问:
Substitution of variable in a second order differential equation
I am trying to make a substitution of a new variable into a second order differential equation. However, I am not sure how to do this, I have seen examples online but they do not seem to be applicable to my case. I would like to know the procedure followed in order to have my original equation be written in terms of t instead of r.

Here is the equation:

This is the new variable I would like the equation to be in terms of:

答:
The new variable is just:
$$t=Cr^2$$
Now we have:
$$\dfrac {dR}{dr}=\dfrac {dR}{dt} \dfrac {dt}{dr}=2Cr\dfrac {dR}{dt}$$
$$\implies \dfrac 2r\dfrac {dR}{dr}=4C\dfrac {dR}{dt}$$
Where $C$ is a constant. Do the same for $\dfrac {d^2R}{dr^2}$

For $\dfrac {d^2R}{dr^2}$
$$\dfrac {d^2R}{dr^2}=\dfrac {dR'}{dr}=\dfrac {d}{dr}\left (2Cr\dfrac {dR}{dt}\right)$$
$$\dfrac {d^2R}{dr^2}=2C\left (\dfrac {dR}{dt}\right)+2Cr\dfrac {d}{dr}\left (\dfrac {dR}{dt}\right)$$
$$\dfrac {d^2R}{dr^2}=2C\left (\dfrac {dR}{dt}\right)+2Cr\left (\dfrac {d^2R}{dt^2}\right)\dfrac {dt}{dr}$$
$$\dfrac {d^2R}{dr^2}=2C\left (\dfrac {dR}{dt}\right)+4C^2r^2\left (\dfrac {d^2R}{dt^2}\right)$$
Finally:
$$\dfrac {d^2R}{dr^2}=2C\dfrac {dR}{dt}+4Ct\dfrac {d^2R}{dt^2}$$


问:
Distribution of determinants of $n\times n$ matrices with entries in $\{0,1,\ldots,q-1\}$

Consider the set $M(n,q)$ of $n\times n$ matrices with entries in $\{0,1,\ldots,q-1\}$, where $q$ is a prime power. What can be said about the distribution of the determinant of matrices in $M(n,q)$? (A 'heuristic' statement of the problem: taking $\{0,1,\ldots,q-1\}$ as a basis for $F=\mathbb{Z}_q$, what do the determinants of matrices over $F$ look like if you don't mod out $q$?)

Obviously $|M(n,q)| = q^{n^2}$. Since $|GL_n(\mathbb{F}_q)| = \prod_{k=0}^{n-1} q^n-q^k$, in $\mathbb{F}_q$ we get a clean answer for how many are divisible by $q$: the values are equally distributed (modulo $q$, there are $\frac{1}{q-1}\prod_{k=0}^{n-1} q^n-q^k$ matrices with determinant $j$, $1\le j\le q-1$). But if we do not look mod $q$, as it were, the question becomes substantially more difficult; to be frank, I'm not sure where to start or if there are any clear patterns. Information about the limiting behavior or any upper bounds on the magnitude of the determinant would be welcome as well.

I computed the distributions for several values of $n=2,3$ and $2\le q\le 5$; the plot labels are of the form $\{n,q\}$.

As expected, determinant zero is the most common option and a determinant of $a$ is just as likely as a determinant of $-a$. Past that, I admit I'm a little out of my league, but it seems like an interesting problem.

答:
In the continuous limit of $n$ fixed, $q \to \infty$, $\frac{1}{q^n}\log |\det(M)|$ is asymptotically normal as $n \to \infty$. See Terry Tao's comments on this MO thread. The linked paper of Nguyen--Vu has a nicely readable intro, see particularly around equations (1.6)-(1.7). The intuition is roughly that the determinant is going to be the (signed) hypervolume, which can be computed as an iterated multidimensional "base times height". Taking the logarithm and fuzzing your eyes, this looks like a sum of i.i.d. random variables. The details are involved, of course, and I have not attempted to digest them. Someone with expertise in this approach may be able to "discretize" it quickly. It's likely more appropriate as an MO question than an MSE question.

Edit: Now that I look at it, Nguyen--Vu's actual main Theorem 1.1 covers this discrete case too (even though the linked MO thread was just after the continuous case), and more generally any distribution with exponentially decaying tails. So, we get that for fixed $q$, $\log |\det(M)|$ is asymptotically normal as $n \to \infty$, with explicit convergence rates.

ContentLicense: CC BY-SA 4.0


问:
Using binomial expansion to derive Pascal's rule
$\displaystyle \binom{n}{k}=\binom{n-1}{k} + \binom{n-1}{k-1}$

$\displaystyle \left(1+x\right)^{n} = \left(1+x\right)\left(1+x\right)^{n-1}$

How do I use binomial expansion on the second equations for the right hand side and convert it to the first equation? The left hand side is obvious, but I'm not sure how to do the right hand side. Please give me some hints

thanks

答: Binomial expansion of both sides of $$\left(1+x\right)^{n} = \left(1+x\right)\left(1+x\right)^{n-1}$$ gives $$\sum_{k=0}^n \binom{n}{k} x^k = \left(1+x\right)\sum_{k=0}^{n-1} \binom{n-1}{k} x^k$$ by distributivity on the right hand side we find $$\left(\sum_{k=0}^{n-1} \binom{n-1}{k} x^k \right)+\left(\sum_{k=0}^{n-1} \binom{n-1}{k} x^{k+1} \right) = \left(\sum_{k=0}^{n} \binom{n-1}{k} x^k \right)+\left(\sum_{k=0}^{n} \binom{n-1}{k-1} x^{k}\right)$$ the limits of the summations do not change the sum because $\binom{n-1}{n} = 0$, $\binom{-1}{n} = 0$. Thus we have $$\sum_{k=0}^n \binom{n}{k} x^k = \sum_{k=0}^{n} \left(\binom{n-1}{k} + \binom{n-1}{k-1}\right) x^k$$ and extracting the $x^k$ coefficients from both sides gives the identity $$\displaystyle \binom{n}{k}=\binom{n-1}{k} + \binom{n-1}{k-1}.$$</p>

ContentLicense: CC BY-SA 3.0


问:
"Deformation" of the kernel of a linear map
It is known that the roots of a monic polynomial of fixed degree vary continuously (smoothly?) with its coefficients, at least over $\mathbb{C}$.  My question is whether there is such a result for linear maps.  To be precise:

Let $U \subseteq \mathbb{R}^k$ be an open set and $A: U \to M_{m \times n}(\mathbb{R}) \cong \mathbb{R}^{mn}$ a smooth map.  Suppose that for all $t \in U$, $\ker(A(t))$ is $l$-dimensional.  Fix $t_0 \in U$.  Is there an open set $\tilde{U} \subseteq U$ containing $t_0$ and smooth function $f: \tilde{U} \times \mathbb{R}^l \to \mathbb{R}^n$ such that $f(t,\mathbb{R}^l) = \ker(A(t))$ for all $t \in \tilde{U}$?

I think this is true.  Here is an example.  Let $A(t) = \begin{pmatrix} \cos t & -\sin t \end{pmatrix}$. Then for $t \in (-\frac{\pi}{2},\frac{\pi}{2})$, we have $\ker A(t) = \langle (\tan t, 1) \rangle$, so we can take $f(t,\lambda) = \lambda (\tan t, 1)$.  Intuitively, it seems that if the rank of a linear map is fixed, nothing else too catastrophic can happen to its level sets - similar to how nothing strange happens with the roots of a polynomial except for their number as the coefficients are varied.

Can anyone come up with a counterexample, or provide a proof of this fact?

答:
Yes. Choose $x_1,\ldots,x_{m-l}\in \mathbb R^n$ such that $y_1(t)=A(t)x_1,\ldots,y_{m-l}(t)=A(t)x_{m-l}$ form a basis for $\mathrm{img}(A(t_0))$ when $t=t_0$, and $y_{m-l+1},\ldots,y_{m}\in \mathbb R^m$ which extend $y_1(t_0),\ldots,y_{m-l}(t_0)$ to a basis for $\mathbb R^m$. Then the determinant
$$\det\begin{pmatrix} y_1(t) | \cdots | y_{m-l}(t) | y_{m-l+1} | \cdots | y_m\end{pmatrix}$$
varies smoothly with $t$, so we have some neighborhood $\tilde U\subseteq U$ of $t_0$ on which it is nonzero. Thus on $\tilde U$ the function
$$Q(t)=P\begin{pmatrix} y_1(t) | \cdots | y_{m-l}(t) | y_{m-l+1} | \cdots | y_m\end{pmatrix}^{-1}$$
is smooth, where $P$ denotes projection onto the first $m-l$ coordinates. Note that $Q(t)$ projects $\mathrm{img}(A(t))$ onto $\mathbb R^{m-l}$.

Define $a:\tilde U\times \mathbb R^n\to \mathbb R^{m-l}$ by $a(t,x)=Q(t)A(t)x$. Rephrasing your question, you want to know whether $\cup_{t\in \tilde U} \ker(A(t))=a^{-1}(0)$ is a smooth manifold. To see this, note that any smooth local parametrization of $a^{-1}(0)$ near $\{t_0\}\times \{0\}$ extends to a smooth parametrization for some neighborhood of $\{t_0\}\times \ker (A(t_0))$ by linearity, and so we have some neighborhood $\tilde U'$ of $t_0$ such that $f(t,\mathbb R^l)=\ker(A(t))$ for $t\in \tilde U'$.  By a standard theorem, it suffices to check that $0$ is a regular value of $a$. But $a$ is a surjective linear map for all fixed $t$, so its differential at any point must be surjective, thus all values of $a$ are regular.


问:
Find the order of each of the elements in the direct product of $Z_3 \times Z_6$
I am stuck in this homework question.

Please help me to find the order of each of the elements in the direct product of $Z_3 \times Z_6$.

This is what I got so far.

elements of $Z_3 = \{0,1,2\}$

elements of $Z_6 = \{0,1,2,3,4,5\}$

$$Z_3 \times Z_6= \{(0,0), (0,1), (0,2), (0,3)(0,4),(0,5),(1,0), (1,1), (1,2), (1,3)(1,4),(1,5),(2,0), (2,1), (2,2), (2,3)(2,4),(2,5)\}$$

Please help me to understand and apply  the theorem to answer this questions.

Let (a1, a2, . . . , an) ∈ Qn

i=1 Gi. If ai is of finite order ri in Gi, then the order of (a1, a2, . . . , an)     in Qn i=1 Gi is equal to the least common multiple of all the ri.

答: Hint:  The order of an element $(a,b)\in G\times H$ is $\rm{lcm}(\rm{ord}(a),\rm{ord}(b))$.


答: Here's a sketch. Working in the subset $W$ of matrices $A$ with rank $r$, we get a continuous map $\rho\colon W\to G(r,n)$ (the Grassmannian of $r$-dimensional subspaces), where $\rho(A)=\text{Row}(A)$. Now, let $C\colon G(r,n)\to G(n-r,n)$ be defined by $C(\xi)=\xi^\perp$. Then $C\circ\rho(A)=\ker(A)$ gives us a continuous (in fact, smooth) map $W\to G(n-r,n)$. ($W$ is in fact a smooth submanifold.)</p>

Title: "Deformation" of the kernel of a linear map

Body:
It is known that the roots of a monic polynomial of fixed degree vary continuously (smoothly?) with its coefficients, at least over $\mathbb{C}$.  My question is whether there is such a result for linear maps.  To be precise:

Let $U \subseteq \mathbb{R}^k$ be an open set and $A: U \to M_{m \times n}(\mathbb{R}) \cong \mathbb{R}^{mn}$ a smooth map.  Suppose that for all $t \in U$, $\ker(A(t))$ is $l$-dimensional.  Fix $t_0 \in U$.  Is there an open set $\tilde{U} \subseteq U$ containing $t_0$ and smooth function $f: \tilde{U} \times \mathbb{R}^l \to \mathbb{R}^n$ such that $f(t,\mathbb{R}^l) = \ker(A(t))$ for all $t \in \tilde{U}$?

I think this is true.  Here is an example.  Let $A(t) = \begin{pmatrix} \cos t & -\sin t \end{pmatrix}$. Then for $t \in (-\frac{\pi}{2},\frac{\pi}{2})$, we have $\ker A(t) = \langle (\tan t, 1) \rangle$, so we can take $f(t,\lambda) = \lambda (\tan t, 1)$.  Intuitively, it seems that if the rank of a linear map is fixed, nothing else too catastrophic can happen to its level sets - similar to how nothing strange happens with the roots of a polynomial except for their number as the coefficients are varied.

Can anyone come up with a counterexample, or provide a proof of this fact?


问:
Why does separation of variable gives the general solution to a PDE
I was reading about physics and came across the method of using separation of variables to solve specific PDEs, but I can't figure out why the specific solutions give rise to the general solution (the book didn't give any explanation for all these).

The specific example in the book was the Laplace Equation in $2$ variables:
$$\frac {\partial^2 V}{\partial x^2}+\frac {\partial^2 V}{\partial y^2}=0$$
For the above example, separation of variable is essentially solving for the eigen-vectors of the operator $\frac {\partial^2 }{\partial x^2}$ and $\frac {\partial^2 }{\partial y^2}$, which are Hermitian and commutes with each other. I know that in the finite dimensional case, such operators are simultaneously diagonalizable, then solving for the eigen-vectors will give all the solution, but I'm not sure does this work for infinite dimension. I'm also not sure does this approach works in the general case, for other PDEs that can be solved by separation of variable.

All the other post I find on here are all explaining how or when separation of variable work, instead of why such techniques will give the general solutions.

Another side question is: What kind of classes will cover these topics? The only undergraduate class that seems relevant at my university is Linear Analysis, which doesn't cover this. The graduate PDE sequence have graduate Real Analysis sequence as pre-requisite, which I don't think I'll be able to take soon.

答:
Separation of variables relies on being able to choose an orthogonal coordinate system in which the Laplace operator separates. That is a rather strong restriction. For example, the 3d Laplacian splits in only a couple dozen different orthogonal coordinate systems. And the solid in which you are solving the Laplace equation must be a cube in the curvilinear coordinate system, so that each surface of the solid is described as a rectangle in two variables of the curvilinear coordinate system. Then, under these conditions, the transformed Laplacian permits the use of separation of variables for solving the Laplace equation.

