问:
Embedding of Tree
Q.

Proof for every Tree can be embedded into the plane.

Conditions.

We cannot use Euler Formula for Planar Graphs. We can use definition of tree, $V-E=1$, no-cycles, every edge is critical, there is a vertex of frequency at most one.

Attempt.
We prove this by induction. True for $V$=1. Trivial. Consider true for $V<n$. Consider a tree $T$ with $n$ edges. Now, we know it has a vertex with at most one edge incident in it. Consider $T$\ $\{v,l\}$ the last vertex and edge. This can be embedded by inductive hypothesis. For the $v$ and $l$, that is left, it can be easily embedded.

The last part seems sloppy. I would appreciate other proofs or improvement on this proof.

答: You're right that there's something very subtle missing.  Suppose we have a tree with uncountably many straight edges, from $(0,0)$ to every point on the circle $x^2+y^2=1$.  Now you can't add another edge to $(0,0)$.  However if your tree is finite your proof is fine.</p>

Title: Embedding of Tree


问:
First and Second Order Derivate of the Function having summation sequence
I have this function having a summation sequence in it
$$ f(x) =\frac {1} { \sum _{k=1}^n \frac{1}{x_k}} $$

How do I determine the first and second derivatives of the function with respect to x?

ContentLicense: CC BY-SA 4.0

Body:
I suppose you mean $f(\boldsymbol x) = \dfrac {1} { \sum _{k=1}^n \frac{1}{x_k}}$, where $\boldsymbol x=(x_1,x_2,\cdots,x_k)$.

Hint. Try small examples:

When $k=2$:
$$
\frac{\partial f(x_1,x_2)}{\partial x_1} = \frac{1}{x_1^2 \left(\frac{1}{x_1}+\frac{1}{x_2}\right)^2}\\
\frac{\partial ^2f(x_1,x_2)}{\partial x_1^2}=\frac{2}{x_1^4 \left(\frac{1}{x_1}+\frac{1}{x_2}\right)^3}-\frac{2}{x_1^3 \left(\frac{1}{x_1}+\frac{1}{x_2}\right)^2}
$$
When $k=3$:
$$
\frac{\partial f(x_1,x_2,x_3)}{\partial x_1}=\frac{1}{x_1^2 \left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right)^2}\\
\frac{\partial ^2f(x_1,x_2,x_3)}{\partial x_1^2}=\frac{2}{x_1^4 \left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right)^3}-\frac{2}{x_1^3 \left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right)^2}
$$

What can you find?


问:
What do we mean when we say "Let $x$ be an element of the set $\mathbb{R}$"?

What do we mean when we say "Let $x$ be an element of the set $\mathbb{R}$"?

Does $x$ represents only a single  element of set $\mathbb{R}$? or does $x$ represent all the element of set $\mathbb{R}$ simultaneously at the same time?

People say that if $x\in\mathbb{R}$ then $x$ is any real number; that means $x$ represents all real numbers. But if $x$ is any real number then let's say $x=1$; so $x$ is one, then how it can represent all real numbers?

Please help me I am very confused.

答:
The set $\mathbb{R}$ is the familiar real number line, including all "decimal expansions." When we say a number is an element of $\mathbb{R}$, we mean that it's a part of the number line. $1 \in \mathbb{R}$, $\sqrt{2} \in \mathbb{R}$, negative fractions that look weird such as $\frac{-1}{\pi}$ are in the set $\mathbb{R}$, just cause they have a decimal expansion.

When we say $x \in \mathbb{R}$, we just mean that there's a number, just like the ones I mentioned above, that is a real number. We just don't know what it is yet? What is $x$? We don't know, but we can give it a name because for whatever reason it's of interest to us. We just know it lies somewhere on the number line. We don't know what $x$ is, but noting that it's in $\mathbb{R}$ for one reason or another has importance. But "variable" in the title of your question is just giving a letter that we don't know its value.


问:
Minimum and Maximum as Random Variables
If we have random variables $A,B,C$, how do we find the density and distribution functions for $\min{\left(A,B,C\right)}$, and $\max{\left(A,B,C\right)}$? I'm asking for the general case, (discrete/absolute continuous, independent/non-independent).

Perhaps, you can illustrate on something simple like $X,Y \sim UNI[0,1]$, for example?

答:
Illustration of one commonly used approach:

Minimum. Let $X_1 \sim \mathsf{Exp}(\lambda_1)$ and independently,
$X_2 \sim\mathsf{Exp}(\lambda_2).$ Then $V = \min(X_1,X_2)$ has reliability function
$$1 - F_V(v) = R_v(V > v) = P(X_1 > v, X_2 > v)\\ =
  P(X_1 > v)P(X_2 > v) = e^{-\lambda_1 v}e^{-\lambda_2 v}
= e^{-(\lambda_1+\lambda_2)v},$$
for $v > 0.$

This is the reliability function of $\mathsf{Exp}(\lambda_1+\lambda_2).$
The method extends to more than two random variables. The product is not always so easily summarized as in this example, but the approach is widely applicable.

Maximum. For $W=\max(X_1 + X_2).$ begin with
the CDF $$F_W(w) = P(W\le w) = P(X_1 < w,X_2 \le w).$$
In the case of two exponential random variables, the maximum is not another exponential random variable. But this method also extends to more than two random variables, and is widely applicable.

