问:
Evaluating power series

How do I evaluate the power series
$$\sum_{n = 1}^{\infty}\dfrac{n}{9^{n}}$$
without using the formula for infinite geometric series?

I am interested in an approach which doesn't make use of infinite geometric series.

Any help would be highly appreciated. New approaches are most welcomed. Thanks.

答: I'm not sure about the exact value, but you can compare it to the integral, which converges since $|\frac{1}{9}|<1$, $\int_{0}^{\infty}x a^x dx$, which will be of the same order, and then derive the difference using Bernoulli polynomials.

Title: Evaluating power series


问:
What is $\lim\limits_{n→∞}(\frac{n-x}{n+x})^{n^2}$?
What is $$\lim_{n\rightarrow\infty}\left(\frac{n-x}{n+x}\right)^{n^2},$$ where $x$ is a real number. Mathematica tells me the limit is $0$ when I put an exact value for $x$ in (Mathematica is inconclusive if I don't substitute for $x$), but using $f(n)=(n-x)^{n^2}$ and $g(n)=(n+x)^{n^2}$, then $$\lim_{n\rightarrow\infty}f(n)=\lim_{n\rightarrow\infty}g(n)=\infty,$$ and by L'Hopital's rule we have
$$\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} = \lim_{n\rightarrow\infty}\frac{(s-z)^{-1+s^2} (s+z)^{1-s^2} (s+2 (s-z) \text{Log}[s-z])}{s+2 (s+z) \text{Log}[s+z]} = 1.$$

I'm not sure which to believe - Mathematica or L'Hopital.

ContentLicense: CC BY-SA 4.0

Body:
The interesting form is
$\left(\frac{n-x}{n+x}\right)^{n}$.
The $n^2$ just blows things up.

Taking the log,
$n \ln \frac{n-x}{n+x}
= n \ln \frac{1-x/n}{1+x/n}
= n \big(\ln (1-x/n)- \ln(1+x/n)\big)
$.
Using $\ln(1+z) = z-z^2/2+z^3/3-z^4/4+...$
and
$\ln(1-z) = -z-z^2/2-z^3/3-z^4/4+...$,
and writing $z$ for $x/n$,
this is
$n((-z-z^2/2-z^3/3-z^4/4+...)-(z-z^2/2+z^3/3-z^4/4+...))
= n(-2z-2z^3/3-...)
=-2n(x/n+x^3/(3n^3))
=-2x-2x^3/(3n^2)-...
$.

So $\left(\frac{n-x}{n+x}\right)^{n}
=\exp(-2x-2x^3/(3n^2)-...)
=\exp(-2x)exp( -2x^3/(3n^2)-...)
=\exp(-2x)(1-2x^3/(3n^2) + O(1/n^4))
$.

If you use the $n^2$,
this behaves like
$\exp(-2xn)$
which $\to 0$ if $x > 0$
and $\to \infty$ if $x < 0$.

ContentLicense: CC BY-SA 3.0


答:
Check with induction that the partial sum is
$$\sum_{k=1}^N \frac{k}{9^k} = \frac{9 - 9^{-N}(8N+9)}{64}$$
Evaluate the limit when $N$ goes to infinity and you get the result.

Title: Evaluating power series

Body:
How do I evaluate the power series
$$\sum_{n = 1}^{\infty}\dfrac{n}{9^{n}}$$
without using the formula for infinite geometric series?

I am interested in an approach which doesn't make use of infinite geometric series.

Any help would be highly appreciated. New approaches are most welcomed. Thanks.


问:
Percentage of the normal area - How to find the area under the normal distribution?
First this is not my homework is an example for learn and solve this little problem.

(Reference online pdf here.)

I have problems only with this section in exercise text:

The middle $60\%$ means that $30\%$ of the area will be on either side of the mean. The corresponding $z$ score is found using $\rm{area} = 0.20$. The $z$ scores are $\pm 0.84$.

How did they find the area is $0.2$ they skipped this part?

答:
I have problems only with this section in exercise text:

The middle $60\%$ means that $30\%$ of the area will be on either side of the mean. The corresponding $z$ score is found using $\rm{area} = 0.20$. The $z$ scores are $\pm 0.84$.

There is an evident typo....it is not 30% but 40% (total) that is on both tails, 20% on the left one and 20% on the right one. Obviously $20\%=0.2$

thus you will claculate the two quantiles corresponding to

$$P(Z<z)=0.2$$

that is

$$\Phi^{-1}(0.2)=-0.84$$

and

$$P(Z>z)=0.2$$

that is

$$\Phi^{-1}(0.8)=0.84$$


答: Notice that $x \frac{d}{dx} \frac{1}{1-x} = \sum_{i=0}^{\infty}ix^i $. So differenciating you get $$x \frac{d}{dx} \frac{1}{1-x} = \frac{x}{(1-x)^2}$$ Substituting $x= \frac{1}{9}$ you get required answer.

Title: Evaluating power series


问:
For any integer $a$, there is an integer $k$ such that $a^2=3k$ or $a^2=3k+1$

Let $a$ be an integer. Prove that there exists an integer $k$ such that $a^2=3k$ or $a^2=3k+1$.

Here is what I did:

I said: $a\in\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$. When I put 2 into $a^2$, I get $4$ but when I put $2=k$, $3k$, I get the number $4$ on the LHS and the number $6$ on the RHS. But $4$ is not equal to $6$. So how can I prove that $a^2=3k$ equivalent?

I've learned what the division algorithm is. However, when I tried to solve this problem, I got confused.

答:
Hint: The division algorithm tells you that, for any given integer $a$, there exist unique integers $q$ and $r$ such that $$a=3q+r\quad\text{and}\quad 0\leq r<3.$$ Note that the only integers greater than or equal to $0$ and less than $3$ are $0$, $1$, and $2$.  Thus, you know that for any given integer $a$, exactly one of three things is true:

$a=3q+0$ for some integer $q$.

$a=3q+1$ for some integer $q$.

$a=3q+2$ for some integer $q$.

In each case, expand $a^2$ and observe which case $a^2$ ends up in. You will see that the last case is impossible. Thus, for any given integer $a$, either

$a^2=3k+0$ for some integer $k$.

$a^2=3k+1$ for some integer $k$.


答:
Denote $A=\sum_{n\geq 1} n9^{-n}$, $B=\sum_{n\geq 2} n9^{-n}$, $C=\sum_{n\geq 1} 9^{-n}$, $D=\sum_{n\geq 2} 9^{-n}$. You have the closed system
$$
A=B+\frac 19,\quad 9B-A=C,\quad C=D+\frac 19,\quad 9D=C,
$$
which gives the value of $A$. Of course, as a byproduct of the computation one also gets the value of the geometric series $C$.

Title: Evaluating power series


问:
Show that two $R$-algebras are not isomorphic.

Consider the $\mathbb{R}$-algebras $A := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{a^2+b^2+c^2+d^2}\right]$ and $B := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{ad-bc}\right]$ (so we are adjoining the inverses of $a^2 + b^2 + c^2 + d^2$ and $ad-bc$, respectively).
I claim that these are not isomorphic as $\mathbb{R}$-algebras, but I'm not sure how to formalize this.

My intuition says that this is because $$a^2 + b^2 + c^2 + d^2 = 0 \iff a=b=c=d=0,$$ while $ad-bc=0$ has nonzero solutions. I tried to formalize this using evaluation homomorphisms, but to no avail. I also tried assuming they were isomorphic and obtaining a contradiction, but haven't been able to.

Does anyone have any pointers on how to approach this?

答:
Question: "Does anyone have any pointers on how to approach this?"

Answer: Your argument is correct: Define

$$A := \mathbb{R}[a,b,c,d]/(a^2+b^2+c^2+d^2), B := \mathbb{R}[a,b,c,d]/(ad-bc).$$

Formally, if there is an isomorphism of affine schemes (over $k$ - the real numbers) $\phi: T:=Spec(A) \cong S:=Spec(B)$ it would follow that $T$ and $S$ have the same $k$-rational points: The map $\phi$ would induce a 1-1 correspondence

$$\phi(k): T(k) \cong S(k)$$

and an equality of sets

$$T(k) \cong Hom_{k-alg}(A,k) \cong Hom_{k-alg}(B,k)\cong S(k),$$

but $T$ has only one $k$-rational point: $x:=(0,0,0,0)$. $S$ has a rational point $x:=(a,b,c,d)$ for any $x$ with $det(x)=0$.

Formally: If $T\cong S$ are isomorphic schemes (over $k$) it follows $h_T\cong h_S$
are isomorphic as functors (over $k$). Hence an isomorphism $\phi$ of affine schemes over $k$ does not exist.


问:
Feedback to Basis and Dimension of $A:= \{x\in\mathbb{R}^4 | x_2 + 3x_3= 0, x_1=x_2\}$
My question: Because of $\mathbb{R}^4$ I assume the vector has to have $x_1,x_2,x_3,x_4$
To calculate the basis I assume that $x_1= 1$ because $x_1=x_2 as$ the problem says and $x_4= 0$ because this was not given. I got this vector $v= (1,1,3,0)$. Are my assumptions wrong? Should $x_1=0$ because it was not given?
My question is about the value of x1. Is it correct that the value of x1 ist 1 because x1 = x2?
I just would like to know if I did something wrong.
Basis and dimension I can do on my own.
Thank you

答:
There are some mistakes, so let me point those out first before giving a solution:

I don't understand why you would assume that $x_1 = 1$, nothing here would suggest that.

"$x_4 = 0$ because it was not given". No, if there are no $x_4$'s in your equations, you can set $x_4$ to whatever you want.

Here's how you should approach it.If we can isolate the variables, it'll be easier to find out what the equations describe. $x_1=x_2$ is fine allready. We can rewrite $x_2 + 3x_3 = 0$ to $x_2 = -3x_3$. From this, we can rewrite your equations as

$x_2=x_1$ and $x_3=-\frac{1}{3}x_2 = -\frac{1}{3}x_1$

So, when you pick some value for $x_1$, the value of $x_2$ and $x_3$ are allready given. Therefore, we get

$A = \{(a,a,-\frac{1}{3}a, b) | a,b \in \mathbb{R}\}$


问:
Expression of $\cos^{-1}\left(4x^3-3x\right)$
The Original question : (translated from French)

Let $f$ be the function
$$
f\left(x\right)=\cos^{-1}\left(4x^3-3x\right)
$$

Find the definition domain of $f$.

Compute $f'\left(x\right)$. Find an expression of $f\left(x\right)$ in terms of $\cos^{-1}\left(x\right)$.

