问:
Inclusion Exclusion Principle Problem

There are 28 people in your family consisting of 18 adults, 13 females, and 11 who have purple hair.  There are 11 adult females, 6 of whom sport purple hair.  There are 10 adults with purple hair.  No female children have purple hair.  How many male children without purple hair are in the family?

I know this problem is an application of the inclusion exclusion principle. However I am not sure how to express one of the statements in the prompt:  "No female children have purple hair. How many male children without purple hair are in the family?"

I'm pretty confident in expressing the rest up until that sentence:

A = adults, F = females, P = purple hair, and $N()$ represents the cardinal number of a set

$N(A) = 18$

$N(F) = 13$

$N(P) = 11$

$N(A \cap F) = 11$

$N(A \cap F \cap P) = 6$

$N(A \cap P) = 10$

I thought maybe it meant $N(F \cap P)=0$, but when I did that I didn't get the right answer which is 7.

Any help is appreciated.  Thank you in advance!

答:
No female children have purple hair.

If $X$ is the set of female children with purple hair, we’re told that $N(X)=0$. The problem, then, is to express $X$ in terms of $F$, $A$, and $P$. If you’re made a Venn diagram, you can see that $X$ is the region inside both $F$ and $P$, but outside $A$. In other words, $X=(F\cap P)\setminus A$. If you’re more accustomed to complements than to set differences, you might be more familiar with that as $X=F\cap P\cap A^c$.

Title: Inclusion Exclusion Principle Problem


问:
Second order Fibonacci numbers
Let's consider a recurrence:
$A_{0}=0$, $A_{1}=1$, $
A_{n}=A_{n-1}+A_{n-2}+F_{n}$, where $F_{n}$ is the $n$th Fibonacci number. How to express $A_{n}$ in a closed form?

Despite the fact that it sounds easy, i got stuck after a number of attempts( i've tried to evaluate the generating function of $A_{n}$)

Any hints that may help?

答:
Let $F(x)$ be the generating function of the Fibonacci sequence, and $A(x)$ be the generating function of your sequence.

We have $$F(x)= \frac{x}{1-x-x^2},$$ so $(1-x-x^2)A(x) = x + F(x)$.
By multiplying both sides by $F(x) $, we get $$A(x) = F(x) + F(x)^2.$$

Now, all you have to do is to compare the coefficient of $x^n$ on both sides.

ContentLicense: CC BY-SA 3.0

Title: Second order Fibonacci numbers

ContentLicense: CC BY-SA 4.0


问:
$\frac{n!}{p^{\varepsilon_{p}(n!)}}$ congruence
Let $n=a_{m}p^{m}+a_{m-1}p^{m-1}+\cdots +a_{0}$, the base $p$ representation of $n$, I have to prove that$$\frac{n!}{p^{\varepsilon_{p}(n!)}} \equiv (-1)^{\varepsilon_{p}(n!)} (a_m)! (a_{m-1})! \cdots (a_{0})! \mod{p}$$
I have done something, but not sure if i am close

Let $n=\sum_{k=0}^{m}a_{k}p^{k}$, where each $0\leq a_{i}<p$ be the radix/base $p$ representation of $n$. If $n=a_{0}$, then note that $p\nmid a_{0}$, therefore$$\frac{n!}{p^{\varepsilon_{p}(n!)}}= \frac{(a_0)!}{1}\equiv (-1)^{0}\cdot a_{0}!\mod{p}$$If $n=a_{1}p+a_{0}$, then we have$$\varepsilon_{p}(n!)=\sum_{k\geq 1}\left\lfloor \frac{n}{p^k}\right\rfloor=\sum_{k\geq 1}\left\lfloor\frac{a_1 p+a_0}{p^k}\right\rfloor=a_1$$Now we have
\begin{align*}
\frac{n!}{p^{\varepsilon_{p}(n!)}}&=\frac{(a_1 p+a_{0})!}{p^{a_1}}
\\ &= \frac{\prod_{i=1}^{p-1} i \cdot \prod_{i=1}^{p-1} (p+i) \cdots \prod_{i=1}^{a_0}(a_{1}p+i) \cdot (p)\cdot(2p)\cdot(3p)\cdots(a_{1}p)}{p^{a_1}}
\\&\equiv \frac{(-1)^{a_1}\cdot (a_{0})!\cdot p^{a_1}\cdot (a_{1})!}{p^{a_1}}\mod{p}
\\ &\equiv (-1)^{\varepsilon_{p}(n!)}\cdot (a_{1})!\cdot (a_{0})!\mod{p}
\end{align*}We have show that the result is true if $n=a_{0}$ and $n=a_{1}p+a_{0}$.

答:
So, you have the following. Consider $n=a_mp^m+\cdots +a_1p+a_0=(a_mp^{m-1}+\cdots +a_1)p+a_0=[\cdot A_1+a_0$ then
$$\frac{n!}{p^{\nu _p(n!)}}=\prod _{i=1}^ni!=\left (\prod _{i=1}^{n-a_0}i\right )\cdot \left (\prod _{i=1}^{a_0}(p\cdot A_1+i)\right )\equiv \prod _{i=1}^ni!=\left (\prod _{i=1}^{n-a_0}i\right )\cdot a_0!\pmod p,$$
now we will have every $p-$residue $A_1$ times so we will get $(p-1)!^{A_1}$ and we will have $p$ times the factorial of $A_1$ but when all the contribution of $p$ is divided, so we will get $\frac{A_1!^p}{p^{\nu _p(A_1!)}}.$

By Wilson's theorem we will get then that
$$\frac{n!}{p^{\nu _p(n!)}}\equiv (-1)^{A_1}\cdot \frac{A_1!}{p^{\nu _p(A_1!)}}\cdot a_0!\pmod p,$$
but if you are doing induction, you probably know something about the middle term. At the end, you will have just to add the exponents of the $-1$ and use Legendre's theorem.


问:
Converting $\sin(A)+\cos(B)$ to a product?
Similar to,

$$ \sin(u) + \sin(v) = 2 \sin \left(\frac{u+v}{2}\right) \cos \left(\frac{u-v}{2}\right)$$

$$ \cos(u) + \cos(v) = 2 \cos \left(\frac{u+v}{2}\right) \cos \left(\frac{v-u}{2}\right)$$

Are there combination laws for expressions such as:
$$ \sin(A) + \cos(B)$$

where $A,B$ are some real angles?

答:
From the comment by @Andrei

$$ \sin(A) + \cos(B) =  \sin(A) + \sin( \frac{\pi}{2} -B)= 2 \sin \left( \frac{A-B}{2} + \frac{\pi}{4} \right) \cos \left( \frac{A+B}{2} - \frac{\pi}{4} \right)$$


问:
Why is this a Riemann surface?
I am reading chapter 3 of Donaldson's
Riemann Surfaces. In the end of this chapter he tries to show that any discrete subgroup $\Gamma$ of Aut(D) can make D/$\Gamma$ into a Riemann surface. I understand that every point in D/$\Gamma$ has a neighborhood homeomorphic to a ball, but what is the overlap map between two charts? I can't see why it is holomorphic.

答:
Roughly speaking, each chart $U \subset D/\Gamma$ is parameterized by an open subset $V$ of $D$ itself.

For two charts $U_1,U_2 \subset D/\Gamma$ parameterized by open subsets $V_1,V_2$ of $D$, the overlap map is the restriction of an element of $\Gamma$.

Since $\Gamma$ is a subgroup of $\text{Aut}(D)$, each element of $\Gamma$ is a holomorphic map, and so its restriction to any open subset is holomorphic.

Caveat: I wrote "roughly speaking", because this answer is so far valid only in the special case that $\Gamma$ is acting freely on $D$. The general case can also be described, but it's complicated by the fact that for a point $z \in D$ which is stabilized by a nontrivial cyclic subgroup of $\Gamma$, the chart around the point of $D/\Gamma$ corresponding to $z$ is not actually parameterized by an open subset of $D$ itself. If you want to know about that case I could describe it too, but I suspect that your question might be mostly about the case of a free action.

Edit: Here's a few more rough details about the case of an action which is not free.

For a point $P \in D$ at which the stabilizer subgroup $\text{stab}(P)$ is a cyclic group of order $n\ge 2$, and letting $p \in D/\Gamma$ be the corresponding point in the quotient, the coordinate chart around $p$ parameterized by first translating $P$ to $0$ by some element of $\text{Aut}(D)$, then taking a small open ball $|z|<r$ in $D$ around $0$, and then taking $w = z^n$. The chart around $p$ is then parameterized by the open ball $|w| < r^n$.

So the overlap maps are a bit more complicated. The most complicated case is one chart centered on a point $p$ with a $w=z^m$ coordinate ($m \ge 2$), and another overlapping chart centered on a different point $p'$ with a $w'=z^n$ coordinate ($n \ge 2$). In this case the overlap map would look like this:
$$w' = f \circ g \circ h(w)
$$
where $h$ is a branch of $\sqrt[m]{z}$, and $g \in \Gamma$, and $f(z)=z^n$. That's a composition of three holomorphic maps, and so the result is holomorphic (I have shoved under the rug the elements of $\text{Aut}(D)$ where we carry out the translation from $P$ to the origin and from $P'$ to the origin; so it's better to think of $g$ as the composition of the inverse of one translation, followed by an element of $\Gamma$, followed by the other translation).

Title: Why is this a Riemann surface?


问:
A possible solution to the ODE $dy+(y\cot(x)-2\cos(x))dx=0$
Given the following ODE
$$dy+(y\cot(x)-2\cos(x))dx=0$$
I found that its integrating factor is
$$\mu(x)=\exp\left(\int\cot(x)dx\right)=\sin(x)$$
and then the solution of the ODE
$$\frac{dy}{dx}+\cot(x)y=2\cos(x)$$
will be
$$y(x)=-\frac{\cos(2x)}{2\sin(x)}+C\cdot\csc(x)$$
where $C\in\mathbb{R}$

I have to choose one alternative that is a possible solution to this equation:

$y=\cos(x)+C\sin(x)$

$y=\cos(x)+C\csc(x)$

$y=-\sin(x)+C\cos(x)$

$y=\sin(x)+C\csc(x)$

$y=\sin(x)+C\sec(x)$

but I couldn't rewrite the solution I found as one of these alternatives. Since
$$\cos(2x)=\cos^{2}(x)-\sin^{2}(x)$$
the best I could get was
$$y=-\frac{1}{2}\cot(x)\cos(x)+\frac{1}{2}\sin(x)+C\csc(x)$$

PS: the 4th alternative actually satisfies the ODE but I don't know how to rewrite my solution as it.

答:
Using
$$ -\frac{\cos(2x)}{\sin(x)}=-\frac{1-2\sin^2(x)}{2\sin(x)}=-\frac{1}{2\sin(x)}+\sin(x) $$
you will get the 4th one.


问:
Prove for Bell numbers
I need to prove the following formula for Bell numbers:

$$ B(n) = \sum_{a_1 + \cdots + a_k = n, a_i \geq 1} \frac{1}{k!} \binom{n}{a_1 \cdots a_k} $$

Do you have any hint for me for getting this done?

Thank you in advance!

答:
To count the number of partitions of $[n]$, you can go through all the ways of carving $n$ up into bin sizes $a_1,\dotsc,a_k$ that add up to $n$ and then count the number of ways of distributing $n$ objects into those bins, which is the multinomial coefficient. However, this overcounts, since if you swap two bins, you get the same partition, but you get a separate count in the sum. So you have to divide by the number of permutations of the $k$ bins, which is $k!$.

While this is the easiest way to get the factor of $k!$, it's also instructive to look at how different permutations of the bins are (over)counted in the sum. If you swap two bins of the same size, you get another distribution of the elements over the same bin sizes, so this overcounting is in the multinomial coefficient, so to speak. On the other hand, if you swap two bins of different sizes, you get a different arrangement of the same bin sizes, so this overcounting is in the sum, so to speak.

Title: Prove for Bell numbers


问:
Complement Finite space is separable
Sow that $(X, )$ with the complement finite topology (the closed sets are the finite subsets of $X$) is separable. If $X$ is countable it's easy but I'm interested in the non-countable infinite case.

答:
Pick any countably infinite subset $D$ of $X$. We claim that this set is dense in $X$. Let $O\in\tau$ be open and non-empty. Then $O^c$ is finite, hence there exists $x\in D\cap O$.

As this holds for arbitrary non-empty $O\in\tau$, the set $D$ is dense.

Title: Complement Finite space is separable


问:
A Banach space isomorphic to his dual is an Hilbert space?
By the Riesz representation theorem, we know that the Hilbert space $\mathcal{H}$ is isomorphic to his dual. Is the converse true ? Does the fact that a Banach space $E$ is isomorphic to his dual implies that $E$ is an Hilbert space ? If it is not the case, does anyone have an example ?

答: No. For any reflexive Banach space $X$ consider $X \oplus X^*$. Then $(X \oplus X^*)^* = X^* \oplus X$, which is of course isomorphic to $X \oplus X^*$ but not necessarily a Hilbert space.


问:
Can a single random variable have a covariance matrix?
This link about the closed-form analytical solution for the Wasserstein distance says:

Let $\mu_1\in \mathbb{R}^N$ be a normally distributed random variable with expected value $m_1$ and covariance matrix $C_1$.

I never knew there could be a covariance matrix for univariate data, only multivariate data. How is it possible for a single random variable to have a covariance matrix?

答: The random variable $\mu_1$ is in $\mathbb{R}^N$, so it's an $N$ dimensional vector.  The covariance matrix for it would be the covariance matrix between each of the components of $\mu_1$.


答: With $Stab(z)$ the subgroup of $\Gamma$ leaving $z$ fixed, which is finite because discrete in some group isomorphic to $PSO_2(\Bbb{R})$, for $u$ close enough to $z$ then $\Gamma u\to \prod_{\gamma \in Stab(z)} (\gamma(u)-\gamma(z))$ looks like a chart from a neighborhood of $\Gamma z$ to a neighborhood of $0$. Thus away from the $z$ with non-trivial $Stab(z)$ we have that $s\to \Gamma s$ is a chart and the transitions functions are the obvious ones, at the remaining $z$ the transitions functions involve some algebraic functions obtained by inverting locally the rational function $u\to \prod_{\gamma \in Stab(z)} (\gamma(u)-\gamma(z))$.

Title: Why is this a Riemann surface?

Body:
I am reading chapter 3 of Donaldson's
Riemann Surfaces. In the end of this chapter he tries to show that any discrete subgroup $\Gamma$ of Aut(D) can make D/$\Gamma$ into a Riemann surface. I understand that every point in D/$\Gamma$ has a neighborhood homeomorphic to a ball, but what is the overlap map between two charts? I can't see why it is holomorphic.


问:
on the commutator subgroup of a special group
Let $G'$ be the commutator subgroup of a group $G$ and $G^*=\langle g^{-1}\alpha(g)\mid g\in G, \alpha\in Aut(G)\rangle$.

We know that always $G'\leq G^*$.

