

Body: Let $\alpha$ be a real number. Define $f = \sin(\alpha x)/ \sin(x)$ if $x \neq 0$ and $\alpha$ at $x = 0.$ How can I prove the derivative of $\sin(\alpha x)/ \sin(x)$ is continuous at $0$? I took the derivative and tried to apply L'Hopital's rule but I kept getting $0/0$ and could not make the argument work... Any explanation is appreciated.


问:
Non-empty intersection.
Let $A_1,A_2,...,A_{2019}$ be sets s.t. $|A_i|$=45 for each $i=\overline{1,2019}$ and $|A_i\cap A_j|=1$ for every $i\neq j$. Is it true that $$\Big|\bigcap _{i=1}^{2019} A_i\Big|=1?$$

What I thought of doing is proving by induction by i, the base is obvious. However I'm stuck on the next step.

答:
If $B:=A_{2019} = \{1,2,...,45\}$ then, by pigeonhole priciple, some $x\in B$ is in at least $45$ sets among sets $A_1,A_2,...A_{2018}$. So $x$ is in at least $46$ sets, say $B,A_1,..,A_{45}$.

Now suppose some set $A_i = \{a_1,...,a_{45}\}$ (for some $i\geq 46$ and $i\ne 2019$) does not contain $x$.

Then some element $a_i$ must be in two sets from the list: $B,A_1,..,A_{45}$. But then these two share $x$ and $a_i$. A contradiction.

Title: Non-empty intersection.


问:
How can I calculate $\log(e^{e^i})$?
I'm studying complex analysis and I'm wondering how to calculate the following multivalued function (using the expression $\log(z)=\ln|z| + i\operatorname{Arg(z)}$):

$$\Large \log(e^{e^i})$$

Thank you!

答: The most general logarithm would be $e^i+2n\pi i=\cos 1+(\sin 1+2n\pi)i$ with $n\in\Bbb Z$, as @AsemAbdelraouf noted. The principal natural logarithm is the case $n=0$ (as @LeonhardEuler discussed), with phase $\in(-\pi,\,\pi]$.

Title: How can I calculate $\log(e^{e^i})$?


答: There are ordered fields which strictly extend the real numbers, there $x>0$ is meaningful, but need not imply $x\in\Bbb R$. </p>


答: $$\begin{align*}f'(0)=&\lim_{h\to0}\frac{\frac{\sin(\alpha h)}{\sin h}-\alpha}h\\=&\lim_{h\to0}\frac{\sin(\alpha h)-\alpha\sin h}{h\sin h}\\=&\lim_{h\to0}\frac{\left[\alpha h-\frac{\alpha^3h^3}{3!}+O(h^5)\right]-\alpha\left[h-\frac{h^3}{3!}+O(h^5)\right]}{h^2}\cdot\lim_{h\to0}\frac h{\sin h}\\=&~0.\end{align*}$$For $x\ne0,f'(x)$ is given by$$\frac d{dx}\left(\frac{\sin(\alpha x)}{\sin x}\right)=\frac{\alpha\cos(\alpha x)}{\sin x}-\frac{\cos x\sin(\alpha x)}{\sin^2x}$$The limit at $0$ is given by$$\lim_{h\to0}\frac{\alpha\sin x\cos(\alpha x)-\cos x\sin(\alpha x)}{\sin^2x}=\lim_{h\to0}\frac{(1-\alpha^2)\sin x\sin(\alpha x)}{\sin 2x}=0$$by L'Hopital. Thus the derivative is continuous.


问:
Prove that $\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\leq\frac14(a+b+c+3)$ if $abc=1$
I found the following exercise in a problem book (with no solutions):

Given $a,b,c>0$ such that $abc=1$ prove that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq\frac{a+b+c+3}{4}$$

I tried AM-GM for the fraction on the LHS but got stuck from there.

答:
let $p=a+b+c,q=ab+bc+ca,r=abc$

Now it is easy to prove the following as $r=1$

$pq\ge 9r=9$

$p^2\ge 3q$

Now we have to prove $$4(3+2p+q)\le (p+3)(2+p+q)$$ or $$p^2+pq\ge 6+3p+q$$ $$p^2+9\ge 6+3p+q$$ using $q\le p^2/3$ $$\frac{2p^2}{3}-3p+3\ge 0$$ which is true as $p\ge 3$


问:
What's the $\lim\limits_{x \to 1} \frac{1+e^\frac{1}{x-1}}{1-e^\frac{1}{x-1}}$
$\lim\limits_{x \to 1} \frac{1+e^\frac{1}{x-1}}{1-e^\frac{1}{x-1}}$

It's always giving me $\infty$ when x approaches 1 from both sides, no matter what I do.
I'm trying to solve it without l'Hopital's Rule or Taylor Series.