The ODEs that result from separation of variables are Sturm-Liouville eigenvalue problems, which is where Sturm-Liouville theory originated. The Sturm-Liouville problems are easier to analyze than the PDE. One can prove that eigenfunction expansions exist for Sturm-Liouville problems. And that gives you enough to solve the Laplace equation by using the eigenfunction expansions coming from the Sturm-Liouville ODEs. You do not necessarily end up with discrete sum expansions of eigenfunctions. If the domain is infinite in one or more coordinates, or if the Jacobian of the orthogonal transformation to curvilinear coordinates vanishes somewhere on the outer surface or at an interior point, then eigenfunction expansions may involve discrete sums and/or integrals of eigenfunctions in the eigenvalue parameter. The theory is not necessarily simple, but it was worked out well before the general theory of Elliptic PDEs, and it remains important because of being able to find explicit solutions for some rather important cases. The method is validated by proving the completeness of eigenfunction expansions associated with Sturm-Liouville problems.

The general theory of Elliptic PDEs is far more general than that required to deal with the problems where separation of variables applies for the Laplace equation. On the other hand, the general theory is not needed when separation of variables applies. Separation of variables is one of the few ways to obtain general, explicit solutions for specific geometries. Even though there are not many cases where explicit solutions are possible, these cases are useful special cases that help reveal the general nature of elliptic PDEs.


问:
Suppose the random variables $X$ and $Y$ have a joint pdf $f(x,y)=6y$, $0<y<x<1$
Find $Cov(X,Y)$

My work:

I found that $E(X) = .75$ and $E(Y)=.50$
$E(XY) = \int_0^1{\int_0^1} xyf(x,y) dydx  = {\int_0^1}\int_0^16xy^2 dy dx = \int_0^1 2x dx = 1$

so $Cov(X,Y) = E(XY) - E(X)\cdot E(Y) = 1 - .75\cdot.5= \frac58$

But my book says that the correct answer is $\cfrac{1}{40}$

Where did I go wrong? I would appreciate any tips.

Body:
$$
E(XY) = \int_0^1 \int_0^1 6xy^2 \mathbb{1}_{0 < y<x<1}dx dy
= \int_0^1 \int_y^1 6xy^2dx dy= \frac{2}{5}
$$


Title: Suppose the random variables $X$ and $Y$ have a joint pdf $f(x,y)=6y$, $0<y<x<1$

Body:
Find $Cov(X,Y)$

My work:

I found that $E(X) = .75$ and $E(Y)=.50$
$E(XY) = \int_0^1{\int_0^1} xyf(x,y) dydx  = {\int_0^1}\int_0^16xy^2 dy dx = \int_0^1 2x dx = 1$

so $Cov(X,Y) = E(XY) - E(X)\cdot E(Y) = 1 - .75\cdot.5= \frac58$

But my book says that the correct answer is $\cfrac{1}{40}$

Where did I go wrong? I would appreciate any tips.

Body: $EXY=\int_0^{1}\int_0^{x} xy (6y)dydx$. This evaluates to $\frac  2 5$ so $cov(X,Y)=\frac  2 5-(0.75)(0.5)=0.025$


问:
Why is the limit $\lim_{x \to c}{(1+x)^{(1/x)}} = e$ only when $c=0$.
I am wondering if the limit
$$
\lim_{x \to c}{(1+x)^{\frac{1}{x}}}
$$
can be $e$ when $c \neq 0$.

I have tried the following:

When $c$ is $\pm \infty$, the limit is $1$, not $e$.

When $c \in (-1,0)\cup(0,\infty)$, the function $(1+x)^{\frac{1}{x}}$ is continuous, and the limit itself equals to $(1+c)^{\frac{1}{c}}$. We can find that the function is decreasing at such a domain, so the value can never be $e$.

But I don't know what to do when $c \in (-\infty,-1)$. One more thing, I haven't learnt anything about complex analysis, so I want the codomain of the function $(1+x)^{\frac{1}{x}}$ be $\mathbb{R}$.

答:
In order that you can work within the real line you have to assume that $1+x >0$ or $x >-1$.

$e^{x} >1+x$ for all $x \neq 0$. Hence we cannot have $(1+c)^{1/c}=e$ or $e^{c}=1+c$ with $-1<c<0$.

Note: $e^{-t} >1-t$ for $t >0$. This follows from the fact that $e^{-t} -(1-t)$ is an increasing function on $[0,\infty)$ and its value at $0$ is  $0$.


问:
Determine $k$ in PDF.

Let $X$ and $Y$ random variables with PDF given by
\begin{eqnarray*}
f(x,y)=\left\{\begin{aligned}kxy, \quad x\geq 0, y\geq 0, x+y\leq 1\\ 0, \quad \text{elsewhere} \end{aligned} \right.
\end{eqnarray*}
$1.$ Calculate $k$.

$2.$ Calculate the marginals density of $X$ and $Y$.

$3.$ Are $X$ and $Y$ independent random variables?

$4.$ Calculate $$\mathbb{P}(X\geq Y), \quad \mathbb{P}(X\geq 1/2 | X+Y\leq 3/4), \quad \text{and} \quad \mathbb{P}(X^{2}+Y^{2}\leq 1)$$

$5.$ Calcule the joint density function of $U=X+Y$ and $V=X-Y$ with their respective marginals.

My approach:

Since that $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)=1 \implies \int_{0}^{1}\int_{0}^{1-x}kxydydx=1 \implies k=24.$$

By definition, we have that $$f_{X}(x)=\int_{0}^{1-x}24xydy=12x(x-1)^{2}, \quad x\in [0,1]$$and $$f_{Y}(y)=\int_{0}^{1-y}24xydx=12y(y-1)^{2}, \quad y \in [0,1]$$

Since that $$f(x,y)\not=f_{X}(x)f_{Y}(y)$$so $X$ and $Y$ are not independents.

Now, here I'm problem in how can I calculate $\mathbb{P}(X\geq 1/2| X+Y\leq 3/4)$ and $\mathbb{P}(X^{2}+Y^{2}\leq 1)$ Can you help me here?. I could find that $\mathbb{P}(X\geq Y)=1/2$.

Here, I don't know how can I solve this part. I was trying to apply the Jacobian transformation, but I don't know how to use this method well. So, I'm not getting a good solution focus.

答:
I have a question, can you explain a little more why $\mathbb{P}(X^{2}+Y^{2}\leq 1)$

Considering that, by assumption of the exercise

$$\mathbb{P}(X,Y \in A)=1$$

I think that the answer is self evident looking at the following drawing

Title: Determine $k$ in PDF.


问:
Compute least hypothenuse of more than n distinct Pythagorean triples
I'm interested in a variation of https://oeis.org/A006339 (Least hypotenuse of n distinct Pythagorean triangles).

Basically, if A006339 is $u(n)$, I want $v(n) := \min_{n \leq i} u(i)$

In other words, $v(n)$ is increasing and for any $k \leq n$, there are less than $v(n)$ triples with hypothenuse $k$.

Are there tables for that sequence? How can I compute it?

A possible way would be to have lower bounds for A006339 and compute it until the lower bound is big enough. I'm not even sure about how I can compute A006339. There is an easy enumeration algorithm for Pythagorean triples due to a result by Hall (1970) and Roberts (1977) (https://stackoverflow.com/a/8263898/5133167), and it allows us to stop the search as any multiplication by $U$, $A$ or $D$ will increase the value of $c$. Is there a more performant way?

Body:
Too big for a comment. These are the factors of the series elements. Perhaps you can find a pattern.

\begin{align*}
1&=5^0\\
5&=5^1\\
25&=5^2\\
125&=5^3\\
65&=5*13\\
3125&=5^5\\
15625&=5^6\\
325&=5^2*13\\
390625&=5^8\\
1953125&=5^9\\
1625&=5^3*13\\
48828125&=5^{11}\\
4225&=5^2*13^2\\
1105&=5*13*17\\
6103515625&=5^{14}\\
 30517578125&=5^{15}\\
 40625&=5^5*13\\
 21125&=5^3*13^2\\
 3814697265625&=5^{18}\\
 203125&=5^6*13\\
 95367431640625&=5^{20}\\
 476837158203125&=5^{21}\\
 5525&=5^2*13*17\\
11920928955078125&=5^{23}\\
274625&=5^3*13^3\\
\end{align*}

If you go here you can find this series taken to $1438$ terms. If you use WolframAlpha as I did, you can perhaps find a pattern in the elements with factors of $13$ and $17$ and $29$ such as

$$2005628955078125 = 5^{11}×13^2×17^2×29^2$$


问:
Complex number equivalency
I'm a bit confused over the solution to a complex ode: $i\alpha y = \beta y''$

The solution to the characteristic polynomial is $r = \pm \sqrt{i\alpha/\beta}$. Somehow my book is getting the solution: $r = \pm (i+1)\sqrt{\alpha/2\beta}$.

Unfortunately, I haven't had a course in complex numbers, so they are a bit of a weak spot for me. Can anyone explain why the two forms are equivalent and how to get from the former to the latter?

This isn't a homework problem or anything, just one step of many in a derivation and I'm trying to fully understand each part.

答:
Note that $(i+1)^{2} = i^{2} + 2i + 1 = -1 + 1 + 2i = 2i$, so in fact, the two roots you've listed are equal.

Just to make it perfectly explicit:

$$r = \pm(i+1)\sqrt{\frac{\alpha}{2\beta}} = \pm\sqrt{\frac{(i+1)^{2}\alpha}{2\beta}} = \pm\sqrt{\frac{2i\alpha}{2\beta}} = \pm\sqrt{\frac{i\alpha}{\beta}}$$

Title: Complex number equivalency


问:
Proving a piecewise function is discontinuous at a point

Give a $\delta,\epsilon$ proof that the following function,
$f(x) = \begin{cases} x,  & \text{if $x \geq 0$} \\ x+3, & \text{if $x
 < 0$} \end{cases}\quad \quad $

is discontinuous at $x=0$.

Attempt:

If $f(x)$ is not continuous at $x=0$, then $\exists \epsilon>0$ such that $\forall \delta > 0$, $\exists x$ such that $0<|x-0|<\delta$ and $|f(x)-f(0)|=|x+3| \geq \epsilon.$

$\quad$Choose $\epsilon=1$. We must show that for a given $\delta$, then $\exists x$ such that $|x+3|\geq1$.

$\quad \quad$Let $a=x+3,b=-x,a+b=3$. By the Triangle Inequality, we have:

$\begin{align}
\quad \quad \quad \quad|a+b|\leq|a|+|b|&\Rightarrow |3|\leq|x+3|+|-x| \\
&\Rightarrow 3-|-x|\leq|x+3|
\end{align}\\$

$\begin{align}
\quad \quad \text{Let } \delta\leq1 &\Rightarrow 0 < x <1\\
&\Rightarrow 3 < x+3 < 4
\end{align}\\$

$\rule{18cm}{.03cm}$

I am unsure as to how to continue with the proof. I believe that after manipulating our $3<x+3<4$, we can use it to create an inequality such that $\epsilon=1<3...\leq|x+3|.$ Which would lead us to the following result:

If $\delta\leq1$, every $x$ where $|x|<\delta$ has $|x+3|\geq \epsilon=1$.

If $\delta >1$, then $(x \bigr ||x|<1)$ is contained in the set of $x$, where $|x|<\delta$. Thus the set of points where $\delta>1$ contains points where $|x+3|\geq \epsilon=1$.

Therefore $f(x)$ is discontinuous at $x=0$.

答:
I only read the first paragraph of the attempt. It doesn’t make sense: it appears to be a misunderstanding of the definition of limit.

The statement “$f(x)$ is discontinuous at $x = 0$” means that $\lim \limits_{x \to 0} f(x)$ does not exist or does exist but is not equal to $f(0)$. In fact, this limit does not exist, as is shown below.

The proof in this answer is not the simplest, but I tried to write it in a way that explains the nature of limits generally.

Intuitive explanation

In order for $\lim \limits_{x \to 0} f(x)$ to exist, $f(x)$ would have to get close to some value (call it $L$) as $x$ got close to $0$. More specifically, we would need to be able to make $f(x)$ get as close as we like to $L$ by making $x$ close enough to $0$.

For this choice of $f$, it is actually the opposite. No matter how close to $0$ we require $x$ to be, we can always find two values of $x$ where the corresponding values of $f(x)$ are far from each other, and therefore can’t both be close to $L$.

Formal proof

By definition, $\lim \limits_{x \to 0} f(x) = L$ is equivalent to $\forall \epsilon > 0$, $\exists \delta > 0$ such that $|x - 0| < \delta \implies |f(x) - L| < \epsilon$. We just need to show that for some $\epsilon$, there is no suitable $\delta$.

Now consider $\epsilon < \frac{3}{2}$ and any value of $\delta$. Let $\delta' = \min\{\frac{\delta}{2}, \frac{3 - 2\epsilon}{2}\}$, $x_1 = -\delta'$ and $x_2 = \delta'$. Then $|x_1 - 0| < \delta$ and $|x_2 - 0| < \delta$, but $|f(x_2) - f(x_1)| \geq 2\epsilon$, so it is impossible for $|f(x_1) - L| < \epsilon$ and $|f(x_2) - L| < \epsilon$ to both be true.


问:
Describing a two person game with a pay-off matrix where i have to change the order to of utlities to find a different action profile?
I understand what weakly dominated strategies are, I understand action profiles, 3 player payoff matrixes and utilities, but I cannot seem to figure out what exactly is asked.

The utilities should be such that, by applying in two different orders the iterated
elimination of weakly dominated strategies, two different and unique action profiles are returned.

Can someone clarify this for me?

答:
This is getting at the fact that the result of the iterative removal of weakly dominated strategies can depend on the order of removal. Consider the game

\begin{array}{|c|c|}
\hline
& L & R \\
\hline
A & (10, 4) & (5, 4) \\ \hline
B & (0, 1) & (4, 6) \\ \hline
\end{array}

Depending on the order of elimination, the set of strategies that remains after iterative removal of weakly dominated strategies can be $\{A, L\}$ or $\{A, R\}$. Note that both of these are Nash equilibrium's, but by using iterative removal of weakly dominated strategies we lose one of them.