It is clear that if $Inn(G)=Aut(G)$, then $G'=G^*$.

Also if $G$ is a non abelian simple group or perfect group, then  $G'=G^*=G$.

Now do there exist a group such that  $Inn(G)\neq Aut(G)$ and $G'=G^*\neq G$?

Thank you

答: Take $G = D_{5} = \langle \sigma, \tau \mid \sigma^5 = \tau^2 = 1, \tau \sigma \tau^{-1} = \sigma^{-1} \rangle$, the dihedral group of order $10$. Clearly, $G' = G^* = \langle \sigma \rangle$. Note that the automorphism $\sigma^i\tau^j \mapsto \sigma^{2i}\tau^j$ is not an inner automorphism.</p>


问:
Can we say anything about $\vert f'(x) \vert$ versus $\vert f''(x) \vert$ if $f$ is concave and goes through the origin?
Suppose we have a function $f(x)$ that is concave, upward sloping, and goes through the origin.

Are we able to say anything about how $\vert f'(x)\vert$ compares to $\vert f''(x)\vert$, such as whether one is greater than/less than another?

答:
No. For any $a > 0$, $f(x) = 1-e^{-ax}$ satisfies those conditions, and $|f''(x)/f'(x)| = a$.

More general, let $h: \Bbb R \to \Bbb R$ be a continuous, positive function. Then
$$
 f(x) = \int_0^x e^{-\int_0^t h(s) ds} dt
$$
satisfies
$$
 f'(x) = e^{-\int_0^x h(s) ds} > 0 \\
 f''(x) = -h(x) e^{-\int_0^x h(s) ds} < 0
$$
so that $f$ is increasing and concave, with $f''(x)/f'(x) = -h(x)$, i.e. that ratio can be equal to an arbitrary prescribed negative continuous function.


问:
Are the mappings linear?
$$
\begin{array}{l}
T_{7}: \mathcal{C}^{2}(\mathbb{R}) \rightarrow \mathcal{C}^{0}(\mathbb{R}),\left(T_{7} f\right)(x):=f^{\prime \prime}(x) \\
T_{8}: \mathcal{C}^{1}(\mathbb{R}) \rightarrow \mathcal{C}^{0}(\mathbb{R}),\left(T_{8} f\right)(x):=f(x) \cdot f^{\prime}(x)
\end{array}
$$

Hello guys, I'm trying to find if the following mappings are Linear or not? But I'm having difficulties understanding how the 2 rules of showing linearity are applied for differential functions as shown in the Image. Any ideas?

答:
You should understand the basic definitión which are

A transformation $T:V \rightarrow W$ over $W,V$ vectorial spaces, then
$T$ is linear if
$$T(u+v)=T(u)+T(v)$$
$$T(\alpha u)=\alpha T(u)$$
Notice that when you need to know if a given aplication is Lineal you need to take
$u,v\in V$ and calculate $T(u+v)$ and $T(\alpha u)$ (according to the rule given by the application) and compare it with $T(u)+T(v)$ and $\alpha T(u)$ and see if the expresions are the same
Remember that $\alpha \in F$ that is $\alpha$ is an escalar in some field $F$

Title: Are the mappings linear?


问:
Let $m, n ∈ \mathbb{N}$ such that $2m^{2} + m = 2n^{2} + n$ , then prove that $m-n$ is a perfect square .
After simply factorizing the equation given in the question I got $\mathbf{m + n = -0.5}$
. But the question mentioned $m, n ∈ \mathbb{N}$ . Then how $m + n = -0.5$ ?

Did I do some mistake or is the question wrong?

答:
As you conclude in the comments, the point is that we must have $m = n$.

Here's an alternative proof: suppose for contradiction that $m \neq n$; in particular take $m > n$. Let $k = 2m^2 + m$. We see that $m$ and $n$ are two solutions to the quadratic equation
$$
2x^2 + x - k = 0.
$$
The sum of the roots of $ax^2 + bx + c = 0$ is given by $-\frac{b}{a} = - \frac 12$, which is to say that $m + n$ is not an integer. This is impossible since $m,n \in \Bbb Z$.


问:
Help solving $2y'-4y\left(\frac{y''}{y'}\right)=\log\left(\frac{y''}{y'}\right)$
I'm at the begginig of a differential equation course, and I'm having trouble finishing this problem. I'm asked to solve the equation
$$2y'-4y\left(\frac{y''}{y'}\right)=\log\left(\frac{y''}{y'}\right). \ \ \ \text{(I)}$$
I'm used the variable change $p=y'$, $\ \dot p=\frac{dp}{dy}$, $ \ y''=\dot pp$, I get the equation:
$$p=2y\dot p+\frac{1}{2}\log \dot p. \ \ \ \text{(II)}$$
That last one is a Lagrange differential equation. I solved it finding a parametric solution using the variable change $\dot p = u$, and using that I got the first order linear equation
$$\frac{dy}{du}+\frac{2y}{u}=\frac{-1}{2u^2}.$$
Solving it by integrating factor, I finally found the parametric values for $y$ and $p$, and they are
$$
\left\{
\begin{array}{rcl}
     y(u,C)&=&\frac{C}{u^2}-\frac{1}{2u}
  \\ p(u,C)&=&\frac{2C}{u^2}-1+\frac{1}{2}\log u
\end{array}
\right.
$$
This two equations are the solution to the equation (II), but I'm asked to solve the equation (I). Since $p=y'(x)$ in the original equation (I), I integrate the expression of $p$ with respect to $x$, and I get
$$y=\int\left(\frac{2C}{u^2}-1+\frac{1}{2}\log u\right)\ dx + K$$
$$\Downarrow$$
$$\boxed{y(x)=x\left(\frac{2C}{u^2}-1+\frac{1}{2}\log u\right) + K}$$
This may be my solution, but something feels wrong about it, since my original $y$ in equation (I) is some $y(x)$, but what I get is a single equation $y(x,u)$ ($C,K$ are constants), and I'm not sure if I'm forgetting another expression to make it have sense. I feel that the solution to (I) must be a parametric one, but I only get one equation for $y$ in function of $x$ and $u$. Is the boxed equation the right solution? If not, where am I wrong? Any help will be appreciated.

答: Differentiating $p=2y\dot p+\log\sqrt{\dot p}$ gives $\dot p=2\dot p+2y\ddot p+\ddot p/(2\dot p)$ so $(1+4yq)\dot q=-2q^2$ where $q=\dot p$. As this is exact, the general solution is $c=f(y)$ where $f_q=1+4yq$ and $f_y=2q^2$. Thus $f(y)=d+q+2yq^2$ so $q=(-1\pm\sqrt{1+Cy})/4y$ and substituting back into $p=y'$ yields $x=\int\frac{dy}{\int q(y)\,dy}$.


问:
Gradient "climbing a mountain"
For the function:

$$f(x,y)=100-x^2-y^3$$

if we take the gradient $\nabla f$ we get $(-2x,{-2y^2})$ so for every positive point in the plane we will get a vector pointing downward? so if we look at the function as a "mountain" and we want to go up in the steepest path, we just take the direction $(2x,{2y^2})$?

答:
Generally speaking the gradient vector itself denotes the direction you have to move in the input-plane (x-y) to get maximum height rise. So, that vector is given in cartesian coordinates as:
$$ \nabla F = \frac{ \partial F}{\partial x} \vec{i} + \frac{\partial F}{\partial y} \vec{j}$$

Title: Gradient "climbing a mountain"


答:
Yes. $m-n$ is a perfect square.

Just observe that your condition implies that
$$ (m-n)(2m+2n+1)=0$$
Since $m$ and $n$ are natural numbers, we must have $m-n = 0$ which is a perfect square

Title: Let $m, n ∈ \mathbb{N}$ such that $2m^{2} + m = 2n^{2} + n$ , then prove that $m-n$ is a perfect square .

Body:
After simply factorizing the equation given in the question I got $\mathbf{m + n = -0.5}$
. But the question mentioned $m, n ∈ \mathbb{N}$ . Then how $m + n = -0.5$ ?

Did I do some mistake or is the question wrong?


问:
$f : \mathbb{R} \to \mathbb{R}$ is continuous iff $\{(x,y):y\lt f(x)\}$ and $\{(x,y):y\gt f(x)\}$ are both open sets in $\mathbb R^2$
Given $f: \mathbb{R} \to \mathbb{R}$ define two sets $A$ and $B$ as $ A = \{(x, y) : y < f(x)\}$ and $B = \{ (x, y) : y > f(x)\}$ in $\mathbb{R^2}$. Show that $f$ is continuous iff  both $A$ and $B$ are open.

I'm not sure if this follows from the definition of continuity. Can someone help?

答:
Okay I think this is how it works.

$=>$ As @bof mentioned, define $g(x, y)=f(x)-y$ which is clearly continuous as $f$ is so. So the sets where $g >0$ and $g<0$ would obviously be open.

$<=$ $A \cup B$ is open, hence graph $f$ is closed. We take a sequence $(x_n, y_n) \in$ Graph $f$ converging to $(x, y)$ in the graph. We have $f(x_n)=y_n$  and $\lim f(x_n) =\lim y_n = y = f(x)$. Hence $f$ is continuous.


问:
How to take derivatives if we re-write a function $f(x,y)$ as $f(x/y)$, $f(x-y)$, etc
For example, consider $f(x,y)=x/y$

Can we instead write this as $f(x/y)=x/y$. Is this valid?

I think it is if we consider it a change of variables (i.e. $u=x/y$ and $g(u)=u)

But what if we do not change the variable and just leave it as $f(x/y)$? Can we even take the partial derivative of this w.r.t $x$? If so how (Would we still use the chain rule?)

A similar example would be writing $f(x,y)=x-y$ as $f(x-y)$

答:
If you as $f(\frac{x}{y})$ that is some sort of notational abuse, doesn't make much sense to me.

Yes, now $u$ is a function of $(x,y)$

If you have a function in the arguments like something of the form:
$$ T(x,y) = F(x-y)$$

Then you can take the partial derivative using chain rule.

$$ \frac{\partial T}{\partial x} = \frac{\partial F}{\partial u}|_{x-y} \frac{ d(x-y)}{d x}$$

Of course, the story is different if $y$ is a function of $x$ then you'd have to use total derivatives.

We can see that $g$ is dependent on $u$ and $u$ is dependent on $(x,y)$. If we are take the derivative of $g$ with $u$ directly, we don't care what caused $u$ to change, we just care of how $g$ would change if some how we could nudge the $u \to u + du$. Hence,

$$ \frac{dg}{du} = 1$$

While say if you took the partial derivative of expression with $x$ that is:

$$ G(x,y) = g(u(x,y))$$

And then took a partial,

$$ \frac{ \partial G}{\partial x} = \frac{ \partial g}{\partial u} \frac{du}{dx}$$

This we can think of 'breaking the connection' between $y$ and $u$/ keeping $y$ as fixed and seeing how the function changes as a whole as we crank $x$


问:
Talyor series representation of $Ln(z)$ centered at $z_0$
This is question from applied complex variables.

Let $P(z)$ : talyor series representation of $Ln(z)$ centered at $z_0 =-1-i$. And let the radius of convergence $R$.

Explain why following is not true : $Ln(z)$=$P(z)$ whenever $|z-z_0|<R$.

My attempt : Since $Ln(z)$ is analytic except {$(x+0i)$ | $x \leq0$ }, radius of convergence = 1.

Then, $Ln$(z) is analytic where $|z-z_0|<1$. So taylor series expansion applies.

But, taylor series implies that series expansion equals $Ln(z)$ where $|z-z_0|<1$.

What did I missed?

答:
You are indeed right that $P(z) = \mathrm{Ln}(z)$ for $|z-z_0| < 1$. But the radius of convergence of the Taylor series is actually $\sqrt{2}$. This is because $\mathrm{Ln}(z)$ can be analytically extended beyond the ray of negative real numbers, but it will be multi-valued. To see this, consider the function $\widetilde{\mathrm{Ln}}$, which we define to be the inverse of
$$\exp: \{z \in \mathbb{C} : \mathrm{Im}(z) \in (-2\pi,0)\} \to \mathbb{C} \setminus \{x \in \mathbb{R} : x \geq 0\}.$$
As the inverse of an analytic function $\widetilde{\mathrm{Ln}}$ will be analytic as well. Moreover, it has to coincide with $\mathrm{Ln}$ on $\{x+iy \in \mathbb{C} : x < 0, y < 0\}$ by the uniqueness of inverses. The Taylor series of $\mathrm{Ln}$ and $\widetilde{\mathrm{Ln}}$ are thus the same in $z_0$. But as $\widetilde{\mathrm{Ln}}$ is analytic on $\{z \in \mathbb{C} : |z-z_0| < \sqrt{2}\}$, the radius of convergence is (at least) $\sqrt{2}$. Since $\mathrm{Ln}$ is discontinuous at the negative real numbers, it follows that the Taylor series does not coincide with $\mathrm{Ln}$ for $1 < |z-z_0| < \sqrt{2}$, $\mathrm{Im}(z) > 0$.

To summarize: The radius of convergence of a Taylor series can be larger than the domain of definition of an analytic function.


问:
On evaluation of $\lim_{h\to0} \sqrt[h^2]\frac{{1+h}}{e^h}$
Evaluate the following limit:$$\lim_{h\to0} \sqrt[h^2]\frac{{1+h}}{e^h}$$

Answer is $\frac{1}{\sqrt{e}}$ as on Wolfram Alpha query and evaluated by sympy (Python). Also here is the math search engine query which returned nothing substantial.

I think the answer is $1$. Why?

If $f(x)\to1$ and $g(x)\to\infty$ then the limit ${f(x)}^{g(x)}=e^{({f(x)-1})\cdot g(x)}$. For those who are curious of the proof:

 $$\lim_{x\to a} [1+(f(x)-1)]^{\frac{1}{f(x)-1} (f(x)-1)g(x)} = \lim_{x\to a} [(1+(f(x)-1))^{\frac{1}{f(x)-1}}]^{\lim_{x\to a} (f(x)-1)g(x)}$$

and then a very popular limit to sum it up $\lim_{x\to0} (1+x)^{\frac{1}{x}}=e$

Applying the limit on the numerator, we get $e^{\frac{1}{h}}$ which is exactly equal to the denominator. So hence, the answer $1$. Only my close friend desmos supports me here:

Whereas Wolfram shows

$$\text{My Questions}$$

$1$. Why does the method I adopted shows a speculatively wrong result?

$2$. Is there any way to correct it?

