问:
Closure of a subset
Show that the closure $\overline E$ of a subset $E \subset \mathbb R^n$ is the smallest closed set containing $E$.

I am new to topology so I was wondering how to establish a solid proof of this statement that seems basic.

Here is my attempt:
First we can show that $\overline E$ is a closed set containing $E$ :

$\overline E = E \cup \partial E $ so it surely contains $E$.

$\overline E^C = E^C \cap (\partial E)^C$. I need to show that $\forall \vec{x} \in \overline E^C, \exists \delta_x > 0 \text{ such that } B(\vec{x},\delta_x) \subset \overline E^C$ but I struggle to show it properly.

Then I need to show that $\overline E$ is the smallest set with the previous conditions. I wanted to do it by contradiction by supposing that $\exists F \subset \overline E$ (so smaller than $\overline E$) and then find a contradiction but it seems too strange to prove.

答:
First show that $\overline{E} = E \cup \partial E$ is closed: so take any $x \notin \overline{E}$. So in particular $x \notin E$ and $x \notin \partial E$. Consider balls $B(x,r)$ around $x$: they will always contain a point of $E^\complement$, namely $x$, so because we know that $x \notin \partial E$, which is defined as the set of all $y$ such that all balls around $y$ contain points from $E$ and also from $E^\complement$, there must be some ball $B(x,r_0)$ that does not contain points from $E$. But the last thing just says that $B(x,r_0) \subseteq E^\complement$, and it also implies $B(x,r_0) \subseteq \partial E^\complement$ (any $y$ in $B(x,r_0)$ has a small ball around it that still sits inside $B(x,r_0)$ so misses $E$ and so $y$ cannot be in $\partial E$ by definition), and it follows that $B(x,r_0) \subseteq \overline{E}^\complement$, and as $x$ was arbitrary there, $\overline{E}^\complement$ is open and $\overline{E}$ is indeed closed.

Now "if $C$ is closed and $E \subseteq C$ then $ \overline{E} \subseteq C$ is the final fact to be shown, which demonstrates the minimality.
So suppose $x \in \overline{E}$ but $x \notin C$. As $C$ is closed and the complement of $C$ is open, we can find some $r_0 >0$ such that $B(x, r_0) \cap C = \emptyset$. It follows that $x \notin E$ (as $ \subseteq C$) and $x \notin \partial E$ (as witnessed directly by this $r_0$) so $x \notin \overline{E}$, contradiction. So $x \in C$ and we're done.

Title: Closure of a subset


问:
Simplify the sum of the square of a polynomial having a fourth degree.
Today I learn about polynomial. Because I want to improve my knowledge. Thank you for your support and time for sharing information and experience.

From question :

If $a, b, c$ and $d$ are the roots of polynomial $Ax^4+Bx^3+Cx^2+Dx+E$ then find the value of
$a^2+b^2+c^2+d^2$

What I know :

$a+b+c+d=\frac{-B}{A}$

$ab+ac+ad+bc+bd+cd=\frac{C}{A}$

$abc+abd+acd+bcd=\frac{-D}{A}$

$abcd=\frac{E}{A}$

What I try :

$$(a+b+c+d)^2=(a+b+c+d)(a+b+c+d)$$
$$\left(\frac{-B}{A}\right)^2=(a^2+ab+ac+ad)+(ab+b^2+bc+bd)+(ac+bc+c^2+cd)+(ad+bd+cd+d^2)$$
$$\left(\frac{-B}{A}\right)^2=2(ab+ac+ad)+2(bc+bd)+2(cd)+a^2+b^2+c^2+d^2$$
$$\left(\frac{-B}{A}\right)^2=2(ab+ac+ad+bc+bd+cd)+a^2+b^2+c^2+d^2$$
$$\left(\frac{-B}{A}\right)^2=2\left(\frac{C}{A}\right)+a^2+b^2+c^2+d^2$$
$$\left(\frac{-B}{A}\right)^2-2\left(\frac{C}{A}\right)=a^2+b^2+c^2+d^2$$
$$\frac{B^2}{A^2}-\frac{2AC}{A^2}=a^2+b^2+c^2+d^2$$
$$a^2+b^2+c^2+d^2=\frac{B^2-2AC}{A^2}$$

My Question:

Is my work correct ?

Is it possible trying from $abcd=\frac{E}{A}$ ?

Thank you for your help and your time. God bless you.

答:
Indeed, we know that

$$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd$$ and you find these sums from Vieta's formulas.

Solving from $abcd$ alone is obviously impossible, because there are infinitely many ways to obtain the same value from different $a,b,c,d$, and these result in different sums $a^2+b^2+c^2+d^2$. To make $a,b,c,d$ uniquely determined, you need more information, i.e. more polynomial coefficients.

(The complete discussion of why all other possible combinations of the polynomial coefficients do not work is out of my reach/time.) Note that you can't retrieve the four roots from less than four coefficients, but we are not looking for the individual roots but for the sum of squares.

ContentLicense: CC BY-SA 4.0


问:
Topology on The space Real of matrices
Consider the real vector space space $\mathrm{M}(2,\Bbb{R})$ of $2\times 2$ real matrices. Identifying $\mathrm{M}(2,\Bbb{R})$ with $\Bbb{R}^4$ with the standard Euclidean metric and the corresponding metric topology, we observe that $\mathrm{GL}(2,\Bbb{R})$ is an open subset of $\mathrm{M}(2,\Bbb{R})$. Let $Y$ be an open subset of $\mathrm{GL}(2,\Bbb{R})$, such that $A,B\in Y$ implies that $AB^{-1}\in Y$. Then prove that $Y$ is closed in $\mathrm{GL}(2,\Bbb{R})$.

Can someone give me some hints about this problem, how to solve it?

Any help will be highly appreciated.

答:
Recall some elementary algebra: $Y$ is a subgroup of $\text{GL}(2,\Bbb R)$. The latter forms a topological group, as the group operations (multiplication and inversion) are continuous.

An open subgroup of a topological group is closed, because its complement is a union of cosets which are also open (being translates of the subgroup).

Title: Topology on The space Real of matrices


问:
Convex function bounded above by linear function
Let $f:\mathbb{R}^{*+}\to \mathbb{R}$ be a convex function and $g:\mathbb{R}^{*+}\to \mathbb{R}$ be a linear function such that $f(x)≤g(x)$ for all $x>0$ and $f(1)=g(1)$. (Linear : $ax+b$)

It's asked to prove that $f=g$

Graphically it looks logical but still can't find a rigorous proof, I tried proof by contradiction assuming $f(a)<g(a) $ for some $a$.

答:
For $x < 1 < y$ we have from the convexity condition:
$$
\begin{align}
 g(1) = f(1) &\le \frac{y-1}{y-x}f(x) + \frac{1-x}{y-x}f(y) \\
&\le \frac{y-1}{y-x}g(x) + \frac{1-x}{y-x}g(y) = g(1)
\end{align}
$$
where the last equality holds because $g$ is linear.

It follows that equality must hold everywhere in that inequality chain, i.e. $f(x) = g(x)$ and $f(y) = g(y)$.


问:
Finding the 'closed' form of this exponential generating function
The exponential generating function of this recurrence relation, $T_n=T_{n-1}\;+(n-1)\cdot T_{n-2}\;$, is

$$f(x)=e^{x + \frac{x^2}{2}}$$

Multiplying the exponential generating functions for each term, we get,
$$=\ \sum_{j\geq0}\;\frac{1}{j!}x^j\cdot\sum_{k\geq0}\;\frac{1}{2^kk!}x^{2k}$$
$$=\ \sum_{j,\ \ k\geq0}\;\frac{1}{j!}\cdot\frac{1}{2^kk!}x^{j+2k}$$

Taking $n\ =\ j\ +\ 2k$,
$$=\ \sum_{n-2k,\ \ k\geq0}\;\frac{1}{(n-2k)!}\cdot\frac{1}{2^kk!}x^n$$

What steps can I take from here to get to the 'closed' form of the exponential generating function, i.e.,

$$\;\sum_{n\ge0}\Big(\sum_{k=0}^{[n/2]}\frac{n!}{(n-2k)!2^kk!}\Big)\frac{x^n}{n!}.$$

答:
If you set
$$
a_{2k}  = \frac{1}{{2^k k!}},\quad a_{2k + 1}  = 0
$$
for $k\geq 0$, then
$$
\!\left( {\sum\limits_{j = 0}^\infty  {\frac{1}{{j!}}x^j } } \right)\!\left( {\sum\limits_{k = 0}^\infty  {\frac{1}{{2^k k!}}x^{2k} } } \right) = \left( {\sum\limits_{j = 0}^\infty  {\frac{1}{{j!}}x^j } } \right)\!\left( {\sum\limits_{k = 0}^\infty  {a_k x^k } } \right) = \sum\limits_{n = 0}^\infty  {\left( {\sum\limits_{k = 0}^n {\frac{1}{{(n - k)!}}a_k } } \right)x^n } .
$$
Now it remains to simplify
$$
\sum\limits_{k = 0}^n {\frac{1}{{(n - k)!}}a_k }.
$$
Since the terms corresponding to even $k$'s contribute only, we put $k=2m$ where $m$ runs between $0$ and $\left[ n/2 \right]$ (in this way we took into account all the even $k$'s between $0$ and $n$). Thus,
$$
\sum\limits_{k = 0}^n {\frac{1}{{(n - k)!}}a_k }  = \sum\limits_{m = 0}^{\left[ {n/2} \right]} {\frac{1}{{(n - 2m)!}}a_{2m} }  = \sum\limits_{m = 0}^{\left[ {n/2} \right]} {\frac{1}{{(n - 2m)!}}\frac{1}{{2^m m!}}} .
$$
Hence, the exponential generating function is
$$
\sum\limits_{n = 0}^\infty  {\left( {\sum\limits_{m = 0}^{\left[ {n/2} \right]} {\frac{1}{{(n - 2m)!2^m m!}}} } \right)x^n }  = \sum\limits_{n = 0}^\infty  {\left( {\sum\limits_{m = 0}^{\left[ {n/2} \right]} {\frac{{n!}}{{(n - 2m)!2^m m!}}} } \right)\frac{{x^n }}{{n!}}} .
$$


问:
Explicit image of $A_5$ inside $PGL(2,\mathbb{C})$
The list of finite subgroups of $PGL(2,\mathbb{C})$ is well known. The cyclic subgroups are easily written down. I was wondering if it is possible to explicitly write down a finite subgroup which is isomorphic to $A_5$. I tried with a presentation of $A_5$ given by

$$
A_5 = \{x,y\ |\ x^5=y^2=(xy)^3=1\},
$$

and tried to find elements of $PGL(2,\mathbb{C})$ which satisfy this relation but I was not successful. What am I missing here? Is there an easier way?

Thank you in advance.

答:
Let $\omega$ be a primitive fifth root of $1$. By a back of the envelope calculation, using traces, I came up with

$$ x \mapsto \left( \begin{array}{cc}\omega&0\\0&\omega^{-1}\end{array}\right),\ \ \ \
x^{-1}y \mapsto \left( \begin{array}{cc}a&b\\1&d\end{array}\right),
$$

where $a = 1/(\omega^2-1)$, $b=-a^2-a-1$, $d=-1-a$, and the images are of course to be understood to be modulo the scalar matrices.


问:
Can I calculate this integral?
I am trying to calculate this integral by DI method but didn't know how to do it (not looking for using gussian formula) :$  \int_{-\infty }^{+\infty} e^{-x^{2}}\,dx $ . Can anyone help me ?

答:
The proof is very simple and well known. It is enough to calculate first $I^2$ where $I$ is you integral passing in polar coordinates. Can  you continue with this hint?

The method using polar coordinates is well explained here

If you are familiar with Gaussian distribution you can by-pass the problem with a substitution

$$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}\underbrace{\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy}_{=1}=\sqrt{\pi}$$

This simply substituting

$$x=\frac{y}{\sqrt{2}}$$

and using the known result because inside the integral you have the Gaussian standard density

Title: Can I calculate this integral?


问:
Relation between R[X] and R[X,Y] as R-modules
In my lecture notes it was left as an exercise to show that for any $R$     $R[X,Y]\simeq R[X]$ as R-modules. To show this you need a bijective function $\phi$ such that $\phi (x+y)=\phi(x)+\phi(y)$ and $\phi(rx)=r\phi(x)$. I don't know how to do this or if its even right since it seems hard to construct phi in such a way not to loose information when going from some $f_0(x,y)$ to $f_1(x)$.

答:
As $R$-modules, $R[X]$ has a simple basis $\{1, X, X^2, \ldots\}$ and $R[X,Y]$ has $\{X^n Y^m \mid n \ge 0, m \ge 0\}$ (where $X^0 = Y^0 =1$ of course etc.) as a basis.

Two $R$-modules with countable bases are isomorphic. Just fix your favourite bijection between the bases and extend $R$-linearly..


问:
How to Taylor expand this function, which is a function of vectors?
Suppose that $a,b,c,d$ are vectors in $\mathbb{R}^2$, so $a_{\alpha}$ where $\alpha = 1,2$. We define a scalar product as $a \cdot b := a_1 b_2 - a_2 b_1$. Suppose also that we have a real-valued function $f(a,b,c)$ which is a function of some $3$ vectors.

What does the Taylor expansion of $f$ look like to first order? For example, I want to Taylor expand $f(a+b,c,d)$ around $(a, c, d)$, so $$f (a+b, c, d) = f(a,c,d) + \ldots $$  but I don't know how to write the remaining terms. Do I get a scalar product of $b$ with some term?

答:
If have only a partial answer, hope this helps:

The difficulty here is definitely your "scalar product". I set this in quotation marks because it is not a scalar product (not even a semi-definite one) because it is skew-symmetric: $a\cdot b = -b \cdot a$

It is merely a skew-symmetric bilinear form, which does not define a metric on your space. Now, I do not know much about exotic topologies, but I am pretty sure you need at least a sense of distance (i.e., a metric) to define derivatives on any space.

This is not given by your scalar product and therefore the Taylor expansion is not defined.

Is it possible that your "scalar product" is independent from the geometry of the space? I.e., it is just a skew-symmetric bilinear form defined in $\mathbb{R}^2$?

Then, the Taylor expansion would be the usual one with the usual scalar product $\langle a, b \rangle = a_1 b_1 + a_2 b_2$.

Let me know if I am wrong with anything!


问:
Möbius transformation preserving polynomial
Let $K$ be an algebraically closed field, and $x$ transcendent over $K$. We know that the group of $K$-automorphism of $K(x)$ is $\operatorname{PGL}(2,K)$, whose elements satisfy $ \alpha (x)= \displaystyle \frac{ax+b}{cx+d}$, with $ad-bc \neq 0$.