答:
Let $t=\frac{1}{x-1} \to \infty$. Then$$\frac{1+e^t}{1-e^t} = \frac{e^{-t} +1}{e^{-t} -1} \to \frac{1}{-1} =-1$$ This is true assuming you mean $x\to 1^+$.

If $x\to 1^{-}$, then $t\to -\infty$ and the limit evaluates to $1$.

Hence, the limit doesn’t exist.


问:
Find the number of integers $n$ of the form $30a0b03$ that are divisible by $13$.

Find the number of integers $n$ of the form $30a0b03$ that are divisible by $13$.

What I Tried: This problem looks simple but I cannot find it's solution. I recalled all the divisible rules of $13$ and tried to put them here, but none of them really worked. One thing which I got was :-
$$\rightarrow 13 | bo3 - 0a0 + 3$$
I only got this by applying one divisibility rule of $13$.
I might try and test for each values of $a$ and $b$ but there will be $100$ configurations, and I don't want to do that.

Can anyone help me?

Edit: After the hints @Daniel gave me, I figured out the solution and found $7$ solutions, can anyone tell me if this is correct or not?

答:
For four or more digit numbers, we do have a divisibility test by $13$ owing to $1001$ being divisible by $13$ ie, $1000$ leaving remainder $-1$. One checks the alternating sum of blocks of three consecutive digits. Eg, $\overline{abcdefgh}$ is divisible by $13$ if $\overline{ab}-\overline{cde}+\overline{fgh}$ is.

So we check how many $n$ there for which following is multiple of $13$ $$\overline{3}-\overline{0a0}+\overline{b03}=\overline{(b-1)(10-a)6}$$

We have $130+26=156$ one such number $\Rightarrow b=2$, $a=5$.

Others can be found by adding/subtracting $130$.


答:
An elementary proof without l'Hôpital nor Landau's big O notation:
for all $x\neq 0$,
\begin{align*}
f'(x)&=\frac{\alpha\cos(\alpha x)\sin(x)-\sin(\alpha x)\cos(x)}{\sin^2(x)}\\
%
&=\left(\frac{(x-0)}{\sin(x)-\sin(0)}\right)^2\frac{\alpha\cos(\alpha x)\sin(x)-\sin(\alpha x)\cos(x)}{(x-0)^2}.
\end{align*}
By continuity of the square and inverse functions, we have:
\begin{align*}
\lim_{x\to0}\left(\frac{(x-0)}{\sin(x)-\sin(0)}\right)^2&=\left(\frac{1}{\lim_{x\to0}\frac{\sin(x)-\sin(0)}{(x-0)}}\right)^2=\left(\frac{1}{\cos(0)}\right)^2=1.
\end{align*}
Let us write:
\begin{align*}
\alpha\cos(\alpha x)\sin(x)-\sin(\alpha x)\cos(x)&=\alpha\big(\cos(\alpha x)-\cos(0)\big)\sin(x)+\alpha\cos(0)\sin(x)\\
&\quad-\sin(\alpha x)\big(\cos(x)-\cos(0)\big)-\sin(\alpha x)\cos(0)
\end{align*}
Notice first that
\begin{align*}
\lim_{x\to0}\frac{\alpha\big(\cos(\alpha x)-\cos(0)\big)\sin(x)}{(x-0)^2}&=\alpha\lim_{x\to0}\frac{\cos(\alpha x)-\cos(0)}{x-0}\frac{\sin(x)-\sin(0)}{x-0}=\alpha(-\alpha\sin(0))\cos(0)=0,\\
%
\lim_{x\to0}\frac{\sin(\alpha x)\big(\cos(x)-\cos(0)\big)}{(x-0)^2}&=\lim_{x\to0}\frac{\sin(\alpha x)-\sin(0)}{x-0}\frac{\cos(x)-\cos(0)}{x-0}=\alpha\cos(0)(-\sin(0))=0
\end{align*}
by definition of the derivative at 0. Next,
\begin{align*}
\lim_{x\to0}\frac{\alpha\cos(0)\sin(x)-\sin(\alpha x)\cos(0)}{(x-0)^2}&=\cos(0)\lim_{x\to0}\frac{\alpha\sin(x)-\sin(\alpha x)}{(x-0)^2}\\
%
&=\cos(0)\lim_{x\to0}\frac{\left[\frac{\alpha\big(\sin(x)-\sin(0)\big)}{x-0}-\alpha\cos(0)\right]-\left[\frac{\sin(\alpha x)-\sin(0)}{x-0}-\alpha\cos(0)\right]}{x-0}\\
%
&=\cos(0)\big(-\alpha\sin(0)+\alpha\sin(0)\big)\\
%
&=0
\end{align*}
by definition of the second derivative at 0. As the limit of products is the products of the limits (provided that all limits exist in $\mathbb{R}$), we therefore have established the following:
\begin{align*}
\lim_{x\to0}f'(x)&=0.
\end{align*}
On the other hand,
\begin{align*}
\lim_{x\to0}\frac{f(x)-f(0)}{x-0}&=\lim_{x\to0}\frac{\frac{\sin(\alpha x)}{\sin(x)}-\alpha}{x-0}\\
%
&=\lim_{x\to0}\frac{x-0}{\sin(x)}\frac{\sin(\alpha x)-\alpha\sin(x)}{(x-0)^2}\\
%
&=\lim_{x\to0}\frac{x-0}{\sin(x)-\sin(0)}\frac{\sin(\alpha x)-\alpha \sin(x)}{(x-0)^2}
\end{align*}
is proved to be 0 as well using the above computations. This shows that $f'(0)=0$ and thus $f'$ is continuous at $x=0$.

