问:
Integral of $\int_{-\infty}^\infty \dfrac{\sigma}{(\sigma^2 + (x-\mu)^2)} \, dx$
I could use some help with this integration and/or an explanation of why Mathematica shows a different output.

The integral is:
$$\int_{-\infty}^\infty \dfrac{\sigma}{(\sigma^2 + (x-\mu)^2)} \,  dx$$

Here is my work:

Take $u = x-\mu$ so that $du = dx$. Then,
\begin{align}
    \int_{-\infty}^\infty \dfrac{\sigma}{\sigma^2 + (x-\mu)^2} \, dx
        &=
            \int_{-\infty}^\infty \dfrac{\sigma}{\sigma^2 + u^2} \, du
        \nonumber
        \\[.5em]
        &=
            \dfrac{1}{\sigma} \int_{-\infty}^\infty
                \dfrac{1}{\dfrac{u^2}{\sigma^2} + 1}
            \,dx
        \nonumber
        \\[.5em]
        &=
            \dfrac{1}{\sigma} \tan^{-1} \left( \dfrac{u}{\sigma} \right) \bigg |_{-\infty}^\infty
        \nonumber
        \\[.5em]
        &=
            \dfrac{1}{\sigma} \left[ \dfrac{\pi}{2\sigma} + \dfrac{\pi}{2\sigma} \right]
        \nonumber
        \\[.5em]
        &=
            \dfrac{\pi}{\sigma^2}
        \nonumber
    \end{align}

When using evaluating the integral in terms of $u$, Mathematica agrees with my results. Although, when expanding $u$ out to be $x-\mu$, Mathematica says the answer is $i(\log(-i/\sigma)-\log(i/\sigma))$, which evaluates to $\pi$ for $\sigma>0$.

I'm a little confused by this. Any insight as to why this is so is appreciated, thank you.

答:
The answer is $\pi$. You made an error at the steps $$\dfrac{1}{\sigma} \int_{-\infty}^\infty \dfrac{1}{\dfrac{u^2}{\sigma^2} + 1} \,dx = \dfrac{1}{\sigma} \tan^{-1} \left( \dfrac{u}{\sigma} \right) \bigg |_{-\infty}^\infty = \dfrac{1}{\sigma} \left[ \dfrac{\pi}{2\sigma} + \dfrac{\pi}{2\sigma} \right]$$

It should be $$\dfrac{1}{\sigma} \int_{-\infty}^\infty \dfrac{1}{\dfrac{u^2}{\sigma^2} + 1} \,dx = \dfrac{1}{\sigma}\color{red}{\cdot \sigma} \tan^{-1} \left( \dfrac{u}{\sigma} \right) \bigg |_{-\infty}^\infty = \color{red}{1} \left[ \dfrac{\pi}{2} + \dfrac{\pi}{2} \right]$$

because the integral of $\frac{1}{\frac{u^2}{\sigma^2}+1}$ is $\sigma\tan^{-1} \left( \dfrac{u}{\sigma} \right)$, not $\tan^{-1} \left( \dfrac{u}{\sigma} \right)$ and because $\tan^{-1} \left( \dfrac{u}{\sigma} \right) \bigg |_{-\infty}^\infty$ doesn't depend on $\sigma$ (it will always be $\pi$).

Edit: As Winther pointed out, this is true only assuming that $\sigma>0$. For $\sigma=0$, the integral is $0$, and for $\sigma<0$, the integral is $-\pi$ instead of $\pi$.


问:
squares of integers, and multiples of 4

Prove that for all $n\in\Bbb Z$ there exists $k\in\Bbb Z$ such that either $n^2=4k$ for $n^2 = 4k + 1$.

A hint given was: What are the possible remainders for n after dividing by 4? Break into the cases where you have each remainder.

答:
Hint:

$k \equiv \pm 1$ or $2(\mod 4) \implies k^2 \equiv 1$ or $0(\mod 4)$

Proof:

When $k \in \text{odd}$, $k=2n+1 \implies k^2=4n^2+4n+1 =4n(n+1)+1$

When $k \in \text{even}, k=2n \implies k^2=4n^2$

ContentLicense: CC BY-SA 3.0

Title: squares of integers, and multiples of 4


问:
one 1-form on the Riemann surface of an algebraic function
Given the many-valued algebraic function $w =\sqrt{(z-1)(z-2)(z-3)}$,  we can get a Riemann surface  $S$ that is topologically equivalent to a torus. I am wondering whether the $1$-form
$ \displaystyle\frac{dz}{w}$ is a holomorphic form on the torus $S$.

On the surface we know that

$$ \frac{dz}{w}=\frac{dz}{\sqrt{(z-1)(z-2)(z-3)}},$$

I think that $z=1,2,3$ are poles of the above 1-form, so it is a meromorphic 1-form on the $z-$plane. Is it a holomorphic 1-form on the torus $S$ determined by the algebraic function $w =\sqrt{(z-1)(z-2)(z-3)}$?

I just know a little about Riemann surface theory. I will appreciate any suggestions and comments. Thanks.

答:
What Riemann surface did you obtain? $C:w^2=(z-1)(z-2)(z-3)$ or $C-(1,0),(2,0),(3,0)$ or the projective closure $E\simeq C\cup (\infty,\infty)$?

Anyway $dz/w$ is holomorphic on them all: near $(1,0)$ we have $z=f(w)w^2$ with $f$ analytic at $0$. At $(\infty,\infty)$ we have $z=(z/w)^{-2} g(z/w),w=(z/w)^{-3} h(z/w)$ with $g,h$ analytic at $0$.

Eventually $dz/w$ is a nowhere vanishing holomorphic 1-form and $p\to \int_{p_0}^p dz/w$ is an isomorphism $E \to \Bbb{C/(uZ+vZ)}$ where $uZ+vZ$ is the lattice obtained by integrating over the closed-loops.

ContentLicense: CC BY-SA 4.0


问:
Solve ODE $y' = \frac{xy}{x^2-y^2}$
How can I solve this ODE: $y' = \frac{xy}{x^2-y^2}$

My approach was to substitute $y(x) \mapsto z(x)x$. Therefore

$$
z'x-z = \frac{x^2z}{x^2(1-z^2)} = \frac{z}{1-z^2}
$$

Hence

$$
z'x = \frac{2z-z^3}{1-z^2} \implies z'\frac{1-z^2}{2z-z^3}=\frac{1}{x}
$$

Integrating both sides with respect to $x$:

$$
\int z'\frac{1-z^2}{2z-z^3} dx=\ln|x|+c
$$

And now I'm stuck.

Any help will be highly appreciated!

EDIT:

As the user 00GB noticed, I started it wrong... Using the substitution $y = zx$ then $y' = z'x + z$ not $z'x - z$. That correction leads to what he stated:
$$
\begin{align*}
\int\frac{1-z^2}{z^3}dz&=\int\frac{1}{x}dx\\
\int\frac{1-z^2}{z^3}dz&=\ln|x| +C\\
\int\frac{1}{z^3}dz-\int\frac{1}{z}dz&=\ln|x| + C\\
-\frac{1}{2z^2}-\ln|z|&=\ln|x| + C
\end{align*}
$$
And substituting back $z = \frac{y}{x}$:
$$
\begin{align*}
-\frac{1}{2\frac{y^2}{x^2}}-\ln|\frac{y}{x}|&=\ln|x| + C\\
-\frac{x^2}{2y^2}&=\ln|\frac{y}{x}|+\ln|x| + C\\
x^2&=y^2(C -2\ln|y|)\\
\end{align*}
$$

答:
Others have already noticed that your equation is incorrect. Anyway, here's how you evaluate this integral
$$
\int \frac{1-z^2}{2z-z^3} dz=\int \frac{2-z^2}{2z-z^3} dz
- \int \frac{1}{2z-z^3} dz=\int \frac{1}{z} dz
- \int \frac{1}{z^3(\frac{2}{z^2}-1)} dz
$$

Put $\frac{2}{z^2}-1=t$ for the second integral and you are done

Title: Solve ODE $y' = \frac{xy}{x^2-y^2}$


问:
How can I simplify this lerp arithmetic to avoid floating point precision errors?
I am making a "cinematic camera" that pans around a scene. Each step has a position for the camera and a point that the camera should be looking at. For instance:

```<code>1. position = (3, 2, 1), facing = (0, 0, 0)
2. position = (6, 5, 4), facing = (9, 9, 9)
</code>```

The camera should start at (3, 2, 1) and be looking at the point (0, 0, 0). By the end of the panning the camera should be positioned at (6, 5, 4) and be looking at the point (9, 9, 9).

I've written some C# to do this:

```<code>public static IEnumerable<(Vector3, Vector3)> CinematicCamera(params (Vector3 location, Vector3 facing, double time)[] points) {
    if (points.Length < 2) {
        throw new ArgumentException("Need at least 2 points.");
    }

    List<(Vector3, Vector3)> returnMe = new List<(Vector3, Vector3)>();

    for (int i = 1; i < points.Length; i++) {
        int numSteps = (int)(points[i].time * 20); // camera runs at 20 frames per second
        double startDistance = Distance(points[i - 1].location, points[i - 1].facing); // distance the camera starts from the point it's looking at
        double endDistance = Distance(points[i].location, points[i].facing); // distance the camera ends from the point it's looking at
        Vector3 startDirection = points[i - 1].location - points[i - 1].facing; // get the start directional vector from the location point and facing point
        Vector3 endDirection = points[i].location - points[i].facing; // get the end directional vector from the location point and facing point

        for (int stepNum = 0; stepNum < numSteps; stepNum++) {
            double progress = (double)stepNum / numSteps; // 0 to 1 based on current step, used for lerp
            double distance = Lerp(startDistance, endDistance, progress); // lerp the distance from the previous point to the next point
            Vector3 currentDirection = Lerp(startDirection, endDirection, progress); // lerp the directional vector

            Vector3 lookAt = Lerp(points[i - 1].facing, points[i].facing, progress); // new point to look at
            Vector3 position = lookAt + currentDirection * distance / Length(currentDirection); // new position

            returnMe.Add((position, lookAt));
        }
    }

    return returnMe;
}

// helper methods
private static double Distance(Vector3 a, Vector3 b) => Math.Sqrt(Math.Pow(a.X - b.X, 2) + Math.Pow(a.Y - b.Y, 2) + Math.Pow(a.Z - b.Z, 2));
private static double Length(Vector3 v) => Math.Sqrt(v.X * v.X + v.Y * v.Y + v.Z * v.Z);
private static double Lerp(double start, double end, double amount) => start + (end - start) * amount;
private static Vector3 Lerp(Vector3 start, Vector3 end, double amount) => new Vector3(Lerp(start.X, end.X, amount), Lerp(start.Y, end.Y, amount), Lerp(start.Z, end.Z, amount));
</code>```

Vector3 is a custom data type that uses double to store x, y, and z. This all works, but the camera is a bit jittery due to floating point imprecision. I graphed some of the points in Excel and you can see the jitter in the graph:

I'm considering switching the entire implementation from double to decimal (double in C# uses 8 bytes while decimal uses 16), but I'm not sure that really addresses the root of the problem. How can I simplify this so that I run into less floating point weirdness?

答:
In prehistoric times using $36$-bit COBOL, I was able to get around this  by two methods.

I took all incoming floating points and multiplied them by a factor that converted them to $36$-bit integers. All calculations were done with integers and the results were then divided by the factor for final output.

For one project, where I began with integers but there would be floating points along the way, I kept numerators and denominators in separate variables and did LCM or GCD calculations as needed to multiply or divide them along the way. This kept the magnitude to $36$ bits and the "division" was used only for the final output.


问:
How do you solve $\cos(θ) = \sin(θ) = a$ non graphically?
If
$\cos(θ) = a$, and
$\sin(θ) = a$,
and if $a$ is between $-1$ and $1$, how could we find $θ$. I could do this graphically by finding the intersection of $\sin(x), \cos(x)$ and $y=a$. But how could I find out with algebra, maybe the sign rules for each function in each quadrant.

答:
A geometric solution:

In the first quadrant, $\cos \theta$ and $\sin \theta$ are complementary angles.

So we have $\theta + \theta = \frac{\pi}{2}$, or $$\theta=\frac{\pi}{4}+2k\pi.$$

The other solution is in the third quadrant (where cosine and sine are both negative).

So the solutions are $$\theta \in \left\{ \frac{\pi}{4}, \frac{5\pi}{4}\right\}+2k\pi.$$


问:
$f(x) = \inf_{k \in \mathbb N} k |x - x_k |$ is continuous if $x_k$ has no accumulation point
I need to show that if $(x_n)$ is a real valued sequence with no accumulation point, then the function $f(x) = \inf_{k \in \mathbb N} k | x - x_k |$ is continuous. I also need to find a sequence that makes $f$ not continuous, but I haven't really worked on that part yet.

My approach so far is the following: fix $y \in \mathbb R$. As $(x_n)$ has no accumulation points, it is unbounded hence $k|y - x_k|$ is unbounded. For every $M \in \mathbb R$, this means the set $\{k|y-x_k|\} \cap [0, M]$ must be finite. If we set $M := |y - x_1|$, the set is inhabited and the minimum of the is equal to $f(y)$. But then $f$ is the minimum of a finite set of continuous functions, hence $f$ is continuous.

I feel very uneasy about the reasoning though, in particular that our choice of $M$ depends on $y$. I would appreciate some guidance on how to solve this problem.

答:
You correctly identified that the goal is to express $f$ as a $\min$ of a finite number of functions.

We note that not only is $x_k$ unbounded, but $|x_k| \to \infty$.

Suppose $|x| \le M$. Note that if we choose some large enough $K$ such that
$|x-x_1| \le 2 |x-x_k|$ for $k \ge K$, then
$f(x) = \min_{k \in \{1,...,K\}} k|x-x_k|$ from which continuity follows on $|x| < M$.

Note that $|x-x_1| \le M+|x_1|$ and $|x_k|-M \le |x-x_k|$, hence if
$M+|x_1| \le 2|x_k| -2M$, or
$|x_k| \ge {1 \over 2} (3M+|x_1|)$ then the inequality is satisfied. Hence we
can choose $K$ such that this is satisfied.


问:
find the equation of the tangents to a quadratic curve through the point $(-7,1)$
Given a quadratic curve $\mathcal{C}: X^2+2XY+4Y^2-2X+6Y+6=0$, determine its type and find the equation of the tangents to $\mathcal{C}$ through the point $(-7,1)$.