Given a polynomial $f(x)\in K[x]$, I want to know which automorphism of K(x) fix $f$, it is, $\alpha (f(x)) = f(\alpha(x)) = f(x)$. They form a finite subgroup of $\operatorname{PGL}(2,K)$. Actually, I am interested in the general case when $f\in K(x)$.

For instance, $f(x)=x^n$, with $\operatorname{char} K \nmid n$, is fixed by the ciclic group generated by $x \mapsto \omega x $, where $ \omega$ is a primitive $n-$root of unity. I'd appreciate any help, even if only works for special cases.

答:
If $f=\sum_{n=0}^N c_n x^n$ is non-constant then it has a pole at $\infty$ and $f(\alpha(x))$ has a pole at $\alpha^{-1}(\infty)$, for $f$ to be fixed we need $\alpha(\infty)=\infty$. This implies that $\alpha(x)=ax+b$.

Looking at the leading coefficient we must have $a = \zeta_d^k$ with $d| N, \gcd(d,k)=1$.

In characteristic $0$, if $a=1$ then $b=0$ for $f$ having finitely many roots. So assume that $a\ne 1,d\ne 1,char(K)\nmid d$.

Whence $f(x)=f(\alpha(x))$ iff $f(x-b/(a-1))=f(ax-b/(a-1))$ iff $f(x-b/(a-1))\in K[x^d]$.

When $char(K)\nmid N$ this implies that $N b/(a-1)=c_{N-1}$: so $b/(a-1)$ is known, then find the $d| N$ such that $f(x-b/(a-1))\in K[x^d]$ from which you know the $a$ that will work, and obtain $b$ from $a$.

When $char(K)| N$ it is the same idea, the coefficients of $f$ give a polynomial equation for $b/(a-1)$.

In characteristic $p$ there is also the case $a=1,b\ne 0$ which is a bit different, though $b$ is still given by a polynomial equation in the coefficients.


问:
If every pair of points is contained in a connected subset, then the space is connected
Suppose for each $a, b \in X$ there exists $A\subset X$ such that $A$ is connected and $a, b\in A$. How to show that $X$ is a connected space?</p>

答:
I will use the following Lemma, the proof of which I leave to you:

Lemma: If $\mathcal C$ is a collection of sets in $X$ with a non-empty intersection, i.e. $\bigcap \mathcal C\neq \varnothing$, then $\bigcup\mathcal C$ is connected.

Theorem: If every pair of points is contained in a connected subset, then the space is connected.

Proof: If $X = \varnothing$, it is trivially connected. Suppose $X\ne \varnothing$. Consider $a\in X$. For every $y\in X$, there exists a connected set $E_{ay}$ such that $a,y\in E_{ay}$. Note that $$X = \bigcup_{y\in X,y\ne a} E_{ay}$$ and $$\bigcap_{y\in X,y\ne a} E_{ay} = \{a\} \neq \varnothing$$ and so $X$ itself is connected. In the last step, I have used the preceding Lemma.

ContentLicense: CC BY-SA 3.0


问:
What is the function for the following plot?
I have the following plot and have huge difficulties finding the corresponding function

$X=1,2,3,6,12,40,100$

$Y=-2,-1,0,1,2,3$

$R=$ the results

$\mathbf{-2}$
$\mathbf{-1}$
$\mathbf{0}$
$\mathbf{1}$
$\mathbf{2}$
$\mathbf{3}$

$1$

$1$
$1.39$
$1.58$
$1.46$

$2$

$0.76$
$1$
$1.2$
$1.26$
$1.11$

$3$
$0.52$
$0.81$
$1$
$1.11$
$1.12$
$0.96$

$6$
$0.85$
$0.94$
$1$
$1$
$0.95$
$0.77$

$12$
$1.02$
$1.02$
$1$
$0.95$
$0.85$
$0.66$

$20$
$1.09$
$1.06$
$1$
$0.92$
$0.8$
$0.61$

$40$
$1.15$
$1.09$
$1$
$0.9$
$0.76$
$0.56$

$100$
$1.2$
$1.11$
$1$
$0.89$
$0.74$
$0.54$

I think it needs to be a combination of a polynomial function and a power function as if I look at the graph for $Y$ vs $R$ by each $X$, these are nice polynomial graphs. If I look at $X$ vs $R$ by each $Y$ they sort of follow a power series graph, but I can not figure out how to "combine them".

答:
So think I found a function and here is how I did it :) First I plottet the two graphs I mentioned above. I then got the following graphs

x vs r by y

y vs r by x

From the first, I see from trendlines that a powerseries on the form of A*x^B should sort of follow the plots.

For the second I find that a polynomial function in the form of A+By+Cy^2 very nicely follows the plots!

A was rather constant at 1.01, so kept it like that. I then looked at B and C pr X. Again I plottet the results ant looked at trendlines.

B and C by x

From here I find that B=-0.2+0.595x^-0.477 and C=-0.0897+0.0213ln(x)

The resulting function is then r=1.01+(-0.2+0.595*x^-0.477)y+(-0.0897+0.0213ln(x))*y^2

So, conclusion: Plot each variable with one series for each other variable. Choose the plots where you find functions that match the plotlines the best. Calculate the constants of these series depending on the second variable. Again, look at the plots for each and find the best matching function. Combine and there you have it :D


问:
advection-diffusion equation with origin-directed drift
Is there a solution (and if yes, what is the solution) to the 3D diffusion equation with a drift velocity directed to the origin $\vec{v}(\vec{x}) = a|\vec{x}|^{-1}\vec{x}$, with $a>0$? Thus a solution to
$$
\partial_t c = \nabla^2c +\nabla \cdot \Big(\vec{v}c\Big) =\nabla^2c +a\nabla \cdot \Big(\frac{\vec{x}}{|\vec{x}|}c\Big)
$$
with initial condition
$$
c(\vec{x},0) = g(\vec{x}-\vec{x}_0)
$$
where $g$ is a function, that has arbitrary small support around $\vec{x}=0$, and might be thought of as an approximation to the delta distribution $g\to\delta$.

To simplify further, the spatial domain of interest is a 3D sphere $S_r(0)\subset\mathbb{R}^3$ with $\vec{x}_0\in S_r(0)$ of small radius $r$ at the origin and a the temporal domain is also small $T\ll1$ in some sense:
$$
c:~~S_r(0)\times[0,T]\ni(\vec{x},t) \mapsto c(\vec{x},t)\in \mathbb{R}_+
$$

Edit: Corrected the sign of the drift.

答:
This is not an answer but rather some non-formal consideration of how the solution will look like.

One can derive many properties from the equation above, giving clear hints to the equations behaviour. To see those more clearly resolve the divergence term:
$$
\partial_t c = \nabla^2c +\frac{2ac}{|\vec{x}|} + a\frac{\vec{x}}{|\vec{x}|}\cdot \nabla c
$$

The overall L1 norm is conserved (since it's a diffusion equation)

The Laplace term decreases the function $c$, wherever it looks gauss-like. Think of 1D and the expansion of the 1D-Laplacian $\partial_x^2c(x) = \lim_{\epsilon\to0} \epsilon^{-2}(c(x+\epsilon) -2 c(x)+c(x-\epsilon))$. Thus $\partial_x^2c$ will be negative when $c$ is concave, positive when convex. This is the known smoothing behaviour.

The second (source) term constantly increases $c$ at the origin leading to a cusp there (seen by the stationary solution).

The last term increases or decreases $c$ depending on the angle of the gradient (vector of steepest ascent) and the vector pointing to the origin.
$$
\vec{x}\cdot \nabla c \propto \cos(\angle\vec{x},\nabla c)
$$

When the gradient points away from the origin, aligning with $\frac{\vec{x}}{|\vec{x}|}$ the last term gives positive contribution and increases $c$

When the gradients points to the origin, opposite to $\frac{\vec{x}}{|\vec{x}|}$, $c$ decreases

This makes any initial gaussian-like concentration package $c(\vec{x},0)$ not centered around the origin move to the origin by lowering its outer flank (gradient pointing to origin) and increasing its inner flank (gradient pointing away from origin).

Finally a stationary normalized solution (but not the general solution) is easy to obtain in spherical coordinates, reading
$$
c(\vec{x}) = \frac{a^3}{8\pi} \exp(-a|\vec{x}|)
$$

Any (smooth) initial state concentrated at some $\vec{x}_0$ would travel to the origin in the time evolution (more precise its center of mass would travel there), forming a cusp as in the stationary solution. The cusp would most likely appear immediately after $t=0$, since it would be very strange if it formed after a finite time?!

What I am missing is the exact form. But I am sure from the considerations above that there should be a nice one.

Edit:
Taylor-expanding the solution for a small time step and assuming the initial condition is a gaussian
$$
g_{\vec{\mu}\sigma}(\vec{x}) \propto \exp\Big(-\frac{|\vec{x}-\vec{\mu}|^2}{2\sigma^2}\Big)
$$
yields a solution
$$
c(\vec{x},t) \propto \exp\Bigg(-\frac{|\vec{x}-\vec{\mu}|^2+2at|\vec{x}|}{2\sigma^2 + 4 t}\Bigg) + \frac{2a}{|\vec{x}|}g_{\vec{\mu}\sigma}(\vec{x})
$$


问:
What is $[\mathbb{Q}(a) : \mathbb{Q}(a+1/a) ]$?
I've been thinking about this for a while and I keep confusing myself, so hopefully someone can help clarify some thoughts.

I'm currently looking at two fields, $\mathbb{Q}(a+1/a)$ and $\mathbb{Q}(a)$, where $a$ is just some algebraic number (in some of the examples I'm looking at, it isn't a number that can be expressed by radicals, which I imagine makes a difference in some cases).

I believe we have that $\mathbb{Q}(a+1/a) \subset \mathbb{Q}(a)$, so we have a field extension $\mathbb{Q}(a) / \mathbb{Q}(a+1/a)$. But what can we say about the degree of this extension?

(In saying this, I'm unsure if this is actually true: every element of $\mathbb{Q}(a+1/a)$ is of the form $u + v(a+1/a) \in \mathbb{Q}(a)$...that is enough to conclude that $\mathbb{Q}(a+1/a) \subset \mathbb{Q}(a)$ in this case, yes?)

I know that in some cases, it's trivial. For example, if $a = \sqrt{2}$, then $\mathbb{Q}(a+1/a) =\mathbb{Q}(a)$. But as I say, in some cases I look at, $a$ cannot be expressed by radicals.

So far, the only real thought I've had is to use Tower Law, with the tower: $\mathbb{Q}(a) / \mathbb{Q}(a+1/a) / \mathbb{Q}$.
This gives $[\mathbb{Q}(a) : \mathbb{Q}(a+1/a)] < [\mathbb{Q}(a) : \mathbb{Q}]$.
But can anything stronger be said?

答:
Note that:
$$a^2 - \left(a+\frac1a\right) a + 1 = 0$$
i.e. $a$ satisfies a quadratic with coefficients in $\Bbb Q\!\left(a+1/a\right)$, so $\Bbb Q(a)/\Bbb Q(a+1/a)$ is at most quadratic, so $[\Bbb Q(a):\Bbb Q(a+1/a)] \le 2$.

It is $1$ iff the equation already has a root in $\Bbb Q(a+1/a)$, which happens iff the discriminant $\left(a+\dfrac1a\right)^2 - 4 = \left(a-\dfrac1a\right)^2$ is a square in $\Bbb Q(a+1/a)$, which occurs iff $a-\dfrac1a \in \Bbb Q(a+1/a)$.

But this criterion is useless because $a-\dfrac1a \in \Bbb Q(a+1/a)$ iff $\dfrac1{2a} \in \Bbb Q(a+1/a)$ iff $a \in \Bbb Q(a+1/a)$.


问:
Compactly generated spaces as quotients of topological sums of compact Hausdorff spaces
I have two questions about Proposition 7.9.2 and Corollary 7.9.3 in tom Dieck's "Algebraic Topology".

Here is the setting, taken from Tammo tom Dieck: Algebraic Topology, European Mathematical Society, 2008 (Section 7.9):

A compact Hausdorff space will be called a ch-space. For the purpose of the
following investigations we also call a ch-space a test space and a continuous map $f: C \to X$ of a test space $C$ a test map. [...]

A subset $A$ of a topological space $(X, \mathcal{T})$ is said to be k-closed (k-open), if for each test map $f: K \to X$ the pre-image $f^{-1}(A)$ is closed (open) in $K$. The k-open sets in $(X, \mathcal{T})$ form a topology $k\mathcal{T}$ on $X$. A closed (open) subset is also k-closed (k-open). Therefore $k\mathcal{T}$ is finer than $\mathcal{T}$ and the identity $\iota=\iota_X: kX \to X$ is continuous. We set $kX = k(X) = (X, k\mathcal{T})$. [...]

A topological space $X$ is called k-space, if the k-closed sets are closed, i.e., if $X = kX$. [...]

The next proposition explains the definition of a k-space. We call a topology $\mathcal{S}$ on $X$ ch-definable, if there exists a family $(f_j: K_j \to X \,|\, j \in J)$ of test maps such that: $A \subset X$ is $\mathcal{S}$-closed $\Leftrightarrow$ for each $j \in J$ the pre-image $f_j^{-1}(A)$ is closed in $K_j$. We can rephrase this condition: The canonical map $\langle f_j \rangle: \bigsqcup_j K_j \to (X,\mathcal{S})$ is a quotient map. A ch-definable topology is finer than $\mathcal{T}$. We define a partial ordering on the set of ch-definable topologies by $\mathcal{S}_1 \leq \mathcal{S}_2 \Leftrightarrow \mathcal{S}_1 \supset \mathcal{S}_2$.

(7.9.2) Proposition. The topology $k\mathcal{T}$ is the maximal ch-definable topology with respect to the partial ordering.

Proof. By Zorn’s Lemma there exists a maximal ch-definable topology $\mathcal{S}$. If this topology is different from $k\mathcal{T}$, then there exists an $\mathcal{S}$-open set $U$, which is not
k-open. Hence there exists a test map $t: K \to X$ such that $t^{-1}(U)$ is not open. If we adjoin this test map to the defining family of $\mathcal{S}$, we see that S is not maximal.

(7.9.3) Corollary. The k-spaces are the spaces which are quotients of a topological sum of ch-spaces.

My questions:

(1) Why do we need Zorn's Lemma? Isn't it true by definition of $k\mathcal{T}$ that this is the coarsest topology on $X$ which is ch-definable? Thus, in particular, it should be a maximal element with respect to the above partial ordering, or what am I missing?

(2) Why is every k-space (homeomorphic to) a quotient of a disjoint union of ch-spaces? Since $k\mathcal{T}$ is a ch-definable topology on $X$ (essentially by definition), there exists a quotient map of the form $\langle f_j \rangle: \bigsqcup_j K_j \to (X,k\mathcal{T})$, so if $X$ is a $k$-space, then $X=kX=(X,k\mathcal{T})$ is bijective to such a quotient. But why is there a homeomorphism? Sorry if this is obvious.

