问:
Calculation on the Power of Means (Probability)
The problem is:

Suppose $X\ge 0$ is a random variable, $p(x)$ is its probability density function. If $\operatorname EX$ exists, $r>0$, prove:

$$\operatorname EX^r = \int_0^\infty rx^{r-1}P(X>x) \, dx.$$

I see $\operatorname E X^r$ as $\int_{-\infty}^\infty x^r p(x) \, dx$, while $$\text{RHS}=\int_0^\infty \left(\int_x^\infty p(t)\,dt \right) dx^r = \left.\left(x^r\int_x^\infty p(t) \, dt\right)\right|_0^\infty-\int_0^\infty x^r p(x) \, dx.$$

However, when calculating $x^r \int_x^\infty p(t) \, dt$ as $x\to \infty$, it comes to a $0 \times \infty$ problem. How can I solve it?

Thanks.

答:
\begin{align}
& \int_0^\infty rx^{r-1}P(X>x) \, dx \\[8pt]
= {} & \int_0^\infty rx^{r-1} \left( \int_x^\infty p(w)\,dw \right) \, dx \\[8pt]
= {} & \iint\limits_{(x,w)\,:\, 0\,<\,x\,<\,w} rx^{r-1} p(w) \, d(x,w) \\[8pt]
= {} & \int_0^\infty \left( p(w) \int_0^w rx^{r-1} \, dx \right) \, dw \\[8pt]
= {} & \int_0^\infty w^r p(w)  \, dw
\end{align}
You said $X\ge0,$ so this last integral is only from $0$ to $+\infty,$ not from $-\infty$ to $+\infty.$


答:
$$E(X^r)=\int_0^{\infty} r x^{r-1} [1-F_X(x)]dx$$

by parts you get

$$E(X^r)=\Big[x^r(1-F_X(x))\Big]_0^{\infty}+\int_0^{\infty} x^r f_X(x)\,dx$$

$$x^r\left[1-F_X(x)\right]_{x=\infty}=\frac{x^r}{\frac{1}{1-F}}=\frac{\infty}{\infty}\rightarrow 0$$

Using the l'hopital $r$ times

Thus you are all set

ContentLicense: CC BY-SA 4.0

Title: Calculation on the Power of Means (Probability)

Body:
The problem is:

Suppose $X\ge 0$ is a random variable, $p(x)$ is its probability density function. If $\operatorname EX$ exists, $r>0$, prove:

$$\operatorname EX^r = \int_0^\infty rx^{r-1}P(X>x) \, dx.$$

I see $\operatorname E X^r$ as $\int_{-\infty}^\infty x^r p(x) \, dx$, while $$\text{RHS}=\int_0^\infty \left(\int_x^\infty p(t)\,dt \right) dx^r = \left.\left(x^r\int_x^\infty p(t) \, dt\right)\right|_0^\infty-\int_0^\infty x^r p(x) \, dx.$$

However, when calculating $x^r \int_x^\infty p(t) \, dt$ as $x\to \infty$, it comes to a $0 \times \infty$ problem. How can I solve it?

Thanks.


问:
Transcendental equation solutions
I'm working on solving a 3d particle in a well for quantum mechanics.

