问:
Integrating an Infinite exponent tower
$$ \int_0^1 x^{2^{x^{2^{x^{\ldots}}}}} ~~ dx =  ~~?$$

What I've tried so far:

$$ x^{2^{x^{2 \ldots}}} \overbrace{=}^{\text{def}} y \\ x^{2^{y}} = y \\ 2^y \ln (x) = \ln(y)$$

Now applying the chain rule:

$$\begin{align} \\&2^y \cdot \frac{dy}{dx} \cdot \ln(2) \cdot \ln(x) + \frac{1}{x} \cdot 2^y \ln(y) =  \frac{dy}{dx} \cdot \frac1y \\ &\implies \frac{dy}{dx} \left (2^y \cdot \ln(x)  \cdot \ln(2) - \frac1y \right) = - \frac1x 2^y \cdot \ln(y) \\ &\implies \frac{dy}{dx} = \left ( \frac{- \frac1x 2^y \cdot \ln(y)}{2^y \cdot \ln(x)  \cdot \ln(2) - \frac1y} \right) \end{align}$$

The problem that this looks like a pretty harsh equation to solve for $dy$ and then substitute it back to the integral, with the fact that we need to find the value of $x$ in terms of $y$.

Any help would be appreciated!

答:
Consider
$$I=\int_0^1 y^{2^{-y}}\,dy=\int_0^\frac 12 y^{2^{-y}}\,dy-\int_1^\frac 12 y^{2^{-y}}\,dy=I_1-I_2$$
Each integrand will be developed as a series expansion around the lower bound to order $(n+1)$ and termwise integrated. This does not lead to very complicated expressions for the integrands and the result of integration is just a polynomial in $\log(2)$.

For example, for $n=3$, the integrands are
$$y-y^2 \log (2) \log (y)+\frac{1}{2} y^3 \left(\log ^2(2) \log ^2(y)+\log ^2(2) \log
   (y)\right)+O\left(y^4\right)$$
$$1+\frac{y-1}{2}+(y-1)^2 \left(-\frac{1}{8}-\frac{\log (2)}{2}\right)+(y-1)^3
   \left(\frac{1}{16}+\frac{\log ^2(2)}{4}\right)+O\left((y-1)^4\right)$$

and the result of integration is
$$\frac{5127+72 \log ^4(2)-36 \log ^3(2)+339 \log ^2(2)-64 \log (2)}{9216}$$
As a function of order $n$, some results
$$\left(
\begin{array}{cc}
 n & I_1-I_2 \\
 1 & 0.56250000 \\
 2 & 0.57249702 \\
 3 & 0.56967702 \\
 4 & 0.56945295 \\
 5 & 0.56962947 \\
 6 & 0.56968187 \\
 7 & 0.56969293 \\
 8 & 0.56969494 \\
 9 & 0.56969522 \\
 10 & 0.56969526 \\
 11 & 0.56969527 \\
 12 & 0.56969528
\end{array}
\right)$$

Title: Integrating an Infinite exponent tower


问:
Correct application of birthday problem
The problem is as follows:

There are 10 shooters at a shooting range. Each shooter is given 5 bullets. They all begin shooting at 9am and end shooting at 10am, They each shoot all 5 of their bullets over that period.

The shooters each choose 5 random time points over the hour and they shoot 1 bullet per time point. The time resolution used is 1 second (3600 possible time points). So the times 09:37:53.100 and 09:37:53.707 are considered to have occurred in the same second.

What is the probability that two shooters fire a bullet at the same time point?

I thought the birthday problem could be used to solve this problem. I replaced the number of days in a year (365) with 3600, then I replaced the number of people in the room with the total number of bullets fired which is 50, resulting in a roughly 29% probability of two shots occurring at the same time.

However when I reduce the problem to just one shooter with 5 bullets, the above construction gives 0.13% probability.

This is wrong because a shooter can only shoot 1 bullet at a time, the probability should be zero, hence the above method is wrong.

I was wondering if some could suggest a way to solve this particular problem.

I think a rewording of the problem could be as follows:

There is a room with MxN people, where M < N. Each person is associated with only one of M clubs and all the people are equally divided amongst the M clubs. What is the probability that two people from different clubs share the same birthday? (day/month combination only)

UPDATE

How would one apply Poisson distribution to this problem? As it seems that the math gets pretty complicated for the exactly 2 collisions problem.

答: Consider the probability that no shots are shot in the same second. The first shooter has $ 3600 $ shots to choose from, the next $ 3595 $, and so forth until the fifth has $ 3580 $ possibilities. Hence, the probability that none of them are in the same second is $$ \prod_{i = 0}^4 \frac{\dbinom{3600 - 5i}{5}}{\dbinom{3600}{5}} \approx 0.9327 $$ Hence, the complementary probability (the one you want) is approximately $ 0.0673 $. </p>

ContentLicense: CC BY-SA 3.0

Title: Correct application of birthday problem


问:
Partial derivatives proof excercise
I need some help in the proof of the next excercise:

Proof that $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}$ in every point of an open set D if and only if there is $g$ derivable such that $f(x,y) = g(x+y)$ $\forall (x,y) \in D$.

Please I'm super lost and I don't know which way to go so I can solve it :(

答: HINT: If $\nabla f$ is a scalar multiple of $(1,1)$ at each points, then the level curves of $f$ are everywhere orthogonal to $(1,1)$, which means they must be lines $x+y=c$. (I'm assuming that $\nabla f \ne 0$ at each point.)

ContentLicense: CC BY-SA 4.0

Title: Partial derivatives proof excercise


问:
Given dihedral angles, find a set of edges
In the paper Space Vectors Forming Rational Angles a special set of tetrahedra is mentioned.

"The remaining three are in the R-orbit of the tetrahedron with dihedral angles ($π/7, 3π/7, π/3, π/3, 4π/7, 4π/7$)."

What is a set of edge lengths or vertices for this tetrahedron? I've written a function that converts edges to angles, but I need the reverse.

I solved it in a very messy way: https://community.wolfram.com/groups/-/m/t/2169279. An elegant solution would still be nice.

答:
The original paper already gives the relative configuration of dihedral angles: if $\alpha_{ij}$ is the dihedral angle at the edge between points $i$ and $j$
$$(\alpha_{12},\alpha_{34},\alpha_{13},\alpha_{24},\alpha_{14},\alpha_{23})=\left(\frac\pi7,\frac{3\pi}7,\frac\pi3,\frac\pi3,\frac{4\pi}7,\frac{4\pi}7\right)$$
so the tetrahedron is rotationally symmetric about the line between the midpoints of edges $12$ and $34$.

Now consider four non-collinear points $p,a,b,c$. Let $\angle bpc=A$ and the dihedral angle between planes $apb$ and $cpa$ be $A'$; define $B,B',C,C'$ cyclically. Then Wikipedia gives the following formula:
$$\cos A'=\frac{\cos A-\cos B\cos C}{\sin B\sin C}$$
Inverting the three formulas obtained from this by cyclic permutation of $A,B,C$ gives, again with cyclic permutations,
$$\boxed{\cos A=\frac{\cos A'+\cos B'\cos C'}{\sin B'\sin C'}}$$
which allows determining the minimal polynomials of the cosines of all face angles:
$$\angle412,\angle123:x^3-3x^2-x+\frac{13}7\ (0.708065\dots)$$
$$\angle421,\angle213:x^3-\frac{\sqrt{21}}3x^2-x+\frac{13\sqrt{21}}{63}\ (0.606682\dots)$$
$$\angle142,\angle231:x^3+\frac{\sqrt{21}}3x^2+\frac x3-\frac{\sqrt{21}}{63}\ (0.131776\dots)$$
$$\angle423,\angle314:x^3-\sqrt{21}x^2+\frac{17}3x-\frac{29\sqrt{21}}{63}\ (0.935327\dots)$$
$$\angle243,\angle431:x^3-\frac{\sqrt{21}}3x^2+\frac x3+\frac{\sqrt{21}}{63}\ (-0.131776\dots)$$
$$\angle234,\angle341:x^3+x^2-x+\frac17\ (0.473952\dots)$$
If edge $12$ has length $1$, the laws of sines and cosines finally give minimal polynomials for edge lengths:
$$24,31:x^3-\sqrt{21}x^2+\frac{3\sqrt{21}}7\ (0.712358\dots)$$
$$14,32:x^3+2x^2-x-1\ (0.801937\dots)$$
$$34:x^3-9x^2-x+1\ (0.286208\dots)$$


问:
How do I factor polynomials with power 3 and above?
All of the videos I've found do things like "let's try -1. oh, it works! okay, now let's divide it out and find the next root".  Is there really no other strategy other than guess and check?  A couple of times I've come across these high power polynomials in contests and I just have no clue what to do with them.  I'm asking this question specifically now because there's a question in my textbook that asks me to find the two imaginary roots of a polynomial of 4th power.  Clearly guess and check is not going to work as well with imaginary numbers, and even with real ones it seems like a pain.  What strategies do you guys use to factor?</p>

答:
There is in fact a Cubic Formula, which can give you the roots, and make factoring trivial...although it is a little more complicated than the famed Quadratic Formula; see here https://math.vanderbilt.edu/schectex/courses/cubic/. And there is also a Quartic Formula, but that is much more complicated https://en.wikipedia.org/wiki/Quartic_function#Solving_a_quartic_equation . And in fact there is $\textbf{not}$ a general equation for 5th and higher order polynomials; a famous theorem known as Abel–Ruffini theorem.

But that is besides the point.

For factoring, particularly in contests; there is normally a trick to see, which is what makes the problems so rich. There are often various symmetries that can be used, or a quantity you can add and subtract which allows you to complete squares, polynomial division, etc.

If we are told roots are rational, then there is a lot that can be accomplished by guess and check, which is often what is taught in high school. However in the case where you have a Quartic Polynomial, I would assume the first step would be to split it into a product of two Quadratic Polynomials and go from there, since if there are 2 imaginary roots, they are probably conjugates of one another which can be expressed in one quadratic polynomial.

There is no general strategy of factoring, just various techniques.


问:
What is the equivalent to $\tan\left(\frac{7\pi}{4}-\theta\right)$?

What is the equivalent to $\tan\left(\frac{7\pi}{4}-\theta\right)$?

I'm very new to the mathjax formatting and the site in general but when I work the problem out it seems as though the answer should be $-1$.

$$\frac{\left(-1\ -\ \tan\left(\theta\right)\right)}{1+\left(-1\cdot\tan\left(\theta\right)\right)}$$

$$\frac{\left(-1\ -\ \tan\left(\theta\right)\right)}{1-\tan\left(\theta\right)}$$

答:
$$x=\tan\left(\frac{7\pi}{4}-\theta\right)
=-\tan\left(θ + \frac{\pi}{4}\right)$$

$$0 \le\theta\le\pi\implies -\infty\le x \le 1$$

Note: $\space
\theta\in\{0,\space\frac{\pi}{2}\}\implies
x=-1\space\land\space
\theta=\pi\implies x=1\space$

and as
$${\theta\rightarrow\frac{\pi}{4}}\implies x\rightarrow - \infty
$$


答:
As with the birthday problem, let's compute the probability that there is no coincidence.
Let the first shooter pick his five seconds.  The second shooter has a $\frac{3595}{3600}$ chance of avoiding a coincidence with his first shot.   He has a $\frac{3594}{3599}$ chance of avoiding a coincidence  with his second shot, and a $\frac{3593}{3598}$ with his third shot.  Hence the desired probability (of avoiding coincidence) is

$$1\cdot \frac{3595^\underline{5}}{3600^{\underline{5}}}\cdot \frac{3590^\underline{5}}{3600^{\underline{5}}}\cdot \frac{3585^\underline{5}}{3600^{\underline{5}}}\cdots \cdot \frac{3555^\underline{5}}{3600^{\underline{5}}}$$

This simplifies to  $$\frac{3595^{\underline{45}}}{(3600^{\underline{5}})^{9}}=\frac{3600^{\underline{50}}}{(3600^{\underline{5}})^{10}}$$

This is in contrast to the birthday problem, whose answer is $$\frac{365^{\underline{n}}}{365^n}$$

Title: Correct application of birthday problem

Body:
The problem is as follows:

There are 10 shooters at a shooting range. Each shooter is given 5 bullets. They all begin shooting at 9am and end shooting at 10am, They each shoot all 5 of their bullets over that period.

The shooters each choose 5 random time points over the hour and they shoot 1 bullet per time point. The time resolution used is 1 second (3600 possible time points). So the times 09:37:53.100 and 09:37:53.707 are considered to have occurred in the same second.

What is the probability that two shooters fire a bullet at the same time point?

I thought the birthday problem could be used to solve this problem. I replaced the number of days in a year (365) with 3600, then I replaced the number of people in the room with the total number of bullets fired which is 50, resulting in a roughly 29% probability of two shots occurring at the same time.

However when I reduce the problem to just one shooter with 5 bullets, the above construction gives 0.13% probability.

This is wrong because a shooter can only shoot 1 bullet at a time, the probability should be zero, hence the above method is wrong.

I was wondering if some could suggest a way to solve this particular problem.

I think a rewording of the problem could be as follows:

There is a room with MxN people, where M < N. Each person is associated with only one of M clubs and all the people are equally divided amongst the M clubs. What is the probability that two people from different clubs share the same birthday? (day/month combination only)

UPDATE

How would one apply Poisson distribution to this problem? As it seems that the math gets pretty complicated for the exactly 2 collisions problem.


问:
Plane through the intersection of two given planes.
If I have two planes $$\mathbf r\cdot\mathbf {\hat n_1}=p_1\\ \mathbf r\cdot\mathbf {\hat n_2}=p_2$$

If they intersect somewhere then the intersection will form a line, if $\mathbf r'$ is the position vector of a point on the intersection line then I have $$\mathbf r'\cdot\mathbf {\hat n_1}=p_1\\ \mathbf r'\cdot\mathbf {\hat n_2}=p_2$$

If I combine these two equations I would get $$\mathbf r'\cdot(\mathbf {\hat n_1}+\mathbf{\hat n_2})=p_1+p_2$$
This looks like a vector equation of a plane.