For a 3x3 example this game has four Nash equilibrium  $(A, L), (B, L), (C, L), (B, R)$. By using iterative removal of weakly dominated strategies we can get $(A, L), (B, L), (C, L)$ depending on the order of removal.

\begin{array}{|c|c|c|}
\hline
& L & M & R \\
\hline
A & (0, 2) & (1, 0) & (0, 0) \\ \hline
B & (0, 1) & (0, 0) & (1, 1) \\ \hline
C & (0, 1) & (0, 1) & (0, 1) \\ \hline
\end{array}


问:
Infinite product involving powers of 2
I have the following infinite product:

$2/1 * 3/2 * 5/4 * 9/8 * 17/16 * 33/32 * 65/64...$

What does it converge to?

I can take its $\ln()$ to get

$\ln(2) + \ln(3/2) + \ln(5/4)....$

Which using the quotient difference rule resolves to

$S(n) = \ln(2^n + 1) - n\ln(2)$ for $n = 0$ to $\infty$

At this point I don't know how to evaluate this: so I took its derivative to see if I could recognize anything:

$S'(n) = 2^n * ln(2) / (2^n + 1) - ln(2)$

Which really looks like it converges to 0...

Integrating $S(n)$ hasn't been much to avail unless its because I lack knowledge.

What's this come out to?

答:
Drop the first term for a moment.

$$
\frac{3}{2}\frac{5}{4}\cdots\frac{2^k+1}{2^k}\cdots = \prod_{k=1}^{\infty} (1 + \frac{1}{2^k})
$$

When the product is expanded, what is the coefficient of $\frac{1}{2^n}$? A term comes from any number of distinct factors of the form $\frac1{2^k}$.  And so, each partition

$$
n = k_1 + k_2 + \cdots + k_m
$$

provides a term in the $\frac{1}{2^n}$.  So, the product is:

$$
\sum_{n=0}^{\infty} \frac{q(n)}{2^n}
$$

where $q(n)$ is the number of partitions of $n$ into distinct integers. I don't think this has a closed form.

Title: Infinite product involving powers of 2


问:
Prove that $\frac{(1-x_1)(1-x_2) \cdots (1-x_n)}{x_1 x_2 \cdots x_n} \geq (n-1)^n$ for $x_i > 0$ and $x_1 + \ldots + x_n = 1$
It is the problem 2.3 in Algebraic Inequalities by Sedrakyan & Sendrakyan.

Solution

I can't figure out how to prove the key inequality in the solution:

$$\frac{(1-x_1)(1-x_2)}{x_1x_2} \geq \frac{(1-\frac{1}{n})(1-x_1 - x_2 + \frac{1}{n})}{\frac{1}{n}(x_1 + x_2 - \frac{1}{n})}$$

The problem 1.17 referenced in the solution had the following inequality:

$$\lambda(x_1 + x_2 - \lambda) \geq x_1x_2  \text{        for    } x_1 \leq \lambda \leq x_2$$

Substituting $\frac{1}{n}$ for $\lambda$ we have $$\frac{1}{n}(x_1 + x_2 - \frac{1}{n}) \geq x_1x_2 \iff \frac{1}{x_1x_2} \geq \frac{1}{\frac{1}{n}(x_1 + x_2 - \frac{1}{n})}$$

The denominators are fine but we still don't have the full inequality and I'm not sure how to get to it. I imagine I must be missing something simple because otherwise the solution would spend more time on this inequality.

答:
By AM-GM $$1-x_i=\sum_{k\neq i}x_k\geq(n-1)\sqrt[n-1]{\prod_{k\neq i}x_k}.$$
Thus, $$\prod_{i=1}^n(1-x_i)\geq(n-1)^n\prod_{i=1}^n x_i$$ and we are done!


问:
If $F_1$ and $F_2$ are both independent of $F_3$ and independent of each other, is $\sigma(F_1\cup F_2)$ independent of $F_3$?
Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space and $\mathcal F_i\subseteq\mathcal A$.

Remember that $\mathcal F_1$ and $\mathcal F_2$ are called ($\operatorname P$)-independent if $$\operatorname P[A_1\cap A_2]=\operatorname P[A_1]\operatorname P[A_2]\;\;\;\text{for all }A_i\in\mathcal F_i\tag1.$$ If $\mathcal F_2$ is a $\sigma$-algebra, then $(1)$ is equivalent to $$\operatorname P\left[A_1\mid\mathcal F_2\right]=\operatorname P[A_1]\;\;\;\text{for all }A_1\in\mathcal F_1\tag2.$$

It's trivial to see that, if

$\mathcal F_1$ and $\mathcal F_3$ are independent; and

$\mathcal F_2$ and $\mathcal F_3$ are independent,

then
3. $\mathcal F_1\cup\mathcal F_2$ and $\mathcal F_3$ are independent.

On the other hand, (1.) and (2.) do not imply that

$\sigma(\mathcal F_1\cup\mathcal F_2)$ and $\mathcal F_3$ are independent.

Question: If we assume (1.), (2.) and additionally

$\mathcal F_1$ and $\mathcal F_2$ are independent,

can we then conclude (4.)? (Maybe, if necessary, assuming that $\mathcal F_i$ is a $\sigma$-algebra).

答:
The answer is NO. There exist events $A,B,C$ such that any two of them are independent but they are not jointly independent. In this case $C$ is not indepdent of $A \cap B$ so we can take $\mathcal F_1=\sigma (A),\mathcal F_2=\sigma (B), \mathcal F_3=\sigma (C)$ for  a counter-example.

In two independent tosses of a  fair coin let $A$ be the event that the first toss results in Heads, $B$ the event that the second one results in Heads and $C$ the even that the outcomes are both heads or both tails. Then $A,B,C$ are pairwise independent but not jointly independent.


问:
Local connectedness of topologist's comb
I'm using Croom's textbook, and came with the following definition of local connectedness:

A space $X$ is locally connected at a point $p$ in $X$ if every open set containing $p$ contains a connected open set which contains $p$.

And the author gives the topologist's comb (https://en.wikipedia.org/wiki/Comb_space) $C$ as an example of a connected but not locally connected space. The reason is that each open ball centered at $p:=(0,t)$ with $0<t\le 1$ contains infinitely many open intervals.

I'm not really sure if this is the case, because the set $B(p,r)\cap C$ contains the interval $(\{0\}\times(t-r,t+r))\cap C$ which is open in $C$, contains $p$ and is connected in $C$. Hence $C$ might be locally connected at $p$. So why is $C$ not locally connected?

答: But $\{0\}\times(t-r,t+r)\cap C$ is not open in $C$: there is no open set in $\Bbb R^2$ whose intersection with $C$ is $\{0\}\times(t-r,t+r)\cap C$.


问:
Stronger than Nesbitt's inequality using convexity and functions
Hi it's a refinement of Nesbitt's inequality  and for that, we introduce the function :

$$f(x)=\frac{x}{a+b}+\frac{b}{x+a}+\frac{a}{b+x}$$

With $a,b,x>0$

Due to homogeneity we assume $a+b=1$ and we introduce the function :

$$g(a)=\frac{a}{1-a+x}$$

Showing that $g(a)$ is convex on $(0,1)$ is not hard so we have :

$$\frac{b}{x+a}+\frac{a}{b+x}\geq 2\frac{a+b}{2(\frac{a+b}{2}+x)}=h(x)$$

So we have :

$$f(x)\geq h(x)+\frac{x}{a+b}$$

Now we put $u=\frac{x}{a+b}$ and we want to show :

$$h(x)+\frac{x}{a+b}=u+\frac{1}{0.5+u}\geq \frac{3}{2}$$

The last inequality is obvious.

My question :

It is correct?

Do you know other refinements?

Thanks in advance!

Ps: I add the tag reference request for the last question.

答:
Stronger than Nesbitts inequality for non-cyclic style

If $a,\,b,\,c$ are positive real numbers, then
$$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$$

If $a,\,b,\,c$ are positive real numbers, then
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}+\frac{27(b-c)^{2}}{16(a+b+c)^{2}}.$$

Let $a,\,b,\,c$ be nonnegative real numbers, no two of which are zero. Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}+\frac{(b-c)^{2}}{2(b+c)^{2}}.$$

Let $a,b,c$ are non-negative real numbers such that $ab+bc+ca>0.$ Prove that
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \geqslant \dfrac{3}{2}+\dfrac{7(b-c)^2}{16a(b+c)+7bc}.$$


问:
Irreducible factors of $x^q-1$ over $\mathbb{Z}_p$
The polynomial $x^3-1$ over $\mathbb{Z}_p$, where $p$ is a prime, factors as $(x-1)(x^2+x+1)$.

The polynomial $x^7-1$ over $\mathbb{Z}_{13}$ factors as, $(x-1)(x^2+3x+1)(x^2+5x+1)(x^2+6x+1)$.

All these $(x-1),(x^2+x+1),(x^2+3x+1),(x^2+5x+1),(x^2+6x+1)$ are irreducible factors.

Does a polynomial $x^q-1$ over $\mathbb{Z}_p$, where $p,q$ are primes ($q<p$), always factorize to $(x-1)$ times several quadratic factors of the form $(x^2+ax+1)$, where $a \in \mathbb{Z}_p$? (a quadratic factor with constant term "1"?)

Thanks a lot in advance.

答:
No. For example, $x^7 - 1 = (x-1)(x^6+x^5+ x^4+x^3+x^2+x+1)$ modulo $19$ as a product of irreducible factors. It is easy to see that $x-1$ is always a factor, but from that point on I think more specific conditions may be required on $p$ and $q$ ($q<p$ seems arbitrary in this respect) for this breakup to occur.

Note : Showing explicitly that the sixth degree polynomial is irreducible requires Galois theory.

We will use the demonstration in the comments for this purpose. We wish to show that $$
x^6+x^5+ x^4+x^3+x^2+x+1
$$

is irreducible modulo $19$. Note that a root of this polynomial can be found in a finite field extension of $\mathbb F_{19}$, which will be of the form $\mathbb F_{19^m}$ for some $m$.

However, any root of this polynomial will also satisfy $x^7 = 1$, therefore we are ask : for which $m$ is there a non-unit $x$ such that $x^7 = 1$ as elements of $\mathbb F_{19^m}$?

If there is such an element $x$, then $7$ must divide $19^m-1$. Conversely, because $7$ is prime, $7$ dividing $19^m-1$ implies the converse by Cauchy's theorem. Therefore, the question now is : which $m$ are such that $7$ divides $19^m-1$?

One can easily check that the set of all such $m$ is of the form $n\mathbb N$ for $n$ being the possible smallest value of $m$. Combining Euler's theorem with a few simple tests, one sees that $m$ can be any multiple of $6$.

So, any root of $x^6+x^5+\ldots+1 = 0$ must then lie in an extension of degree at least $6$. However, because the degree of this polynomial is also at most $6$, it follows that the root lies in an extension of degree $6$, and therefore $x^6+x^5+\ldots+1$ is its minimal polynomial, hence irreducible over $\mathbb F_{19}$.


问:
What is the difference between a random variable and an event?
In layman terms, what is the difference between a random variable and an event? To my understanding a random variable is a function outputting a real number. And an event is an outcome or a set of outcomes.</p>

答:
This may be an instance where the most possible general definition gives more insight. Consider a pair $(\Omega, \Sigma)$ where $\Omega$ is the sample space and $\Sigma \subseteq 2^\Omega$ is a $\sigma$-algebra. The measurable sets (i.e., elements of $\Sigma$) are called events. A random variable is a measurable function $X\colon \Omega \to \Bbb R$. By measurable function we mean that for every open interval $]a,b[\subseteq \Bbb R$, we have that $X^{-1}\big(]a,b[\big) \in \Sigma$ is measurable. To put it simply:

A random variable is a nice function from the sample space to the real line (and a random vector takes values in some $\Bbb R^k$, etc.). An event is a certain subset of the sample space, to which a probability may be assigned.

This does not depend on any choice of probability. More precisely, a probability space is a triple $(\Omega, \Sigma, \Bbb P)$ where $(\Omega, \Sigma)$ is as above and $\Bbb P\colon \Sigma \to \Bbb R_{\geq 0}$ is a measure with $\Bbb P(\Omega) = 1$. Then finding $\Bbb P(X = x)$ means finding the "measure" of the event $\{\omega \in \Omega \mid X(\omega) = x\}\in \Sigma$, inside of the sample space $\Omega$.

Example: consider the experiment of tossing a fair coin $n$ times. The sample space is $\Omega = \{{\rm heads}, {\rm tails}\}^n$. The $\sigma$-algebra of events will be all possible subsets of $\Omega$. Say that we write $0$ for heads and $1$ for tails, and the random variable $X \colon \Omega \to \Bbb R$ gives the result of the second toss. That is: $X\colon \Omega \to \Bbb R$ is given by $X(\omega_1,\ldots, \omega_n)= \omega_2$. The event "getting heads on the second toss" is $X^{-1}(0) = \{  (\omega_1,\ldots, \omega_n) \in \Omega \mid \omega_2 = 0  \}$, and the probability of this happening is $\Bbb P(X = 0) = 1/2$.


问:
probability density function of $c\times y$
I have been given a pdf by $f(y)=2y$ for $0<y<1$. Now I am asked to find the pdf for $c\times y$ where $c$ is some constant $c\in \mathbb{R}_{+}$.

My initial thought process was that $f(c\times y)=c\times f(y)$. But this is wrong, because when I evaluate the integral

$\int_{0}^{1}(c\times f(y))dy=c$ instead of $1$.

So can anyone help me with this?
Thanks in Advance.

答:
Start by finding the CDF of $Z=cY$. Then $Z\in(0,c)$.

$P(Z\le z)=P(cY\le z)=P(Y\le z/c)=F_Y(z/c)$

Now differentiate it with respect to $z$ to obtain PDF,$$f_Z(z)=\frac{d}{dz}[F_Y(z/c)]=\frac1cf_Y(z/c)$$


问:
$T(x)=T(y)$, whenever $|x-y|=1/n$
$\textbf{Question.}$ Let $T:[0.1]\to\mathbb{R}$ be a continuous function such that $T(0)=T(1)$. Then

For each $n\in\mathbb{N}$ show that there exists, $x,y\in[0,1]$ satisfying $|x-y|=\frac{1}{n}$ and $T(x)=T(y)$.