I have the transcendental equation
$$ - \cot(z) = \sqrt{(z_0/z)^2-1} $$

A graph of this looks like

```<code>Plot[{Sqrt[(5*\[Pi]/z)^2 - 1], -Cot[z]}, {z, 0, 5*\[Pi]}]
</code>```

A solution manual for Griffiths QM book states that the solutions ONLY occur at multiples of pi, and supplies the following graph

why are those other non-intervals of pi intersections not considered?

答:
Consider that you are looking for the zero's of function
$$f(z)=\sqrt{\frac{a^2}{z^2}-1}+\cot (z)$$ You cannot have any solution if $z \gt a$.

Concerning the possible roots, using Taylor expansion around $z=k \pi$ and series reversion, they are given by
$$z_{(k)}=k \pi \left(1-\frac{1}{\left(a^2-\pi ^2 k^2\right)^{1/2}}+\frac{\pi ^2 k^2}{2 \left(a^2-\pi ^2 k^2\right)^{3/2}}+\frac{a^2}{\left(a^2-\pi ^2 k^2\right)^2} +\cdots\right)$$

Title: Transcendental equation solutions


问:
From range [0,1] to range [-117,117]
Do you know the formula to get from a range of values [0,1] to a range of values [-117,117]?

For example if I had the 0.1 value, I'd like to know the corrisponding of the other range.

Thanks.

答:
The initial range has length $1$, the final one length $334$. Using this observation, dilate and shift:
$$x\mapsto 334x-117.$$

Title: From range [0,1] to range [-117,117]


问:
A geometry problem with the reflection of the incenter
Consider $\triangle ABC$ and its incenter $I$. Let $M$ be the midpoint of arc $BC$ (not containing $A$) and $P$ be the second intersection point of the circumcircle of $\triangle ABC$ and the circle with center $I$ and radius $IA$. Prove that if $I'$ is the reflection of $I$ with respect to $BC$, then $I', P, M$ are collinear (lie on the same line).

Please help, I am lost, I don't have any idea! Thanks a lot in advance! Anyway, this problem is based on a problem connected to the reflection of the incenter, which appeared in Mathematical Excalibur.

For help, here is a figure:

答:
After a suggestion by @brainjam I found an offical solution:

 This was posted initial and it is not a solution, just partial result if someone find it usefull.
Let $PM$ meet smaller circle at $D$. Then $AD\bot BC$.

Proof: It is easy to see that angle between altitude from $A$ and angle bisector from is $|\beta-\gamma|$.
 Now we see that $\angle MBC = \angle MAC = \alpha $ so $$\angle DPA = \angle MPA = \angle MBA = \alpha+2\beta$$ From here we get (notice: $\alpha +\beta+\gamma = 90^{\circ}$ and that $ADI$ is isosceles) $$\angle DJA = 360^{\circ}-2(\alpha+2\beta) \implies \angle  DAI = \beta - \gamma$$ and we are done.
Let perpendicular from $I$ to $BC$ meet $PM$ at $J$. If we prove that $BC$ bisect $IJ$ we are done, i.e. $J=I'$. But no idea how to finish.


问:
Existence and Uniqueness of First-Order ODE System
Consider the following ODE-system
\begin{align}
\dot{x}(t)&=f(x(t),z(t))\\
\dot{y}(t)&=g(x(t),y(t),z(t)) \\
\dot{z}(t)&=h((x(t),y(t),z(t)))
\end{align}
with the following initial/terminal conditions: $\lim\limits_{t\rightarrow \infty} x(t) = x_\infty, y(0)=y_0, z(0)=0$. Moreover, it is known that $\lim_{t\rightarrow \infty}y(t)=0$ and $\lim_{t\rightarrow \infty} z(t)=z_\infty>0$. The solution of the system must therefore converge to a point $(x_\infty,0,z_\infty)$, where $x(\infty)$ is known but $z(\infty)$ is not. Assume that $f,g$, and $h$ are Lipschitz continuous.

Under which conditions does this system have a unique solution? I understand that if I had an initial condition for $x(0)$, standard theorems like Picard-Lindeloef would apply. However, without this initial condition but a boundary condition at infinity I do not know whether there exist theorems to apply.

答:
I'll assume that $f,g,h$ are Lipschitz continuous so you have local existence and uniqueness.

In general it is unlikely that there are any solutions to your problem, but there are some
cases where this will happen.  Since you don't say what $y$ and $z$ are supposed to be doing as $t \to \infty$ while $x \to x_\infty$, there are many scenarios, but the simplest is that there is an isolated fixed point $P = (x_\infty, y_\infty, z_\infty)$ that is the limit of
your solution as $t \to \infty$.  The next question is: what is the dimension of the stable manifold for that fixed point?

If that dimension is $0$, the only way to approach it is to start there, i.e. you must have $y_0 = y_\infty$ and $0 = z_\infty$, and the unique solution is the constant solution $x = x_\infty, y=y_\infty, z = z_\infty$.

If the dimension is $1$, there are two trajectories approaching $P$ from opposite directions, and your problem has a unique solution if one of these happens to intersect the line
$y=y_\infty, z= 0$ in exactly one point and the other does not intersect it.

If the dimension is $2$, i.e. the stable manifold is a surface, you want that surface to intersect the line $y=y_\infty, z=0$ in exactly one point.  This may be the most likely
scenario.

If the dimension is $3$, you won't have a unique solution: the set of initial conditions that end up approaching the fixed point is an open set, and if it intersects your line
$y=y_\infty, z=0$ it does so in an open interval.


问:
Finding Grad of G,
OK, so I'm out of ideas. The question is:
Let  f(x,y) be a function with continous partial derivatives that upholds:

$$
\bigtriangledown f (0,-8) = (-3\widehat{i}, 5\widehat{j})
$$
let g(x,y) be the function :
$$
g(x,y) = f(xy+x^2 , xy-y^2)
$$

Calculate: (a Vector)
$$
\bigtriangledown g(-2,2) = ?
$$

My take on this:
$$
\bigtriangledown g = \left(\frac{\partial \:g}{\partial \:x},\:\frac{\partial \:g}{\partial \:y}\right)
$$
let u,v be:
$$
u\left(x,y\right)\:=\:xy+x^2\:\:,\:v\left(x,y\right)\:=\:xy-y^{2\:}
$$

and then By the chain Rule:
$$
\frac{\partial g}{\partial x}\:=\:\frac{\partial g}{\partial u}\:\cdot \:\frac{\partial u}{\partial x}\:+\:\frac{\partial g}{\partial v}\cdot \frac{\partial v}{\partial x}
$$
$$
\frac{\partial g}{\partial y}\:=\:\frac{\partial g}{\partial u}\:\cdot \:\frac{\partial u}{\partial y}\:+\:\frac{\partial g}{\partial v}\cdot \frac{\partial v}{\partial y}
$$

Placing du/dx du/dy and dv/dx dv/dy , and evaluating at the point (-2,2)
$$
\frac{\partial g}{\partial y}\:=\:+2\frac{\partial g}{\partial u}\:+6\:\frac{\partial g}{\partial v}
$$
$$
\frac{\partial g}{\partial x}\:=\:+2\frac{\partial g}{\partial u}\:-2\:\frac{\partial g}{\partial v}
$$
My question is how to procceed further?, What do i do with The partial derivatives of g with respect to u and v.

答:
I would think that you could use the chain rule on its matrix form. Let

$$h(x,y)=(xy+x^2, xy-y^2)$$

and observe that $h(-2, 2)=(0,-8)$. Now $g(x,y)=(f \circ h)(x,y)$ and according to the chain rule

$$(\nabla (f \circ h)(x,y))^T = (\nabla f(h(x,y))^T J_h (x,y).$$

So we get

\begin{align}
(\nabla (f \circ h)(-2,2))^T &= (\nabla f(0,-8))^T J_h (-2,2) \\
&= [-3 \quad 5] \begin{bmatrix} -2 & -2 \\
2 & -6
\end{bmatrix} = [16 \quad -24]
\end{align}

Now, I don't think this is the most elegant way of using the chain rule here, so I would like to see other solutions.

Title: Finding Grad of G,


问:
Normal and invertible operator's polar decomposition does not belong to $C^* (A)$.
I'm studying Conway's Functional analysis, solving problem 8.3.24.

Give an example of a normal operator $A$ such that the partial isometry in the polar decomposition of $A$ does not belong to $C^*(A)$.

In the above problem, they state that, if $A$ is normal and invertible, then it's polar decomposition belongs to $C^*(A)$. This is easily shown because $A = W(A^*A)^{1/2}$ and $(A^*A)^{1/2} $ is invertible since $A$ is invertible. So, I'm trying that $A$ is not invertible matrix, but cannot find the example. Can you help me?

答:
THE FINITE DIMENSIONAL CASE

Let $f$ be the function defined on $\mathbb R$ by
$$
  f(x) = \left\{\matrix{
  \displaystyle \frac{1}{\sqrt x}, & \text{ if } x\neq 0,\cr
  0, & \text{ if } x=0.\cr
  }\right.
  $$
Clearly $f$ is discontinuous, but its restriction to the finite spectrum of $A^*A$ is continuous.  Therefore
$$
  W:= Af(A^*A)
  $$
lies in the C*-algebra generated by $A$ and it is not hard to see that $W$ is indeed the polar part of the polar decomposition of $A$.

THE INFINITE DIMENSIONAL CASE

Consider the diagonal operator $A$ on $\ell ^2$ defined on the standard orthonormal basis by $A(e_n)=e_n/n$.  Then $A\geq 0$, so
$A=(A^*A)^{1/2}$, and one has that $W$ is the identity operator.

Since $A$ is a compact operator we have that $C^*(A)\subseteq \mathcal K(\ell ^2)$, so $W$ is not in $C^*(A)$.


问:
Is it that these two polynomials doesn't have a common zero?
Consider $x$ a vector of $n$ components (complex). Now consider the following two polynomial equations.

$x^TPx-1=0$ where $P$ is an $n\times n$ squared matrix.

$q(x)=0$ with $q(x)$ a polynomial of degree greater than 2 and homogenenous of degree $\alpha$, i.e. $q(tx)=t^\alpha q(x), t>0$.

I wanted to study the common zeros of these two equations, however I noticed something strange.

Write these two polynomials as

$x^TPx- 1 = p(x)r(x)$

$q(x) = s(x)r(x)$

where $p,r,s$ are polynomials. $r(x)$ is a polynomial which vanish at the common zeros of $x^TPx-1$ and $q(x)$. Moreover and $p,s$ doesn't vanish when $r(x)$ vanishes. Now pick any $y$ such that $r(y)=0$ and choose $x = ty$ for $t>0$ in order to study the rational function $q(x)/(x^TPx-1)$. Then,
$$
\lim_{t\to 1}\frac{q(x)}{x^TPx-1}=\lim_{t\to 1}\frac{s(x)}{p(x)} = \lim_{t\to 1}\frac{s(ty)}{p(ty)}=L
$$
where $L\neq 0$ since $p(ty)<\infty$ for any $t$ near 1 (since $p(y)$ its a polynomial) and $s(y)\neq 0$. However,
$$
\frac{q(x)}{x^TPx-1}=\frac{q(ty)}{t^2y^TPy-1} = \frac{t^\alpha }{t^2-1}q(y)=0
$$
for any $t\neq 1$ since $q(y)=0$ and $y^TPy=1$. Hence, I arrive at a contradiction:  inspecting $q(x)/(x^TPx-1)$ in terms of $s/p$ I obtain that it must be different to 0 near $x=y$. However, by inspecting $q(x)/(x^TPx-1)$ directly I obtain that it must be zero near $x=y$. Hence, I cannot write them as $x^TPx-1=p(x)r(x)$ and $q(x) = p(x)r(x)$ so they do not have the common factor $r(x)$. Therefore they don't share any common solution.

However, I thought polynomial systems of equations always had (complex) solutions. I never assumed that the common factor $r(x)$ vanished somewhere at the real line, so I don't know where my procedure failed.

So, is it that these two polynomial equations share a common zero? If so, whats the flaw in my argument? Is my procedure correct?

Hopefully, I made some stupid error, but I cannot find it.

答:
It's only for polynomials in a single variable that solutions must lead to factors of the polynomial.  It certainly doesn't happen in this case.

If $q(x)$ is homogeneous, the factor $r(x)$ must also be homogeneous of its degree.  So either $r$ is constant or $r(0)=0$, and therefore can't be a factor of $x^T P x - 1$.

It may help to look at a simple example:
$x^2 + y^2 - 1 = 0$ and $x^2 - y^2 = 0$.  The common solutions are $x=\pm 1/\sqrt{2}$, $y=\pm 1/\sqrt{2}$.  But $x^2-y^2$ and $x^2 + y^2 - 1$ have no common factors.


问:
Let $X$ and $Y$ be topological spaces and $f: X \to Y$ a function....
Let $X$ and $Y$ be topological spaces and $f: X \to Y$ a function.

a) Show that if $f$ is continuous, then for each $ x \in  X$ and each sequence $ (x_n) _ {n \in \mathbb N}$  in $X$ such that $ x_n \to x $ we have to  $f( x_n) \to f(x) $.

b) Show that the reciprocal is true in the case where $X$ has a fundamental countable neighborhood system.

===================================================================

Let $U$ be an open neighborhood of $f (x)$, we have that $ f^{- 1} (U)$ is a neighborhood of $x$, so there is $V$ such that $ x_n \in f^{- 1}(U) $ , for $n  \ge \mathbb N $, and therefore $f (x_n) \in  V$ for $n \ge \mathbb N $.

I know that the exercise is a variation of the Theorem 30.1 in the book Topology by James R. Munkres, 2nd edition, but I am confused, mainly if my answer to item a) is acceptable, because I have the impression that I cannot use what I used in item a) without first proving item b ) what I'm not able to do for now.

Thank in advance for any assistance.

答:
Your proof is almost there, except for the part where you say that $n \geq \mathbb{N}$ which doesn't make much sense. ($\mathbb{N}$ isn't even a number).

Suppose that $U$ is some arbitrary open neighborhood of $f(x)$. Our goal is to show that there is some natural number $N \in \mathbb{N}$ such that for all $n \geq N$, $f(x_n) \in U$.

Since $f$ is continuous, $f^{-1}(U)$ is an open neighborhood of $x$, and since $x_n \rightarrow x$, there is some $N \in \mathbb{N}$ such that for all $n \geq N$, $x_n \in f^{-1}(U) \implies f(x_n) \in U$. In summation for $n \geq N$, $f(x_n) \in U$, which is exactly what we needed.

Part (b) is a tad more complicated. Suppose we have sequential continuity, that is, if $x_n \rightarrow x$, then $f(x_n) \rightarrow f(x)$ for every sequence and point. Suppose that $f$ is not continuous at some point $x \in X$, i.e, there is some $U \subseteq Y$ open such that $f(x) \in U$ but for every open $V \subseteq X$ with $x \in V$, we have that $V \not \subseteq f^{-1}(U)$ which means $f(V) \not \subseteq U$. Now let $\{B_n : n \in \mathbb{N} \}$ be a countable local base at $x$. Notice that we may assume $B_{n+1} \subseteq B_{n}$ for every $n$ (why?). Then by our assumption $f(B_n) \not \subseteq U$ for every $n$, meaning there is some point $x_n \in B_n$ such that $f(x_n) \not \in U$. Notice that the $x_n$ form a sequence such that $x_n \rightarrow x$. By sequential continuity this means $f(x_n) \rightarrow f(x)$. But $U$ is an open neighborhood of $f(x)$ and for every $n$, we have that $f(x_n) \not \in U$, which is a contradiction. So $f$ is continuous at $x \in X$, and $x$ was chosen arbitrarily.