Simulation. Sometimes, especially for more difficult cases than exponential, it is helpful to have approximate descriptions of the distribution of minimum or maximum, which can often
be obtained easily by simulation. For example, for the minimum of independent $X_1 \sim \mathsf{Exp}(3)$
and $X_2 \sim \mathsf{Exp}(4),$ a simulation in R might
go as follows:

```<code>set.seed(2020)
x1 = rexp(10^6, 3);  x2 = rexp(10^6, 4)
v = pmin(x1, x2)
summary(v);  sd(v)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
0.0000003 0.0412125 0.0990582 0.1429672 0.1981739 1.9674862
[1] 0.1429503  # aprx SD(V) = 1/7 =  0.1428571

hist(v, prob=T, br=50, col="skyblue2", main="EXP(7)")
 curve(dexp(x, 7), add=T, col="orange", lwd=2)
</code>```

Title: Minimum and Maximum as Random Variables

Body:
Illustration of one commonly used approach:

Minimum. Let $X_1 \sim \mathsf{Exp}(\lambda_1)$ and independently,
$X_2 \sim\mathsf{Exp}(\lambda_2).$ Then $V = \min(X_1,X_2)$ has reliability function
$$1 - F_V(v) = R_v(V > v) = P(X_1 > v, X_2 > v)\\ =
  P(X_1 > v)P(X_2 > v) = e^{-\lambda_1 v}e^{-\lambda_2 v}
= e^{-(\lambda_1+\lambda_2)v},$$
for $v > 0.$

This is the reliability function of $\mathsf{Exp}(\lambda_1+\lambda_2).$
The method extends to more than two random variables. The product is not always so easily summarized as in this example, but the approach is widely applicable.

Maximum. For $W=\max(X_1 + X_2).$ begin with
the CDF $$F_W(w) = P(W\le w) = P(X_1 < w,X_2 \le w).$$
In the case of two exponential random variables, the maximum is not another exponential random variable. But this method also extends to more than two random variables, and is widely applicable.

Simulation. Sometimes, especially for more difficult cases than exponential, it is helpful to have approximate descriptions of the distribution of minimum or maximum, which can often
be obtained easily by simulation. For example, for the minimum of independent $X_1 \sim \mathsf{Exp}(3)$
and $X_2 \sim \mathsf{Exp}(4),$ a simulation in R might
go as follows:

set.seed(2020)
x1 = rexp(10^6, 3);  x2 = rexp(10^6, 4)
v = pmin(x1, x2)
summary(v);  sd(v)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
0.0000003 0.0412125 0.0990582 0.1429672 0.1981739 1.9674862
[1] 0.1429503  # aprx SD(V) = 1/7 =  0.1428571

hist(v, prob=T, br=50, col="skyblue2", main="EXP(7)")
 curve(dexp(x, 7), add=T, col="orange", lwd=2)


问:
Find maximum size of set family, where at least one member has at least one of any three points, but do not contains all those three points
Let $n$ and $s$ be non-negative integers such that  $n\geq 3$ and $2s\log2>3\log n$. Prove that there exists sets $A_1,A_2,\ldots,A_s \subseteq [n]$ such that

for every $B\in \binom {[n]} 3$ there exists $1 \leq j \leq s$ with both $A_j \cap B \neq \emptyset$ and $A_j \cap B \neq B$.

Should I start by counting the sets that do not either not intersect $B$ or contains $B$?

答:
Roughly speaking: If we increase $s$ by $3$, we allow $n$ to grow by a factor of four.
This suggests the following recursive procedure:

If $n=4k-r$ with $k\ge 1$ and $0\le r\le 3$, we may assume that $s-3$ sets suffice to solve that problem for three-element subsets of $[k]$ (and also for $[k-1]$).
Thus let $A'_1,\ldots,A'_{s-3}\subseteq[k]$ and $A''_1,\ldots,A''_{s-3}\subseteq[k-1]$ be such sets. Let
$$A_i=\begin{cases}4A'_i\cup(4A'_i+1)\cup(4A'_i+2)\cup(4A'_i+3)&\text{if }r=0,\\
4A'_i\cup(4A'_i+1)\cup(4A'_i+2)\cup(4A''_i+3)&\text{if }r=1,\\
4A'_i\cup(4A'_i+1)\cup(4A''_i+2)\cup(4A''_i+3)&\text{if }r=2,\\
4A'_i\cup(4A''_i+1)\cup(4A''_i+2)\cup(4A''_i+3)&\text{if }r=3.\\
\end{cases}$$
Then the sets $A_1,\ldots, A_{s-3}$ solve the problem for all sets $B$ consisting only of numbers pairwise congruent modulo $4$.
Add $A_{s-2}=4\mathbb Z\cap [n]$, $A_{s-1}=(4\mathbb Z+1)\cap [n]$, $A_{s}=(4\mathbb Z+2)\cap [n]$.
Any three-element set $B$ that has elements from at least two resudue classes mod $4$, has one element in one of these three sets and another not in it.

To complete the proof, we need to verify the claim for some mall $n$. For $n\le 2$, the claim is vacuously true.
From $n=3$ on, the condition implies $s\ge 3$, and hence the reduction works fine.

ContentLicense: CC BY-SA 3.0


问:
Show that $4x^2-yz$ is a perfect square
Here is my problem.

$A=xy+yz+zx$, where $x,y,z\in\mathbb{Z}$. It is known that if we add $1$ to $x$, and subtract $2$ from both $y$ and $z$, the value $A$ won't change. Prove that $-A$ is a square of whole number.

My attempt:

\begin{align}
A=xy+yz+zx
\end{align}

and

\begin{align}
A=(x+1)(y-2)+(y-2)(z-2)+(z-2)(x+1)
\end{align}

comparing them we get

\begin{align}
4x+y+z=0.
\end{align}

Now we insert last equation to first equation:

\begin{align}
-A&=-xy-yz-zx\\
&=-x(y+z)-yz\\
&=4x^2-yz
\end{align}

From here I don't know how to show that $4x^2-yz$ is a perfect square.

答:
If we expand, we get
$$A=(x+1)(y-2)+(y-2)(z-2)+(z-2)(x+1)\quad =\quad x y + x z - 4 x + y z - y - z$$
Subtracting the original from this equation should be zero
$$(x y + x z - 4 x + y z - y - z)-(xy+yz+xz)=0\\
\implies 4x+y+z=0\implies z = -4 x - y$$
Substituting $z$ now allows the equation to be viewed more simply
$$xy+y(-4 x - y)+x(-4 x - y)= -4 x^2 - 4 x y - y^2=A\\
\implies A=-(4x+4xy+y^2)$$
$$\therefore -A=(2x+y)^2$$

\begin{align*}
\end{align*}

Title: Show that $4x^2-yz$ is a perfect square


问:
Find the area under curve using the given data

Consider the curve $y=f(x)$ which satisfies the DE $(1+x^2)\frac{dy}{dx} +2xy=4x^2$ and passes through the origin. Find area enclosed by $f^{-1}x$, x axis, and $x=2/3$

After some calculation which I don’t think is necessary to show here, I got
$$y=\frac{4x^3}{3(1+x^2)}$$

Now the inverse for this function can’t be found directly, and I don’t know how else to do it. Can I get a hint?

Body: we have $$I=\int_{0}^{2/3} f^{-1}(x)dx$$ take $x=f(t),dx=f'(t)dt$ then $$I=\int_{0}^1 tf'(t)dt=1\cdot f(1)-0\cdot f(0)-\int_0^1 f(t) \, dt$$  can you finsish?


答:
Notice that $f(1) = \frac{2}{3}$ and $f(0) = 0$

Thus the integrals of the function and its inverse will add up to the rectangle of area $\frac{2}{3}\cdot 1$ as we can see on this graph

$$\int_0^{\frac{2}{3}}f^{-1}(x)dx+\int_0^1f(x)dx = \frac{2}{3}$$

The antiderivative of $f$ is given by

$$\frac{2}{3}\int 2x\left(1-\frac{1}{1+x^2}\right)dx = \frac{2}{3}x^2-\frac{2}{3}\ln(1+x^2)+C$$

which means the final answer is

$$\frac{2}{3}\ln 2$$

Title: Find the area under curve using the given data

Body:
Consider the curve $y=f(x)$ which satisfies the DE $(1+x^2)\frac{dy}{dx} +2xy=4x^2$ and passes through the origin. Find area enclosed by $f^{-1}x$, x axis, and $x=2/3$

After some calculation which I don’t think is necessary to show here, I got
$$y=\frac{4x^3}{3(1+x^2)}$$

Now the inverse for this function can’t be found directly, and I don’t know how else to do it. Can I get a hint?


问:
Find the equation of the equilateral hyperbola given one of its foci points $(1;1)$ and an asymptote $x+y=0$
If the asymptote is $y=-x$, then doesn't it mean that the other one is $y=x$?

But does't that also mean that the foci points are $(c_1,0)$, $(c_2,0)$?

But then how is it $(1,1)$?

答: Hope the following shifted rectangular hyperbola addresses all questions:


问:
bound of standard normal random variable
Let $X$ be a standard normal random variable, $h>1$ is a constant. How can I get an estimation of $M$, if $$\mathbb{P}[-M\leq X\leq M] =1-\frac{1}{h^2}.$$ For example, an upper bound and a lower bound of $M$ in terms of $h$, if a closed form solution for $M$ is not possible.

答:
$$M = \sqrt{2} \,\mathrm{erf}^{-1}\left(1 - \frac{1}{h^2}\right)  \text{,}  $$
where $\mathrm{erf}^{-1}$ is the inverse error function.  Depending on your definition of "closed form", we are done.  (For instance, see the Python implementation scipy.special.erfinv().)

A series expansion centered at $\infty$ simplifies to
$$  M \approx \sqrt{-\log \left(\pi  \log \left(\frac{2 h^4}{\pi }\right)\right) + 4 \log(h) + \log(2)}  \text{.}  $$
It is an upper bound for $h$ up to about $3.8$.
It is egregiously terrible for $h \in [1,3/2]$, but the error is less than $0.066$ for $h \geq 2$ and less than $0.011$ for $h \geq 3$.

Using the Taylor series for the inverse error function centered at $0$ (and the argument $1-1/h^2$), we can get a lower bound
$$ M \geq \sqrt{\frac{\pi }{2}} \left(
\left(1-\frac{1}{h^2}\right) +
    \frac{\pi}{12}
   \left(1-\frac{1}{h^2}\right)^3 +
    \frac{7\pi^2}{480}
   \left(1-\frac{1}{h^2}\right)^5 +
    \frac{127 \pi ^3}{40320}
   \left(1-\frac{1}{h^2}\right)^7
\right)  \text{,}  $$
which can be improved by using more terms of the series.  The error is less than $0.13$ on $[1,2]$.

So a set of bounds is...  Let
\begin{align*}
a_1 &= \sqrt{-\log \left(\pi  \log \left(\frac{2 h^4}{\pi }\right)\right) + 4 \log(h) + \log(2)}  \\
a_2 &= \sqrt{\frac{\pi }{2}} \left(
\left(1-\frac{1}{h^2}\right) +
    \frac{\pi}{12}
   \left(1-\frac{1}{h^2}\right)^3 +
    \frac{7\pi^2}{480}
   \left(1-\frac{1}{h^2}\right)^5 +
    \frac{127 \pi ^3}{40320}
   \left(1-\frac{1}{h^2}\right)^7
\right)  \text{.}
\end{align*}
Then
$$  \min\{ a_1 - 0.066, a_2\} \leq M \leq a_1 + 0.066  \text{.}  $$

If you want tighter bounds than this, I recommend using bisection to polish (since you know the forward function, erf, and you know that erf is monotonic).


问:
How to find coordinates of tangent point on circle, given center coordinates, radius, and end point of tangent line
I'm working on a hobby programming project and my mathematic thinking is feeling a bit slow at the moment. Basically, in situations similar to the image I made, given the information I mentioned in the title, I need to be able to calculate the coordinates of the tangent point. (Dimensions are from the origin.) Could you all help me out with finding a straightforward way of doing so? Thanks in advance.

(P.S. In the program, this will actually be in 3D, but once I have the 2D solution I should be able to figure it out in 3D.)

答:
The point at which the tangent line touches the circle (call it $A$), the endpoint of the tangent line (call it $B$), and the center of the circle (call it $C$) form a right triangle. We know the radius of the circle (length of $AC$) and the distance from the center of the circle to the endpoint of the tangent line (length of $BC$).

All that remains is to use the Pythagorean theorem and a bit of trigonometry to find the length of the remaining side, as well as the angle between the remaining side ($AB$) and side $BC$. That should be enough information to find the coordinates of point $A$.


问:
Give an example of a vector field that satisfies the given line integral
Given the path C which is a circle with equation $(x-9)^2 +y^2 =36$ in the anticlockwise direction.
Give an example of a vector field $F(x,y)$ that satisfies $\oint_C \vec F \cdot dr = 11\pi$

I used the parametrization $(x,y)=(9+6cos(t),6sin(t))$ and I think I need to use green's theorem to find a function (for example $F(x, y) = x^2\mathbf{i} + xy\mathbf{j} $ ) that satisfies $\oint_C F(r(t))\cdot r'(t)\,dt =11\pi$

Is my parametrization correct since the circle is centered off-origin?

Is this only solvable by trial and error of multiple functions or is there something I'm missing?

答:
Your parametrization is correct.

If the vector field is $(P, Q)$ then as per Green's theorem,

$\oint_C P \, dx + Q \, dy = \iint_D (Q_x - P_y) \,dA$

Given area of the circle is $36 \pi$, some easy examples of the vector field resulting in line integral of $11\pi$ will be where $Q_x - P_y = \frac{11}{36}$.

We can pick $\vec{F} = (c, \frac{11x}{36}) \,$ where $c$ is any constant or you can pick $(y, \frac{47x}{36})$.

For the first vector field, the line integral will turn out to -

$\int_0^{2\pi} (-6 c \sin t + 11 \cos^2t + \frac{33}{2} \cos t) \, dt = 11 \pi$

For the second vector field, the line integral will turn out to -

$\int_0^{2\pi} (47 \cos^2t - 36 \sin^2t + \frac{141}{2} \cos t) dt = 11 \pi$


问:
Calculate $X_A(x) $ and $m_A(x) $ of a matrix $A\in \mathbb{C}^{n\times n}:a_{ij}=i\cdot j$

Caclulate the characteristic & the minimal polynomial of the matrix:

$$A\in\mathbb C^{n\times n}:a_{ij}=i\cdot j ,\forall i,j=1,..,n$$

$$\text{i.e for $n=3$,  } $$
$$A=\left[\begin{matrix}1 & 2 & 3\\2 & 4 & 6\\3 & 6 & 9\end{matrix}\right]$$

$$\text{I've calculate $X_A(x),m_A(x)$ for the following cases:}$$

$$ \left[\begin{matrix}1 & 2\\2 & 4\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3\\2 & 4 & 6\\3 & 6 & 9\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3 & 4\\2 & 4 & 6 & 8\\3 & 6 & 9 & 12\\4 & 8 & 12 & 16\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3 & 4 & 5\\2 & 4 & 6 & 8 & 10\\3 & 6 & 9 & 12 & 15\\4 & 8 & 12 & 16 & 20\\5 & 10 & 15 & 20 & 25\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3 & 4 & 5 & 6\\2 & 4 & 6 & 8 & 10 & 12\\3 & 6 & 9 & 12 & 15 & 18\\4 & 8 & 12 & 16 & 20 & 24\\5 & 10 & 15 & 20 & 25 & 30\\6 & 12 & 18 & 24 & 30 & 36\end{matrix}\right]$$

$$\xrightarrow{X_A(x)} \left [ x \left(x - 5\right), \quad - x^{2} \left(x - 14\right), \quad x^{3} \left(x - 30\right), \quad - x^{4} \left(x - 55\right), \quad x^{5} \left(x - 91\right)\right ]$$

$$\xrightarrow{m_A(x)} \left [ x \left(x - 5\right), \quad x \left(x - 14\right), \quad x \left(x - 30\right), \quad x \left(x - 55\right), \quad x \left(x - 91\right)\right ]$$

$$\text{Hence, i assumed that if }A\in\mathbb{C}^{n\times n}:$$
$$\fbox{$X_A(x)=(-1)^nx^{n-1}\big( x-\operatorname{tr}(A)\big)$},$$
$$\fbox{$m_A(x)=x\big( x-\operatorname{tr}(A)\big)$} .$$

How can i prove this in general case?

any suggestions?

答:
Let $\mathbf{v}:=(1,2,\ldots,n)$ and $\mathbf{w}_i:=(i,0,\ldots,0,-1,0,\ldots)$ for $i=2,\ldots,n$. These vectors are non-zero orthogonal and hence form a basis of $\mathbb{R}^n$. They are also eigenvectors of $A$ since $$A\mathbf{v}=|\mathbf{v}|^2\mathbf{v},\qquad A\mathbf{w}_i=\mathbf{0}$$ Note that $|\mathbf{v}|^2=\mathrm{trace}(A)$.

Hence $A$ has eigenvalues $|\mathbf{v}|^2$ and $0$ repeated $(n-1)$ times. The characteristic polynomial is thus $$X_A(x)=(-1)^n(x-\lambda_1)\cdots(x-\lambda_n)=(-1)^nx^{n-1}(x-|\mathbf{v}|^2)$$
The minimal polynomial can be found by multiplying $A$ with itself: $$A^2=\begin{bmatrix}\mathbf{v}\\ 2\mathbf{v}\\ \vdots\\n\mathbf{v}\end{bmatrix}\begin{bmatrix}\mathbf{v}&2\mathbf{v}&\cdots&n\mathbf{v}\end{bmatrix}=|\mathbf{v}|^2A$$ $$\therefore\ m_A(x)=x^2-|\mathbf{v}|^2x=x(x-|\mathbf{v}|^2)$$ $m_A$ is the minimal polynomial since it is the least degree polynomial containing both eigenvalues.

These results for $X_A$ and $m_A$ generalize to any matrix of the type $A={\mathbf{v}\mathbf{v}}^\top$.


答:
You do not actually need trigonometry. Let $AC = r$ be the radius, let point $A$ be at $(x_1, y_1)$, and point $B$ be at $(x_2, y_2)$. You know the values of $r, x_2$, and $y_2$.

First find the length of $AB$ using Pythagoras. Then using the distance formula (Pythagoras again) and squaring both sides, $(y_2 - y_1)^2 + (x_2 - x_1)^2 = AB^2$.

In addition, $(x_1, y_1)$ must lie on the circle, so $x_1^2 + y_1^2 = r^2$. Solving these two equations will give you $x_1, y_1$ in terms of the other variables. There will be two solutions, but you can plot each of them and determine which one is correct for the general setup.

Title: How to find coordinates of tangent point on circle, given center coordinates, radius, and end point of tangent line

Body:
I'm working on a hobby programming project and my mathematic thinking is feeling a bit slow at the moment. Basically, in situations similar to the image I made, given the information I mentioned in the title, I need to be able to calculate the coordinates of the tangent point. (Dimensions are from the origin.) Could you all help me out with finding a straightforward way of doing so? Thanks in advance.

(P.S. In the program, this will actually be in 3D, but once I have the 2D solution I should be able to figure it out in 3D.)


答:
Looking at your data, you generated random numbers from a geometric distribution counting the failures before first success

$$P(X=x)=0.8^x\cdot 0.2$$

$x=0;1;2...$

If you do the average of the generated numbers you have to find 4 and a variance of 20. Thus it is difficult that your sample comes from a $Geo(0.2)$

No, there is no order, correlation or dependence in those numbers (I hope, if the random generating algorithm works well)

You can use particular statistical methods to generate this sample by your own

Title: What's the meaning of these numbers in geometric distribution?

Body:
For a geometric distribution, I know there's a certain pattern, for example, given the probability of success $0.2$ for a event, you can have:

$$P(X = r) = {q}^{(r-1)} \times p$$

When r = 1, probability is 0.2

When r = 2, probability is 0.32

When r = 3, probability is 0.128

So on..

This tells me the probability of trial 1, 2, 3 .. and I can see it decreases for every trial.

I tried to do this same example using code to generate data for geometric distribution using python programming language:

```<code>s = np.random.geometric(p, 100000)
</code>```

It means for 'p' = 0.20, give me 100,000 samples.

Here are 500 samples from these and what they look like:

[1, 6, 1, 2, 3, 2, 3, 1, 5, 3, 9, 12, 6, 2, 7, 3, 7, 7, 4, 2, 4, 4, 1, 2, 9, 3, 9, 1, 5, 9, 1, 10, 7, 8, 2, 1, 6, 1, 8, 1, 5, 2, 2, 7, 4, 3, 7, 3, 6, 4, 5, 4, 1, 9, 1, 7, 6, 2, 7, 13, 24, 1, 3, 2, 2, 8, 2, 1, 2, 8, 3, 2, 1, 2, 3, 12, 7, 5, 14, 6, 2, 3, 1, 1, 4, 3, 1, 4, 1, 12, 2, 7, 18, 8, 6, 5, 5, 6, 2, 12, 4, 1, 1, 2, 5, 3, 2, 2, 2, 4, 3, 12, 6, 7, 5, 9, 3, 12, 3, 5, 9, 7, 3, 8, 3, 12, 2, 27, 2, 7, 2, 14, 3, 2, 5, 5, 9, 1, 1, 12, 6, 1, 12, 14, 4, 2, 7, 25, 7, 1, 3, 1, 3, 3, 10, 2, 4, 1, 6, 11, 2, 12, 5, 5, 14, 3, 1, 1, 1, 2, 1, 10, 8, 6, 4, 1, 2, 2, 3, 14, 13, 2, 3, 2, 8, 8, 5, 6, 2, 6, 1, 5, 2, 12, 1, 6, 1, 3, 12, 9, 6, 4, 1, 4, 6, 16, 8, 3, 3, 4, 1, 7, 2, 1, 1, 4, 4, 7, 2, 5, 3, 7, 8, 6, 3, 7, 2, 8, 2, 1, 1, 4, 9, 2, 5, 4, 2, 10, 1, 10, 6, 10, 1, 6, 2, 3, 14, 2, 1, 10, 4, 2, 11, 6, 10, 6, 1, 3, 3, 4, 2, 1, 1, 1, 2, 9, 19, 1, 2, 17, 1, 6, 7, 12, 4, 15, 11, 1, 3, 3, 4, 2, 31, 2, 22, 6, 2, 4, 10, 1, 1, 3, 9, 3, 4, 4, 15, 8, 5, 1, 1, 18, 2, 3, 2, 1, 11, 5, 8, 8, 2, 8, 4, 7, 3, 2, 5, 1, 8, 2, 20, 1, 8, 2, 1, 4, 4, 3, 5, 2, 5, 2, 1, 1, 9, 3, 6, 11, 2, 5, 7, 1, 1, 10, 1, 1, 1, 3, 7, 2, 6, 4, 4, 9, 3, 1, 3, 3, 1, 1, 7, 7, 1, 3, 2, 2, 21, 4, 1, 12, 7, 4, 7, 5, 3, 4, 1, 2, 4, 2, 3, 1, 16, 21, 7, 2, 9, 3, 4, 2, 6, 9, 2, 1, 1, 32, 5, 3, 5, 2, 5, 4, 5, 1, 2, 7, 4, 18, 12, 3, 2, 4, 12, 10, 5, 8, 3, 1, 7, 7, 7, 8, 9, 10, 1, 22, 7, 2, 16, 5, 9, 1, 2, 2, 2, 4, 1, 3, 7, 7, 4, 14, 9, 3, 4, 4, 5, 4, 1, 7, 3, 1, 4, 5, 3, 2, 19, 2, 2, 17, 4, 4, 5, 3, 9, 6, 3, 5, 5, 2, 6, 1, 8, 6, 1, 1, 2, 6, 11, 3, 3, 6, 9, 4, 8, 1, 9, 6, 3, 8, 9, 5, 11, 2, 1, 3, 1, 6, 3, 1]

What I want to know is that how these numbers represent geometric distribution? Is there a certain order? The probability of say, 0, 1, 2, or 3 appearing as next number? Does the order matter at all?


