问:
Embedding $\beta X$ in $\mathbb{N}^{\kappa}$ for some cardinal $\kappa$.
I found an interesting exercise but I don't know how to solve it.

Given a topological space $X$, suposse that there exist $\lambda$ a cardinal such that $X$ is homeomorphic to a closed subset of $\mathbb{N}^{\lambda}$ ($X$ can be viewed, in fact, as a closed subset of $\mathbb{N}^{\lambda}$). Is there $\kappa$ a cardinal such that $\beta X$ is homeomorphic to a subspace of $\mathbb{N}^{\kappa}$? Here, the subspace is not necesarilly closed nor open. It's only a subspace of $\mathbb{N}^{\kappa}$.

I really don't know how to solve it. Is it true? My first attempt was take $X\subseteq\mathbb{N}^{\lambda}$ with $X$ closed in $\mathbb{N}^{\lambda}$. By the Tychonoffness of $\mathbb{N}^{\lambda}$ then such space can be viewed as a subspace of $[0,1]^{\alpha}$ for some cardinal $\alpha$, i.e., $\mathbb{N}^{\lambda}\subseteq [0,1]^{\alpha}$. Then $X\subseteq \mathbb{N}^{\lambda}\subseteq [0,1]^{\alpha}$. If we take the closure respect to $Z=[0,1]^{\alpha}$, i.e., $$\text{cl}_{Z}(X)\subseteq\text{cl}_{Z}(\mathbb{N}^{\lambda})\subseteq [0,1]^{\alpha}$$
Can be $\text{cl}_{Z}(X)=\beta X$? From here I don't know how to go because $\text{cl}_{Z}(\mathbb{N}^{\lambda})$, in general, is not of the form $\mathbb{N}^{\kappa}$. Any hint? Thanks for your help.

答:
Recall that a space is called $\mathbb{N}$-compact, if it can be embedded as a closed subset of $\mathbb{N}^{\lambda}$ for some cardinal $\lambda$.
Note that a compact subset of $\mathbb{N}^{\lambda}$ is closed in $\mathbb{N}^{\lambda}$, hence $\mathbb{N}$-compact.

With this notation the question is, when $\beta X$ is $\mathbb{N}$-compact.
In fact, Tyrone already almost pointed out the answer, namely:

Let X be completely regular. Then $\beta X$ is $\mathbb{N}$-compact, iff $X$ is strongly $0$-dimensional.

Proof:
"$\Rightarrow$":
Obviously, $\beta X$ is $0$-dimensional. Hence, $X$ is strongly $0$-dimensional.

"$\Leftarrow$": Since $X$ is strongly $0$-dimensional, $\beta X$ is strongly $0$-dimensional. Since it is compact, in particular, realcompact, it is
$\mathbb{N}$-compact. (This is a well-known result proven in 1967 by H. Herrlich. I couldn't find an online version of the proof, but it is cited, for instance, here, and also mentioned in the paper of Mysior mentioned by Tyrone.)

The example given in Mysior is $\mathbb{N}$-compact, but not strongly $0$-dimensional, hence $\beta X$ is not $\mathbb{N}$-compact, thus answering the assumption in the negative.


问:
$\mathbb{Q}/\mathbb{Z}$ has cyclic subgroup of every positive integer $n$?
I would like to know whether $(\mathbb{Q}/\mathbb{Z},+)$ has

$1$. Cyclic subgroup of every positive integer $n$?

$2$. Yes, unique one.

$3$. Yes, but not necessarily unique one.

$4$. Does not have cyclic subgroup of every positive integer $n$.

What I know about the given group is infinite group but every element has finite order. Please help how to proceed.

答:
Under the identification
$$
\frac{\Bbb R}{\Bbb Z}\stackrel{\sim}\longrightarrow S^1\simeq
\{\text{$z\in\Bbb C$ such that $||z||=1 $}\},\quad
r+\Bbb Z\mapsto e^{2\pi i r}
$$
the group ${\Bbb Q}/{\Bbb Z}$ is identified to the group $\mu(\Bbb C)$ of the complex roots of 1. Thus, a primitive $n$-th root $\zeta_n$ will correspond to an element in ${\Bbb Q}/{\Bbb Z}$ of exact order $n$.

Moreover, we know that every finite subgroup of the multiplicative group $\Bbb C^\times$ is cyclic. Thus there's just one subgroup of $\mu(\Bbb C)$ of order $n$, and same is true for ${\Bbb Q}/{\Bbb Z}$.

ContentLicense: CC BY-SA 3.0


问:
Showing $||k_{\lambda} * f - f||_{\infty} \to 0$ as $\lambda \to \infty$
Let $f$ be bounded and uniformly continuous on $\mathbb{R}$ and let $(k_{\lambda})_{\lambda \in \mathbb{N}}$ be a Dirac family on $\mathbb{R}$. I would like to show that $||k_{\lambda} * f - f||_{\infty} \to 0$ as $\lambda \to \infty$. Here, $*$ is a convolution operator and $||g||_{\infty} = \operatorname{ess sup}_{x \in \mathbb{R}}|g(x)|$.

My attempt:

We have, by triangle inequality for integral,
$$|k_{\lambda} * f - f| \leq \int_{\mathbb{R}}|f(x - t)- f(x)||k_{\lambda}(t)|\ dt.$$
By uniform continuity of $f$, we know that $\forall \varepsilon > 0, \exists \delta > 0$, such that $|f(x-t) - f(x)| < \varepsilon$. Also, by boundedness of $f$, we can find $M > 0$ such that $|f(x)| \leq M$ for any $x \in \mathbb{R}$. So,
\begin{align*}
|k_{\lambda} * f - f|  &\leq \int_{|t|< \delta}|f(x - t)- f(x)||k_{\lambda}(t)|\ dt + \int_{|t| \geq \delta}|f(x - t)- f(x)||k_{\lambda}(t)|\ dt\\
&< \varepsilon\int_{|t|< \delta}|k_{\lambda}(t)|\ dt + 2M\int_{|t| \geq \delta}|k_{\lambda}(t)|\ dt.
\end{align*}
Since $(k_{\lambda})_{\lambda \in \mathbb{N}}$ is a Dirac family, we know that for any $\delta > 0$, $\lim_{\lambda \to \infty}\int_{|x| > \delta}|k_{\lambda}(x)|dx = 0$. So $2M\int_{|t| \geq \delta}|k_{\lambda}(t)|\ dt \to 0$ as $\lambda \to \infty$. What is left is $\varepsilon\int_{|t|< \delta}|k_{\lambda}(t)|\ dt$. I am not sure what I can do with it. I know that, a Dirac family on $\mathbb{R}$ is a sequence $(k_{\lambda})_{\lambda \in \mathbb{N}}$ of continuous functions in $L^1(\mathbb{R})$, which satisfies

$\forall \lambda \in \mathbb{N} : \int_{\mathbb{R}}k_{\lambda}(x)\ dx = 1,$

$\limsup_{\lambda \to \infty}||k_{\lambda}|||_1 < \infty,$

$\forall \delta > 0: \int_{|t| \geq \delta}|k_{\lambda}(t)|\ dt \to 0$ as $\lambda \to \infty$.

I have used property 3. I don't know how to use the other properties, so I can show that $||k_{\lambda} * f - f||_{\infty} \to 0$ as $\lambda \to \infty$. Any idea would be appreciated.

答: You are almost done. By 2) $ \lim \sup_{\lambda \to \infty} \int_{|t|<\delta} |k_{\lambda} (t)| dt \leq \lim \sup_{\lambda \to \infty}\|k_{\lambda}\|_1 <\infty$.

ContentLicense: CC BY-SA 4.0


问:
How to integrate $\frac{1}{3+\sin x+\cos x}$
I can easily do it with $\tan {\frac{x}{2}}$ substitution, but my problem requires $x\in(0,2\pi)$.

$\int \frac{1}{3+\sin x+\cos x}dx, x\in(0,2\pi)$

How do i solve it now ?

Body:
Take $\sin x=2t/(1+t^2), \cos x=(1-t^2)/(1+t^2), tan(x/2)=t \implies 2 \sec^2(x/2) dx=dt$. Then
$$I=\int \frac{dx}{3+\sin x+ \cos x}=\frac{1}{4}\int \frac{dt}{t^2+t+2}=\frac{1}{4}\int \frac{dt}{(t+1/2)^2+7/4}==\frac{1}{2\sqrt{7}} \tan^{-1} \frac{2t+1}{\sqrt{7}}+C.$$
where $t=\tan(x/2)$.


问:
Supremum and infimum of $\{\frac{m}{n}+\frac{1}{m}+\frac{1}{n}+\frac{n}{m}: m,n\in\mathbb{N}\}$
Let $$S=\left\{\frac{m}{n}+\frac{1}{m}+\frac{1}{n}+\frac{n}{m}: m,n\in\mathbb{N}\right\}.$$ I think that $\sup S =\infty$ since the set is unbounded from above. By AG inequality, $\inf S =2$.

Am I right?

Any help is welcome.

答:
Your answer is correct. However using the AM-GM Inequality, you have got a lower bound only. You cannot determine the $\operatorname{inf}$ using AM-GM Inequality

To determine the $\operatorname{inf}$ use the following.

When $m=n$, then $$\frac{m}{n}+\frac{1}{m}+\frac{1}{n}+\frac{n}{m}=2+\frac{2}{n}$$

Now use the fact that $$\lim_{n \to \infty} 2+\frac{2}{n}=2$$


问:
$\mathbb{C}$ and $\mathbb{R}$ vector spaces.
$\mathbb{C}^n$ is a $\mathbb{C}$-vector space with a natural structure of $\mathbb{R}$-vector space. Then, what is the dimension of $\mathbb{C}^n$ as $\mathbb{R}$-vector space? I'm struggling so hard with this question, I do not even know how to start. Can somebody help me please.

答:
If $\{e_1,\dots,e_n\}$ is the standard basis of $\mathbb{C}^n$, consider
$$
\{e_1,\dots,e_n,ie_1,\dots,ie_n\}
$$


答: The dimension is $2n$. Think of $\mathbb{C}$. It can be generated by just one non-zero complex number (if multiplication by complex scalars is allowed) or, alternatively, by two independent complex numbers like $1$ and $i$ if only multiplication by reals is allowed (any complex number is $a+bi$ where $a,b$ are reals). You can easily generalize this to the $n$-dimensional case.

Title: $\mathbb{C}$ and $\mathbb{R}$ vector spaces.

Body: $\mathbb{C}^n$ is a $\mathbb{C}$-vector space with a natural structure of $\mathbb{R}$-vector space. Then, what is the dimension of $\mathbb{C}^n$ as $\mathbb{R}$-vector space? I'm struggling so hard with this question, I do not even know how to start. Can somebody help me please.


问:
$\sum_{n=0}^{\infty}3^{-n} (z-1)^{2n}$ converges when
$\sum_{n=0}^{\infty}3^{-n} (z-1)^{2n}$ converges when,

$1.|z|\le 3$

$2. |z|<\sqrt{3}$

$3.|z-1|<\sqrt{3}$

$4.|z-1|\le \sqrt{3}$

The radius of convergence can be found by applying the root test to the terms of the series. The root test uses the number $$C=\lim\sup|3^{-n}(z-1)^{2n}|^{1\over n}$$ for convergence $C<1$ and I get $|z-1|<\sqrt{3}$, Is it okay?

答:
Do the root test.

Set $$\limsup_{n\to\infty} \left(3^{-n} (z -1)^{2n}\right)^{1/n} < 1.$$
Cancelling exponents leaves
$$|z - 1|^2/3 < 1$$
or $$|z - 1| < \sqrt{3}.$$
Note that this fails to converge at both endpoints.  It is, in fact, a geometric series.  You are lookin' good.


答:
Hint: Note that $$\dim(\mathbb{C}^{n},\mathbb{C},\oplus, \odot)=n$$
and $$\dim(\mathbb{C}^{n},\mathbb{R},\oplus,\odot)=2n$$

This is very intuitive if you take $\mathbb{C}^{1}$ or $\mathbb{C}^{2}$ examples on $\mathbb{R}$ and $\mathbb{C}$ scalar fields, for example.


问:
The people can be seated around a circular table so that everyone is seated between two friends.
Suppose $n \ge 4$ people gathered for a dinner. Of those $n$ people any two of them together knows the remaining $n-2$. To show that the people can be seated around a circular table so that everyone is seated between two friends.

How I tried to approach the problem:

Consider the $n$ points as vertices of a graph $G$. There is an edge between two vertices if they know each other. By the condition that "any two of them together knows the remaining $n-2$," we have $$deg(u) + deg(v) = n-2.$$ My aim was to show that a Hamiltonian Circuit exists in the given graph so that I can define the sitting arrangement using the Hamiltonian Circuit.

But by $\textbf{Ore's theorem}$ we have:

If $deg(u) + deg(v) \ge n$ for every pair of distinct non-adjacent vertices $v$ and $w$ of $G$, then $G$ is Hamiltonian.

答:
The condition "any two of them together knows the remaining $n−2$" is much stronger than

$$deg(u) + deg(v) = n-2$$

in fact it implies that for any vertex $v$,

$$deg(v) \ge n-2$$

i.e. every person must know everyone else except possibly one other person

This is because if there were a person who didn't know two other people, then those two people would not know him, contradicting the given condition. Thus

$$
\begin{align}
deg(u) + deg(v) &\ge 2n-4 \\
\\
&\ge n
\end{align}$$

(the last step following since $n \ge 4$) and Ore's Theorem applies. Of course, as @Steven mentioned in a comment below, since $deg(v) \ge n-2$, Hamiltonicity can be more directly established by Dirac's Theorem.


答:
Making the substitution $t=\tan(x/2)$ gives \begin{align}I(x)&=\frac{1}{2}\int\frac{1+t^2}{2+t+t^2}\frac{2}{1+t^2}\,dt=\int\frac{1}{2+t+t^2}\,dt=\frac{4}{7}\int\frac{1}{1+(\frac{2t+1}{\sqrt7})^2}\,dt\\&=\frac{2}{\sqrt7}\arctan(\frac{2t+1}{\sqrt7})=\frac{2}{\sqrt7}\arctan(\frac{2\tan(x/2)+1}{\sqrt7})\end{align}

The problem with this is that this substitution is only valid for $0\le x<\pi$. In fact, at $x=\pi^+$, the above formula become negative, but there's nothing amiss in the original integrand at $x=\pi$.

