问:
Question on Probability and Bayes' theorem
You roll a fair six-sided die. If the die shows an odd number then you toss a fair coin that number of times. If the die shows an even number you do not toss the coin at all. What is the probability that you see at least two Tails?

I kind of know where to start - you find the probability of getting 0 and 1 tails and take that away from 1. Don't know what to do next

答:
$T: \text{tossing tails}$

$$\begin{aligned}
P({T\ge2})
&= P(\texttt{roll 3}\texttt{ AND } T\in\{2,3\})+ P(\texttt{roll 5} \texttt{ AND }T\in\{2,3,4,5\}) \\
&=P(\texttt{roll 3})\cdot P(T\in\{2,3\}|\texttt{Flip coin thrice})+
P(\texttt{roll 5})\cdot P(T\in\{2,3,4,5\}|\texttt{Flip coin five times})\\
&=\frac{1}{6}\cdot P(T\in\{2,3\}|\texttt{Flip coin 3 times})+
\frac{1}{6}\cdot P(T\in\{2,3,4,5\}|\texttt{Flip coin 5 times})\\
&=\frac{1}{6}\cdot \frac{a}{8} + \frac{1}{6}\cdot \frac{b}{32}
\end{aligned}$$

We'll leave $a$ and $b$ for you to calculate.

UPDATE:

It's been a while.

$$P(T\in\{2,3\}|\texttt{Flip coin 3 times})=P(TTH)+P(HTT)+P(THT)+P(TTT)=\frac{4}{8}=\frac{1}{2}.$$

$$P(T\in\{2,3,4,5\}|\texttt{Flip coin 5 times})=1-P(T\in\{0,1\}|\texttt{Flip coin 5 times})=1-P(HHHHH)-5P(THHHH)=1-\frac{6}{32}=\frac{13}{16}.$$

$$P(T \ge 2)=\frac{1}{6}\left(\frac{1}{2}+\frac{13}{16}\right)=\frac{7}{32}.$$


问:
Evaluating stochastic integral with exponential function
After solving for the Ornstein-Uhlenbeck process, i'm still stuck with a stochastic integral that I can't solve.

The Ornstein-Uhlenbeck process:

$X_{t} = X_{0} e^{-\theta t}+ \mu(1-e^{-\theta t}) + \sigma e^{-\theta t} \int_{0}^{t} e^{\theta s}dW_{s} $

Does anyone have an idea on how to solve this stochastic integral? ($\int_{0}^{t} e^{\theta s}dW_{s} $)

*Note that i'm trying to evaluate this expression for a Monte-Carlo simulation. I have already tried discretizing the integral but I would like to improve my results by using the exact solution.

Thanks in advance!

答:
If you only want to simulate that integral, then nothing is mysterious here.

It's just a normal variable of law $ \mathcal{N}\left( 0, \int_{0}^t e^{2 \theta s}.ds \right)$.

If you want to so look at $(W_s)$ as a stochastic process then you can see easily that $ (U_t ):= \left(  \int_{0}^t e^{ \theta s}.dW_s \right) $ is just a change time Brownian process. To be more precise,

Letting $g(t)=  \int_{0}^t e^{2 \theta s}.ds$.

Roughly speaking, there is a Brownian process $\tilde{W}$ such that:

$(U_t)=\left( \tilde{W}_{g(t)} \right)$

ContentLicense: CC BY-SA 4.0


问:
Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$
Here is my attempt:

Let $(x, y)$ be any such point.

Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we have the equalities
$$
\lvert x \rvert = \lvert y \rvert = \sqrt{ (x-4)^2+(y-2)^2}.
$$
From $\lvert x \rvert = \lvert y \rvert$, we obtain $y = \pm x$.

Thus we have the equations
$$
\sqrt{ (x-4)^2 + (\pm x -2)^2} = \lvert x \rvert,
$$
which implies
$$
(x-4)^2 + ( \pm x - 2)^2 = x^2,
$$
which simplifies to
$$
x^2 - 2(4 \pm 2)x + 20 = 0.
$$

Thus we have the following two quadratic equations
$$
x^2 - 12x + 20 = 0 \qquad \mbox{ and } \qquad x^2 -4x + 20 = 0,
$$
and the solutions of these quadratic equations are
$$
x = \frac{12 \pm 8 }{ 2 } \qquad \mbox{ and } \qquad x = \frac{ 4 \pm 8 \iota }{ 2 },
$$
that is,
$$
x = 10, 2 \qquad \mbox{ and } \qquad x = 2 \pm 4 \iota.
$$
We will of course only need the real values for our $x$.

Thus there are eight possible points satisfying the condition given in the problem, namely
$$
(10, 10), (10, -10), (-10, 10), (-10, -10), (2, 2), (2, -2), (-2, 2), (-2, -2).
$$

Is my solution correct in terms of the technique employed as well as the answers obtained? Or, are there any mistakes?

答:
$$\dfrac{x-4}{\cos t}=\dfrac{y-2}{\sin t}=r\ge0$$

So, the point $P(4+r\cos t,2+r\sin t)$$

We need $$(4+r\cos t)^2=(2+r\sin t)^2=r^2$$

$$r(1+\cos t)=4$$ as $r(1-\cos t)\ge0$

Similarly, $r(\sin t+1)=2$

Using https://en.m.wikipedia.org/wiki/Weierstrass_substitution#The_substitution, $$\dfrac42=\dfrac2{(1+p)^2}$$

where $p=\tan\dfrac t2$

$\implies(1+p)^2=1\implies p=0,-2$

If $p=0,r=2$

If $p=-2,r=?$


问:
What does "$A \leq B : \Longleftrightarrow A \subseteq B$ is an order relation of $\mathcal{P}(N)$" mean?
I have read the following in some exercise for discrete mathematics.
Let $N$ be a set and $\mathcal{P}(N)$ be its power set. Then $A \leq B : \Longleftrightarrow A \subseteq B$ is an order relation of $\mathcal{P}(N)$.

I am not really sure what is asked here. I know how to show that $\subseteq$ is a partial order on the power set but I am not sure what exactly is asked here. What order relation do I need to show and what should I do with $A \leq B$?

答: A "ordered set" is a shortcut for "partially ordered set", so you must prove the 3 axioms of a ordered relation. See here.</p>


问:
How can I show if $AA^T+A^TA=A+A^T$, then $Col(A)=Col(A^T)$?
Question: For a square matrix $A$, if $$AA^T+A^TA=A+A^T,$$ prove that $$\text{Col}(A)=\text{Col}(A^T).$$

I want to prove this statement, but it's quite difficult...

I think I should use "$Null(A)$ is orthogonal to $Col(A^T)$"

Any good ideas?

答:
We prove that the null-spaces of $A$ and $A^T$ are equal by showing
$$
Av=0\implies A^Tv=0
\\A^Tv=0\implies Av=0
$$
First, to show $Av=0\implies A^Tv=0$, we assume $Av=0$ for some vector $v$. Then by multiplying both sides of $AA^T+A^TA=A+A^T$ in $v^T$ from left and $v$ from right we obtain $${
v^TAA^Tv+v^TA^TAv=v^TAv+v^TA^Tv\implies
\\
v^TAA^Tv+v^TA^TAv=2v^TAv\implies
\\
v^TAA^Tv+0=0\implies
\\
||A^Tv||^2=0\implies
\\
A^Tv=0
}$$
Similarly the proof holds conversely and the statement is proved $\blacksquare$.


问:
Method of Characteristic for PDE
Given the following partial differential equation with initial condition
$$tu_x+u_t=2t\cdot u,\;IV: u(x,0)=x.$$
So I apply the method of characteristic to solve this problem as following

Forming three characteristic equation

$$\frac{dx}{ds}=t,\;\frac{dt}{ds}=1,\; and\;\frac{du}{ds}=2t\cdot u $$
$$\frac{dt}{ds}=1\rightarrow dt = ds\rightarrow t = s+t_0\rightarrow t=s\;since\; t_0=0\; from\; IV$$
then
$$\frac{dx}{ds}=t\rightarrow \frac{dx}{ds}=s\rightarrow x=\frac{s^2}{2}+x_0$$
also for the last equation we have
$$\frac{du}{ds}=2t\cdot u\rightarrow \frac{du}{u}=2s\;dt\rightarrow u=e^{s^2+u_0}$$
Then we substitute the $s=t$, and $x_0=x-\frac{s^2}{2}$ to initial condition
$$ u(x_0,0)=e^{0+u_0}=x\rightarrow u_0=lnx$$
Thus the finial solution I have should be $u(x,t)=e^{t^2+lnx}=xe^{t^2}$

However this is not same as the answer given, could someone tell me where I make the mistake in this problem?

答:
$$tu_x-u_t=tu$$
Charpit-Lagrange system of characteristic ODEs :
$$\frac{dx}{t}=\frac{dt}{1}=\frac{du}{tu}=ds$$
This is what you correctly found, but presented on an equivalent form.

A first characteristic equation comes from solving $\frac{dx}{t}=\frac{dt}{1}$ :
$$t^2-2x=c_1$$
A second characteristic equation comes from solving $\frac{dt}{1}=\frac{du}{tu}$ :
$$ue^{-x}=c_2$$
The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is :
$$ue^{-x}=F(t^2-2x)$$
$F$ is an arbitrary function (to be determined later in order to satisfy the initial condition).
$$\boxed{u(x,t)=e^xF(t^2-2x)}$$

CONDITION :

$$u(x,0)=x=e^xF(-2x)$$
$$F(-2x)=xe^{-x}$$
Let $x=-\frac12 X$
$$F(X)=-\frac12 X  e^{\frac12 X}$$
Now the function $F(X)$ is known. We put it into the above general solution where $X=t^2-2x$

$F(t^2-2x)=-\frac12 (t^2-2x)  e^{\frac12 (t^2-2x)}$
$$u(x,t)=e^x\left(-\frac12 (t^2-2x)  e^{\frac12 (t^2-2x)}\right)$$
After simplification, the particular solution which satisfies both the PDE and the condition is :
$$\boxed{u(x,t)=(x-\frac{t^2}{2})e^{\frac12 t^2}}$$

Title: Method of Characteristic for PDE


问:
Check that $\mathbb{Q}[x]/(x^4+1)$ is a field
$\mathbb{Q}[x]/(x^4 + 1)$ is a field?

I understand, that in order to $\mathbb{Q}[x]/(x^4 + 1)$ be a field, polynomial $(x^4 + 1)$ must be irreducible. And I checked its roots, they are not in $\mathbb{Q}$: $\pm \frac{1\pm i}{\sqrt{2}}$. But how I can simply check that in $\mathbb{Q}[x]$ there are no divisors of $x^4+1$? There are infinitely many polynomials of degree less than $4$ in $\mathbb{Q}[x]$.

答:
Since  $\mathbb{Q}[x]$ is a PID, it is sufficient to prove that $x^4+1$ is irreducible as element of $\mathbb{Q}[x]$

(Observe that the same polynomial is not irreducible over $\mathbb{R}[x]$, it can be useful to find various counterexamples!)

Now $x^4+1$ is reducible if and only if $(x+1)^4+1$ is reducible, but the latter polynomial is irreducible by Heisenstein criterion with $p=2$.


答:
A binary relation $\leq$ over a set $X$ is an order if the following hold:

$\leq$ is reflexive, i.e. $n \leq n$ for all $n \in X$;

$\leq$ is antisymmetric, i.e. if $n \leq m$ and $m \leq n$ then $n = m$, for all $n,m \in X$;

$\leq$ is transitive, i.e. if $n \leq m$ and $m \leq \ell$ then $n \leq \ell$, for all $n, m, \ell \in X$.

Often an order relation on $X$ is called partial to distinguish it from a total order, which is an order on $X$ such that, for all $n,m \in X$, one has $n \leq m$ or $m \leq n$.

You are asked to prove that $\subseteq$ is a reflexive, antisymmetric and transitive relation on $\mathcal{P}(N)$.

Title: What does "$A \leq B : \Longleftrightarrow A \subseteq B$ is an order relation of $\mathcal{P}(N)$" mean?

Body:
I have read the following in some exercise for discrete mathematics.
Let $N$ be a set and $\mathcal{P}(N)$ be its power set. Then $A \leq B : \Longleftrightarrow A \subseteq B$ is an order relation of $\mathcal{P}(N)$.

I am not really sure what is asked here. I know how to show that $\subseteq$ is a partial order on the power set but I am not sure what exactly is asked here. What order relation do I need to show and what should I do with $A \leq B$?


问:
How to solve $x^2 \equiv 12 \pmod {13}$?

Solve $x^2 \equiv 12 \pmod {13}$

By guessing I can say that the solutions are $5$ and $8$, but is there another way to find the solution besides guessing?

答:
Hint :

As $x^2 \equiv a^2 \pmod {n}$ is easiest to handle, and $12\equiv25 \pmod {13}$, we have

$$ x^2 \equiv (\pm 5)^2 \pmod {13}$$


问:
If $f(a) < g(a)$ for certain $a$ then exists $\delta > 0$ so $f(x) \leq g(x)$ por every $x \in (a-\delta, a+\delta).$
If $f,g:\mathbb{R} \to \mathbb{R}$ are continuous, $a \in \mathbb{R}$ and $f(a) < g(a)$ then exists $\delta > 0$ so $f(x) \leq g(x)$ por every $x \in (a-\delta, a+\delta).$
I'm studying real analysis and I'm not sure on how to apply the continuity definition to solve this problem. Any hint would be appreciated.