问:
Markov Chain problem with first passage time
Let $X$ a Markov Chain with space state $S$, and transition matrix $P$. Let $A \subset S$ and $\tau_A = \inf\{n \ge 0: X_n \in A\}$ . Suppose that exists $n \ge 1$ and $\alpha > 0$ that for all $x \in A^c$ (the complement of $A$), $p^n_{x,A} \ge \alpha$. i) Show that for all $k \in \mathbb{N}$ and $x \in S$, $\mathbb{P}_x(\tau_A > kn) \le (1 - \alpha)^k$. ii) Show that $\mathbb{E}_x[\tau_A] \le \frac{n}{\alpha}$, and in particular, $\mathbb{P}_x(\tau_A < \infty) = 1$.

This is how I tried to solve it:

If $x \in A$, we have
$$\mathbb P _x(\tau_A = 0) = 1 \implies \mathbb P _x(\tau_A > kn) = 0 \le (1-\alpha)^k.$$

Let $x \in A^c$. If $\lambda = (\lambda_i : i \in S)$ is a probability distribution of $X$, so $\lambda_x = \mathbb P_x(X_0 = x)$. Moreover, it's known that

By definition of $\lambda$, we have $\lambda_x \le 1$;

Since $X$ is Markov Chain, we have (from Chapman-Kolmogorov)
$$p^n_{x,A^c} = \sum_{i_1, \dots, i_{n-1} \in S}{p_{x,i_1} \cdots p_{i_{n-1},A^c} }$$

Since $A^c \subset S$,
$$p_{x,A^c}\underbrace{p_{A^c,A^c} \cdots p_{A^c,A^c} }_{n-1 \text{ times}} \le \sum_{i_1, \dots, i_{n-1} \in S}{p_{x,i_1} \cdots p_{i_{n-1},A^c} } = p^n_{x,A^c}$$

Similarly,
$$\underbrace{p_{A^c,A^c} \cdots p_{A^c,A^c} }_{n \text{ times}} \le \sum_{i_1, \dots, i_{n-1} \in S}{p_{A^c,i_1} \cdots p_{i_{n-1},A^c} } = p^n_{A^c,A^c} $$

It's easy to see, that
$$p^n_{A^c,A^c} = 1 - p^n_{A^c,A^c}$$

For $x \in A^c$,
$$p^n_{A^c,A} \ge p^n_{x,A}$$

And finally,
$$p^n_{x,A^c} = 1 - p^n_{x,A} \le 1 - \alpha$$

Using some properties about Markov Chains and all these information, it follows that

$$\begin{aligned}\mathbb P_x(\tau_A > kn) &= \mathbb P (X_0 \not \in A,X_1 \not \in A, \dots, X_n \not \in A, X_{n+1} \not \in A, \dots, X_{2n} \not \in A,\dots, X_{kn - 1} \not \in A,X_{kn} \not \in A\mid X_0 = x)\\
&=\mathbb P(X_0 = x, X_1 \not \in A, \dots, X_n \not \in A, \dots, X_{2n} \not \in A,\dots, X_{kn} \not \in A )\\
&= \lambda_x p_{x,A^c}\underbrace{p_{A^c,A^c}\cdots p_{A^c,A^c}}_{n-1 \text{ times}}\underbrace{\underbrace{p_{A^c,A^c}\cdots p_{A^c,A^c}}_{n \text{ times}}\cdots \underbrace{p_{A^c,A^c}\cdots p_{A^c,A^c}}_{n \text{ times}}}_{k - 1 \text{ times}}\\
&\le p_{x,A^c}\underbrace{p_{A^c,A^c}\cdots p_{A^c,A^c}}_{n-1 \text{ times}}\underbrace{\underbrace{p_{A^c,A^c}\cdots p_{A^c,A^c}}_{n \text{ times}}\cdots \underbrace{p_{A^c,A^c}\cdots p_{A^c,A^c}}_{n \text{ times}}}_{k - 1 \text{ times}} \text{ (from 1)}\\
&\le \left(\sum_{i_1,\dots, i_{n-1} \in S}{p_{x,i_1}p_{i_1,i_2}\cdots p_{i_{n-1},A^c}}\right) \cdot \left(\sum_{i_1,\dots, i_{n-1} \in S}{p_{A^c,i_1}p_{i_1,i_2}\cdots p_{i_{n-1},A^c}}\right)^{k-1} \text{ (from 3 and 4)}\\
&= \left(p^n_{x,A^c}\right)\left(p^n_{A^c,A^c}\right)^{k-1} \text{ (from 2 and 4)}\\
&\le \left(p^n_{x,A^c}\right)\left(1 - p^n_{x,A}\right)^{k-1} \text{ (from 5 and 6)}\\
&\le (1-\alpha)^k \text{ ( from 7)}\end{aligned}$$

This is how I tried to solve item i). Note that,for $i \ge 1$
$$\{\tau_A > i + 1\} \subseteq \{\tau_A > i\}$$
It means that
$$\mathbb P_x(\tau_A = \infty) = \mathbb P_x(\lim_{k \to \infty}{\tau_A > kn}) = \lim_{k \to \infty} {\mathbb P_x({\tau_A > kn})} \le \lim_{k \to \infty}{(1-\alpha)^k} = 0$$
It implies that,
$$\mathbb P_x(\tau_A < \infty) = 1 - \mathbb P_x(\tau_A = \infty) = 1$$

Finally,

$$
\begin{aligned}
\mathbb{E}_x[\tau_A] &= \frac{n}{n}\mathbb E_x[\tau_A]\\
&= n\mathbb E_x\left[\frac{\tau_A}{n}\right]\\
&= n\left(\sum_{i \ge 1}{\mathbb P_x\left(\frac{\tau_A}{n} > i\right)}\right)\\
&= n\left(\sum_{i \ge 1}{\mathbb P_x(\tau_A > ni)}\right)\\
&\le n\left(\sum_{i \ge 1}{\mathbb (1-\alpha)^i}\right)\\
&=n \cdot \frac{1}{1 - (1-\alpha)} = \frac{n}{\alpha}
\end{aligned}
$$
Please, help me to find any mistakes. Thanks for the help!

答: Note that if $x\in A$ then $\tau_A=0$ and so $$0=\mathbb{P}_x(\tau_A>kn)\leq (1-\alpha)^k$$ On the other hand, if $x\notin A$ we can say that $$\mathbb{P}_x(\tau_A>kn)=\mathbb{P}(X_1\notin A,X_2 \notin A,\ldots,X_{kn}\notin A|X_0=x)$$ The right hand side is bounded above by $$\mathbb{P}(X_n\notin A,X_{2n}\notin A,\ldots,X_{kn}\notin A|X_0=x)$$ Using the chain rule for conditional probability along with Markov this equals $$\mathbb{P}(X_n\notin A|X_0=x)\times \mathbb{P}(X_{2n}\notin A|X_n\notin A)\times \dots \times\mathbb{P}(X_{kn}\notin A|X_{(k-1)n}\notin A)$$ which is bounded above by $(1-\alpha)^k$. Clearly if $x\in A$ then $\mathbb{E}_x(\tau_A)=0$ so for $x \notin A$ we have that $$\mathbb{E}_x(\tau_A)=\sum_{j=0}^{\infty}\mathbb{P}_x(\tau_A > j)=1+\sum_{j=1}^{n-1}\mathbb{P}_x(\tau_A>j)+\sum_{k=1}^{\infty}\sum_{j=kn}^{kn+n-1}\mathbb{P}_x(\tau_A>j)$$ Because $$\sum_{j=1}^{n-1}\mathbb{P}_x(\tau_A>j)\leq(n-1)\mathbb{P}_x(\tau_A>1)\leq(n-1)(1-\alpha)$$ $$\sum_{j=kn}^{kn+n-1}\mathbb{P_x}(\tau_A>j)\leq n\mathbb{P}_x(\tau_A>kn)\leq n(1-\alpha)^k$$ it follows that $$\mathbb{E}_x(\tau_A)\leq1+(n-1)(1-\alpha)+n\sum_{k=1}^{\infty}(1-\alpha)^k=\alpha(1-n)+\frac{n}{\alpha}\leq \frac{n}{\alpha}$$ as desired. Now we'll prove $\mathbb{P}_x(\tau_A<\infty)=1$ for $x\notin A$. For $j\geq n$ we have $$\mathbb{P}(X_j\notin A,\ldots ,X_1 \notin A|X_0=x)\leq \mathbb{P}(X_{n\left \lfloor{j/n}\right \rfloor }\notin A,\ldots,X_n \notin A|X_0=x)$$ The right hand side is bounded above by $(1-\alpha)^{\left \lfloor{j/n}\right \rfloor }$ and finally $$\mathbb{P}_x(\tau_A=\infty)\leq \lim_{j\rightarrow \infty}(1-\alpha)^{\left \lfloor{j/n}\right \rfloor}=0 $$


问:
Let $G$ be a finite group with $|G|>2$. Prove that ${\rm Aut}(G)$ contains at least two elements.

Let $G$ be a finite group with $|G|>2$. Prove that ${\rm Aut}(G)$ contains at least two elements.

We know that ${\rm Aut}(G)$ contains the identity function $f: G \to G: x \mapsto x$.

If $G$ is non-abelian, look at $g : G \to G: x \mapsto gxg^{-1}$, for $g\neq e$. This is an inner automorphism unequal to the identity function, so we have at least two elements in ${\rm Aut}(G).$

Now assume $G$ is abelian. Then the only inner automorphism is the identity function. Now look at the mapping $\varphi: G \to G : x \mapsto x^{-1}$. This is an homomorphism because $\varphi (xy) = (xy)^{-1} = y^{-1} x^{-1} = x^{-1} y^{-1} = \varphi (x) \varphi (y)$. Here we use the fact that $G$ is abelian. This mapping is clearly bijective, and thus an automorphism.

This automorphism is unequal to the identity function only if there exists an element $x \in G$ such that $x \neq x^{-1}$. In other words, there must be an element of order greater than $2$.

Now assume $G$ is abelian and every non-identity element has order $2$. By Cauchy's theorem we know that the group must have order $2^n$.

I got stuck at this point. I've looked at this other post, $|G|>2$ implies $G$ has non trivial automorphism, but I don't know what they do in the last part (when they start talking about vector spaces). How should this prove be finished, without resorting to vector spaces if possible?

Thanks in advance

答:
Building off of OP's work, we now consider Abelian groups, where each non-identity element has order 2.

Let $\{g_1, g_2, \ldots g_n \}$ be any minimal generating set of the group. Since $|G| > 2 $, thus $n\geq 2$.

Each element of the group $g$ can be uniquely written as $\prod_{i \in S_g} g_i$. (Exponents are no needed, since each element has order 2.)

Then, consider the the map that replaces $g_1$ with $g_2$ and vice versa.

Show that this is an automorphism.


答:
Starting with the first derivative
$$\frac {\partial}{\partial x_p} \Bigg[ \dfrac {1} { \sum _{k=1}^n \frac{1}{x_k}}\Bigg]=\frac {\partial}{\partial x_p} \Bigg[ \dfrac {1} { \frac{1}{x_p}+C}\Bigg]=\frac {\partial}{\partial x_p}\Bigg[ \dfrac {x_p} { 1+Cx_p}\Bigg] $$since $$C=\sum _{k=1}^{p-1} \frac{1}{x_k}+\sum _{k=p+1}^{n} \frac{1}{x_k}$$ does not depend on $x_p$

$$\frac {\partial}{\partial x_p} \Bigg[ \dfrac {1} { \sum _{k=1}^n \frac{1}{x_k}}\Bigg]=\frac{1}{(1+C x)^2}$$
$$\frac {\partial^2}{\partial x_p^2} \Bigg[ \dfrac {1} { \sum _{k=1}^n \frac{1}{x_k}}\Bigg]=-\frac{2 C}{(1+C x)^3}$$
$$\frac {\partial^3}{\partial x_p^3} \Bigg[ \dfrac {1} { \sum _{k=1}^n \frac{1}{x_k}}\Bigg]=\frac{6 C^2}{(1+C x)^4}$$
$$\frac {\partial^4}{\partial x_p^4} \Bigg[ \dfrac {1} { \sum _{k=1}^n \frac{1}{x_k}}\Bigg]=-\frac{24 C^3}{(1+C x)^5}$$ I am sure that you see the very clear pattern.

Title: First and Second Order Derivate of the Function having summation sequence

Body:
I have this function having a summation sequence in it
$$ f(x) =\frac {1} { \sum _{k=1}^n \frac{1}{x_k}} $$

How do I determine the first and second derivatives of the function with respect to x?


问:
Get the right vectors in a differential equation system
I'm stuck with this differential equation system:

$
Y'(x) =
    \begin{pmatrix}
    2 & 1 & 3 \\
    0 & 2 & -1 \\
    0 & 0 & 2 \\
    \end{pmatrix}
Y(x)
$

I get only one eigenvalue, $\lambda = 2$ (with algebraic multiplicity 3).

For this eigenvalue, I obtain:

$
Y'(x) =
    \begin{pmatrix}
    0 & 1 & 3 \\
    0 & 0 & -1 \\
    0 & 0 & 0 \\
    \end{pmatrix}
    \begin{pmatrix}
    v_{x} \\
    v_{y} \\
    v_{z} \\
    \end{pmatrix}
=
    \begin{pmatrix}
    0 \\
    0 \\
    0 \\
    \end{pmatrix}
    \Rightarrow
    \begin{pmatrix}
    v_{x} \\
    v_{y} \\
    v_{z} \\
    \end{pmatrix}
=
v_{x}
    \begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
$

I just got only one eigenvector, so that I've to get two more.