Remark. The above proof works as well for any function $f(x)=\frac{g(\alpha x)}{g(x)}$ where $g$ is smooth and satisfies $g(0)=0$ and $g'(0)=1$.


问:
When two polynomials $f(x),g(x)$ over a field $F$ are said to be relatively prime?
When two polynomials $f(x),g(x)$ over a field $F$ are said to be relatively prime?

Following the definition given for the integers I guess when two of them have no factors in common other than $1.$ But if we follow such definition it implies $-1$ and $x+1$ are not relatively prime.

My second question is when does $f,g$ are relatively prime provided one of them is $0?$

答:
The relevant property here is that any field is a GCD domain, which means that any two nonzero elements have a GCD (ordered by divisibility).  And once $F$ is a GCD domain, $F[x]$ is as well.  "Relatively prime" means that their GCD is 1 (up to units, of course).

In this context $\gcd(-1,x+1)=1$, up to units; so in some sense $F\setminus \{0\}$ is the gcd.  Also for nonzero $\alpha$, $\gcd(\alpha,0)=\gcd(\alpha,\alpha)=\alpha$, again up to units.


问:
Showing that $X = int(A) \cup Fr(A)\cup int(X-A)$
Let $X$ be a topological space and $A\subset X$. Show that $X = int(A) \cup Fr(A)\cup int(X-A)$, this being a union disjointed.

To show this equality I must show the inclusions:
\begin{equation}
X \subseteq int(A) \cup Fr(A)\cup int(X-A)
\end{equation}
and
\begin{equation}
int(A) \cup Fr(A)\cup int(X-A)\subseteq X
\end{equation}

Picking up a point $x$ from $X$ and showing that it's in $int(A) \cup Fr(A)\cup int(X-A)$. And vice versa?
Is that what it was?

答:
Yes.

The second inclusion comes for free, since $X$ is the topological space you're working in, so every set you're treating is a subset of $X$.

The other needs a little more work.  If you don't know where to start, try drawing $A$ on a piece of paper, and delimiting clearly what are $\text{int}(A)$, $\text{fr}(A)$, and $\text{int}(X-A)$ :).


问:
Finding a plane containing $(-1,2,1)$, parallel to $\frac{x}{2}=-\frac{y}{3}=-z$, and making a $60^\circ$ angle with $x=y, z=0$
I have to find the equation of the plane that passes through $$(-1,2,1),$$ is parallel to the line: $$\frac{x}{2}=-\frac{y}{3}=-z$$ and makes $60^\circ$ angle with $$x=y, z=0$$

I thought that I could take two points from the line and make a vector with them that would be parallel to the plane, but that means I'm just using one of the given conditions and also getting the wrong answer.

Isn't that last condition extra? If I have a point and a parallel line why do I need anything else? And why am I getting the wrong answer? I don't know what I'm doing wrong.

Could you please help me with this? Thanks a lot

答:
Hint.