For $x>\pi$, one needs to add a constant to keep the integral continuous, namely $$I(x)=\frac{2}{\sqrt7}\arctan(\frac{2\tan(x/2)+1}{\sqrt7})+\frac{2\pi}{\sqrt7}$$

Title: How to integrate $\frac{1}{3+\sin x+\cos x}$

Body:
I can easily do it with $\tan {\frac{x}{2}}$ substitution, but my problem requires $x\in(0,2\pi)$.

$\int \frac{1}{3+\sin x+\cos x}dx, x\in(0,2\pi)$

How do i solve it now ?


答:
Every element $v\in\mathbb C^n$ is uniquely expressed using $2n$ real numbers $a_1,\dots,a_n,b_1,\dots,b_n\in\mathbb R$ as
$$
v=
\begin{pmatrix} a_1+\mathrm i b_1 \\ a_2+\mathrm i b_2 \\ \vdots \\ a_n+\mathrm i b_n\end{pmatrix}
=
a_1\begin{pmatrix}1 \\ 0 \\ \vdots \\ 0\end{pmatrix}
+ \dots +
a_n\begin{pmatrix}0 \\ \vdots \\ 0 \\ 1\end{pmatrix}
+
b_1\begin{pmatrix}\mathrm i\\ 0 \\ \vdots \\ 0\end{pmatrix}
+ \dots +
b_n\begin{pmatrix}0 \\ \vdots \\ 0 \\ \mathrm i\end{pmatrix}.
$$
Hence, the $2n$ vectors
$$
\begin{pmatrix}1 \\ 0 \\ \vdots \\ 0\end{pmatrix}
,\dots,
\begin{pmatrix}0 \\ \vdots \\ 0 \\ 1\end{pmatrix}
,
\begin{pmatrix}\mathrm i\\ 0 \\ \vdots \\ 0\end{pmatrix}
,\dots,
\begin{pmatrix}0 \\ \vdots \\ 0 \\ \mathrm i\end{pmatrix}
$$
form a basis of $\mathbb C^n$ as an $\mathbb R$-vector space.


问:
Show that $x(t) = e^{- \gamma t} \cos(\omega t) x(0)$ is the solution a diffusion equation.
I am working through Kurt Jacob's Stochastic Processes for Physicists and came across a question I am unsure about on linear diffusions.

The equation for the damped harmonic oscillator is

\begin{align} m \frac{\mathrm{d}^2 x}{\mathrm{d}^2 t}  + 2 \gamma m
 \frac{\mathrm{d} x}{\mathrm{d} t} + kx = 0 . \end{align}

Assume that $\gamma < \sqrt{k/m}$, in which the resonator is said to
be "under-damped". Show that

$$ x(t) = e^{- \gamma t} \cos(\omega t) x(0)$$

is the solution to the equation above, and find the expression for the
frequency $\omega$ in terms of $m,k$ and $\gamma$.

(pg 24 Question 1)

答:
Let the solution be $ce^{zt}$ where $c$ and $z$ are complex numbers. The equation turns to

$$c(mz^2+2\gamma mz+k)e^{zt}=0.$$

By solving the quadratic we have

$$z=-\gamma\pm i \sqrt{\frac km-\gamma^2},$$ which gives you $\omega$.

Note that the initial condition $x(0)$ is not sufficient to make the answer unique. You also need a second one, such as $x'(0)$.


问:
Does anyone know of any good ways to get good at algebra without as much"grind" as doing hundreds of questions a night
I'm coming to a point in college where I can't avoid my math classes any longer. I need to get better at algebra so I don't flunk out of the class when I take it, however I've never been able to get good at doing it without taking forever on one problem. The main problem I always seem to have is just memorizing the rules for various things such as inequalities. Does anyone know of any good methods to learn and get better at remembering the rules of these equations without having to do hundreds of questions a night. Please note that I know homework will involve doing hundreds of problems a night, and I can deal with that, however my main issue is that I want to at least get to a point over the summer so that I can get good enough to not take ten minutes on one question while not taking all day for multiple days a week doing math questions.

Also, if anyone can provide me with ideas that involve computer programming in the process, that would make it more fun for me, as I do enjoy computer programming, and it would make learning how to do the equations more fun.

Thanks,
Flyboy

答:
A certain amount of repetition is necessary for proficiency. It seems to me that it is better to have a good understanding of how to solve a few problems of a given type rather than to mindlessly solve billions and billions of problems. By this I mean: How does each step in the solution help you get closer to a solution for the problem? Is there a reason the steps are done in this order or could they be done in some other order?

It does not matter so much how long the first problem of a given type take. This supposes we are talking about minutes, not years. What matters is that once you start on a new type of problem you have the means to get proficient in a reasonable length of time.


问:
Expectation Values
Suppose that $\{X_n\}_{n\ge1}$ take values $-\dfrac{1}{2}$ and $\dfrac{1}{2}$ with probability $a$ and $1-a$ respectively and $0$ otherwise. Suppose further that they are independent and discrete and random.

Then, how do I calculate the limit of $E\left[\dfrac{X_1+X_2+...+X_n}{3^n}\right]$ as $n$ tends to infinity.

If I had just variables and probability, I would use $$E[X]=\sum_{n=1}^\infty x_ip_i$$. But I am not sure about distribution of these values or whatsoever.
I would appreciate any help.

答:
Since $E\left[\sum{\dfrac{x}{a}}\right] = \dfrac{\sum{E[x]}}{a}$, you have that

$$
E\left[\frac{X_1+X_2 + ...+X_n}{n^3}\right] = \dfrac{n*E[X]}{n^3} = \dfrac{n*(-1/2*\alpha + 1/2*(1-\alpha))}{n^3}=\dfrac{n*(1/2-\alpha)}{n^3}=\dfrac{1/2-\alpha}{n^2}
$$

The rest is simple.

Title: Expectation Values


问:
Paradox in definite integrals
I don't know whether this question is suitable or not for this forum because it is not exactly rigorous mathematics.

We generally descibe geometrically absolute value of a definite integral as area under the curve(exception comes when the integrand changes its sign but even in that case also we can describe it geometrically and in this question it is irrelevant)

Now consider $$A = \int_{0}^{1} e^x \, dx$$

If we take our $x$ in meters or if we measure our coordinate axis in meters($m$) we see that $A$ has the unit of $m^2$ because it is area. Now obviously $dx$ has the unit of $m$ and $e^x$ is unitless. So RHS of the above equation has its unit in meters where as LHS has $m^2$. This perplexing inconsistency is haunting me. Can any one explain this paradox?

答: $e^x$ is not unitless. It is the distance from the $x$ axis measured in $m$ as well. Think of velocity, if an object moves with velocity $e^t$, this means that at time $t$ its velocity is $e^t$ something. Define the units of this something, do an integral and you'll know how far it went in these units.

Title: Paradox in definite integrals


答:
First make the substitution $x=\pi + y$ with $y\in (-\pi, +\pi)$. The integral becomes
\begin{equation}
\int \frac{d y}{3 - \sin(y)-\cos(y)} = \frac{2}{\sqrt{7}}\arctan\left(
\frac{4\tan\left(\frac{y}{2}\right) - 1}{\sqrt{7}}\right) + C
\end{equation}
after integration with the $\tan(y/2)$ substitution. Now replace $y$ with $x-\pi$ and the solution is defined in $(0, 2\pi)$.


问:
Show there exists $z \in \mathbb{U}$ such that $\prod_{k=1}^n (z - a_k) \in \mathbb{U}$
The exercise

Let $n\geq 3$ and $a_1, \dots, a_n \in \mathbb{U}$. Show there exists $z \in \mathbb{U}$ such that $\prod_{k=1}^n (z - a_k) \in \mathbb{U}$.

My try

Writing $z = e^{i \theta}$ and $a_k = e^{i \theta_k}$, I get that the module of the product is $2^n\prod_{k=1}^n | \sin(\frac{\theta - \theta_k}{2}) |$, using the fact that $e^{i \theta} - e^{i \theta_k} = 2 i \sin(\frac{\theta - \theta_k}{2}) e^{i \frac{\theta+\theta_k}{2}}$. I then need to find a value of $\theta$ that would make this number equal to $1$. Clearly, the function $f : \theta \mapsto 2^n \prod_{k=1}^n | \sin(\frac{\theta - \theta_k}{2}) |$ is continuous over $\mathbb{R}$. I would like to use the intermediate value theorem, but I have to show first $f$ takes at least one value $<1$ and another $> 1$.

Your help is welcome!

答:
$f(\theta_k) = 0$ for each $k$, so that it remains to show that $f(\theta) \ge 1$ for some $\theta$. Here helps the complex analysis:
$$
 g(z) = (z-a_1) \cdots (z-a_n)
$$
is holomorphic in $\Bbb C$. The maximum modulus principle implies that $|g|$ attains its maximal value on the closed unit disk at a point on the boundary:
$$
 1 = |g(0)| \le \max_{|z|=1} |g(z)| = |g(e^{i\theta})|
$$
for some $\theta \in [0, 2 \pi]$. (Actually the inequality is strict because $g$ is not constant, but that is not needed here.)


问:
Volume of a pyramid with a spherical base
Given the attached figure, we are interested in finding the volume of this red pyramid ($ABCV$) with the spherical base. We assume that the Cartesian coordinates of the points $A, B, C, V$ are known and consequently,

$$A = (x_A, y_A, z_A),  B= (x_B, y_B, z_B),  C = (x_C, y_C, z_C), V = (x_V, y_V, z_V)$$

The radius $r$ of the sphere is also assumed to be known.

Edit: Since one can easily obtain the plane equation of $ABC$, the desired volume of the red pyramidoid is obtained as: volume of large pyramid $ABCO$ + volume of small pyramid $ABCV$ - volume of the "spherical triangle - like" $ABCO$.

So the new question is: how can we compute the volume of this "spherical pyramidoid" $ABCO$?

Any help would be useful.

答:
You need not worry about the pyramidoid outside the sphere.

Area of spherical triangle  $ A_{sph\;triangle} $ is found by spherical excess times r^2.

$$( \alpha + \beta + \gamma - \pi) r^2$$

These three are the dihedral angles at the three vertices.The three equitorial planes have three intersections with the sphere as great circles.

The volume of the convex "pyramidoid" is

$$ V= A_{sph\;triangle} \cdot \dfrac{r}{3}$$
It can be found by integrating thin differential shell elements.

Check for pyramid with very small solid angle at vertex $\to 0$

$$  V=(\pi/3+\pi/3+\pi/3-\pi) r^2 \cdot r/3 \to 0 $$

Check for hemisphere

$$  V=(\pi+\pi+\pi-\pi) r^2 \cdot r/3 = \frac23 \pi r^3 $$

Check for full sphere

$$  V=(\pi+\pi+\pi-\pi) r^2 \cdot r/3 = \frac43 \pi r^3 $$


问:
Calculate Size of Pixel in X and Y Direction
we have analytically calculated distance between the centers of a big circle and a small circle in mm.

We also have coordinates of the centers of the mentioned circles - yellow and blue - identified from an the image with image processing algorithm in pixels. Now we have to calculate the size of a pixel in mm in $x$-direction and $y$-direction based on the following information;

The analytical distance between center of the two circles in mm - $RDistance$

The center of the two circles identified with image processing in pixels, i.e. $(x_1, y_1)$ and $(x_2, y_2)$

I can calculate a single pixel size with the formula
$$\frac{RDistance}{\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}.$$

However, the requirement is to calculate the pixel size in both $x$- and $y$-direction differently. How can we calculate the pixel size $x$-direction and pixel size $y$-direction separately?

答:
I would make an approximate solution like so: Let's assume that the number of pixels in each direction is large, so that the accuracy is sufficient. We don't know where the exact point lands within one pixel, so we have to approximate that the end points of the line are at the centre of a pixel.

Consider the line that connects the two points. It makes some angle with the horizontal axis, let's call that angle $\theta$. We get
$$
\tan \theta \approx \frac{y_2-y_1 }{x_2-x_1} = Q
$$
Calculate this auxiliary quantity $Q$, as it will be used later. If the width of one pixel is $\Delta x$ and height is $\Delta y$, we get
$$
\cos \theta = \frac{1}{\sqrt{1+Q^2}} \approx \Delta x \cdot \frac{x_2 - x_1}{RDistance}
$$
$$
\sin \theta= \frac{Q}{\sqrt{1+Q^2}} \approx \Delta y \cdot \frac{y_2 - y_1}{RDistance}
$$
from which you can solve for $\Delta x$ and $\Delta y$. Can you continue from here?


问:
Integrating $\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^n \, d\theta$, for $n=2$ and $n=3$
How do you integrate the following functions:
$$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^2 \, d\theta$$ and $$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^3 \, d\theta
$$
respectively?

Note: Initially, I tried integrating the function without the power and obtained the result below.
$$
\int \frac{\cos\theta}{1+\sin^2\theta} \, d\theta = \arctan(\sin\theta)+ C
$$
However, from here it is difficult to proceed. Integration by substitution doesn't seem to work.

How should I go on from here? Any pointers would be greatly appreciated.

答:
See question Evaluating $\int P(\sin x, \cos x) \text{d}x$.