Source code for sympy evaluation.

```<code>#Python 3.8
import math
from sympy import *

x = symbols('x')
expr = ((1+x)**(1/x**2))/(math.e)**(1/x);

print("Expression : {}".format(expr))

# Use sympy.limit() method
limit_expr = limit(expr, x, 0)

print("Limit of the expression tends to 0 : {}".format(limit_expr))
</code>```

Run the code:

```<code>Expression : 2.71828182845905**(-1/x)*(x + 1)**(x**(-2))
Limit of the expression tends to 0 : 0.606530659712633
</code>```

答: For your first question see Daniel's comment. To get W|A's answer we just need L'Hopital:$$\lim_{h\to0}\left(\frac{1+h}{e^h}\right)^{1/h^2}=\exp\lim_{h\to0}\frac{\log((1+h)/e^h)}{h^2}=\exp\lim_{h\to0}\frac{\frac{e^h}{1+h}\cdot\frac{-h}{e^h}}{2h}=\frac1{\sqrt e}.$$


答:
$f(z)=Ln(z)$ is defined on the open set $U=\mathbb C -\mathbb R_-$ while $P(z)$, the Taylor series of it, at $z_0=-1-i$ is defined on the open set $V$ given by $|z-z_0| < \sqrt 2$ as one can easily see by taking derivatives which are just $\pm z_0^{-n}(n-1)!$ and finding the radius of convergence to be $|z_0|=\sqrt 2$.

By analytic continuation $f(z)=P(z)$ whenver $z \in W$ the connected component of $U \cap V$ containing $z_0$ hence for $|z-z_0| < \sqrt 2, \Im z <0$ which is the description of $W$ seen by drawing a picture.

However, there is no guarantee that $f=P$ outside $W$ even if both are defined there, only that $f-P$ is constant (multiple of $2\pi i$ of course) on $W_1$ = $U \cap V -W$ (which has description$|z-z_0| < \sqrt 2, \Im z >0$) and indeed by continuity $\Im P$ is $-\pi$ on the negative imaginary axis hence slighly above where defined it will be $-\pi -\epsilon$, while $\Im f$ is $\pi -\epsilon$ there so on $W_1$ we have $f-P=2\pi i$

Title: Talyor series representation of $Ln(z)$ centered at $z_0$

Body:
This is question from applied complex variables.

Let $P(z)$ : talyor series representation of $Ln(z)$ centered at $z_0 =-1-i$. And let the radius of convergence $R$.

Explain why following is not true : $Ln(z)$=$P(z)$ whenever $|z-z_0|<R$.

My attempt : Since $Ln(z)$ is analytic except {$(x+0i)$ | $x \leq0$ }, radius of convergence = 1.

Then, $Ln$(z) is analytic where $|z-z_0|<1$. So taylor series expansion applies.

But, taylor series implies that series expansion equals $Ln(z)$ where $|z-z_0|<1$.

What did I missed?


问:
Equivalent of definite integral in fourier space
I would like to know the equivalent in Fourier space of an double integral over a circular domain:

$$\int\int_C f(x,y) =\int_0^l \rho d\rho\int_0^{2\pi}d\theta f(\rho,\theta)\rightarrow ????????$$

I know the $\hat{f}(k)$ but there's no closed form for its inverse Fourier transform.

Thanks

答:
If your function $f$ is radially symmetric, then the equivalent of the FT is

$$\hat{f}(k) = 2 \pi \int_0^{\infty} dr \, r \, f(r)\, J_0(k r) $$

where $J_0$ is the Bessel function of the first kind of zeroth order and is equal to

$$J_0(x) = \frac{1}{2 \pi} \int_0^{2 \pi} d\theta \, e^{i x \cos{\theta}}$$

If not, then you may write $f(r,\theta)$ as a Fourier series in $\theta$

$$f(r,\theta) = \sum_{m=-\infty}^{\infty} c_m(r) e^{i k \theta}$$

and

express in terms of Bessels of higher order.


Title: On evaluation of $\lim_{h\to0} \sqrt[h^2]\frac{{1+h}}{e^h}$

Body:
Evaluate the following limit:$$\lim_{h\to0} \sqrt[h^2]\frac{{1+h}}{e^h}$$

Answer is $\frac{1}{\sqrt{e}}$ as on Wolfram Alpha query and evaluated by sympy (Python). Also here is the math search engine query which returned nothing substantial.

I think the answer is $1$. Why?

If $f(x)\to1$ and $g(x)\to\infty$ then the limit ${f(x)}^{g(x)}=e^{({f(x)-1})\cdot g(x)}$. For those who are curious of the proof:

 $$\lim_{x\to a} [1+(f(x)-1)]^{\frac{1}{f(x)-1} (f(x)-1)g(x)} = \lim_{x\to a} [(1+(f(x)-1))^{\frac{1}{f(x)-1}}]^{\lim_{x\to a} (f(x)-1)g(x)}$$

and then a very popular limit to sum it up $\lim_{x\to0} (1+x)^{\frac{1}{x}}=e$

Applying the limit on the numerator, we get $e^{\frac{1}{h}}$ which is exactly equal to the denominator. So hence, the answer $1$. Only my close friend desmos supports me here:

Whereas Wolfram shows

$$\text{My Questions}$$

$1$. Why does the method I adopted shows a speculatively wrong result?

$2$. Is there any way to correct it?

Source code for sympy evaluation.

```<code>#Python 3.8
import math
from sympy import *

x = symbols('x')
expr = ((1+x)**(1/x**2))/(math.e)**(1/x);

print("Expression : {}".format(expr))

# Use sympy.limit() method
limit_expr = limit(expr, x, 0)

print("Limit of the expression tends to 0 : {}".format(limit_expr))
</code>```

Run the code:

```<code>Expression : 2.71828182845905**(-1/x)*(x + 1)**(x**(-2))
Limit of the expression tends to 0 : 0.606530659712633
</code>```

Body:
In this case we have

$$\left(\frac{{1+h}}{e^h}\right)^\frac1{h^2}=\left[\left(1+\left(\frac{{1+h}}{e^h}-1\right)\right)^\frac{1}{\frac{{1+h}}{e^h}-1}\right]^{\frac{\frac{{1+h}}{e^h}-1}{h^2}}$$

with

$$\left(1+\left(\frac{{1+h}}{e^h}-1\right)\right)^\frac{1}{\frac{{1+h}}{e^h}-1} \to e$$

and

$$\frac{\frac{{1+h}}{e^h}-1}{h^2}=\frac1{e^h}\frac{1+h-e^h}{h^2} \to -\frac12$$


答:
I assume that you are familiar with L'Hospital's Rule for evaluating limits.

Applying the formula which you mentioned in the question,
$$lim_{h\to0}\frac{e^{-h}(1+h) - 1}{h^2} = \frac{-1}{2}$$
by the application of L'Hospital's Rule.

So the final limit is $\frac{1}{\sqrt(e)}$


问:
How to solve integrand that diverge?
In the following Eq. 1, the integral diverges. However, it is part of my equation for the Expectation of rate and it should not diverge. I will really appreciate any help as I am stuck on this for over 4 days now.

$$\begin{align}
E = \int_r^\infty\dfrac{\mathrm{e}^{-x}}{1+ \mathrm{e}^{b-cx}}dx \tag{1} \label{1}
\end{align} $$

In the following, I found I can solve it using the My Take below, but even that diverges.

My Take:

--> following the properly $(1+z)^{-1} =  \sum_{j=1}^{\infty} (-1)^{j-1}(z)^{j-1}$

$$E = \int_r^\infty\mathrm{e}^{-x}\sum_{j=1}^{\infty} (-1)^{j-1}(e^{b-cx})^{j-1}dx$$

$$E = \sum_{j=1}^{\infty} (-1)^{j-1} e^{b(j-1)}\int_r^\infty\mathrm{e}^{-x}.e^{-cx(j-1)} dx$$

$$E = \sum_{j=1}^{\infty} (-1)^{j-1} e^{b(j-1)}\int_r^\infty\mathrm e^{-x-cx(j-1)}dx$$

$E$ diverge even now.

How can I solve this?

答:
Provided that $b,c>0$ we have
$$\frac{e^{-x}}{1+e^{b-c x}}=\frac{1}{e^x+e^{b+(1-c) x}}<\frac{1}{e^b\left(e^{x-b}+e^{x-cx}\right)}<\frac{1}{ke^{x}};\;k>0$$
Therefore the integral converges for the comparison test.

Title: How to solve integrand that diverge?


问:
Simple proof showing a subset has smaller cardinal
In Shoenfield, page 254, I read a proof of the following fact: $x \subseteq y \implies Card(x) \le Card(y)$.

Recall

$Card(x)$ is the least ordinal $\alpha$ that is bijective to $x$.

Thus the above inequality just means that $Card(x) = Card(y) \lor Card(x) \in Card(y)$.

Simple proof

The proof uses a remark issued from the proof that the axiom of choice implies Zermelo's theorem. I'm looking for a direct proof which assumes already Zermelo's theorem to be equivalent to the axiom of choice and simple facts about ordinals such as the facts gathered here.

答:
The thing here is that not everyone have their copy of Shoenfield's open, or even at reach (maybe it's in the office), or even a copy of the book.

In general, in $\sf ZF$ we say that for two arbitrary sets, the cardinal of $x$ is less or equal the cardinal of $y$ if and only if there is an injection from $x$ into $y$. So if $x\subseteq y$, the identity function is such injection.

Shoenfield's book, in contrast, defines $Card(x)$ as the smallest initial ordinal which is in bijection with $x$. Now the axiom of choice is equivalent to the statement "For every $x$, $Card(x)$ exists".

But you don't need the axiom of choice for most basic cardinal arithmetic results when involving ordinals, and certainly not for this one. Shoenfield's argument is as follows (and I apologise in advance if my notation is different):

$x\subseteq y$, so if $f\colon y\to\sigma=Card(y)$ is a bijection, $f\restriction x$ is a bijection between $x$ and a subset of $\sigma$, say $f``x$.

Looking at the function mapping $f``x$ to its order-type, say $\chi$, then $\chi$ is an ordinal and $\chi\leq\sigma$.

Therefore, $Card(x)=Card(\chi)\leq\chi\leq\sigma=Card(y)$.

The axiom of choice is only used to claim that if $y$ is any arbitrary set, then $Card(y)$ exists. But the proof does not use any more choice than that. And if we consider a broader notion of cardinals which allow for sets which are not well-orderable to have a cardinal (which is not an ordinal, in this case), then the definition of $Card(x)\leq Card(y)$ is aligned with making this statement trivial (although we still need to prove that for initial ordinals the cardinal ordering coincides with the ordinal ordering which coincides with the inclusion ordering).


问:
Prove $\sin x\ge x-\frac{x^3}{3!}$

Prove $$\sin x\ge x-\frac{x^3}{3!}$$ for $x\ge 0$.

I know that a calculus/continuity proof exists, but I am curious if this can be proved without that. Here is a sketch I have made.;

we will use for $x\ge 0$;

$$\sin x\le x\le \tan x$$

proof of this see here

I got a weaker bound :
$$\tan\left(\frac{x}{2}\right)\ge \frac{x}{2}$$
$$\sin\left(\frac{x}2\right)\cos\left(\frac{x}2\right)\ge \frac{x}{2}\cos^2\left(\frac{x}{2}\right)$$
$$\sin x= x\left[1-\sin^2\left(\frac{x}{2}\right)\right]\ge x\left(1-\frac{x^2}{4}\right)= x-\frac{x^3}{4}$$
Thus $$\sin x \ge x-\frac{x^3}{4}$$
As you can see is there any way I can take this to strengthen the inequality, to get $$\sin x\ge x-\frac{x^3}{6}.$$
Also could this possibly be extended by strengthening inequality further to get the Taylor series. I think I am going too far!

答:
Using $\sin^2(\frac t2)\le (\frac{t}{2})^2$, one has
$$ \sin(x)=\int_0^x\cos(t)dt=\int_0^x\bigg(1-2\sin^2(\frac t2)\bigg)dt\ge \int_0^x(1-\frac12t^2)dt=x-\frac{x^3}{3!}.$$

Title: Prove $\sin x\ge x-\frac{x^3}{3!}$


问:
Unit vector in $ds$ substitution (while changing coordinate systems)
I had to calculate the line integral of a homogenous vector field $E=E.e_y$ from an angle $\phi=o$ to $\phi=\phi_{e}$ having a radius $\rho$. the formula normally is $\varphi=\int E \space ds$ but since the coordinate system used is cylindrical I converted $ds$ to $R\ dR\ d\phi$. However, the answer I got was wrong and in the provided solution. $ds$ was substituted by $Rd\varpThe session negotiation failed.

Error: Failed to authenticate user '' with PAMhi_e\varphi$ below I have added a picture of the problem.

答:
The issue is with the notation $d\mathbf{s}$ is not a surface element, so it's not $Rdrd\phi$. It is a length, tangent to the curve. You need to decompose it into $ds_x$ and $ds_y$. The one that you care about is $ds_y$, since $ds_x\mathbf{e}_x\cdot\mathbf{E}=0$.

You can write $ds_y$ in terms of the angle:
$$ds_y=\rho\cos\phi d\phi$$
See if this solves the problem.


问:
Probability that a larger random subset contains a smaller random subset.

Let $A$ be a subset of $\{1,..,n\}$ such that $|A| = k$ and let $B$ be another random subset of $\{1,..,n\}$ such that $|B| = t < k$.

What is the probability that $A$ contains $B$? (both $A$ and $B$ are chosen uniformly)

I need to show that this probability is $\Theta\left(\left(\frac{k}{n}\right)^t\right)$. Any advice?

答:
Hints:

How many sets $B$ are there with $|B|=t$ and $B \subset \{1,2,\cdots,n\}$?

Given $A$, how many sets $B_A$ are there with $|B_A| =t$ and $B_A \subset A$?


问:
Composition of $\mathrm H^p$ function with Möbius transform
Let $f:\mathbb D\rightarrow \mathbb C$ be a function in $\mathrm{H}^p$, i.e. $$\exists M>0,\text{ such that }\int_0^{2\pi}|f(re^{it})|^pdt\leq M<\infty,\forall r\in[o,1)$$
Consider a Möbius transform of the disk $\varphi :\mathbb D\rightarrow\mathbb D$, which generally does not fix $0$.

(or more generally consider a holomorphic function $g:\mathbb D \rightarrow \mathbb D$)

Is the composition $f\circ\varphi $ in $\mathrm H^p$?

If $\varphi $ fixes $0$ then we can apply Littlewood subordination theorem and derive the result. However, what can we say if $\varphi$ does not fix zero? Does it hold $f\circ\varphi $ in $\mathrm H^p$, or is there a counterexample?

答:
Yes. If $\varphi$ is a holomorphic map of unit disk into itself, the composition operator $f\mapsto f\circ \varphi$ is bounded on $H^p$. In fact,
$$\|f\circ \varphi\|_{H^p}\le \left(\frac{1+|\varphi(0)|}{1-|\varphi(0)|}\right)^{1/p}\|f \|_{H^p} \tag1$$

Original source:

John V. Ryff, Subordinate $H^p$ functions, Duke Math. J. Vol. 33, no. 2 (1966), 347-354.

