问:
Fourier Transform Dirac Delta
I have recently learnt about tempered distributions, and how one can define the Fourier transform of a tempered distribution $v$ to be $\hat v$ so that
$$\langle\hat v,\varphi\rangle=\langle v,\hat \varphi\rangle.$$
for all Schwartz function $\varphi$. Here, let's say I take the convention that the Fourier transform is
$$\hat\varphi(\xi)=\frac{1}{\sqrt{2\pi}}\int_\mathbb R e^{-i x\xi}\varphi(x) \, dx.$$
 In particular, one can show that the Fourier transform of $1$ is the Dirac delta $\delta_0$. Here, $1$ is in the sense of tempered distribution, defined by
$$\langle 1,\varphi\rangle=\int\varphi(x)\,dx.$$
No problem there.

However, I have come across a confusing usage of this fact: I was reading a proof of some estimate, the author claimed that ($g$ is just some function of $x$)

$$\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\int_\mathbb{R} e^{-ixt}g(x)\,dt\,dx=\int_\mathbb{R}\delta(x)g(x)dx.$$
It looks like the author took the Fourier transform of $1$ "at the point $x$". But this is confusing to me because I don't understand what it means to take Fourier transform "at a point". I also don't understand what $\delta(x)$ means, and what it means to "integrate" it.

If the context is helpful, I am reading this set of seminar notes.
https://www.math.ucla.edu/~visan/Oberwolfach2012.pdf
Namely, page 11 Lemma 2.11 (Local Smoothing). I simplified the notation above the make the question less complicated.

答:
Why so complicated?  The inner integral is  FT(g)(-t) and so we have to
look at the FT(FT(g))(x) for x = 0, but  FT(FT(g))(x) = g(-x) for test functions.</p>

Title: Fourier Transform Dirac Delta


问:
Probability that $X < Y$ for bivariate normal variable
For the following probability questions, I'm a bit lost on how to proceed.

Let $X,Y$ be bivariate normal random variables with $\mathbb{E}(X)=10$, $\mathbb{E}(Y)=-7$, $Var(X)=30$, $Var(Y)=100$, and $\rho(X,Y)=-0.2$. Determine:

a) $P(X<\mathbb{E}(Y)-Y)$,

b) $P(X+10Y<0)$.

For a), I have been thinking about integrating the joint distribution to get the unconditional distribution of $X$ but since both R.V.'s are in the inequality this doesn't seem to be the way to go. Any ideas?

答: Hint: $X+Y$ and $X+10 Y$ are normal random variables.  What are their means and variances?

ContentLicense: CC BY-SA 4.0


问:
Integral $\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$ vanishes
Come across

$$I=\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$$

and break the integrand as

$$\frac{1-x(2-\sqrt x)}{1-x^3}=\frac{1-x}{1-x^3}- \frac{x(1-\sqrt x)}{1-x^3}
$$
The first term simplifies and the second term transforms with $\sqrt x\to x$. Then, the integral becomes

$$I=\int_0^\infty \frac{1}{1+x+x^2}dx
- 2\int_0^\infty \frac{x^3}{1+x+x^2+x^3+x^4+x^5}dx
$$
The first integration is straightforward $\frac{2\pi}{3\sqrt3}$ and the second is carried out via partial fractional decomposition. After some tedious and lengthy evaluation, it surprisingly yields the same value $\frac{2\pi}{3\sqrt3}$.

Given its vanishing value, there may exist a shorter and more direct derivation, without evaluation of any explicit intermediate values.

答:
Note
\begin{align}
&\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx=\int_0^\infty \frac{1+x^{\frac32}-2x}{1-x^3}dx\\= &\int_0^\infty \frac{1+x^{\frac32}}{1-x^3}dx
-\int_0^\infty \underset{x\to x^{\frac12} }{\frac{2x}{1-x^3}}dx= \int_0^\infty \frac{dx}{1-x^{\frac32}}
-\int_0^\infty \frac{dx}{1-x^{\frac32}}=0
\end{align}


问:
Limit with variable in integral boundary
I have the following integral

$$\int_{\vert x\vert\leq\varepsilon}\frac{1}{\varepsilon}\frac{1}{\vert x\vert^2}\,\mathrm{d}x,$$

where the integral should be understood as a Lebesgue integral and where $x\in\mathbb{R}^{n}$. My question is now, can we say what happens if $\varepsilon\to 0$? I don't know how to approach this question in a mathematical way. The problem here is that on the one hand our integral region becomes smaller and smaller, which means that in the end we will end up with a Lebesgue-zero set, but in the same time our integrand becomes larger and larger.  Any ideas?

答:
You can definitely ask about what happens when $\epsilon \to 0$ because after all for each (fixed) $\epsilon>0$ your integral is a number -- one real number. So, there is an association: $\epsilon \to I(\epsilon)$, where $I(\epsilon)$ stands for the value of the integral. Of course, one needs to show that the integral is finite.

So, your question is
$$
\lim_{\epsilon \to 0^+} I(\epsilon)= ?
$$
As you pointed out the answer is not trivial, becasue although your set shrinks down to a null set, the interand grows to infinity. So, how do they balance each other out?

The answer can be found by explicitly computing the integral, probably using spherical coordinates. Notice that $1/\epsilon$ is a constant and can be pulled out of the integral sign.


问:
For which $n$ does $p_n(x)=\sum\limits_{{k=1,(k,n)=1}}^n o(k) x^k $ have exactly two real roots?
Let $n\in\mathbb{N}$ be fixed and denote by $o(k)$ the multiplicative order of $k$ modulo $n$. Define $$p_n(x)=\sum_{\substack{k=1 \\ (k,n)=1}}^n o(k) x^k ;$$here the sum is taken over $k$ that are coprime to $n$, so $o(k)$ is well-defined. Here are two simple observations:

If $n$ is even, $p_n(x)$ is odd. Further, in this case it only has one real root.

If $n$ is odd, aside from $n=1$ it seems that $n$ has an even number of real roots. Here is a picture of the number of roots for $1\le n \le 1000$ and $n$ odd.

I checked OEIS and it didn't return any matches. Perhaps this problem is very difficult, as I don't know how easy it is to compute $o(k)$ for arbitrary $n$. If we denote $R(n)$ as the number of real roots of $p_n(x)$ for odd $n$, here are some questions I think are worth pursuing about $R(n$):

Is there an estimate or asymptotic for $R(n)$? Is there an upper bound for $R(n)$?

When does $R(n)=2$? For $1\le n\le 1000$, this occured when $n=\{3, 5, 7, 11, 31, 89, 127, 257, 331, 635\}$; OEIS didn't recognize this sequence either.

As a follow-up, does $R(n)=2$ infinitely often?

答:
$\deg(p_n) = n-1$.

$p_n$ has $2d_n$ non-real roots thus $n-1-2d_n$ real roots counted with multiplicity, maybe you just didn't find the rare cases where it has some double real roots.

For $n$ even then $p_n$ is odd and it is $> 0$ on $(0,\infty)$ thus its only one real root is at $0$.

The same holds when replacing $order(k\bmod n)$ by any non-negative function such that $f(n-1)>0$.

Not convinced that we can say anything about $R(n)$, do more simulations and see if some non-random estimate pops up.


问:
What could be better than base 10?
Most people use base 10; it's obviously the common notation in the modern world.

However, if we could change what became the common notation, would there be a better choice?

I'm aware that it very well may be that there is no intrinsically superior base, but for the purposes of humans, is there a better one?

I've heard from sources such as this and this that base 12 is better, from here that base 8 is better, and, being into computer science, I would say that base 16 is the most handy.

Base 12 does seem to be the most supported non-base 10 number system, mainly due to the following reason pointed out by George Dvorsky:

First and foremost, 12 is a highly composite number — the smallest
  number with exactly four divisors: 2, 3, 4, and 6 (six if you count 1
  and 12). As noted, 10 has only two. Consequently, 12 is much more
  practical when using fractions — it's easier to divide units of
  weights and measures into 12 parts, namely halves, thirds, and
  quarters.

And, on top of that, previous societies considered very advanced used other systems, such as the Mayans using base 20, and the Babylonians using base 60.

So, summarized, my question is: Is there an intrinsically superior base? If not, is there one that would be best for society's purposes? Or does the best base depend on the context it is being used in?

答:
Not all bases are 10 in their own notation.  There are a group of alternating bases where the base is not 'ten' but a 'hundred'.  The most elegant of these is the long-hundred of the proto-germanics and their decendents.  Reckoning in the six-score long-hundred (ie 120), was still common enough in 1350 to pass without comment.

Yes, i have used this base for some thirty years.  It's truly elegant, being more efficient than either 10 or 12.  It's the first base, for which the (number of proper divisors)/(ln base) is greater than 3.

Also 120 is the smallest multiply perfect number, and has the same features as the perfect numbers.  For example, 120 = 1+2+4+8+15+30+60 = 3+5+6+10+12+20+24+40, all of these numbers make the total divisors of 120.  The second set corresponds to a set of weights, eg

1 oz, 2 oz, 4 oz, 8 oz, 1 lb, 2 lb, 4 lb.  15 oz = 1 lb , 120 oz = 1 clove.

1 ct, 2 ct, 4 ct, 8 ct, 1 dr, 2 dr, 4 dr : 15 ct = 1dr ,  120 ct = 1 oz

1 lb, 2 lb, 4 lb, 8 lb, 1 st, 2 st, 4 st  : 15 lb = 1 st,  120 lb = 1 cwt

When one considers not just integers, but also fractions x/y and y/z, arranged by xy, one finds that the expressions of these are very short for the first sixty or so, even in things like 56 (8/7 = 1:17.17.17, vs 7/8 = V5), and this unusual pair at 96 (3/32 = 11:30, 32/3 = 10:80.

ContentLicense: CC BY-SA 3.0

Title: What could be better than base 10?


问:
Shorthand for quantifying two variables?
Say I have two real variables $x$ and $y$. Can I universally quantify them with $\forall x,y\in\mathbb{R}$ instead of using $\forall x\in\mathbb{R},\forall y\in\mathbb{R}$? Is this correct notation, or is it deemed sloppy by mathematicians?

答:
This is entirely standard shorthand notation. It could be argued that it is slightly unclear whether $x\in\Bbb R$, but mostly this is apparent from context.

If you want to remedy this minor issue without being too verbose, you could, for instance, say $$\forall (x,y)\in\Bbb R^2$$and there is absolutely no more room for any reasonable ambiguity.


问:
Is it possible to know if the feasible region is unbounded without drawing it?
Given the following set:

$$ -x_1 + x_2 = 4$$

$$ x_1 - 2x_2 + x_3 <= 6 $$
$$ x_3 >= 1 $$
$$ x_1,x_2,x_3 >= 0 $$

Without drawing the feasible region can I know if it is bounded or unbounded?

答: A polyhedron is unbounded if there exists a point $x$ and a direction $d$ such that $x + \alpha d$ is in the polyhedron for all $\alpha \geq 0$. Denote the polyhedron as $Ax \leq b$, then $x$ and $d$ need to satisfy $A(x+\alpha d) \leq b$ for all $\alpha \geq 0$. That is equivalent to finding $x$ and $d$ such that $Ax \leq b$ and $Ad \leq 0$.


答:
Here is the joint distribution.

The regions are colored according to your problem b).  Just integrate.

Title: Probability that $X < Y$ for bivariate normal variable

Body:
For the following probability questions, I'm a bit lost on how to proceed.

Let $X,Y$ be bivariate normal random variables with $\mathbb{E}(X)=10$, $\mathbb{E}(Y)=-7$, $Var(X)=30$, $Var(Y)=100$, and $\rho(X,Y)=-0.2$. Determine:

a) $P(X<\mathbb{E}(Y)-Y)$,

b) $P(X+10Y<0)$.

For a), I have been thinking about integrating the joint distribution to get the unconditional distribution of $X$ but since both R.V.'s are in the inequality this doesn't seem to be the way to go. Any ideas?


问:
Particular solution of $t^2y''-2y=t^2$ using the method of variation of parameters
Using the Anzats $y(t) = t^r$, we find $r(r-1)-2=0$ and thus the fundamental set of solutions $y_1(t)=t^2$ and $y_2(t)=t^{-1}$.

We have $W[y_1,y_2](t) = y_1(t)y'_2(t)-y_1'(t)y_2(t) = -t^2 t^{-2} - 2t t^{-1} = -3$.

The method of variation of parameters now gives a particular solution of the form $y_p(t) = u_1(t)y_1(t)+u_2(t)y_2(t)$ with $$u_1'(t) = -\frac{t^2 y_2(t)}{W[y_1,y_2](t)} = \frac{1}{3} t$$ and $$u_1'(t) = \frac{t^2 y_1(t)}{W[y_1,y_2](t)} = -\frac{1}{3}t^4.$$

Hence, we have $u_1(t) = \frac{1}{6}t^2$ and $u_2(t) = -\frac{1}{15}t^5$. This gives $$y_p=u_1(t)y_1(t)+u_2(t)y_2(t) = \bigg( \frac{1}{6}-\frac{1}{15} \bigg)t^4  = \frac{1}{10}t^4\ .$$

Inserting this into the ODE $L[y] = t^2y''-2y$, results in $L[y_p]=t^4$, not $t^2$.

Wolfram Alpha gives the particular solution $y_p(t) = \frac{1}{3}t^2 \log (t)$

Question: why does my application of the method of variation of parameters give a wrong particular solution? Where have I made a mistake?

Context: this is problem 2.4.9 of Braun's Differential Equations and Their Applications.

答:
Put the differential equation in the classical form before to apply the method of variation of parameters:
$$y''+b(t)y'+c(t)y=f(t)$$
$$y''-\dfrac 2{t^2}y=\color{red}{1}$$

So now you have that:
$$u_1'(t) = -\frac{ y_2(t)}{W[y_1,y_2](t)} = \frac{1}{3t} $$
And
$$u_1'(t) = \frac{ y_1(t)}{W[y_1,y_2](t)} = -\frac{1}{3}t^2$$


问:
Algebraic equation concerning the Jacobi sn function
What are the solutions $u$ to the equation $\sqrt{1-m} \operatorname{sn} (u \mid \frac{1-m}{1+m}) = 1$ for a given $m$? Because $\mbox{sn}$ is an elliptic function of order 2, we know it attains any value two times in the fundamental parallellogram, so that there are two solutions. Let's first assume $0 < m < 1$, in particular that $m$ is real, and denote $M = \frac{1-m}{1+m}$. Numerics suggest that indeed there are 2 solutions in the fundamental parallellogram which are related by symmetry, of the form $u^{\pm} = K(M) + \left[ 1 \pm \gamma(m) \right] i K'(M)$ for some $\gamma \in (0,1)$. A plot of the numerically obtained $\gamma$ may be found here (note: near $m=1$ the resolution becomes bad, but I believe $\gamma \rightarrow 0$ as $m \rightarrow 1$. also, it seems that $\gamma \rightarrow 1/2$ as $m \rightarrow 0$). We may identify the potential solutions because $\mbox{sn}$ is symmetric about the point $K(M) + i K'(M)$. Does an algebraic expression for $u$ exist?

