问:
Balls and Spheres in $(n; k; c)$-expander.
Let $G = (V;E)$ be an $(n; k; c)$-expander. Let $v\in V$ be a vertex of $X$, and $r$ a non-negative integer. The ball of radius $r$ centered at $v$ is define to be
$$B_r(v) = \{w \in V \mid d(v;w) \leq r\}$$

Let $v_1$ and $v_2$ be some vertices in $G$, and let $d(v_1,v_2)=d$; we define 2 balls, $B_r(v_1)$ and $B_r(v_2)$ such that $r=d/2$.

I know that $$B_r(v_1) \cap B_r(v_2) = 1$$

but I try to find the number of vertex that are in $B_{r+1}(v_1)$ and $B_{r+1}(v_2)$,
meaning $$ B_{r+1}(v_1) ∩ B_{r+1}(v_2) = ??$$

Definition ($c$–expander)- we say that a graph $G= (V;E)~$ is an $c$–expander if for every set $|S|\leq n/2$, we have that $\frac{|\partial(S)|}{|S|}\ge c$ for a real number $c >0~$. An $(n,c,k)$-expander is a graph such that each vertex has degree $k$ and it is a $c$–expander.

答:
There is an issue when $d$ is odd. Then $r$ is not an integer and the two balls do not intersect. For example, in the simple graph $v_1 - \cdot - \cdot - v_2$, $d = 3, r = 1.5$ and each of the balls is just the center and immediate neighbor.

If $d$ is even, then it is true that $$\left|B_r(v_1) \cap B_r(v_2)\right| \ge 1$$
But I see nothing in your conditions that would require it to actually be $1$. There could be multiple paths of length $d$ connecting $v_1$ and $v_2$.

Let $N_r = B_r(v_1) \cap B_r(v_2)$.  $N_r \subseteq N_{r+1}$, so $|N_{r+1}| \ge |N_r|$. Also for any $v \in N_r$ all $k$ of its neighbors must be in $N_{r+1}$. But some of those neighbors can be other vertices in $N_r$, so they doesn't necessarily increase the size of $N_{r+1}$. Any single vertex in $N_r$ guarantees $k+1$ vertices (itself and its $k$ neighbors) in $N_{r+1}$, so we are guaranteed that $|N_{r+1}| \ge {k+1}$, but those two requirements is all you can guarantee.
$$|N_{r+1}| \ge \max\{k+1, |N_r|\}$$

But that is the best you can do for a lower bound. I do not see how being a $c$-expander would change this, as the requirements could be met elsewhere than in the intersection fo the two balls.

You can get an upper limit by supposing that no two elements of $N_r$ share any neighbors. This assures that the $k+1$ points of intersection in $N_{r+1}$ guaranteed by each element of $N_r$ do not overlap. But there can be other new points of intersection. If there were non-intersection points on the boundary of the $r$-radius balls that were within a distance $\le 2$ of each other, those points generate new intersections for the balls of radius $r+1$. And assuming that every point on the boundary of $B_r(v_1)$ that is not in $N_r$ has only one connection to the interior, and the other $k-1$ neighbors are all in $N_{r+1}$, this gives a maximum of $(k-1)|\partial B_r(v_1) \setminus N_r|$ new intersection points. And similarly for $v_2$, except that the new intersection points for each must be the same. This means
$$\begin{align}\max\{k+1, |N_r|\} \le &|N_{r+1}| \le (k+1)|N_r| + (k-1)\min\{|\partial B_r(v_1)|-|N_r|, |\partial B_r(v_2)|-|N_r|\}\\\\\max\{k+1, |N_r|\} \le &|N_{r+1}| \le 2|N_r|+ (k-1)\min\{|\partial B_r(v_1)|, |\partial B_r(v_2)|\}\end{align}$$

In this, I haven't made any use of the $c$-expander property, but I don't think it will make much of a difference.


问:
how to tell whether x and y are independent or not
Suppose that $f_{x,y}(x,y) = \lambda^2 e^{\displaystyle-\lambda(x+y)}, 0\leq x , 0\leq y.$ Find $\operatorname{Var(X+Y)}$.

I'm having trouble with this problem the way to find $\operatorname{Var(X+Y)} = \operatorname{Var(X)}+\operatorname{Var(Y)}+2\operatorname{Cov(X,Y)}$, however if $X$ and $Y$ are independent, then $\operatorname{Cov(X, Y)}=0$, the answers indicated that $X$ and $Y$ are independent since they just used $\operatorname{Var(X+Y)} = \operatorname{Var(X)}+\operatorname{Var(Y)}+0$, my question is how do I tell whether $X$ and $Y$ are independent or not, based on looking only at $f_{x,y}(x,y) = \lambda^2 e^{\displaystyle-\lambda(x+y)}, 0\leq x , 0\leq y.$

ContentLicense: CC BY-SA 3.0

Body: $$f_{xy}(x,y)\neq f_X(x)f_Y(y)$$


问:
Do closed, connected subsets of manifolds always admit open neighborhoods to which they are homotopy equivalent?
Let $M$ be a topological manifold and $C$ a closed, connected subset of $M$. Can we always find an open neighborhood $U$ of $C$ such that the inclusion $C \to U$ is a homotopy equivalence?

My thought was to take $U$ as a union of small neighborhoods around each point in $C$ and show that $U$ deformation retracts onto $C$. However, I don't know how to make this rigorous. If we're working with smooth manifolds and $C$ a submanifold, we can just take a disk bundle of the normal bundle. But in general, $C$ may not have such a tubular neighborhood.

答: This fails miserably already for $M=\mathbb{R}^2$. The inclusion $C\rightarrow U$ being a homotopy equivalence would imply that $\pi_1(C)\cong\pi_1(U)$, but $U$, being an open subset of $\mathbb{R}^2$, is a noncompact surface and these are known to have free fundamental group (see here). On the other hand, there are plenty of closed, connected subsets of the plane with non-free fundamental group, the most famous one probably being the Hawaiian earring.

ContentLicense: CC BY-SA 4.0


问:
How to prove that every number of the form $245\cdot 22..11-239$ is congruent to $10^m \mod 41$?
Consider Numbers:

\begin{align}
245&\cdot 211-239\\
245&\cdot 2211-239\\
245&\cdot 22211-239\\
245&\cdot 222211-239{}{}{}\\
245&\cdot 2222\ldots11-239
\end{align}
I am stuck, how to prove that every number of this form is congruent to $10^m \bmod 41$ for some $m$?

答:
Let $a_n = 245\cdot 2222\cdots11-239$ with $n$ $2$'s. Then $a_n \equiv 10^{n-1}$. (All congruences are mod $41$.)

Indeed, $a_{n+1}-a_n = 245 \cdot (2 \cdot 10^{n+2}) \equiv 9 \cdot 10^{n-1} $.

Now let $b_n = 10^{n-1}$. Then $b_{n+1}-b_n \equiv 9 \cdot 10^{n-1} \equiv a_{n+1}-a_n$.

Since $a_1 \equiv 1 \equiv b_1 \bmod 41$, we have $a_n \equiv b_n$ for all $n$.


问:
Solving a Cauchy Problem using Method of Characteristics
I came across a partial differential equation (IVP PDE) that I would like to solve:

$$\{u_y+\cos(ky)u_x=ax^2 | u(x,0)=0\}$$

This should be a quasilinear PDE, and is in the format of a Cauchy Problem, in the form of:

$$au_x+bu_y=c$$
Such that a, b, and c are constants.

In my particular case, I have:

$$
\left\{
\begin{array}{c}
a=\cos(ky) \\
b=1 \\
c=ax^2
\end{array}
\right.
$$
Using the Lagrange-Carpit Equations: $$\frac{dx}{\cos(ky)}=\frac{dy}{1}=\frac{du}{ax^2}$$
Rearranging gives: $$\frac{dx}{dy}=\cos(ky)$$
$$dy=\frac{du}{ax^2}$$
I then canclulated $dx-dy$:
$$du=(ax^2\cos(ky)+1)dy-dx$$
Integrate to obtain $u(x,y,a,k)$:
$$u(x,y,a,k)=\frac{1}{k}ax^2\sin(ky)+y-x+C_1$$
I applied the initial condition of $u(x,0)=0$ and made $x=0$, which gave $C_1=0$.
The question is thus finalised.

This however does not look correct. Am I doing the right thing?

I would also like to see what the characteristic curves look like, what are these? Are they essentially the terms $a, b, c$ I obtained above? Or rather, I simply differentiate those three variables in terms of $s$. Essentially calculating $\frac{dy}{ds}$, $\frac{dx}{ds}$ and $\frac{dz}{ds}$.

答:
This is not just a quasilinear PDE - it is in fact simply linear, as it is of the form
$$a(x,y)\partial_x u(x,y)+b(x,y)\partial_yu(x,y)=c(x,y)$$
In particular,
$$\begin{cases}
\cos( ky) \partial _{x} u( x,y) +\partial _{y} u( x,y) =ax^{2}\\
u( r,0) =0
\end{cases}$$
So the method of characteristics is fairly routine here. We have a triple of what are technically speaking partial differential equations, but can be solved as ordinary differential equations.
$$\begin{cases}
\partial _{s} x( r,s) =\cos( k\cdotp y( r,s)) & x( r,0) =r\\
\partial _{s} y( r,s) =1 & y( r,0) =0\\
\partial _{x} z( r,s) =a\cdotp x( r,s)^{2} & z( r,0) =0
\end{cases}$$
Fortunately, these equations can be decoupled. One can notice immediately that $y=s$, giving
$$\partial_sx(r,s)=\cos(ks)\implies x(r,s)=\frac{1}{k}\sin(ks)+f_1(r)$$
And our initial condition gives
$$x(r,s)=\frac{1}{k}\sin(ks)+r.$$
Hence,
$$\partial_s z(r,s)=a\left(\frac{\sin^2(ks)}{k^2}+\frac{2r\sin(ks)}{k}+r^2\right)$$
Which we can integrate directly as
$$z(r,s)=a\left(\frac{1}{k^2}\left(\frac{s}{2}-\frac{\sin(2ks)}{4k}\right)-\frac{2r\cos(ks)}{k^2}+sr^2\right)+f_3(r)$$
Since our initial condition is $z(r,0)=0$ we plug $s=0$ in:
$$a\left(\frac{1}{k^2}\left(\frac{0}{2}-\frac{\sin(2k\cdot 0)}{4k}\right)-\frac{2r\cos(k\cdot 0)}{k^2}+0\cdot r^2\right)+f_3(r)=0$$
Giving
$$f_3(r)=\frac{2ar}{k^2}$$
This gives us the implicit relations
$$\begin{cases}
x=\frac{\sin( ks)}{k} +r\\
y=s\\
z=a\left(\frac{1}{k^{2}}\left(\frac{s}{2} -\frac{\sin( 2ks)}{4k}\right) -\frac{2r\cos( ks)}{k^{2}} +sr^{2}\right) +\frac{2ar}{k^{2}}
\end{cases}$$
The second line is rather fortunate and allows us to say
$$s(x,y)=y$$
Which gives
$$r(x,y)=x-\frac{\sin(ky)}{k}$$
Finally,
$$u(x,y)=z(r(x,y),s(x,y))$$
So,
$$u(x,y)=\frac{a}{4k^3}\left(4k^3x^2y-8\sin(ky)(k^2xy+1)-8kx\cos(ky)+4k(2x+y)+3\sin(2ky)-2ky\cos(2ky)\right)$$
You can check the solution with Mathematica, Maple, etc to confirm it is correct.

BONUS: Here is a plot of the solution for $a=1,k=1$ in the range $x,y\in[-10,10]$.


问:
A question about winding numbers and the Global Version of Cauchy's Theorem
In his book, Introduction to Complex Analysis at a Graduate Level, Serge Lang starts the chapter on the Global Version of Cauchy's Theorem with the following discussion

However, I fail to follow what he means. He talks about how if a path winds around some point outside of $U$ then we will have an integral not equal to zero, but this can't be true. Take the integral along the unit circle of $\frac{1}{z-5}$. And furthermore, he mentions that the integrals of functions such as $f(z)=\frac{1}{z - \alpha}$ where $\alpha$ is a pint not in $U$ have an integral =/0 for closed paths in $U$, which is again false (for them to have values not equal to zero doesn't do point need to be in $U$? For example $f(z)=\frac{1}{z}$ along the unit circle has a value of $2\pi i$ and the point 0 and the closed path are both in the open set, say the circle of radius 2 centered at the origen.

I think I am probably missing something. Is whats written in the discussion by Serge Lang correct? And if yes could someone please help me understand where I have gone wrong?

All help is really appriciated.

答:
Perhaps that you are missing that $U$ is the domain of $f$. So, in your example with $f(z)=\frac1z$, the natural domain of $f$ is $U=\Bbb C\setminus\{0\}$. And so the only point outside $U$ is $0$. And the integral of $f$ along $\gamma\colon[0,2\pi]\longrightarrow\Bbb C$ defined by $\gamma(t)=Re^{it}$ $(R>0$) is indeed $2\pi i$, which is different from $0$.

On the other hand, note that Lang wrote that if the path winds around some point outside of $U$, then we can find functions whose integral is not equal to $0$. He does not claim that the integral is always different from $0$.


问:
Show that for all nonnegative sequences $\{a_k:1 \leq k \leq n\}$, one has $(\sum_{k=1}^n a_k^{1/2} )^2 \leq (\sum_{k=1}^n a_k^{1/3} )^3.$
Problem: Show that for all nonnegative sequences $\{a_k:1 \leq k \leq n\}$, one has $$\left(\sum_{k=1}^n a_k^{1/2} \right)^2 \leq \left(\sum_{k=1}^n a_k^{1/3} \right)^3.$$
This is a problem from the book "The Cauchy Schwarz Master Class".

Question:  In the solution, it says that the inequality is homogeneous of order one, so we can assume that $\sum\limits_{k=1}^n a_k^{1/3} = 1$. But I'm not quite sure how can we assume that is the case here. Can anyone explain to me what this sentence mean?