The universal standard substitution to evaluate an integral of a rational fraction in $\sin \theta,\cos \theta$, i.e. a rational fraction of the form

$$R(\sin \theta,\cos \theta)=\frac{P(\sin \theta,\cos \theta)}{Q(\sin \theta,\cos \theta)},$$

where $P,Q$  are polynomials in $\sin \theta,\cos \theta$ is a trigonometric substitution known as the Weierstrass substitution

$$
\begin{equation*}
\tan \frac{\theta }{2}=t,\qquad\theta =2\arctan t,\qquad d\theta =\frac{2}{1+t^{2}}dt
\end{equation*},
$$

which converts the integrand into a rational function in $t$. We know from trigonometry (see this answer) that

$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{
\theta}{2}}=\frac{1-t^2}{1+t^2},\qquad \sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}
\frac{\theta }{2}}=\frac{2t}{1+t^2}.$$

Applying this substitution to e.g. the first integral, we get
$$
\begin{eqnarray*}
I &=&\int \left( \frac{\cos \theta }{1+\sin ^{2}\theta }\right) ^{2}d\theta
=\int \left( \frac{\cos \theta }{2-\cos ^{2}\theta }\right) ^{2}d\theta  \\
&=&\int \frac{2\left( 1+t^{2}\right) \left( t^{2}-1\right) ^{2}}{\left(
1+6t^{2}+t^{4}\right) ^{2}}dt.
\end{eqnarray*}
$$

[Edited and added.] The integrand can be expanded as
$$
\begin{eqnarray*}
\frac{2\left( 1+t^{2}\right) \left( t^{2}-1\right) ^{2}}{\left(
1+6t^{2}+t^{4}\right) ^{2}} &=&\frac{16+80t^{2}}{\left(
1+6t^{2}+t^{4}\right) ^{2}}+\frac{2t^{2}-14}{1+6t^{2}+t^{4}} \\
&=&\frac{1}{4}\frac{4-3\sqrt{2}}{t^{2}+3-2\sqrt{2}}+\frac{5\sqrt{2}-7}{
\left( t^{2}+3-2\sqrt{2}\right) ^{2}}\\&+&\frac{1}{4}\frac{4+3\sqrt{2}}{t^{2}+3+2
\sqrt{2}}-\frac{7+5\sqrt{2}}{\left( t^{2}+3+2\sqrt{2}\right) ^{2}}.
\end{eqnarray*}
$$

With the help of SWP I evaluated
$$
\begin{eqnarray*}
\int \frac{1}{4}\frac{4\mp 3\sqrt{2}}{t^{2}+3-2\sqrt{2}}dt &=&\frac{1}{4}
\frac{4\pm 3\sqrt{2}}{\sqrt{2}-1}\arctan \frac{t}{\sqrt{2}-1}, \\
\int \frac{-7+5\sqrt{2}}{\left( t^{2}+3-2\sqrt{2}\right) ^{2}}dt &=&\frac{
(7-5\sqrt{2})t}{2\left( -3+2\sqrt{2}\right) \left( t^{2}+3-2\sqrt{2}\right) }\\
&+&\frac{7-5\sqrt{2}}{2\left( -3+2\sqrt{2}\right) \left( \sqrt{2}-1\right) }
\arctan \frac{t}{\sqrt{2}-1}, \\
-\int \frac{7+5\sqrt{2}}{\left( t^{2}+3+2\sqrt{2}\right) ^{2}}dt &=&-\frac{
(7+5\sqrt{2})t}{2\left( 3+2\sqrt{2}\right) \left( t^{2}+3+2\sqrt{2}\right) }\\&-&
\frac{7+5\sqrt{2}}{2\left( 3+2\sqrt{2}\right) \left( \sqrt{2}+1\right) }
\arctan \frac{t}{\sqrt{2}+1}.
\end{eqnarray*}
$$

As a consequence the integral $I$ can be written as
$$
\begin{eqnarray*}
I &=&\frac{1}{4}\frac{-10+7\sqrt{2}}{-5\sqrt{2}+7}\arctan \frac{t}{\sqrt{2}-1}+\frac{1}{4}\frac{10+7\sqrt{2}}{7+5\sqrt{2}}\arctan \frac{t}{\sqrt{2}+1}\\&-&\frac{t^{3}-t}{t^{4}+6t^{2}+1}+C, \\
&=&\frac{1}{4}\frac{-10+7\sqrt{2}}{-5\sqrt{2}+7}\arctan \frac{\tan \frac{
\theta }{2}}{\sqrt{2}-1}+\frac{1}{4}\frac{10+7\sqrt{2}}{7+5\sqrt{2}}\arctan
\frac{\tan \frac{\theta }{2}}{\sqrt{2}+1}\\&-&\frac{ \tan ^{3}\frac{\theta }{2}
-\tan \frac{\theta }{2}}{ \tan^{4} \frac{\theta }{2}
+6 \tan ^{2}\frac{\theta }{2} +1}+C.
\end{eqnarray*}
$$


问:
Normal ditribution and proofs
Can someone check my proof please?

Let $X~ N(\mu, \sigma^2 )$; use the symmetry of the density of X to suggest a transformation g such that Y = g(X) has the same density as X. (The identity transformation, $Y = X$, is not permitted).

X follows Normal distribution with mean µ and variance $\sigma^2$ X is symmetric about the mean $\mu$.
Because of this symmetry, the transformation $Y = -X $ will also have the same density as that of X.

Proof:

By Jacobian of transformations,

$$f(y)=f(x)\left | \frac{dx}{dy} \right |$$
$$\left | \frac{dx}{dy} \right | = 1 $$

The domain of X is -infinity to infinity.
Since Y=-X, the domain of Y is also -infinity to infinity.
So, Y also follow the same density of X.
Y follows Normal distribution with mean $\mu$ and variance $\sigma^2$.

答: You have mis-understood the question. You have to come up with $g$ not equal to the identity function such that $g(X)$ has the same distribution as $X$. For this use the fact that $\frac  {X-\mu} {\sigma}$ has the same  distribution as $-\frac  {X-\mu} {\sigma}$ since $N(0,1)$ is symmetric. No add $-\frac {\mu} {\sigma} $ to both sides and multiply by $-\sigma$ to conclude that $2\mu-X$ has the same distributoion as $X$ so  $g(x)=2\mu -x$ does the job.

Title: Normal ditribution and proofs


问:
lasso approach to compute lasso estimate
Consider the following LASSO problem
$min_{\beta} \sum\limits_{i=1}^n(y_{i}-\sum\limits_{j=1}^p x_{ij}\beta_{j})^2$ , subject to $\sum\limits_{j=1}^p|\beta_{j}|\leq t$

where $t \geq 0$  is a constant.

(a) If $t = 0$, compute $\hat{\beta}_{j}^{lasso}$ , for $j = 1, . . . , p$.

(b) Define $t_{0}= \sum\limits_{j=1}^p \hat |{\beta}_{j}^{ols}| $. Prove that, if $t \geq t_{0}$, then

$\hat{\beta}_{j}^{lasso} $=$ \hat{\beta}_{j}^{ols}$

My approach:
for (b) $ \hat{\beta}_{j}^{ols}=\sum\limits_{i=1}^n(y_{i}-\hat{y})^2$
and $\hat{\beta}_{j}^{lasso} = \sum\limits_{i=1}^n(y_{i}-\hat{y})^2 + \lambda \sum\limits_{j=1}^p |\beta_{j}|$,

when $\lambda =0 $
we can get get    $\hat{\beta}_{j}^{lasso} $=$ \hat{\beta}_{j}^{ols}$

I am not sure about the approach, also don't know how to do part (a).
Please guide.

答:
Are you a student at UA? Let me give you some hint.

Consider the problem below.

Find real number x and $y$ which minimize $x^2 + y^2$ subject to $|x|+|y|\le t$.

(a)If t = 0, find the x and $y$.

(b)If t = 5, find x and $y$.

For part (a), the restriction is $|x|+|y|\le 0$, so x must be 0 and $y$ must be 0. Otherwise $|x|+|y|> 0$.

For part (b), the global solution for minimizing $x^2 + y^2$ without any restriction is x=0, $y$=0.  This global solution without any restriction satisfied the restriction $|x|+|y|\le 5$. Thus, the solution for minimize $x^2 + y^2$ subject to $|x|+|y|\le 5$ is also x=0, $y$=0.

Please work on your question with the hint.


问:
Pinter's Set Theory Exercise 1.7.6: Let G and H be graphs. Prove that if G and H are sets, then G inverse and GoH are sets.
I've managed to do a proof of this by using the previous exercise that shows the domain and range of G are both sets. Then {y} is a set as it is a subset of the range for any y in the range, and similarly for any x in the domain. So for all x and y, {y,x} is a set by being a union of {x} and {y}, and then by Axiom 5 {{y}, {y,x}} = (y,x) is a set for all x and y and the union of these is G inverse so this is also a set.

I can also do GoH similarly by breaking the sets down and reconstructing with unions. But... It's so ugly. It's all I've got but I feel like I'm not even in the right ballpark of what the intended proof is. Is there a more attractive path to this result?

答:
Hint

I assume that with "$G$ is a graph" you mean $G = \{ (x,y) \mid x \in X \text { and } y \in Y \}$ where $X,Y$ are sets.

Thus, what is $G^{-1}$ ? It is $\{ (y,x) \mid x \in X \text { and } y \in Y \}$ where $X,Y$ are sets.

So, if the condition "$x \in X \text { and } y \in Y$" allows us to use Specification axiom schema to prove that $G$ is a set, the same holds for $G^{-1}$.

With more details: $G^{-1} = \{ z \mid \exists y \exists x ( z= \{ \{ y \} \{ y,x \} \} \text { and } x \in X \text { and } y \in Y \}$.

But browsing on the textbook, it seems that the author follows a difefrent approach.

If so, IMO you can use Axiom A.3: Every subclass of a set is a set, as well as Th.1.51: If $A,B$ are sets, then $A \times B$ is a set.

Thus, if $G \subseteq X \times Y$ is a set, also $G^{-1} \subseteq Y \times X$ will be.


问:
A problem related to $2x^3-3x^2-x+\frac{3}{2}=0$

let $f(x)=2x^3-3x^2-x+\frac{3}{2}$.Then prove that
$$\int_{1/8}^{7/8} f(f(x)) \text d x\neq \frac{3}{4}$$

Factorising$$f(x)=2(x-1.5)(x-\frac{1}{\sqrt{2}})(x+\frac{1}{\sqrt{2}})$$
$$f(f(x))=2(f(x)-1.5)(f^2(x)-0.5)$$
But i don't see anything nice from this.
Maybe if i could prove that $$f(f(x+\frac{1}{2}))=-f(f(x-\frac{1}{2}))$$ then the integral would turn out to be zero.

also i dont think we have to find the exact value of integral for second part if we could just set up an inequality or prove that the integral is negative we are done!

Please note that i am intersted in a proof without actually finding $f(f(x))$

答:
Too long for the comment.

Taking in account the brilliant condition
$$f(f(1-x)) = 1 - f(f(x))$$
by JimmyK4542, one can write
$$I=\int\limits_{\large^1/_8}^{\large^1/_2}f(f(x))\text dx +\int\limits_{\large^1/_2}^{\large^7/_8}f(f(x))\text dx
=\int\limits_{\large^1/_8}^{\large^1/_2} \big(f(f(x))+f(f(1-x))\big)\text dx =\int\limits_{\large^1/_8}^{\large^1/_2}\text dx,$$
$$\color{brown}{\mathbf{I = \dfrac38.}}$$


问:
Evans PDE Ch. 3.5 exercise with subdifferential of convex function
I am very stuck on this question in Evans PDE book. The question is as follows:

Here  $L(q) = \sup_{s \in \mathbb{R}^n}\left\{  s \cdot q - H(q) \right\}. $

Let $(1)$ be the statement that

$$ v \in \partial H(p), $$

$(2)$ be the statement that

$$  p \in \partial L(v), $$

and $(3)$ be the statement that

$$ L(v) + H(p) = p \cdot v.  $$

I was able to show $$ (3) \Rightarrow (1) \Rightarrow (2), $$

but I can't seem to show $(2) \Rightarrow (3)$, or find any other way to complete the chain.

Can someone please help me?

Thanks!

答:
I think it is easier to prove $(3) \iff (1) \iff (2).$

I'll try to give you some hints because it all proofs of these kind are based on a similar way of reasoning and it is useful to assimilate it.

Let's see $(1) \Rightarrow (3)$.

A good way to prove $$L(v) + H(p) = p \cdot v$$
is to prove both inequalities (usually one of the $2$ is trivial). In this case $\leq$ is trivial as $L$ is a $\sup$.
The other inequality is "trickier" as you have the $\sup$ on one of the $2$ sides and this doesn't allow you to really replace $L(v)$ with something equal to it. But you don't need to replace $L$ with an equality, as an inequality is sufficient for your purposes. Using the definition of $\sup$, you know that $\forall\varepsilon$ there is at least one $r$ such that $$L(v) - \varepsilon \leq v \cdot r - H(r),$$ or equivalently $$L(v) \leq v \cdot r - H(r) + \varepsilon.$$ Now if you use $(1)$ you can play with the r.h.s. to obtain an inequality that holds for every $\varepsilon$ (and with no $r$).

Let's see $(2) \Rightarrow (1)$.

You want to prove $$H(r) \geq H(p) + v\cdot(r -p)\ \ \ \ \forall r.$$
Using the fact that $H$ is convex you know that $H^{*}=L$, thus
$$ H(r) \geq t\cdot r -L(t),$$
for every $t$, in particular for $t=v$, i.e. $H(r) \geq v\cdot r - L(v)$. Now if you use $p \in \partial L(v)$ to play with $- L(v)$ you should be able to get the result.


问:
Deriving the 37-percent rule for dating
I am trying to prove the theoretical "37-percent rule" for dating.  The setup, if I remember correctly, is this.  Suppose that you will meet exactly $N$ potential mates in your life, and you will meet them one at a time, in a perfectly random order.  The potential mates rank from best to worst (in a total ordering), and you want to maximize the probability that you end up with the best one.  However, you can only tell how good the mates are relative to each other, so while you can fully rank the people you've already met, you can't say anything about the ones you have yet to meet.  Also, for each potential mate, you can either stay with them forever or leave forever, i.e. there is no divorce or post-breakup dating.

The result I have heard, and which I am trying to prove, is that your best strategy is to wait and reject the first 37% of them ($1/e$, to be precise), and then marry the next one that is better than all you have previously met.  The $1/e$ number presumably arises as the limit as $N \to \infty$.

Obviously, you should never marry someone who isn't strictly better than all the previous ones, because then your chances of picking the right one are $0$.  Also, given a strategy that you wait through the first $K$ partners and then marry the next one that is the best so far, I calculate your expected chances of succeeding as
\begin{equation}
\frac{\displaystyle \sum_{M = K}^{N-1} \frac{M - 1 \choose K-1}{N - M}}{N \choose K}
\end{equation}

(Let $M$ be the maximum value among the first $K$ people you meet, where $1$ is the value of the worst person, $2$ is the next, and so on, with your desired mate having value $N$.  Given $M$, your chances of winning are $\frac{1}{N - M}$, because the value $N$ must be the first to appear out of the highest $N - M$ values.  The chances of the maximum being exactly $M$ are ${M - 1\choose K - 1}/{N \choose K}$.)

(The above formula doesn't technically work for $K = 0$, but the reasonable convention ${-1 \choose -1} = 1$ gives the desired value $\frac1N$.)

The two things I am unable to prove, and which I would like to see ideas for, are:

Given that you never pick someone unless they are the best so far, how do you prove further that the best strategy must involve waiting for some $K$ people and then going for anyone else after that $K$?

Why is the formula above optomized at $K = N / e$, and how could one show this?