Does the above result hold true if $|x-y|=h_n$ for some $h_n$ which is not of the form $\frac{1}{n}$?

It looks like a basic application of the definition of limit, but I am not really sure, where to start or what to look at.
Would be grateful in case of a detailed answer. Thank you.

答:
Let $n \in \mathbb{N}$.

The question is equivalent to showing that the function $f: [0,1-\frac{1}{n}] \to \mathbb{R}$ defined by $f(x) = T(x+\frac{1}{n}) - T(x)$ has a zero.

Suppose $f$ has no zeroes. Then convince yourself that $f$ must be either positive everywhere or negative everywhere (hint: intermediate value theorem).

Suppose $f$ is positive everywhere. Then
$$0<\sum_{i=0}^{n-1}f\left(\frac{i}{n}\right) = \cdots = -T(0) +T(1) = 0 $$

Title: $T(x)=T(y)$, whenever $|x-y|=1/n$


问:
help negating statements (exclusive or)
I) All flowers are pink or have a stem.

(There is at least one flower which isn't pink and doesn't have a stem?)

II) All flowers are either pink or have a stem (exclusive or)

(isn't logical equivalence the negation of the exclusive or?, so would it then be A flower is pink iff it has a stem?)

I have posted my thoughts regarding the solutions, would be thankful for advice and correction if needed.

答:
"Exclusive or" $(p\oplus q)$ can be written as $(p\land\lnot q)\lor(\lnot p\land q)$, where as "if and only if" $(p\leftrightarrow q)$ can be written as $(p\land q)\lor(\lnot p\land\lnot q)$.

And so, it can be seen that $(p\oplus q)$ is equivalent to $(p\leftrightarrow\lnot q)$.

And so $\neg(p\oplus q)$ is equivalent to $(p\leftrightarrow q)$.

[But, be careful using a biconditional in a existential statement.]

Title: help negating statements (exclusive or)


问:
Find the infimum of the set $\{x+\frac 1x\}$
I am pretty new to the website, so I apologize in advance if I ask my question incorrectly.

I am trying to find the infimum of the Set
$$A=\left\{x+\frac{1}{x} \text{ for all }x \in \mathbb N\right\}.$$

Here's how I started: Using the definition of infimum, there is a $u$ in the set $A$ such that $u<x+\frac{1}{x}$.

From the Archimedean property, $\epsilon>0$ implies $\frac{1}{\epsilon}>0$.

I am not sure how to continue. Any advice?

答: Observe that the sequence $(a)_n = \{n + \frac{1}{n}\}_{n \in \mathbb{N}}$ is strictly increasing, and that $2 = 1 + \frac{1}{1} \in (a)_n$ is the first element in the sequence, thus a lower bound for it, by our previous comment. Indeed, $2$ is the greatest lower bound for $(a)_n -$ if $L$ is a greater lower bound, then $L \not< 2$.


问:
flow generates a vector field
In a smooth manifold M every flow generates a vector field. How can I prove this.

if X is a tangent vector at (0,p) in $\mathbb{R}\times M$ then the differential of the flow map generates a tangent vector at p in M. In this way we can always assign a tangent vector at each point p. But how this assignment is smooth?...or any other way to prove?

答: If $\Phi\colon \Bbb R\times M \to M$ is a flow, we fix $x \in M$ and look at the curve $$\Bbb R \ni t \mapsto \Phi(t,x) \in M.$$For $t = 0$, we see that it starts at $\Phi(0,x) = x$. The velocity of this curve gives us a vector $$X_x= \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \Phi(t,x) \in T_xM.$$So $x \mapsto X_x$ defines $X \in \mathfrak{X}(M)$. It is smooth because $\Phi$ is (more precisely, you're taking derivatives of something you assume smooth, hence the result is smooth as well).

Title: flow generates a vector field


问:
Derivatives or Integrals of Factorials
I am wondering, why is it so hard to take the derivative or integral of a factorial function defined as $f(x) = x!$.

I know that approximations do exist, and it wouldn't be so hard to subject them to derivation or integration. I doubt that the reason is that factorials are defined only for whole numbers...

Is my thought right? What other things make this kind of a derivation seemingly hard (or impossible) to get ?

One thing I did is like this:
$$ (n+1)! - n! = n\times n!$$

This is a commonly known thing.

Now, derivatives are defined as (I am very much of a noob in calculus, even if I know a little bit of it):
$$\lim_{h\to 0}\dfrac{f(x+h) - f(x)}{h}$$
Taking $f(x) = x!$, when we try to take the derivative:
$$\lim_{h \to 0} \dfrac{(x+h)! - x!}h$$

In this case, simple substitution won't work, and since I don't know many other rules well, maybe I won't be able to work it out well.

But if I try to use Lagrange's mean value theorem to see if there is a place where the $f(x)$ here has a minima or a maxima (I only have a small knowledge about this) here, I'd have $f'(c) = \dfrac{a! - b!}{a-b}$  ($a$ and $b$ are distinct) and this can't be zero anywhere, since factorials are defined only for positive integers and $a = b$ means that the $f'(c)$ we get enters indeterminate forms.

Anything more that I can go through to understand why $\dfrac d{dx}{x!}$ is hard to evaluate ?

答:
I am wondering, why is it so hard to take the derivative or integral of a factorial function

I doubt that the reason is that factorials are defined only for whole numbers.

That's exactly the reason. Calculus tools are made to handle functions defined on (most of) the real number line. That's their raison d'être. They can't really handle a function that is defined only on the integers, without either a massive reworking of calculus itself, or redefining the function to have a value everywhere.

Title: Derivatives or Integrals of Factorials


问:
Why must a and p be relatively prime in Fermat's Little Theorem?
A variant of Fermat's Little Theorem states that $a^{p - 1} \equiv 1~mod~p$ if $a$ is not divisible by $p$.

Why is this last condition important? Why must $a$ and $p$ be relatively prime?

答:
If $p$ divides $a$, then $a\equiv 0\bmod p$, so you will have
$$a^{p-1}\equiv 0^{p-1}\equiv 0\not\equiv 1\bmod p.$$
If you prefer a statement of Fermat's little theorem that doesn't have this restriction:
$$a^p\equiv a\bmod p$$
for any integer $a$ (regardless of whether it's relatively prime to $p$).</p>


问:
Show that the rank of a symmetric matrix is the maximum order of a principal sub-matrix which is invertible
The question is this-

Show that the rank of a symmetric matrix is the maximum order of a principal sub-matrix which is invertible.

I can show that there cannot exist a sub-matrix with rank more than the actual rank of the matrix. But I cannot show the other way around, i.e when the rank of the actual matrix is $r$, then there exist a principal sub-matrix of the same rank (though I can prove that there exist a sub-matrix with rank $r$).

I was thinking like this: if the rank of the matrix is $r$, we can find r linearly independent rows of the matrix, say $a_1, a_2, ..., a_r$ -th rows are linearly independent. Then the corresposding columns $a_1^t, a_2^t, ..., a_r^t$ are also linearly independent. But how to show that the submatrix that they produce is of rank r?

I hope my question is clear. Any hints or help would be highly appreciated.

答:
let $A\in \mathbb R^{n\times n}$ be symmetric with rank $r$. Note: there is nothing to do if $r\in\big\{0,n\big\} $ so assume $r\in\big\{1,2,...,,n-1\big\}$.

Arguing by contrapositive:

Suppose that the maximal sized invertible principal submatrix is $r-k\times r-k$ for integer $k\geq 1$.

Then the sum over all $r\times r$ principal minors is zero (because each minor is zero), and so is the sum over all $r+1\times r+1$ principal minors (again each is zero by hypothesis) and so on up to the $n\times n$ principal minor given by $\det\big(A\big)=0$.

Up to a sign, these sums over principal minors give specific coefficients of $A$'s characteristic polynomial -- in particular the last $n-r+1$ coefficients in the characteristic polynomial of $A$, so

$p_A(x)= (x-0)^{n-r+1}\cdot q(x)$.  or

$\text{algebraic multiplicity of zero}\big(p_A\big) \geq n-r+1$.

Finally apply rank-nullity and the fact that real symmetric matrices are diagonalizable so algebraic multiplicities equal geometric multiplicities

$\text{rank}\big(A\big) = n -\dim\ker\big(A\big)= n -\text{algebraic multiplicity of zero}\big(p_A\big)\leq n - (n-r+1) = r-1$

$\implies k \not \geq 1$

and you've already shown  $k\not \leq -1$, thus $k=0$ as desired.


问:
Singularities of complex functions
I am trying to find the type of singularities of
$f(z)=e^{z+\frac{1}{z}}$ at $0,\infty$.
At $z=0$, if we consider the expansion of $f$ around 0, we see
$e^{z+\frac{1}{z}}=1+(z+\frac{1}{z})+ \cdots + \frac{1}{n!}(z+\frac{1}{z})^n+\cdots$
we see that the principal part has infinitely many terms. Hence, I can conclude that $z=0$ is an isolated essential singularity of $f$.
Again, to find the singularity at $\infty$, we have to consider the singularity of $f(\frac{1}{z})$ at 0. But we notice that $f(\frac{1}{z})=f(z)$. Thus,  we conclude that $f$ has an essential singularity at $\infty$ also. Am I doing it right?

Further, they have even asked to calculate the integral of $f$ around the circle $|z|=1$ taken in the anticlockwise sense. Well, if there is essential singularity at 0, then we can't apply the residue theorem. Then,  how can I calculate this integral?
Thank you in advance.

答:
You are right about the type of singularity.

On the other hand,\begin{align}e^{z+1/z}&=e^ze^{1/z}\\&=\left(1+z+\frac12z^2+\frac1{3!}z^3+\cdots\right)\left(1+\frac1z+\frac12\frac1{z^2}+\frac1{3!}\frac1{z^3}+\cdots\right)\end{align}and therefore the coefficient of $\frac1z$ in the Laurent series of $f$ centered at $0$ is$$1+\frac12+\frac1{2\times3!}+\frac1{3\times4!}+\cdots\tag1$$So, the value of the integral that you're after is $2\pi i$ times $(1)$.

Title: Singularities of complex functions


问:
Projection to Intersection of Two Hyperplanes
Let $x \in \mathbb{R}^n$ and $V$ a vector space.

We know that if we have a hyperplane $S = \{x \in V | \langle a \,, x \rangle = b\}$, then the projection of $y$ onto $S$, i.e. $\arg\min_{x\in S} ||x-y||$ is $$z = y-\left(\frac{\langle a \,, x \rangle - b}{||a||^2}\right)a$$

Now consider the intersection of two hyperplanes $S_1 = \{x \in V | \langle a_1 \,, x \rangle = b_1\}$ and $S_2 = \{x \in V | \langle a_2 \,, x \rangle = b_2\}$. How do we generalize the above result to find the explicit form of the projection of $y$ onto $S_1 \cap S_2$, i.e. $\arg\min_{x\in S1\cap S2} ||x-y||$? I know that it's in the form $$y-c_1a_1 - c_2a_2$$

How do we find $c_1$ and $c_2$?

答:
Let $V$ be any finite dimensional real inner product space. Given a (linear) subspace $U\subseteq V$, the set $U^\perp = \{\, y\in V \mid \forall x\in U\colon \langle x,y\rangle =0 \,\}$ is the orthogonal complement of $U$ and $V$ decomposes as the direct sum $V=U\oplus U^\perp$. Any vector $v\in V$ uniquely decomposes as $v=u+w$ with $u\in U$ and $w\in W$. In this situation, $u$ is the orthogonal projection of $v$ onto $U$.

By picking an orthonormal basis $u_1,\dots,u_k,w_1,\dots,w_l$ of $V$ such that the $u_i$ span $U$ and the $w_i$ span $U^\perp$, we obtain the projections as
$$
v = \underbrace{\sum_{i=1}^k \langle v, u_i\rangle u_i}_{\in U} + \underbrace{\sum_{i=1}^l \langle v, w_i\rangle w_i}_{\in W}.
$$

Hence, all you need to obtain orthogonal projection onto $U$ is either an orthonormal basis of $U$ to obtain $u$ directly, or an orthonormal basis of $U^\perp$ to obtain $u$ as $u=v-w$.

Now we consider the case where $U=S_1\cap S_2$ for central hyperplanes $S_i=\{\,x\in V\mid \langle x, a_i\rangle =0\,\}$ with $a_1, a_2$ linearly independent. In this case $U^\perp$ is spanned by $a_1, a_2$ and an orthonormal basis can be obtained from $a_1, a_2$ using the Gram–Schmidt process:
\begin{align*}
w_1 &= \frac{1}{\|a_1\|} a_1, \\
w_2 &= \frac{1}{\|w_2'\|} w_2' \quad\text{where}\quad w_2' = a_2 - \frac{\langle a_2, a_1\rangle}{\langle a_1, a_1\rangle } a_1.
\end{align*}
The orthogonal projection of $v\in V$ onto $U=S_1\cap S_2$ is then given by
$$
u = v-\sum_{i=1}^2 \langle v, w_i\rangle w_i.
$$
When the hyperplanes $S_1$ and $S_2$ are not central as is the case in your question, you can find an arbitrary point $p$ in $S_1\cap S_2$ first and then shift everything so that $p$ becomes the origin. Now the situation is the same as described above.


问:
Given joint pdf $f(x,y,z) = \frac{2}{3}(x+y+z)$, how to find $E(X+Y|Z)$?
As the title says, how do you find $E(X+Y|Z)$ given $f(x,y,z) = \frac{2}{3}(x+y+z)$?

Not sure how to go about approaching this problem. The furthest I got was:

$$
E(X+Y|Z) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x+y)f(x+y|z)dxdy
$$

And now I'm stuck because I'm not sure how to get $f(x+y|z)$.

答: Are you sure $f(x,y,z)$ does not have some bound on its domain? If I integrate $f(x,y,z)$ from $-\infty$ to $\infty$ I definitely don't get 1.


答: If $a$ is divisible by $p$, then $a^{p-1} \equiv 0$ (mod $p)$. </p>

Title: Why must a and p be relatively prime in Fermat's Little Theorem?