But if we see in internet or in books we would find there is scalar $\lambda$  multiplied.

$$\mathbf r'\cdot(\mathbf {\hat n_1}+\lambda\mathbf{\hat n_2})=p_1+\lambda p_2$$

So my question is why there is $\lambda$ there? Please also give me the geometrical image of the plane which is passing through the intersection of those two planes. I'm not able to imagine how would that plane look like.

答:
Lambda is used for direction here
We can see the plane passes through there intersection can be perpendicular to them or at 120 degrees, so its purpose is to give different planes for different values of lambda.</p>


问:
Alternative approaches to prove the following inequality
For $a,b,c \in \mathbb{R^+},$ prove that
$$\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3.$$
I managed to prove this problem using the technique of isolated fudging. In particular, one can prove that $\dfrac{1}{3}\left(\frac{2a}{b+c} \right)^{\frac{2}{3}} \geq  \dfrac{a}{a+b+c} $ (motivation is explained below) using AM-GM, as follows:
\begin{align*}
\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} \geq  \dfrac{a}{a+b+c} & \iff \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} \geq  \dfrac{3a}{a+b+c} \\
& \iff  \left(\dfrac{2a}{b+c} \right)^{2} \geq  \left(\dfrac{3a}{a+b+c}\right)^3  \\
& \iff 4a^2(a+b+c)^3 \geq 27a^3(b+c)^2.
\end{align*}
The preceding inequality is homogeneous, hence W.L.O.G. set $a+b+c=3$. It thus suffices for us to prove that $4 \geq a(3-a)^2 \iff 8 \geq 2a(3-a)^2$. But this is obvious from AM-GM: $2a(3-a)^2 \leq \left(\dfrac{2a+(3-a)+(3-a)}{3}\right)^3=8.$ Similarly, we have  $\dfrac{1}{3}\left(\dfrac{2b}{a+c} \right)^{\frac{2}{3}} \geq  \dfrac{b}{a+b+c}  $ and $\dfrac{1}{3}\left(\dfrac{2c}{a+b} \right)^{\frac{2}{3}} \geq  \dfrac{c}{a+b+c} $. Thus, summing cyclically, we obtain:
$$\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \dfrac{1}{3}\left(\dfrac{2b}{a+c} \right)^{\frac{2}{3}} + \dfrac{1}{3}\left(\dfrac{2c}{a+b} \right)^{\frac{2}{3}} \geq 1 \Rightarrow \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3 .$$
And we are done.

Some motivation:

Let $f(a,b,c)=\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} - \dfrac{a^r}{a^r+b^r+c^r}$, for some $r \in \mathbb{R}$. Ideally, we want $f(a,b,c) \geq 0$ for all  values of $a,b,c$. Note that $f(1,1,1)=0$, which suggests that we should set $(1,1,1)$ as a local minimum of $f$. Hence, take the partial derivative of $f$ with respect to $a$, and set it to zero at $(1,1,1)$. By solving the resulting equation, we find a corresponding value of $r$:

$$\dfrac{\partial f}{\partial a}=  \dfrac{1}{3} \cdot \sqrt[^3]{4} \cdot \left(\dfrac{1}{b+c} \right)^{\frac{2}{3}} \cdot \dfrac{2}{3} \cdot a^{\frac{-1}{3}} - \dfrac{ra^{r-1}(a^r+b^r+c^r)-a^r(ra^{r-1})}{(a^r+b^r+c^r)^2}$$
Hence,
$$\frac{\partial f}{\partial a }\Bigr|_{(1,1,1)} =0 \Rightarrow \dfrac{2}{9}- \dfrac{2r}{9} =0 \Rightarrow r=1.$$

My question is, is there any other way to solve this inequality besides isolated fudging? I.e. would methods such as Cauchy / Holder / Jensen also work here? I would love to see any other alternative approach.

答:
Let $p=a+b+c,\,q=ab+bc+ca,\,r=abc.$ Using the Holder inequality, we have
$$\left[\sum \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}}\right]^3 \sum a^2(b+c)^2 \geqslant 4(a+b+c)^4.$$
It remain to show that
$$4(a+b+c)^4 \geqslant 27 \sum a^2(b+c)^2,$$
or
$$2p^4+27pr \geqslant 27q^2.$$
If $p^2 > 9q,$ then $$2p^4 > 2(9q)^2 > 27q^2.$$
If $p^2 \leqslant 9q,$ by the Schur inequality, we have
$$r \ge \frac{4pq-p^3}{9}.$$
We will show that
$$2p^4+3p^2(4q-p^2) \geqslant 27q^2,$$
equivalent to
$$(9q - p^2)(p^2-3q) \geqslant 0.$$
This is true. The proof is completed.


问:
Solve by using method of variation of constants
$y''=sin(x)-cos(x)$

I tried integrating twice and got the general solution, but I need to do it with the method of variation of constants, does anyone know how to do this?

答:
First we solve the homogenous equation:

$$y_h''=0 \rightarrow y=C_1x+C_2$$

Then we solve for a particular solution of the specific equation:

$$y_p=C(x) y_h$$
$$y_p''=C''(x)y_h+2C'(x)y'_h+C(x)y''_h=\sin(x)-\cos(x)$$

From here we can simplify and find:

$$C(x)=\frac{-\sin(x)}{C_1x+C_2}+\frac{\cos(x)}{C_1x+C_2}$$

So the general solution is:

$$y=y_p+y_h= \left(\frac{-\sin(x)}{C_1x+C_2}+\frac{\cos(x)}{C_1x+C_2}\right)(C_1x+C_2) + C_1x+C_2$$
$$y = -\sin(x)+\cos(x)+ C_1x+C_2$$


问:
What is the rank of $Q+Q^T-Q\circ I$ if $Q = qq^T$
I have a question about the rank of $$Q+Q^T-Q\circ I,$$ where

$Q = qq^T$, $q\in \mathbb{R}^n$. Suppose $q^Tq=1$.

$A\circ B:$ Hadmard (elementwise) product

$Q\circ I$ just takes the diagonal entries of $Q$.

I know

$Q$ is rank $1$, positive-semidefinite.

$Q+Q^T$ is still rank $1$.

However, when considering the last term, what the rank will be?   full rank?

Thanks!

答:
First, let's assume all the entries of $q \in \mathbb{R}^n$ are non-zero. It is easy to check that $Q+Q^T = 2qq^T$ and $Q \circ I = D$ where $D$ is a diagonal matrix with entries $D_{n,n} = q_n^2$.

Hence, $Q+Q^T+Q\circ I = D+2qq^T$ where $D$ is an invertible matrix (since it is diagonal with non-zero diagonal entries). So by the Sherman-Morrison formula, we have that $Q+Q^T+Q\circ I = D+2qq^T$ is invertible iff $1+2q^TD^{-1}q \neq 0$. But this holds trivially since$$1+2q^TD^{-1}q = 1+2\sum_{k = 1}^{n}q_k \cdot (D^{-1})_{k,k} \cdot q_k = 1+2\sum_{k = 1}^{n}q_k \cdot q_k^{-2} \cdot q_k = 1+2\sum_{k = 1}^{n}1 = 2n+1 \neq 0.$$

Thus, if all the entries of $q$ are non-zero, then $Q+Q^T+Q\circ I = D+2qq^T$ is invertible, and thus, has rank-$n$.

Now, suppose $q$ has exactly $r$ entries which are non-zero. You can check that $$(Q+Q^T+Q \circ I)_{i,j} = (D+2qq^T)_{i,j} = \begin{cases} 2q_iq_j & \text{if} \ j \neq i \\ 3q_i^2 & \text{if} \ j = i\end{cases}.$$ Hence, $Q+Q^T+Q\circ I$ will have $n-r$ rows and columns which are all zero. If you look at the $r \times r$ submatrix of $Q+Q^T+Q\circ I$ formed by deleting all the zero rows/columns, you can use the same Sherman-Morrison formula based argument to show that this $r \times r$ submatrix is invertible, and thus, full rank. This is enough to conclude that $Q+Q^T+Q \circ I$ has rank-$r$.

Therefore, the rank of $Q+Q^T+Q \circ I$ will be exactly the number of non-zero entries in $q$.


问:
Book of exercises on sequences and/or series
I'm looking for a dedicated problem/exercise book or pdf to work on limits of sequences and/or convergence of series (real analysis). Something like 1000 exercises with or without solution depending on the difficulty. A new year resolution of mine is studying one a day.</p>

答:
Problems in Mathematical Analysis I: Real Numbers, Sequences and Series

Problems in Real Analysis: Advanced Calculus on the Real Axis by Teodora-Liliana Radulescu, Titu Andreescu, and Vicentiu Radulescu

Knopp's Theory and Applications of Infinite Series

Infinite Series by Earl David Rainville

Bonar and Khoury's Real Infinite Series

An Introduction to the Theory of Infinite Series by Thomas John I'Anson Bromwich


问:
Find the general solution of the next differential equation
Im trying to solve this problem but have no idea how start

$$y''-\frac{2x}{x^{2}+1}y'+\frac{2}{x^{2}+1}y=(x^2+1)^2$$

答:
$$y''-\dfrac{2x}{x^{2}+1} y'+\dfrac{2}{x^{2}+1}y=0$$

As @Satyendra answered, $y=x$ is a solution. There is another I found almost by accident which is $y=(x-i)^2$.

In both cases, you can reduce the order but the particular solution given by @Satyendra is from far away the simplest.

I suppose that tha particular solution is just a polynomial with only even powers (find why).


问:
Cauchy's Theorem- Trigonometric application
any help would be very much appreciated. The question asks to evaluate the given integral using Cauchy's formula. I plugged in the formulas for $\sin$ and $\cos$ ($\sin= \frac{1}{2i}(z-1/z)$ and $\cos= \frac12(z+1/z)$) but did not know how to proceed from there.<br>
$$\int_0^{2\pi} \frac{dθ}{3+\sinθ+\cosθ}$$
Thanks. </p>

答:
Real Method

Often, this type of integral is workable using the substitution
$$
z=\tan(\theta/2)\quad\text{and}\quad\mathrm{d}\theta=\frac{\mathrm{2\,d}z}{1+z^2}\\
\sin(\theta)=\frac{2z}{1+z^2}\quad\text{and}\quad\cos(\theta)=\frac{1-z^2}{1+z^2}
$$
then
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}\theta}{3+\sin(\theta)+\cos(\theta)}
&=\int_{-\pi}^\pi\frac{\mathrm{d}\theta}{3+\sin(\theta)+\cos(\theta)}\\
&=\int_{-\infty}^\infty\frac{2\mathrm{d}z}{3(1+z^2)+2z+(1-z^2)}\\
&=\int_{-\infty}^\infty\frac{\mathrm{d}z}{z^2+z+2}\\
&=\int_{-\infty}^\infty\frac{\mathrm{d}z}{(z+1/2)^2+7/4}\\
&=\frac{2\pi}{\sqrt7}
\end{align}
$$

Complex Method
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}\theta}{3+\sin(\theta)+\cos(\theta)}
&=\oint\frac1{3+\frac1{2i}(z-\frac1z)+\frac12(z+\frac1z)}\frac{\mathrm{d}z}{iz}\\
&=\oint\frac{2z}{6z-i(z^2-1)+(z^2+1)}\frac{\mathrm{d}z}{iz}\\
&=\oint\frac{-2i}{(1-i)z^2+6z+(1+i)}\mathrm{d}z\\
\end{align}
$$
The singularities of the integrand are at $\frac{-3+\sqrt7}{2}(1+i)$ and $\frac{-3-\sqrt7}{2}(1+i)$. The second is outside the unit circle, so we only need to compute the residue at the first, which is
$$
\frac{-2i}{2(1-i)z+6}=\frac{-2i}{2\sqrt7}
$$
Thus, the integral is $2\pi i$ times the residue:
$$
\int_0^{2\pi}\frac{\mathrm{d}\theta}{3+\sin(\theta)+\cos(\theta)}=\frac{2\pi}{\sqrt7}
$$


问:
Why is $\frac{\frac{1}{x}}{1+\frac{1}{x^2}}=\frac{x}{1+x^2}$ true?
While trying to adapt Euler's series for $\tan^{-1}{x}$ to be used for $\cot^{-1}{x}$ so I could more efficiently compute $\pi$ with Machin-like formulae, I discovered the following identity:
$$\frac{\frac{1}{x}}{1+\frac{1}{x^2}}=\frac{x}{1+x^2}$$
I was surprised to discover that it's true because it seems counterintuitive. Why should the expression $\frac{x}{1+x^2}$ be the exact same value for both $x$ and its reciprocal, regardless of the value of $x$? It's not exactly the first example that comes to mind when you consider when you see an identity like
$$f\left(\frac{1}{x}\right)=f(x)$$
What's the principle behind it?

答:
Just multiply the numerator and denominator by $x^2$.

$$\frac{\frac{1}{x}}{1+\frac{1}{x^2}} = \frac{\frac{1}{x}}{1+\frac{1}{x^2}} \cdot \underbrace{\frac{x^2}{x^2}}_{=1} = \frac{\frac{1}{x} \cdot x^2}{\left(1+\frac{1}{x^2}\right) \cdot x^2} = \frac{x}{x^2+1}$$

Nothing particularly special going on here.