So far I've written the matrix of $\mathcal{C}$ as
$$A = \begin{bmatrix}6 & -1 & 3 \\ -1 & 1 & 1 \\ 3 &1 &4\end{bmatrix}.$$

I used the fact that $\det A\neq 0$ and $\begin{vmatrix}1 & 1 \\ 1 & 4\end{vmatrix}>0$ to say that this is an ellipse, by some theorems we did in class. Now I know that if a point $P(x,y)$ is on the curve, then the unique tangent at this point is the polar line which would be calculated by taking the homogeneous vector of $P$ and multiplying it by A as follows $(1:x:y)A$. However $(-7,1)$ is not on the curve although I calculated $x=\frac{16}{7}$ to be its tangent if it was. so I am stuck here. What could I use to find the tangents since the point is not on the curve? If we write $t_P: y=ax+b$ then $t_P$ must contain $P$ and intersect $\mathcal{C}$ at a point, but I am not sure what point that would be.

答:
The first example has been treated.

Let us consider the second example you have given in one of your comments:

$$x^2-xy-y^2-2x+2y+1=0 \tag{1}$$

(which is a hyperbola: see figure below) with point $P(4,-2)$.

The generic parametric equations of lines passing through point $P$ are:

$$\begin{cases}x&=& \ \ \ 4+at\\y&=&-2+bt\end{cases}$$

If we substitute them in (1), we obtain the following quadratic equation in variable $t$:

$$(a^2 - ab - b^2)t^2 + 2(4a + b)t + 9=0$$

Its discriminant is, up to an unimportant factor $4$, equal to :

$$\Delta=(4a + b)^2-9(a^2 - ab - b^2)$$

$$\Delta=7a^2 + 17ab + 10b^2\tag{2}$$

Setting $\Delta=0$ is expressing that there is a double root, a fact characterizing a tangent line.

Let us take arbitrarily $a=1$ in (2). The two possible roots of quadratic equation obtained from (2)

$$7 + 17b + 10b^2=0$$

are $b=-1$ and $b=-0.7$ ($b$ is in fact plainly the slope of the straight lines)

Therefore the two parametric equations of tangent lines issued from point $P(4,-2)$ are:

$$\begin{cases}x&=& \ \ \ 4+t\\y&=&-2-t\end{cases} \ \text{and} \ \begin{cases}x&=&\ \ \ 4+t\\y&=&-2-0.7t\end{cases}$$

otherwise said, with resp. cartesian equations:

$$x+y=2 \ \ \ \ \text{and} \ \ \ \ 7x+10y=8$$


答:
I started from your solution
$$
z'x+z = \frac{x^2z}{x^2(1-z^2)} = \frac{z}{1-z^2}
$$
then
$$xz'=\frac{z^3}{1-z^2}$$
$$\frac{1-z^2}{z^3} \text{dz}=\frac{1}{x}\text{dz}$$
Integrate both side and then you are done

Title: Solve ODE $y' = \frac{xy}{x^2-y^2}$

Body:
How can I solve this ODE: $y' = \frac{xy}{x^2-y^2}$

My approach was to substitute $y(x) \mapsto z(x)x$. Therefore

$$
z'x-z = \frac{x^2z}{x^2(1-z^2)} = \frac{z}{1-z^2}
$$

Hence

$$
z'x = \frac{2z-z^3}{1-z^2} \implies z'\frac{1-z^2}{2z-z^3}=\frac{1}{x}
$$

Integrating both sides with respect to $x$:

$$
\int z'\frac{1-z^2}{2z-z^3} dx=\ln|x|+c
$$

And now I'm stuck.

Any help will be highly appreciated!

EDIT:

As the user 00GB noticed, I started it wrong... Using the substitution $y = zx$ then $y' = z'x + z$ not $z'x - z$. That correction leads to what he stated:
$$
\begin{align*}
\int\frac{1-z^2}{z^3}dz&=\int\frac{1}{x}dx\\
\int\frac{1-z^2}{z^3}dz&=\ln|x| +C\\
\int\frac{1}{z^3}dz-\int\frac{1}{z}dz&=\ln|x| + C\\
-\frac{1}{2z^2}-\ln|z|&=\ln|x| + C
\end{align*}
$$
And substituting back $z = \frac{y}{x}$:
$$
\begin{align*}
-\frac{1}{2\frac{y^2}{x^2}}-\ln|\frac{y}{x}|&=\ln|x| + C\\
-\frac{x^2}{2y^2}&=\ln|\frac{y}{x}|+\ln|x| + C\\
x^2&=y^2(C -2\ln|y|)\\
\end{align*}
$$


答:
Try to consider two different cases:

If $n=2k$ when $n$ is even, so $n=4k^2=4k', k,k'\in\mathbb Z$

If $n=2k+1$ when $n$ is odd, so $n^2=4k'+1, k'\in\mathbb Z$

Title: squares of integers, and multiples of 4

Body:
Prove that for all $n\in\Bbb Z$ there exists $k\in\Bbb Z$ such that either $n^2=4k$ for $n^2 = 4k + 1$.

A hint given was: What are the possible remainders for n after dividing by 4? Break into the cases where you have each remainder.


问:
Finding the middle of n points
Sorry for my possible bad English, I have a problem that I spent a bit of time on and I have been blocked on it for a couple of hours, I'll try to translate it as best as I can:

Given an integer $n\ge3$, and given $A_1,\ldots,A_n$ points on a plane, under what condition(s) can we find points $P_1,\ldots,P_n$ such that $A_1$ is the midpoint of $[P_1,P_2]$, $A_2$ is the midpoint of $[P_2,P_3]$, ... and $A_n$ is the midpoint of $[P_n,P_1]$?

Please tell me if I wasn't clear enough in my translation, and thank you very much in advance!

答:
A remark; not an answer.

If you had posed the question for the line rather than the plane, then
$A$ needs to be the Voronoi Diagram for $P$.
For example, for
$$
A = \left( \frac{1}{2}, 3, 6 \right) \;,
$$
we need
$$
P = ( 0, 1, 5, 7 ) \;.
$$
So $A$ must represent a possible $1$-dimensional Voronoi Diagram.

I have ignored the wrap-around requirement that $A_n=(P_n+P_1)/2$,
and instead used the more natural (to me) $A_n=(P_n+P_{n+1})/2$.
So $|P|=|A|+1$.

Title: Finding the middle of n points


问:
Proof Verification - divisibility through induction
I'm studying for a final and was wondering if this proof is correct. I've seen several variants but they all seem less logical, so I want to make sure I'm not overlooking something.

The proof is essentially:
Prove, using induction that $5 | 11^{n} - 6$, for all positive integers n

Here is my proof:
Base case, n = 1

$11^1 - 6 = 5$ which is divisible by 5.

Inductive hypothesis:: Assume the proposition is true for some $k \in Z^{+}$, or 5|$11^k - 6$

Inductive step: Prove, using the assumption made in the previous step that $5 | 11^{k+1} - 6$
From our inductive hypothesis we know that there exists some positive integer a such that 5a = $11^{k} - 6$

$5 | 11^{k+1} - 6 $

$11^{k+1} - 6 $ = $11^k * 11^1 - 6$
but, we know that 5a = $11^k - 6$, so substituting, we get:
$11*5a = 55a$, which is a factor of 5.

Thus, the proposition holds.

答:
An easier solution for the induction step may be that if $11^k-6=5m$ for some integer $m$,

i.e. $11^k=5m+6$

then $11^{k+1}=11(5m+6)=55m+66 = 5(11m+12)+6$

i.e. $11^{k+1}-6=5(11m+12)$

or in other words, if $11^k-6$ is a multiple of $5$ then $11^{k+1}-6$ is also a multiple of $5$


答:
So you'd like to see $A_1 = P_1/2 + P_2/2, A_2 = P_2/2 + P_3/2, \dots, A_n = P_n/2 + P_1/2$. What if we try this in $p$ dimensions (you asked about $p = 2$) and represented the components of $A_1$ as $A_{11}, \dots, A_{1p}$, and so forth for all $A$s, then the same thing for the $P$s. You'd have

\begin{equation}
A := \begin{bmatrix}
A_{11} & A_{12} & \cdots & A_{1p} \\
\vdots &        &        & \vdots \\
A_{n1} & A_{n2} & \cdots & A_{np}
\end{bmatrix}.
\end{equation}
and
\begin{equation}
P := \begin{bmatrix}
P_{11} & P_{12} & \cdots & P_{1p} \\
\vdots &        &        & \vdots \\
P_{n1} & P_{n2} & \cdots & P_{np}
\end{bmatrix}.
\end{equation}

The coordinates of $A_i$ are represented by row $i$ of $A$, etc.

Looking at it this way, the question you're asking is whether it's possible
to find $P$ satisfying $A = WP$ where $W$ is the $n \times n$ matrix

\begin{equation}
W := \begin{bmatrix}
1/2 & 1/2 & 0 & \dots & 0 \\
0   & 1/2 & 1/2 & \dots & 0 \\
\vdots & & & & \vdots \\
1/2 & 0 & 0 & \dots & 1/2
\end{bmatrix}.
\end{equation}

You'd be home-free if you could show the determinant of $W$ is nonzero.

Title: Finding the middle of n points

Body:
Sorry for my possible bad English, I have a problem that I spent a bit of time on and I have been blocked on it for a couple of hours, I'll try to translate it as best as I can:

Given an integer $n\ge3$, and given $A_1,\ldots,A_n$ points on a plane, under what condition(s) can we find points $P_1,\ldots,P_n$ such that $A_1$ is the midpoint of $[P_1,P_2]$, $A_2$ is the midpoint of $[P_2,P_3]$, ... and $A_n$ is the midpoint of $[P_n,P_1]$?

Please tell me if I wasn't clear enough in my translation, and thank you very much in advance!


问:
Ideal of a group
The ideal is defined in the ring theory;

In ring theory, a branch of abstract algebra, an ideal of a ring is a special subset of its elements

In the answer to this question What is the exponent of a group?, the term ideal is used for groups as;

The exponent of a group $G$ is the non-negative generator of the ideal $\{z \in \mathbb{Z} : \forall g \in G (g^z=1)\}$.

Is this usage of the term ideal correct? If not, what is the correct term?

答:
This can be verified directly from the definition of an ideal.

Let $\mathbf I = \{z \in \mathbb{Z} : \forall g \in G (g^z=1)\} \subset \mathbb Z.$

Suppose $z_1, z_2 \in \mathbf I$. Then $\forall g \in G, g^{z_1} = g^{z_2} =1$. So $\forall g \in G, g^{(z_1+z_2)} =  g^{z_1} g^{z_2} =1$. Hence $z_1+z_2 \in \mathbf I$.

Suppose $k \in \mathbb{Z}$ and $z \in \mathbf I$. Then $\forall g \in G,g^{z} = 1$. So $g^{kz} = (g^{z})^k =1$. Hence $kz \in \mathbf I$.

It follows that $\mathbf I$ is an ideal of $\mathbf Z$.

Title: Ideal of a group


问:
Christoffel symbols manipulation help
We have $\sum_l\Gamma^{l}_{ij}g_{lk}=\frac{1}{2}(\frac{\partial(g_{jk})}{\partial{x}_i}$+$\frac{\partial(g_{ki})}{\partial{x_j}}$-$\frac{\partial(g_{ij})}{\partial{x}_k})$

Since $(g^{lk})$ is the inverse of $(g_{lk})$,

$\Gamma^{m}_{ij}=\frac{1}{2}\sum_k(\frac{\partial(g_{jk})}{\partial{x_i}}$+$\frac{\partial(g_{ki})}{\partial{x_j}}$-$\frac{\partial(g_{ij})}{\partial_{x_j}})g^{km}$

May someone please supply the linear algebra argument above ? the indices are slightly confusing me every time I attempt to understand the linear algebra argument.

答:
The Christoffel symbols are the components of the connection $\nabla$. They have a proper definition for any suitable connection, not just the Levi-Civita. $\Gamma_{ij}$ is a vector that describes how much the $i$th basis vector varies along the coordinate lines of the $j$th component. It is what we get out on the other side when we take the covariant derivative on our basis vectors:
$$\nabla_i \mathrm{e}_j=\Gamma^k_{ij}\mathrm{e}_k$$
In a more intuitive way, $\Gamma_{ij}$ is a vector that describes how much the $\mathrm{e}_i$ basis vector changes when we nudge it in the $x^j$ direction. ( See this video ). We can use the equation above as well as the required properties of the connection to write expressions for the covariant derivatives of vectors, covectors, and eventually, tensors. I'll omit the details here, but it turns out that the components of the covariant derivative of a $(0,2)$ tensor are
$$\nabla_k T_{ij}=\partial_k T_{ij}-\Gamma^l_{ki}T_{lj}-\Gamma^l_{kj}T_{il}$$
The connection $\nabla$ is precisely the Levi-Civita connection if $\nabla g=0$ and if $\Gamma$ is symmetric in its lower two indices. This means
$$\nabla_k g_{ij}=\partial_kg_{ij}-\Gamma^l_{ki}g_{lj}-\Gamma^l_{kj}g_{il}=0\tag{1}$$
But equivalently,
$$\nabla_i g_{kj}=\partial_i g_{kj}-\Gamma^l_{ki}g_{lj}-\Gamma^l_{ji}g_{kl}=0\tag{2}$$
And,
$$\nabla_j g_{ik}=\partial_j g_{ik}-\Gamma^l_{ij}g_{lk}-\Gamma^l_{kj}g_{il}=0\tag{3}$$
A nice trick is that we can add $(1)+(2)-(3)$:
$$\partial_k g_{ij}+\partial_i g_{kj}-\partial_j g_{ik}\color{red}{-\Gamma^l_{ki}g_{lj}}\color{blue}{-\Gamma^l_{kj}g_{il}}\color{red}{-\Gamma^l_{ki}g_{lj}}\color{green}{-\Gamma^l_{ji}g_{kl}+\Gamma^l_{ij}g_{lk}}\color{blue}{+\Gamma^l_{kj}g_{il}}=0$$
But both $\Gamma$ and $g$ are  symmetric, so the blue and green terms vanish. In all,
$$\partial_k g_{ij}+\partial_i g_{kj}-\partial_j g_{ik}=2\Gamma^l_{ik}g_{lj}$$
Divide both sides by $2$ and contract both sides with the inverse metric tensor:
$$\frac{1}{2}g^{pj}\left(\partial_k g_{ij}+\partial_i g_{kj}-\partial_j g_{ik}\right)=\Gamma^l_{ik}g_{lj}g^{pj}=\Gamma^p_{ik}$$
Renaming the indices into the order we typically see them,
$$\Gamma^k_{ij}=\frac{1}{2}g^{kl}\left(\partial_i g_{il}+\partial_j g_{lj}-\partial_l g_{ij}\right)$$

Title: Christoffel symbols manipulation help


问:
Homotopy problem
Let $X$ and $Y$ metric space with $X$ compact and $f,g: X \rightarrow Y$ homotopic. Then for each $\epsilon>0$, exists continuous functions $f_0,\ldots,f_k: X \rightarrow Y$ such that:

$$ d(f_i(x),f_{i-1}(x))<\epsilon, \quad \forall x\in X $$

and $f_0=f$, $f_k=g$.