Sketch of proof. The function $|f|^p$ is subharmonic and has uniformly bounded circular means (i.e., mean values on  $|z|=r$, $r<1$). Let $u$ be its the smallest harmonic majorant in $\mathbb D$. Then $u\circ \varphi$ is a harmonic majorant of $|f\circ \varphi|^p$. This implies that the circular means of $|f\circ \varphi|^p$ do not exceed $u(\varphi(0))$. In particular, $f\circ \varphi\in H^p$. Furthermore, the Harnack inequality for the disk yields $$u(\varphi(0))\le u(0)\frac{1+|\varphi(0)|}{1-|\varphi(0)|}$$ proving (1).


问:
Whole number roots of a polynomial equation with a prime constant term
We have quadratic equation $x^2-4x-21-q=0$ and we know that $q$ is a prime number. The question is to find every value of $q$ where the equation has whole solutions. I found one $q=11$ but I don't understand how to prove that no other prime will satisfy this equation. Thanks in advance!

答:
Solved it with a hint from Mr. Peter. Taking the $D$1$=(\frac{b}{2})^2-ac=q+25$, for the solutions to be whole, $D$1 must be a perfect square, so take some $k^2=q+25$. $$q=k^2-25$$ $$q=(k+5)(k-5)$$ For $q$ to be prime $k-5$ needs to be equal to $1$ since it is one of the two factors of a prime number. $$k-5=1$$ $$k=6$$ plug in $q=6^2-25=11$.
Check that value of $11$ for $q$ yields whole roots, which it does.


答:
There are $\binom{n}{t}$ ways to choose a $B$, and there $\binom{k}{t}$ ways to choose a $B$ that is a subset of any given $A$. Thus:

$$P(B\subset A) = \frac{\binom{k}{t}}{\binom{n}{t}} = \frac{\frac{k!}{t!(k-t)!}}{\frac{n!}{t!(n-t)!}} = \frac{k!(n-t)!}{n!(k-t)!} = \overbrace{\underbrace{\frac{k(k-1)(k-2)...(k-t + 1)}{n(n-1)(n-2)...(n-t + 1)}}_{t\text{ factors}}}^{t\text{ factors}}$$

Can you take it from here?

Title: Probability that a larger random subset contains a smaller random subset.

Body:
Let $A$ be a subset of $\{1,..,n\}$ such that $|A| = k$ and let $B$ be another random subset of $\{1,..,n\}$ such that $|B| = t < k$.

What is the probability that $A$ contains $B$? (both $A$ and $B$ are chosen uniformly)

I need to show that this probability is $\Theta\left(\left(\frac{k}{n}\right)^t\right)$. Any advice?


答:
Given a function $f:\mathbb R\to\mathbb R$ define
$$A_f=\{(x,y)\in\mathbb R\times\mathbb R:f(x)\gt y\},$$
$$B_f=\{(x,y)\in\mathbb R\times\mathbb R:f(x)\lt y\}.$$

Lemma 1. The function $f:\mathbb R\to\mathbb R$ is upper semicontinuous if and only if $B_f$ is open in $\mathbb R\times\mathbb R$.

Proof. First, suppose $f$ is USC. Let $(x_0,y_0)$ be any point in $B_f$. Let $\varepsilon=\frac{y_0-f(x_0)}2\gt0.$ Choose $\delta\gt0$ so that $|x-x_0|\lt\delta\implies f(x)\lt f(x_0)+\varepsilon$. Then
$$|x-x_0|\lt\delta,\ |y-y_0|\lt\varepsilon\implies f(x)\lt f(x_0)+\varepsilon=y_0-\varepsilon\lt y\implies(x,y)\in B_f.$$
This shows that $B_f$ is open.

Conversely, suppose $B_f$ is open. Let $x_0\in\mathbb R$ and $\varepsilon\gt0$ be given. Since $(x_0,f(x_0)+\varepsilon)\in B_f$, we can find $\delta\gt0$ such that
$$|x-x_0|\lt\delta\implies(x,f(x_0)+\varepsilon)\in B_f\implies f(x)\lt f(x_0)+\varepsilon.$$
This shows that $f$ is USC.

Lemma 2. The function $f:\mathbb R\to\mathbb R$ is lower semicontinuous if and only if $A_f$ is open in $\mathbb R\times\mathbb R$.

Proof. Similar to Lemma 1.

Theorem. The function $f:\mathbb R\to\mathbb R$ is continuous if and only if both of the sets $A_f$ and $B_f$ are open in $\mathbb R\times\mathbb R$.

Proof. Lemmas 1 and 2.

Example. The function $f=\{(x,0):x\le0\}\cup\{(x,\frac1x):x\gt0\}$ is lower semicontinuous but not upper semicontinuous; the sets $A_f$ and $A_f\cup B_f$ are open, but $B_f$ is not open.

Example. The function $f=\{(x,y): xy=1\}\cup\{(0,0)\}$ is neither upper semicontinuous nor lower semicontinuous; $A_f\cup B_f$ is open, but neither $A_f$ nor $B_f$ is open.

Title: $f : \mathbb{R} \to \mathbb{R}$ is continuous iff $\{(x,y):y\lt f(x)\}$ and $\{(x,y):y\gt f(x)\}$ are both open sets in $\mathbb R^2$

Body:
Given $f: \mathbb{R} \to \mathbb{R}$ define two sets $A$ and $B$ as $ A = \{(x, y) : y < f(x)\}$ and $B = \{ (x, y) : y > f(x)\}$ in $\mathbb{R^2}$. Show that $f$ is continuous iff  both $A$ and $B$ are open.

I'm not sure if this follows from the definition of continuity. Can someone help?


问:
Uniform continuity and sequences negation
Show that if $f$ is not uniformly continuous on an interval $[a,b]$ then there are sequences $\{x_n\}$ and $\{y_n\}$ chosen from $[a,b]$ so that $x_n-y_n\to0$ but $|f(x_n)-f(y_n)|>c$ for some $c>0$

I've only barely started the proof for this one.

Proof

Suppose $f:[a,b]\to \mathbb{R}$ is not uniformly continuous

Then $\exists c >0: \forall \delta >0,$ there are $x_ ,y \in [a,b]: |x-y|\leq \delta \quad \text{and}\quad |f(x)-f(y)|\geq c$

Let $\{x_n\}, \{y_n\} \subseteq \mathbb{R}$

...

I'm not sure where to go from here. I get stuck because we don't know what $\{x_n\}$ and $\{y_n\}$ converge to, which makes using the definition of convergence tricky here. Any ideas?

答:
The negation of $$\forall \varepsilon >0, \exists \delta >0: \forall x,y\in [a,b]: |x-y|\leq \delta \implies |f(x)-f(y)|<\varepsilon $$

Is $$\exists \varepsilon >0: \forall \delta >0, \exists x_\delta ,y_\delta \in [a,b]: |x-y|\leq \delta \quad \text{and}\quad |f(x)-f(y)|\geq \varepsilon .$$
In particular, there is $\varepsilon >0$ s.t. for all $n\in\mathbb N$, there are $x_n,y_n\in[a,b]$ s.t. $$|x_n-y_n|\leq \frac{1}{n}\quad \text{and}\quad |f(x_n)-f(y_n)|>\varepsilon .$$


问:
How to compare points in homogeneous coordinates universally?

Problem

Given a plane, I have lines in the canonical form (described by 3D vectors). Using a cross product, I can find an intersection of any two lines. I want to check, whether two pairs of lines intersect at the same point, taking noisiness of the data into account. This means that I should compare them with respect to a chosen threshold

$$
p_1 = \ell_1 \times \ell_2 \\
p_2 = \ell_3 \times \ell_4 \\
\left( l_1 \nparallel l_2 \right) \land \left( l_3 \nparallel l_4 \right) \implies
\left( p_1^z \neq 0 \right) \land \left( p_2^z \neq 0 \right) \implies
\left\lVert \frac{p_1}{p_1^z} - \frac{p_2}{p_2^z} \right\rVert < \varepsilon?
$$

If two given lines are not parallel, their intersection point has a non-zero third coordinate. So, I can divide by it to get ordinary 2D coordinates of the point and compare it with another one in the usual way.
Otherwise, if the lines are parallel, the third coordinate of their intersection point is close to zero (meaning it's the point at infinity), so I cannot divide by it and should compare the directions of two points at infinity

$$
p_1 = \ell_1 \times \ell_2 \\
p_2 = \ell_3 \times \ell_4 \\
\left( l_1 \parallel l_2 \right) \land \left( l_3 \parallel l_4 \right) \implies
\left( p_1^z \approx 0 \approx p_2^z \right) \implies
\left\lVert
\frac{p_1}{\left\lVert p_1 \right\rVert} - \frac{p_2}{\left\lVert p_2 \right\rVert}
\right\rVert < \varepsilon?
$$

Question

Is there a universal way to compare points in a homogeneous coordinate system to avoid the separate check of the third coordinate?

答:
You can "normalize" as you've done: replace $p_1$ by $q_1 = \frac{p_1}{\|p_1\|}$, and similarly for $p_2$ to get a pair of unit vectors in the 2-sphere in 3-space. There's a small problem here, which is that even if $p_1$ and $p_2$ represent the same point, $q_1$ and $q_2$ might point in opposite directions (i.e., $p_1 = -q_2$).

So the next step is to compute $d = q_1 \cdot q_2$, and if $d < 0$, you negate $q_2$. (I'm going to assume you've done this, and now call the replaced vector $q_2$.)

Now you've got two vectors in the same hemisphere of $S^2$, so to measure how far apart they are, you can compute the angle between them:
$$
\theta = \arccos(q_1 \cdot q_2).
$$

(Noting that $\cos$ is an even function, you'll realize that the whole "negate $q_2$ isn't actually necessary...)

This value $\theta$ gives you a "distance" between the computed intersections.

Is $\theta$ invariant with respect to the various transformations that matter to you? You might, for instance, say "If I changed coordinates by translating everything by $(a,b)$ for any part of numbers $(a,b)$, then the 'distance' should be unchanged". If you want that, then you have to check whether $\theta$ has that property. Since I don't know what properties you wish to have, I can't really go any farther.


问:
proof with induction $2^n + 4 > n^2 + 2n$
prove w/ full induction : $2^n + 4 > n^2 + 2n$.
Let's skip the other steps for a moment, I'm just getting stuck with the proof
the claim is: $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$
and now the proof  $$2^{n+1} + 4 > 2 * 2^n + 4$$ $$> 2 * (n^2 + 2n)$$ $$> 2n^2 + 4n$$
i am stuck here

Body:
It is true that for $n\geq5$

$$2^n+4>n^2+2n$$

Assume that for $n=k$
$$2^k+4>k^2+2k$$

Let's show for $n=k+1$

$$2^{k+1}+4>(k+1)^2+2(k+1)$$

Now

$$2(2^k+4)>2(k^2+2k)$$

$$2^{k+1}+4>2k^2+2k-4$$

$$2^{k+1}+4>2k^2+2k-4>(k+1)^2+2(k+1)$$

$$2k^2+2k-4>k^2+4k+3$$

$$k^2-2k-7>0$$

for $k\geq5$

$$(k-1)^2-8>0$$


答:
$$A= \sqrt[h^2]\frac{{1+h}}{e^h}\implies \log(A)=\frac 1{h^2}\left( \log(1+h)-h\right)$$ Now, by Taylor
$$\log(A)=\frac 1{h^2}\left(-\frac{h^2}{2}+\frac{h^3}{3}+O\left(h^4\right) \right)=-\frac{1}{2}+\frac{h}{3}+O\left(h^2\right)$$
$$A=e^{\log(A)}=\frac{1}{\sqrt{e}}+\frac{h}{3 \sqrt{e}}+O\left(h^2\right)$$


问:
Are there any functions that satisfy these properties?
You're given a function $f$ : $\mathbb{R} \rightarrow \mathbb{R}$ that satisfy these properties
$$ f(x +1)=f(x), f(n + \dfrac{1}{2})=1, f(n)=0  \ \forall x \in \mathbb{R}, n \in \mathbb{N}$$
Are there any functions $m,n$ : $\mathbb{R} \rightarrow \mathbb{R}$ satisfy

$$ m(x) \neq 0 \ \ \text{with all $x$ satisfy} \ \ f(x)=m(x)(n(x) + 1) $$

答:
Note that $n(x) = f(x) - 1$, $m(x) = 1$ satisfy the conditions of the problem. Here is a specific example of $f(x), n(x), m(x)$that work for $x\in \mathbb{R}$:

$$f(x) = \frac{-\cos(2\pi x) + 1}{2}$$

$$n(x) = \frac{-\cos(2\pi x) - 1}{2}$$

$$m(x) = 1$$


问:
$\lim\limits_{t\to\infty}x(t)=0$ in a kind linear ODE in these conditions but I think I didn't need one of the conditions of the exercise to prove it
The exercise is: Let $a,b:\Bbb R \to\Bbb R $ be continuous functions and $M\in \Bbb R, M>0,$ such that $a(t)\ge M$ for all $t\in\Bbb R$ and $b(t)\ge \frac{1}{t^{100}}$ for all $t\ge 1$ and $\lim\limits_{t \to \infty}b(t)=0$. Proof that all the solutions of $x'(t)=-a(t)x(t)+b(t)$ verify $\lim\limits_{t\to\infty}x(t)=0$.

This is what I did:

The solutions of the ODE are  $$ x_{_C}(t)=e^{-\int a(t)dt}\int (e^{\int a(t)dt}b(t))dt+Ce^{-\int a(t)dt}, C\in \Bbb R$$
First, I proved that given $t_0\in \Bbb R$, then $\lim\limits_{t\to \infty} e^{-\int_{t_0}^t a(s)ds}=0$. I shall explain if someone asks me.

Using $\lim\limits_{s \to \infty}b(s)=0$, given $\varepsilon>0$ exists $s_0$ such that $s>s_0 \implies |b(s)|<M\varepsilon$, so

$|\int_{t_0}^t (e^{\int_{t_0}^s a(r)dr}b(s))ds|=_{t>s_0}\le M\varepsilon |\int_{t_0}^t (e^{\int_{t_0}^s a(r)dr})ds|=$
$M\varepsilon|(\frac{1}{a(s)}e^{\int_{t_0}^s a(r)dr}]_{t_0}^t|\le \varepsilon|e^{\int_{t_0}^t a(s)ds}-1|$,

Multiplying by $|e^{-\int_{t_0}^t a(s)ds}|$ in the inequality:

$|e^{-\int_{t_0}^t a(s)ds}\int_{t_0}^t (e^{\int_{t_0}^s a(r)dr}b(s))ds|\le \varepsilon|1-e^{-\int_{t_0}^t a(s)ds}|<\varepsilon$.

With these we can say that $\lim\limits_{t\to \infty}x_c(t)=0$, for all $C\in \Bbb R$.

You can see that $b(t)\ge \frac{1}{t^{100}}$ wasn't needed if I made it well. Is this reasoning correct?

答:
First, I would define the general solution with limits of integration since that's what you use later:
$$ x_{C}(t)=e^{-\int_{t_0}^t a(t')dt'}\int_{t_1}^t e^{\int_{t_0}^{t''}a(t')dt'}b(t'')dt''+Ce^{-\int_{t_0}^t a(t')dt'}, C\in \Bbb R.$$

When you bound your integral with $M\epsilon$ to get rid of the $b(t)$, it's not correct because the inequality $b(t)<M\epsilon$ must be satisfied on the whole interval of integration.