答:
Let $\sum_{k=1}^n a_k^{1/3} = A^{1/3}$ for some $A,$ and define $b_k = \dfrac{a_k}A$.

If the inequality holds for $b_k,$ then:
$$
\left(\sum_{k=1}^n b_k^{1/2} \right)^2 \leq \left(\sum_{k=1}^n b_k^{1/3} \right)^3 \\
\implies 1/A \left(\sum_{k=1}^n a_k^{1/2} \right)^2 \leq 1/A \left(\sum_{k=1}^n a_k^{1/3} \right)^3 \\
$$


问:
How to calculate the volume of a rectangular section of a sphere?
I may not be using the correct terminology; please correct me if so.

There is an existing question How to calculate the area covered by any spherical rectangle? which is similar to mine, but I am looking for the volume of the pyramid-like shape bounded by a spherical rectangle and the four planes formed by each of its sides when taken together with the center of the sphere. That is, if the corners of the spherical rectangle are labeled A, B, C, and D, and the center of the sphere is labeled S, I am looking for the volume bounded by the spherical rectangle ABCD and the four planes ABS, BCS, CDS, and DAS.

答:
If the spherical rectangle $ABCD$ has area $S$, covering a $\frac{S}{4\pi r^2}$ fraction of the surface of the sphere, then the pyramid-like region covers the same fraction of the volume. From $\frac{S}{4\pi r^2} = \frac{V}{\frac43 \pi r^3}$ we get $V = \frac r3 \cdot S$.

The same holds for any shape on the surface of a sphere (and the spherical section we get if we extend every point of that shape to the center of the sphere).


问:
Present Value of Perpetuity
Really really struggling with the following question:

Suppose you receive $\$44000$ per year, but get a $\$1000$ raise every $k = 9$ years. That is, you are paid $\$44000$ in years $1, 2, ..., k$, and paid $\$44000 + \$1000$ in years $k+1, k+2, ..., 2k$, and paid $\$44000+\$2000$ in years $2k+1, 2k+2, ..., 3k$, and so on into perpetuity.  Given an interest rate $2.7\%$, what is the present value of these cash flows?  (to nearest $\$0.01$) (Hint: work out a formula rather than just calculating numerically. Can you convert this question to the PV of an annuity of length 9 years?)

The answer should be $\$1,766,315.72$.

I tried calculating annuity for $n=9$ first and get the annuity of $\$44,000+\$1000$ and $\$44,000+\$2000$ separately. Is that what I do? If yes, how?

Any help would be appreciated! Thanks.

答:
This is an increasing perpetuity that has the following cash flow, in thousands:  $$PV = 44 (v + v^2 + \cdots + v^9) + 45 v^{9} (v + v^2 + \cdots + v^9) + 46 v^{18} (v + v^2 + \cdots + v^9) + \cdots$$ where $v = 1/(1+i)$ is the annual present value discount factor.  Note we are assuming that payments are made in arrears, so this is a perpetuity-immediate and the present value is calculated with respect to one year before the first payment of $44$.  Clearly, we may perform a factorization:
$$PV = (v + v^2 + \cdots + v^9)(44 + 45 v^{9} + 46 v^{18} + \cdots ),$$ and the first factor is simply $$\require{enclose} a_{\enclose{actuarial}{9}i} = \frac{1 - v^9}{i} \approx 7.89619.$$  The second factor is $$44(1 + v^9 + v^{18} + \cdots) + (v^9 + 2v^{18} + 3v^{27} + \cdots) = 44 \ddot a_{\enclose{actuarial}{\infty} j} + (Ia)_{\enclose{actuarial}{\infty} j},$$ where $j = (1+i)^{9} - 1$ is the effective nine-year rate of interest.  Since a perpetuity-due has a present value of $$\ddot a_{\enclose{actuarial}{\infty} j} = 1 + \frac{1}{j},$$ and an increasing perpetuity-immediate has a present value of $$(Ia)_{\enclose{actuarial}{\infty} j} = \frac{1}{j} + \frac{1}{j^2},$$ the second factor in our equation of value is $$44(4.6905) + 17.3103 \approx 223.692.$$  Therefore the present value is $$PV \approx (7.89619) ( 223.692) = 1766.31472367$$ in units of thousands of dollars.

As an addendum, the formulas for the present values of the perpetuities can be found as follows.  As usual, $i$ is some periodic interest rate and $v = 1/(1+i)$.  We assume $i > 0$, otherwise the present value is infinite.  Then $0 < v < 1$ and

First, $$\ddot a_{\enclose{actuarial}{\infty} i} = 1 + v + v^2 + \cdots = \frac{1}{1-v} = \frac{1}{1 - \frac{1}{1+i}} = \frac{1+i}{i} = 1 + \frac{1}{i},$$ from the formula for the sum of an infinite geometric series.
Hence, $$a_{\enclose{actuarial}{\infty} i} = v + v^2 + v^3 + \cdots = \frac{1}{i},$$ which is the present value of a perpetuity-immediate.
Finally, let $$x = (Ia)_{\enclose{actuarial}{\infty} i} = v + 2v^2 + 3v^3 + \cdots.$$  Then $$vx = v^2 + 2v^3 + 3v^4 + \cdots,$$ and $$
vx + a_{\enclose{actuarial}{\infty} i} = (v^2 + 2v^3 + 3v^4 + \cdots) + (v + v^2 + v^3 + \cdots) = v + 2v^2 + 3v^3 + \cdots = x.$$  Solving for $x$ yields $$x = (Ia)_{\enclose{actuarial}{\infty} i} = a_{\enclose{actuarial}{\infty}i} \cdot \frac{1}{1-v} = (a_{\enclose{actuarial}{\infty}i}) (\ddot a_{\enclose{actuarial}{\infty}i}) = \frac{1}{i} + \frac{1}{i^2},$$ as claimed.

Title: Present Value of Perpetuity


问:
Concentration of coordinatewise Lipschitz function
Let $f:\mathbb R^n \to \mathbb R$ be coordinatewise $1$-Lipschitz, i.e $|f(x')-f(x)| \le |x'_k-x_k|$ whenever $x=(x_1,\ldots,x_n)$ and $x' = (x'_1,\ldots,x_n')$ are two vectors in $\mathbb R^n$ which only differ in their $k$th coordinates. Let $X=(X_1,\ldots,X_n)$ be a random vector in $\mathbb R^n$ with coordinates drawn iid from $\mathcal N(0,1)$ and define $Y:=f(X)$

Question. What is a non-trivial concentration inequality for $Y$ ?

Note. The particular function I have in mind is $f(x) := \max_{i \ne j} |x_i - x_j|$.

Observation

If one is too lazy, one can observe that any such $f$ is $n$-Lipschitz w.r.t the euclidean norm, and so by a standard result, we get
$\mathbb P(|Y - \mathbb E[Y]| \ge t) \le 2e^{-t^2/(2n^2)}$. I'm hoping for a concentration inequality with a much better dependence on the dimension $n$. It is in this sense that I mean "nontrivial concentration inequality".

答:
You can get a tighter dependence on $n$ by using McDiarmid for Rademacher r.v.'s then using a standard "tensorization trick" to have the result carry over to the Gaussian measure: essentially, looking at $(\{-1,1\}^m)^n$ and define your new function $g\colon \{-1,1\}^{mn} \to \mathbb{R}$ by
$$
g(x) = f\!\left( \frac{1}{\sqrt{m}}\sum_{k=1}^m x_k, \frac{1}{\sqrt{m}}\sum_{k=m1}^{2m} x_k, \dots, \frac{1}{\sqrt{m}}\sum_{k=(n-1)m+1}^{nm} x_k\right)
$$
and applying McDiarmid to $g$ (which is $1/\sqrt{m}$-Lipschitz in each coordinate), before letting $m\to\infty$. See, e.g., Corollary 2.1 in these lecture notes.

This gives that, under your assumption of coordinate-wise $1$-Lipschitzness of $f\colon\mathbb{R}\to\mathbb{R}$ and under the standard Gaussian distribution,
$$
\mathbb{P}\{ |f(X)-\mathbb{E}[f(X)] | \geq t \} \leq 2e^{-2t^2/n} \tag{$\dagger$}
$$

Moreover, this bound is tight, as one can check by taking $f(x) = \sum_{k=1}^n x_k$ (as then $f(X)$ is itself Gaussian with mean $0$ and variance $n$).


问:
Applications of the Fibonacci sequence
The Fibonacci sequence is very well known, and is often explained with a story about how many rabbits there are after $n$ generations if they each produce a new pair every generation. Is there any other reason you would care about the Fibonacci sequence?

答: Suppose you're writing a computer program to search a sorted array for a particular value. Usually the best method to use is a binary search. But binary search assumes it's the same cost to read from anywhere in the array. If it costs something to move from one array element to another, and the cost is proportional to how many array elements you need to skip over to get from one element you read to the next, then Fibonacci search works better. This can apply to situations like searching through arrays that don't fit entirely in your computer's cache so it's generally cheaper to read nearby elements that distant ones.</p>

Title: Applications of the Fibonacci sequence

ContentLicense: CC BY-SA 2.5


问:
Minimize Boolean function
I have got some silly task, but I am quite confused.

Need to minimize function.

$$f(x_1,x_2,x_3,x_4)=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4.$$

Thanks.

Sorry for my English. Minimize Boolean function

Title: Minimize Boolean function

Body:
Do you mean to reduce the form? i.e. something like
$$ \begin{eqnarray}
 f(x_1,x_2,x_3,x_4) &=& x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4 \nonumber \\
                    &=& x_1(x_2+x_3+x_4)+x_2(x_3+x_4) \nonumber \\
                    &=& x_1x_2+(x_1+x_2)(x_3+x_4)
\end{eqnarray}$$


问:
Extending the Lebesgue Measure
In D.Williams Probability with Martingales, we define the Lebesgue Measure as follows:

We have a set function $\mu_0 = \sum_{k \leq r} (b_k-a_k)$ where the sigma algebra over (0,1] is defined to be the collection of intervals of the form $F = (a_1,b_1] \text{ } \cup \text{ } ... \text{ } \cup \text{ } (a_r,b_r]$.

We then use caratheodory's extension theorem to extend this set function to the Lebesgue measure on $((0,1], \mathcal{B}(0,1])$.

Then, Williams says, to obtain the Lebesgue measure on $([0,1], \mathcal{B}[0,1])$, we need to define the Lebesgue measure of the set {0} as zero. My question is why do we have to do this?

答:
The algebra $\Sigma_0$ is defined in the following way. $F \in \Sigma_0$ if $F$ can be written as $F = (a_1,b_1] \cap ... \cap (a_r,b_r].$ $\Sigma_0$ defined like that is an algebra on (0,1] and this allows  to use later the theorem 1.7 to extend $(\Sigma_0,\mu_0)$ to $(\Sigma,\mu).$

Sets of the type $F = [a_1,b_1] \cap ... \cap [a_r,b_r]$ don't form an algebra on $[0,1]$. For example, if we have $E = [0,1/2],$ then $E^c = (1/2,1] \notin \Sigma_0.$ So you will not be able to use theorem 1.7 and define $\mu$ on (the $\sigma$-algebra) $\Sigma,$ which is the goal.

Title: Extending the Lebesgue Measure


问:
On a log-gamma definite integral
A very famous log-gamma integral due to Raabe is
$$\int_0^1 \log \Gamma (x) \, dx = \frac{1}{2} \log (2\pi).$$
Several proofs of this result can be found here.

I would like to known about the evaluation of the similar looking log-gamma integral

$$I = \int_0^1 \frac{\log \Gamma (x)}{1 - x} \, dx.$$

I very much doubt a closed-form solution for this integral can be found, though I would really like to see one if it could be found (we live in hope you see). Here is what I have managed to come up with so far. Integrating by parts gives
$$I = \int_0^1 \log (1 - x) \psi (x) \, dx.$$
Here $\psi (x)$ is the digamma function. Expanding the log term, after swapping the order of the summation with the integration we have
$$I = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^1 x^n \psi (x) \, dx.$$
Now amazingly, a closed form solution for $\int_0^1 x^n \psi (x) \, dx$ exists. It is
$$\int_0^1 x^n \psi (x) \, dx = \sum_{k = 0}^{n - 1} (-1)^k \binom{n}{k} \left (\zeta'(-k) - \frac{B_{k + 1} H_k}{k + 1} \right ).$$
Thus
$$I = \sum_{n = 1}^\infty \sum_{k = 0}^{n - 1} \frac{(-1)^{k + 1}}{n} \binom{n}{k} \left (\zeta'(-k) - \frac{B_{k + 1} H_k}{k + 1} \right ).$$
Here $H_n$ is the $n$th harmonic number, $B_n$ is the $n$th Bernoulli number, while $\zeta' (x)$ denotes the derivative of the Riemann zeta function. I am guessing not much more can be done with these sums.

Any other alternative approaches one could try?

答:
By changing $x=1-t$, the integral reads
\begin{align}
 I &= \int_0^1 \frac{\log \Gamma (x)}{1 - x} \, dx\\
 &=\int_0^1\frac{\log \Gamma (1-t)}t \, dt
\end{align}
Then, using the identity (DLMF)
\begin{equation}
 \sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k}t^{k}=-\gamma t+\ln\Gamma\left(1-t\right)
\end{equation}
one obtains, by swapping summation and integral,
\begin{align}
 I&=\int_0^\infty\left[\gamma t+\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k}t^{k}\right]\frac{dt}{t}\\
 &=\gamma+\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k^2}
\end{align}
The series can be made more rapidly convergent by introducing the Basel series:
\begin{equation}
 I=\frac{\pi^2}{6}-1+\gamma+\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)-1}{k^2}
\end{equation}

Title: On a log-gamma definite integral


问:
Correct use of $\rightarrow$ "tends to" relation.
Which of the following two examples is the correct use of the $\rightarrow$ "tends to" relation?

Example 1

$$ \lim_{n \rightarrow \infty} \frac{\pi(n)}{n/\ln(n)} = 1 $$

Example 2
$$ \lim_{n \rightarrow \infty} \frac{\pi(n)}{n/\ln(n)} \rightarrow 1 $$

If both are valid, or conventional, please do state in which contexts the traditions exist.