I was thinking of using the fact that $H: X \times [0,1] \rightarrow Y$ is continuous in a compact, so is uniformly continuous. But I can't prove what it asks for

答:
You can use the Lebesgue number lemma.

Let $\cal V$ be the open cover of $Y$ by open balls of radius $\epsilon/2$ around each point of $Y$.

Let $\mathcal U = \{H^{-1}(V) \mid V \in \mathcal V\}$, which is an open cover of the compact space $X \times [0,1]$.

Applying the Lebesgue number lemma, and using the ordinary product metric on $X \times [0,1]$, let $\lambda > 0$ be a Lebesgue number for $\mathcal U$, and so every subset of $X \times [0,1]$ of diameter $< \lambda$ is contained in some element of $\mathcal U$.

Let $N$ be a natural number with $N > \frac{1}{\lambda}$.

For each $x \in X$ and each $i=1,\ldots,n$, note that $\{x\} \times [i-1/n,i/n]$ has diameter $<\lambda$ and so $H\left(\{x\} \times [(i-1)/n,i/n]\right)$ is contained in some element of $\cal V$, and hence $f(x,(i-1)/n)$ and $f(x,i/n)$ have distance $\le 2 \cdot (\epsilon/2) = \epsilon$.

Now take $f_i(x) = f(x,i/n)$ for $i=0,\ldots,N$.

Title: Homotopy problem


问:
Indefinite Integral $\int\sqrt[3]{\tan(x)}~dx$
For calculating $\int\sqrt{\tan(x)}~dx$, I used this easy method
$$\begin{align}\int\sqrt{\tan(x)}~dx&=\frac{1}{2}\int\left(\sqrt{\tan(x)}+\sqrt{\cot(x)}\right)dx+\frac{1}{2}\int\left(\sqrt{\tan(x)}-\sqrt{\cot(x)}\right)dx\\&=\frac{1}{2}\int\frac{\sin(x)+\cos(x)}{\sqrt{\sin(x)\cos(x)}}~dx-\frac{1}{2}\int\frac{\cos(x)-\sin(x)}{\sqrt{\sin(x)\cos(x)}}~dx\\&=\frac{\sqrt{2}}{2}\int\frac{du}{\sqrt{1-u^2}}-\frac{\sqrt{2}}{2}\int\frac{dv}{\sqrt{v^2-1}}.\end{align}$$$$u=\sin(x)-\cos(x), v=\sin(x)+\cos(x)$$

Does there exist an easy method for $\int\sqrt[3]{\tan(x)}~dx$?

Body:
For $\int\sqrt[3]{\tan x}~dx$ ,

Let $u=\sqrt[3]{\tan x}$ ,

Then $x=\tan^{-1}u^3$

$dx=\dfrac{3u^2}{u^6+1}~du$

$\therefore\int\sqrt[3]{\tan x}~dx$

$=\int\dfrac{3u^3}{u^6+1}~du$

$=\dfrac{3}{2}\int\dfrac{u^2}{u^6+1}~d(u^2)$

$=\dfrac{3}{2}\int\dfrac{v}{v^3+1}~dv$ (Let $v=u^2$)

With reference to Finding $\int \frac{1}{1+x^3}dx$ without partial fractions,

Let $v=-\dfrac{w-1}{w+1}$ ,

Then $dv=-\dfrac{2}{(w+1)^2}dw$

$\therefore\dfrac{3}{2}\int\dfrac{v}{v^3+1}~dv$

$=\dfrac{3}{2}\int\dfrac{-\dfrac{w-1}{w+1}}{-\dfrac{(w-1)^3}{(w+1)^3}+1}\left(-\dfrac{2}{(w+1)^2}\right)~dw$

$=3\int\dfrac{w-1}{(w+1)^3-(w-1)^3}~dw$

$=\int\dfrac{3(w-1)}{2(3w^2+1)}~dw$

$=\int\dfrac{3w}{2(3w^2+1)}~dw-\int\dfrac{3}{2(3w^2+1)}~dw$

$=\dfrac{\ln(3w^2+1)}{4}-2\sqrt3\tan^{-1}(\sqrt3w)+C$

$=\dfrac{1}{4}\ln\dfrac{3(v-1)^2+(v+1)^2}{(v+1)^2}+2\sqrt3\tan^{-1}\dfrac{\sqrt3(v-1)}{v+1}+C$

$=\dfrac{1}{4}\ln\dfrac{3(u^2-1)^2+(u^2+1)^2}{(u^2+1)^2}+2\sqrt3\tan^{-1}\dfrac{\sqrt3(u^2-1)}{u^2+1}+C$

$=\dfrac{1}{4}\ln\dfrac{3(\tan^\frac{3}{2}x-1)^2+(\tan^\frac{3}{2}x+1)^2}{(\tan^\frac{3}{2}x+1)^2}+2\sqrt3\tan^{-1}\dfrac{\sqrt3(\tan^\frac{3}{2}x-1)}{\tan^\frac{3}{2}x+1}+C$

For $\int\sqrt[4]{\tan x}~dx$ ,

Let $t=\sqrt[4]{\tan x}$ ,

Then $x=\tan^{-1}t^4$

$dx=\dfrac{4t^3}{t^8+1}~dt$

$\therefore\int\sqrt[4]{\tan x}~dx$

$=\int\dfrac{4t^4}{t^8+1}~dt$

$=\int\dfrac{4t^4}{(t^4-\sqrt2t^2+1)(t^4+\sqrt2t^2+1)}~dt$

$=\int\dfrac{\sqrt2t^2}{t^4-\sqrt2t^2+1}~dt-\int\dfrac{\sqrt2t^2}{t^4+\sqrt2t^2+1}~dt$ (with reference to https://www.wolframalpha.com/input/?i=4x%5E2%2F%28x%5E4%2B1%29)

$=\int\dfrac{\sqrt2}{t^2-\sqrt2+\dfrac{1}{t^2}}~dt-\int\dfrac{\sqrt2}{t^2+\sqrt2+\dfrac{1}{t^2}}~dt$

$=\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}-\sqrt2}~dt+\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}-\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}+\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}+\sqrt2}~dt$

$=\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+2-\sqrt2}~dt+\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{\left(t+\dfrac{1}{t}\right)^2-2-\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+2+\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{\left(t+\dfrac{1}{t}\right)^2-2+\sqrt2}~dt$

$=\dfrac{1}{\sqrt2}\int\dfrac{d\left(t-\dfrac{1}{t}\right)}{\left(t-\dfrac{1}{t}\right)^2+2-\sqrt2}+\dfrac{1}{\sqrt2}\int\dfrac{d\left(t+\dfrac{1}{t}\right)}{\left(t+\dfrac{1}{t}\right)^2-2-\sqrt2}-\dfrac{1}{\sqrt2}\int\dfrac{d\left(t-\dfrac{1}{t}\right)}{\left(t-\dfrac{1}{t}\right)^2+2+\sqrt2}-\dfrac{1}{\sqrt2}\int\dfrac{d\left(t+\dfrac{1}{t}\right)}{\left(t+\dfrac{1}{t}\right)^2-2+\sqrt2}$

$=\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tan^{-1}\dfrac{t-\dfrac{1}{t}}{2-\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tanh^{-1}\dfrac{t+\dfrac{1}{t}}{2+\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tan^{-1}\dfrac{t-\dfrac{1}{t}}{2+\sqrt2}+\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tanh^{-1}\dfrac{t+\dfrac{1}{t}}{2-\sqrt2}+C$

$=\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tan^{-1}\dfrac{\sqrt[4]{\tan x}-\dfrac{1}{\sqrt[4]{\tan x}}}{2-\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tanh^{-1}\dfrac{\sqrt[4]{\tan x}+\dfrac{1}{\sqrt[4]{\tan x}}}{2+\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tan^{-1}\dfrac{\sqrt[4]{\tan x}-\dfrac{1}{\sqrt[4]{\tan x}}}{2+\sqrt2}+\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tanh^{-1}\dfrac{\sqrt[4]{\tan x}+\dfrac{1}{\sqrt[4]{\tan x}}}{2-\sqrt2}+C$


问:
Fourier Transform Symmetry Misunderstanding
I learned Fourier Transform about 2 years ago, but recently I found that I can't understand a simple property of that.

there is a lot of proof that shows Fourier Transform of even/odd signal is even/odd, but when I saw the proofs they missed a minus in the differential. here is my proof : (for even)

if $f(t) = f(-t)$

$F(-w) = \int_{-\infty}^{\infty} f(t)\:e^{jw2\pi t}\:dt = \int_{-\infty}^{\infty} f(-u)\:e^{-jw2\pi u}\:(-du) = \int_{-\infty}^{\infty} -f(u)\:e^{-jw2\pi u}\:du = -F(w)\:\:\:(!!)$

I use changing variable like all other proofs ( $u = -t$ )
but in other proofs they didn't do $du = -dt$ and they use $du = dt$ and get to the $F(-w) = F(w)$ easily.( I cant get why they did that)

please tell my what's my mistake.

Thanks!

答:
When you change variables $u=-t$ you do, indeed, get $du = -dt$. But you also change the limits on the integral:
$$
\int_{-\infty}^{+\infty} \phi(t)\;dt =
\int_{+\infty}^{-\infty} \phi(-u)\;(-du)
$$
then the convention says
$$
-\int_{+\infty}^{-\infty} \phi(-u)\;du = +\int_{-\infty}^{+\infty} \phi(-u)\;du .
$$


答:
The problem is that you have an incorrect change of variables.

Let's do it in the level of Riemann integrals.

Assuming absolute integrability and recall the definition
$$
\int_{-\infty}^\infty f(x)dx=\lim_{N\to\infty}\int_{-N}^N f(x)\,dx
$$

If you do the change of variables $u=-x$, then
$$
\int_{-N}^N f(x)\,dx=\int_{N}^{-N}f(-u)\cdot(-1)du=\int_{-N}^N f(-u)\,du
$$
which implies that
$$
\int_{-\infty}^\infty f(x)dx
=\int_{-\infty}^\infty f(-u)du
=\int_{-\infty}^\infty f(-x)dx
.\tag{1}
$$

Note that we do not use any parity assumptions on $f$ here.

So in your calculation, you should have
$$
\begin{align}
F(-w)
 &= \int_{-\infty}^{\infty} f(t)\:e^{jw2\pi t}\:dt\\
 &= \int_{-\infty}^{\infty} \color{red}{f(-u)\:e^{-jw2\pi u}}\:du\\
 &= \int_{-\infty}^{\infty} \color{red}{f(u)\:e^{-jw2\pi u}}\:du \\
 &= F(w)
\end{align}
$$

Title: Fourier Transform Symmetry Misunderstanding

Body:
I learned Fourier Transform about 2 years ago, but recently I found that I can't understand a simple property of that.

there is a lot of proof that shows Fourier Transform of even/odd signal is even/odd, but when I saw the proofs they missed a minus in the differential. here is my proof : (for even)

if $f(t) = f(-t)$

$F(-w) = \int_{-\infty}^{\infty} f(t)\:e^{jw2\pi t}\:dt = \int_{-\infty}^{\infty} f(-u)\:e^{-jw2\pi u}\:(-du) = \int_{-\infty}^{\infty} -f(u)\:e^{-jw2\pi u}\:du = -F(w)\:\:\:(!!)$

I use changing variable like all other proofs ( $u = -t$ )
but in other proofs they didn't do $du = -dt$ and they use $du = dt$ and get to the $F(-w) = F(w)$ easily.( I cant get why they did that)

please tell my what's my mistake.

Thanks!


问:
How to read $\frac{d^2y}{dx^2}$
Is it correct to say 'dee squared y over dee x squared'?

I am just trying to make sure because I have to teach this to someone.

I learnt that $f''(x)$ is pronounced 'f double prime of x' but I couldn't find how to pronounce Leibniz notation. Thanks!

答:
Three ways of reading $\dfrac {d^2 y}{dx^2}$...

"dee squared y over dee x squared" (your way)

"dee two y over dee x two" (informal)

The second derivative of $y$ with respect to $x$ (formal)

Title: How to read $\frac{d^2y}{dx^2}$


问:
How to prove the use of boundedness may be the key to this problem
Let $f$ be a bounded convex function on $(0,+\infty)$,prove that $f$  is decreasing and
$$\displaystyle\lim_{x\to+\infty}xf'_+(x)=0.$$
Geometrically speaking, the significance of the first problem is obvious, but I don't know how to analyze and prove it.I try to use the following inequality$$\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\leqslant\dfrac{f(x_3)-f(x_1)}{x_3-x_1}\leqslant\dfrac{f(x_3)-f(x_2)}{x_3-x_2}.$$I don't know how to take advantage of boundedness...

答:
The use of boundedness

Suppose $f$ is not decreasing. Then there are numbers $0\le a<b$ such that $f(a)<f(b)$. Let $c=b-a$.

By convexity, $f(a+2c)+f(a)\ge 2f(a+c)$ and so $$f(a+2c)-f(a+c)\ge f(a+c)-f(c).$$

Similarly, $$f(a+3c)-f(a+2c)\ge f(a+2c)-f(a+c)$$ and so on. Therefore, summing these inequalities, $$f(a+nc)-f(a)\ge n(f(a+c)-f(a))$$ and so $f(a+nc)$ is unbounded as $n$ tends to infinity.

This contradiction proves that $f$ is decreasing.


问:
Why is a locally finite measure defined on Hausdorff spaces only?
Recall the definition of a locally finite measure 1.