I would do it like that: if the integral
$$
\int_{t_1}^\infty e^{\int_{t_0}^{t''}a(t')dt'}b(t'')dt''
$$
exists, then you are done. I believe I have to use the condition $b(t)>\frac{1}{t^{100}}$ to claim that $b(t)$ is positive (at least starting at $t=1$) and thus, if the integral above does not exists, it diverges to $\infty$. If the integral diverges, then use L'Hopital rule on
$$
\lim_{t\rightarrow \infty} e^{-\int_{t_0}^t a(t')dt'}\int_{t_1}^t e^{\int_{t_0}^{t''}a(t')dt'}b(t'')dt''=
\lim_{t\rightarrow \infty} \frac{\int_{t_1}^t (e^{\int_{t_0}^{t''}a(t')dt'}b(t''))dt''}{e^{\int_{t_0}^t a(t')dt'}}=\lim_{t\rightarrow\infty}\frac{b(t)}{a(t)}=0,
$$
where the next to last inequality was obtained by using L'Hopital rule and differentiating the top and bottom.


问:
Convergence in $\mathcal{S}'$ and $L^p$ spaces
I expect that the following statement is true:

Let $p \in (0, \infty], f \in L^p(\mathbb{R}^n), \{f_j\} \subset L^1(\mathbb{R}^n)$ and $f_j \to f \in \mathcal{S}'(\mathbb{R}^n)$. In addition, if $\{f_j\}$ is $L^1$-bounded, then, $f \in L^1(\mathbb{R}^n)$.

I can neither prove nor disprove this claim.
I would be grateful if anyone could give me some advice.

My English is not very good. Please forgive me if there are any grammatical errors or confusing parts.

答:
Lets assume $p≥1$. $f_j\to f$ in $\mathcal S'$ means that $\int_{\Bbb R} f_j s\to\int_{\Bbb R} f s$ for all $s\in\mathcal S$, this condition can be weakened a bit, fix some $n$ and then you have:

$$\int_{[-n,n]}f_j(x) s(x)\,dx\to\int_{[-n,n]}f(x)s(x)\,dx$$
for all $s\in \mathcal S$. In particular the left-hand side is bounded above by $C\cdot \|s\|_\infty$ for $C:=\sup_{j}\|f_j\|_1$, so the right-hand side also is bounded above by this, ie:

$$\int_{[-n,n]}f(x) s(x)\,dx≤ C\,\|s\|_\infty$$

Three things:

$L^1 ([-n,n])\supseteq L^p([-n,n])$ for $p\in[1,\infty)$, so $f\lvert_{[-n,n]}\in L^1([-n,n])$.

The dual of $C([-n,n])$ includes $L^1([-n,n])$, so you can get the norm $\|f\lvert_{[-n,n]}\|_1$ by taking the supremum of $f[s]$ over all unit vectors $s$ in $C([-n,n])$.

The restriction of $\mathcal S$ to $[-n,n]$ is dense in $C([-n,n])$ if you give it the norm $\|\cdot\|_\infty$. So in the above consideration you can just take the norm over the norm $≤1$ vectors in $\mathcal S$ rather than $C([-n,n])$.

Written out:
$$\|f\lvert_{[-n,n]}\|_1 = \sup_{s\in C[-n,n], \|s\|_\infty ≤1} f[s] =\sup_{s\in \mathcal S, \|s\|_\infty ≤1}\ \int_{[-n,n]}f(x)s(x)\,dx≤ C$$
And then:
$$\lim_n \int_{[-n,n]}|f(x)|\,dx ≤ C$$
giving that $f\in L^1$.


问:
Are solvable groups Howson?
A finitely generated group $G$ has the Howson property if the intersection of any two finitely generated subgroups is again finitely generated. (Finitely generated) free groups, nilpotent, and polycyclic groups all have the Howson property.

Are solvable groups Howson?

I cannot find a reference to this. I expect the answer to be negative, but can't think of any counterexample at the moment.

答:
Let $H = {\rm BS}(1,k) = \langle  a,b \mid b^{-1}ab=a^k \rangle$ for some $k>1$.

Then $H$ is solvable, and the normal subgroup $\langle a^H \rangle$ is abelian and not finitely generated.

Now let $G = \langle  a,b,c \mid b^{-1}ab=a^k, [c,b] = [c,a] = 1\rangle \cong H \times {\mathbb Z}$.

Then the subgroups $\langle a,b \rangle$ and $\langle a,bc \rangle$ of $G$ are finitely generated, but their intersection $\langle a^H \rangle$ is not.

Title: Are solvable groups Howson?


问:
$\pi$-base of topological space
I've recently came across the $\pi$-weight which is the cardinality of the smallest $\pi$-base of topological space. Unfortunately I've never heard of such thing as $\pi$-base and can't find any article which has the definition of it. What exactly is a $\pi$-base?

答:
Well, one I know for sure is

Stover, Derrick. "On $\pi $-metrizable spaces, their continuous images and products." Commentationes Mathematicae Universitatis Carolinae 50.1 (2009): 153-162.
link

From the second sentence of the intro:

Recall that for a space $X$ a collection of nonempty open sets $\Theta$ is called a $\pi$-base if for every nonempty open set $\mathcal O$, there exists a $\mathcal U\in\Theta$ such that $\mathcal U\subset \mathcal O$.

This is confirmed also in another solution on this site.

It also appears in Wikipedia's glossary of topology with the citation:
Hart, Nagata, Vaughan Sect. d-22, page 227 which seems to refer to

Hart, Klaas Pieter, Jun-iti Nagata, and Jerry E. Vaughan. Encyclopedia of general topology. Elsevier, 2003.

Title: $\pi$-base of topological space


答:
Your curve is parametrized as

$$\mathbf{s}(\phi) = \rho(\cos\phi,\sin\phi,0),$$

hence

$$\mathrm{d}\mathbf{s} = \rho(-\sin\phi,\cos\phi,0)\,\mathrm{d}\phi.$$

Combined with $\mathbf{E} = (0,E,0)$, we get

$$\int\mathbf{E}\cdot\mathrm{d}\mathbf{s} = \int_0^{\phi_e}E\rho\cos\phi\,\mathrm{d}\phi = E\rho \sin\phi_e.$$

Title: Unit vector in $ds$ substitution (while changing coordinate systems)

Body:
I had to calculate the line integral of a homogenous vector field $E=E.e_y$ from an angle $\phi=o$ to $\phi=\phi_{e}$ having a radius $\rho$. the formula normally is $\varphi=\int E \space ds$ but since the coordinate system used is cylindrical I converted $ds$ to $R\ dR\ d\phi$. However, the answer I got was wrong and in the provided solution. $ds$ was substituted by $Rd\varpThe session negotiation failed.

Error: Failed to authenticate user '' with PAMhi_e\varphi$ below I have added a picture of the problem.


Title: proof with induction $2^n + 4 > n^2 + 2n$

Body:
prove w/ full induction : $2^n + 4 > n^2 + 2n$.
Let's skip the other steps for a moment, I'm just getting stuck with the proof
the claim is: $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$
and now the proof  $$2^{n+1} + 4 > 2 * 2^n + 4$$ $$> 2 * (n^2 + 2n)$$ $$> 2n^2 + 4n$$
i am stuck here

Body:
Please note this is not true for low values of $n \in \mathbb{Z+}$ up to $4$.

Just your induction step -

If $2^{n} + 4 > n^2 + 2n$ is true

we need to prove $\, 2^{n+1} + 4 > (n + 1)^2 + 2(n+1)$

$2^{n+1} + 4 = 2 \times 2^{n} + 8 - 4 = 2 (2^{n} + 4) - 4$

$ \gt 2n^2 + 4n - 4 = n^2 + (n+1)^2 + 2(n+1) - 7 = (n+1)^2 + 2(n+1) + (n^2 - 7)$

As $(n^2 - 7) \gt 0$ for $n \geq 3$, this is true.


问:
Entire function $f$ such that $f(n) = \frac{(-1)^n}{n!}$
I know that there exists an entire function $f$ s.t. $f(n) = \frac{(-1)^{n}}{n!}$ for $n \in \mathbb{N}$. (Using Weierstrass theorem or Mittag-Leffler theorem)

Is there any simple formula to express $f$? The value $\frac{(-1)^n}{n!}$ comes from the residue of $\Gamma(z)$, so I guess that $f$ would be expressed in terms of $\Gamma(z)$, but I'm not sure.

答:
The simplest thing to try would be, writing $(-1)^n = \exp(i\pi n)$ and $n! = \Gamma(n+1)$.  But that fails because $\Gamma(-1) = \infty$.  However,
$f(z) := 1/\Gamma(z)$ is an entire function (removable singularities), so an answer is
$$
e^{i\pi x} f(x+1) = \frac{e^{i\pi x}}{\Gamma(x+1)}
$$
Once we have this, we can get other solutions by adding any entire function that vanishes on $\mathbb Z$.  Thus, adding anything of the form $(e^{2i\pi x}-1)G(x)$ with $G$ entire.


问:
Sample vs Population Covariance Matrices
Let $P(X)$ be a probability density function over $\mathbb{R}^{n}$, and assume its random variable $X$ to have a zero mean and a covariance matrix:

$\Sigma = cov(X, X) = \mathbb{E}[XX^T]$

In case of kernel PCA methods, $X$ is actually the feature vector of the original data point, and $n$ may be finite or infinite depending on the associated kernel.

Let $\{ X_i \}_{i = 1}^{N}$ to be i.i.d. samples from $P(X)$, and define estimator of $\Sigma$ as:

$\hat{\Sigma} = \frac{1}{N} \sum_{i = 1}^{N} X_i X_i^T$

Let $\{ \lambda_j, v_j \}$ and $\{ \hat{\lambda}_j, \hat{v}_j \}$ be eigenpairs of $\Sigma$ and $\hat{\Sigma}$ respectively, where eigenvalues are sorted in decreasing order.

How close are $n \times n$ matrices $\Sigma$ and $\hat{\Sigma}$? How close are their eigenvalues $\{ \lambda_j \}$ and $\{ \hat{\lambda}_j \}$, and eigenvectors $\{ v_j \}$ and $\{ \hat{v}_j \}$? Specifically, can we lower-upper bound the inner product $<v_k, \hat{v}_m>^2$ more tightly than the obvious $[0, 1]$ interval?

Below I summarize several papers that answer some parts of my question. If anyone has more suggestions, please share them below.

答:
Here I add short summary of papers I found to be relevant for the above questions.

Denote by $V_k$ and $\hat{V}_k$ sub-spaces that correspond to spans of $\{ v_j \}_{j = 1}^{k}$ and $\{ \hat{v}_j \}_{j = 1}^{k}$ respectively. Denote by $P_k$ and $\hat{P}_k$ the orthogonal projectors on these sub-spaces. Finally, denote by $V_k^{\perp}$ and $\hat{V}_k^{\perp}$ the orthogonal complements of $V_k$ and $\hat{V}_k$.

John Shawe-Taylor,2005 produces probabilistic bounds on eigenvalue sums, i.e. $\sum_{j = 1}^{k} \lambda_j$. Additionally, there are bounds on projection of new data onto $\hat{V}_k$ and onto $\hat{V}_k^{\perp}$. Finally, it gives bounds on $\hat{v}_j^T \Sigma \hat{v}_j$.

Laurent Zwald, 2006 bounds Hilbert-Schmidt norm of a differences $\Sigma - \hat{\Sigma}$ and $P_k - \hat{P}_k$. Likewise, there is bound on $<v_k, \hat{v}_k>^2$.

Gilles Blanchard, 2007 there are bounds on projection of new data onto $V_k^{\perp}$ and onto $\hat{V}_k^{\perp}$. Likewise, bounds produced to relate sums $\sum_{j = k}^{n} \lambda_j$ and $\sum_{j = k}^{n} \hat{\lambda}_j$. This work refines some of the results of 1.

Roman Vershynin, 2010 bounds operator norm of $\Sigma - \hat{\Sigma}$, showing that this norm is small with high probability when $N = O(n)$ and the distribution has finite fourth moment.

Lorenzo Rosasco, 2010 bounds $\sum_{j \geq 1} (\lambda_j - \hat{\lambda}_j)^2$, $\sup_{j \geq 1} (\lambda_j - \hat{\lambda}_j)^2$, and $|tr(\Sigma) - tr(\hat{\Sigma})|$.

Andreas Louka, 2017 bounds $|<v_k, \hat{v}_m>|$ for any $k$ and $m$, showing its bound to be inverse proportional to $|\lambda_k - \lambda_m|$. Further, a probabilistic version of this bound is proposed relating expressions $|<v_k, \hat{v}_m>|$ and $|\lambda_k - \hat{\lambda}_k|$ with a measure of the concentration of the distribution in
the direction of $v_k$.

Wainwright Martin J, book, 2019 bounds 2-norm of $\Sigma - \hat{\Sigma}$ (Section 6), and $||v_1 - \hat{v}_1||_2$ (Section 8). Also, some theorems are provided especially for sparse covariance matrices.


问:
A question regarding the joint density of two random variables
Please take a look at my working. I would like to know if it is correct. Thank you for your time.

Let $X$ and $Y$ be independent exponentially distributed random variables with parameters $\lambda$ and $\mu$. Find the joint density of $X$ and $Y$ and calculate $P(X\lt Y)$.

Attempt

The joint density of $X$ and $Y$ is as follows.

$$f(x,y)=\begin{cases}
\lambda e^{-\lambda x} \mu e^{-\mu y} & \text{ if } x \ge 0, y \ge 0, \\
0 & \text{ otherwise }.
\end{cases}$$

We have $$P(X\lt Y)= \int_{0}^{\infty}\int_{0}^{y}\lambda e^{-\lambda x} \mu e^{-\mu y}dx dy$$

Then

$$\lambda \mu\int_{0}^{\infty}\int_{0}^{y}e^{-(\lambda x + \mu y)}dx dy=\lambda \mu \int_{0}^{\infty}\left[\dfrac{e^{-(\lambda x + \mu y)}}{-\lambda}\right]_{0}^{y} dy$$

Continuing, we have

$$ \dfrac{\lambda \mu}{-\lambda} \int_{0}^{\infty} \left(e^{-(\lambda + \mu )y}-e^{-\mu y}\right)dy=-\mu \left [ \dfrac{e^{-(\lambda + \mu )y}}{-(\lambda + \mu)} - \dfrac{e^{-\mu y}}{(-\mu)} \right]_{0}^{\infty}=\dfrac{\lambda}{\lambda + \mu}.$$

答:
I think that is correct.  I'd have found your working slightly easier to follow if you had made it clearer that you were doing the x-integration first, e.g. you could have written something like

$$
\lambda \mu \int_{0}^{\infty}\left[\dfrac{e^{-(\lambda x + \mu y)}}{-\lambda}\right]_{x=0}^{x=y} dy
$$

but no big deal!