My attempt :
$f$ is defined on $\left[-1;1\right]$ and
$$
f'\left(x\right)=\frac{3-12x^2}{\sqrt{1-\left(4x^3-3x\right)^2}}
$$
To answer the question I thought i could express it as
$$
f'\left(x\right)=\frac{3-12x^2}{\sqrt{-1+3x-4x^3}\sqrt{1-3x+4x^3}}
$$
and try a kind of decomposition to find it as
$$
f'\left(x\right)=a\frac{3-12x^2}{\sqrt{-1+3x-4x^3}}+b\frac{3-12x^2}{\sqrt{1-3x+4x^3}}
$$
What am I missing ?

答:
Since the domain is $[-1,1]$, you can let $t=\cos^{-1} x \implies x=\cos t$. Why? Because $4\cos^3 t -3\cos t = \cos 3t $. $$f(t) = \cos^{-1} (\cos 3t) $$

The range of $\cos^{-1} x$ is $[0,\pi] $, and so you need to be careful and divide $f(t)$ into suitable pieces as follows:

$$f(t) =\begin{cases} 3t, & 0\le t\le \pi/3 \\ 2\pi-3t, & \pi/3 \le t \le 2\pi/3 \\ 3t -2\pi, & 2\pi/3 \le t \le \pi \end{cases} $$

Now just replace $t$ while noting that $0\le \cos^{-1} x \le \pi/3 \iff \frac 12 \le x\le 1$ and $\pi/3 \le \cos^{-1} x \le 2\pi/3 \iff -\frac 12 \le x \le \frac 12$ and $2\pi/3 \le \cos^{-1} x \le \pi \iff -1\le x\le -\frac 12$.

$$f(x)  =\begin{cases} 3\cos^{-1} x, & \frac 12 \le x \le 1 \\ 2\pi-3\cos^{-1} x, & -\frac 12  \le x \le \frac 12 \\ 3\cos^{-1} x -2\pi, & -1 \le x \le  -\frac 12 \end{cases} $$


问:
Cyclic vector space
In class we defined what it means that there is a creating element $v$ of a vector space, such that for an endomophism $A$ on $V$ we have:
${\rm span}(v,Av,...,A^{n-1}v)=V$.

Also we said that if the minimal polynomial of the endomorphism is $p^k$ where $k$ is a natural number and $\deg(p)=\dim(V)$ then there exists such an element. Unfortunately, we did not make an example how to find such an element for a given endomorphism. Is there any chance, that someone could give me a hint, how one would try to find such a $v \in V$?

答:
This time the question is more clearly posted:

Claim: With the notation used in the OP, if $\,A^k=0\;$ but $\;A^m\neq 0\;,\;\forall m<k\,$ , then for $\,v\in V\;s.t.\; A^{k-1}v\neq 0\;,$ , we have that $\;\{v,Av,\ldots,A^{k-1}\}\,$ is linearly independent.

Proof: Suppose $\,a_i\in\Bbb F\,$ are s.t. (with $\,A^0=I\,$) :

$$0=\sum_{i=0}^{k-1}a_iA^iv\implies 0=A^{k-1}(0)=A^{k-1}\left(\sum_{i=0}^{k-1}a_iA^iv\right)=\sum_{i=0}^{k-1}a_iA^{k+i-1}v=$$

$$=a_0A^{k-1}v+\underbrace{a_1A^kv+\ldots+a_{k-2}A^{2k-2}v+a_{k-1}A^kv}_{\text{All this is zero}}=a_0A^{k-1}v$$

and since $\,A^{k-1}v\neq 0\;$ we get $\,a_0=0\,$ . So delete the first summand from the first equality above (i.e., begin the sum from $\,i=1\,$ ) and this time apply $\;A^{k-2}\;$ and etc., getting $\,a_i=0\;\;\forall\,i\,\;\;\;\;\;\square$

Title: Cyclic vector space


问:
solve $|\sin(a)|=\cos(3a)$
$|\sin(a)|=\cos(3a)$ is an alternative version of an equation $\sqrt{1-x^2}=4x^3-3x$, where I made a substitution $x=\cos(a)$ for $x \in [-1, 1]$. Unfortunately, I have no idea how to solve trigonometric equations.

答:
Hint

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

Title: solve $|\sin(a)|=\cos(3a)$


答:
Hint:

One possibility is $$\sin a=\cos 3a \implies \cos(\frac{\pi}{2} - a) =\cos 3a$$ and the other is $$-\sin a = \cos 3a \implies \cos(\frac{\pi}{2} +a ) =\cos 3 a$$

Now, $$\cos x=\cos y \iff x\pm y = 2k\pi$$ Find all possibilities for $a$, and hence $x=\cos a$.

Title: solve $|\sin(a)|=\cos(3a)$

Body: $|\sin(a)|=\cos(3a)$ is an alternative version of an equation $\sqrt{1-x^2}=4x^3-3x$, where I made a substitution $x=\cos(a)$ for $x \in [-1, 1]$. Unfortunately, I have no idea how to solve trigonometric equations.


问:
Which Slope is larger?
If a line has $\tan(120^\circ)$ as gradient or slope and another line has $\tan(130^\circ) $as slope, which slope is larger? I mean numerically $\tan(120^\circ)=-1.7320$ and $\tan(130^\circ)=-1.1917$. Since $-1.1917>-1.7320$ so this slope seems grater to me. But the steepness of slope $\tan(120^\circ)$ is higher. So which slope is larger?

答: Without additional context or information, the first argument seems to be correct. Since $\tan(120^\circ) < \tan(130^\circ)$, the second slope is larger.

Title: Which Slope is larger?


答: The steepness of the slope is the absolute value of the gradient. $y=-10x$ looks steeper than $y=x$ even though $-10 < 1$.

Title: Which Slope is larger?

Body: If a line has $\tan(120^\circ)$ as gradient or slope and another line has $\tan(130^\circ) $as slope, which slope is larger? I mean numerically $\tan(120^\circ)=-1.7320$ and $\tan(130^\circ)=-1.1917$. Since $-1.1917>-1.7320$ so this slope seems grater to me. But the steepness of slope $\tan(120^\circ)$ is higher. So which slope is larger?


问:
Monotone Properties of Random Graphs
I am learning about the monotone properties of random graphs. I came across this question which I am unable to prove. Let $s = s(N) ∈ \{0, 1, . . . , N\}$ where ${s(N) → ∞}$ and ${N − s(N) → ∞ }$ as ${N → ∞}$.
Let
$ξ ∼ Bin(N, s/N)$. Prove that $P(ξ = s) → 0$ as $N → ∞$. Is it true for constant 's'? I want to know how to approach this type of numerical (I do have a bit of understanding of probability and measure theory but I am very new to random graph). Any help is appreciated.

Moreover, can anyone please suggest some reference or literature for the Random graph? I am following Random Graphs - Svante, Łuczak, Rucinski and Random Graphs by Béla Bollobásbook (which is generally instructed to follow) but I am finding it a bit advance for my level. What I am looking for is something that shows how probability methods are used in Random graphs and have vast examples for the same (since I am not able to get the intuition of solving various random graphs related numerical even though I can understand various solved lemmas).

Thanks in advance.

答:
You want to evaluate
$$
P(\xi=s)={N \choose s}\left(\frac{s}{N}\right)^{s}\left(1-\frac{s}{N}\right)^{N-s}
$$

Now you know
$$
{N \choose s} \leq \left(\frac{eN}{s}\right)^s,
$$
and
$$
\left(1-\frac{s}{N}\right)^{N-s} = \exp\left( (N-s) \ln{\left(1-\frac{s}{N}\right)}\right) \leq \exp\left( -(N-s) \frac{s}{N}\right),
$$
last inequality follows from $\ln(1-x) \leq -x$ for $x\in (0,1)$. Note that you will be done if $(N-s) \frac{s}{N}\to \infty$ as $N\to \infty$. Now there can be two cases:

$s/N \to 0$. Then $(1-\frac{s}{N}) {s}\to \infty$ since  $s\to \infty$.

$s/N \not\to 0$. Then $(N-s) \frac{s}{N}\to \infty$ since $N-s \to \infty$.

For the second part of your question. Generally, the two books that you referred to staple texts in this field. Another excellent reference is the book by Remco van der Hofstad, RANDOM GRAPHS AND COMPLEX NETWORKS Volume I (it is available on his webpage). You may read the past few chapters to review the basic materials. Hope this helps you to grasp the material better. Happy reading :)

Title: Monotone Properties of Random Graphs


问:
Tricky probability bound
Here's an easy-looking probability theory problem that so far has defied my attempts at solving it:

Let $X$ and $Y$ be two real independent identically distributed random variables. Prove that

$$
Pr(|X-Y| \le 2) \leq 3Pr(|X-Y| \le 1)
$$

答:
This is certainly not recreational-mathematics, is called the 123 theorem and is due to Noga Alon and Raphael Yuster. The full statement is:

For every $b> a> 0$ and for every two independent identically distributed real random variables $X$ and $Y$, $P[|X − Y | ≤ b] < (2⌈b/a ⌉ − 1)P[|X − Y | ≤ a]$. This is tight for all admissible pairs $(a, b)$.

Title: Tricky probability bound


问:
The meaning of $dx$ in an indefinite integral
This semester I'm taking integral calculus for the first time. We started with the differential (i.e. $dy=f'(x)\,dx$) and right after that with the indefinite integral. Since then, I've been trying to make sense of the $dx$ when it's part of an indefinite integral (i.e. $\int f(x) \, \boldsymbol{dx}$). I know there are already a bazillion answers regarding this question, but all of them refer to the definite integral and the ones that do touch on indefinite integrals only say things like "it's just a syntactical device to tell you the variable to differentiate with respect to or the integration variable" (Ihf, 2012, web). I don't like this answer, especially because I was taught that given two functions $f(x)$, $g(x)$ and the antiderivative of their product $\int f(x)g(x)\,dx$, if I were to assign $f(x)$ to $u$ and $g(x)\,dx$ to $dv$ in order to integrate by parts, then I would have to integrate $dv$ to find $v$. This only makes sense if $dx$ means something by itself and is not just a quirk of the notation, otherwise we would end up with something like the following: $$\int g(x)\,dx\space dx$$

After thinking about it a lot, I believe I finally found a way to make sense of the $dx$ as something that isn't purely and simply a notation device. My reasoning is as follows:

Given a function $f(x)$, let $y=f(x)$. Then $$\frac{dy}{dx}=f'(x)$$

Then from the fundamental theorem of calculus, we know that $$\int\frac{dy}{dx}\,dx=y$$

Let $dy=f'(x)\,dx$. Then the equation $$dy=\frac{dy}{dx}\,dx$$ holds and $$\int\frac{dy}{dx}\,dx=\int dy=y$$

Finally, integrate both sides of the equation $dy=f'(x)\,dx$ in order that $$\int dy=\int f'(x)\,dx\implies y=f(x)$$

From this I conclude that, by integrating a function, what we are really doing is integrating the differential of that function. This makes perfect sense to me, although I'm aware that seemingly logical things aren't necessarily logical. That's why I would appreciate it if someone could tell me whether the above is mathematically correct or pure gibberish.