答: Use example $1$, as the limit is a number by definition. However, there is another option: $\tfrac{\pi(n)}{n/\ln n}\to1$ as $n\to\infty$. These clauses may be listed in the other order, especially when describing several $n\to\infty$ limits.


答:
The correct notation is the first one. The second is wrong since a limit cannot “tend” to anything. The limit is a real number (in this case).

The limit equals the value, while the sequence of numbers tends to the value. We say that a sequence of numbers “tends” to a number when we mean that its limit “equals” 1.

Title: Correct use of $\rightarrow$ "tends to" relation.

Body:
Which of the following two examples is the correct use of the $\rightarrow$ "tends to" relation?

Example 1

$$ \lim_{n \rightarrow \infty} \frac{\pi(n)}{n/\ln(n)} = 1 $$

Example 2
$$ \lim_{n \rightarrow \infty} \frac{\pi(n)}{n/\ln(n)} \rightarrow 1 $$

If both are valid, or conventional, please do state in which contexts the traditions exist.


问:
How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$
I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}$$
and
$$I_n=\int_0^1\frac{x-1}{x^{n+1}-1}dx$$
It makes sense to me that, for all values of $n$, $I_n$ is convergent since the bottom of the function is always above zero and the integral exists for $n\to\infty$ however I cannot seem to find a nice closed form for this.

One thought I did have was using:
$$\sum\ln(x_i)=\ln\left(\prod x_i\right)$$
but I cannot seem to make it work. Does anyone have any hints for this type of problem as I would like to try and complete it myself. Thanks :)

Body:
If you wanna find a closed form for that integral then you must do some partial fractions.

Let $ n\in\mathbb{N}^{*} $,

\begin{aligned}\frac{x-1}{x^{n}-1}=\frac{1}{\prod\limits_{k=1}^{n-1}\left(x-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{n}}\right)}=\frac{2\mathrm{i}}{n}\sum_{k=1}^{n-1}{\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{n}}\sin{\left(\frac{k\pi}{n}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{n}}}}\end{aligned}

If working with an even integer :

\begin{aligned}\small\frac{x-1}{x^{2n}-1}=\frac{\mathrm{i}}{n}\sum_{k=1}^{2n-1}{\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{2n}}\sin{\left(\frac{k\pi}{2n}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}}}&\small=\frac{\mathrm{i}}{n}\left(\sum_{k=1}^{n-1}{\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{2n}}\sin{\left(\frac{k\pi}{2n}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}}}-\frac{\mathrm{i}}{x+1}+\sum_{k=n+1}^{2n-1}{\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{2n}}\sin{\left(\frac{k\pi}{2n}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}}}\right)\\ &\small=\frac{\mathrm{i}}{n}\left(\sum_{k=1}^{n-1}{\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{2n}}\sin{\left(\frac{k\pi}{2n}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}}}-\frac{\mathrm{i}}{x+1}+\sum_{k=1}^{n-1}{\frac{\mathrm{e}^{\mathrm{i}\frac{3\left(2n-k\right)\pi}{2n}}\sin{\left(\frac{\left(2n-k\right)\pi}{2n}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{\left(2n-k\right)\pi}{n}}}}\right)\\ &\small=\frac{1}{n\left(x+1\right)}+\frac{\mathrm{i}}{n}\sum_{k=1}^{n-1}{\left(\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{2n}}\sin{\left(\frac{k\pi}{2n}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}}-\frac{\mathrm{e}^{-\mathrm{i}\frac{3k\pi}{2n}}\sin{\left(\frac{k\pi}{2n}\right)}}{x-\mathrm{e}^{-\mathrm{i}\frac{k\pi}{n}}}\right)}\\ \small\frac{x-1}{x^{2n}-1}&\small=\frac{1}{n\left(x+1\right)}+\frac{2}{n}\sum_{k=1}^{n-1}{\frac{\sin{\left(\frac{k\pi}{2n}\right)}-x\sin{\left(\frac{3k\pi}{2n}\right)}}{x^{2}-2x\cos{\left(\frac{k\pi}{n}\right)}+1}\sin{\left(\frac{k\pi}{2n}\right)}}\end{aligned}

If our integer is odd :

\begin{aligned}\small\frac{x-1}{x^{2n+1}-1}=\frac{2\mathrm{i}}{2n+1}\sum_{k=1}^{2n}{\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{2n+1}}\sin{\left(\frac{k\pi}{2n+1}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}}}&\small=\frac{2\mathrm{i}}{2n+1}\left(\sum_{k=1}^{n}{\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{2n+1}}\sin{\left(\frac{k\pi}{2n+1}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}}}+\sum_{k=n+1}^{2n}{\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{2n+1}}\sin{\left(\frac{k\pi}{2n+1}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}}}\right)\\ &\small=\frac{2\mathrm{i}}{2n+1}\left(\sum_{k=1}^{n}{\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{2n+1}}\sin{\left(\frac{k\pi}{2n+1}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}}}+\sum_{k=1}^{n}{\frac{\mathrm{e}^{\mathrm{i}\frac{3\left(2n+1-k\right)\pi}{2n+1}}\sin{\left(\frac{\left(2n+1-k\right)\pi}{2n+1}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{2\left(2n+1-k\right)\pi}{2n+1}}}}\right)\\ &\small=\frac{2\mathrm{i}}{2n+1}\sum_{k=1}^{n}{\left(\frac{\mathrm{e}^{\mathrm{i}\frac{3k\pi}{2n+1}}\sin{\left(\frac{k\pi}{2n+1}\right)}}{x-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}}-\frac{\mathrm{e}^{-\mathrm{i}\frac{3k\pi}{2n+1}}\sin{\left(\frac{k\pi}{2n+1}\right)}}{x-\mathrm{e}^{-\mathrm{i}\frac{2k\pi}{2n+1}}}\right)}\\ \small\frac{x-1}{x^{2n+1}-1}&\small=\frac{4}{2n+1}\sum_{k=1}^{n}{\frac{\sin{\left(\frac{k\pi}{2n+1}\right)}-x\sin{\left(\frac{3k\pi}{2n+1}\right)}}{x^{2}-2x\cos{\left(\frac{2k\pi}{2n+1}\right)}+1}\sin{\left(\frac{k\pi}{2n+1}\right)}}\end{aligned}

Thus, we have in general : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{x-1}{x^{n}-1}=\frac{1+\left(-1\right)^{n}}{n\left(x+1\right)}+\frac{4}{n}\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}{\frac{\sin{\left(\frac{k\pi}{n}\right)}-x\sin{\left(\frac{3k\pi}{n}\right)}}{x^{2}-2x\cos{\left(\frac{2k\pi}{n}\right)}+1}\sin{\left(\frac{k\pi}{n}\right)}}\end{array}$} $$

If you know how to solve an integral like $ \int{\frac{cx+d}{x^{2}-ax+1}\,\mathrm{d}x} $, then it would be simple to figure out what the closed form would be.

If you just want to find the limit, then we have for any $ n\in\mathbb{N} $ : $$ \int_{0}^{1}{\frac{1-x}{1-x^{n}}\,\mathrm{d}x}=\frac{1}{2}+\int_{0}^{1}{\frac{x^{n}\left(1-x\right)}{1-x^{n}}\,\mathrm{d}x}=\frac{1}{2}+\frac{1}{n}\int_{0}^{1}{\frac{x^{\frac{1}{n}}\left(1-x^{\frac{1}{n}}\right)}{1-x}\,\mathrm{d}x} $$

Since : $$ \left|\frac{1}{n}\int_{0}^{1}{\frac{x^{\frac{1}{n}}\left(1-x^{\frac{1}{n}}\right)}{1-x}\,\mathrm{d}x}\right|\leq\frac{1}{n}\int_{0}^{1}{\frac{x^{\frac{1}{n}}\left(1-x\right)}{1-x}\,\mathrm{d}x}=\frac{1}{n+1}\underset{n\to +\infty}{\longrightarrow}0 $$

The limit would be $ 1/2 $.


问:
Function which is strongly convex and L-smooth
Let $f:\mathbb{R}^{n}\to \mathbb{R}$. We say that $f$ is $\mu$ strongly convex if
$$ f(y)\ge f(x)+\nabla f(x)^T(y-x)+\frac{\mu}{2}\lVert x-y \rVert^2$$
and we say that $f$ is $\mu$ smooth if
$$ f(y)\le f(x)+\nabla f(x)^T(y-x)+\frac{\mu}{2}\lVert x-y \rVert^2$$
If $f$ is both of them, how can I prove that
$$f(x)=\frac{\mu}{2}\lVert x-b\rVert^2+c$$
It is obvius that
$$f(y)=f(x)+\nabla f(x)^T(y-x)+\frac{\mu}{2}\lVert x-y \rVert^2$$
but I can not do any more. Any help will be appreciated.

答: You also don't have to do more, $x$ is fix. In the case $n=1$ it is easiest to see. $$f(x)+ f'(x)(y-x) + \frac{\mu}{2} (x-y)^2 = \frac{\mu}{2}y^2 + (f'(x)+\mu x)y +(f(x)+\frac{\mu}{2}x^2)$$ now complete the square to get something of the form $$\frac{\mu}{2}(y-b)^2 + c.$$ In higher dimensions this is the same (coordinatewise).


问:
I can't find the pattern while evaluating $\int_0^{\pi/2}\sin^n(x)\,dx$
While integrating:
$$
\int_0^{\frac{\pi}{2}}\sin^nx \, dx
$$
I have noticed that when $n$ is even the value is a multiple of $\pi$, but when $n$ is odd it is rational.

My results are:

n=1, 1

n=2, $\frac{\pi}{4}$

n=3, $\frac{2}{3}$

n=4, $\frac{3\pi}{16}$

n=5, $\frac{8}{15}$

n=6, $\frac{5\pi}{32}$

n=7, $\frac{16}{35}$

n=8, $\frac{35\pi}{256}$

and so on...

Is there a pattern? And if so what is it?

答:
These are Wallis' integrals : https://en.wikipedia.org/wiki/Wallis%27_integrals

An integration by parts leads to $W_{n+2}=\frac{n+1}{n+2}W_n$ therefore $W_{2n}=\frac{(2n)!\pi}{2^{2n+1}(n!)^2}$ and $W_{2n+1}=\frac{4^n(n!)^2}{(2n+1)!}$.


问:
Does a function with two frequencies have a special name?
Consider a real function of the form $f(x) = A \cos(x) + B \cos(2x)$.   Does this type of function have a special name? At first, I thought it could be bi-periodic, or bi-harmonic, or doubly-periodic, but I see that these terms are already used for other kinds of functions.

答: The function is still just periodic. With period $2\pi$.


答: As @StefanLafon notes, using integration by parts twice viz.$$\int_0^{\pi/2}uv^{\prime\prime}dx=[uv^\prime]_0^{\pi/2}-\int_0^{\pi/2}u^\prime v^\prime \,dx = [uv^\prime-u^\prime v]_0^{\pi/2} + \int_0^{\pi/2} u^{\prime\prime} v \, dx$$with $u=\sin^{n+1}x,\,v=-\sin x$ gives$$\int_0^{\pi/2}\sin^{n+2}x\,dx=[n\sin^{n+1}x\cos x]_0^{\pi/2}-(n+1)\int_0^{\pi/2}(n\sin^nx-(n+1) \sin^{n+2}x) \, dx,$$where we've used $\cos^2x=1-\sin^2x$. Rearranging,$$\int_0^{\pi/2}\sin^{n+2}x \, dx=\frac{n+1}{n+2}\int_0^{\pi/2}\sin^nx \, dx.$$You can also write your integral in terms of the Beta function:$$\int_0^{\pi/2}\sin^ndx=\tfrac12\operatorname{B}\left(\tfrac12,\,\tfrac{n+1}{2}\right)=\tfrac{\sqrt{\pi}\Gamma\left(\tfrac{n+1}{2}\right)}{2\Gamma\left(\tfrac{n+2}{2}\right)}.$$

Title: I can't find the pattern while evaluating $\int_0^{\pi/2}\sin^n(x)\,dx$

Body:
While integrating:
$$
\int_0^{\frac{\pi}{2}}\sin^nx \, dx
$$
I have noticed that when $n$ is even the value is a multiple of $\pi$, but when $n$ is odd it is rational.

My results are:

n=1, 1

n=2, $\frac{\pi}{4}$

n=3, $\frac{2}{3}$

n=4, $\frac{3\pi}{16}$

n=5, $\frac{8}{15}$

n=6, $\frac{5\pi}{32}$

n=7, $\frac{16}{35}$

n=8, $\frac{35\pi}{256}$

and so on...

Is there a pattern? And if so what is it?


问:
Evaluate the following definite integral $\int_{0}^t\cos(\sin s)\, ds$?
Evaluate
$$\int_{0}^t\cos(\sin s)\, ds\ \ \ \text{or}\ \ \ \int_0^{\sin t}\frac{\cos p \, dp}{\sqrt{1-p^2}}.$$
Please help how to start.

答:
As proposed in the comments, we can use Jacobi-Anger expansion:
$$\cos(\sin(s)) = J_0(1)+2\sum_{n=1}^\infty J_{2n}(1)\cos(2ns).$$
Integrating with respect to $s$, we obtain:
$$I = \int_0^t\cos(\sin(s))ds = t\left(J_0(1)+2\sum_{n=1}^\infty J_{2n}(1)\frac{\sin(2nt)}{2nt}\right).$$

Maybe the only cases with closed form solution are for $t = k\pi$, $k\in\mathbb{R}^+$, when $\sum_{n=1}^\infty J_{2n}(1)\frac{\sin(2\pi n)}{2\pi n}=0$, in which case $I=k\pi J_0(1)\approx 2.40393943k$.