Given two lines $L_1,L_2$ and a point $p_0\in \Pi$ such that

$$
\cases{
L_1\to p = \lambda_1\vec v_1\\
L_2\to p = \lambda_2\vec v_2\\
\Pi\to (p-p_0)\cdot\vec n=0
}
$$

we need

$$
\cases{
\vec n_\cdot \vec v_1 = 0\\
\vec n\cdot\vec v_2 = |\vec n||\vec v_2|\cos\left(\frac{\pi}{3}\right)\\
|\vec n|= 1
}
$$

or

$$
\cases{
\vec n_\cdot \vec v_1 = 0\\
\vec n\cdot\vec v_2 = |\vec v_2|\cos\left(\frac{\pi}{3}\right)\\
|\vec n|= 1
}
$$

three equations and three unknowns $\vec n = (n_x,n_y,n_z)$. Here

$$
\cases{
\vec v_1 = (2,-3,-1)\\
\vec v_2 = (1,1,0)
}
$$


问:
Show that such $x$ and $y$ exist

Let $n$ be a positive integer which is not a square. Prove that for every integer $a$ relatively prime to $n$, there exist integers $x$ and $y$ satisfying
$$ax \equiv y \pmod{n}\text{ with } 0<x<\sqrt n \text{ and } 0<|y|<\sqrt n$$

I'm unable to make any progress with this problem. Any help will be appreciated.

答: Just look at the expression $ax-y$ for $x,y$ $\in \{0,...,\lfloor\sqrt n\rfloor \}$. Notice that there are $(\lfloor\sqrt n\rfloor \ +1)^2 > n$ possibilities for $(x,y)$. Since there are $n$ possible values for $ax-y \pmod n$, by pigeonhole principle, there exists distinct $(x_1,y_1), (x_2,y_2)$ in the previous range with $ax_1-y_1 \equiv ax_2-y_2 \pmod n$. So, $a(x_1-x_2) \equiv y_1-y_2 \pmod n$. Now, take $x=\lvert x_1-x_2 \rvert$, and, $y=y_1-y_2$ or $y=-(y_1-y_2)$ depending on the sign of $x$. Clearly, $x,\lvert y \rvert \in \{0,...,\sqrt n\}$. We also have $x\neq \sqrt n$, and, $\lvert y \rvert \neq \sqrt n$, since $n$ is not a square. All it remains is to show that $x\neq 0$, and, $\lvert y \rvert \neq 0$, which can be done easily using the fact that $(x_1,y_1)$ and $(x_2,y_2)$ are distinct and $(a,n)=1$.

Title: Show that such $x$ and $y$ exist


答:
The last condition is not extra. Parallel to the given line and passing through the given point, one may have infinitely many planes. These can be generated by rotating any one such plane about an axis passing through $(-1,2,1)$ and parallel to the line.

Let the normal vector of the desired plane be $(a,b,c)$ with $a^2+b^2+c^2=1$. The plane is parallel to $l_1$, so its normal vector is perpendicular to $l_1$, i.e.$$(a,b,c)\cdot(2,-3,-1)=0\iff c=2a-3b$$$l_2$ makes $60^\circ$ with the plane and thus $30^\circ$ with the normal vector. So$$\frac{(a,b,c)\cdot(1,1,0)}{\sqrt2}=\cos(30^\circ)\iff a+b=\sqrt\frac32$$Can you solve these to get the normal vector? Once that is done, the equation of the plane is given by$$((x,y,z)-(-1,2,1))\cdot(a,b,c)=0$$

Title: Finding a plane containing $(-1,2,1)$, parallel to $\frac{x}{2}=-\frac{y}{3}=-z$, and making a $60^\circ$ angle with $x=y, z=0$

Body:
I have to find the equation of the plane that passes through $$(-1,2,1),$$ is parallel to the line: $$\frac{x}{2}=-\frac{y}{3}=-z$$ and makes $60^\circ$ angle with $$x=y, z=0$$

I thought that I could take two points from the line and make a vector with them that would be parallel to the plane, but that means I'm just using one of the given conditions and also getting the wrong answer.

Isn't that last condition extra? If I have a point and a parallel line why do I need anything else? And why am I getting the wrong answer? I don't know what I'm doing wrong.

Could you please help me with this? Thanks a lot


答:
Of course, everything depends on context. I usually prefer to say things like

For any $x\in\mathbb{R}$ with $x>0$, ...

instead of

For any $x>0$, ...

to remove ambiguity, but I'm not insistent on it; I might be willing to sacrifice the "$x\in\mathbb{R}$" if it's making an orphan at the end of a paragraph, for example.

In contrast, everyone knows what

For any $\epsilon>0$, ...

almost always means, and it doesn't really add anything to say $\epsilon\in\mathbb{R}$. Of course, if you want to write $\epsilon>0$ and $\epsilon$ is not an element of $\mathbb{R}$, then it is all the more incumbent upon you to warn the reader of this non-standard use.