Side notes: further numerical investigation of $\gamma$ leads me to conjecture that it has the form $\gamma(m) = \frac{1}{2} - \frac{1}{2 \pi} \left( m + \frac{11}{48}m^3 + \frac{599}{5120}m^5 + \mathcal{O}(m^7) \right)$ as $m \rightarrow 0$. If this is true and such an analytic expansion of $\gamma$ exists it also gives us the answer $\forall m \in \mathbb{C} \setminus \{0,1\}$ by analytic continuation. Because $\sqrt{1-m} \rightarrow 0$ when $m \rightarrow 1$, I believe $\gamma \rightarrow 0$ in this limit because $\operatorname{sn} (u \mid M)$ must become singular to satisfy the equation. So this places the constraint $1 + \frac{11}{48} + \frac{599}{5120} + \dots = \pi$ on the coefficients of the (hypothetical) expansion of the (hypothetical) $\gamma$. The burden of proof remains...

答:
By definition, $\newcommand{sn}{\operatorname{sn}}\sn$ and the first-kind incomplete elliptic integral $F$ are inverses of each other.
$$\sqrt{1-m}\sn\left(u,\frac{1-m}{1+m}\right)=1$$
$$\sn\left(u,\frac{1-m}{1+m}\right)=\frac1{\sqrt{1-m}}$$
$$u=F\left(\sin^{-1}\frac1{\sqrt{1-m}},\frac{1-m}{1+m}\right)$$
By the transformation given here this may be separated into real and imaginary parts when $m\in[0,1]$:
$$u=K\left(\frac{1-m}{1+m}\right)+iF\left(\sin^{-1}\sqrt{\frac{1+m}2},\frac{2m}{1+m}\right)$$


问:
Calculation of permutation in tensor of rank 4
I am trying to compute the symmetric part of a 4th order tensor $A_{ijkl}$

From a previous post (Symmetric Part of Product of 2 tank 2 tensors), I saw that I need to compute the permutations of $A_{ijkl}$ (4!=24 permutations).

I would like to know if anyone can give some hints on how to compute these permutations.

If $$A_{ij}=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 & 0 &1 \end{bmatrix}\quad\text{and}\quad A_{kl}=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 & 0 &1 \end{bmatrix}$$

Then

$$A_{ijkl}=\begin{bmatrix}
1 &0 &0 \\
0 &1 &0 \\
0 & 0 &1
\end{bmatrix}
 \begin{bmatrix}
1 &0 &0 \\
0 &1 &0 \\
0 & 0 &1 \end{bmatrix} \\
= \begin{bmatrix}
1.& 0.& 0.& 0.& 1.& 0.& 0.& 0.& 1. \\
 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
1.& 0.& 0.& 0.& 1.& 0.& 0.& 0.& 1.\\
0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
 1.& 0.& 0.& 0.& 1.& 0.& 0.& 0.& 1.\\
\end{bmatrix}
$$

How can I compute the remainder, e.g., $A_{ikjl},A_{jikl}, A_{lkij}, \ldots$

Kind Regards!

答:
Solved in python with:</p>
```<code>import numpy as np
from itertools import permutations
# create a 2nd order tensor
a=np.eye(3)
a[0][0]=2
a[0][-1]=3
a[-1][-1]=4
# create a 2nd order tensor
b=np.eye(3)

# Function that will permutate the indices
def permutation(str1,tmp):
    store=np.zeros([3,3,3,3]) # variable to store the results

    #function to split the string with the indices
    def split(word):
       return [char for char in word]

    a,b,c,d=split(str1) # gets indices from string

    for i in [0,1,2]:
        for j in [0,1,2]:
            for k in [0,1,2]:
                for l in [0,1,2]:
                    store[i,j,k,l]=tmp[vars()[a],vars()[b],vars()[c],vars()[d]]
    return store

result=np.einsum('ij,kl->ijkl',a,b) # gets the result of A_ij*B_kl = P_ijkl

permutation_set=[''.join(p) for p in permutations('ijkl')] # generate a list of permutations with the indices ijkl

for p in permutation_set:

    permutation_wanted=np.einsum(p,result) # permutates the result for the with the intended permutation with the funciton einsum

    check=permutation(p,result) # gets the permutation with the permutation function defined above
    print("Current permutation is: %s" %(p))
    if(np.count_nonzero(check-permutation_wanted)==0):
        # if the difference between check and permutation_wanted is 0 (all the elements are equal)
        print("Script is working")
    else:
        print("Script is not working!")
</code>```

Body:
Solved in python with:

import numpy as np
from itertools import permutations
# create a 2nd order tensor
a=np.eye(3)
a[0][0]=2
a[0][-1]=3
a[-1][-1]=4
# create a 2nd order tensor
b=np.eye(3)

# Function that will permutate the indices
def permutation(str1,tmp):
    store=np.zeros([3,3,3,3]) # variable to store the results

    #function to split the string with the indices
    def split(word):
       return [char for char in word]

    a,b,c,d=split(str1) # gets indices from string

    for i in [0,1,2]:
        for j in [0,1,2]:
            for k in [0,1,2]:
                for l in [0,1,2]:
                    store[i,j,k,l]=tmp[vars()[a],vars()[b],vars()[c],vars()[d]]
    return store

result=np.einsum('ij,kl->ijkl',a,b) # gets the result of A_ij*B_kl = P_ijkl

permutation_set=[''.join(p) for p in permutations('ijkl')] # generate a list of permutations with the indices ijkl

for p in permutation_set:

    permutation_wanted=np.einsum(p,result) # permutates the result for the with the intended permutation with the funciton einsum

    check=permutation(p,result) # gets the permutation with the permutation function defined above
    print("Current permutation is: %s" %(p))
    if(np.count_nonzero(check-permutation_wanted)==0):
        # if the difference between check and permutation_wanted is 0 (all the elements are equal)
        print("Script is working")
    else:
        print("Script is not working!")


问:
Proof of $\operatorname{arccot}(x)= \arctan(\frac{1}{x})$
$\newcommand{\ac}{\operatorname{arccot}}$
Here is my proof:

$ \theta = \ac(x)$

$ \cot \theta = x $

$ \displaystyle \tan \theta = \frac{1}{\cot \theta} = \frac{1}{x}$

$ \displaystyle \theta = \arctan \frac{1}{x}$

But based on the graph, $\ac(x) = \arctan(\frac{1}{x})$ is only satisfied when $ x > 0$. What is wrong with my proof? What is the error occurred inside a proof?

答:
There are two conventions for principal value:

$\operatorname{arccot} x \in (-\pi/2, \pi/2]$ with a discontinuity at zero.

$\operatorname{arccot} x \in (0, \pi)$

See https://mathworld.wolfram.com/InverseCotangent.html

It turns out that $\operatorname{arccot} x = \arctan(1/x)$ for all $x\ne 0$ for one convention of principal value.

If you want to adapt or enforce a particular range for $\operatorname{arccot} x$, then $\operatorname{arccot} x = \arctan(1/x) + k \pi$ for some integer $k$.

Here is what Mathematica thinks:

```<code> Plot[{ArcTan[1/x] + .01, ArcCot[x]}, {x, -7 Pi, 0}, Frame -> True,
  PlotRange -> All]
</code>```

You can rotate this second plot $90^\circ$ and define your range anyway you want up to a multiple of $\pi.$

Body:
There are two conventions for principal value:

$\operatorname{arccot} x \in (-\pi/2, \pi/2]$ with a discontinuity at zero.

$\operatorname{arccot} x \in (0, \pi)$

See https://mathworld.wolfram.com/InverseCotangent.html

It turns out that $\operatorname{arccot} x = \arctan(1/x)$ for all $x\ne 0$ for one convention of principal value.

If you want to adapt or enforce a particular range for $\operatorname{arccot} x$, then $\operatorname{arccot} x = \arctan(1/x) + k \pi$ for some integer $k$.

Here is what Mathematica thinks:

 Plot[{ArcTan[1/x] + .01, ArcCot[x]}, {x, -7 Pi, 0}, Frame -> True,
  PlotRange -> All]

You can rotate this second plot $90^\circ$ and define your range anyway you want up to a multiple of $\pi.$


答:
There's notation and there's notation!

If you are using notation in the sense of shorthand symbolism helpfully added to mathematicians' English, with "$\forall$" shorthand for "for all", then "$\forall x, y \in \mathbb{R}$" is perfectly acceptable shorthand "for all $x$ and $y$ in the real numbers". There are no strict rules about this sort of thing, no one "correct" way of doing things --  whatever works, works.

If you are (differently) aiming to use notation in the sense of writing formulas of a formalized language, then all bets are off. There will be a determinate answer to what is correct, depending on the precise rules of the language in question. (Something like "$\forall x, y$" might be allowed; usually not, though.)

Title: Shorthand for quantifying two variables?

Body: Say I have two real variables $x$ and $y$. Can I universally quantify them with $\forall x,y\in\mathbb{R}$ instead of using $\forall x\in\mathbb{R},\forall y\in\mathbb{R}$? Is this correct notation, or is it deemed sloppy by mathematicians?


问:
Prove that there is a matrix $B$ so that $\det(M + tB) \neq 0$
Let $M$ denote an $n\times n$ matrix with entries in a field $\mathbb{F}$. Prove that there is a $n\times n$ matrix $B$ with entries in $\mathbb{F}$ so that $\det(M + tB)\neq 0$ for every non-zero $t\in\mathbb{F}$.

答:
First of all, we note that we can take $B = I$ for the case that $M = 0$ and $B = 0$ in the case that $\det(B) \neq 0$.

For the remaining cases: with a rank-decomposition, we can find invertible matrices $P,Q$ such that
$$
M = P\pmatrix{I_r&0\\0&0_{(n-r)\times (n-r)}}Q.
$$
If we take
$$
B = P \pmatrix{0_{r \times r} & 0\\0 & I_{n-r}} Q,
$$
then we find that
$$
M + tB = P \pmatrix{I_r & 0\\0 & t I_{n-r}} Q,
$$
and we see that for all non-zero $t$ we have
$$
\det(M + tB) = \det(P)\det(Q) t^{n-r} \neq 0.
$$


问:
Find a polynomial in $\mathbb Q[x]$ such that $2^{1/2} + 3^{1/3}$ is a root
What would be a way to find a polynomial in  $\mathbb Q[x]$ such that $2^{1/2} + 3^{1/3}$ is a root.

答: Think in terms of conjugates.  $2^{1/2} + 3^{1/3}$ lives in the field $L := \Bbb{Q}(2^{1/2}, 3^{1/3}, \frac{-1+i\sqrt{3}}{2})$, so we could just take the product of all six linear factors of the form$$(x - a2^{1/2} - b3^{1/3}),$$ where $a = \pm 1$, $b = (\frac{-1 + i\sqrt{3}}{2})^k$ for $k = 0, 1, 2$. Since this polynomial's coefficients will be invariant under any automorphism $\sigma \in \operatorname{Gal}(L : \Bbb{Q})$, all the coefficients must actually live in $\Bbb{Q}$.


问:
Is it okay to multiply both sides of an inequality with a positive number if one side is 0?
I was wondering if it would be okay to multiply both sides of an inequality by a positive number $n$ if the inequality is the following:

$$-2/n < 0$$

And thus getting that $-2<0$?

答: Yes. It is always valid to do the same thing to both sides of an inequality, providing that it preserves ordering. Multiplication by positive numbers preserves ordering.</p>


答:
You have first to put the equation in the form
$$
y''+P(t)y'+Q(t)y=R(t)
$$

Title: Particular solution of $t^2y''-2y=t^2$ using the method of variation of parameters

Body:
Using the Anzats $y(t) = t^r$, we find $r(r-1)-2=0$ and thus the fundamental set of solutions $y_1(t)=t^2$ and $y_2(t)=t^{-1}$.

We have $W[y_1,y_2](t) = y_1(t)y'_2(t)-y_1'(t)y_2(t) = -t^2 t^{-2} - 2t t^{-1} = -3$.

The method of variation of parameters now gives a particular solution of the form $y_p(t) = u_1(t)y_1(t)+u_2(t)y_2(t)$ with $$u_1'(t) = -\frac{t^2 y_2(t)}{W[y_1,y_2](t)} = \frac{1}{3} t$$ and $$u_1'(t) = \frac{t^2 y_1(t)}{W[y_1,y_2](t)} = -\frac{1}{3}t^4.$$

Hence, we have $u_1(t) = \frac{1}{6}t^2$ and $u_2(t) = -\frac{1}{15}t^5$. This gives $$y_p=u_1(t)y_1(t)+u_2(t)y_2(t) = \bigg( \frac{1}{6}-\frac{1}{15} \bigg)t^4  = \frac{1}{10}t^4\ .$$

Inserting this into the ODE $L[y] = t^2y''-2y$, results in $L[y_p]=t^4$, not $t^2$.

Wolfram Alpha gives the particular solution $y_p(t) = \frac{1}{3}t^2 \log (t)$

Question: why does my application of the method of variation of parameters give a wrong particular solution? Where have I made a mistake?

Context: this is problem 2.4.9 of Braun's Differential Equations and Their Applications.


答:
Usually what you want to do to find a polynomial such that a given $\alpha$ is a root, you take powers of $x-\alpha$ and play around with both sides of the equation. In this case:

$$ x-\alpha=0 \Longleftrightarrow x-\sqrt2=\sqrt[3]{3} $$
$$ (x-\sqrt2)^3 = 3 $$
You'll still have some radicals left here, but all you have to do is to move all of them on one side of the equation and take powers again of both sides, i.e.
$$ x^3-3\sqrt2 x^2+6x-2\sqrt2 = 3 $$
$$ x^3+6x-3=\sqrt2 (3x^2+2) $$
$$ (x^3+6x-3)^2= 2(3x^2+2)^2, $$
which will finally give the degree-6 polynomial that you were looking for.

Title: Find a polynomial in $\mathbb Q[x]$ such that $2^{1/2} + 3^{1/3}$ is a root

Body: What would be a way to find a polynomial in  $\mathbb Q[x]$ such that $2^{1/2} + 3^{1/3}$ is a root.


答:
There are a few cases to consider. If $n = 0$, then the inequality is nonsense.

If $n > 0$, then we may multiply both sides by $n$ to obtain
$$
-2 < 0
$$
If $n < 0$, then we may multiply both sides by $n$ to obtain
$$
-2 > 0
$$
The fact that $-2 > 0$ is not true tells us that the case $n < 0$ has no solutions (supposing there were solutions with $n < 0$ produced a contradiction).