答:
WLOG you can assume that $f\equiv0$ (if not, take $G(x)=g(x)-f(x)$).

Applying continuity at $g(a)$, given $\varepsilon=g(a)>0$, there is $\delta>0$ such that for $x\in(a-\delta,a+\delta)$ $|g(x)-g(a)|<\varepsilon=g(a)$. Hence
$-g(a)<g(x)-g(a)<g(a)$ and, in particular, $g(x)>0$ for $x\in (a-\delta,a+\delta)$.


答:
As cosmo5 mentioned in his comment. You can do the following which is a little bit better than guessing since you probably can instantly recognise square numbers.
For every natural number $n$ we have
$$
x^2 \equiv 12 \mod 13
\quad\Leftrightarrow\quad
x^2 \equiv 12+13n \mod 13.
$$
Hence, you can start
$$
\begin{array}{c|c}
n & 12+13n \\
\hline
1 & 25 \\
2 & 38 \\
3 & 51 \\
4 & 64
\end{array}
$$
Then you will see that for $n=1$ you have $5^2$ and for $n=4$ you have $8^2$, which gives you the solutions $x_1=5$ and $x_2=8$.

Alternative you can make a polynomial division after you found one solution, because
$$
x^2-12 \equiv (x-x_1)(x-x_2) \mod 13.
$$
However, this might take longer than guessing for small numbers.

Title: How to solve $x^2 \equiv 12 \pmod {13}$?

Body:
Solve $x^2 \equiv 12 \pmod {13}$

By guessing I can say that the solutions are $5$ and $8$, but is there another way to find the solution besides guessing?


问:
Spherical coordinates of a sphere above the origin
I have to evaluate the volume of the sphere
$$x^2+y^2+(z-2)^{2}=4$$
using spherical coordinates

I found that $0\leq\theta\leq2\pi$ and $\displaystyle0\leq\phi\leq\frac{\pi}{2}$, but I can't get an appropriate interval for $\rho$.

What I thought was: since
$$z-2=\rho\cos\phi$$
and
$$0\leq z\leq4$$ then
$$z-2\leq2\Rightarrow\rho\cos\phi\leq2$$
Therefore
$$0\leq\rho\leq\frac{2}{\cos\phi}$$
Am I right?

答:
Note that if you have defined $z-2=\rho\cos\phi$, you have actually violated the standard coordination transformation since the correct transformation from Cartesian to Spherical coordinates is
$$
{x=\rho \sin\theta\cos\phi
\\
y=\rho \sin\theta\sin\phi
\\
z=\rho \cos\theta}
$$
regardless of what problem you manage to solve. The shifted-from-origin sphere then, has the following equivalent representation:
$$
{x^2+y^2+z^2=4z
\\
\rho^2=4\rho\cos\theta
\\
\rho=4\cos\theta\quad,\quad 0\le \theta\le{\pi\over 2}
}
$$


答:
$z = \rho \cos \phi, 0 \leq z \leq 4 \implies 0 \leq \rho \leq 4 \cos \phi$

$0 \leq \phi \leq \frac{\pi}{2}$

Title: Spherical coordinates of a sphere above the origin

Body:
I have to evaluate the volume of the sphere
$$x^2+y^2+(z-2)^{2}=4$$
using spherical coordinates

I found that $0\leq\theta\leq2\pi$ and $\displaystyle0\leq\phi\leq\frac{\pi}{2}$, but I can't get an appropriate interval for $\rho$.

What I thought was: since
$$z-2=\rho\cos\phi$$
and
$$0\leq z\leq4$$ then
$$z-2\leq2\Rightarrow\rho\cos\phi\leq2$$
Therefore
$$0\leq\rho\leq\frac{2}{\cos\phi}$$
Am I right?


问:
Question using Rouché's Theorem
This is from Stein Complex Analysis textbook page 106

Suppose $f$ and $g$ are holomorphic in a region containing the disc
$|z|\leq1$. Suppose that $f$ has a simple zero at $z=0$ and vanishes
nowhere else in $|z|\leq 1$. Let $f_\epsilon(z)=f(z)+\epsilon g(z)$.
Show that if $\epsilon$ is sufficiently small, then $f_\epsilon (z)$
has a unique zero in $|z|\leq 1$.

My approach:

Note that $|f(z)|>0$ if $z\in B_1(0)$ since $f(z)$ does not vanish on the circle. Also, $f(B_1(0))$  and $g(B_1(0))$ are compact and so they attain maximum and minimum on the circle.(Is my reasoning correct?)

Therefore, there exists $\epsilon>0$ such that $|\epsilon g(z)|<|f(z)|$ on $B_1(0)$. Using Rouché's theorem $f(z)+\epsilon g(z)$ and $f(z)$ have the same number of zero inside the circle, and it remains to show that $f(z)+\epsilon g(z)$ does not vanish on the circle. At this point, I don't know how to show it.
Can anybody give an insight?..

答:
That is the idea.

let $m$ be the mininum of $|f(z)|$ on the circle $|z| = 1$

let $M$ be the maximum of $|g(z)|$ on the circle.

Since the functions are holomorphic on the disk, these extrema must exist.

Let $\epsilon < \frac {m}{M}$

Then by Roche's theorem as $|f(z)|>|\epsilon g(z)|$ everywhere on the countour, $f(z) + \epsilon g(z)$ has the same number of zeros as $f(z)$ inside the contour.

Title: Question using Rouché's Theorem


问:
Vector spaces and subsets of Matrices

Hi, I was wondering if anyone could help me with this linear algebra problem.

I have solved a) by using matrix multiplying by scalars and matrix addition/subtraction. And hence concluded that I should get the return of

Row 1: (3 3) Row 2: (-2 -5)

I have also solved b) by creating a matrix from the 2x2 matrices and used row reducing operations. From this I found each line to have a pivot variable and hence having a unique solution. Which thus means that they are linearly independent.

However I'm now stuck on parts c) and d). I know to prove that a subset is a subspace it must meet the three conditions:

It is closed under addition

It is closed under scalar multiplication

The 0 vector is in the subset.

But I'm unsure how to prove that these do/do not hold true.

Any help is appreciated.

答:
Recall our basic definitions of matrices and arithmetic surrounding matrices...

Very loosely defined, a matrix is a rectangular array of elements with some number of rows and some number of columns (even possibly zero).  They can be pictured either with square brackets or with parentheses, both are common and mean the same thing.

We can add two matrices so long as they have the same number of rows and have the same number of columns.  To do so, we just perform the elementwise addition of each element.  For example, we have:

$$\begin{bmatrix}1&2\\3&4\end{bmatrix}+\begin{bmatrix}50&60\\70&80\end{bmatrix}=\begin{bmatrix}51&62\\73&84\end{bmatrix}$$

We can multiply a matrix by a scalar.  To do so, we just multiply each element of the matrix individually by the scalar.  For example, we have:

$$5\cdot \begin{bmatrix}1&2\\3&4\end{bmatrix}=\begin{bmatrix}5&10\\15&20\end{bmatrix}$$

This understanding of how to add matrices and multiply matrices by scalars is a critical first step in adequately understanding and defining the vector space in your question.  Make sure you are comfortable with matrices as objects.

Now, in your question you are interested very specifically at the set of $2\times 2$ matrices (matrices with two rows and two columns) with real entries whose top left, top right, and bottom right entries add up to be equal to zero.  That is to say, the set:

$$\left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}~:~a+b+d=0\right\}$$

This includes for instance the matrices $\begin{bmatrix}1&2\\3&-3\end{bmatrix}$ since $1+2+(-3)=0$ as well as $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ since $1+0+(-1)=0$ and so on...

We ask whether or not this set (coupled with the scalar field of $\Bbb R$ and arithmetic operations as described earlier) satisfies the properties necessary to be called a vector space.

We should already know that the space of all $2\times 2$ real matrices is in fact a vector space over $\Bbb R$ and so we do not need to check here all of the mundane properties such as associativity of addition of vectors, distributivity of scalars over vectors, that the scalar field is actually a field and what not... we can instead focus all of our attention on proving that our set is closed under addition, is closed under scalar multiplication, and contains the "zero" vector (the additive identity of our set, in this case the $2\times 2$ matrix who contains only zeroes).

To check that the set is closed under addition, we need to check that any two matrices we take from our set such as $\begin{bmatrix}1&2\\3&-3\end{bmatrix}$ and $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, their sum will also wind up in the set.  Indeed, here we have $\begin{bmatrix}1&2\\3&-3\end{bmatrix}+\begin{bmatrix}1&0\\0&-1\end{bmatrix} = \begin{bmatrix}2&2\\3&-4\end{bmatrix}$ is also an element of the set since $2+2+(-4)=0$, the defining trait of what it means to be in our set.

Of course, there are infinitely many vectors in our set, we can't check every addition possible individually... so instead we choose to be general.

Suppose that $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and $\begin{bmatrix}a'&b'\\c'&d'\end{bmatrix}$ are two elements in our set of matrices.

 (the apostrophes here have no mathematical meaning and are just used in order to give the second set of variables similar but different names to tell them apart.  They could just as easily have been called $e,f,g,h$ or $\heartsuit,\clubsuit,\diamondsuit,\spadesuit$ or any other four symbols)

By the fact that they are elements of our set, that means by definition of our set that they satisfy $a+b+d=0$ and $a'+b'+d'=0$ respectively.

We ask whether their sum also belongs to the set, that is whether $\begin{bmatrix}(a+a')&(b+b')\\(c+c')&(d+d')\end{bmatrix}$ satisfies the property that the top left, top right, and bottom right entries add up to equal zero... that is whether or not $(a+a')+(b+b')+(d+d')=0$

Indeed, by a bit of algebraic rearranging, we have $$(a+a')+(b+b')+(d+d')=(a+b+d)+(a'+b'+d')=0+0=0$$

Here, we used what we knew about the original pair of matrices, that they were in the set as well and so individually had their entries satisfying the property.  This shows that regardless what two matrices we picked, so long as they were from our set their sum will also be in our set.

As for scalar multiplication, this follows similarly.  We let $k$ be any element from our scalar field and $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ be an element from our set of matrices.  We ask if $k\cdot \begin{bmatrix}a&b\\c&d\end{bmatrix}$ is also a matrix in our set.

Indeed, $k\cdot \begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}ka&kb\\kc&kd\end{bmatrix}$ and we have the sum of the top left, top right, and bottom right entries happens to be:

$$ka+kb+kd = k(a+b+d)=k(0)=0$$

Thus, our set is indeed closed under scalar multiplication.

We move on to asking about "if zero is an element of our set."  It is important here to understand what we mean by "zero."  There are many different mathematical objects, all of whom might be given the name "zero."  In this context, we mean very specifically that element from our proposed vector space who acts as an additive identity, an element who when added to any other will not change the result.  In our specific problem, that "zero" is very specifically the $2\times 2$ matrix whose entries are all zero, $\begin{bmatrix}0&0\\0&0\end{bmatrix}$.

So... we ask... is $\begin{bmatrix}0&0\\0&0\end{bmatrix}$ a matrix in our set?  Well, the sum of the top left, top right, and bottom right entries are $0+0+0=0$ so yes, it is an element of our set.

 It is worth pointing out that this condition can be replaced with the condition "Our set is non-empty" and the set of definitions would have been equivalent.  After all, a set containing "zero" is nonempty since it contains at least zero, and a set who is nonempty and closed under scalar multiplication would include some $V$ and also contain $0\cdot V$, the result of the scalar zero multiplied by the vector $V$ which results in the vector "zero."

By showing these three properties to hold true, we have accomplished what we wanted in showing the set to indeed be a vector subspace of $\mathcal M_{2\times 2}(\Bbb R)$ (what is written in your screenshot simply as $M$)

Now, on to the question of finding a basis and a dimension for this, the common technique is to try to think in terms of "free-variables."  If we can change a variable to a "free-variable" and allow it to freely change without affecting its ability to stay in our set, we do so, but keeping in mind that some variables are dependent on one another so not all can be simultaneously turned into free-variables.  Think of it like the process of parametrizing a curve.

In your case, $c$ is a completely free variable, not relying on anything else.  We could set $c$ as anything and it would never have any impact on whether or not it is an element of our set.  Let us replace $c$ by a free variable, what I will label as $t$.

Next, we can set some (but not all) of the others to free variables.  Letting $a$ be replaced by the free variable $r$ and letting $b$ be replaced by the free variable $s$ we see that $d$ is then forced to be equal to $-r-s$ in order to satisfy the defining property of our set that $a+b+d=0$.

That is to say, every element in our space can be written in the form:

$$\left\{\begin{bmatrix}r&s\\t&-r-s\end{bmatrix}~:~r,s,t\in\Bbb R\right\}$$

We can split this up and factor this as well, factoring the free variables outside, giving us:

$$\left\{r\cdot \begin{bmatrix}1&0\\0&-1\end{bmatrix} + s\cdot \begin{bmatrix}0&1\\0&-1\end{bmatrix}+t\cdot\begin{bmatrix}0&0\\1&0\end{bmatrix}~:~r,s,t\in\Bbb R\right\}$$

Well... this is precisely what is meant when we talk about the span of the vectors $\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\0&-1\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix}$.

These vectors, which can be trivially shown to be linearly independent and span our space, then satisfy the definition of what it means to be a basis of our space.

Finally, we note that there are three vectors in our basis for our space and so the dimension of the space is simply the number of vectors in the basis and is then equal to $3$.