I look for:

$
    \begin{pmatrix}
    0 & 1 & 3 \\
    0 & 0 & -1 \\
    0 & 0 & 0 \\
    \end{pmatrix}
    \begin{pmatrix}
    v_{x} \\
    v_{y} \\
    v_{z} \\
    \end{pmatrix}
=
    \begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
    \Rightarrow
    \begin{pmatrix}
    v_{x} \\
    v_{y} \\
    v_{z} \\
    \end{pmatrix}
=
    \begin{pmatrix}
    v_{x} \\
    1 \\
    0 \\
    \end{pmatrix}
$

From this, I choose the vector $(0,1,0)$

To get the last vector, I do:

$
    \begin{pmatrix}
    0 & 1 & 3 \\
    0 & 0 & -1 \\
    0 & 0 & 0 \\
    \end{pmatrix}^2
    \begin{pmatrix}
    v_{x} \\
    v_{y} \\
    v_{z} \\
    \end{pmatrix}
=
    \begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
$

$
    \begin{pmatrix}
    0 & 0 & -1 \\
    0 & 0 & 0 \\
    0 & 0 & 0 \\
    \end{pmatrix}
    \begin{pmatrix}
    v_{x} \\
    v_{y} \\
    v_{z} \\
    \end{pmatrix}
=
    \begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
    \Rightarrow
    \begin{pmatrix}
    v_{x} \\
    v_{y} \\
    v_{z} \\
    \end{pmatrix}
=
    \begin{pmatrix}
    v_{x} \\
    v_{y} \\
    1 \\
    \end{pmatrix}
$

From this, I choose the vector $(0,0,1)$

So that, I think with these three (and linearly independent) vectors I should get the general solution like this:

$
y(x) =
    C_{1}
    \begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{2}x
    \begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{2}
    \begin{pmatrix}
    0 \\
    1 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    \frac{C_{3}x^2}{2}
    \begin{pmatrix}
    1 \\
    0 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{3}x
    \begin{pmatrix}
    0 \\
    1 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{3}
    \begin{pmatrix}
    0 \\
    0 \\
    1 \\
    \end{pmatrix}
    e^{2x}
$

But this is not correct since it doesn't proove the initial relation. I get for both sides of the initial relation:

$
Y'(x) =
    C_{1}
    \begin{pmatrix}
    2 \\
    0 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{2}x
    \begin{pmatrix}
    2 \\
    0 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{2}
    \begin{pmatrix}
    1 \\
    2 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    \frac{C_{3}x^2}{2}
    \begin{pmatrix}
    2 \\
    0 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{3}x
    \begin{pmatrix}
    1 \\
    2 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{3}
    \begin{pmatrix}
    0 \\
    1 \\
    2 \\
    \end{pmatrix}
    e^{2x}
$

And

$
    \begin{pmatrix}
    2 & 1 & 3 \\
    0 & 2 & -1 \\
    0 & 0 & 2 \\
    \end{pmatrix}
    Y(x) =
    C_{1}
    \begin{pmatrix}
    2 \\
    0 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{2}x
    \begin{pmatrix}
    2 \\
    0 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{2}
    \begin{pmatrix}
    1 \\
    2 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    \frac{C_{3}x^2}{2}
    \begin{pmatrix}
    2 \\
    0 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{3}x
    \begin{pmatrix}
    1 \\
    2 \\
    0 \\
    \end{pmatrix}
    e^{2x}
    +
    C_{3}
    \begin{pmatrix}
    3 \\
    -1 \\
    2 \\
    \end{pmatrix}
    e^{2x}
$

I'm almost sure that I get the eigenvectors wrongly but I don't know how to get it right...

Please, anybody knows where is the error and what should I do in order to get the right vectors?

Thank you very much in advance.

答: The solutions of your differential equation are of the form$$Y(t)=k\exp\left(t\begin{bmatrix}2&1&3\\0&2&-1\\0&0&2\end{bmatrix}\right).\tag1$$Now, note that$$\begin{bmatrix}2&1&3\\0&2&-1\\0&0&2\end{bmatrix}=2\operatorname{Id}+\begin{bmatrix}0&1&3\\0&0&-1\\0&0&0\end{bmatrix}$$and that this expresses$$\begin{bmatrix}2&1&3\\0&2&-1\\0&0&2\end{bmatrix}$$as the sum of two commuting matrices. Therefore, $(1)$ is equal to$$k\exp(2t\operatorname{Id})\exp\left(t\begin{bmatrix}0&1&3\\0&0&-1\\0&0&0\end{bmatrix}\right).$$But\begin{align}\exp\left(t\begin{bmatrix}0&1&3\\0&0&-1\\0&0&0\end{bmatrix}\right)&=\operatorname{Id}+t\begin{bmatrix}0&1&3\\0&0&-1\\0&0&0\end{bmatrix}+\frac1{2!}\left(t\begin{bmatrix}0&1&3\\0&0&-1\\0&0&0\end{bmatrix}\right)^2+\cdots\\&=\operatorname{Id}+t\begin{bmatrix}0&1&3\\0&0&-1\\0&0&0\end{bmatrix}+\frac{t^2}2\begin{bmatrix}0&0&-1\\0&0&0\\0&0&0\end{bmatrix}\\&=\begin{bmatrix}1&t&3t-\frac{t^2}2\\0&1&-t\\0&0&1\end{bmatrix}.\end{align}Can you take it from here?


问:
Orthogonal complement in $ L_2 $
Find the orthogonal complement of a subspace
$$ M = \{ x \in L_2(-1, 1):x(t)=-x(-t), \int_0^1 x(t)t^2dt=0 \} $$
in $L_2(-1, 1).$

As I understand M can be described as all odd functions which are orthogonal to $ \lambda t^2 $ on $(0, 1)$. But I don`t know how to find the orthogonal complement

答: Let $f(t)=t^{2}\chi_{(0,1)}(t)$. Then $M$ consists of functions which are odd and orthognal to $f$. This means $M$ is precisely the orthogonal complement of the span of even functions and $f$. Thus, $M^{\perp} =\{g+cf:c \in \mathbb R, g \, \text {is even} \}$

Title: Orthogonal complement in $ L_2 $


问:
A limit on binomial coefficients
Let $$x_n=\frac{1}{n^2}\sum_{k=0}^n \ln\left(n\atop k\right).$$ Find the limit of $x_n$.

What I can do is just use Stolz formula. But I could not proceed.

Title: A limit on binomial coefficients

Body:
I've managed to reduce it somewhat.
$$x_n=\frac{1}{n^2}\ln(\prod_{k=0}^n \binom{n}{k})$$

$$x_n=\frac{1}{n^2}\ln \left ( \frac{n!}{(n-0)!0!}\frac{n!}{(n-1)!1!}\frac{n!}{(n-2)!2!}...\frac{n!}{(n-n)!n!}\right )$$

$$x_n=\frac{1}{n^2}\ln \left ( \frac{n!^{(n+1)}}{(0!1!2!...n!)^2} \right )$$

$$x_n=\frac{n+1}{n^2}\ln \left ( n! \right )-\frac{2}{n^2}\ln \left ( 0!1!2!...n! \right )$$

$$x_n=\frac{n+1}{n^2}\ln \left ( n! \right )-\frac{2}{n^2}\sum_{k=0}^n \ln(k!)$$

Using Stirling's approximation for large $n$ ($ \ln(n!)=n \ln(n)-n$):

$$x_n=\frac{n+1}{n^2}(n \ln(n)-n)-\frac{2}{n^2}\sum_{k=0}^n \ln(k!)$$

$$x_n=\frac{n \ln(n)-n}{n}+\frac{n \ln(n)-n}{n^2}-\frac{2}{n^2}\sum_{k=0}^n \ln(k!)$$

$$x_n=\ln(n)-1+\frac{ \ln(n)-1}{n}-\frac{2}{n^2}\sum_{k=0}^n \ln(k!)$$

Noting that $\lim_{n \rightarrow \infty} \frac{\ln(n)}{n}=0$, for $n \rightarrow \infty$:

$$x_n=\ln(n)-1-\frac{2}{n^2}\sum_{k=0}^n \ln(k!)$$


问:
Suppose that $n \in \mathbb{Z}$. Prove that if $n^2 + 1$ is a perfect square, then $n$ is even.
This is a homework problem that I cannot figure out. I have figured out that if $n^2 + 1$ is a perfect square it can be written as such:

$n^2 + 1 = k^2$.

and if $n$ is even it can be written as such:

$n = 2m$

I believe I'm supposed to use the fact that if $n \pmod{4} \equiv 0$ or $1$ then it's a perfect square (maybe that's wrong).

I cannot figure this out.

答: Suppose n is odd, write $n=2m+1$. Then $n^2+1=(2m+1)^2+1=4m^2+4m+2$, which is even. Therefore  if this is a perfect square, it must be divisible by $4$. But clearly $n^2+1\equiv 2\pmod 4$, a contradiction.


问:
I am using it in my m.phil research so it is very vital for me to Evaluating $\sum_{n=0}^\infty\frac1{\Gamma(n/2+2)}$
I want to sum the following:
$$\sum_{n=0}^\infty\frac1{\Gamma(n/2+2)}$$

答:
If you are aware of the incomplete gamma function,
$$\sum_{x=0}^n \frac{1}{\Gamma \left(\frac{x}{2}+2\right)}=-\left(1+\frac{2}{\sqrt{\pi }}+e\, \text{erfc}(1)\right)+\frac{e \,\Gamma \left(\frac{n+4}{2},1\right)}{\Gamma
   \left(\frac{n+4}{2}\right)}+\frac{e\, \Gamma \left(\frac{n+3}{2},1\right)}{\Gamma
   \left(\frac{n+3}{2}\right)}$$ When $n\to \infty$, the sum of the last two terms tend to $2e$ and this makes
$$\sum_{x=0}^\infty \frac{1}{\Gamma \left(\frac{x}{2}+2\right)}=e (1+\text{erf}(1))-\left(1+\frac{2}{\sqrt{\pi }}\right)\sim 2.88060$$

Now, you need to prove that this is correct.

Such an answer is of absolutely of no use if you do not know or understand the steps.


问:
partial fraction of $\prod_{i=1}^{n}1/(\lambda_{i}+x)$ and its integration
For positive scalars $\lambda_{i}$, the partial fraction of $\prod_{i=1}^{n} \frac{1}{\lambda_{i}+x}$ can be obtained easily for $n=2,3$. Then the integration $\int_{0}^{\infty}\prod_{i=1}^{n} \frac{1}{\lambda_{i}+x}dx$  can be computed correspondingly for $n=2,3$.

The question is that what is the partial fraction and its integration for larger $n$?

PS: The integration correponds to the first and second divided difference of the logarithm function $f(x) =\ln(x)$ for $n=2,3$, respectively. So it is natural to think the correpondence may be correct for larger $n$.

答:
The partial fraction decomposition is
\begin{equation}
f(x) = \prod_{i=1}^n\frac{1}{\lambda_i+x} = \sum_{i=1}^n\frac{\alpha_i}{\lambda_i+x}, \qquad \alpha_i=\prod_{j\neq i}\frac{1}{\lambda_j-\lambda_i}
\end{equation}
because $g_i(x) = (\lambda_i+x)f(x)$ satisfies $g_i(-\lambda_i) = \alpha_i$.


问:
Proving the equivalence of norms is an equivalence relation
Two norms $\Vert -\Vert _1 $, $\Vert -\Vert _2$ are equivalent if:

for two constants $a,b$ and $x$ from $V$ a vector space over a field it holds that:  $$a\Vert x\Vert _1\leqslant \Vert x\Vert _2\leqslant  b\Vert x\Vert _1.$$

This is a equivalence relation because:

$$a\Vert x\Vert _1\leqslant \Vert x\Vert _2\leqslant  b\Vert x\Vert _1$$ and $$c\Vert x\Vert _2\leqslant  \Vert x\Vert _3 \leqslant  d\Vert x\Vert _2$$

it follows that there are constants such that (transitivity): $$e\Vert x\Vert _1 \leqslant  \Vert x\Vert _3 \leqslant  f\Vert x\Vert _1$$

Reflexivity:  $$a\Vert x\Vert _1\leqslant  \Vert x\Vert _1\leqslant  b\Vert x\Vert _1$$ with $a,b = 1$; this is true.

Symmetry: $$a\Vert x\Vert _1 \leqslant  \Vert x\Vert _2 \leqslant  b\Vert x\Vert _1$$

if we take: $$-\frac{1}{b}\Vert x\Vert _2\leqslant  \Vert x \Vert_1 \leqslant  \frac{-1}{a} \Vert x\Vert _2.$$

Is this a valid proof  that the equivalence of two norms is truly a equivalence relationship?

Attempt 2:

Symmetry:

$$a\Vert x\Vert _1 \leqslant  \Vert x\Vert _2 \leqslant  b\Vert x\Vert _1$$

$\Rightarrow $: $$\frac{1}{b} \Vert x\Vert _2 \leqslant  \Vert x\Vert _1 \leqslant  \frac{1}{a}\Vert x\Vert _2.$$

Transitivity: given

$$a\Vert x\Vert _1\leqslant \Vert x\Vert _2\leqslant  b\Vert x\Vert _1$$ and $$c\Vert x\Vert _2\leqslant  \Vert x\Vert _3 \leqslant  d\Vert x\Vert _2$$

$\Rightarrow$ : $$ac \Vert x\Vert _1 \leqslant  c\Vert x\Vert _2\leqslant  \Vert x\Vert _3 \leqslant d\Vert x\Vert _2\leqslant  db\Vert x\Vert _1.$$

答:
$
\newcommand{\norm}[2]{\left \Vert #1 \right \Vert_{#2}}
\newcommand{\N}{\mathcal{N}(V)}
\newcommand{\a}{\alpha}
\newcommand{\b}{\beta}
\newcommand{\g}{\gamma}
\newcommand{\l}[1]{\stackrel{(#1)}{\le}}
$
I feel like your proof could be better phrased to make it clearer what the conditions imply and require, where they come from, and your understanding thereof, at least for your scratch work. I come away from your proof not fully sure you even know what needs to be verified. (Granted, this could entirely be a language barrier problem.) I will try to rewrite the proof in such a way to make these notions and your writing clearer, if just for future readers if not to enlighten you.

First though, some more specific criticisms:

It should be specified that the constants in your field are positive. You use negatives in particular in your first attempt at symmetry.

You didn't include any real mention of transitivity in your original proof. Or rather you kinda did, but even then it wasn't even clear where your constants came from. After all, equivalence relations need all three conditions.

Your second attempt at symmetry doesn't make it clear where you obtained the constants from, or why they indeed work.

Your second (?) attempt at transitivity suffers from a similar problem, but how it follows is a bit less out of the way in this case. I still feel it should be noted where, though, for the sake of completeness and clarity.

Referencing $1,2,3$ for the norms feels a bit off to me; it makes it look like you're referencing specific norms, e.g. $p$-norms. This is why I'll be using more "arbitrary-looking" indices $\a,\b,\g$ in my write-up, to help avoid the potential confusion. Also, I would use $\norm \cdot \a$ as opposed to $\norm - \a$ (since that's the more familiar notation in my experience). Granted, this bullet point in particular might all be a personal issue, so take that for what you will.

Defining the Relation: So we define the relation as so: for norms $\norm \cdot \a, \norm \cdot \b$ on $V$ a vector space over an ordered field $F$,

$$\norm \cdot \a \sim \norm \cdot \b \iff (\forall x \in V)(\exists a,b \in F^+) \left(a \norm x \a \le \norm x \b \le b \norm x \a\right)$$

To show $\sim$ is an equivalence, we need to verify the three necessary conditions. Note that $\sim$ is a relation on the set $\N$ of norms $\norm \cdot \a : V \to F$. Once verified, we will say that $\sim$ denotes the equivalence of norms.

Reflexivity: The requirement here is that $\norm \cdot \a \sim \norm \cdot \a$ for every $\norm \cdot \a \in \N$. This holds by taking $a=b=1$, trivially.