Any help is much appreciated. Many thanks in advance!

答:
Using the comments, I would like to answer my first question:

The class of all test maps to $X$ is not in general a set, so it is not clear from the definition of $k\mathcal{T}$ that this topology is ch-definable. Tom Dieck uses Zorn's Lemma in order to obtain a maximal ch-definable topology on $X$ with respect to the partial ordering he has defined, and then he shows that this topology is $k\mathcal{T}$. In particular, this implies that $k\mathcal{T}$ is in fact ch-definable.

And I think I have found an answer to my second question:

If $X=(X,\mathcal{T})$ is a k-space, that is, $\mathcal{T}=k\mathcal{T}$, then by the proposition, its topology is ch-definable, so there exists a quotient map $q: Z\to X$ for some topological sum $Z$ of ch-spaces. We define an equivalence relation $\sim$ on $Z$ by $z_1\sim z_2 \Leftrightarrow q(z_1)=q(z_2)$. Then $q$ induces a well-defined bijective continuous map $f:Z/\!\sim\, \to X$ satisfying $f\circ p = q$ where $p:Z\to Z/\!\sim$ is the canonical projection.
This map $f$ is a homeomorphism: Given an open subspace $U\subset Z/\!\sim$, we need to show that the image $f(U)$ is open in $X$, or equivalently, that $q^{-1}(f(U))$ is open in $Z$ (since $q:Z\to X$ is a quotient map). This is true because $q^{-1}(f(U)) = (f\circ p)^{-1}(f(U)) = p^{-1}(f^{-1}(f(U))) = p^{-1}(U)$ (by injectivity of $f$), which is open in $Z$ (by definition of the quotient topology on $Z/\!\sim$).
Hence, $X$ is homeomorphic to the quotient $Z/\!\sim$.


问:
How to use integral to get the marginal mean.

The Question

Suppose that the conditional distribution of $Y(t)$ given $X$ is

$$Y(t)|X \sim N(t+X \exp(-tX),1).$$

The conditional mean of $Y(t)$ given $X$ is $t+X \exp(-tX).$ Suppose also that the marginal distribution of $X$ is unit exponential. The marginal mean of $Y(t)$ is obtained by integrating out the covariate to get

$$\mu(t)=E[t+X \exp(-tX)]=t+\frac{1}{(t+1)^2}.$$

My Understanding

As stated in the figure below, we want to figure the expected value of $Y$ via integrating out the $X$. I followed the law of the unconscious statistician but cannot get the result provided by the author. Also I m not sure if the unit exponential means the pdf of $x: f(x) = e^{-x}$. Any suggestions would be appreciated.

答:
You can use the law of total Expectation

$$\mathbb{E}[Y]=[\mathbb{E}[\mathbb{E}[Y|X]]=\mathbb{E}[t+xe^{-xt}]=t+\mathbb{E}[xe^{-xt}]=t+\int_0^{\infty}xe^{-xt}\cdot e^{-x}dx$$

to solve the integral without a lot of calculations observe that

$$\int_0^{\infty}x e^{-x(t+1)}dx=\frac{1}{(t+1)^2}\underbrace{\int_0^{\infty}[x(t+1)] e^{-x(t+1)}d[x(t+1)] }_{=\Gamma(2)=1}$$


问:
Independence of $T$ and $B_T$
Let $\{B_t:t\ge0\}$ be a real brownian motion such that $B_0=0$. Let $T=\inf \{t:B_t \notin (-a,a)\}$ with $a>0$. Are $T$ and $B_T$ independent?

I tried the following and I would like your opinion. Let $t>0$,
$P(T<t)=P(T<t, B_T=a)+P(T<t, B_T=-a)=2 P(T<t, B_T=a)$.

答:
Your approach looks good.  You might want to think about why it is that $P(T < t, B_T = -a) = P(T < t, B_T = a)$ (you used this in your last equality).  One way to see it is to use the fact that $-B_t$ is also a Brownian motion.

The last thing to note is that $P(B_T = a) = 1/2$, and thus you have shown $$P(T < t, B_T = a) = P(T < t) P(B_T = a)$$
and so the events $\{T < t\}$ and $\{B_T = a\}$ are independent.  The same argument shows this also holds with $-a$ in place of $a$.  Since $B_T$ can only take the values $a$ and $-a$ (by continuity), this shows that $T$ and $B_T$ are independent.

Title: Independence of $T$ and $B_T$


问:
Find $f_{xy}(0,0)$ and $f_{yx}(0,0)$ for the following function
Find $f_{xy}(0,0)$ and $f_{yx}(0,0)$ for the following function:

$$f(x,y) =
  \begin{cases}
                                   \frac{xy}{x^{2}+y^{2}} & (x,y) \ne (0,0) \\
  0 & (x,y) = (0,0)
  \end{cases}$$

My try:

$$f_x =
  \begin{cases}
                                   \frac{y(y^2-x^2)}{(x^2+y^2)^2} & (x,y) \ne (0,0) \\
  0 & (x,y) = (0,0)
  \end{cases}$$

$$f_y =
  \begin{cases}
                                   \frac{x(x^2-y^2)}{(x^2+y^2)^2} & (x,y) \ne (0,0) \\
  0 & (x,y) = (0,0)
  \end{cases}$$
$$\color{blue}{f_x(0,0)}=\lim_{\Delta x \to 0}\frac{f\left(\Delta x,0\right)-f\left(0,0\right)}{ \Delta x}=\lim_{\Delta x \to 0}\frac{0-0}{\Delta x}=0$$
$$f_{xy}(0,0)=\lim_{\Delta y \to 0} \frac{f_x\left(0,\Delta y\right)-\color{blue}{f_x(0,0)}}{ \Delta y}=\lim_{\Delta y \to 0}\frac{
\frac{1}{\Delta y}-0}{\Delta y}=0$$
$$\color{red}{f_y(0,0)}=\lim_{\Delta y \to 0}\frac{f\left(0,\Delta y\right)-f\left(0,0\right)}{ \Delta y}=\lim_{\Delta y \to 0} \frac{0-0}{\Delta y}=0$$

$$f_{yx}(0,0)=\lim_{\Delta x \to 0} \frac{f_y\left(\Delta x,0\right)-\color{red}{f_y(0,0)}}{ \Delta x}=\lim_{\Delta x \to 0}\frac{\frac{1}{\Delta x}-0}{\Delta x}=0$$

If you have an alternative solution please first check mine.

答:
Your only mistake here is
\begin{equation}
f_{xy}(0,0) = \lim_{h\rightarrow0}\frac{\frac{1}{h}}{h} = \lim_{h\rightarrow0}\frac{1}{h^2} = +\infty,
\end{equation}
and similarly $f_{yx}(0,0) =+\infty$. That being said, I think some remarks are well in order.

The function you are dealing with is a classical example illustrating some of the subtleties that arise when passing from single-variable calculus to multi-variate calculus. Embarassingly enough, I missed a lot of them myself.

Specifically, by setting $f(0,0)=0$ all 'horizontal' and 'vertical' sections of $f$ are continuous. By this I mean that for all fixed $y\in\mathbb{R}$, the single-variable function $f(\,\cdot\,,y)$ is continuous, and so is $f(x,\,\cdot\,)$ for any fixed $x\in\mathbb{R}$. Nonetheless $f$ is not continuous in $(0,0)$ as a multi-variate function, as you can see by examining its value on the diagonals $y=x$ and $y=-x$:
\begin{equation}
f(x,x) = \frac{x^2}{2x^2} = \frac{1}{2} \quad \forall x\neq0, \qquad f(x,-x) = \frac{-x^2}{2x^2} = -\frac{1}{2} \quad \forall x\neq0.
\end{equation}

This is important because continuity is a necessary condition for differentiability. Thus this is a function which is derivable in the origin, but not differentiable. In particular, the partial derivatives are not continuous in the origin as, for example,
\begin{equation}
f_x(x,0) = 0, \qquad f_x(0,y) = \frac{1}{4y} \quad \forall y\neq0,
\end{equation}
hence $f_x(x,0)\rightarrow0$ for $x\rightarrow0$ whereas $f_x(0,y)\rightarrow+\infty$ for $y\rightarrow0^+$.


问:
Name of "divided difference" transform $\frac{f(x)-f(x_0)}{x-x_0}$ and special case $\frac{e^x - 1}{x}$?
Given an analytic function / formal power series

$$\displaystyle f(x)=\sum _{n=0}^{\infty }\frac{f^{(n)}(x_0)}{n!}\left(x-x_{0}\right)^{n}=f(x_0)+f'(x_0)(x-x_{0})+ \tfrac{1}{2}f''(x_0)(x-x_{0})^{2}+\ldots$$

we can construct another analytical function via the "divided difference" transformation

$$ R(f)(x) := \frac{f(x)-f(x_0)}{x-x_0} := f'(x_0) + \tfrac{1}{2}f''(x_0)(x-x_0) +  \tfrac{1}{6}f'''(x_0)(x-x_0)^2 + \ldots $$

Note that this is similar but not equal to the derivative operator

$$ D(f)(x) = f'(x_0) + f''(x_0)(x-x_0) +  \tfrac{1}{2}f'''(x_0)(x-x_0)^2 + \ldots $$

Does this transformation have a name in the literature? What are its properties? In the discrete case it is known as divided differences, however here I am explicitly interested in this transform as a map $$T\colon A(D)\to A(D),\, f\mapsto R(f)$$ between the space of analytic functions $A(D)$ on an open interval $D\subset \mathbb R$.

Does the function we get by applying this transformation to the exponential function, i.e. $\frac{e^x-1}{x}$ have a name in the literature?

The latter appears in the integrand of the exponential integral; is apparently related to the derivative of the exponential map from Lie groups and occurs in the solution of inhomogeneous linear ODEs, e.g. $\dot x = a x+b, x(t_0)=x_0$ has the solution $  x^*(t) = e^{a(t-t_0)}x_0 + \frac{e^{a(t-t_0)} -1}{a} b$; and in the multivariate case $\dot x(t) = A \cdot x(t) + b ,  x(t_0) = x_0$ we get the analogous $  x^*(t) = e^{A(t-t_0)}x_0 + \frac{e^{A(t-t_0)}-I}{A}b$. Note that $\frac{e^{A(t-t_0)}-I}{A}$ exists even when $A$ is singular, hence it would be useful to drop this notation and give the function a name instead.

答:
I found the answer I was looking for. It appears that these are known as the $\varphi$-functions in numerical mathematics and they appear in the context of exponential integrators:

$\varphi_0(z) = e^{z}$

$\varphi_1(z)=\frac{e^{z}-1}{z}$

$\varphi_1(z)=\frac{e^{z}-1 -z}{z^2}$

$\ldots$

$z\varphi_{n+1}(z) = \varphi_{n}(z)-\frac{1}{(z-1)!}$

$\varphi_{S}(z)=\frac{1}{(s-1) !} \int_{0}^{1} e^{(1-\theta) z} \theta^{s-1} d \theta$

And they can be computed efficiently using Saad's Trick:

$$\exp\left(\begin{bmatrix} z & 1 \\ 0 & 0 \end{bmatrix}\right) = \begin{bmatrix} \varphi_0(z) & \varphi_1(z) \\ 0 & 1\end{bmatrix}$$

and in particular

$$\exp\left(\begin{bmatrix} A & b \\ 0 & 0 \end{bmatrix}\right) = \begin{bmatrix} \varphi_0(A) & \varphi_1(A)b \\ 0 & 1\end{bmatrix}$$

References:

Computing the Action of the Matrix Exponential, with an Application to Exponential Integrators, Al-Mohy and Higham (SIAM 2011)

Matrix Exponential Based Integration Algorithms for Circuit Transient Simulation


问:
Find point nearest to the origin
Find the points on the curve $5x^2 - 6xy + 5y^2 = 4$ that are nearest the origin.

The first method I've tried is I've taken the derivative of the equation to optimize (Pythagorean Theorem) and also the function of the curve using implicit differentiation and plugged stuff in (the way I've done it so far). I got $y = x$ which I tried to plug into the function of the curve but didn't match the answers at the back of the textbook. I stared at it and figured I made a sign error somewhere since $-y = x$ worked but couldn't find it.

The second method I tried was solving for the function of the curve in terms of $y$ and tried both completing the square and using the quadratic formula to find $x$ which gave the same failed answer where $y$ subtracted to nonexistence when I plugged them into the function of the curve.

I don't know what lagrange multipliers are, which was the only solution on the web. I hope it could be done just with simple algebra (or calculus) since the textbook expects that (I think).

答:
Let $x=r\cos\theta$ and $y=r\sin\theta$. Then
$$4=5x^2-6xy+5y^2=r^2-6r^2\cos\theta\sin\theta=r^2(5-3\sin 2\theta).$$
To minimize $r^2$, we maximize $5-3\sin 2\theta$. The maximum value of $5-3\sin 2\theta$ is $8$. </p>

Title: Find point nearest to the origin


问:
Convergence of power series at the boundary
Consider the following complex power series
$$
\sum_{n\geq 1}{\frac{ni^n}{2^n}{z^{n-1}}}
$$
By the root test, I have concluded that the disc of convergence is $D:=D(0,2)$. Then, I would like to study the convergence of the series in $\partial D$. Considering $z=2e^{i\theta}\in \partial D$, $\theta \in \mathbb{R}$, the series becomes
$$
\frac{1}{2}\sum_{n\geq 1}{n i^n e^{i(n-1)\theta}}
$$
which is not absolutely convergent and, indeed,
$$
\lim_{n} n i^n e^{i(n-1)\theta} \neq 0
$$
for every $\theta \in \mathbb{R}$, so the series cannot be convergent at the boundary. Is this reasoning true, or am I missing some aspect of the boundary convergence of complex power series? Thanks in advance for your answers.

答:
You are correct.
To be more precise: let $a_n:=n i^n e^{i(n-1)\theta}$, then $|a_n|=n$ for all $n$, thus $(a_n)$ does not converge to $0.$


问:
Algebra projective over subalgebra
Let $A$ be an algebra, $S \subset A$ a subalgebra.*
I was wondering:

Is $A$ projective as an $S$-module?

I have a feeling that this cannot be true in this generality (the feeling comes from the statement being false on the level of rings), hence I should actually be asking

When is $A$ projective as an $S$-module?

When is it not aka what are some counterexamples?

It's true for group algebras because of Lagrange's theorem.
More generally, the Nichols-Zoeller theorem shows that it's true for Hopf algebras and Hopf subalgebras.

*EDIT: I should add "over some ring", and the answers can/should also involve conditions on the ring. Also, in my examples I'm really thinking "finite-dimensional and over a field" just to be safe.

答:
No it is not true. I don't think one can get a satisfactory answer to your question, because it is too general.

For example, if $A$ is a ring with prime characteristic $p$, it contains $\mathbb{F}_p$, and your question  translate as:let $S$ be a subring. Is $A$ a projective $S$-module ?