Let $(S,\,{\cal{T}})$ be a Hausdorff topological space and let $\Sigma$ be a $\sigma$-algebra containing the topology $\cal{T}$. A measure $\mu$ on $\Sigma$ is locally finite if $\forall$ point $p\in S$ lies in an open set of finite measure.

${\underline{\mbox{Question 1.}}}~~~~$ Why is it necessary for the topological space to be Hausdorff?

${\underline{\mbox{Question 2.}}}~~~~$ Does this definition imply that a locally finite measure is finite on any open set?

Now recall the definition of compactness:

A set $A\subseteq S$ is compact if any of its open covers contains a finite subcover.

Wikipedia 2 says that

(a)  A locally finite measure is finite on compact sets.

(b)  The converse also holds, for locally compact Hausdorff spaces.

Can you please help me to prove these two statements?

Specifically, will this proof require the Hausdorffness?

答:
Question 2.  No, Lebesgue measure on $\mathbb R$ is locally finite, and $\mathbb R$ is an open set with infinite measure.

a. Suppose $\mu$ is a locally finite measure.  Let $K$ be a compact set.

The collection $\mathscr U = \{U \subseteq S : U \text{ open and } \mu(U) < +\infty\}$ is an open cover of $K$, so it has a finite subcover.  Thus $K$ is contained in a finite union of open sets of finite measure, and then $\mu(K) < \infty$.

b. Use definition of locally compact.


问:
Are Jordan Domains Locally Connected?
I have been trying to answer this question, to which I do not know the answer. Are Jordan Domains locally connected?

In this context, a Jordan Domain is just the bounded interior of a Jordan curve.

Edit: In fact, I know the answer. It is true due to the Caratheodory Extension theorem. However, could it be proven using exclusively topological arguments?

答: Yes. The bounded interior is actually homeomorphic to an open ball, by the Schönflies Theorem. Hence it is locally connected. The closure is also locally connected, being homeomorphic to a closed disc, by the same theorem.</p>

Title: Are Jordan Domains Locally Connected?


Title: Solve ODE $y' = \frac{xy}{x^2-y^2}$

Body:
$$y' = \frac{xy}{x^2-y^2}$$
$$x^2-y^2 = {xx'y}$$
Substitute $u=x^2$:
$$u-y^2=\dfrac y 2u'$$
$$yu'-2u=-2y^2$$
$$ \left (\dfrac u {y^2} \right)'=-\dfrac 2y$$
Integrate.
$$x^2=y^2(C-2 \ln y)$$


问:
How to Axiomize the Notion of "Continuous Space"?
EDIT (to clear up controversy and misunderstandings caused by my poor wording): Historically, Riesz's efforts to try and make rigorous a notion of a "continuous space" (as opposed to "discrete ones") were part of some of the movements in mathematics which lead to the establishment of the axioms of topology.

Some people tout topologies as being "the axiomization of the notion of a continuous space". However, they are clearly too general, as I argue below.

What is an appropriate level of axiomization to make rigorous a notion of "continuous space"?

Why do we allow either discrete or trivial topologies in the definition of topology? (if we want it to define "continuous spaces", otherwise there is no issue)

More generally, why do we allow points to be open (the existence of isolated points) or the failure of the T0 axiom (existence of topologically indistinguishable points)? (if we want it to define "continuous spaces", otherwise there is no issue)

Both scenarios could be prevented by addition of the following axioms to the definition of topology:

No point is open

Every point has a unique neighborhood system

This question is a follow-up to my previous question: Why study non-T1 topological spaces?

The historical motivation for topology, as far as I have read, was to give a rigorous notion to the intuition of a "continuous space". The three axioms commonly used are equivalent to those proposed by Kuratowski, Riesz, and Hausdorff, who were trying to axiomatize "continuous space".

But discrete spaces are the exact opposite of that, and in general allowing isolated points allows the possibility of spaces with "discrete components". It also makes the definition of "limit" unnecessarily complicated. The discrete topology is also equivalent to the power set, so is in effect not a new notion. Also every function from a discrete space is continuous, and any definition of a function at an isolated point is continuous, which goes counter to the expectation that the morphisms of a given category should be "special" functions. Basically discrete topologies gains us no new insights compared to elementary set theory.

Moreover, points not having unique neighborhood systems lead to all sorts of pathologies (for example here: Non-T1 Space: Is the set of limit points closed?), and even the Zariski topology, which is the most pathological commonly used topology of which I am aware, is still in general $T0$.

Finally, any non-$T0$ space should be homeomorphic (I think) to the quotient of the space under the equivalence relation " $x \sim y$ if and only if $x$ and $y$ are topologically indistinguishable".

This question is similar in spirit to this one, but to me it is very clear why we would want to consider non-metric spaces, at least as someone who is interested non-metricizable spaces like those which can occur in functional analysis. I am aware of the notion of continuum, but since the definition requires the space to be metric, and focuses more on compactness than separation axioms and connectedness rather than path-connectedness, I believe it to be the incorrect axiomization of a "continuous space". The existence of indecomposable continua lends credence to this notion, in my view.

However, it seems clear to me that the proper axiomization of a "continuous space" must lie somewhere in between the notion of "metric space" and "topological space", as the former is far too restrictive, and the latter far too general. I would enjoy your thoughts on the matter.

(I want to better axiomize the notion of "continuous space" so as to better facilitate the study of stochastic processes, not that this is directly relevant to the immediate question at hand.)

EDIT:

At least Riesz appeared to have been clearly interested in defining some notion of "continuous space" or "continuum" (excerpt from "History of Topology" on google books (https://books.google.com/books?id=7iRijkz0rrUC&pg=PA212&lpg=PA212&dq=riesz+topology+axioms&source=bl&ots=B_xcnG6StL&sig=EajpOCr3XRUtA9hwqpLwArKLmoY&hl=ru&sa=X&ved=0ahUKEwj3pMPs9ZvNAhVBGFIKHcXmB28Q6AEIJzAC#v=onepage&q&f=false):

In a footnote Riesz criticisms the way in which philosophers have dealt with notions like continuous and discrete and he repeats Russell’s remark about the followers of Hegel: «the Hegelian dictum (that everything discrete is also continuous and vice versa) has been tamely repeated by all of his followers. But as to what they meant by continuity and discreteness, they preserved a discrete and continuous silence; […]» (Riesz [147]). The relation of our subjective experience of space and time and mathematical continua is described by Riesz as follows. Mathematical continua possess certain properties of continuity, coherence and condensation. On the other hand, our subjective experience of time is discrete and consists of countable sequences of moments. Systems of subsets of a mathematical continuum can be interpreted as a physical continuum when subsets with common elements are interpreted as undistinguishable and subsets without common elements as distinguishable. Rise [147, p. 111] is an interesting paper in which Riesz, who had read Frechet’s work and appreciated it, developed a different theory of abstract spaces, based on the notion of «Verdichtungsstelle», i.e. «condensation point», or as we will translate «limit point». In his theory Riesz succeeded in deriving the Bolzano-Weierstrass Theorem and the Heine-Borel Theorem. We will not discuss this paper. We will restrict ourselves to a shorter paper that was presented by Riesz in 1908 at the International Congress of Mathematicians in Rome. In that paper, «Stetigkeit und Abstract Mengenlehre» (Riesz [148]), concentrates on the characterization of mathematical continua. We will briefly describe some of the ideas the Riesz describes in the paper. As we said, Riesz’s basic notion is the notion of a limit point (Verdichtungsstelle).

I'm not asking anyone to agree with Riesz's basic goal (to get a rigorous distinction between "continuous" and "discrete" spaces or time).

To those who might object that the standard three axioms are all that are necessary to define the notions of connectedness and compactness, I have several responses:

Connectedness actually isn't that nice of a property. We could look at the topologist's sine curve or the infinite broom for examples, but for me the fact that there exists a countable, Hausdorff, and connected set implies that topological connectedness is not quite the intuitive connectedness from Euclidean spaces which we want to generalize for an arbitrary "continuous space". In my mind, path connectedness better satisfies this criterion, since every non-trivial T1 path-connected space is uncountable. Path connectedness does provide reasonable properties, although one can object that the definition is somewhat of a tautology ("a space behaves like the real line if it behaves like the real line") and requires the prior construction of the real line in order to define, rather than being constructible from first principles.

Compactness also isn't very nice unless we are in a Hausdorff space, since otherwise compact sets aren't even closed. Sure $T0$ isn't Hausdorff, but it's certainly a step in the right direction. (Hence why the distinction between "compactness" and "quasicompactness" is so prevalent in algebraic geometry where the non-Hausdorff Zariski topology so frequently comes into play).

For non-$T0$ spaces, the set of limit points isn't even closed, limits aren't unique or well-defined... the notion of limit so intrinsic to the idea of a "continuous space" clearly requires at least $T0$, if not even $T1$ or $T2$, to function even remotely similarly to intuition.

Related but different: Why the axioms for a topological space are those axioms?

The definition of metric space,topological space

What concept does an open set axiomatise?

Why do we require a topological space to be closed under finite intersection?

https://xorshammer.com/2011/07/09/a-logical-interpretation-of-some-bits-of-topology/

https://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets

https://en.wikipedia.org/wiki/Decidability_(logic)#Semidecidability

meaning of topology and topological space

Motivation behind topology

Why is discrete space ``discrete"

Discretizations of Differential, Geometric and Topological Notions

Origins of the modern definition of topology

答:
This might be a little late to the party, but I highly recommend reading about the Theory of Linear Structures from Tim Maudlin.

Rather than using the notion of an open set as primitive, this theory uses the notion of a Line (essential to this notion is the linear ordering of points). A linear structure is a point set S along with a set of lines L, satisfying certain axioms. The lines contain the information of which points "surround" which, specifying the neighborhood structure, which specifies open set structure, and so on (this theory does employ different ways of expressing concepts using the available info of lines, as opposed to the usual definitions with open sets).

A space is called a continuum if the linear order of every line is dense and complete. Does that help?


问:
What does it mean geometrically that $H^*(B\mathbb{Z}_2) = \mathbb{Z}[a]$?
I found cited in a physics paper that $H^*(B\mathbb{Z}_2) = \mathbb{Z}_2[a]$ and I wanted to know what geometric shapes this corresponded to.  Notice the answer is a polynomial ring.  Here $\mathbb{Z}2 = \mathbb{Z}/2\mathbb{Z}$.

Wikipedia's article on classifying space refers to homotopy theory and gives the example that the circle $S^1$ is the classifying space for the infinite cycle group $\mathbb{Z}$.  This is written as $E\mathbb{Z} = \mathbb{R}$ and they are discussing the "universal bundle".

I'd like a more down-to-earth explication.  What kind of objects are being represented by the classifying space of $\mathbb{Z}_2$ and what do the equivalences classes in the cohomology ring look like?

What could mod-2 cohomology be measuring about the shape of the space?  Here it says $M_3$ could we could have $M = S^3 / i(S^1)$ with some reasonable map $i : S^1 \to S^3 = \mathbb{R}^3 \cup \{ pt\}$  (a "knot").  This is taken from a Physics paper from arXiv which I am still looking at [1].

Just to collect from the comments:

$B\mathbb{Z}_2$ is the classifying space of principal $\mathbb{Z}_2$-bundles over paracompact spaces.  What I am seeing is an perfectly good abstract definition:

For a discrete group $G = \mathbb{Z}/2\mathbb{Z}$ is ... a path-connected topological space $X$ such that the fundamental group of $X$ is isomorphic to $G$ and the higher homotopy groups are trivial.

Since I have only read through Munkres and Hatcher the last parts of that sentence are challenging.
It sounds like we could just say "two-sheet cover".

The cohomology ring of real projective space $H^*(\mathbb{R}P^\infty, \mathbb{Z}_2) = \mathbb{Z}_2[a]$ which is thes same as the ring in question. Here they do mention that for the configuration space of three points on a sphere $H^*(S^{m-1}) \simeq \mathbb{Z}[a]/(a^2) $ so this is already a much smaller ring than the answer that we have.

Reading through the definition of paracompact space we have, at least that metric spaces are paracompact and CW complexes are paracompact.

Here "geometric" just means "shape".  Here's another possible question... why are 2-sheeted covers of CW complexes classified by a polynoial ring?

答:
It's unclear how to respond to your question without knowing more about your background in topology. In any case, this answer will mostly repeat things that have already been said in the comments.

$B \mathbb{Z}_2$ can be thought of as either the classifying space of principal $\mathbb{Z}_2$-bundles (which correspond to double covers) or as the classifying space $BO(1)$ of real line bundles. Maybe the most "geometric" description of it is that it is an infinite real projective space $\mathbb{RP}^{\infty}$, so you can get some intuition for it by studying the finite real projective spaces.

If $M$ is path-connected and maybe satisfies some other mild hypotheses then homotopy classes of maps $M \to B \mathbb{Z}_2$ are computed by

$$H^1(M, \mathbb{Z}_2) \cong \text{Hom}(\pi_1(M), \mathbb{Z}_2) \cong \text{Hom}(H_1(M), \mathbb{Z}_2).$$

As far as the cohomology calculation goes, the generating class $a$ corresponds to the first Stiefel-Whitney class $w_1 \in H^1(-, \mathbb{Z}_2)$ of a real line bundle. Loosely speaking it measures how the line bundle "twists" (more formally, it is measuring monodromy, which the identification with $\text{Hom}(\pi_1(M), \mathbb{Z}_2)$ also suggests), and is zero iff the line bundle is trivial(izable). The fact that the cohomology (over $\mathbb{Z}_2$) is the polynomial algebra on $w_1$ says that every characteristic class of real line bundles is a polynomial in $w_1$ and that $w_1$ is free in the sense that it satisfies no polynomial relations. This reflects the fact that, by the computation above, it can take on any value in $H^1(M, \mathbb{Z}_2)$ for any $M$.

The fact that $BO(2) \cong B \mathbb{Z}_2$ says that $w_1$ is a complete invariant of real line bundles: two real line bundles $L, L'$ are isomorphic iff $w_1(L) = w_1(L')$.

It's unclear whether any of this constitutes an answer to your question; it would help if you were more specific about which of this most sounds like what you're looking for.


问:
number of ways to make $2.00
How many different ways can you make $2.00 using only 1 cent, 5 cent, 10 cent, and 25 cent pieces, and 1 and 2 dollar bills (there are 100 cents in a dollar)? I have worked out an equation:

$$p + 5n + 10d + 25q + 100l + 200t = 200$$

Where all variables are non-negative integers. But I don't know what to do from here to figure out how many different sets of non-negative integers there are that satisfy this equation. What do I do next?