问:
What does $U(e^{j \lambda})$ mean?
I am reading “Signal processing for communications” by Paolo Prandoni and Martin Vetterli. In the section 4.4 The DTFT (Discrete-Time Fourier Transform) (p. 72 of 2008 ed.) they write:

The somewhat odd notation $X (e ^{j \omega}) $ is quite standard in the signal
processing literature and offers several advantages:
<...>
regardless of context, it immediately identifies a function as the
Fourier transform of a discrete-time sequence: for example $U(e^{j \lambda})$
is just as readily recognizable ...

My question: what $U(e^{j \lambda})$ is?

Because I can’t “readily recognize” it. And authors don't use it anywhere else in the text.

I suspect, that it is a common notation in English, related to complex numbers in general. But I had education in different language and didn’t have enough exposure to math in English. Or I simply don’t know. And, unfortunately, it is hard to create a relevant google search.

答: What they mean is just that $U(e^{j\lambda})$ is recognizable as a DTFT of some sequence $u[n]$, no matter what names one chooses for the function and for the frequency variable. I agree that they could have made their point a bit clearer.

Title: What does $U(e^{j \lambda})$ mean?


问:
How do we prove this equality involving pseudoinverse?
Given a rectangular matrix $A$, how do we prove that $A^{\dagger} \left(A^{\dagger}\right)^{H} = \left(A^{H}A\right)^{-1}$, where $A^{\dagger} = \left(A^{H}A\right)^{-1}A^{H}$ is the right pseudo inverse operation?

答: Just substitute the definition in $A^\dagger(A^\dagger)^H$ using $(XY)^H=Y^HX^H$, and that $A^HA$ is self-adjoint hence so is its inverse.


答: It's probably best to say something like "there is a real number $x > 0$ such that...," or "let $x > 0$ be a real number," etc. depending on what you want to say about $x$.  This avoids ambiguity, and also using words to describe the new symbol being introduced makes life easier for the reader.  The goal of writing (even mathematical writing) should not be to avoid redundancy.


问:
A class has 6 men and 4 women and they are ranked among themselves then how many different rankings are possible?
A class has $6$ men and $4$ women and they are ranked among themselves; how many different rankings are possible?

My solution:

Possible rankings among men = $6!$

Possible rankings among women = $4!$

Then the total number of possible rankings is $6! + 4!$

But the book in which I am reading says that the total number of possible rankings is $6!\cdot 4!$

Which one is the right answer? And if second one is right then how exactly?

答:
As you write, there are $6!$ possible rankings for men and $4!$ for women.
To get the total you can't just add the two results.

Think about how you would count one by one.

You first choose one possible ranking for the men, and they count the $4!$ possibilities for women. You already have $4!$ possibilities.

Then you take the next possible ranking for the men, and you do the same : you have already $2.4!$ possilities.

...

And you do this $6!$ times.

So you have $(4!)(6!)$ possibilities.

Hope it helped


问:
Proving Poincare disk lines are orthogonal to the unit circle.
I'm doing self study with the book "Continuous Symmetry," and I had one question about proving the incidence axioms for the Poincare disk.

I'm to prove,  using analytic geometry, that the circle $C$ meets the unit circle orthogonally iff
$$
C := \{(x,y) : (x - a)^2 + (y - b)^2 = a^2 + b^2 - 1\}
$$

I was able to prove the necessary portion, yet the authors say I need to use some possibly messy algebra to get the orthogonality from
$$
r^2 = a^2 + b^2 -1.
$$
I see I could use the converse of the pythagorean theorem to get the right angles, but I feel like I'm missing an easier solution that the authors intended.

Any hints would be appreciated.

答:
I don't know where the "possibly messy algebra" comes in. This seems pretty straightforward:

$$\overline{OT}\perp\overline{PT} \quad\iff\quad r^2 + 1^2  = a^2 + b^2$$


问:
Polynomials - The sum of two roots
If the sum of two roots of
$$x^4  + 2x^3 - 8x^2 - 18x - 9 = 0$$ is
$0$, find
the roots of the equation</p>

答:
I'd do it this way. If $x=\pm a$ are roots of the equation, then $f(x)=x^2-a^2$ is a factor of the polynomial. Note that this contains only even powers of $x$.