I would be surprised if there was a reasonable answer to that.

However, one can answer (more or less) if you pick $S$ wisely:

Let $A$ be a finitely generated (commutative associative unital) $k$-algebra, where $k$ is a field.

Noether normalization lemma states that there is a polynomial subalgebra $S$ (that is $S\simeq k[X_1,\ldots,X_n]$) such that $A$ is integral over $S$, hence in particular a finitely generated $S$-module.

By Quillen -Suslin theorem (https://en.wikipedia.org/wiki/Quillen%E2%80%93Suslin_theorem), such a module is $S$-projective if and only if it is $S$-free.

This should give you plenty of examples/counterexamples. For an explicit counter-example, you can take  $S=k[x^4,y^4]$ and $A=k[x^4,x^3y,xy^3,y^4]$ (see When an integral extension of integral domains is flat?)

Title: Algebra projective over subalgebra


问:
How come modulo gives same result while applying on each step or applying once in end?
(abc)%n = (((ab)%n)c)%n = (((ac)%n)b)%n

Means, result is always same even if I apply on each step or once at end.

For Ex. a=3, b=2, c=3, n=5, always gives answer 4.

Why it feels weird to me?

If I assume modulo as clock with numbers from 0-4, we walk on this clock from 0-∞ and always remains on clock. But distance traveled in above cases is very different but still final position is same.

36=729 and final position is 4

32=9, and 43=64, now distance is 9+64=73, much lesser than previous but final position still 4.

33=27, 22=4, here distance is only 31, but final position still 4.

I don't know if I am using this clock process right, but it feels weird and confusing. Can anyone tell me why they gives same result or point me to some resource.

答: This is true because taking the residue modulo $n$ is a ring morphism from $\Bbb Z$ to $\Bbb Z{/}n\Bbb Z$. So we can interchange ring operations (like $+,-,\times$ (and their derived operations, like powers) and taking residues modulo $n$. You could formally show it using some recursion on arithmetical expressions, I suppose.


问:
What sequence is this $\prod_{i=1}^n \frac{2i}{2i + 1} = $?
I'm trying to find a closed form solution for $\prod_{i=1}^n \frac{2i}{2i + 1} $, but I'm confused what kind of sequence this is. Does it have a closed form?

答:
Closed Form
$$
\begin{align}
a_n
&=\prod_{k=1}^n\frac{\color{#C00}{2k}}{\color{#090}{2k+1}}\tag{1a}\\
&=\frac{\color{#C00}{2^nn!}\,\color{#090}{2^nn!}}{\color{#090}{(2n+1)!}}\tag{1b}\\[3pt]
&=\frac{4^n}{(2n+1)\binom{2n}{n}}\tag{1c}
\end{align}
$$

Bounds
$$
\begin{align}
\frac{\sqrt{n+1}\ a_n}{\sqrt{n}\ a_{n-1}}
&=\sqrt{\frac{n+1}{n}}\frac{2n}{2n+1}\tag{2a}\\
&=\sqrt{\frac{4n^3+4n^2}{4n^3+4n^2+n}}\tag{2b}\\[9pt]
&\le1\tag{2c}
\end{align}
$$
Therefore, $\sqrt{n+1}\ a_n$ is decreasing.
$$
\begin{align}
\frac{\sqrt{n+\frac34}\ a_n}{\sqrt{n-\frac14}\ a_{n-1}}
&=\sqrt{\frac{4n+3}{4n-1}}\frac{2n}{2n+1}\tag{3a}\\
&=\sqrt{\frac{16n^3+12n^2}{16n^3+12n^2-1}}\tag{3b}\\[12pt]
&\ge1\tag{3c}
\end{align}
$$
Therefore, $\sqrt{n+\frac34}\ a_n$ is increasing.

Since $\sqrt{n+1}\ a_n$ is decreasing and greater than $\sqrt{n+\frac34}\ a_n$, which is increasing, and their ratio tends to $1$, they tend to a common limit, $L$.
$$
\begin{align}
L
&=\lim_{n\to\infty}\sqrt{n+1}\prod_{k=1}^n\frac{2k}{2k+1}\tag{4a}\\
&=\lim_{n\to\infty}\frac{\sqrt{n+1}}{2n+1}\frac{4^n}{\binom{2n}{n}}\tag{4b}\\
&=\frac{\sqrt\pi}2\tag{4c}
\end{align}
$$
Explanation:

$\text{(4a)}$: definition of $L$

$\text{(4b)}$: apply $(1)$

$\text{(4c)}$: Theorem $1$ from this answer

Thus $(2)$, $(3)$, and $(4)$ yield
$$
\bbox[5px,border:2px solid #C0A000]{\sqrt{\frac{\pi}{4n+4}}\le\prod_{k=1}^n\frac{2k}{2k+1}\le\sqrt{\frac{\pi}{4n+3}}}\tag5
$$


问:
Inequality for a nested family of convex sets.
We consider a family of bounded open convex sets in $\mathbf{R}^n$ which we'll denote $\{G_h\}_{h \in (0,1]}$. They are non-decreasing in the sense $h_1 > h_2$ implies $G_{h_1} \supseteq G_{h_2}$. Put $G:= \bigcap_{h \in (0,1]}G_h$. Let $x_0 \in G$ and $r>0$ be given. I want to prove (or alternatively find a counterexample) to the implication
$$ h_1 > h_2 \implies \frac{|G_{h_1} \cap B_r(x_0)|}{|G_{h_1}|} \leq \frac{|G_{h_2} \cap B_r(x_0)|}{|G_{h_2}|}.$$

I think the implication is true (though am not certain). The intuition being that as the sets decrease the portion contained in the ball increases.

答: I think something like this should be a counterexample:


问:
Approximating $\int \log\big[1+\sin^2(t)\big]\,dt$
On yesterday, a now closed question asked for a while how to compute the antiderivative
$$I=\int \log\big[1+\sin^2(t)\big]\,dt$$ which, as given by a CAS, is just awful (see @Forester's comment and @Turing's answer). During the time the question was open, I tried on my side and gave up since being unable to find anything simpler.

Then, I tried to approximate the integrand since we just need a good representation of it for $0 \leq t \leq \frac \pi 2$ because of the symmetries and periodicity. What I am proposing is a polynomial function matching the function and its first and second derivative values at a minimum number of data points. For the time being, I chose $t=0$, $t=\frac \pi 4$ and $t=\frac \pi 2$. The final result is
$$f(t)=\log\big[1+\sin^2(t)\big]\sim t^2+\frac 19 \sum_{n=3}^8 \frac{a_n}{\pi^n} t^n=g(t)$$ the $a_n$'s being
$$\left(
\begin{array}{ccc}
n & a_n & \frac{a_n}{9\pi^n}\\
 3 & -768 \pi -217 \pi ^2-22608 \log (2)+18432 \log (3) &         +0.087806356 \\
 4 & 7680 \pi +1940 \pi ^2+286416 \log (2)-221184 \log (3) &      -1.360303642 \\
 5 & -27648 \pi -8500 \pi ^2-1501632 \log (2)+1105920 \log (3) &  +1.225383826 \\
 6 & 43008 \pi +19568 \pi ^2+3960576 \log (2)-2801664 \log (3) &  -0.512958585 \\
 7 & -24576 \pi -22720 \pi ^2-5170176 \log (2)+3538944 \log (3) & +0.102611702 \\
 8 & 10496 \pi ^2+2654208 \log (2)-1769472 \log (3) &             -0.007207813
\end{array}
\right)$$

About the norm
$$\Phi=\int_0^{\frac \pi 2} \big[f(t)-g(t)\big]^2 \, dt=1.79 \times 10^{-8}$$ while the beautiful approximation
$$\sin(t) \simeq \frac{16 (\pi -t) t}{5 \pi ^2-4 (\pi -t) t}\qquad \text{for} \qquad 0\leq t\leq\pi$$ would lead to  $\Phi=8.74 \times 10^{-7}$ (almost $50$ times larger).

From this
$$J(x)=\int_0^x \log\big[1+\sin^2(t)\big]\,dt\sim \frac 13 x^3+\frac 19 \sum_{n=3}^8 \frac{a_n}{(n+1)\,\pi^n} x^{n+1}$$

As a test
$$J\left(\frac{\pi }{2}\right)=\frac{\pi }{181440}\left(432 (128 \log (3)-87 \log (2))-55 \pi ^2 \right)\approx 0.591383$$ to be compared to the exact value
$$ \pi \log\left(\frac{1+\sqrt{2}}{2}\right)\approx 0.591331$$ recalled by @projectilemotion in comments.

My questions

Without adding more terms, could it be possible to improve the approximation ? Would changes of base points could have a significant impact (if yes, how to optimize them ?)

Would, by chance, be known (in a simple form) the exact values of the integrals
$$K_n=\int_{-\frac \pi n}^{\frac \pi n} \log\big[1+\sin^2(t)\big]\,dt$$ beside the case of $n=2$ ? If there were, I suppose that much better approximations could be done.

答:
Just for fun. Better than an approximate, the nice exact solution :

Of course this isn't from a normal man. Thanks to WolframAlpha.


问:
Conditional Probability, balls in bag.
There are 5 white balls and 4 black balls in a box. Three random balls are taken out.
What's the probability that out of the 3 balls the number of black balls is odd provided that the last taken ball was black?

I understand that this is an extremely straight forward problem, but I not able to match the answer with the solution. So, I needed a second pair of eyes to check if I am correct.

Here is what I did.

Set A (# 1 or 3 black balls) = {('W', 'B', 'W') ('W', 'W', 'B') ('B', 'B', 'B') ('B', 'W', 'W')}

Set B (last black ball) = {('W', 'B', 'B') ('W', 'W', 'B') ('B', 'B', 'B') ('B', 'W', 'B')}

Hence the probability that out of the 3 balls the number of black balls is odd provided that the last taken ball was black will be

$P(A | B) = \frac{\#(A \cap B)}{\#(B)}$

$A \cap B = {('W', 'W', 'B'), ('B', 'B', 'B')}$

$\Rightarrow P(A|B) = \frac{2}{4} = 0.5$

Thanks for helping.

答:
Your sets do not consider the probability of them occurring which can be different for different sets.

Say the event of third ball being black is $B$ and event of odd number of black balls is $A$.

Outcomes where third ball is black - $\small\{2B, B\}, \{2W, B\}, \{1B 1 W, B\}$

Outcomes where we have odd number of black balls and the last ball is black - $\small \{2B, B\}, \{2W, B\}$

So, $\small P(B) = \displaystyle \frac{4 \cdot 3 \cdot 2}{9 \cdot 8 \cdot 7} + \frac{5 \cdot 4 \cdot 4}{9 \cdot 8 \cdot 7} + 2 \cdot\frac{5 \cdot 4 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{4}{9}$

$\small P(A\cap B) = \displaystyle \frac{4 \cdot 3 \cdot 2}{9 \cdot 8 \cdot 7} + \frac{5 \cdot 4 \cdot 4}{9 \cdot 8 \cdot 7} = \frac{13}{63}$

So, $\small P(A|B) = \displaystyle \frac{13}{28}$

Title: Conditional Probability, balls in bag.


问:
What is the probability of winning this game?

A tennis player has a 60% chance of winning any given point in a tennis game. Calculate the probability that she will win the game within the first 6 points, stating any assumptions you make.

(A game is won when a player has won at least 4 points and won at least 2 more points than their opponent.)

So there are 3 choices:

she wins 4 points with the probability 60%$^{4}$

she wins 4 points and lose 1 point with probability 4 $\times$ 60%$^{4}$ $\times$ 40%

she wins 4 points and lose 2 points 15 $\times$ 60%$^{4}$ $\times$ 40%$^{2}$

So the total probability of her winning the game is 0.69984. Is this right?

答:
To win in $\leq 6$ points, she must lose at most $2$ points in $6$, so

P(win) = $\binom 6 0\cdot 0.4^0\cdot 0.6^6 + \binom 6 1\cdot 0.4^1\cdot 0.6^5  + \binom 6 2\cdot  0.4^2 \cdot 0.6^4 = 0.54432$

PS

The above formulation is the simplest way to get the answer. However, just for corroboration, let us also solve it by summing up

P(win in exactly $4,5,\;or\; 6$) points

To win in exactly $4$ points, she must win all: $0.6^4 =0.1296$

To win in $5$ points, she must win the fifth point and three of the first four: $\left(\binom 4 3 \cdot 0.6^3\cdot0.4\right)\cdot0.6 = 0.20736$

To win in $6$ points, she must win the sixth point, and three of the first five:$\left(\binom 5 3 \cdot 0.6^3\cdot0.4^2\right)\cdot0.6 = 0.20736$

$0.1296+ 0.20736 +0.20736 = 0.54432$, as before

But error prone method with possibly confusing binomial coefficients


问:
harmonic function in a 2d disc.
So my professor gave as the following question: Given the function u which is continuous in $\bar{D}$ , where $D = \left \{ (x,y) \in \mathbb{R}^{2} : x^{2}+y^{2}<1 \right \}$, and harmonic in $D$.

Find $u(0,0)$, if $u(x,y)=x^{2}y^{2}$ for $(x,y) \in \partial D$.

Now I know from the mean value theorem that the value of $u$ at the center of the disc is the mean of its values at the border of the disc $\partial D $. So we get $u(0,0)=\frac{1}{\left | \partial D \right |} \int_{(x,y) \in \partial D} u(x,y)ds$, but how do we calculate this?
Please help me, thank you in advance.

Title: harmonic function in a 2d disc.

Body: $$u(0,0)= \frac{1}{2 \pi}\int_0^{2 \pi}u( \cos t, \sin t) dt=\frac{1}{2 \pi}\int_0^{2 \pi} \cos^2 t \sin^2 t dt.$$


问:
Generalized "Worm on the rubber band " problem
I found this « Worm on the rubber band » problem in Concrete Mathematics book.

A slow worm $W$ starts at one end of a meter-long rubber band and crawls one centimetre per minute toward the other end.

At the end of each minute, a keeper of the band $K$ stretches it one meter.

Does the worm ever reach the finish?

The given solution is: $W$ reaches the finish if and when $H(n)$ ever surpasses 100, where $H(n)$ is the $n$th Harmonic number.

How to solve this generalized problem with continuous data and with $W$ crawling with a velocity $u=f(t)$ and $K$ crawling with velocity $v=g(t)$  where $u(t)$ and $v(t)$ are both arbitrary functions of time.

For example, I want to find if and when the worm will reach the end of a rubber band of length $L$ if (with $a$ and $b$ constants) $u(t)=a*t$ and $v(t)=b*(t)$ ?

答:
Bonjour, Jean-Pierre, the relative position of the insect on the rubber band is
$$r(0)=x_W(0) / x_K(0)$$
and you will have to know it to solve the problem. It will be a number between 0 and 1.