It should be emphasized that this is just one among many possible choices of basis.  It is a well known theorem however that a vector space will always have the same number of basis elements regardless which basis is used to describe it and so the dimension is unambiguous and does not rely on any specific basis.  For us, we could have had as a basis $\begin{bmatrix}1&2\\3&-3\end{bmatrix},\begin{bmatrix}2&3\\4&-5\end{bmatrix},\begin{bmatrix}0&0\\42\pi&0\end{bmatrix}$ for instance and been just as correct.  It would not have necessarily been what was written in a solutions manual or what the majority of people would have come up with (like the earlier choice I gave would be) but would still have been just as mathematically correct.

Title: Vector spaces and subsets of Matrices


问:
Given a unit segment, construct one of length $4+\sqrt{3}$, and one of length $7+\sqrt{4+\sqrt{3}}$

Given a unit segment, construct a segment for the number $4 + \sqrt3$.

Given a unit segment, construct a segment for the number $7 + \sqrt{4 + \sqrt3}$.

I get how to use a unit segment and create a segment $\sqrt2$ and then use another unit segment at a right angle and get a segment $\sqrt3$. Then do I just tack on $4$ unit lengths to that line to construct a long segment that is $4 + \sqrt3$? And then it seems the next part builds on the last, but I’m not sure how.

答:
From the segment $4+\sqrt 3$ build a rectangle having height $1$ so its area is $4+\sqrt 3$

Now we build an equivalent square, so its side will be $\sqrt{4+\sqrt 3}$

With center in the midpoint $M$ of the segment $AB$ build a circle and the perpendicular to $AB$ in the point $H$. The intersection $C$ is the vertex of a triangle were

$CH^2=AH\cdot HB\to CH =\sqrt{4+\sqrt 3}$


问:
How do we define the $L^p$ norm of a tempered distribution?
I am finishing up a class on function theory and I am trying to reconcile
a few statements that I have seen.

Let us define $L^p(\mathbb R^n)$ to be the set of measurable functions $f$ so that
$\int_{\mathbb R^n} |f|^p dx < \infty$, with the norm $||f||_p = \Big(\int_{\mathbb R^n
} |f|^p dx\Big)^{1/p}$.

I have also seen it written that $L^p(\mathbb R^n)$ is the set of all tempered distributions
$f \in S'$ satisfying $ ||f||_p < \infty$. Here $S$ is the collection of rapidly decreasing complex valued functions on $\mathbb R^n$, and $S'$ is the space of continuous linear functionals on $S$ (tempered distributions).

I would like to show that these two definitions "are the same". However, I am at a disadvantage, since I don't even see how to define the $||\cdot||_p$ norm on
the space of tempered distributions.

Thus, how do we define the $L_p$ norm of a tempered distribution?

答:
You cannot define $\|\cdot\|_p$ in all space of tempered distributions. In general, if $T$ is a distribution, for the value $\|T\|_p$ make sense, we need to identify $T$ with a function defined in $\mathbb{R}^n$. There are some distributions that can not be identified with functions. On the other hand, the distributions that can be identified with functions and satisfies $\|T\|_p<\infty$, constitute the space $L^p$.

Consider for example the Dirac Distribution $\delta$. If you were to identify $\delta$ with some function $f$, then $f(x)=0$ in $\mathbb{R}\setminus\{0\}$, $f(0)=\infty$ and $\int_\mathbb{R} f(x)dx=\infty$. So, you can see that the notion of distribution, generalizes the notion of function.

ContentLicense: CC BY-SA 3.0


问:
What happens to the Stone-Cech compactification if you change “compact Hausdorff” to “T1 compact”?
I have asked a similar question elsewhere:

what happens to the Stone-Cech compactification if you change “compact Hausdorff” to “$T_1$ compact”? Here, I've added $T_1$
as opposed to this. Is $K$ there $T_1$ or at least $T_0$ ? In other words, does this modified thing always exist ?

答:
The answer is negative and has essentially already been established by Eric Wofsey. Following his construction almost exactly we assume that $X$ is a noncompact $T_1$ space and $f:X\rightarrow Y$ is a map into a compact $T_1$ space $Y$ through which any map $X\rightarrow Z$ into a compact $T_1$ space factors uniquely. Note that $X$ necessarily has infinitely many points.

Let $\widetilde K=X\sqcup\{a,b\}$ and topologise it by giving it the base of open sets generated by $i)$ the open subsets $U\subset X$, $ii)$ the sets $U=(X\setminus F)\cup\{a\}$ where $F\subseteq X$ is finite, $iii)$ the sets $U=(X\setminus G)\cup\{b\}$ where $G\subseteq X$ is finite, and $iv)$ the sets $U=(X\setminus H)\cup\{a,b\}$, where $H\subseteq X$ is finite.

Then $\widetilde K$ is compact $T_1$ so accepts a map $g:Y\rightarrow \widetilde K$ which is uniquely determined by the inclusion $i:X\hookrightarrow \widetilde K$. Let $\theta:\widetilde K\rightarrow\widetilde K$ be the map
$$\theta(x)=\begin{cases}x&x\in X\\b&x=a\\a&x=b.\end{cases}$$
Then $\theta$ is a homeomorphism satisfying $\theta i=i$. On the other hand the two maps $\theta g$ and $g$ are necessarily distict extensions of $i$. The full argument for this is that of Wofsey already cited above.

Replacing $T_1$ with $T_0$ everywhere above we see also that there are no initial maps into compact $T_0$ spaces. Similarly $T_1$ can be replaced by $T_D$ to arrive at the same conclusion for compact $T_D$ spaces.

Here is some discussion. Our argument hinged on two assumptions; $1)$ That $X$ is $T_1$, $2)$ that $f:X\rightarrow Y$ is a closed embedding.

It is easy to see that the second assumption causes no harm, and we can even assume without loss of generality that this embedding is dense. More subtle is the first assumption.

Now the full subcategory of $T_1$ spaces is extremal epireflective in $Top$. Thus for each space $X$ there is a $T_1$ space $X_1$ and a quotient surjection $X\rightarrow X_1$ through which any map $X\rightarrow Y$ into a $T_1$ space $Y$ will factor uniquely. This is all true for abstract reasons. Namely because the property of being $T_1$ is both productive and hereditary. The space $X_1$ can be realised as the quotient $X/\sim$, where $\sim$ is the intersection of all the closed equivalence relations on $X$.

Clearly if $X$ is compact, then so is $X_1$.

Each compact space has a reflection in compact $T_1$ spaces.

Of course by the same line of reasoning any space $X$, compact or not, for which $X_1$ is compact will have a reflection in compact $T_1$ spaces. As an example let $X=[0,\infty)$ topologised with the base of open sets $\{[0,x)\}_{x\in[0,\infty)}$. Then $X$ is a noncompact $T_0$ space which fails to be $T_1$, and its $T_1$ reflection $X_1=\ast$ is a singleton.

There are noncompact spaces with reflections in compact $T_1$ spaces.

This is all elementary and included only to indicate that the situation is more interesting than a blanket negative answer. It is also the setup required to verify that our restriction to $T_1$ spaces did no harm.

Question: Can the spaces which do admit reflections in compact $T_1$ spaces be characterised?

Of course if I knew how to answer this I would not pose it as a question.


问:
If $x^2 + ax - b$ is a factor of $x^3 - 2bx^2 + ax -6$, show that $a = -2b -6/b$.

If $x^2 + ax - b$ is a factor of $x^3 - 2bx^2 + ax -6$, show that $a = -2b -6/b$.

答:
Suppose $x^3-2bx^2+ax-6$ has the factor $x^2+ax-b$. Suppose the other factor is $x-q$. Then:

$$(x-q)(x^2+ax+b)=x^3-2bx^2+ax-6$$

But by expanding:

$$(x-q)(x^2+ax+b)=x^3-qx^2+ax^2-qax+bx-qb=x^3+(a-q)x^2+(b-aq)x-bq$$

By comparing coefficients we have the equations:

$$a-q=-2b, \quad b-aq = a, \quad -bq=6$$

Hence $q=a+2b$ and $6 = -b(a+2b)$. This will reduce to $a = -2b-6/b$.


问:
A question about $\mathfrak{su}(n)$
Let the $\{T^a\}$ be the generators of the adjoint representation of $\mathfrak{su}(n)$, and $f^{abc}$ the structure constants (therefore, $[T^a,T^b]=if^{abc}T^c$ with an implicit sum on the repeated index $c$). I am looking for sets of $n^2-1$ Hermitian traceless matrices $\{X^a\}$ of size $(n^2-1)\times(n^2-1)$ that satisfy
$$[T^a,X^b] = i f^{abc} X^c \tag{1}$$

Obviously, $X^a\equiv const\times T^a$ (with the same constant prefactor for all $a$'s) is a solution of this problem.

The matrices $\{D^a\}$ defined from the symmetric structure constants (sorry for the physicist's jargon; I do not know their proper name - these are the constants that appear, e.g., in the anticommutators of generators in the fundamental representation, $\{t^a,t^b\}=\delta^{ab}/n + d^{abc}t^c$)
by  $(D^a)_{bc}\equiv d^{abc}$ provide another solution.

Does (1) have other solutions (Hermitian traceless), besides linear combinations of the above two?

Is the extra assumption that the $X^a$ should be antisymmetric matrices sufficient for the $\{T^a\}$ to be the unique solution up to a rescaling?

答:
(1) describes tensors transforming under the adjoint representation, Hermitian traceless $(^2−1)×(^2−1)$ dimensional matrices; as a representation the adjoint, your T, is unique, but your Xs need not be a representation, just tensors.  Your T (call them F to comport with standard notation, e.g. Macfarlen et al 1968 ), and D     imaginary antisymmetric and real symmetric, respectively, are very sparse matrices.

But the D s are constructed out of the F, as per equation (2.27) of that reference,
$$
D_i= 2 d_{ijk}F_jF_k ~/n.
$$

The only isotropic tensors of su(n) are, I believe, $f_{ijk}$ and $d_{ijk}$, and you may combine them with the Fs and the Ds to construct adjoint tensors, like your solutions sought. However, bilinears of $f_{ijk}$ and $d_{ijk}$ correlate, equations (2.8-10), and trilinears collapse/reduce, equations (2.15-2.18), so  I have a  sense you will not find new solutions through longer strings of them, but I could be wrong. Recall the quadratic Casimir $F\cdot F$ , the cubic $F\cdot D$, etc..., trivializing the resulting contractions.

Your answer might be in the appendix of  Kaplan and  Resnikoff, Jou  Math  Phys  8 2194 (1967), but it could take work to tease it out...

Title: A question about $\mathfrak{su}(n)$


问:
Is it true that any 3-regular bipartite graph containing $K_{3,2}$ as subgraph must be non-planar?
I asked a similar question previously at Is it true that any 3-regular graph containing K_{3,2} as subgraph must be non-planar? The requirement can be relaxed so that the $K_{3,2}$ "locally" respect the bipartite structure, namely there can't be edge connecting two vertices on the dangling side. Many thanks in advance.

答:
Yes, it is true.

Combinatorially speaking, there is only one plane embedding of $K_{3,2}$: the one shown below.

To complete this graph to a $3$-regular bipartite graph, there must be at least one more vertex in the component containing copy of $K_{3,2}$. That vertex must be in the interior of one of the length-$4$ faces of this plane embedding. It doesn't matter which one, but let's put it in the interior of the face with corners $v_1, w_1, v_2, w_2$.

That vertex must be part of a fragment which is bipartite and nearly $3$-regular, except for an edge to one or both of $v_1, v_2$. Crucially, since $v_1$ and $v_2$ are in the same part of the bipartition, the edges to $v_1$ and/or $v_2$ must also come from the same side of the bipartition of the fragment.

But now, suppose that the fragment has $k$ vertices in one part and $\ell$ vertices in the other part (with the edges to $v_1$ and/or $v_2$ coming from the part of size $\ell$). Then there are $3k$ edges from the first part to the second, but either $3\ell-1$ or $3\ell-2$ edges from the second part to the first. This is a contradiction, so no bipartite $3$-regular planar graph can contain a $K_{3,2}$.

However, if we relax the condition "bipartite" to "triangle-free", then there are many counterexamples. Take your favorite bipartite, planar, $3$-regular graph $G$; delete an edge $u_1 u_2$ on the exterior boundary of a plane embedding of $G$, and splice it into the embedding above by adding edges $u_1v_1$ and $u_2v_2$ instead. (The result is no longer bipartite, because $u_1$ and $u_2$ were in different parts of the bipartition of $G$.)

We now have a planar triangle-free graph which is $3$-regular except for $v_3$. Take two copies of this graph and join them by an edge connecting $v_3$ and its copy, and we get a planar $3$-regular triangle-free graph that contains a copy of $K_{3,2}$.


问:
$ \sum_{a \mid n} \Bigl( f(a) \sum_{b \mid \frac{n}{a}} g(b) \Bigr) = \sum_{a \mid n} \Bigl( g(a) \sum_{b \mid \frac{n}{a}} f(b) \Bigr) $
Is the following equation true?

$$
\sum_{a \mid n} \Bigl( f(a) \sum_{b \mid \frac{n}{a}} g(b) \Bigr) =
\sum_{a \mid n} \Bigl( g(a) \sum_{b \mid \frac{n}{a}} f(b) \Bigr)
$$

If this equation is true, how can I prove it?