答:
The decision to marry or not need only be made when the current result is better than all previous results (otherwise the current choice is definitely suboptimal).
The choice, at point $K$ to marry or not (provided the $K$th date is better than all before) does not depend on the previous dates in anay ways (aprt from them being worse than the $K$th) because all relatiev orders are equally likely.
Therefore, any optimal strategy must consist of probabilities $p_k$ for $1\le k\le N$ such that when the $k$th date is better than all before, we marry (and stop) with probability $p_k$ and continue otherwise.
Let $q_k$ be the probability of picking the best provided we picked none of the first $k$. Then the probability of picking the best provided we picked none of the first $k-1$ is
$$ q_{k-1}=\frac{\frac1Np_k+\frac{N-k}{N}\frac1k(1-p_k)q_k+\frac{N-k}{N}(1-\frac1k)q_k}{P(\text{none of the first $k-1$ selected})}=\alpha_kp_k+\beta_k$$
($k$th is best and we pick, or best is after $k$ and $k$ is best of first and we don't pick it, or best is after $k$ and $k$ is not best of first)
and hence is clearly maximized at $p_k=0$ ($\alpha<0$) or $p_k=1$ ($\alpha>0$) or in a rare possibility $p_k$ does not matter at all ($\alpha=0$) and can wlog. be selected to be $0$ or $1$.

You can make yourself clear that haveing $p_k=1$ and $p_{k+1}=0$ cannot be optimal.

OwnerDisplayName: anon

Title: Deriving the 37-percent rule for dating


问:
Definition of projectively normal in Harshorne Ex II.5.14
In Hartshorne, Ex II.5.14, he defines that a closed subscheme $X \in \mathbb{P}^r_A$ is called projectively normal if the homogeneous coordinate ring $S(X)$ is integrally closed. This is also given by many notes and in wikipedia. But the denominators of elements of the fraction field of $S(X)$ may not be homogeneous, which makes things funny and the problem difficult to solve.

My question is whether this definition is the right one. If yes, how does it make sense? If no, what is the common way to fix the definition?

答:
This definition is fine and part (d) gives you a definition which is equivalent and doesn't encounter this issue of your. Your concern can be fixed with a little ingenuity.

Suppose $R=\bigoplus_d R_d$ is a graded domain, and let $K$ be its quotient field. Assume there exists a nonzero element $y\in R_1$. Let $K_d$ denote the subgroup of elements which can be represented as $\frac{r_1}{r_2}$ for $r_1,r_2$ homogeneous and $\deg r_1-\deg r_2=d$. Then $K'=\bigoplus_{d\in\Bbb Z} K_d$ is a subring of $K$, it's isomorphic to $K_0[y,1/y]$, and $K=K_0(y)=\operatorname{Frac} K'$. We observe that $K_0[y,1/y]$ is integrally closed in its field of fractions, being the localization of a polynomial ring over a field, so the integral closure of $R$ must be a subring of $K'$, and in particular consist of sums of homogeneous fractions. So we've eliminated your difficulty.


问:
Determine the number of these subsets for which the sum of the three elements is a multiple of 3. [Read Full Q. Below]

Consider the collection of all three-element subsets drawn from the set ${1,2,3,4,..., 299, 300}$. > Determine the number of these subsets for which the sum of the three elements is a multiple of 3.

This is a question from the Chapter Combinatorics but from what I understood of the question, I tried to find a pattern between subsets from which I found that the sum of all consecutive 3 positive integers is a multiple of 3, meaning if repetition isn't allowed, the three-element subsets would be $100$

I can't find a solution if repetition is allowed so please verify my answer if repetition isn't allowed and answer the question thinking it is allowed.

EDIT/UPDATE: IF repetition is allowed and we take the fact that the sum of all 3 consecutive integers is a multiple of 3 then each consecutive set {1, 2, 3},
{4, 5, 6}, will have 3 more sets for example the first will also be viable for
{2, 3, 4}, {3, 4, 5}; also, a set with the same number repeating 3 times would also be a multiple of 3 so the final answer would be $3 * 100 + 300 = 600$
Would this be correct?

答:
I'm considering the problem on top of your question. There is no "consecutive" involved, and "repetition" does not occur in sets.

We have $x_1+x_2+x_3=0$ modulo $3$ iff either all $x_i$ have different remainders modulo $3$, or all $x_i$ have the same remainder modulo $3$. It follows that there are
$${100\choose1}^3+3{100\choose3}=1\,485\,100$$
three-element subsets of $[300]$ whose sum is divisible by $3$.


问:
A random inequality
Let $A \in M(n,\mathbb{C})$ be abitrary and $B\in M(n,\mathbb{C})$ be symmetric (not sure if necessary). Then is the following inequality true?
$$
|(AB)_{ii}|^2+|(A^*B)_{ii}|^2 +2\Re(A_{ii} (B^*A^*B)_{ii})\ge0
$$
for all $i=1,2,...,n$.

This probably seems random, but I stumbled across this during some calculations. If the following were true, it would greatly simplify my calculations. A quick test using random matrices on a computer seems to always give me a $\ge 0$ value, but I can't think of why it would be true in general.

答:
Counterexample: $A=\begin{pmatrix}1&-1\\-1&0\end{pmatrix}$, $B=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$, then $AB=A^*B=\begin{pmatrix}0&2\\-1&-1\end{pmatrix}$, $B^*A^*B=BAB=\begin{pmatrix}-1&1\\1&3\end{pmatrix}$, so the inequality is false for $i=1$.

What one can show is the following:

$$0\le|\langle x,ABx\rangle+\langle x,A^*Bx\rangle|^2=|\langle x,ABx\rangle|^2+|\langle x,A^*Bx\rangle|^2+2\mathrm{Re}\langle ABx,x\rangle\langle x,A^*Bx\rangle$$ in particular for $x=e_i$, so that $\langle e_i,ABe_i\rangle=(AB)_{ii}$, etc.

Moreover, if $A$ is positive self-adjoint, then by Cauchy-Schwarz, $$|\langle x,Ay\rangle|^2\le\langle x,Ax\rangle\langle y,Ay\rangle$$
Hence putting $y=Bx$ gives $$|\langle x,ABx\rangle|^2\le\langle x,Ax\rangle \langle x,B^*ABx\rangle$$ so the given inequality is again (trivially) true.

Title: A random inequality


问:
infinite series expansion and rational functions
Consider the following function
$$f(x) = \sum_{k=0}^{\infty}a_kx^k.$$

Let us say I know $a_k$ belongs to a strict subset $F\subset\mathbb{R}$.

I want to characterize the set of sequences $\{a_k\}_{k=0}^\infty$ for which $f(x)$ is a rational function in $x$.

Can someone point me to relevant results in literature?

答:
As

$$f(x)=\frac{p(x)}{q(x)},$$ past a certain degree the coefficients of $f(x)q(x)$ vanish.

Hence except for the first few, the $a_k$ are constrained by a linear recurrence on $\text{deg}(q)+1$ terms.


问:
PDEs in biology
I am student who mostly heard lectures on partial differential equations and homogenization. But I really like the idea of working in biology or with biologists - but (with my lack of overview) it seems to me it's either statistics or some ODE problems.
Is there anything more PDE releated in biology?</p>

答: There is a great book about this: Leah Edelstein-Keshet, Mathematical models in biology chapter 10. </p>

Title: PDEs in biology


答:
If the aspect ratio, which @MattiP mentioned in his comment above, is equal to 1, you have,
$$\mathrm{pixel\space size\space in}\space y\mathrm{-direction} \space =\mathrm{pixel\space size\space in}\space x\mathrm{-direction} \space=\frac{RDistance}{\sqrt{2\left(\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2\right)}}.$$

If the aspect ratio is not equal to 1, then you have to use your image processing algorithm to find out,

$(A).\space$ the coordinates of the center of the $\mathrm{\color{blue}{blue\space circle}}$ directly above the $\mathrm{\color{brown}{yellow\space circle}}$ and

$(B).\space$ the coordinates of the center of the $\mathrm{\color{blue}{blue\space circle}}$ on the left hand side of the $\mathrm{\color{brown}{yellow\space circle}}.$

Denote the coordinates of the yellow circle and that of the blue circle above it as $\left(x_{yel}, y_{yel}\right)$ and $\left(x_{bluA}, y_{bluA}\right)$  respectively. Your image processing algorithm must give $x_{yel}=x_{bluA}$. We also assume that the analytical distance between these two circles is $RDistance_A$. Then, you shall write,
$$\mathrm{pixel\space size\space in}\space y\mathrm{-direction}=\frac{RDistance_A}{\left|y_{bluA}-y_{yel}\right|}.$$

Now, denote the coordinates of the blue circle on the left hand side of the yellow circle as $\left(x_{bluB}, y_{bluB}\right)$. In this case, the image processing algorithm must give $y_{yel}=y_{bluB}$. We also assume that the analytical distance between these two circles is $RDistance_B$. Then, you shall write,
$$\mathrm{pixel\space size\space in}\space x\mathrm{-direction}=\frac{RDistance_B}{\left|x_{bluB}-x_{yel}\right|}.$$

Title: Calculate Size of Pixel in X and Y Direction

Body:
we have analytically calculated distance between the centers of a big circle and a small circle in mm.

We also have coordinates of the centers of the mentioned circles - yellow and blue - identified from an the image with image processing algorithm in pixels. Now we have to calculate the size of a pixel in mm in $x$-direction and $y$-direction based on the following information;

The analytical distance between center of the two circles in mm - $RDistance$

The center of the two circles identified with image processing in pixels, i.e. $(x_1, y_1)$ and $(x_2, y_2)$

I can calculate a single pixel size with the formula
$$\frac{RDistance}{\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}.$$

However, the requirement is to calculate the pixel size in both $x$- and $y$-direction differently. How can we calculate the pixel size $x$-direction and pixel size $y$-direction separately?


问:
how to solve this probability exercise -selecting perfect parts from 2 boxes?
there are two boxes of parts, there are 50 parts in box A and 10 are perfect, 30 parts in box B and 18 are perfect. A box is selected randomly and two parts are taken out from it at a time without replacement.

1- find the probability that the first part is perfect.
2- suppose that the first part is found to be perfect. Find the probability that the second part is still perfect

I have resolved the first question and I got 0.4, but I still trying to resolve the second one, I would appreciate it If I get help or at least some hints

答:
The probability to get the second part perfect given the first is perfect can be calculated using conditional probability definition

$$\frac{\mathbb{P}[\text{both parts perfect}]}{\mathbb{P}[\text{first part perfect}]}=\frac{\frac{\binom{10}{2}}{\binom{50}{2}}+\frac{\binom{18}{2}}{\binom{30}{2}}}{\frac{10}{50}+\frac{18}{30}}\approx48.56\%$$


问:
Criterion for the linear independence for a family of vectors
How does one prove the following lemma?

Let $E$ and $F$ two nontrivial vector spaces and let $(u_i)_{i \in I}$ be a family of vectors in $E$.
The family $(u_i)_{i \in I}$ is linearly independent iff for every family $(v_i)_{i \in I}$, of vectors in $F$, there exists some linear map $f: E \rightarrow F$ such that $f(u_i) = v_i$.

答:
Here are some hints: the family is linearly independent if $\sum_i \alpha_i u_i=0$ iff $\alpha_i=0$ for all $i$.

Suppose there exists $f$ such that $f(u_i)=v_i$ and the $v_i$ are linearly independent, what can you say about $f(\sum_i \alpha_i u_i)$?

Now suppose that $u_i$ is linearly independent, simply define $f$ by its action $f(u_i)=v_i$. Is this a well defined linear function?


问:
Is there anything interesting about the limit $\lim_{h\rightarrow 0}\Big|\frac{f(a+h)-f(a)}{|h|}\Big|$?
I was wondering if there is anything interesting to say about the following limit:

Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$, and $a\in \mathbb{R}^n$. Consider

$$\lim_{h\rightarrow 0}\Big|\frac{f(a+h)-f(a)}{|h|}\Big|$$

If $f$ is differentiable I don't think the limit needs to converge. Suppose $f$ is differentiable at $a$, then

$$\lim_{h\rightarrow 0}\Big|\frac{f(a+h)-f(a)}{|h|}\Big|=\lim_{h\rightarrow 0}\Big|\frac{f(a+h)-f(a)-T(h)}{|h|}-\frac{|h|T(u)}{|h|}\Big|=\Big|T(u)\Big|$$
where $u$ is the unit vector in the direction of $h$. So the direction at which the limit is taken matters.

答:
If f is from Rn it needs to take n values, so the equation doesnt make sense for n not equal 1, for n = 1 this is a derivative which is think is pretty intresting.

Edit:
if you mean that h is being added to one element of a this is a partial derivative, and if it is added to all elements of a than without deviding by h this is something known as the total diffrential


问:
Independent, Pairwise Independent and Mutually Independent events
I want to understand the difference between Independent, Pairwise Independent and Mutually Independent events. I have read multiple answers related to this like in here, here and here. Most of them talk with 3 events and explain the difference between pairwise independent and mutually independent. I understand that. But what happens when there are $n$ events?

Suppose, $A_1$, $A_2$,....$A_n$ are n events, if

$$P(A_1 \cap A_2 \cap .... A_n) = P(A_1)P(A_2)....P(A_n)$$
but they are neither pairwise independent nor mutually independent. Only the above statement holds. Now, are these events still called Independent Events? Or is there any separate nomenclature for that?

答:
We will take the following definitions

Suppose $A_1,A_2,\ldots,A_n$ are $n$ events.

Definition 1: They are pairwise independet if
$$P(A_i\cap A_j)=P(A_i)P(A_j)\; \forall 1\leq i,j\leq n\;,i\not=j$$

Definition 2: They are mutually independet if
$$P(A_{i_1}\cap A_{i_2}\cap\ldots\cap A_{i_m})=P(A_{i_1})P(A_{i_2})\ldots P(A_{i_m})$$
$\forall 1\leq i_1<i_2<\ldots<i_m\leq n$ , $\forall m=2,3\ldots,n$, that is, for any combination of events you choose, satisfy the product rule.

Definition 3 They are independent if
$$P(A_1\cap A_2\cap \ldots \cap A_n)=P(A_1)P(A_2)\ldots P(A_n)$$

Remark :

Pairwise independent doesn't imply mutually independent but mutually implies pairwise.

Mutually independent implies independent, but not the converse because it is possible to create a three-event example in which

$${\displaystyle \mathrm {P} (A\cap B\cap C)=\mathrm {P} (A)\mathrm {P} (B)\mathrm {P} (C)}$$
and yet no two of the three events are pairwise independent (and hence the set of events are not mutually independent) George, Glyn, "Testing for the independence of three events


问:
Writing function $f(x,y)$ along a line as the value at a point plus integral of partial derivatives?
Suppose We have a function $f(x,y)$. As an example lets consider $f(x,y)=x^2y^2$

Suppose we know the following

The value of the point at a point, $(a,b)$. As an example, consider $(a,b)=(5,1)$

The partial derivatives of the function, $f_1$ and $f_2$.