Now, you want to find $t$ such that $x_W(t)=x_K(t)$ i.e. $r(t)=1$. To find it, you must calculate the time derivative of $r(t)$.
$$r'(t) = \frac{u(t)}{x_K(t)} . $$
It's because the relative position of the insect is determined essentially by its velocity $u(t)$ but the longer the rubber band is, i.e. the greater is $x_K(t)$, the less the insect manages to change the relative position.

Given your data, $x_K(t)$ is the indefinite integral (that you have to calculate) of the function $v(t)$ you have mentioned and $u(t)$ is something you mentioned. So both the numerator and denominator are, in principle, known functions of $t$, so you solve the differential equation above.

One may also write the general compact formula for a given $r(0),u(t),v(t)$. But it's more interesting to know a major example. If $v(t)$ is constant, $x_K(t)$ goes like $Kt+x_0$ and if $u(t)$ is constant, the equation above says $r'(t)\sim 1/(t+t_0)$. This is solved by $r(t)\sim \ln((t+t_0)/T)$ and when you invert it, you will find out that the $t$ for which $r(t)=1$ is an exponential function of the parameters. It's a straightforward exercise, and given the risk that this is a homework, I don't want to write every detail and the final answer here.

So the ant will finally get there but you will need an exponentially long time! For more general functions, the computation is more complicated.


问:
Help with a Vardi Integral
I ran into some trouble trying to evaluate this integral:

$$\mathcal{I}=\int_{0}^{1}\frac{x}{x^{2}+1}\log\log\left(\frac{1}{x}\right)\mathrm dx$$

Basically, what I have at the moment is a generalized expansion

$$\int_{0}^{1}\frac{x^{p-1}}{\delta + x^{w}}\log\log\left(\frac{1}{x}\right)\mathrm dx = -\gamma\int_{0}^{1}\frac{x^{p-1}}{\delta+x^{w}}\mathrm dx-\frac{1}{\delta}\sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(wk+p)}{\delta^{k}(wk+p)}$$

Hence,

$$\mathcal{I} = -\gamma\int_{0}^{1}\frac{x}{1+x^{2}}\mathrm dx - \sum\limits_{k\geq 0}(-1)^{k}\cdot\frac{\log(2k+2)}{(2k+2)}$$

I can't figure out how to solve the sum. I'm positively certain there's some trick with digamma functions or Hurwitz-Zeta, but I'm too sleep-deprived to make sense of it. Maybe I'm going about this all wrong and there exists a more straightforward method to solve $\mathcal{I}$? Please let me know your opinion!

答:
Using
$$\frac x{x^2+1}=\sum_{n=0}^\infty (-1)^n x^{2n+1}$$ and, after a simple change of variable
$$\int  x^{2n+1} \log (-\log (x))\,dx=-\frac{\text{Ei}(2 (n+1) \log (x))-x^{2 n+2} \log (-\log (x))}{2 (n+1)}$$
$$\int_0^1  x^{2n+1} \log (-\log (x))\,dx=-\frac{\log \left(4 (n+1)^2\right)-2 i \pi +2 \gamma }{4 (n+1)}$$ you should arrive at the result
$$\mathcal{I}=\int_{0}^{1}\frac{x}{x^{2}+1}\log\log\left(\frac{1}{x}\right)\mathrm dx=-\frac{3}{4} \log ^2(2)$$

Title: Help with a Vardi Integral


问:
Solving differential equation ${\rm d}y/{\rm d}x = (4x - y)^2$
Let $u = 4x - y$

${\rm d}u = 4{\rm d}x - {\rm d}y$

$4{\rm d}x - {\rm d}u = {\rm d}y$

$4{\rm d}x - {\rm d}u = (4x - y)^2$

$4{\rm d}x - {\rm d}u = u^2$

I was stuck at $-{\rm d}u = (u - 4){\rm d}x$

by separation of variables (SOV), so what's next?

答:
$$u=4x-y \implies \frac{du}{dx}=4-\frac{dy}{dx},$$ the the ODE becomes
$$\frac{du}{dx}=4-u^2 \implies \int \frac{du}{4-u^2}= \int dx \implies \frac{1}{2} \tanh^{-1}(u/2)=x+C.$$


问:
Is the derivative of a continuous function integrable on a closed interval?
Say, if f', derivative of a continuous function f, is defined on [a,b], will it be Riemann Integrable on [a,b]?</p>

答:
Example. Let $f: [0,1] \to \mathbb R$ be defined by

$f(x)=x^{3/2} \sin (1/x)$ if $x \ne 0$ and $f(0)=0.$

Then $f$ is continuous and differentiable on $[0,1].$

Compute $f'(x)$ for $0 \le x \le 1$ and show that $f'$ is unbounded on $[0,1]$. Hence $f$ is not Riemann interable.


问:
Using implicit function theorem without using the inverse function theorem.
Let $f:U\rightarrow \mathbb{R}$ defined on the open set $U \subset \mathbb{R}^m$. It the function $g(x):U\rightarrow \mathbb{R}$, given by the expression $$g(x)= \int_{0}^{f(x)} (t^2 + 1)dt,$$ of class $C^{\infty}$, then f is $C^{\infty}$.Explicitly using the implicit function theorem, knowing that the inverse function is equivalent.</p>

答:
The function
$$\phi:\quad {\mathbb R}\to{\mathbb R},\qquad y\mapsto y+{y^3\over3}$$
is $C^\infty$ and has first derivative $\geq1$; therefore it maps ${\mathbb R}$ diffeomorphically onto  ${\mathbb R}$. It follows that $\phi$ has a $C^\infty$ inverse $\psi:\  {\mathbb R}\to{\mathbb R}$.

Now $$g(x)=\phi\bigl(f(x)\bigr)\qquad(x\in U)$$
and therefore
$$\psi\bigl(g(x)\bigr)= f(x)\qquad(x\in U)\ .$$
This proves that the function $f$ is $C^\infty$ in $U$.


问:
Time to return to initial position given $v(t)$
Question:

A particle moves in a straight line such that its velocity, $v$ in m/s, at time $t$ seconds, is given by:

$$v(t)=\left\{\begin{array}{cc} 5-(t-2)^2, & 0\leq t\leq 4\\ 3-\frac{t}{2},& t>4\end{array}\right.$$

the particle returns to its initial position at $t=T$. Find the value of $T$.

Doubt/Problem:

The mark scheme suggests that the initial position is equal to the distance travelled before coming to rest. Whereas I believe the initial position is the displacement at $t=0$, which I found out to be $4.33$ meters.

I am unable to identify what's wrong with my thinking pattern.

Mark scheme:

答:
Let's call $x(t)$ to the particle position at time $t$. From the definition of velocity,
$$v(t)=\frac{\mathrm{d}x}{\mathrm{d}t}\Longrightarrow\mathrm{d}x=v(t)\mathrm{d}t\Longrightarrow x(t)-x(0)=\int_0^tv(s)\mathrm{d}s\Longrightarrow x(t)=x(0)+\int_0^tv(s)\mathrm{d}s.$$

If $0\leq t\leq4$, the position is
\begin{align*}
x(t) & =x(0)+\int_0^t\bigl(5-(s-2)^2\bigr)\mathrm{d}s=x(0)+\int_0^t(-s^2+4s+1)\mathrm{d}s=\\
& =x(0)+\left[-\frac{s^3}{3}+2s^2+s\right]_0^t=x(0)-\frac{t^3}{3}+2t^2+t.
\end{align*}

And, if $t>4$,
\begin{align*}
x(t) & =x(0)+\int_0^4\bigl(5-(s-2)^2)\mathrm{d}s+\int_4^t\left(3-\frac{s}{2}\right)\mathrm{d}s=\\
& =x(0)-\frac{4^3}{3}+2\cdot4^2+4+\left[3s-\frac{s^2}{4}\right]_4^t=\\
& =x(0)+\frac{44}{3}+3t-\frac{t^2}{4}-\left(12-\frac{16}{4}\right)=\\
& =x(0)+\frac{t^2}{4}+3t+\frac{20}{3}.
\end{align*}

Now, let's see if the particle can get back to its initial position at $t\leq4$:
$$x(0)-\frac{t^3}{3}+2t^2+t=x(0)\Longrightarrow t\left(-\frac{t^2}{2}+2t+1\right)=0.$$
The solutions of the above equation are $t_1=0$, $t_2=2+\sqrt{6}>4$ and $t_3=2-\sqrt{6}<0$  and none of them is valid for this case.

So, now we look for $t>4$. We have
$$x(0)+\frac{t^2}{4}+3t+\frac{20}{3}=x(0)\Longrightarrow\frac{t^2}{4}+3t+\frac{20}{3}=0.$$
The solutions are
$$t_4=6-2\sqrt{\frac{47}{3}}<0,\quad t_5=6+2\sqrt{\frac{47}{3}}\approx13.9.$$
So $T=t_5\approx 13.9$.


问:
Nonlinear System of ODES
I have been trying to solve this nonlinear system of ODEs analytically with no luck:

$$\frac{d\psi}{dt}=-cot \theta cos \psi$$

$$\frac{d\theta}{dt}=-sin\psi$$

$$\frac{d\phi}{dt}=\frac{cos\psi}{sin\theta}$$

Where of course $\psi$, $\theta$ and $\phi$ are functions of $t$.

Any hints would be appreciated!

答:
$$\frac{d\psi}{dt}=-\cot (\theta) cos (\psi)$$

$$\frac{d\theta}{dt}=-\sin(\psi)$$

$$\frac{d\phi}{dt}=\frac{cos(\psi)}{sin(\theta)}$$

Hint: notice how there are the same arguments in the first and third equation.

$$\frac{d\psi}{dt}=-\dfrac{cos(\theta)}{\sin(\theta)} \cos (\psi)$$

Let's solve the 3rd equation to $\sin(\theta)$

$$\sin(\theta)=\dfrac{dt}{d\phi}\cos(\psi)$$

And plug it in the 1st one:

$$\frac{d\psi}{dt}=-\dfrac{cos(\theta)}{\dfrac{dt}{d\phi}\cos(\psi)} \cos (\psi)$$

Then $\cos(\psi)$ falls out as long as it doesn't equal zero.

Edit: I made a mistake as i accidentally wrote $d\psi/dt$ instead of $d\phi$ which is fixed now. Further simplification is required, later i'll try to make it simpler.

$$\frac{d\theta}{dt}=-\sin(\psi)$$

$$\frac{d\psi}{dt}=-\dfrac{cos(\theta)}{\dfrac{dt}{d\phi}}$$

Don't forget that $y'\cdot y'\neq y''$ but  $(y')^2$

Title: Nonlinear System of ODES


问:
Gomori algorithm: Stuck at second constrain because it gives me the same solution as when I applied the first costrain.
Here is the function to be maximised: $f = x_1 + 2x_2$.

Constrains: $$-5x_1+2x_2\le 10 \\ 3x_1+2x_2\le12 \\ x_1-4x_2\le 3$$
I need to find a solution to this problem with $x_1,x_2 \in Z$ using Gomori algorithm.
First I applied simplex algorithm to find the solution to the problem, where $x_1,x_2$ are not necessarily integers:
Last symplex table and solution

From there, I choose to change the row of symplex table with basal variable $x_2$:
$x_6 = \frac{-5}{8}+\frac{3}{16}x_3+\frac{5}{16}x_4$. Because $x_6 \ge 0 \Rightarrow 3x_3 +5x_4 \ge 10$. I express $x_3, x_4$ from LPP canonical:
$$x_3 = 5x_1-2x_2+10 \\ x_4 = -3x_1-2x_2+12$$

Then $$10 \le 3(5x_1-2x_2+10) +5(-3x_1-2x_2+12) \Rightarrow x_2\le 5$$

New problem with the additional constrain rewritten in canonical form $LPP_1$:
$$f=x_1+2x_2 \rightarrow \max \\ -5x_1+2x_2= 10 \\ 3x_1+2x_2=12 \\ x_1-4x_2= 3 \\ x_2+x_6 = 5$$
The last Symplex table and solution:

So, now I have $x* = (\frac{2}{3},5)^T$. The first component is non integer, so I will make changes co a Symplex table row with basal variable $x_1$:

$$x_7 = -\frac{2}{3}+\frac{1}{3}x_4 -\frac{2}{3}x_6 \Rightarrow \frac{1}{3}x_4 -\frac{2}{3}x_6 \ge \frac{2}{3}$$. Then from the $LPP_1$ I will express $x_4, x_6:$

$$x_4 = -3x_1-2x_2+12 \\ x_6 = -x_2+5$$

From there: $$2 \le (-3x_1-2x_2+12) -2(-x_2+5) \Rightarrow x_1 \le 0 \Rightarrow x_1 = 0$$
How I go from there and isn't there any mistakes in my computations?

答: Maximum/minimum of the linear function for example $z=x_1+2x_2$, will occur at the vertices of the closed region indicated by these three lines. This is due to the theorem of linear programming. The max value is $23/2$ and it occurs at $x=1/4,y=45/8.$


问:
Express $(\theta, \phi)$ of a torus using $(x,y,z)$
I know how to write the $(x,y,z)$ coordinates of a torus using $(\theta, \phi)$:
$$
\begin{align}
x(\theta, \phi) &= (R+r\cos(\theta))\cos(\phi) \\
y(\theta, \phi) &= (R+r\cos(\theta))\sin(\phi) \\
z(\theta, \phi) &= r\sin(\theta)
\end{align}
$$

For $\theta, \phi\in [0, 2\pi]$. How can I find the inverse relation? I would like to find expressions
$$
\theta(x,y,z) = \ldots \\
\phi(x,y,z)=\ldots
$$

Given that I have a point on the torus in cartesian coordinates $(x, y,z)$. How can I find its expression in terms of angles $(\theta, \phi)$?

From the last expression I can get
$$
\theta = \arcsin\left(\frac{z}{r}\right)
$$
and then I guess from the first equation one could find
$$
\phi = \arccos\left(\frac{x}{R + r\cos(\theta)}\right)
$$
which works when the denominator is not zero. To avoid having the denominator being zero we need
$$
R+r\cos(\theta) \neq 0 \implies \theta \neq \arccos\left(-\frac{R}{r}\right)
$$

答:
If you remember what $r$ and $R$ are (inner and outer radius), you see that $R > r$ and so your additional assumption is automatically satisfied:
$$
R > r
\quad \Longrightarrow \quad
R > r \cos \theta
\quad \Longrightarrow \quad
R - r \cos \theta \neq 0
\quad \text{for any } \theta.
$$

To be precise, these formulas only work in some region, in the same way that $\alpha = \arcsin(\sin \alpha)$ is only sometimes true. To fix this, you can either treat each case separately or compute both $\sin$ and $\cos$ for the angles you need (which is better - see the explanation below and Adam Latosiński's answer).