答:
We can conveniently prove this identity between arithmetical functions $f$ and $g$ using the Dirichlet convolution
\begin{align*}
(f\ast g)(n)=\sum_{d|n}f(d)g\left(\frac{n}{d}\right)
\end{align*}
The Dirichlet convolution is commutative $f\ast g=g\ast f$ and associative  $f\ast(g\ast h)=(f\ast g)\ast h$. We also use the constant function  $\zeta$ which is defined as
\begin{align*}
\zeta(n)=1\qquad\qquad n\geq 1
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{a|n}f(a)\sum_{b|\frac{n}{a}}g(b)}&=\sum_{a|n}f(a)\sum_{b|\frac{n}{a}}g(b)\zeta\left(\frac{\frac{n}{a}}{b}\right)\\
&=\sum_{a|n}f(a)\left(g\ast\zeta\right)\left(\frac{n}{a}\right)\\
&=\left(f\ast\left(g\ast\zeta\right)\right)(n)\\
&=\left(\left(g \ast\zeta\right)\ast f\right)(n)\\
&=\left(g \ast\left(\zeta\ast f\right)\right)(n)\\
&=\left(g \ast\left(f\ast \zeta\right)\right)(n)\\
&=\sum_{a|n}g(a)\left(f\ast \zeta\right)\left(\frac{n}{a}\right)\\
&=\sum_{a|n}g(a)\sum_{b|\frac{n}{a}}f(b)\zeta\left(\frac{\frac{n}{a}}{b}\right)\\
&\,\,\color{blue}{=\sum_{a|n}g(a)\sum_{b|\frac{n}{a}}f(b)}\\
\end{align*}
and the claim follows.


问:
When the primitive of a function is the function itself, then $f(x)=0$ (using the FTC)?
Let f: R → R be a continuous function such that:

$f(x)=\int_{0}^{x}f(t)\,dt$

Show that $f(x)=0$; for all $x$.

I did it like this:
According to the statement, $ f (x) $ is already the primitive of $ f (x) $, so $ f (x) $ is equal to $ F (x) $. By the Fundamental Theorem of Calculus we already know that $ F´ (x) = f (x) $ just like $ f (x) = F (x) $ we have $ F´ (x) = f´ (x) $ and therefore $ f (x) = f´ (x) $. I don't know how to show that it's zero.

答:
If $f(x)=f'(x)$ then $\frac{f'(x)}{f(x)}=1$. Integrating, you get that
$\ln f(x)=x+K$ whence $f(x)=ce^x$ for some $c\in\Bbb R$.

But you also know that $f(x)=\int_0^xf(t)dt$, so $f(0)=\int_0^0f(t)dt=0$.

Putting altogether $0=f(0)=ce^0=c$ and you'll get that $f(x)=0$.

EDIT: Thh above argument works whenever $f\ne0$. Observe that, in fact, $c\ne0$ but $f(x)=0$ is a trivial solution of $f'(x)=f(x)$. So we can write that all solutions of $f'(x)=f(x)$ are $f(x)=ce^x$ with $c\in \Bbb R$.


问:
prove that the unit circle $x^2+y^2=1$ is a closed set
I need to prove that the unit circle $x^2+y^2=1$ is a closed set in $\mathbb{R}^2$ is closed using convergent sequence method.</p>

答:
The euclidean norm $$(x,y)\mapsto ||(x,y)||$$
is continuous then if $(x_n,y_n)\in S^1\to(x,y)$ we have $$||(x_n,y_n)||=1\to ||(x,y)||=1$$
so $$(x,y)\in S^1\quad \text{hence} \quad S^1\quad \text{is closed subset}$$</p>


问:
Help with Derivation of: $-(\exists x) -Lx \vdash (\forall x) Lx$
I am trying to derive:  $-(\exists x) -Lx \vdash (\forall x) Lx$
As part of a broader understanding of relationships between universal affirmative; universal negative; particular affirmative; and particular negative.

But I am having trouble getting started with this one because of the negated Lx in the premise.

With a similar one, I was able to derive along the lines of:

$
1 (1)\hspace{10 mm} -(\exists x) -Lx \hspace{10 mm}A \\
2 (2)\hspace{10 mm} Lx \hspace{29 mm}A \\
2 (2)\hspace{10 mm} (\exists x) Lx \hspace{21 mm}2,\exists I\\
$

And then introduce a contraditction between $-(\exists x)Lx \wedge (\exists x)Lx)$, but again, the issue with the negated Lx in the premise is causing me some difficulty.

Does anyone have any suggestions on how I might approach this?
The logic system is restricted to primitive rules, so some of the obvious equivalencies are not allowed.

答: Your idea of deriving a contradiction with the premise is correct. To provoke a contradiction with $\neg \exists x \neg Lx$, you need to get $\exists x \neg Lx$. So just change your assumption on line 2 to $\neg Lx$. That's what you later want to apply the contradiction rule on.


问:
Grp and Ab are not equivalent as categories
I am trying to understand an example leading to the notion of adjoint.

Let $\mathbf{Grp}$ and $\textbf{Ab}$ denote the category of groups and abelian groups, respectively. Further, suppose that $U: \textbf{Ab} \to \textbf{Grp}$ denotes the forgetful functor and $F: \textbf{Grp} \to \textbf{Ab}$ be the abelization functor, i.e. for $G \in \text{Ob}(\textbf{Grp})$, we set $F(G) := G/G'$ with $G' = [G,G]$ being the commutator subgroup of G.

The claim is that (1) $F \circ U \cong \text{id}_{\textbf{Ab}}$, but (2) $U \circ F \not \cong \text{id}_{\textbf{Grp}}$.

Can you explain why (1) and (2) holds?

答:
Claim 1 holds, because for every group you can find a (natural) isomorphism between $FU(G)$ (the abelianization of $G$, that was already abelian and has just been regarded as a mere group, forgetting abelianity: the commutator subgroup $[G,G]$ is trivial if $G$ is abelian; actually, if and only if). If you're new to the concept of "natural isomorphism", this is a good first instance of how they work.

So, (somewhat sloppily, but this is the main idea) claim 2 holds by virtue of the fact that there exists at least one group that is not abelian. For example, the free group of two elements.


答:
As $f$ is continuous, all its antiderivatives are differentiable and the differential equation

$$f'(x)=f(x)$$ holds.

We know that the general solution of this ODE is $f(x)=ce^x$ (including $c=0$, from $f'(x)=f(x)=0$).

Now as $f(0)=0$, only the solution with $c=0$ is possible.

Title: When the primitive of a function is the function itself, then $f(x)=0$ (using the FTC)?

Body:
Let f: R → R be a continuous function such that:

$f(x)=\int_{0}^{x}f(t)\,dt$

Show that $f(x)=0$; for all $x$.

I did it like this:
According to the statement, $ f (x) $ is already the primitive of $ f (x) $, so $ f (x) $ is equal to $ F (x) $. By the Fundamental Theorem of Calculus we already know that $ F´ (x) = f (x) $ just like $ f (x) = F (x) $ we have $ F´ (x) = f´ (x) $ and therefore $ f (x) = f´ (x) $. I don't know how to show that it's zero.


问:
Proof that the method of Gauss/Jordan yields the inverse of a matrix
I have trouble in solving the following exercise: let A be an invertible matrix. Consider the matrix A|I where I is the identity matrix.
Prove that the matrix obtained by transforming A|I in it's row-echelon form is equal to I|B where B is the inverse matrix of A. Any help would be appreciated.</p>

答:
Hints:

(1) Any elementary row (column) operation on  $\,A\,$ is like multiplying $\,A\,$ from the left (from the right) by an elementary matrix $\,E_j\,$ ($\,F_j\;$)

(2) We know (I hope...) all the elementary matrices are invertible

(3) After you're done with the algorithm, you got

$$A\to E_rE_{r-1}\cdot\ldots\cdot E_1AF_1\cdot\ldots F_s=I$$

(4) If $\,A,B,P,Q\,$ are square matrices of the same order,  with $\,P,Q\,$ invertible, then

$$PAQ=B\iff A=P^{-1}BQ^{-1}$$

Edit: As noted in the comments by Sami, you can not mix row/columns operations, so you must choose either $\,E_j$'s or $\,F_j$'s but don't mix them.


答:
Let $A$ be abelian group. Apply $U$ to get $U(A)=A$. Now apply $F$ to get $FU(A) = A/A'$. Now, since $A$ is abelian, $A' = \{e\}$ is trivial. Thus, $FU(A)=A/A' \cong A$.

Now, choose $G$ nonabelian. Apply $F$ to get $F(G) = G/G'$. Now apply $U$ to get $UF(G) = G/G'$. Since $G$ was nonabelian, $G' \neq \{e\}$ and therefore $UF(G) = G/G' \not\cong G$.

Title: Grp and Ab are not equivalent as categories

Body:
I am trying to understand an example leading to the notion of adjoint.

Let $\mathbf{Grp}$ and $\textbf{Ab}$ denote the category of groups and abelian groups, respectively. Further, suppose that $U: \textbf{Ab} \to \textbf{Grp}$ denotes the forgetful functor and $F: \textbf{Grp} \to \textbf{Ab}$ be the abelization functor, i.e. for $G \in \text{Ob}(\textbf{Grp})$, we set $F(G) := G/G'$ with $G' = [G,G]$ being the commutator subgroup of G.

The claim is that (1) $F \circ U \cong \text{id}_{\textbf{Ab}}$, but (2) $U \circ F \not \cong \text{id}_{\textbf{Grp}}$.

Can you explain why (1) and (2) holds?


问:
Connection between Gauss and Poisson distribution?
I know that these distributions are connected by the central limit theorem.

But as it is written here: The central limit theorem says that the distribution of the mean of N draws from a probability distribution approaches a Gaussian of width $\frac{\sigma}{\sqrt{N}}$ for large N.

In the same script, in Eq(70) he says the standard deviation of a poisson distribution is $\sigma = \sqrt{\lambda t}$

Below Eq(79) he says the standard deviation of the gaussian (as limit of the poisson) is also  $\sigma = \sqrt{\lambda t}$.

For me this is confusing, where is with the $\frac{1}{\sqrt{N}}$ ?

In our case, the central limit theorem says that the distribution of the mean of N draws from the poisson distribution approaches a Gaussian.

As in the example of the radioactive decay:

I measure N times, the number of decays during the time t (this is the mean) and consider the distribution. This should be a Gaussian? But in this script, he do not sample over anything, he just says that he consider m >> 1 with m as the number of events during t. Is this right? I mean it should be right - he is professor in Harvard - but I don't get it.

答:
If you have $N$ independent Poisson random variables, each with parameter $\lambda$, their sum is Poisson with parameter $N\lambda$.  So a Poisson random
variable with large parameter can be considered as a sum of independent Poisson random variables, and you can apply the Central Limit Theorem to that.


问:
Computability -- prove the program does or does not exist
I know that we can prove a program is uncountable by basing it off the halting program, which does not exist. To clarify, we can use the theorem

The Halting Problem is uncomputable; i.e., there does not exist a
computer program TestHalt with the behavior specified above on all
inputs (P, x).

But I don't really get how to put it into practice.

How would you say a program P(F, x, y) that returns true if the program F outputs y when given x as input (i.e. F(x) = y) and false otherwise can't exist?

答: Suppose such program $P$ existed, but change it a little to output "yes" or "no" instead of true and false. Let $G$ be the program that outputs 1 on any input, and for any program $F$, let $F_1 = G\circ F$. Then $F_1$ is computable, so the function $H(x) =  P(F_1, x, 1)$ is also computable. But this function is exactly $\text{TestHalt}(F,x)$, a contradiction.


问:
Combinatorics identity sum of
Prove that:

$$\sum^{n}_{k=0}\binom{k}{2n-k}2^k = 2^{2n}$$

By using only combinatorics identities.

答:
From this result we have

$$2^n=\sum_{k=0}^n2^{-k}\binom{n+k}k=\frac1{2^n}\sum_{k=0}^n2^{n-k}\binom{n+k}k=\frac1{2^n}\sum_{k=0}^n2^k\binom{2n-k}{n-k}=\frac1{2^n}\sum_{k=0}^n2^k\binom{2n-k}n\;,$$

so

$$\sum_{k=0}^n2^k\binom{2n-k}n=2^{2n}\;,$$

which I suspect is the intended result.

Title: Combinatorics identity sum of


问:
How to find the number of marbles which someone has when it is a multiple of an unknown number?
The problem is as follows:

Mark has $\overline{ab}$ marbles and Ferdinand has $\overline{cde}$
marbles. Find the total of marbles which both have knowing that the
product of the number of marbles they have is the ninth part of
$\overline{abcde}$.

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&\textrm{237 marbles}\\
2.&\textrm{133 marbles}\\
3.&\textrm{126 marbles}\\
4.&\textrm{128 marbles}\\
5.&\textrm{132 marbles}\\
\end{array}$

How exactly are these numbers being asked found?. What I've attempted to do was to write the question into an equation.

Since it mentions that the product between the numbers is:

$$\overline{ab} \times \overline{cde}= \frac{1}{9} \overline{abcde}$$

And they ask to find $E=\overline{ab}+\overline{cde}$

This is assuming that the intended number system is on base 10.

But that's it. There's where I'm stuck. The only other thing which I could spot was that $c$
cannot be zero as well $a$. But from there I don't know how to use the hint which they have referring to being the ninth part of $\overline{abcde}$ does it exist some sort of identity or thing which to use to use this clue?.

Since I am lost into this question, can someone explain this in a step by step approach and the most detailed as possible to find the answer?. The part most confusing to me was to reduce the available options within the numbers to get the right combination to match what it is being asked.

More importantly, does it exist an approach to get the answer by finding individually first the digits involved instead of getting the requested answer first?.