PS, I had never written a mathematical proof before and I have taken no proofs courses yet, so any suggestions are welcome.

Reference: lhf. (2012, May 9). What does $dx$ mean?. Mathematics Stack Exchange. https://math.stackexchange.com/q/143262

答:
Remark

In mathematics it may be that "$\int$" and "$dx$" are considered to be brackets, inclosing the integrand:
$$
\int f(x)\;dx
$$
In physics it may be that "$\int dx$" is used as an operator, so it precedes the integrand:
$$
\int dx \;f(x)
$$


问:
Calculating the variance of the ratio of random variables
I want to calculate $\newcommand{\var}{\mathrm{var}}\var(X/Y)$. I know:
$$\var(X - Y) = \var(X) + \var(Y) - 2 \var(X) \var(Y) \mathrm{corr}(X,Y) \>,$$
What is the equivalent for $\newcommand{\var}{\mathrm{var}}\var(X/Y)$?

答:
The ratio of two random variables does not in general have a well-defined variance, even when the numerator and denominator do.  A simple example is the Cauchy distribution which is the ratio of two independent normal random variables.  As Sivaram has pointed out in the comments, the formula you have given (with the correction noted by Henry) is for the variance of the difference of two random variables.

Henry's wiki link includes a formula for approximating the ratio distribution with a normal distribution.  It seems like this can be used to create the sort of formula you are looking for.


Title: For any integer $a$, there is an integer $k$ such that $a^2=3k$ or $a^2=3k+1$

Body:
Let $a$ be an integer. Prove that there exists an integer $k$ such that $a^2=3k$ or $a^2=3k+1$.

Here is what I did:

I said: $a\in\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$. When I put 2 into $a^2$, I get $4$ but when I put $2=k$, $3k$, I get the number $4$ on the LHS and the number $6$ on the RHS. But $4$ is not equal to $6$. So how can I prove that $a^2=3k$ equivalent?

I've learned what the division algorithm is. However, when I tried to solve this problem, I got confused.

Body:
By division, $\ a = 3j\!+\!r,\ r\in \{0,\color{#c00}{\pm1}\},\,$ so $\,(3j\!+\!r)^2 = 3(3j^2\!+\!2jr)\!+\!r^2,\ r^2\in\{0,\pm1\}^2\! = \{0,1\}$

Note: we used $\ 3j\!+\!2 = 3(j\!+\!1)\!\color{#c00}{-1}$.


问:
Let $c \in \mathbb{R}$ and $X=(c, \infty)$, $Y=(-\infty,c).$ Show that $X$ is homeomorphic to $Y$.

Let $c \in \mathbb{R}$ and $X=(c, \infty)$, $Y=(-\infty,c).$ Show that $X$ is homeomorphic to $Y$.

To show that $X$ is homeomorphic to $Y$ I would need to find a bicontinuous function $f :X \to Y$?

Define $f(x)= 2c-x$. If $x \in X$, then $f(x) < c$ thus $f(x) \in Y$. Define $g(y) =2c-y$ if $y \in Y$, then $g(y) > c$ so $g(y) \in X$. So I have that $f : X \to Y$ and $g : Y \to X$ also $$g \circ f=g(f(x))=g(2c-x)=2c-(2c-x)=x \\ f\circ g=f(g(y)) = f(2c-y)=2c-(2c-y)=y$$ which implies that $g = f^{-1}$. To show that this is indeed bicontinuous I have that $$|f(x)-f(y)|=|(2c-x)-(2c-y)|=|x-y|, x,y\in X$$ which would imply that $f$ is bi-lipschitz and thus homeomorphic. Is this a valid way to show this?

答: Sure, this proof is fine. Instead of Lipschitz you could also use the order-reversing property (and the fact that these sets have the order topology).</p>


问:
How to prove a "generalization" of the "Clarkson inequality"?
So let $s\leq r$ and $s\leq2$, and $x,y\in\mathbb{R}$. I am interested in how to prove the following inequality:
\begin{equation}
\left( \lvert x+y \rvert^r + \lvert x-y \rvert^r \right)^{1/r} \leq 2^{1-1/s}\left( \lvert x \rvert^s + \lvert y \rvert^s \right)^{1/s}.
\end{equation}

In fact, the original problem is to prove that for required $r$ and $s$ ($s\leq p\leq r$ and $s\leq2$), every $f$ and $g$ in $L_p(X)$ satisfy the inequality above. I realized that since every normed value is real, the inequality for $\mathbb{R}$ may be a reasonable reduction, but then I am stuck on proving it. Thank you and I appreciate any insights!

答:
The left-hand side is increasing in $r$, see for example How do you show monotonicity of the $\ell^p$ norms?. Therefore it suffices to prove the inequality for the smallest possible value of $r$, that is for $r=s$:
$$
\left( \lvert x+y \rvert^s + \lvert x-y \rvert^s \right)^{1/s} \leq 2^{1-1/s}\left( \lvert x \rvert^s + \lvert y \rvert^s \right)^{1/s}.
$$

This Clarkson type inequality holds for $s \ge 2$, see for example:

Showing $\left|\frac{a+b}{2}\right|^p+\left|\frac{a-b}{2}\right|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p$

Three related inequalities (the first being $2(|a|^p + |b|^p) \leq |a + b|^p + |a - b|^p \leq 2^{p-1}(|a|^p + |b|^p)$)

On the other hand, the inequality does not hold in general for $0 < s < 2$, as can be seen by setting $x=1$ and $y=0$.


问:
Given a parabolic shape with maximum height $OC=8m$ and maximum width $AB=20m$. If $M$ is the middle of $OB$, then what is the height $MK$, from $M$?
Given a parabolic shape with maximum height $OC=8m$ and maximum width $AB=20m$. If $M$ is the middle of $OB$, then what is the height $MK$, from $M$?

I attempted to solve the question as follows:

$OB=10m$

Hence, $OM=5m$

$OC=8m$

I state that $O$ is point $(0,0)$ on the Cartesian coordinate system.

$\implies M(5, 0), C(0, 8), B(0, 10), A(-10, 0)$. I then attempted to find the coordinates of K, but I don't know how to. Obviously, $\triangle AKB$ is isosceles, but I couldn't see anything else further than that. Could you please explain to me how to solve this question?

答:
For the given question, you can simply solve using basic properties of parabola and without using coordinate geometry.

Draw a perpendicular to $CO$ from point $K$ and say it meets $CO$ at $H$.

We observe that axis of the parabola is along $OC$ and $C$ is the vertex of the parabola. Second, we know that the perpendicular distance from the vertex to an ordinate is directly proportional to the square of the ordinate. So we have,

$CO = k \cdot OB^2$ and $CH = k \cdot HK^2$ where $k$ is some constant.

As $ \ CH = CO - HO = CO - MK$ and $HK = OM$,

$\displaystyle \frac{CO}{CO - MK} = \frac{OB^2}{OM^2} = 4 \ $ (as $OB = 2 \cdot OM$)

Since $CO = 8, \ MK = 6$


问:
Why is do no complete graphs have genus 7?
My teacher mentioned this in class, but he rushed his explanation due to time restraints. Why is it that complete graphs ($K_n$) never have genus $7$?</p>

答:
It is a theorem of Ringel and Youngs that for all $n \geq 3$, the genus of the complete graph $K_n$ -- i.e., the minimal genus of a compact orientable surface into which $K_n$ embeds -- is $\lceil \frac{(n-3)(n-4)}{12} \rceil$.  Granting this result, what you ask about is very straightforward: the given function is weakly increasing.  For $n = 12$ it takes the value $6$.  For $n = 13$ it takes the value $8$.  Thus it never takes the value $7$ (the first of infinitely many values that it skips).

Not being a graph theorist, I confess that I don't know the proof of the Ringel-Youngs Theorem, and the paper was not freely available to me on the internet.  For all I know it may be possible to avoid this result and give an argument more specific to your precise question...but I will guess that it will probably be more fruitful to learn the proof of the general result.

Ringel, G. and Youngs, J. W. T. "Solution of the Heawood Map-Coloring Problem." Proc. Nat. Acad. Sci. USA 60, 438-445, 1968.


Title: solve $|\sin(a)|=\cos(3a)$

Body:
Hint

$1-x^2=x^2(3-4x^2)^2\iff -16x^6+24x^4-10x^2+1=0$

If we set $w=x^2$ we get $-16w^3+24w^2-10w+1=0$ which has $\frac{1}{2}$ as a solution


问:
Is there a probability measure $\mu$ on $\mathbb R^n$, with all Borel sets having a measure in $\{0,1\}$ and all singleton sets having measure 0?
My guess would be no, as I'm failing to find an example of such a measure. I tried the following to prove that there is no such measure:

Attempt: We can partition $\mathbb R^n$ into countably many subsets. By sigma-additivity, we find that only a single set will have measure 1, and all other will have measure 0.

I want to use this to prove that by partitioning $\mathbb R^n$ in a "finer and finer" way, we can eventually find smaller and smaller sets, of which only a single one can have nonzero measure, so eventually we find a singleton set that would have nonzero measure.

I'm not sure if this is the right approach to this problem, nor how it would be possible to put my idea in rigorous terms.

答:
The answer is no. The following proof readily generalizes to arbitrary dimension but becomes slightly more tedious to write down:

Proposition
Let $\mu$ be a $\sigma$-finite Borel measure on $\mathbb{R}$ which takes values in $\mathbb{N}\cup \{0\}$. Then, $\mu$ is a linear combination of Dirac-masses.

Proof: Let $I_{n,k}=[k2^{-n}, (k+1)2^{-n}]$ and notice that for every sequence of intervals $J_m:=I_{m,k_m}$ such that $J_m\supseteq J_{m+1}$, we have that $\mu(J_m)$ is decreasing and hence, eventually constant. Furthermore, every such $(J_m)_{m\in \mathbb{N}}$ defines an $x\in \mathbb{R}$ as the unique element of $\cap_{m\in \mathbb{N}} J_m$. Every $x$ is defined in this way by at most two different such sequences. As such, let $J_m^x$ denote the unique such sequence for which $x<\sup J_m^x$ eventually.

If there is an interval $[-K,K]$ containing infinitely many $x_n$ such that the corresponding $J_m^{x_n}$ all have positive measure, then by compactness, these $x_n$ have a point of condensation point $x_{\infty}$ and so, for every $\varepsilon>0$, we have $\mu([x_{\infty}-\varepsilon,x_{\infty}+\varepsilon])=\infty$ (since it contains arbitrarily many distinct $J_m^{x_n}$ that all have measure at least $1$). This contradicts the assumption of $\sigma$-finiteness.