Thus, all we've managed to demonstrate with this approach is that $n > 0$.

It turns out that $n > 0$ precisely describes the solutions to this inequality. To see this, observe that $-2$ divided by any positive number is indeed less than $0$.

The moral of the story is that we can say nothing about what multiplying both sides of the inequality by $n$ does to the inequality without first establishing that $n$ is strictly positive or strictly negative.

Title: Is it okay to multiply both sides of an inequality with a positive number if one side is 0?

Body:
I was wondering if it would be okay to multiply both sides of an inequality by a positive number $n$ if the inequality is the following:

$$-2/n < 0$$

And thus getting that $-2<0$?


Body:
We can say that $x=\sqrt[3]{2}+\sqrt[3]{3}$ and proceed from there.

$$x^3 = 5+ 3\sqrt[3]{12}+3\sqrt[3]{12}$$
$$x^3 -5 = 3(\sqrt[3]{12}+\sqrt[3]{18})$$
$$x^3 -5 = 3\sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3})$$
Now, since we already defined that $x=\sqrt[3]{2}+\sqrt[3]{3}$, we say that
$$x^3 -5 = 3\sqrt[3]{6}(x)$$
$$x^3-3\sqrt[3]{6}x-5 = 0$$
And cubing,
$$x^9-15x^3-87x^3-125$$


问:
How to solve this integral $\int \frac{1+2x^2}{x^2(1+x^2)}dx$
How to solve this integral

$$\int \frac{1+2x^2}{x^2(1+x^2)}dx$$

I thought it should be $ x + 3x^2$ in the numerator so that I will take $x+x^3 = u$ then taking derivative both sides and it comes; $1+3x^2$  so this is wrong.

Please suggest how to proceed.

答:
Hint:

Use partial fractions.

$\dfrac{1+2x^2}{x^2(1+x^2)}=\dfrac{1}{1+x^2}+\dfrac{1}{x^2}$


问:
Monte carlo methods and the curse of dimensionality for solving PDEs
Do Monte Carlo methods suffer from the curse of dimensionality when solving PDEs? I was under the impression that they do not, since the variance scales like $\sigma /\sqrt{N}$, and thus it only depends on the number of samples and not on the number of dimensions.

However, I've read in several references (for example in this one), that direct Monte Carlo methods for simulating the Fokker-Planck equation also suffer from the curse of dimensionality, hence my question. I'm not sure I understand where this dependence is coming from, are there any good references about this topic? I would like to know how this type of Monte Carlo algorithms to solve Fokker-Planck equations scale with the dimensions.

Also, is there any Monte Carlo method to solve this type of equations that doesn't suffer from the curse of dimensionality?

Thank you very much.

答:
I checked your reference, and that claim does not have any additional detail or any reference. I am not sure you should take it too seriously.

Monte-Carlo does indeed beat the curse of dimensionality, but it also changes the problem a bit. As Robert Israel pointed out, if you goal is to solve the PDE on a grid with $N^d$ points in dimension $d$, then you cannot beat the curse of dimensionality. So something has to give.

Monte-Carlo methods are used when the goal is to compute the value of the solution at some given point $(x,t)$ (and only that point, or at some small collection of points). The idea is to reformulate the equation in terms of a Brownian motion or stochastic ODE, so that you can write $u(x,t)$ as the expectation against this stochastic process, in the form:

$$u(x,t) = \mathbb{E}[\cdots].$$

Then you sample some Brownian motion paths, or simulate the stochastic ODE, and replace the expectation by an empirical mean. As you said, the variance scales like $1/\sqrt{N}$ where $N$ is the number of sample paths, which is independent of dimension. This means the number of sample paths needed to obtain a particular error is independent of dimension.


问:
Recursive sequence with first term defined in interval
I have to prove that the recursive sequence $$a_{n+1}=2a_n(a_n+1)$$ converges for $-1 \le a_0 \lt 0$.

I began by trying to prove that the sequence is decreasing:
$$ a_n - a_{n+1}>0, $$
$$ a_n-2a_n(a_n+1)>0, $$
$$ a_n-2a_n^2 - 2a_n>0. $$
I don't know how to prove that inequality since $a_0$ is defined on an interval.
What should I do next?

答: $a_0=-1$ means $a_1=0$ and $a_n=0$ so assume that $-1<a_n<0$ by induction $0< a_n<1$ since,  if $-1<a_n<0$ then by adding one  $0< a_n+1 < 1$. $a_{n+1}=a_n(a_n+1)$ is the product of $0< a_n+1 < 1$ and $-1< a_n < 0$ so the product will be strictly between -1 and 0. For example $\frac{-1}{2}\times\frac{1}{2}=\frac{-1}{4}<0$ and $\frac{-1}{4}>-1$. Okay now showing it is actually increasing is possible $a_{n+1}=a_n(a_n+1)>a_n$ since $a_n<0$ and $0<a_n+1<1$. Bounded increasing sequences converge.


问:
Three isosceles triangles △ABC', △AB'C, and △A'BC were constructed on the sides of triangle ABC.
Three isosceles triangles $△ABC'$, $△AB'C$, and $△A'BC$ were constructed on the sides of triangle $ABC$. The vertex angles have the following measurements; $∠A'=α, ∠B'=β, ∠C'=γ$. Prove that if $α+β+γ=360°$ then the measurements of the angles of $△A'B'C'$ do not depend on the measurements of the angles of $△ABC$.

$D$, $E$, and $F$ are the midpoints of the sides of $△ABC$, making $A'E$, $B'F$, and $C'D$ the perpendicular bisectors of their respective triangles.

答:
Let us consider the circles $\Gamma_A,\Gamma_B,\Gamma_C$ centered at $A',B',C'$ with radii $A'B,B'C,C'A$.

The constraint on the angles ensures that these circles meet at a point $P$.

We also have that $PB'AC', PC'BA'$ and $PA'CB'$ are kites, therefore $\widehat{B'A' C'}=\frac{1}{2}\alpha$ and so on.


问:
Does almost sure convergence and $L^1$-convergence imply almost sure convergence of the conditional expectation?

Question. Let $ X_{n}, X  $ be random variables on some probability space  $ ( \Omega, \mathcal{F},\mathbb{P} ) $ and let $ \mathcal{G} \subset \mathcal{F} $ be a sub-$\sigma$-algebra. Moreover suppose that $ X_{n } \to X $ a.s. and in $ \mathrm{L}^{ 1 } ( \Omega, \mathcal{F}, \mathbb{P}) $. Does this already imply that $ \mathbb{ E}  ( X_{ n } | \mathcal{G} ) \to \mathbb{ E}  ( X | \mathcal{G} ) $ a.s.?

We know that the $ \mathrm{L}^{1} $-convergence implies that $\mathbb{ E}  ( X_{ n } | \mathcal{G} ) \to \mathbb{ E}  ( X | \mathcal{G} ) $ in $\mathrm{L}^{ 1 } $. However, we can in general only expect a subsequence to converge a.s. This can be seen by looking at $\mathcal{ G } = \mathcal{ F } $ and taking the usual counterexample that $ \mathrm{L}^{ 1 } $ convergence does not always imply a.s. convergence.
But what happens if we additionally assume the a.s. convergence of the original sequence?

答:
No, in general this is not true.

The ingredients: Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of non-negative random variables such that $Y_n \to 0$ in $L^1$ but $Y_n$ does not converge almost surely to $0$ (e.g. the classical example of a sequence converging in probability to $0$ but not almost surely; see Example 1 here). Let $(\xi_n)_{n \in \mathbb{N}}$ be a sequence of independent random variables such that $(Y_n)_{n \in \mathbb{N}}$ and $(\xi_n)_{n \in \mathbb{N}}$ are independent and $\mathbb{P}(\xi_n=2^n) = 2^{-n}$, $\mathbb{P}(\xi_n=0) = 1-2^{-n}$. If we set $$Z_n := \prod_{j=1}^n \xi_j,$$ then the Borel Cantelli lemma shows that $Z_n(\omega)= 0$ for $n\geq N(\omega)$ sufficiently large. Moreover, $\mathbb{E}(\xi_n)=1$ implies $\mathbb{E}(Z_n)=1$ for all $n \in \mathbb{N}$.

The counterexample: Define $X_n := Y_n \cdot Z_n$. Since $Z_n(\omega)=0$ for $n \geq N(\omega)$, we clearly have $X_n \to 0$ almost surely. Moreover, $X_n \geq 0$, and so

$$\mathbb{E}(|X_n|) = \mathbb{E}(X_n) = \mathbb{E}(Y_n) \underbrace{\mathbb{E}(Z_n)}_{=1} \to 0,$$

shows that $X_n \to 0$ in $L^1$. Hence, $X_n \to 0=:X$ almost surely in $L^1$. Now consider the sub $\sigma$-algebra $\mathcal{G} := \sigma(Y_n; n \geq 1)$. Then, by the independence,

$$\mathbb{E}(X_n \mid \mathcal{G}) = \mathbb{E}(Y_n Z_n \mid \mathcal{G}) = Y_n \mathbb{E}(Z_n \mid \mathcal{G}) = Y_n \underbrace{\mathbb{E}(Z_n)}_{=1}.$$

Since we have chosen $(Y_n)_{n \in \mathbb{N}}$ such that $Y_n$ does not converge almost surely to $0$, we conclude that $\mathbb{E}(X_n \mid \mathcal{G})=Y_n$ does not converge almost surely to $\mathbb{E}(X \mid \mathcal{G})=0$.


问:
A quotient group over a commutator of a normal subgroup is in the center.
Suppose $H \triangleleft G$  is a normal subgroup, $K = [G, H]$ is ф commutator subgroup. Then K is the smallest normal subgroup such that $H/K \leqslant Z(G/K)$, where Z(G/K) is the center of G/K.

Since $K \geqslant H'$ then $K \triangleleft H$ and $H/K$ is an abelian group. But why does it commute with other elements of $G/K$? If I realize that, I think I will prove its smallness.

答:
$gKhK = ghK = [g, h]hgK = [g, h]KhKgK = hKgK$

since $[g, h] \in K$


问:
Elliptic Deformations of Multiplication
Consider the squared absolute value function on complex numbers $a+bi \rightarrow a^2+b^2$.

It's easy to verify that the squared absolute value is a homomorphism over multiplication. That is
$$ |(a+bi)(c+di)|^2 =  (ac - bd) + (ad+bc)i = (ac-bd)^2 + (ad+bc)^2 = (ac)^2-2acbd+(bd)^2 + (ad)^2 + 2abcd + (bc)^2 = (ac)^2+(bd)^2+(ad)^2+(bc)^2= (a^2+b^2)(c^2+d^2)= |(a+bi)|^2|(c+di)|^2$$

If we define an elliptic absolute value as the map given a choice of $(u,v)$ and a complex number $a+bi \rightarrow \frac{a^2}{u^2} + \frac{b^2}{v^2}$ we can easily see that unless we have $u=v=1$ this function almost surely doesn't respect multiplication. (We can try some difference choices of $u,v$ to see this for ourselves).

So now we get to a very subjective question, what if we insist there ought to be an operation like multiplication, what we shall call a $(u,v)$-deformation of multiplication, which has the property that the $(u,v)$-elliptic absolute value respects this operation, and this operation is in some sense as "natural" as possible. We could probably say that the foci of the ellipse generated by $(u,v)$ would replace the function of 0 w.r.t this multiplication.

What would be some candidates for such a multiplication?

Some Preliminary thoughts:

If we have a complex function $\Psi$ that maps the unit circle to an ellipse centered at 0. Then we can define multiplication on the ellipse by taking two points $\rho_1, \rho_2$ and computing $\Psi(\Psi^{-1}(\rho_1)\Psi^{-1}(\rho_2))$.

This new multiplication operation at the least would "respect" our elliptic squared absolute values for the unit ellipse of a particular $(u,v)$ class.

It's meaningful then to ask, given a $(u,v)$ pair what is the space of meromorphic functions that map complex circles of radius $r$ to the ellipse $x^2/u^2 + y^2/v^2 = r^2$.

These would create can candidates for making an elliptical multiplication.

答:
By your assumptions,

$$\Psi(re^{i\theta})=r\Psi(e^{i\theta})=r\big(u\cos\psi(\theta)+iv\sin\psi(\theta)\big)$$

for some real function $\psi$.

The Cauchy-Riemann equations in polar coordinates are

$$ir\frac{\partial\Psi}{\partial r}=\frac{\partial\Psi}{\partial\theta};$$

that is,

$$ir(u\cos\psi+iv\sin\psi)=r(-u\sin\psi+iv\cos\psi)\psi'.$$

Cancelling $r$ and separating into real and imaginary parts,

$$-v\sin\psi=-u\sin\psi\,\psi',\quad u\cos\psi=v\cos\psi\,\psi'.$$

Since $\Psi$ is invertible, $\psi$ is not constantly a multiple of $\pi/2$, so we can cancel $\cos\psi$ and $\sin\psi$:

$$v=u\psi',\quad u=v\psi'$$

$$\psi'=\frac vu=\frac uv$$

$$u=\pm v$$

$$\psi=\pm\theta+C$$

$$\Psi(re^{i\theta})=ru\,e^{\pm i\psi}=re^{i\theta}\cdot ue^{\pm iC}$$

Thus $\Psi$ must be multiplication by a constant, and the ellipse must be a circle.

Title: Elliptic Deformations of Multiplication


答: Hint: write the numerator as $1+x^2 + x^2$ and write the whole expression as two fractions. One with numerator $1+x^2$ and the other one with $x^2$</p>

Title: How to solve this integral $\int \frac{1+2x^2}{x^2(1+x^2)}dx$

Body:
How to solve this integral

$$\int \frac{1+2x^2}{x^2(1+x^2)}dx$$

I thought it should be $ x + 3x^2$ in the numerator so that I will take $x+x^3 = u$ then taking derivative both sides and it comes; $1+3x^2$  so this is wrong.

Please suggest how to proceed.


答:
Consider the function $f(x)=2x^2(x+1)$. Basic analysis allows you to show that this function sends $[-1,0]$ onto $[-1/2,0]$,  so that if $a_0 \in [-1,0]$, then $a_n$ also for all $n$.

We now seperate the values in $[-1,0)$ into $3$ cases : either $a_0<-1/2$, either $a_0=-1/2$, either $a_0>-1/2$.

The second case is easy as $-1/2$ is fixed under the function $f$. In the first case, we notice that if $a_0=-1$, the sequence gets stuck on $0$ and stays there. If $-1<a_0<-1/2$, then $a_1 >-1/2$ and we are in the third case. We now only need to show that if $a_0>-1/2$ then the sequence converges. This is simple, for you can show that for all $x \in (-1/2,0)$, $f(x)<x$. The sequence is thus decreasing and bounded, and so converges.