Body:
The claim is $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$

Assuming it is true for $n$, that is $2^n + 4 > n^2 + 2n\to 2^n>n^2+2n-4$ we have
$$2^{n+1} + 4=2\cdot 2^n+4>2\left(n^2+2n-4\right)+4=2n^2+4n-4=(n^2+2n+1)+(2n+2)+n^2-6$$
$$>(n+1)^2+2(n+1);\;n>2$$
Proved


问:
What is the formula for Cross partials after change of variable?
Suppose we have a function $f(x,y) = 2x^2+2y^2$

Next we change the variables with $u(x,y)=2x^2+2y^2$ so we have $f(x,y)=g(u(x,y))=u(x,y)$

Since these functions are equal everywhere, we can write $f(x,y)=g(u(x,y))$ and take partials of both sides.

Question: What is the formula for $\frac{\partial}{\partial y}\frac{\partial}{\partial x} g(u(x,y))$ (in general, not for the specific above example)

For example, we can say that
$$ \tag{1}
\frac{\partial}{\partial x} f(x,y) =
\frac{\partial}{\partial x} g(u(x,y)) = \frac{\partial g }{\partial u} \frac{\partial u }{\partial x}
$$

But I want such a a relationship for cross partials.

(note, both the LHS and RHS of (1) both are $4x$ for the example I have given)

Note: I am using $\frac{\partial}{\partial y}\frac{\partial}{\partial x}$ to mean first take the partial w.r.t to $x$ and then the partial of that partial w.r.t $y$. I may have the order backwards; I always forget what is the correct order to write the partials in.

Body: Sincce $f_x=g^\prime(u)u_x$, $f_{xy}=g^{\prime\prime}(u)u_xu_y+g^\prime(u)u_{xy}$.


问:
Find the Frequency Components of a Time Series Graph
For a periodic (and not so periodic) function, it is always possible to use Fourier series to find out the frequencies contained in the function.

But what about function that cannot be expressed in mathematical terms? For example, this graph (accelerogram):

Is there anyway to apply a sort of Fourier series, in order to find out the frequencies contained in it?

答: What you are after is known in statistics as spectral analysis. There is a whole literature on extracting frequency components from noisy data.  See Shumway and Stoffer (2006, chapter 4) for example.</p>

ContentLicense: CC BY-SA 2.5


问:
Number of common elements in $A \times B $ and $B \times A$
This question was part of maths worksheet of my brother and I was unable to solve it. He studies in High Shcool.

So, I am asking for help here.

Let there are 5 common elements between A and B set. Then what is the number of common elements in $A \times B $ and $B \times A$ ?

Although this question seems to be of basic Sets and relation but I am clueless on how to approach this despite being an bachelor student of mathematics myself .

So, CAn you pleaese help me on how should I aproach this question .

Thanks!!

答: Hint Show that $(x,y) \in A \times B \cap B \times A$ if and only if $x,y \in A \cap B$.


问:
Show that a 2x2 matrix A is symmetric positive definite if and only if A is symmetric, trace(A) > 0 and det(A) > 0
I need to show two parts of the implication are true.
First: if $A$ is $2\times 2$ and is symmetric positive definite then $trace(A)>0$ and $\det(A)>0$.

Second: if $trace(A)>0$ and $\det(A)>0$ then $A$ is symmetric positive definite.

For the first part I was thinking: $A$ is symmetric and positive definite then $A$ has its eigenvalues positive. If $A$ is $2\times 2$  then characteristic polynomial of $A$ is
 $x^2−x.trace(A)+\det(A)=0$  If we compute the discriminant we get $(tr(A))^2 ≥ 4.\det(A)$
Now $tr(A)$ is squared so it is positive. How do I know that $\det(A)$ is also positive?

After copper.hat's response we argue that the eigenvalues of A are all positive because $A$ is spsd and the $\det(A)$ is the product of its eigenvalues. Thus $\det(A)$ is strictly positive.

Now, how do I verify the second part?

Thanks.

答:
What isn't quite proved here is 2nd part in number 3 of Jesus RS' answer, I think?

Anyhoo:

Proof:

Real symmetric is orthogonally diagonalisable. (I think we use just diagonalisable._ Let $A=QDQ^T$.

We want to show $x^TAx=v^TDv > 0$ if $v=Q^Tx \ne 0$ (iff $x \ne0$ since $Q^T$ is invertible (since $Q^T$ is orthogonal. again, I think just diagonalisable is fine))'.

For $D=diag[d_1,d_2]$, we have $v^TDv=d_1v_1^2+d_2v_2^2$.

We have $0 < tr(A)=tr(D) = d_1 + d_2$ and $0 < \det(A)=\det(D) = d_1 d_2$, i.e. $d_1,d_2>0$.

By (3) and (4), we have $v^TDv > 0$ if $v \ne 0$. QED


问:
The matrix function $\frac{e^x - 1}{x}$ for non-invertible matrices
I am trying to evaluate the matrix function
$$
f(X) = \frac{e^X - I}{X}, \tag{1}
$$
where $e^X$ is the usual matrix exponential, defined by
$$
e^X = \sum_{n=0}^{\infty} \frac{1}{n!} X^n.
$$
The problem is that all matrices I am working with are singular (if it is of relevance, $X \in \mathfrak{so}(3)$). According to this Wikipedia article, one may consider the function $f$ to be a real function, find its Maclaurin series, and use this to evaluate the function for a matrix. This seems reasonable enough, and it also seems to be how the matrix exponential is defined. The Maclaurin series for $(1)$ is given by
$$
f(x) = \sum_{n=0}^{\infty} \frac{X^n}{(n+1)!} = I + \frac{1}{2!}X + \frac{1}{3!}X^2 + \ldots,
$$
indicating that the invertibility of $X$ shouldn't matter when it comes to finding a meaningful value for $(1)$.

Now for my question:
Is there a closed form of $(1)$ that do not involve the inverse? To be clear, my goal is to implement the function in Python, and I would rather do it without computing a large partial sum and use that as the approximation. I know a closed form of the matrix exponential for matrices in $\mathfrak{so}(3)$.

答:
$\def\mymatrix#1{ \left[\begin{array}{c}#1\end{array}\right] }$
Nick Higham has a recommendation for calculating such matrix functions.

Write a simple function, applicable to non-singular matrices.

For example, here is some Julia code

```<code>function phi(X)
   return (exp(X)-I) / X
end
</code>```

For every singular matrix there is a nearby non-singular matrix. So evaluate the simple function by perturbing the singular matrix with a random matrix $(X+R)\,$ where
$\,\|R\|\approx\sqrt{\varepsilon}\;$ where $\varepsilon$ is the machine epsilon. The function itself will be accurate to $\sqrt{\varepsilon}\approx10^{-8}$.

So for singular matrices, call the function like so

```<code>ε,X = 1e-8, rand(4,2)*rand(2,4)  # create a singular matrix
P = phi(X + ε*randn(4,4))        # evaluate the function
norm(I+X*P - exp(X))             # validate the function
</code>```

If this simple method isn't good enough, there are rational (Pade) approximations for the exponential and phi-functions. Here is a fairly comprehensive paper.

Another suggestion is to evaluate the function using quad precision and a perturbation $\|R\|\approx10^{-16}.\;$ Then the function will be accurate to double precision (for singular matrix arguments).

Update

Another approach is to calculate the exponential of a block-triangular matrix
$$\eqalign{
\exp\left(\mymatrix{X&I\\0&0}\right)
 &= \mymatrix{e^X&\frac{e^X-e^0}{X}\\0&e^0}
 = \mymatrix{e^X&\frac{e^X-I}{X}\\0&I}
}$$
then extract the upper-right block. I'm not sure of the numerical accuracy of this method, but it should be as accurate as the underlying $\,\exp\,$ function. Note that this result holds for any analytic function, not just the exponential, i.e.
$$\eqalign{
f\left(\mymatrix{X&I\\0&0}\right)
 &= \mymatrix{f(X)&\frac{f(X)-f(0)}{X}\\0&f(0)}
}$$

Body:
$\def\mymatrix#1{ \left[\begin{array}{c}#1\end{array}\right] }$
Nick Higham has a recommendation for calculating such matrix functions.

Write a simple function, applicable to non-singular matrices.

For example, here is some Julia code

function phi(X)
   return (exp(X)-I) / X
end

For every singular matrix there is a nearby non-singular matrix. So evaluate the simple function by perturbing the singular matrix with a random matrix $(X+R)\,$ where
$\,\|R\|\approx\sqrt{\varepsilon}\;$ where $\varepsilon$ is the machine epsilon. The function itself will be accurate to $\sqrt{\varepsilon}\approx10^{-8}$.

So for singular matrices, call the function like so

ε,X = 1e-8, rand(4,2)*rand(2,4)  # create a singular matrix
P = phi(X + ε*randn(4,4))        # evaluate the function
norm(I+X*P - exp(X))             # validate the function

If this simple method isn't good enough, there are rational (Pade) approximations for the exponential and phi-functions. Here is a fairly comprehensive paper.

Another suggestion is to evaluate the function using quad precision and a perturbation $\|R\|\approx10^{-16}.\;$ Then the function will be accurate to double precision (for singular matrix arguments).

Update

Another approach is to calculate the exponential of a block-triangular matrix
$$\eqalign{
\exp\left(\mymatrix{X&I\\0&0}\right)
 &= \mymatrix{e^X&\frac{e^X-e^0}{X}\\0&e^0}
 = \mymatrix{e^X&\frac{e^X-I}{X}\\0&I}
}$$
then extract the upper-right block. I'm not sure of the numerical accuracy of this method, but it should be as accurate as the underlying $\,\exp\,$ function. Note that this result holds for any analytic function, not just the exponential, i.e.
$$\eqalign{
f\left(\mymatrix{X&I\\0&0}\right)
 &= \mymatrix{f(X)&\frac{f(X)-f(0)}{X}\\0&f(0)}
}$$


问:
Complexification of realification: Unnecessary computation of eigenvalues?
I started studying the book of Daniel Huybrechts, Complex Geometry An Introduction. I tried studying backwards as much as possible, but I have been stuck on the concepts of almost complex structures and complexification. I have studied several books and articles on the matter including ones by Keith Conrad, Jordan Bell, Gregory W. Moore, Steven Roman, Suetin, Kostrikin and Mainin, Gauthier.

I have several questions on the concepts of almost complex structures and complexification. Here is one:

Question: Below, I describe what I understand is happening in Suetin, Kostrikin and Mainin (12.13 of Part I), where the authors prove for any $\mathbb C$-vector space $L$, $L_{\mathbb R}^{\mathbb C} \cong L \bigoplus \overline L$ (I also asked about this isomorphism here.) Also, I base my understanding on Daniel Huybrechts, Complex Geometry An Introduction (Chapter 1.2). Is my understanding, given in the two parts as follows, correct?

Part 0. Assumptions, definitions, notations:

Complex structure on map definition : See here.

On $L_{\mathbb R}^2$, we have almost complex structure $J(l,m):=(-m,l)$. $J$ is canonical in the sense that we define complexification $(L_{\mathbb R})^{\mathbb C}$ of $L_{\mathbb R}$ with respect to $J$: $(L_{\mathbb R})^{\mathbb C} := (L_{\mathbb R}^2,J)$. Similarly, we have complexification $f^{\mathbb C}$ of $f \in End_{\mathbb R}(L_{\mathbb R})$ defined with respect to the canonical $J$: $f^{\mathbb C} := (f \oplus f)^J$.

2.1. For every $f \in End_{\mathbb R}(L_{\mathbb R})$, $f \oplus f$ commutes with $J$  i.e. $f^{\mathbb C} := (f \oplus f)^J$ is $\mathbb C$-linear.

Let $i^{\sharp}$ be the unique almost complex structure on $L_{\mathbb R}$ such that $L=(L_{\mathbb R},i^{\sharp})$.

Let $\hat i := i^{\sharp} \oplus i^{\sharp}$ such that $(\hat i)^J = (i^{\sharp})^{\mathbb C}$. Then $\hat i$ is another almost complex structure on $L_{\mathbb R}^2$.

4.1. By (2.1), $\hat i$ and $J$ commute, i.e. both $(\hat i)^J$ and $J^{\hat i}$ are $\mathbb C$-linear. (See here for related question.)

Part I of my understanding:

The authors compute the eigenvalues of $J^{\hat i}$ and not $(\hat i)^J$. Then, they compute the corresponding eigenspaces.

We know 'eigenspaces are subspaces', so if someone were to ask

Why do the authors have to explain why $L^{1,0}$ and $L^{0,1}$ are $\mathbb C$-subspaces of $(L_{\mathbb R})^{\mathbb C}$?,

then the answer would be:

We know $L^{1,0}$ and $L^{0,1}$ are $\mathbb C$-subspaces of $(L_{\mathbb R}^2,\hat i)$ by "eigenspaces are subspaces", but we want to also show that $L^{1,0}$ and $L^{0,1}$ are $\mathbb C$-subspaces of $(L_{\mathbb R}^2,J) = (L_{\mathbb R})^{\mathbb C}$.

Part II of my understanding: It is unnecessary for the authors to compute the eigenvalues of $J^{\hat i}$ and then show that the eigenspaces are $\mathbb C$-subspaces of $(L_{\mathbb R}^2,J) = (L_{\mathbb R})^{\mathbb C}$.

The eigenvalues of $J^{\hat i}$ are the same as the eigenvalues of $(\hat i)^J$. The corresponding eigenspaces also have the same underlying sets. (See here for related question.)

By (7), the authors could have directly computed eigenvalues $(\hat i)^J$ and corresponding eigenspaces. Then, there is no need to explain why said eigenspaces would be $\mathbb C$-subspaces of $(L_{\mathbb R}^2,J) = (L_{\mathbb R})^{\mathbb C}$.

8.1. Note: An $\mathbb R$-vector space $A$ has an almost complex structure $H$ if and only if $A=B_{\mathbb R}$ for some non-unique $\mathbb C$-vector space $B$, such as $B=(A,H)$.

8.2. By (8.1), I believe (8) is precisely what Huybrechts (Chapter 1.2) does except Huybrechts uses arbitrary almost complex structure '$I$' on '$V$' instead of specifically what would be $i^{\sharp}(v): = iv$ on $V$, viewed as the realification of some $\mathbb C$-vector space $L$, i.e. viewed as $V = L_{\mathbb R}$. Furthermore, the 'i', '$I$' and '$I^{\mathbb C}$' of Huybrechts correspond, respectively, to the $J$, $i^{\sharp}$ and $(i^{\sharp})^{\mathbb C}$ of Suetin, Kostrikin and Mainin.

答:
Just so this has an answer:

Yes, it is correct.


问:
Connes cyclic set - construction explaination.
This is from a paper by Scholze and Nikolaus, THH, p145, second equation.

The author defines two categories.