问:
Given two positive real numbers $A$ and $l$, does there exist a plane, closed, simple curve with area $A$ and length $l$?
We are given two real positive numbers $A$ and $l$, such that $4\pi A \leqslant l^2$ (so the Isoperimetric inequality holds). Is it true we can always find a plane, simple, closed curve so that curve has length $l$ and region that it encloses has area $A$ ? If not, give a counterexample.

Intuitively, I think it is true, but I'm not sure have to define the curve with given properties.

答:
The area of an ellipse with radii $a$ and $\frac A{\pi a}$ is $A$. If we let $a$ vary continuously over the interval $[\sqrt{\frac A\pi},\infty)$, i.e. from  a circle to an arbitrarily stretched almost-line, the perimeter varies accordingly (and as a continuous function of $a$) over $[2\sqrt{\pi A},\infty)$.

If you want something more constructive, as shape like below (with a part of a circle pushed up by some distance $a$), can give you curves with area $A=\pi r^2$ and perimeter $\ell = 2\pi r+4a$ where you can easily pick $a$ arbitrarily in $[0,2r)$, i.e., this covers the cases
$$ 4\pi \le \frac{\ell^2}{A}<\frac{(2\pi+8)^2}{\pi}=64.9382\ldots$$
So for $\frac{\ell^2}A$ in that range, simple pick $r$, compute the desired $a$, and construct the figure accordingly - with the middle part thin enough to avoid problems, namely if the width of the middle part is $b$, you want $a^2+b^2<4r^2$.

On the other hand, $x\times y$ rectangles with $x\le y$ cover the cases $\frac{\ell^2}A\ge 16$. for given $\frac{\ell^2}A$, it is straightforward to compute $y$ for given $x$, say, from a quadratic equaiton.


问:
Application of Fermat's little theorem?

Question: Let $p$ and $q$ be distinct primes and let $n=pq$.  Prove that if $a=p+n\mathbb{Z}$ and $b=q+n\mathbb{Z}$, then $a^{q-1}+b^{p-1}=1.$

I'm not entirely convinced that the above is true.  I have applied Fermat's little theorem.  Since $p$ and $q$ are distinct primes - $a^{q-1}\equiv 1 \pmod q$ and $b^{p-1}\equiv 1 \pmod p.$  Where I am getting stuck is how to apply that second given: $a=p+n\mathbb{Z}$ and $b=q+n\mathbb{Z}.$

Added due to player3236:

$a^{q-1}+b^{p-1}\equiv 1 \pmod q$ since $b^{q-1}\equiv 0\pmod q$.  Similarly for $a$ and $p$.  This shows that $a^{q-1}+b^{p-1}-1\equiv 0 \pmod {pq}$,or that $a^{q-1}+b^{p-1}-1$ is divisible by $pq$ since $p$ and $q$ are distinct primes.

答:
Well  $a \equiv p\pmod q$ and $b\equiv q\equiv 0 \pmod q$ so $a^{q-1}+b^{p-1}\equiv p^{q-1}\equiv 1 \pmod q$.

Likewise $a\equiv p\equiv 0\pmod p$ and $b\equiv q\pmod p$ so  $a^{q-1}+b^{p-1}\equiv q^{p-1}\equiv 1 \pmod p$.

Now by Chinese Remainder theorem we know any mutual solution $a^{q-1}+b^{p-1}\equiv 1 \pmod p$ and $a^{q-1}+b^{p-1}\equiv 1 \pmod q$ is unique $\mod pq$.  And $a^{q-1} + b^{p-1}\equiv 1 \pmod {pq}$ is a solution.  So it is the only solution.

Title: Application of Fermat's little theorem?


问:
If sum of triangle angles is $180$ degrees, how $\sin(270)$ is possible?
I'm not new to trigonometry, but this question always bothers me.

As it is in Wolfram MathWorld-

$$
\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}
$$

We know that the sum of the angles in a triangle is 180 degrees, which means each angle is less than 180 degrees, so how, for example, $\sin 270$  is possible?

Edit

After Joe posted his answer, I went back to MathWorld and read it more carefully. I noticed what Brian Drake mentioned in his answer. MathWorld has already explained both definitions. I kept the question because of the brilliant answers below.

答:
Sinus is a function defined on all of ${\mathbb R}$, even on all of ${\mathbb C}$. The graph of $$t\mapsto \sin t\qquad(-\infty<t<\infty)\tag{1}$$
is the infinite wavy periodic curve you have seen many times. It so happens that in right triangles with legs $a$, $b$, and hypothenuse $c$, one has $${a\over c}=\sin\alpha\ ,$$
when the angle $\alpha$ (opposite the leg $a$) is measured in "radians".

Note that the "time process" $(1)$ can be viewed at a thousand places in nature; whereas right triangles are a tiny matter in mathematics. It is a wonder that there is one large concept of sine that describes all of this. To get hold of this concept you have to follow the course Calculus 101, or something similar.


问:
Euler Lagrange equation, example
can anyone help me solve the following minimisation problem ( w.r.t the matrix $A\in \mathbb{R}^{n\times n}$) using the Euler Lagrange equations?

$$ \inf \Big\{ \int_0^1\sqrt{\langle A\dot{\sigma},\dot{\sigma} \rangle},~~ \sigma:[0,1]\to \mathbb{R}^n ~\text{absolutely continuous on $[0,1]$},\sigma_0=x,\sigma_1=y \Big\} $$

答:
We have $L(t, \sigma, \dot \sigma) = \sqrt{\langle A \dot \sigma, \dot \sigma \rangle}$. We see that $\frac{\partial L}{\partial \sigma} = 0$. To find $\frac{\partial L}{\partial \dot\sigma}$, we need a formula for the derivative of the function
$$
f(x) = \sqrt{\langle A x,x \rangle}.
$$
To do this, write $f(x) = \sqrt{g(x)}$ with $g(x) = \langle Ax,x \rangle$ and apply the chain rule. We have
$$
g(x+h) = \langle A (x+h),(x+h) \rangle = \langle Ax,x \rangle + \langle (A + A^T)x, h \rangle + \langle Ah,h \rangle,
$$
from which it follows that $\frac{\partial g(x)}{\partial x} = (A + A^T)x$. With that, the chain rule gives us
$$
\frac{\partial f}{\partial x} = \frac 1{2 \sqrt{g(x)}}\cdot (A + A^T)x.
$$
With that, we find that the Euler-Lagrange equation becomes
$$
\frac d{dt} \frac{\partial L}{\partial \dot \sigma} = 0 \implies
\frac d{dt} \left[\frac{(A + A^T)\dot \sigma}{2 \sqrt{\langle A\dot \sigma,\dot \sigma\rangle}} \right] = 0.
$$
If something has derivative zero, then it must be constant. That is, there is a vector $c$ for which
$$
\frac{(A + A^T)\dot \sigma}{2 \sqrt{\langle A\dot \sigma,\dot \sigma\rangle}} = c \implies
\frac{A + A^T}{2} \cdot \frac{\dot \sigma}{\sqrt{\langle A \dot \sigma, \dot\sigma \rangle}} = c.
$$
Without further assumptions on the matrix $A$ (e.g. that $A$ is positive definite), I see no way to move forward from here.

If we are given that $A$ is indeed (symmetric and) positive definite, then we can quickly see that $\dot \sigma$ must be a multiple of some constant vector, which to say that $\sigma$ moves within a straight line.

We might also have made the entire problem more straightforward with the substitution $\tau := A^{1/2}\sigma$, which would have led to the equivalent minimization of
$$
\int_0^1 \sqrt{\langle \dot \tau, \dot \tau\rangle}\,dt,
$$
which is the integral defining arc length (with respect to the usual metric).

Title: Euler Lagrange equation, example


问:
Let $x,y,a,b\in\mathbb{Z}$. If $x+y\equiv 0\ (\mathrm{mod}\ 3)$ and $a\equiv b\ (\mathrm{mod}\ 3)$, then show that $ax+by\equiv 0\ (\mathrm{mod}\ 3)$.
Stuck on this homework problem. At this point I know the basic definitions of what it means to be congruent, I just can't seem to link them together to solve this problem, and I'm not really sure how to use the $x+y\equiv 0\ (\mathrm{mod}\ 3)$

I assume I must expand/alter something and then plug it into the $ax+by\equiv 0\ (\mathrm{mod}\ 3)$.

So far for this proof, I know to $a\equiv b\ (\mathrm{mod}\ 3)$, which then implies that $a = b + 3k$, for some $k\in\mathbb{Z}$. Similarly, $b= a + 3j$ for some $j\in\mathbb{Z}$.

Any help on how to link these theorems would be appreciated.

答:
We can substitute $b=a+3k$ for some $k$ since $a\equiv b\text{ mod }3$

Then $ax+by=ax+(a+3k)y=ax+ay+3yk=a(x+y)+3yk$.

Since $x+y$ is a multiple of $3$, both terms of the sum are divisible by $3$ so $$ax+by\equiv 0\text{ mod }3$$


问:
Showing the limit of given function does not exist
Given $f:\mathbb{R}\to \mathbb{R}$ such that $\mid f(x)\mid\geq M$ for some $M > 0$, $\forall x\in \mathbb{R}$.

I need to show that $\lim_{x\to 0} \frac{f(x)}{\sqrt[3]{x}}$ does not exist.

My try :

Assume limit exist, say L.

$\lim_{x\to 0} \frac{f(x)}{\sqrt[3]{x}}=\frac{\lim_{x\to 0} f(x)}{\lim_{x\to 0} \sqrt[3]{x}}=L$

$\Rightarrow$  $\lim_{x\to 0} f(x)=L.\lim_{x\to 0} \sqrt[3]{x}=0$ , contradiction to the fact that $\mid f(x)\mid \geq M, \forall x$

答:
In order to prove this rigorously, you need to use the definition or prove by contradiction. Suppose that $$\lim_{x\rightarrow 0}\frac{f(x)}{x^{1/3}}=l.$$

Take $\epsilon =1$,  $\exists\delta>0$, for all $x$, such that $0<|x|<\delta \Rightarrow |\frac{f(x)}{x^{1/3}}-l|<\epsilon$. Using triangle inequality, we have $|f(x)|<|x^{1/3}|\cdot (|l|+1)$. Now take $x=\min \{\delta/2, \frac{M^3}{(|l|+1)^3}\}$ and got a contradiction.


问:
Weak convergence of measure
Suppose that the measure $\mu_n$ weakly convergences to the measure $\mu$. Does the following hold?

For any measurable set $A$, $\limsup_{n\to \infty} \mu_n(A)\ge \mu(A)$?

For any measurable set $A$, $\liminf_{n\to \infty} \mu_n(A)\le \mu(A)$?

答:
In general, no.  Let $\mu_n$ be a point mass at $1/n$ and $\mu$ be a point mass at $0$.  A counterexample to (1) is the singleton $A = \{0\}$ and a counterexample to (2) is $A = (0,1]$.

For some more detailed information you can read about the portmanteau theorem

Title: Weak convergence of measure


问:
Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$
Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$.

I tried using substitution as follows:
$$ a = x^2 $$
$$ b = (m + 1)^2$$
Using the substitution:
$$ a^2 - (1 + b)a + b \geq 0$$

Substituting in the second degree equation:
$$ \frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
Solving the binomial coefficients $(m + 1)^4$ and $(m + 1)^2$, I got:
$$ \frac{-(1 + (m + 1)^2) + \sqrt{m^4 + 4m^3 + 8m^2 + 8m + 8}}{2} $$

From here I don't know how to proceed...

答:
Your method might work, but I couldn't really see how to proceed. A substitution that I am sure would work is as follows:

First expand your polynomial: $x^4-(m^2+2m+2)x^2+m^2+2m+1$.

Next, you can use your $a=x^2$ substitution: $a^2-(m^2+2m+2)a+m^2+2m+1$. Then, you use essentially the same $b$ substitution, but expanded: $b=m^2+2m+1$. Once you do this you get $a^2-(b+1)a+b$, you solve and get $a=1, a=b$ (verify yourself that they work). Then substitute back $a$ and $b$: $x^2=1, x^2=m^2+2m+1$, so your solutions for $x$ are: $x=1$, $x=-1$, $x=m+1$, $x=-m-1$. Therefore, since an extremely large or small value of $x$ will make the equation positive, your range of x-values are $(-\infty, -|m|]\cap[-1, 1]\cap[|m|, \infty)$. Note when $m=-2$, your range of x-values are $(-\infty, \infty)$

Edit: My steps are essentially your steps (expanded out), but I didn't use quadratic formula like on the fourth step, but factored it.

Using what you have:

$ a = x^2 $

$ b = (m + 1)^2$

$ a^2 - (1 + b)a + b \geq 0$

$a=b, a=1$ are solutions to the inequality, and solve for bounds

The rest is the same.


问:
Laplace transform:$\int_0^\infty \frac{\sin^4 x}{x^3} \, dx $
I have a trouble with a integral:
Using this Laplace trasform equation:
$$\begin{align}
 \int_0^\infty F(u)g(u) \, du  & = \int_0^\infty f(u)G(u) \, du \\[6pt]
 L[f(t)] & = F(s) \\[6pt]
 L[g(t)] & = G(s)
 \end{align}$$

Applying to compute this integral:

$$I = \int_0^\infty \frac{\sin^4 x}{x^3} \, dx $$

答:
Related problem. Recalling the Laplace transform of a function $f(x)$

$$ F(s)=\int_{0}^{\infty} f(x)e^{-sx}dx .$$

Let

$$ G(u)=\frac{1}{u^3} \implies g(u)=\frac{u^2}{2!}, $$

and

$$ f(u)= \sin(u)^4  \implies F(u)=\frac{24}{u(u^2+4)(u^2+16)}. $$

Now,

$$ \int_0^\infty \frac{\sin^4 x}{x^3} \, dx= \frac{24}{2}\int_0^\infty \frac{u^2}{u(u^2+4)(u^2+16)} \, dx = \ln(2)$$

Note:

1) To evaluate the last integral, you can write the integrand as

$$\frac{u}{(u^2+4)(u^2+16)}= \frac{1}{12}{\frac {u}{{u}^{2}+4}}-\frac{1}{12}{\frac {u}{{u}^{2}+16}} $$

2) To find Laplace transform of $\sin^4(x)$, first, write

$$ \sin^4(x) = \frac{1}{(2 i)^4} ( e^{i x}-e^{- i x} )^{4}, $$

then use the binomial theorem to expand the above expression, and finally use the Laplace transform of $e^{ax}$

$$ F(s)=\frac{1}{s-a}. $$

ContentLicense: CC BY-SA 3.0


问:
Expected number of tries
Say there are 400 marbles in a bag. I want 4 specific marbles and those marbles have a 1/100, 1/200, 1/300 and 1/400 chance of being drawn. I keep drawing marbles 1 at a time with replacement until I have drawn all 4 marbles that I wanted.

Two questions:

What is the expected number of marbles I pick before I get at least 1 of each of the 4 specific marbles I was looking for?

What is the probability of drawing all 4 marbles at least once by the nth draw?

Thanks

答:
We approach by recursion and markov-chains.  Let our states be represented by binary 4-tuples where a $1$ in a position corresponds to us having drawn a marble of the corresponding type and $0$ if we have not drawn a marble of that type.  For examples $(1,0,0,0)$ corresponds to having drawn at least one marble of the first type and no other marbles of the other desired types (and any number of undesired marbles).

The "game" ends when we have drawn the final missing desired marbletype.

Let $f~:~\{0,1\}^4\to\Bbb R$ be the function which maps a state to the expected number of remaining draws required to reach the final end-state of $(1,1,1,1)$.

You should be able to see that these states form a tesseract (a $4$-dimensional cube) and on each draw of a new marble one can travel from one state to another higher state by changing a $0$-bit into a $1$-bit or you may stay at the current state.

(Tesseract image taken from wikipedia)