Added later. A general remark on using $\arcsin$ and $\arccos$, especially in coding context. Knowing $\cos \alpha$ or $\sin \alpha$ is not enough to recover $\alpha$. You could artificially divide the domain into several parts and use a different formula (involving $\arcsin$ or $\arccos$) in each region, but even then, the method is numerically bad.

In contrast, if you know both $\sin \alpha$ and $\cos \alpha$, the angle $\alpha$ is uniquely determined (up to a multiple of $2\pi$). Programming languages provide two-variable functions that allow you to recover $\alpha$ from these two numbers, see here for Python.


问:
how can we prove that this function is homomorphism?
If $G_1 , ... , G_n$ are groups
let $G=G_1\times G_2\times\dotsb\times G_n$.

$E_p\colon G\to G_{p^{-1}(1)}\times G_{p^{-1}(2)}\times\dotsb\times G_{p^{-1}(n)}$
where $p \in S_n$ is defined as
$(g_1 , g_2 , \dotsc , g_n )\mapsto(g_{p^{-1}(1)} , g_{p^{-1}(2)} , \dotsc ,g_{p^{-1}(n)})$.

Now in the case $G_1 = G_2 = \dotsb = G_n$, show that the function $A\colon S_n\to\operatorname{Aut}(G)$ defined by $p\mapsto E_p$ is homomorphism.

I tried to solve it "the idea is not difficult" but every time, I found it's not a homomorphism!

So I need help

thanx

This is an exercise #8 in Dummit and Foote's abstract algebra book
page 156 but i have changed the symbols because I can't use the greek symbols in this site.

Body:
On the one hand,

$E_{p_1p_2}(g_1,\ldots,g_n) = (g_{(p_1p_2)^{-1}(1)},\ldots,g_{(p_1p_2)^{-1}(n)})=(g_{p_2^{-1}(p_1^{-1}(1))},\ldots,g_{p_2^{-1}(p_1^{-1}(n))})$

On the other hand,

$E_{p_1}(E_{p_2}(g_1,\ldots,g_n)) = E_{p_1}(g_{p_2^{-1}(1)},\ldots,g_{p_2^{-1}(n)})$

Now define $g'_n:=g_{p_1^{-1}(n)}$. Then we have

$E_{p_1}(E_{p_2}(g_1,\ldots,g_n)) = E_{p_1}(g'_1,\ldots,g'_n) = (g'_{p_1^{-1}(1)},\ldots,g'_{p_1^{-1}(n)}) = (g_{p_2^{-1}(p_1^{-1}(1))},\ldots,g_{p_2^{-1}(p_1^{-1}(n))})$

So we get $E_{p_1p_2}=E_{p_1}E_{p_2}$.


问:
Difficulty with Po-shen Loh Quadratic Equation
I am a math tutor looking for a better way to explain quadratics to my students.

I came across the so-called "Po-shen Loh" method on YouTube, and I tried to apply it to a quadratic equation that one of my students was having trouble with. (Here is the video that I learned it from, skipped to the example that is most similar to my problem: https://youtu.be/XKBX0r3J-9Y?t=1663)

The equation that I want to solve is $5x^2 - 4x - 6 = 0$

Here is the correct result, as confirmed by symbolab (https://www.symbolab.com/solver/quadratic-equation-calculator/solve%20for%20x%2C%205x%5E%7B2%20%7D-4x-6%3D0):

$$
x_{1,2} = \frac{2 \pm \sqrt{34}}{5}
$$

However, when I try to work it out using the "Po-shen Loh" method, I keep getting this result:

$$
x_{1,2} = \frac{2 \pm \sqrt{40}}{5}
$$

Here are my calculation steps, please help me if you can:

答:
(I am eternally grateful to player3236)

My mistake was in the line:

$\frac{2}{5} - u^2=-\frac{6}{5}$

However, this should have been:

$\frac{4}{25} - u^2=-\frac{6}{5}$

Why? In the immortal words of player3236:

$(2/5 - u)(2/5 + u) \neq 2/5 - u^2$

The "Poh-shen Loh" method involves a difference of squares at that stage in the calculation.


问:
Angle between the curves without differentiation
$y^2=4(x+1)$ and $x^2=4(y+1)$

Are the two curves, the answer online uses calculus and we have not been taught that.

can this be solved any other way?

Thanks in advance.

答: Write the product of the equations of the parabolas around the intersection points $(2\pm 2\sqrt 2, 2\pm 2\sqrt2):$ $-16[(x-(2\pm 2\sqrt{2}))^2\pm \sqrt{2}(x-(2\pm 2\sqrt{2}))(y-(2\pm 2\sqrt{2}))+(y-(2\pm 2\sqrt{2}))^2]+$ higher order terms, and use the $\tan \theta = {2 \sqrt{h^2-ab} \over a+b},ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0:$ $$\tan \theta={2 \sqrt{(\sqrt{\pm 2})^2-1} \over 1+1}=1.$$


答:
To solve
$$dy/dx = (4x-y)^2$$
is to find a function $y(x)$, that satisfies this equation.

Substitute
$$u=4x-y$$
It follows
$$\Rightarrow du/dx = 4 - dy/dx \Leftrightarrow dy/dx = 4-du/dx  $$
Plug in the $u$ and the $du/dx$ in the original equation
$$\begin{align} & dy/dx = 4-du/dx = u^2 = (4x-y)^2 \\ \Rightarrow & \ du/dx = 4-u^2 \\  \Rightarrow & \ dx/du = 1/(4-u^2) \\  \Rightarrow &  \ dx = du/(4-u^2)\end{align}$$

It follows
$$\begin{align}x &= \int du/(4-u^2) +C\\
&= (1/4) \cdot  \int du/(1-(u/2)^2) + C\\
&= (2/4) \cdot  \int da/(1-a^2) + C && a = u/2 \ \ \text{and} \ \ da/du = 1/2\\
&= (1/2) \cdot \text{arctan}(a) + C \\
&= (1/2) \cdot \text{arctan}\left(u/2\right) + C \\
&= (1/2) \cdot \text{arctan}\left((4x-y)/2\right) + C && u = 4x-y \\
\end{align}$$

This means that

$$ x = (1/2) \cdot \text{arctan}\left((4x-y)/2\right) + C$$

which has to be solved for $y$. Maybe this way:

Substract C from both sides
$$ x - C = (1/2) \cdot \text{arctan}\left((4x-y)/2\right) $$

Multiply by $2$
$$ 2(x - C) =  \text{arctan}\left((4x-y)/2\right) $$

Take $\tan$ on both sides
$$ \tan(2(x - C)) =  (4x-y)/2 $$

Multiply by $2$
$$ 2\tan(2(x - C)) =  4x-y $$

Rearrange
$$ y =  4x-2\tan(2(x - C)) $$

Not so sure if $ y =  4x-2\tan(2(x - C)) $ satisfies $dy/dx = (4x-y)^2$ ... ?

Title: Solving differential equation ${\rm d}y/{\rm d}x = (4x - y)^2$

Body:
Let $u = 4x - y$

${\rm d}u = 4{\rm d}x - {\rm d}y$

$4{\rm d}x - {\rm d}u = {\rm d}y$

$4{\rm d}x - {\rm d}u = (4x - y)^2$

$4{\rm d}x - {\rm d}u = u^2$

I was stuck at $-{\rm d}u = (u - 4){\rm d}x$

by separation of variables (SOV), so what's next?


问:
Zorn's lemma implies the well-ordering principle
I am little confused about the proof given here http://euclid.colorado.edu/~monkd/m6730/gradsets05.pdf

On the second page, when defining $P$, the author says that $B\subset A$ and $(B,<)$ is a well-ordering structure. Isn't this exactly what we want to prove? How do we know that $B$ can be well-ordered? What happens if $P$ is empty?

答: If $A$ is non-empty then every finite subset is well-orderable, by definition of finite. So $P$ is never empty if $A$ is non-empty.</p>


问:
existence of Riemann metric on smooth manifold
Let $M$ be the smooth manifold,prove there exist some Riemann metric on it.

This is the standard application of partition of unity thoerem:First find the local $g_\alpha$ on each chart by pullback the standard metric on Euclidean space.

since the chart of the manifold covers $M$,appling partition of unity to glue those $g_\alpha$ togother as :

$$g = \sum\rho_\alpha g_\alpha$$

To prove it's a Riemann metric,we need to check two things :

first it's smooth,second at each point it satisfy the inner product axioms.

Here is a bit detail that confuses me is how to define $g_\alpha$ outside original domain $U_\alpha$?It may depend on the existence of smooth extension of $g_\alpha$ outside $U_\alpha$.Does this result holds?To make it more clear I make the proposition below:

Let $M$ be a smooth manifold,prove given closed subset $A \subset M$,and a smooth tensor field on $A$ ,there exist a smooth extension of this tensor field to $M$.

答: It doesn't matter because $\rho_\alpha$ is supported on $U_\alpha$, which means that $\rho_\alpha g_\alpha$ is zero outside $U_\alpha$.


问:
99% of all readings are within WHAT temperature range?
I am a bit stumped.

"The temperature reading from a thermocouple placed in a constant-temperature medium is
normally distributed with mean of 82.5C and standard deviation .1C. 99% of all readings
are within what temperature range?"

The question states that multiple readings are taken - no sample size is given. The mean is 82.5 with standard deviation of 0.1.
I am not sure whether the question is badly phrased or some other error in it.

I attempted is as a Confidence Interval, however a N would be necessary.

I would appreciate at least a hint because I feel like I am going mad.

答:
Here is a different phrasing of the same problem:

Find an interval such that the probability that a reading lies within that interval is 99%.

Your thinking is too advanced and complicated. Do the simple and immediate thing.

Of course, there are many possible answers. Presumably you are asked to find the interval which is centered on the mean.


答:
simply you have to calculate which is the minimum range covering 99% of the distribution. Due to the symmetry of the Gaussian you just have to waste 0.5% per tail

Can you complete the problem?

Title: 99% of all readings are within WHAT temperature range?

Body:
I am a bit stumped.

"The temperature reading from a thermocouple placed in a constant-temperature medium is
normally distributed with mean of 82.5C and standard deviation .1C. 99% of all readings
are within what temperature range?"

The question states that multiple readings are taken - no sample size is given. The mean is 82.5 with standard deviation of 0.1.
I am not sure whether the question is badly phrased or some other error in it.

I attempted is as a Confidence Interval, however a N would be necessary.

I would appreciate at least a hint because I feel like I am going mad.


答:
Note that you have $$ \sqrt{x^2+y^2} = R+r\cos\theta$$
We have then
$$ \cos\phi = \frac{x}{\sqrt{x^2+y^2}} $$
$$ \sin\phi = \frac{y}{\sqrt{x^2+y^2}} $$
$$ \cos\theta = \frac{\sqrt{x^2+y^2}-R}{r}$$
$$ \sin\theta = \frac{z}{r}$$
Which can be used to recover $\theta$ and $\phi$. You should be careful with using $\arcsin$ though, at it will give you the correct value of the angle only if said angle is in the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$.

To deal with that you need to consider different cases. That is because the equation $\sin\alpha = t$ has many solutions; one is $\alpha=\arcsin t$, but $\alpha=2\pi n+\arcsin t$, $n\in\mathbb{Z}$ is also a solution, as is $\alpha=2\pi n+\pi-\arcsin t$, $n\in\mathbb{Z}$. Out of all of these solutions the correct one is the one that gives the correct sign of $\cos \alpha$ and gives angle $\alpha$ in the correct predefined interval. For example, if you want $\phi$ to be in the interval $[0,2\pi)$, the full formula for $\phi$ will be
$$ \phi = \left\{\begin{array}{ll}\arcsin\frac{y}{\sqrt{x^2+y^2}} & \text{if } x\ge 0,y\ge 0 \\ \pi-\arcsin\frac{y}{\sqrt{x^2+y^2}} &\text{if } x \le 0\\ 2\pi+\arcsin\frac{y}{\sqrt{x^2+y^2}} &\text{if } x\ge 0, y<0\end{array}\right. $$
On the other hand, if you wanted the values of $\phi$ to be in the range $(-\pi,\pi]$, the formula would be
$$ \phi = \left\{\begin{array}{ll}\arcsin\frac{y}{\sqrt{x^2+y^2}} & \text{if } x\ge 0 \\ \pi-\arcsin\frac{y}{\sqrt{x^2+y^2}} &\text{if } x \le 0, y\ge 0\\ -\pi-\arcsin\frac{y}{\sqrt{x^2+y^2}} &\text{if } x\le 0, y<0\end{array}\right. $$
It can be checked that if $\phi$ is definied like this, then $\sin\phi$ and $\cos\phi$ satisfy the previous equations regardless of the signs of $x$ and $y$. $\theta$ can be properly defined in an analogous way, with different formulas depending on the signs of $z$ and $\sqrt{x^2+y^2}-R$.

Title: Express $(\theta, \phi)$ of a torus using $(x,y,z)$

Body:
I know how to write the $(x,y,z)$ coordinates of a torus using $(\theta, \phi)$:
$$
\begin{align}
x(\theta, \phi) &= (R+r\cos(\theta))\cos(\phi) \\
y(\theta, \phi) &= (R+r\cos(\theta))\sin(\phi) \\
z(\theta, \phi) &= r\sin(\theta)
\end{align}
$$

For $\theta, \phi\in [0, 2\pi]$. How can I find the inverse relation? I would like to find expressions
$$
\theta(x,y,z) = \ldots \\
\phi(x,y,z)=\ldots
$$

Given that I have a point on the torus in cartesian coordinates $(x, y,z)$. How can I find its expression in terms of angles $(\theta, \phi)$?

From the last expression I can get
$$
\theta = \arcsin\left(\frac{z}{r}\right)
$$
and then I guess from the first equation one could find
$$
\phi = \arccos\left(\frac{x}{R + r\cos(\theta)}\right)
$$
which works when the denominator is not zero. To avoid having the denominator being zero we need
$$
R+r\cos(\theta) \neq 0 \implies \theta \neq \arccos\left(-\frac{R}{r}\right)
$$


问:
Prove that if $\lim_{n\to\infty} \frac{x_n}{y_n} = 0$, then $\lim_{n\to\infty} x_n \div \frac{x_n + y_n}{2} = 0$
Let $\{x_n\}$ and $\{y_n\}$ be positive sequences. Assume $\lim_{n\to\infty} \frac{x_n}{y_n} = 0$. I have to prove if the claim $\lim_{n\to\infty} x_n \div \frac{x_n + y_n}{2} = 0$ is always true, always false, or sometimes true and sometimes false depending on the sequences.