To finish, let $(x_n)_{n\in I}$ denote the discrete set of points such that $(\mu(J^{x_n}_m))_{m\in \mathbb{N}}$ is not eventually $0$. Then, applying the uniqueness theorem for $\sigma$-finite measures, we get
$$
\mu=\sum_{n\in I} a_n \delta_{x_n},
$$
where $a_n=\lim_{m\to\infty} \mu(J^{x_n}_m)$.


答:
Suppose that such measure exists. Then taking countable cover by open sets of diameter $\leq 1$, one of those sets, say $U_1$, must have measure $1$. Cover $U_1$ by countable amount of open sets of diameter $\leq 1/2$ and take the one having measure $1$.

This process gives us a decreasing sequence of open sets $U_k$ with diameter $\leq 1/k$ and $\mu(U_k) = 1$.

Then $\mu(\bigcap_{k = 1}^\infty U_k) = \lim_{k\to \infty} \mu(U_k) = 1$ from continuity of measure, but $\bigcap_{k = 1}^\infty U_k$ has at most one element.

Title: Is there a probability measure $\mu$ on $\mathbb R^n$, with all Borel sets having a measure in $\{0,1\}$ and all singleton sets having measure 0?

Body:
My guess would be no, as I'm failing to find an example of such a measure. I tried the following to prove that there is no such measure:

Attempt: We can partition $\mathbb R^n$ into countably many subsets. By sigma-additivity, we find that only a single set will have measure 1, and all other will have measure 0.

I want to use this to prove that by partitioning $\mathbb R^n$ in a "finer and finer" way, we can eventually find smaller and smaller sets, of which only a single one can have nonzero measure, so eventually we find a singleton set that would have nonzero measure.

I'm not sure if this is the right approach to this problem, nor how it would be possible to put my idea in rigorous terms.


问:
Distance of point on ellipse from focii

Let $P( \alpha, \beta)$ be a point on the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ where foci are $F_1$ and $F_2$. Then $(PF_1 - PF_2)^2= ?$

I went for a brute for approach, the focii are located at $(\mp ae,0)$ so the distance of point from the general focus is given as:

$$ d = \sqrt{ (\alpha  \pm ae )^2 + \beta^2} = \sqrt{\alpha^2 +(ae)^2 \pm 2 \alpha a e + \beta^2}$$

Now, I use the parameterization $ \alpha = a \cos \theta $ and $ \beta = b \sin \theta$, this leads to :

$$ \sqrt{a^2 + b^2 + (ae)^2 \pm 2 a^2 \cos \theta e} \ \  \color{red} *$$

Further I use the property that $(ae)^2 = a^2 -b^2$,

$$ \sqrt{2a^2 \pm 2a^2 \cos \theta e} =a \sqrt{2} \sqrt{1 \pm e \cos \theta}$$

Substituting $ \cos \theta = \frac{\alpha}{a}$

$$ a \sqrt{2} \sqrt{ 1 \pm e \cos \theta} =  \sqrt{2a} \sqrt{a \pm e \alpha}$$

I'm not quite sure how to proceed anymore...

In the book it was given that the distance of point from focus is given $(a \pm e \alpha)$, did I make mistake? How do I correctly proceed?

答: After the parameterization, the expression becomes $$\sqrt{a^2 \cos^2 \theta +(a^2-a^2 e^2) \sin^2 \theta +(ae)^2 \pm 2ea^2 \cos \theta} \\ = \sqrt{a^2 +a^2e^2 \cos^2 \theta   \pm 2ea^2 \cos \theta   } \\ = a\sqrt{1 +(e\cos \theta)^2 \pm 2e\cos\theta} \\ = a(1\pm e\cos\theta) \\ = a \pm e\alpha$$

Title: Distance of point on ellipse from focii


问:
Choosing $a$ s.t. $\frac{a^k - 1}{a-1}$ is not a prime power

Let us suppose that we are presented with a positive integer $k$ and asked to come up with a positive integer $a$ such that $\frac{a^k - 1}{a-1}$ is not a prime power, or just to prove in an elementary way that this can be done. How do we proceed?

Of course, in practice this is essentially trivial. But is there a cannonical choice? For that matter, is there an elementary proof that shows that this is always possible?

This came up in a discussion with another Brown mathie about interesting things to talk about in an elementary number theory class I will be teaching this summer. To be more specific, we were trying to prove the infinitude of primes in arithmetic progressions of the form $1, 1+p^n, 1+2p^n, ...$ using a naive a completely elementary approach (hopefully to be replicated by my students this summer), and this side topic came up.

答:
take instead of $a$ $a^2$ then we want to prove
$$\frac{a^{2k}-1}{a^2-1}$$ is not a prime.

but $\frac{a^{2k}-1}{a^2-1}=\frac{(a^k-1)(a^k+1)}{a^2-1}$

So if we suppose $k>3$, then choose and $a$ such that $a^k-1>a^2-1$ and $a^k+1>a^2-1$

Hence there is at least one prime factor from each $a^k-1$ and $a^k+1$ in the expression $$\frac{a^{2k}-1}{a^2-1}$$


问:
Existence of successive Picard approximations, $y'=3y +1, \enspace y(0)=2$
I'm trying to show that the successive approximations for $$y'=3y +1, \enspace y(0)=2$$

exist for all real $x$. I think the general formula for approximation $\phi_k$ is

$$\phi_k=2+7x+\frac{21}{2}x^2+\frac{21}{2}x^3+\cdots+\frac{k-1}{k}\bigg(\frac{21}{2}x^k\biggr)$$

which I have inferred  by computing some of the approximations, so it is not a real solution. My book says to use induction to show the approximations exist. I'm confused about what this means. What would be my inductive statement in this case?

答:
I think I came up with a way to prove this. There is no need to derive a general formula for $\phi_k$. Proceed to proof by induction. The base case $$\phi_1 = 2$$

clearly exists ( is defined ) for all $x$. Now suppose that $$\phi_k = 2 + \int_0^{x}3(\phi_{k-1}(t))+1dt$$

exists for all $x$. By Picard ( successive approximation method ) we have that $$\phi_{k+1} = 2+ \int_0^{x} 3\bigg(2+\int_0^{x}3(\phi_{k-1}(t))+1dt\bigg)+1 dt$$

exists for all $x$. Why? Because it will be a polynomial of degree $k+1$, and as such is continuous everywhere and defined for all $x$. Thus all approximations exists by induction? Does this make sense?


问:
If $A$ and $B$ are subgroups of $G$ such that $[[A, B], B] = 1$, then $[A, B]$ is abelian.
I am stuck on the following question:

If $A$ and $B$ are subgroups of $G$ such that $[[A, B], B] = 1$, then $[A, B]$ is abelian, where $[X, Y]$ denotes the commutator subgroup of $X$ and $Y$ as usual.

I have seen that $[A, B]$ is in the intersection of all the centralizer of each $b\in B$. And I am trying to show that $B$ is a normal subgroup of $G$, then it follows that $[A, B] \subseteq Z(B)$, where $Z(B)$ denotes for the centre of $B$. Then the result follows.

But I have no idea how to show that $B$ is normal in $G$ or maybe I am completely on the wrong way.

Anyone can help?

答:
Lemma Let $A,B$ be subgroups a of group $G$, then both $A$ and $B$ normalize $[A,B]$.

Proof By symmetry we only prove this for $A$: let $a,a_1 \in A, b_1 \in B$, then
$$a^{-1}[a_1,b_1]a=(a_1a)^{-1}b_1^{-1}(a_1a)b_1.b_1^{-1}a^{-1}a_1^{-1}a_1b_1a=[a_1a,b_1][b_1,a] \in [A,B].$$

Proposition Let $A,B$ subgroups of a group $G$ and let $A$ or $B$ centralize $[A,B]$, then $[A,B]$ is abelian.

Proof Again, by symmetry, we only need to prove this if $[[A,B],B]=1$. Let $a_1,a_2 \in A$ and $b_1,b_2 \in B$, it is enough to show that the commutators $[a_1,b_1]$ and $[a_2,b_2]$ commute. Observe that by the Lemma $[a_2,b_2]^{a_1^{-1}} \in [A,B]$:
$$[a_1,b_1][a_2,b_2]=a_1^{-1}b_1^{-1}a_1b_1[a_2,b_2]\underset{b_1 \text { commutes with } [a_2,b_2]}=a_1^{-1}b_1^{-1}a_1[a_2,b_2]b_1=a_1^{-1}b_1^{-1}[a_2,b_2]^{a_1^{-1}}a_1b_1 =\underset{\text{ using the Lemma and } b_1^{-1} \text{ centralizes }[A,B]}=\\ a_1^{-1}[a_2,b_2]^{a_1^{-1}}b_1^{-1}a_1b_1=[a_2,b_2][a_1,b_1].$$

Note (added March 24th 2021) There is still another elegant way to show that $[A,B]$ is abelian, and that is by an appeal to the Three Subgroups Lemma:

Lemma Let $X,Y,Z$ be subgroups of a group $G$. If $[X,[Y,Z]]=1$ and $[Z,[X,Y]]=1$ then $[Y,[Z,X]]=1$.

Note that the subgroups/letters are shifted cyclically. For a proof and a slight generalization (replacing $1$ by containment in a normal subgroup $N$) see for example here. Proofs usually apply the so-called Hall-Witt Identity.

Now let's show that if $[A,B]$ is centralized by $B$, $[A,B]$ must be abelian. By the first Lemma, $A$ normalizes $[A,B]$, so $[A,[A,B]] \subseteq [A,B]$, implying $[B,[[A,B],A]]=1$. Further, $[A,[B,[A,B]]]=[A,1]=1$. The Three Subgroups Lemma now yields, $[[A,B],[A,B]]=1$, that is $[A,B]$ is abelian.


问:
Drawbacks of gradient descent
In the method of gradient descent, which in each iteration we reduce the problem to a single variable optimization problem in order to find $\alpha$ (called the learning rate?) using the update rule:

$$\mathbf{x}_{t+1} = \mathbf{x}_t-\alpha\nabla f(\mathbf{x}_t)$$

Drawbacks:

computational: we must evaluate the gradient, of at least estimate it using finite differences.

order of convergence, due to the fact that $$\nabla f (\mathbf{x}_{t+1})^T\nabla f(\mathbf{x}_t) = 0$$ we will obtain a "zig-zag" pattern which will increase the number of iterations.

When will the gradient descent do a $90$ degrees turn? On the one hand we see that

$$\nabla f (\mathbf{x}_{t+1})^T\nabla f(\mathbf{x}_t) = 0$$

On the other hand we know that the gradient is perpendicular to the level set?

I see that in some case we start with a random starting point and set a learning rate $\alpha$ and in other cases, we just start with staring point, and find $\alpha$ using line search methods, is this the same optimization method?