We can start filling in the values for $f$ by starting at the top and working our way down.  We begin by noting that if we are in the final end-state we require no additional draws to finish since we are already done.  For all others, when pulling this adds $1$ to our overall expected wait time and we will then need an additional amount of time based on what state we arrive at which varies based on what state it is and will occur with what probability it is to travel to that state.  From there, we can with algebra and earlier found values calculate the wait time for each respective state if we had worked on finding these values from top-down.

No missing marbletypes

$f(1,1,1,1)=0$

One missing marbletype

$f(0,1,1,1) = 1 + \frac{1}{100}f(1,1,1,1)+\frac{99}{100}f(0,1,1,1) = 100$

$f(1,0,1,1) = 1 + \frac{1}{200}f(1,1,1,1)+\frac{199}{200}f(1,0,1,1) = 200$

$f(1,1,0,1) = 1 + \frac{1}{300}f(1,1,1,1)+\frac{299}{300}f(1,1,0,1) = 300$

$f(1,1,1,0) = 1 + \frac{1}{400}f(1,1,1,1)+\frac{399}{400}f(1,1,1,0) = 400$

Two missing marbletypes

$f(0,0,1,1) = 1 + \frac{1}{100}f(1,0,1,1) + \frac{1}{200}f(0,1,1,1) + \frac{197}{200}f(0,0,1,1) = \frac{700}{3}$

 Working through this line in more detail, we draw a marble and then from $(0,0,1,1)$ we could travel up to $(1,0,1,1)$ by having drawn a marble of the first type which occurs with probability $\frac{1}{100}$ or traveled to $(0,1,1,1)$ by having drawn a marble of the second type which occurs with probability $\frac{1}{200}$.  Having drawn a marble of any other type we remain in the current state which occurs with probability $1 - \frac{1}{100}-\frac{1}{200}$.  We previously calculated $f(1,0,1,1)$ and $f(0,1,1,1)$ as being $200$ and $100$ respectively.  We have then $f(0,0,1,1)=1+\frac{1}{100}\cdot 200 + \frac{1}{200}\cdot 100 + \frac{197}{200}f(0,0,1,1)$ and after rearranging we have $\frac{3}{200}f(0,0,1,1)=\frac{7}{2}$.  Multiplying both sides by $\frac{200}{3}$ gives $f(0,0,1,1)=\frac{700}{3}$

$f(0,1,0,1) = 1 + \frac{1}{100}f(1,1,0,1) + \frac{1}{300}f(0,1,1,1) + \frac{296}{300}f(0,1,0,1) = 325$

$f(0,1,1,0) = 1 + \frac{1}{100}f(1,1,1,0)+\frac{1}{400}f(0,1,1,1)+\frac{395}{400}f(0,1,1,0) = \dots$

$\vdots$

$f(1,1,0,0) = 1 + \frac{1}{300}f(1,1,1,0)+\frac{1}{400}f(1,1,0,1)+\frac{1193}{1200}f(1,1,0,0)=\dots$

Three missing marbletypes

$f(0,0,0,1) = 1 + \frac{1}{100}f(1,0,0,1) + \frac{1}{200}f(0,1,0,1)+\frac{1}{300}f(0,0,1,1) + \frac{589}{600}f(0,0,0,1) = \dots$

$\vdots$

Four missing marbletypes

$f(0,0,0,0) = 1+\frac{1}{100}f(1,0,0,0)+\frac{1}{200}f(0,1,0,0)+\frac{1}{300}f(0,0,1,0)+\frac{1}{400}f(0,0,0,1) + \frac{1175}{1200}f(0,0,0,0) = \dots$

This final value is what you are after.  It is going to be highly tedious and so I leave it to you to complete.  With some programming skill, you could automate the calculations (though be careful about floating arithmetic errors).

Title: Expected number of tries


答:
Summarizing a set of comments below the OP, the only mistake in your proof is in the problematic expression $\lim_{x\to0}f(x)\over\lim_{x\to0}\sqrt[3]x$, which is of the utterly meaningless form $\text{possibly undefined}\over 0$. But it's an expression you don't need at all; it's enough to say $\lim_{x\to0}{f(x)\over\sqrt[3]x}=L$. That equation is all it takes to express the assumption that the limit, call it $L$, exists.  As for the denouement, what you've written is basically correct, but it might be clearer to write it as

$$\lim_{x\to0}f(x)=\lim_{x\to0}\left({f(x)\over\sqrt[3]x}\sqrt[3]x\right)=\left(\lim_{x\to0}{f(x)\over\sqrt[3]x}\right)\cdot\left(\lim_{x\to0}\sqrt[3]x\right)=L\cdot0=0$$

The take home message here, in terms of learning to write proofs, is to look at each and every assertion you make and ask yourself if it's both meaningful and true. If an assertion fails on either criterion, you then need to ask yourself what's it doing there? In this case, the answer is Nothing!

Title: Showing the limit of given function does not exist

Body:
Given $f:\mathbb{R}\to \mathbb{R}$ such that $\mid f(x)\mid\geq M$ for some $M > 0$, $\forall x\in \mathbb{R}$.

I need to show that $\lim_{x\to 0} \frac{f(x)}{\sqrt[3]{x}}$ does not exist.

My try :

Assume limit exist, say L.

$\lim_{x\to 0} \frac{f(x)}{\sqrt[3]{x}}=\frac{\lim_{x\to 0} f(x)}{\lim_{x\to 0} \sqrt[3]{x}}=L$

$\Rightarrow$  $\lim_{x\to 0} f(x)=L.\lim_{x\to 0} \sqrt[3]{x}=0$ , contradiction to the fact that $\mid f(x)\mid \geq M, \forall x$


答:
You are correct. If by $\sin \theta$ we mean $\frac{\text{opposite}}{\text{hypotenuse}}$ then $\sin 270$ is indeed meaningless. Therefore, it is a little disingenuous to say that
$$
\sin\theta:=\frac{\text{opposite}}{\text{hypotenuse}} \, .
$$
(The $:=$ sign means 'defined to be equal to'.)

Mathematicians rarely use the above definition for this exact reason: the definition only applies when $0 < \theta < 90$. It is much more useful to have a definition that applies to all values of $\theta$, even negative values. To create this new, extended definition of sine we use a unit circle:

A unit circle is a circle with a radius of one. If we place a triangle in the unit circle, as pictured in the diagram, then we know that
$$
\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\text{opposite}}{1}=\text{opposite} = y\text{-coordinate} \, .
$$
So far, we have just used the old definition of sine, but if we want $\sin \theta$ to have a meaningful value when $\theta$ is not between $0$ and $90$ degrees, then we can say that $\sin \theta$ is simply the $y$-coordinate when you go $\theta$ degrees counterclockwise around the unit circle, starting from the point $(1,0)$. In this context, we are no longer talking about right-angled triangles, and the definition applies regardless of how big or small $\theta$ is. That's a lot to take in, and it might be unclear to you as to why we would define sine in this way. There are two reasons why we do this:

This definition makes sense when $\theta$ is any number, whereas the right-angled triangle definition of sine only makes sense when $0 < \theta < 90$.*

When $\theta$ is between $0$ and $90$ degrees, then the two definitions of sine—the old and the new one—'agree'. For example, $\sin 30$ is both the ratio of the opposite side to the hypotenuse, and the $y-$coordinate when you go $30$ degrees anticlockwise around a unit circle. This is not a coincidence. We literally defined sine in such a way that the old definition still applies when $0 < \theta < 90$.

Extending the definition of something is very common in mathematics. You have probably seen it before. For example, when $n$ is a positive integer, we can say that
$$
a^n = \underbrace{a \times a \times a \times a \times \cdots \times a}_{n \text{ times}} \, .
$$
But you probably know what $9^{1/2}$ is. Again, $9^{1/2}$ is meaningless when you define it in the above way. Instead, we say that
$$
a^{p/q}=(\sqrt[q]a)^p \, .
$$
This means that $9^{1/2}=\sqrt{9}=3$. Why do we define exponents in this way? It's because the essential properties of exponentiation are preserved under the new definition. It's very obvious that when $m$ and $n$ are positive integers, that
$$
a^{m+n}=a^m \times a^n \, .
$$
If we define $a^{p/q}$ as $(\sqrt[q]a)^p$, then it is still true that
$$
a^{m+n}=a^m \times a^n
$$
even when $m$ and $n$ are not positive integers. In the case of trigonometry, people noticed that if we place a triangle in a unit circle, then the $y-$coordinate of the triangle is equal to $\sin x$. Extending the definition of sine means that this property still holds, even when $\theta$ is not between $0$ and $90$ degrees. It also means that the sine graph has many nice symmetries, such as $\sin (\theta)=-\sin(-\theta)$.

Credit to congusbongus for sharing this delightful animation. Source: Visually stunning math concepts which are easy to explain.

*When $\theta$ is negative, instead of moving counterclockwise, we move clockwise around the unit circle. When you think of negative numbers as being 'opposite' to positive numbers, this seems logical.

In the interests of keeping this post accessible, I did not mention complex numbers. However, it might be worth noting that we can extend the definition of sine even further. If we consider the Taylor series of sine,
$$
\sin \theta = \theta - \frac{\theta ^ 3}{3!} + \frac{\theta ^ 5}{5!} - \frac{\theta ^ 7}{7!} + \cdots
$$
then the RHS is meaningful even when $\theta \in \mathbb{C}$. So rather than treating the Taylor series of sine as a result that you can derive, you can instead use it as the definition of sine. You most certainly cannot go $i$ radians counterclockwise around a unit circle. However, you can plug $i$ into the formula
$$
\theta - \frac{\theta ^ 3}{3!} + \frac{\theta ^ 5}{5!} - \frac{\theta ^ 7}{7!} + \cdots \, .
$$
Therefore, we may define $\sin z$ as
$$
z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots
$$
for all $z \in \mathbb{C}$. So once again, sine becomes more abstract, but more powerful.

Title: If sum of triangle angles is $180$ degrees, how $\sin(270)$ is possible?

Body:
I'm not new to trigonometry, but this question always bothers me.

As it is in Wolfram MathWorld-

$$
\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}
$$

We know that the sum of the angles in a triangle is 180 degrees, which means each angle is less than 180 degrees, so how, for example, $\sin 270$  is possible?

Edit

After Joe posted his answer, I went back to MathWorld and read it more carefully. I noticed what Brian Drake mentioned in his answer. MathWorld has already explained both definitions. I kept the question because of the brilliant answers below.


答:
Although the word ‘trigonometry’ originally meant ‹triangle measurement›, trigonometry is really about much more than just triangles.  The formula $ \operatorname { sine } = \operatorname { opposite } ÷ \operatorname { hypotenuse } $ is correct when talking about angles in a right triangle, but that is only a small part of the ways in which angles appear in geometry, and geometry itself is only part of what the sine is good for.  So it's a place to start, but there is much more to the sine than that.

The two comments that have appeared under your question so far (by Deepak and runway44) will introduce you to the uses of sines in geometry beyond right triangles.  The one other answer so far (by Christian Blatter) hints at a non-geometric application: cyclic processes through time.  (ETA: And now there is another answer, by Joe, that talks about both.)  These are all things that you will want to learn about to understand trigonometry.  But the simple answer to your question is just that your formula only applies to the special case of right triangles.


答:
The definition of function $\sin(\alpha)$ that you mentioned is correct, but is applicable only for $90 > \alpha > 0$

Much better definition of this function would be: $ \sin(\alpha) := y $ coordinate of a unit vector with origin in $(0, 0)$, such that angle between the vector and $x$ axis is $\alpha$. It gives the same result as the first definition if $90 > \alpha > 0$, but works for any $\alpha$.


问:
Inapproximability of the $k$-center problem
The $k$-center problem:

Given:

Undirected, complete graph $G=(V,E)$,

a distance function $d:E\rightarrow\mathbb{R}$ such that $d_{ii}=0, d_{ij}=d_{ji}$ for each pair for vertices $i,j\in V$, and that the distances obey triangle inequality,

A positive integer $k$.

Find a set $S\subseteq V, |S|=k$ of $k$ cluster centers with the objective to minimize the maximum distance of a vertex to its cluster center.

We know that the problem is NP-hard and a simple greedy algorithm gives 2-approximation. I was going through a proof from the book "The Design of Approximation Algorithms" for a theorem that states that an approximation algorithm with factor less than 2 would imply P=NP. This is done using a reduction from the dominating set problem.

Theorem: There is no $\alpha$-approximation algorithm for the $k$-center problem with $\alpha<2$ unless P=NP.

Proof: Given an instance of the dominating set problem, we can define an instance of the $k$-center problem by setting the distance between adjacent vertices to 1 and non-adjacent vertices to 2. Then there is a dominating set of size $k$ if and only if the optimal radius for this $k$-center instance is 1. Furthermore, any $\alpha$-approximation with $\alpha<2$ must always produce a solution of radius 1 if such a solution exists.

Question: I think I understand the proof. However, I am confused with the setting of distance between non-adjacent vertices to 2. What if we set it to 3 instead of 2? Would that mean the $\alpha<3$ is not possible unless P=NP? This is not true as we have a 2-approximation algorithm. The proof should not work for values greater than 2. However, I am not able to figure out which step of the proof breaks when we set distance between non-adjacent vertices to 3.

Dominating Set Problem: Given an undirected graph $G=(V,E)$ and an integer $k$, we must decide if there exists a set $S\subseteq V$ of size $k$ such that each vertex is either in $S$ or adjacent to a vertex in $S$.

答:
If the weight for adjacent vertices is $w_{a}$, using any other $w_{na} > 2 w_{a}$ for non-adjacent ones would break the triangle equality, which is a fundamental assumption to prove the 2-approximation algorithm.
In fact, think about three vertices $i,j,k$ in the $k$-center instance such that $i$ and $j$ are adjacent, $j$ and $k$ are adjacent but $i$ and $k$ are not. This would break $d_{ij}+d_{jk}=2w_{a}\geq d_{ik}= w_{na}$ because $w_{na}\geq 2 w_{a}$.

The constant numbers 1 and 2 are just two example of constants $w_{a}$ and $w_{na}$ that satisfy $0 < w_{a} < w_{na} \leq 2 w_{a}$.


问:
Homology of pair