答:
Use generating functions.
All you have to do is find the coefficient of $x^{200}$ in the formula
$$
(1 + x + x^2 + \cdots)(1 + x^5 + x^{10} + \cdots)(1 + x^{10} + x^{20} + \cdots)
$$
$$
\cdot (1 + x^{25} + x^{50} + \cdots)(1 + x^{100} + x^{200} + \cdots)(1 + x^{200} + x^{400} + \cdots)
$$

The problem is inherently quite computation-heavy, but reducing it to the question above makes the computation much more straightforward, and also very easy to do with a computer.

Title: number of ways to make $2.00


问:
Find surface area of cylinder $y^2 + z^2 = 1$ between two planes: $x + y - 2 = 0$ and $x-z+4 = 0$.
Find surface area of a cylinder $y^2 + z^2 = 1$ between two planes: $x + y - 2 = 0$,  $x-z+4 = 0$.

This was my approach:

I drew the picture and projected the figure onto $xOy$ plane, and after that I found partial detivatives $\frac{\partial z}{\partial x} $ and $ \frac{\partial z}{\partial y}$ where $z = + \sqrt{1-y^2}$. I choose the positive sqare root because of the symmetry, and that's why I will multiply following integral by $2$.

Eventually, my integral looks like this:

$P(S) = 2\begin{gather*}
    \iint_D \sqrt{1+ \frac{y^2}{1-y^2} + 0^2}\,dx\,dy =
 2\iint_D \frac{1}{\sqrt{1-y^2}}\,dx\,dy =
... = 9\pi
\end{gather*}
$ .

$D$ is $\{ -4 \leq x \leq 1 \land -1\leq y \leq 1  \} \cup \{ 1 \leq x \leq 3 \land  -1 \leq y \leq 2-x \}$

I wonder if this is correct and if anyone could tell me if I am wrong on this, I would really appreciate it.

答:
Solving this problem is easier if you parameterize this surface. Consider $$\vec{r}(u,v)=\big$$ for $\sin(v)-4 \leq u \leq -\cos(v)+2$ and $0\leq v \leq 2\pi$. The surface area in your question equals $$\int_0^{2\pi} \int_{\sin(v)-4}^{-\cos(v)+2}||\vec{r}_u \times \vec{r}_v||dudv=12\pi$$

Remark: If you wanted to solve it your way, you wouldn't be able exploit the symmetry of the cylinder $y^2+z^2=1$ by the way it intersects the plane $x-z+4=0$. Calculating the surface area with $\int \int _{D}\sqrt{1+z_x^2+z_y^2}dA$ gives $$\int_{-1}^{1} \int_{-4-\sqrt{1-y^2}}^{-y+2}\frac{dxdy}{\sqrt{1-y^2}}+\int_{-1}^{1} \int_{-4+\sqrt{1-y^2}}^{-y+2}\frac{dxdy}{\sqrt{1-y^2}}=12\pi$$

Body:
Solving this problem is easier if you parameterize this surface. Consider $$\vec{r}(u,v)=\big<u,\cos(v),\sin(v)\big>$$ for $\sin(v)-4 \leq u \leq -\cos(v)+2$ and $0\leq v \leq 2\pi$. The surface area in your question equals $$\int_0^{2\pi} \int_{\sin(v)-4}^{-\cos(v)+2}||\vec{r}_u \times \vec{r}_v||dudv=12\pi$$

Remark: If you wanted to solve it your way, you wouldn't be able exploit the symmetry of the cylinder $y^2+z^2=1$ by the way it intersects the plane $x-z+4=0$. Calculating the surface area with $\int \int _{D}\sqrt{1+z_x^2+z_y^2}dA$ gives $$\int_{-1}^{1} \int_{-4-\sqrt{1-y^2}}^{-y+2}\frac{dxdy}{\sqrt{1-y^2}}+\int_{-1}^{1} \int_{-4+\sqrt{1-y^2}}^{-y+2}\frac{dxdy}{\sqrt{1-y^2}}=12\pi$$


问:
How to integrate $\int_0^{\infty} \frac{x}{e^x+1} dx$
I want to integrate it explicitly. I looked up if this had an exact solution, and I got that it is $\frac{\pi^2}{12}$ here: https://www.integral-calculator.com/#expr=x%2F%28e%5Ex%2B1%29&lbound=0&ubound=inf

I thought it could be solved using geometric series, so I tried to turn my integral into something similar:
$$\int_0^{\infty} \frac{x}{e^x+1} dx= \int_0^{\infty} \frac{x}{e^x+1} \frac{e^x-1}{e^x-1} dx=\int_0^{\infty} \frac{x e^x }{e^{2x}-1}dx -\int_0^{\infty} \frac{x}{e^{2x}-1} dx= \int_0^{\infty} \frac{x e^x }{e^{2x}-1}dx -\frac{1}{4}\int_0^{\infty}  \frac{x}{e^{x}-1} dx=\int_0^{\infty} \frac{x e^x}{e^{2x}-1} - \frac{1}{4}\frac{\pi^2}{6}$$

Where the second term is easy to solve with geometric series, and it's also defined with the Riemann zeta:

$$\int_0^{\infty} \frac{x}{e^x-1}=\zeta(2)=\frac{\pi^2}{6}$$

I think I can use geometric series also with the other integral, I've tried to solve it this way and I got:
$$\int_0^{\infty}\frac{xe^x}{e^x-1}=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{\left(n-\frac{1}{2}\right)^2} $$

And I don't know how to obtain a value from this infinite sum, although if I'm not wrong it should be $\pi^2/2$.

Anyone knows how to do this? Also, I think I might be looping the loop and there's some easier way to proceed from the beggining, so I'd appreciate another point of view to solve this.

Thanks.

答:
Alternatively, let $t=e^{-x}$ to express the integral as

$$\int_0^{\infty} \frac{x}{e^x+1} dx= - \int_0^1 \frac{\ln t}{1+t}dt
\overset{IBP} = \int_0^1 \frac{\ln (1+t)}{t}dt
 = \frac{\pi^2}{12}
$$
where the result
Finding $ \int^1_0 \frac{\ln(1+x)}{x}dx$ is used.


问:
Probability that a quadratic polynomial with random coefficients has real roots
The following is a homework question for which I am asking guidance.

Let $A$, $B$, $C$ be independent random variables uniformly distributed between $(0,1)$. What is the probability that the polynomial $Ax^2 + Bx + C$ has real roots?

That means I need $P(B^2 -4AC \geq 0$). I've tried calling $X=B^2 -4AC$ and finding $1-F_X(0)$, where $F$ is the cumulative distribution function.

I have two problems with this approach. First, I'm having trouble determining the product of two uniform random variables. We haven't been taught anything like this in class, and couldn't find anything like it on Sheldon Ross' Introduction to Probability Models.

Second, this strategy just seems wrong, because it involves so many steps and subjects we haven't seen in class. Even if I calculate the product of $A$ and $C$, I'll still have to square $B$, multiply $AC$ by four and then subtract those results. It's too much for a homework question. I'm hoping there might be an easier way.

答:
We want the volume of the region in the unit cube satisfying the property $AC < Z \equiv B^2 / 4$. Separate into two regions: $A < Z$ and $A > Z$. For the former, all $C$ up to $1$ satisfy the property, while for the latter, all $C$ up to $Z / A$ satisfy the property. We then have

$$
\int_0^1 dB \int_0^Z dA \int_0^1 dC + \int_0^1 dB \int_Z^1 dA \int_0^{Z/A} dC = \frac{1}{12} + \frac{1 + 3 \ln 2}{18} \approx 0.2544
$$

ContentLicense: CC BY-SA 2.5


问:
Why are singletons of Real Numbers an interval according to this definition?
"Definition: A subset S of R is said to be an interval if it has the following property: if x∈S, z∈S, and y∈R are such that x < y < z, then y∈S"

According to the book, singletons are intervals from this definition, but since singletons only have one element, why would that be the case?

答:
Consider the singleton $S=[x,z]$ with $z=x$; i.e., $S=[x,x]=\{x\}$.

For this $S$, there is no $y\in\mathbb R$ such that $x<y<z=x$,

so the property in the definition of an interval is satisfied vacuously.


问:
Possible darts scores and probability thereof
As an intellectual exercise only, I am attempting to solve the following:
by throwing 3 darts at a standard dartboard, it is possible to score a maximum of 180 (3 x treble 20), and a minimum of zero (all darts miss the scoring area).
1. Which total scores within this range are not possible to achieve? (eg 179)
2. For each of the possible scores, how many ways are there to achieve that score? (eg a total of 5 could be
0, 0, 5, or
0, 3, 2, or
0, treble 1, double 1, or
0, 1, double 2

etc etc)

Remember that we are looking at combinations, not permutations, since the order of the 3 darts is irrelevant.

I am working on an Excel solution to this at present and would be interested to see if anyone else has a solution to compare with what I hope will be my solution.

Regards

答:
You are correct. There are 9 scores that cannot be achieved and they are in fact 163, 166, 169, 172, 173, 175, 175, 178, 179. I did this with python, here is my code:

```<code>possibleShotScore = []

for i in range(0, 21):
    possibleShotScore.append(i)
    possibleShotScore.append(i * 2)
    possibleShotScore.append(i * 3)

# Special cases:
possibleShotScore.append(25)
possibleShotScore.append(50)

# Only save unique scores:
possibleShotScore = set(possibleShotScore)

possibleFinalScores = []
for first in possibleShotScore:
    for second in possibleShotScore:
        for third in possibleShotScore:
            possibleFinalScores.append(first + second + third)
            # Uncomment following to see possible partitions for a number if present.
            # Here, I have it set to 149 but you can change this value to your liking.
            # if first + second + third == 149: print(first, second, third)

counter = 0
unattainablePartitions = []
for x in range(0, 181):
    if x not in possibleFinalScores:
        unattainablePartitions.append(x)
        counter += 1

print("No. of unattainable scores: ", counter)
print("Unattainable Scores: ", unattainablePartitions)
</code>```

Output:

```<code>$ python3 dart.py
No. of unattainable scores:  9
Unattainable Scores:  [163, 166, 169, 172, 173, 175, 176, 178, 179]
</code>```

Body:
You are correct. There are 9 scores that cannot be achieved and they are in fact 163, 166, 169, 172, 173, 175, 175, 178, 179. I did this with python, here is my code:

possibleShotScore = []

for i in range(0, 21):
    possibleShotScore.append(i)
    possibleShotScore.append(i * 2)
    possibleShotScore.append(i * 3)

# Special cases:
possibleShotScore.append(25)
possibleShotScore.append(50)

# Only save unique scores:
possibleShotScore = set(possibleShotScore)

possibleFinalScores = []
for first in possibleShotScore:
    for second in possibleShotScore:
        for third in possibleShotScore:
            possibleFinalScores.append(first + second + third)
            # Uncomment following to see possible partitions for a number if present.
            # Here, I have it set to 149 but you can change this value to your liking.
            # if first + second + third == 149: print(first, second, third)

counter = 0
unattainablePartitions = []
for x in range(0, 181):
    if x not in possibleFinalScores:
        unattainablePartitions.append(x)
        counter += 1

print("No. of unattainable scores: ", counter)
print("Unattainable Scores: ", unattainablePartitions)

Output:

$ python3 dart.py
No. of unattainable scores:  9
Unattainable Scores:  [163, 166, 169, 172, 173, 175, 176, 178, 179]


问:
exponential generating functions containing an even/odd number of cycles
How to derive the exponential generating functions that having an even/odd number of cycles?
And how to define a bijection between them? Is there any example of this? Thanks in advance!</p>

答:
[Edit: corrected the silly mistake pointed out by Marc]

The parity of the number of cycles of a permutation corresponds to the parity of the permutation itself. For $n\gt1$ a bijection is given by multiplying every even permutation by a fixed transposition. Thus there are $n!/2$ permutations of each kind. This doesn't work for $n\le1$ since in this case there's no transposition to multiply by. There's one permutation with an even number ($0)$ of cycles for $n=0$ and one permutation with an odd number ($1$) of cycles for $n=1$, so the exponential generating functions are

$$
1+\sum_{n=2}^\infty\frac12x^n=1+\frac12\left(\sum_{n=0}^\infty x^n-1-x\right)=\frac12\left(1-x+\frac1{1-x}\right)
$$

for an even number of cycles and

$$
x+\sum_{n=2}^\infty\frac12x^n=x+\frac12\left(\sum_{n=0}^\infty x^n-1-x\right)=\frac12\left(-1+x+\frac1{1-x}\right)
$$

for an odd number of cycles.


问:
L'Hospital's rule doesn't converge for a function with square root
I was trying to find the limit of a function of the form $\frac{x}{\sqrt{(x+a)(x+b)}}$. When I apply L'Hospital's rule and differentiate the numerator and denominator, after simplification I end up with the same form as I started with:

$$L = \lim_{x \to \infty} \frac{x}{\sqrt{(x+a)(x+b)}} \\
= \lim_{x \to \infty} \frac{1}{\frac{1}{2} \frac{(2x+a+b)}{\sqrt{(x+a)(x+b)}}} = \frac{2\sqrt{(x+a)(x+b)}}{2x+a+b} = \frac{\frac{2x+a+b}{\sqrt{(x+a)(x+b)}}}{2} \\
= \lim_{x \to \infty} \frac{x}{\sqrt{(x+a)(x+b)}} \textrm{(because $\lim_{x \to \infty} \frac{a+b}{\sqrt{...}}$ is zero)}$$

I noticed if I solve it by computing the limit of the squared value, the solution is easy:
$$L^2 = \lim_{x \to \infty} \frac{x^2}{(x+a)(x+b)} \\
L^2 = \lim_{x \to \infty} \frac{2x}{2x+b+c} = 1 \\
L = \sqrt{1} = 1$$

Before I found this solution, I was searching online for different ways of evaluating limits and applying L'Hospital's rule, but none of the resources I came across seem to cover this case. Am I missing another straightforward way of solving this (whether using L'Hospital's or not)? When does L'Hospital's rule fail to converge, and what does it mean when it does?