Symmetry: Suppose $\norm \cdot \a \sim \norm \cdot \b$. Then to show symmetry we will want to show $\norm \cdot \b \sim \norm \cdot \a$. So, we know that

$$(\forall x \in V)(\exists a,b \in F^+)\left(a \norm x \a \le \norm x \b \le b \norm x \a\right) \tag 1$$

What we want to show is the similar condition

$$(\forall x \in V)(\exists a',b' \in F^+)\left(a' \norm x \b \le \norm x \a \le b' \norm x \b\right)$$

Consider the inequality in $(1)$. Divide by $a$ to obtain $(2a)$, and by $b$ to obtain $(2b)$. Note that inequality is maintained since $a,b > 0$.

\begin{align*}
&\norm x \a \le \frac 1 a \norm x \b \le \frac b a \norm x \a \tag{2a} \\
\frac a b &\norm x \a \le \frac 1 b \norm x \b \le  \norm x \a \tag{2b}
\end{align*}

Combine the two about $\norm x \a$ and we have the chain of inequalities $(3)$, with the boxed portion being most relevant:

$$\frac a b \norm x \a \le \boxed{\frac 1 b \norm x \b \le \norm x \a \le \frac 1 a \norm x \b} \le \frac b a \norm x \a \tag{3}$$

Since $a,b > 0$, then $1/a,1/b>0$ as well. Thus, take $a' = 1/a$ and $b' = 1/b$ to obtain symmetry!

Transitivity: So, suppose $\norm \cdot \a \sim \norm \cdot \b$ and $\norm \cdot \b \sim \norm \cdot \g$. We want to show $\norm \cdot \a \sim \norm \cdot \g$. So, what we know is that

\begin{align*}
(\forall x \in V)(\exists a,b \in F^+) &\left(a \norm x \a \le \norm x \b \le b \norm x \a\right) \tag 4\\
(\forall x \in V)(\exists c,d \in F^+) &\left(c \norm x \b \le \norm x \g \le d \norm x \b\right) \tag 5
\end{align*}

where the first known relation is in $(4)$, and the second is in $(5)$. On the basis of these, we want to demonstrate $(6)$:

$$(\forall x \in V)(\exists e,f \in F^+) \left(e \norm x \a \le \norm x \g \le f \norm x \a\right) \tag 6$$

We now pass around between the inequalities in $(4)$ and $(5)$, using $\norm x \g$ as a starting point, and constructing inequalities each way. We obtain the following (where the numbers above each inequality show which one implies the inequality):

$$
a c \norm x \a
\l 4 c \norm x \b
\l 5  \norm x \g
\l 5 d \norm x \b
\l 4 bd \norm x \a
$$

For emphasis, focus on the first, third, and fifth terms:

$$
a c \norm x \a
\le  \norm x \g
\le bd \norm x \a
$$

Thus, take $e = ac$ and $f=bd$ in $(6)$ to obtain transitivity! And, with all three properties verified, $\sim$ is an equivalence relation!

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


答:
$x_n=\frac{1}{n^2}\sum_{k=0}^{n}\ln{n\choose k}=\frac{1}{n^2}\ln(\prod {n\choose k})=\frac{1}{n^2}\ln\left(\frac{n!^n}{n!^2.(n-1)!^2(n-2)!^2...0!^2}\right)$ since ${n\choose k}=\frac{n!}{k!(n-k)!}$

$e^{n^2x_n}=\left(\frac{n^n(n-1)!}{n!^2}\right)e^{(n-1)^2x_{n-1}}=\left(\frac{n^{n-1}}{n!}\right)e^{(n-1)^2x_{n-1}}$

By Stirling's approximation, $n! \sim n^ne^{-n}\sqrt{2\pi n}$

$e^{n^2x_n}\sim \left(\frac{e^n}{n\sqrt{2\pi n}}\right)e^{(n-1)^2x_{n-1}}$

$x_n \sim \frac{(n-1)^2}{n^2}x_{n-1}+\frac{1}{n}-\frac{1}{n^2}\ln(n\sqrt{2\pi n})$

The $\frac{1}{n}$ term forces $x_n$ to tend to infinity, because $\frac{1}{n^2}\ln(n\sqrt{2\pi n})$ does not grow fast enough to stop it diverging.

Title: A limit on binomial coefficients

Body:
Let $$x_n=\frac{1}{n^2}\sum_{k=0}^n \ln\left(n\atop k\right).$$ Find the limit of $x_n$.

What I can do is just use Stolz formula. But I could not proceed.


问:
Find the locus of points on a complex plane that satisfy |z-3| - |z+3| = 4
Can you find the locus of points on the complex plane represented by the equation below:

|z-3| - |z+3| = 4

答:
Hint:

Given two points called the foci (plural of focus), which are $(c,0)$ and $(-c,0)$ in the picture below, a hyperbola is precisely the locus of points $(x,y)$ such that the difference between $(x,y)$ and the foci is constant.

Since the complex plane can be visualized like an $xy$-plane, this should help. Say $z=x+iy$. Then your equation says that the distance between $(x,y)$ and $(3,0)$ minus the distance between $(x,y)$ and $(-3,0)$ is constant, equal to 4. This is exactly half of the hyperbola you see below with $c=3$. The equation of this hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, where $a$ is the real number that's the solution of the equation that you should try to find, and $a^2+b^2=c^2$.

Reference and more info: https://courses.lumenlearning.com/waymakercollegealgebra/chapter/equations-of-hyperbolas/#:~:text=Reviewing%20the%20standard%20forms%20given,%3D%20a%202%20%2B%20b%202%20.


答:
The partial fractions of $\prod_{i=1}^n 1/(x+\lambda_i)$ can be easily calculated by writing this as

$$\sum_{i=1}^n \frac{A_i}{x+\lambda_i} $$

and solving it to get

$$A_i = -\prod_{j=1,j\neq i}^{n} \frac{1}{\lambda_i - \lambda_j}$$

since the function $\phi_i(x)$ defined by$$\phi_i(x) = (x+\lambda_i)\prod_{k=1}^{n} \frac{1}{x+\lambda_k}$$ satisfies the equation $\phi_i(-\lambda_i) = A_i$.

Title: partial fraction of $\prod_{i=1}^{n}1/(\lambda_{i}+x)$ and its integration

Body:
For positive scalars $\lambda_{i}$, the partial fraction of $\prod_{i=1}^{n} \frac{1}{\lambda_{i}+x}$ can be obtained easily for $n=2,3$. Then the integration $\int_{0}^{\infty}\prod_{i=1}^{n} \frac{1}{\lambda_{i}+x}dx$  can be computed correspondingly for $n=2,3$.

The question is that what is the partial fraction and its integration for larger $n$?

PS: The integration correponds to the first and second divided difference of the logarithm function $f(x) =\ln(x)$ for $n=2,3$, respectively. So it is natural to think the correpondence may be correct for larger $n$.


问:
Boundary operator: what does it mean?
I'm getting some confusion in simplicial homology...Take a very simple example, a (solid) tetrahedron:

Following the well known property that "the bounday of a boundary is zero", we would end up with $\partial\partial=0$.

Instead, using the "chain complex" concept, the boundary operator seems to map the solid tetrahedron to the "void tetrahedron", and the latter to its edges.

So it seems to me that "taking the boundary" has at leat two different meanings...where am I wrong?

答:
Simply because when you take one of these triangle and you apply once again the boundary operator you obtain the (oriented) segments of the triangle.
To put it "simply" the boundary operator decompose a figure into pieces that compose him.

The accuracy of my answer is maybe not too high, because I also begin in the study of such object.

Title: Boundary operator: what does it mean?


问:
Proof that limits are unique
I have been working through a few calculus proofs, and to prove that a limit was unique i assumed the fact that if $c,d\in\mathbb{R}$ if given any $\epsilon>0$ we have $|c-d|<\epsilon$, then $c=d$.  However, is there an actual proof of this statement, or is this more of an obvious reality?

答: This is actually not provable without really getting deep into the definition/construction of $\mathbb R$. We need the fact that if $a \neq b$, then $|a-b|>0$, and proving that, rather than taking it as an axiom, requires rather complicated mathematics.

Title: Proof that limits are unique


答:
The reason why that last case is done separately is because in that case we cannot describe a non-trivial automorphism with operations intrinsic to the group. In the non-abelian case we could use an inner automorphism, and in other abelian cases we can use negation. Using vector space structure is the "next best thing" in the sense that then everybody will immediately accept multiplication by an invertible matrix as an automorphism.

That last case is about groups that are isomorphic to a direct product of finitely many copies of $C_2$
$$
G=C_2\times C_2\times\cdots C_2.
$$
Any permutations of components is then obviously an automorphism. But proving that such a group has this direct product structure, while not too tricky, is something the answerers did not want to do, because it is much more convenient to observe that a finite abelian group where all the non-trivial elements have order two is a vector space over $GF(2)$.

Here's another way of doing this step. We inductively find elements $x_1,x_2,\ldots,x_\ell\in G$ such that at all steps $x_k$ is not in the subgroup generated by $x_1,x_2,\ldots,x_{k-1}$. Because $G$ is finite, at some point we will have exhausted all of $G$, and thus $G=\langle x_1,x_2,\ldots,x_\ell\rangle$. The claim is then that
$$
G=\langle x_1\rangle\times\langle x_2\rangle\times\cdots\langle x_\ell\rangle
$$
is a way of writing $G$ as a direct product of subgroups of order two. This is more or less obvious by construction.

But you should notice that this is essentially the same argument proving that a finitely generated vector space over a field has a finite basis.

Body:
Let $G$ be a finite group with $|G|>2$. Prove that ${\rm Aut}(G)$ contains at least two elements.

We know that ${\rm Aut}(G)$ contains the identity function $f: G \to G: x \mapsto x$.

If $G$ is non-abelian, look at $g : G \to G: x \mapsto gxg^{-1}$, for $g\neq e$. This is an inner automorphism unequal to the identity function, so we have at least two elements in ${\rm Aut}(G).$

Now assume $G$ is abelian. Then the only inner automorphism is the identity function. Now look at the mapping $\varphi: G \to G : x \mapsto x^{-1}$. This is an homomorphism because $\varphi (xy) = (xy)^{-1} = y^{-1} x^{-1} = x^{-1} y^{-1} = \varphi (x) \varphi (y)$. Here we use the fact that $G$ is abelian. This mapping is clearly bijective, and thus an automorphism.

This automorphism is unequal to the identity function only if there exists an element $x \in G$ such that $x \neq x^{-1}$. In other words, there must be an element of order greater than $2$.

Now assume $G$ is abelian and every non-identity element has order $2$. By Cauchy's theorem we know that the group must have order $2^n$.

I got stuck at this point. I've looked at this other post, $|G|>2$ implies $G$ has non trivial automorphism, but I don't know what they do in the last part (when they start talking about vector spaces). How should this prove be finished, without resorting to vector spaces if possible?

Thanks in advance


问:
Is any algebraic extension a subfield of a spitting field?
Let $F \subset K$ be an algebraic field extension, is $K$ a subfield of a splitting field of $S=\{{f_i}\}_{i \in I} $ over $F$? I tried: Let $\{{u_i}\}$ be a $F$-basis of K, $\{{p_i}\}$ be its corresponding minimal polynomials ,can we deduce that $K$ is  a subfield of the splitting field of $\{{p_i}\}$ over $F$? I got somewhat confused about the underlying sets. Thanks in advance.

答:
Zorn's lemma (see Existence of extension of an embedding for an algebraic extension of a field) implies that there is an embedding $K \rightarrow \overline{F}$.

Now, $\overline{F}$ is the splitting field for $S = \text{every polynomial in $F[x]$}$ , and $K$ is isomorphic to the image of the embedding in $\overline{F}$.


问:
How to make a plot of function in MATLAB
I have a function ${\dot{\varphi } = \gamma - F(\varphi )}$ (where $F(\varphi)$ - is 2${\pi }$-periodic function) and graph of function $F(\varphi)$.

So I need to create a program that outputs 6 plots of φ(t) for different meaning of $\gamma (\gamma= 0.1, 0.5, 0.95, 1.05, 2, 5)$, and t ∈ [0, 100]

I've got a following F(φ) function:
$$F(\phi)=\begin{cases}
 & {-\frac{\phi}{a}-\frac{\pi}{a}},  &\text{if } {\phi \in [-\pi, -\pi + a]},\\
 & -1, &\text{if } {\phi \in [-\pi + a, - a]}, \\
 & -\frac{\phi}{a},  &\text{if }  {\phi \in [- a,  a]}, \\
 & 1, &\text{if } {\phi \in [a, \pi- a]},\\
 & {-\frac{\phi}{a}+\frac{\pi}{a}},&\text{if }  {\phi \in [\pi-a, \pi]}.
\end{cases}$$

```<code>                 ^
                 |
                 |1   ______
                 |   /|     \
                 |  / |      \
                 | /  |       \
__-π_______-a____|/___|________\π____>
   \        |   /|0    a
    \       |  / |
     \      | /  |
      \     |/   |
       ¯¯¯¯¯¯    |-1
</code>```

I've tried to do it in MATLAB but I faced with some problems. I don't know what to put in ode45 function (on which section should each part of the graph be displayed and what is the initial value to take). Because the evolution of  $\phi(t)$ must be continuous.
It's a code for $\gamma= 0.1$

```<code>hold on;
df1dt=@(t,f1) 0.1 - f1 - 3.14;
df2dt= @(t,f2)- 1;
df3dt=@(t,f3) 0.1 + f3;
df4dt= @(t,f4)+1;
df5dt=@(t,f5) 0.1- f5 + 3.14;
[T1,Y1] = ode45(df1dt, ...);
[T2,Y2] = ode45(df2dt, ...);
[T3,Y3] = ode45(df3dt, ...);
[T4,Y4] = ode45(df4dt, ...);
[T5,Y5] = ode45(df5dt, ...);
plot(T1,Y1);
plot(T2,Y2);
plot(T3,Y3);
plot(T4,Y4);
plot(T5,Y5);
hold off;
title('\gamma  = 0.1')
</code>```