答:
Let $\overline{ab} = x$ and $\overline{cde} = y$. Then, we want to find $x + y$ given $xy = \frac{1000x + y}{9}$. We have:

$$9xy = 1000x + y$$

$$9xy - 1000x - y = 0$$

$$81xy - 9000x - 9y = 0$$

$$81xy - 9000x - 9y + 1000 = 1000$$

$$(9x - 1)(9y - 1000) = 1000$$

Now, since $9x-1$ and $9y-1$ are integers, we can find factor pairs of $1000$ and solve for $x$ and $y$. (Keep in mind here that $x < 100$ and $y < 1000$.) We find:

$$(9x - 1)(9y -1000) = 125(8)$$

$$9x -1 = 125, 9y - 1000 = 8$$

$$9x = 126, 9y = 1008$$

$$x = 14, y = 112$$

$$x + y = \boxed{126}$$

An alternate, shorter solution (that may not work with different answer choices):

Note that $\overline{abcde}\equiv a + b + c + d +e\equiv 0\pmod{9}$ because we are working in base $10$. Then, $\overline{ab} + \overline{cde} = 10a + b + 100c + 10d + e \equiv a + b + c + d+e \equiv 0\pmod{9}$. Thus, the correct answer must be divisible by $9$. Note that the only answer choice divisible by $9$ is $126$, so this must be the correct answer.


问:
Tail behavior of series when means converge to zero.
Let $\{d_t\}_{t \geq 1}$ be a sequence of real numbers with each $d_t \in [0, 1]$. Suppose that we know that:
$$
\lim_{T \to \infty} \frac{1}{T} \sum_{t=1}^{T} d_t = 0 \:. \quad\quad (1)
$$

Fix a constant $\alpha \in (0, 1)$.
Is it necessarily true that there must exist an $\beta \in (0, 1)$ such that:
$$
\limsup_{T \to \infty} \max_{k=\lceil \alpha T \rceil, ..., T} \frac{1}{(T-k+1)^\beta} \sum_{t=k}^{T} d_t < \infty \:. \quad\quad (2)
$$

For example, consider the sequence with $d_t = 1/\sqrt{t}$.
Then:
$$
\sum_{t=1}^{T} d_t = \sum_{t=1}^{T} \frac{1}{\sqrt{t}} \leq 2 \sqrt{T} \:,
$$
and so condition (1) holds.
On the other hand:
$$
\sum_{t=k}^{T} d_t = \sum_{t=k}^{T} \frac{1}{\sqrt{t}} \leq 2 (\sqrt{T} - \sqrt{k-1}) \:.
$$
By the concavity of $x \mapsto \sqrt{x}$, we can upper bound for $k \geq 2$:
$$
  \sqrt{T} - \sqrt{k-1} \leq \frac{1}{2\sqrt{k-1}} (T - k + 1) \:.
$$
Hence when $k \geq \alpha T$:
$$
  \frac{1}{\sqrt{T-k+1}} \sum_{t=k}^{T} d_t \leq \frac{\sqrt{T-k+1}}{\sqrt{k-1}} = \sqrt{\frac{T}{k-1} - 1} \leq \sqrt{\frac{1}{\alpha} \frac{T}{T-1/\alpha} - 1} \:.
$$
It is clear that this bound is finite as $T \to \infty$.
Hence (2) holds with $\beta=1/2$.

My question is if this type of behavior is general or is there a counter-example when this does not occur?

答:
Here is the counter example.

We generate a sequence $\{d_t\}$ according to the following procedure. Initially set all $d_t$'s equal to zero. Now for $k=1, 2, ...$, we set $d_t = 1$ for all $t \in [2^k - (k-1), 2^k]$.

First, (1) holds since $\sum_{t=k}^{T} d_t$ is bounded by roughly $\log_2^2(T)$, since at position $T$ there are at most $\lceil{\log_2(T)\rceil}$ subsequences of consecutive 1's in $\{d_t\}_{t=1}^{T}$, each of length at most $\log_2(T)$.

Now (2) clearly does not hold, since when $T=2^i$ for some $i$, then for $k=T - \log_2(T) + 1$ the tail sum $\sum_{t=k}^{T} d_t = T - k$.


问:
Help with iteratively solving an overconstrained system of non-linear equations
I'm working on a problem at my job, where I am trying to estimate the locations of N points in 3D space, given the distances between pairs of points. Ideally all $N(N+1)/2$ distances would be known, although it could be less, though I don't expect it would be by much. Each point is guaranteed to know the distance to at least 4 other points.

Here's what I have so far:

In order to remove the ambiguity of global orientation (which can be solved for later using a method external to this problem), the first point, P1, is fixed at the origin, and the next 3 are placed in the first octant. For simplicity, P2 is placed on the positive x-axis, P3 in the positive x-y plane, and P4 in the positive x-y-z octant. This fixes 6 variables (x1, y1, z1, y2, z2, z3 all equal zero), and constrains 6 more (x2, x3, y3, x4, y4, z4 all greater than zero). The points P1-P4 could be solved exactly for using multilateration, but I don't know if that is better or worse than including them in the system of equations.

If I know $M \le \frac{N(N+1)}{2}$ distances, then I can set up a system of M x 3N equations of the form $f_{mn} = (x_m - x_n)^2 + (y_m - y_n)^2 + (z_m - z_n)^2 - d_{mn}^2$ . If $M = 3N$, then I am familiar with solving this through the Newton-Raphson method using the Jacobian. However, M is expected to almost always be greater than 3N. Using the constraints mentioned in the previous point, I can reduce this down to something like $M - 6 \times 3N - 12$, or somewhere in between, but again I'm not certain whether or not that is helpful.

Here's what I am trying to figure out:

Is this the sort of problem I could solve with an iterative method? I am aware of the Newton-Raphson method for an exactly constrained system, but I do not know what to do about an overconstrained system or one with variable constraints. I am an engineer more than a mathematician, so it is slightly hard to find the right sorts of resources on these things. If an iterative method isn't appropriate, there is a search method I could use, where I would perform multilateration on successive points, but I would prefer not to resort to this, as it can be tricky to programmatically determine points that are least likely to be coplanar, which would propagate large errors.

Is there information on what sort of errors I can expect? Ideally, is there information on positional error as a function of uncertainty in the distance between points? This is something I could determine after the fact, but if there is literature already, then that would be great.

Thanks for any help! I really appreciate it

答:
The typical approach to overconstraint systems of equations is least squares approach. Instead of solving $f_{nm} = 0$ the least squares method minimizes
$$
J = \sum_{(n, m) \in \mathcal M} f_{nm}^2 \to \min_{x_i, y_i, z_i}. (*)
$$
Here I've denoted by $\mathcal M$ all the point pairs with known distance between them.
You might take a look at Levenberg–Marquardt algorithm, it was designed for generic nonlinear least-squares problem.

Sadly, this might not work for big N, or may require very close initial guess. The method might find a local optimum (just like in case of $M = 3N$ the Newton's method might hit a singular jacobian and terminate without a solution).

If you know most of the pairwise distances, then you may try to solve the problem in a flood fill manner. Let $K$ be the set of points with known coordinates. On the start $K = \{1, 2, 3, 4\}$. Then you repeatedly choose a point $q$ that has most known distances to points from $K$. If there are many of such points, take the closest (e.g. by sum of the distances) to $K$. Solve the problem for $(x_q, y_q, z_q)$ (in a least-squares fashion) and add $q$ to $K$. Repeat until no more points left.

If this algorithms succeeds to find all points' coordinates you may refine them further by taking them as initial guess for global optimization problem (*).

I made a proof-of-concept Jupyter notebook with this algorithm. I'm not sure if it works all the time, but in various random cases it recovers original points pretty well (up to rigid motion and distance perturbation, of course). Sometimes, locating point by known points gave incorrect result (local optimum problems, as I said before) but choosing initial guess as average of known points seemed to fix that.

Here's the link to the code
https://gist.github.com/uranix/c5e5e6db687eda0d12ffa4fa570bc3b4


问:
Find the differential equation that solves a certain problem
I'm at the beggining of a differential equation's course, and I need to solve the following problem:

Find every curve in the plane $XY$ so that the middle point of the
normal's segment between each point $(x,y)$ of the curve and the
abscissa axis ($OX$ axis) travels the $y^2=px$ parabola.

I tried to translate the problem the best I can, so I apologise if it's not the best way to express it (You can tell me in the comments if it's not clear). I just want help finding the equation. I don't need help solving the equation, my problem is translating this problem into a differential equation I can solve. I want the $y$ variable to be a function of $x$, to be said, the variables of my equation will be $x$ and $y(x)$ (I'm not sure if it's easier doing it any other way or it just doesn't matter using $y$ and $x(y)$ instead). Any help or hint will be appreciated.

答:
At $(X,Y)$, the normal to the curve is given by$$\frac{y-Y}{x-X}=-\frac 1{y'}$$The normal's $x$ intercept is $(X+Yy',0)$ and the midpoint of the segment between $(X,Y)$ and $(X+Yy',0)$ is $(X+Yy'/2,Y/2)$. This point lies on $y^2=px$, giving you the differential equation$$Y^2/4=p(X+Yy'/2)$$

We get$$y^2=4px+2pyy'\implies yy'-y^2/2p=-2x$$Keep $y^2=v,2yy'=v'$, so$$v'/2-v/2p=-2x\implies v'-v/p=-4x$$which is a first order linear ODE with solution$$\begin{align*}ve^{-x/p}&=-\int4xe^{-x/p}dx\\&=-4p^2\int me^mdm~~[\because m=-x/p]\\&=-4p^2(m-1)e^m+k\end{align*}$$Finally we get $y^2=4p^2+4px+ke^{x/p}$.


问:
Arrange books on a shelf- combinatorics
I wonder if I'm correct at this question:

You have $20$ different books and $3$ of them are books about tennis. In how many ways can you put $12$ books on a shelf if order matters and the $3$ tennis books must be on the shelf?

My attempt was: $3! \cdot P(17,9) \cdot C(12,3)$. There are $3!$ ways to arrange the tennis books, $P(17,9)$ ways to arrange the other books and out of $12$ positions three is supposed to be for tennis. Am I correct?

答: Just 9 book form 17, that you can do on $${17\choose 9}$$ ways and then arrange all 12 book on a shelf that is $12!$ ways. Multiply both and you are done.

Title: Arrange books on a shelf- combinatorics


问:
Probability of choosing a cell in a $ N\times M$ grid with specific conditions
I have this problem :
Given a $N\times M$ grid.
You cover $2$ cells by a $1\times2$ tile.
If you pick a cell,what is the probability for it to be covered by the tile?

It's not hard to evaluate the number of the tile possible positions, which is $N(M-1)$.
But how could I find the answer ?

答:
For any fixed tile placement, the probability is $\frac{2}{MN}$ that a randomly chosen cell is covered by the tile. There are $2$ cells covered, out of $MN$ cells total.

Since this is true no matter where the tile is placed, it is still true when the tile is placed randomly.

(Note, though, that once you pick a cell, the probability that a randomly placed tile will cover it can vary. Corner and edge cells are less likely to be covered by a tile. That's not relevant for this problem, but I'm mentioning it for completeness.)


问:
The inf and sup of sets
Do the following identitities hold:
Inf(S+T)=inf(S)+inf(T), where S and T are sets of nonnegative real numbers
Also sup(S+T)=sup(S)+sup(T)</p>

答:
It's true: Let $m = \inf \{S+T\}$. i.e.

$$m = \inf_{s \in S, t \in T}s + t $$

which implies that for $\epsilon > 0$, $\exists s \in S, t \in T$ s.t

$$m \leq s + t \leq  m + \epsilon$$
$$\Rightarrow \inf_{s \in S} s +t \leq  m + \epsilon$$
$$\Rightarrow \inf_{s \in S} s +\inf_{t \in T}t \leq  m + \epsilon$$

As $\epsilon$ is arbitrary now, we have
$$\Rightarrow \inf_{s \in S} s +\inf_{t \in T}t \leq  m $$

Now for the reverse, let $m_1 = \inf_{s \in S} s$, $m_2 = \inf_{t \in T} t$

Then by similar arguments, $\forall \epsilon > 0$, $\exists s \in S, t \in T$ s.t
$$ s \leq m_1 + \epsilon/2, t \leq m_2 + \epsilon/2$$

$$\Rightarrow s+t \leq m_1 + m_2 + \epsilon$$

$$\Rightarrow \inf_{s \in S, t \in T}s + t \leq m_1 + m_2 + \epsilon$$
Take $\epsilon$ to 0 to get the reverse.
QED

Title: The inf and sup of sets


问:
Proof using derivative definition?

Let $\varphi(h)$ be a function such that $\varphi (h)\neq 0$ for all
$h \neq 0$ and $\lim_{h \rightarrow 0} \varphi(h)=0$. Prove that for
any differentiable function $f(x)$ that

$$ \lim_{h \rightarrow 0}\frac{f(x+\varphi(h))-f(x)}{\varphi(h)} =
f'(x) $$

This seems obviously true to me but I'm having trouble writing the proof. I tried using sequence criteria like so:

$$ Let \space h = \frac{1}{n} $$

$$ f'(x) = \lim_{n \rightarrow
 \infty}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}} $$

$$ f'(x) = \lim_{n \rightarrow
 \infty}\frac{f(x+\varphi(\frac{1}{n}))-f(x)}{\varphi(\frac{1}{n})} $$

$$ f'(x) = \lim_{h \rightarrow
 0}\frac{f(x+\varphi(h))-f(x)}{\varphi(h)} $$

But this isn't convincing and I'm not sure if it's accurate. I would appreciate a nudge in the right direction.