问:
linearization on Matlab for Optimal control
I am studying a tutorial about pendulum-on-a-cart control on youtube and its purpose is to stabilise the pendulum at the upper position where $\theta=\pi$. The state-vector is
$$\mathbf{x}=\begin{bmatrix}
\chi \\
\dot{\chi}\\
\theta \\
\dot{\theta}\\
\end{bmatrix}$$
Therefore, we linearise about
$$\mathbf{x}=\begin{bmatrix}
0 \\
0\\
\pi\\
0\\
\end{bmatrix}$$
, and use the pole-placement command on Matlab to choose arbitrarily some set of admissible eigenvalues.

```<code>m = 1;M = 5;L = 2;g = 10;d = 1;
A = [0 1 0 0;
     0 -d/M m*g/M 0;
     0 0 0 1;
     0 -d/(M*L) (m+M)*g/(M*L) 0];
B = [0; 1/M; 0; 1/(M*L)];
p = [-1.0;-1.1;-1.2;-1.3];
K = place(A,B,p);
</code>```

To verify that the specific K does indeed do the job, we cal introduce the $u=K(y-r)$ back to the system and see how angle $\theta$ changes in time:

```<code>tspan = 0:0.001:10;
reference = [0; 0; pi; 0];
y0 = [0; 0; pi+.01; 0];
[t,y] = ode45(@(t,y)model(y,m,M,L,g,d,-K*(y-reference)),tspan,y0);
x = y(:,1);
x_dot = y(:,2);
theta = y(:,3);
theta_dot = y(:,4);

figure()
plot(t,theta);axis equal;xlabel('time');ylabel('theta');
</code>```

If we introduce it to the full nonlinear system:

```<code>function dydt = model(y,m,M,L,g,d,u)
  a = 1/(M+m-m*(cos(y(3)))^2);
  dydt(1,1) = y(2);
  dydt(2,1) = a*(g*sin(y(3))*cos(y(3)) + m*L*(y(4)^2)*sin(y(3)) - y(2)*d + u);
  dydt(3,1) = y(4);
  dydt(4,1) = (a/L)*(-(m+M)*g*sin(y(3)) - m*L*(y(4)^2)*sin(y(3))*cos(y(3)) + y(2)*cos(y(3))*d - u*cos(y(3)));
end
</code>```

the result is this:

If instead we introduce it to the linearized version:

```<code>function dydt = model(y,m,M,L,g,d,u)
    dydt(1,1) = y(2);
    dydt(2,1) = (-d/M)*y(2) + (m*g/M)*y(3) + (1/M)*u;
    dydt(3,1) = y(4);
    dydt(4,1) = (-d/(M*L))*y(2) + ((m+M)*g/(M*L))*y(3) + (1/(M*L))*u;
end
</code>```

, the result is this:

The pendulum never returns close to the reference, and practically goes all the way down to the stable equilibrium. Why does this happen? I would expect similar results, since the linearized version is valid and since the chosen eigenvalues and initial conditions are not unrealistic.

答:
The dynamics of a nonlinear system $\dot{x}=f(x,u)$ can be approximated with a linearization using

$$
\dot{x} \approx  \left.\frac{\partial\,f(x,u)}{\partial\,x}\right|_{\bar{x},\bar{u}} (x - \bar{x}) + \left.\frac{\partial\,f(x,u)}{\partial\,u}\right|_{\bar{x},\bar{u}} (u - \bar{u}),
$$

with $\bar{x}$ and $\bar{u}$ the values of $x$ and $u$ respectively at the operating point, such that $f(\bar{x},\bar{u})=0$.

In your implementation for the linearized system you multiplied the partial derivatives with $x$ (and $u$) instead of $x-\bar{x}$ (and $u-\bar{u}$).


问:
Uniform convergence of sequence of functions on unbounded intervals
I am new to this topic(sequence and series of functions) so I have a doubt.All theorems related to uniform convergence which I am studying are given on closed interval for example

(1) If a sequence of functions $f_n(x)$ converges pointwise to a function $f$ on $[a,b]$ then $$Sup_{[a,b]}\vert f_n(x)-f(x)\vert= 0\Rightarrow f_n(x)  \text{converges uniformly on [a,b]}$$
(2) Dini's theorem: For an increasing sequence $F={f_n}$ of continuous functions on an interval $I=[a,b]$ which converges pointwise on $I$ to a continuous function $f$ on $I$, Dini's theorem states that $F$ converges to $f$ uniformly on $I$

My questions:(1)Is the study of of uniform convergence not important for unbounded intervals?if it is not so then why we are only studying the convergence on bounded intervals(in particular closed intervals).

2)Are the methods to test the uniform convergence remains same if we work on unbounded intervals,if it is not so then what is the best method to check uniform convergence on unbounded intervals?

Any help would be appreciable

Thanks in advance

答: There is a lot of difference between bounded and unbounded domains. Dini's Theorem is stated for compact intervals because it is not valid for unbounded intervals. A very simple example to illustrate the difference between the two casees is $f_n(x)=\frac x n$. This sequence converges to $0$ uniformly on any bounded interval but not on $\mathbb R$.


问:
Prove that $x^2 + 1$ divides $f(x) - \frac{(f(i) - f(-i))}{2i} x - \frac{(f(i) + f(-i))}{2}$ in the ring $\mathbb{Q}(i)[x]$.
Let $\mathbb{Q}(i) = \{a + bi ~|~ a, b \in \mathbb{Q}\}$. With addition and multiplication as in $\mathbb{C}$, $\mathbb{Q}(i)$ is a field.

(Note: You are not being asked to prove that $\mathbb{Q}(i)$ is a field.)

Let $f(x) \in \mathbb{Q}(i)[x]$. Prove that $x^2 + 1$ divides
\begin{align*}
    f(x) - \frac{(f(i) - f(-i))}{2i} x - \frac{(f(i) + f(-i))}{2}
\end{align*}
in the ring $\mathbb{Q}(i)[x]$.

$\textbf{My Attempt:}$

Since, $i, i^2, i^3, i^4 \in \mathbb{Q}(i)$. Also, $i^5 = i \in \mathbb{Q}(i)$.

So, for some $n \in \mathbb{N}$, $i^n \in \mathbb{Q}(i)$.

Since, we let $f(x) \in \mathbb{Q}(i)[x]$.

Let $f(x)$ has degree of $n$ where $n \in \mathbb{N} \cup \{ 0 \}$.

Then, $f(i) = a_n i^n + \cdots + a_0$ and $f(-i) = a_n (-i)^n + \cdots + a_0$. for some $a_j\in \mathbb{Q}(i)$.

Since, $\mathbb{Q}(i)$ is a field with addition and multiplication as in $\mathbb{C}$.

Then, $f(i) \in \mathbb{Q}(i)$ and $f(-i) \in \mathbb{Q}(i)$.

Also, $(f(i) - f(-i)) \in \mathbb{Q}(i)$ and $(f(i) + f(-i)) \in \mathbb{Q}(i)$.

Since, notice that $-\frac{1}{2i} = \frac{1}{2}i \in \mathbb{Q}(i)$.

Then, $-\frac{(f(i) - f(-i))}{2i} = (\frac{1}{2}i)(f(i) - f(-i)) \in \mathbb{Q}(i)$ and $-\frac{(f(i) + f(-i))}{2} = (-\frac{1}{2})(f(i) + f(-i)) \in \mathbb{Q}(i)$.

Let $\alpha = -\frac{(f(i) - f(-i))}{2i}$ and $\beta = -\frac{(f(i) + f(-i))}{2}$. Where $\alpha, \beta \in \mathbb{Q}(i)$.

Then, $f(x) - \frac{(f(i) - f(-i))}{2i}x - \frac{(f(i) + f(-i))}{2} = f(x) + \alpha x + \beta$.

Since, we let $f(x) \in \mathbb{Q}(i)[x]$ and we set $f(x)$ has a degree of $n$.

So, $f(x) = b_n x^n + \cdots + b_1 x + b_0$ where $b_j \in \mathbb{Q}(i)$.

Then, $f(x) + \alpha x + \beta = (b_n x^n + \cdots + b_1 x + b_0) + \alpha x + \beta = b_n x^n + \cdots + (b_1 + \alpha)x + (b_0 + \beta)$.

Since, $b_1, b_0, \alpha, \beta \in \mathbb{Q}(i)$. So, $(b_1 + \alpha), (b_1 + \beta) \in \mathbb{Q}(i)$.

Then, we set $g(x) = b_n x^n + \cdots + (b_1 + \alpha)x + (b_0 + \beta) = c_n x^n + \cdots + c_1 x + c_0$.

$~\hspace{10mm}$ Which $c_n = b_n$, ..., $c_1 = (b_1 + \alpha)$, $c_0 = (b_0 + \beta)$. Where each $c_j \in \mathbb{Q}(i)$.

Then, $g(x)$ has degree of $n$ as well.

Also, $g(x) = c_n x^n + \cdots + c_1 x + c_0 = f(x) - \frac{(f(i) - f(-i))}{2i}x - \frac{(f(i) + f(-i))}{2}$.

$~\hspace{10mm}$ Where $g(x)$ is in the ring $\mathbb{Q}(i)[x]$

Next, I want to prove that $x^2 + 1$ divides g(x) in the ring $\mathbb{Q}(i)[x]$:
Which I need to find a $h(x) \in \mathbb{Q}(i)$ such that $g(x) = (x^2 + 1)h(x)$.

But I don't know how to find it. And I don't think it always work since if $g(x) = x + 1$. Then, I cannot find a $h(x)$.

答:
Your solution is very hard to read. Note that $x+i$ and $x-i$ are coprime and so for a polynomial to be divisible by $(x+i)(x-i) = x^2+1$ if suffices for it to have $i$ and $-i$ as roots.

A direct computation shows that

$$
\begin{align*}
    f(i) - \frac{(f(i) - f(-i))}{2i} i - \frac{(f(i) + f(-i))}{2} = f(i)-f(i) = 0
\end{align*}
$$

and

\begin{align*}
    f(-i) - \frac{(f(i) - f(-i))}{2i} (-i) - \frac{(f(i) + f(-i))}{2} = f(-i)-f(-i) = 0.
\end{align*}


问:
What is the FTC of a function from the perspective of Stokes' theorem?
In the case of $\int_a^b f(x) dx=F(b) - F(a)$ this can be written as $\int_a^b \frac d {dx} F(x) dx=F(b)- F(a).$

Comparing to Stokes' theorem, $\int_{\partial\Omega} \omega=\int_\Omega d\omega$ where $d\omega$ is a differential form, I guess it would be something like

$$\int_{\Omega=[a,b]} dF=\int_{\partial\Omega=\{a,b\}} F$$?

So the manifold $\Omega$ is the compact interval $[a,b]$ over which the exterior derivative $dF$ is integrated, whereas the boundary consists of the set of the two limit points $\partial\Omega=\{a,b\}$ over which the differential form $F$ is integrated.

PS: Great explanation:

答:
By the choice of F, dF/dx = f(x). In the vocabulary of differential forms, this is saying that f(x) dx is the exterior derivative of the 0-form, namely the function, F. In other words, that dF = f dx. The general Stokes theorem applies to higher differential forms ω instead of just 0-forms such as F.

A closed interval [a, b] is a simple example of a one-dimensional manifold with boundary. Its boundary is the set consisting of the two points a and b. Integrating f over the interval may be generalized to integrating forms on a higher-dimensional manifold. In the more general case, the manifold has to be orientable, and the form has to be compactly supported in order to give a well-defined integral.

The Orientation of the interval can be thought of as the ordering of the elements, the closed bounded interval [a,b] is compact as you learn in analysis.

The two points a and b form the boundary of the closed interval. More generally, Stokes' theorem applies to oriented manifolds M with boundary. The boundary ∂M of M is itself a manifold and inherits a natural orientation from M. Intuitively, a inherits the opposite orientation as b, as they are at opposite ends of the interval. So, "integrating" F over two boundary points a, b is taking the difference F(b) − F(a).

This integrating over two points is really more of a degenerate case and maybe it is confusing as to why you are just plugging in the values.

You can understand it as integrating over a singleton using an atomic measure with the two point boundary. An atomic measure gives a non-zero measure to the singleton set so that taking an integral over a singleton makes sense in a way that is more than just hand waving.


问:
Find the limit of $\lim\limits_{x \to 1/2}\frac{\tan(\pi x)}{2x - 1}+\frac{2}{\pi(2x - 1)^2}$
Find the following limit:

$$\lim_{x \to 1/2}\frac{\tan(\pi x)}{2x - 1}+\frac{2}{\pi(2x - 1)^2}$$

Using a calculator, I know the function is divergent, but I don't know how to show that it is divergent. I tried to simplify it first, but that just made the function more complicated.

答:
As other did show, it is almost always easier to work around $0$. So
$$\lim_{x \to \frac 12}\frac{\tan(\pi x)}{2x - 1}+\frac{2}{\pi(2x - 1)^2}=\lim_{y \to 0}\frac{1-\pi  y \cot (\pi  y)}{2 \pi  y^2}=\pi\lim_{z \to 0}\frac{1-z \cot (z)}{2 z^2}$$

Now, using Taylor expansions
$$\frac{1-z \cot (z)}{2 z^2}=\frac {1-z\left(\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}+O\left(z^5\right) \right) } {2z^2}=\frac{1}{6}+\frac{z^2}{90}+O\left(z^4\right)$$ So, your limit is $\frac \pi 6$.


答:
You can follow the below references:

A Problem Book in Real Analysis by Asuman G. Aksoy & Mohamed A. Khamsi

Advanced Calculus by Robert Wrede, Murray Spiegel

Methods of Solving Sequence and Series Problems by  Ellina Grigorieva

Sequences and Series by Edward A. Azof

All the above mentioned books contains various problems with solutions (not all). I think these will help you a lot.

Title: Book of exercises on sequences and/or series

Body: I'm looking for a dedicated problem/exercise book or pdf to work on limits of sequences and/or convergence of series (real analysis). Something like 1000 exercises with or without solution depending on the difficulty. A new year resolution of mine is studying one a day.</p>


问:
Are there any random variable $X$ and measurable function $g$ such that $\mathbb{E} [g(X)]=\mathbb{E}[Xg(X)]=0$ while $g(X)\ne 0$?
As titled. I was going through some problems related to the relation between uncorrelatedness and indepedence, and reached a situation where the existence of such RV and function would be nice</p>

答: Let $X$ be any symmetric random variable with finite third moment such that $X^{2}$ is not  a constant and $g(x)=x^{2}+a$ where $a =-EX^{2}$. Then $Eg(X)=0$ and $EXg(X)=EX^{3}+aEX=0+0=0$. Specific example: $X \sim N(0,1)$.


问:
Solving an AAA triangle using circles
I was asked to find the angles of an isosceles triangle by circumscribing a circle around it, then drawing another circle with the tip of the triangle as its center, and a radius of half of the leg.

I can see that the ratio of AD to DE is the same as AB to BC, but how does this help me know the angles of the triangle? The lengths are 10 for the legs and 6 for the base. I know I can use other methods to find the angles, but how do I use this method - meaning, how can I specifically use the circles to find the angles? I know the radius of the upper circle to be $\frac{1}{2} AB$, and that $DE=\frac12 BC$.

答:
Is this what they were asking for?

$\alpha = m \angle BAC = m \operatorname{arc} DE$

$\dfrac{\pi - \alpha}{2} = m \angle ABC = \dfrac 12(m \operatorname{arc} GE)$

\end{document}

Title: Solving an AAA triangle using circles


问:
How to write $\sin(x_1).\sin(x_2)$ as an integral?
We know that $\sin(x)$ is the same as writing $\int_{0}^3 \cos(x)dx$.