Somehow interesting, since $\frac{\sin(2nt)}{2nt}\leq \frac{1}{2nt}$, and $\sum_{n=1}^\infty J_{2n}(1) = \frac{1-J_0(1)}{2}$, we can upper bound the integral as
$$I = \color{red}{\int_0^t\cos(\sin(s))ds} = tJ_0(1)+2t\sum_{n=1}^\infty J_{2n}(1)\frac{\sin(2nt)}{2nt}\leq tJ_0(1)+2\sum_{n=1}^\infty\frac{J_{2n}(1)}{2n}<\color{blue}{(t-1)J_0(1)+1},$$

thus the dominant term is $tJ_0(1)$ (see the picture).


问:
Evaluate $\lim\limits_{n\to +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$
How to evaluate:
$$\lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$$

Personally, i always have trouble to evaluate sum of trignometric series, if you know a paper to recommend, or part of a book, about this, please let me know.

Anyway

$$\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right) = \frac{1}{n^2}\left(\sum_{k=1}^{n} \frac{1}{\sin^2\left(\frac{\pi k}{2n+1}\right)} - \sum_{k=1}^{n} 1\right)$$

What now?

答:
Let $P_n(X)=\frac{(X+i)^{2n+1}-(X-i)^{2n+1}}{2i}$, then
$$ \begin{aligned} P_n(z)=0&\iff \left(\frac{z+i}{z-i}\right)^{2n+1}=1 \\
&\iff \exists k\in[\![1,2n+1]\!],\frac{z+i}{z-i}=e^{\frac{2ik\pi}{2n+1}} \\
&\iff\exists k\in[\![1,2n+1]\!],z=i\frac{e^{\frac{2ik\pi}{2n+1}}+1}{e^{\frac{2ik\pi}{2n+1}}-1} \\
&\iff\exists k\in[\![1,2n+1]\!],z=\cot\left(\frac{k\pi}{2n+1}\right)
\end{aligned}$$
Therefore the roots of $P_n$ are $\alpha_k:=\cot\left(\frac{k\pi}{2n+1}\right)$. Moreover, using the binomial theorem, we have
$$ P_n(X)=\frac{1}{2i}\sum_{k=0}^{2n+1}\binom{2n+1}{k}\left(i^k-(-i)^k\right)X^{2n+1-k}=\sum_{k=0}^n\binom{2n+1}{2k+1}(-1)^k X^{2n-2k}$$
because the even terms cancel each other. Let $\displaystyle Q_n(X)=\sum_{k=0}^n\binom{2n+1}{2k+1}(-1)^k X^{n-k}$, then $Q_n(X^2)=P_n(X)$ and the roots of $Q_n$ are the $\alpha_k^2=\cot^2\left(\frac{k\pi}{2n+1}\right)$. But since $\alpha_{2n+1-k}=-\alpha_k$ it remains only $n=\deg Q_n$ roots which are the $\alpha_k^2$ for $k\in[\![1,n]\!]$. Therefore, the lead coefficient of $Q_n$ being $2n+1$, we have
$$ Q_n(X)=(2n+1)\prod_{k=1}^n(X-\alpha_k^2)$$
Thus, the sum $\sum_{k=1}^n\alpha_k^2$ is the coefficient of degree $n-1$ in $Q_n$ divided by $-\frac{1}{2n+1}$, that is
$$\sum_{k=1}^n\alpha_k^2=\frac{1}{2n+1}\binom{2n+1}{3}=\frac{n(2n-1)}{3}$$
Finally,
$$ \lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^n\cot^2\left(\frac{k\pi}{2n+1}\right)=\frac{2}{3} $$


问:
Number of ways for two people to select k items each such that every item occurs even number of times
Select $a_1,\dots,a_n, b_1,\dots, b_n \in \mathbb{N}_{\ge 0}$ (non-negative integers) such that $|a| = \sum_{i=1}^n a_i = k$ and $|b| = \sum_{i=1}^n b_i = k$, and further $a_i + b_i$ is even for each $i$.
How many ways are there?

答:
Hint: So you want $a_i$ to be even iff $b_i$ is even. So select the indices where the $a_i$ and $b_i$ are odd. If you select say $\ell$ of them, then you will have $2x+\ell=k,$ where $x$ now you want to partition it however you like, no? So you will need $\ell$ to have the same parity as $k.$ Add over all possibilities and use Stars and Bars when you have fixed the indices.

I get something like:

 $$\sum _{\ell \equiv k \pmod 2}\binom{n}{\ell}\binom{\frac{k-\ell}{2}+n-1}{n-1}^2$$


问:
Find constant term in the expansion of $\left (1+\frac x2 -\frac 2x \right)^4$
Finding the constant term for

$$\left (1+\frac x2 -\frac 2x \right)^4$$

is easy, but that would require converting the expression into a binomial.

However, I have no idea about how to do that.   Completing squares doesn’t work because we have a negative sign on $\frac 2x$.

答:
And Now for Something Completely Different:

Let $f_k= \begin{cases} -2, & k=-1, \\
1,& k=0,\\
{1 \over 2},& k=1,\\
0, & \text{otherwise}\end{cases}$,

Formally we have $1+{x \over 2}-{2 \over x}  = \sum_k f_k x^k$ and so
the constant term of $(1+{x \over 2}-{2 \over x})^4 $ is found from the convolution
$\sum_{k_1+...+k_4 = 0} f_{k_1} ... f_{k_4}$.

Note that if any of $k_1,...,k_4$ are not in $\{-1,0,1\}$ then $f_{k_1} ... f_{k_4} = 0$.

Let $K= \{ (k_1,...,k_4) | k_1+...+k_4 = 0, k_i \in \{-1,0,1\} \}$.

It is not hard to see that the $K$ contains 1 element with no $-1$, $12 = 4 \cdot 3$ elements with exactly one $-1$ and $\binom{4}{2}$ elements with exactly two $-1$s. Computing the product $f_{k_1} ... f_{k_4}$ for each gives
$1 \cdot 1^4 + 12 \cdot (-2 \cdot {1 \over 2}) +6 ((-2)^2 \cdot ({1 \over 2})^2 ) = -5$.


问:
Markov's Chain stationary matrix determination
I have the following transition matrix $P$

\begin{equation}
P=
\begin{pmatrix}
1/2 & 1/3 & 1/6\\
1/4 & 3/4 & 0\\
1/5 & 2/5 & 2/5
\end{pmatrix}
\end{equation}

I have tried to calculate the stationary matrix as
\begin{equation}
P^n= C D^n C^{-1}
\end{equation}

With $D=C^{-1} P C$ the diagonal matrix. Calculating $C$ with the eigenvalues and eigenvectors

\begin{equation}
C=
\begin{pmatrix}
v_1 & w_1 & z_1\\
v_2 & w_2 & z_2\\
v_3 & w_3 & z_3
\end{pmatrix}
\end{equation}

where the eigenvectors are

\begin{align}
\vec{v} =\langle (v_1,v_2,v_3)\rangle &= \langle(1,1,1)\rangle \\
\vec{w} =\langle(w_1,w_2,w_3)\rangle &= \langle\left(\frac{2i\sqrt{39}+3}{6},\frac{-3i\sqrt{39}-7}{16},1\right)\rangle \\
\vec{v} =\langle(v_1,v_2,v_3)\rangle &= \langle\left(\frac{-2i\sqrt{39}+3}{6},\frac{3i\sqrt{39}-7}{16},1\right)\rangle \\
\end{align}

so the diagonal matrix is
\begin{equation}
D=
\begin{pmatrix}
1 & 0 & 0\\
0 & \frac{-i\sqrt{39}+39}{120} & 0\\
0 & 0 & \frac{i\sqrt{39}+39}{120}
\end{pmatrix}
\end{equation}

how can i calculate $D^{n}$? is the diagonalization correct?

I tried by other way using $BP=B$ where B is the final vector state, then

\begin{align}
[p_1 \quad p_2 \quad p_3]  \begin{pmatrix}
1/2 & 1/3 & 1/6\\
1/4 & 3/4 & 0\\
1/5 & 2/5 & 2/5
\end{pmatrix} = [p_1 \quad p_2 \quad p_3]
\end{align}

using Gauss-Jordan elimination for solving this system it gives me infinite solutions.

I dont know if it is a problem with the transition matrix.

答:
Since $1$ is the dominant eigenvalue of the Markov Chain (the derivation of that fact can be found elsewhere, thus I leave that up to the reader to investigate if desired), all we need to solve is

$$ \begin{bmatrix} 1/2 & 1/4 & 1/5 \\1/3 & 3/4 & 2/5 \\ 1/6 & 0 & 2/5 \end{bmatrix} \begin{bmatrix} x  \\y \\ z \end{bmatrix}= \begin{bmatrix} x  \\y \\ z \end{bmatrix}$$
Solving gives infinite (why?) solutions as follows: $x=3.6z,y=6.4z$  with $z$ free. Using the constraint $x+y+z=1$ we find $z=18/55,y=32/55,z=1/11$ and thus we find for the Stationary matrix: $$ \begin{bmatrix} 18/55 & 18/55 & 15/55 \\32/55 & 32/55 & 32/55 \\ 1/11 & 1/11 & 1/11 \end{bmatrix}$$ The rudiments of the calculations is nothing but row operations (Gauss Elimination)


问:
How to find the generators for $y^2 = x^3 + 5x$
This is a problem (3.9b on pg. 113) from "Rational Points on Elliptic Curves 2nd Edition" by Silverman and Tate.

Let $C : y^2 = x^3 + 5x$ and $\overline{C} : y^2 = x^3 -20x $. Let $\Gamma = C(\mathbb{Q})$ and $\overline{\Gamma} = \overline{C}(\mathbb{Q})$. In the textbook, they define the homomorphisms $\alpha : \Gamma\to \mathbb{Q}^*/{\mathbb{Q}^*}^2$ and $\overline{\alpha}: \overline{\Gamma}\to \mathbb{Q}^*/{\mathbb{Q}^*}^2$ by:

$$ \alpha((x,y)) = x \mod{{\mathbb{Q}^*}^2} $$
$$ \alpha((0,0)) = 5 \mod{{\mathbb{Q}^*}^2} $$
$$ \overline{\alpha}((x,y)) = x \mod{{\mathbb{Q}^*}^2} $$
$$ \overline{\alpha}((0,0)) = -20 \mod{{\mathbb{Q}^*}^2} $$
The methodology presented in the book tells us that the rank $r$ of $C$ is equal to:
$$ 2^r = \frac{\left|\alpha(\Gamma)\right|\left|\overline{\alpha}(\overline{\Gamma})\right|}{4} $$
I have found that:
$$ \alpha(\Gamma) = \{ 1,5 \} \subset \mathbb{Q}^*/{\mathbb{Q}^*}^2 $$
$$ \overline{\alpha}(\overline{\Gamma}) = \{ \pm 1, \pm 5 \} \subset \mathbb{Q}^*/{\mathbb{Q}^*}^2 $$
Therefore, the rank of $C$ is $1$. Now, for the second part of the problem it's asking us to find generators for $C(\mathbb{Q})/2C(\mathbb{Q})$.

$$ C(\mathbb{Q}) = \mathbb{Z}\times (\text{torsion subgroup}) $$
$$ C(\mathbb{Q})/2C(\mathbb{Q}) = \mathbb{Z}/2\mathbb{Z}\times (\text{half of torsion subgroup}) $$

The generators for "half of torsion subgroup" are easy to find via Nagell-Lutz. I was wondering how I can find a generator for $\mathbb{Z}/2\mathbb{Z}$ in order to finish the question.

答:
I just figured it out. I will reference notation and results in the book without going into too much detail. But for anyone else reading the book, I hope this answer will be helpful. Following the notation in the book, there are homomorphisms $\psi : \overline{\Gamma} \to \Gamma$ and $\phi: \Gamma \to \overline{\Gamma}$ such that:
$$ \psi(\phi(\Gamma)) = 2\Gamma $$
By a result in the book, we know that $\ker(\alpha) =\psi(\overline{\Gamma})$. Therefore, we simply have to find an element $(x,y)\in \Gamma$ with infinite order such that $\alpha((x,y))\neq 1$. In other words, this will ensure that $(x,y)\notin 2\Gamma$ and hence will indeed be the non-zero element in $\mathbb{Z}/2\mathbb{Z}$.

Explicitly:

Using the method developed in Chapter 3.6, we know that $5\in \alpha(\Gamma)$ if and only if there is a solution in $M,e,N$ with $e > 0$, $M\neq 0$ to the equation $N^2 = 5M^4 + e^4$, and moreover
$$ (x_0,y_0) = \left( 5 \cdot\frac{M^2}{e^2}, 5\cdot \frac{MN}{e^3} \right) $$
will be an affine point on the curve $C$ satisfying $\alpha((x_0,y_0)) = 5$. Thus, with a little bit of luck, I found that $M=2,N=9,e=1$ works, and therefore we have a candidate
$$ \left( 20,90 \right) \notin 2\Gamma $$
It suffices to check that this element has infinite order, which is easy by taking twice the point and checking that its coefficients are non-integer.


问:
If $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

Claim: if $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.

Please, see if I made some mistake in the proof below. I mention some theorems in the proof:

The condition to $f(x)$ be continuous at $x=x_0$ is $\lim\limits_{x\to x_0} f(x)=f(x_0)$.

(1) If $f(x)$ is differentiable at $x-x_0$, then $f'(x)=\lim\limits_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists and the function is defined at $x=x_0$.

(2) Therefore, by the Limit Linearity Theorem, $\lim\limits_{x\to x_0} f(x)$ exists and we'll show it is equals $f(x_0)$.

(3) We'll do this by the Precise Limit Definion: given $ \epsilon>0, \exists\delta|0<|x-x_0|<\delta$, then $0<|f(x)-f(x_0)|<\epsilon$. As this limit exists by (2), we can make $f(x)$ as close to  $f(x_0)$ as one wishes, therefore $\lim\limits_{x\to x_0} f(x)=f(x_0)$, what satisfies the condition for $f(x)$ be differentiable at  $x=x_0$. The end.