Now suppose we want the value of the function along a line that has $(a,b)$ as an endpoint.

Question: Can we write $f(x,y)$ along every point on this line in terms of $f(a,b)$ and $f_1$, and $f_2$?

For example, suppose I want the values of $f$ along the line $6-x$ for $x∈ [1,5]$, and suppose we know that $f(5,1)= 25$.

Then can I write $$f(x,6-x) = f(5,1) + \int_x^5 \left(-f_1(x,6-x) + f_2 (x,6-x)\right)dx $$

A quick simulation suggests that this works for $f(x,y)=x^2y^2$ and the line $y=6-x$, but I don't know if such a result always holds

Also the (-) sign on $f_1$ is important.

If we instead consider the line $y=x-4, x>4$ then we need a (+) sign. Is there a way to write the formula regardless the direction we are moving?

Note: Assume we are considering an $f$ that is continuous and infinitely differentiable. (and at points where the partial derivatives are all defined) If other smoothness conditions are needed to make the result true feel free assume them, but please state what they are.

答:
Given a line $y=Mx+c$, we can parametrise points along this line in terms of a new variable $t$ by something like $x=t, y=Mt+c$. Then we can consider $f$ as a function on this line by defining
$$F(t)=f(t,Mt+c)$$

By the fundamental theorem of calculus
$$F(t) -F(t_0) =\int_{t_0}^t F'(t)dt$$

By the multivariable chain rule
$$F'(t)= \frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$

Using this, you get a formula that works for any line, once you fix a parametrisation.

In this case
$$F'(t) = 2(t)(Mt+c)^2(1) + 2(t^2)(Mt+c)(M)$$

and it is the sign of your M that determines the sign in the sum.

Note that this works for any path from $t_0$ to $t$, but the expression for the integral depends explicitly on the parametrisation. Unfortunately, there is no path independent approach that I know of.

(Indeed, if $f$ is not sufficiently smooth, the value of the integral will depend explicitly on the parametrisation, much as in integration of functions with poles in complex analysis)


问:
Property of series of independent random variables that have same distribution.
Let $X_1, X_2, X_3 ...$ be a sequence of independent random variables that have same distribution with bounded variance.
Let $S_n$ be random variable that is a sum of first $n$ of $X_i$.

I need to prove that $lim\; \mathbb{P}(S_n \le b)$  exists and has only $0, 1, 0.5$ as possible values.

Because of $0-1$ Kolmogorov's law this series either converge or not.
If $\mathbb{E}X_i > 0$ then $\mathbb{E}S_n = n\mathbb{E}X_1 \to \infty$ and I guess(Not sure how to prove it precisely) this is the case of $0$ for every $b$. And if $\mathbb{E}X_1<0$ this should be case of $1$ for every b.

Now going to case of $\mathbb{E}X_i = 0$ we can see that $\mathbb{E}S_n = 0$ too. Now the series converges to $0$ but I don't understand how to get this $0.5$ for $b = 0$ in case of my intuition not being wrong.

答:
By SLLN's $\frac {S_n} n \to m$ a.s., hence also in probability,  where $m$ is the mean.

Case 1)

$m>0$

Take $\epsilon \in ( 0, m)$ and note that $P(S_n \leq b)\leq P(|\frac {S_n} n-m| >\epsilon) \to 0$.

Case 2)

$m <0$.

Take $0<\epsilon <-m$ and note that $P(S_n > b)\leq  P(|\frac {S_n} n-m| >\epsilon) \to 0$. Hence $P(S_n \leq b) \to 1$.

Case 3)

$m=0$.

Here $P(S_n \leq b)= P(\frac {S_n} {\sqrt n} \leq \frac b {\sqrt n}) \to \Phi(0)=\frac 1 2$ (by CLT) where $\Phi$ is the standard normal distribution function .


问:
Computing expectation using Binomial Expansion
I'm solving for a problem. I got this probability mass function
$$
P(X=x) = \begin{pmatrix} x-1 \\ n-1 \end{pmatrix} p^n(1-p)^{x-n} \hspace{1cm}; x=n,n+1,...
$$

I should find the expectation

$$
E(\frac{n-1}{x-1})
$$

I started solving it

$
E(\frac{n-1}{x-1}) = \sum_{x=n}^{\infty} \frac{n-1}{x-1} \begin{pmatrix} x-1 \\ n-1 \end{pmatrix} p^n(1-p)^{x-n}
$

$
\hspace{1.6cm} = \sum_{x=n}^{\infty} \begin{pmatrix} x-2 \\ n-2 \end{pmatrix} p^n(1-p)^{x-n}
$

$
\hspace{1.6cm} = p^n\sum_{x=n}^{\infty} \begin{pmatrix} x-2 \\ n-2 \end{pmatrix}(1-p)^{x-n}
$

$
\hspace{1.6cm} = p^n\sum_{x=n}^{\infty} \frac{(x-2)!}{[(x-2)-(n-2)]!(n-2)!}(1-p)^{x-n}
$

$
\hspace{1.6cm} = p^n\sum_{x=n}^{\infty} \frac{(x-2)!}{(x-n)!(n-2)!}(1-p)^{x-n}
$

I'm stuck at this point. I don't know how to proceed. Can someone help me on how to solve or proceed further. I think Binomial Expansion might be useful but I don't know how to apply it in this case.

答:
Taking from your las step which can be re-written as
$$E=p^n\sum_{x=n}^{\infty} {x-2 \choose n-2} (1-p)^{x-n}=p^n\sum_{j=0}^{\infty} {j+n-2 \choose j} (1-p)^j=p^n[1-(1-p)]^{-(n-2)-1}=p.$$
$j=x-n$ used in above.
Here we have also used the binomial series for negative integer index as:
$$\sum_{J=0}^{\infty} {J+N \choose J} Z^{J}=(1-Z)^{-N-1}$$


问:
$\int_0^1 x^{2n} \sqrt{1-x^2 +\alpha(1-x^4)}\: d x, \quad \mathrm{for} \: \alpha>0 \: \mathrm{and} \: n \in \mathbb N$
I am trying to compute integral
\begin{equation}
I(n):=\int_0^1 x^{2n} \sqrt{1-x^2 +\alpha(1-x^4)}\: d x, \quad \mathrm{for} \: \alpha>0
\: \mathrm{and} \: n \in \mathbb N.
\end{equation}
This integral can be done by Mathematica when I take $n \leq 11$. For example, when $n = 0$ the integral turns out to be
\begin{equation}
I(0) = \frac{\sqrt{\alpha +1}}{3 \alpha } \left((2 \alpha +1) K\left(\frac{1}{\alpha
   +1}-1\right)-E\left(\frac{1}{\alpha +1}-1\right)\right),
\end{equation}
where $K(x)$ and $E(x)$ are the complete elliptic integrals of the first kind and second kind, respectively. Similarly, when $n=2$ Mathematica gives
\begin{equation}
I(1) = \frac{\sqrt{\alpha +1}}{15
   \alpha ^2} \left(-(\alpha  (6 \alpha +7)+2) K\left(\frac{1}{\alpha +1}-1\right) +(6 \alpha  (\alpha +1)+2) E\left(\frac{1}{\alpha
   +1}-1\right)\right).
\end{equation}
When I increase $n$ I find that Mathematica again gives the linear combinations of these two elliptic integrals. Therefore I guess that $I(n)$ should be the form
\begin{equation}
I(n) = A(n) K\left(\frac{1}{\alpha +1}-1\right) + B(n) E\left(\frac{1}{\alpha
   +1}-1\right).
\end{equation}
I find that Mathematica cannot give a result if I take $n>12$. Is it possible to compute this integral for arbitrary $n$? Or equivalently, is it possible to determine the coefficients $A(n)$ and $B(n)$ for arbitrary $n$?

答:
There is a general simpler formulation
$$I_n=\frac{\sqrt{(\alpha +1)\pi} }4 \,\,\frac{\Gamma \left(n+\frac{1}{2}\right)}{\Gamma(n+2)}\,\,\, _2F_1\left(-\frac{1}{2},n+\frac{1}{2};n+2;-\frac{\alpha}{\alpha +1}\right)$$

$$I_n=\frac{\sqrt{a+1}}{8}  \Gamma (n+2)\sum_{k=0}^\infty (-1)^{k+1} \frac{\Gamma \left(k-\frac{1}{2}\right) \Gamma
   \left(k+n+\frac{1}{2}\right)}{\Gamma (k+1) \Gamma (k+n+2)}\left(\frac{\alpha }{\alpha +1}\right)^k$$

Using this formula, there is no problem for the generation in terms of elliptic integrals (I computed $I_{50}$ instantly).

Now, what are the polynomials, that is the question.


问:
Find the derivative of the root of cubic equation
Consider the following cubic equation:

\begin{align}
\alpha ^3-2 \alpha ^2- \alpha(6+4x^2) +8 x^3-4x&=0
\end{align}
Where $x\in \mathbb{R}$.

The solutions $\alpha_1,\alpha_2,\alpha_3$ will depend on $x$. Solving this equation on Mathematica or Wolfram Alpha gives me complicated third roots.

I am not interested in the actual value of the roots, I am only looking for the value of $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$. Does a method exist to obtain a "nice" expression for $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$ without having to actually solve the cubic equation?

Edit:
Applying the advice from the answer, I obtain:
$$\frac{\mathrm{d}\alpha}{\mathrm{d}x}=\frac{(4 + 8 \alpha x - 24 x^2)}{(-6 - 4 \alpha + 3 \alpha^2 - 4 x^2)}$$
Is there a way to eliminate the $\alpha$ terms and have an expression $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$ uniquely depending on $x$?

答:
Implicit function theorem to the resccue!

We have :

$$ F(x, \alpha) = 0$$

Say that $\alpha(x)$ that is alpha is some function of x around some nice local interval which keeps us on the zero set of F.. and then take derivative

$$ \frac{dF}{dx} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial \alpha} \frac{d \alpha}{dx}$$

Hence,

$$ \frac{d \alpha}{dx} = \frac{-\frac{\partial F}{\partial x} }{ \frac{\partial F}{\partial \alpha}}$$

With,

$$ F(x, \alpha) = \alpha ^3-2 \alpha ^2- \alpha(6+4x^2) +8 x^3-4x$$

I'll leave the computation up to you :-)

Going further, suppose we have a point on the curve $(x_o,\alpha_o)$ , then we can write a taylor expansion of $\alpha$ locally as:

$$ \alpha(x) = \alpha_o + \frac{d \alpha}{dx}|_{x_o,\alpha_o} (x-x_o) + \frac{1}{2!} \frac{ d^2 \alpha}{dx^2}|_{x_o,\alpha_o} (x-x_o)^2 + O \left( (x-x_o)^3 \right)$$

The catch is that this polynomial only behaves like a curve for some small segment of it (geometrically thinking). Also note that taking derivative of above polynomial yields $\frac{d \alpha}{dx}$ as a polnyomial in terms $(x-x_o)$


问:
Problems whose first solutions had been using Calculus but later was shown to be done by non-Calculus methods
I was wondering about mathematical problems whose first published solutions  was obtained by using methods of Calculus but later was shown  (or known) to be solvable by using non-Calculus methods.

Are there really any such kind of problems?

Note:-

By the words "shown (or known)" I wanted to mean that the later non-Calculus solution were either published (and didn't existed before the solution using Calculus was published) or was later revealed to have existed even before the published solution.

答:
Semi-examples of what you ask for might be relevant for your purposes, and are easier to come up with. For example, Visual Calculus obtains results so eerily easily as to rival non-Calculus proofs. Another semi-example is when a non-Calculus solution is known, but the problem is given anyway as an exercise in Calculus. Perhaps the canonical example along this line is Regiomontanus’ angle-maximization problem. Here is the link to the Wikipedia article in which the Calculus-based solution is mentioned as being popular, but gratuitous, but with the happy result that this nifty problem has become more widely known: https://en.wikipedia.org/wiki/Regiomontanus%27_angle_maximization_problem

Also, the article, by Murray S. Klamkin, ‘On Two Classes of Extremum Problems without Calculus’ might be of interest. Here is the link to it: https://www.researchgate.net/publication/266192671_On_Two_Classes_of_Extremum_Problems_without_Calculus


问:
Weak convergence and tensors
Let:

$\Omega$ a bounded domain of $\mathbb{R}^d$ with a Lipshitz boundary $\partial\Omega =\Gamma_1 \cup\Gamma_2\cup\Gamma_3.$

$S_{d}$ is the space of second order symetric tensors on $\mathbb{R}^d \left(d=2,3\right)$

$\sigma.\tau=\sigma_{ij}.\tau_{ij} \forall \sigma,\tau\in S_{d} $

$W=\{\sigma=\sigma_{ij}; \sigma_{ij}=\sigma_{ji}\in
L^2\left(\Omega\right)/1\leq i\leq d\}=L^2\left(\Omega; S_{d}\right)$

$H=\{u=\left(u_{i}\right)\ u_{i}\in L^2\left(\Omega\right)\leq i\leq d \}=L^2\left(\Omega\right)^{d}$

$H_{1}=\{u=\left(u_{i}\right)\ u_{i} \in H^1\left(\Omega\right)\ 1\leq i\leq d\}=H^1\left(\Omega\right)^d$

$W_{1}=\{\sigma \in W ;Div \sigma \in H\}$

$  V=\{ v\in H_{1}\left(\Omega\right) ;v=0$ in $\Gamma_{1}\}$

$\left(\sigma,\tau\right)_W=\int_{\Omega}\tau_{ij}\sigma_{ij}dx \forall \sigma,\tau \in W$

$\left(u,v\right)_H=\int_{\Omega}u_iv_idx \forall u,v \in H$

$\left(u,v\right)_{H_{1}}=\left(u,v\right)_H+\left(\epsilon\left(u\right),\epsilon\left(v\right)\right)_W \forall u,v \in H_{1}$

$\left(\sigma,\tau\right)_{W_{1}}=\left(\tau,\sigma \right)_W+\left(Div \tau ,Div \sigma \right)_H$

$\left(u,v\right)_V=\left(\epsilon\left(u\right),\epsilon\left(v\right)\right)_W \forall u,v \in V$

$\epsilon\left(u\right)=\epsilon_{ij}\left(u\right)=\frac{1}{2}\left(\frac{\partial u_{i}}{\partial x_{j}}+\frac{\partial u_{i}}{\partial x_{j}} \right)$

There exists $C$ such that $\lvert \epsilon\left(v\right)\rvert_{W}\geq C\lvert v \rvert_{H_1\left(\Omega\right)}$

The non linear operator of viscosity$ A:\Omega \times S^{d}\rightarrow S^{d} $satisfy:

1-There exists a constant $L_{A}$ such that:
$\lvert A\left(x,\epsilon_1\right)-A\left(x,\epsilon_2\right)\rvert\leq L_{A} \lvert \epsilon_{1}-\epsilon_{2} \rvert$

2-There exists $m_A\geq 0$ such that:
$\left(A\left(x,\epsilon_1\right)-A\left(x,\epsilon_2\right).\left( \epsilon_{1}-\epsilon_{2}\right)\right)\geq m_{A}\lvert\epsilon_{1}-\epsilon_{2}\rvert^2 \forall \epsilon_1,\epsilon_2 \in S^d, a.e  x \in \Omega $

3-$x\rightarrow \left(A\left(x,\epsilon\right)\right)$ is lebesgue measurable in $\Omega$.