However, I am interested if there is a similar way to write $\sin(3). \sin(4) $ as an integral.

One way I think is $$\sin(3).\sin(4) \Longrightarrow  \int_{0}^3 \cos(x)dx.\int_{0}^4 \cos(y)dy $$
However, it is not so satisfying.
Please help me to know some other way to write $\sin(3).\sin(4)$ (or, in general, $\sin(x_1).\sin(x_2)$) as an integral.
 Thanks for your help.

Body:
We have:

$$\sin x_1 \sin x_2 = \frac12(\cos (x_1-x_2) - \cos (x_1+x_2))$$

Hence we have:

$$\int_{x_1-x_2}^{x_1+x_2}\frac{\sin x}2 dx = \frac12(\cos (x_1-x_2) - \cos (x_1+x_2))=\sin x_1 \sin x_2$$


Title: How to write $\sin(x_1).\sin(x_2)$ as an integral?

Body:
We know that $\sin(x)$ is the same as writing $\int_{0}^3 \cos(x)dx$.

However, I am interested if there is a similar way to write $\sin(3). \sin(4) $ as an integral.

One way I think is $$\sin(3).\sin(4) \Longrightarrow  \int_{0}^3 \cos(x)dx.\int_{0}^4 \cos(y)dy $$
However, it is not so satisfying.
Please help me to know some other way to write $\sin(3).\sin(4)$ (or, in general, $\sin(x_1).\sin(x_2)$) as an integral.
 Thanks for your help.

Body:
\begin{align}
\sin(x_1)\sin(x_2) &= \frac12 \left( \cos (x_1-x_2) - \cos(x_1+x_2\right)\\
&=\frac12 \int_{x_1+x_2}^{x_1-x_2} -\sin(t) \, dt \\
&=\frac12 \int_{x_1-x_2}^{x_1+x_2} \sin(t) \, dt \\
\end{align}


问:
Show that $\left(2n\right)!\leq \left(n\left(n+1\right)\right)^{n}$.
Show that $\left(2n\right)!\leq \left(n\left(n+1\right)\right)^{n}$.

My attempt so far:

$\left(2n \right)!=1\cdot2\cdot...\cdot\left(2n-1\right)\cdot2n\leq\left(n\left(n+1\right)\right)^{n}=\left(n^{2}+n\right)^{n}$

This is where I am stuck. I know that his holds for some values of $n$, such as 2, but I am having a hard time proceeding from here. I am not sure if induction would be the best approach or if there would be a more elegant solution. Thanks in advance for any hints.

答: You are done, but should probably put some words in to support the $\le$ sign.  Break the product on the left up into pairs of numbers, taking the highest and lowest off the run to make the pairs.  There are $n$ pairs and each one is less than $n(n+1)$


问:
Determine : $\lim _{n \rightarrow \infty} \left[\frac{1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !}{(n+1) !}\right]^{(n+1)!}$
What is the limit of the following :
$$
\lim _{n \rightarrow \infty} \left[\frac{1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !}{(n+1) !}\right]^{(n+1)!}
$$
I think it is clear that the numerator approaches infinity faster than the denominator so the result should be $+\infty$?

答:
[Note] : Notice that $n\cdot n! = (n+1)!-n!$, $\forall n\in\mathbb{N}$ go ahead and try to prove that using induction. It follows that :
\begin{align*}
\lim _{n \to \infty} \left[\displaystyle\frac{1 \cdot 1 !+2 \cdot 2 !+\ldots+n \cdot n !}{(n+1) !}\right]^{(n+1)!}&=\lim_{n\to\infty}\left[\frac{(n+1)!-1}{(n+1)!}\right]^{(n+1)!}\\
&=\lim_{n\to\infty}\left(1-\frac{1}{(n+1)!}\right)^{(n+1)!}\\
&=\boxed{\frac{1}{e}}
\end{align*}


问:
Show that a function $f(x,y) = \phi(x) + \psi(y)$ is monotonically decreasing if and only if $\phi(x)$ and $\psi(y)$ are decreasing.
I have come to a problem in a multivariate calculus book that I'm having trouble solving.

Let $I$ and $J$ be two non-empty intervals in $\mathbb{R}$. Given any
$\phi : I \rightarrow \mathbb{R}$ and $\psi : J \rightarrow \mathbb{R}$, define
$f,g : I \times J \rightarrow \mathbb{R}$ by $f(x,y) := \phi(x) + \psi(y)$ and
$g(x,y) := \phi(x)\psi(y)$.

The problem asks the reader to prove that $f$ is monotonically decreasing on $I \times J$
if and only if $\phi$ is decreasing on $I$ and $\psi$ is decreasing on $J$.

Here $f$ is defined as monotonically decreasing on $I \times J$ if for all
$(x_{1},y_{1}),(x_{2},y_{2}) \in I \times J$ we have :
\begin{equation}
(x_{1},y_{1}) \leq (x_{2},y_{2}) \Rightarrow f(x_{1},y_{1}) \geq f(x_{2},y_{2})
\end{equation}
Where here $\leq$ is a partial order on $\mathbb{R}^{2}$ s.t. :
\begin{equation}
(x_{1},y_{1}) \leq (x_{2},y_{2}) \Leftrightarrow x_{1} \leq x_{2} \text{ and } y_{1} \leq y_{2}
\end{equation}

Can someone help solve this ?

答:
Let first $\phi, \psi$ be decreasing and $(x_1,y_1)\leq (x_2,y_2)$. Then $x_1\leq x_2$ and $y_1\leq y_2$. Thus $\phi(x_1)\geq\phi(x_2)$ and $\psi(y_1)\geq\psi(y_2)$ implying $\phi(x_1)+\psi(y_1)\geq \phi(x_2)+\psi(y_2)$ or $f(x_{1},y_{1}) \geq f(x_{2},y_{2})$.

Now let $x_1\leq x_2$ and choose $y\in J$. Then $f(x_{1},y) \geq f(x_{2},y)$. That is $\phi(x_1)+\psi(y)\geq \phi(x_2)+\psi(y)$. So $\phi(x_1)+\geq \phi(x_2)$ as desired. For $\psi$ the same idea works.


Title: Determine : $\lim _{n \rightarrow \infty} \left[\frac{1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !}{(n+1) !}\right]^{(n+1)!}$

Body:
What is the limit of the following :
$$
\lim _{n \rightarrow \infty} \left[\frac{1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !}{(n+1) !}\right]^{(n+1)!}
$$
I think it is clear that the numerator approaches infinity faster than the denominator so the result should be $+\infty$?

Body:
You have
$$a_n=\Bigg[\frac {\sum_{i=1}^n i\,i!}{(n+1)!}\Bigg]^{(n+1)!}=\Bigg[\frac {(n+1)!-1}{(n+1)!}\Bigg]^{(n+1)!}=\Bigg[1-\frac {1}{(n+1)!}\Bigg]^{(n+1)!} $$
$$\log(a_n)=(n+1)! \log\Bigg[1-\frac {1}{(n+1)!}\Bigg]\sim -1 $$
$$a_n \sim \frac 1e$$


问:
Metric assuming the value infinity
If we instead define a metric as $d:X\times X \rightarrow [0,\infty]$, do we lose any nice properties of metric spaces?

The reason I ask is that I saw this theorem: Given a finite measure space $(X,\mathcal{M},\mu )$, define equivalence relation $A \sim B$ if $\mu(A \mathbin{\Delta} B)=0$ for $A,B\in \mathcal{M}$. Then $d: \mathcal{M} /{\sim} \rightarrow [0,\infty)  $ defined by $d(A,B)=\mu(A \mathbin{\Delta}  B)$ is a metric.

But the same argument would go through if $X$ wasn't a finite measure space, the only place where it seems to fail is because metrics are defined not to take the value $\infty$.

答:
Instead, think about this. Many measure spaces are $\sigma$-finite; this just means that they are countable unions of subsets of finite measure.

Suppose $(\Omega, \mathcal{S}, \mu)$ is a $\sigma$-finite measure space.  Then you can partition it with $\{E_n\}_{n=1}^\infty$ so $\mu(E_n)<\infty$ for all n.

Then for all $n$, $A\mapsto \mu(A\cap E_n)$ is a pseudometric on $\mathcal{S}$.
Now put

$$d(A, B) = \sum_{n=1}^\infty {\mu(A \mathbin{\Delta} B)\cap E_n)\over 2^n(1+ \mu(A\ \mathbin{\Delta} B)\cap E_n))}.\qquad A, B \in \mathcal{S}.$$

This is a pseudometric with many of the qualities you are seeking.  In fact, it can be seen to be a metric on $\mathcal{M}/\sim$.

Title: Metric assuming the value infinity


答:
For $1\leq i \leq n$ we have $i(2n-i+1) \leq n^{2}+n$. This is because  $n^{2}-(2i-1)n+i^{2}-i =(n-(i-\frac1  2))^{2}+i^{2}-i- (i-\frac1  2)^{2}\geq 0$. [ I have used the fact that $|n-(i-\frac  12)| \geq \frac1  2$ which gives $(n-(i-\frac  12))^{2} \geq \frac1  4$].

Now just mulptly the inequalities $i(2n-i+1) \leq n^{2}+n$ for $i=1,2,...,n$.

Body:
Show that $\left(2n\right)!\leq \left(n\left(n+1\right)\right)^{n}$.

My attempt so far:

$\left(2n \right)!=1\cdot2\cdot...\cdot\left(2n-1\right)\cdot2n\leq\left(n\left(n+1\right)\right)^{n}=\left(n^{2}+n\right)^{n}$

This is where I am stuck. I know that his holds for some values of $n$, such as 2, but I am having a hard time proceeding from here. I am not sure if induction would be the best approach or if there would be a more elegant solution. Thanks in advance for any hints.


问:
Convergence of sequence and serie
Let $\{a_n\}\subset (0, +\infty)$ a sequence such that $\sum_{n = 1}^{\infty} a_n < \infty$. Define:
$$\rho_1= \frac{1}{1 + a_1} \ \ and \ \ \rho_{n + 1} = \frac{\rho_n}{1 + a_n} $$
Show that $\rho_n = \prod_{k = 1}^{n} \frac{1}{1 + a_k}$ and $\rho_n$ is monotone and decreasing, bounded and $\rho = \lim \rho_n > 0$.
The first part is easy, my trouble is about the convergence of $\rho$, I'm trying with this:
$$ \rho_n = \frac{1}{1 + a_1}\times\frac{1}{1+a_2}\times \cdots \times \frac{1}{1 + a_n} $$
$$ln(\rho_n) = -[ln(1 + a_1) + ln(1 + a_2) + \cdots +ln(1 + a_n)] $$
Finally:
$$ln(\rho_n) < -ln(a_1 a_2\cdots a_n) $$

答:
The sequence ${\rho_{n}}$ is bounded below by 0 and above by 1. Also $\rho_{n}$ is monotone decreasing. therefore $\rho_{n}$ converges to some limit $\rho$ as a consequence of the monotone convergence theorem for sequences. You aren't asked to find $\rho$ so you are done.

So you also want to show $\rho$ is not zero.

As you have done take the natural log:

$ln(\prod \frac{1}{1+a_{n}}) = - \sum ln(1+a_{n}) $

For all finite sums.

Looking at a series expansion of ln(1+x) around zero you can obtain bounds that you can use to bound your series.

So expanding as a series about x=0 we have $$ln(1+x)= x - \frac{x^{2}}{2} + \frac{x^3}{3} + ...$$

So near zero $ln(1+x) < x$ and a tail of $a_{n}$ will be arbitrarily close to 0 eventually. So $\sum ln(1+a_{n}) < \sum a_{n}$ where here the sum is taken over a tail of the sequence.

So $ln(\prod \frac{1}{1+a_{n}}) > -M$ for some M since the sum of $a_{n}$ is less than infinity. So $\rho > e^{-M}$.

Title: Convergence of sequence and serie


答:
Ross Millikan has perhaps the slickest approach, but if one doesn’t see it, one can also prove the result by induction on $n$. We can check the cases $n=0,1,2$ by hand. Suppose that $n\ge 2$, and $(2n)!\le n^n(n+1)^n$.

$$\frac{(2(n+1))!}{(2n)!}=2(n+1)(2n+1)\,,$$

and

$$\frac{(n+1)^{n+1}(n+2)^{n+1}}{n^n(n+1)^n}=\frac{(n+1)(n+2)^{n+1}}{n^n}\,,$$

so we’re done if we can show that

$$2(n+1)(2n+1)\le\frac{(n+1)(n+2)^{n+1}}{n^n}\,,$$

i.e., that

$$\frac{4n+2}{n+2}\le\left(\frac{n+2}n\right)^n\,.$$

And this follows from the binomial theorem:

$$\begin{align*}
\left(\frac{n+2}n\right)^n&=\left(1+\frac2n\right)^n=\sum_{k=0}^n\binom{n}k\left(\frac2n\right)^k\\
&\ge 1+n\cdot\frac2n+\frac{n(n-1)}2\cdot\frac4{n^2}\\
&=3+\frac{2n-2}n\\
&=5-\frac2n\\
&\ge 4\\
&\ge 4-\frac6{n+2}\\
&=\frac{4n+2}{n+2}
\end{align*}$$

Title: Show that $\left(2n\right)!\leq \left(n\left(n+1\right)\right)^{n}$.


答:
$$\left(2n\right)!\leq \left(n\left(n+1\right)\right)^{n}$$
$$\log((2n)!) \leq n \log(n(n+1))$$ Using Stirling approximation for the lhs and Taylor for the rhs
$$\text{lhs}=n (2 \log (n)-2+2\log (2))+\frac{1}{2} \log (4 \pi  n)+\frac{1}{24
   n}+O\left(\frac{1}{n^3}\right)$$
$$\text{rhs}=2 n \log (n)+1-\frac{1}{2 n}+\frac{1}{3
   n^2}+O\left(\frac{1}{n^3}\right)$$
$$\text{rhs- lhs}= 2(1-\log (2))n+\left(1-\frac{1}{2} \log (4 \pi  n)\right)-\frac{13}{24
   n}+O\left(\frac{1}{n^2}\right)$$


答:
Consider the following function of $x$:

$$ f(x) = \frac{1}{x + \frac 1x}. \tag1$$

But since $\frac{1}{1/x} = x,$ we find that

$$ f\left(\frac1x\right) = \frac{1}{\frac 1x + x} = f(x). $$

In other words, we have a formula that is symmetric in $x$ and $\frac1x,$ so of course we can interchange $x$ and $\frac1x$ and get the same result.