答:
In general these are considered different algorithms. For gradient based algorithms, typically, the differences arise in conditions under which line searches are terminated.

For example, if an algorithm performs an exact line search on a smooth objective function $f$, then at each iteration, the following sub-problem is solved

$$\begin{equation}\label{1} \min_{\alpha} \quad \phi(\alpha) := f(x_k - \alpha g_k) \tag{1}.\end{equation}$$

Here $g_k = \nabla f(x_k)$. If $\alpha^*$ is the solution of \ref{1}, then  the next iterate is $x_{k+1} = x_k - \alpha^*g_k $ and it follows that $\phi^{\prime}(\alpha^*) = g_{k+1}^T g_k = 0$, but this is not true if the sub-problem is not solved exactly.

As solving the subproblem is, in general, not an easy thing to do, it is solved approximately using some variant of bracketing or interpolation schemes. More practical approaches are to use line search termination conditions such as Armijo or Wolfe or Goldstein condition, which take a step whenever sufficient decrease is guaranteed. These conditions are primarily designed so that the algorithm terminates.

Title: Drawbacks of gradient descent


问:
Compute $Df^{-1}(2,5)$ given $f(u,v)=(uv,u^2+v^2)$ using the Inverse Function Theorem (IFT)
I need help figuring out where my solution below goes wrong, because the question was phrased as if the derivative would exist.
$$f(u,v)=(uv,u^2+v^2)$$
We prove $D {f}^{-1}(2,5)$ does not exist. Clearly $ {f}$ is $\mathcal{C}^1$ on $\mathbb{R}^2$. Note that $ {f}(1,2)=(2,5)= {f}(2,1)$. We consider the Jacobian of $ {f}$ at $(1,2)$ and at $(2,1)$:
$$\Delta_{ {f}}(1,2)=\begin{vmatrix} 2 & 1 \\ 2 & 4 \end{vmatrix}=2(4-1)=6.$$
$$\Delta_{ {f}}(2,1)=\begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix}=2(1-4)=-6.$$
Since $\Delta_{ {f}}(1,2)\neq 0$ and $\Delta_{ {f}}(2,1)\neq 0$, by the IFT there exists an open $V$ and $W$ respectively in $\mathbb{R}^2$ such that $ {f} ^{-1} $ is $\mathcal{C}^1$ on $ {f}(V)$ and $ {f}(W)$. We know $ {f}$ is $\mathcal{C}^1$ on $\mathbb{R}^2$ and $V$, $W$ are open in $\mathbb{R}^2$, thus $ {f}(V)$, $ {f}(W)$ are open in $\mathbb{R}^2$. Note that $ {f}( {f} ^{-1}(2,5))= {f}(1,2)= {f}(2,1)$. Since $(2,5)= {f}(1,2)\in {f}(V)$ and $(2,5)= {f}(2,1)\in {f}(W)$ by the IFT we have:
$$D {f} ^{-1}(2,5)  =  [D {f}( {f} ^{-1}(2,5))] ^{-1}  =  [D {f}(1,2)] ^{-1} $$
and
$$D {f} ^{-1}(2,5)  =  [D {f}( {f} ^{-1}(2,5))] ^{-1}  =  [D {f}(2,1)] ^{-1}.$$
Thus $[D {f}(1,2)] ^{-1} =[D {f}(2,1)] ^{-1} $. We know $[D {f}(u,v)] ^{-1} =\begin{bmatrix} v & u \\ 2u & 2v \end{bmatrix} ^{-1} $ so for $|u|\neq|v|$ we have:
$$\displaystyle[D {f}(u,v)] ^{-1} =\frac{1}{2v^2-2u^2}\cdot\begin{bmatrix} 2v & -u \\ -2u & v \end{bmatrix}.$$ Since $1\neq 2$ we have:
$$D {f} ^{-1}(2,5)=[D {f}(1,2)] ^{-1} =\frac{1}{6}\cdot\begin{bmatrix}
    4 & -1 \\ -2 & 2
\end{bmatrix}$$
$$D {f} ^{-1}(2,5)=[D {f}(2,1)] ^{-1} =-\frac{1}{6}\cdot\begin{bmatrix}
    2 & -2 \\ -4 & 1
\end{bmatrix}$$
However, it is clear that $\displaystyle \frac{1}{6}\cdot\begin{bmatrix}
    4 & -1 \\ -2 & 2
\end{bmatrix}\neq -\frac{1}{6}\cdot\begin{bmatrix}
    2 & -2 \\ -4 & 1
\end{bmatrix}$. Thus $D {f} ^{-1}(2,5)\neq D {f} ^{-1}(2,5)$ a contradiction, so $D {f} ^{-1}(2,5)$ does not exist.

What I really don’t understand is that this contradiction implies that $Df^{-1}(2,5)$ does not exist, thus $f^{-1}$ is not differentiable. This would contradict the IFT since one of the implications of the theorem is that $f^{-1}$ must be differentiable on the open intervals $f(V)$/$f(W)$.

答:
Notice that you're applying the IVT twice here. Once at the point $(1,2)$ and another times around $(2,1).$

By applying the IFT around the point $(1,2)$ you get an open set $W_1$ containing $(1,2)$ and a second open set $V_1 = f(W_1)$ containing $f(1,2) = (2,5)$. The IFT tells you that the restriction of $f$ to $W_1$ is invertible and that it's inverse $f_{W_1}^{-1}$ is also differentiable.

By applying this theorem again at the point $(2,1)$ you will find open sets $W_2$ and $V_2 = f(W_2)$ and the IVF tells you that these the restriction of $f$ to $W_2$ , $f_{W_2}$ is invertible and that its inverse $f_{W_2}$ and that it's inverse $f_{W_2}^{-1}$ is also differentiable.

We are therefore dealing with two different functions here : $f_{W_1}$ and $f_{W_2}$ !

In your solution you do not take this into account and are giving these functions the same name $f$. The apparent contradiction that you found isn't one because you're calling two different matrices $D{f_{W_1}^{-1}}(2,5)$ and $Df_{W_2}^{-1}(2,5)$ by the same name $Df^{-1}(2,5)$.

It might be surprising to you that
$$ D{f_{W_1}^{-1}}(2,5) \neq Df_{W_2}^{-1}(2,5)$$

but when you think about it $f_{W_1}$ and $f_{W_2}$ are just different functions so one shouldn't always expect the above quantities to be same. That doesn't mean that they aren't related however !

I suspect that we have $f_{W_1} = f_{W_2} \circ \pi_{W_1}$ where $\pi_{W_1}$ is  the smooth invertible function

$$ \pi_{W_1} : W_1 \rightarrow W_2 : x,y \mapsto \pi_{W_1}(x,y) = (y,x)$$

We would then have $f_{W_1}^{-1} = \pi_{W_1}^{-1}\circ f_{W_2}^{-1}$ and using the chain rule you will find that

$$ Df_{W_1}^{-1}(2,5) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \times Df_{W_2}^{-1}(2,5).$$


问:
How many paths are there in graph $K_n$?
We know In $K_4$ there is $(4*3)+(4*3*2)+(4*3*2*1)$ paths in graph.

How many paths are there in graph $K_n$?

I want a simple formula...

Body:
As you given for $K_4$, the answer is simply $n!(1+\frac{1}{2!}+\dots+\frac{1}{(n-2)!})= [n!e]-n-1$ for $n\ge 2$. The sum is computed as follows
$$
\begin{align}
&n!e-n!(1+\frac{1}{2!}+\dots+\frac{1}{(n-2)!}+\frac{1}{(n-1)!}+\frac{1}{n!})\\
=&\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\dots\\
<&\frac{1}{n+1}e\\
<&1.
\end{align}
$$


问:
Question about criterion for the existence of holomorphic logarithm
Theorem 6.2 of Stein's Complex Analysis says that: if $f$ is a nowhere vanishing holomorphic function on a simply connected region $\Omega$, then f has a holomorphic logarithm on $\Omega$.

A question here on stackexchange also states the same theorem:
Existence of logarithm of holomorphic function

However, if I directly use the principle branch of logarithm (That is, for $z=re^{i\theta}$ we define the logarithm $log(z)=log(r)+i\theta$, with |$\theta|<\pi$) to define a function $g$ so that $g(z)=log(f(z))$ provided that the image of $f$ on $\Omega$ is contained in $\mathbb{C} \backslash (-\infty,0]$, then we would have $g'(z)=\frac{f'(z)}{f(z)}$ by chain rule and we know that $g$ is holomorphic. It seems that $\Omega$ need not be simply connected or even connected.

Perhaps in the process above some step is wrong but I didn't find any. Could anyone help?

答:
A comment that got too long:

Note that $\mathbb{C} \backslash (-\infty,0]$ is a simply connected set avoiding zero so $\log w$ is defined and analytic there, hence one can do a composition $\log \circ f$ which is analytic; same proof applies if the image of $f$ is contained in a simply connected set $U$ not containing zero.

However, such a condition on the image of $f$ is very restrictive as for example it means that the image of $f$ cannot contain any full circle (or for that matter any Jordan curve) that encircles the origin

The result quoted in the original post is much more general as one needs only that the domain of definition of $f$ is simply connected but the image can be any (open) set that avoids zero (eg full punctured plane) and $f$ can be very complicated. As we saw in the beginning applying this theorem to $g(w)=w$ and the domain  $\mathbb{C} \backslash (-\infty,0]$ or any other simply connected open $0 \notin U$ containing the image of $f$ immediately gives us by composition that $\log f$ exists in those cases


问:
If f(n) = o(g(n)), is h(f(n)) = o(h(g(n))?
Definition: $o(g(n)) = \{f(n) | \forall c>0, \exists N\in \mathbb{N}:0\leq f(n)<c\cdot g(n)\}$ as in CLRS book

Is the following statement true?

If $f(n) = o(g(n))$
and $f,g,h:\mathbb{N}\rightarrow\mathbb{N}$ are increasing sequences that approach $\infty$,
then
$h(f(n)) = o(h(g(n))$

答:
No, in fact both of the statements are false:

If $f(n) = o(g(n))$ then $h(f(n)) = o(h(g(n))$

If $f(n) = O(g(n))$ then $h(f(n)) = O(h(g(n))$

(Assume $f,g,h:\mathbb{N}\rightarrow\mathbb{N}$ are increasing sequences that approach $\infty$)

Example for $h(f(n)) \neq o(h(g(n))$:

$f(n) = 2^{n/2}$

$g(n) = 2^n$

$h(n) = lg(n)$

If the statement was true, then we get $n/2 = o(n)$, which is false.

Example for $h(f(n)) \neq O(h(g(n))$:

$f(n) = 2n$

$g(n) = n$

$h(n) = e^n$

If the statement was true, then $e^{2n} = O(e^n)$, which is false.