Now suppose we multiply $f(x)$ by another polynomial $p(x)$ so that $f(x)p(x)=P(x)$. Let e(x) be the polynomial obtained by picking out the even powers of $p(x)$ with $o(x)$ as the odd powers, so that $p(x)=e(x)+o(x)$ and $$f(x)p(x)=f(x)e(x)+f(x)o(x)=E(x)+O(x)=P(x)$$ gives a decomposition of $P(x)$ into the sum of even powers of $x$ plus the sum of odd powers of $x$, with $f(x)$ a factor of both $E(x)$ and $O(x)$.

In brief, we know that, in the case presented $$x^4-8x^2-9$$ and$$2x^3-18x$$

have a common factor of the kind we are seeking. In a more complex example we could use the Euclidean algorithm to pin this down further (there might be other common factors). But here the even and odd parts are easy to factorise as $$(x^2-9)(x^2+1)$$ (treat this as a quadratic in $x^2$) and$$2x(x^2-9)$$

From there we can easily finish the factorisation.

Note: this is an entirely systematic approach, which simplifies the problem without having to guess.

Note: this is essentially the approach suggested by Erick Wong in a comment on another answer, which I spotted after I had written this all out.

Title: Polynomials - The sum of two roots


问:
Find probability with the help of Poisson distribution
There are $3$ cyclone on average in a year in russia .

What is the probability that there will be  cyclone  in the next TWO years  ?

I just want to know the value of $\lambda$ in Poisson distribution! As we know $\lambda$ actually works on the same time interval.
My calculations: as it's Two years, $\lambda = 3\cdot 2 = 6$.

But my confusion is should we calculate the present year too? I mean if we don't calculate this year we can calculate as time = $2$ years otherwise $3$ ...

Is my calculation ok?

答:
$\lambda=6$ thus your probability is

$$P[X>0]=1-e^{-6}\approx 99.75\%$$


答:
Let $x\in X$ be arbitrary and let $\mathcal N_x$ denote its set of neighborhoods.

Purely on logical grounds we can say that exactly one of the following options is correct:

$\exists N\in\mathcal N_x [N\subseteq A]$

$\exists N_x\in\mathcal N_x [N\subseteq A^\complement]$

$\forall N\in\mathcal N_x [N\cap A\neq\varnothing\text{ and }N\cap A^{\complement}\neq\varnothing]$

This can be rephrased as:

$x\in\mathsf{int}(A)$

$x\in\mathsf{int}(A^{\complement})$

$x\in\mathsf{Fr}(A)$

Title: Showing that $X = int(A) \cup Fr(A)\cup int(X-A)$

Body:
Let $X$ be a topological space and $A\subset X$. Show that $X = int(A) \cup Fr(A)\cup int(X-A)$, this being a union disjointed.

To show this equality I must show the inclusions:
\begin{equation}
X \subseteq int(A) \cup Fr(A)\cup int(X-A)
\end{equation}
and
\begin{equation}
int(A) \cup Fr(A)\cup int(X-A)\subseteq X
\end{equation}

Picking up a point $x$ from $X$ and showing that it's in $int(A) \cup Fr(A)\cup int(X-A)$. And vice versa?
Is that what it was?


答:
Hint: $\operatorname{Int}(A)\sqcup\operatorname{Fr(A)}=\operatorname{cl}(A)$, which is (by definition? I guess it depends on your definition of closure and boundary of a subset) the smallest closed subset of $X$ containing $A$.

This way, you just have to show that $X=\operatorname{cl}(A)\,\sqcup\,\operatorname{Int}(X\smallsetminus A)$; but again, $\operatorname{Int}(X\smallsetminus A)$ is (by definition) the largest open subset of $X$ contained in $X\smallsetminus A$.

Considering this, you can easily find that $\operatorname{Int}(X\smallsetminus A)=X\smallsetminus\operatorname{cl}(A)$, which leads us to what you're trying to prove. (Notice that all unions here are disjoint!)


问:
Confusion in module structure of indecomposable injectives over quiver path algebra
I got very confused while working out the explicit module structure on the indecomposable injective modules $I_i$ over a finite-dimensional quiver path algebra $kQ/I$. I know that the right injective indecomposable $I_i$, where $i$ is a vertex of $Q$, may be described as

$$I_i=\{\text{paths starting at }i \text{ modulo relations}\}^*,$$

where $^*$ denotes the $k$-dual module. But what does it mean? Does this mean that in this module a path $p$ corresponds to a linear map to $k$ denoted $p^*$ which takes $p$ to 1 and other paths to 0? If so, am I right in understanding that multiplication acts by "deletion", i.e. if $p^*\cdot q$ is $s^*$ such that in the quiver $qs=p$ (I am composing paths as maps, i.e. $q$ must start where $s$ begins)?