答:
Discussion of L'Hopital's Rule aside, the straightforward way to solve the limit would be to multiply both numerator and denominator by $\frac1x$:

$$L = \lim_{x \to \infty} \frac{x}{\sqrt{(x+a)(x+b)}} = \lim_{x \to \infty} \frac{1}{\sqrt{\frac {(x+a)(x+b)}{x^2}}}=\lim_{x \to \infty} \frac{1}{\sqrt{\left(1+\frac ax\right)\left(1+\frac bx\right)}}=1$$


问:
Finding Solutions to a System of Diophantine Equations
I'm trying to find triplets of integer $(x, y)$ pairs - $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ - that satisfy the following equations:

$$
{x_1}^2 + {y_1}^2 = {x_2}^2 + {y_2}^2 = {x_3}^2 + {y_3}^2 \\
x_1 + x_2 + x_3 = 5 \\
y_1 + y_2 + y_3 = 0 \\
(x_1, y_1) \neq (x_2, y_2) \\
(x_1, y_1) \neq (x_3, y_3) \\
(x_2, y_2) \neq (x_3, y_3)
$$
Currently, for each integer $c$ that can be written as the sum of two squares, I create a list of all possible integer pairs $(a, b)$ for which $a^2 + b^2 = c$ and then check all possible triplets from each list for validity (actually I only need to check all possible combinations of 2 pairs, but the idea is the same). The only constraints I've found are $c$ must be divisible by 5 and $a$ and $b$ must have different parity. This is pretty inefficient, as only a very small subset of possible values of $c$ produce valid triplets. Checking the validity of any given triplet is relatively easy, so I'm hoping someone can help me find further constraints on the values of $c$. It is also entirely possible that there is some other, more efficient method of finding these triplets. Either way, help would be much appreciated.

答:
We can use functions derived from the Pythagorean theorem where $A^2+B^2=C^2.\quad$ One generator is Euclid's formula, shown here as
$ \quad A=m^2-k^2\quad B=2mk \quad C=m^2+k^2$

All $C$-values are of the form
$4n+1$ but not all $4n+1$ are valid $C$-values. A short list is shown here. You will notice, for $C\in \{65,85,145,185,205,221,265,305,325,365\}$, there are $2$ entries because those have $2$ different primitive Pythagorean triples each. For $C=1105$, there would be $4$.

We can find the $(m,k)$ pairs for triples for any given $C$-value by solving the $C$-function for $k$ and testing a defined range of $m$-values to see which yield integers. If none are found, there is no primitive or $2\times$ or square-multiple of a primitive for that $C$-value. Here is how we find triples.

\begin{equation}
C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\\
\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.

$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\\
\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$

The number of "primitive" triples will be $\quad 2^{n-1}$ where $n$ is the number of prime factors of $C$ meaning there will never be $3$ and only $3$ primitive triples for any $C$-value. For example, with $C=65=5\times13$, there are $2^1=2$ primitive triples. There $are$ values that also have non-primitive corresponding triples such as $C=325$. The first is non-primitive but those triples are

$$f(15,10)=(125,300,325)\quad f(17,6)=(253,204,325)\quad f(18,1)=(323,36,325)$$
So we have $\qquad125^2+300^2\space =\space 253^2+204\space =\space 323^2+36^2\space =\space 325^2$

The combination $125+253-323\ne5\space$ but $\space 125+253-323=55$ if that helps (kidding). Dividing both $x$ and $y$ by $11$ would be valid but $300-204-36\ne0$ so this triple of triple will not work in any case.

There are $67$ C-values where
$\quad C=4n+1\space\text{ for }\space 81\le n\le 11925\quad$ that have $3$ triples each. You would have to find them programmatically as I did but perhaps one of them satisfies your requirement
$\quad x_2+x_2+x_3=5\quad\land\quad y_2+y_2+y_3=0$


问:
find the condition on A for the summation to be convergent
The summation is:
$$\sum_{n=1}^\infty \frac{ \sqrt { n + 1 } - \sqrt n }{n^A}$$

I don't know how to even begin. Hints??

答:
Hint: A standard beginning is to multiply top and bottom by $\sqrt{n+1}+\sqrt{n}$.

We end up with
$$\sum_{n=1}^\infty \frac{1}{n^A(\sqrt{n+1}+\sqrt{n})}.$$
Now note that $\sqrt{n}\lt \sqrt{n+1}\le 2\sqrt{n}$. From this we obtain
$$     \frac{1}{3n^{A+1/2}}       \lt    \frac{1}{n^A(\sqrt{n+1}+\sqrt{n})}\lt \frac{1}{2n^{A+1/2}}.$$
These inequalities should be enough to do the appropriate comparisons with familiar series.


答: Here's another approach which utilizes the total law of probability directly. $$P(B^2\geq 4AC)=\int_{0}^{1}P(B^2\geq4AC|C=c)f_{C}(c)dc$$ Using independence the above simplifies to $$P(B^2\geq 4AC)=\int_{0}^{1}P(B^2\geq4Ac)dc$$ When $c\in\Big(0,\frac{1}{4}\Big)$ we have $$P(B^2 \geq 4Ac)=\int_0^1 \int_{\sqrt{4ac}}^1dbda=1-\frac{4\sqrt{c}}{3}$$ When $c\in \Big[\frac{1}{4},1\Big)$ we have $$P(B^2 \geq 4Ac)=\int_0^1 \int_0^{b^2/4c}dadb=\frac{1}{12c}$$ Putting everything together $$P(B^2\geq 4AC)=\int_0^{1/4}\Bigg(1-\frac{4\sqrt{c}}{3}\Bigg)dc+\int_{1/4}^1\frac{dc}{12c}=\frac{5+3\ln(4)}{36}$$

Title: Probability that a quadratic polynomial with random coefficients has real roots

Body:
The following is a homework question for which I am asking guidance.

Let $A$, $B$, $C$ be independent random variables uniformly distributed between $(0,1)$. What is the probability that the polynomial $Ax^2 + Bx + C$ has real roots?

That means I need $P(B^2 -4AC \geq 0$). I've tried calling $X=B^2 -4AC$ and finding $1-F_X(0)$, where $F$ is the cumulative distribution function.

I have two problems with this approach. First, I'm having trouble determining the product of two uniform random variables. We haven't been taught anything like this in class, and couldn't find anything like it on Sheldon Ross' Introduction to Probability Models.

Second, this strategy just seems wrong, because it involves so many steps and subjects we haven't seen in class. Even if I calculate the product of $A$ and $C$, I'll still have to square $B$, multiply $AC$ by four and then subtract those results. It's too much for a homework question. I'm hoping there might be an easier way.


问:
Decay of time derivative
If a function $f(\vec{x}, t)$ goes to zero faster than $\frac{1}{|\vec{x}|}$, will $\frac{\partial f}{\partial t}$ always decay faster than $\frac{1}{|\vec{x}|}$ too? Assume that $f$ is differentiable.

答:
Seems reasonable and desirable but without other constraints, such as satisfying a differential equation, the answer is No; here's a counterexample.

In one spatial dimension take $f(x, t) = \sin(x/t) / x^{3/2}$.
For $t > 0$, $\vert f(x,t) \vert \leq 1 / \vert x \vert^{3/2}$ but
$f_t(x, t) = -\cos(x/t)/(x^{1/2} t^2)$, which, as $x \rightarrow \infty$, doesn't decay at the rate you'd like it to.

Title: Decay of time derivative


Title: find the condition on A for the summation to be convergent

Body:
The summation is:
$$\sum_{n=1}^\infty \frac{ \sqrt { n + 1 } - \sqrt n }{n^A}$$

I don't know how to even begin. Hints??

Body:
HINT:

$$\frac{\sqrt{n+1}-\sqrt{n}}{n^A}=\frac{\sqrt{n+1}-\sqrt{n}}{n^A}\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac1{n^A\left(\sqrt{n+1}+\sqrt{n}\right)}$$


问:
Product in the Category of $L$-structures with Homomorphisms.
let $\scr {A}_1$, and $\scr {A}_2$ be $L$-structures. Define an L-structure $\scr A_1 \times \scr A_2$ with universe $A_1 \times A_2$, such that the natural epimorphism $\pi_i :\scr A_1 \times \scr A_2 \to \scr A_i $ for $i=1,2$ satisfy the following universal property:

given any $L$-structure $\scr D$ and homomorphisms $\phi _i : \scr D \to \scr A_i$ for $i=1,2$, there is a unique homomorphism $\psi : \scr D \to \scr A_1 \times \scr A_2 $ such that $\pi_i(\psi)=\phi_i$ for $i=1,2$

This notion also popped up in a category theory book I stumbled upon. However, it was taken as an axiom without any further explanation.

Construction:

Let $\scr A_1,A_2$ be two $L$-structures with $f_i^{\scr A_i} \in \scr A_i$.

Define $\scr A_1 \times \scr A_2$ as $\langle A_1 \times A_2, \pi_i \rangle$ including constants induced by $f_i$ and relation on each $A_i$

Define
$\pi_i: A_1 \times A_2 \to A_i$, as follows,

$\pi_i=\{(a,f(a))| a,f(a)\in \scr A_i\}$ with $i=1,2$

Will this work?

答:
What you've written in the construction section makes no sense to me. Most of it doesn't even parse correctly.

The task is to define an $L$-structure $\mathscr{A}_1\times \mathscr{A}_2$. You are told that the universe of this structure should be the cartesian product $A_1\times A_2$. It remains to define, for each symbol in the language, an interpretation of that symbol on $A_1\times A_2$.

You don't need to define $\pi_1$ and $\pi_2$: the "natural epimorphisms" referred to in the question are just the projections onto each coordinate. So $\pi_1\colon A_1\times A_2\to A_1$ is defined by $\pi_1(a_1,a_2) = a_1$, and $\pi_2\colon A_1\times A_2\to A_2$ is defined by $\pi_2(a_1,a_2) = a_2$.

Ok, let's say we have a constant symbol $c$ in the language. Then we have elements $c^{\mathscr{A}_1}\in A_1$ and $c^{\mathscr{A}_2}\in A_2$. We want to interpret $c^{\mathscr{A}_1\times \mathscr{A}_2}$ as an element of $A_1\times A_2$. Which element shall we pick? There's really only one reasonable choice: $c^{\mathscr{A}_1\times \mathscr{A}_2} = (c^{\mathscr{A}_1},c^{\mathscr{A}_2})$.

Now let's say we have an $n$-ary function symbol $f$ in the language. Then we have functions $f^{\mathscr{A}_1}\colon A_1^n\to A_1$ and $f^{\mathscr{A}_2}\colon A_2^n\to A_2$. We want to interpret $f^{\mathscr{A}_1\times \mathscr{A}_2}$ as a function $(A_1\times A_2)^n\to (A_1\times A_2)$. Again, there's really only one reasonable choice: $f^{\mathscr{A}_1\times \mathscr{A}_2}((a_1,a_1'),\dots,(a_n,a_n')) = (f^{\mathscr{A}_1}(a_1,\dots,a_n),f^{\mathscr{A}_2}(a_1',\dots,a_n'))$.

Finally, let's say we have an $n$-ary relation symbol $R$ in the language. Then we have relations $R^{\mathscr{A}_1}\subseteq A_1^n$ and $R^{\mathscr{A}_2}\subseteq  A_2^n$. We want to interpret $R^{\mathscr{A}_1\times \mathscr{A}_2}$ as a subset of $(A_1\times A_2)^n$. This time it's not so obvious what to choose, but the right choice is still the most natural one: $((a_1,a_1'),\dots,(a_n,a_n'))\in R^{\mathscr{A}_1\times \mathscr{A}_2}$ if and only if $(a_1,\dots,a_n)\in R^{\mathscr{A}_1}$ and $(a_1',\dots,a_n')\in R^{\mathscr{A}_2}$.

The above defines the $L$-structure $\mathscr{A}_1\times \mathscr{A}_2$. Having given you the construction, I'll leave it to you to try to verify the universal property.


问:
Prove $\sin(\pi/2-x)\cot(x+\pi/2) = -\sin(x)$
I've managed to use cofunction identities to get the left side of the equation to (pi/2-x)-tanx. From here, I keep ending up at -cotx? I'm pretty sure that's wrong, and I have no idea how to to go about proving this in another way.</p>

答:
Hint: Show that $$\sin\left(\frac{\pi}{2} -x\right) = \cos(x)$$ and $$\cot\left(x +\frac{\pi}{2}\right) = -\tan(x)$$

Both of the above can be shown using
$$
\sin(x +y) = \sin(x) \cos(y) + \cos(x) \sin(y)
$$
and
$$
\cos(x +y) = \cos(x) \cos(y) - \sin(x) \sin(y)
$$
combined with $\cot(x) = \frac{\cos(x)}{\sin(x)}$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$.


问:
Suppose that typically 3/10 of the customers order “grande” mocha. Assume that customers are independent.
Suppose that typically $\frac3{10}$ of the customers order “grande” mocha. Assume that customers are independent.

a. What is the probability that exactly $2$ of the next $5$ customers will order
a “grande” mocha?

b. What is the probability that the $1$st $3$ customers do not order “grande”
mochas and the $4$th and $5$th customers do order “grande” mochas?

My work:

a) I used the binomial distribution for this: $P(X=2) = {5 \choose 2}.3^2.7^3 = .3087$

b) I need help with b) I know that the  $1$st $3$ customers do not order “grande”
mochas and the $4$th and $5$th customers do order “grande” mochas is one specific outcome. But I'm not sure how what the denominator (all total outcomes) would be in order to calculate probability. Can someone give me a clue?

答: Since the customers are independent, the probability in (b) can be found just by multiplying the probabilities in each slot together. The answer is found by removing the $\binom52$ factor from (a): $.3^2.7^3=0.03087$.


问:
How to read $\frac{d^2y}{dx^2}$
Is it correct to say 'dee squared y over dee x squared'?

I am just trying to make sure because I have to teach this to someone.

I learnt that $f''(x)$ is pronounced 'f double prime of x' but I couldn't find how to pronounce Leibniz notation. Thanks!

答: Formally, you should say : second order derivative of y w.r.t. x

Title: How to read $\frac{d^2y}{dx^2}$


问:
Show that every homeomorphism is a local homeomorphism
This is the definition of local homeomorphism I was given:

" Let $(X,\tau)$ and $(Y,\tau_1)$ be topological spaces. A map $f:X \to Y$ is said to be a local homeomorphism if each point $x \in X$ has an open neighbourhood $U$ such that the restriction of $f$ to $U$ maps $U$ homeomorphically onto an open subspace $V$ of $(Y,\tau_1)$; that is, if the topology induced on $U$ by $\tau$ is $\tau_2$ and the topology induced on $V=f(U)$ by $\tau_1$ is $\tau_3$, then $f$ is a homeomorphism of $ (U,\tau_2)$ onto $(V,\tau_3)$. The topological space $(X,\tau)$ is said to be locally homeomorphic to $(Y,\tau_1)$ if there exists a local homeomorphism of $(X,\tau)$ into $(Y,\tau_1)$."