If you graph $y = f(x)$ on log-log graph paper (or even semi-log graph paper using the logarithmic axis for $x$) you will get a kind of bell-shaped symmetric curve.
In general, any function $\phi$ whose graph on such paper is left-to-right symmetric will satisfy $\phi\left(\frac1x\right) = \phi(x).$

Your two formulas, $\frac{1/x}{1+1/x^2}$ and $\frac{x}{x^2+1},$
are just the right-hand side of Equation $(1)$ multiplied top and bottom by either $\frac1x$ or $x.$ That is, you have two alternative formulas for $f(x).$
And as we know that $f\left(\frac1x\right) = f(x),$ we know you can substitute $\frac1x$ for $x$ (or $x$ for $\frac1x$) in any formula of $f(x)$ and get back another formula of $f(x).$

For some more fun, consider

$$ \DeclareMathOperator{\sech}{sech}
   g(x) = \frac12 \sech x = \frac{1}{e^x + e^{-x}}. $$

This function is a kind of bell-shaped curve that is symmetric left-to-right
across the $y$ axis.
It also happens to be related to the function $f$ defined in Equation $(1)$ by the relationship

$$ f(x) = g(\ln x). $$

Also note that the property $\phi\left(\frac1x\right) = \phi(x)$
is a property of any formula that uses $x$ and $\frac1x$ in a completely symmetric fashion.
For example:

$$ \phi_1(x) = x + 1 + \frac1x, \qquad \phi_1\left(\frac1x\right) = \phi_1(x). $$
$$ \phi_2(x) = x^3 + x + \frac1x + \frac1{x^3}, \qquad
   \phi_2\left(\frac1x\right) = \phi_2(x). $$
$$ \phi_3(x) = \sqrt x + \sqrt{\frac1x},
   \qquad \phi_3\left(\frac1x\right) = \phi_3(x). $$

On the other hand, we can write a formula that uses $x$ and $\frac1x$
in an antisymmetric way; for example:

$$ \phi_4(x) = \frac{1}{x - \frac1x}, \qquad
    \phi_4\left(\frac1x\right) = -\phi_4(x). $$

Title: Why is $\frac{\frac{1}{x}}{1+\frac{1}{x^2}}=\frac{x}{1+x^2}$ true?

Body:
While trying to adapt Euler's series for $\tan^{-1}{x}$ to be used for $\cot^{-1}{x}$ so I could more efficiently compute $\pi$ with Machin-like formulae, I discovered the following identity:
$$\frac{\frac{1}{x}}{1+\frac{1}{x^2}}=\frac{x}{1+x^2}$$
I was surprised to discover that it's true because it seems counterintuitive. Why should the expression $\frac{x}{1+x^2}$ be the exact same value for both $x$ and its reciprocal, regardless of the value of $x$? It's not exactly the first example that comes to mind when you consider when you see an identity like
$$f\left(\frac{1}{x}\right)=f(x)$$
What's the principle behind it?


问:
Prove that $(ab = -b )\land (a\ne 0)\implies a+1=0$
I am having some troubles proving the implication that if $ab=-b$, and $b$ is not equal to $0$, then $a=-1$. I am only supposed to use axioms of integers to prove the proposition. I have written a proof, though I am not sure if it achieves its purpose. I am looking for some help to improve my proof where it may be weak.

Let $a,b$ be elements of integers. If $ab = -b$, then $a+1 = 0$

$ab = -b \implies ab+b = -b+b \implies ab+b = 0$

If $ab+b = 0$, then $ab+b = 0 = a+1$.

$a+1 = 0 \implies b(a+1) = 0 \implies ba+b = 0 \implies ab+b = 0$.

$ab+b = a+1 \implies b(a+1) = a+1 $

Here is where I get stuck. I need to prove $ab+b = a+1 \implies b(a+1) = a+1 \implies b=1$ Is there a way I can do so without division? So far I have this: $\implies (a+b)|(ba+b)=(a+b)|(a+b) \implies b=1$ Though I would like to avoid division if possible?

If $b=1$, then $ab=-b \implies a(1)=-(1) \implies a=-1$

As proven, $ab = -b$, then $a+1=0$.

If you have another approach to proving the implication, I'd really appreciate it if you could share it with me. I'm looking for some alternate methods to prove it.

答:
$$ ab = -b $$
$$ ab + b = 0$$

Using converse of distributive property,

$$ b (a+1) = 0 $$

Now for $b(a+1)$ to be zero either $b$ or $a+1$ has to be zero. Since it is given that b is not zero so $a+1$ will be zero

$$ a+1=0\implies a = -1 $$

Hence proved that $a+1$ is zero as well as $a = -1$


问:
Why does my book use $\subseteq$ instead of $\subset$ to describe a transitive set?
The following FOL statement is used in my book to describe a transitive set (which ultimately sets the stage for the definition of ordinals):

$z$ is a transitive set iff $\forall y \in z [ y \subseteq z]$

Given that $y\subseteq z \leftrightarrow y \subset z \ \lor y=z $, I am a little confused as to why $y \subseteq z$ is used.

In the context of ZFC set theory, it is my understanding that $z \in z$ is always false.

Therefore if $y \in z$, then clearly $y \neq z$. So why not just use $y \subset z$?

Am I missing something? (There is clearly no logical issue either way...it just seemed odd).

Edit: After realizing that, perhaps, there are different definitions for the various symbols $\subset$, $\subseteq$, and $\subseteqq$, I figured I would add that my book (The Foundations of Mathematics by Kenneth Kunen) asserts the following definition:

$x \subseteq y \iff \forall z ( z \in x \rightarrow z \in y)$

Also, it appears as though the only two subset symbols that are used in my book are $\subseteq$ and $\subsetneqq$ (the latter has no formal definition)...there is also $\nsubseteq$, but that is obviously not applicable here.

答:
In the presence of the axiom of foundation, it is true as you indicate that no set belongs to itself, and so the definition of transitive set can be written with $\subset$ (or $\subsetneq$, whichever symbol you prefer).

However, one may study also set theories where foundation fails, and then it is natural to define transitive sets in a way that allows self-membership. Indeed, any set $\Omega$ such that $\Omega=\{\Omega\}$, for instance, should be considered transitive.

One minor thing is that the natural translation of the natural language statement describing transitive sets gives us the rendering used by Kunen: A set $t$ is transitive if and only if any element of $t$ is a subset of $t$. This description is agnostic as to whether containment is proper or not. A somewhat more mathematical observation is that, even if Kunen is working here in a context where foundation is assumed, the definition he gives can be ported as is to a setting where foundation fails, there is no need to redefine things in this new context. This "portability" is particularly useful if one finds oneself in a situation where only fragments of ZFC hold, which is typically the case when studying independence proofs (or using countable structures in combinatorial arguments).


问:
Retrying Leray: new solutions to Navier-Stokes
This is an attempt to revisit Leray's self-similar solutions of the 3-dimensional incompressible Navier-Stokes equation (NSE) and to show, that they are an isolated case of the general set of self-similar solutions. I'm seeking constructive feedback about this proof. It would be really helpful to see a detailed answer addressing the results.

This question is part of a study which I published on arXiv. You can read it in full detail here: Polihronov, "Charles Bouton and the Navier Stokes Global Regularity Conjecture" arXiv:1902.01985 [math.GM].

Update: This article lists the definitions and properties of differential invariants in a concise form. For full details see C. L. Bouton, Am. J. Math. vol21, p25 (1899). The reader is encouraged to familiarize themselves with the following differential invariants:

absolute invariant

absolute covariant

relative invariant

relative covariant.

and how they behave under ${\it finite}$ (not infinitesimal) transformation.
Without this knowledge, it would be very difficult to grasp how Lie groups apply to the NSE.

This question is built mostly on the scaling properties of the NSE. Why is scaling so important? Because it determines the criticality of the NSE. Detailed explanation on this was written by T. Tao and can be found here.

Why are criticality and scaling important? Because the NSE was written originally for a continuum. This concept allows infinite rescalings and may lead to blow up in the solutions. More detailed explanation on this is again found in T. Tao's article referenced above.

The NSE is scale-invariant, and the scaling transform that preserves it has been discovered only recently by A. Ercan and M. L. Kavvas, Chaos vol25, p123126 (2015):

$$\vec{u}^\prime=k^{\alpha_x-\alpha_t}\vec{u}$$
$$(x,y,z)^\prime=k^{\alpha_x} (x,y,z)$$
$$t^\prime=k^{\alpha_t} t$$
$$p^\prime = k^{2(\alpha_x-\alpha_t)} p$$
$$\nu^\prime = k^{2\alpha_x-\alpha_t} \nu$$

where "$p$" stands for "$p/\rho$".

Regarding the viscosity $\nu$: it is a constant with a fixed value, or a constant parameter. In the dimensionalized NSE, $\nu$ has dimension, and is thus a scalable constant because its dimensions are scalable. Note that $\nu$ is a fixed constant at a fixed scale, however this does not mean that $\nu$ is non-scalable.

In the nondimensionalized NSE, all variables $x,y,z,t$ are dimensionless. Written in nondimensionalized form, the NSE retains its form and is still considered to have scaling invariance. It is for this reason that Leray's solutions (see below) were written for the nondimensionalized NSE as "self-similar" solutions, due to their scalability.  Is then the fixed constant $\nu$ (or, the Reynolds number) non-scalable within the nondimensionalized NSE? Is scalability a property of $x,y,z,t$ only? Why should we set the viscosity to be non-scalable? Hard to answer, there is no unified viscosity theory, since the molecular origins of viscosity are unclear.
Note that scalability is not to be confused with fixed constant values at a fixed scale. Scalability can be the property of a fixed constant to take fixed, but different values at different scales.

The viscosity is scalable in the dimensionalized NSE, it is an undeniable fact. We cannot hand pick $\nu$ as non-scalable in the nondimensionalized NSE because after  all, it is still the same equation. Any nondimensionalized solution can be dimensionalized and must be a solution of both dimensionalized and nondimensionalized NSE.

Scalability of the Reynolds number/viscosity in the nondimensionalized NSE is known (discussions on this are found even in this forum). For this reason, in this work the fixed constant $\nu$ is considered scalable within the nondimensionalized NSE (when initially posted, this was considered obvious, or well-known).

We see at once that the classic NSE scaling trasform.
$$\vec{u}^\prime = \frac{1}{k} \vec{u}$$
$$p^\prime = \frac{1}{k^2} p$$
$$(x,y,z)^\prime=k(x,y,z)$$
$$t^\prime=k^2 t$$
is an isolated case of the general trasform. when $\alpha_x=1$, $\alpha_t=2$.

Let's define isobaric weights as follows:
$W(x^a) =W(y^a) =W(z^a) =a \alpha_x$;

$W(t^b) =b\alpha_t$;

when variables are multiplied, their weights are added.

To write the self-similar solutions of the NSE, one can use Bouton's theorem (see C. L. Bouton, Am. J. Math. vol21, p25 (1899)).

About Bouton's work: While it is true that his article is more than 100 years old and may use older notation, his first theorem is very important. In essence, it states that the self-similar solutions of the NSE must be isobaric functions. This is the main reason why his name is the title of my article, because through his theorem I was able to construct the general form of all NSE's self-similar solutions. Bouton's work on differential invariants deserves much credit and recognition alongside its modern counterparts.

According to Bouton's theorem, scaling invariance of the self-similar solutions means that they are isobaric functions (polynomials or rational functions or representations thereof) of $x,y,z,t$. More precisely, $\vec{u}$ must be isobaric in $x,y,z,t$ of weight $W=\alpha_x-\alpha_t$, while $p$ must be isobaric in $x,y,z,t$ of weight $2(\alpha_x-\alpha_t)$.

In the literature, Leray is credited with proposing the form of NSE self-similar solutions as

$$\vec{u} = \frac{1}{\sqrt{t}} \mathbf{F} \left( \frac{x}{\sqrt{t}}, \frac{y}{\sqrt{t}}, \frac{z}{\sqrt{t}} \right)$$

(analogously for $p$)

but these are an isolated case $\alpha_x=1$, $\alpha_t=2$ of the general self-similar solution of the NSE. To find it, write Bouton's theorem above as

$$c\left( x\frac{\partial \vec{u}}{\partial x} + y\frac{\partial \vec{u}}{\partial y} + z\frac{\partial \vec{u}}{\partial z} \right) +at \frac{\partial \vec{u}}{\partial t} = b\vec{u}$$

and find (I used Maple)

$$\vec{u} = t^{\frac{b}{a}} \mathbf{F} \left( \frac{x}{t^\frac{c}{a}}, \frac{y}{t^\frac{c}{a}}, \frac{z}{t^\frac{c}{a}} \right)$$

then seek

$$\frac{b}{a} = \frac{\alpha_x-\alpha_t}{\alpha_t}$$
$$\frac{c}{a} = \frac{\alpha_x}{\alpha_t}$$

which gives Bouton's self-similar solutions

$$\vec{u} = t^{\frac{\alpha_x-\alpha_t}{\alpha_t}} \mathbf{F} \left( \frac{x}{t^{\frac{\alpha_x}{\alpha_t}}} , \frac{y}{t^{\frac{\alpha_x}{\alpha_t}}}, \frac{z}{t^{\frac{\alpha_x}{\alpha_t}}}\right)$$

These are identical with Leray's solutions as long as $\alpha_t=2\alpha_x$.

The expression for $p$ is analogous. It is a scalar function with isobaric weight of zero multiplied by $t^{2(\alpha_x-\alpha_t)/\alpha_t}$.

Could you kindly provide constructive feedback about this proof. It would be really helpful to see a detailed answer addressing the results, rather than argumentative comments.

So far, one opinion was given (I have not voted on it yet), however we would like to see constructive reasoning on the mathematics of viscosity scaling. The facts that viscosity traditionally was considered non-scalable; or that Charles Bouton's article was published more than 100 years ago do not contribute to the constructive discussion that is needed.