For example, consider a quiver $Q$ with 2 vertices and one edge $x: 1\to2$. I would say from the above description that in $I_i$ we have $x^*\cdot x=e_1^*$, since $(x^*\cdot x)(e_1)=x^*(x\cdot e_1)=1$?

By the way, I got lost in this because I was trying to understand the criteria for a quiver path algebra to be self-injective, for which I believe it is enough to have a permutation of vertices $\pi$ such that $P_i\cong I_{\pi(i)}$, but I also saw that it is enough to have $\text{soc}(P_i)\cong\text{top}(I_{\pi(i)})$. I would be thankful for any references on this topic!

答:
There are several things to unpack here.  Let's review the notions that already appear in your post, and we'll get to a nice description of injective modules at the end of this answer.

First, let's make our conventions explicit.  From your post, I see you are considering right modules, and also that you are composing arrows in a quiver from right to left (that is, if the target of the arrow $x$ is the source of the arrow $y$, then $yx$ is their composition).  Also, to simplify notations, let  $A=kQ/I$.

Before looking at the injective modules, let's describe the indecomposable projective modules.  For each vertex $i$, there is an idempotent $e_i$ in $A$ (corresponding to the path of length zero at $i$); the corresponding indecomposable projective right module is $e_iA$, which is the set of linear combinations of paths of $Q$ ending in $i$ (modulo $I$).  Note that the indecomposable projective left module is $Ae_i$, which is the set of linear combinations of paths of $Q$ starting in $i$ (modulo $I$).

Now, to get injective right modules, we will use the duality functor $(?)^* = \operatorname{Hom}_k(?,k)$  from the category of left $A$-modules to the category of right $A$-modules.  The duality functor sends projective modules to injective modules, and preserves indecomposability.  Thus the indecomposable injective right module corresponding to the vertex $i$ is
$$
  I_i \cong (Ae_i)^* = \operatorname{Hom}_k(Ae_i,k).
$$

The $A$-module structure of $\operatorname{Hom}_k(Ae_i,k)$ is defined as follows: for any $\phi\in \operatorname{Hom}_k(Ae_i,k)$ and any $a\in A$,  $\phi\cdot a$ is the map sending $x$ to $\phi(a\cdot x)$.

In order to obtain a more explicit description of $I_i$, we turn to quiver representations.  There is an equivalence of categories between the category of finite-dimensional right $A$-modules and the category of finite-dimensional representations of the opposite quiver $Q^{op}$ which satisfy the relations in $I$.  This equivalence is fairly straightforward to describe on objects: given a right $A$-module $M$, the corresponding representation is defined by

putting the vector space $Me_i$ at each vertex $i$, and

putting the linear map $Me_j \to Me_i: m \mapsto ma$ at each arrow $a:i\to j$.

Let's apply this to the injective module $I_i$.  For any vertex $j$, we have that
$$
I_ie_j = \operatorname{Hom}_k(Ae_i, k)e_j \cong \operatorname{Hom}_k(e_jAe_i, k).
$$

A basis of $I_ie_j$ is thus given by a dual basis of $e_jAe_i$, the space generated by paths from $i$ to $j$ in $Q$ (modulo $I$).

Finally (and perhaps most importantly in this discussion), for an arrow $a:h\to j$, the corresponding linear map $I_ie_j \to I_ie_h$ is dual to the map $\lambda_a:e_hAe_i \to e_jAe_i$ given by left multiplication by $a$.  In particular, if you fix a basis of $e_hAe_i$ and $e_jAe_i$ and let $[\lambda_a]$ be the matrix of the map $\lambda_a$ in those bases, then the matrix of the dual map $I_ie_j \to I_ie_h$ in the dual bases is the transpose of $[\lambda_a]$.

In particular, if you know a representation $V$ isomorphic to the left projective $A$-module $Ae_i$, then to get a representation isomorphic to the right injective $A$-module $I_i$, all you need to do is take the transpose of all the matrices defining $V$.


问:
JAM 2005 Mathematics Question 7

This is a multiple-choice question having only a single correct option that appeared in JAM 2005 (an Exam in India). Now according to me options B, C, and D all should be correct but only option D is marked as correct. My reasoning behind B, C and D is that:

$$ T(x) = cx, I(x) = x$$
$$ \therefore (T -cI)x = T(x) -cI(x)=cx-cx=0 $$ for all $x \in \mathbb{R}^3 $. Therefore rank of the transformation $(T-cI) $ should be the dimension of its image, which is the vector $0$. Hence the rank of the transformation is $0$. So B, C and D should all be correct. Where am I wrong?