You can also show that it always converge to $-1/2$ except when $a_0=-1$. Give it a try!

Title: Recursive sequence with first term defined in interval

Body:
I have to prove that the recursive sequence $$a_{n+1}=2a_n(a_n+1)$$ converges for $-1 \le a_0 \lt 0$.

I began by trying to prove that the sequence is decreasing:
$$ a_n - a_{n+1}>0, $$
$$ a_n-2a_n(a_n+1)>0, $$
$$ a_n-2a_n^2 - 2a_n>0. $$
I don't know how to prove that inequality since $a_0$ is defined on an interval.
What should I do next?


问:
Why does $P(X=E(X)) = P(X=E(X)-1)$ in the Poisson distribution?
When a discrete random variable follows the Poisson distribution and if $E(X)$ is a natural number, $E(X)$ has the greatest probability, but why does $E(X)-1$ have the same probability?

答: Fleshing out the first part of Brian Moehring's comment: If $X$ is Poisson with mean $E[X]=\lambda$ where $\lambda$ is an integer, then $P(X=\lambda) = e^{-\lambda} \frac{\lambda^\lambda}{\lambda!}$ while $P(X=\lambda-1) = e^{-\lambda} \frac{\lambda^{\lambda - 1}}{(\lambda-1)!}$.


问:
what is a coloring in set theory?
I've been interested in the Open Coloring Axiom,  but I couldn't find a formal definition for Coloring. Is there an actual definition for Coloring? Is a coloring a function?</p>

答:
A coloring of a set $A$ with a set $C$ of colors is a function $f:A\to C$.  Usually (and in particular in OCA) the set $A$ being colored is obtained from some other set $X$, for example as the set of $2$-element subsets of $X$.

A coloring $f$ as above is almost equivalent to a partition of $A$ into pieces indexed by $C$, namely the pieces $f^{-1}(\{c\})$ for $c\in C$. Here "almost" refers to the detail that pieces in a partition are supposed to be nonempty, whereas $f$ is not required to be surjective.

Title: what is a coloring in set theory?


问:
Linearly independent vectors coefficients
Suppose that the set $\{\vec u_1, \vec u_2, \ldots \vec u_n\}$ is linearly independent and that $\vec w$ is a linear combination of these vectors. Is it true that the coefficients of the linear combination are unique? How so?

I do not understand how to prove that the coefficients are unique.

答:
$w=\displaystyle\sum_{k=1}^n\alpha_kv_k$ where $v_1,...,v_n$ is a set of linearly independent vectors and $\alpha_1,...,\alpha_n$ are its correcponding coefficients.

We will prove this statement using proof by contradiction.

We start by assuming that the choice of $\alpha_1,...,\alpha_n$ is not unique. i.e.

$w=\displaystyle\sum_{k=1}^n\alpha_kv_k=\sum_{k=1}^n\beta_kv_k$ where at least one $\alpha_k\neq\beta_k$

Here,

$\displaystyle\sum_{k=1}^n\alpha_kv_k=\sum_{k=1}^n\beta_kv_k\\
\displaystyle\sum_{k=1}^{n-1}\alpha_kv_k+\alpha_nv_n=\displaystyle\sum_{k=1}^{n-1}\beta_kv_k+\beta_nv_n\\
v_n=\displaystyle\frac{\displaystyle\sum_{k=1}^{n-1}\beta_kv_k-\displaystyle\sum_{k=1}^{n-1}\alpha_kv_k}{\alpha_n-\beta_n}$

However, this contradicts our assumption that the vectors $v_1,...,v_n$ are linearly independent.

So, our assumption that the choice of $\alpha_1,...,\alpha_n$ is not unique is false.
So, the choice of $\alpha_1,...,\alpha_n$ is unique.


问:
Tensor product of separating vectors in von Neumann algebra.
Let $\mathcal{M}_1\subseteq \mathcal{B}(\mathcal{H}_1)$ and $\mathcal{M}_2\subseteq\mathcal{B}(\mathcal{H}_2)$ be von Neumann algebras with separating vectors $\xi_1\in \mathcal{H}_1$ and $\xi_2\in\mathcal{H}_2$ respectively. Then how can we prove that $\xi_1 \otimes \xi_2\in\mathcal{H}_1\overline\otimes\mathcal{H}_2$ is a separating vector of the von Neumann algebra $\mathcal{M}_1\overline\otimes\mathcal{M}_2$?

答: As $\xi_1\otimes\xi_2$ is cyclic for $\mathcal{M}_1'\overline\otimes\mathcal{M}_2'$ and we have $\mathcal{M}_1'\overline\otimes\mathcal{M}_2'\subseteq (\mathcal{M}_1\overline\otimes\mathcal{M}_2)'$, therefore $\xi_1\otimes\xi_2$ is cyclic for $(\mathcal{M}_1\overline\otimes\mathcal{M}_2)'$. Hence $\xi_1\otimes\xi_2$ is separating for $\mathcal{M}_1\overline\otimes\mathcal{M}_2$.


问:
Second order differential equation from physics
Trying to use Desmos to curve-fit some positional data of a ball bearing falling through water, and I want that curve to be relevant to $F_D=-kv^2$. So by mixing that with Newton's 2nd law, I got this equation:
$$
mg-k\left(\frac{dx}{dt}\right)^2=m\frac{d^2x}{dt^2}
$$
Where $m$, $g$, and $k$ are positive, non-zero constants, and for all $t\geq0$, $x(t)\geq0$, $x'(t)>0$

But I can't see how to solve the thing for $x$.

Any ideas?

答:
Hint: Start by solving a simpler equation:$$m\frac{dv}{dt}=mg-kv^2$$
Then integrate to get $x$. Let me know if you still have problem with this.

Edit:
$$\frac{dv}{dt}=g\left(1-\frac k{mg}v^2\right)$$
Note that you can define the terminal velocity when the acceleration is zero, so $$v_T^2=\frac{mg}k$$
Then $$\int\frac{dv}{1-\frac{v^2}{v_T^2}}=g\int dt$$
Integrating, you get $$v_T\mathrm{arctanh}\frac{v}{v_T}=gt+C$$
From here, I would need the initial velocity to get $C$. Let's assume that $v(0)=0$. Then $C=0$ and $$v=v_T\tanh\frac{gt}{v_T}$$
Integrating $\tanh$ is easy (it's $\sinh$ divided by $\cosh$). Assuming $x(0)=0$, you get $$x(t)=\frac{v_T^2}g\ln\cosh\frac{gt}{v_T}$$


问:
Evaluating $\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$
This is the sequel of my previous question

$$I(a)=\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$$ I want to use differentiation under the integral sign with respect to parameter "a" but so far without success.

Any hint?

答:
Good evening,

I found a simpler faster way to deal with this integral :

1/ The numerator being a pair function of Sin(x) one can easily show that the integral is equal to :

F(a)=Integral from 0 to Pi/2 Arctan(a.(Sin(x)^2))) / (Sin(x)^2) dx

2/ Then we apply Feynman’trick to F(a) and get :

F’(a)=Integral from 0 to Pi/2 dx / (1+ a^2. (Sin(x))^4

3/ Then we use t = Tan(x) and get :

F’(a)=Integral ftom 0 to inf (1+t^2) dt / (t^4(1+a^2) + 2t^2 +1)

4/ Then we use Schlomilch master formula and get :

F’(a)=Pi / (2^(1/2)) . (1+(1+a^2)^(1/2)) / (1+a^2)^(1/2)

5/ Then we integrate F’(a) and get :