My attempt:
\begin{align*}
x_n + y_n &\geq y_n \\
0 < \frac{1}{x_n + y_n} &\leq \frac{1}{y_n} \\
0 < \frac{x_n}{x_n + y_n} &\leq \frac{x_n}{y_n}
\end{align*}
Then by Squeeze Theorem, $\lim_{n\to\infty} \frac{x_n}{x_n + y_n} = 0 \implies \lim_{n\to\infty} \frac{2x_n}{x_n + y_n} = 0$ which is what we wanted. Could you check if my proof makes sense. Is there any cases of sequences of x_n and y_n such that this statement is not true?

答:
Alternative approach

$\frac{y_n}{x_n} \to \infty \implies$

$\frac{1}{2} \times \frac{x_n + y_n}{x_n} = \frac{1}{2}
\times \left[1 + \frac{y_n}{x_n}\right] \to \infty.$


答:
$$\frac{x_n}{x_n+y_n\over 2}=2\frac{x_n \over y_n}{{x_n\over y_n}+1}$$

Therefore $$\lim_{n\to \infty}\frac{x_n}{x_n+y_n\over 2}=2 \frac{0}{{0}+{1}}=0 $$

Title: Prove that if $\lim_{n\to\infty} \frac{x_n}{y_n} = 0$, then $\lim_{n\to\infty} x_n \div \frac{x_n + y_n}{2} = 0$

Body:
Let $\{x_n\}$ and $\{y_n\}$ be positive sequences. Assume $\lim_{n\to\infty} \frac{x_n}{y_n} = 0$. I have to prove if the claim $\lim_{n\to\infty} x_n \div \frac{x_n + y_n}{2} = 0$ is always true, always false, or sometimes true and sometimes false depending on the sequences.

My attempt:
\begin{align*}
x_n + y_n &\geq y_n \\
0 < \frac{1}{x_n + y_n} &\leq \frac{1}{y_n} \\
0 < \frac{x_n}{x_n + y_n} &\leq \frac{x_n}{y_n}
\end{align*}
Then by Squeeze Theorem, $\lim_{n\to\infty} \frac{x_n}{x_n + y_n} = 0 \implies \lim_{n\to\infty} \frac{2x_n}{x_n + y_n} = 0$ which is what we wanted. Could you check if my proof makes sense. Is there any cases of sequences of x_n and y_n such that this statement is not true?


问:
What is the variance of residual of regression (matrix form)?
I am confused at the variance of residual in regression (matrix form)

Given $Y = X*\beta \hat{}+ \epsilon$

Y is a nx1 predicted vector

X is a nxp matrix

$\beta\hat{}$ is a px1 coefficient vector = $(X^TX)^{-1}X^TY$

$\epsilon$ is a nx1 residual vector

Then $var(\epsilon)=var(Y-X\beta\hat{}$) = var($Y$ - $X(X^TX)^{-1}X^TY$) (1)

Let projection matrix $P$ = $X(X^TX)^{-1}X^T$ . It is idempotent $(P^n=P)$

(1) becomes $var(Y - PY)$ (2)

Method 1: (2) = $var ((I-P)Y) = (I-P)Var(Y)(I-P)^T = \sigma^2(I-P)$

Method 2: (2) = $var(Y)+var(PY) = \sigma^2I + P\sigma^2P^T = \sigma^2(I+P)$

I expect that both methods should result in the same formula but they did not. I am not sure what is missed here

答:
The second formula you have assumed independence (or at least zero covariance) of $Y$ and $PY$, that does not hold.

See here, you forgot two important terms.

$$var (Y - PY) = var (Y) + var(PY) + cov(Y, -PY) + cov(-PY, Y)  $$


问:
Invert "matrix of polynomials" for a change of presentation of an Ideal
If I start with a polynomial ideal generated by a set of polynomials $f_i$, which is not a standard basis for the ideal, and then I obtain a standard basis for the ideal to be a set of polynomials $g_j$, how can I figure out how to write each $g_j$ as a linear combination of $f_i$?

Because $g_j$ is a standard basis, without much difficulty we can write each $f_i$ as a linear combination of $f_i = \sum_j M_{ij} g_j$, where each $M_{ij}$ is a polynomial. So we have a "matrix of polynomials" describing the "change of presentation" from $g_j$ to $f_i$ (not sure the correct phrasing, as $f_i$ is not a basis, so this is not a change of "basis"... maybe change of generators?).

With this matrix $M_{ij}$ in hand, is there some way I can "invert" it to get the matrix $N_{ij}$, such that $g_j = \sum_i N_{ji} f_i$?

If I knew the proper phrases to describe this process, I could probably search better. So that would be helpful as well.

This question is very related:

Use / take advantage of Gröbner basis $ G = (g_j)$ to write $f \in I = \langle f_i \rangle$ as a $k$-linear combination of the polynomials in $I$?

It shows how one could, in principle, calculate this by carrying extra information along while calculating the standard basis. But I am not calculating the standard basis by hand. The example I am working with is large, and while computer packages can calculate the basis fairly easily (in about 10 minutes with a few GB of memory), they fail if I try to calculate the matrix simultaneously: it quickly runs through all the memory on my computer trying to carry around extra information in case it is needed for the matrix. That is why I want to know if there is some way to use just the basis itself after it is calculated (and the $M_{ij}$ which is not hard to generate from it).

答:
Writing $g$ as a linear combination (with polynomial coefficients) of $f_1,
\ldots,f_r$ is sometimes called "lifting", e.g. lift in Singular and lift in SageMath (which uses Singular). The resulting coefficients can be called witnesses to (or a certificate of) the membership of $g$ in the ideal generated by the $f_i$.

I'm not sure exactly how the computation goes, but I think it amounts to finding syzygies of $f_1,\ldots,f_r,g$ where the coefficient of $g$ is a unit (or more generally syzygies of $f_1,\ldots,f_r,g_1,\ldots,g_s$ where the coefficients of $g_j$ are units).
The internal implementation in Singular is found in idLift, which someone might be able to decipher.


问:
Prove $\nabla \cdot (X \times Y) = (\nabla \times X) \cdot Y - X \cdot (\nabla \times Y)$ by converting to the language of differential forms
I wish to prove the identity $$\nabla \cdot (X \times Y) = (\nabla \times X) \cdot Y - X \cdot (\nabla \times Y)$$ where $X, Y$ are vector fields in $\mathbb{R}^3$ and $\times, \cdot$ are the cross, dot product respectively by converting it to the language of differential forms. Letting $X, Y$ correspond to the 1-forms $\omega_1, \omega_2,$ we have $X \times Y = *(\omega_1 \wedge \omega_2), \nabla \cdot = * d *, \nabla \times = * d,$ so $$\nabla \cdot (X \times Y) = *(d(*(*(\omega_1 \wedge \omega_2)) = *d(\omega_1 \wedge \omega_2) = *(d\omega_1 \wedge \omega_2 - \omega_1 \wedge d\omega_2).$$

Thus, we just need to show $(\nabla \times X) \cdot Y$ corresponds to $*(d\omega_1 \wedge \omega_2)$ and the other term will be similar. The best I could do was get it to $(*d\omega_1) \cdot Y = (*d\omega_1)(Y)$ using the fact that $A \cdot B = \omega(B)$ if $A, B$ are vector fields and $\omega$ is the 1-form corresponding to $A.$

Letting $\omega = d\omega_1,$ the claim we need to prove is that $*(\omega \wedge \omega')$ and $(*\omega)(A)$ are equivalent if $A, \omega'$ are. Because both sides are linear, we can just check the case $\omega' = dx_1, A = e_1$ WLOG. But even this seems like too much work because we have to write $\omega = a_1 dx_1 \wedge dx_2 + a_2 dx_2 \wedge dx_3 + a_3 dx_3 \wedge dx_1$ and expand both sides. Is there a faster way to finish the problem that relies on expressing the dot product purely in differential forms (no vector fields)?

答: It turns out that $*(\omega_1 \wedge *\omega_2) = *(*\omega_1 \wedge \omega_2)$ corresponds to $X \cdot Y.$ Using this on $\nabla \times X = * d \omega_1, Y = \omega_2,$ we get $$(\nabla \times X) \cdot Y = *(**d\omega_1 \wedge \omega_2) = *(d\omega_1 \wedge \omega_2).$$


答: No, the author doesn't assume what he want to prove, we just looks at the set $P$ of all $(B, <)$, where $B$ is any subset of $A$ and $<$ is a well-order on it. There are such pairs, for sure, as for example $(\emptyset, \emptyset)\in P$.</p>

Title: Zorn's lemma implies the well-ordering principle

Body:
I am little confused about the proof given here http://euclid.colorado.edu/~monkd/m6730/gradsets05.pdf

On the second page, when defining $P$, the author says that $B\subset A$ and $(B,<)$ is a well-ordering structure. Isn't this exactly what we want to prove? How do we know that $B$ can be well-ordered? What happens if $P$ is empty?


问:
What is the probability of winning this game?

A tennis player has a 60% chance of winning any given point in a tennis game. Calculate the probability that she will win the game within the first 6 points, stating any assumptions you make.

(A game is won when a player has won at least 4 points and won at least 2 more points than their opponent.)

So there are 3 choices:

she wins 4 points with the probability 60%$^{4}$

she wins 4 points and lose 1 point with probability 4 $\times$ 60%$^{4}$ $\times$ 40%

she wins 4 points and lose 2 points 15 $\times$ 60%$^{4}$ $\times$ 40%$^{2}$

So the total probability of her winning the game is 0.69984. Is this right?

答:
Using Stars and Bars, the correct answer is:

$$P(\text{win})=\binom{3}{0}0.6^4+\binom{4}{1}0.6^4\times0.4+\binom{5}{2}0.6^4\times0.4^2$$
$$=0.54432$$

She must win the last point, and so the sums represent 3 wins no losses, 3 wins 1 loss and 3 wins 2 losses in the preceding points of the game.


问:
How to find the inverse of this matrix? (in order to solve the question).
I am trying to solve this question and things got really complicated while trying to find the inverse of the matrix, here's the question:

Given $
A = \left[\begin{array}{l}a&1&-1\\1&a&-1\\1&-1&a\end{array}\right]$, $a\in \mathbb{R}$, if $A$ is invertible, then $(A)_{11}*(A^{-1})_{11}=?$

I started to doubt myself since I reached $\left[\begin{array}{l}a&1&-1\\0&a-\frac{1}{a}&-1+\frac{1}{a}\\0&-1-\frac{1}{a}&a+\frac{1}{a}\end{array}\right]$.

And I couldn't figure out how to keep going, and started to doubt I'm missing something (or I'm just weak at finding inverse matrix).

Note I checked the determinant and got that it's invertible if and only if $a \ne 0,1,2$ so there's not problem with dividing by $a$.

Any help would be really appreciated, thanks in advance.

答:
Since you are asked only to find the product $(A)_{11} \cdot \left( A^{-1} \right)_{11}$, you only need to find the $(1,1)$-entry of $A^{-1}$ and not the whole matrix. For this purpose, since the matrix is $3 \times 3$ it is useful to use the formula for the adjugate matrix. We have

$$ \det(A) = a \cdot (a^2 - 1) - (a - 1) + (-1 + a) = a \cdot (a^2 - 1), \\
\operatorname{adj}(A)_{11} = a^2 - 1 $$
and hence
$$ \left( A^{-1} \right)_{11} = \frac{1}{\det(A)} \left( \operatorname{adj}(A) \right)_{11} = \frac{a^2 - 1}{a \cdot (a^2 - 1)} = \frac{1}{a}.$$

Alternatively, let's write

$$ A^{-1} = \begin{pmatrix} x & ? & ? \\ y & ? & ? \\ z & ? & ? \end{pmatrix}. $$

We need to find $x$. Since $A \cdot A^{-1} = I$, we must have
$$ ax + y - z = 1, \\
 x + ay - z = 0,\\
 x - y + az = 0. $$

There are three equations for three unknowns (where $a$ is a parameter) and the augmented matrix associated to this system is
$$ \left( \begin{array}{ccc|c}
a & 1 & -1 & 1 \\
1 & a & -1 & 0 \\
1 & -1 & a & 0 \end{array} \right). $$

Performing elementary operations, we get the equivalent system
$$ \left( \begin{array}{ccc|c}
a & 1 & -1 & 1 \\
1 & a & -1 & 0 \\
1 & -1 & a & 0 \end{array} \right) \xrightarrow{R_1 \leftrightarrow R_2}
\left( \begin{array}{ccc|c}
1 & a & -1 & 0 \\
a & 1 & -1 & 1 \\
1 & -1 & a & 0 \end{array}  \right) \xrightarrow{R_2 = R_2 - aR_1, R_3 = R_3
- R_1}
\left( \begin{array}{ccc|c}
1 & a & -1 & 0 \\
0 & 1 - a^2 & a - 1 & 1 \\
0 & -1 - a & a + 1 & 0 \end{array}  \right)
\xrightarrow{ R_2 = R_2 + (1 - a) R_3}
\left( \begin{array}{ccc|c}
1 & a & -1 & 0 \\
0 & 0 & a(1-a) & 1 \\
0 & -1 - a & a + 1 & 0 \end{array}  \right)
$$
which translates into the system
$$
x + ay - z = 0, \\
a(1-a)z = 1, \\
(a+1)z = (a+1)y.
$$

Hence, $z = y  = \frac{1}{a(1-a)}$  and $x = z - ay = \frac{1}{a(1-a)} - \frac{1}{1-a} = \frac{1}{a}$.


问:
How to take the root of a complex number when the right side of the equation is zero?
so I know that when wanting to take the root of the complex number in a case like this

$$z^4=1+\sqrt3i$$

it's really just computing the polar coordinates and finding the solutions for $z_{0}, z_{1},...,z_{3}$ with De Moivre's formula etc.

but when we have something like this instead...

$$z^7+7i=0$$

how am I supposed to compute the $\Re$ and $\Im$ parts of this? With the first exercise I can easily come up with the polar form, but the second one is just horrendous. I don't see where that number has its real and imaginary parts which makes computing $r$ and especially the angle $\varphi$ a mystery.

I would really appreciate any help ;-; thank you so much!!

答:
You can write it as $$z^7 = -7 \iota$$and then take modulus to get $r^7 = 7$

Here $\iota = \sqrt{-1}$


问:
Second Linear ODE
Confused on how I can approach this ODE, and I just need some help on figuring out what to do

$y''-4y'-21=5e ^{-t}$ with initial conditions $y(0)=0$ and  $y'(0)=4$

After Laplace transforming, I got:
$(\frac{(5)}{(s+1)}{+ 4+\frac{(21)}{(s+1)}}) /( s^2-4s) $

And now I am stuck on what to do here

答:
I don't know about Laplace transformation but I would solve it a little differently. Look at it the following way:
$$y''-4y' = 5e^{-t} + 21$$
Now solving the homogenous equation $y''-4y'=0$ is not too complicated. It has the characteristic polynomial $t^2-4t$ with zeros at $t=0$ and $t=4$. Therefore the solution to the homogenous part equation is $c_1 e^{4t} + c_2e^{0t} = c_1e^{4t} + c_2$.