答:
Here is a proof that uses the $(\varepsilon,\delta)$ definition of a limit.

It was assumed that $f$ is differentiable, so $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ exists for every $x\in\text{dom}[f]$. We deduce from the $(\varepsilon,\delta)$ definition of a limit that given $\varepsilon>0$, there exists a $\delta_1>0$ such that

$$\left|\frac{f(x+h)-f(x)}{h}-f'(x)\right|<\varepsilon$$

for all $0<|h-0|<\delta_1$. It was also assumed that $\lim_{h\to0}\varphi(h)=0$, so there exists a $\delta>0$ such that

$$|\varphi(h)-0|<\delta_1$$

for all $0<|h-0|<\delta$.

Notice that we must have $0<|\varphi(h)-0|$ because $\varphi(h)\neq 0$ for every $h\neq 0$. Therefore, if $0<|h-0|<\delta$, then $0<|\varphi(h)-0|<\delta_1$, and since

$$\left|\frac{f(x+h)-f(x)}{h}-f'(x)\right|<\varepsilon$$

is true for every $h$ with $0<|h-0|<\delta_1$, it will also be true for $\varphi(h)$. Thus,

$$\left|\frac{f(x+\varphi(h))-f(x)}{\varphi(h)}-f'(x)\right|<\varepsilon$$

Since $\varepsilon$ was given arbitrarily, the same argument applies to every positive real number. Thus,

$$\lim\limits_{h\to 0}\frac{f(x+\varphi(h))-f(x)}{\varphi(h)}=f'(x)$$

This style of argument can be used to prove a wide variety of limits.

Title: Proof using derivative definition?


问:
Double series and series
Let me start with a simpler version of my actual problem. Say I have a partition $\mathbb{N} = \cup_k I_k$ where the $I_k$ are disjoint, and an absolutely convergent series $\sum_n x_n$ in $\mathbb{R}^d$. I want to show the identity $\sum_n x_n = \sum_k \sum_{i\in I_k} x_i$. In this simple situation I can show this using Fubini's theorem, but actually I'm interested in the situation of Banach space valued series that converge unconditionally, so I'd appreciate a more elementary approach that is better suited for generalization.

答:
Thanks to Daniel Fischer's hint, I was able to solve the question on my own. I will focus on my specific issue with Banach space valued series, but the scheme is probably always the same.

I recall the definition of summability in a Banach space from "Analysis in Banach spaces" by Hytönen, van Nerven, Veraar and Weis, Definition 4.1.2. They also have a proposition included that shows equivalence to several other definitions, including invariance under reordering, see Proposition 4.1.6 in-there.

Def: An indexed family $(x_j)_j$ in a Banach space $X$ is called summable, if there is $x\in X$ such that for all $\varepsilon > 0$ there is a finite set $F_\varepsilon \subseteq J$ such that if $F\subseteq J$ is a finite set containing $F_\varepsilon$, then $\| x - \sum_{j\in F} x_j \| \leq \varepsilon$. Write $\sum_j x_j = x$.

Proof: Let $(x_j)_j$ be a summable family in a Banach space. Note that also its subfamilies are summable, see the comment of Daniel Fischer below and Lemma 4.1.4 in the aforementioned book for a proof of the used characterization. Also, let $\varepsilon > 0$. Moreover, let $J^k$ be a partition of $J$ indexed by the natural numbers.

Put $x_k := \sum_{j\in J^k} x_j$, which is well defined by summability over subcollections. Also, there are finite sets $J^k_0 \subseteq J^k$ such that all finite subsets $J_k' \subseteq J^k$ containing $J^k_0$ satisfy $\| x_k - \sum_{j\in J_k'} x_j \| \leq \varepsilon 2^{-k}$. Similarly, with $x:= \sum_j x_j$ there is a finite set $J_0 \subseteq J$ such that any finite set $J' \subseteq J$ containing $J_0$ verifies $\| x - \sum_{j\in J'} x_j \| \leq \varepsilon$.

Define $J_k' := J^k_0 \cup (J_0 \cap J^k)$ and let $N \geq 0$ be such that $J' := \cup_{k=1}^N J_k' \supseteq J_0$, which is possible by finiteness of $J_0$. Note that $\sum_{k=1}^N \sum_{j\in J_k'} x_j = \sum_{j\in J'} x_j$ by finiteness of the sums. Then

$$\| x - \sum_{k=1}^N x_k \| \leq \| x - \sum_{k=1}^N \sum_{j\in J_k'} x_j \| + \sum_{k=1}^N \| x_k - \sum_{j\in J_k'} x_j \| \leq \varepsilon + \varepsilon \sum_{k=1}^N 2^{-k} \leq 3\varepsilon.$$

Note that the bound is independent of $N$. Hence, we can take the limit $N\to \infty$ to conclude.

Title: Double series and series


问:
Finding all nonzero integers $k$ so that $\sqrt{k^2 - pk}$ is a positive integer
Let $p$ be a prime. Find all nonzero integers $k$ such that $\sqrt{k^2 − pk}$ is a positive integer.

I first let $k^2 - pk = x^2,$ where $x$ is a positive integer. However, I got stuck from here as I wasn't quite sure what to do with the condition that $p$ is a prime. Can someone help me please?

答:
hint

Consider the condition
$$k^2-pk-x^2=0$$

as a quadratic in $ k $ with discriminant

$$\Delta= p^2+4x^2$$

which must be a perfect square

$$p^2=y^2-4x^2=(y+2x)(y-2x)$$


问:
What is the number of $s-t$ cuts on $n$ vertices?
I am trying to know why exactly the number of $s-t$ cuts (cuts that have $s$ in the tail and $t$ in the head) is $2^{n-2}$?

I tried to assume that $s$ has to be on the first set and $t$ has to be in the second set:

$=<\{s,.... \},\{t,.... \}>$

Then since the order is relevant, so we have to use permutation. And we have to fix $s$ and $t$, so could you please tell me to proceed from here to get the final answer $2^{n-2}$?

答: For each of the $n-2$ vertices other than $s$ and $t$, you have a choice between $2$ options: put the vertex in the first set or the second set. These choices are independent, giving $2^{n-2}$ total outcomes.


问:
If $x^2+ky^2$ represents $a$ and $ab$ must it also represent $b$?
Consider the binary quadratic form $f(x,y) = x^2 + k y^2$ over the integers.

If there exists integers $u,v$ such that $f(u,v) = a$, we say that the form can represent $a$.

I can show that if $f$ represents $a$ and $b$ that it must also represent $ab$.

Is it possible to partially invert this and show that if $f$ represents $a$ and $ab$ it must also represent $b$?

If not, what are some necessary conditions which would allow this conclusions to go through? Maybe some restriction on the class number $h(-4k)$? I may have misunderstood, but I saw a lecture that seemed to assume this is obvious when the class number is odd. I'm hoping there is a simple reason it is true in general, but am having trouble proving anything.

My attempt

First note:
$$
\begin{aligned}
(r^2 + k s^2) (t^2 + k u^2)
&= (r t)^2 + (k s u)^2 + k((s t)^2 + (r u)^2) \\
&= (r t + k s u)^2 - 2k r u s t + k (s t - r u)^2 + k 2 r u s t \\
&= (rt+ksu)^2 + k (st-ru)^2 \\
\end{aligned}
$$

Therefore if $f$ represents $a$ and $b$ that it must also represent $ab$.

Now focusing on whether representation of $a$ and $ab$ implies representation of $b$, I'll start with some given representation of $a$ and $ab$ and try to construct a representation of $b$.
$$
\begin{aligned}
a &= r^2 + k s^2 \\
ab &= x^2 + k y^2 \\
\end{aligned}
$$
Then using the identity above, try to solve for $t,u$ from
$$
\begin{aligned}
x &= (rt+ksu) \\
y &= (st-ru) \\
\end{aligned}
$$
as this would give $t^2 + k u^2 = b$.

Combining the equations we find:
$$
\begin{aligned}
sx - ry &= u (k s^2 + r^2) = u a \\
rx + ksy &= t (r^2 + k s^2) = t a \\
\end{aligned}
$$
We will get the desired conclusion if we can show
$$
\begin{aligned}
sx - ry &= 0 \pmod{a} \\
rx + ksy &= 0 \pmod{a} \\
\end{aligned}
$$
so that our solutions for $t$ and $u$ are integers.

I'm stuck here, and not sure if this is even the appropriate path to try to prove this.

答:
Inspired by the conditions @WhatsUp found experimentally, I can now prove they are necessary.

That is:
If $k>1$ then it must be square free, and for $k \ge 4$ it must be 1 or 2 mod 4.

Note: These conditions are not sufficient, as noted by the $k=14$ case tested by WhatsUp.

Step 1. Proving that $k>1$ cannot be a square.

If $k=n^2 > 1$, then

$f(0,1) = k$

$f(n,1) = 2k$

but $f$ cannot represent 2 since it is less than $k$ and not a square.

So to have this property, if $k>1$ it cannot be a square.

Step 2. Proving that $k>1$ must be square free.

If $k=an^2$ with $a,n>1$ and $a$ is square free, then

$f(0,1) = an^2$

$f(n,0) = n^2$

but $f$ cannot represent $a$ because it is less than $k$ and not a square.

So combined with the previous step we can conclude that for the quadratic form to have this property, if $k>1$ it must be square free.

Step 3. Proving the $\pmod 4$ conditions.

If $k=4n$ then it is not square free, and so if $k>1$ and $k \equiv 0 \pmod 4$ the quadratic form will not have the desired property.

For $k=4n+3$ it is helpful to split into a case analysis of when $(n+1)$ is a square or not.

If $k=4n+3$ with $n>0$ and $(n+1)$ not a square, then

$f(1,1) = 1 + k = 4(n+1)$

$f(2,0) = 4$

but $f$ cannot represent $(n+1)$ because it is less than $k$ and not a square.

If $k=4n+3$ with $n>0$ and $(n+1)=m^2$, then

$k = 4(m^2-1)+3 = 4m^2 - 1 = (2m+1)(2m-1)$

$f(2m + 1, 1) = (2m+1)^2 + (2m+1)(2m-1) = (2m+1)(4m)$

$f(2m - 1, 1) = (2m-1)^2 + (2m+1)(2m-1) = (2m-1)(4m)$

$f(2,0) = 4$

As $(2m+1)m$ and $(2m-1)m$ are less than $k$, they must both be squares to be representable. Since $\gcd(2m+1,m)=1$, and similarly $\gcd(2m-1,m)=1$, this also requires $2m+1$, $2m-1$, and $m$ to all be squares for these values to be representable. However there are no integer squares that differ by 2, so this is impossible.

All together this allows us to conclude that for the quadratic form to have the desired property, if $k \ge 4$ then $k$ must be 1 or 2 mod 4.

Other Notes

I don't quite see what is going on yet, but the "bad" values of k seem to be related to almost all values that can be taken by a similar "form" of quadratic forms.

$$k = m^2 n - r^2$$

The main part that doesn't look nice in this is that the $n=1$ case seems to limit $r \in \{0,1\}$.

Consider $k = m^2$ with $m>1$

We've already shown these values of $k$ do not work.

Consider $k = m^2 - 1 > 4$

I couldn't find a quick example, but using previous work:

If $m$ is even, then $m^2 = 4t$, so $k = m^2 - 1 = 4(t-1) + 3$.

If $m$ is odd, then $m^2 = 4t + 1$, so $k = m^2 -1 = 4t + 0$.

So these $k$ will not work.

This appear to be the limit of general arguments against the difference of squares, as WhatsUp tests found that $k = 3^2 - 2^2 = 5$ was probably okay.

Consider $k = m^2n - 1$ with $m,n > 1$ and $n$ square-free

$f(1,1) = m^2n$

$f(m,0) = m^2$

cannot represent $n$, so this value of $k$ will no work.

More generally,

$k = m^2 n - r^2$ with $m>1$, $k>n>1$, and $n$ square free

$f(r,1) = m^2 n$

$f(m,0) = m^2$

Cannot represent $n$, so this value of $k$ will not work.

Unlike the $n=1$ case, here the values of $r$ are much less restricted.

Since another quadratic form seems to be rearing its head, there is probably something simpler connecting all this. Hopefully someone else can see the thread I am missing, and unravel the rest.


答:
Proceed with your notation. We have $k^2-x^2 = pk \iff (k-x)(k+x) = pk$. Now since $p$ is a prime, it must be the case that at least one of the $k-x$ and $k+x$ is a multiple of $p$.

(1) If $x$ and $k$ are both multiple of $p$: let $k$ = $mp$ and $x=np$, thus the equation simplifies to $m(m-1)=n^2$. Since $\operatorname{gcd}(m,m-1) = 1$, it must be the case that $m$ and $m-1$ are squares. The only possibility is $m=1$.

(2) Otherwise, $k$ and $x$ can't be multiple of $p$ (why?). Also observe that $(k-x)<p$. Can you continue from here?

Title: Finding all nonzero integers $k$ so that $\sqrt{k^2 - pk}$ is a positive integer

Body:
Let $p$ be a prime. Find all nonzero integers $k$ such that $\sqrt{k^2 − pk}$ is a positive integer.

I first let $k^2 - pk = x^2,$ where $x$ is a positive integer. However, I got stuck from here as I wasn't quite sure what to do with the condition that $p$ is a prime. Can someone help me please?