F(a) = Pi / (2^(1/2)) . a / (1+(1+a^2)^(1/2))^(1/2)

```<code>     = Pi / (2^(1/2) . ((1+a^2)^(1/2) - 1)^(1/2)
</code>```

fjaclot

Body:
Good evening,

I found a simpler faster way to deal with this integral :

1/ The numerator being a pair function of Sin(x) one can easily show that the integral is equal to :

F(a)=Integral from 0 to Pi/2 Arctan(a.(Sin(x)^2))) / (Sin(x)^2) dx

2/ Then we apply Feynman’trick to F(a) and get :

F’(a)=Integral from 0 to Pi/2 dx / (1+ a^2. (Sin(x))^4

3/ Then we use t = Tan(x) and get :

F’(a)=Integral ftom 0 to inf (1+t^2) dt / (t^4(1+a^2) + 2t^2 +1)

4/ Then we use Schlomilch master formula and get :

F’(a)=Pi / (2^(1/2)) . (1+(1+a^2)^(1/2)) / (1+a^2)^(1/2)

5/ Then we integrate F’(a) and get :

F(a) = Pi / (2^(1/2)) . a / (1+(1+a^2)^(1/2))^(1/2)

     = Pi / (2^(1/2) . ((1+a^2)^(1/2) - 1)^(1/2)

fjaclot


问:
Proving a sequence has a finite Limit
I was wondering how to prove that a sequence is converging to a finite limit.

I know that if a sequence is bounded and decreasing / increasing then it has a finite limit.

although I have 2 questions where I don't know how to prove them :

the sequence : $a_n = \frac {1}{n} + \frac {1}{n+1} + ... \frac {1}{3n}$

This sequence is bounded between 1 and 3 , I tried to prove its decreasing but failed to.

the sequence : $a_n = \frac {1}{\sqrt{1}} + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} + ...+ \frac {1}{\sqrt{n}}$
is not bounded and has no finite limit ($\infty$)

what is the way to prove?
Thank you.

Title: Proving a sequence has a finite Limit

Body:
hint

$$a_n=\frac 1n+\frac{1}{n+1}+...+\frac{1}{3n}$$

$$a_{n+1}=\frac{1}{n+1}+...+\frac{1}{3n}+\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3(n+1)}$$

then

$$a_{n+1}-a_n=$$
$$\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}-\frac 1n=$$
$$\frac{.......}{n(3n+1)(3n+2)(3n+3)}$$

For the second sequence, use the fact that for any $ k\in\{1,2,...n\}$,

$$ \frac{1}{\sqrt{k}}\ge \frac{1}{\sqrt{n}}$$
to get
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\ge \frac{n}{\sqrt{n}}$$

Observe that the RHS goes to infinity and conclude.


问:
Evaluating $\int_0^4 \int_{\sqrt {x}}^2 \frac{x^2 e^{y^2}}{y^5}\mathrm dy\,\mathrm dx$
I need help starting with this. I can't find an example like this anywhere in my book

$$\int_0^4 \int_{\sqrt {x}}^2 \frac{x^2 e^{y^2}}{y^5}\mathrm dy\,\mathrm dx$$

Body:
Hint:
$$
\int_{x = 0}^4 {\int_{y = \sqrt x }^2 {\frac{{x^2 e^{y^2 } }}{{y^5 }}\,dy} \,dx}  = \int_{y = 0}^? {\int_{x = 0}^? {\frac{{x^2 e^{y^2 } }}{{y^5 }}\,dx} \,dy}.
$$


问:
A quotient space of a closed annulus is homeomorphic to a Klein bottle
I was told that a quotient space of a closed annulus centered at the origin obtained with a relation $x \sim - x$ for $x$ in the boundary is homeomorphic to a Klein bottle, which is a connected sum of two real projective planes.

So if my quotient map is $q$ and $U$ is the annulus without the inner boundary and $V$ is the annulus without the outer boundary, I was told $q(U)$ and $q(V)$ are projective planes missing a point, where the missing point is $q(\text{missing boundary})$.

However, I do not see this directly, and not sure how to show this.

Also, once I show this, I can see what is the fundamental group of the space by the consequence of the Van Kampen theorem, but what would be the generators of the fundamental group?

答:
If you can appeal to the classification of connected, compact surfaces, then you can look at a cell decomposition: vertices are the 4 (really only 2) points where the boundaries meet the $x$-axis, edges are the 4 arcs (really only 2) plus the 2 segments along the $x$-axis, and 2-cells are the upper and lower regions these enclose.  You can compute the Euler characteristic and can directly find an embedded copy of a Möbius band.

For the statements about the quotient maps, it might be easier to understand what's going on if you replace "are" by "are homeomorphic to".  A surface minus a point is homeomorphic to a surface minus a closed disk centered at this point.  This point of view reveals the boundary.

For the fundamental group, since you are thinking about the Van Kampen theorem, a natural choice would be to look at a generator for the fundamental group of the intersection of $U$ and $V$ and generators for the fundamental groups of $U$ and $V$.  This will give three generators, but you can eliminate one using the relations that arise from Van Kampen's Theorem.  A little bit of care is needed in choosing a base point.


问:
Prove this integration by parts formula
I have been struggling on this problem :

We work in $([0,\infty[,B(\mathbb{R}),\nu_1)$.  Let $f,g\in \mathcal{L^1}$, and
\begin{align}
F(x)&=\int_0^{x}f(t)\,dt\\
G(x)&=\int_0^{x}g(t)\,dt
\end{align}
Show:
$$\int_0^{\infty}F(x)g(x)\,dx=\int_0^{\infty}f(x)\,dx\int_0^{\infty}g(x)\,dx-\int_0^{\infty}f(x)G(x)\,dx$$

The first question is about showing that $F,G$ are bounded (that's because $f,g\in\mathcal{L^1}$), and that they are continuous (by using the theorem about continuity of integrals with parameters with $f*1_{[0,x]}$ as function of $x$ and $t$).

But when it comes to proving the equation, I really am stuck. I've tried a couple of things, but none that's worth sharing. Any help would be great, thanks in advance.

答:
Proof 1 ( Using the notion of absolute continuity and its consequences)

Fix some $t>0$. Since, $f,g$ are integrable, it follows that $F,G$ are absolutely continuous on the closed and bounded interval $[0,t]$. This means, that $F,G$ are differentiable everywhere on $[0,t]$ and $F'(x)=f(x),\,G'(x)=g(x)$ almost everywhere. Now, $F\cdot G$ is also absolutely continuous on $[0,t]$ hence it is differentiable almost everywhere and
$$\tag{1} [F(x)\cdot G(x)]'=f(x)G(x)+g(x)F(x)$$
almost everywhere. Also, by the absolute continuity of $F\cdot G$ we also have
\begin{align}
\int_{0}^{t}[F(x)\cdot G(x)]'&=F(t)\cdot G(t)-F(0)\cdot G(0)\\
&=\biggl(\int_0^tf(x)\,dx\biggr)\cdot \biggl(\int_0^tg(x)\,dx\biggr)\tag{2}
\end{align}
Integrating $(1)$ and using $(2)$ we obtain
$$\int_0^t F(x)g(x)\,dx=\biggl(\int_0^tf(x)\,dx\biggr)\cdot \biggl(\int_0^tg(x)\,dx\biggr)-\int_0^tf(x)G(x)\,dx\tag{3}$$
Now, let $s_t(x)=F(x)g(x)\cdot \mathbb{1}_{[0,t]}$. Then, $s_t(x)\to F(x)g(x)\cdot \mathbb{1}_{[0,\infty)}$ as $t\to \infty$. Also,
$$|s_t(x)|\leq M|g(x)|\cdot \mathbb{1}_{[0,\infty)}$$
where $M$ is the bound of $F(x)$. Since $g$ is integrable it follows by the dominated convergence theorem that
$$\lim_{t\to \infty}\int_{0}^tF(x)g(x)\,dx=\int_0^\infty F(x)g(x)\,dx
$$
Similarly, $\lim_{t\to \infty}\int_0^tf(x)G(x)\,dx=\int_{0}^\infty f(x)G(x)\,dx$ and $$\lim_{t\to\infty}\int_0^t f(x)\,dx=\int_0^\infty f(x)\,dx,\ \lim_{t\to \infty}\int_{0}^t g(x)\,dx=\int_0^\infty g(x)\,dx$$
So, take $t\to \infty$ in $(3)$ to conclude
$$\int_{0}^\infty F(x)g(x)\,dx=\biggl(\int_0^\infty f(x)\,dx\biggr)\cdot \biggl(\int_0^\infty g(x)\,dx\biggr)-\int_0^\infty f(x)G(x)\,dx$$

Proof 2 (Approach with indicator functions and approximation arguments)

I think you can prove $(2)$ (in Proof 1) for indicator functions, say $f=\mathbb{1}_A,\,g=\mathbb{1}_B$ with $\mu(A)<\infty,\,\mu(B)<\infty$ and then use the classic method by approximating general measurable functions $f,\,g\in L^1$ with simple functions and use the convergence theorems (Monotone for positive $f,g$, dominated for general $f,g$). Although, i find this approach quite ugly but is the only think i could think off. So suppose $f=\mathbb{1}_A,\,g=\mathbb{1}_B$ with $\mu(A),\mu(B)<\infty$. Then, fix $t>0$ and observe that
$$F(x)=\mu\bigl(A\cap[0,x]\bigr),\, G(x)=\mu\bigl(B\cap [0,x]\bigr)$$
for every $0\leq x\leq t$. Write,
\begin{align}
\int_0^t F(x)g(x)\,dx&=\int_0^t\mu\bigl(A\cap [0,x]\bigr)g(x)\,dx\\
&=\int_0^t\bigg(\mu\bigl(A\cap [0,t]\bigr)-\mu\bigl(A\cap [x,t]\bigr)\biggr)\cdot g(x)\,dx\\
&=\mu\bigl(A\cap [0,t]\bigr)\int_0^t g(x)\,dx-\int_0^t\mu\bigl(A\cap [x,t]\bigr)g(x)\,dx\\
&=\biggl(\int_0^t f(x)\,dx\biggr)\cdot \biggl(\int_0^t g(x)\,dx\biggr)-\int_0^t\mu\bigl(A\cap [x,t]\bigr)g(x)\,dx
\end{align}
And we claim that
$$\tag{4} \int_0^t\mu\bigl(A\cap [x,t]\bigr)g(x)\,dx=\int_0^t G(x)f(x)\,dx$$
Writing $\mu\bigl(A\cap [x,t]\bigr)$ as $\int_x^t \mathbb{1}_A(y)\,dy$ and using Tonelli's theorem to interchange the order of integration we obtain
\begin{align}
\int_0^t\mu\bigl(A\cap [x,t]\bigr)g(x)&=\int_0^t\biggl(\int_x^t \mathbb{1}_A(y)\,dy\biggr)g(x)\,dx\\
&=\int_0^t\biggl(\int_x^t \mathbb{1}_A(y)\,dy\biggr)\cdot \mathbb{1}_B(x)\,dx\\
&=\int_0^t\biggl(\int_0^t \mathbb{1}_{A\cap [x,t]}(y)\cdot \mathbb{1}_B(x)\biggr)dy\,dx\\
&=\int_0^t\biggl(\int_0^t \mathbb{1}_{A\cap [x,t]}(y)\cdot \mathbb{1}_B(x)\biggr)dx\,dy \tag{5}
\end{align}
Now, fix $0\leq y\leq t$ and observe that
$$\tag{6}\mathbb{1}_B(x)\cdot \mathbb{1}_{A\cap [x,t]}(y)=\mathbb{1}_A(y)\cdot \mathbb{1}_{B\cap [0,y]}(x)$$
for every $0\leq x\leq t$. Using $(6)$ in $(5)$ we obtain
\begin{align}
\int_0^t \mu\bigl(A\cap [x,t]\bigr)\cdot \mathbb{1}_B(x)&=\int_0^t\biggl(\int_0^t \mathbb{1}_A(y)\cdot \mathbb{1}_{B\cap[0,y]}(x)\,dx\biggr)\,dy\\
&=\int_0^t \mathbb{1}_A(y)\biggl(\mathbb{1}_{B\cap [0,y]}(x)\,dx\biggr)\,dy\\
&=\int_0^t \mathbb{1}_A(y)\cdot \mu\bigl(B\cap [0,y]\bigr)\,dy\\
&=\int_0^t f(y)\cdot G(y)\,dy
\end{align}
and $(4)$ is proven! (phew!)

Ps: I didnt check the details for the approximation arguments for general $f,g$ i expect that it can work.

Title: Prove this integration by parts formula


问:
Proof in Circle geometry question

Let Γ be a semicircle with diameter . The point  lies on the diameter  and points  and  lie on the arc , with  between  and . Let the tangents to Γ at  and  meet at . Suppose that ∠ = ∠. Prove that ∠ = ∠ + ∠.

I got this question from a sample competition paper aimed at 16-18 year olds.
I think that I can prove this by proving that the point C must be at the circle's centre by contradiction; if it wasn't then the 2 angles which are supposedly equal wouldn't in fact be equal.

I have 3 questions:

Is my argument correct?

If my argument is correct, what would be a rigorous proof for my argument?

Is there an alternative method? If so please hint to it.

Thank you for your help.

答:
It is certainly not the case that $C$ must be the center. Consider the diagram below:

Here $C$ is not the center. However $\angle EFD = \angle ACD + \angle ECB$ still holds.

To show that equality, it suffices to show that $CEFD$ is cyclic.

This can be accomplished by proving the orange angles below are equal, and that is my hint. (There could be alternative methods but I'm not sure)

We will need to show that $CE = CG$ given that $\angle ECB = \angle GCB$.

Consider the center of the circle, $O$. $OE = OG$ as radii, with $OC$ as common sides, leads to $\triangle COE \cong \triangle COG$ by "long side-short side-angle".

We can also see this by dropping perpendiculars from $O$. Then $\triangle COP \cong \triangle COP'$ by AAS, and $\triangle POE \cong \triangle P'OE$ by RHS. Either way, we have $CE = EG$, and we can proceed.

Title: Proof in Circle geometry question


问:
Integrating $f(x,y,z)$ within the region of a cylinder and a hyperboloid in cartesian coordinates
I'm confident with solving this question in cylindrical coordinates (for which there are few answers) but I'm doubtful about cartesian coordinates and can't seem to figure out the limits. $z$ needs to be the outer integral in the below question:

Right now, I've figured that:

$z$ must vary from $-1$ to $1$ (since it describes the radius of the cylinder).

If we make $x$ the inner integral and dependent on $y$ we could say that it varies from $-2y^2$ to $(1-2y^2)$ (obtained by substituting the cylinder equation and adding in $a=0$ & $a=1$

I'm not sure how to evaluate the middle integral and how exactly there are 2 regions over which we must integrate.

答:
Hyperboloid $z = \sqrt{a^2+x^2+y^2} \,$ where $a \in (0,1)$ is upper portion of the hyperboloid (above $XY$ plane). Please note $z = -\sqrt{a^2+x^2+y^2}$ gives you a similar region below $XY$ plane.

Given $a \lt 1,$ it does intersect with cylinder $y^2 + z^2 = 1$.

If you visualize the region is bound below and on sides by hyperboloid and bound above by the cylinder.

In cartesian coordinates,

Bounds of $z$ is $ \,\sqrt{a^2+x^2+y^2} \leq z \leq \sqrt{1-y^2} \,$ (bound below by hyperboloid and above by the cylinder)

To find bounds of $x,y$ as you wrote, $1 - a^2 = x^2 + 2y^2 \implies y = \pm \sqrt{\frac{1 - a^2 - x^2}{2}}$

Bounds of $y$, $-\sqrt{\frac{1 - a^2 - x^2}{2}} \leq y \leq \sqrt{\frac{1 - a^2 - x^2}{2}}$

From above equation $1 - a^2 = x^2 + 2y^2$, the lower and upper bounds of $x$ (occurs when $y = 0$) is $\pm \sqrt{1-a^2}$.

Bounds of $x, \, \, -\sqrt{1-a^2} \leq x \leq \sqrt{1-a^2}$

If you need to integrate over both regions above $XY$ plane and below $XY$ plane, just multiply the integral by $2$.


问:
Asymptotics for convolutions of arithmetic functions
I would like to know the asymptotics for $n \to \infty$ of the two functions
$$\sum_{d \mid n} \frac{\mu(d)}{d^2}n^2$$

and
$$\sum_{d \mid n} \frac{\mu^2(d)}{d} n.$$

I see nothing else to do than to write the expanded sum and... this is not enough to "see" the main term. Is there any general method to estimate asymptotics for such convolutions?

答:
Those things don't have a simple asymptotic because

$$\sum_{d |n} \frac{\mu(d)^r}{ d^2}n =n\prod_{p^k \| n} \sum_{d |p^k}\frac{\mu(d)^r}{ d^2}=n\prod_{p | n} (1+ \frac{(-1)^r}{p^2})$$