问:
Given two natural numbers $m, n$ so that $m> n.$ Prove that there exist any real numbers $x$ so that $2\sin nx\cos mx\geq 1$

Given two natural numbers $m, n$ so that $m> n.$ Prove that there exist any real numbers $x$ so that (unsolved in a case..)
$$2\sin nx\cos mx\geq 1$$
Source: StackMath/@RiverLi_ Prove: $(\forall m, n\in\Bbb N_{>0})(\exists x\in\Bbb R)$ s. t. $2\sin n x \cos m x \ge 1$ .

I couldn't find a concrete value of $x,$ but I succeeded at proving the problem by using Intermediate value theorem (My friend has taught me a solving plan..). Let $x$ on the interval $I= \left [ -\frac{5}{6}\pi, \frac{\pi}{6} \right ],$ we call $f\left ( x \right ):=\frac{1}{2\sin x}, g\left ( x \right ):=\cos\alpha x, \alpha:=\frac{m}{n}> 1,$ I devide my problem into 2 cases as followings

When $1< \alpha< \frac{6}{5},$ we have $f, g$ continuous both and $f\left ( -\frac{3}{4}\pi \right )\geq g\left ( -\frac{3}{4}\pi \right ),$ by Intermediate value theorem $\exists x\in I$ so that $f\left ( x \right )= g\left ( x \right )$

When $\alpha\geq\frac{6}{5},$ we have $f, g$ continuous and $\exists x\in I$ so $g\left ( x \right )= -1\Rightarrow\exists x\in I$ so that $f\left ( x \right )= g\left ( x \right )$

Conclusion.
$$\forall\alpha= \frac{m}{n}> 1, \exists x\in I: f\left ( x \right )= g\left ( x \right )\Leftrightarrow\forall m, n, \exists x\in I: 2\sin x\cos\alpha x= 1$$
$$\Rightarrow\forall m, n, \exists x\in \mathbb{R}: 2\sin nx\cos mx\geq 1$$
I'm waiting for @RiverLi's complete solution and want to see such a value of $x$ for my proof, thanks

答:
Extended comment ($a=\alpha$).

Your proof is hard to follow.
First, you have not specified a point $x\in I$, s.t. $f(x)<g(x)$. Second one cannot apply IVT on the interval $I=\left[-\frac{5\pi}6, \frac{\pi}{6}\right]$ because the function $f(x)$ is not continuous on this interval. Third the claim that there exists a point with $g(x)=-1$ in the chosen interval is not proven.

Below I will try to remedy these shortcomings. First we define the interval to be $I=\left[-\frac{3\pi}4, -\frac{\pi}{3a}\right]$. Clearly both functions are continuous on this interval and
$f\left(-\frac{\pi}{3a}\right)<g\left(-\frac{\pi}{3a}\right)=\frac12.$ So we need only to show that there is a point $x\in I$, s.t. $f(x)>g(x)$.

Similarly as in you proof we let the point be $x=-\frac{3\pi}4$ for $1<a<\frac43$. For larger $a$ we can choose any point $x$ such that $\cos(a x)=-1$, $-\frac{3\pi}4\le x<-\frac{\pi}4$ (this choice is somewhat artificial and done only due to some symmetry).
To find $x$ we need to resolve the following inequality for $k\in\mathbb Z$:
$$
-\frac{3\pi}4\le -\frac\pi a(2k+1)<-\frac{\pi}4\iff \frac a4<2k+1\le\frac{3a}4.\tag1
$$
The essence here is: for $a>\frac43$ the inequality $(1)$ always has at least one solution. Observe that $k=0$ is the only solution of $(1)$ for $a<4$ and cancels to be a solution for $a>4$. This however does not harm  as we can take $k=1$ for $4<a<\frac{20}3$. For higher values of $a$ multiple solutions for $k$ are possible (though smaller $k$ repeatedly cancel to be the solutions).


问:
Analytic Solution for 1D Wave Equation

Im trying to solve the 1D Wave Equation on $x \in \mathbb{R}, t > 0$: $$u_{tt} = c^2u_{xx}, \hspace{5mm} u(x,0) = \exp(-100(x-0.5)^2), \hspace{5mm} u_t(x,0) = 0$$ with $c = 1$ and a periodic boundary condition $$u(x,t) = u(x+1,t)$$ From what I know the general solution to this type of wave equation is $$u(x,t) = \frac{1}{2}(g(x+ct) + g(x-ct))$$ where $g(x) = u(x,0)$, which in this case is $$u(x,t) = \frac{1}{2} (\exp(-100(x-t-0.5)^2) + \exp(-100(x+t-0.5)^2)$$   however this does not satisfy the above wave equation.

Does anyone know what the correct closed form analytical solution is of this wave equation?

答:
Because you have periodic boundary conditions the solution is different from what you stated.
With this starting condition you cannot have a solution from the basic fact that it is not periodic.

In order to get the general solution you will need to do separation of variables and account for the periodic boundary conditions like follows:

Let's assume $u(x,t)=X(x)T(t)$ you can see that the equation becomes:
$$X_{xx}=-k^2X, T_{tt}=-k^2T$$ (The minus and the square is just for simplicity and because k can be complex doesn't eliminate potential solutions)

The general solutions for these equations are:
$$X(x)=Ae^{ikx}+Be^{-ikx}$$
From periodic boundary conditions you can get the value of k and see it is $2\pi$.

Title: Analytic Solution for 1D Wave Equation


问:
Number of possible solutions to $2 \le |x-1||y+3| \le 5$, where $x$ and $y$ are negative integers

If $$2 \le |x-1||y+3| \le 5$$ and both $x$ and $y$ are negative integers, find the number of possible combinations of $x$ and $y$ .

Below is my solution approach :-

As $x$ is a negative integer, hence $|x-1|$ in the $2 \le |x-1||y+3| \le 5$ will be come $-(x-1)$ or $(1-x)$ and the main equation will transform into $2 \le (1-x)|y+3| \le 5$.

$1st$ case when $y+3 \ge 0 \Rightarrow y \ge -3 \Rightarrow y \in \{{-3,-2,-1}\} $ as $y$ is a negative integer :

For $y=-3$ we can see that there is no valid solution for $x$ as $|y+3|$ part will become $0$, hence this case is invalid.

For $y=-2$ we get the solution set for $x$ to be $x \in \{{-4,-3,-2,-1}\} $  and total number of solutions possible in this case is $4$.

For $y=-1$ we get the solution set for $x$ to be $x \in \{{-1}\} $ and total number of solutions possible in this case is $1$.

So for this $1st$ case when $y+3 \ge 0$ we have in total 5 solutions.

$2nd$ case when $y+3 \lt 0 \Rightarrow y \lt -3$ and in this case $y$ will have infinite values and I am not able to proceed from here.

The answer for the total number of solutions provided is $10$ and you can see that I've been able to find out the $5$ solutions in my $1st$ case. Can someone please guide or help me about how to proceed in the 2nd case?

Thanks in advance !

答:
The possible values of the product are $2,3,4,5$.

Hence the possible factorizations,

$$1\cdot2,2\cdot1,1\cdot3,3\cdot1,1\cdot4,2\cdot2,4\cdot1,1\cdot5,5\cdot1.$$

But $|x-1|\ge2$, which leaves

$$2\cdot1,3\cdot1,2\cdot2,4\cdot1,5\cdot1,$$

while $|y+3|$ can achieve $1$ and $2$ in two ways.

$$10.$$


问:
Difficulty in solving the following integral
How to solve the following definite integral:

$$I= \int_{\pi/4}^{3\pi/4} e^{-e^{\frac{1}{x}}}dx$$

I have tried solving this using Feynman's trick but to me, it turns out there is no obvious way to parametrize this function. I am very weak in contour integration and hence couldn't apply it here. I couldn't think of any obvious substitution as well.

The only thing I can see is that $1/x$ is the derivative of $log$ $x$ which is the inverse of exponential. So probably this integral requires exploitation of that but I can't see how.

Also, I prefer answers that do not go beyond undergraduate knowledge in mathematics (if possible) since I encountered it in an undergraduate exam.

Body:
Try this: Let $t= e^{\frac1x}$
$$\int_{\pi/4}^{3\pi/4} e^{-e^{\frac{1}{x}}}dx
= - \int_{e^{\frac4\pi}}^{e^{\frac4{3\pi}}}\frac{e^{-t}}{t\ln^2t}dt= J(0)
$$
where
$J(a)= - \int_{e^{\frac4\pi}}^{e^{\frac4{3\pi}}}\frac{t^{a-1} e^{-t}}{\ln^2t}dt
$
$$J’’(a) =  - \int_{e^{\frac4\pi}}^{e^{\frac4{3\pi}}}{t^{a-1} e^{-t}}dt = \Gamma(a, e^{\frac4{3\pi}}) -\Gamma(a, e^{\frac4\pi})
$$


问:
In the figure below, the side length of of the smaller square is 4, and the area of the red region is 1. What is the area of the larger square?
I already did some progress and labeled some similar segments.

答: The red triangle has area $1$ but it has base and height $x$ so the area will be $x^2/2=1$ so $x^2=2$ and finally $x=\sqrt 2$. By the Pythagorean theorem we also have that $y^2=x^2+4^2=4^2+\sqrt 2^2=18$. But $y^2$ is just the area of the larger square since it's the side lenght squared. So the area will be $18$


问:
Exercise in Tarski's Introduction to Logic
In Chapter 6 Exercise 9 (p.145) of Introduction to Logic, Tarski asks the reader to prove that the following statements, taken as axioms, are independent:

Theorem III. For any elements $x$, $y$, and $z$ of the set $S$, if $x\cong y$ and $x\cong z$, then $y\cong z$.

Theorem IV. For any elements $x$, $y$, and $z$ of the set $S$, if $x\cong y$ and $y\cong z$, then $z\cong x$.

Theorem V. For any elements $x$, $y$, $z$, and $t$ of the set $S$, if $x\cong y$, $y\cong z$, and $z\cong t$, then $x\cong t$.

No other axioms appear to be assumed in the question, and Tarski seems to be using the $\cong $  symbol here to designate any relation whatsoever, not necessary an equivalence relation (for an example of this usage, see pp.121-124).

My problem: I have not yet found a model where (III) and (IV) hold while (V) fails. In fact, I seem to be able to prove that (III) and (IV) imply (V). Here is my argument (Edit: thanks to @DanielV for providing a much simpler argument, reproduced here):

Suppose $x\cong y\cong z\cong t$. Then $z\cong x$ follows from (IV). By (III), $z\cong x$ and $z \cong t$ give us $x\cong t$, as needed. $\square$

My question: What am I missing? Assuming I am wrong, I would appreciate learning the error in my argument and, ideally, seeing an example of a model in which (III) and (IV) hold but (V) fails.