Thank-you

答:
It seems that scaling is an important topic for you, c.f. your prior questions 1
2
3
4.

In view of your persistence, I am confused as to why you ignored me asking in link 1 above if you could demonstrate the scaling from Kavvas and Ercan is indeed as you said. The computation would have revealed the below error.

In fact, I see no evidence in the paper that K&E  claim to have a new scaling for Navier--Stokes that was just missed by everyone else until 2015.

Instead it seems  that (among other things) they talk about relaxing the scalings of other things like the density and forcing term, which allows you to compare different physical scenarios.  This is not the fixed viscosity scenario considered by Leray / Tao etc. Obviously if the viscosity is fixed then from your expression for $\nu'$, one immediately gets  $\alpha_t=2\alpha_x$ which you seem to accept, makes it collapse to the same one used by other people.

Being a refutation of the premises, I see no need to address the rest of the post. Which is fortunate for me, because as in link 3 above, it seems we will not agree on the relevant definitions, nor do I want to check a paper for you that's over 100 years old...


问:
Volume of a Sphere with Cylinder in the Middle
Question: Let $b$ be a positive constant. Suppose a sphere with radius $2b$, has a cylindrical hole with radius $b$. Find the volume of the sphere using the shell method.

My Attempt:

I tried to think about the general formula for the volume of the sphere, and the cylinder. Knowing that it would have to be $$V = V_{\text{sphere}} - V_{\text{cylinder}}$$

The height of the cylinder would be $h = 2r_{\text{sphere}}$, which will be $h = 4b$. And so, I computed this formula,
\begin{align*}
V &= V_{\text{sphere}} - V_{\text{cylinder}} \\
&= \frac{4}{3}\pi r_{\text{sphere}}^3 - \pi r_{\text{cylinder}}^2h \\
&= \frac{4}{3}\pi (2b)^3 - \pi (b)^2(4b) \\
&= \frac{20\pi b^3}{3}
\end{align*}

If this is the volume of the sphere, I am not sure how I am supposed to show this using the shell method. Any assistance would help! A picture below depicts the problem.

答:
You have note mentioned which axis cylinder is along but let's say it is parallel to $z-$axis. Then in cylindrical coordinates, we have

Equation of sphere: $r^2+z^2 = 4b^2$

Equation of cylindrical hole: $r = b$

To find volume using cylindrical shell method, please take shells along the axis of the cylinder.

At any given radius, $-\sqrt{4b^2 - r^2} \leq z \leq \sqrt{4b^2 - r^2}$

So the height of the shell is $2 \sqrt{4b^2 - r^2}$

Also $b \leq r \leq 2b, \ $ as we are bound by radius of the cylinder and radius of the sphere.

So integration to find volume of the given sphere with cylindrical hole using shell method is,

$\displaystyle \int_b^{2b} 2 \pi r \cdot 2\sqrt{4b^2 - r^2} \ dr $

As far as your calculation without the integration,

at the intersection of cylinder and sphere, we have $z = \pm \sqrt3b$.

So the cylindrical hole is a cylinder of height $2\sqrt3b$ and two spherical caps on top and bottom,

Height of each spherical cap $ = (2-\sqrt3)b$ and radius of its base = $b$.

So, volume of cylinder along with spherical caps

$V_{hole} = 2b\sqrt3 \cdot \pi b^2 + 2 \cdot \frac{\pi b^3}{6} (2-\sqrt3) \big(3 + (2-\sqrt3)^2\big)$

$ = \frac{32 \pi b^3}{3} - 4 \sqrt3 \pi b^3$

So $V_{sphere} - V_{hole} = 4 \sqrt3 \pi b^3$


问:
find the formula for the following expression
$$S(n)=\sum _{d=3}^n6^{-d+n}d\,,\qquad n\ge 3$$

I have attempted to take out the $6^n$ out of the sigma as it will be a constant. I'm not sure how to form a sum formula to represent the left over in the sigma.

For example, when $n=5$, we can calculate the sum by $3 \cdot6^2 + 4\cdot 6 + 5 \cdot 6^0.$

答:
Consider
$$S(n)=6^n x\sum _{d=3}^n d\,x^{d-1}=6^n x\left(\sum _{d=3}^n x^{d}\right)'$$

When done, let $x=\frac 16$.


问:
Confusion regarding the domain of an inverse function?
In my textbook, the following example is given. I am confused as to why the domain for $f^-1(x) = x^2+5$ is "all real $x$ greater than $0$". Doesn't this function, take all real x-values? Or is it something to do with this inverse function ($x^2+5$) not passing the horizontal one-to-one test when plotted? Any explanation would be appreciated.

]1

答:
The  inverse of $f:A\to B$ is a function $f^{-1}:B\to A$, where $B$ is the range of $f$.

The range of $f(x)=\sqrt{x-5}$ is $[0,\infty)$ hence the domain of its inverse is $[0,\infty)$.

The domain of $f(x)$ is $[5,\infty)$ hence the range of its inverse is $[5,\infty)$.

It just so happens that we can extend $f^{-1}$ over $\Bbb R^-$. But that will mean it is no longer the inverse of $f$ since for no value of $x$ does $f(x)\in\Bbb R^-$.


问:
To prove differentiability of a function at a point
I am studying Denlinger's Elements of Real Analysis, there a problem is given as-

Suppose $f:\mathbb{R}\to\mathbb{R}$ is differentiable at $x_o$, and define $g:\mathbb{R}\to\mathbb{R}$ by $g(x)=x^3f(x^2)$. Prove that g is differentiable at $x_o$ and find $g'(x_o)$.

I have attached the pic of my work.

I have a doubt that we are just given that f is differentiable at $x_o$. Please tell is my way incorrect or I am missimg something.

Please clarify the doubt.

答:
You are right. The problem statement is wrong. For example, $$f(x)=\begin{cases}(x+1)^2&x\in\Bbb Q\\0&0>x\notin \Bbb Q\\\frac1{x^2}&0<x\notin\Bbb Q\end{cases}$$
is differentiable at $x_0=-1$, but $g(x)=x^3f(x^2)$ is nowhere differentiable (in fact, nowhere continuous).

The problem statement should ask for differentiability at $\sqrt{x_0}$ (hoping that $x_0\ge0$) or specify $x_0=0$ or $=1$.


问:
Ordered combinations
If I have 2 sets of elements, say $\{A, B, C, D\}$ and $\{P, Q, R, S\}$, how can I calculate combinations of combined set $$\{A, B, C, D, P, Q, R, S\}$$ such that order $A\to B\to C\to D$ and $P\to Q\to R\to S$ is maintained in each combination.

For example $$\{A, B, C, D, P, Q, R, S\},\\\{A, P, Q, R, B, S, C, D\}, \\\{P, Q, R, S, A, B, C, D\}\hphantom{,}$$ are valid combinations whereas
$$\{A, B, C, D, Q, P, R, S\}, \\ \{A, B, D, C, P, Q, R, S\}\hphantom{,}$$ are invalid.

答: Hint: Once you know which 4 positions $A, B, C, D$ are in, how many possible ordered combinations are there?

Title: Ordered combinations


答:
The Simplest way is as fallows

You can write down the inequality as:
$$ (2n)! \le (n^2+n)^n $$
Since $ n \ge 1 $:
$$ 2n \le n^2+n $$
$$ (2n)^n \le  (n^2+n)^n $$
Also:
$$ (2n)! \le (2n)^n $$
So:
$$ (2n)! \le (n^2+n)^n $$


Body:
Since
$n! < n(n/e)^n$,
$(2n)! < 2n(2n/e)^{2n}$,
so we want
$2n(2n/e)^{2n}
\lt (n^2+n)^n
$
or
$(2n)^{1/n}(2n/e)^2
\lt n^2+n
$.

This is true if
$(2n)^{1/n}(2n/e)^2
\lt n^2
$
or
$(2n)^{1/n}
\lt e^2/4
$.

Since
$(2x)^{1/x}$
is decreasing for
$x > e/2$
and
$(2n)^{1/n}
\lt e^2/4$
for $n = 3$,
(since
$(e^2/4)^3
=e^6/64
\gt 6.3$)
we are done.


问:
an question about compact space
Theorem:

Let $(X, \tau)$ be a  compact  space and $A$ a subset of $X$. Then $(X, \tau(A))$ is
a compact  space if and only if $\overline{A} \setminus A$ is a $\tau$-closed subset of $X$ and $ A$ is  a compact  subspace of $(X, \tau)$.

I can only prove one side of the theorem, but I do not know how to use closeness of $\overline{A} \setminus A$ and compactness of $A$  to show $(X, \tau(A))$ is a compact  space. Can you help me?

If $(X, \tau)$  is a topological space, $A \subset X$,
and $A\not\in \tau$, then the topology with subbase $\tau \cup \{A \}$ is called the simple expansion of $\tau$ by $A$ and is denoted $\tau(A)$.

$\tau(A) =\{ U \cup ( V \cap A): U , V \in \tau  \} $

答: The result is false. Let $X=[0,1]$, let $\tau$ be its usual topology, and let $A=\{0\}$. Then $\operatorname{cl}_\tau A\setminus A=\varnothing$, which is certainly $\tau$-closed, and $A$ is certainly $\tau$-compact. Let $U_0=\{0\}$, for $n\in\Bbb Z^+$ let $U_n=(2^{-n},1]$, and let $\mathscr{U}=\{U_n:n\in\Bbb N\}$; then $\mathscr{U}$ is a $\tau(A)$-open cover of $X$ with no finite subcover.

Title: an question about compact space


问:
How to prove the inequality $\Theta(x,y)\le \Theta(x,z)+\Theta(z,y)$?
Let $x, y$ be two complex vectors, $$\cos\Theta(x,y):=\operatorname{Re} \frac{y^*x}{\|x\|\|y\|} .$$ Then I want to prove that $$\Theta(x,y)\le \Theta(x,z)+\Theta(z,y) .$$  </p>

答: You should think about distances on the unit sphere.</p>


问:
How to solve these elementary algebra questions?

Lionel is thinking of two numbers. The sum of twice the larger number and 4 times the smaller number is 40. Twice the smaller number is 4 more than larger number. Find the larger number.

Mrs tan spent $140 on buying 4 toys and 6 shirts. If a toy cost twice as much as a shirt, find the cost of a toy.

A bakery sells pancakes which come in two flavors,peanut or strawberry. Each peanut pancake cost \$1.10 and each strawberry pancake costs \$1.30. The number of strawberry pancakes sold was twice that of peanut pancakes. The total sale is \$400. Find the minimum number of pancakes sold.

答:
1) Let the two number be $x$ and $Y$ with $x<y$. Then we have

$2y + 4x=40$; $2x=4+y$

So $y=8$ which is the larger number.

2) Let the toy cost be $x$ and shirt cost be $y$. Then we have

$4x+6y=140$; $x=2y$

So $x=20$, which is the toy cost.


问:
Definition of local ring (unique maximal left/right ideals)
Wikipedia lists a few equivalent definitions of local rings, the first two of which are

$R$ has a unique maximal left ideal.

$R$ has a unique maximal right ideal.

However, it does not list this condition:

$R$ has a unique maximal two-sided ideal.

I feel that both conditions in Wikipedia imply $R\backslash R^\times$ is an additive subgroup, thus implies 3. Then why is 3 not listed? Is it because 3 is not strong enough in general? What is a counterexample? Or is it simply "too easy to be listed?"

答: Yes, condition 3 is much weaker.  For instance, if $R=M_n(k)$ for a field $k$, then $0$ is the unique maximal two-sided ideal, but one-sided maximal ideals in $R$ are not unique if $n>1$.


问:
Probability of an event given conditional and unconditional probabilities
Let's assume the following:

a) the conditional probability of $B$ given $A$ is 0.8

b) the conditional probability of $B$ given $\text{not }A$ is 0.4

c) the unconditional probability of $B$ is 0.5

What is the probability of $A$?

EDIT

I'm really struggling with this.  Here is what I'm thinking and could really use some help.

$$P(B) - P(B|\text{not }A) \cdot P(\text{not }A) = P(B|A) \cdot P(A) \implies P(\text{not }A)=1-P(A)$$

Is this correct?

答:
The law of total probability states $P(A)=\displaystyle \sum_{j=1}^k P(B_j)P(A|B_j)$ with $B_1,...,B_k$ a partition of the sample space.

Here, $A$ and $A^c$ form a partition of the sample space, and also A and B are switched around from the statement of the law. We have $P(B)=P(B|A)P(A)+P(B|A^c)P(A^c)$

$.5=.8P(A)+.4(1-P(A))$

$P(A)=.25$


问:
If sum of LI vectors is Cauchy, do the vectors themselves have to be Cauchy?
Say I have a normed space $X$ and a Cauchy sequence $(x_n) \in X$. We can define $W$ to be an arbitrary subspace of X, with the Hamel Basis $\{e_1, e_2, e_3 ...\}.$

Given that $W$ is a subspace of $X$, we can define a Hamel Basis of $X$ as $$\{e_1, e_2, e_3 ... \}\cup\{e'_1,e'_2,e'_3...\}$$
From this, I think we can define unique sequences $(y_n)$ and $(w_n)$ such that $x_n = y_n + w_n$ for all $n \in \mathbb{N}$, and that $w_n \in W$ and $y_n \in \text{Span}(\{e'_1,e'_2,e'_3...\})$ for all $n$. Is it true then that for all normed spaces $X$, that $(y_n)$ and $(w_n)$ have to be Cauchy? $X$ and $W$ are not necessarily complete and not necessarily of finite dimension.

I think it's true, but in my attempts, I can't think of a way to get to where I need to go without using the assumption that $||y+w||\ge||y||,||w||$ for $y \in \text{Span}(\{e'_1,e'_2,e'_3...\})$ and $w \in W$. I don't think that inequality is true in general.

Either pointers or solutions would both be appreciated.

答: The answer is no. Let $X=(\Bbb R^2)^\infty$ with the supremum norm inherited from $\Bbb R^2$, and define $W=W_1\times W_2\times\cdots$ where each $W_n$ in the $n$th coordinate is spanned by $(n,1)$. Let $(x_n)\subset X$ be the sequence where all coordinates of $x_n$ equal $(0,0)$ except that the $n$th coordinate equals $(0,\frac1n)$. Then $(w_n)\subset W$ is the sequence where all coordinates of $w_n$ equal $(0,0)$ except that the $n$th coordinate equals $(1,\frac1n)$, while $(y_n)\subset X\setminus W$ is the sequence where all coordinates of $y_n$ equal $(0,0)$ except that the $n$th coordinate equals $(-1,0)$; neither of these sequences will be Cauchy (since the nonzero coordinate keeps changing indices).