Now to show that every  homeomorphism is a local homeomorphism, the main idea is that each open neighborhood is an open set and the image is also an open set. Then the induced subspace of both are actually just open sets from the original topologies and we take the same function from the original homeomorphism.

答: What you wrote out is a proof but another (and in my opinion simpler proof) is for each point to take the entire space, $X$, as your open neighborhood. The restriction of $f$ to $X$ is just $f$ which is a homeomorphism by assumption.


问:
The Legendre family of elliptic curves
The Legendre family of elliptic curves over the $\lambda$ line is given by the equation
$$E_{\lambda}:y^2=x(x-1)(x-\lambda),\lambda \in P^1_{\mathbb{C}},$$
which has three singular points, $\lambda=0,1,\infty$. I remember that I read from a reference, which I could not find now, that the Legendre family is the family of elliptic curve over $\Gamma(2) \backslash \mathbb{H}$ (or $\Gamma_0(2) \backslash \mathbb{H}$, which I cannot remember clearly), where $\mathbb{H}$ is the upper half plane of $\mathbb{C}$. Could anyone explain why? How to construct an explicit "isomorphism" between $P^1_{\mathbb{C}}$ and $\Gamma(2) \backslash \mathbb{H}$ (or $\Gamma_0(2) \backslash \mathbb{H}$)?

答:
The Legendre family is isomorphic to $X(2) = H/\Gamma(2)$.  The latter space is easily seem to be the same as the space of complex tori E with the choice of an isomorphism between $Z/2 \times Z/2$ and the 2-torsion $E[2]$.  It is a covering space of $M_1 = H/SL_2(Z)$ with deck group $SL_2(Z/2)$.  In the Legendre family, the origin of $E_\lambda$ is at infinity, and the points of order exactly 2 are the points with $x=0,1$ and $\lambda$ and $y=0$.  (Note that the tangent line to $E$ at a point $P$ passes through $E$ at the point $Q = -2P$ in the group law on $E$; and the points of order 2 just listed are where the cubic curve has a vertical tangent, passing through the flex at infinity.)

In concrete terms, if $E$ in $X(2)$ has marked 2-torsion $0,a,b,c$, then there is a unique isomorphism between $E/(z \rightarrow -z)$ and the projective line that sends $0$ to $\infty$ and $(a, b, c)$ to $(0,1,\lambda)$ for some $\lambda$.  In other words $\lambda$ is the cross-ratio of the branch points of the Wierstrass $p$--function, ordered by the marking.  This allows one to write down an explicit formula for $\lambda$ as a function of $\tau$ in $ H$.  See Ahlfors, Complex Analysis, Chapter 7:  Elliptic functions, section 3.4: The modular function $\lambda(\tau)$.

More geometrically, a fundamental domain for $\Gamma(2)$ is given by
the ideal hyperbolic quadrilateral with vertices $(0,1,-1,\infty)$.  The side identifications yield as quotient a sphere with 3 points removed.  This sphere is naturally identified with the $\lambda$-line with $0,1,\infty$ removed.

Title: The Legendre family of elliptic curves


问:
What is the inverse of a square matrix with orthogonal columns?
I know that for an orthogonal matrix, i.e. a square matrix with orthonormal columns, the inverse is the transpose. What about the general case when the columns of a square matrix are non-zero and orthogonal, but not necessarily of length $1$? (The matrix is invertible since non-zero orthogonal vectors are linearly independent.) Is there a simple characterization/formula for the inverse, or a simple way to compute it?

What if we additionally require that the columns are orthogonal, and the rows are orthogonal? (I know there exist invertible matrices with orthogonal columns, but not orthogonal rows.)

答: A matrix with orthogonal columns (which is easily seen to be invertible) may be written as $QD$ where $Q$ is orthogonal and $D$ is invertible diagonal. Its inverse is then $D^{-1}Q^T$.


问:
Prove that the variance of the MLE of $f(x\mid\theta)=θx^{θ−1}$ tends to $0$ for large $n$
I am given that $X_1,\ldots,X_n$ are randomly sampled from the following distribution:
$f(x\mid\theta)=\theta x^{\theta−1}$, $0<x<1,$ $0<θ<∞.$

I need to show that the variance of the MLE of $θ$ tends to zero as $n$ tends to infinity.

I know that the MLE of $θ$ is $\hat{\theta} =\frac{-1}{(1/n)\sum_{i=1}^n \log X_i}$.

I am stuck on finding the variance of the MLE, however.

I've been thinking about solving $E(X^2) = \int x^2f(x)\,dx$ and $E(X) = \int xf(x) \, dx$ with the MLE as the argument, and rearranging to solve for $\operatorname{var}(\widehat{\theta})$, but I'm not sure of what $f(x)$ would look like using the MLE. It's a really hairy looking integral.

For instance,  $E(\,\widehat{\theta^2}\,) = \int \left( \frac{-1}{(1/n) \sum_{i=1}^n\log X_i} \right)^2 \theta x^{\theta−1}\,dx$ would require that I uniformly replace $\theta$ with the MLE $\hat{\theta}$ ?
Does this look right? Am I on the right/wrong track?

答:
If $X \sim f(x|\theta)$ it's pretty easy to show that $-\ln(X)\sim \exp(\theta)$ which makes $$\frac{1}{\hat{\theta}}=\frac{1}{n}\sum_{j=1}^n-\ln(X_j)\sim \text{Erlang}(n,n\theta)$$ From the law of the unconscious statistician,  $$\mathbb{E}\Big(\hat{\theta}\Big)=\int_0^{\infty}\frac{1}{t}\cdot \frac{(n\theta)^nt^{n-1}e^{-n\theta t}}{(n-1)!}dt=\frac{n\theta}{n-1}$$

$$\mathbb{E}\Big(\hat{\theta}^2\Big)=\int_0^{\infty}\frac{1}{t^2}\cdot \frac{(n\theta)^nt^{n-1}e^{-n\theta t}}{(n-1)!}dt=\frac{n^2 \theta^2}{(n-1)(n-2)}$$ So finally, $$\mathbb{V}\Big(\hat{\theta}\Big)=\mathbb{E}\Big(\hat{\theta}^2\Big)-\Big(\mathbb{E}\big(\hat{\theta}\big)\Big)^2=\frac{n^2 \theta^2}{(n-1)(n-2)}-\Bigg(\frac{n\theta}{n-1}\Bigg)^2$$ which tends to $0$ as $n \longrightarrow \infty$.


问:
Coefficients of characteristic polynomial of a matrix
For a given $n \times n$-matrix $A$, and $J\subseteq\{1,...,n\}$ let us denote by $A[J]$ its principal minor formed by the columns and rows with indices from $J$.

If the characteristic polynomial of $A$ is $x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, then why $$a_k=(-1)^{n-k}\sum_{|J|=n-k}A[J],$$
that is, why is each coefficient the sum of the appropriately sized principal minors of $A$?

答:
$\newcommand\sgn{\operatorname{sgn}}$
I learned of the following proof from @J_P's answer to what effectively is the same question. It arises from expanding the usual definition $\det A=\sum_{\sigma\in S_n}\sgn\sigma\prod_{1\le k\le n}A_{k,\sigma(k)}$, and deserves to be more well-known than it currently is.

Let $[n]:=\{1,\dots,n\}$, and write $\delta_{i,j}$ for the Kronecker delta, which is equal to $1$ if $i=j$, and is $0$ otherwise. Note that $\prod_{1\le k\le n}(a_k+b_k)=\sum_{C\subseteq[n]}\prod_{i\in C}a_i\prod_{j\in[n]-C}b_j$, since every term in the expansion on the left hand side will choose from each expression $(a_k+b_k)$ either $a_k$ or $b_k$, and so we may sum over all possible ways $C$ of choosing the $a_k$ terms.
We compute
\begin{align*}
\det(tI-A)
&=\sum_{\sigma\in S_n}\sgn\sigma\prod_{1\le k\le n}
(t\delta_{k,\sigma(k)}-A_{k,\sigma(k)})\\
&=\sum_{\sigma\in S_n}\sgn\sigma\sum_{C\subseteq[n]}
\prod_{i\in C}(-A_{i,\sigma(i)})\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}\\
&=\sum_{C\subseteq[n]}(-1)^{|C|}\sum_{\sigma\in S_n}\sgn\sigma
\prod_{i\in C}A_{i,\sigma(i)}\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}.
\end{align*}
For fixed $C\subseteq[n]$ and $\sigma\in S_n$, the last product $\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}$ vanishes unless $\sigma$ fixes the elements of $[n]-C$, in which case the product is just $t^{n-|C|}$. So we need only consider the contributions of the permutations of $C$ in our sum, by thinking of a permutation $\sigma\in S_n$ that fixes $[n]-C$ as a permutation in $S_C$. The sign of this permutation considered as an element of $S_C$ remains the same, as can be seen if we consider the sign as $(-1)^{T(\sigma)}$, where $T(\sigma)$ is the number of transpositions of $\sigma$. We thus have
\begin{align*}
\sum_{C\subseteq[n]}(-1)^{|C|}\sum_{\sigma\in S_n}\sgn\sigma
\prod_{i\in C}A_{i,\sigma(i)}\prod_{j\in[n]-C}t\delta_{j,\sigma(j)}
&=\sum_{C\subseteq[n]}(-1)^{|C|}\sum_{\sigma\in S_C}\sgn\sigma
\prod_{i\in C}A_{i,\sigma(i)}t^{n-|C|}\\
&=\sum_{C\subseteq[n]}(-1)^{|C|}t^{n-|C|}\sum_{\sigma\in S_C}\sgn\sigma
\prod_{i\in C}A_{i,\sigma(i)}.
\end{align*}
The term $\sum_{\sigma\in S_C}\sgn\sigma\prod_{i\in C}A_{i,\sigma(i)}$ is precisely the determinant of the principal submatrix $A_{C\times C}$, which is the $|C|\times|C|$ matrix with rows and columns indexed by $C$, and so
\begin{align*}
\sum_{C\subseteq[n]}(-1)^{|C|}t^{n-|C|}\sum_{\sigma\in S_C}\sgn\sigma
\prod_{i\in C}A_{i,\sigma(i)}
&=\sum_{C\subseteq[n]}(-1)^{|C|}t^{n-|C|}\det(A_{C\times C})\\
&=\sum_{0\le k\le n}\sum_{\substack{C\subseteq[n]\\|C|=k}}(-1)^kt^{n-k}
\det(A_{C\times C})\\
&=\sum_{0\le k\le n}t^{n-k}\left((-1)^k
\sum_{\substack{C\subseteq[n]\\|C|=k}}
\det(A_{C\times C})\right)\\
&=\sum_{0\le k\le n}t^k\left((-1)^{n-k}
\sum_{\substack{C\subseteq[n]\\|C|=n-k}}
\det(A_{C\times C})\right).
\end{align*}


问:
Yitang Zhang: Prime Gaps
Has anybody read Yitang Zhang's paper on prime gaps? Wired reports "$70$ million" at most, but I was wondering if the number was actually more specific.

*EDIT*$^1$:

Are there any experts here who can explain the proof? Is the outline in the annals the preprint or the full accepted paper?

答:
As to the "idea" of the proof, I would suggest looking at the following.

As mentioned by Mark Lewko, according to WolframAlpha, the proof gives a gap size of $63, 374, 610.$

Title: Yitang Zhang: Prime Gaps


答:
Take LHS
First , Sin(π/2 - x) = cosx
Second , cot(π/2+ x) = -tanx because in 2nd quadrant cot is negative so when cot changes to tanx we add extra negative sign.

Now Put -tanx = - sinx / cosx and cancel out cos x from numerator and denominator
hence you got -six in LHS which is equal to RHS.

Title: Prove $\sin(\pi/2-x)\cot(x+\pi/2) = -\sin(x)$

Body: I've managed to use cofunction identities to get the left side of the equation to (pi/2-x)-tanx. From here, I keep ending up at -cotx? I'm pretty sure that's wrong, and I have no idea how to to go about proving this in another way.</p>


答:
I am answering my own question, to build on the other answers and comments, in case anyone needs more details.

In general, if $A$ is a square matrix, and $D$ a diagonal matrix with diagonal entries $d_1, d_2, ..., d_n$, then it is readily seen that the $i$th column of $AD$ equals $d_i$ times the $i$th column of $A$.

In our case, if $A$ has orthogonal columns, let $Q$ be the orthonormal matrix that is created from $A$ by dividing each $i$th column of $A$ by its length $a_i := \lVert A_{\text{col $i$}} \rVert$. Put these $a_i$ as the entries of a diagonal matrix $D$. Then $A = QD$, and so $A^{-1} = D^{-1}Q^{-1} = D^{-1}Q^T$.

Here $D$ is indeed invertible because the norms $a_i$ of the columns of $A$ are nonzero. The inverse of a diagonal matrix is, of course, the diagonal matrix formed by taking the reciprocals.

Title: What is the inverse of a square matrix with orthogonal columns?

Body:
I know that for an orthogonal matrix, i.e. a square matrix with orthonormal columns, the inverse is the transpose. What about the general case when the columns of a square matrix are non-zero and orthogonal, but not necessarily of length $1$? (The matrix is invertible since non-zero orthogonal vectors are linearly independent.) Is there a simple characterization/formula for the inverse, or a simple way to compute it?

What if we additionally require that the columns are orthogonal, and the rows are orthogonal? (I know there exist invertible matrices with orthogonal columns, but not orthogonal rows.)


答:
Here's a hint,

Try using the following identities
\begin{align*}
\cot(x) &= \frac{\cos(x)}{\sin(x)}\\
\sin(-x) &= -\sin(x)\\
\cos(-x) &= \cos(x)\\
\sin(x+\pi) &= -\sin(x)\\
\cos(x+\pi) &= -\cos(x)
\end{align*}

Expand the LHS with the first one, and then try to change some terms (and maybe get them to cancel) by using the phase shift identities


问:
Suppose $M$ and $N$ are smooth manifolds then $T(M \times N)$ is diffeomorphic to $ TM \times TN $
I think I found a map but I cannot prove that is a diffeomorphism.
That is the map :

$F : T(M \times N) \to T(M) \times T(N)$ defined by $F((m,n),X) = ((m,(\pi_{1})_{\star,(m,n)}(X)),(n,(\pi_{2})_{\star,(m,n)}(X)))$

where $\pi_1 : M \times N \to M$ and $ \pi_2 : M \times N \to N$ are the projections and the "little star" represent the push-forward.