答:
I have never answered a MATLAB question here. Please ask such type of questions somewhere else.

```<code>clear
stp=0.001;
fi=-pi:stp:pi;
a=1;
for s=1:length(fi)
    x=fi(s);
    if x<-pi+a
        f(s)=-(x/a+pi/a);
    elseif x<-a
        f(s)=-1;
    elseif x<a
        f(s)=x/a;
    elseif x<pi-a
        f(s)=1;
    else
        f(s)=-x/a+pi/a;
    end
end
plot(fi,f,'linewidth',1.5)
xlabel('$\phi$','interpreter','latex')
ylabel('$F(\phi)$','interpreter','latex')
title('$a=1$','interpreter','latex')
set(gca,'fontsize',14)
axis([-pi-.5 pi+.5 -1.5 1.5])
grid on
</code>```

PS: your graph shows a positive slope around zero, but you function doesn't.

Body:
I have never answered a MATLAB question here. Please ask such type of questions somewhere else.

clear
stp=0.001;
fi=-pi:stp:pi;
a=1;
for s=1:length(fi)
    x=fi(s);
    if x<-pi+a
        f(s)=-(x/a+pi/a);
    elseif x<-a
        f(s)=-1;
    elseif x<a
        f(s)=x/a;
    elseif x<pi-a
        f(s)=1;
    else
        f(s)=-x/a+pi/a;
    end
end
plot(fi,f,'linewidth',1.5)
xlabel('$\phi$','interpreter','latex')
ylabel('$F(\phi)$','interpreter','latex')
title('$a=1$','interpreter','latex')
set(gca,'fontsize',14)
axis([-pi-.5 pi+.5 -1.5 1.5])
grid on

PS: your graph shows a positive slope around zero, but you function doesn't.


问:
How to find coordinates of tangent point on circle, given center coordinates, radius, and end point of tangent line
I'm working on a hobby programming project and my mathematic thinking is feeling a bit slow at the moment. Basically, in situations similar to the image I made, given the information I mentioned in the title, I need to be able to calculate the coordinates of the tangent point. (Dimensions are from the origin.) Could you all help me out with finding a straightforward way of doing so? Thanks in advance.

(P.S. In the program, this will actually be in 3D, but once I have the 2D solution I should be able to figure it out in 3D.)

答:
The following includes all.

$$ \text{ Power=} T^2, \; C^2 = T^2 +R^2 =  h^2+k^2$$

$$ \tan \gamma =\dfrac{R}{T};\tan \beta =\dfrac{k}{h} $$

$$  \text{Tangent points  } [\;T  \cos (\beta \pm \gamma), T \sin (\beta \pm \gamma)\;] $$

$$(h,k)= (-30, -70).$$

Without Trigonometry  the formulas for tangent point coordinates plug-in and calculations are:

$$\boxed{ \dfrac{hT\mp kR}{C},\dfrac{kT\pm hR }{C}}$$


问:
Proving manifold dimension theory using homology
The text I am using stated that any manifold with different dimension cannot be homeomorphic, but did not prove it (only proved the case when $n = 2$) until now, and it is given as an exercise in the homology chapter, but I am not sure how to approach this problem. Can anyone provide a solution or hint?

答: Just check that being locally homeomorphic to $\mathbb{R}^n$ is a property that is invariant under homeomorphism; the conclusion follows from the fact that $\mathbb{R}^n$ and $\mathbb{R}^m$ are homeomorphic if and only if $m=n$ (and to prove that you need homology).


答: By structure theorem for finite abelian groups, we have $$G\cong(\Bbb Z_2)^n$$ for some $n$. Consider the function on $(\Bbb Z_2)^n$ that swaps the $1$st and $n$th coordinates. Show that that is an automorphism.</p>

Title: Let $G$ be a finite group with $|G|>2$. Prove that ${\rm Aut}(G)$ contains at least two elements.


问:
Oriented manifolds and the intermediate value theorem
Working through a proof that any connected orientable manifold has two orientations (From Tu's manifolds), I'm having a difficulty with a small nuance in the proof. Consider a manifold $ M $ with two orientations $ \mu $ and $ \nu $, and some arbitrary point $ p \in M $. $ \mu_p $ and $ \nu_p $ are the same or opposite orientations of $ T_p M $. The strategy of the proof is to show that the function $ f:M \rightarrow \{1,-1\} $ (where $ f(p) = 1 $ if $ \mu_p = \nu_p $ and $ f(p) = -1 $ if $ \mu_p = - \nu_p $) is locally constant, and therefore globally constant since $ M $ is connected.

Okay, all well and good. Now, the orientations are continuous, so for any point there is a connected open neighborhood of $ p $, $ U $, where $ \mu = [(X_1,...,X_n)] $ and $ \nu = [(Y_1,...,Y_n)] $ are equivalence classes of continuous frames on $ U $. Of course, there is a transition matrix between these frames, $ A:U \rightarrow GL(n,\mathbb{R}) $ with continuous entries, meaning the nonsingular $ \text{det} A:U \rightarrow \mathbb{R}^\times $ is continuous.

Here's the kicker, though: Tu says that because of the Intermediate Value Theorem, $ (\text{det} A)(p) $ is always positive or always negative on $ U $, being unable to change signs without smoothly crossing over zero---thus showing that $ f $ is constant on U. However, I have no idea what the intermediate value theorem means when not talking about functions from some closed interval of $ \mathbb{R} $ to $ \mathbb{R} $. Naturally, I'd say $ \text{det} A $ being continuous means that $ \text{det} A \circ \phi^{-1}:\phi(U) \rightarrow \mathbb{R} $ is continuous for a chart $ (U,\phi) $. But, how is the intermediate value theorem extended to a function $ \mathbb{R}^n \rightarrow \mathbb{R} $? My intuition is that, like in one dimension, the connectedness of $ U $ is key to an analogous theorem, but can a generalization of the IVT be made without introducing notions of curves along $ U $ and other differential geometry machinery into the mix? Can I apply the 1-D IVT to all possible curves between two points in $ U $ or something? I get the intuitive sense of what he means by it, but how can I put it on more mathematical footing?

答: The determinant is a continuous function from a connected open set to $\mathbb{R} -\{0\}$, so its image is a connected set.


问:
If P is the total of points on the plane. Find all functions $f:P\rightarrow P$, which fullfil some conditions.
If P is the total of points on the plane. Find all functions $f:P\rightarrow P$ such that for every two points $A, B$ of $P$, the points $A, B, f(A), f(B)$ are either collinear or con-cyclic.

I attempted to do it in the following way:

I state $S=(A:f(a)\ne A)$. Then for every three different points $X,Y,Z$ we will write $C(X,Y,Z)$ for the one radius circle or the line which passes through these points.

Afterwards I started taking cases like:

If $S$ is empty, then $f$ is an identity function.

If $|S|=1$, then $f$ is an identity function, in all places apart from one point.

I did not know how to continue it from here on, but I believe that this method is extremely tedious. Could you please explain to me an intuitive solution?

答:
Let $A,B$ be points and $T$ a set of points with $A\notin T$. Define
$$f(x)=\begin{cases}B&x=A\\A&x\in T\\x&\text{otherwise}\end{cases}$$
Then $f$ has the desired property because four pints $x,y,f(x),f(y)$ are actually just three points.

Or pick any circle-or-line (or CoL for short) and define $f$ as any map of this CoL to itself while being the identity outside that CoL. Then such $f$ also has this property.

Assume there is $f$ not matching the above. Then we can find two non-fixpoints $A,B$  points such that $A,B,f(A),f(B)$ are four distinct points and are on a unique CoL. We also know that there is another non- fixpoint $C$ not on That COL. Then the CoL through $A,C,f(A)$ and the CoL through $B,C,f(B)$ intersect exactly in $C$ and $f(C)$.

This is not complete yet, but I suspect that one cannot have many non- fuxpoints in such constellations.


问:
Polynomial equation root
I need to show whether:

$$\frac{1}{x}\times A - \left(\frac{1}{(x+b)(x-b)}-\frac{1}{(x+c)(c-x)}\right)$$

has a single root between $b$ and $c$, where $c>b>0$ and $A>0$. My numerical plots suggest so, but not 100% sure.

答:
The argumentation in you're answer seems to be correct, although I haven't calculated it all.

You should add the possibility, that P only "touches" the x-axis in $x^*$. Consider for example the square function, where $f'(0)=0$. That means, P hasn't to cross the root. But I think, that's easily solved by adding some $\leq$.

Title: Polynomial equation root


问:
Multivariate normal distribution density function
I was just reading the wikipedia article about Multivariate normal distribution: http://en.wikipedia.org/wiki/Multivariate_normal_distribution

I use a little bit different notation. If $X_1,\ldots,X_n$ are independent $\mathcal{N}(0,1)$ random variables, $X=(X_1,\ldots,X_n)$ and $m=(m_1, \ldots, m_n)$ are $n$-dimensional vectors and $B$ is a $n\times n$ matrix, then $Y=m+BX$ has density function

$$f_Y(x)=\frac{1}{\sqrt{(2\pi)^n|\boldsymbol\Sigma|}}
\exp\left(-\frac{1}{2}({x}-{m})^T{\boldsymbol\Sigma}^{-1}({x}-{m})
\right)$$

where $\Sigma=BB^T$

Question: In the wikipedia article its part of the definition that $\Sigma$ is the covariance matric from $Y$, i.e $\sum_{ij}=\operatorname{Cov}(Y_i,Y_j)$, but why? How can this be proved.

答:
By the definition, any covariance matrix $\Sigma$ consists of $\Sigma_{i,j} = Cov(X_i, X_j) = E[(X_i - \mu_i)(X_j - \mu_j)]$, where randoms $X_i$ and $X_j$ have expectations $E[X_i] = \mu_i$ and $E[X_j] = \mu_j$ accordingly. So your problem is to prove $Cov(m_i + \mathbf{b}_i \mathbf{X}, m_j + \mathbf{b}_j \mathbf{X}) = (B \cdot \Sigma \cdot B^\top)_{i,j}$, where $B$ is non-singular matrix with rows $\mathbf{b}_i = (b_{i,1}, \ldots, b_{i,n})$, $1 \le i \le n$ and $\mathbf{X} = (X_1, \ldots, X_n)$ is $n$-dimensional random, $(m_1, \ldots, m_n)$ is vector of scalars.

Let $\mathbf{X}$ has the covariance $n \times n$-matrix $\Sigma$, which is symmetric.

Note, $E[m_i + \mathbf{b}_i \mathbf{X}] = m_i + \mathbf{b}_i \mathbf{\mu}$, where $\mathbf{\mu} = (\mu_1, \ldots, \mu_n)$. Expectation operator $E$ is linear. So, the proof:

$Cov(m_i + \mathbf{b}_i \mathbf{X}, m_j + \mathbf{b}_j \mathbf{X}) = E[(\mathbf{b}_i (\mathbf{X} - \mathbf{\mu}))(\mathbf{b}_j (\mathbf{X} - \mathbf{\mu}))] = E[\sum_{t=1}^n \sum_{p=1}^n b_{i,t} b_{j,p} (X_t - \mu_t)(X_p - \mu_p)] = \sum_{t=1}^n \sum_{p=1}^n b_{i,t} b_{j,p} E[(X_t - \mu_t)(X_p - \mu_p)] = \sum_{t=1}^n b_{i,t} \left(\sum_{p=1}^n b_{j,p} \Sigma_{p,t}\right) = \sum_{t=1}^n b_{i,t} \left( B \cdot \Sigma \right)_{j,t} = \left(B \cdot \Sigma \cdot B^\top\right)_{i,j}$

So the covariance matrix of $\mathbf{m} + B \cdot \mathbf{X}$ is $B \cdot \Sigma \cdot B^\top$, where $Cov(\mathbf{X}) = \Sigma$.


问:
Find the area under curve using the given data

Consider the curve $y=f(x)$ which satisfies the DE $(1+x^2)\frac{dy}{dx} +2xy=4x^2$ and passes through the origin. Find area enclosed by $f^{-1}x$, x axis, and $x=2/3$

After some calculation which I don’t think is necessary to show here, I got
$$y=\frac{4x^3}{3(1+x^2)}$$

Now the inverse for this function can’t be found directly, and I don’t know how else to do it. Can I get a hint?

答:
Since you received very good and simple answers, let me be stupid and work with the inverse function.

$$y=\frac{4x^3}{3(1+x^2)}\implies 4 x^3-3  y\,x^2-3 y=0$$ Using the hyperbolic method for solving cubic equations when only one real root,
$$x=\frac{1}{4} y \left(1+2 \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(1+\frac{24}{y^2}\right)\right)\right)$$ which does not show any antiderivative !


问:
Solve the linear congruences
To solve $\,26x\equiv 6\pmod{\!110},\,$ note $\,\gcd(26,110)=2\,$ so there are $\,2\,$ solutions.

Next, $\ 104x\equiv 24\pmod{\!110}\ $ follows by scaling the above by $\,4.$

i.e. $\,\ \ \ \ \ {-}6x\equiv 24\pmod{\!110}$

Hence $\ \ \ \ \,x\equiv -4\pmod{\!55}\ $ by dividing prior by $\,6$.

What is the final answer?

答: You could first solve $13x\equiv 3\mod 110$ by the extended Euclidean algorithm which is possible since $gcd(13,110)=1$ and then multiply the congruence by $2$ giving a solution of the original congruence.

Title: Solve the linear congruences


问:
Embedding of Tree
Q.

Proof for every Tree can be embedded into the plane.

Conditions.

We cannot use Euler Formula for Planar Graphs. We can use definition of tree, $V-E=1$, no-cycles, every edge is critical, there is a vertex of frequency at most one.

Attempt.
We prove this by induction. True for $V$=1. Trivial. Consider true for $V<n$. Consider a tree $T$ with $n$ edges. Now, we know it has a vertex with at most one edge incident in it. Consider $T$\ $\{v,l\}$ the last vertex and edge. This can be embedded by inductive hypothesis. For the $v$ and $l$, that is left, it can be easily embedded.

The last part seems sloppy. I would appreciate other proofs or improvement on this proof.

答:
You may prefer to chose an embedding where all edges are mapped to straight line segments.

As you did, let $v$ be a vertex of degree $1$, let $w$ be its only neighbour.
As you are not allowed to use very much, I assume you already know/can use that $v\ne w$ and removing vertex $v$ and edge $vw$ from $T$ does indeed produce a tree $T'$.
By induction hypothesis, there is an embedding $f\colon T\to\mathbb R^2$ of $T'$ where all edges are mapped to straight line segments.
Especially, the image of any edge $xy$ where $w\ne\{x,y\}$ is compact, hence has positive distance to $f(w)$.
Since there are only finitely many such edges, for some $r>0$ the open ball $B_r(f(w))$ interscts only the finitely many straight line segments $f(wx)$ for edges incident with $w$.
Select $f(v)$ in $B_r(f(w))\setminus\bigcup_{x\text{ neigbour of } w}f(wx)$ and map $vw$ to the straight line segment from $f(v)$ to $f(w)$.

Title: Embedding of Tree


问:
Is this braced heptagon a rigid graph?
In Mathematica, GraphData[{"UnitDistance", {21, 2}}]

Is this 42-edge graph rigid?  It has chromatic number 4. If it was floppy, that might make it an interesting tool in high-chromatic graph studies.

The current record for heptagon bracing is 59, so this would beat that.

答:
The graph is rigid. In fact I will show that the following $35$-edge graph formed by deleting the two interior vertices of one triangle is (minimally) rigid:

We first need to find a parametrisation of the vertices of the graph with as few parameters as possible (that relate to metrics like lengths and angles); three parameters suffice. Fix $A$ and $B$ in the diagram below, set as parameters the angles $\alpha_1,\alpha_2,\alpha_3$ that the vectors $\vec{A0},\vec{B1},\vec{B2}$ make with $\vec{AB}$, then construct each of the remaining numbered points in order as the point that is a distance $1$ away from two already-determined points. (See the program at the end of this answer for the exact construction sequence.)

Define the function $f:\mathbb R^3\to\mathbb R^3$ as
$$f(\alpha_1,\alpha_2,\alpha_3)=(d(11,12),d(13,14),d(15,2))$$
i.e. it returns the lengths of the three thick edges, which are not created as part of the construction but are in the graph. Let the parameters corresponding to the regular braced heptagon be $\alpha_1^*,\alpha_2^*,\alpha_3^*$; clearly $f(\alpha_1^*,\alpha_2^*,\alpha_3^*)=(1,1,1)$. To check the rigidity of the ($35$-edge) graph, we construct the Jacobian matrix $J$ of $f$ at $(\alpha_1^*,\alpha_2^*,\alpha_3^*)$; if it has full rank we can immediately conclude that it is (infinitesimally or first-order) rigid.

It turns out that $J$ is a $3×3$ matrix with rank $2$. This does not mean that the graph is not rigid; we still have to analyse the behaviour of $f$ along the infinitesimal motion associated with this rank-deficient $J$. Such a motion is given by an element of the nullspace of $J$; suppose this element is $(\delta_1,\delta_2,\delta_3)$. Define
$$g(t)=f(\alpha_1^*+t\delta_1,\alpha_2^*+t\delta_2,\alpha_3^*+t\delta_3)$$
When we plot $g$ for $t$ around zero we get the following graph (blue, red and green lines represent the three coordinates in order):

The lengths of all three thick edges are individually on the same side of $1$. Together with the result from the Jacobian matrix, we infer that said lengths are not $1$ in a punctured neighbourhood of $(\alpha_1^*,\alpha_2^*,\alpha_3^*)$, so the $35$-edge graph is (second-order) rigid.

As a final note, it can be easily shown that the $35$-edge graph is a Laman graph by means of a Henneberg construction, so no edges can be removed without making the graph flexible.