Let $X$ and $A$ two path connected spaces, $A$ subset of $X$. My question is how to commute the group homolgy $H_{0}(X,A)$?
Here is what I have tried: For the pair $(X, A)$ we have long sequence
$$\cdots\longrightarrow H_{q}(A)\longrightarrow H_{q}(X)\longrightarrow H_{q}(X, A)\longrightarrow H_{q-}(A)\longrightarrow\cdots$$
For q=0, we obtain
$$H_{0}(A)\longrightarrow H_{0}(X)\longrightarrow H_{0}(X, A)\longrightarrow 0$$ By the connectedness, $H_{0}(A)=H_{0}(X)=\mathbb{Z}$:
$$\mathbb{Z}\longrightarrow \mathbb{Z}\longrightarrow H_{0}(X, A)\longrightarrow 0$$

From Here I dont know what to do. Could you help me please? And what is the result in the general case ( $A$ and $X$ are not path connected)?

答:
At this point, it helps to look at the cycle level. Letting $u \in X \subset X$ be a point, the singular $0$-chain
$$
\sigma = 1 \cdot u
$$
is a cycle, and in fact a generator of $H_0(A)$. Under the inclusion map, it becomes a generator of $H_0(X)$. Hence the map from $\Bbb Z \to \Bbb Z$ turns out to be an isomorphism; in particular, it's surjective, so that last bit of that long-exact sequence can be replaced by
$$
0 \to H_0(X,A) \to 0
$$
whence we conclude that $H_0(X,A) = 0$.

Why does $\sigma$ generate $H_0(A)$? Suppose that
$$
\tau = \sum c_i P_i
$$
is an integer linear combination of points of $A$, i.e., a $0$-chain. Because the boundary in dimension $0$ is just the $0$-map, $\tau$ is also a $0$-cycle. For each $P_i$, let $\gamma_i$ be a path from $u$ to $P_i$ (which exists because $A$ is path-connected). Note that on the chain level, we have $\partial \gamma_i = 1 \cdot p_i - 1 \cdot u$.  Then
$$
Q = \sum c_i \gamma_i
$$
has the property that
\begin{align}
\partial Q
&= \sum c_i \cdot (\partial \gamma_i)\\
&= \sum c_i \cdot (P_i - u) \\
&= \sum (c_i \cdot P_i)  - (\sum c_i) u)
\end{align}
so that the chain $\tau$ is homologous to the chain $(\sum c_i) u$.

Thus every chain is homologous to an integer multipl of $\sigma$, so $\sigma$ is a generator of the quotient group.

Exactly the same argument works for $H_0(X)$, except that the paths (and points) must be taken in $X$ rather than just $A$.

Title: Homology of pair


问:
Prove or disprove: $\forall p\in[0,1),\,\int_0^1 x\oplus p\,dx=\frac12$, where $\oplus$ is the bitwise XOR operation.
For $x\in[0,1)$, let $(x_k)\in\{0,1\}^{\mathbb N^*}$ be its bits coefficients, i.e.
$$x=\sum_{k=1}^\infty x_k2^{-k}.$$

I want to prove (or disprove) that

$$\forall p\in[0,1),\,\int_0^1\sum_{k=1}^\infty\left(x_k\oplus p_k\right)2^{-k}\,dx=\frac12,$$

where $\oplus$ is the XOR operation, i.e.

$$x_k\oplus p_k=
\begin{cases}
1\quad\text{if $x_k\neq p_k$}\\
0\quad\text{if $x_k=p_k$}
\end{cases}=(x_k+p_k) \mod2.$$

The idea is very simple, when we do the XOR with $p$ for all numbers in $[0,1]$, the results are kind of uniformly distributed in $[0,1]$, thus we'll finally have $\frac12$.

The problem here is how to prove this observation using formal mathematical language, noticing that the function $f:x\mapsto x\oplus p$ is very strange, not continue, not Riemann integrable (but measurable and bounded).

答: You want to break your interval up into $2^k$ intervals, and put upper and lower bounds on each interval. Then take limit as k goes to infinity, using bounded convergence theorem, bounded by the constant function 1. For instance when you break it into one interval, the upper bound is 1 and lower bound is 0. When you break it up into two intervals, the upper bound is 1/2(1+1/4) and lower bound is 1/2(0+1/2). When you break it into 4 intervals the upper bound is 1/4(1+1/2+1/2+1/4) and the lower bound is 1/4(3/4+1/2+1/4)=6/16 and so on. I will leave it to you to workout the details.


问:
How many Dyck paths start from $(x, y)$?
Dyck paths count paths from $(0,0)$ to $(n,n)$ in steps going east $(1,0)$ or north $(0,1)$ and that remain below the diagonal.

How many of these pass starts from given point $(x,y)$ with $x \leq y$?

Note that, I found the similar question, here, but my question is starting from $(x,y)$.

答:
In the question that you are citing, they find the number of Dyck paths that go from $(0,0)$ to $(n,n)$ that pass thru the point $(y,x).$(is backwards because they are doing it there above the diagonal). They (Brian M. Scott) find  that this is
$$\begin{align*}
&\frac{x+1-y}{x+1}\binom{x+y}{x} \cdot \frac{x+1-y}{n+1-y}\binom{2n-x-y}{n-y} =\\
&\frac{(x+1-y)^2}{(x+1)(n+1-y)} \binom{2n-(x+y)}{n-x} \binom{x+y}{x}.
\end{align*}.$$
Notice that a path here is any path going from $(0,0)$ to $(y,x)$ and then a path from $(y,x)$ to $(n,n).$ It is also shown there that the number of paths that go from $(0,0)$ to $(y,x)$ is $\frac{x+1-y}{x+1}\binom{x+y}{x}$ and hence if you divide the answer by this, you will get the number of paths that start at $(y,x)$ to $(n,n).$

Namely, $$\frac{(y-x+1)}{(n+1-x)}\binom{2n-(x+y)}{n-y}$$
is the answer to your problem.


答:
Here's how it's done in high school, using Vieta's relations and results on the sign of a quadratic polynomial:

First set $y=x^2$ and solve for $y$: you have the quadratic inequation
$$y^2-\bigl(1+(m+1)^2\bigr)y+(m+1)^2 \ge0,\qquad y\ge 0.$$
The discriminant is
$$\Delta=\bigl(1+(m+1)^2\bigr)^2-4(m+1)^2=\bigl(1-(m+1)^2\bigr)^2,$$
therefore the quadratic polynomial in $y$ has two roots $y_0$ and $y_1\;(y_0\le y_1)$.

Furthermore these roots are nonnegative, since $y_0+y_1=\bigl(1+(m+1)^2\bigr)^2>0\:$ and $\:y_0y_1=(+1)^2\ge 0$.

Conclusion; the original inequation is equivalent  to

$$0\le y\le y_0\;\text{ or }\;y\ge y_1\iff \begin{cases}-\sqrt{y_0\mathstrut}\le x\le \sqrt{y_0\mathstrut},\\
\quad\text{ or }\\ x\le -\sqrt{y_1\mathstrut},\enspace x\ge\sqrt{y_1\mathstrut}. \end{cases}$$

Body:
Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$.

I tried using substitution as follows:
$$ a = x^2 $$
$$ b = (m + 1)^2$$
Using the substitution:
$$ a^2 - (1 + b)a + b \geq 0$$

Substituting in the second degree equation:
$$ \frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
Solving the binomial coefficients $(m + 1)^4$ and $(m + 1)^2$, I got:
$$ \frac{-(1 + (m + 1)^2) + \sqrt{m^4 + 4m^3 + 8m^2 + 8m + 8}}{2} $$

From here I don't know how to proceed...


问:
Markov Chain: Calculating conditional probability
For this problem there are 3 states zonal flow $(R1)$, intermediate $(R2)$ and blocked regime$ (R3).$ Let $A$ denote the event that more than three consecutive days are classified as blocked
regime. Calculate the probability that more than three consecutive days are classified
as blocked starting from day n which is already blocked, $P(A|Xn = 3)$. Assume the
Markov property:

I defined the state space $S= \{{1,2,3}\}$ with$ 1$ corresponding  to $R1$, $2$ corresponding to $R2 $ and $3$ corresponding to $R3$.  I let $A =\{{(X_{n+1}, X_{n+2}, X_{n+3}) = R3} \}$  which i'm not sure is correct.
Then $P(A|X_n=3) = P(X_{n+3}, X_{n+2}=3,X_{n+1}=3 | X_n =3)$. I am not sure if i'm right at all.
Would greatly appreciate the help or reassurance

答: The expression should be $\mathbb{P}(X_{n+1} = 3 | X_n = 3) \cdot \mathbb{P}(X_{n+2} = 3 | X_{n+1} = 3) \cdot \mathbb{P}(X_{n+3} = 3 | X_{n+2} = 3)$


问:
Prove that the process is a standard 2-dim brownian motion.
I have this exercise:

Let $(B_t^1,B_t^1)_{t \in (0,\infty)}$ be a standard two-dimensional brownian motion.
Prove that the process $((X_t^1, X_t^2))_{t \in (0,\infty)}$ defined by $$ X_t^1 = \int_0^t cos(B_s^1)dB_s^1 - \int_0^t sin(B_s^1)dB_s^2$$  $$X_t^2 = \int_0^t sin(B_s^1)dB_s^1 + \int_0^t cos(B_s^1)dB_s^2$$
is a standard two-dimensional brownian motion.

I was thinking to solve it by using a two dimensional levy characterization of brownian motion. However, I just now the 1 dimensional version, so in general I would check that that given a martingale $M_t$, it is a standard brownian motion if $\langle M \rangle_t = t$. Is it the right way of proving this ? Maybe by considering a covariation ?

答:
Yes, you can prove it using the multidimensional Levy theorem. This is, checking that the process $((X_t^1, X_t^2))_{t \in (0,\infty)}$ is a 2-dimensional local martingale and that

$[X^{1}]_t=[X^{2}]_t = t;$

$[X^{1},X^{2}]_t = 0.$


问:
Determine the values of k that make the equation a circle.
Given the equation: $$(2x^2) + (2y^2) + 4kx - 2y = 2k - 8$$

Find the values of k that make the equation a circle.

My question is what is the procedure for reaching this conclusion:

Answer: $k \in (-\infty, -5/2) \cup (3/2, +\infty)$

答:
Hint:

Complete the squares to get

$2x^2+4kx+2k^2+2y^2-2y+\frac12=2k-8+2k^2+\frac12=\frac12(2k-3)(2k+5)$,

and, to get a circle, the right side must be positive.


问:
Frechet derivative of trace of matrix expression
I am new to the concept of Frechet Derivatives. I have encountered a problem where I am supposed to find the Frechet derivative of $\operatorname{trace}(XAX+AXA^T)$ where my $X,A \in \Bbb R^{n\times n}$ and $X=X^T$.

答: Let $f(X) = {\rm tr}(XAX+AXA^T)$. Since ${\rm tr}$ is a linear functional, the total derivative of ${\rm tr}$ is ${\rm tr}$ itself, so the product rule gives $$Df(X)(H) = {\rm tr}(HAX+XAH+AHA^T).$$


问:
How to compute Frechet Derivatives that involve matrices.
Suppose I have a linear operator which maps $S^n\to\mathbb R^{n \times n}$. It is represented as $LX:= XA^T + AX.$ How can I find the Frechet derivative of $f(X) = XA^T + AX.$

答: The best linear approximation to a function that is already linear is the function itself. So $DL(X) = L$.


Title: Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$

Body:
You made a mistake. From here
$$ a^2 - (1 + b)a + b \geq 0$$

you went here
$$ a=\frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
While the correct solution is
$$a=\frac{-(1 + b) + \sqrt{ (1 + b)^2-4b}}{2} =\frac{-(1 + b) + \sqrt{ (1 - b)^2}}{2}$$
$$a_1=1;\;a_2=b$$

$$a\le 1\lor a\ge b$$
that is
$$x^2\le 1 \lor x^2\ge (m+1)^2$$
or
$$-1\le x \le 1 \lor  x\ge| m+1|$$


问:
What is the cardinality of set of functions $F:\mathbb{N} \longrightarrow \{0,1,2\}$?
Actually I have tried it but I am unable to create a mapping to see the reality as I wants to know that what is the value of 3 to the power aleph null?</p>

答:
Let $F_n$ be set of all functions $\mathbb N \to \{0, 1, \ldots, n\}$.

Cardinality of $F_2$ is continuum (either definition or should be already known).

$|F_3| \geq |F_2|$, as $F_2 \subset F_3$. Analogously, $|F_3| \leq |F_4|$.

To show that $|F_2| = |F_4|$, we can construct bijection between them: for $f \in F_2$ take $g \in F_4$ s.t. $g(k) = 2\cdot f(2k) + f(2k + 1)$ (essentially, $f$ is binary sequence, and if we group bits by $2$, we can think about this as a sequence of quaternary sequence).

So, $|F_2| \leq |F_3| \leq |F_4| = |F_2|$. From Cantor-Bernstein theorem it follows $|F_2| = |F_3|$.


问:
For each of the following relations, determine whether it is reflexive, symmetric or transitive
a) $X$ is the set of real numbers, $n$ is a natural number
 $$R =  \{(x, y) \mid x, y \in X, x^n = y^n\}.$$

b) $X$ is the set of people in the world
$$R = \{ (x, y) \mid x, y \in X, x\text{ and }y\text{ share a parent}\}$$

I believe both relations are reflexive, symmetric and transitive, aren't they?

I mean, for a) you can have $x^n = x^n \rightarrow (x,x)$, you can have $(x,y)$ and $(y,x) \rightarrow$ symmetric and you can have transitivity by $\leq.$

And for b) it's pretty much the same. As it doesn't say $x$ mustn't be equal to $y$, then you can have $(x,x)$ which is true. You can also have $(y,x) \rightarrow$ symmetry and you can also get transitivity for "if $(x,y)$ is true and $(y,z)$ is true, then $(x,z)$ is true," right?

答:
Relation (b) isn't transitive: consider half-siblings.

In detail: let $M(x)$ and $F(x)$ be the mother and father of person $x$. If $M(a)=M(b)$ and $F(b)=F(c)$ but $M(a)\ne M(c)$ and $F(a)\ne F(c)$ then we have $(a,b),(b,c)\in R$ but $(a,c)\not\in R$.


问:
The proof that a permutation and its conjugate have the same cycle structure from Cooper's notes
I've been idly reading "Cooper's Notes" on graph theory the past few days.

Following one example, he states (see $\S2.10$ Conjugates) that

"the permutation and its conjugate have the same cycle structure."

where a conjugate of a permutation $a$ is defined as $b^{-1}ab$ for an arbitrary permutation $b$ on the same set.

How interesting! I think. It's followed by a theorem:

Theorem 2: The conjugate of $a=(x_1\:x_2\:\dots\:x_n)\dots$ by $b$ is $c=a^b=(b(x_1)\:b(x_2)\dots\:b(x_n))\dots$

no problem yet. Then the proof begins:

Proof: We'll show that $ab=bc$ from which it follows that $b^{-1}ab=c$.

Wait a minute. First of all, $c$ is defined as $b^{-1}ab$. Shouldn't it be the other way around? Show that $b^{-1}ab=c$ means $ab=bc$?

Second, and more importantly, all that this shows is that $ab$ has the same cycle structure as $bc$. Why should $ab=bc$ mean that $c$ has the same cycle structure as $a$?

The proof itself is short, which I reproduce here:

Now $ab(x_1)=b(a(x_1))=b(x_2)$ and $bc(x_1)=c(b(x_1))=b(x_2)$ Thus
$ab$ and $bc$ have the same effect on the symbol $x_1$. Similarly they
have the same effect on any symbol in the cycle notation for $a$. If
$z$ is any other symbol then it's fixed by $a$ so
$ab(z)=b(a(z))=b(z)$. Since $z$ is not present in the cycle notation
for $c$ it's fixed by $c$ and so $bc(z)=c(b(z))=b(z)$. We've thus
shown that $ab$ and $bc$ behave identically on all symbols so $ab=bc$.

What am I missing? Why does this proof have anything at all to do with the Theorem? Note that while I'd be interested in seeing a different proof, I'm more interested in why this specific proof works.

答:
Let's call the theorem:

"the permutation and its conjugate have the same cycle structure."