Now we need to find a special solution to the inhomogenous problem. To be honest I don't know how to do that systematically, but a little "good guessing" can get us there: How do we get the $+21$? Let us put a $\frac{21}{4} t$ in the solution, so that the $4y'$ produces our $21$. How do we get $5e^{-t}$? Because of the fact that the second derivative of $e^{-t}$ is $e^{-t}$ and the first derivative is $-e^{-t}$ we can find the special solution
$$u_{sp} = e^{-t} - \frac{21}{4} t$$

Then the whole solution to this equation is:
$$u= c_1e^{4t}+c_2 + e^{-t} - \frac{21}{4} t$$

Now we try to find $c_1, c_2$ via the initial conditions. Note $0=u(0) = c_1 + c_2+1$ and $4=u'(0) = 4c_1 -1- \frac{21}{4}$ Solving this we get $c_1 = \frac{41}{16}$ and $c_2 = \frac{-57}{16}$.

If anybody knows how we could have gotten the special solution in a more systematic way feel free to let me know. If the Laplace transform method would be easier than my approach to solving linear ODEs with constant coefficients I am also happy if you let me know because I don't know how to do Laplace transformation.

Title: Second Linear ODE


问:
Divergence of magnetic field $B = \frac{\mu_0 I}{2\pi r}$
I have to show that the divergence of this magnetic field is 0. I can do this pretty easily using the divergence theorem; however, if I try using try computing the divergence directly $\nabla B$ does not equal $0$.

To solve it indirectly I used the definition that defines the divergence as the limit of a surface integral. I chose a cylindrical surface and then showed that the flux through that surface is zero as the field lines form circles.

答:
The answer is quite simple. If I understand correctly, then you are solving the problem of a magnetic field around a infinite conductor with current. Writing down the equation for the magnetic field circulation, you got the $\phi$-th component of the magnetic field (only it is nonzero), i.e.
\begin{equation}
B_{\phi} = \frac{\mu_0 I}{2\pi r},\\
B_{z} = 0,\\
B_{r} = 0.
\end{equation}
The divergence in cylindrical coordinates is as follows
\begin{equation}
\vec{\nabla}\cdot\vec{B} =\frac{1}{r}\frac{\partial(r B_{r})}{\partial r} + \frac{\partial B_z}{\partial z} + \frac{1}{r}\frac{\partial B_{\phi}}{\partial\phi},
\end{equation}
which is in your case is
\begin{equation}
\vec{\nabla}\cdot\left(\frac{\mu_0 I}{2\pi r}\vec{e}_{\phi}\right) = \frac{1}{r}\frac{\partial\left(\frac{\mu_0 I}{2\pi r}\right)}{\partial\phi} = 0,
\end{equation}
where $\vec{e}_{\phi}$ is the unit vector along $\phi$.


答:
HINT

Simply rearrange into the form
$$z^7=-7i$$
We can let $z=r(\cos(\theta+2k\pi)+i\sin(\theta+2k\pi))$ and proceed like usual. If you need any more help please don't hesitate to ask.

Title: How to take the root of a complex number when the right side of the equation is zero?

Body:
so I know that when wanting to take the root of the complex number in a case like this

$$z^4=1+\sqrt3i$$

it's really just computing the polar coordinates and finding the solutions for $z_{0}, z_{1},...,z_{3}$ with De Moivre's formula etc.

but when we have something like this instead...

$$z^7+7i=0$$

how am I supposed to compute the $\Re$ and $\Im$ parts of this? With the first exercise I can easily come up with the polar form, but the second one is just horrendous. I don't see where that number has its real and imaginary parts which makes computing $r$ and especially the angle $\varphi$ a mystery.

I would really appreciate any help ;-; thank you so much!!


问:
Probability of a Coin Toss
I am stuck with this question, I thought $(\frac{1}{2})^7$ but it is wrong. Any help would be much appreciated.

```<code>In a football game, the sides in the soccer field are determined on the
basis of a single coin flip by the referee and the team which correctly
guesses the flip result gets its desired side.

a) What is the probability that a team will have to wait until its 7th
game to get its desired side?
</code>```

答:
your result looks correct to me. It is a geometric distribution

$$\mathbb{P}[X=x]=(1-p)^{x-1}\cdot p=\left(\frac{1}{2}\right)^6\cdot\frac{1}{2}$$

Title: Probability of a Coin Toss


问:
Can someone explain why this simplification works?
https://i.stack.imgur.com/bOLuF.png

I don't understand wouldn't -x+x cancel out to become 0?

答: Just check that $$x-\dfrac{x}{2}=\dfrac{\color{red}{2}x}{\color{red}{2}}-\dfrac{x}{2}=\dfrac{x}{2}.$$


答:
$\theta = (-\pi/2) \implies -i = e^{i\theta} \implies
z^7 = -7i = 7e^{i\theta}$.

Therefore, one of the roots of the equation is

$z_r = (7)^{(1/7)}e^{i(\theta/7)}.$

Having found one of the roots, you find all 7 roots, by computing

$w_k = z_r \times z_k : k \in \{0,1,2,3,4,5,6\}$

where $z_k$ is the $k$-th root of $z^7 = 1 \implies $

$z_k = e^{(i2k\pi/7)}.$

Edit

To elaborate:

$e^{i\theta}$ is shorthand for $[\cos(\theta) + i\sin(\theta)].$

Although (for example) $\cos(-\pi/14)$ is probably expressible by radicals, this is almost certainly not the point of the problem.  Therefore, expressing the roots by their polar coordinates should be fine.


问:
A convergence in probability associated to order statistics
Consider $n$ independent random variables $U_1,\dots,U_n$ uniformly distributed on $[0,1]$. Order them like so :
$$U_{(1)}\leq U_{(2)}\leq\cdots\leq U_{(n)}$$

Let $x>0$ be given. How does one prove the following assertion :
$$J_n=\frac1n\sum_{k=1}^n\mathbf{1}_{\{n(U_{(k)}-U_{(k-1)})>x\}}\xrightarrow[n~\to~\infty]{~\text{in probability}~}e^{-x}~?$$
(we set $U_{(0)}=0$).

What I have tried / observations :

Since we are trying to establish convergence in probability to a constant, it suffices to show convergence in distribution (see wikipedia)

We can thus try to show either (a) pointwise convergence of the characteristic functions $\Phi_n=\Phi_{J_n}$ to $\Phi_{e^{-x}}$, or (b) pointwise convergence of the cumulative distribution functions $F_n=F_{J_n}$ to $F_{e^{-x}}$ at all points of continuity of the cumulative distribution function of the constant random variable $e^{-x}$, (that is, on $\mathbb{R}\setminus\{e^{-x}\}$)

The $J_n$ only take on finitely many values that are non negative integers, and thus the $F_n$ only take on discrete values.

The problem asks one to establish the fact that
$$U_{(1)}\leq U_{(2)}\leq\cdots\leq U_{(n)})\sim\Big(\frac{T_{1}}{T_{n+1}},\dots,\frac{T_{n}}{T_{n+1}}\Big)$$
with $T_k=\mathscr{E}_1+\cdots+\mathscr{E}_k$ with the $\mathscr{E}_i$, $i=1,\dots,n+1$ iid of exponential law of parameter $1$, aswell as the fact that
$$nU_{(1)}\xrightarrow[n~\to~\infty]{~\text{in distribution}~}\mathrm{Exp. Law}(1)$$

Another observation is the fact that convergence in probability is equivalent to convergence of $\mathbb{E}[|J_n-e^{-x}|\wedge 1]\to 0$, that is, since $-1\leq J_n-e^{-x}\leq +1$, $J_n\to e^{-x}$ in $L^1$.

The problem I don't know how to get rid of is fact that the $n(U_{(k)}-U_{(k-1)}$ aren't independent ... The facts I quote in the last bullet point entail that each $\mathbf{1}_{\{n(U_{(k)}-U_{(k-1)})>x\}}$ has the same expectation and they tend to $=e^{-x}$ as $n\to\infty$. Unfortunately the law of large numbers cannot be applied (and in any case, we are only asked to show convergence in probability anyways).

答:
Instead of studying the order statistics let us have a look on the lengths of the intervals between the order statistics: let
$$ L_i=U_{(i)}- U_{(i-1)} \qquad \text{for } 1\leq i\leq n+1,$$
where we use the convention that $U_{(0)}=0$ and $U_{(n+1)}=1$.
The random vector $(L_1,\dots,L_{n+1})$ has the uniform distribution on the simplex  $$ \left\{ (x_1,\dots,x_{n+1}) : x_1+\cdots+x_{n+1}=1, \quad x_1,\dots,x_{n+1}\geq 0 \right\}. $$

An alternative way to generate a random vector with this probability distribution is to consider the vector
$$ \left( \frac{Z_1}{Z_1+\cdots+Z_{n+1}}, \dots, \frac{Z_{n+1}}{Z_1+\cdots+Z_{n+1}} \right), $$
where $Z_1,\dots,Z_{n+1}$ are independent random variables with the exponential distribution $\operatorname{Exp}(1)$. (TODO: missing reference for the proof of this result.)

Our goal is to study the probability distribution of the number of indexes $k\in\{1,\dots,n\}$
with the property that $n L_k > x$ or, equivalently, the number $N$ of the indexes such that
$$ Z_k > x \cdot \frac{Z_1+\cdots+Z_{n+1}}{n}.$$
By Bienaymé-Chebyshev inequality the right-hand side converges in probability to $x$.

Take $\epsilon>0$ and denote by $N'$ the number of the indexes such that
$$ Z_k > x +\epsilon.$$
We have
$$ \mathbb{P}\left\{ N < N' \right\} \leq \mathbb{P}\left\{ x \cdot \frac{Z_1+\cdots+Z_{n+1}}{n} > x+\epsilon \right\} \xrightarrow{n\to\infty} 0.$$

Since the random events counted in $N'$ are independent, the law of large numbers implies that
$$ \frac{N'}{n} \xrightarrow{P} e^{-x-\epsilon}. $$

The union bound implies that
$$ \mathbb{P}\left\{ \frac{N}{n} < e^{-x -\epsilon} - \epsilon \right\} \leq  \mathbb{P}\left\{ N < N'\right\} +
 \mathbb{P}\left\{ \frac{N'}{n} < e^{-x -\epsilon} - \epsilon \right\}   $$
converges to zero for each $\epsilon>0$.

Analogously one can show the upper bound that the probability of the event
$$ \frac{N}{n} > e^{-x +\epsilon} + \epsilon $$
converges to zero for each $\epsilon>0$.


问:
How to find sum of $\sum_{n=1}^{\infty }\frac{(n)\bmod(k)}{n(n+1)}$?

How to find sum of $\sum_{n=1}^{\infty }\frac{(n)\bmod(k)}{n(n+1)}$?

I know that if $n \bmod k \le k-1$ then this sum is converge then it has finite sum, I just guess it's $\ln(k)$ because when $k=1$ sum is $0=\ln(1)$. I really don't know how to find it. Please help me.

答:
Using the same ideas as in a recent answer (https://math.stackexchange.com/a/4024522/198592) here is a solution which is very close to that of @xpaul and @Gary but more elementary in the final step (without polygamma functions).

We will prove that

$$s(k):=\sum _{n=1}^{\infty } \frac{n \bmod k}{n (n+1)}=\log (k)\tag{1}$$

simplify the mod function letting

$$n = m k +j, m=0,1,2,..., j=1, 2, ..., k-1\tag{2}$$

notice that $j$ runs only up to $k-1$ because $(m k + k) mod(k) = 0$.

Then  $n \bmod k = j$ and defining

$$u(k,j) = \sum_{m=0}^{\infty} \frac{j}{(m k+j)(m k+j+1)}\tag{3}$$

we get

$$s(k) = \sum_{j=1}^{k-1} u(k,j)\tag{4}$$

Calculate $u$

Instead of doing the $m$-sum directly which leads to polygamma functions we prefer to represent denominators by integrals using the formula $\frac{1}{a}=\int_{0}^{\infty} e^{-a t}\,dt$

After partial fraction decomposition

$$\frac{j}{(j+k m) (j+k m+1)}=\frac{j}{j+k m}-\frac{j}{j+k m+1}$$

this gives after also doing the $m$-sum

$$u(k,j) = \int_{0}^{\infty} j\sum_{m=0}^{\infty}\left(e^{-t(j+m k)}-e^{-t(1+j+m k)}  \right)\,dt=\int_{0}^{\infty} j\left( \frac{e^{k t-j t}}{e^{k t}-1}-\frac{e^{k t-(j+1) t}}{e^{k t}-1}\right)\,dt\tag{5}$$

do the $j$-sum under the integral gives for the integrand

$$\sum_{j=1}^{k-1} j\left( \frac{e^{k t-j t}}{e^{k t}-1}-\frac{e^{k t-(j+1) t}}{e^{k t}-1}\right) = \frac{k \left(-e^t\right)+e^{k t}+k-1}{\left(e^t-1\right) \left(e^{k t}-1\right)}=\frac{1}{e^t-1}-\frac{k}{e^{k t}-1}\tag{6}$$

the final step is to calculate the $t$-integral

$$s(k) = i(k) := \int_{0}^{\infty} \left(\frac{1}{e^t-1}-\frac{k}{e^{k t}-1}\right)\,dt\tag{7}$$

The indefinte integral (antderivative) is

$$a(t) = \int \left(\frac{1}{e^t-1}-\frac{k}{e^{k t}-1}\right)\,dt\\
=-t+k t+\log \left(\frac{e^t-1}{e^{k t}-1}\right)\tag{8}$$

and for the limits at the integration boundaries we find

$$a(t \to \infty) = 0, a(t \to 0) = \log(\frac{1}{k})\tag{9}$$

Hence the integral $(7)$ becomes $i(k)=\log(k)$ which completes the proof.

Details of the limits

$$\begin{align}
a(t \to 0) :
& -t + k t+\log \left(\frac{e^t-1}{e^{k t}-1}\right)\\
& \to  \log \left(\frac{\left(1+t+...\right)-1}{\left(1+k t +...\right)-1}\right)\\
& \to \log \left(\frac{1}{k}\right)
\end{align}\tag{10a}$$

$$\begin{align}
a(t \to \infty) :
& -t + k t+\log \left(\frac{e^t-1}{e^{k t}-1}\right)\\
& \to -t +k t+\log \left(\frac{e^t \left(1-e^{-t}\right)}{e^{k t} \left(1-e^{-k t}\right)}\right)\\
& \to -t+k t+\log \left(\frac{e^t}{e^{k t}}\right)+\log \left(\frac{1-e^{-t}}{1-e^{-k t}}\right)\\
& = \log \left(\frac{1-e^{-t}}{1-e^{-k t}}\right) \to 0
\end{align}\tag{10b}$$

Discussion

The integral $i(k)$ extends the domain of $s(k)$ to real values of $k$.