Body:
A variant of Fermat's Little Theorem states that $a^{p - 1} \equiv 1~mod~p$ if $a$ is not divisible by $p$.

Why is this last condition important? Why must $a$ and $p$ be relatively prime?


问:
A problema about probability and exponential distribution.

A machine has four identical components, whose duration times in years are independent and identically distributed random variables, with exponential distribution of the parameter $1/4$. Determine the life of the machine, assuming it fails

$1.$ When all components have failed.

$2.$ When one of the components fails.

$3.$ When only one component remains in operation.

In the three cases above, find the probability that the machine will last at least $5$ years.

My approach: Let $Y_{1},Y_{2},Y_{3},Y_{4}$ the indentical components in the machine. Now, by hypothesis we know that $Y_{i} \overset{i.i.d}{\sim}\textbf{Exp}(1/4)$.

We need to know when all components have failed, so taking $U_{4}=\max\{Y_{1},Y_{2},Y_{3},Y_{4}\}$ and since $$f_{U_n}(y)=nf(y)[F(y)]^{n-1}$$ because $Y_{i}\overset{i.i.d}{\sim}\textbf{Exp}(1/4)$ and since that here $n=4$, we have that $$f_{U_n}(y)=4\left(\frac{1}{4}e^{-y/4} \right)(1-e^{-y/4})=e^{-y/4}(1-e^{-y/4})^{3}$$
Note that $F(y)=\int_{-\infty}^{y} \frac{1}{4}e^{-t/4}dt=\int_{0}^{y}\frac{1}{4}e^{-t/4}dt=1-e^{-y/4}$
Also, $$\mathbb{P}(U_{n}\geq 5)=\int_{5}^{\infty}e^{-y/4}(1-e^{-y/4})^{3}dy\approx 0.7408$$.

We need to know when one of the components fails, so taking $U_{1}=\min\{Y_{1},Y_{2},Y_{3},Y_{4}\}$ and since the $Y_{i}$ are i.i.d, so we have that $$f_{U_1}(y)=nf(y)[1-F(y)]^{n-1}$$ and here $n=4$, so we have that $$f_{U_{1}}(y)=e^{-y}, \quad y\geq 0$$ also, $$\mathbb{P}(U_{1}\geq 5)=\int_{5}^{\infty}e^{-y}dy\approx 0.006738$$.

3.But, I don't know how can I solve this part. Can you help me in this part?

答:
Hint: Writing the serving times of components in a variational series

$$
\min\{Y_1, Y_2, Y_3, Y_4\} := Y_{(1)} \le Y_{(2)} \le Y_{(3)} \le Y_{(4)} := \max\{Y_1, Y_2, Y_3, Y_4\}
$$

one notices that $U_3 = Y_{(3)}$ is the quantity you are asked in the question. The latter is the $3$-rd order statistics of $Y_1, \dots, Y_4$.


问:
Suppose $f$ satisfies $m(\{x \in [0,1]:f(x)>y\}) \leq \frac{y}{(\ln y)^2}$ for $y \geq 2$. Show that $f$ is integrable.
Suppose $f:[0,1] \to [0,\infty]$ is measurable and satisfies $$m(\{x \in [0,1]:f(x)>y\}) \leq \frac{1}{y\,(\ln y)^2}$$ for $y \geq 2$. Show that $f$ is integrable.

We want to show that $\int f\, dm<\infty$. Let $E=\{x \in [0,1]:f(x)>y\}$. Then, note that
$$\int f\, dm=\int_{E} f\, dm+\int_{E^c} f\, dm\, .$$

Now, I want to find an upper bound for the integrals. I am stuck as $$\int_{E^c} f\, dm \leq \int_{E^c} y\, dm \leq \, ?$$

答:
By a well known application of Fubini's Theorem we have $\int f(x)dx=\int_0^{\infty} m\{x: \in [0,1]: f(x) >y\}dy$. Hence $$\int f(x)dx$$ $$ \leq \int_0^{2} m\{x: \in [0,1]: f(x) >y\}dy$$ $$+\int_2^{\infty} m\{x\in [0,1]: f(x) >y\}dy$$ $$ \leq 2+ \int_2^{\infty}\frac  1 {y (\ln y)^{2}}dy$$ make the substitution $\ln y=z$ to show that the second term is finite.

The first step is justified as follows:

$$\begin{align}\int_0^{\infty} m \{x: f(x) >y\}dy &= \int_0^{\infty} \int \chi _{\{x: f(x)>y\}} dm(x)dy \\
&=\int_0^{\infty} \int_0^{f(x)} dy dm(x)\\
&=\int f(x)dm(x)\end{align}$$


问:
How many ways you can distribute 8 distinct items in 2 different boxes?
Given that there are $8$ items $m, n, p, q, r, s, t,$ and $u$ will be distributed in box X and box Y, answer the following questions:

a.  How many ways can you distribute the given items if either Box $X$ or Box $Y$ can have no item?

This is how I answered this problem.

Consider the equation $X + Y = 8$ where $X$ and $Y$ are non-negative integers from $0$ to $8$. We're looking for the number of solutions this equation has. Thus, we can use the start and bars method.

In this case, the total number of ways is $_9C_1$.

But accoding to my partner, the answer must $2^8$.

b.  How many ways you can distribute the given items such that there must be at least $1$ item in both Box $X$ and Box $Y$?

In this case, I expressed $X$ as $X' + 1$ and $Y$ as $Y'+ 1$

Hence, $X'+1 + Y'+1 = 8$.
$X'+Y' = 6$
This follows that the number of ways to distribute is $_7C_1$.

But the answer of my partner is $2^8-2$ ways.

I am a bit confused on our answers.
Any comments or suggestions will be much appreciated. Thank you in advance.

答:
a) $2^8$ is correct. Note that we have $2$ choices for a box, and we have to make this choice $8$ times.

b) From part a), we know that there are $2^8$ ways to distribute $8$ unique items among $2$ unique boxes. We see the word “at least”; this implies that we may be able to use complementary counting. That is, $$\text{The number of ways at least one item is in both boxes } = \text{ The number of ways we can place distribute the items without restrictions } - \text{The number of ways a box has all of the items}.$$ This is a common technique used in counting. We see that the number of ways we can have each box have all $8$ items is $2$, since either box $X$ or box $Y$ has it. Indeed, $2^8 - 2$ is the correct answer.


问:
Geometric Progression Question with terms to infinity
I'm having a problem with the question stated below. I stumbled upon it during my revision And I was hoping one of you guys could help me solve it and better yet Understand how to go about it.

**A geometric Progression has the first term a, common ratio r and sum to infinity 6. A second geometric Progression has the first term 2a, common ratio r^2 and sum to infinity 7. What are the values of a and r

答:
so using the formula of sum of infinity...
we will get 2 equation. i) 6 = a/1-r      ii) 7 = 2a/1-r^2

then can do simultaneous equation but i did a much easier way by dividing equation ii) with equation i).....then 'a' will cancel each other and we will get an algebraic expression of 'r'---> r will have 2 values r=5/7 , r=1(ignored coz doesnt make sense XD)
then substitute r to either of the i) or ii) to find a, there you goo. r = 5/7 , a = 12/7 ;D


答: the number of solutions to $X+Y=8$ does not  take into account that they are distinct items. For example, having 3 items in X and 5 items in Y does not count one configuration because $X = \{m,n,p\}, Y = \{q,r,s,t,u\}$ is a different configuration you need to count separately from $X = \{q,n,p\}, Y = \{m,r,s,t,u\}$ (switched $q$ and $m$)

Title: How many ways you can distribute 8 distinct items in 2 different boxes?

Body:
Given that there are $8$ items $m, n, p, q, r, s, t,$ and $u$ will be distributed in box X and box Y, answer the following questions:

a.  How many ways can you distribute the given items if either Box $X$ or Box $Y$ can have no item?

This is how I answered this problem.

Consider the equation $X + Y = 8$ where $X$ and $Y$ are non-negative integers from $0$ to $8$. We're looking for the number of solutions this equation has. Thus, we can use the start and bars method.

In this case, the total number of ways is $_9C_1$.

But accoding to my partner, the answer must $2^8$.

b.  How many ways you can distribute the given items such that there must be at least $1$ item in both Box $X$ and Box $Y$?

In this case, I expressed $X$ as $X' + 1$ and $Y$ as $Y'+ 1$

Hence, $X'+1 + Y'+1 = 8$.
$X'+Y' = 6$
This follows that the number of ways to distribute is $_7C_1$.

But the answer of my partner is $2^8-2$ ways.

I am a bit confused on our answers.
Any comments or suggestions will be much appreciated. Thank you in advance.


问:
Using De Moivre to show $\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$

Use the De Moivre Theorem to show that
$$\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$$

I got this question on my exam today and had no idea how to do it.
There must be a connection to complex numbers since its the topic its connected to. I only learnt how to use the De Moivre theorem with complex numbers in polar form so I was completely lost here.

Sorry if its a dumb question :<.

答: From $$(\cos{\theta}+i\sin{\theta})^6=\cos{6\theta}+i\sin{6\theta},$$ by using binomial theorem $$(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^k b^{n-k},$$ you can get $\cos{6\theta}$ as $Re\{(\cos{\theta}+i\sin{\theta})^6\}$ and $\sin{6\theta}$ as $Im\{(\cos{\theta}+i\sin{\theta})^6\}$. Now, $$\tan{6\theta}=\frac{\sin{6\theta}}{\cos{6\theta}}.$$


答:
By De Moivre's theorem we have
$$\cos(6\theta)+i\sin(6\theta)=(\cos(\theta)+i\sin(\theta))^6$$
$$=-\sin^6(\theta) + \cos^6(\theta) + 6 i \sin(\theta) \cos^5(\theta) - 15 \sin^2(\theta) \cos^4(\theta) - 20 i \sin^3(\theta) \cos^3(\theta) + 15 \sin^4(\theta) \cos^2(\theta) + 6 i \sin^5(\theta) \cos(\theta)$$ by the Binomial expansion. Then comparing the real and imaginary parts we have:

$$\cos(6\theta)=-\sin^6(\theta) + \cos^6(\theta)- 15 \sin^2(\theta) \cos^4(\theta) + 15 \sin^4(\theta) \cos^2(\theta)$$
$$\sin(6\theta)= 6  \sin(\theta) \cos^5(\theta) - 20 \sin^3(\theta) \cos^3(\theta) + 6  \sin^5(\theta) \cos(\theta)$$

So we have $$\tan(5\theta)=\frac{\sin(6\theta)}{\cos(6\theta)}$$
$$=\frac{6  \sin(\theta) \cos^5(\theta) - 20 \sin^3(\theta) \cos^3(\theta) + 6  \sin^5(\theta) \cos(\theta)}{-\sin^6(\theta) + \cos^6(\theta)- 15 \sin^2(\theta) \cos^4(\theta) + 15 \sin^4(\theta) \cos^2(\theta)}$$
$$=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$$
after dividing through by $\cos^6(\theta)$

Title: Using De Moivre to show $\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$

Body:
Use the De Moivre Theorem to show that
$$\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$$

I got this question on my exam today and had no idea how to do it.
There must be a connection to complex numbers since its the topic its connected to. I only learnt how to use the De Moivre theorem with complex numbers in polar form so I was completely lost here.

Sorry if its a dumb question :<.


答:
So you begin by writing

$$
\tan 6\theta \equiv {\sin 6\theta \over \cos 6\theta}
$$

From there you apply De Moivre's Theorem which states:

$$
(\cos \theta +\iota\sin\theta)^n \equiv (\cos n\theta + \iota\sin n\theta)
$$

In this case we have $n = 6$, so:
$$
(\cos \theta +\iota\sin\theta)^6 \equiv (\cos 6\theta + \iota\sin 6\theta)
$$

Expanding the left side with the binomial theorem gives:

$$
\cos6\theta + \iota \sin6\theta = C^6 + 6 i C^5 S - 15 C^4 S^2 - 20 i C^3 S^3 + 15 C^2 S^4 + 6 i C S^5 - S^6
$$

Where $C = \cos\theta$ & $S = \sin\theta$

Separating the real and imaginary parts, we get:

$$
\cos6\theta = C^6 - 15 C^4 S^2 + 15 C^2 S^4 - S^6 \\
\sin6\theta = 6  C^5 S - 20  C^3 S^3 + 6  C S^5
$$

From here

$$
\tan6\theta = {6  C^5 S - 20  C^3 S^3 + 6  C S^5 \over C^6 - 15 C^4 S^2 + 15 C^2 S^4 - S^6}
$$

Dividing both the numerator and the denominator by $\cos^6\theta$ or $C^6$

$$
\tan6\theta = {{[6  C^5 S - 20  C^3 S^3 + 6  C S^5]\over C^6} \over {[C^6 - 15 C^4 S^2 + 15 C^2 S^4 - S^6]\over C^6}}
$$

Which simplifies to:

$$
\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}
$$


问:
Rigorous proof that $\lim_{z \to -a}$ Arg$(z)$ does not converge for $a > 0$
I am required to prove that $\lim_{z \to -a}$ Arg$(z)$ does not converge for $a > 0$. I've broken the question down into showing that $\lim_{z \to -a}$ Arg$(z) = \pi$ for $\Im (z) \geq 0$ and $\lim_{z \to -a}$ Arg$(z)=-\pi$ for $\Im(z) < 0$ (which is pretty obvious).

In order to prove these limits, I will need to use the epsilon-delta definition of a limit. Since I am new to these proofs in complex analysis, I'm unsure where to go after setting it out. For the first limit, $0 < |\text{Arg}(z) - \pi| < \varepsilon$. I am unsure where to go from here. Any guidance is greatly appreciated!

答:
Let $z_1 = (-a + i[0]).$

It is assumed that Arg$(z_1) = \pi.$

In order to try to show that the Arg function is continuous at $z_1$

you would have to show that for any $\epsilon > 0$ there exists a $\delta > 0$

such that $0 < |z - z_1| < \delta \implies |Arg (z) - Arg (z_1)| < \epsilon.$

In order to refute the assertion of continuity of the Arg function at $z_1$, you
have to show the contradiction.