With $n=m!$ and $m$ large then $\prod_{p | n} (1+ \frac{(-1)^r}{p^2})$ will be close to $1/\zeta(2)$ if $r$ is odd or $\zeta(2)/\zeta(4)$ if $r$ is even, whereas for $n$ prime $\prod_{p | n} (1+ \frac{(-1)^r}{p^2})$ will be close to $1$.


问:
How do I calculate the annihilator of series with fraction?
How do I calculate the annihilator of this:
$$s_n = s_{n-1} + \frac{1}{(n + 1)(n + 2)(n + 3)}$$

I know that $E-1$ is the annihilator part for $s_n - s_{n-1}$ but I don't know how to get the part for the fraction.

I need to find the closed form of $s_n$ and the first thought I had was to use annihilators.

答:
$$s_n-s_{n-1}=\frac{1}{(n+1) (n+2) (n+3)}\tag{1}$$
The homogeneous recurrence
$$s_n-s_{n-1}=0$$
has the trivial solution $s_n=s_0$

To solve $(1)$ we guess the particular solution to be
$$s_n^*=\frac{a}{n+2}+\frac{b}{n+3}$$
plug this in the recurrence and get
$$\frac{a}{n+2}+\frac{b}{n+3}-\frac{a}{n+1}-\frac{b}{n+2}=\frac{1}{(n+1) (n+2) (n+3)}$$
and then
$$\frac{n (-a-b)+(-3 a-b)}{(n+1) (n+2) (n+3)}=\frac{1}{(n+1) (n+2) (n+3)}$$
so we have to solve the system
$$\begin{cases}
-a-b=0\\
-3a-b=1\\
\end{cases}\to a=-\frac{1}{2};\;b=\frac12
$$
The particular solution is
$$s_n^*=-\frac{1}{2(n+2)}+\frac{1}{2(n+3)}=-\frac{1}{2 (n+2) (n+3)}$$
and
$$s_n=s_0+s_n^*\to s_n=s_0-\frac{1}{2 (n+2) (n+3)}$$
for $n=0$ we have $s_0=\frac{1}{12}$
Finally the solution of the recurrence is
$$s_n=s_0+\frac{1}{12}-\frac{1}{2 (n+2) (n+3)}\to s_n=s_0+\frac{n (n+5)}{12 (n+2) (n+3)}$$


答:
A comment has made clear the problem is not understanding factorials well. Note that$$(k+1)!=\underbrace{1\cdot2\cdot\dots\cdot k}_{k!}\cdot(k+1)=k!(k+1),$$so$$\frac{P(X=k+1)}{P(X=k)}=\frac{\frac{\lambda^{k+1}}{(k+1)!}e^{-\lambda}}{\frac{\lambda^k}{k!}e^{-\lambda}}=\lambda\frac{k!}{(k+1)!}=\frac{\lambda}{k+1}.$$At first, the ratio of consecutive probabilities is high, but it quickly diminishes. (That makes sense: the probabilities haev to shrink eventually for infinitely many of them to sum to $1$, but we expect very small $k$ to be improbable if $\lambda$ is very large, so probabilities have to rise at first.) In particular:

$P(X=k+1)>P(X=k)$ if $k<\lambda-1$;

$P(X=k+1)=P(X=k)$ if $k=\lambda-1$; and

$P(X=k+1)<P(X=k)$ if $k>\lambda-1$.

Title: Why does $P(X=E(X)) = P(X=E(X)-1)$ in the Poisson distribution?

Body: When a discrete random variable follows the Poisson distribution and if $E(X)$ is a natural number, $E(X)$ has the greatest probability, but why does $E(X)-1$ have the same probability?


问:
Homotopy classes of a subset of $\mathbb{R}^2$ with two holes.
I’m working through some self-devised examples to better understand homotopy classes. In a convex subset of $\mathbb{R}^2$, for example, the homotopy class is trivial. In a convex subset of $\mathbb{R}^2$ with one hole in it, I can see that there are two classes of loops: the trivial one, and the class of loops going “around” the hole (there’s no formal proofs here, just my intuition to see which loops can or cannot be deformed into other loops).

I’m not sure about convex subsets of $\mathbb{R}^2$ with two holes in them, so I want to get my answer checked. As far as I can see, there’s still the trivial class, the class of loops that go around one hole, the class of loops that go around both holes, and (possibly) the “figure-eight” loops. Am I missing something?

I know there are machinery that could help computing these in a formal way, but that will come later. For now I’d appreciate if the problem could be approached at an intuitive level.

答:
You're not missing anything that comes to my mind, the only thing you should be aware of is that these classes come with a natural structure of group, given by the "composition" of loops.

This group structure however is generally not commutative: in your case, $\mathbb R^2$ with two holes makes space for all possible composition of loops around each of the two holes (and, of course, the trivial class, given by the identity of the group, or the class of the constant loop if you will); up to homotopy, every loop in your space is the composition of loops around each of the two holes! However, you have to account not only for "how many times" you go around a specific hole, but also for the order of these compositions.

In a more algebraic fashion, you say that the class of loops in this space [i.e., its fundamental group, $\pi_1(\mathbb R^2\smallsetminus\{p,q\})$] is the free group generated by the two classes of the two "simple" loops around each hole. The simplest way you can think about this concept is as "the group of words in an alphabet of two letters": let's call the two "simple" loop classes (winding only one time in a direction of your preference) around each hole $f$ and $g$; let's also denote as $\tilde{f},\tilde{g}$ their inverses, i.e. the same loops but "running in the opposite direction".

Then, up to homotopy, every loop in your space is equal to some combination of $f,g,\tilde{f},\tilde{g}$; for instance, $ff\tilde{g}\tilde{f}ggf$.
All you have to take into account at this point is that, clearly, if a loop and its inverse come one after the other, they cancel out (always "up to homotopy", of course): for example, $gf\tilde{f}fg=gfg$.


Title: Proving a sequence has a finite Limit

Body:
I was wondering how to prove that a sequence is converging to a finite limit.

I know that if a sequence is bounded and decreasing / increasing then it has a finite limit.

although I have 2 questions where I don't know how to prove them :

the sequence : $a_n = \frac {1}{n} + \frac {1}{n+1} + ... \frac {1}{3n}$

This sequence is bounded between 1 and 3 , I tried to prove its decreasing but failed to.

the sequence : $a_n = \frac {1}{\sqrt{1}} + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} + ...+ \frac {1}{\sqrt{n}}$
is not bounded and has no finite limit ($\infty$)

what is the way to prove?
Thank you.

Body:
The following shows that $a_{n}\geq a_{n+1}$ for all $n$, that is, $(a_{n})$ is decreasing:

$\frac {1}{n}  + ... + \frac {1}{3n} > \frac {1}{n+1}+...+\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3} \Longleftrightarrow \frac {1}{n}>\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}$

Since $\frac {1}{n} > \frac {1}{n+\frac {1}{3}}=\frac {3}{3n+1}= \frac {1}{3n+1}+\frac {1}{3n+1}+\frac {1}{3n+1} > \frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}$, the result follows.


问:
Union of finite connected subsets which intersect pairwise is connected proof verification
$X$ is a topological space and $A_1 ,\ldots ,A_n \subset X$ connected subsets such that $A_k  \cap  A_{k+1} \neq \emptyset$ for all $k = 1 ,\ldots ,n-1$. I must show that $A_1 \cap \ldots \cap A_n$ is connected. I am using proof by induction along with the fact that a space $(X,\tau)$ is connected $\iff$ when $f: X \longrightarrow \{0,1\}$ is continuous, it is constant. My attempt at a proof is as follows:

Assume $A_1 \cup A_2$ disconnected. Then, there exist two non-empty, disjoint and open subsets, say $B$ and $C$, such that $A_1 \cup A_2 = B \cup C$. Define $f: U \cup V \longrightarrow \{0,1\}$ by $$f(x)=\begin{cases}1& \; ; x\in U\\0&\; ; x\in V\end{cases}$$To show it is continuous, consider some open subset of $\{0,1\}$, say $W$. It is a simple case by case verification: $$\begin{array}{cc} 1) & 0 \notin W, 1 \notin W, f^{-1}(W) = \emptyset \\ 2) & 0 \in W, 1 \in W, f^{-1}(W) = A_1 \cup A_2 \\ 3) & 0 \in W, 1 \notin W, f^{-1}(W) = V \\ 4) & 0 \notin W, 1 \in W, f^{-1}(W) = U \end{array}$$
Thus, $f$ is continuous since $f^{-1}(W)$ is open whenever $W$ is open. Now, having proved our $f$ is continuous, we need to show it is constant. Indeed, since $A_1$ is connected, $f(A_1) = 0$ or $f(A_1) = 1.$ Say it equal to 0. Then, $f(A_1 \cap A_2) = 0$ since $A_1 \cap A_2 \subset A_1$ but since $A_2$ is also connected, this means that $f(A_2) = f(A_1 \cap A_2) = f(A_1) = 0$. It is the exact same if $f(A_1) = 1$. This gives us a contradiction to our initial assumption that $A_1 \cup A_2$ is disconnected which implies that it is indeed connected.

Assume the case is true for $A_k \cup A_{k+1}$. We will show it is true for $A_{k+1} \cup A_{(k+1)+1}$. Namely, if we proceed in the same fashion as for the trivial case, we will arrive at the point that $$f(A_{(k+1)+1}) = f(A_{k+1} \cap A_{(k+1)+1}) = f(A_{k+1}) = \text{constant}$$ where the constant is either $0$ or $1$. We may now disregard the intersection and apply our hypothesis, namely that $$f(A_{(k+1)+1}) = f(A_{k+1}) = f(A_k) = f(A_{k-1}) = \cdots = f(A_1) = \text{constant}$$ which shows that $A_1 \cup \cdots \cup A_n$ is indeed connected.

This does make sense to me but if there is anything wrong with it at all, please do let me know where I went wrong as I would like to try figure this out on my own. I appreciate any help given!

答:
The theorem that you want to prove is that for every $n\in\Bbb Z^+$, $\bigcup_{k=1}^nA_k$ is connected if each of the sets $A_k$ is connected, and $A_k\cap A_{k+1}\ne\varnothing$ for $k=1,\ldots,n-1$. Since you’re trying to prove that something is true for each positive integer $n$, you should be inducting on $n$. The result is trivial for $n=1$. For the induction step we assume that the result is true for $n$ and try to prove it for $n+1$, so let $A_1,\ldots,A_{n+1}$ be connected subsets of $X$ such that $A_k\cap A_{k+1}\ne\varnothing$ for $k=1,\ldots,n$. Then

$$\bigcup_{k=1}^{n+1}A_k=\left(\bigcup_{k=1}^nA_k\right)\cup A_{n+1}\,.$$

For convenience let $A=\bigcup_{k=1}^nA_k$; by the induction hypothesis $A$ is connected, and clearly $A\cap A_{n+1}\supseteq A_n\cap A_{n+1}\ne\varnothing$. The argument that you used to show that $A_1\cup A_2$ is connected can easily be adapted to show that $A\cup A_{n+1}$ is connected, and that completes the induction step.


问:
Which functions are spherical derivatives?
Let us define the differential operator
$$
Z=x_1 \partial_{x_2} - x_2 \partial_{x_1}, $$
where $(x_1, x_2, x_3)$ are the standard Cartesian coordinates on $\mathbb R^3$. I would like to characterize the functions $h\in C^\infty(\mathbb S^{2})$ such that
$$\tag{1}
h=Zf, \qquad \text{for some }f\in C^\infty(\mathbb S^{2}).$$

The operator $Z$ is one of the generators of the rotation group $SO(3)$, in the sense that
$$\tag{2}
Zf(x)=\lim_{\epsilon\to 0}\frac1\epsilon \big( f(R_\epsilon x)-f(x)\big), $$
where $R_\epsilon$ is the matrix
$$
R_\epsilon=\begin{bmatrix}
\cos \epsilon & -\sin \epsilon & 0  \\
\sin \epsilon & \cos \epsilon & 0   \\
0 & 0 & 1
\end{bmatrix}.$$
The formula (2) implies that
$$\int_{\mathbb S^{2}} Zf(x)\, dS(x)=0, $$
where $dS$ denotes the standard Lebesgue measure on the sphere. Thus, a necessary condition for (1) to hold is that
$$\tag{3}
\int_{\mathbb S^{2}} h\, dS = 0.$$
Also, since $Z$ vanishes at $(0,0, \pm 1)$, another necessary condition is
$$
\tag{4} h(0,0,\pm 1)=0.$$
Are these two last conditions also sufficient?

答:
No. A stronger condition is needed. Since the orbints of $Z$ are the circles on constant lattitude, $Zf$ must integrate to zero over each such circle. This can be seen by working in standard spherical coordinates $(\theta,\varphi):[0,2\pi)\times(0,\pi)\to S^2$. In these coordinates, $Z=\partial_\theta$, so we can apply the fundamential theorem of calculus along the lattitudes, obtaining
$$
f(\theta,\varphi)=f(0,\varphi)+\int_0^\theta Zf(\theta',\varphi)d\theta'
$$
Since $f(0,\varphi)=\lim_{\theta\to 2\pi^-}f(\theta,\varphi)$, we must have $\int_0^{2\pi} Zf(\theta',\varphi)d\theta'=0$. If you modify your set of conditions, you can check sufficiency by checking if the above expression gives a smooth function on $S^2$ for some choice of $f(0,\varphi)$.


问:
Proof of $\operatorname{arccot}(x)= \arctan(\frac{1}{x})$
$\newcommand{\ac}{\operatorname{arccot}}$
Here is my proof:

$ \theta = \ac(x)$

$ \cot \theta = x $

$ \displaystyle \tan \theta = \frac{1}{\cot \theta} = \frac{1}{x}$

$ \displaystyle \theta = \arctan \frac{1}{x}$

But based on the graph, $\ac(x) = \arctan(\frac{1}{x})$ is only satisfied when $ x > 0$. What is wrong with my proof? What is the error occurred inside a proof?

答:
HINT.-Using a figure of a right triangle, it is evident that for $0\lt \theta\lt \dfrac{\pi}{2}$ one has the proposed equality.

from which the equality is true for all $x\gt 0$.

However, for $x\lt0$ the equality is not true because, for example,

$$\operatorname{arccot}(-2)\approx2.6779\\\arctan(-\dfrac 12)\approx-1.1071$$


问:
Finding parameters in the intersection of two sets
I'm having trouble with the following problem. I have two sets $A=\{(x,y)\in\mathbb{R}^{2}: 3x+y\le 0 \}$ and $B=\{(x,y)\in\mathbb{R}^{2}: ax+by < 0 \}$ and I am supposed to find all real parameters $a,b$ such that $A\cap B=\emptyset$. How to find them?

答:
The elements in $A$ are $(x,y) \in \mathbb{R}^2$ such that $y \leq -3x.$ Therefore, any $(z,w) \in \mathbb{R}^2$ such that $y>-3x$ are not in $A.$

But

$$y>-3x \iff 3x +y > 0 \iff -3x - y < 0.$$

So consider $a = -3$ and $b = -1.$

Next, observe that $y + 3x = 0$ and $ax + by = 0$ are lines that passes at $(0,0).$ So these are the only real values of $a$ and $b$ such that $A \cap B = \emptyset.$


问:
bivariate Yoneda lemma
$\newcommand{\Hom}{\operatorname{Hom}}$Any category is equipped with a covariant hom-functor $\Hom(A,-)$, by letting the second argument vary. The covariant Yoneda lemma says $\operatorname{Nat}(\Hom(A,-),F)\cong F(A)$ for any covariant functor $F$.