4-The mapping $x\rightarrow \left(A\left(x,0\right)\right)$ belongs to $W$.

If $ u^{'}_n$ is a bounded sequence in $L^2\left(0,T;V\right)$ such that $u_n^{'}$ converges weakly to $u^{'}$ in $L^2\left(0,T;V\right)$ , how can we show that $\left(A\epsilon\left(u^{'} _n\right),\epsilon\left(v\right)\right)_{W}$ converges weakly  to $\left(A\epsilon\left(u^{'}\right),\epsilon\left(v\right)\right)_{W}$ in $L^2\left(0,T\right)$?

I tried to do it but I have found a difficulty because of the nonlinearity of the operator $A$.

NB: $u$ depend of $x$ and $t$  with $x\in \Omega$ and $t \in \left[0,T\right]$and $u^{'}$ is the derivative of $u$ with  respect to $t$.

答:
I don't think that this will be true.
Let us ignore the time-dependence and replace $\varepsilon(u_n)$ by $f_n$.

Then, you essentially ask whether the weak convergence $f_n \rightharpoonup f$ implies
$$A(f_n) \rightharpoonup A(f)$$
for some non-linear, pointwise defined mapping $A$. This is, however, not true.

Let us take
$$A(f) = \begin{cases} f & \text{for } f < 0 \\ 2 \, f & \text{for } f \ge 0.\end{cases}$$
This $A$ satisfies your assumption.

Now, take a standard oscillating sequence $f_n$ which takes only the values $-1$ and $1$ with $f_n \rightharpoonup 0$.
Then, $A(f_n)$ takes the values $-1$ and $2$ and will converge weakly to $(-1+2)/2 = 1/2 \ne 0 = A(0)$.

Title: Weak convergence and tensors


问:
Finding the subgroup of $(\mathbb Z_{56},+)$ which is isomorphic with $(\mathbb Z_{14},+)$ by GAP
I am sorting some easy questions for the students in Group Theory I. One of them is:

Is $(\mathbb Z_{14},+)$ isomorphic to a subgroup of $(\mathbb Z_{35},+)$? What about $(\mathbb Z_{56},+)$?

I know the first claim is false because if it is true then I have $14\nmid 35$ which is a contradiction. For the second one, I see the function $f:(\mathbb Z_{14},+)\to(\mathbb Z_{56},+),  f(g)=4g$ is a nice one-one homomorphism.

My problem is to find $f(\mathbb Z_{14})$ by using GAP. I know how to define Cyclic Groups or Free Groups in GAP's environment also. Thanks for the ANY help.