答:
AS J.W. Tanner said, we don't have $T(x)=cx$ for all $x$, but only for one.

Here the only thing you can say is $(T-cI)$ is not inversible because it is not injective, so its rank cannot be $3$.

But you can find $T$ such that its rank is $0$ (for instance $T=0$), or $1$ (for instance $T(x,y,z) = (x,y,0)$), or $2$ (for instance $T(x,y,z) = (x,0,0)$).

The concept behind this question is the one of eigenvalues.
cf https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors.

Title: JAM 2005 Mathematics Question 7


答:
If $\text{rank}(T-cI) = 3$, i.e. full rank, $(T-cI)$ is invertible and the only solution of $(T-cI)x = 0$ is $x = 0$. However, we are given a non-zero $x\in\mathbb{R}^3$ such that $x\in \ker(T-cI)$, i.e. $Tx = cx$ so $\text{rank}(T-cI) \neq 3$.

The rank may be $0,1$ or $2$ - we don't know what it is exactly! This is the best we can say when we are given the existence of at least one $x$ such that $Tx=cx$. If $Tx = cx$ happens for all $x$, then $(T-cI) = 0$ (the zero operator on $\mathbb{R}^3$) - but this isn't the case!

Title: JAM 2005 Mathematics Question 7

Body:
This is a multiple-choice question having only a single correct option that appeared in JAM 2005 (an Exam in India). Now according to me options B, C, and D all should be correct but only option D is marked as correct. My reasoning behind B, C and D is that:

$$ T(x) = cx, I(x) = x$$
$$ \therefore (T -cI)x = T(x) -cI(x)=cx-cx=0 $$ for all $x \in \mathbb{R}^3 $. Therefore rank of the transformation $(T-cI) $ should be the dimension of its image, which is the vector $0$. Hence the rank of the transformation is $0$. So B, C and D should all be correct. Where am I wrong?


问:
Prove that every open set in $\mathbb{R}$ is the union of an at most countable collection of disjoint segments.
My attempt:

We know that $\mathbb{Q}$ is dense in $\mathbb{R}$.Now, we consider an open set $S$ in $\mathbb{R}$ then let $x \in S$, $x$ is an interior point of $S$ and we can get an open ball $(x-r,x+r)$ ,$r>0$ contained in $S$. Since $\mathbb{Q}$ is dense, let $q_1 \in (x-r,x+r)$.Now, $q_1 \in (x-r,x+r)$ and let $r_1=|q_1-x|$.We consider the rational number $ > r_1$ (which is possible to choose as $\mathbb{Q}$ is dense ) and let us call it $d_1$ and we consider the open ball $(q_1-d_1,q_1 +d_1)$ which will contain $x$.

We pick $x' \in S$ , and consider $r_2>0$ such that $(x'-r_2,x'+r_2) \cap (q_1-d_1,q_1+d_1) = \phi$ , if such an $r_2$ doesn't exist then we can conclude that $x' \in (q_1-d_1,q_1+d_1)$. Repeating this process we can conclude that the open set $S $ can be written as the union of disjoint segments.

Since $\mathbb{Q}$ is countable then it can be shown that union of this disjoint segment is also countable.

I am trying to do Rudin on my own and need someone to cross check my proofs.I know this might be a duplicate but if someone goes through my proof and picks up my mistake and a probable way out that would be helpful.

答:
Well being technically correct the fact that "reapeting this process we can conclude something" doesn't prove anything unless you show that in fact we can conclude it.

The basic idea of the proof is that we can approximate every open interval in $\mathbb{R}$ using open intervals with rational ends.

Take any open set $U\subseteq\mathbb{R}$. Now take $x\in U$ - by the definition there exist an open interval s.t $x\in I\subseteq U$. For such interval $I$ we can find an interval $J$ s.t $I\subseteq J\subseteq U$ and there's no "larger" interval in $U$ containing $J$. Take a family of all such maximal intervals in $U$. Every segment contains rational so that solves the countability. Can you see that every interval in such family is disjoint?


问:
RSA: what is $k$ in the formula to calculate $d$?
In RSA, to calculate $d$, when given $\phi(n)$ and $e$, I stumbled upon this formula:

$$d = \dfrac{k \phi(n) + 1}{e}$$

But what does $k$ stand for? How to obtain the value for $k$?

Thank you,

Chris

答:
$d$ in an integer which satisfies

$ed\equiv 1\mod \phi(n)$,

which means that there exists an integer $k$ such that

$ed = 1 + k\cdot \phi(n).$