$\Lambda_\infty$ a fullsubcaregory of Posets with $\Bbb Z$ action: Its objects are  $\frac{1}{n}\Bbb Z$, with its natural ordering.

$$\Lambda_p := \Lambda_\infty/ Bp\Bbb Z$$
They said a description of this category is given by:

i)  objects are the same as $\Lambda_\infty$

ii) the morphism space are quotient out by the relation.
$$ f \sim \sigma^pf$$
where $\sigma$ is a generator of the action of $\Bbb Z$ on morphism space. Explicitly $\sigma^pf=f+p$.

I get the latter description. However I don't understand what this quotient $\Lambda_\infty /Bp\Bbb Z$ means:

a) where is this taking placing?

b) how is this computation done?

Some elaborationi wuold be nice!

答:
Since you already know what the answer should be, it's maybe not so important where the operation is taking place.  But here is one possible context that makes sense.

As explained in the paper, there is an action of $B\mathbb{Z}$ on $\Lambda_\infty$, i.e., we have a <s>functor</s> $\alpha: B\mathbb{Z} \times \Lambda_\infty \to \Lambda_\infty$, which restricts to an action $\alpha: B(p\mathbb{Z}) \times \Lambda_\infty \to \Lambda_\infty$.  The notation $\Lambda_\infty / B(p\mathbb{Z})$ simply means to take the quotient of $\Lambda_\infty$ by this action, which can be modeled for example as the coequalizer of $$B(p\mathbb{Z}) \times \Lambda_\infty \overset{\alpha}{\underset{\mathrm{pr}}{\rightrightarrows}} \Lambda_\infty$$ in the category of (small) categories, where $\mathrm{pr}$ is the projection onto the second factor.

There's not really a "computation" to do here - everything is just a matter of unpacking definitions.  Since $\alpha$ preserves objects in $\Lambda_\infty$, the coequalizer can just be computed as a quotient category of $\Lambda_\infty$ by the congruence generated by identifying $\operatorname{pr}(p\sigma, f) = f$ with $\alpha(p\sigma, f) = f + p$.  In other words the objects $\Lambda_\infty / B(p\mathbb{Z})$ are the same as $\Lambda_\infty$, and the morphisms are equivalence classes of morphisms in $\Lambda_\infty$ under the relation $f \sim f + p$.

EDIT: What I wrote above is slightly off: $\alpha$ doesn't seem to actually be a functor.  There may be some way to salvage this, but I don't see an easy way right now, so I'm offering another perspective that hopefully has fewer mistakes.

We think of the action of $B\mathbb{Z}$ (really $\mathbb{Z}$) on $\Lambda_\infty$ as a functor $$F: B\mathbb{Z} \to \mathsf{Cat}$$ that sends the object $*$ of $B\mathbb{Z}$ to $\Lambda_\infty$, and a generator $\sigma$ of $\mathbb{Z} = \operatorname{mor}(B\mathbb{Z})$ to the functor $\sigma: \Lambda_\infty \to \Lambda_\infty$ that is the identity on objects and $f \mapsto f + 1$ on morphisms. Then, we set $\Lambda_\infty / B(p\mathbb{Z})$ to be the colimit of the composite $B(p\mathbb{Z}) \to B\mathbb{Z} \xrightarrow{F} \mathsf{Cat}$.


问:
Bound on the shortest non-zero vector in any full rank n-dimensional lattice $\Lambda \subseteq \mathbb{R}^n$ with respect to the $1$-norm.
How can i prove
$$\lambda_1 \; \leq \; (n! \; det(\Lambda))^{\frac{1}{n}} \approx \frac{n \; det{(\Lambda)}^\frac{1}{n}}{e}.$$

Here $\Lambda_1$ is shortest non-zero vector. My initial thought was using Minkowski theorem (choosing $S =$ n-Ball of radius $\sqrt n \frac{\lambda_1}{n}$) and proof by contradiction (assuming $\lambda_1 \; > \; (n! \; det(\Lambda))^{\frac{1}{n}}$ and contracting with minimality of $\lambda_1$).

[Minkowski’s convex body theorem] : Let $\Lambda$ be a full dimensional lattice.  If $S \subset \mathbb{R}^n$ is a symmetric convex body of volume $vol(S) > 2^n det(\Lambda)$ ,  then $S$ contains a  non-zero lattice point.

$1-$norm for a vector $x$ : $\sum{}{}{|x_i|}$.

Body:
$$C_n=Vol(\{ x, \|x\|_1\le 1\})= \int_{-1}^1 Vol(\{ y, \|y\|_1\le 1-x_1\})dx_1$$ $$=\int_{-1}^1 (1-|x_1|)^{n-1}C_{n-1}dx_1=\frac{2}n C_{n-1}=\frac{2^n}{n!}$$
For all $r<\lambda_1/2$ then $ \{x, \|x\|_1\le r\}$ doesn't intersect its $\Lambda$-translates. Thus $$\det(\Lambda)\ge Vol(\{ x, \|x\|_1\le r\})=r^n\frac{2^n}{n!}$$ and hence $$\lambda_1^n\le  n! \det(\Lambda)$$


问:
Number of reflexive relations on the set {1,2,...,n}
I am solving questions of a bachelors exam and was unable to solve this question and i am looking for help!!

Find Number of reflexive relations on the set {1,2,...,n}.

I know that R is called reflexive if for every a $\in $R a R a. But when I used it here 1 got that there would be only 1 reflexive relation ie each element goes to itself but that's wrong according to answers.

I don't have any idea on how too approach the question .

Waiting for your reply!

答: But you can have whatever relation you want for $aRb$ with $a\neq b.$ Show that there are $2\binom{n}{2}$ of them. And either they are there or not. Giving a total of $2^{2\binom{n}{2}}.$


答:
As the set $S=\{1,2,\ldots, n\}$ has $n$ elements, the set of all pairs of elements in $S$ (i.e. $S\times S$) has $n^2$ elements.

Out of those, $n$ pairs are of the form $(x,x)$ for $x\in S$. Those pairs must belong to a reflexive relation $R$. As for the remaining $n^2-n$ pairs, each of them may or may not belong to the relation $R$, so the number of potential candidates for $R$ is $2^{n^2-n}$.

Body:
I am solving questions of a bachelors exam and was unable to solve this question and i am looking for help!!

Find Number of reflexive relations on the set {1,2,...,n}.

I know that R is called reflexive if for every a $\in $R a R a. But when I used it here 1 got that there would be only 1 reflexive relation ie each element goes to itself but that's wrong according to answers.

I don't have any idea on how too approach the question .

Waiting for your reply!


问:
Holomorphic series with its real part positive $f(z)=1+\sum_{n=1}^\infty a_n z^n$
Let $$f(z)=1+\sum_{n=1}^\infty a_n z^n,$$

$f \in H(B(0,1))$, and $\operatorname{Re} f(z)\ge 0$,
$\forall z \in B(0,1) $.

Prove:

(1) $| a_n | \le2$;

(2) $|a_1^2-a_2| \le 2, |2a_1a_2-a_1^3-a_3|\le2$.

答:
Part (1) follows from the Herglotz representation theorem.

For part (2):

$$ \frac{1}{f(z)} = 1 - a_1 z + (a_1^2-a_2) z^2 - (a_1^3 - 2 a_1a_2 + a_3) z^3 +  \dotsc
$$


问:
All forms of integers
I had a question, just a beginner so are all integers represented by the forms $$4n \\ 4n+1 \\ 4n+2 \\ 4n+3$$

Thanks and regards
Sorry for this question, if it seems too easy...

答:
Yes. When you divide any integer by $4$, your remainder will be one of $0,1,2,3$.

Potentially you may be curious about dividing negative numbers by $4$ (e.g. what is $-23\div 4$ ?), but even there you can take the remainder to be positive. You just need to round your initial number down to the first smaller number divisible by $4$. For example, when dividing $-23$ by $4$, notice $-24$ is divisible by $4$ and so the quotient can be taken to be $-6$ and remainder to be $1$ (rather than quotient $-5$ and remainder $-3$).

Title: All forms of integers


问:
Derivative of a Functional. The Chain Rule? Whats going on here.
Hi suppose I have a functional $I_{x,y}$ acting on continuous paths

$$\{\xi~:~\xi:[0,1]\to \mathbb{R}^n,~\xi(0)=x,~\xi(1)=y\}$$ .

In my case $I_{x,y}[\xi]=\int_0^1 L(\xi(t),\dot{\xi}(t))dt$ where $L(\xi(t),\dot{\xi}(t))\in \mathbb{R}$ is some function.

How can I write

$$ \partial_\epsilon I_{x,y}[\xi+\epsilon \eta] $$

in terms of

$$ \nabla_y I(x,y)\cdot \eta ~~~?~\text{or something like this }. $$

I think this has something to do with directional or functional derivative but im very confused.

答:
The mapping from $\eta$ to the directional derivative that you are asking to evaluate is exactly what we call the functional derivative of $I_{x,y}$. For notation let's say the first argument of $L$ is denoted by $q$ and the second is denoted by $p$ then if you replace $\xi$ by $\xi + \epsilon \eta$ then the argument of $L$ becomes $L(\xi+\epsilon \eta,\dot{\xi}+\epsilon \dot{\eta})$. Thus linearization of $L$ (which is just an ordinary scalar function of two scalar variables) yields $L(\xi,\dot{\xi})+L_q(\xi,\dot{\xi}) \epsilon \eta + L_p(\xi,\dot{\xi}) \epsilon \dot{\eta}$, using the ordinary multivariable chain rule (nothing functional-analytic going on here). Thus the functional derivative of your $I_{x,y}$ is the mapping

$$\xi \mapsto \left ( \eta \mapsto \left ( \int_0^1 L_q(\xi(t),\dot{\xi}(t)) \eta(t) + L_p(\xi(t),\dot{\xi}(t)) \dot{\eta}(t) dt \right ) \right )$$

Now if you are looking for extrema of $I_{x,y}$, as you probably are,  this is not entirely useful because not all $\eta$ are admissible with your boundary condition. Specifically if you want to find a critical point of $I_{x,y}$ subject to the boundary condition, only $\eta$ that vanish on the boundaries are admissible. This can be incorporated into this formalism by integration by parts:

$$\int_0^1 L_p(\xi(t),\dot{\xi}(t)) \dot{\eta}(t) dt = -\int_0^1 \frac{d}{dt} \left [ L_p(\xi(t),\dot{\xi}(t)) \right ] \eta(t) dt$$

where the boundary term has vanished because $\eta(0)=\eta(1)=0$.

Now your functional derivative at a fixed point takes the form $\eta \mapsto \int_0^1 F[\xi(t),\dot{\xi}(t)] \eta(t) dt$. The $\eta$ here are "sufficiently general" that, for this to vanish for all $\eta$, one must have the differential equation

$$\frac{d}{dt} \left [ L_p(\xi(t),\dot{\xi}(t)) \right ] = L_q(\xi(t),\dot{\xi}(t)).$$

This is called the Euler-Lagrange equation. This very last step takes some theoretical machinery to prove, but the rest of the derivation is just ordinary calculus.


问:
In $\Delta ABC, AB = 13, AC = 5, BC = 12.$ Points $M,N$ lie on $AC,BC$ respectively with $CM = CN = 4$.

In $\Delta ABC, AB = 13, AC = 5, BC = 12.$ Points $M,N$ lie on $AC,BC$ respectively with $CM = CN = 4$. Points $J,K$ are on $AB$ such that  $MJ$ and $NK$ are perpendicular to $AB$. Find the area of the pentagon $(CMJKN)$ to the nearest whole number.

What I Tried: Here is a picture :-

Almost everything of what I did is shown in the picture, I am typing the rest.

First of all, $\Delta ABC \sim \Delta AMJ \sim \Delta NBK$.
So we have $(AB = 13 , BC = 12 , AC = 5)$ as well as $(AM = 1 , BN = 8)$ .

Hence :- $$AJ = \frac{5}{13} , MJ = \frac{12}{13}$$ and :-
$$NK = \frac{40}{13} , BK = \frac{96}{13}$$
Also $JK = 13 - \frac{101}{13} = \frac{68}{13}$ .

After joining $MN$ and $MK$ , I find $MN = 4\sqrt{2}$ and $MK = \frac{4\sqrt{298}}{13}$ .

With all these lengths known, I will be able to find $[\Delta MCN] , [\Delta JMK]$ by using normal area formula, and $[\Delta MNK]$ using Heron's Formula and then simply add them up to get my required solution.

The Problem is, finding $[\Delta MNK]$ using Heron's Formula will be a lot difficult for me, because the side lengths are all complicated.

Can anyone give me a solution to this problem which I am facing? Thank You. (Alternate Solutions are also welcome.)

答:
$$[\triangle ABC]=30$$

since by A.A.A similarity theorem $$\triangle ABC \sim \triangle AMJ \sim \triangle NBK$$

and the ratio between area of two similar triangles is equal to the square of the ratio of length of corresponding sides

$$\frac{[\triangle ABC]}{[\triangle AMJ]}=(\frac{13}{1})^2=\frac{30}{[\triangle AMJ]}$$

$$[\triangle AMJ]=\frac{30}{169}$$

$$\frac{[\triangle ABC]}{[\triangle NBK]}=(\frac{13}{8})^2=\frac{30}{[\triangle NBK]}$$

$$[\triangle NBK]=\frac{1920}{169}$$

$$[CMJKN]=[\triangle ABC]-([\triangle AMJ]+[\triangle NBK])$$


答:
Because $\triangle AJM\sim\triangle ACB\sim\triangle KNB$, note that $[AJM] =
 \frac{[ACB]}{169}$ and $[KNB] = \frac{64[ACB]}{169}$. Thus, $[CMJKN] = [ACB] -
[AJM] - [KNB] = \frac{104[ACB]}{169} = \frac{8[ACB]}{13}$. However, $[ACB] = 30$,
so $[CMJKN] = \frac{240}{13}\approx 18.46$, and your answer is $\boxed{18}$.

Body:
In $\Delta ABC, AB = 13, AC = 5, BC = 12.$ Points $M,N$ lie on $AC,BC$ respectively with $CM = CN = 4$. Points $J,K$ are on $AB$ such that  $MJ$ and $NK$ are perpendicular to $AB$. Find the area of the pentagon $(CMJKN)$ to the nearest whole number.

What I Tried: Here is a picture :-

Almost everything of what I did is shown in the picture, I am typing the rest.

First of all, $\Delta ABC \sim \Delta AMJ \sim \Delta NBK$.
So we have $(AB = 13 , BC = 12 , AC = 5)$ as well as $(AM = 1 , BN = 8)$ .

Hence :- $$AJ = \frac{5}{13} , MJ = \frac{12}{13}$$ and :-
$$NK = \frac{40}{13} , BK = \frac{96}{13}$$
Also $JK = 13 - \frac{101}{13} = \frac{68}{13}$ .

After joining $MN$ and $MK$ , I find $MN = 4\sqrt{2}$ and $MK = \frac{4\sqrt{298}}{13}$ .