Now take a permutation $\sigma$ written as a product of disjoint cycles

$$\sigma = c_1 \dots c_l$$

The conjugate is

$$a \sigma a^{-1} = (a c_1 a^{-1}) \dots (a c_l a^{-1}) \tag{1}$$ which according to theorem 2 is a product of same length cycles. That are disjoint because the conjugation is bijective. This proves the theorem.

Remember that conjugation is a group homomorphism: this is what equation $(1)$ states.


问:
Properties of vector multiplication: why is $(a^Tb)b = (bb^T)a$?
Suppose $a, b \in \mathbb{R}^{n}$. Why is it the case that $(a^Tb)b = (bb^T)a$?

I know that $a^Tb = b^Ta$, but I'm not sure why the above holds.

答:
Note that $a^Tb$ is a scalar. We have
$$
(a^Tb)b = b(a^Tb) = b(b^Ta) = (bb^T)a.
$$


问:
splitting a definite integral over an interval into infinitely many sub-intervals
if I break an interval $[a,b]$ into an infinite sequence of intervals $I_1, I_2, \dots, I_n$ and prove that $\forall i$ the function $f$ is Reimann integrable over $I_i$ can I say that $f$ is Reimann integrable over $[a,b]$?

答:
Not necessarily. Imagine splitting $[0,1]$ into subintervals $I_n = [1/(n+1),1/n]$ for $n \in \mathbb{N}$. Just because the function is integrable over $I_n$, resulting in $\int_{1/(n+1)}^{1/n} f = a_n$, that does not mean than $\sum a_n < \infty$.

But if you restrict to Riemann integration perhaps you can argue that $a_n \to 0$ since the interval width is shrinking...

UPDATE

Here is a definite counter-example. In the above scheme, let $f(x) = 1/x$ to get that
$$
\int_{1/(n+1)}^{1/n} \frac{dx}{x} = \ln(1/n) - \ln(1/(n+1) = \ln(1+1/n)
$$
which is well defined for any $n \in \mathbb{N}$ but
$$
\int_0^1 \frac{dx}{x} = \sum_{n=1}^\infty \ln(1+1/n),
$$
and of course both sides diverge.


问:
Given two positive real numbers $A$ and $l$, does there exist a plane, closed, simple curve with area $A$ and length $l$?
We are given two real positive numbers $A$ and $l$, such that $4\pi A \leqslant l^2$ (so the Isoperimetric inequality holds). Is it true we can always find a plane, simple, closed curve so that curve has length $l$ and region that it encloses has area $A$ ? If not, give a counterexample.

Intuitively, I think it is true, but I'm not sure have to define the curve with given properties.

答:
Let $s = \frac12\sqrt{\ell^2-4\pi A}$ and $r = \frac1{2\pi}(\ell - 2s)$.

Start from a circle of radius $r$, cut it along a diameter into two semicircles. Insert a $(2r) \times s$ rectangle between the "cut" and glue the two semi-circles to the two sides of rectangle with length $2r$. You obtain a "pill"-like figure of area $A$ and perimeter $\ell$.

The algebra is left as an exercise.


问:
$z^3 = i * \frac{|z^5|}{z* \bar z}$ Did I do it right?
I just want to check whether I understand the basic algebra of complex numbers.

I have to find solution to: $z^3 = i * \frac{|z^5|}{z* \bar z}$.

So I transform that expression into: $z^3 = i * \frac{|z^5|}{|z^2|} \iff z^3 = i * |z^3|$.

Then I take trigonometric form of $z^3$: $|z^3|[\cos(3\alpha) + \sin(3\alpha) i] = i * |z^3| \iff \cos(3\alpha) + \sin(3\alpha) i = i \iff 3\alpha = \frac{\pi}{2} \iff \alpha = \frac{\pi}{6}$

So ultimately I get $|z|(\frac{\sqrt{3}}{2}+\frac{1}{2}i)$. Is that the final solution? Did I get it right?

答:
Watch our for more solutions (there are lots of them)!
Note that, if
$$z^3 = i |z^3|$$
then $\alpha = z/|z|$ satisfies $\alpha^3 = i$. Hence $\alpha$ is a cube root of $i$, i.e. $\alpha$ could be any of
$$ \alpha_k = e^{(\pi/2 + 2\pi i k)/3}$$
for $k= 0, 1,2$. Therefore $z$ is of the form $z = |z|\alpha_k$ for some $k=0, 1,2$. Since this equation does not force any restriction on the size of $|z|$ we see that there are infinitely many solutions each of the form $r \alpha_k$ for some $k=0,1,2$ and $r \in \mathbb{R}$.

Edit: The single solution you have found is correct but there are more (so it's not `bad' but just missing all of the solutions). In your calculation you have $\cos(3\alpha) + \sin(3\alpha) i = i  \iff 3\alpha = \frac{\pi}{2}$. This is not correct as it misses out the other two possibilities $3\alpha = \frac{5\pi}{2}$ and $3\alpha = \frac{7\pi}{2}$ - you should really check all of the possible solutions to $3\alpha = \frac{\pi}{2} + 2\pi  k$ for $k \in \mathbb{N}$ and see that there are $3$.


问:
Application of Peter-Weyl theorem: groups with no small-subgroups admit a faithful finite-dimensional representation
My version of the Peter-Weyl theorem says that if $G$ is a compact group, then the matrix coefficients of $G$ are uniformly dense in $C(G)$. Consequently, the matrix coefficients are also dense in $L^2(G).$

Consider the following fragment from Bump's book "Lie groups":

I have two questions:

(1) Why does the Peter-Weyl theorem imply that such a matrix coefficient $f$ exists?

I tried to construct some continuous function with Urysohn's lemma and then use the Peter-Weyl theorem to approximate this but I do not succeed in forcing $f(1)=0$. Maybe I should use a translation? How should I construct $f$?

(2) Why if $f$ is constant on the kernel of $\pi$, we have $\ker \pi \subseteq U$?

Maybe we can show that $f < 1$ on $\ker \pi$?

答:
Building upon the OP's idea of using Urysohn's lemma the argument can be completed as follows.

Because $G$ is normal as a topological space, Urysohn's lemma implies the existence of a continuous function $g:G\to\Bbb{R}$ such that $g(x)=2$ for all $x\notin U$ and $g(1)=0$.

By Peter-Weyl theorem, there exists a matrix coefficient $f$ of $G$ such that the sup-norm $||f-g||<1/2$.

So for the matrix coefficient $f$ from step 2 we have $f(1)<1/2$ and $f(x)>3/2$ for all $x\in U$.

Consider the function $h:G\to\Bbb{C}$ defined by $h(g)=f(g)-f(1)$. The space of matrix coefficients of $G$ is closed under linear combinations, and contains the constant functions. Hence $H$ is a matrix coefficient of some representation $\pi$ of $G$.

By definitiion $h(1)=f(1)-f(1)=0$ and for all $x\notin U$ we have $h(x)=f(x)-f(1)>1$.

If $g\in G$ is in the kernel of the representation $\pi$, then $\pi(g)=\pi(1)$. Because $h$ is a matrix coefficient of $\pi$, this implies that $h(g)=h(1)=0$. Therefore $g\in U$, and thus $\operatorname{ker}(\pi)\subseteq U$.

As $U$ contains no non-trivial subgroups, we can conclude that $\operatorname{ker}(\pi)$ is trivial.


问:
Is it ever proper to say that the limit of a function equals infinity?
If I calculate a limit and get the value $\infty$, what is the proper way to communicate this? Can I say that the $\lim_{n\to\infty}a_n=\infty$ and therefore the sequence $\{a_n\}$ diverges, or do I say that the limit does not exist?</p>

答:
I would say the limit "tends to" $\infty$ and therefore your sequence diverges.

When working out limits, what you're doing is looking at what happens as you get very close to some point. So when working out the limit as $n \to \infty$, you're looking to see what happens as $n$ gets very big and as you're not working (whatever) out at a specific point, I wouldn't say its correct to say it "equals". "Tends to" means that it gets close to that point.


问:
Solve differential equation (Double integral)

Problem

Consider the following differential equation, where u = $\sqrt{x^2 + y^2 + z^2}$
$$\frac{d^2{f}}{du^2} + \frac{2}{u}\frac{df}{du} = \frac{d}{u^2}(u^2 \frac{df}{du}) = 0$$
And we are to determine a general solution of this ordinary differential equation for $u > 0$.
In short solve for $f$

My attempt

$$\frac{1}{u^2}\frac{d}{du}(u^2\frac{df}{du}) = 0$$
$$\frac{d}{du}(u^2\frac{df}{du}) = 0$$
Now I'm stuck at integrating it twice to solve for $f$, because when you integrate you have to integrate with regards to a variable, being $u$. So you get:
$$\int\int\frac{d}{du}(u^2\frac{df}{du}) du? = \int\int0 du ?$$
Doesn't the $du$'s take each other out? Leaving us with an integral which we do not know with regards to what variable to integrate from?

Body: $$u^2f''+2uf'=(u^2f')'=0\to u^2f'=c_0\to f'=\frac{c_0}{u^2}\to f=\frac{c_1}u+c_2.$$


问:
Coefficient of $x^3$ in $(1+2x-x^2+3x^3)^4$ and $(1+2x-x^2+3x^3+4x^4)^4$
Let $a$ and $b$ be the coefficient of $x^3$ in $(1+2x-x^2+3x^3)^4$ and $(1+2x-x^2+3x^3+4x^4)^4$ respectively. Then find the value of $a-b$.

My approach is as follow

$T=2x-x^2+3x^3$

$T^2=x^2(2-x+3x^2)^2$

$T^3=x^3(2-x+3x^2)^3$

$T^4=x^4(2-x+3x^2)^4$

$(1+T)^4=^4C_0+^4C_1T+^4C_2T^2+^4C_3T^3+^4C_4T^4$

After this step not able to approach as it is getting complicated

答:
Let $A = 1 + 2x - x^2 + 3x^3$ and $B = 4x^4$ then the difference of your both polynomials is:

$$
\begin{align}
A^4 - (A+B)^4 &= A^4 - (A^4 + 4A^3B + 6A^2B^2 + 4AB^3 + B^4) \\
&= -4A^3B - 6A^2B^2 - 4AB^3 - B^4
\end{align}
$$

Because $B$ has $x^4$, all polynomial terms with a multiple of $B$ have no coefficient for $x^c$ with $c < 4$. Therefore all the coefficients $0$ to $3$ of the difference polynomial are garanteed to be $0$.

Therefore $a - b = 0$.


问:
Let $O$ be any point inside $\Delta ABC$. Extend $AO,BO,CO$ to meet $BC,CA,AB$ at $D,E,F$ respectively.

Let $O$ be any point inside $\Delta ABC$. Extend $AO,BO,CO$ to meet $BC,CA,AB$ at $D,E,F$ respectively. If $AO = 30, FO = 20, BO = 60, DO = 10, CO = 20$, find $EO$.

What I Tried: Here is a picture in Geogebra :-

The statement of the problem is really simple, every length surrounding point $O$ is given except $EO$ , which we have to find. My question is, how?

I am only given the lengths, which I don't even know how to use them. They neither help in finding lengths of other unknown lengths, nor help in angle-chasing, and so on. I don't find any similar triangles where I could have used the lengths. The only thing which I find interesting is $FO = CO$. Maybe we can use them in the areas of the triangles considering them as bases, but I can't see to find a way out there. Also that dosen't make the point $O$ special, does it, or am I missing something?

Can anyone help me or give me some hints? Thank You.

答: Hint: Apply Melanaus theorem in $\Delta ABD,\Delta CBF,\Delta ADC,\Delta AFC,\Delta CFA,\Delta BEA$


问:
$y_1(t) = t^2$ and $y_2(t) = t|t|$ linearly independent on [-1,1]
For $t<0$, we have $t|t|= -t^2$. For $t>0$, we have $t|t|=t^2$.

Hence the Wronskian $$W[y_1,y_2](t) = y_1 y_2' -y_1' y_2  \\
= t^2 \cdot (-2t) - (2t)\cdot(-t^2) =  0 \quad t<0 \\ t^2\cdot (2t)-(2t)\cdot t^2 = 0 \quad t>0\ .
$$
In either case, the Wronskian is zero. How are these functions then linearly independent on [-1,1]?

Context: this is problem 2.1.11 of Braun's Differential Equations and Their Applications, 4th edition.

(a) Show that $y_1$ and $y_2$ are linearly dependent on the interval $0\leq t \leq 1$.

(b) Show that $y_1$ and $y_2$ are linearly independent on the interval $-1\leq t \leq 1$.

答:
The Wronskian test seems inconclusive in this context. As you have pointed  out, $t\lvert t \rvert = -t^{2}$ for $t < 0 $ and $t \lvert t \rvert = t^{2} $ for $t \ge 0$  ( "$\ge$" since the absolute value has $0$ inclusive) so this translates to the system of equations
\begin{align*}
c_{1} t^{2} + c_{2}t^{2} &= 0 \quad ,t \ge 0  \\
c_{1}t^{2} - c_{2}t^{2} &= 0 \quad ,t <0
\end{align*}
Solving this system of equation yields
$$
\frac{t^{2}}{t^{2}}\begin{pmatrix}
1 & 1 & | \ \ \ 0 \\ 1 & -1  &| \ \  \ 0
\end{pmatrix} \to \begin{pmatrix} 1 & 0 & |\ \ \  0  \\ 0 & 1 & |\ \ \  0 \end{pmatrix} \implies c_{1} = c_{2} \equiv 0,$$
the system is linearly independent for arbitrary $t$ so it is linearly independent for $t \in (0,1)$.


问:
Prove or disprove: $\forall p\in[0,1),\,\int_0^1 x\oplus p\,dx=\frac12$, where $\oplus$ is the bitwise XOR operation.
For $x\in[0,1)$, let $(x_k)\in\{0,1\}^{\mathbb N^*}$ be its bits coefficients, i.e.
$$x=\sum_{k=1}^\infty x_k2^{-k}.$$

I want to prove (or disprove) that

$$\forall p\in[0,1),\,\int_0^1\sum_{k=1}^\infty\left(x_k\oplus p_k\right)2^{-k}\,dx=\frac12,$$

where $\oplus$ is the XOR operation, i.e.

$$x_k\oplus p_k=
\begin{cases}
1\quad\text{if $x_k\neq p_k$}\\
0\quad\text{if $x_k=p_k$}
\end{cases}=(x_k+p_k) \mod2.$$

The idea is very simple, when we do the XOR with $p$ for all numbers in $[0,1]$, the results are kind of uniformly distributed in $[0,1]$, thus we'll finally have $\frac12$.

The problem here is how to prove this observation using formal mathematical language, noticing that the function $f:x\mapsto x\oplus p$ is very strange, not continue, not Riemann integrable (but measurable and bounded).

答: Use the King property $\int_0^1f(x)dx=\tfrac12\int_0^1(f(x)+f(1-x))dx$. Since$$(x\oplus p)+((1-x)\oplus p)=\sum_k2^{-k}=1$$(with countably many exceptions, namely the dyadic fraction values of $x$), the original integral is $\tfrac12\int_0^1dx=\tfrac12$.


问:
A left $K$-module is topologically free if and only if it is separated complete and torsion free.
Following Christian Kassel's "Quantum Groups", page 390.

Prop: Let $K= \mathbb{C}[[h]]$, the algebra of formal power series in one indeterminant over the complex numbers. A left $K$-module is topologically free if and only if it is separated complete and torision free.

$(\rightarrow$) is trivial

Proof of $(\leftarrow$): Let $M$ be a separated, complete, and torsion-free module. We have to show that $M$ is of the form $V[[h]]$ for some complex vector space $V$. Choose a vector subspace of $M$ which is suplementary to $hM$. Because of the torision-free assumption, we get $h^nM = h^nV \oplus h^{n+1}M$ for all $n \geq 0$.

Hence: $$ \frac{M}{h^nM}=V \oplus hV \oplus \cdots \oplus h^{n-1}V = \frac{V[[h]]}{h^nV[[h]]}$$

Taking inverse limits and using the fact that $M$ and $V[[h]]$ are separated and complete, we get

$$M \cong V[[h]]$$

Can somebody explain the bold part to me? Thanks!

答:
View $M$ as a vector space over $\mathbb{C}$ and $hM$ as its subspace. Then there is a subspace $V \subseteq M$ such that $M = V \oplus hM$ (direct sum of $\mathbb{C}$-vector spaces).

The mapping $\varphi \colon M \longrightarrow M$, given by $\varphi(m) = hm$, is a linear transformation, because $M$ is a $K$-module. Applying it $n$ times to the decomposition above we get $h^n M = h^n V \oplus h^{n+1} M$. This allows us to write

$$ M/hM = V \oplus hV \oplus \dots \oplus h^{n-1}V \ \ \text{(again as $\mathbb{C}$-vector spaces)}.$$

I think the only place where we use torsion freeness of $M$ is in the following identification

$$ V \oplus hV \oplus \dots \oplus h^{n-1}V \cong V[[h]]/h^nV[[h]] $$

because we want $V \cong h^n V$ for all $n \in \mathbb{Z}_+$.

Viewing $M/hM \cong V[[h]]/h^nV[[h]]$ again as $K$-modules and taking completions we get the desired topological freeness.


问:
Show that these two integrals are the same
I am trying to prove this. Where do I start?
$$ \int_{0}^{x}\Biggl[\int_{0}^{t} F(u)\,du\Biggr]dt = \int_{0}^{x}(x-u)F(u)\,du$$

答:
Integration by parts using  $u=\int_{0}^{t}{F}(u)du\quad ,dv=dt$
$$=\left. t\int_{0}^{t}{F}(u)du \right|_{0}^{x}-\int_{0}^{x}{tF\left( t \right)dt}=x\int_{0}^{x}{F}(u)du-\int_{0}^{x}{tF\left( t \right)dt}=\int_{0}^{x}{\left( x-t \right)F\left( t \right)dt}$$


问:
Complex analysis - inequalities
Prove
$|e^z-1|\leq |e^x-1|+e^x|z|$
Where z=x+iy

I tried using $|e^z|=e^x$ and $e^x\geq1+x$ with no success

答:
Hint: By the triangle inequality,
$$|e^{z} -1 | = |e^{x+iy} - e^{x} + e^{x} - 1| \le |e^{x}-1|+ e^x|e^{iy} -1|.$$
Now sketch the unit circle in $\mathbb{C}$ and note that the distance between $e^{iy}$ and $1$ is less than the length of the arc of the circle joining $e^{iy}$ and $1$. The length of this arc is $y$ (modulo $2\pi$) and so....?

Title: Complex analysis - inequalities


问:
Calculate the area of the sphere $x^2+y^2+z^2=2Rz$ where $R>0$ inside of the cone $x^2+y^2=z^2$ where $z \geq 0$

Calculate the area of the sphere $x^2+y^2+z^2=2Rz$ where $R>0$ inside of the cone $x^2+y^2=z^2$ where $z \geq 0$

Attempt

First we should use the elemental area formula given by
$$\int_{S}f \dot dS=\int_{D} ||T_{u} \times T_{v}|| dudv$$ where $T_{u}$ and $T_{v}$ are the tangent vectors to the surface  $S$ which is the surface parametrized .

Notice that these area is the upper semi sphere, since the sphere has center $C=(0,0,R)$ and radius $R$ and it is given by $$x^2+y^2+(z-R)^2=R^2$$  we only need to parametrize the sphere and get the upper area.

Let the parametrization
$$x=R\cos \theta \sin \phi$$
$$y=R\sin\theta \sin \phi$$
$$z=R \cos \phi$$
notice that the upper sphere begin in $z=R$ this is one projection of the cone $x^2+y^2=z^2$ in $XY$ plane.
since
$$R\cos \phi=\sqrt{(R\cos \theta \sin \phi)^2+(R\sin\theta \sin \phi)^2}$$
$$R\cos \phi=\sqrt{R^2\cos^2 \theta \sin^2 \phi+R^2\sin^2\theta \sin^2 \phi}$$
$$Rcos \phi=\sqrt{R^2sin^2\phi}$$
$$Rcos \phi=R\sin\phi$$
$$\cos \phi= \sin \phi$$
since $\phi\in [0, \pi]$ we deduce that $0 \leq \phi \leq \frac{\pi}{4}$.

Now for calculating the tangent vectors to the surface we get
$$T_{\theta}=(-R\sin \theta \sin \phi)i+(R\sin \phi \cos \theta )j+0k$$
$$T_{\phi}=(R\cos \phi \cos \theta )i+(R\cos \phi \sin \theta)j+(-R \sin \phi)k$$
Now calculating the cross product
$$T_{\theta} \times T_{\phi}=(-R^2\sin^2 \phi \cos \theta)i+(-R^2\sin \phi \sin \theta)j+(-R^2 \cos \phi \sin \phi)k$$
$$||T_{\theta} \times T_{\phi}||=R^2\sin \phi$$
Finally applying the formula of the elemental área we get
$$\int_{0}^{2\pi} \int_{0}^{\pi/4}R^2 \sin \phi d\phi d\theta=\pi R^2(2-\sqrt{2})$$

Is my answer right or have I done a mistake with the interpretation of the problem.

答:
Please note this is a sphere with center at $(0, 0, R)$ and radius $R$.

So the correct parametrization is $ \, \rho = 2R \cos \phi$

$x = \rho \cos \theta \sin \phi = 2R \cos \theta \sin \phi \cos \phi = R \cos \theta \sin2\phi$

Similarly,

$y = R \sin \theta \sin2\phi$

$z = \rho \cos \phi = 2R \cos^2\phi$

$T_{\theta}=(-R\sin \theta \sin 2\phi, R\cos \theta \sin 2\phi,0)$

$T_{\phi} = (2R\cos \theta \cos 2\phi, 2R\sin \theta \cos 2\phi, -2R \sin 2\phi)$

You can now find $|T_{\theta} \times T_{\phi}|$.

You should see the integral become -

$\displaystyle \int_{0}^{2\pi} \int_{0}^{\pi/4} 2R^2 \sin 2\phi \, d\phi \, d\theta$


问:
Low rank approximation using the SVD
I have been looking at some LSI implementation which assumes that the following property always hold:

Given the Singular Value Decomposition of a real matrix $A$, $$A = U\Sigma V^T$$
and its approximation at the first $k$ singular values:
$$\hat{A_k} = U_k \Sigma_k V_k^T$$
it holds:
$$\hat{A_k} = U_k \Sigma_k V_k^T=AV_kV_k^T$$

Can you help me proving that this is true (or false) ?

Edit:
I have found this proof.

The $V$ matrix is orthonormal, so $V^{-1} = V^T$, therefore:
$$AV = U\Sigma$$
and then truncating at the same level:
$$AV_k = U_k \Sigma_k$$

However, I'm not very convinced by this last step, do you think that it is correct?

答:
The last step is fine, just break $AV=U\Sigma$ into blocks and carry out the matrix multiplication. In particular, suppose $A$ is $n\times m$ and of rank $r$ and that $k\leq r$. Then,

$$\underbrace{U}_{n\times r}=\begin{bmatrix}U_k|\tilde{U}\end{bmatrix},\quad \underbrace{\Sigma}_{r\times r}=\begin{bmatrix}\Sigma_k&|&0\\\hline0&|&\tilde{\Sigma}\end{bmatrix},\quad \underbrace{V^T}_{r\times m}=\begin{bmatrix}V_k^T\\\hline \tilde{V}^T\end{bmatrix}$$

where $U_k$ is $n\times k$, $\tilde{U}$ is $n\times r-k$, $\Sigma_k$ is $k\times k$, $\tilde{\Sigma}$ is $r-k\times r-k$, $V_k$ is $m\times k$ and $\tilde{V}$ is $m\times r-k$. So,

$$AV=A\begin{bmatrix} V_k |\tilde{V}\end{bmatrix}=\begin{bmatrix} AV_k |A\tilde{V}\end{bmatrix},$$

$$U\Sigma=\begin{bmatrix}U_k|\hat{U}\end{bmatrix}\begin{bmatrix}\Sigma_k&|&0\\\hline0&|&\tilde{\Sigma}\end{bmatrix}=\begin{bmatrix}U_k\Sigma_k|\tilde{U}\tilde{\Sigma}\end{bmatrix}.$$

Thus,

$$AV=U\Sigma\Leftrightarrow \begin{bmatrix} AV_k |A\tilde{V}\end{bmatrix}=\begin{bmatrix}U_k\Sigma_k|\hat{U}\tilde{\Sigma}\end{bmatrix}\Rightarrow AV_k=U_k\Sigma_k.$$

Title: Low rank approximation using the SVD


问:
Are these proofs regarding the infimum and supremum correct?
I would like to know if I solved correctly the following (well known/easy) problems about infimums and supremums. I couldn't find these precise proofs anywhere and I wish to know if they are all right.

Problem 1. Let $A\subset R$ be a bounded non-empty set. Show that $\sup(-A)=-\inf(A)$, where $-A=\{-a | a\in A\}$.

Solution. For any $a\in A$ we have that $a\ge \inf A \iff -a\le -\inf A$ and from here it follows that $x\le -\inf A, \forall x\in -A$, so $-\inf A$ is an upper bound for $-A$.

We have a sequence $(x_n)_{n\ge 1}\subset A$ such that $x_n \to \inf A$. This means that $-x_n\to -\inf A$ and $(-x_n)_{n\ge 1}\subset -A$ so now we may say that $\sup(-A)=-\inf A$.

Problem 2. Find $\inf B$ and $\sup B$ if $B=\{\frac{m}{n} | m,n \in \mathbb{N}, m<2n\}$.

Solution. Obviously, $0<x<2$ for any $x$ in $B$, so $0$ is a lower bound and $2$ is an upper bound.

We have $(x_n)_{n\ge 1} \subset B$, $x_n=\frac{1}{n}\to 0$, so we may say that $\inf B=0$, and we also have that $(y_n)_{n\ge 1}\subset B$, $y_n=\frac{2n-1}{n}\to 2$, so we may say that $\sup B=2$.

I think that these proofs are all right, but most of the infimum supremum proofs I have seen used epislons. I just want to make sure that my solutions are correct too.

答: The proofs are correct. However, the first proof has a useless part. That occurs after you prove that, if $a\in A$, then $-a\leqslant-\inf A$. For some reason, you decided to deduce from this that, if $x\in A$, then $-x\leqslant-\inf A$. Why? It's the same assertion as the one that you had just proved. You simply replaced $a$ with $x$, but that changes nothing.


问:
Does the Fundamental Theorem of Algebra follow from the real Spectral Theorem?
I found a slick way to prove that every real normal matrix is orthogonally equivalent to a block diagonal form consisting of diagonal parts and $2 \times 2$ parts of the form $$\begin{pmatrix}a & b \\ -b & a\end{pmatrix}.$$

This seems awfully close to proving FTA - if every real polynomial is the characteristic polynomial of a normal matrix, then every real polynomial consequently splits over $\mathbb C$. So my question is:

Is there a  slick way to find a real normal matrix given a characteristic polynomial?

Note that the companion matrix is not normal in general.

答:
An algebraic solution cannot exist; you can adjoin the eigenvalues of rational symmetric matrices to $\mathbb Q$ to obtain a field which satisfies the spectral theorem (symmetric matrices diagonalize orthogonally) but is not a real closed field.

Definition. Let $F \subseteq K$ be an extension of ordered fields. Then $\alpha \in K$ is symmetric over $F$ if $\alpha$ is an eigenvalue of a symmetric matrix with entries in $F$. $K$ is a symmetric extension of $F$ if every element of $K$ is symmetric over $F$.

Lemma. Let $A$ be an $n \times n$ symmetric matrix over an ordered field $F$, and let $p, q \in F[x]$ be coprime polynomials. Then $\ker p(A)$ and $\ker q(A)$ are orthogonal subspaces of $F^n$, and therefore intersect trivially.

Proof. Let $ap+bq=1$ and use the substitution $a(A)p(A)$ to assume WLOG that $p(x) = x$ and $q(x) = 1-x$. Then use the fact that the kernel of a symmetric matrix is orthogonal to its image.

Lemma. A symmetric matrix over an ordered field is semisimple.

Proof. Apply the spectral theorem to a real closure of the field of coefficients.

Theorem. Let $A$ be a symmetric matrix over an ordered field $F$. Then $A$ is orthogonally equivalent to a block diagonal matrix in which each block has irreducible characteristic polynomial.

Lemma Let $A, B$ be symmetric matrices over an ordered field $F$, of size $n \times n$ and $m \times m$ respectively. Then polynomials in $A \otimes I_m$ and $I_n \otimes B$ witness the fact that the symmetric elements of $K$ in an extension $F \subseteq K$ form a subfield of $K$.

Lemma Let $F \subseteq K \subseteq L$ be a chain of extensions of ordered fields. If $K/F$ is symmetric and $L/K$ is symmetric, then $L/F$ is symmetric.

Thus we can speak of a "symmetric closure" of $\mathbb Q$ in $\mathbb R$, consisting of roots of irreducible polynomials over $\mathbb Q$ which are the characteristic polynomials of rational symmetric matrices. In particular, every root of such a polynomial must be real, so $x^4-2$ does not have a root in the symmetric closure of $\mathbb Q$. Since $\sqrt[4]{2}$ must exist in a real closed field, this proves that there are fields satisfying the spectral theorem which are not real closed.

This answer suggests that it is an unsolved problem to determine which rational polynomials are the characteristic polynomial of a symmetric matrix.


问:
Two sided Ideals of a simple ring
Suppose $R$ is semisimple and all left simple ideals of $R$ are isomorphic.

I want to show that except $0$ and $R$, there is no two sided ideal in $R$.

Let $J$ be a two sided ideal. So it is a left ideal of $R$, hence it is a left $R$-module semisimple. So $J$ is a sum of simple left ideals of $J$.

My goal is to show that all simple left ideals of $J$, are also simple left ideals of $R$.

答: Let $I$ be the unique, up to isomorphism, simple left ideal of $R$. Let $D$ be the division ring $\operatorname{End}_R(I)$. By semi-simplicity of $R$ there is a positive integer $n$ with ${}_RR\simeq\bigoplus_{i=1}^nI$ and thus $R^\mathrm{op}\simeq\operatorname{End}({}_RR)\simeq M_n(D)$ i.e. $R\simeq M_n(D^\mathrm{op})$. Since for any ring $S$ one has a one-to-one correspondence between $\{$two-sided ideals of $S\}$ and $\{$two-sided ideals of $M_n(S)\}$ via $J\longrightarrow M_n(J)$, $R$, which is isomorphic to $M_n(D^\mathrm{op})$, has only two two-sided ideals: $0$ and itself.

Title: Two sided Ideals of a simple ring


问:
Compute derivative of real valued tensor function (second invariant)
In the context of continuum mechanics, given a tensor $A$, the second invariant $i_2(A) = \frac{\text{tr}(A)^2 - \text{tr}(A^2)}{2} $. I need to compute its derivative $Di_2(A)[S]$ (I'm using the notation from Gurtin's book - An introduction to continuum mechanics), where $S$ is a tensor.
(see here for better comprehension)
I will analyize one term at a time:

$D\text{tr}(A)^2[S] = 2\text{tr}(A)[D \text{tr}(A)[S]] = 2 \text{tr}(A)\text{tr}(S)$

$D\text{tr}(A^2)[S] = \text{tr}[DA^2[S]] = \text{tr}(AS+ SA)$

where I used the fact that $D \text{tr}(A)[S] = \text{tr}(S)$ because the trace is a linear operator.

So I would obtain $$D i_2(A)[S] = \text{tr}(A)\text{tr}(S)  - \frac{1}{2} \text{tr}(AS+SA)$$

Is this correct?

答:
Yes, your answer is right, though for the second term, your notation is misleading. Taken literally, $D\text{tr}(A^2)[S]$ means the derivative of trace at the point $A^2$ evaluated on $S$. This would just equal $\text{tr}(S)$, since trace is linear. Though, what you meant is that
\begin{align}
D\left(A\mapsto \text{tr}(A^2)\right)_A[S] &= [(D\text{tr}_{A^2})\circ \left(D(A\mapsto A^2)_A\right)][S]\tag{chain rule}\\
&= \text{tr}(AS + SA).
\end{align}
Here, I put the point of evaluation of derivatives in subscript to avoid excessive bracketing.

This is just one of those issues which comes up over and over in differential calculus, about functions vs function values.


问:
How does the Kronecker delta select coodinates?
I'm reading A Gentle Introduction to Tensors. In introducing the Kronecker delta it says it "[...] it may be used as a coordinate selector, in the following sense:"

$\delta_j^i x^j = x^i$