答:
Here, a solution I've found at "Introduction to Analysis" byt Arthur Mattuck:

$$\lim_{x\to x_0} (f(x)-f(x_0)) = \lim_{x\to x_0} (\dfrac{f(x)-f(x_0)}{x-x_0})(x-x_0)\\ =  \lim_{x\to x_0} (\dfrac{f(x)-f(x_0)}{x-x_0}) \cdot \lim_{x\to x_0} (x-x_0)\\
=  f'(x)\cdot 0 \\
=0$$


问:
Relative homology group of cone of space with five path components
I am working on Exercise 5.21 from Rotman's algebraic topology book:

Assume that $X$ has five path components. If $CX$ is the cone on $X$, what is $H_1(CX,X)$?

Here is my solution:

We know that there is a long exact sequence $$\dots\to \tilde H_1(CX)\to H_1(CX,X)\to\tilde H_0(X)\to\tilde H_0(CX).$$ Since $CX$ is contractible, we know that $\tilde H_1(CX)=H_1(CX)=0$. This means that the image of the map $\tilde H_1(CX)\to H_1(CX,X)$ is $0$, and so the map $H_1(CX,X)\to\tilde H_0(X)$ has $\ker=0$.

We also know that $\tilde H_0(X)$ is the free abelian group with rank equal to one less than the number of path components of $X$. Thus $\tilde H_0(X)=\mathbb Z^4$ and $\tilde H_0(CX)=0$. Hence the map $H_1(CX,X)\to H_0(X)$ has image isomorphic to $\mathbb Z^4$ (since the map $\tilde H_0(X)\to\tilde H_0(CX)$ is everywhere $0$), from which we conclude that $H_1(CX,X)\cong\mathbb Z^4$.

I don't see any mistakes, but this is one of my first relative homology group computations, so I'd just like to check if the solution is actually correct.

EDIT: I just edited my solution (I messed up a bit by working with regular homology, instead of reduced homology groups; I think the solution is basically the same either way, but I think this one is better).

答:
This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your proof is correct.  If you have an exact sequence
$$0 \to A \to B \to 0,$$
then $A \to B$ is an isomorphism. Take $A = H_1(CX,X)$ and $B = \tilde H_0(X)$.


问:
Matrix Polynomial with Non-Zero Coefficients
Let $A ∈ M_{n \times n}(F)$ for some field $F$. Prove there is some non-zero polynomial
$p(x)$ such that $p(A) = 0$.

I approached this question by dealing with each entry of the matrix and equating it to the respective $0$ on the $0$ matrix but I am stuck now as to what I should do next. Also the question does not allow the use of Cayley Hamilton Theorem.

答: $I,A,A^{2},...,A^{n^{2}+1}$ cannot all be linearly independent because the dimension of the space of $n \times n$ matrices is $n^{2}$. Hence, some non-trivial linear combnation of these must vanish.


问:
What are collinear lines?
I know about collinear points (points lying on a same straight line).

I know about collinear vector (parallel vectors).

But, coming to collinear lines, I am not even able to imagine them, and I can't find any definition.

Exact Question I am dealing with is:
$$\text{ [ The three straight lines ax+by=c,bx+cy=a and cx+ay=b are collinear, if ]}$$

One possible approach I did,
Collinear lines have collinear points, which will have zero area, so while calculating area, I get $0=0$ identity.

So, kindly shed knowledge upon collinear lines.

PS: Googling the exact question gives the same answer on every site. They basically add the equations, and compare the coefficient.

答:
First, I've not seen this usage in 65+ years of looking at mathematics... which leads me to suspect it's either a literal typo, or is a sort of word-typo, where the word that was intended was replaced by something similar.

Second, literally, "collinear" would tend to mean something like "on the same line", or "parallel", maybe, the latter for lines themselves?

But, we should admit/note that it is too ambiguous, and is not a good (=helpful) usage.

Title: What are collinear lines?


问:
Given $N(\alpha) = pq$, $p$ ≡ $3$ mod $4$ and $q$ ≡ $3$ mod $4$, prove that $p = q$.
Suppose that $p,q \in \mathbb{N}$ are prime numbers and $\alpha \in \mathbb{Z[i]}$ satisfy
$N(\alpha) = pq$, $p$ ≡ $3$ mod $4$ and $q$ ≡ $3$ mod $4$.
Using the fact that $p$ and $q$ remain irreducible in $\mathbb{Z[i]}$, prove that $p = q$.

Proof attempt:
Assume false, that is, $p \neq q$. As $N(\alpha) = \alpha\alpha^* = pq$, we have $p \mid \alpha\alpha^*$ in $\mathbb{Z[i]}$. Given that $p$ is congruent to $3$ modulo $4$, we have $p$ is irreducible in $\mathbb{Z[i]}$. Then $p$ is prime implying $p \mid \alpha$ or $p \mid \alpha^*$ in $\mathbb{Z[i]}$. Taking norms, we get in both cases that $p^2 \mid pq$ in $\mathbb{Z}$. As $p$ and $q$ are primes in $\mathbb{Z}$, we get $p = q$. Hence a contradiction so $p = q$.

This is my attempt. I'm not certain it's correct. The reason being I haven't used the fact that $q$ is congruent to $3$ modulo $4$.

答:
I think a proof is okay and a more general fact holds: since the norm is multiplicative, then $s | x$ in $\mathbb{Z}[i]$ implies $x = sy$ for some $y$ and

$$
N(x) = N(s)N(y),
$$

so that $N(x)| N(y)$.

If $p$ is a Gauß prime, then $p \mid N(x)= xx^\ast$ implies $p \mid x$ and thus $p^2 = N(p) \mid N(x)$. In your case, we have $p^2 \mid N(x) = pq$, hence $q = p$ by uniqueness of prime factorization (in $\mathbb{Z}$).

In general, gaussian primes are of the form $pi$ and $p$ with $p \in \mathbb{4\mathbb{N}_0}+3$ or $a+ib$ such that $ab \neq 0$ and $N(a+ib)$ is prime in $\mathbb{Z}$. Thus, if $x$ has a certain prime factorization

$$
x = p_1^{n_1} \cdots p_s^{n_s}f_1^{m_1} \cdots f_r^{m_r}
$$

in $\mathbb{Z}[i]$, with each $p_j \in 4\mathbb{Z}_0
+3$ and $q_i = f_i^2 \in \mathbb{N}$ primes, taking norms we get

$$
N(x) = N(p_1)^{n_1} \cdots N(p_s)^{n_s}N(q_1^{m_1})\cdots N(q_r^{m_r}) = p_1^{2m_1} \cdots p_s^{2n_s}f_1^{m_1} \cdots f_r^{m_r}.
$$

This shows that gaussian primes appearing in the factorization of $N(x)$ for some $x \in \mathbb{Z}[i]$ always have even exponent.

Going back to the problem of the question, since $N(x) = pq$, either $p = q$ or no Gaussian primes divide $x$. But since $p \equiv 3 \pmod{4}$, the former is true. No extra hypotheses on $q$ are needed.


问:
How good can a "near-miss" polyomino packing be?
Given a polyomino $P$ with $n$ cells, we can ask about its maximal packing density $\delta_P$ in the plane (perhaps the limsup if we are concerned about convergence issues, though I don't think this actually comes up).

If $\delta_P=1$, then $P$ tiles the plane: having arbitrarily good packings implies that $P$ can cover arbitrarily large $N\times N$ squares, and from there the result follows via a compactness argument.

We can then ask: among polyominoes on $n\ge 7$ cells that do not tile the plane, which one achieves the highest density? Call this maximal value $\Delta_n$. For each $n$, $\Delta_n<1$, though of course in the limit for large $n$ it approaches $1$.

As a simple lower bound, we always have $\Delta_n\ge n/(n+1)$, as exhibited by the polyomino given by taking all but the first square in a spiral of length $n+1$ around the origin; once we plug the hole, it tiles the plane without gaps, but the hole cannot be filled. For instance, with $n=7$:

In contrast, putting remotely nontrivial upper bounds on $\Delta_n$ in general should be exceedingly difficult or impossible, since any computable upper bound less than 1 would yield an algorithm to decide the tiling problem for polyominoes, which is suspected not to exist. (Non-computable upper bounds on the order of $1-1/\text{BB}(k\cdot n)$ can be done, of course, but these are rather silly.)

However, improved lower bounds and exact values for small $n$ seem pretty tractable, and I'm curious what is known in this direction. Some questions:

Is the sequence $\Delta_7,\Delta_8,\ldots$ monotonically increasing? I suspect not.

When does $\Delta_n$ first exceed $n/(n+1)$? I don't actually know that it does, but I strongly suspect so for reasons described above. When $n=7$ I have not found any packings of density greater than $7/8$, although only for one of the four non-tiling heptominoes (the one pictured above) have I proven this is optimal. (All four obtain $7/8$ by adding one square to yield a tiling octomino.)

Can $\Delta_7$ be proven to equal $7/8$, if indeed it does so? Exhausting all tilings of an $N\times N$ square by each of the four non-tiling polyominoes and counting the max density therein would at least yield an upper bound; if $\Delta_7$ exceeds $7/8$, I would guess it does so via finding a tiling $15$-omino which is the union of two copies of a non-tiling heptomino and an additional cell.

[Subjective] What are some examples of interesting "near-miss" tilings?

答: Here's a very simple heptomino tiling that is indeed "a tiling 15-omino which is the union of two copies of a non-tiling heptomino and an additional cell."


问:
Stationary distribution of a DTMC that has recurrent and transient states
I need to construct a stationary distribution for $\{X_n, n\in \mathcal{N}\}$ with state space $E=\{0,1,2,3,4,5\}$ and transition matrix:
\begin{bmatrix}
 0.8 & 0.2 & 0&0&0&0\\
0&0.1&0.4&0.5&0&0\\
0&0.3&0.3&0.4&0&0\\
0&0.6&0.3&0.1&0&0\\
0&0&0&0&0.7&0.3\\
0&0&0&0&0.5&0.5
\end{bmatrix}

In the previous steps of this problem I am asked to find the stationary distribution of all positive recurrent classes. For the positive recurrent class $\{1,2,3\}$ I was able to find the stationary distribution to be $1/3$ for each state, and for $\{4,5\}$ to be $5/8$ and $3/8$, respectively. However, I need to put these distributions together to find the stationary distribution for $X_n$ and I am not sure how to do so. This is just a practice problem as I study for my exam.  Thank you for any help

答: States $\{0,1,2,3\}$ are isolated from $\{4,5\}$, so the stationary distribution isn't unique: any weighted combination of the stationary distributions for the subsystems will be stationary. The probabilities will evolve to a combination of the two, with weights equal to the total probability each subsystem started with. For instance, if the initial probabilities are $(1/3,1/3,0,0,0,1/3),$ the limiting probabilities are $(0,2/9,2/9,2/9,5/24,3/24).$


问:
Show that for all nonnegative sequences $\{a_k:1 \leq k \leq n\}$, one has $(\sum_{k=1}^n a_k^{1/2} )^2 \leq (\sum_{k=1}^n a_k^{1/3} )^3.$
Problem: Show that for all nonnegative sequences $\{a_k:1 \leq k \leq n\}$, one has $$\left(\sum_{k=1}^n a_k^{1/2} \right)^2 \leq \left(\sum_{k=1}^n a_k^{1/3} \right)^3.$$
This is a problem from the book "The Cauchy Schwarz Master Class".

Question:  In the solution, it says that the inequality is homogeneous of order one, so we can assume that $\sum\limits_{k=1}^n a_k^{1/3} = 1$. But I'm not quite sure how can we assume that is the case here. Can anyone explain to me what this sentence mean?

答:
The inequality is homogeneous because if you replace $a_1,\ldots,a_n$ by $ca_1,\ldots, ca_n$ (for some constant $c>0$), the inequality becomes, after some algebra,
$$c\left(\sum_{k=1}^n a_k^{1/2} \right)^2 \leq c\left(\sum_{k=1}^n a_k^{1/3} \right)^3.
$$
So the inequality is unchanged. Notice that choosing
$$
c:=\frac1{\left(\sum a_k^{1/3}\right)^3}
$$
makes the RHS equal to $1$. So WLOG we may assume RHS of the original equality equals 1.


问:
For each positive integer $n \geq 3$, determine the centre of $D_{2n}$ (help for understand)
For each positive integer $n \geq 3$, determine the centre of $D_{2n}$.

Proof. Let $F$ be any flip and $R$ any rotation. Drawing out the effects of each operation,
we find that $FR = R^{−1}F$. This is $RF$ if and only if $R= R^{−1}$. Letting $R = R_{360/n}$,
we find that $R\neq R^{−1}$. Thus, no flip is central. In fact, $R = R^{−1}$ if and only if $R = R_0$ or $R = R_{180}$. If n is odd, there is no $R_{180}$, so $Z(D_{2n}) = {R_0}$. If $n$ is
even, we see that $R_{180}$ commutes with every flip, and surely with every rotation, so
$Z(D_{2n}) = \{R_{0}, R_{180}\}$.

About this proof I have a couple of doubts.

First:

Drawing out the effects of each operation,
we find that $FR = R^{−1}F$. This is $RF$ if and only if $R= R^{−1}$

How can I determine that this is indeed generally true?

Second:

$R = R^{−1}$ if and only if $R = R_0$ or $R = R_{180}$

This is clear, but how could it be proven?

It must be said that it is not a very advanced material that I am studying. I hardly go by subgroups. Thanks in advance

答:
For the first one:
$$
R=R^{-1}\Longrightarrow R\color{blue}F=R^{-1}\color{blue}F=FR
\\
RF=R^{-1}F\Longrightarrow RF\color{blue}{F^{-1}}=R^{-1}F\color{blue}{F^{-1}}=R\left( FF^{-1} \right) =R^{-1}\left( FF^{-1} \right) \Longrightarrow R=R^{-1}
$$
For the second one:
Assume $R_x$ satisfies $R_x=R^{-1}_x$, then $R_x=R_{(360-x)\text{ mod } 360}$, then it is obvious that $x=0$ or $x=180$ (Note $0\leq x<360$).


问:
Resources for self-study of general topology
Could you point me to good resources for self-study of general topology. I want to learn the basics and how to prove theorems about structures like polyhedra by myself.</p>

答:
Fred H . Croom,
Principles of Topology

See: Croom’s:Principles of Topology

Here are partial list of the chapters:

Introduction

Line and the Plane

Metric Spaces

Topological Spaces

Connectednes

Compactness

Looking at the book,the layout is beautiful and the proofs
are clearly written showing you how to do the exercises.