This means that you have to exhibit a specific value for $\epsilon > 0$ such that

no matter how small you make $\delta > 0$ you will always be able to find

a $z$ such that $0 < |z - z_1| < \delta$ and $|Arg (z) - Arg (z_1)| \geq \epsilon.$

Choose $\epsilon = 1.$

For any $\delta > 0,$ choose $z = -a - i[\delta/2]$.

Then $0 < |z - z_1| < \delta$ and $-(\pi/2) > Arg(z) > -\pi.$

This means that $|Arg (z) - Arg (z_1)| \geq \epsilon.$

Therefore, you have just exhibited a specific value for $\epsilon$ such that

no satisfying value for $\delta$ can be found, no matter how small you set $\delta.$

You are done.

Note: showing that the Arg function is not continuous at $z_1$ is equivalent to showing that it is not the case that

the $\lim_{z \to z_1} ~Arg(z) = Arg (z_1).$


答:
Please note that in star and bar method, all stars are alike. For your question, take a case where you put $7$ items in box A and $1$ item in box B. It matters which is that one item in box $B$. There are $8$ ways to do just that whereas star and bar method will count it as $1$.

So the way you have to look at is that you have two independent (box) choices for each item and so for two distinguisable items $2^2$ choices and so on...

In part ($b$), minus $2$ is to take out $2$ cases where one of the boxes is empty - there are only two cases - when all items get into either box $A$ or box $B$.


问:
Isomorphism of direct products
If $G, A, B$ are groups, it is not true in general that $G\times A\cong G\times B \implies A\cong B$. For instance, $G=\mathbb{Z}\times\mathbb{Z}\times\cdots, A=\mathbb{Z}, B=\{1\}$.

What happens when $G,A,B$ are restricted to be finite? Are there finite groups $G,A,B$ such that $G\times A\cong G\times B$, but $A\not\cong B$?

答:
If $G$ satisfies ACC and DCC (ascending and descending chain conditions) on normal subgroups, which applies when $G$ is finite, then $A \cong B$.

Thais follows from the Krull-Schmidt Theorem, which is moderately difficult to prove.

Title: Isomorphism of direct products


问:
How can I approximate $\int_{0}^{\infty}x^{7/6}\ln(\frac{1}{1-e^{-x}})dx$?
My approximation here needs to be valid for large values of $x$ here.
$$\int_{0}^{\infty}x^{7/6}\ln\Bigg(\frac{1}{1-e^{-x}}\Bigg)dx$$
I can't seem to get around this natural log no matter what I do. How can this integral be approximated without going to $0$? I know there is a solution in an integral table if the natural log wasn't there.

Body:
This is not a full answer.

Considering
$$I(a)=\int_{0}^{\infty}x^{(1+a)}\log\Bigg(\frac{1}{1-e^{-x}}\Bigg)\,dx $$ by Taylor
$$I(a)=\zeta (3)+\sum_{n=1}^\infty \frac {a^n}{n!}\int_{0}^{\infty} x \log ^n(x)\log\Bigg(\frac{1}{1-e^{-x}}\Bigg)\,dx $$

The remaining integrals

$$J_n=\int_{0}^{\infty} x \log ^n(x)\log\Bigg(\frac{1}{1-e^{-x}}\Bigg)\,dx$$
do not show closed form but they can be tabulated and, as long as $a$ is small, the summation seems to converge quite fast. Summing up to $p$, for $a=\frac 16$ the results would be
$$\left(
\begin{array}{cc}
p & \text{approximation} \\
 1 & 1.253737667 \\
 2 & 1.268492231 \\
 3 & 1.268514471 \\
 4 & 1.268612760 \\
 5 & 1.268608473 \\
 6 & 1.268609145 \\
 7 & 1.268609093 \\
 8 & 1.268609098 \\
 9 & 1.268609097
\end{array}
\right)$$

Tested for $a=\frac 13$, we have the same kind of convergence
$$\left(
\begin{array}{cc}
p & \text{approximation} \\
 1 & 1.305418432 \\
 2 & 1.364436687 \\
 3 & 1.364614603 \\
 4 & 1.366187225 \\
 5 & 1.366050056 \\
 6 & 1.366093078 \\
 7 & 1.366086316 \\
 8 & 1.366087690 \\
 9 & 1.366087444 \\
 10 & 1.366087490 \\
 11 & 1.366087481 \\
 12 & 1.366087483
\end{array}
\right)$$


问:
Show that orbitally equivalent autonomous systems do have same orbits
According to Wikipedia,

Two dynamical systems on the same state space, defined by $\dot {x}=f(x)$ and $\dot {x}=g(x)$, are said to be orbitally equivalent if there is a positive function, $\mu :X\to \mathbb {R}$, such that $g(x)=\mu (x)f(x)$. Orbitally equivalent system differ only in the time parametrization.

I wonder why do orbitally equivalent have same orbits. Intuitively, the two systems would have the same "direction of velocity" at any given point, but this is not a mathematical proof, right?

I can not found any relating information except this question where lhf implicitly uses this property.

答:
If $x=\xi(t)$ is a solution of $dx/dt=f(x)$, let $T(s)$ be a solution of $dT/ds=\mu(\xi(T(s)))$. Then $t=T(s)$ is a time reparametrization that does the trick for that particular orbit, i.e., $x=\xi(T(s))$ solves $dx/ds=g(x)$:
$$
\frac{dx}{ds}
=\frac{d}{ds} \Bigl( \xi(T(s)) \Bigr)
=\frac{d\xi}{dt}(T(s)) \cdot \frac{dT}{ds}(s)
= f(\xi(T(s))) \cdot \mu(\xi(T(s)))
=g(\xi(T(s)))
=g(x)
.
$$


问:
Prove the following inequality for non-negative Borel measurable functions defined on $\Bbb R.$

Let $f : \Bbb R \longrightarrow [0,\infty)$ be a Borel measurable function. Show that $$\displaystyle{\sum\limits_{n = 1}^{\infty} \text {m} \left (\left \{f \gt n \right \} \right ) \leq \int f\ \text {dm} \leq \sum\limits_{n = 1}^{\infty} \text {m} \left (\left \{f \geq n \right \} \right ) },$$ where $\text {m}$ denotes the Lebesgue measure.

Approximating $f$ by a non-negative sequence of increasing simple measurable functions I find that $$\begin{align*} \int f\ \text {dm} & = \lim\limits_{n \to \infty} \left [\sum\limits_{k = 0}^{n 2^n - 1} \dfrac {k} {2^n}\ \text {m} \left (\left \{\dfrac {k} {2^n} \leq f \lt \dfrac {k + 1} {2^n} \right \} \right ) + n\ \text {m} \left (\left \{f \geq n \right \} \right ) \right ]. \end{align*}$$

How do I proceed now? Any help will be highly appreciated.

Thanks in advance.

Source $:$ ISI (Indian Statistical Institute) PhD entrance test in Mathematics held in $20$th September this year, TEST CODE $:$ MTA (FORENOON SESSION), Question No. $7.$

Body:
Since
$$
\int f\,dm=\int_{0}^{\infty}m(\{f>x\})\,dx=\sum_{n\ge 0}\int_{n}^{n+1}m(\{f>x\})\,dx
$$
and
$$
m(\{f>n+1\})\le\int_{n}^{n+1}m(\{f>x\})\,dx\le m(\{f>n\}),
$$
one gets
$$
\sum_{n\ge 1}m(\{f>n\})\le\int_{0}^{\infty}m(\{f>x\})\,dx\le \sum_{n\ge 0}m(\{f>n\}).
$$


问:
Detect numbers surrounded by equal numbers in a matrix
I need an algorithm that extracts from a matrix those fields that are surrounded by equal values. I'll provide you an example to see it clearer.

This is my intitial matrix.

$$
        \begin{matrix}
        1 & 1 & 1 & 1 & 1 & 0 & 1 \\
        1 & 0 & 1 & 0 & 0 & 0 & 1 \\
        1 & 0 & 1 & 1 & 1 & 0 & 1 \\
        0 & 0 & 1 & 0 & 1 & 0 & 1 \\
        1 & 1 & 1 & 0 & 1 & 1 & 1 \\
        0 & 0 & 1 & 0 & 0 & 0 & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 & 1 \\
        \end{matrix}
$$

As you can see, the spaces marked with an "X" are those fields that
are surrounded by equal values, in this case, surrounded by value 1.

$$
        \begin{matrix}
        1 & 1 & 1 & 1 & 1 &   & 1 \\
        1 &   & 1 &   &   &   & 1 \\
        1 &   & 1 & 1 & 1 &   & 1 \\
          &   & 1 & X & 1 &   & 1 \\
        1 & 1 & 1 & X & 1 & 1 & 1 \\
          &   & 1 & X & X & X & 1 \\
        1 & 1 & 1 & 1 & 1 & 1 & 1 \\
        \end{matrix}
$$

The question is if there exists any mathematic formula that can return which are those fields. If not, I'd be grateful if you gave me some theorems that can help me developing the algorithm.

Thanks.

答:
I’m afraid that no theorems or formulas can help you much in the developing of the algorithm, and therefore  you should concentrate on the algorithm itself. I wrote a simple Pascal program to illustrate one of the simplest of such algorithms.

```<code>program SurNum;
const
  m=7;
  n=7;
  var
    IFi,OFi:Text;
    tC:Char;
    i,j,k:Byte;
    a:array[1..m,1..n]of Byte;
    Found:Boolean;
begin

{matrix input}
assign(IFi,'input.txt');
reset(IFi);
for i:=1 to m do begin
  for j:=1 to n do begin
    read(IFi,tC);
    if tC='1' then a[i,j]:=1 else a[i,j]:=0;
  end;
  readln(IFi);
end;
close(IFi);

{main part}

{find initial unisolated cells}
for j:=1 to n do if a[1,j]=0 then a[1,j]:=2;
for j:=1 to n do if a[m,j]=0 then a[m,j]:=2;
for i:=1 to m do if a[i,1]=0 then a[i,1]:=2;
for i:=1 to m do if a[i,n]=0 then a[i,n]:=2;

{try to find additional unisolated cells}

repeat
Found:=False;
for i:=2 to m-1 do for j:=2 to n-1 do
  if (a[i,j]=0) and ((a[i-1,j]=2) or (a[i+1,j]=2) or (a[i,j-1]=2) or (a[i,j+1]=2))
   then begin a[i,j]:=2; Found:=True end;
until not Found;

{matrix output}

assign(OFi,'output.txt');
rewrite(OFi);
for i:=1 to m do begin
  for j:=1 to n do begin
    if a[i,j]=0 then write(OFi,'X');
    if a[i,j]=1 then write(OFi,'1');
    if a[i,j]=2 then write(OFi,' ');
  end;
  writeln(OFi);
end;
close(OFi);

end.
</code>```

The contents of the file “input.txt” is:

```<code>1111101
1010001
1011101
0010101
1110111
0010001
1111111
</code>```

The contents of the file “output.txt” should be:

```<code>11111 1
1 1   1
1 111 1
  1X1 1
111X111
  1XXX1
1111111
</code>```

Body:
I’m afraid that no theorems or formulas can help you much in the developing of the algorithm, and therefore  you should concentrate on the algorithm itself. I wrote a simple Pascal program to illustrate one of the simplest of such algorithms.

program SurNum;
const
  m=7;
  n=7;
  var
    IFi,OFi:Text;
    tC:Char;
    i,j,k:Byte;
    a:array[1..m,1..n]of Byte;
    Found:Boolean;
begin

{matrix input}
assign(IFi,'input.txt');
reset(IFi);
for i:=1 to m do begin
  for j:=1 to n do begin
    read(IFi,tC);
    if tC='1' then a[i,j]:=1 else a[i,j]:=0;
  end;
  readln(IFi);
end;
close(IFi);

{main part}

{find initial unisolated cells}
for j:=1 to n do if a[1,j]=0 then a[1,j]:=2;
for j:=1 to n do if a[m,j]=0 then a[m,j]:=2;
for i:=1 to m do if a[i,1]=0 then a[i,1]:=2;
for i:=1 to m do if a[i,n]=0 then a[i,n]:=2;

{try to find additional unisolated cells}

repeat
Found:=False;
for i:=2 to m-1 do for j:=2 to n-1 do
  if (a[i,j]=0) and ((a[i-1,j]=2) or (a[i+1,j]=2) or (a[i,j-1]=2) or (a[i,j+1]=2))
   then begin a[i,j]:=2; Found:=True end;
until not Found;

{matrix output}

assign(OFi,'output.txt');
rewrite(OFi);
for i:=1 to m do begin
  for j:=1 to n do begin
    if a[i,j]=0 then write(OFi,'X');
    if a[i,j]=1 then write(OFi,'1');
    if a[i,j]=2 then write(OFi,' ');
  end;
  writeln(OFi);
end;
close(OFi);

end.

The contents of the file “input.txt” is:

1111101
1010001
1011101
0010101
1110111
0010001
1111111

The contents of the file “output.txt” should be:

11111 1
1 1   1
1 111 1
  1X1 1
111X111
  1XXX1
1111111


问:
Converting $C_1e^{i\omega t} + C_2 e^{-i\omega t} $ to $A\cos(\omega t +\phi) $
It seems intuitive to be able to replace:
$$C_1e^{i\omega t} + C_2 e^{-i\omega t} $$

by:
$$ A\cos(\omega t +\phi) $$

And not lose any "information". But I was wondering if there is any algebraic manipulation I can do to show that it is indeed the case.

答:
No, this won't work for  general $C_1$ and $C_2$. Supposing $C_1$ and $C_2$ are real (the following argument can also be extended - below - to complex $C_1$ and $C_2$) you have the identity
$$C_1e^{i\omega t} + C_2 e^{-i\omega t} = \sqrt{C_1 \cdot C_2} (e^{\phi + i\omega t} + e^{- \phi -i\omega t} )
$$
where $\phi = \frac12 (\log C_1 - \log C_2)$

So you don't receive a $\cos$-function with a real phase shift.

For complex $C_1$ and $C_2$, write w.l.o.g. $C_1 = |C_1|e^{i \alpha_1}$ and  $C_2 = |C_2|e^{i \alpha_2}$. Then the above identity generalizes to
$$C_1e^{i\omega t} + C_2 e^{-i\omega t} = e^{i \frac12 (\alpha_1 + \alpha_2)}\sqrt{|C_1| \cdot |C_2|} (e^{\phi + i(\omega t + \gamma)} + e^{- \phi -i(\omega t + \gamma)} )
$$
where $\phi = \frac12 (\log |C_1| - \log |C_2|)$ and $\gamma = \frac12 (\alpha_1 - \alpha_2)$. Hence $\alpha_1$ and $\alpha_2$ lead to a real phase shift of $\omega t$ which is what OP was intending.

In general,  we identify $A = e^{i \frac12 (\alpha_1 + \alpha_2)}\sqrt{|C_1| \cdot |C_2|} $ and we see that it is $\phi$ which prevents us from writing the desired $\cos$-function with real phase shift, which, in turn, is possible only if $\phi=0$, i.e. iff $|C_1|=|C_2|$.


问:
A problem about probability and unit square
How can I solve this problem?

A point $P = (X, Y)$ of the unit square $(0,1) \times  (0,1)$ is chosen at random. Let $\theta$ be the angle formed between the $x-$axis and the segment joining the origin and $P$. Find the density of $\theta$.

I know for example that $\theta=\tan^{-1}\left( \frac{Y}{X}\right)$ and since that $0<x<1$ and $0<y<1$ so $X\sim \textbf{U}(0,1)$ and $Y\sim \textbf{U}(0,1)$ and $X$ and $Y$ are independent random variables.
But, I don't know how can I continue. In fact I don't quite understand the problem.

In my attempt, I think I could use the Jacobian transformation theorem, but I don't know how to approach the problem that way.

答:
First of all you can calculate the distribution of $Z=\frac{Y}{X}$

Without integrals, but only with a simple drawing you get

$$ F_Z(z) =
\begin{cases}
0,  & \text{if $z<0$ } \\
\frac{z}{2},  & \text{if $0\leq z<1$ } \\
1-\frac{1}{2z}, & \text{if $z\geq 1$}
\end{cases}$$

Then easy find the distribution of $\theta=arctan(Z)$ with the fundamental transformation theorem

$$ f_\Theta(\theta) =
\begin{cases}
\frac{1}{2}+\frac{tan^2\theta}{2},  & \text{if $0\leq \theta<\frac{\pi}{4}$ } \\
\frac{1}{2}+\frac{1}{2 tan^2\theta}, & \text{if $\frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}$} \\
0, & \text{elsewhere}
\end{cases}$$

Solution details & explanation

How to derivate $F_Z(z)$?

Frist of all observe that $Z \in [0;\infty)$.

Using the definition we have

$$F_Z(z)=\mathbb{P}[Z\leq z]=\mathbb{P}[Y\leq zX]$$

Let's do a drawing:

Z-CDF is the purple area...

Immeditately find $f_Z(z)$ derivating F

and also easy find the density of $\Theta=tan^{-1}(Z)$ simply applying the known formula

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$


问:
Provable soundness of finite fragments of ZFC
Fix some reasonable computable enumeration of the axioms of ZFC, and let $ZFC_n$ be the theory consisting of the first n axioms.

Is it the case that, for each natural number n, and each sentence $\phi$, ZFC proves the sentence

$(ZFC_n \vdash \phi ) \rightarrow \phi$?

(Where we formalize provability in ZFC in some reasonable way.)

Please note that this is distinct from asking whether ZFC proves

$\forall n \forall \phi \ : \ (ZFC_n \vdash \phi ) \rightarrow \phi$

(Which ZFC trivially does not, if it is consistent.)

答:
Consider any finite list $\Phi$ of axioms of ZFC and any other sentence $\phi$. By the Lévy-Montague reflection theorem, there is some rank-initial segment $V_\theta$ of the universe for which all the sentences in $\Phi$ and also $\phi$ are absolute between $V_\theta$ and $V$. Since the sentences of $\Phi$ are part of ZFC, they are true in $V$ and hence also in $V_\theta$. In particular, $V$ looks upon $V_\theta$ as a model of $\Phi$, according to the truth predicate that it can define for this set structure. Therefore, if $V$ thinks that $\Phi\vdash\phi$, then it will think that $V_\theta\models\phi$. Since $\theta$ was chosen so that this sentence is absolute, this implies $\phi$ holds in $V$, as desired. So we've established any instance of the implication.

As you noted in the question, we get this implication only as a scheme, a separate statement for each instance, because we have the reflection theorem also only as a scheme.

Addendum. Let me explain that one can also strengthen the conclusion somewhat, by assuming not that the sentences of $\Phi$ are part of ZFC, but rather merely that they are true. In other words, I claim that ZFC proves every instance of the scheme:
$$(\psi\vdash\phi)\to(\psi\to\phi).$$
If we take $\psi$ to be the conjunction of the sentences in $\Phi$, this generalizes your scheme. But the same proof works here. By the Lévy-Montague reflection theorem, there is $V_\theta$ for which both $\psi$ and $\phi$ are absolute between $V_\theta$ and $V$. Now, if $\psi\vdash\phi$ and $\psi$ is true (in $V$), then $\psi$ is true in $V_\theta$, and so $\phi$ also is true there, and so $\phi$ is true in $V$, as desired.


答:
This is possible for complex $A$ iff $|C_1|=|C_2|$, and is possible for real $A$ iff $C_2=\overline{C_1}$.

(I assume $\phi$ is restricted to real values, too.)

The "only if"s are because if we rewrite $A\cos(\omega t+\phi)$ in terms of complex exponentials, the coefficients $C_1=\frac{1}{2}Ae^{i\phi}$ and $C_2=\frac{1}{2}Ae^{-i\phi}$ have the same size, and indeed if $A$ is real then the coefficients $C_1$ and $C_2$ must be complex conjugates.

Conversely, if $|C_1|=|C_2|$ then $C_1=Ce^{i(\theta+\phi)}$ and $C_2=Ce^{i(\theta-\phi)}$ for some angles $\theta,\phi$ and

$$ C_1e^{i\omega t}+C_2e^{-i\omega t}= Ce^{i\theta}(e^{i(\omega t+\phi)}+e^{-i(\omega t+\phi)})=A\cos(\omega t+\phi) \tag{$\ast$} $$

where $A=2Ce^{i\theta}$. If $C_2=\overline{C_1}$ then $\phi=0$ and $A$ is real of course.

If you want to allow complex values for $\phi$ then you can write $C_1=e^{\theta+i\phi}$ and $C_2=e^{\theta-i\phi}$ for complex values of $\theta$ and $\phi$ and do the same algebraic manipulation as $(\ast)$.

Title: Converting $C_1e^{i\omega t} + C_2 e^{-i\omega t} $ to $A\cos(\omega t +\phi) $

Body:
It seems intuitive to be able to replace:
$$C_1e^{i\omega t} + C_2 e^{-i\omega t} $$

by:
$$ A\cos(\omega t +\phi) $$

And not lose any "information". But I was wondering if there is any algebraic manipulation I can do to show that it is indeed the case.


问:
Show that the matrix $A^2 + I$ is invertible for all matrices $A$, where $A$ is an $n \times n$ symmetric matrix.
I'm a little stuck on this problem. I know that since $A$ is symmetric, $A=A^{T}$. I'm also pretty sure that $AA^{T}$ is invertible. Therefore $A^2$ would be invertible. I'm not really sure how to account for the identity matrix (In).

Thanks in advance!

答:
For an arbitrary real valued matrix $A$, $A^TA$ is positive-semidefinite, so all its eigenvalues are nonnegative. It follows that all the eigenvalues of $A^TA+I$ are no less than $1$, so $A^TA+I$ is invertible(and positive definite). In particular, when $A$ is symmetric, $A^2+I$ is invertible(and positive definite).

A more direct way is to show that for any nonzero column vector $v$, $(A^TA+I)v\ne 0$. To see this, note that for every column vector $w$, $w^Tw\ge 0$ and the equality holds if and only if $w=0$. Then
$$v^T(A^TA+I)v=(Av)^T(Av)+v^Tv>0.\tag{1}$$
$(1)$ implies that $(A^TA+I)v\ne 0$, which completes the proof. In fact, $(1)$ also shows that $A^TA+I$ is positive definite.


答:
There is a very simple way to get rid of the logarithm: integration by parts. So we have
$$\int^\infty_0x^a\ln\left(\frac1{1-e^{-x}}\right)\,dx=\frac1{a+1}\int^\infty_0x^{a+1}\frac{e^{-x}}{1-e^{-x}}\,dx.$$ The fraction can be written as a geometric series:
$$\frac{e^{-x}}{1-e^{-x}}=\sum^\infty_{k=1}e^{-kx},$$ and since it's a positive series, we can exchange summation and integral:
$$\int^\infty_0x^{a+1}\frac{e^{-x}}{1-e^{-x}}\,dx=\sum^\infty_{k=1}\int^\infty_0x^{a+1}e^{-kx}\,dx.$$ Substituting $kx=t$, we see that
$$\int^\infty_0x^{a+1}e^{-kx}\,dx=\frac1{k^{a+2}}\int^\infty_0t^{a+1}e^{-t}\,dt=\frac{\Gamma(a+2)}{k^{a+2}},$$ using the well-known integral representation of the Gamma function. So our integral becomes
$$\int^\infty_0x^{a+1}\frac{e^{-x}}{1-e^{-x}}\,dx=\Gamma(a+2)\,\sum^\infty_{k=1}\frac1{k^{a+2}}=\Gamma(a+2)\,\zeta(a+2).$$ Putting all of that together, we obtain
$$\int^\infty_0x^a\ln\left(\frac1{1-e^{-x}}\right)\,dx=\frac{\Gamma(a+2)\,\zeta(a+2)}{a+1}=\Gamma(a+1)\,\zeta(a+2)$$ (many thanks to @Gary for the last simplification).

For $a=7/6$, that's approximately $1.2686090974543140685705195862758272411$.

Title: How can I approximate $\int_{0}^{\infty}x^{7/6}\ln(\frac{1}{1-e^{-x}})dx$?

Body:
My approximation here needs to be valid for large values of $x$ here.
$$\int_{0}^{\infty}x^{7/6}\ln\Bigg(\frac{1}{1-e^{-x}}\Bigg)dx$$
I can't seem to get around this natural log no matter what I do. How can this integral be approximated without going to $0$? I know there is a solution in an integral table if the natural log wasn't there.


问:
Compute the limit $\lim\limits_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin tx} $
I have been working on this limit for days, but I am not getting it. The question is

Compute the limit $$\lim_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (tx)}$$

Note that the integral is well defined and convergent for every $t >0$. Indeed the integrand function is a positive function for every $t >0$ since
$$e^x + \sin tx > e^x-1 > x>0$$
And as $x \to + \infty$ the integrand function behaves like $e^{-x}$.

WHAT I TRIED:

I consider $t=2n \pi$ a multiple of $2 \pi$, and see what happens:
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} = \sum_{k=0}^\infty \int_{k /n}^{(k+1) /n} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)}$$
Making the change of variables $u = 2n \pi x$ I get
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{u/2n \pi}+ \sin (u)} &\ge
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(2k+2) \pi/2n \pi}+ \sin (u)} \\&=
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)}\end{align} where I write the lower bound with the minimum of the function at $u=(2k+2) \pi$. Now I use the fact that the integrand function does is integrated over a period of $2 \pi$, and using the result for $C>1$
$$\int_0^{2 \pi} \frac{ \mathrm d u}{C+ \sin (u)} = \frac{2 \pi}{\sqrt{C^2-1}}$$
I get the estimate
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)} &= \sum_{k=0}^\infty \frac{1}{2n \pi} \frac{2 \pi}{\sqrt{e^{2(k+1)/n} -1 }} \\&= \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}\end{align}
Summing all up, I got that
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} \ge \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}$$
As $n \to \infty$ the series converges to the Riemann integral
$$\int_0^{+ \infty} \frac{\mathrm d y}{\sqrt{e^{2y}-1}} = \frac{\pi}{2}$$

Hence the limit should be a number larger than $\pi/2$, or $+ \infty$.

Using WA I got for large values of $t$ that the integral is between $1$ and $2$, thus $\pi/2$ could be the actual limit.

答:
Claim: We have $\lim\limits_{t \to + \infty} \int_0^{+ \infty}(e^x+ \sin tx)^{-1}\,dx=\pi/2$.

Proof. We start by writing