```<code>#!/usr/bin/env python3
from mpmath import *
mp.dps = 100

def cu(z1, z2):
    # Constructs the point at a distance of 1 from z1 and z2,
    # left of the line from z1 to z2.
    m = (z1 + z2) / 2
    d, theta = polar(z2 - z1)
    return m + sqrt(1 - d * d / 4) * expj(theta + pi / 2)

def f(a1, a2, a3):
    # Progressive construction of the graph's vertices.
    # Returns the distances between three pairs of vertices that are
    # not directly related in the construction but are still linked
    # by an edge.
    A = mpf(0)
    B = mpf(1)
    p0 = expj(a1)
    p1 = 1+expj(a2)
    p2 = 1+expj(a3)
    p3 = cu(p1, p0)
    p4 = cu(B, p1)
    p5 = cu(p0, p3)
    p6 = cu(p2, p5)
    p7 = cu(p4, p6)
    p8 = cu(p6, p2)
    p9 = cu(p8, A)
    p10 = cu(A, p9)
    p11 = cu(p0, p7)
    p12 = cu(p9, p7)
    p13 = cu(p12, p1)
    p14 = cu(p5, p10)
    p15 = cu(p13, p14)
    return abs(p11-p12), abs(p13-p14), abs(p15-p2)

# Known parameters at the intended solution. a2 is exact, 1.43262130059...
a1 = 5*pi/7
a2 = phase(polyroots([7, 28, 70, 133, 203, 259, 281, 259, 203, 133, 70, 28, 7])[10])
a3 = 2*pi/7

# Jacobian matrix of f
def fderiv(i, j):
    ff = lambda x, y, z: f(x,y,z)[i]
    dvec = tuple(int(n == j) for n in range(3))
    return diff(ff, (a1,a2,a3), dvec)
J = matrix([[fderiv(i, j) for j in range(3)] for i in range(3)])

# J is rank-deficient at the reference solution,
# i.e. the framework is not first-order (infinitesimally) rigid.
# Not to worry. Analyse f's behaviour along the one infinitesimal motion...
_, Sigma, V = svd(J)
nprint(chop(Sigma))
da1, da2, da3 = V[2,:]
def g(z):
    return f(a1+z*da1, a2+z*da2, a3+z*da3)

# The lengths of the three "test rods" are individually on the same side of 1
# for sufficiently small multiples of the infinitesimal motion.
# This shows that the graph is (second-order) rigid.
plot([lambda z: g(z)[0], lambda z: g(z)[1], lambda z: g(z)[2]], [-0.1, 0.1])
</code>```

The graphs above were drawn with the help of a script in my Dounreay repository. Run ./drawgraph.py heptagon in the folder containing drawgraph.py to reproduce the full $42$-edge graph; the above pictures are the result of tweaking in Inkscape.

Removing two consecutive vertices of one of the inner stars leads to a first-order rigid graph in either case (the determinant of $J$ is non-zero):

```<code>A = mpf(0)
B = mpf(1)
p0 = expj(a1)
p1 = 1+expj(a2)
p2 = 1+expj(a3)
p3 = cu(p1, p0)
p4 = cu(B, p1)
p5 = cu(p0, p3)
p6 = cu(p2, p5)
p7 = cu(p4, p6)
p8 = cu(p6, p2)
p9 = cu(p8, A)
p10 = cu(p3, p8)
p11 = cu(p9, p7)
p12 = cu(p11, p1)
p13 = cu(p12, p5)
p14 = cu(p0, p7)
p15 = cu(p13, p2)
p16 = cu(p14, p10)
return abs(p11-p14), abs(p12-p15), abs(p15-p16)
</code>```

```<code>A = mpf(0)
B = mpf(1)
p0 = expj(a1)
p1 = 1+expj(a2)
p2 = 1+expj(a3)
p3 = cu(p1, p0)
p4 = cu(B, p1)
p5 = cu(p0, p3)
p6 = cu(p2, p5)
p7 = cu(p4, p6)
p8 = cu(p6, p2)
p9 = cu(p8, A)
p10 = cu(A, p9)
p11 = cu(p0, p7)
p12 = cu(p3, p8)
p13 = cu(p10, p4)
p14 = cu(p5, p10)
p15 = cu(p14, p2)
p16 = cu(p11, p12)
return abs(p12-p13), abs(p13-p16), abs(p16-p15)
</code>```

Title: Is this braced heptagon a rigid graph?

Body:
The graph is rigid. In fact I will show that the following $35$-edge graph formed by deleting the two interior vertices of one triangle is (minimally) rigid:

We first need to find a parametrisation of the vertices of the graph with as few parameters as possible (that relate to metrics like lengths and angles); three parameters suffice. Fix $A$ and $B$ in the diagram below, set as parameters the angles $\alpha_1,\alpha_2,\alpha_3$ that the vectors $\vec{A0},\vec{B1},\vec{B2}$ make with $\vec{AB}$, then construct each of the remaining numbered points in order as the point that is a distance $1$ away from two already-determined points. (See the program at the end of this answer for the exact construction sequence.)

Define the function $f:\mathbb R^3\to\mathbb R^3$ as
$$f(\alpha_1,\alpha_2,\alpha_3)=(d(11,12),d(13,14),d(15,2))$$
i.e. it returns the lengths of the three thick edges, which are not created as part of the construction but are in the graph. Let the parameters corresponding to the regular braced heptagon be $\alpha_1^*,\alpha_2^*,\alpha_3^*$; clearly $f(\alpha_1^*,\alpha_2^*,\alpha_3^*)=(1,1,1)$. To check the rigidity of the ($35$-edge) graph, we construct the Jacobian matrix $J$ of $f$ at $(\alpha_1^*,\alpha_2^*,\alpha_3^*)$; if it has full rank we can immediately conclude that it is (infinitesimally or first-order) rigid.

It turns out that $J$ is a $3×3$ matrix with rank $2$. This does not mean that the graph is not rigid; we still have to analyse the behaviour of $f$ along the infinitesimal motion associated with this rank-deficient $J$. Such a motion is given by an element of the nullspace of $J$; suppose this element is $(\delta_1,\delta_2,\delta_3)$. Define
$$g(t)=f(\alpha_1^*+t\delta_1,\alpha_2^*+t\delta_2,\alpha_3^*+t\delta_3)$$
When we plot $g$ for $t$ around zero we get the following graph (blue, red and green lines represent the three coordinates in order):

The lengths of all three thick edges are individually on the same side of $1$. Together with the result from the Jacobian matrix, we infer that said lengths are not $1$ in a punctured neighbourhood of $(\alpha_1^*,\alpha_2^*,\alpha_3^*)$, so the $35$-edge graph is (second-order) rigid.

As a final note, it can be easily shown that the $35$-edge graph is a Laman graph by means of a Henneberg construction, so no edges can be removed without making the graph flexible.

#!/usr/bin/env python3
from mpmath import *
mp.dps = 100

def cu(z1, z2):
    # Constructs the point at a distance of 1 from z1 and z2,
    # left of the line from z1 to z2.
    m = (z1 + z2) / 2
    d, theta = polar(z2 - z1)
    return m + sqrt(1 - d * d / 4) * expj(theta + pi / 2)

def f(a1, a2, a3):
    # Progressive construction of the graph's vertices.
    # Returns the distances between three pairs of vertices that are
    # not directly related in the construction but are still linked
    # by an edge.
    A = mpf(0)
    B = mpf(1)
    p0 = expj(a1)
    p1 = 1+expj(a2)
    p2 = 1+expj(a3)
    p3 = cu(p1, p0)
    p4 = cu(B, p1)
    p5 = cu(p0, p3)
    p6 = cu(p2, p5)
    p7 = cu(p4, p6)
    p8 = cu(p6, p2)
    p9 = cu(p8, A)
    p10 = cu(A, p9)
    p11 = cu(p0, p7)
    p12 = cu(p9, p7)
    p13 = cu(p12, p1)
    p14 = cu(p5, p10)
    p15 = cu(p13, p14)
    return abs(p11-p12), abs(p13-p14), abs(p15-p2)

# Known parameters at the intended solution. a2 is exact, 1.43262130059...
a1 = 5*pi/7
a2 = phase(polyroots([7, 28, 70, 133, 203, 259, 281, 259, 203, 133, 70, 28, 7])[10])
a3 = 2*pi/7

# Jacobian matrix of f
def fderiv(i, j):
    ff = lambda x, y, z: f(x,y,z)[i]
    dvec = tuple(int(n == j) for n in range(3))
    return diff(ff, (a1,a2,a3), dvec)
J = matrix([[fderiv(i, j) for j in range(3)] for i in range(3)])

# J is rank-deficient at the reference solution,
# i.e. the framework is not first-order (infinitesimally) rigid.
# Not to worry. Analyse f's behaviour along the one infinitesimal motion...
_, Sigma, V = svd(J)
nprint(chop(Sigma))
da1, da2, da3 = V[2,:]
def g(z):
    return f(a1+z*da1, a2+z*da2, a3+z*da3)

# The lengths of the three "test rods" are individually on the same side of 1
# for sufficiently small multiples of the infinitesimal motion.
# This shows that the graph is (second-order) rigid.
plot([lambda z: g(z)[0], lambda z: g(z)[1], lambda z: g(z)[2]], [-0.1, 0.1])

The graphs above were drawn with the help of a script in my Dounreay repository. Run ./drawgraph.py heptagon in the folder containing drawgraph.py to reproduce the full $42$-edge graph; the above pictures are the result of tweaking in Inkscape.

Removing two consecutive vertices of one of the inner stars leads to a first-order rigid graph in either case (the determinant of $J$ is non-zero):

A = mpf(0)
B = mpf(1)
p0 = expj(a1)
p1 = 1+expj(a2)
p2 = 1+expj(a3)
p3 = cu(p1, p0)
p4 = cu(B, p1)
p5 = cu(p0, p3)
p6 = cu(p2, p5)
p7 = cu(p4, p6)
p8 = cu(p6, p2)
p9 = cu(p8, A)
p10 = cu(p3, p8)
p11 = cu(p9, p7)
p12 = cu(p11, p1)
p13 = cu(p12, p5)
p14 = cu(p0, p7)
p15 = cu(p13, p2)
p16 = cu(p14, p10)
return abs(p11-p14), abs(p12-p15), abs(p15-p16)

A = mpf(0)
B = mpf(1)
p0 = expj(a1)
p1 = 1+expj(a2)
p2 = 1+expj(a3)
p3 = cu(p1, p0)
p4 = cu(B, p1)
p5 = cu(p0, p3)
p6 = cu(p2, p5)
p7 = cu(p4, p6)
p8 = cu(p6, p2)
p9 = cu(p8, A)
p10 = cu(A, p9)
p11 = cu(p0, p7)
p12 = cu(p3, p8)
p13 = cu(p10, p4)
p14 = cu(p5, p10)
p15 = cu(p14, p2)
p16 = cu(p11, p12)
return abs(p12-p13), abs(p13-p16), abs(p16-p15)


问:
Probability of a random variable defined as the sum of two independent random variables
Let $X,Y,Z$ be independent random variables taking values $0$ and $1$ each with probability $1/2$.
Define the random variable $U$ as
$U =  X$  if $Z=0$ and $U = Y$ if $Z=1$

Let $A$ be another random variable that takes $0$ with probability $p$ and $1$ with probability $1-p$.
Define $C = U+A$.
Find the probabilities $P(C=X), P(C=Y)$ and $P(C=Z)$.

I have done the first two and have a confusion with the third. I feel it should be zero.

答:
You can make a table of outcomes.  Note that I put $Z$ first since its outcome determines whether $U = X$ or $U = Y$.
$$\begin{array}{c|c|c|c|c}
Z & X & Y & C = U & C = U + 1 \\
\hline
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 1 & 0 & 1 & 2 \\
0 & 1 & 1 & 1 & 2 \\
1 & 0 & 0 & 0 & 1 \\
1 & 0 & 1 & 1 & 2 \\
1 & 1 & 0 & 0 & 1 \\
1 & 1 & 1 & 1 & 2 \\
\end{array}$$
The fourth column, $C = U$, is taken with probability $p$.  The fifth column, $C = U + 1$, is taken with probability $1-p$.  Then $\Pr[C = Z]$ is $p$ times the number of rows in which $C = Z$ from the fourth column, plus $1-p$ times the number of rows in which $C = Z$ from the fifth column, divided by $8$, since each triplet $(Z, X, Y)$ has equal probability $1/8$ of occurring.  Since there are $4$ rows in the fourth column that match, and $2$ rows in the fifth column that match, the desired probability is $$\Pr[C = Z] = \frac{4p + 2(1-p)}{8} = \frac{p+1}{4}.$$


问:
Formal power series with coefficient in $\{0,1 \}$ which is not rational
Recall that $f \in \mathbb{C}[[x]]$ is rational if $f = ab^{-1}$ where $a, b \in \mathbb{Q}[x]$ (the inclusion  $ \mathbb{Q}[x] \to \mathbb{C}[[x]]$ is obvious).