问:
Use of completeness theorem in decidability of monadic predicate logic
One way of proving the decidability of monadic predicate logic is via the finite model lemma:

If a monadic formula F with $k$ unary predicates is satisfiable, then it is satisfiable in a domain with at most $2^k$ elements.

To prove this lemma we assume that F is satisfiable and construct a model with size $|2^k|$. Most proofs assume that decidability follows immediately from this lemma. However, isn't it also necessary to also show that if F is $not$ satisfiable, we can demonstrate this unsatisfiability by a finite procedure?

My guess is that this requires an implicit (but essential) appeal to the completeness theorem: if ~F is unsatisfiable, then ~F is refutable.

答:
No, the lemma on its own is enough.

Here's the decision procedure to tell if a monadic second-order sentence $\varphi$ using only unary relation symbols is satisfiable:

Search through all structures in the language of $\varphi$ of size $\le 2^{\vert\varphi\vert}$ for a model of $\varphi$. If you find one, then $\varphi$ is satisfiable; if you don't find one, then $\varphi$ is not satisfiable.

We have to prove four things about this process:

It's computable.

It always halts and gives an answer.

It never gives a false positive: if the algorithm says "$\varphi$ is satisfiable" then $\varphi$ is actually satisfiable.

It never gives a false negative: if the algorithm says "$\varphi$ is not satisfiable" then $\varphi$ is actually not satisfiable.

The first two bulletpoints are immediate. The fourth bulletpoint (or rather, its contrapositive) follows from the lemma in the OP: by the lemma, if $\varphi$ is satisfiable then the algorithm above will indeed report $\varphi$ to be satisfiable.

The third bulletpoint meanwhile is trivial: the only way the algorithm declares $\varphi$ to be satisfiable is if it actually finds an actual model of $\varphi$, so by definition it can't falsely conclude that $\varphi$ is satisfiable.


问:
Largest prime factor of 600851475143
I'm trying to use a program to find the largest prime factor of 600851475143.
This is for Project Euler here: http://projecteuler.net/problem=3

I first attempted this with the code that goes through every number up to 600851475143, tests its divisibility, and adds it to an array of prime factors, printing out the largest.

This is great for small numbers, but for large numbers it would take a VERY long time (and a lot of memory).

Now I took university calculus a while ago, but I'm pretty rusty and haven't kept up on my math since.

I don't want a straight up answer, but I'd like to be pointed toward resources or told what I need to learn to implement some of the algorithms I've seen around in my program.

答:
The thing with Project Euler is that there is usually an obvious brute-force method to do the problem, which will take just about forever. As the questions become more difficult, you will need to implement clever solutions.

One way you can solve this problem is to use a loop that always finds the smallest (positive integer) factor of a number. When the smallest factor of a number is that number, then you've found the greatest prime factor!

Detailed Algorithm description:

You can do this by keeping three variables:

The number you are trying to factor (A)

A current divisor store (B)

A largest divisor store (C)

Initially, let (A) be the number you are interested in - in this case, it is 600851475143. Then let (B) be 2. Have a conditional that checks if (A) is divisible by (B). If it is divisible, divide (A) by (B), reset (B) to 2, and go back to checking if (A) is divisible by (B). Else, if (A) is not divisible by (B), increment (B) by +1 and then check if (A) is divisible by (B). Run the loop until (A) is 1. The (3) you return will be the largest prime divisor of 600851475143.

There are numerous ways you could make this more effective - instead of incrementing to the next integer, you could increment to the next necessarily prime integer, and instead of keeping a largest divisor store, you could just return the current number when its only divisor is itself. However, the algorithm I described above will run in seconds regardless.

You said you don't want a straight up answer. Regardless, if you want to see one implementation of it, I've uploaded my code for this problem on Pastebin: here (C) and here (Python).

Example: Let's find the largest prime factor of 105 using the method described above.

Let (A) = 105. (B) = 2 (we always start with 2), and we don't have a value for (C) yet.

Is (A) divisible  by (B)? No. Increment (B) by +1: (B) = 3. Is Is (A) divisible  by (B)? Yes. (105/3 = 35). The largest divisor  found so far is 3. Let (C) = 3. Update (A) = 35. Reset (B) = 2.

Now, is (A) divisible  by (B)? No. Increment (B) by +1: (B) = 3. Is (A) divisible  by (B)? No. Increment (B) by +1: (B) = 4. Is (A) divisible  by (B)? No. Increment (B) by +1: (B) = 5. Is (A) divisible  by (B)? Yes. (35/5 = 7). The largest divisor we found previously is stored in (C). (C) is currently 3. 5 is larger than 3, so we update (C) = 5. We update (A)=7. We reset (B)=2.

Then we repeat the process for (A), but we will just keep incrementing (B) until (B)=(A), because 7 is prime and has no divisors other than itself and 1. (We could already stop when (B)>((A)/2), as you cannot have integer divisors greater than half of a number - the smallest possible divisor (other than 1) of any number is 2!)

So at that point we return (A) = 7.

Try doing a few of these by hand, and you'll get the hang of the idea. Or just play with the code I've provided. Insert some print statements, see how it iterates through the numbers.

Title: Largest prime factor of 600851475143


问:
Can an open interval be bounded?
I have a question which is as follows:

Let $A = \{(-\cos x, \sin x)\} \in \mathbb{R^2} : x \in (-\pi, \pi)$

I have to justify what kind of set it is. I said it's an open and bounded set. I would say it is bounded since cos and sin are both bounded functions. However, the only thing that I'm unsure of is the fact that its only defined at $x \in (-\pi, \pi)$. I'm guessing that it doesn't really matter if its not defined at $-\pi$ or $\pi$ because the values at that points are irrelevant for concluding whether A is bounded or not. And as for whether the set is open or not, I said that it's open since the domain is open since the limit points $-\pi$ and $\pi$ are not included in the set.

答:
$A$ is simply the unit circle minus the point $\langle 1,0\rangle$, so it is indeed bounded: every point of $A$ is exactly $1$ unit from the origin. It is not, however, open: it does not contain an open ball around any of its points. It is also not closed, since $\langle 1,0\rangle$ is in its closure.

It is true $(-\pi,\pi)$ is open in $\Bbb R$ and that $A$ is the image of that open set under the continuous function

$$f:\Bbb R\to\Bbb R^2:t\mapsto\langle-\cos t,\sin t\rangle\,,$$

but that function is not open: it does not take open subsets of $\Bbb R$ to open subsets of $\Bbb R^2$.

Title: Can an open interval be bounded?


问:
If the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ intersects the axes at points $A,B,C$ then Area of Triangle $= \sqrt{b^2c^2+c^2a^2+a^2b^2}$
I'm working through this problem,

Compute a surface area by integration to show that if the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ intersects the axes at points $A,B,C$ then Area of Triangle $= \sqrt{b^2c^2+c^2a^2+a^2b^2}$

and keep hitting a bump along the way I'm not sure how to overcome.. So far I have:

Let $z=f(x,y)=c\left(1-\frac{x}{a}-\frac{y}{b}\right)$ and thus used:

$$A(S)=\iint_{S}dS=\iint_{D}\sqrt{1+\left(\frac{\partial{z}}{\partial{x}}\right)^2+\left(\frac{\partial{z}}{\partial{y}}\right)^2}dA=\iint_{D}\sqrt{b^2c^2+c^2a^2+a^2b^2}dA$$

Now when considering D (the projection of $S$ onto the $xy$ plane, I have found a triangle with vertices $(0,0), (a,0), (0,b)$. Thus, I had limits of integration as:
$$0\leq x\leq a$$$$0 \leq y \leq b\left(1-\frac{x}{a}\right)$$

My issue is that evaluating this I am obtaining
$$\int_{0}^{a}\int_{0}^{b\left(1-\frac{x}{a}\right)}\sqrt{b^2c^2+c^2a^2+a^2b^2} dydx = \frac{ab}{2}\sqrt{b^2c^2+c^2a^2+a^2b^2}$$

I can see if $a=b=1$ then the desired result of $\frac{1}{2}\sqrt{b^2c^2+c^2a^2+a^2b^2}$ is obtained, but I'm not sure how to state my final result from this, or if I have made an error in the integral itself.

Any help massively appreciated!

答:
The three coordinate planes and the given plane $\pi$ determine a simplex $T$ in the first octant. The volume of $T$ can be computed in two ways:
$${\rm vol}(T)={1\over6}abc={1\over3}A(S)h\ ,$$
where $h$ is the height of  $T$ with respect to $S$. It follows that
$${\rm area}(S)={abc\over 2h}\ .$$
In order to determine $h$ we intersect the plane normal $t\mapsto t\left({1\over a},{1\over b},{1\over c}\right)$
with $\pi$. When $P$ is the resulting point we have
$$h=|OP|={ab c\over\sqrt{b^2c^2+c^2a^2+a^2b^2}}\ ,$$
so that
$${\rm area}(S)={1\over2}\sqrt{b^2c^2+c^2a^2+a^2b^2}\ .$$


问:
About possible signatures of a non-degenerate symmetric bilinear form.
Say, I have a non-degenerate symmetric bilinear form on a vector space with Dim = $4$. Suppose I have a set of basis $\{v_1,v_2,v_3,v_4\}$ and I know for any $i$, $<v_i,v_i>$ is positive. So what are the possible signatures?

I know $(4,0)$ is trivially a possible signature, just consider an Identity matrix.

However, about $(0,4)$, I do not know how to prove it is not possible. (I guess it is not)

So generally how to solve such kind of question, are there any useful tools?

答:
It should be obvious $(0,4)$ isn't possible, since every vector has negative norm in that signature.

For the rest, it suffices to check $\mathbb{R}^{p,q}$ for every other possible signature $(p,q)$. (In theory, this might be useless idea if this is hard to do, but it turns out not to be hard to do.) Indeed, if you play around a bit you should be able to generalize to the following fact:

Proposition. A pseudo-Euclidean vector space admits a basis of vectors with positive quadratic form so long as it is not negative-definite signature.

Proof. Without loss of generality (by Sylvester's Theorem), we may as well consider $\mathbb{R}^{p,q}$ with $p>0$. If we pick any $v\in\mathbb{R}^p$ with $v\cdot v>1$ then there is such a basis

$$ \{e_1,\cdots,e_p,v+e_{p+1},\cdots,v+e_{p+q}\}. $$


问:
How to solve $x^2 \equiv 12 \pmod {13}$?

Solve $x^2 \equiv 12 \pmod {13}$

By guessing I can say that the solutions are $5$ and $8$, but is there another way to find the solution besides guessing?

答:
In the special case of trying to find a square root of $-1$ modulo a prime $p \equiv 1 \pmod{4}$, we can use the following algorithm: first, select some random integer $a$ with $1 < a < p - 1$.  Now, if $p-1 = 2^k \cdot q$ with $q$ odd, then form $a^q \operatorname{mod} p$.  If you get something other than $1$ or $-1$ at this point, then repeatedly square until you get $-1$.  Then, the number just before $-1$ will be a square root of $-1$.  (The reason this works: by Fermat's little theorem, $a^{p-1} \equiv 1 \pmod{p}$, so after taking the square $k$ times, you get to $1$.  Since $p$ is prime, the number just before you reach $1$ for the first time is forced to be $-1$.)

On the other hand, if $a^q \equiv \pm 1 \pmod{p}$ already, then go back and choose another random value of $a$ to try.  Since $q < \frac{p - 1}{2}$, it is impossible for every value of $a$ to satisfy $a^q \equiv \pm 1 \pmod{p} \Leftrightarrow a^{2q} \equiv 1 \pmod{p}$.

So, in the case of $p = 13$, where $p - 1 = 2^2 \cdot 3$ and so $q = 3$, let us first start with a trial of $a = 2$.  Then, $a^q = 2^3 = 8$ and $8 \not\equiv \pm 1 \pmod{13}$, so we're good to go.  From here, $8^2 = 64 \equiv -1 \pmod{13}$, and so we find 8 as a square root of $-1 \pmod{13}$.


问:
Why is the argument function defined in Complex Analysis discontinuous?
I've been working through Spivak's Calculus and I'm currently working through Ch. 26 on Complex Functions. One of the functions they define is the argument function. Recall that the argument of a nonzero complex number $z$ is a (real) number $\theta$ such that:

$$z = |z|(\cos \theta + i \sin \theta)$$

There are infinitely many arguments for $z$m but just one which satisfies $0 \leq \theta < 2\pi$. If we call this unique argument $\theta(z)$, then $\theta$ is a (real-valued) function (the "argument function") on $\{z \in \mathbb{C}: z \neq 0\}$.

Now there is a claim in the text that the argument function is discontinuous at all nonnegative real numbers. Attached is two screenshots that correspond to one another.

In the text the example they give is after shifting and redefining the argument function $\theta(z)$. Even after redefining it I still don't see how the function is discontinuous. At the current moment I'm not concerned with the formal proof of how it is discontinuous, but an overall picture of why.

To see what I'm trying to get at, when first reading that the argument function is discontinuous I envisioned what would happen is that when solving for the argument I would end up with an undefined function. So if we attempt to solve for the argument of a complex number $z = x+iy$, recalling the trigonometric definition of a complex number $z$, we arrive at:

$$\arctan\bigg(\frac{y}{x}\bigg) = \theta$$

Restricting $\theta$ to the first quadrant of the complex plane. Using $\arccos$ or $\arcsin$ will have similar requirements for their individual functions.