\begin{align}\int_0^{+ \infty} \frac{dx}{e^x+ \sin (tx)}&=\int_0^1\frac{du}{1-u\sin(t\log u)}\tag1\\&=\int_0^1\sum_{n=0}^\infty u^n\left(\frac{u^{it}-u^{-it}}{2i}\right)^n\,du\tag2\\&=\int_0^1\sum_{n=0}^\infty\frac1{(2i)^n}\sum_{k=0}^n\binom nku^{n+(2k-n)it}(-1)^{n-k}\,du\tag3\\&=\sum_{n=0}^\infty\left(\frac i2\right)^n\sum_{k=0}^n\binom nk\frac{(-1)^k}{n+1+(2k-n)it}\tag4.\end{align} Uniform convergence means we can interchange limit and sum. On interchanging the order of summation, we can see that the only nonzero term that contributes to the integral as $t\to+\infty$ occurs when $n=2k$ so that the coefficient of $t$ is zero. Thus \begin{align}\lim_{t\to+\infty}\int_0^{+ \infty} \frac{dx}{e^x+ \sin (tx)}&=\sum_{k=0}^\infty\left(-\frac14\right)^k\binom{2k}k\frac{(-1)^k}{2k+1}=\frac\pi2\tag5.\end{align}

$(1):$ substitute $x=-\log u$

$(2):$ geometric series with radius of convergence $|u\sin(t\log u)|<1$ for $u\in(0,1)$

$(3):$ binomial theorem

$(4):$ interchange of integral and sum by Fubini

$(5):$ taylor series of $\arcsin1$


答:
If we use the Maclaurin series of the logarithm, we obtain that the integral is
$$
\sum\limits_{n = 1}^\infty  {\frac{1}{n}\int_0^{ + \infty } {x^{7/6} e^{ - nx} dx} }  = \Gamma \left( {2 + \tfrac{1}{6}} \right)\sum\limits_{n = 1}^\infty  {\frac{1}{{n^{3 + 1/6} }}}  = \Gamma \left( {2 + \tfrac{1}{6}} \right)\zeta \left( {3 + \tfrac{1}{6}} \right),
$$
where $\Gamma$ is the gamma function and $\zeta$ is the Riemann zeta function.


问:
Investigating convergence of integral for certain parameters
Suppose we have an integral of the form,

$$ \int_0^\infty \frac{x^{b-1}[\log(1+x)]^{1+a}}{(1+x^2)^{c+b/2}}dx $$

I want investigate values of $a,b,c > 0$ for which this integral converges. Using the basic inequality, $\log(1+x) \leq x$ for $x > 0$, the integral is bounded by,

$$ \int_0^\infty \frac{x^{b-1}[\log(1+x)]^{1+a}}{(1+x^2)^{c+b/2}}dx \leq \int_0^\infty \frac{x^{a+b}}{(1+x^2)^{c+b/2}}dx = \frac{\Gamma\left(\frac{a+b+1}{2}\right)\Gamma\left(c-\frac{a+1}{2}\right)}{2\Gamma\left(c+\frac{b}{2}\right)}$$

(this integral was computed in Mathematica). So we can see that convergence holds if $c - \frac{a+1}{2} > 0$ or $a < 2c - 1$. Unfortunately this forces $c > \frac{1}{2}$. Is there a better bound than $\log(1+x) \leq x$ I can use? Ideally, I'd like to fix $b,c > 0$ and choose $a$ alone.

答:
First, clearly,
$\int_0^1 \frac{x^{b-1}[\log(1+x)]^{1+a}}{(1+x^2)^{c+b/2}}\mathrm{d} x$ is convergent.

Second, since $x(1+x^2)^c \ge \frac{1+x}{2}(\frac{(1 + x)^2}{2})^c$ for all $x$ in $[1, \infty)$, we have
$$\int_1^\infty \frac{x^{b-1}[\log(1+x)]^{1+a}}{(1+x^2)^{c+b/2}}\mathrm{d} x
\le \int_1^\infty \frac{[\log(1+x)]^{1+a}}{\frac{1+x}{2}(\frac{(1 + x)^2}{2})^c}\mathrm{d} x$$
It is easy to prove that RHS is convergent. We are done.


问:
proof involving Cauchy sequence
Is there a proof of the following statement?

A sequence $(x_n)$ has a Cauchy subsequence if and only if it has a subsequence $(x_{n_k})$ for which $d(x_{n_k}, x_{n_{k+1}}) < \frac{1}{2^k}$ for all $k$.

答:
Let $(x_n)$ be a sequence such that a subsequence $(x_{n_k})$ has the property you mention. Choose $\epsilon>0$ and let $k_0$ be such that
\begin{equation}
\sum_{k=k_0}^{+\infty}2^{-k}<\epsilon.
\end{equation}
If $i,j\geq k_0$ and $j>i$, then by triangular inequality
\begin{equation}
d(x_{n_i},x_{n_j})\leq\sum_{k=i}^{j-1}d(x_{n_k},x_{n_{k+1}})<\sum_{k=i}^{j-1}2^{-k}<\sum_{k=k_0}^{+\infty}2^{-k}<\epsilon.
\end{equation}
That is: $(x_{n_k})$ is the Cauchy subsequence you were looking for.

Conversely, assume that $(x_n)$ has a Cauchy subsequence and, without loss of generality, assume that this subsequence is $(x_n)$ itself. We construct $(x_{n_k})$ inductively. First, let $n_1$ be such that $d(x_n,x_m)<2^{-1}$ for all $n,m\geq n_1$. Suppose you have already determined $n_1,\dots,n_k$. Let $n_{k+1}$ be such that $n_{k+1}\geq n_k$ and $d(x_n,x_m)<2^{-(k+1)}$ for all $n,m\geq n_{k+1}$. Since both $n_k$ and $n_{k+1}$ are greater than or equal to $n_k$, one clearly has $d(x_{n_k},x_{n_{k+1}})<2^{-k}$ and the proof is finished.

Title: proof involving Cauchy sequence


问:
Suggesting a form of the equation of the given curve
I am really struggling with this problem for a long time.

I think it is easy to understand, but difficult to solve;

We have a curve that has a lower limit as $x \rightarrow -\infty$, and it has another lower limit as $x \rightarrow \infty$. This curve increases only before the peak and decreases only after the peak. The rate at which this curve increases may not be the same rate at which this curve decreases (apart from the sign). What I mean is: the function may increases rapidly but decreases slowly, or vice versa.

Also, it is not a piecewise function. Moreover, it should be integerable at any domain.

If the form of that function contains something like this: $ae^{-x^2}$ (Gaussian Distribution), then it will not be integerable.

I was trying to combine two expression (piecewise) to make it (not piecewise), but no luck: like combining

$y=a-be^{cx}$ for $-\infty <x \le j$, and $y=A-Be^{Cx}$ for $j < x < \infty$; where $a,b,c,A,B,C,j$ are constants.

The following curve is plotted (using piecewise function) just to illustrate the problem.

So the goal is to find the form of the equation of the curve with the following criteria:

has a lower limit before the peak.

has another lower limit after the peak.

The rate of increasing may be different from the rate of decreasing (apart from the sign).

Must be integerable in any domain.

Must not be a piecewise function.

Your help would be appreciated. THANKS!

答:
The piecewise function :
$$y(x)=\begin{cases}
a+b\:e^{c\:x}\qquad x<\chi \\
A+B\:e^{C\:x}\qquad x>\chi
\end{cases}$$
is defined on an other form :
$$y(x)=\left(a+b\:e^{c\:x}\right)H(\chi-x)+\left(A+B\:e^{C\:x}\right)H(x-\chi)$$
in which $H$ is the Heaviside step function.

If one want to avoid the picewise form one can replace the Heaviside function by a continuous function which approximates the Heaviside function. They are several possibilities.
For example a function of tbe logistic kind can be used :
$$H(X)\simeq\frac{1}{1+e^{-K\:X}}$$
$K$ is an arbitrary positive large number. Doesn't mater the value of $K$ insofar it is very large. For example :

In order to fit the equation to data one have to adjust the parameters $a,b,c,A,B,C,\chi$ using a regression method.

Since there is no scale on the graph published by the OP, arbitrary nomalized scales are added on the above graph. The values of parameters were adjusted in order to fit the blue curve to the given red points.

NOTE :

If one want a smoother transition between the two exponentials, a smaller value of $K$ has to be adjusted. So, one parameter more for the regression.

Even more general, one can introduce more paramerers such as slightly different $K_1,K_2$ and slightly different $\chi_1,\chi_2$ in order to have a wide range of possible shapes of the curve.
$$y(x)=(a+be^{cx})\frac{1}{1+e^{K_1(x-\chi_1)}} +(A+Be^{Cx})\frac{1}{1+e^{K_2(\chi_2-x)}}$$
On the other hand, a so large number of parameters $a,b,c,\chi_1,K_1,A,B,C,\chi_2,K_2$ would increases a lot the difficulty for a robust regression process. With so many parameres I doubt that the available softwares will converge to reliable result.


问:
Relationship between $\left\Vert A-B \right\Vert$ and $\left\Vert A_s- B_s \right\Vert$ where $A$ is similar to $A_s$ and $B$ is similar to $B_s$?
I'm looking for a bound that relates $\left\Vert A-B \right\Vert_{op}$ and $\left\Vert A_s- B_s \right\Vert_{op}$ where $A$ and $A_s$ are similar matrices and $B$ and $B_s$ are similar matrices. If it helps one can assume that all four matrices are real and that $A$ and $B$ are symmetric (but $A_s$ and $B_s$ may not be). I define the matrix norm here as the operator norm induced by the Euclidian norm.

It is easy to check these quantities are not equal. Is it possible to construct a bound of the form $$\left\Vert A-B \right\Vert_{op} \leq f\left\Vert A_s- B_s \right\Vert_{op}$$
where $f$ is a variable that depends on the matrices in question?

答:
Let $A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $B=B_s=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. Then $A$ is similar to $$A_s:=B=\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}^{-1}A\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}$$

As $\|A-B\|=\|\begin{pmatrix}-1&1\\1&1\end{pmatrix}\|=\sqrt2$ but $A_s-B_s=0$, there is no number $c$ such that $\|A-B\|\le c\|A_s-B_s\|$.


问:
Property of an Operator.
Let $f\in C^\infty_c(\mathbb{R}) $ given.

Let $T (f(x))=-\partial_t v(x,0)$. Where $v:\mathbb{R}^2\to \mathbb{R}$.

$$
\partial_x^2v(x,t)+\partial_t^2v(x,t)=0 ~\in\mathbb{R^2},~~\text{and}~~ v(x,0)= f(x)\in \mathbb{R}.
$$

How can I show $ \int_{\mathbb{R}} T(f(x)) f(x) dx\geq 0$, any hints?

答:
The initial value problem
$$
v_{xx}+v_{tt}=0, \quad v(x,0)=f(x) \tag{1}
$$
does not enjoy uniqueness, and hence the operator $Tf=v_t(x,0)$ is not well-defined.

If $v$ is a solution of $(1)$, then so is $\tilde v(x,t)=v(x,t)+at$.

Title: Property of an Operator.


问:
Calculating $\lim_{x\to 0}(f(x)+f(5x))$ and $\lim_{x\to 0}f(x)$
I apologize for the lengthy question in advance. As you can see in the title, I'm trying to calculate the limits of the two functions.

They are defined the following:

(1) If $\lim_{x\to 0}f(x) = 2$ , what is $\lim_{x\to 0}(f(x)+f(5x))$ ?

(2) If $\lim_{x\to 0}(f(x)+1/f(x)) = 2$, what is $\lim_{x\to 0}f(x)$?

I've never worked with limits like these before. But these are my thoughts so far:

We know that $\lim_{x\to 0}f(x) = 2$ and since $\lim_{x\to 0}(f(x)+f(5x))$ = $\lim_{x\to 0}f(x)$ + $\lim_{x\to 0}f(5x)$ (At least I hope this is allowed), this results in $2$ + $\lim_{x\to 0}f(5x)$

However, I do not quite know how to go on from this. Is there a way I can manipulate $\lim_{x\to 0}f(5x)$ so I can use the given property to calculate a limit?

For (2), it's quite similar.I'm not quite sure if I can rewrite it again to $\lim_{x\to 0}f(x) + \lim_{x\to 0}1/f(x)$. I'm not sure if this is relevant to this task, but we're taking a limit of a function and it's inverse (not the inverse function, but the inverse of the value). Is this something I can use for this exercise or is this just an unnecessary observation in this case?

答:
For the first one we can use that

$$\lim_{x\to 0} 5x \to 0$$

and then by $y=5x \to 0$

$$\lim_{x\to 0}f(5x)=\lim_{y\to 0}f(y) =0$$

by this theorem

General proof of limit composition theorem on continuous function

For the second one refer to

If $\lim_{x\to 0}\left(f(x)+\frac{1}{f(x)}\right)=2,$ show that $\lim_{x\to 0}f(x)=1$.


问:
Word to describe "semi-optimal" solutions
I have an algorithm (based on heuristics) that solves for a quantity $x$, producing a practically good result but not guaranteed to be optimal.

What adjective can I use to describe this "solution" $x$? "Suboptimal" has a negative connotation. I want to communicate that $x$ is good and useful although not the best possible.

答: This is not really a mathematical question. You could try "near optimal" or "satisficing".</p>


问:
Can we wrap a square onto itself with constant singular values?
I have now cross-posted this on mathoverflow.

Let $0<\sigma_1<\sigma_2$ satisfy $\sigma_1\sigma_2=1$, and let $D=[-1,1]^2$.

Does there exist a Lipschitz bijective* map $f:D \to D$ such that $df$ has almost everywhere the fixed singular values $\sigma_1,\sigma_2$?

Is there such a diffeomorphism of $D$? (thinking of $D$ as a manifold with corners, or requiring smoothness only on the interior etc.)

*I am fine with requiring only $|f^{-1}(y)|=1$ for a.e. $y \in D$; the Area formula then implies that $f$ is surjective.

Clearly, no affine map would be suitable. We somehow need a map whose singular vectors are 'rotating' from point to point.

Comment: If we replace $D$ with a disk then we have $ f_t:(r,\theta) \to (r,\theta+t \log r)$, which
is the flow of $\log r \frac{\partial}{\partial \theta}$. $f_t$ has constant singular values (which depend on $t$.)

答:
Here are some elements of reflection, not at all a full answer.

Let us think into the complex plane, using conformal mappings (i.e. bi-holomorphic bijections). [seeing your $\theta+c\log(r)$, we are not far anyway from  complex numbers...]

There exists explicit conformal mappings from the square $D$ (any square in fact) onto unit disk $U$. Let for example $\varphi$ be the one given here using complex function  $$\varphi(u):=\int_0^u \dfrac{dz}{(z^4-1)^{1/2}}\tag{1}$$

(itself issued from Schwarz-Christoffel transformation).

Caution: in this case the vertices are $\{1,i,-1,-i\}$ instead of $\pm1\pm i$.

There exists an infinity of conformal mappings: $\mu:U \to U$, under the form of these Möbius transformations:

$$\mu_{\theta,a}(z)=\mu(z)=e^{i\theta }{\frac{z+a}{\bar{a}z+1}}, \ \ |a|<1. \tag{2}$$

Then, you can take $$\varphi^{-1}\circ \mu \circ \varphi: D \to D \tag{3}$$

Turning (3) into its equivalent in terms of real functions ($x+iy \to (x,y)$), it may give you, using chain rule, a way to attack your issue.

But I am fully conscious that this does not bring a direct answer to your query...


问:
How to solve this non linear and second order ODE
$$y'' + ay(y')² + by = 0$$

Here, $a$ and $b$ are constants. I can't solve this!

答:
Write it as
$$
\frac{2ay'y''}{ay'^2+b}+2ayy'=0
$$
and integrate to get
$$
ay'^2+b=ce^{-ay^2}.
$$
At this point you can transform it into a quadrature problem, but I do not believe that there will be a symbolic solution.