问:
Gradient direction / descent
it was a while ago I read multivariable calculus so I need to refresh certain results.

Given $ f:R^n\to R $, at a local stationary point $ x $ the gradient is $ \nabla f(x) = 0 $. However, given the fact that the gradient points at the direction which $f$ increases the most, how come the gradient is zero at a local minima?

Also, about Gradient descent we use that fact to find a local minima, as per saying that if $ \nabla f$ points in the direction with maximum increase, $-\nabla f $ points in the direction of maximum decrease.

How is that equivalent?

答:
Main Answer

the gradient points at the direction which $f$ increases the most

…

$-\nabla f $ points in the direction of maximum decrease

Those statements are a good rule of thumb, but they are only guaranteed to hold under four conditions:

$\nabla f$ is not zero at the point.

You're looking "close enough" to the point.

$f$ is differentiable at the point.

You're sort of "averaging" $f$'s behavior near the point (or the derivative of $f$ is continuous at the point).

If $f$ has a nice formula, then typically the last two conditions hold. And you might have already had the intuition about the second condition. But the first condition is the crux of your question.

1. Nonzero Gradient

Single Variable Functions

In order to understand the basic idea of why we need $\nabla f\ne\boldsymbol 0$, it suffices to look at $f:\mathbb R\to\mathbb R$. The derivative at a point then tells you the slope of a line that the function looks approximately like as you zoom in arbitrarily far. (See, for instance How is the derivative truly, literally the "best linear approximation" near a point?.)

If $f'(a)>0$, then the line has positive slope, and the value of $f$ would generally increase if you move to the right (the direction of $\nabla f=\langle f'(a)\rangle$) from $a$ a small amount. Similarly, if $f'(a)<0$, then the line has negative slope, and the value of $f$ would generally increase if you move to the left from $a$ a small amount.

But if $f'(a)=0$ a lot of things could happen. For example, $f(x)=x^4-3x^2$ has a local maximum at $x=0$ that's not a global maximum. $f(x)=7$ has no direction of increase at all. $f(x)=x^3$ always has a direction of increase to the right, but still looks approximately like a horizontal line when you zoom in to the origin, so $f'(0)=0$.

Even if the function is given by a nice formula, the idea is that "looks approximately like a horizontal line" doesn't tell you which of these things the function acts like. To determine the direction of increase of a nice function with $f'(a)=0$, you need more information than the first derivative provides.

If the second derivative were a nonzero value, you'd know both directions increase or decrease, according to the second derivative test. If $f''(a)=0$ as well, it might be the case that one direction increases and the other does not. If we then have, say, $f'''(a)<0$, then Taylor's theorem tells us that the function is approximately $f'''(a)(x-a)^3+f(a)$, which we know increases to the left of $a$ since it basically looks like $-x^3$.

Higher Dimensions

In $2\text{D}$, the gradient isn't just a number, so there are many directions it can point. But (assuming differentiability), it still tells us about a good local approximation: $f(x,y)$ is approximately $\nabla f(\mathbf a)\bullet\left(\langle x,y\rangle-\mathbf a\right)+f(\mathbf a)$ near $\mathbf a$.

If $\nabla f(\mathbf a)\ne\boldsymbol 0$, then this dot product is maximized when $\langle x,y\rangle-\mathbf a$ goes in the same direction as $\nabla f(\mathbf a)$. So that explains why it would be the "direction of greatest increase".

But if $\nabla f(\mathbf a)=\boldsymbol 0$, then the best local linear approximation is the horizontal plane through $f(\mathbf a)$. And just like in $1\text{D}$, this doesn't really tell you anything about where nearby the function might increase/decrease.

The $n\text{D}$ case completely analogous.

2. "Close Enough" and Gradient Ascent/Descent

As mentioned above,  $f(x)=x^4-3x^2$ has a local maximum at the origin. So if you start in the interval $\left(-\sqrt{\dfrac32},\sqrt{\dfrac32}\right)$ and follow the gradient/sign of the derivative, you'll end up at the local maximum value $0$. But if you start outside of that interval, you'll reach values with $|x|>\sqrt3$ where the function is positive, and $f$ (and $|f'|$) will increase without bound. So "greatest increase" has to be interpreted in a local sense for this sort of reason.

3. Differentiability

A lot of the intuition here works out if $f$ is differentiable. But things can break down quite badly if that doesn't hold.

Directions of Ascent

For example, consider the function $f(x,y)=\sqrt[3]{xy}+x$. The gradient at the origin is $\langle1,0\rangle$. But the directional derivative doesn't even exist in any direction not aligned with the axes.

The Wrong Direction

We can still have problems even if we require that the function is continuous around the point and all directional derivatives exist. Consider:

$$f(x,y)=\cases{\dfrac{x^2y}{x^2+y^2} + \dfrac{y}2 & if $(x,y)\ne(0,0)$\\0 & if $(x,y)=(0,0)$}$$

The gradient at the origin is $\langle0,\dfrac12\rangle$. We might hope the direction of greatest increase is $\langle0,1\rangle$, with directional derivative $\dfrac12$. But the directional derivative is attained in, for instance, the direction $\left\langle\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right\rangle$. That directional derivative is $\left(\dfrac{\left(\dfrac{h}{\sqrt{2}}\right)^{3}}{2\left(\dfrac{h}{\sqrt{2}}\right)^{2}}+\dfrac{\dfrac{h}{\sqrt{2}}}{2}\right)/h=\dfrac{1}{\sqrt2}>\dfrac12$

4. "Averaging"

Even if the first three conditions hold, it needn't be the case that $f$ is actually increasing on a tiny line segment with one end at the point in question. $f$ is just well-approximated by something that is increasing.

Scattered Points

Consider the function $f(x)=\cases{x^2+x & if $x\in\mathbb Q$\\x & if $x\not\in\mathbb Q$}$ and note that $f'(0)=1$. For $x$ near zero, the points on the graph of $f$ jump up and down between the parabola $x^2+x$ and the line $x$. Because of this, the sense in which $f$ is "increasing to the right of $0$" is a bit strained.

Infinite Oscillation

For a function with a more connected graph, consider $f(x)=\cases{x^2\sin\left(\dfrac1x\right) - \dfrac{x}2 & if $x\ne0$\\0 & if $x=0$}$. We have $f'(0)=-\dfrac12$. However, at points of the form $\dfrac{1}{(2k+1)\pi}$, the derivative is $\dfrac12>0$. So even though the function "should" be increasing as we move leftwards from zero, it actually decreases near zero on infinitely many intervals.

Title: Gradient direction / descent


问:
Homomorphisms from $\mathbb{Z}[x] \to R$

Problem: Let $R$ be a ring with identity. Prove that the number of ring homomorphisms $\mathbb{Z}[x] \to R$ is equal to the number of elements of $R.$

I can't seem to find any approach for this. There are no generators of $\mathbb{Z}[x]$. Can we try something else? Any hint is highly appreciated.

Thank you.

答:
For each $r \in R$ consider the ring homomorphism $\operatorname{ev}_r : \mathbb Z[x] \to R$ given by $\operatorname{ev}_r(p(x)) = p(r)$. Then the map $\operatorname{ev}_{(-)} : R \to \operatorname{Hom}_{\textsf{Ring}}(\mathbb Z[x],R)$ is bijective.

Proof. It is injective since if $\operatorname{ev}_r = \operatorname{ev}_s$, then $r=\operatorname{ev}_r(x) = \operatorname{ev}_s(x) = s$. It if surjective since if $f : \mathbb Z[x] \to R$ is a ring homomorphism, then for any non-negative integer $n$ and any $a_0,\dots,a_n \in \mathbb Z$ we have
$$\begin{align}
f \bigg( \sum_{k=0}^n a_kx^k \bigg) &= \sum_{k=0}^n f(a_k)f(x)^k \\ &=^{\color{blue}1} \sum_{k=0}^n (a_k1_R)f(x)^k \\ &= \sum_{k=0}^n a_kf(x)^k \\ &= \operatorname{ev}_{f(x)} \bigg( \sum_{k=0}^n a_kx^k \bigg) \end{align}$$
and so $f = \operatorname{ev}_{f(x)}$.

${}^{\color{blue}1}$ Recall that there is only one ring homomorphism $\mathbb Z \to R$, namely $m \mapsto m1_R$.


问:
$X_1,..., X_n \sim$ Uniform$(0,\theta)$ and $\theta$ has prior distribution Pareto$(\alpha,\theta_0)$. Posterior expectations and variances?
We have IID $X_1,..., X_n \sim$ Uniform$(0,\theta)$ and we have a prior distribution $p(\theta)=\frac{\alpha \theta_0^\alpha}{\theta^{\alpha +1}}\mathbb{1}\{\theta_0\leq\theta\}$ where $\alpha>2$. Let $\textbf{X}=(X_1,...,X_n)$ and $\textbf{x}=(x_1,...,x_n)$. My questions are:

Am I correct in saying that the posterior distribution of $\theta$ is a Pareto$(n+\alpha, \theta_0) $ distribution? And hence is $\mathbb{E}(\theta|\textbf{x})=\frac{(n+\alpha)\theta_0}{n+\alpha-1}$ and $\text{Var}(\theta|\textbf{x})=\frac{(n+\alpha)\theta_0^2}{(n+\alpha-1)^2(n+\alpha-2)}$?

Let $X_1,...,X_n \sim \text{Uniform}(0,\theta^*)$ be IID. Let $Y=\max\{X_1,...,X_n\}$ and assume $\max\{Y,\theta\}\to_{\mathbb{P}(\cdot;\theta^*)}\max\{\theta,\theta^*\}$. I need to find the limit that $\mathbb{E}(\theta|\textbf{X})$ converges to in probablility, and the limit that $\text{Var}(\theta|\textbf{X})$ converges to in probability? I have no idea how to proceed with this one.

For $1$, my reasoning is $$p(\theta|\textbf{x})\propto f(\textbf{x};\theta)p(\theta)=\big(\frac{1}{\theta^n}\big)\frac{\alpha \theta_0^\alpha}{\theta^{\alpha +1}}\mathbb{1}\{\theta_0\leq\theta\}\mathbb{1}\{x_1,...,x_n\in[0,\theta]\}=\frac{\alpha \theta_0^\alpha}{\theta^{n+\alpha +1}}\mathbb{1}\{\theta_0\leq\theta\}\mathbb{1}\{x_1,...,x_n\in[0,\theta]\}$$
And since expectation of a Pareto distribution for $\alpha>1$ is $\frac{\alpha\theta_0}{\alpha-1}$ and its variance for $\alpha>2$ is $\frac{\alpha\theta_0^2}{(\alpha-1)^2(\alpha-2)}$ the result follows?

答:
Setting $Y=\max(X_1,\dots,X_n)$ as your text did, the likelihood is

$$L(\theta)=\frac{1}{\theta^n}\cdot \mathbb{1}_{[Y;\infty)}(\theta)$$

The prior is

$$p(\theta)\propto \frac{1}{\theta^{\alpha+1}}\cdot \mathbb{1}_{[\theta_0;\infty)}(\theta)$$

Thus the posterior is

$$p(\theta|\mathbf{x})\propto\frac{1}{\theta^{n+\alpha+1}} \cdot \mathbb{1}_{[\max(Y;\theta_0);\infty)}(\theta)$$

This means that

$$(\theta|\mathbf{x})\sim Pareto\Big[\max(Y;\theta_0);n+\alpha\Big]$$

With mean

$$\mathbb{E}[\theta|\mathbf{x}]=\frac{(\alpha+n)\max(Y;\theta_0)}{\alpha+n-1}$$

In the exercise there is a nice assumption about the convergence of $max(Y;\theta)$ so I think you can conclude by yourself...

Similar reasoning for the variance


问:
why is the "not" operator considered an operator
an assertion as i understand it is, a claim. so, something like "i like jello". It can be false or true. The "not" operator switches the boolean value of a statement. so if statement x was true it switches it to false and vise-versa. i think this is done by claiming statement x is false. so, if it was false to begin with the claim becomes "statement x is false" which is true. however, if it was true the claim becomes "statement x is false" which is false. so, my question is, isn't the not operator an assertion?

thanks in advance :)

答:
I think what is tripping you up may be that "not" is unary while "and" and "or", the other main logical operators, are binary. The "not" operator creates an assertion to the same extent that "and" can create an assertion from two other ones. If we have two statements $X$ and $Y$, the statement $X \wedge Y$ is asserting that both $X$ and $Y$ are true, while $\neg X$ is asserting that $X$ is false. At the end of the day, these are just the building blocks of statements that may or may not be true.

Edit. As an extra, I found it useful early on to think of these operators as functions. The "not" function takes a proposition and returns another proposition (its negation). The "or" function takes two propositions $X$ and $Y$ and returns the proposition "$X$ or $Y$".