My try:

Now the idea using local charts I think is the following :

Let $(U,\phi)$ be a local chart for $M$ , $(V,\psi)$ be a local chart for $N$, and so $(U\times V,\Phi:= \phi\times \psi)$ is a local chart for $M\times N$ by definition, and $(TU, \tilde \phi)$ $(TV,\tilde \psi)$ $(T(U\times V), \tilde\Phi)$ will be the induced local charts in $TM, TN $ and $T(M\times N)$. Now the only problem is to check that this maps are in fact smooth. Now I will do this for $F$.

So in these local charts we have that $(\tilde \phi\times \tilde \psi)\circ F\circ(\tilde \Phi)^{-1}(x_1,...x_m,y_1,...,y_n,\epsilon_1,...\epsilon_m,\eta_1,...,\eta_n)=\\= (\tilde \phi\times \tilde \psi)\circ F(\phi^{-1}(x_1,...x_m),\psi^{-1}(y_1,...,y_n),\epsilon_1,...,\epsilon_m,\eta_1,...,\eta_n)=\\=(\tilde\phi\times \tilde \psi)(\phi^{-1}(x_1,...x_m),\epsilon_1,...,\epsilon_m,\psi^{-1}(y_1,...,y_n),\eta_1,...,\eta_n)=\\
=(x_1,...,x_m,\epsilon_1,...\epsilon_m,y_1,...,y_n,\eta_1,...,\eta_n)$

Now to clarify what is happening when we apply $F$  at the tangent vector that this $v=\epsilon_1 \frac{\partial}{\partial x_1}+...+\epsilon_n \frac{\partial}{\partial x_n}+...+\eta_1 \frac{\partial}{\partial y_1}+...+\eta_n \frac{\partial}{\partial y_n}$, we need to look at the diferentials $d\pi_1$ and $d\pi_2$. I will only look at one of them since the other is analogous. So for example at $d\pi_1(\frac{\partial}{\partial x_i})=\sum_{j=1}^{m}\frac{(\phi\circ \pi_1 \circ (\phi\times \psi)^{-1} )^j}{\partial x_i}\frac{\partial }{\partial x_j}=\frac{\partial}{\partial x_i}$, and $d\pi_1(\frac{\partial}{\partial y_i})=\sum_{j=1}^{m}\frac{(\psi\circ \pi_1 \circ (\phi\times \psi)^{-1} )^j}{\partial y_i}\frac{\partial }{\partial x_j}=0$.

My question : How could I construct $F^{-1}$? could anyone help?

答:
Adding to @Bargabbiati's answer. In proposition 3.14, proved in problem 3-2 of this same book, it is proven that
\begin{equation*}
\alpha (v) = (d {\pi _{1}} _{p} (v), ..., d {\pi _k}_p (v))
\end{equation*}
for all $v \in T_p (M_1 \times \cdots \times M_k)$ is an isomorphism between this tangent space and the direct sum
$$\bigoplus T_{p_i} M_i.$$

That being said, such $F$, as you defined it, is an isomorphism (that is almost clear).


问:
Calculating an angle of incidence of a ray on a curved surface
So I'm dealing with a ray travelling inside a hemispherical shell. The angle of entry into the shell theta2 is known, as is the inner and outer radii r1 and r2 respectively. The angle of incidence with the inner interface theta3 needs to be found. I derived something myself using the law of sines, but it doesn't match something I've found in pre-existing literature. I've triple checked my work and I'm unsure if I've done something wrong or if the expression I've found is incorrect. Help is much appreciated.

答:
You have actually gotten the right equation, just in a different form. In an earlier step, you found:

$$\frac{r_{2}}{r_{1}}\sin(\theta_{2}) =\sin(\pi -\theta_{3})$$

However, $\sin(\pi - x) = \sin(x)$ for all real $x$. (You can derive this fact using either the angle subtraction formula for $\sin$ or just by observing that the graph of $\sin(x)$ is symmetric about $x = \frac{\pi}{2}$). Thus, what you have found is equivalent to:

$$\frac{r_{2}}{r_{1}}\sin(\theta_{2}) = \sin(\theta_{3})$$


问:
Bertrand's box paradox
I have $3$ coins, $1$ coin has $2$ heads (HH), 1 coin has $2$ tails (TT), $1$ coin has $1$ head and $1$ tail (HT). I toss the coin, it fells on my hand, and the side i see is a tail. What's the chance that the other side is also a tail?

I got this as a teaser from a friend, possible from here, as you can see he is insisting on $\frac12$ as not being the correct answer, I got $\frac13$ as my answer, am I right?

答:
A very basic solution will be as follows. Since we know that a tail has appeared, there can be three outcomes.
We can be looking at the
1). Tail side of the HT coin
2). Face 1 of the TT coin
3). Face 2 of the TT coin
So 2 of the 3 outcomes possible are in favour of the coin being a 2 tailed coin.</p>

Title: Bertrand's box paradox


问:
Functions through & from a point
Suppose there are two functions f(x) & g(x) and there is a point A. One of the given functions passes via A and the other function originates at A; I am interested to find out which function passes via A and which originates at A.
Is there a method / way to find this out without knowing the Domain & Range of the functions ?

答: If you are referring to continuous real-valued functions $f$ and $g$, since you specify that one function (say $f(x)$) passes through A, that would imply there exist values for both $f(x+\epsilon)$ and $f(x-\epsilon)$, whereas $g(x)$ would only be defined for either $g(x+\epsilon)$ or $g(x-\epsilon)$ but not both, for some small number $\epsilon > 0$.

Title: Functions through & from a point


问:
A combinatorial formula for different sums $a\cdot( k_1+ k_2+...+k_{n-1})+k_n$
I am looking for a combinatorial rule for the following: let $n \in \mathbb{N}$ and $k_1,k_2,...,k_n \in \mathbb{N}$ (these are known). Since we know what the sum $\sum_{j=1}^{n}k_i$ is, let's say that $\sum_{j=1}^{n}k_i=B$.

Now, let us define a new combinatorial arrangement of $k_i$'s, let us label this as $k_{x_1},k_{x_2},...,k_{x_n}$. Also, let $a \in \mathbb{N}$ be known. Now, we want to determine how many different values may the following sum take:

$a\cdot k_{x_1}+a\cdot k_{x_2}+...+k_{x_n}=a\cdot(k_{x_1}+...+k_{x_{n-1}})+k_{x_n}$?

Here is an example: let $k_1=2$, $k_2=3$ and $k_3=1$ (i.e. $n=3$ and $k_1+k_2+k_3=6$). Now we would have the following arrangements (let $A_i$ stand for arrangement):

$\{k_{x_1},k_{x_2},k_{x_3}\}:$ $A_1=\{2,3,1\}$, $A_2=\{3,2,1\}$, $A_3=\{2,1,3\}$, $A_4=\{3,1,2\}$, $A_5=\{1,2,3\}$ and $A_6=\{1,3,2\}$.

Now, if we want to determine different described sums for these arrangements, we may see the following:

For $A_1$ we have $a\cdot(2+3)+1=5a+1$.

For $A_2$ we have $a\cdot (3+2)+1=5a+1$.

For $A_3$ we have $a\cdot (2+1)+3=3a+3$.

For $A_4$ we have $a\cdot (3+1)+2=4a+2$.

For $A_5$ we have $a\cdot (1+2)+3=3a+3$.

For $A_6$ we have $a\cdot (1+3)+2=4a+2$.

Thus, we have the total of $3$ different formulas for $a\cdot (k_{x_1}+...+k_{x_n})+k_{x_n}$. However, we do not know the exact value of $a\in \mathbb{N}$. For example, if we have $a=1$ then we would have to have $3a+3=4a+2=5a+1$, thus reducing the number of different sums to exactly $1$. However, if $a=2$ then $5a+1=11$, $3a+3=9$ and $4a+2=10$ and we would have $3$ different sums. Would there be a combinatorial formula that takes into consideration all different possibilities and would determine the exact number of different possible sums given the original $k_i$'s (in the example $k_1=2$, $k_2=3$ and $k_3=1$) and an arbitrary choice of $a \in \mathbb{N}$?

答:
You have the sum of all of the $k_i$, or any combinatorial arrangement of them, equal to a fixed and known sum, e.g., as you stated,

$$\sum_{j=1}^{n}k_i = \sum_{j=1}^{n}k_{x_i} = B \tag{1}\label{eq1A}$$

Next, for some $a \in \mathbb{N}$, your requested sum is

$$\begin{equation}\begin{aligned}
a\sum_{j=1}^{n-1}k_{x_i} + k_{x_n} & = a\sum_{j=1}^{n-1}k_{x_i} + ak_{x_n} - ak_{x_n} + k_{x_n} \\
& = a\sum_{j=1}^{n}k_{x_i} - ak_{x_n} + k_{x_n} \\
& = aB + (1 - a)k_{x_n}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

As you noticed, if $a = 1$, then the sum is always $aB$, so there's always only one possible value. Otherwise for $a \neq 1$, the number of possible values of \eqref{eq2A} is the number of distinct values of $k_i$, e.g., it will be $n$ if they are all distinct.


答: If the columns of $A$ are orthogonal then $$A^TA=D^2$$ where the diagonal elements of $D$ are the lengths of the columns. Thus $$A^{-1}=D^{-2}A^T=(AD^{-2})^T$$ The recipe is therefore to divide the column vectors of $A$ by their lengths squared and then take the transpose.


问:
extreme point of convex function
For twice differentiable convex function $f(x)$ on $\mathbb{R}^n$, with $x=\langle x_1,x_2,...,x_n \rangle$,

prove that, $\forall x \in \mathbb{R}^n$ and $\forall i \in \{1,2,...,n\}$, if $\frac{\partial f(x) }{\partial{} ~x_i} < 0$, then

$$x^*_i \ge
 x_i$$

where $x^*=\langle x^*_1,x^*_2,...,x^*_n \rangle$ is the extreme point (global minimum) of convex function $f$.

答:
A counterexample for two dimension convex function, picture shows the contour lines, where all rings are perfect circle!

A concrete example would be $f(x,y)=x^{2}+\left(y+x\right)^{2}$

Title: extreme point of convex function


问:
Integral using the Riemann Sum
Here I have a question to to calculate $$\int_{0}^{2}e^{x}dx$$ using Riemann sums. I got to the part where $$\int_{0}^{2}e^{x}dx=\lim....(e^{\frac{2i}{n}})\frac{2}{n}=\lim....(e^{\frac{2(\frac{r-r^{n+1}}{1-r})}{n}})\frac{2}{n}$$
What should I do now? Can I expand this equation further? It doesnt seem like it. Thank you for helping.

答:
Yea, so we are partitioning $[0,2]$ into $n$ equal pieces each of length $\frac{2}{n}$. In particular:

$$x_i = 0+\frac{2i}{n} = \frac{2i}{n}$$

Then, we pick our evaluation set $T = \{x_i: i \in \{1,2,\ldots,n\}\}$. So, consider the Riemann Sum:

$$R(f,P,T) = \sum_{i=1}^{n} f(x_i) \Delta x_i = \sum_{i=1}^{n} e^{\frac{2i}{n}} \frac{2}{n}$$

Observe that $\sum_{i=1}^{n} \left(e^{\frac{2}{n}} \right)^i$ is a geometric sum with common ratio $e^{\frac{2}{n}}$. So, we have:

$$\sum_{i=1}^{n} \left(e^{\frac{2}{n}} \right)^i = \frac{1-\left(e^{\frac{2}{n}}\right)^{n+1}}{1-e^{\frac{2}{n}}}$$

Observe that the numerator of this expression simplifies to $1-e^{2+\frac{2}{n}}$. It is easy to see that this expression goes to $1-e^2$ as $n \to \infty$. Then, study the expression:

$$\frac{\frac{2}{n}}{1-e^{\frac{2}{n}}}$$

It is relatively easy to see that as $n \to \infty$, this expression goes to $-1$ (try proving this). So, it is the case that:

$$\lim_{n \to \infty} R(f,P,T) = e^2-1$$

and we are done. Now, alternatively, if you don't want to go that route with it, then there's a faster way to do the problem without having to do this sort of gymnastics.

Since $x \mapsto e^x$ is continuous on $[0,2]$, it is Riemann Integrable there. Let $P = \{0 = x_0 < x_1 < \ldots < x_{n-1} < x_n = 2\}$ be any partition of $[0,2]$. Since $x \mapsto e^x$ is continuous and differentiable on every interval of $\mathbb{R}$:

$$\forall i: \exists t_i \in (x_{i-1},x_i): e^{t_i} = \frac{e^{x_i}-e^{x_{i-1}}}{\Delta x_i}$$

This is just a consequence of the MVT. Then, pick $T = \{t_i: i \in \{1,2,\ldots,n\}\}$ as your evaluation set and compute the Riemann Sum:

$$R(f,P,T) = \sum_{i=1}^{n} f(t_i) \Delta x_i = \sum_{i=1}^{n} \left(e^{x_i}-e^{x_{i-1}} \right) = e^2-e^0 = e^2-1$$

where we have made use of the fact that we have a telescoping sum. This approach is nice because it actually generalizes and gives us a pretty neat proof of the Fundamental Theorem of Calculus :D

Title: Integral using the Riemann Sum


Title: Integral using the Riemann Sum

Body:
Here I have a question to to calculate $$\int_{0}^{2}e^{x}dx$$ using Riemann sums. I got to the part where $$\int_{0}^{2}e^{x}dx=\lim....(e^{\frac{2i}{n}})\frac{2}{n}=\lim....(e^{\frac{2(\frac{r-r^{n+1}}{1-r})}{n}})\frac{2}{n}$$
What should I do now? Can I expand this equation further? It doesnt seem like it. Thank you for helping.

Body:
\begin{align}
\int_0^2e^xdx&=\lim_{n\to\infty}\sum_{i=0}^{n-1}\frac2{n}e^{2i/n}\\
&=\lim_{n\to\infty}\frac2{n}\left(\frac{e^{2}-1}{e^{2/n}-1}\right)\\
&=e^2-1.
\end{align}