Alternatively any category is equipped with a contravariant hom-functor, by instead letting the first argument vary. Then for contravariant functors $F\colon C^\mathrm{op}\to \operatorname{Set},$ we have the contravariant Yoneda lemma $\operatorname{Nat}(\Hom(-,X),F)\cong F(X).$

Instead of choosing one or the other argument to vary, we may let both vary. Giving the hom functor as a bivariate functor $\operatorname{Hom}(-,-)\colon C^\mathrm{op}\times C\to \operatorname{Set}$. Is there a Yoneda lemma for this functor as well?

答:
To extend on the answer and comments of Derek Elkins.

Indeed, applying e.g. the covariant Yoneda lemma $\left[\,{\rm Nat}(\hom_{\mathcal C}(C,-),\,F)\,\simeq\,F(C)\right.$ for $\left. F:\mathcal C\to\mathcal Set\,\right]$ with $\mathcal C=\mathcal A^{op}\times\mathcal B$ and an arbitrary profunctor $F:\mathcal A^{op}\times\mathcal B\to\mathcal Set$ will produce
$$ {\rm Nat}(\hom_{\mathcal A^{op}\times\mathcal B}((A,B),\,-),\,F)\,\simeq\,F(A,B) $$
where $\hom_{\mathcal A^{op}\times\mathcal B}((A,B),\, (X,Y))=\hom_{\mathcal A}(X,A)\times\hom_{\mathcal B}(B,Y)$.

Let $H_{A,B}$ denote this profunctor $H_{A,B}(X,Y)=\hom_{\mathcal A}(X,A)\times\hom_{\mathcal B}(B,Y)$.

Observe that this is basically the same as considering the disjoint union $\mathcal A+\mathcal B$ and freely joining an arrow $h_{A,B}:A\to B$:

the set of arrows $X\to Y$ arising this way, for $X\in Ob\,\mathcal A,\ Y\in Ob\,\mathcal B$, consists exactly of formal compositions $\beta\circ h_{A,B}\circ\alpha$ for arbitrary $\alpha:X\to A$ and $\beta:B\to Y$.

(In the above setting, we have $h_{A,B}=(1_A,1_B)\in H_{A,B}(A,B)$.)

We can also view any profunctor $F:\mathcal A^{op}\times\mathcal B\to\mathcal Set$ as a category containing $\mathcal A+\mathcal B$ by considering $F(A,B)$ as the set of (hetero-)morphisms 'from $A$ to $B$'. Compositions are given by the action of $F$ on morphisms.

Also, natural transformations $F\to G$ simply correspond to functors that are identical on $\mathcal A+\mathcal B$.

From this perspective, it's clear that ${\rm Nat}(H_{A,B},\,F)\cong F(A,B)$, since any morphism of profunctors $\Phi:H_{A,B}\to F$ is indeed uniquely determined by its image on the freely joined arrow $h_{A,B}$ which can be any arrow $A\to B$ in  $F$, i.e. any element of $F(A,B)$.

It's easy to see that both the covariant and the contravariant Yoneda lemmas are consequences of this instance, by letting either $\mathcal A$ or $\mathcal B$ be the trivial, one element category.

To answer your original question, the bivariate hom functor has less to do with Yoneda, nevertheless we can have a similar statement:

For any 'endoprofunctor' $F:\mathcal A^{op}\times\mathcal A\to\mathcal Set$, we have
$${\rm Nat}(\hom_{\mathcal A}(-,-),\,F)\,\simeq\,{\rm Nat}(I_1,I_2)\,\simeq\,\int_{\mathcal A}F$$
where $I_1,I_2:\mathcal A\to\mathcal F$ are the two embeddings of $\mathcal A$ into the category $\mathcal F$ constructed as above from $F$, and the integral denotes the end of $F$.

For the correspondence, observe that any natural transformation $\Phi:\hom_{\mathcal A}(-,-)\to F$ is determined by the images $t_A:=\Phi_{A,A}(1_A)\in F(A,A)$ of $1_A\in\hom_{\mathcal A}(A,A)$, and these elements $t_A$ need to satisfy a commutativity condition for each arrow $\alpha:A\to A'$.

Title: bivariate Yoneda lemma


问:
Description of a set in the Baire space.
I have the following exercise.

In the Baire space, $\omega ^ \omega$ ,describe a set $A$ such that $int(A)\neq \emptyset$ and $A\neq \bar{A}$

I recently just got to know the Baire space, it is difficult for me to understand it a bit and I have come across this exercise. Could you please help me with an example. Thank you.

Edit:

I have considered the following set: $A=\{ x\in \omega ^ \omega: x\nleq f\}$ And I have shown that it is open. But I have the doubt, that for example in the discrete metric all open is closed at the same time, and I cannot see how to show that this set is not closed

答: HINT: Consider the point $f$ itself: clearly $f\notin A$, but you should be able to show that $f\in\operatorname{cl}A$. It may help to prove that the sets $$B_n=\{x\in\omega^\omega:x(k)=f(k)\text{ for all }k\le n\}$$ for $n\in\omega$ are a local base at $f$, if you don’t already know this fact.


问:
A comparison principle for Eikonal equation with viscosity solution
Given $u\in \mathrm{C}(\Omega)$ where $\Omega$ is an open subset of $\mathbb{R}^n$, then the followings are equivalent:

$u$ is a viscosity solution of $|Du|\geq 1$ in $\Omega$.

If $V$ is an open bounded subset of $\Omega$ and $\varphi\in \mathrm{C}^\infty(V)$ such that $|D\varphi| \leq 1$ in $V$ and $\varphi \leq u$ on $\partial V$ then $\varphi\leq u$ in $\overline{V}$.

答:
For (i) implies (ii): If $u-\phi$ has a min at $x_0\in \overline{V}$ and
$$ \min_{\partial V}(u-\phi) > \min_{V}(u-\phi)$$
then we must have $x_0\in V$, then we can find $r>0$ such that $B(x_0,r)\subset V$ and thus we can actually make it as a strict local minimum by replacing $\phi(x)$ with $\phi(x)-\eta(x-x_0)|x-x_0|^2$ where $\eta\in \mathrm{C}^\infty_c$ and $\eta$ is compactly supported in $B(0,r)$ and $\eta = 1$ near $0$.

The sequence $\phi_\lambda(x) = \lambda \phi(x)$ with $\lambda<1$ satisfies $\phi_\lambda \rightarrow \phi$ uniformly, thus $u-\phi_\lambda$ has a nearby local minimum at $x_\lambda$ and $x_\lambda \rightarrow x_0$ (up to subsequence), therefore the supersolution test of $u$ implies that
$$|D\phi_\lambda(x_\lambda)| = \lambda |D\phi(x_\lambda)| \geq 1$$
which is a contradiction since by assumption $\lambda|D\varphi(x_\lambda)|\leq \lambda < 1$. Therefore the asumption is wrong and thus
$$ 0\leq \min_{\partial V}(u-\phi) \leq \min_{V}(u-\phi).$$

For (ii) implies (i): Assume that $u$ is not a super-solution, then there exists $x_0\in \Omega$ such that there exists $\varphi\in \mathrm{C}^\infty(\Omega)$ such that $u-\varphi$ has a strict minimum at $x_0\in \Omega$ and $|D\varphi(x_0)| < 1$. By replacing $\varphi(x)$ by $\varphi(x) - |x-x_0|^2 - C$ where $C = (u-\varphi)(x_0)$ we can assume $u(x_0) = \phi(x_0)$ and thus $(u-\phi)(x)>0$ for all $x$ near $x_0$, say $x\in B(x_0,3r)$. By continuity we can also assume that
$$ |D\varphi(x)| < 1 \qquad\text{for all}\qquad x\in B(x_0,2r).$$

Let $\varphi^\varepsilon(x) = (1+\varepsilon\eta(x))\varphi(x)$ where $\eta\in \mathrm{C}_c^\infty$ with $\eta = 1$ on $B(x_0, r)$ and $\mathrm{supp}(\eta)\subset B(x_0,2r)$, we have
$$ |D\varphi^\varepsilon(x)|\leq (1+\varepsilon)|D\varphi(x)| + C\varepsilon,\qquad x\in B(x_0,2r)$$
where $C = \sup_{\overline{B(x_0,2r)}}|D\eta(x)\varphi(x)|$. Since $|D\varphi(x)|<1$ in $B(x_0,2r)$, we can find $\varepsilon$ small such that
$$|D\varphi^\varepsilon(x)|<1, \qquad x\in B(x_0,2r).$$
Applying (ii) with $V = B(x_0,2r)$, we see that on the boundary $\varphi^\varepsilon = \varphi<u$ on $\partial V$, therefore
$$\varphi^\varepsilon(x) \leq u(x) \qquad
\text{for all}\qquad x\in V$$
which is a contradiction since
$$\varphi^\varepsilon(x_0) = (1+\varepsilon)\varphi(x_0) > u(x_0).$$


问:
Prove or disprove statements about the greatest common divisor
Help with prove or disproving either of these statements would be really appreciated, one or the other is fine, I just need a start or a solution to one and I'm sure I could probably figure the other out.

(a) For all $a, b, c \in \mathbb{Z}$, $\gcd(ac, bc) = c \gcd(a, b)$

(b) For all $a, b, c \in \mathbb{Z}$, $\gcd(ab, c) = \gcd(a, b) \gcd(b, c)$

答:
If the highest power of prime $p$ in $a,b,c$ are $r_a,r_b,r_c$ respectively,

the highest power of $p$ in gcd$(a,b)=$min $(r_a,r_b)$

the highest power of $p$ in $c\cdot$ gcd$(a,b)=r_c+$min $(r_a,r_b)$

the highest power of $p$ in  gcd$(a\cdot c,b \cdot c)=$min $(r_a+r_c,r_b+r_c)=r_c$+min $(r_a,r_b)$

the highest power of $p$ in gcd$(a\cdot b,c)=$min $(r_a+r_b,r_c)$

the highest power of $p$ in gcd$(a,b)\cdot $gcd $(b,c)=$min $(r_a,r_b)$ + min$(r_b+r_c)$

Can you prove min $(r_a+r_b,r_c)=$min $(r_a,r_b)$ + min$(r_b+r_c)?$


问:
A question about floor functions and series
The sequence $\{a_n\}^{∞}_{n=1} = \{2,3,5,6,7,8,10,...\}$
consists of all the positive integers that are not perfect squares.

Prove that $a_n= n+ [\sqrt{n} + \frac{1}{2}]$.

Well, I managed to prove that
$ m^2 < n+ [\sqrt{n} + \frac{1}{2}] < (m+1)^2 $,
where $[\sqrt{n} + \frac{1}{2}] = m$.

But is this enough to answer the question? Or do we also need to prove that $n+ [\sqrt{n} + \frac{1}{2}] $ can take all non-perfect square values. If yes, then how?

Any help is appreciated, thanks!

答:
The following is just a first step toward the answer.

From
$$
\eqalign{
  & a_{\,n}  = n + \left\lfloor {\sqrt n  + {1 \over 2}} \right\rfloor   \cr
  & a_{\,n + 1}  - a_{\,n}
  = 1 + \left\lfloor {\sqrt {n + 1}  + {1 \over 2}} \right\rfloor
  - \left\lfloor {\sqrt n  + {1 \over 2}} \right\rfloor  \cr}
$$
since we have
$$
\eqalign{
  & \left\lfloor {x - y} \right\rfloor  =  - \left\lceil {y - x} \right\rceil  =   \cr
  &  = \left\lfloor x \right\rfloor  - \left\lfloor y \right\rfloor
  + \left\lfloor {\left\{ x \right\} - \left\{ y \right\}} \right\rfloor  =   \cr
  &  = \left\lfloor x \right\rfloor  - \left\lfloor y \right\rfloor
  - \left[ {\left\{ x \right\} < \left\{ y \right\}} \right] \cr}
$$
where $[P]$ denotes the Iverson bracket, then
$$
\eqalign{
  & a_{\,n + 1}  - a_{\,n}  = 1 + \left\lfloor {\sqrt {n + 1}  + {1 \over 2}} \right\rfloor
  - \left\lfloor {\sqrt n  + {1 \over 2}} \right\rfloor  =   \cr
  &  = 1 + \left\lfloor {\sqrt {n + 1}  - \sqrt n } \right\rfloor
  + \left[ {\left\{ {\sqrt {n + 1}  + {1 \over 2}} \right\}
 < \left\{ {\sqrt n  + {1 \over 2}} \right\}} \right] =   \cr
  &  = 1 + \left[ {\left\{ {\sqrt {n + 1}  + {1 \over 2}} \right\}
 < \left\{ {\sqrt n  + {1 \over 2}} \right\}} \right] \cr}
$$

Now the Iverson bracket will be one, and thus we will have a jump in $a_{\,n}$, when
$$
\left\{ {\sqrt {n + 1}  + {1 \over 2}} \right\} < \left\{ {\sqrt n  + {1 \over 2}} \right\}
$$
which for $1 \le n$ , which means $sqrt{n+1} -\sqrt{n} < 1/2$, can occur only when
$$
\sqrt n  + {1 \over 2} < m < \sqrt {n + 1}  + {1 \over 2}
$$
with $m$ an integer, that is
$$
\eqalign{
  & \sqrt n  < m - {1 \over 2} < \sqrt {n + 1} \quad  \Rightarrow   \cr
  &  \Rightarrow \quad n < m^{\,2}  + {1 \over 4} - m < n + 1\quad  \Rightarrow   \cr
\quad  \Rightarrow   \cr
  &  \Rightarrow \quad n - {1 \over 4} < m\left( {m - 1} \right) < n + {3 \over 4}\quad  \Rightarrow   \cr
  &  \Rightarrow \quad n_ *   = m\left( {m - 1} \right) \cr}
$$

Thus
$$
a_{\,n + 1}  - a_{\,n}  = 2\quad {\rm iff}\quad n = m\left( {m - 1} \right)
$$

It remains then to demonstrate that
$$
\forall q\;\exists m:\quad a_{\,n_{\, * } }  + 1
 = m\left( {m - 1} \right) + \left\lfloor {\sqrt {m\left( {m - 1} \right)}
  + {1 \over 2}} \right\rfloor  + 1 = q^{\,2}
$$


问:
Evaluate the following integral :$\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$
Evaluate the following integral :

$$I=\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$$

I was tried use change variable ,

If I use $x=y^2$ integral becomes :

$$I=2\int\limits_0^{\infty}\frac{\log (1+x^{8})}{1+x^{2}}dx$$

From here I have one idea  the derivative under sing integral but I got I difficult integration :

$$I=2\int\limits_0^{\infty}\frac{x^{8}}{(1+ax^{8})(1+x)}dx$$

I already to see you hints or solution!

答:
Using the principal branch of the logarithm, we have

$$ \begin{align} I &= \int_{0}^{\infty} \frac{\ln(1+t^{4})}{\sqrt{t}(1+t)} \, \mathrm dt = 2\int_{0}^{\infty} \frac{\ln(1+x^8)}{1+x^2} \, \mathrm dx  = \int_{-\infty}^{\infty} \frac{\ln(1+x^8)}{1+x^2} \, \mathrm dx \\ &= \sum_{n=0}^{7} \int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \\ &= \sum_{n=0}^{3} \left(\int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx + \int_{-\infty}^{\infty} \frac{\ln \left(1 \,  {\color{red}{+}} \, xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \right)\\ &= \sum_{n=0}^{3} \left(\int_{-\infty}^{\infty} \frac{\ln \left(1-xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx+ \int_{\infty}^{-\infty} \frac{\ln \left(1 - ue^{i \pi(2n+1)/8}\right)}{1+u^2} \,  (- \mathrm du) \right) \\&= 2 \sum_{n=0}^{3} \int_{-\infty}^{\infty} \frac{\ln \left(1 - xe^{i \pi(2n+1)/8}\right)}{1+x^2} \, \mathrm dx \\ & \stackrel{(1)}= 2\sum_{n=0}^{3}    2 \pi i \operatorname{Res} \left[ \frac{\ln \left(1-ze^{i \pi(2n+1)/8}\right)}{1+z^2}, i  \right] \\ &=  2\sum_{n=0}^{3} 2 \pi i  \, \frac{\ln \left(1-ie^{i \pi(2n+1)/8}\right)}{2i} \\ &= 2\pi \sum_{n=0}^{3} \ln \left(1-ie^{i \pi(2n+1)/8}\right)  \\ &= 2 \pi \ln   \left[ \left(1-ie^{i \pi/8} \right) \left(1-ie^{3 \pi i/8} \right) \left(1+ie^{-3 \pi i/8} \right) \left(1+ie^{- i \pi/8} \right) \right]\\ &= 2 \pi \ln \left[\left(2+ 2 \sin \left(\frac{\pi}{8} \right) \right) \left(2+ 2 \sin \left(\frac{3\pi}{8} \right) \right)\right] \\ &= 2 \pi \ln \left(4 +2 \sqrt{2-\sqrt{2}} +2 \sqrt{2+\sqrt{2}} + \sqrt{2} \right) \\ &= 2 \pi \ln \left(4 + 2 \sqrt{4 +2\sqrt{2}} + \sqrt{2}\right) \\ &= 2 \pi \ln \left(\left(\sqrt{2}+\sqrt{2+\sqrt{2}} \right)^{2}\right) \\ &= 4 \pi \ln \left(\sqrt{2}+\sqrt{2+\sqrt{2}} \right) \end{align}$$

$(1)$ The branch cut for $\ln \left(1-ze^{i \pi(2n+1)/8}\right) $ is in the lower half of the complex plane.


问:
one example on game playing with some details, how this average step was calculated?
I have one example in my TA notes:

We have a two players games. two players will be agree on number $n$. game start with $x=2$. first player select by Coin flipping and start game. after that each of players arbitrary do one of the following: $x=x^2$ or $x=x^3$.

for example if "David" start game and choose $x^3$ then $x$ will be equals to $8$. after playing of each players that condition $x>n$ occurred, the game will be stopped and the last player announced as winner. at average this game ended after $O(\log \log n)$ steps.

This is very nice example, but how we can detect this average order can be achieved?

答:
If we take $\log_2$ of everything, we end up with a simpler game that is equivalent to the original ($\log$ is monotonically increasing, so $x>n$ iff $\log x > \log n$). I will denote $\log_2x$ by $y$ and $\log_2n$ by $m$. When we take the log, squaring becomes doubling, and cubing becomes tripling.

Then this game is equivalent to the following:

Start with $y=1$. The players take turns either doubling or tripling $y$. The game ends when $y >m$.

We can take the log again. When we take the log of a product, that's the sum of the log of the multiplicands, so doubling becomes adding the log of $2$, and tripling becomes adding the log of $3$ (side note: this means that this is a nim game, as adding until you reach some number is equivalent to starting at that number and subtracting until you get to zero).  So then the game becomes:

Start with $z=0$. Each turn, players add $\log_22=1$ or $\log_23$ to $z$. The game ends when $z > \log_2m$.

Clearly, this game will end the fastest if each players increases $z$ by $\log_23$ each turn, in which case it will (with rounding) end after $\frac{\log_2m}{\log_23 }= \log_3m = \log_2\log_3n$ turns. The longest it can take is $\log_2\log_2n$. It's not clear that "average" is well defined, but it regardless of how it's defined, it will be between these two numbers. And if we're dealing with Big-O notation, the bases of the logarithms don't matter, so we're left with $\log\log n$


问:
Not sure why integral of $\sqrt{a^2-x^2}$ result includes absolute value
I worked out the integral of $\sqrt{a^2-x^2}dx$ and found it to be $\frac{x\sqrt{a^2-x^2}+a^2arcsin(\frac{x}a)}2 + C$ via the trig sub $x=asin(\theta)$

However, integral-calculator.com returns the result shown below:

This agrees with my answer up until the last step where an absolute value is introduced and for some reason $a^2-x^2$ is factorised. Please can anyone explain why this is done and why my answer is incorrect?

答:
Your answer is correct. the other one is just factorized out by $ a $.

The difference  comes from the fact that

$$\sqrt{1-\frac{x^2}{a^2}}=$$

$$\sqrt{\frac{a^2-x^2}{a^2}}=$$

$$\frac{\sqrt{(a+x)(a-x)}}{\sqrt{a^2}}=$$

$$\frac{\sqrt{-(x-a)(x+a)}}{|a|}$$

because
$$\sqrt{a^2}=|a|$$


问:
Limit with variable in integral boundary
I have the following integral

$$\int_{\vert x\vert\leq\varepsilon}\frac{1}{\varepsilon}\frac{1}{\vert x\vert^2}\,\mathrm{d}x,$$

where the integral should be understood as a Lebesgue integral and where $x\in\mathbb{R}^{n}$. My question is now, can we say what happens if $\varepsilon\to 0$? I don't know how to approach this question in a mathematical way. The problem here is that on the one hand our integral region becomes smaller and smaller, which means that in the end we will end up with a Lebesgue-zero set, but in the same time our integrand becomes larger and larger.  Any ideas?

答: Hint: Different answers for different $n.$ For example, if $n=1,2,$ you should be able to see that for any $\epsilon>0,$ the integrals equal $\infty.$ That's not true in higher dimensions however.


问:
Evaluating $\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$
This is the sequel of my previous question

$$I(a)=\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$$ I want to use differentiation under the integral sign with respect to parameter "a" but so far without success.

Any hint?

Body: One more solution: using $\sum_{n=-\infty}^{\infty}\left(x+2\pi n\right)^{-2}=\frac{1}{4}\csc^{2}\frac{x}{2}$ and$$\sum_{n\ge0}C_{n}x^{n}=\frac{1-\sqrt{1-4x}}{2x}\implies\sum_{n\ge0}C_{2n}x^{2n}=\frac{\sqrt{1+4x}-\sqrt{1-4x}}{4x}$$(by taking the even part) in terms of the Catalan numbers $C_{n}=\frac{\left(2n\right)!}{n!\left(n+1\right)!}$,$$\begin{align}I\left(a\right)&=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\arctan\left(a\sin^{2}x\right)}{x^{2}}dx\\&=\frac{1}{2}\int_{0}^{2\pi}\sum_{n=-\infty}^{\infty}\frac{\arctan\left(a\sin^{2}x\right)}{\left(x+2\pi n\right)^{2}}dx\\&=\frac{1}{8}\int_{0}^{2\pi}\arctan\left(a\sin^{2}x\right)\csc^{2}\frac{x}{2}dx\\&=\frac{1}{4}\int_{0}^{\pi/2}\arctan\left(a\sin^{2}x\right)\left(\csc^{2}\frac{x}{2}+\sec^{2}\frac{x}{2}\right)dx\\&=\int_{0}^{\pi/2}\arctan\left(a\sin^{2}x\right)\csc^{2}xdx\\&=\sum_{n\ge0}\frac{\left(-1\right)^{n}}{2n+1}a^{2n+1}\int_{0}^{\pi/2}\sin^{4n}xdx\\&=\pi a\sum_{n\ge0}C_{2n}\left(-\frac{a^{2}}{16}\right)^{n}\\&=\frac{\pi}{i}\left(\sqrt{1+ia}-\sqrt{1-ia}\right)\\&=\frac{\pi}{\sqrt{2}}\frac{a}{\sqrt{1+\sqrt{1+a^{2}}}}.\end{align}$$


问:
Solution verification, complex inequality
$$\{z\in \text{complex numbers :}\quad|z| + Re(z) \leq Im \}$$
$Re...\text{real component}$

$Im... \text{imaginary component}$

$$|x + iy| + x \leq y$$
$$\sqrt{x^2 + y^2} + x \leq y $$
$$\sqrt{x^2 + y^2} \leq y -x $$
$$1.1: \text{for } y \leq x $$
$$ 0 \leq -2xy \leftrightarrow 0\geq xy$$
$$\text{solution 1 : }\, y \geq 0 \,\text{ and }\, x\leq 0$$
$$1.2 \text{ for }\,y\geq x $$
$$ xy \geq 0$$
$$\text{solution 2: }x\geq 0 \, \text{ and }y\geq 0\text{ and } y \leq x  $$
$$\text{Final solution: Solution 1 union Solution 2}$$

答:
The condition is $$\sqrt{x^2+y^2}\le y-x\Leftrightarrow\begin{cases}x^2+y^2\le (y-x)^2\\ y\ge x\end{cases}\Leftrightarrow \begin{cases} xy\le 0\\ y\ge x\end{cases}\Leftrightarrow\\\Leftrightarrow\begin{cases}x=0\\ y\ge x\end{cases}\lor \begin{cases}x<0\\ y>0\\ y\ge x\end{cases}\lor \begin{cases}x>0\\ y<0\\ y\ge x\end{cases}\lor \begin{cases}y=0\\ y\ge x\end{cases}\Leftrightarrow\begin{cases}x=0\\ y\ge 0\end{cases}\lor \begin{cases}x<0\\ y>0\end{cases}\lor \begin{cases}y=0\\ x\le0\end{cases}$$

So the solution set is the closed II quadrant, i.e. the set defined by the condition $[x\le0\land y\ge 0]$.


问:
Alternative definition of refinement
Are the following interchangable ways of defining the definition of refinement of a partition on a set?

$P$ is a refinement of $S$ if and only if both sets are partitions of a set $U$ and every $P$-cell is included in a $S$-cell. (this implies that for every $S$-cell, there is a $P$-cell included in it, while the converse is false).

$P$ is a refinement of $S$ if and only if both sets are partitions and every $S$-cell is a union of $P$-cells.

I do not think they are actually interchangable.
2 does not imply 1, in fact we could have {a,b,c,d} as a partition and {x,y} as a partition in which x is the union of a with b and y is the union of a with c, but d is not included in any member of {x,y}.

Would an equivalent way to express 2 be:

$P$ is a refinement of $S$ if and only if both sets are partitions of a set $U$ and every $S$-cell includes a $P$-cell?

From 2 to 3 it is clear, but how can I prove that from 3 it follows that every $S$-cell is the union of some $P$-cells?

答:
If $x=a\cup b$ and $y=a\cup c$, then $\{x,y\}$ isn’t a partition of anything unless $a=\varnothing$, in which case $\{a,b,c,d\}$ isn’t a partition of anything: the members of a partition must be pairwise disjoint and non-empty.

Assuming that $P$ and $S$ are both partitions of the same set $U$, $(1)$ and $(2)$ are equivalent. That $(2)$ implies $(1)$ is clear, so assume $(1)$, i.e., that for each $p\in P$ there is an $s_p\in S$ such that $p\subseteq s_p$.

First observe that $s_p$ is unique: if $s,s'\in S$, $p\subseteq s$, and $p\subseteq s'$, then $s\cap s'\ne\varnothing$, so $s=s'$. Next, for each $s\in S$ there is at least one $p\in P$ such that $s=s_p$: if not, $U=\bigcup P\subseteq U\setminus s\subsetneqq U$, which is absurd. For each $s\in S$ let $$P_s=\{p\in P:s_p=s\}\,;$$ we just saw that $P_s\ne\varnothing$, and I claim that in fact $s=\bigcup P_s$, so that every member of $S$ is indeed a union of members of $P$.

To see this, let $u\in s$. There is some $p\in P$ such that $u\in p$, so $$s_p\cap s\supseteq p\cap s\ne\varnothing\,,$$ and therefore $s=s_p$. But then $u\in p\in P_s$, so $u\in\bigcup P_s$. And $u$ was an arbitrary element of $s$, so $\bigcup P_s\subseteq s\subseteq\bigcup P_s$, i.e., $s=\bigcup P_s$.

Title: Alternative definition of refinement


问:
Why is $2f(x)$ in the same function?
My SAT Math essentials book says:
$$f(x) = 8x - 2$$
becomes
$$f(3) = 8(3) - 2f(3) = 24 - 2 = 22$$
So, when $x = 3$, the value of the function is $22$.

Why did the author insert $2f(3)$ if that line equals $44$?

答: You are correct that $2f(3)=44$.  I think your book author meant “$f(3)=8(3)-2\color{red}; f(3)=24-2$,” which is correct.

Title: Why is $2f(x)$ in the same function?


问:
Cross section is a regular hexagon.Is it a cube?
One of the cross sections in a rectangular box is a regular hexagon.Prove that the box is a cube

I tried to prove that certain lengths were equal by showing that certain triangles are congruent but that was of no use.Could someone help me with this one?

答:
Since the lines on opposite faces are parallel, and every diagonal is the same length, we have $b^2+c^2 = d^2+e^2 = f^2+a^2$.

For the hexagon to be in a plane, we need
$det \begin{pmatrix}b&0&-a\\c&d&0\\0&e&f\end{pmatrix} = 0$,
hence $bdf = ace$

Finally, we need the angles of the hexagon to be the same, so looking at the scalar products, you get $cd=ef=ab$.
This makes $5$ equations. So proving $a=b=c=d=e=f$ might be possible from there :

$(b-a)((a+b)bd+abc) = bd(b^2-a^2)+(ab)bc-(ab)ac = bd(b^2-a^2)+(cd)bc-(ef)ac \\= bd(b^2+c^2-a^2) - (ace)f = bdf^2 - bdf^2 = 0$

Since $(a+b)bd+abc > 0$, we get $a=b$.

Similarly, $c=d$ and $e=f$, and $cd = ef = ab$ becomes $c^2 = e^2 = a^2$, hence $a=b=c=d=e=f$


问:
If $x,y\in (0,1]$ and $xy$ is rational, are $x$ and $y$ algebraic
I'm working on the problem already specified in the title.
As the "easy" counterexamples - $x= 0$ or $y = 0$ - are excluded by the definition of the problem, I'm not sure how to approach this problem.

答:
No. Let $a=\frac{4}{5}$ say. Then for every $x \in (0,1)$ there is a $y$ such that $x^y =a$. Now the set of such $x$ that are transcendental is uncountable and has measure 1, while the set of alegraic $x, y \in (0,1)$ has measure 0. So the set of $x \in (0,1)$ s.t. either $x$ or $y=\frac{\ln a}{\ln x}$ is algebraic has measure 0. Thus there has to be an transcendental $x$ and a transcandental $y$ such that $x^y=4/5$.

ETA this line of reasoning holds for $x × y=a$ as well; the set of such $x$ that either $x$ or $y$ is transcendental has measure 0, so almost all pairs $x,y$ s.t. $x×y=a$ are s.t. both $x$ and $y$ are transcendental.


问:
Show the convergence of $\sum_{n=0}^{\infty}\frac{1}{2}\cdot \frac{(2n)!}{n!(n+1)!}\cdot \left(\frac{1}{4}\right)^n$
As stated by the prompt, I'm looking to show the convergence of the series $$\sum_{n=0}^{\infty}\frac{1}{2}\cdot \frac{(2n)!}{n!(n+1)!}\cdot \left(\frac{1}{4}\right)^n$$
I've tried using the ratio, root, and Limit Comparison test (w/ a geometric series), and all have given me inconclusive answers.  If anyone has some insights for how I might be able to show this series converges, your help would be much appreciated.

答: In fact, if $|x|\le\tfrac14$ then $\sum_{n\ge0}\frac{(2n)!}{n!(n+1)!}x^n=\frac{1-\sqrt{1-4x}}{2x}$, so the given sum is $1$.


问:
Cauchy and Convergent Sequence
Let X = {3, −4,3, −4,3, −4, ... }.
a) Show that X is not a Cauchy sequence in R.
b) Is X a convergent sequence in R? Justify your answer.

I am solving this exercise and i want to show that the sequence is not Cauchy. I put that the two subsequences of X { 3, 3, 3} and {-4, -4, -4} converge to different limits, so the sequence is not convergent therefore it is not Cauchy.
but in part 2 they asked about the convergence so I think what I did was wrong can any body help me to find other proof
thank you

答: What you are doing is ok. In the real numbers convergent iff Cauchy. If you want to use the definitions, without using the equivalence above, then, for part a), you can assume that the sequence is Cauchy, then for $\varepsilon=1>0$, there exists $N\in\mathbb{N}$ such that, for  $n,m>N$ you have $|x_n-x_m|<1$. You get a contradiction if you take $n=N+1,\,m=N+2$, since you have $|x_{_{N+1}}-x_{_{N+2}}|=7>1$. For part b), you can use a similar argument.

Title: Cauchy and Convergent Sequence


问:
Show that $f$ is periodic from an integral property
Let $f$ be a continuous function such that for all $t\in\mathbb{R}$,
$$
\int_a^bf(x)\,\mathrm{d}x=\int_{a+t}^{b+t}f(x)\,\mathrm{d}x
$$

How can I show that $f$ has to be a periodic function? I was able to prove the statement in the other way, but in this question I end up with
$$
\int_a^b(f(x+t)-f(x))\,\mathrm{d}x=0
$$
and from that I don't know what else to do. :(

答:
Define $F(t)=\int_{a+t}^{b+t} f(x) dx = \int_{a}^{b} f(x+t) dx$. You know that $F$ is constant. So $\frac{d}{dt} F(t) = 0$.

On the other hand
\begin{align*}
\frac{d}{dt} F(t) &= \frac{d}{dt} \int_{a}^{b} f(x+t) dx \newline
&=  \int_{a}^{b} \frac{d}{dt} f(x+t) dx  \newline
&=  f(b+t)-f(a+t)  \newline
\end{align*}

Finally $f(b+t)=f(a+t)$  $\forall t\in \mathbb{R}$. You can even use $u=a+t$ so that $f((b-a)+u)=f(u)$ $\forall u\in \mathbb{R}$.