问:
Fixing a Calculation Mistake on p.51 in "Mathematical Analysis I" by Vladimir A. Zorich?
While reading "Mathematical Analysis I" by Vladimir A. Zorich. I came across this proof regarding the irrationality of $\sqrt{2}$ :

Let $X$ and $Y$ be the sets of positive real numbers such that $\forall x \in X\left(x^{2}<2\right)$ $\forall y \in Y\left(2<y^{2}\right) .$ Since $1 \in X$ and $2 \in Y,$ it follows that $X$ and $Y$ are nonempty sets. Further, since $(x<y) \Leftrightarrow\left(x^{2}<y^{2}\right)$ for positive numbers $x$ and $y,$ every element of $X$ is less than every element of $Y .$ By the completeness axiom there exists $s \in \mathbb{R}$ such that $x \leq s \leq y$ for all $x \in X$ and all $y \in Y$ We shall show that $s^{2}=2$ If $s^{2}<2,$ then, for example, the number $s+\frac{2-s^{2}}{3 s},$ which is larger than $s,$ would have a square less than $2 .$ Indeed, we know that $1 \in X,$ so that $1^{2} \leq s^{2}<2,$ and $0<\Delta:=2-s^{2} \leq 1 .$ It follows that :

$$
\boxed{\left(s+\frac{\Delta}{3 s}\right)^{2}=s^{2}+2 \cdot \frac{\Delta}{3 s}+\left(\frac{\Delta}{3 s}\right)^{2}<s^{2}+3 \cdot \frac{\Delta}{3 s}<s^{2}+3 \cdot \frac{\Delta}{3 s}=s^{2}+\Delta=2}
$$

The author made a simple mistake during the expansion as seen in the box above. Therefore, I attempt to fix the proof as follow :

Since $1 \leq s^2$ and $\frac{\Delta}{3}<1$ hold, then :
\begin{align*}
\left(s+\frac{\Delta}{3s}\right)^2&=s^2+2\cdot\frac{\Delta}{3}+\left(\frac{\Delta}{3s}\right)^2\\
&\leq s^2+2\cdot\frac{\Delta}{3}+\left(\frac{\Delta}{3}\right)^2\\
&<s^2+2\cdot\frac{\Delta}{3}+\frac{\Delta}{3}\\
&= s^2+3\cdot\frac{\Delta}{3}\\
&=s^2+\Delta \\
&= 2
\end{align*}

Hence, $(s+\frac{\Delta}{3s})^2<2$ is true as required.

Are my steps correct in my attempt to fix this part of the proof?

答:
Alternate approach:

Worth looking at the concrete equation $x^2 - 2 = 0$ we are dealing with. We know that there are methods to solve this equation. Among them, there is the Newton  approximation (going on the tangent), and the secant method.
In this case, since we want values less than the solution, we use the secant method.

So consider the points on the graph $(s, s^2-2)$ below the horizontal,  axis, and $(2, 2^2 - 2)$ above the horizontal axis. Join them and consider the point of intersection with the horizontal axis $(s', 0)$.
From the picture, we should have $s'> s$, and $s'^2-2 < 0$.

Now, a simple calculation gives
$$s' = \frac{2(s+1)}{s+2}$$
Check that
\begin{eqnarray}
s' - s &=& \frac{2-s^2}{s+2}\\
2 - s'^2 &=& \frac{2(2-s^2)}{(s+2)^2}
\end{eqnarray}

$\bf{Added:}$ Pleased to notice  this matches the calculation  in Rudin's Principles (exp 1.1).  The fog is rising..


答:
Let $T_4$ be the Chebyshev polynomial of degree $4$. Then $p(x)=\tfrac18 T_4(x) = x^4-x^2 + \tfrac18$ is monic and oscillates between $-\tfrac18$ and $ \tfrac18$ on $[-1,1]$. In particular, there are numbers $$-1 = x_0 < x_1 < x_2 < x_3 < x_4 = 1$$ such that $$p(x_0)=p(x_2)=p(x_4)=\tfrac18$$ and $$ p(x_1)=p(x_3)=-\tfrac18.$$ Now suppose that $q(x)$ is a polynomial of degree $4$ such that $\lvert q(x) \rvert < \tfrac18$ on $[-1, 1]$.  Then $q\neq p$ and $p - q$ must have a root in each of the four intervals $(x_0, x_1), \ldots, (x_3, x_4)$ since it switches sign on each of these.  In particular $p-q$ must be of degree $4$ and therefore $q$ cannot be monic.

In other words for any monic $q$ of degree $4$ there is some $x \in [-1, 1]$ such that $\lvert q(x) \rvert \geq \tfrac18$.

In the same way (using $T_n$) it follows that this bound is $2^{1-n}$ for monic polynomials of degree $n \geq 1$.

Title: How to prove $|x^{4}+ ax^{3}+ bx^{2}+ cx+ d|\geq 1/8$ for some $x\in[-1, 1]$

Body:
I am trying to prove $|x^{4}+ ax^{3}+ bx^{2}+ cx+ d|\geq 1/8$ for some $x\in[-1, 1].$

I found one case where $\max|x^{4}+ ax^{3}+ bx^{2}+ cx+ d|= 1/8$ for some $x\in[-1, 1],$ which is
$$x^{4}- x^{2}+ \frac{1}{8}$$


问:
Is it possible for the pullback of an ample line bundle under projection to be big?
Given an ample line bundle on a curve is there any chance for its pullback to the projective bundle of some vector bundle to be big? It is known that first cohomology of inverse of nef and big line bundles can be killed by Frobenius if the dimension of variety $\geq 2$. I was wondering whether such a vanishing can hold for pullbacks of ample line bundles to projective bundles.

答: No. Just use the projection formula to compute the cohomology of powers of the pullback.</p>


问:
Why is it important that a basis be orthonormal?
Why is it important that a basis be orthonormal? The requirement for Orthonormal basis is so often repeated in linear algebra that it seems linear algebra depends on it as requirement. What is gained (or lost) if the basis are not orthonormal? Is it just for convention and convenience or is it just to restrict to the type of problems that assumption of orthonormality is beneficial? </p>

答:
Calculating scalar product -- hence length and angle -- of vectors given with coordinates, is much simpler w.r.t. an orthonormal basis.

If $b_1,\ldots,b_n$ is any basis, and a scalar product $\langle-,-\rangle$ is given in the vector space (else it wouldn't even make sense to speak about 'orthogonality'), then for the general formula for scalar products with given coordinates, we have to form the so called Gram-matrix of the basis:
$$\Gamma:=(\langle b_i,b_j\rangle)_{i,j}$$
And then
$$\left\langle\sum_i\alpha_ib_i,\,\sum_i\beta_ib_i\right\rangle=\sum_{i,j}\alpha_i\,\langle b_i,b_j\rangle\,\beta_j={\bf\alpha}^T\Gamma{\bf\beta}\,,$$
where ${\bf\alpha}^T=(\alpha_1,\ldots,\alpha_n)$.

If $(b_i)$ is an orthonormal basis, then $\Gamma$ is the identity matrix.

Why would one carry $\Gamma$ in all formulas which uses coordinates and scalar products, if we can always construct (by Gram-Schmidt process) an orthonormal basis starting out from arbitrary basis?


问:
Integral of $1/x^2$ without power rule
I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents

\begin{equation*}
  \int \frac{1}{x^2} \; dx
\end{equation*}

When using integration by parts, you end up with the same integral in the rhs so it seems out of luck

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}

This question is inspired by this blackpenredpen's video

Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}

you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.

The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that

\begin{equation*}
    \int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}

and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)

Title: Integral of $1/x^2$ without power rule

Body:
General way:

$$ x^n \cdot x^{-n} = 1$$

$$ nx^{n-1} x^{-n} + x^n \frac{d}{dx} x^{-n}=0$$

$$ \frac{n}{x^{n+1} } + \frac{d}{dx} x^{-n} = 0$$

Or,

$$ \frac{d}{dx}x^{-n} = \frac{-n}{x^{n+1} }$$

integrating both sides with $x$,

$$ x^{-n} +C = \int \frac{-n}{x^{n+1} } dx$$

Or,

$$ \frac{-n}{x^n} +C'  = \int \frac{1}{x^{n+1} } dx$$


问:
Solving definite integral from -1 to 1
Please help me with this. I can't prove the result. Tried integral by parts or notations, nothing working

$$\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx$$

Title: Solving definite integral from -1 to 1

Body:
Let $t=-x$, then
\begin{align*}
\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx
&=\int_{1}^{-1}{\frac{t^2}{e^{-t}+1}}(-dt)\\
&=\int_{-1}^{1}{\frac{t^2}{e^{-t}+1}}dt\\
&=\int_{-1}^{1}{\frac{x^2e^x}{e^x+1}}dx.
\end{align*}
\begin{align*}
\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx
&=\frac{1}{2}\left(\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx
+\int_{-1}^{1}{\frac{x^2e^x}{e^x+1}}dx\right)\\
&=\frac{1}{2}\int_{-1}^{1}x^2dx=\int_{0}^{1}x^2dx\\
&=\frac{1}{3}.
\end{align*}


问:
Strictly increasing sequence in $\mathcal P(\omega)/fin$?
Is it possible to find a sequence $(A_\alpha)_{\alpha < \omega_1}$ in $\mathcal P(\omega)$ with the following property:

(1) If $\alpha  < \beta$, then $|A_\alpha \setminus A_\beta|$ is finite but $|A_\beta \setminus A_\alpha|$ is infinite;

$\mathcal P(\omega)$ is the powerset of $\omega$. I'm really at a loss as to how to approach this, so any hints are appreciated. For what it's worth, an affirmative answer to this question was assumed in an answer to another question of mine at MO.

答:
I would recursively construct sets $B_\xi$ for $\xi<\omega_1$ so that $B_\eta\subset^*B_\xi$ whenever $\xi<\eta<\omega_1$ and then let $A_\xi=\omega\setminus B_\xi$ for each $\xi<\omega_1$. (Here $X\subset^*Y$ means that $|X\setminus Y|<\omega=|Y\setminus X|$.)

Suppose that $\eta<\omega_1$, and we have infinite subsets $B_\xi$ of $\omega$ for $\xi<\eta$ such that $B_\xi\subset^*B_\zeta$ whenever $\zeta<\xi<\eta$. Enumerate $\{B_\xi:\xi<\eta\}$ as $\{C_k:k\in\omega\}$, recursively choose distinct $m_k,n_k\in\omega$ so that $m_k,n_k\in\bigcap_{i\le k}C_i$ for each $k\in\omega$, let $B_\eta=\{n_k:k\in\omega\}$, and check that $B_\eta\subset^*B_\xi$ for each $\xi<\eta$.


问:
Computing degree of map $f: S^n\to S^n$

Compute the degree of map $f: S^n\to S^n$, where $f$ is defined by$$f(x_0,\ldots x_i,\ldots x_j,\ldots x_n) =(x_0,\ldots x_j,\ldots x_i\ldots x_n)$$
  which is an exchange of $x_i$ and $x_j$.

Can we find this general formula? Considering the case where $n=2$ or $3$, $f$ has degree $-1$ since it is a composition of rotation and reflection. (Am I right?) However is it generally true for $S^n$?

Thank you!

答: It's always a reflection ... across the hyperplane $x_i=x_j$. (This is conjugate to reflection across a "standard" hyperplane.)</p>

Title: Computing degree of map $f: S^n\to S^n$


问:
Prove that for a convex set C in the Euclidean space E, a convex function has bounded level sets if and only if it satisfies the growth condition.
Requiring the function $f$ to have bounded level sets is a "growth condition". Another example is the stronger condition
\begin{equation}
\liminf_{\left\|x\right\|\to\infty}\frac{f(x)}{\left\|x\right\|}>0,
\end{equation}
where we define
$$\liminf_{\left\|x\right\|\to\infty}\frac{f(x)}{\left\|x\right\|}=\lim_{r\to +\infty}\inf\left\{\frac{f(x)}{\left\|x\right\|}\ \middle|\ x\in C\cap rB^c\right\}.$$

The proof is said to be outlined as follows:

Find a function with bounded level sets which does does not satisfy the growth condition.

Prove that any function satisfying the growth condition has bounded level sets.

Suppose the convex function $f:C\to\mathbb{R}$ has bounded level sets but the growth condition fails. Deduce the existence of a sequence $(x^m)$ in $C$ with
$$f(x^m)\leq\frac{\left\|x^m\right\|}{m}\to +\infty.$$
For a fixed point $\bar{x}$ in $C$, derive a contradiction by considering the sequence
$$\bar{x}+\frac{m}{\left\|x^m\right\|}(x^m -\bar{x}).$$
Hence complete the proof of the proposition.

答:
If $C$ is bounded then the level sets are bounded and the growth condition vacuous, so I am presuming that $C$ is unbounded.

Let $L_\alpha = \{x \in C| f(x) \le \alpha \}$.

Suppose $\liminf_{\|x\|\to\infty}\frac{f(x)}{\|x\|} = r > 0$.
Then for some $M$ if $x \in C$ and $\|x\| \ge M$ we have ${f(x) \over \|x\| } \ge {1 \over 2}r$
or $f(x) \ge {1 \over 2} r \|x\|$.

In particular, if $\|x\| \ge M'=\max(M, {2 \over r} (\alpha+1))$, then $f(x) > \alpha$ and so $L_\alpha \subset B(0,M')$.

Now suppose the level sets are bounded. Pick some $x_0$ then there is some $M$ such that $L_{f(x_0)+1} \subset B(0,M)$.

Choose some $x$ with $\|x-x_0 \| \ge M$ and let $x' = x_0+ M {x - x_0 \over \| x - x_0 \|}$. Then
${f(x)-f(x_0) \over \| x-x_0\||} \ge {f(x')-f(x_0) \over \|x'-x_0\|} \ge {1 \over M}$ and so
\begin{eqnarray}
{f(x) \over \|x\|} & \ge & {f(x) -f(x_0)\over \|x\|} + {f(x_0) \over \|x\|} \\
&=& {f(x) -f(x_0)\over \|x-x_0\|} { \|x-x_0\| \over \|x\|} + {f(x_0) \over \|x\|}
\end{eqnarray}
and so
$\liminf_{\|x\| \to \infty} {f(x) \over \|x\|} \ge {1 \over M} > 0$.


问:
Introductory/Intuitive Functional Analysis Book
Can you recommend a gentle introduction to the abstract thinking and motivation of functional analysis? I'm looking for a book that holds you by the hand and shows the details of exercises, etc.

Thanks.

答: I would recommend Introduction to Functional Analysis by S.C. Bose. The text is really good and simple to understand; there is a considerable number of solved exercises and examples. I'm getting a desire to re-read!


问:
What is the probability that L does not intersect C1?
Let C1 and C2 be two concentric circles with radius r and 2r respectively. Given a fixed point Q on the circumference of C2, a line L is drawn at random going through Q. What is the probability that L does not intersect C1?

I tried using geometry to calculate probability whereby ratio of segment area to the overall area of C2 should be the answer. However I am not getting the right solution. It seems to be straightforward question unless I am missing something obvious.

答:
Note that we have to consider angles as probability and not length.

Angle subtended by APB is $\pi/6$. (from trigonometry, sin 30=0.5). So the probability will be $(\pi-\pi/6)/pi)$ (as it is line, we need to consider both ends, hence total possible angles of rotation  are $\pi$ and not $2\pi$)

so answer must be $5/6$=0.8334

Hope you will be able to carry on