With all these lengths known, I will be able to find $[\Delta MCN] , [\Delta JMK]$ by using normal area formula, and $[\Delta MNK]$ using Heron's Formula and then simply add them up to get my required solution.

The Problem is, finding $[\Delta MNK]$ using Heron's Formula will be a lot difficult for me, because the side lengths are all complicated.

Can anyone give me a solution to this problem which I am facing? Thank You. (Alternate Solutions are also welcome.)


问:
Order of an element and its inverse in a group
Let order of element $a$ be $m$, then prove that order of $a'$ ( inverse of $a$) is also $m$ in a group.
For proving this I used the fact-
Inverse of inverse element is again the same element.
$(a')'=a$,
then taking the power $-m$ to both sides for m in set of positive integers,
we get
$((a')')^{-m} =a^{-m}$ ,
implies that $(a')^m=(a^m)^{-1}$,
implies that
= $e^{-1}$
= $e$
we get the proof finally. I think the proof is not systematic. Need systematic proof.</p>

答:
Here is a slightly different take.

We have $\langle a \rangle = \{ a^n : n \in \mathbb Z \} = \{ a^{-n} : n \in \mathbb Z \} =  \{ (a^{-1})^n : n \in \mathbb Z \} = \langle a^{-1} \rangle$.

Therefore, $o(a) = |\langle a \rangle| = |\langle a^{-1} \rangle| = o(a^{-1})$.


问:
Finding two numbers when having their sum and product
I have two numbers, their sum is 41 and their product is 238. What are the numbers?

I got during this far in my calculations:

$a+b=41,\quad  ab=238,\quad  238=41-b.$

I appreciate answers or tips to help me complete this. Thanks in advance

答:
Your first two equations are correct. One way to continue is to solve one of them for a variable. For example, you could find
$$
a = 41 - b
$$
and substitute into the other equation to get
$$
(41 - b)b = 238.
$$
The equation in $b$ can be rewritten $b^2 - 41b + 238 = 0$, which has solutions $7$ and $34$ (you can use the quadratic formula to find these).

If we use this information in $a + b = 41$, we also get solutions of $7$ and $34$ for $a$. As far as we are concerned in this problem, there is no difference between the solution $a = 7$ and $b = 34$ and the solution $a = 34$ and $b = 7$. We may simply say: "The two numbers with sum $41$ and product $238$ are $7$ and $34$."


问:
Prove that two random variates must be negatively correlated

Problem statement

Given the probability density
$$ p(s,t|a,b,c) = \cases{Z(a,b,c)^{-1} \exp(-as-bt-cst) &if $s \geq 1$ and $t \geq 1$ \\ 0 &otherwise} \label{1} \tag{1}$$
where
$$ \begin{align}
Z(a,b,c) &\equiv \int \int \exp(-as-bt-cst) \ ds \ dt = \frac{1}{c} \exp\Big(\frac{ab}{c}\Big) \Gamma\Big(0,\frac{(a+c)(b+c)}{c}\Big) \\
\Gamma(0,x) &= \int_x^\infty \exp(-t) \frac{dt}{t}
\end{align} $$
and $(a,b,c)$ are "admissible" such that $Z(a,b,c)$ exists, i.e.,
$$ c > 0, \quad a + c > 0, \quad b + c > 0.$$

Prove (or disprove) that for any such admissible $(a,b,c)$:
$$ \langle st \rangle \leq \langle s \rangle \langle t \rangle. \label{2} \tag{2}$$
In other words, the random variates $(s,t)$ must be negatively correlated.

Comments

Lower bound on $\langle st \rangle$

For all $(s,t)$ in the support of \eqref{1} we have $1 \leq (s,t) \leq st$. So $1 \leq \max(\langle s \rangle, \langle t \rangle) \leq \langle st \rangle$.

The case $c=0$

If $c=0$ the random variates $(s,t)$ are independent and thus uncorrelated, such that \eqref{2} reduces to $\langle st \rangle = \langle s \rangle \langle t \rangle.$

Background

Given the constraints:
$$ 1 \leq (s,t) < L, \quad \langle s \rangle = S, \quad \langle t \rangle = T, \quad \langle st \rangle = U $$
the density \eqref{1} is the maximum entropy distribution with respect to Lebesgue measure for $L \rightarrow \infty$ where $(a,b,c)$ are admissible Lagrange multipliers which solve the simultaneous equations:
$$ -\nabla_{a,b,c} \log Z \equiv (\langle s \rangle, \langle t \rangle, \langle st \rangle) = (S,T,U). \label{3} \tag{3}$$
Thus by construction \eqref{1} is in the exponential family.

I implemented a numerical solver that finds admissible $(a,b,c)$ for given $(S,T,U)$ but it fails badly when $U$ is moderately larger than $ST$: it drives $c \rightarrow 0$ but cannot accurately satisfy \eqref{3}. Further numerical investigation led me to believe (95% sure) that \eqref{3} can only be solved when $U \leq ST$, but after much effort I could not prove this.

I tried several approaches to prove \eqref{2} (e.g. directly attacking \eqref{2}; bounding $\langle st \rangle$ with the moment generating function of \eqref{1}) but the only noteworthy result I could get is the following:

For $c \approx 0$, the solution to \eqref{3} is roughly:
$$ a = \frac{1}{S-1}, \quad b = \frac{1}{T-1}, \quad c = -\frac{U - ST}{S^2(T-1)^2 + T^2(S-1)^2 + (S-1)^2 (T-1)^2}. $$
This can be obtained by a first-order expansion of \eqref{3} at $c = 0$ and solving for $(a,b,c)$.
Since $c > 0$, we must have that $U < ST$, such that \eqref{2} holds.

I'm unsure what the significance of this result in relation to proving \eqref{2} might be.

答:
Short answer

Your conjecture is true. Long answer is follows.

Long answer

I'll start by giving two simple inequalities, one is a special case of an important inequality, one is an inequality I crafted just for my work.

Preliminaries

Theorem 1 (Special case of FKG inequality)

Let $X$ be real random variable, $f$ and $g$ be two real increasing functions such that $f(X),g(X) \in L^2$. Then,
$$ \mathbb{E}( f(X)g(X)) \ge \mathbb{E}(f(X))\mathbb{E}(g(X)) $$
Demonstration 2

We extend our probability space to have another random variable $X_1$ which independent of and identically distributed to $X$.

By the monotonicity of $f,g$, we have:
$$ \mathbb{E}\left[  (f(X)-f(X_1))(g(X)-g(X_1) \right] \ge 0 $$

Our desired inequality is just the our expression after expanding.

Done.

Theorem 3 ( Logdensity and the morphing of distribution functions)

Let $X$ and $Y$ be two real random variables, $h$ and $g$ are their respective pseudo densities (definitions below).
Suppose that :
$$ \frac{h(t)}{h(s)} \ge \frac{g(t)}{g(s)} \forall t\ge s$$

Let $f$ be any increasing function such that $\mathbb{E}( f(X))$ and $\mathbb{E}( f(Y))$  are well defined, then

$$ \mathbb{E}(f(X)) \ge \mathbb{E}(f(Y))$$

Remark 4 the above inequality is taked in the sense that $\frac{a}{b} \ge \frac{c}{d} \leftrightarrow ad-bc \ge 0$ for all $b,d \ge 0$ ) .

Remark 5 all the conditions about $L^2$ about the well-definedness are just there to make all the statements valid, i.e, they are just technical, just forget it. The essence is the inequalities

Remark 6: (Definition of pseudo density) A measurable function $f$ is called a pseudo density if $f$ is nonnegative and $0< \int f < +\infty$. Clearly, by then $\tilde{f}:= \dfrac{f}{|f|_1}$ is a density function.

Demonstration 6

Under our assumption on the monotonicity and  the inequality for logdensity, we have:

$$ \int_{\mathbb{R}^2} \left[ f(s)- f(t)\right] [ h(s)g(t)-h(t)g(s) ]dsdt \ge 0$$
Thus,
$$ \Vert g \Vert_1 \int_{\mathbb{R}} f(s)h(s)ds \ge \Vert h \Vert_1 \int_{\mathbb{R}} f(s)g(s)ds  $$
Done.

Remark 7 FKG's inequality has some interesting generalizations, but they are not within our scope.

Remark 8 The intuition behind those inequalities is that if we assign something big with other big things, something small with other small things, we should have some kind of increment.

Remark 9 If $\log h, \log g$ is well defined and have derivatives, the above inequality condition case be rephrased as  $ \dfrac{\partial \log h }{\partial t} \ge \dfrac{\partial \log g }{\partial t}$

Back to our main question

Main question

Let $h_s := p( s , \cdot |a,b,c)$

If $(a,b,c)$ is an admissble choice, we have:
$$ \dfrac{ \partial^2 \log h_s(t)}{\partial s \partial t} = -c < 0 $$
Thus for all $s_1< s_2$, we have:
$$ \dfrac{\partial \log h_{s_1} }{\partial t} \ge \dfrac{\partial \log h_{s_2} }{\partial t}$$

Besides, we see that $h_s$ is the peusdo density of the conditional law of $T$, given $\{S=s\}$
.
As $f(x):=x$ is an increasing function, we hence imply that :
$$ g(s):= \mathbb{E}\left( T| S=s \right) \text{ is a decreasing function}$$

So, we see:

$$\langle ST \rangle = \mathbb{E}(ST) = \mathbb{E}\left( S \mathbb{E}(T|S) \right)= \mathbb{E}\left( f(S)g(S) \right) \underbrace{\le}_{FKG's} \mathbb{E}(S) \mathbb{E}(g(S)) = \langle S \rangle \langle T \rangle $$
(Note that $f$ is increasing, while $g$ is decreasing)

So the answer is Affirmative.

Comments

So you see the sign of $\dfrac{ \partial^2 \log p}{\partial s \partial t}$ implies the sign of correlation.

The theorem 3 gives some sense of the behaviour of $\mathcal{L}(T|S=s)$ when $s$ varies.


答:
Let $S=\frac 12CA\cdot CB=30=[ABC]$ be the area of $\Delta ABC$. (We use brackets to denote the area of a convex polygon.)

Note the similitudes $\Delta ABC\sim \Delta AMJ\sim\Delta NBK$. We know the proportions for the three hypothenuse values $13, 1,8$. This implies:
$$
\begin{aligned}{}
[CMJKN]
&=[ABC]-[AMJ]-[NBK]\\
&=S-\left(\frac 1{13}\right)^2S-\left(\frac 8{13}\right)^2S
=\frac{13^2-1^2-8^2}{13^2}\cdot S
\\
&=\frac{104}{169}\cdot S\approx 18.4615384615385\dots
\end{aligned}
$$

Title: In $\Delta ABC, AB = 13, AC = 5, BC = 12.$ Points $M,N$ lie on $AC,BC$ respectively with $CM = CN = 4$.


问:
Is $|X|$ always a random variable given that $X$ is one?
Is it true that $|X|$ is always a random variable given that $X$ is a random variable?

答: A random variable is a measurable quantity of our choice. For example, we can say X is a random variable that represents outcome of a dice roll. Similarly we can say Y is a random variable representing if the outcome of a dice roll is even. Its all about what you want to measure. As long as X is measurable Y=|X| is also a random variable measuring some other quantity (in this case absolute value of what X measures).


问:
Rewrite $\int\limits_{x=0}^\infty{2\sqrt{a -\frac{b}{x}}{K_1}({2\sqrt{a-\frac{b}{x}}}){x^{M-1}}\exp({-\frac{x}{c}})dx}$ as non-elementary function?
Is it possible to rewrite this integral $I=\int\limits_{x = 0}^\infty  {2\sqrt {a - \frac{b}{x}} {K_1}\left( {2\sqrt {a - \frac{b}{x}} } \right){x^{M - 1}}\exp \left( { - \frac{x}{c}} \right)dx}$ as non elementary function (For exaple $Ei(x)$, $Li(x)$) ?

${K_1}\left( x \right)$ is the modified Bessel function of the second kind.

$a,b,c$ are positive real number and $M$ is a positive integer

It is ok if someone can help me express the integral $I$ as an infinite series. I have also think of using the expansion
${e^x} = \sum\limits_{k = 0}^\infty  {\frac{{{x^k}}}{{k!}}}$ to deal with the exponential term but I cannot proceed.

For clarification, I am an engineer working in Telecommunications. This integral is something come out of the Probability the signal fall below some decoding threshold.

Thank you for your enthusiasm !

答:
No way to find a closed form.

I found an approximation which uses $K_2$, too
$$\int_0^{\infty } 2 x^{M-1} \sqrt{a-\frac{b}{x}} K_1\left(2 \sqrt{a-\frac{b}{x}}\right)e^{-x/c} dx\approx \\2 c^{M-1} \Gamma (M-1) \left(b K_2\left(2 \sqrt{a}\right)-\frac{K_1\left(2 \sqrt{a}\right) (b-a c (M-1))}{\sqrt{a}}\right)$$

Mathematica code

```<code>Integrate[
  Series[2 E^(-(x/c)) Sqrt[a - b/x] x^(-1 + M)
      BesselK[1, 2 Sqrt[a - b/x]], {x, Infinity, 1}] // Normal, {x, 0,
    Infinity}] // FullSimplify
</code>```

At $x\to\infty$

$$2 x^{M-1} \sqrt{a-\frac{b}{x}} K_1\left(2 \sqrt{a-\frac{b}{x}}\right)e^{-x/c}\sim \\e^{-\frac{x}{c}} x^{M-1} \left(\frac{2 b \left(\sqrt{a} K_2\left(2 \sqrt{a}\right)-K_1\left(2 \sqrt{a}\right)\right)}{\sqrt{a} x}+2 \sqrt{a} K_1\left(2 \sqrt{a}\right)\right)$$

Body:
No way to find a closed form.

I found an approximation which uses $K_2$, too
$$\int_0^{\infty } 2 x^{M-1} \sqrt{a-\frac{b}{x}} K_1\left(2 \sqrt{a-\frac{b}{x}}\right)e^{-x/c} dx\approx \\2 c^{M-1} \Gamma (M-1) \left(b K_2\left(2 \sqrt{a}\right)-\frac{K_1\left(2 \sqrt{a}\right) (b-a c (M-1))}{\sqrt{a}}\right)$$

Mathematica code

Integrate[
  Series[2 E^(-(x/c)) Sqrt[a - b/x] x^(-1 + M)
      BesselK[1, 2 Sqrt[a - b/x]], {x, Infinity, 1}] // Normal, {x, 0,
    Infinity}] // FullSimplify

At $x\to\infty$

$$2 x^{M-1} \sqrt{a-\frac{b}{x}} K_1\left(2 \sqrt{a-\frac{b}{x}}\right)e^{-x/c}\sim \\e^{-\frac{x}{c}} x^{M-1} \left(\frac{2 b \left(\sqrt{a} K_2\left(2 \sqrt{a}\right)-K_1\left(2 \sqrt{a}\right)\right)}{\sqrt{a} x}+2 \sqrt{a} K_1\left(2 \sqrt{a}\right)\right)$$