How do I obtain an example of  $f \in \mathbb{C}[[x]]$ with coefficients $0,1$ which is not rational?  I have an experience of proving that some series is not rational (say, $exp(x)$).I could't come up with an example myself, though.

答: Since $f = \sum c_nz_n$ has coefficients $0$ and $1$, $f(1/2)$ is the binary number $c_0.c_1c_2c_3 \dotsm$. If $f$ is rational this number is rational and hence ultimately periodic. Thus just choose a non ultimately periodic sequence to get a non-rational series.


问:
Substitute $u=\exp\left(\mathrm{i}t\right)$ in an integral where $t\in\left[-\pi,+\pi\right]$
Consider the following integral.

$$\int_{-\pi}^{\pi}f\left(\exp\left(\mathrm{i}xt\right)\right)\mathrm{d}t$$

Where
$$u=\exp\left(\mathrm{i}t\right),t=\frac{1}{\mathrm{i}}\ln u,\mathrm{d}t=\frac{1}{\mathrm{i}u}\mathrm{d}u$$
makes the integrand
$$\frac{f\left(u^{x}\right)}{\mathrm{i}u}$$

How should we get the new range of integration given $t\in\left[-\pi,+\pi\right]$, and $u=\exp\left(\mathrm{i}t\right),
t=\frac{1}{\mathrm{i}}\ln u$?

TL;DR: How to use u-substitution on an integral using $u=\exp\left(\mathrm{i}t\right)$ where $t\in\left[-\pi,+\pi\right]$?

答:
$
\newcommand{\sint}[1]{ \int \limits_{#1}  }
\newcommand{\soint}[1]{\oint \limits_{#1}  }
\newcommand{\dd}{\mathrm{d}}
$
Naively, the new range of integration will be $u$ (recall, $u$ is a function of $t$, so $u$ is a shorthand for $u(t)$) evaluated at the old bounds. Thus,

$$u(-\pi) = e^{-i \pi} = -1 \qquad u(\pi) = e^{i \pi} = -1$$

But then the integral is obviously zero, but that couldn't be right. The question is -- what went wrong?

The problem is that since we're now working with complex numbers, we need to work with a contour, not just the bounds of integration. So what your original integral is, in the complex analytic sense, over the contour $[-\pi,\pi]$. Then the transformation $u$ provides turns $[-\pi,\pi]$ into a circle. In other words, you get the contour $C$ parameterized by $u(t) = e^{i t}$ for $t \in [-\pi,\pi]$. So, now you have to integrate over the unit circle in the complex plane. We can't naively just focus on the bounds anymore.

(Also, another thing of note: bear in mind that this parameterization ensures we only go around the loop once. This is important in the use of some calculations and techniques.)

Intuitively, you can think of this like your substitution mapping every point of the original interval elsewhere, and then you're integrating over that. Try to think of it in the familiar, one-dimensional case you think of from calculus - your $u$ substitution "stretches" and "squishes" the interval and the function, not just the endpoints. So in a sense, this extends very naturally to how things work in multiple dimensions, and in particular is why we actually are integrating over a circle here. The entire path matters, not just the bounds of integration.

That is, then,

$$\sint{[-\pi,\pi]} f(e^{ixt}) \, \dd t = \frac 1 i \soint{|z|=1} \frac{f(z^x)}{z} \, \dd z$$

Integrals of this form can be handled through techniques such as Cauchy's integral theorem, Cauchy's integral formula, or, the method encompassing both of these and more, the residue theorem. Though I think it ultimately depends on the function $f$ we're working with.


问:
Differences between $L^p$ and $\ell^p$ spaces
Could someone explain some differences between the $L^p$ and $\ell^p$ spaces? Thanks a lot.</p>

答:
Either L or ℓ denotes the first alphabetic letter of Lebesgue. According to the definitions, there are interpretations for both L^p spaces and ℓ^p space. Please note that Frigyes Riesz named L^p spaces with adding the pulral "s" to the word "space". Mr. Riesz proved the L^2 spaces and ℓ^2 space are isomorphic. Of course, ℓ^2 space is the Hilbert space.

1.L^p (function) spaces

L^p spaces are function spaces defined using a natural generalization of the p-norm for finite-dimensional vector spaces. L^p spaces is called Lebesgue spaces, named after Henri Lebesgue.

2.ℓ^p (sequence) space

The above-mentioned p-norm can be extended to vectors that have an infinite number of components (sequences), which yields ℓ^p space. ℓ^p space is the sequence space consisting of the p-power summable sequences with the p-norm. In addition, it has bounded sequences.

However, some mathematicians used both L^p spaces and ℓ^p spaces to mention the Banach space, and both L^2 spaces and ℓ^2 space for the Hilbert space interchangeably (because infinite-dimensional vectors are used in L^p function spaces sometimes even though the usage is not strict in the math). The mathematicians have avoided the subtle difference of the two concepts. It makes people confused sometimes.


Title: Solve the linear congruences

Body:
To solve $\,26x\equiv 6\pmod{\!110},\,$ note $\,\gcd(26,110)=2\,$ so there are $\,2\,$ solutions.

Next, $\ 104x\equiv 24\pmod{\!110}\ $ follows by scaling the above by $\,4.$

i.e. $\,\ \ \ \ \ {-}6x\equiv 24\pmod{\!110}$

Hence $\ \ \ \ \,x\equiv -4\pmod{\!55}\ $ by dividing prior by $\,6$.

What is the final answer?

Body:
$26x \equiv 6 \pmod{110} \iff 110 | (26x - 6) \iff $

$55 | (13x - 3) \iff 13x \equiv 3 \pmod{55}.$

$55 = (4)13 + 3.$

$13 = 4(3) + 1.$

Therefore,

$1 = 13 - 4[55 - 4(13)] = 17(13) - 4(55)$

Therefore, $17 \times 13 \equiv 1 \pmod{55}.$

Therefore, $3 \times (17 \times 13) \equiv 3 \equiv 13 x \pmod{55}.$

Therefore, $x \equiv 51 \pmod{55}.$

Therefore $55 | \{(13[51]) - 3\}$.

Therefore $110 | \{(26[51]) - 6\}$.

Therefore $\{26[51]\} \equiv 6 \pmod{110}.$

Therefore $x \equiv 51 \pmod{110}.$


问:
How to intuitively understand the $n$-dimensional cube as the dimension grows large
So I read* that for the convex body, i.e. the cube $[-1,1]^n$ in $\mathbb{R}^n$, the smallest ball containing it has radius $\sqrt{n}$, while the largest ball inside the cube has radius $1$.

Also,

"...as the dimension grows, the cube resembles a ball less and less."

How do I visualize these things when $n\geq 4$? I just can't see it!

It'd be great if I could get some help with the intuition involved here. Thanks!

*See page 2 of

Keith Ball, "An elementary introduction to modern convex geometry" in Flavors of Geometry, Silvio Levy ed., Cambridge 1997.

Edit: While the suggested answers are very good, I don't think they address the particular geometric structure I am concerned with in my question.

答:
What makes you think that we can visualize higher cubes and spheres? For $n=4$ you might play games like using some sort of time slider to draw the intersection of your object with the $xyz$-hyperplane, but for $n>4$ those kind of hacks will become unavailable very quickly.

The intuition behind facts like the ones you quote, isn’t intuition but computation. In some sense mathematics builds around our intuition for 2, 3 or maybe even 4-dimensional space, by which I mean that most definitions mimick something in this low dimensional worlds. Yet the definitions are vastly more general in the way that the dimension is inessential, so we might as well try to find out what they do in higher dimensions (thinking of manifolds). It is a pity for sure that we can’t see what is happening there, because sure enough things start to break down. Manifolds become unsmoothable or have multiple distinct smooth structures, classification results are impossible to obtain and spheres become pointy and computationally start to look and behave rather alien. To state one example: The Poincare-conjecture was one of the millenium problems (ie was on the same level of difficulty than the Riemann hypothesis or $P$ vs $NP$) and was about $3$-spheres. Higher geometry is hard.

On the other hand this is the whole fun about abstract mathematics. Intuitive definitions derived from a small collection of examples soon enough turn out to have more exotic but interesting instances, which makes the definition even more interesting and worthy of studying.


问:
Definition(s) of rational numbers
The definitions of rational numbers are somewhat confusing for me. The definition of rational numbers on wikipedia and most other sites is:

In mathematics, a rational number is any number that can be expressed as the quotient or fraction $\frac pq$ of two integers, with the denominator $q$ not equal to zero.

All integers can be expressed as the quotient $\frac pq$. So if they are not written in the fraction form, will they still be called rational numbers?

Another definition which I've found is:

A rational number is a quotient $\frac mn$ where $m$ and $n$ are integers and $n \neq 0$.

These definitions might be the same. However, I'm not able to figure out how! Are rational numbers those numbers which can be expressed as the quotient $\frac mn$ of two integers or those which are directly the quotient of two integers? For example, if $n$ is an integer, so is $n$ a rational number? Or is $\frac n1$ a rational number?

答:
The integers are indeed rational, because each integer can be expressed as such a quotient as you describe.

The second definition seems different, at first glance, but it isn't. If a number $x$ can be expressed as such a quotient $\frac{m}{n}$, then it is such a quotient, in the sense that $\frac{m}{n}=x.$ On the other hand, such a quotient can immediately be expressed as such a quotient.

Title: Definition(s) of rational numbers


问:
Noether‘s Theorem and Moment Maps
Noether‘s Theorem says that every continuous symmetry of a physical system (i.e., a Lie group action on phase space ${\bf R}^{2n}$ preserving a Hamiltonian $H$) leads to a conservation law (i.e., a function $I$ satisfying $\left\{I,H\right\}=0$).

In the literature one finds this as an application of the method of moment maps. The general result is that for a group action of a Lie group $G$ on a symplectic manifold $(M,\omega)$ one has the associated moment map $\mu\colon M\to {\mathfrak g}^*$, in particular for fixed $X\in {\mathfrak g}$ a map $\mu^X\colon M\to{\bf R}$, and then $\left\{\mu^X,H\right\}=\omega(X_H,X)$ for the Hamiltonian vector field $X_H$. The right hand side equals $\iota_X dH$, which vanishes because the Lie group action is assumed to preserve $H$.

I do not understand, why the literature makes the assumption that the Lie group preserves the symplectic form and, more seriously, I do not understand why this assumption should be true in the setting of Noether‘s Theorem. That theorem is just about continuous symmetries preserving the given Hamiltonian on ${\bf R}^{2n}$, so why should they preserve the symplectic form $\sum_k dx_kdy_k$?

答:
I invite you to read this book, especially chapter 4 and 5:
Jean-Louis Koszul, Introduction to Symplectic Geometry
https://link.springer.com/book/10.1007%2F978-981-13-3987-5</p>

Title: Noether‘s Theorem and Moment Maps


问:
Zeros of elements of $C\{M_n\}$.
Let $M_0,M_1,\ldots$ be positive constants and $C\{M_n\}$ be the class of all infinitely differentiable functions $f$ on the real axis such that the inequalities
\begin{equation}
|f^{(n)}(x)| \le \beta_fB_f^nM_n.
\end{equation} holds for all $x \in R$ and $n=01,2,\ldots$, where $\beta_f$ and $B_f$ are positive constants. We say $C\{M_n\}$ is quasi-analytic if the condition $f \in C\{M_n\}$ satisfies $f^{(n)}(0)=0$ for all $n=0,1,2,\ldots$ will imply $f(x)=0$ on the real axis. I wonder the existence of a function $f$ of this class which has infinitely many zeros on a bounded interval. Any comments? Thanks!

答:
If $f$ has infinitely many zeroes on a bounded interval $I$, then by Bolzano-Weierstrass theorem the set of zeroes has an accumulation point $x_0\in I$. I claim that $f^{(n)}(x_0) = 0$ for all nonnegative integers $n$. Indeed, if there is a sequence of distinct points $x_n\to x_0$ with $f(x_n)=0$, then by Rolle's theorem we can obtain a sequence $c_n$ of distinct points between the $x_n$'s such that $c_n\to x_0$ as $n\to\infty$ and also $f'(c_n)=0$. Thus $f'(x_0)=0$ by continuity. Inductively replacing $f$ with $f^{(k)}$ in this argument proves the claim.

Since $C\{M_n\}$ is a quasianalytic class, $f^{(n)}(x_0)=0$ for all $n$ implies $f\equiv 0$ identically. Hence the only function in $C\{M_n\}$ with infinitely many zeroes in a bounded interval is the constant $0$ function.

Title: Zeros of elements of $C\{M_n\}$.


问:
What do we mean when we say "Let $x$ be an element of the set $\mathbb{R}$"?

What do we mean when we say "Let $x$ be an element of the set $\mathbb{R}$"?

Does $x$ represents only a single  element of set $\mathbb{R}$? or does $x$ represent all the element of set $\mathbb{R}$ simultaneously at the same time?

People say that if $x\in\mathbb{R}$ then $x$ is any real number; that means $x$ represents all real numbers. But if $x$ is any real number then let's say $x=1$; so $x$ is one, then how it can represent all real numbers?

Please help me I am very confused.

答:
Suppose I say : if x // y and y // z then x // z.

Is this sentence meaningful? In order this sentence to make sense, I should first say : Let x, y  and z be straight lines in a given plane D.

So, saying " let x belong to R" is a way to state the domain in which a sentence will have meaning, and, therefore, will have a truth value . ( Maybe, allowing x to be some non-real number would yield a meaningless sentence.)

Second reason: it might happen that a sentence has  meaning in some large domain D, but is false for some values of this domain. In case your goal is to establish a universally true sentence, you restrict the possible values of x to a subset D* of D.

Third reason: sometimes , you use " Let x belong to some domain D" as an hypothesis in a conditional proof.

Let x belong to R  $\space \space $ (Hypothesis for conditional proof).

x² = 4

$\sqrt{x²} = \sqrt 4$

$|x| = 2$

$x = 2$ OR $x = -2$

$x² =4 \rightarrow (x= 2$ OR $x = -2)$

x belongs to R $\rightarrow [x² =4 \rightarrow (x= 2$ OR $x = -2) ] $

For all $x_{\in R} \space ,  \space x² =4 \rightarrow (x= 2$ OR $x = -2)$

Note : the conclusion ( for all x belonging to R ... ) is allowed on the ground that, in the hypothesis, x was arbitrary. If the concluson holds for any number in R , it also holds for all numbers in R.