Also another professor on another site suggested it over
Conover’s A First Course on Topology which I currently use
because Conover’s thesis was not shown in American Geneology of Mathematicians

Another excellent book you can get by hard back is

Bert Mendelson,Introduction to Topology

See:
Mendleson’s Introduction to Topology

It is heavy on metric spaces.

Here is a list of the chapters:

1.Theory of Sets

2.Metric Spaces

3.Topological Spaces

4.Connectedness

5.Compactness

His proofs of many of the theorems are easily understood and
the problems he gives should not be to hard to do.

Chapter 4 has a section on homotopic paths.Homeomorphism is mentioned in Chapter 3.

This is a basic text in Topology.


问:
Good concentration inequality for $\max_{i,j} X_j- X_i$ where $X_1,\ldots,X_n \overset{iid}{\sim} N(0,1)$.
Let $n$ be a large positive integer and define $\Delta:\mathbb R^n \to \mathbb R$ by $\Delta(x) := \max_{i,j} x_j - x_i$. Let $X_1,\ldots,X_n$ be iid from $N(0,1)$

Question. What is a good concentration inequality for $\Delta(X)$ ?

Observation

Let $x'$ and $x$ be vectors in $\mathbb R^n$ which only dependend on their $k$th coordinate. Then

$$
\begin{split}
|\Delta(x') - \Delta(x)| &= |(|\max_{i,j} x'_j-x'_i|) - (|\max_{i,j} x'_j-x'_i)| \le \max_{i,j} \left||x'_j-x'_i|-|x_j-x_i|\right|\\
&\le \max_{i,j}|(x_j'-x_j)-(x_i'-x_i)| = \begin{cases}0, &\mbox{ if }k \not\in\{i,j\},\\
|x'_i-x_i|,&\mbox{ if }i = k \ne j,\\ |x'_j-x_j|,&\mbox{ if }j = k \ne i.\end{cases}
\end{split}
$$
Therefore $|\Delta(x') - \Delta(x)| \le |x'_k-x_k|$, and so $\Delta$ is $1$-Lipschitz in every coordinate and it follows from Corollary 2.1 of this monograph that

$$
\mathbb P(|\Delta(X) - \mathbb E[\Delta(X)]| \ge t) \le 2e^{-t^2/(2n)}.
$$

I hope for an inequality with a better dependence on the dimension $n$.

答:
The functions $v \mapsto \max_i v_i$ and $v \mapsto \min_i v_i$ are $1$-Lipschitz with respect to the Euclidean metric. Indeed, take any $x,y \in \mathbb{R}^n.$ In the case that $\max_i x_i > \max_j y_j,$ $$ | \max_i x_i - \max_j y_j| = \max_i x_i - \max_j y_j \le \max_i (x_i - y_i) \le \max_i |x_i - y_i| \le  \|x-y\|,$$ and the same argument reversed works if $\max_i y_i > \max_i x_i.$ The claim follows for $\min$ because $\min_i v_i = - \max_i (-v_i).$

The upshot of this is that $\max_{i \neq j}  x_i -x_j = \max_i x_i - \min_j x_j $ is $2$-Lipschitz with respect to the Euclidean metric. Now using theorem 2.2 from the notes linked to in the question gives a concentration radius independent of $n$.


问:
how to tell whether x and y are independent or not
Suppose that $f_{x,y}(x,y) = \lambda^2 e^{\displaystyle-\lambda(x+y)}, 0\leq x , 0\leq y.$ Find $\operatorname{Var(X+Y)}$.

I'm having trouble with this problem the way to find $\operatorname{Var(X+Y)} = \operatorname{Var(X)}+\operatorname{Var(Y)}+2\operatorname{Cov(X,Y)}$, however if $X$ and $Y$ are independent, then $\operatorname{Cov(X, Y)}=0$, the answers indicated that $X$ and $Y$ are independent since they just used $\operatorname{Var(X+Y)} = \operatorname{Var(X)}+\operatorname{Var(Y)}+0$, my question is how do I tell whether $X$ and $Y$ are independent or not, based on looking only at $f_{x,y}(x,y) = \lambda^2 e^{\displaystyle-\lambda(x+y)}, 0\leq x , 0\leq y.$

答: You need to find the marginal densities and show that the joint is the product of the marginals in order to show that they are independent.</p>


问:
How did they came up with this simplification of given differential equation? (kepler's first law proof)
So for the proof of Kepler's first law they make a simplification of a differential equation, but I don't get how they come up with that simplification. It goes like this

$$2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}=0$$
once simplified (which I don't understand) they came up with this answer
$$\frac{1}{2r}\frac{d}{dt}\left(r\left(\frac{d\theta}{dt}\right)^2\right)=0$$
could someone please explain to me what they did?

thanks in advance.

答:
$$\frac{dr}{dt}\frac{d\theta}{dt}+\color {red} {2r}\frac{d^2\theta}{dt^2}=0$$
More simply:
$$r'\theta '+2r\theta ''=0$$
Multiply by $\theta ':$
$$r'(\theta ')^2+2 r \theta '\theta ''=0$$
Then you have:
$$r' (\theta ')^2+r ((\theta ')^2)'$$
Finally:
$$(r((\theta ')^2))'=0$$

But this
$$2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}=0$$
Is :
$$2r'\theta ' +r \theta ''=0$$
$$2r'r\theta ' +r^2 \theta ''=0$$
$$(r^2 \theta ')'=0$$
The two equations you wrote are not giving the same result.


问:
Inverse of $A+b^{T}b$ and $A+B^{T}+B$?
Let $A$ be a $n$ $\times$ $n$ positive-definite matrix, $b$ is a $1$ $\times$ $n$ vector. Assume that $A+b^{T}b$ is invertible.
Is there a formula to compute the inverse of $A+b^{T}b$, where $T$ is transponse?

Thanks!

Another questions:
What is the inverse of $A+B^{T}+B$, where $B$ is a $n$ $\times$ $n$ matrix, given that it is invertible?

答:
The Sherman-Morrison forumla states that if $A$ is an invertible $n \times n$ matrix and $u,v$ are $n \times 1$ vectors with $1+v^TA^{-1}u \neq 0$, then $A+uv^T$ is invertible and $$(A+uv^T)^{-1} = A^{-1} - \dfrac{A^{-1}uv^TA^{-1}}{1+v^TA^{-1}u}.$$

Applying that formula here with $u = v = b^T$ yields $$(A+b^Tb)^{-1} = A^{-1} - \dfrac{A^{-1}b^TbA^{-1}}{1+bA^{-1}b^T}.$$ Note that since $A$ is positive definite in your problem, $A^{-1}$ is also positive definite, and thus, $1+bA^{-1}b^T \ge 1 > 0$.

Title: Inverse of $A+b^{T}b$ and $A+B^{T}+B$?


问:
Is there a solid reason why some people assume the fundamental theorem of calculus should still hold for divergent integrals with improper bounds?
I asked this question on Mathoverflow and received 6 downvotes so far and no answer.

In the comment section of this question several people seemingly freely assume that fundamental theorem of calculus should still hold for divergent integrals with improper bounds.

That is, they assume

$$\int_a^b f(\varphi(x))\varphi'(x) dx = \int_{\varphi(a)}^{\varphi(b)} f(u)du$$

holds even if $a$ or $b$ are $\pm\infty$ and the integral in the left-hand side is divergent to infinity.

I wonder, what can justify this assumption. In my view it is totally unjustified. Why do I think so? Because it allows the following:

$$I=\int_0^\infty1dx=2\int_0^\infty1du=2I$$
or (with substitution $u=2x$)
$$\int_1^{+\infty}\frac1x dx=\int_2^{+\infty}\frac1u du$$

In the second case the indegrals even have different regularized values, the left-hand side integral has the regularized value $0$ while the right-hand side integral has the regularized value $-\ln2$. In other words, they are different divergent integrals with different properties.

In the comment section the user Johannes Hahn justified the relation with words "because it's true" but I do not see in what sense this equality even can be true for divergent integrals?

Is it equality of the values? If so, what value can have a divergent integral without extension of real numbers? Or an extension is assumed?

Is it equality of some other set of properties? In that case we see that the regularized values of these two integrals before and after substitution are different.

Something else?

答:
Taken from MathOverflow's help section:

MathOverflow's primary goal is for users to ask and answer mathematical questions related to current research in mathematics.

So the downvotes are definitely warranted as this is not related to current research in mathematics.

Now let's address the crux of your argument. First note that improper integrals are limits and
$$\int_0^\infty\mathrm dx = \lim_{b\to\infty}\int_0^b\mathrm dx=\infty$$
Which means that the limit does not exist and the improper integral diverges. So what does it mean when an integral or a sum diverges? It basically means that neither of the two classical summability methods worked. Now if a different summability method is applied, then it must be explicitly stated. Note that different summability methods can report different values for divergent integrals or sums. To my surprise, all of this was already explained to you here. Let me know if you have any other questions.


答:
A quick and dirty way to get a dimension-independent bound without using the nice observation of stochasticboy321:

$$\begin{align}
\mathbb{P}\{ |\max_{i,j} (X_i- X_j) &- \mathbb{E}\max_{i,j} (X_i- X_j) |\geq t \}\\
&= \mathbb{P}\{ |(\max_i X_i- \mathbb{E} \max_{i} X_i)- (\min_j X_j - \mathbb{E} \min_{i} X_i)| \geq t \} \\
&\leq\mathbb{P}\{ |\max_i X_i- \mathbb{E} \max_{i} X_i|\geq t/2 \}
+ \mathbb{P}\{ |\min_i X_i- \mathbb{E} \min_{i} X_i|\geq t/2 \} \\
&=2\mathbb{P}\{ |\max_i X_i- \mathbb{E} \max_{i} X_i|\geq t/2 \}\\
\end{align}$$
and we have a dimension-independent bound for the concentration of the max of $n$ i.i.d. Gaussians, so
$$\begin{align}
\mathbb{P}\{ |\max_{i,j} (X_i- X_j) - \mathbb{E}\max_{i,j} (X_i- X_j) |\geq t \}
&\leq Ce^{-C't^2}
\end{align}$$
for some explicit constants $C,C'>0$ I don't recall at the moment.

Title: Good concentration inequality for $\max_{i,j} X_j- X_i$ where $X_1,\ldots,X_n \overset{iid}{\sim} N(0,1)$.

Body:
Let $n$ be a large positive integer and define $\Delta:\mathbb R^n \to \mathbb R$ by $\Delta(x) := \max_{i,j} x_j - x_i$. Let $X_1,\ldots,X_n$ be iid from $N(0,1)$

Question. What is a good concentration inequality for $\Delta(X)$ ?

Observation

Let $x'$ and $x$ be vectors in $\mathbb R^n$ which only dependend on their $k$th coordinate. Then

$$
\begin{split}
|\Delta(x') - \Delta(x)| &= |(|\max_{i,j} x'_j-x'_i|) - (|\max_{i,j} x'_j-x'_i)| \le \max_{i,j} \left||x'_j-x'_i|-|x_j-x_i|\right|\\
&\le \max_{i,j}|(x_j'-x_j)-(x_i'-x_i)| = \begin{cases}0, &\mbox{ if }k \not\in\{i,j\},\\
|x'_i-x_i|,&\mbox{ if }i = k \ne j,\\ |x'_j-x_j|,&\mbox{ if }j = k \ne i.\end{cases}
\end{split}
$$
Therefore $|\Delta(x') - \Delta(x)| \le |x'_k-x_k|$, and so $\Delta$ is $1$-Lipschitz in every coordinate and it follows from Corollary 2.1 of this monograph that

$$
\mathbb P(|\Delta(X) - \mathbb E[\Delta(X)]| \ge t) \le 2e^{-t^2/(2n)}.
$$

I hope for an inequality with a better dependence on the dimension $n$.


问:
How can we say that if $a+bi=c+di$ then $a=c$ and $b=d$, where $a,b,c,d$ are one-variable functions?
I have been reading Paul Nahin's book and in the last chapter (page number 213), he writes
$$\frac{\pi}{b}e^{-ab}=\int_{-\infty}^{\infty}\frac{\cos(ax)}{b^2+x^2}\mathrm{d}x+i\int_{-\infty}^{\infty}\frac{\sin(ax)}{b^2+x^2}\mathrm{d}x$$
implies that
$$\int_{-\infty}^{\infty}\frac{\cos(ax)}{b^2+x^2}\mathrm{d}x=\frac{\pi}{b}e^{-ab}$$ and $$\int_{-\infty}^{\infty}\frac{\sin(ax)}{b^2+x^2}\mathrm{d}x=0$$
But earlier in the same chapter he shows that $$\int_{-1}^{1}\frac{\mathrm{d}x}{x}=i\pi$$
So, how can we compare the real and imaginary parts in the first equation, when we don't know whether the integrals are real or imaginary?

答:
The integrals in the first equation are integrals of real functions over the real axis. So (if they converge) they converge to real numbers.

(That the second integral is $0$ is clear since it's the integral of an odd function.)


问:
Zero probability and impossibility
I read a comment under this question:

There are plenty of events that can occur that have zero probability.

This reminds me that I have seen similar saying before elsewhere, and have never been able to make sense out of it. So I was wondering

if zero probability and impossibility mean the same?

if an event with zero probability doesn't mean that the event is impossible to occur, how probability theory represents/describes impossibility?

Thanks and regards!

答:
I believe the root of this confusion is all because Math(in its nature) is in general(including probabilities) always more serious towards its concepts, than we -the humans- are. And this seriousness is exactly what brings Math its clarificationability(the power of turning vague into non-vague) and simplificationability(the power of turning complicated into simple), two of its core ultimate goals!

An instance of this amount of seriousness: When Math says "probability" it really means "probability"! I.e, mathematical probabilities are ALWAYS about "probable" expectations, and so even its 1 is not certainty, and even its 0 is not impossibility! Rather indeed, its 1 is absolute probability and its 0 is absolute improbability!