From here since the claim was that the argument function is discontinuous at every positive real number and keeping in mind that the $x$ or $y$ are defined as real numbers I thought of taking the limit, that would mean something along the lines of if $a,b \in \mathbb{R}$:

$$\lim_{(a,b) \to (x,y)}\arctan\bigg(\frac{b}{a}\bigg) = \theta(x,y) = \arctan\bigg(\frac{y}{x}\bigg)$$

Writing this out now as I think about it and the behaviour of multivariable functions, I believe that the discontinuities can occur because approaching the limiting value (in this case $(x,y)$) happens over an infinite amount of different paths and some of these will possibly have instances where the argument function will produce different values compared to the expected result.

Is this interpretation correct as to why the argument function is discontinuous? at least in an informal way?

答:
The argument function must be discontinous along a ray from the origin to the point at infinity.  Here, with the definition you give, the ray is the positive real axis.  Since the argument function is multi-valued, there has to be somewhere where it jumps by $2\pi$.

Let $\epsilon>0$.  Consider $z_1=x_0+i\epsilon$ and $z_2=x_0-i\epsilon$.  Then $\theta(x_0+i\epsilon)=\tan^{-1} \frac{\epsilon}{x_0}$ and $\theta(x_0-i\epsilon)=2\pi-\tan^{-1} \frac{\epsilon}{x_0}$.  Let $\epsilon\to 0$.  We see that the limits $\lim_{\epsilon\to 0} z_1 =\lim_{\epsilon\to 0} z_2 = x_0$ but $\lim \theta(z_1) \ne \lim \theta(z_2).$


问:
How to use the modular function of a locally compact group?
Let $G$ be a locally compact group with left Haar measure $\mu$.

Let $\delta:G\to(0,\infty)$ the unique modular function such that $\delta(x)\mu(E) = \mu(Ex)$ for all Borel sets $E$ and $x\in G$.

How is this definition used in integration?

Are the following steps correct?  (Please alert me of even the slightest mistake!)

For fixed $x\in G$, and Borel set $E$, we have $\delta(x)\mu(E) = \mu(Ex)$.

$\int_{G}f(y)\mu(dy) = \int_{G}f(yx)\mu(dy\cdot x) = \int_{G}\delta(x)f(yx)\mu(dy)$

For a $\mu$-integrable function $f$, I write $\int_{G}f(g)\mu(dy)$ to be mean the same as $\int_{G}fd\mu$, with the indication that $y$ is my variable of integration.

By $\mu(dy\cdot x)$ in the middle step, I mean I am integrating with respect to the measure $\lambda$ defined by $\lambda(E) = \mu(Ex)$.

My motivation is that I am trying to solve this problem: Verifying Convolution Identities

$\bf{\text{Next}}$:

When trying to verify the Convolution identity

$$(f*g)(x) := \int_{G}f(y)g(y^{-1}x)\mu(dy) = \int_{G}\delta(y)^{-1}f(y^{-1})g(yx)\mu(dy)$$

How do I deal with the fact that the argument of $\delta$ is the variable of integration??

答: Yes, this looks correct to me.  It's how you handle right translation when you've got a left Haar measure. Usually the next step would be to pull the $\delta(x)$ out of the integral. </p>


问:
Integrating partial Weierstrass products of $\sin(x)$ divided by $\sin(x)$
Recall the Weierstrass factorization of $\sin(z)$, valid for any complex $z$:
$$
\sin(z) = z \prod_{k=1}^{\infty}\left(1-\frac{z^2}{k^2\pi^2}\right)
$$Apropos of nothing, I thought, 'what if I integrated these partial products divided by sine over integer multiples of $\pi$?' The results Mathematica gave were somewhat striking:
\begin{align}
\int_{-\pi}^{\pi} \frac{x\left(1-\frac{x^2}{\pi^2}\right)}{\sin(x)}\,dx &= \frac{-21 \zeta(3)}{\pi}\\
\int_{-2\pi}^{2\pi} \frac{x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)}{\sin(x)}\,dx &= \frac{5 \left(35 \pi ^2 \zeta (3)-93 \zeta (5)\right)}{2 \pi ^3}\\
\int_{-3\pi}^{3\pi} \frac{x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)}{\sin(x)}\,dx &= \frac{7 \left(392 \pi ^4 \zeta (3)-2170 \pi ^2 \zeta (5)+1905 \zeta (7)\right)}{8 \pi ^5}\\
\int_{-4\pi}^{4\pi} \frac{x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\left(1-\frac{x^2}{16\pi^2}\right)}{\sin(x)}\,dx &= \frac{7 \left(3044 \pi ^6 \zeta (3)-24831 \pi ^4 \zeta (5)+51435 \pi ^2 \zeta (7)-22995 \zeta
   (9)\right)}{16 \pi ^7}\\ &
\end{align}
Here as usual, $\zeta(s)$ is the Riemann zeta function, the analytic continuation of $\sum_{n=1}^{\infty} n^{-s}$; the odd positive-integer values are presently unknown. The results had a similar form with different integer coefficients if the range of integration was restricted to $(-\pi,\pi)$. I am seeking an explanation for these identities, as well as an explanation for how they were derived. I don't think computing antiderivatives is the best way to go; perhaps there is a transform that makes these integrals more manageable?

Body:
Use integration by parts and induction in $k$ to find $$\int_{-\pi}^\pi x^{k}e^{-inx}dx$$
then
$$\int_{-\pi}^{\pi} \frac{x^{k+1}(1-x^2/\pi^2)}{\sin(x)}d=\lim_{r\to 1^-} \int_{-\pi}^\pi \frac{x^{k+1}(1-x^2/\pi^2) 2i e^{-ix}}{1-re^{-2ix}}dx$$ $$=\lim_{r\to 1^-} \int_{-\pi}^\pi \sum_{n=0}^\infty 2i x^{k+1}(1-x^2/\pi^2) r^n e^{-(2n+1)ix}dx$$ $$= 2i\lim_{r\to 1^-} \sum_{n=0}^\infty r^n \int_{-\pi}^\pi x^{k+1}(1-x^2/\pi^2) e^{-(2n+1)ix}dx$$


答:
By the way, the definition of $\theta$ is NOT $\arctan\left(\frac{y}{x}\right)$. This is only valid if you restrict $\theta$ to the first quadrant of the complex plane. The full correct definition requires a bunch of case work.

Let $a\geq 0$ be any real number. Then, as Spivak mentions, $\theta$ is not continuous at the point $a$; the definition of continuity says that given a small $\epsilon$ ball around the point $\theta(a)$, you must be able to find a small $\delta$ ball around the point $a$, such that $\theta$ maps the $\delta$ ball INTO the $\epsilon$ ball (see page 535, figure 3). The crux of the argument is the following: if $\alpha$ is close to $a$, but lies "above" (i.e has positive imaginary part, for example $\alpha = a+ i\frac{\delta}{2}$) then $\theta(\alpha)$ is a "small" positive number; i.e it is close to $\theta(a)=0$.

On the other hand, a point like $\beta= a-i\frac{\delta}{2}$ is also close to $a$; in the sense that $|\beta-a|<\delta$, however, its argument $\theta(\beta)$ is very close to $2\pi$; i.e $\theta(\beta)$ lies outside the $\epsilon$ ball drawn around $\theta(a)$.

So, to summarize, the trouble is that no matter how small of a ball you draw around the point $a$, you can always find points above and below which map close to $0$ and $2\pi$ respectively. It is this large discrepancy which is the reason for the discontinuity.

Another way of phrasing it (more computational, less geometric) is to say that you have two paths along which we get different values for the limit: if we consider the paths $\gamma_1(t) = a+it$ and $\gamma_2(t) = a-it$, then
\begin{align}
\lim_{t\to 0^+}\theta(\gamma_1(t)) = 0 \quad \text{but} \quad
\lim_{t\to 0^+}\theta(\gamma_2(t)) = 2\pi.
\end{align}
The differences in limits show that $\theta$ is discontinuous at the point $a$. Similar line of reasoning works to show any of the other argument functions you define will also be discontinuous.

Title: Why is the argument function defined in Complex Analysis discontinuous?

Body:
I've been working through Spivak's Calculus and I'm currently working through Ch. 26 on Complex Functions. One of the functions they define is the argument function. Recall that the argument of a nonzero complex number $z$ is a (real) number $\theta$ such that:

$$z = |z|(\cos \theta + i \sin \theta)$$

There are infinitely many arguments for $z$m but just one which satisfies $0 \leq \theta < 2\pi$. If we call this unique argument $\theta(z)$, then $\theta$ is a (real-valued) function (the "argument function") on $\{z \in \mathbb{C}: z \neq 0\}$.

Now there is a claim in the text that the argument function is discontinuous at all nonnegative real numbers. Attached is two screenshots that correspond to one another.

In the text the example they give is after shifting and redefining the argument function $\theta(z)$. Even after redefining it I still don't see how the function is discontinuous. At the current moment I'm not concerned with the formal proof of how it is discontinuous, but an overall picture of why.

To see what I'm trying to get at, when first reading that the argument function is discontinuous I envisioned what would happen is that when solving for the argument I would end up with an undefined function. So if we attempt to solve for the argument of a complex number $z = x+iy$, recalling the trigonometric definition of a complex number $z$, we arrive at:

$$\arctan\bigg(\frac{y}{x}\bigg) = \theta$$

Restricting $\theta$ to the first quadrant of the complex plane. Using $\arccos$ or $\arcsin$ will have similar requirements for their individual functions.

From here since the claim was that the argument function is discontinuous at every positive real number and keeping in mind that the $x$ or $y$ are defined as real numbers I thought of taking the limit, that would mean something along the lines of if $a,b \in \mathbb{R}$:

$$\lim_{(a,b) \to (x,y)}\arctan\bigg(\frac{b}{a}\bigg) = \theta(x,y) = \arctan\bigg(\frac{y}{x}\bigg)$$

Writing this out now as I think about it and the behaviour of multivariable functions, I believe that the discontinuities can occur because approaching the limiting value (in this case $(x,y)$) happens over an infinite amount of different paths and some of these will possibly have instances where the argument function will produce different values compared to the expected result.

Is this interpretation correct as to why the argument function is discontinuous? at least in an informal way?


问:
Find the generating function for $ a_n = n^4$ .
Find the generating function for $a_n = n^4$.

How should I approach this question? I have tried to go $1,16,81\dots$ and generate the function with the series but no idea how to do so.

答:
Notice

$$ \sum_{n=0}^\infty x^n = \frac{1}{1-x}.$$

Take a derivative to get

$$\sum_{n=0}^\infty (n+1) x^{n} = \frac{1}{(1-x)^2}.$$

Now we get

$$ \frac{1}{(1-x)^2} - \frac{1}{1-x} = \sum_{n=0}^\infty n x^n + \sum_{n=0}^\infty x^n - \sum_{n=0}^\infty x^n = \sum_{n=0}^\infty n x^n.$$

Keep going like this.


问:
Proof that the method of Gauss/Jordan yields the inverse of a matrix
I have trouble in solving the following exercise: let A be an invertible matrix. Consider the matrix A|I where I is the identity matrix.
Prove that the matrix obtained by transforming A|I in it's row-echelon form is equal to I|B where B is the inverse matrix of A. Any help would be appreciated.</p>

答:
Let $e_1$, $e_2$, ..., $e_n$ denote the columns of the identity matrix. When you solve the system $Ax=e_i$, you find a unique column $v_i$ such that $Av_i=e_i$. What's this column? When you consider the matrix $[A\;|\;e_i]$ and bring it to row echelon form, you get $[I\;|\;v_i]$.

Thus when you do the row echelon form reduction of $[A\;|\;I]$ you're doing $n$ simultaneous reductions; there's no difference in the steps you perform in each reduction $[A\;|\;e_i]\to[I\;|\;v_i]$ because you find the same pivots in each row, so in the first $n$ columns, and never use the last column for the elimination.

Thus the row echelon form you get is exactly $[I\;|\;v_1\; v_2\; \dots\; v_n]$ and
$$
A[v_1\; v_2\; \dots\; v_n]=[e_1\;e_2\;\dots\;e_n]=I
$$
by definition of $v_1$, $v_2$, ..., $v_n$.


问:
Is a complex-valued function on an interval constant iff its derivative is zero?
I know that this is true for real-valued functions. However, the proof is based on the mean value theorem and the latter is not valid for complex-valued functions (here is a counter-example).

That being said, the following generalization was proven in one of my lectures:

Let $V$ and $W$ be normed vector spaces over the same field $F$. Let $U\subset V$ be open and connected, then a map $f\colon U\to W$ is constant iff $Df(x)=0$ for all $x\in U$, where $Df(x)\in L(V,W)$ is the Fréchet-derivative.

In the title of my question, I consider the case $V=\mathbb{R}$ and $W=\mathbb{C}$. The problem is that the Fréchet derivative is not defined when $V$ is real vector space and $W$ is a complex vector space, since the set $L(V,W)$ is only defined when $V$ and $W$ are vector spaces over the same field.

Is there another definition of the derivative in this situation and does it allow a similar theorem?

答:
In case $V=\mathbb R$ and $W \in \mathbb C$, write
$$
f(x) = u(x) + iv(x)
$$
where $u(x)$ and $v(x)$ are real.  Then the derivative is
$$
f'(x) = u'(x) + iv'(x) .
$$

In finite-dimensional spaces, the Fréchet derivative is the same as the ordinary derivative.

For general $V$ and $W$, we want
$$
f(x+h) = f(x) + f'(x)(h) +o\big(\|h\|\big)
$$
as $h \to 0$.  Here $x$ and $h$ are in $V$, $f'(x) \in L(V,W)$,
and $h \in V$, so $f'(x)(h) \in W$ makes sense.