答:
Since we know it's cyclic, we just have to find an element of order $14$:</p>

```<code>G:=CyclicGroup(56);
gen:=Filtered(G,g->Order(g)=14);
Group(gen[1]);
</code>```

Body:
Since we know it's cyclic, we just have to find an element of order $14$:

G:=CyclicGroup(56);
gen:=Filtered(G,g->Order(g)=14);
Group(gen[1]);


问:
Evaluate $\lim_{n\to\infty}\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k}}$ & $\lim_{n\to\infty}n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$.
This is a homework question. I have to find two limits:

i. $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}$$

ii. $$\lim_{n\to \infty} n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$$

I was able to solve the first one:

$\sqrt{n^8+1}\le \sqrt{n^8+k}\le \sqrt{n^8+n}$

$\implies \displaystyle\dfrac{1}{\sqrt{n^8+1}}\ge \dfrac{1}{\sqrt{n^8+k}}\ge \dfrac{1}{\sqrt{n^8+n}}$

$\implies \displaystyle\dfrac{k^3}{\sqrt{n^8+1}}\ge \dfrac{k^3}{\sqrt{n^8+k}}\ge \dfrac{k^3}{\sqrt{n^8+n}}$

$\implies \displaystyle\sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+1}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+k}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+n}}$

$\implies \displaystyle\dfrac{n^2(n+1)^2}{4\sqrt{n^8+1}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+k}}\ge \dfrac{n^2(n+1)^2}{4\sqrt{n^8+n}}$

The upper fraction goes to $\frac{1}{4}$ and the lower as well. From the Squeeze theorem:

$$\lim\limits_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}} = \frac{1}{4}$$

The same method did not work for the second limit. Can I get a clue or a hint, please?

答:
Note first that $\sum_{k=1}^n k^3=\frac14n^2(n+1)^2$. Now note that $\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}=\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+k/n^8}}$ and hence $$\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+n/n^8}}\leq\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}\leq\frac1n\sum_{k=1}^n (k/n)^3$$ implying that $\lim_{n\to\infty}\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}=\frac14$.

For part 2 again see that
$$n\left(\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+n/n^8}}-\frac14\right)\leq n\left(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac14\right)\leq n\left(\frac1n\sum_{k=1}^n (k/n)^3-\frac14\right)$$ which directly implies the second limit your are after is $\frac12$.


问:
How to prove that x $\longrightarrow$ x$_B$ gives a bijective linear map between V and $\mathbb{R}^n$?
How to prove that x $\longrightarrow$ x$_B$ gives a bijective linear map between V and $\mathbb{R}^n$ ?

Given:

Let V and W be vector space and $ A: V \longrightarrow W $ be a linear map.

x$_B $ is the column consisting of the coordinates of x with respect to B, where x $\in$ V can be uniquely written as x = $ x_1 $$ $b$ _1 $ + ... + $ x_n $$ $b$ _n $ and where B is an ordered basis, B = {b _1, ..., b _n }.

I got this assignment for school and I think I should start proving injectivity and bijectivity, and I know that for a linear map that is bijective that ker(A) = {0} and A(V) = W. However, I really don't know how to start with the proof and how to work with x $\longrightarrow$ x$_B$.

Thanks in advance for your help!

答:
Let $V$ is a vector space with a ordered basis $B=\{v_{1},v_{2},\ldots,v_{n}\}$, also let
\begin{eqnarray*}
[\cdot]_{B}: (V,F,+,\cdot) &\to & \mathbb{R}^{n}\\
v\mapsto [v]_{B}&=&(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})
\end{eqnarray*}
where $v=\alpha_{1}v_{1}+\alpha_{2}v_{2}+\cdots \alpha_{n}v_{n}$, so $[\cdot]_{B}$ is a bijection from $V$ in $\mathbb{R}^{n}$.

Statement: $[\cdot]_{B}$ is a linear transformation. Indeed, you can prove that

$[\alpha v]_{B}=\alpha[v]_{B}$

$[v+w]_{B}=[v]_{B}+[w]_{B}$

Now, you know that $$[\cdot]_{B} \quad \text{is bijection} \quad \iff [\cdot]_{B} \quad \text{injective} \quad \wedge \quad [\cdot]_{B} \quad \text{surjective}$$

Now, note that
\begin{eqnarray*}
\mathbf{ker}([v]_{B})=\left\{v \in V: [v]_{B}=\vec{0}_{\mathbb{R}^{n}} \right\} \overbrace{\implies}^{solving} \mathbf{ker}([v]_{B})=\vec{0}_{V} \implies [v]_{B} \quad \text{is injective}.
\end{eqnarray*}
Also, you can see that
\begin{eqnarray*}
\mathbf{im}([v]_{B})=\left\{w \in \mathbb{R}^{n}:\exists v \in V: [v]_{B}=w \right\} \overbrace{\implies}^{solving} \mathbf{im}([v]_{B})=\mathbb{R}^{n} \implies [v]_{B} \quad \text{is surjective}.
\end{eqnarray*}

So, $[v]_{B}$ is a bijection from $V$ to $\mathbb{R}^{n}$.


问:
Why do we say A given B in Bayes theorem?
I enjoyed this blog on visualizing Bayes theorem but I have a question on the wording for .

$p(A\mid B) = p(A,B)/p(B)$

In words I understand this to be "probability of A given B"

if we are "given B" why do we need to divide by p(B) ?   Surely if I am "given" something then its probability is $1$?

Why do we say it this way?

[Update]

I am seeing that p(B) = 1 (because we are "given" it )

thus p(A,B)/p(B) = p(A,B)

I am looking for the right words to articulate that we still need to divide by p(B) and that it is not 1

If we were saying "given we know the probability of B" instead of "given B" I would not be so confused.

答:
One way it becomes more clear (for me at least), is when you look at the Metropolis-Hastings algorithm https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm

It uses Bayes theorem to generate samples from a probability distribution from which direct sampling is difficult. So it generates a posterior distribution from a variable from which we only can give a prior distribution and we can calculate the likelihood of observing the data given that the random variable has a certain value (so we treat it as a given).

Then a posterior distribution can be found by updating the prior distribution with the likelihood of observing the random variable as that given value and the probability it actually takes on that value. By keep doing random sampling of 'B' and observing 'A' and accepting a new 'B' with a certain probability, we get closer and closer to the posterior distribution.


问:
Evaluate the line integral of the given vector field
The formula I'm using is $\int_C \mathbf F \bullet d\mathbf r$

The given vector field is
$$F(x,y) = \frac{x}{(x^2 + y^2)^{3/2}}\vec i +\frac{y}{(x^2 + y^2)^{3/2}} \vec j$$
and the curve C given parametrically by
$$r(t)=e^{2t}\cos(3t)\vec i + e^{2t}\sin(3t)\vec j,\space 0\leq t\leq 2\pi$$
$$r^{'}(t)= -3e^{2t}\sin(3t)+2e^{2t}\cos(3t)\vec i + 2e^{2t}\sin(3t)+3e^{2t}\cos(3t)\vec j$$

The integral will then be
$$\int_0^{2\pi}\left<\frac{e^{2t}\cos(3t)}{(e^{2t}\cos(3t))^2 + (e^{2t}\sin(3t))^2)^{3/2}}+ \frac{e^{2t}\sin(3t)}{(e^{2t}\cos(3t))^2 + (e^{2t}\sin(3t))^2)^{3/2}}\right> \bullet \left<-3e^{2t}\sin(3t)+2e^{2t}\cos(3t), \space2e^{2t}\sin(3t)+3e^{2t}\cos(3t)\right>dt$$
After working that whole thing out I got
$$\int_0^{2π} \frac{2}{e^{2t}}dt=1-\frac{1}{e^{4\pi}}$$
Am I correct?

答:
Your working is correct. However I am not sure if you are aware that for conservative vector field, there is a much easier way for you to get to the answer. The line integral will only depend on the end points and not on the chosen path. Here is details for this problem -

$F(x,y) = \frac{x}{(x^2 + y^2)^{3/2}}\vec i +\frac{y}{(x^2 + y^2)^{3/2}} \vec j$ is a conservative vector field.

Please note the potential function for the vector field is

$f(x,y) = -\frac{1}{\sqrt{x^2 + y^2}} + c \,$ where $c$ is a constant.

Just to confirm, you can see  $\frac{\partial f}{\partial x}i + \frac{\partial f}{\partial y}j \,$ is your vector field.

So $\int_C \vec{F}  \cdot{d\vec{r}} = \int_C \nabla {f} \cdot {d\vec{r}} = f(\vec{r}(2\pi)) - f(\vec{r}(0))$.

Now $\vec{r}(0) = (1, 0) \,$ and $\vec{r}(2\pi) = (e^{4\pi}, 0) \,$

So our line integral is simply $ = f(e^{4\pi}, 0) - f(1,0) = \displaystyle 1 - \frac{1}{e^{4\pi}}$.


问:
Lifting a map to the total space of a circle bundle
Let $\pi:P \to M $ be a smooth circle bundle, so $S^1$ is the fibre, $f:N \to M$ a smooth map. I would like to know what are the necessary and sufficient conditions for $f$ to lift to a map $g:N \to P$ covering $f$.

There are related questions reading which I think the answer is that $f$ lifts if and only if $f^*:H^2(M,\mathbb{Z})\to H^2(N,\mathbb{Z})$ is the zero map. However my understanding of algebraic topology is very basic and I cannot quite follow the arguments presented, so I would like for someone to confirm or  correct what I wrote. Incidentally, advice for the most accessible introduction to obstruction theory you know of would be appreciated.

In fact I am interested in a related but slightly different question: with $\pi:P\to M$ as above when does a diffeomorphism $f:M\to M$ lifts to a map $h:P\to P$? If the above criterium is correct $f$ lifts if and only if $\pi^*: H^2(M) \to H^2(P)$ is the zero map. Can anything be said about when that happens given that $P$ is a circle bundle?

答:
You do not need obstruction theory for that: $f$ lift to $g$ if and only if $[v]=f^*([u])=0$ where $u\in H^2(M,\mathbb{Z})$ is the Euler class of $P\rightarrow M$. To see that:

Consider the pullback $f':P'\rightarrow N$ of $P\rightarrow M$. It is classified by $[v]$.  $[v]=0$, if and only if $P'\rightarrow N$ is trivial, This is equivalent to $P'$ is isomorphic to $N\times S^1$ which induces a section $g':N\rightarrow P'$ that can be composed with $P'\rightarrow P$ (the projection of the pullback). On the other hand, if the map $g$ is a lift of $f$, $g$ defines a section of the pullback of $P\rightarrow M$ by $f$, which is thus trivial since it is a principal bundle.


问:
The right "weigh" to do integrals
Back in the day, before approximation methods like splines became vogue in my line of work, one way of computing the area under an empirically drawn curve was to painstakingly sketch it on a piece of graphing paper (usually with the assistance of a French curve) along with the axes, painstakingly cut along the curve, weigh the cut pieces, cut out a square equivalent to one square unit of the graphing paper, weigh that one as well, and reckon the area from the information now available.

One time, when we were faced with determining the area of a curve that crossed the horizontal axis thrice, I did the careful cutting of the paper, and made sure to separate the pieces above the horizontal axis from the pieces below the horizontal axis. Then, my boss suddenly scooped up all the pieces and weighed them all.

I argued that the grouped pieces should have been weighed separately, and then subtract the weights of the "below" group from the "above" group, while my boss argued that we were calculating the total area, and thus, not separating the pieces was justifiable.

Which one of us was correct?

答:
you are correct. Actually, the integral is equal to: "(total above weight) minus (total below weight)", which I'm sure is what you meant. What your boss is calculating is actually
$$
\int_a^b |f(x)| dx
$$</p>

ContentLicense: CC BY-SA 2.5

Title: The right "weigh" to do integrals


问:
Prove using an example that there is no plane on $\mathbb{R}^3$ that contains every group of 4 points
Well, this is a homewrok question (which I know I should not be asking, but I cannot find an answer to this anywhere):

The exercise is as follows:
i) Find the equation of the plane of $\mathbb{R}^3$ that contains the following three points:
 (0,0,1),(-1,3,5) and (1,-1,0)
ii) Prove using an example that there is no plane on $\mathbb{R}^3$ that contains every group of 4 points

So I have managed to solve (i) using a lot of Google and a third party tool (my notes are not all that good), but I cannot find (ii) anywhere. Can someone prove this providing an example and explaining it properly so that I can understand the reason this is impossible?

Thanks in advance!

P.S: This is an exercise from a course on Matrixes and how to solve Systems using matrices, in case anyone wants to know.

答:
From the phrasing of the question, it sounds like you understand that there is only one plane through those three points in part i). Knowing that, then we can say this:

Hint: If you find a point $P$ that does not satisfy the equation you found in i), then that point, along with the original three points, is an example of a set of four points that doesn't fit in a single plane.


问:
Determine the minimal polynomial of $\alpha = 1 + 3^{1/3} + 9^{1/3}$ over $\mathbb{Q}$. What is $[\mathbb{Q}(\alpha):\mathbb{Q}]$?

Determine the minimal polynomial of $\alpha = 1 + 3^{1/3} + 9^{1/3}$
over $\mathbb{Q}$. What is $[\mathbb{Q}(\alpha):\mathbb{Q}]$?

I tried to rearrange $\alpha$ in a way that $f(\alpha) = 0$ but can't figure out this part.
If I take $(a-1)^3 = (3^{1/3} + 3^{2/3})^3$, it doesn't end. I can't get rid of power of $1/3$.

I also tried $\alpha = (1+3^{1/3})^2 - 3^{1/3}$ but doesn't work either.

Is my approach wrong?

Body:
Hint:

$$(\alpha-1)^3=(3^{1/3}+3^{2/3})^3$$

$$\implies\alpha^3-3\alpha^2+3\alpha-1=3+3^2+3(3^{1/3}\cdot3^{2/3})(\alpha-1)$$

Alternatively,

$$\alpha=\dfrac{(3^{1/3})^3-1}{3^{1/3}-1}$$

$$\iff3^{1/3}=?$$

Now take cube in both sides


问:
Proof of the Brouwer fixed-point theorem for intervals in $\mathbb{R}$
Proof of the Brouwer fixed-point theorem for intervals in $\mathbb{R}$. I'm looking for feedback and corrections.

Lemma. If $\overline{B}_1(0)=\{x\in\mathbb{R}:\left|x\right|\leq1\}=[-1,1]\subset \mathbb{R}$ and $f\colon \overline{B}_1(0)\to \overline{B}_1(0)$ is continuous, then there exists $x_0\in \overline{B}_1(0)$ such that $f(x_0)=x_0$.

Proof of lemma. Let $g\colon \overline{B}_1(0)\to \overline{B}_1(0)$ be defined as $g(x)=f(x)-x$. As $f(x)$ and $x\mapsto x$ are continuous functions, then $g$ is also continuous.

Because $\overline{B}_1(0)$ is closed, we define $g(-1)$ and $g(1)$, and without loss of generality let us say that $g(-1) < 0< g(1)$. From the intermediate value theorem, we know that for $0\in \overline{B}_1(0)$, there is a $x_0\in \overline{B}_1(0)$ such that $g(x_0)=0$, but as $g(x)=f(x)-x$, we know that $g(x_0)=0=f(x_0)-x_0 \implies f(x_0)=x_0$. $\blacksquare$

(Brouwer fixed-point theorem for intervals in $\mathbb{R}$.) If $f\colon [a,b]\to [a,b]$ is continuous, then there is a $x_0\in [a,b]$ such that $f(x_0)=x_0$.

Proof. Consider $h\colon [a,b]\to [-1,1]$ a continuous bijection with $h(a)=-1$ and $h(b)=1$.

Define $g(x)=h(f(x))-h(x)$ with $g\colon [a,b]\to [-1,1]$. By the lemma there's a $x_0\in [a,b]$ such that $h(f(x_0))=h(x_0)$, and because both $h$ and $f$ are bijections (and therefore admit inverses), $f(x_0)=x_0$. $\blacksquare$

答:
Your idea is correct, but your proof of the lemma does not work as it is. In fact, you cannot be sure that $g(x) \in [-1,1]$. Take for example $f(x) = -x$. But that does not matter, for your IVT-argument you do not need $g([-1,1]) \subset [-1,1]$. Next, you cannot be sure that $g(1), g(-1)$ are non-zero with opposite sign. Look at $f(x) = x$.

I suggest to proceed as follows.

Consider $f : [a,b] \to [a,b]$ and define $g : [a,b] \to \mathbb R, g(x) = f(x) - x$. Then $g(a) = f(a) - a \ge a - a = 0$ and $g(b) = f(b) - b \le b - b = 0$. Now the IVT tells us that $g(x_0) = 0$ for some $x_0 \in [a,b]$.


问:
Eigenvalues of a diagonalizable stochastic matrix
Suppose $T$ is a stochastic diagonalizable matrix. Then we can write $T\cdot 1=P^{-1}DP\cdot 1=1$ where $1$ is the column vector with all entries equal to one. Then we have $DP\cdot 1=P\cdot 1$ where $D=diag\left(d_1,d_2,...,d_n\right)$. This means that
$d_k\cdot\Sigma_j\left(p_{k,j}\right)= \Sigma_j\left(p_{k,j}\right)$ and eventually $d_k=1$ for any $k$. But this implies that all eigenvalues of $T$ are equal to one so also its determinant.
Is this correct?

答:
No, that only means the logical disjunction "$d_k=1$ or $\sum_jp_{kj}=0$" is true. E.g. consider
$$
T=\frac14\pmatrix{1&3\\ 2&2}=\pmatrix{-\frac32&1\\ 1&1}\pmatrix{-\frac14\\ &1}
\pmatrix{-\frac25&\frac25\\ \frac25&\frac35}=P^{-1}DP,
\quad P\mathbf1=\pmatrix{0\\ 1}.
$$
The first diagonal entry of $D$ is $-\frac14\,(\ne1)$ and the first entry of $P\mathbf1$ is zero.

In fact, by Perron-Frobenius theorem, $1$ is a simple eigenvalue of $T$ whenever $T$ is irreducible.


问:
Integration over a unit circle
If I have to integrate a function in the region of the unit circle...Is there a way without even evaluating the integral I might get the result using the symmetry of a unit circle.

For example, if I'm integrating the following function over a unit circle.
$$\int\int_R e^x dxdy$$
Can I solve without really solving the integral or get to a point where the integral might be not as difficult as this?

答:
Using polar coordinates, the integral of $x^n$ over the unit circle is given by $$\int_0^{2\pi}\int_0^1 (r\cos\theta)^{n}r\,dr\,d\theta=\frac{1}{n+2}\int_0^{2\pi}\cos^{n}\theta\,d\theta=2\pi\frac{(2n-1)!!}{(2n)!!(n+2)}$$ for even $n$, and $0$ otherwise (using a table of definite integrals or establishing a recurrence relation by parts).

Hence the integral of $e^x$ yields

$$2\pi\sum_{k=0}^\infty\frac{(4k-1)!!}{(4k)!!(2k+2)(2k)!}$$

and this will work for other functions with a known Taylor development.

Title: Integration over a unit circle


问:
Evaluate the line integral with the given closed curve

Evaluate the line integral of the closed curve C oriented counterclockwise

答:
Hint: Applying Green's Theorem

$ \int_C (Pdx + Qdy) = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial x}) dA$

When you find $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial x}$, you will realize you get a number and your integral is that number multiplied by $\iint_D dA$. So simply calculate the area of the given closed path splitting into rectangles and squares, without integral. That should lead you to the answer.


问:
Convergence by using Cauchy Criterion
this is the sequence:

$(a_n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdot\cdot\cdot+\frac{1}{2n}$

And this is what I tried to do so far:

$|a_{n+1} - a_{n} | = \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1} = \frac{1}{2n+1}-\frac{1}{2n+2}$
in order to show (hopefully?) that the sequence IS convergent, since it is (hopefully) a Cauchy-Sequence.

However, as you can see, I don't know if a) I'm on the right way, b) what to do next.

答:
You have
$${1\over n + k} \ge \int_{n+k}^{n+k+1}{dx\over x}.$$
Hence
$$\sum_{k = 0}^n {1\over n + k} \ge \sum_{k=0}^n\int_{n+k}^{n+k+1}{dx\over x}
= \int_{n}^{2n}{dx\over x} = \log(2).$$
Another similar inequality lurks.  You can show this, in fact, converges, and that it is therefore Cauchy.</p>

Title: Convergence by using Cauchy Criterion


问:
Showing that Equality Holds for $\mathbb{P} ( X + Y + Z \leq 15) = \mathbb{P}(X+Y+Z \geq 26)$ for R.V.s $X, Y, Z$
I am a bit struggling to prove the following:

Let $X$ be the outcome of a standard die roll, $Y$ the outcome of an icosahedral die roll - 20 faces - and $Z$ the outcome of a dodecahedral die roll - 12 faces. If the three dice are rolled simultaneously, how can I show that$\mathbb{P} ( X + Y + Z \leq 15) = \mathbb{P}(X+Y+Z \geq 26)$?

So far, I have been trying to algebraically manipulate the probabilities, but I got stuck. This is what I have:
\begin{align*}
\mathbb{P} (X+Y+Z \leq 15) &= \sum_{k=3}^{15} \mathbb{P}(X+Y+Z = k) \\
&= \sum_{k=3}^{15} \sum_{i=1}^{k} \sum_{j=1}^{k} \mathbb{P}(X=i, Y=j, Z=k-(i+j))
\end{align*}
With this expression, I am not entirely sure how to proceed to show that the inequality holds. I would appreciate any help on how to prove this!

答:
This answer is essentially a special case of that provided by Masacroso, but I feel it provides a distinct enough perspective to warrant exposition.

You are rolling a D6, a D20 and a D12. The outcome of your experiment is a triple $(x,y,z)$ of numbers, with $1\leq x \leq6,$ $1\leq y\leq 20,$ and $1\leq z\leq 12.$ Hence the sample space is just $\Omega=\{1,\ldots,6\}\times\{1,\ldots,20\}\times\{1,\ldots,12\}.$ The random variable you are interested in is $X+Y+Z\colon\Omega\to\mathbb{R}$ given by $(x,y,z)\mapsto x+y+z.$

You want to show that $\mathbb{P}(X+Y+Z\leq 15) = \mathbb{P}(X+Y+Z\geq26).$ By definition,
$$\mathbb{P}(X+Y+Z\leq k) = \frac{\lvert\{(x,y,z)\in\Omega:x+y+z\leq k\}\rvert}{\lvert \Omega\rvert} = \frac{\lvert\{(x,y,z)\in\Omega:x+y+z\leq k\}\rvert}{1440}.$$
Now note that if $(x,y,z)\in\Omega$ satisfies $x+y+z\leq 15,$ then $(7-x,21-y,13-z)\in\Omega$ also, and $(7-x)+(21-y)+(13-z)\geq26.$ Moreover, the map $\Omega\to\Omega$ given by $(x,y,z)\mapsto(7-x,21-y,13-z)$ is a bijection. Therefore,
$$\lvert\{(x,y,z)\in\Omega:x+y+z\leq 15\}\rvert = \lvert\{(x,y,z)\in\Omega:x+y+z\geq 26\}\rvert.$$
Hence the result.

To relate this argument back to your own attempt, note that what I have essentially shown is that $\mathbb{P}(X+Y+Z=k) = \mathbb{P}(X+Y+Z=7+21+13-k) = \mathbb{P}(X+Y+Z=41-k).$ Plugging this into your attempt gives the answer, and indeed, this is essentially what Masacroso does in their answer (but in more high-falutin' language).


问:
Find $\dot{x}$ and $\ddot{x}$ of $x(t) = e^{-\gamma t}\:\cos \left(\omega \:t\right) x(0)$
I am trying to find the first and second derivative of a function
$$
x(t) = e^{-\gamma t}\:\cos \left(\omega \:t\right) x(0)
$$

i.e.
$$
\frac{\partial}{\partial \:t}\left(e^{-\gamma \:t\:}\:\cos \left(\omega \:t\right)\right) \ \ \ \& \\ \frac{\partial ^2}{\partial \:t^2}\left(e^{-\gamma \:t\:}\:\cos \left(\omega \:t\right)\right)
$$
I have some solutions below and would like to check if they are correct and if not what is the correct solution?

My solution
$$
\dot{x} = - \gamma e^{- \gamma t} \cos{\left (\omega t \right )} - \omega e^{- \gamma t} \sin{\left (\omega t \right )}
$$

$$
\ddot{x} = \gamma^{2} e^{- \gamma t} \cos{\left (\omega t \right )} - 2 \gamma \omega e^{- \gamma t} \sin{\left (\omega t \right )} - \omega^{2} e^{- \gamma t} \cos{\left (\omega t \right )}
$$

答:
$$x=e^{-gt} \cos wt \implies x'=-g e^{-gt} \cos w t- e^{-gt} w \sin wt=-(g\cos wt+w \sin wt)e^{-gt}$$
$$x''=-(-g w \sin wt+w^2\cos wt)e^{-gt}+g(g \cos w t+w \sin wt)e^{-gt}$$
$$x''=[(g^2-w^2) \cos wt +2gw \sin wt]e^{-gt}.$$


答:
Using the lemma is good, but not the way you're trying to.

Consider $g(y)=h(f(h^{-1}(y)))$; this is a continuous function $g\colon[-1,1]\to[1,1]$, so it has a fixed point $y_0\in[-1,1]$. Then $x_0=h^{-1}(y_0)$ is a fixed point for $f$, because
$$
y_0=g(y_0)=h(f(x_0))
$$
and so $x_0=h^{-1}(y_0)=f(x_0)$.

Of course, using $g(x)=f(x)-x$ and applying the intermediate value theorem to $g$ is much easier.

Body:
Proof of the Brouwer fixed-point theorem for intervals in $\mathbb{R}$. I'm looking for feedback and corrections.

Lemma. If $\overline{B}_1(0)=\{x\in\mathbb{R}:\left|x\right|\leq1\}=[-1,1]\subset \mathbb{R}$ and $f\colon \overline{B}_1(0)\to \overline{B}_1(0)$ is continuous, then there exists $x_0\in \overline{B}_1(0)$ such that $f(x_0)=x_0$.

Proof of lemma. Let $g\colon \overline{B}_1(0)\to \overline{B}_1(0)$ be defined as $g(x)=f(x)-x$. As $f(x)$ and $x\mapsto x$ are continuous functions, then $g$ is also continuous.

Because $\overline{B}_1(0)$ is closed, we define $g(-1)$ and $g(1)$, and without loss of generality let us say that $g(-1) < 0< g(1)$. From the intermediate value theorem, we know that for $0\in \overline{B}_1(0)$, there is a $x_0\in \overline{B}_1(0)$ such that $g(x_0)=0$, but as $g(x)=f(x)-x$, we know that $g(x_0)=0=f(x_0)-x_0 \implies f(x_0)=x_0$. $\blacksquare$

(Brouwer fixed-point theorem for intervals in $\mathbb{R}$.) If $f\colon [a,b]\to [a,b]$ is continuous, then there is a $x_0\in [a,b]$ such that $f(x_0)=x_0$.

Proof. Consider $h\colon [a,b]\to [-1,1]$ a continuous bijection with $h(a)=-1$ and $h(b)=1$.

Define $g(x)=h(f(x))-h(x)$ with $g\colon [a,b]\to [-1,1]$. By the lemma there's a $x_0\in [a,b]$ such that $h(f(x_0))=h(x_0)$, and because both $h$ and $f$ are bijections (and therefore admit inverses), $f(x_0)=x_0$. $\blacksquare$


答:
Here is a different take that does not rely on algebraic manipulations.

$\mathbb{Q}(3^{1/3})$ has degree $3$ over $\mathbb{Q}$, with basis $\{1, 3^{1/3},9^{1/3}\}$.

The minimal polynomial of $\alpha$ is the minimal polynomial of the linear transformation $x \mapsto \alpha x$. This polynomial  can be computed using its matrix with respect to the basis above:
$$
A
=
\begin{pmatrix} 1 & 3 & 3  \\ 1 & 1 & 3  \\ 1 & 1 & 1  \end{pmatrix}
$$
The characteristic polynomial of this matrix is $x^3 - 3 x^2 - 6 x - 4$.
This polynomial is irreducible over $\mathbb Q$ because it has degree $3$ but no rational root (*), and so is the minimal polynomial of $A$ and hence of $\alpha$. Therefore, $\alpha$ has degree $3$ and so $\mathbb{Q}(\alpha)=\mathbb{Q}(3^{1/3})$.

(*) use the rational root theorem here.

Body:
Determine the minimal polynomial of $\alpha = 1 + 3^{1/3} + 9^{1/3}$
over $\mathbb{Q}$. What is $[\mathbb{Q}(\alpha):\mathbb{Q}]$?

I tried to rearrange $\alpha$ in a way that $f(\alpha) = 0$ but can't figure out this part.
If I take $(a-1)^3 = (3^{1/3} + 3^{2/3})^3$, it doesn't end. I can't get rid of power of $1/3$.

I also tried $\alpha = (1+3^{1/3})^2 - 3^{1/3}$ but doesn't work either.

Is my approach wrong?


问:
Examples of functions with a natural domain $\Bbb R\setminus 2\Bbb N$

I would like to find a few examples of a function $f$ with a natural domain $\Bbb R\setminus 2\Bbb N$ which can be defined in a single expression.

Thoughts:

Since $\lfloor x\rfloor=\lceil x\rceil\iff x\in\Bbb Z$, I took
$$f(x)=\begin{cases}\frac1{\left|\left\lfloor\frac{x}2\right\rfloor\right|-\left\lceil\frac{x}2\right\rceil},&x\ne 0\\0,&x=0\end{cases}\space$$
Similarly, it can be defined as:
$$f(x)=\begin{cases}\ln\left(\left|\left|\left\lfloor\frac{x}2\right\rfloor\right|-\left\lceil\frac{x}2\right\rceil\right|\right),&x\ne 0\\0,&x=0\end{cases}$$
Another attempt is:

$$f(x)=\begin{cases}\tan\left(\left\lfloor\left(\frac32\right)^{\operatorname{sgn}(x)}\right\rfloor\frac{x+1}2\pi\right),&x\ne0\\0,&x=0\end{cases}$$

but $0\notin\Bbb N$, so I had to define $f(0)$ separately (however, $\tan\left(\operatorname{sgn}(x)\left\lfloor\left(\frac32\right)^{\operatorname{sgn}(x)}\right\rfloor\frac{x+1}2\pi\right)$ works if one takes $\operatorname{sgn}(0)=0$).
Furthermore, it is constant on $(-\infty,0]$, which isn't quite interesting.

How can I fix this problem if I want $f$ to be composed of $\tan(x)$? Are there any other examples of such functions defined by one expression?

Thank you in advance!

答:
In the meantime, I found a function that isn't defined piece-wisely and I figured out what to do with the problematic $x=0$ in my previous attempts where I assumed $\boxed{\operatorname{sgn}(0)=1}$ as in some literature.

Let's consider a bijection $f:(a,b)\to\Bbb R$ defined by: $$f:=\tan\left(\frac\pi{b-a}\left(x-\frac{a+b}2\right)\right)$$

with a prime period $P=b-a$ and shifted by $x_0=\frac{a+b}2$ on the $x$-axis.

Since we don't want even natural numbers (I don't include $0$ in $\Bbb N$), let $a=2k, b=2(k+1),k\in\Bbb N$.

My idea was to make the desired function constant on $(-\infty,2)$ and use the above pattern on $(2,+\infty)$.

Let's take $a=2,b=4$ and proceed as follows:

$$f(x)=\tan\left(\left\lfloor2^{\operatorname{sgn(x-2)}}\right\rfloor\frac\pi4(x-3)\right)$$

this works because $\left\lfloor 2^{\operatorname{sgn}(x-2)}\right\rfloor=\begin{cases}0,&x<2,\\2,&x\ge 2\end{cases}$ if one takes $\operatorname{sgn}(0)=1$.


问:
Starting mathematical education late
(Math)degreeless people who are self taught are doubtless viewed with at best suspicion, having countless holes in their mathematical knowledge, questionable methods, whose lack of academic qualifications understandably inspire little confidence in their ability. As someone who did well in the subject at school, but chose another path (degrees in other subjects), approaching middle age, I find myself drawn back to mathematics. Is this viable, or am I too old to start (I am 38). Is it necessary to pursue the academic path (masters, phd, etc.), or could I get anywhere "going it alone", submitting papers to journals, and writing to others re collaboration. What is the protocol regarding the latter?</p>

答:
The answer to your question is given by the following classic exchange:

A: Who does research?

B: Those that can, do. Those that can’t teach.

Also, something to bear in mind is Derek Sivers’ advice to keep your goal(s) to yourself. Here’s the link to his TED talk on this: https://www.ted.com/talks/derek_sivers_keep_your_goals_to_yourself?language=en

Also, you might want to consider the ‘Mathematics Educators’ Stack Exchange site:
https://matheducators.stackexchange.com/

Title: Starting mathematical education late


问:
Give an unbiased estimators for the third central moment of the underlying distribution.
Let $X_{i}, i=1,\ldots, n$ denote a random simple of size $n$ from a population with mean $3$. Assume that $\hat{\theta}_{2}$ is an unbiased estimator of $\mathbb{E}[X^{2}]$ and that $\hat{\theta}_{3}$ is an unbiased estimator for $\mathbb{E}[X^3]$. Give an unbiased estimator for the third central moment of the underlying distribution.

My approach:

I know that $$\mu_{3}=\mathbb{E}[(X-3)^{3}]=\mathbb{E}[X^{3}]-9\mathbb{E}[X^{2}]+54$$
so, we can see that $\hat{\mu_{3}}=\hat{\theta}_{3}-9\hat{\theta}_{2}+54$ is an unbiased estimator for
$\mu_{3}$ of the underlying distribution.

Is it correct?

答:
Yes it is unbiased and so correct.  To check:

$$\mathbb E[\hat{\mu_{3}}]\\=\mathbb E[\hat{\theta}_{3}]-9\mathbb E[\hat{\theta}_{2}]+54  \\ = \mathbb E[X^3]-3\cdot3\cdot\mathbb E[X^2]+3 \cdot3^2\cdot\mathbb E[X] - 3^3 \\ = \mathbb E[(X-3)^3]$$

It is not the only unbiased estimator.


问:
Find the derivative of the root of cubic equation
Consider the following cubic equation:

\begin{align}
\alpha ^3-2 \alpha ^2- \alpha(6+4x^2) +8 x^3-4x&=0
\end{align}
Where $x\in \mathbb{R}$.

The solutions $\alpha_1,\alpha_2,\alpha_3$ will depend on $x$. Solving this equation on Mathematica or Wolfram Alpha gives me complicated third roots.

I am not interested in the actual value of the roots, I am only looking for the value of $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$. Does a method exist to obtain a "nice" expression for $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$ without having to actually solve the cubic equation?

Edit:
Applying the advice from the answer, I obtain:
$$\frac{\mathrm{d}\alpha}{\mathrm{d}x}=\frac{(4 + 8 \alpha x - 24 x^2)}{(-6 - 4 \alpha + 3 \alpha^2 - 4 x^2)}$$
Is there a way to eliminate the $\alpha$ terms and have an expression $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$ uniquely depending on $x$?

答:
If you look at the determinant
$$\Delta=-1472 x^6+1152 x^5+2944 x^4+1408 x^3+1488 x^2-992 x+1008$$ it is negative except for $-1.3< x < 2.1$.

So, if you exclude this region, using the hyperbolic method for one real root, you have
$$\alpha=\frac{2}{3} \left(1+\sqrt{12 x^2+22} \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\frac{-54 x^3+18 x^2+27 x+31}{\sqrt{2} \left(6
   x^2+11\right)^{3/2}}\right)\right)\right)$$

Write it as
$$\alpha=\frac{2}{3} \left(1+f(x) \cosh \left(\frac{1}{3} \cosh ^{-1}(g(x))\right)\right)$$ from which
$$\frac 32\frac{d\alpha}{dx}=f'(x) \cosh \left(\frac{1}{3} \cosh ^{-1}(g(x))\right)+$$ $$\frac {f(x)\,g'(x)}{3\sqrt{g^2(x)-1}} \sinh \left(\frac{1}{3} \cosh ^{-1}(g(x))\right)$$ Just a bit (!!) of patience.

Have fun. But, back to serious, enjoy the implicit function theorem.


问:
Find the degree of the differential equation of the family of circles touching the lines $y^2-x^2=0$ and lying in the first and second quad
The centre obvious lies on the $y$ axis

So
$$x^2 + (y-y_1)^2 = r^2$$

Where $r^2 = \frac{y_1^2}{2}$

Then
$$x^2 + y^2+y_1^2 -2yy_1 = \frac{y_1^2}{2}$$

Since the degree of the equation is determined by the number of constants, the degree here should be 2, because $y_1$ and $y_1^2$ are separate constants

Or maybe they are not. Both constants are dependent on the value of $y_1$, so I am confused whether to have the degree 1 or 2

答:
This family of circle is given by ONE parameter family of circles
$$x^2+(y-k)^2=k^2/2~~~~(1)$$
We need to get a FIRST order ODE only.
D. (1) w.r.t. $x$, to get
$$2x+2(y-k)y'=0 \implies k-y=\frac{x}{y'} \implies k=(x/y'+y)$$
Putting these in (1) we get
$$x^2+(x/y')^2=\frac{1}{2}(x/y'+y)^2,$$
which may be simplified further.


答:
Hint :

As  pointed out by user Albus Dumbledore we can just consider the case where all the variable are positive now try to show that with this constraint of positivity :

$$(2a+b)(2b+c)(2c+a)\leq 1$$

And :

$$(1+a+2b)(1+b+2c)(1+c+2a)\leq 8$$

Title: If $a,b,c \in \mathbb{R}$ and $a+b+c = 1$, prove that $(2a+b)(2b+c)(2c+a)+(1+a+2b)(1+b+2c)(1+c+2a) \leq 9$


问:
Theorem or just a change of varibles?
I have a formula in my text:

$$\int \int_{S} F \cdot n dA= \int \int_{w} F(G(u,v)) \cdot (dG_{u}\times dG_{v}) du dv$$

I am really lazy and hate remembering formulas  to me this looks like a change of variables from x,y to u,v with a new normal vector to this surface can someone derive this? id much rather just remember the derivation that and if im write this should apply to higher dimensions i just need to know how to apply it.

Edit: the only reason i can't do a normal change of variables  is the n the formula makes u sub in the values of n from the change of varibles then cross product them out and im not sure why, i can blindly follow the formula but well this isnt physics id rather just know how to do it.

答:
This is the definition of a surface integral. If you were integrating over two different regions in the plane, it would be a change of variables. But you're not far off: The unit normal is
$$n =\frac{\frac{\partial G}{\partial u}\times\frac{\partial G}{\partial v}}{\|{\frac{\partial G}{\partial u}\times\frac{\partial G}{\partial v}}\|}\,,$$
and you can think of the element of surface area on the surface as
$$dA = \Big\|{\frac{\partial G}{\partial u}\times\frac{\partial G}{\partial v}}\Big\|\,du\,dv\,.$$
To see this, remember that the magnitude of the cross-product of two vectors is the area of the parallelogram they span, and a little bit of surface is approximated by the tangential parallelogram with sides $\dfrac{\partial G}{\partial u}\Delta u$ and $\dfrac{\partial G}{\partial v}\Delta v$.</p>

Title: Theorem or just a change of varibles?