It's just an intentional neglection of all the odd possibilities. For instance, when I throw a die, it is possible that I will get a 7...! Why?! Maybe someone has added a dot to the 6 side! But we neglect that possibility on purpose for the sake of simplification, because it's an odd(read improbable) possibility.

Another Example: Yes, it is possible (even absolutely possible!) that the exact number 3 might occur in the continuous interval [1, 4], but it is absolutely improbable(notice how you can't disagree), thus the P=0. (Notice: Computers are never really continuous, so don't even try doing it programmatically, it won't be a counterexample).

Title: Zero probability and impossibility


问:
Convergence of composition of power series
I am doing a project, and I have to find an article/book/ something with a certain proof. It is about function/power series. My starting point is this:

$$f(z)=\sum_{n=0}^\infty a_nz^n$$ with radius r>0

$$g(z)=\sum_{n=1}^\infty b_nz^n$$ with radius R>0

$$\lbrace g(z) : |z|<r \rbrace \subset D(0,r)$$

I have to prove that f o g (composition) has a power series expansion in a neighbourhood of zero, and say what we can say about the radius.

Also, I have to prove that analytic function composition is analytic, but I think I solved that using chain's rule and seeing that it is holomorphic.

答:
For $T$ small enough we have $\sum_n |b_n| |z|^n<r$ for $|z|<T$.

So that $$\sum_k |a_k| (\sum_n |b_n| |z|^n)^k$$ converges for $|z|<T$. The absolute convergence means that we can change the order of summation in $$f(g(z))=\sum_k a_k(\sum_n b_n z^n)^k=\sum_k a_k \sum_m B_{k,m} z^m=\sum_m z^m \sum_k a_k B_{k,m}$$
The same argument shows that for $G$ analytic on $U$ and $F$ analytic on $G(U)$ then $F(G(z))$ is analytic on $U$ (given by a power series around every point).


问:
Is it correct to define $\operatorname{cl}(\Omega) := \Omega \cup\partial \Omega$?
In my differential geometry class the professor wrote $$\operatorname{cl}(\Omega) := \Omega \cup\partial \Omega, \tag{1}$$
where $\operatorname{cl}$ is closure.
From Topology I recall it was $$\operatorname{cl}(\Omega) := \operatorname{Int}(\Omega) \cup \partial \Omega \tag{2},$$
where $\operatorname{Int}$ is Interior.

Is $(1)$ also correct? If it is not , what is missing in that union?
I feel the isolated points may not be taken into consideration in $(1)$.

Furthermore he said that historically the $\partial \Omega$ notation comes from the topological concept of derived set.  But I don't think  that the boundary is the same as the derived set, is it?

Edit: I am attempting a proof, taking $(2)$ as definition:

$$\operatorname{Int}(\Omega) \subseteq \Omega \implies \operatorname{Int}(\Omega) \cup\partial\Omega \subseteq \Omega\cup \partial\Omega  $$

$$\implies \operatorname{cl}(\Omega) \subseteq\Omega\cup \partial\Omega $$

But don't know how to prove the oposite inclusion, any idea?

答:
It's the same thing. If you require $\Omega$ to be an open set (or you use $\operatorname{Int}(\Omega)$ instead) you just end up with a disjoint union, but the same equality holds anyways:
$$ \operatorname{cl}(\Omega) = \operatorname{Int}(\Omega)\,\sqcup\,\partial\Omega$$

The derived set (the set of all limit points) is a concept similar to closure, but not the same: the derived set of $\Omega$ doesn't take isolated points into account. In fact, the closure of any set $\Omega$ in a topological space is the disjoint union of limit points AND isolated points:
$$ \operatorname{cl}(\Omega)=\operatorname{der}(\Omega)\,\sqcup\,\operatorname{Is(\Omega)}$$
Notice that all isolated points end up to be part of the boundary $\partial\Omega$. In general, you have:
$$ \operatorname{Int}(\Omega)\subseteq \operatorname{der}(\Omega) \subseteq \operatorname{cl}(\Omega) $$
$$ \operatorname{Is(\Omega)} \subseteq \partial\Omega \subseteq \operatorname{cl}(\Omega) $$

The difference between $\Omega$ and $\operatorname{Int}(\Omega)$ has not necessarily to do with isolated points.

Take $\Omega=[0,1)\subset\mathbb R$ as an example; $\Omega$ (and its closure, $[0,1]$) have no isolated points, and still $\Omega\smallsetminus\operatorname{Int}(\Omega)\neq\varnothing$. You can "decompose" $\operatorname{cl}(\Omega)$ as $\operatorname{Int}\sqcup\partial=(0,1)\sqcup\,\{0,1\}$, and these are all limit points.

You might try experimenting with $\Omega=[0,1)\cup\{2\}$ to better understand what happens to closure and boundaries when isolated points are included.


答:
I've just had a look at the book, it says

if we change variables to $x = -e^{i\theta}$, with $0\leq \theta \leq \pi$ radians, then the integral becomes
$$
I=\int_{0}^{\pi} \frac{-i e^{i \theta} d \theta}{-e^{i \theta}}=\int_{0}^{\pi} i d \theta=i \pi
$$

And I doubt the correctness of the variable substitution. Say, the substitution can lead to the extra imaginary part, and it doesn't really equal to the originial one. For example, we take $\theta=\dfrac \pi4$, then $e^{-i\frac{ \pi}{4} }=\dfrac{1-i}{\sqrt{2}}$ is not a real number, while $x$ in the originial integral must be real ($x\in[-1,-1]$).

Generally speaking, the integral $$\int_{-1}^{1}\frac{\mathrm{d}x}{x}$$ does not converge. If we talk about cauchy principal value, we can have it as
$$
V.P. \int_{-1}^1\dfrac1xdx=0
$$

Return to your question,

when can we compare the real and imaginary parts

The answer is when all the integrals converge, that is, integrals of real functions over the real axis converges to real numbers.

Title: How can we say that if $a+bi=c+di$ then $a=c$ and $b=d$, where $a,b,c,d$ are one-variable functions?

Body:
I have been reading Paul Nahin's book and in the last chapter (page number 213), he writes
$$\frac{\pi}{b}e^{-ab}=\int_{-\infty}^{\infty}\frac{\cos(ax)}{b^2+x^2}\mathrm{d}x+i\int_{-\infty}^{\infty}\frac{\sin(ax)}{b^2+x^2}\mathrm{d}x$$
implies that
$$\int_{-\infty}^{\infty}\frac{\cos(ax)}{b^2+x^2}\mathrm{d}x=\frac{\pi}{b}e^{-ab}$$ and $$\int_{-\infty}^{\infty}\frac{\sin(ax)}{b^2+x^2}\mathrm{d}x=0$$
But earlier in the same chapter he shows that $$\int_{-1}^{1}\frac{\mathrm{d}x}{x}=i\pi$$
So, how can we compare the real and imaginary parts in the first equation, when we don't know whether the integrals are real or imaginary?


问:
Solve three-variable system $x^2 - yz = 1,\>y^2 - xz = 2,\>z^2 - xy = 3$
I am going around in circles on this system

$$x^2 - yz = 1\\
y^2 - xz = 2\\
z^2 - xy = 3$$

I have tried a few things (below) but keep hitting a wall. I know that by adding the three equations we get:
$x^2 + y^2 + z^2 - yz - xz - xy =6$. And also that:

$(x-y)^2 + (x-z)^2 + (y-z)^2 = 2x^2+2y^2+2z^2-2xy-2xz-2yz=12$

I have tried the following:

$(x-y)(x+y+z)=(x^2-yz)-(y^2-xz)=-1$

$(x-z)(x+y+z)=(x^2-yz)-(z^2-xy)=-2$

$(y-z)(x+y+z)=(y^2-xz)-(z^2-xy)=-1$

So I think it follows that $x-y=y-z$ and therefore $x-z=2(x-y)$

Substituting in:

$(x-y)^2+[2(x-y)]^2+(x-y)^2=12$

$6(x-y)^2=12$

$x-y=\sqrt{2}$

Therefore: $y-z=\sqrt{2}$ and $x-z=2\sqrt{2}$

But when I rearrange and substitute into one of the original equations, the solutions I get for x, y and z don't actually work.

$y=x-\sqrt{2}$ and $z=x-2\sqrt{2}$

Substituting into the first equation:

$x^2-(x-\sqrt{2})(x-2\sqrt{2})=1$

$x=\frac{3\sqrt{2}}{2}$

Giving $y=\frac{\sqrt{2}}{2}, z=-\frac{\sqrt{2}}{2}$

Unfortunately, substituting these values into the three original equations only satisfies the second one. Can anyone please help?

答:
You're on the right track but made an algebra error (cf. $y - z$).  These are quadratic equations, so there are two solutions:

$$\left\{x\to -\frac{5}{3 \sqrt{2}},y\to \frac{1}{3 \sqrt{2}},z\to \frac{7}{3
   \sqrt{2}}\right\}$$

$$\left\{x\to \frac{5}{3 \sqrt{2}},y\to -\frac{1}{3 \sqrt{2}},z\to -\frac{7}{3
   \sqrt{2}}\right\}$$


问:
Disprove if $a_n$ converges then $a_{n+1}/a_n$ converges

Disprove: if $a_n$ converges then $a_{n+1}/a_n$ converges.

I want to disprove this statement but I couldn't find a sequence that disproves it.

Further more it will be really helpful to know what are the things to look for in those kind of questions.

答:
Hint: We could try $ 1, , 1, -1, -1, 1, 1, -1, -1, \ldots$ but the sequence doesn't converge.

The ratio is $  1, -1, 1, -1, \ldots$ which doesn't converge.

Modify it so that the sequence converges, but the ratio doesn't.

Things to look for

What could make the ratio not converge? Could we make it arbitrarily large?

If $ a_n \rightarrow L \neq 0$, prove that $ \frac{ a_n } { a_{n-1}} \rightarrow 1$. (This should be obvious after playing around with it for a while.)

Hence, if we want a counter example, what must be true of $ L$?

So we know now that $ a_n \rightarrow 0$. Can the ratio of small terms be arbitraily large?


问:
Complex-number method to minimize equilateral-triangle area inside right triangle of side lengths $2\sqrt3$, $5$, and $\sqrt{37}$

The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt3$, $5$, and $\sqrt{37}$, as shown, is $\tfrac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.

I am trying to solve this problem just with complex numbers. I let the vertices of the large triangle to be $a,b,c$ for $b=0,a=2\sqrt{3}i,c=5.$ I let the vertices of the equilateral triangle to be $z_1,z_2,z_3$ for $z_1$ along $AB$, $z_2$ along $AC$, and $z_3$ along $BC$. In the first place, it is well-known that $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1.$

It is futile to consider the equations for $A,Z_1,B$ and $B,Z_3,C$ to be collinear because we set those as the imaginary and real axes. With the equation for $A,Z_2,C$ to be collinear, we have
$$\frac{z_2-a}{z_2-c}=\frac{\overline{z_2}-\overline{a}}{\overline{z_2}-\overline{c}}=\frac{\overline{z_2}+a}{\overline{z_2}-c},$$

with that step from $\overline{a}=-a$ and $\overline{c}=c,$ as these are along the imaginary and real axes respectively. With some manipulation, we get to
$$a(z_2+\overline{z_2})+c(z_2-\overline{z_2})-2ac=0$$
$$\Longleftrightarrow a\Re(z_2)+c\Im(z_2)i=10\sqrt{3}$$
$$\Longleftrightarrow 2\sqrt{3}x_2+5y_2=10\sqrt{3},$$
for $\Re(z)=x,\Im(z)=y.$

But I'm not quite sure what to do now, because I can't think of any synthetic simplifications.

答:
Rewrite the equilateral condition $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$ as
$$z_3^2 -(z_1+z_2)z_3 + z_2^2+z_3^2-z_1z_2=0$$
which is quadratic in $z_3$, yielding
$$z_3= e^{i\frac\pi3}z_1 + e^{-i\frac\pi3}z_2$$
Substitute $z_3$ into the line equation below for the hypotenuse

$$\frac{z+\bar z}{10}+ \frac{z-\bar z}{4\sqrt3 i}=1$$
obtained from the given side lengths, leading to
$$\frac7{20}|z_1 |+ \frac{11\sqrt3}{60}|z_2 |=1
$$
where $z_1=\bar z_1$ and $z_2= -\bar z_2$ are used. Then, the area of the equilateral triangle is

$$A= \frac{\sqrt3}{4} |z_2-z_1|^2
= \frac{\sqrt3}{4}( |z_2|^2 +|z_1|^2 )\ge \frac{\sqrt3}{4} \frac{\left(\frac7{20}|z_1 |+ \frac{11\sqrt3}{60}|z_2 |\right)^2 }{\left(\frac7{20}\right)^2
+ \left(\frac{11\sqrt3}{60}\right)^2 }= \frac{75\sqrt3}{67}
$$
where the C-S inequality is applied.


答:
Let
$$a_n = \begin{cases}
\frac{1}{n} & \text{if }n\text{ is even} \\
\frac{1}{n^2} & \text{if }n\text{ is odd} \\
\end{cases}$$
Then $a_n \to 0$, but
$$\frac{a_{n+1}}{a_n} = \begin{cases}
\frac{n}{(n+1)^2} < \frac{1}{n} & \text{if }n\text{ is even} \\
\frac{n^2}{n+1} \geq \frac{n}{2} & \text{if }n\text{ is odd} \\
\end{cases}$$
Hence $\lim a_{n+1}/a_n$ does not exist.

Title: Disprove if $a_n$ converges then $a_{n+1}/a_n$ converges

Body:
Disprove: if $a_n$ converges then $a_{n+1}/a_n$ converges.

I want to disprove this statement but I couldn't find a sequence that disproves it.

Further more it will be really helpful to know what are the things to look for in those kind of questions.


答:
If a sequence, $b_n$, is equidistributed in $[-1, 1]$ (or a smaller interval), then $a_n = \prod_{k=1}^n b_k$ would converge to $0$, while $\frac{a_{n+1}}{a_n} = b_{n+1}$ does not converge.