Where superscript indices are rows, and subscript indices are columns. My current understanding is that this does nothing to $x$ besides change $i$ to $j$, which doesn't seem to be meaningful in this context, as it is still the same column vector. How does the Kronecker delta select coordinates? Is this just a tool to change tensor indices in equations?

答:
Here is the intuition. Recall that repeated indices imply summation. To put an example consider 3 dimensions only. Hence
$$
\delta^i_jx^j = \delta^i_1x^1+\delta^i_2x^2+\delta^i_3x^3
$$

However, remember that $\delta_j^i=0$ for $i\neq j$ and $\delta_i^i=1$. Therefore, among $\delta^i_1,\delta^i_2,\delta^i_3$ only one of those 3 quantities is nonzero. For example if $i=2$ then $\delta^i_1=0,\delta^i_2=1,\delta^i_3=0$
Therefore, $\delta^2_jx^j=\delta^2_1x^1+\delta^2_2x^2+\delta^2_3x^3 = x^2$ and this is how the delta selects coordinates.

So in general among all terms involved in $\delta^i_jx^j$ just the $i$-th terms survives: the one with $x^i$. Therefore $\delta^i_jx^j=x^i$.

Moreover $\delta^i_jx^j$ is an operation over the whole vector $x$, and the result $x^j$ is just a component. This technique is one of the most useful rules to "play" with indices. Here is an example where this identity is useful apart from just "interchanging" indices in another answer of mine here.


问:
Isomorphism between $\mathbb{Z}[i]/{p}$ and $\mathbb{F}^2_p$
Say $p$ is a prime number such that $p=4k+1$. I need to prove that $\mathbb{Z}[i]/p$ (quotient ring) and $\mathbb{F}^2_{p}$ are isomorphic. I know that since $p \equiv 1 \mod4$ I can say that $p=xy$ where $x,y$ are conjugate prime numbers in $\mathbb{Z}[i]$, so I tried using the CRT (Chinese remainder theorem) - so I know that $\mathbb{Z}[i]/p$ is isomorphic to $\mathbb{Z}[i]/x\times\mathbb{Z}[i]/y$, but I wasn't able to take it from there. Any hint would be helpful.

答:
More precisely we have
$$\mathbb{Z}[i]/(p) \cong \mathbb{F}_p[X]/(X^2+1) \cong
\begin{cases}
\mathbb{F}_{p^2} &\text{if } p \equiv 3 \pmod 4\\
\mathbb{F}_{p} \times \mathbb{F}_{p} &\text{if } p \equiv 1 \pmod 4
\end{cases}$$

So the comment on whether it is written $\Bbb F_{p^2}$ or $\Bbb F_p^2$ was really meaningful.

Reference: Quotient ring of Gaussian integers $\mathbb{Z}[i]/(a+bi)$ when $a$ and $b$ are NOT coprime


问:
Generating $S_7$ only by $7$-cycles
Do all the $7$-cycles in $S_7$ generate $S_7$? How?

Can we generalize the idea?

(Edited)

I found a new idea:

Assume $$\sigma = (x_1,x_2,x_3,x_4,x_5,x_6,x_7)$$ and $$\tau=(x_7,x_6,x_5,x_4,x_3,x_1,x_2)$$ Now: $$\sigma\tau=(x_1,x_2,x_3)$$
So, we can generate all $3$-cycles.

Theorem: Let $n ≥ 3$. Prove that $A_n$ can be generated by all the $3$-cycles in $S_n$.

What should I do for $S_n - A_n$?

Thank you for your answers.

答: No. A $7-$cycle has parity $+1$, therefore you cannot reach a transposition just using $7-$cycles.

Title: Generating $S_7$ only by $7$-cycles


问:
Is it ever proper to say that the limit of a function equals infinity?
If I calculate a limit and get the value $\infty$, what is the proper way to communicate this? Can I say that the $\lim_{n\to\infty}a_n=\infty$ and therefore the sequence $\{a_n\}$ diverges, or do I say that the limit does not exist?</p>

答:
Saying that the limit is infinity has the same precise meaning as writing $\lim_{n\to\infty}a_n=\infty$. Saying that the sequence diverges has not the same precise meaning, because the limit could also be $-\infty$.

I'd avoid phrases such as "the limit tends to"; the limit either doesn't exist or, if it exists, it is a real number, $\infty$ or $-\infty$.

In any case, it's a "local" question: terminology is not fixed; somebody uses "divergent" as synonymous to "non convergent", for instance. But what you're allowed to write you should be also allowed to say.


问:
Function $f(x,y)$ analytic with respect to both $x$ and $y$, is $\frac{\partial}{\partial y} f(x,y)$ analytic with respect to $x$?
Assume $U$ is some open set in $\mathbb{R}$. Let $f: U \times U \rightarrow \mathbb{R}$ be a function such that for a fixed $y$ $$x \mapsto f(x,y)$$ is analytic, and for a fixed $x$ $$y \mapsto f(x,y)$$ is analytic. Is it true that for a fixed $y$ $$x \mapsto \frac{\partial}{\partial y} f(x,y)$$ is also analytic?

答:
See edit before reading.

The answer is yes but requires some work. If $x \mapsto f(x,y)$ is analytic for any fixed $y \in U$ and $y \mapsto f(x,y)$ is analytic for any fixed $x\in U$ then by Hartogs' Theorem:

https://en.wikipedia.org/wiki/Hartogs%27s_theorem

$f(x,y)$ is analytic in the 2-variable sense, i.e. it admits a power series expansion in 2 variables for $(x,y) \in U  \times U$. In particular, you can differentiate this power series with respect to $y$ and obtain a power series expansion for $x \mapsto \frac{\partial}{\partial y} f(x,y)$ for $x \in U$. Hence your desired map is analytic.

Edit: I misread the question thinking that $U$ was a subset of $\mathbb{C}$. The answer to the question is in fact no and a counter example can be found in the link to the Wikipedia article above.


问:
Probability inequalities with indicator variables
Consider a non-negative random variable $X$ such that $\mathbb{E}(X^2)>0$ is finite, and let $t\in(0,1)$. I need to prove that
$\mathbb{E}\left[X\mathbb{I}_{X>t\mathbb{E}(X)}\right]\geq(1-t)\mathbb{E}(X)$ and $\mathbb{P}\left[X>t\mathbb{E}(X)\right]\geq(1-t)^2\frac{\mathbb{E}(X)^2}{\mathbb{E}(X^2)}$. I know that $\mathbb{E}\left[\mathbb{I}_{X>t\mathbb{E}(X)}\right]=\mathbb{P}[X>t\mathbb{E}(X)]$, and I feel like I should use the tower law somehow to deal with the product with the indicator variable, but I am not sure what to condition on. For the second inequality, I suspect Cauchy-Schwarz will come into play, but I can't really see a path to success. Could someone please guide me on this? Thanks!

答:
For the first inequality, write $X = X \cdot \mathbb{I}_{X > t\mathbb{E}(X)} + X \cdot \mathbb{I}_{X \leq t \mathbb{E}(X)}$ and take expectations of both sides:
$$
\mathbb{E}(X) = \mathbb{E}(X \cdot \mathbb{I}_{X > t \mathbb{E}(X)}) + \mathbb{E}(X \cdot \mathbb{I}_{X \leq t \mathbb{E}(X)}) \leq \mathbb{E}(X \cdot \mathbb{I}_{X > t \mathbb{E}(X)}) + t \mathbb{E}(X).
$$
Then rearranging gives $\mathbb{E}(X \cdot \mathbb{I}_{X > t \mathbb{E}(X)}) \geq (1-t) \mathbb{E}(X)$.

Now for the second inequality, use the first part and apply Cauchy-Schwarz to $\mathbb{E}(X \cdot \mathbb{I}_{X > t \mathbb{E}(X)})$:
$$
(1-t)\mathbb{E}(X) \leq \mathbb{E}(X \cdot \mathbb{I}_{X > t \mathbb{E}(X)}) \leq \mathbb{E}(X^2)^{1/2} \cdot \mathbb{E}(\mathbb{I}^2_{X > t\mathbb{E}(X)})^{1/2} = \mathbb{E}(X^2)^{1/2} \cdot \mathbb{P}(X > t\mathbb{E}(X))^{1/2}.
$$
Now square both sides and rearrange and you get the second inequality.


问:
Transpositions and idempotents in matrix algebras
Let $A$ be a semi-simple algebra over $\mathbb{C}$. An idempotent $x$ in $A$ is an element of $A$ such that $x^2 = x$. A minimal non-zero idempotent $y$ of $A$ is a non-zero idempotent of $A$ such that for each non-zero idempotent $x \in A$ satisfying $xy = x$ we have $x = y$. We have a linear 'transposition' map on $A$ that sends an element $x \in A$ to $x^T \in A$ with the following properties:

$1^T = 1$;

$(x^T)^T = x$  for all  $x \in A$;

$(x y)^T = y^Tx^T$ for all $x,y \in A$.

Is it true that there exists a set of minimal idempotents $x_1,\dots,x_n$ in $A$ such that

$x_i^T = x_i$ for each $i \in 1,\dots,n$;

$\sum_{i=1}^n x_i= 1$;

each idempotent $y$ of $A$ such that $y^T = y$ can be written as $\sum_{j=1}^k x_{i_j}$ for some $k \leq n$ with $i_j \in 1,...,n$ for each $j \in 1,...,k$.

If $A$ is a matrix algebra and $T$ is the normal transpose on $A$, then the diagonal matrices with exactly one $1$ and $0$ elsewhere do satisfy this condition.

答:
The product algebra $A=\Bbb{C}\times\Bbb{C}$ with $(x,y)^T:=(y,x)$ is a counter example since $A$ has only two minimal idempotents, $(1,0)$ and $(0,1)$ which are interchanged by $T$.

I expect the answer to be "yes" if you further imposed that $A\simeq\prod_{i=1}^rM_{n_i}(\Bbb{C})$ with distinct $n_i$: the "transpose map" ought to be of the form $x\mapsto g\cdot{}^tx\cdot g^{-1}$ for some $g\in A^\times$ and the usual "term by term transposition" $(x_1,\dots,x_r)\mapsto (~{}^tx_1,\dots,{}^tx_r) $.


问:
Limit as n approaches infinity involving roots
$$\lim_{n\to\infty}\frac{n}{1+2\sqrt{n}}$$

Given my understanding of how to solve these problems, I need to take the highest power of $n$ in the denominator and then divide both the numerator and denominator by that power. But the highest power in the denominator is lower than the one in the numerator, so I will still be left with $n^{1/2}$ in the numerator, which gives $\frac{\infty}{2}=\infty$. Is that the correct answer or am I missing something? I would have thought that since the power of the denominator is smaller than the numerator that you could factor it by long division, but I'm not sure I know how to do that.

答: $\forall$  $M >0$ exist $n_0$  such that $\forall n >n_0$ $|\frac{n}{1+2\sqrt n}|>|\frac{ n}{4\sqrt n}|>|\sqrt n|>\sqrt {n_0}>\sqrt{M}$ so$ |\frac{n}{1+2\sqrt n}|$ will go to infinite for n enough large</p>


问:
Independent and Identically distributed, conditional independent and Naive bayes
I'm reading about Naive Bayes classification concept, noting that we make the conditionally independence assumption. But isn't this the general assumption that is always made dealing with machine learning algorithms?

Suppose we have a supervised binary classification problem setup, with a dataset
$\mathcal{D} = \{(x_1,t_1), \dots, (x_n,t_n)\}$ where $x_i \in \mathbb{R}^D$ and $t_i \in\{0,1\} \, \, \forall i=1, \dots n$.

I've read everywhere that we always make the assumption that data are iid (independent and with the same probability dstribution, this would mean that $p((x_i, t_i),(x_j,t_j)) = p((x_i,t_i))p((x_j,t_j))$..right?). At this point it is reasonable to think of a Bernoulli distribution to model the data. Let $p(\mathcal{D}|\theta)$ the likelihood function: then we want to find

$$\hat{\theta} = arg\max_{\theta}p(\mathcal{D}|\theta)$$

where $p(\mathcal{D}|\theta) = p((x_1,t_1), \dots ,(x_n,t_n)|\theta) = p((x_1,t_1)|\theta)\, , \dots , \, p((x_n,t_n)|\theta)$ Here we should use a conditional independence hypothesis in order to go on. So in every situation we use the naive bayes hypothesis? I'm having troubles trying to distinguish..

答:
In Naive Bayes, we assume that all features in
$x$ are mutually independent, conditional on the category
$C_k$.

The features can be height and weight and the category, $C$ can be whether your BMI is higher than certain quantity.

Here given $C$, we assume that height and weight are independent. The conditional independence is described for the features.


问:
Understanding proof that $C_c^{\infty}$ is dense in $C^{\infty}$ for the convergence in $C^{\infty}$
Prior notations: $\mathcal{E} = C^{\infty}(\mathbb{R}^n)$ and $\mathcal{D}=C^{\infty}_c(\mathbb{R}^n)\subsetneq \mathcal{E}.$

Definition 'convergence in $\mathcal{E}$': $\psi_j\to \psi$ in $\mathcal{E}$ iff for each multi-index $\alpha$ we have that $D^{\alpha}\psi_j \to D^{\alpha}\psi$ uniformly on every compact subset of $\mathbb{R}^n$ (here: $D = \partial_{x_1}\dots \partial_{x_n}$).

The subspace $\mathcal{D}$ is dense in $\mathcal{E}$ for convergence in $\mathcal{E}$.

Proof. Let $\phi\in\mathcal{E}$. For each $j\in\mathbb{N}$, consider $\psi_j\in\mathcal{D}$ such that $\psi_j=1$ on $B(0,j)$. Then $\psi_j\phi \to \phi$ in $\mathcal{E}$.

Questions: why do such test functions $\psi_j$ exist?

As for the convergence, is this reasoning correct: Let $K$ be a compact subset of $\mathbb{R}^n$ and suppose $K\subseteq B(0,j)$ for sufficiently large $j$. Then $\psi_j\phi = \phi$ on $K$, so it's clear that $D^{\alpha}(\psi_j\phi)\to D^{\alpha}\phi$ uniformly (since we will have $\forall k\ge j: \forall x\in K: 0=|(\psi_k\phi)(x)-\phi(x)|<\varepsilon, \forall \varepsilon >0$ and similarly for all possible derivatives $D^{\alpha}$).

Thanks in advance.

答:
The convergence proof looks good to me.

Existence of test functions with prescribed support

First consider
$$f(t)=\left\{\begin{array}{ll}
0 & \text{ if }t\leq 0\\
e^{-1/t} & \text{if }t>0\end{array}\right.$$
You can show that $f$ is $C^{\infty}$ by induction. Now consider
$$g(t)=\dfrac{f(2-t)}{f(2-t)+f(t-1)}$$
It is $C^{\infty}$, $g(t)=1$ if $t\leq 1$ and $g(t)=0$ if $x\geq 2$. Finally
$$\psi(x)=g(\|x\|)$$
is a $C^{\infty}$ function in $\mathbb{R}^n$ such that $\psi(x)=1$ when $\|x\|\leq 1$ and $\psi(x)=0$ when $\|x\|>2$. It can be scaled with
$$\psi_j(x)=g\left(\dfrac{1}{j}\|x\|\right)$$
so that $\psi_j(x)=1$ on $B(0,j)$.

Note: This is the constructive approach. You can also construct such test functions by convolution methods.


问:
Can someone explain the isomorphism $\frac{V[[h]]}{h^nV[[h]]}\cong V \otimes \frac{\mathbb{C}[h]}{h^n\mathbb{C}[h]}$
Let $V$ be a complex vector space, $V[[h]]$ the ring of former power series over $V$, $\mathbb{C}[h]$ the complex polynomial ring over $h$. Can somebody explain the isomorphism: $$\frac{V[[h]]}{h^nV[[h]]}\cong V \otimes \frac{\mathbb{C}[h]}{h^n\mathbb{C}[h]}$$

Thanks!

答:
The isomorphism

$$ V[[h]]/h^n V[[h]] \cong V[h]/h^nV[h] $$

is given by

$$ v_1 + v_2 h + \dots + v_{n-1} h^{n-1} + h^n V[[h]] \longmapsto v_1 + v_2 h + \dots + v_{n-1} h^{n-1} + h^n V[h].$$

And clearly $ V[h]/h^nV[h] \cong V\otimes_k k[h]/h^n k[h]$.

Note: I don't think that the suggested in the comments approach through exact sequences will work because $V \otimes_k k[[h]]$ is not isomorphic to $V[[h]]$, when $V$ is infinite dimensional.