For example, take $b_n = \sin(n) \to a_n = \prod_{k=1}^n \sin(k)$. Then $a_n \to 0$, but $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}$ does not exist.


问:
Matrix multiplication related to complex numbers?
Evaluate and simplify the product

$\begin{bmatrix} r\cos(\alpha) & -r\sin(\alpha) \\ r\sin(\alpha) & r\cos(\alpha)\\ \end{bmatrix}$ $\begin{bmatrix} s\cos(\beta) & -s\sin(\beta) \\ s\sin(\beta) & s\cos(\beta)\\ \end{bmatrix}$

where $\alpha$, $\beta$ are any real numbers and $r$, $s$ are nonnegative real numbers. Relate your answer to multiplication of complex numbers. Is there any connection with addition of complex numbers?

I was able to multiply through and I will post my answer below, but I can't see the connection to complex numbers. Could someone please explain? Thanks.

答: $\begin{bmatrix} rs\cos(\alpha+\beta) & -rs\sin(\alpha+\beta) \\ rs\sin(\alpha+\beta) & rs\cos(\alpha+\beta)\\ \end{bmatrix}$ </p>


问:
A painter has three different jars of paint colors available, in colors green, yellow, and purple.
A painter has three different jars of paint colors
available, in colors green, yellow, and purple. She wants to paint something
abstract, so she blindfolds herself, randomly dips her brush, and paints on the
canvas. She continues trying paint jars until she finally gets some purple onto
the canvas (her assistant will tell her when this happens) and then she stops. Assume that she does not repeat any of the jars because her assistant removes
a jar once it has been used. So the sample space is
$S = \{(P),(G, P),(Y, P),(Y, G, P),(G, Y, P)\}$

Find the Probabilities of each of the following events:

$\{(P)\}$

$\{(G, P),(Y, P)\}$

$\{(G, P),(G, Y, P)\}$

$\{(Y, G, P),(G, Y, P)\}$

$ \{(P),(Y, P)\}$

What I did was this:

$P\{(P)\} = \cfrac{1}{5}$ since $(P)$ is just one of the five outcomes in the sample space

$P\{(G, P),(Y, P)\} = \cfrac{2}{5}$ since $(G,P)$ and $(Y,P)$ are two of the five outcomes in the sample space

$P\{(G, P),(G, Y, P)\} = \cfrac{2}{5}$ since $(G,P)$ and $(G,Y,P)$ are two of the five outcomes in the sample space

$P\{(Y, G, P),(G, Y, P)\} = \cfrac{2}{5}$ since $(Y,G,P)$ and $(G,Y,P)$ are two of the five outcomes in the sample space

$P\{(P),(Y, P)\} = \cfrac{2}{5}$ since $(P)$ and $(Y,P)$ are two of the five outcomes in the sample space

But the final answer in the back of the book says this:

$P\{(P)\} = 1/3;$

$P\{(G, P),(Y, P)\} = 1/3;$

$P\{(G, P),(G, Y, P)\} = 1/3;$

$P\{(Y, G, P),(G, Y, P)\} = 1/3;$

$P\{(P),(Y, P)\} = 1/2;$

I have no idea where they got these values from. Where did I go wrong in my reasoning?

答:
The problem is that the events in your sample space are not all equally likely.

For instance, she has a $1/3$ chance to pick purple at the start, so the probability of event $(P)$ is $1/3$. (Hopefully, that much is intuitively obvious: three pots to pick from, one is chosen at random.)

However, on the other hand, to get $(Y,P)$ for another example, she has to pick yellow, and then purple: that is a probability of $1/3 \cdot 1/2 = 1/6$ ($1/3$ for yellow, then $1/2$ for purple once yellow is removed).

The notion of probability being equal to

$$\text{probability} = \frac{\text{# of desired events}}{\text{total events in sample space}}$$

only applies whenever each event in the sample space is equally likely.


问:
Proof of rule for multiplying numbers with uncertainties
So suppose you have a value $A$ with an absolute uncertainty $\pm a$ and another value $B$ with absolute uncertainty $b$, it is easy to prove that the rule for dealing with the uncertainties when adding these values like so:
$$(A\pm a)+(B\pm b)=(A+B)\pm (a+b)$$
However the rule for dealing with uncertainties when multiplying the values is:
$$(A\pm a)\times(B\pm b)=(A\times B)\pm \left[\left(\frac{a}{A}\cdot100\right)+\left(\frac{b}{B}\cdot100\right)\right]\%$$
my question is how would one go about proving this fact or is it just useful convention with no logical proof. If it is the latter why is useful to deal with uncertainties this way. Thanks in advance :)</p>

答:
Suppose $A, B$ are the variables and $a, b$ are their absolute uncertainties respectively.
We could then calculate absolute uncertainty of $AB$ as follows:

$$\max(AB)= (A+a)(B+b)=AB + Ab + Ba + ab$$
$$\min(AB)= (A-a)(B-b)=AB - Ab - Ba + ab$$

So absolute uncertainty of $AB = \frac{\max(AB) - \min(AB)}2 =
\frac{2Ab + 2Ba}2 = Ab + Ba$

% uncertainty of $AB$ = absolute / $AB = b/B + a/A =$ %uncertainty of $A$ + % uncertainty of $B$.


问:
Investigating whether a given relation is reflexive, symmetric, and transitive
Let $X = \{0, 1, 2, ... , 10\}$, Define the relation $R$ on $X$ by, for all $a, b \in X$, $aRb$ if and only if $a + b = 10$.

Is $R$ reflexive? symmetric, transitive? Give reasons.

Here are my answers, please see if I made any mistakes?

$R$ is not reflexive, because there exists $a \in X$ such that $a$ does not relate $a$. For example, let $a=1$, $1+1=2$ which is not equal to $10$.

$R$ is symmetric, because $a + b = b + a$. sum of integers are symmetric. So $4R6$ and $6R4$.

$R$ is not transitive, because there are $a, b, c$ integers such that $aRb$ $bRc$ but $a$ does NOT relate $c$. Let $a = 4, b = 6$ and $c = 4$. Then, $4R6$ and $6R4$ but $4$ does NOT relate $4$.

答:
Your overarching ideas are correct, but there are some minor mistakes along the way. So some minor clarifications and comments:

For reflexivity, your work is fine. If $R$ were reflexive, then we would have $1R1$, but $1+1 \ne 10$, so that doesn't hold.

For symmetry, you have the right idea, but you should bear in mind that a simple example does not mean that it holds all of the time. (For instance, that $5R5$ holds doesn't mean the relation as a whole is reflexive!) What you need to see is that symmetry holds if, whenever $aRb$ holds, then $bRa$ holds. This follows from commutativity of addition:

$$aRb \iff a+b = 10 \iff b+a = 10 \iff bRa$$

For transitivity, you have the right idea: a counterexample is always a nice means of disproof when possible. If it were transitive, then we would have $4R6$ and $6R4$ implies $4R4$, but $4+4=8 \ne 10$, so we don't have transitivity.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
Gaussian distribution of crest factor of Gaussian noise
I have a confusing problem.

I've produced a collection of waveforms, comprised of $n$ samples from a Gaussian distribution. All waveforms are composed of the same number of samples, picked from a Gaussian with the same values for $\mu$ and $\sigma$ (it's actually a natural source, however it's fairly consistent).

So, when I plot the sample values on a histogram for a single waveform, I get a Gaussian distribution (horizontal axis being sample value, vertical axis being # of samples). Similarly, when I plot the SNR for each sample (horizontal axis being SNR), where noise is defined as the RMS of the waveform, I get a Gaussian distribution.

Now, I'm looking at the SNR of whole waveforms. I define the SNR of a waveform to be the maximum value of the waveform, minus the minimum value of the waveform, quantity divided by the RMS of the waveform. So, $\frac{max-min}{RMS}$. I compute this value for each waveform, and make a histogram with the horizontal axis being # of waveforms, and the vertical axis being SNR, as defined.

Somewhat suprisingly, I get a very clear Gaussian distribution here, too. I guess I was expecting that $\frac{max-min}{RMS}$ would be roughly the same for all waveforms, and therefore I'd get a nice sharp peak somewhere in the histogram.

Why is this? I can't make up my mind if it's surprising, or expected.

答:
I'm no stats expert, but here's some insight from what I've learned in the world of signal processing for random vibration:

The distributions of the maximum and minimum of the waveform will each follow an Extreme Value Distribution--namely the Gumbel distribution when the random variable is drawn from a Gaussian distribution. The location and scale parameters of the Gumbel distribution will be related to the length of waveform $n$, process mean $\mu$, and process standard deviation $\sigma$.

The RMS of the waveform is also a random variable. For a gaussian process distribution, it will follow a chi distribution, and is related to the true standard deviation of the Gaussian process.

Putting it all together, the maximum and minimum values for a single waveform are both randomly chosen from independent Gumbel distributions mirrored about the process mean, and the RMS from a chi distribution, so your $\frac{max-min}{RMS}$ statistic is another random variable, resulting from the combination of each of these independent statistics. While the parameters of the Gumbels and chi are dependent, the samples drawn from them are independent.

The result you are producing is the range of the waveform (positive) divided by the RMS (positive), so this quantity is strictly positive and would not be Gaussian.


答:
If I change variable/u-substitute in a divergent improper integral, I get another divergent improper integral. Is that not intuitively obvious? A change of variable shouldn’t be able to turn divergent improper integrals into convergent ones, or vice versa.

Changes of variable should also preserve the manner of divergence -- whether the original improper integral approaches $+\infty$, approaches $-\infty$, or oscillates forever (or some combination), the $u$-substituted version is going to be doing the exact same thing.

Edit: Let's take the example from the original post, where we change variable in $\int_1^\infty \frac{1}{x} dx$ with $u = 2x$. We're supposed to get $du = 2x dx$, $du/2 = dx$, and $$\int_1^\infty \frac{1}{x} dx = \int_2^\infty \frac{1}{u/2} \frac{du}{2} = \int_2^\infty \frac{1}{u} du.$$

It seems that OP doesn't like when people say, informally, "The value of the improper integral here is $+ \infty$." In fact, that's widely understood to mean a certain kind of limiting behavior which gives more specificity over just saying, "The value of the improper integral does not exist", or, "The improper integral does not converge to a value." Let's go back to high school calculus and remember how these improper integrals are computed: $$"\int_1^\infty \frac{1}{x} dx" \text{ means/equals/is defined as } "\lim_{b \to \infty} \int_1^b \frac{1}{x} dx, \text{ if it exists."}$$

To say that the improper integral "is" $+ \infty$, is just to say that that limit "is" $+ \infty$, right? That is, as $b \to \infty$, the value of the integral is increasing without bound. (I'm putting "is" in quotes here to remind us that the $+ \infty$ is really describing a limiting behavior and not a numerical value.)

So let's look at what that means for our $u$-substituted integral:

$$\int_1^\infty \frac{1}{x} dx = \lim_{b \to \infty} \int_1^b \frac{1}{x} dx = \lim_{b \to \infty} \int_2^{2b} \frac{1}{u} du = \int_2^\infty \frac{1}{u} du,$$

and clearly if one of those limits "is" $+ \infty$, then the other limit "is" $+ \infty$ too. The same would hold true regardless of which specific $u$-substitution I made in the original--the corresponding limit is going off to $+ \infty$.

Note that I haven't said anything about the "regularized values" of these integrals. The "regularized value" is not to be confused with the integral itself. In fact, if the integral diverges but it has a "regularized value", then the value is already, by definition, different from the integral itself! There's no reason a change of variable should preserve whatever "regularized" number is assigned to a divergent integral, much less to think that this casts fundamental doubt on $u$-substitution itself.

Title: Is there a solid reason why some people assume the fundamental theorem of calculus should still hold for divergent integrals with improper bounds?

Body:
I asked this question on Mathoverflow and received 6 downvotes so far and no answer.

In the comment section of this question several people seemingly freely assume that fundamental theorem of calculus should still hold for divergent integrals with improper bounds.

That is, they assume

$$\int_a^b f(\varphi(x))\varphi'(x) dx = \int_{\varphi(a)}^{\varphi(b)} f(u)du$$

holds even if $a$ or $b$ are $\pm\infty$ and the integral in the left-hand side is divergent to infinity.

I wonder, what can justify this assumption. In my view it is totally unjustified. Why do I think so? Because it allows the following:

$$I=\int_0^\infty1dx=2\int_0^\infty1du=2I$$
or (with substitution $u=2x$)
$$\int_1^{+\infty}\frac1x dx=\int_2^{+\infty}\frac1u du$$

In the second case the indegrals even have different regularized values, the left-hand side integral has the regularized value $0$ while the right-hand side integral has the regularized value $-\ln2$. In other words, they are different divergent integrals with different properties.

In the comment section the user Johannes Hahn justified the relation with words "because it's true" but I do not see in what sense this equality even can be true for divergent integrals?

Is it equality of the values? If so, what value can have a divergent integral without extension of real numbers? Or an extension is assumed?

Is it equality of some other set of properties? In that case we see that the regularized values of these two integrals before and after substitution are different.

Something else?


问:
Prove $\gcd(n, n + 1) = 1$ for any $n$
Let $n \in \mathbb Z$ be even. Then $n + 1$ is odd. So, $2$ doesn't divide $n + 1$. Thus there's no even number for which $\gcd(n, n+1)$ is not $1$. I am not sure how to show it for odd numbers. Is there a better way to prove the statement?</p>

答: Suppose $n$ $\in \mathbb{Z}$ $s.t$ $n = 1$. Obviosuly, $gcd(n,n+1) = gcd(1,n+1) = 1$. Now, suppose $n$ $\in$ $\mathbb{Z}$ $s.t$ $n$ $\gt$ $1$. By the division algorithm, we have $$\ (n+1) = nq + r$$ $$0 \le r \lt n$$In particular $$\ (n+1) = n(1) +1$$ which gives us $$ gcd(n+1,n) = gcd(n,1) =1$$ $\mathbb{Q.E.D}$

Title: Prove $\gcd(n, n + 1) = 1$ for any $n$