答:
IMO, the problem is stated diffrently

Show that the three following "axioms" are independent: Axiom I and Theorems I and II of Sect.37, that are:

$\text {Ax.I } x \cong x \text { (Reflex)}$

$\text {Th.I } \text { if } y \cong z, \text { then } z \cong y \text { (Symm)}$

$\text {Th.II } \text { if } x \cong y \text { and } y \cong z, \text { then } x \cong z \text { (Trans)}$


问:
Graded version of Quillen and Suslin's theorem (graded projectives of graded-free and finite rank modules are graded-free)
I'm trying to solve the following exercice about graded $R$-modules, when $R=\mathbb{R}[x_1,\ldots,x_n]$ is a polynomial ring :

Let $M$ be a graded-free $R$-module of finite rank. Show that any graded summand $N$ of $M$ is graded-free.

Note that everything should be understood in the graded sense, eg graded-free means free with a homogeneous basis.

I haven't got far: let ${m_k}$ be a homogeneous basis of $M$, and $p\colon M\to N$ the projection map (graded homogeneous, as $N$ is a graded summand). Then $\{p(m_k)\}$ is a homogeneous generating set of $N$, and we can extract a minimal generating set $\{p(m_{k_i})\}$. We must then show that if $\sum_i \lambda_{k_i} p(m_{k_i})=0$, $\lambda_{k_i}=0$ for all $i$. I guess I should use that $R$ is a polynomial ring, but I don't know where.

Any hint?

答:
Thanks to @lhl73's comment, I found a reference which shows this result using Nakayama's lemma. I showed it in a difference way again using this lemma, which I describe here in case it helps anyone (any confirmation that this is indeed correct is welcome).

Polynomial rings have the special feature that their lowest grading space is a field $k$. For $R=k[x_1,\ldots,x_n]$, any $R$-module $M$ is in a particular a $k$-module, that is a vector space, and we can often use linear algebra to solve questions about $R$-module. Nakayama's lemma tells us how. We denote $R_+ = \oplus_{i>0} R_i$ and $[m]$ the projection of $m\in M$ in $M/R_+ M$.

(one version of the) Graded Nakayama's lemma. Let $M$ be a finitely generated $R$-module, and $m_1,\ldots,m_r\in M$ homogeneous elements such that $[m_1],\ldots,[m_r]$ is a basis of the $k$-vector space $M/R_+ M$. Then $m_1,\ldots,m_r$ is a minimal generating set of $M$.

Let $M=N\oplus P$ and $m_1,\ldots,m_k$ such a minimal generating set (it always exists). As $[m_1],\ldots,[m_r]$ is a basis of $N/R_+ N\oplus P/R_+ P$, we can modify it into the union of a basis of $N/R_+ N$ and a basis $P/R_+ P$. Such a linear change can directly be applied to the elements $m_1,\ldots,m_k$, so that we now assume we have elements $n_1,\ldots,n_s$ and $p_1,\ldots,p_t$ defining a minimal generating set (ie a basis as $M$ is free) such that $[n_1],\ldots,[n_s]\in N/R_+ N$ and $[p_1],\ldots,[p_t]\in P/R_+ P$.

Let $\pi_N$ and $\pi_P$ be the projections. $\{\pi_N(n_1),\ldots,\pi_N(n_s),\pi_P(p_1),\ldots,\pi_P(p_t)\}$ is still a basis of $M$ by Nakayama's lemma, as the values of the projections onto $M/R_+ M$ are left unchanged. This shows that $\{\pi_N(n_1),\ldots,\pi_N(n_s)\}$ is a basis of $N$, and hence $N$ is free.


问:
Convex quadrilateral geometry and circumradius
ABCD is a convex quadrilateral in which AC and BD meet at P. Given PA =1, PB = 2 , PC = 6 and PD = 3 . Let O be the circumcenter of triangle PBC. If OA is perpendicular to AD, find the circumradius of triangle PBC.

I have tried using similarity of the four triangles formed by the two diagonals by utilizing the dimensions given. However I am not able to proceed further.

答:
Let circumradius of $\triangle PBC$ be $R$. Since $OAD$ is right triangle,
$$AD^2=OD^2-OA^2=(OD^2-R^2)-(OA^2-R^2)$$
$$=\text{Pow}_{(PBC)}(D)-\text{Pow}_{(PBC)}(A)$$
$$=DP\cdot DB-AP\cdot AC=15-7=8$$

But $PD^2=9$ and $PA^2=1$. Therefore $AD^2=PD^2-PA^2 $ implies $\triangle PAD$ is right with $\angle PAD=90^{\circ}$. This means $P$ lies on $OA$.

Hence $PC$ is the diameter and
$$R=PC/2=3$$


问:
A negative binomial distribution for two random variables?
In a sequence of Bernoulli trials with success chance $p$, the probability that we will see $k$ failures before the $r$-th success is given by the negative binomial distribution (NBD) with pmf:
$$ Pr(k) = \binom{k+r-1}{r-1} p^r (1-p)^k $$
But what if we had two sequences of simultaneous Bernoulli trials and we wanted to know the probability of how many trials we need, before we see at least $r_1$ successes in the first sequence and $r_2$ successes in the second, with success chances $p_1$ and $p_2$, respectively?

For simplicity, consider the case where $r_1 = r_2 = 1$. The NBD pmf simplifies to:
$$ Pr(k) = p (1-p)^k $$
Let's distinguish two independent random variables $X$ and $Y$ with their respective pmf's:
$$ Pr_x(k_x) = p_x (1-p_x)^{k_x} \quad \text{and} \quad Pr_y(k_y) = p_y(1-p_y)^{k_y}$$

If we perform a sequence of trials for both $X$ and $Y$ simultaneously, and it takes $n$ trials before we see at least one success in each sequence, then at least one sequence must have had its first success exactly after $n$ trials (or $n-1$ failures). Suppose that this was the $X$-sequence. Then the $Y$-sequence could also have had its first success after $n-1$ failures. The probability of this happening is:
$$ Pr_x(n-1) \cdot Pr_y(n-1) $$
But the $Y$-sequence could also have seen its first success before the $n$-th trial. The probability of that happening is:
$$ Pr_x(n-1) \cdot Pr_y(< n-1) = Pr_x(n-1) \cdot \Bigg(\sum_{j = 0}^{n-2} Pr_y(j)\Bigg) $$
Lastly, there is the case where the $Y$-sequence has its first success after exactly $n-1$ failures, while the $X$-sequence had its first success before the $n$-th trial. The probability of this occurring is:
$$ Pr_x(< n-1) \cdot Pr_y(n-1) = \Bigg(\sum_{j = 0}^{n-2} Pr_x(j)\Bigg) \cdot Pr_y(n-1) $$
The summations in the two previous expressions are geometric. As long as we assume that $p_x, p_y \neq 0$, we can simplify them. Let $(q_x, q_y) := (1-p_x, 1-p_y)$ and by adding all three probabilities together we find a nice, symmetric expression:
$$ Pr(n) = p_x q_x^{n-1} p_y q_y^{n-1} + p_x q_x^{n-1} (1 - q_y^{n-1}) + p_y q_y^{n-1} (1 - q_x^{n-1}) $$
I've verified this expression with a few experiments in Python. Is the constructive argument I've given above enough to prove that the pmf is correct?

So, what do we do in the more general case where we're looking for more than one success? I've already struggled quite a bit with this simple case. And more importantly, is this sort of distribution well-known and if so, what is it called?

答:
An alternative approach to your calculation for $r_1 = r_2 = 1$ is to say the probability of both being successful at least once by the $n$th attempt is $(1-q_x^{n})(1-q_y^{n})$ and of both being successful at least once by the $n-1$th attempt is $(1-q_x^{n-1})(1-q_y^{n-1})$ so the probability the second is initially successful after $n$ attempts or $n-1$ failures is the difference: $$(1-q_x^{n})(1-q_y^{n}) - (1-q_x^{n-1})(1-q_y^{n-1})$$

If you multiply this out and rearrange than you get your result.

For more general $r_1,r_2$ you could produce a similar expression involving regularised incomplete beta functions or something like

$$\left(1-\sum_{i=0}^{r_1-1}{n \choose i}p_x^{i}q_x^{n-i}\right)\left(1-\sum_{i=0}^{r_2-1}{n \choose i}p_y^{i}q_y^{n-i}\right) \\-\left(1-\sum_{i=0}^{r_1-1}{n-1 \choose i}p_x^{i}q_x^{n-1-i}\right)\left(1-\sum_{i=0}^{r_2-1}{n-1 \choose i}p_y^{i}q_y^{n-1-i}\right)$$


问:
ideals of ring of Gaussian integers modulo $n$
In the ring of integers modulo $n$, $\mathbb{Z}_n$, we have all the ideals of $\mathbb{Z}_n$ are principal and they are $\langle d\rangle$, where $d$ is a divisor of $n$ and prime ideals are $\langle p\rangle$ where $p$ is a prime divisor of $n$.

What about the ring of Gaussian integers modulo $n$, $\mathbb{Z}_n[i]$, In this ring, whether all the ideals are principal and which are the ideals of $\mathbb{Z}_n[i]$ and which of them are principal?

Since these rings are not integral domains in general, I cannot use results about PID or UFD.

答:
Your ring $\mathbb{Z}_n[i]$ can be seen as the quotient $$\Bbb Z_n [i] \cong \Bbb Z[i] / (n).$$ One of the isomorphism theorems, which can be found on every algebra book, says that the ideals of a quotient $R/A$ are in bijection with the ideals of $R$ that contain $A$, and the bijection is given by $$B \mapsto B/A = \{b+A : b \in B\}.$$Because $\Bbb Z[i]$ is a PID, every ideal that contains $(n)$ is of the form $(m)$, where $m \mid n$. If $$n=p_1^{k_1} \ldots p_r^{k_r}$$is the decomposition of $n$ as product of primes (in $\Bbb Z [i]$), then any ideal containing $(n)$ is of the form $(p_1^{s_1} \ldots p_r^{s_r})$, where $s_j \leq k_j$ and, in particular, there are $(k_1+1) \ldots (k_r+1)$ different ideals, which, in turn, correspond to different ideals of $\mathbb{Z}_n[i]$.\
Note that there is nothing particular of $\Bbb Z [i]$ apart from being a PID that we are using here. The same method proves how to find all the ideals in the quotient of a PID by one of its ideals, as you would do with $\Bbb Z _n$, but now taking into account that you need to factorise in $\Bbb Z[i]$ so, for example, $5$ is not prime because $5=(2+i)(2-i)$.

As it has been said in the comments, every ideal of $\mathbb{Z}_n[i]$ is principal. This is a general result. If you look back to the correspondence of ideals in a quotient, if $B$ is generated by $b$, then $B/A$ is generated by $b+A$.


问:
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.

Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
\lim\limits_{n \to \infty} \frac{\ln\left(\cos \frac{1}{n} \right)}{\frac{1}{n^2}} &= -\frac{1}{2} \cdot \lim\limits_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\cos \frac{1}{n}} \\
&= -\frac{1}{2}.
\end{align*}
Hence, the limit is equal to $-\frac{1}{2} \ne 0$. Since the $p-$ series is convergent with $p=2$, by the limit comparison test, the series is convergent.

Am I true?

答:
Alternative way using Taylor series:
$$\ln\left(\cos \frac{1}{n}\right) = \ln\left(1-O\left(\frac{1}{n^2}\right) \right) = O\left(\frac{1}{n^2}\right)$$


问:
Proof of $ind(\underline X) = 0 \Rightarrow Ind(\underline X) = 0$ if $\underline X$ is $T_1$ with a countable basis.
In H. Herrlich's Topologie I book there is a proof of the fact that if $\underline X$ is $T_1$ and has a countable basis, then it holds that $\operatorname{ind}(\underline X) = 0 \Rightarrow \operatorname{Ind}(\underline X) = 0$.

Here $\underline X$ is said to have $\operatorname{ind}(\underline X) = 0$ if it has a basis $\mathscr{B}$ consisting of clopen sets.

Whereas $\operatorname{Ind}(\underline X) = 0$ if for any neighborhood $U_A$ of some closed set $A$ there is a clopen set $B$ in $\underline X$, such that $A \subset B \subset U_A$.

The author starts by defining:

$$C = \{ B \in  \mathscr{B} \ | \ B \subset U  \}$$
$$D = \{ B \in  \mathscr{B} \ | \ B \subset X \setminus A \}$$

Then proceeds to state that $C$ and $D$ can be written as:

$$C = \{ C_n \ | \ n \in N\}$$
$$D = \{ D_n \ | \ n \in N\}$$

Then also defines:

$$E_n = C_n \setminus \bigcup\{D_m \ | \ m \le n\}$$
$$F_n = D_n \setminus \bigcup\{C_m \ | \ m \lt n\}$$
$$E = \bigcup \{ E_n \ | \ n \in N\}$$
$$F = \bigcup \{ F_n \ | \ n \in N\}$$

Saying that $E$ and $F$ are open sets in $\underline X$ with $F = X \setminus E$.

He then concludes with the fact that $E$ is therefore a clopen set with $A \subset E \subset U$ and thus it holds that $\operatorname{Ind}(\underline X) = 0$. $\blacksquare$

I have a difficulty understanding how the author rewrites $C$ and $D$ as:

$$C = \{ C_n \ | \ n \in N\}$$
$$D = \{ D_n \ | \ n \in N\}$$

such that he can later conclude $F = X \setminus E$. While I get that they can be represented as countable sets and thus as some sequences of sets, I can't understand how one goes to the conclusion from that. Especially what $E_n$ and $F_n$, $E$, $F$ are supposed to represent. Seems to me that it's somehow about the area formed by $U \setminus A$.

Does anyone have an idea and could explain the steps in more detail?

答:
The rewriting is a pure formality for notational reasons: $\mathcal{B}$ is countable (given) so $\mathcal{C} = \{B \in \mathcal{B}: B \subseteq U\}$ is also a countable family of clopen sets (being a subfamily of $\mathcal{B}$) so we fix some enumeration of it and write it as $\{C_n \mid n \in \Bbb N\}$. The same applies to $\mathcal{D} = \{B \in \mathcal{B}\mid B \subseteq X\setminus A\}$ which we can write as $\{D_n\mid n \in \Bbb N\}$. Having these indices makes it easy to define

$$E_n = C_n \setminus \bigcup \{D_m \mid m \le n\}$$

and $$F_n = D_n \setminus \bigcup \{C_m \mid m < n\}$$

which is a standard way (in other proofs this also occurs) to "disjointify" a family of sets.

If we just took the union of the $C_n$ as given, that union would be just $U$ because of the fact that $\mathcal{B}$ is a base for $X$, so $U$ as an open set is exactly a union of members of $\mathcal{B}$ and we take all such $C_n$. Likewise, the union of all $D_n$ is just $X\setminus A$, which is also open. So these unions would intersect exactly in the gap $U\setminus A$.

A nice thing about the definition of the $E_n$ and $F_n$ is their (almost) symmetry: and the fact that we can use that all finite unions of clopen sets are clopen and their difference is also clopen, so all $E_n$ and $F_n$ are still clopen, like the $C_n$ and $D_n$.

Let's check that $E \cap F = \emptyset$:

Suppose $x \in E \cap F$. Define $n_0 = \min \{n\mid x \in E_n\}$ and
$m_0 = \min \{m \mid x \in F_m\}$. So $x \in E_{n_0}$, so $x \in C_{n_0}$ and $x \notin \bigcup_{m \le n_0} D_m$ which implies $m_0 > n_0$. On the other hand we have $x \in F_{m_0}$, so $x \in D_{m_0}$ and $x \notin \bigcup_{m < m_0} C_m$ which in turn implies $n_0 \ge m_0$. We cannot have both $m_0 > n_0$ and $n_0 \ge m_0$ so this cannot happen.

Also, $X = E \cup F$: if $x \in A$, $x \in U$ so $x \in C_m$ for some $m$ and $x$ is not in any $D_n$ as these are all disjoint from $A$, so $x \in E_m \subseteq E$ is immediate. Similarly we get that $x \notin U$ implies $x \in X\setminus A$ so $x \in D_m$ for some $m$ and in no $C_n$ (as these sit inside $U$) and hence $x \in F_m \subseteq F$ for such $m$. The interesting case is when $x \in U\setminus A$ when we can again define $n_0=\min \{n\mid x \in C_n\}$ and $m_0=\min \{m\mid x\in D_m\}$. Now, If $m_0 > n_0$ we have that $x \in C_{n_0}$ but $x \notin D_m$ for any $m \le n_0$ (or that would contradict minimality of $m_0$) and so $x \in E_{n_0} \subseteq E$. If however $m_0 \le n_0$, we have that $x \in D_{m_0}$ but $x \notin C_m$ for any $m < n_0$ (or we'd contradict minimality of $n_0$), so $x \in F_{m_0} \subseteq F$.

So in all cases $E \cup F = X$ and so $E,F$ are clopen and it's easy to see that $A \subseteq E \subseteq U$. So $\operatorname{Ind}(X) = 0$.


问:
Van der Waerden Theorem - Deducing Ackermann function Upper bound
According to the book "Ramsey Theory on the Integers, Landman", the original proof of Van der Waerden's Theorem yields the upper bound ack(n) for w(k;2).

I didn't quite understand the proof that book gave but I read this one and it was quite clear: http://www.sfu.ca/~vjungic/RamseyNotes/sec_proof_vdW.html

Is there a way to deduce this ackermann function upperbound (or a similar one) from the proof in this link?

答:
It's more that there's a bound which grows roughly like the Ackermann function, with some other things happening that are less relevant. (Although I did see the actually-literally-Ackermann bound in Landman and Robertson's book, it's not clear to me if that's meant to be taken literally.)

From the prove you give, let's define $F(\ell, k, r)$ to be the least $M$ such that if $[1,M]$ is $k$-colored, there is either

a monochromatic $\ell$-term arithmetic progression, or

$r$ color-focused $(\ell-1)$-term arithmetic progressions.

The proof gives us three inequalities:

$F(\ell,k,1) \le 2W(\ell-1,k)$.

$F(\ell,k,r) \le 2 F(\ell, k, r-1) \cdot W(\ell-1, k^{2 F(\ell,k,r-1)})$.

$W(\ell,k) \le 2 F(\ell,k,k)$.

Define two related functions $w(\ell,k)$ and $f(\ell,k,r)$ by:

$f(\ell,k,1) = w(\ell-1,k)$.

$f(\ell,k,r) = w(\ell-1, f(\ell,k,r-1))$.

$w(\ell,k) = f(\ell,k,k)$.

These are a sort of "lower bound on the upper bound": we have stripped away the less-relevant parts of the recursions in the upper bounds on $W$ and $F$. In one sense, $w$ and $f$ will end up much smaller than $W$ and $F$: where $F(\ell,k,r)$ has a $k^{2 F(\ell,k,r-1)}$ in its recursion, $f(\ell,k,r)$ just has $f(\ell,k,r-1)$. But when we get to Ackermann-sized bounds, the difference between these two things is insignificant, so $w$ and $f$ still capture the essence of the growth rate of $W$ and $F$.

So where does the Ackermann function come in? Well, the recurrence for $f$ and $w$ unpacks into
$$
   w(\ell,k) = w(\ell-1, w(\ell-1, w(\ell-1, \dots w(\ell-1,k)\dots)))
$$
with $k$ applications of $w$, and the Ackermann function satisfies $A(m,n) = A(m-1,A(m,n-1))$ and $A(m,0) = A(m-1,1)$ from which we get
$$
   A(m,n) = A(m-1, A(m-1, A(m-1, \dots A(m-1,1)\dots)))
$$
with $n$ applications of $A$. These aren't quite the same, and may not have the same base cases (actually, I haven't said what the base case for $w$ is). But they will still follow the same pattern that $A(m,\bullet)$ is a nested application of $A(m-1,\bullet)$ and $w(\ell,\bullet)$ is a nested application of $w(\ell-1,\bullet)$. If at one "level" (one value of $m$ or of $\ell$ respectively) the function grows like addition, then at the next it will grow like multiplication, at the next like exponentiation, at the next like power towers, and so on through the Knuth up-arrows.


问:
If a family has two children, what is the probability that at least one is a boy?
If a family has two children, what is the probability that at least one is a boy, assuming there are only two sexes?

One way can be : The sample space is $\{BB, BG, GB, GG\}$ so that $P = \frac{1}{4}$.

This might also be a way : The sample space is $\{0 \ B, 1 \ B, 2 \ B\}$ so that $P = \frac{1}{3}$.

Are both models correct? Is there a fallacy? I usually see the answers to be $\frac{1}{4}$ so I am confused.

Edit : As I have been pointed out, I did mean $\frac{3}{4}$ and $\frac{2}{3}$.

答:
It's $\frac{3}{4}$.

The mistake in your second example

{0B, 1B, 2B}

is that these 3 possibilities are not equally probable.

I'm assuming that there's an equal probability of a child being a boy or a girl. With that in mind, if a family has 2 children then we have 4 possible outcomes each with equal probability:

GG

GB

BG

BB

Each of these outcomes has a probability of $\frac{1}{4}$. Now we can just count how many of them correspond to the outcome we're interested in, namely that at least one is a boy. The answer is clearly that 3 of them correspond to that, so the probability is $\frac{3}{4}$.


问:
show $\lim_{x\rightarrow\infty} x^n\exp(-x^2)=0$
I want to show for all $n\in\mathbb N$
$$\lim_{x\rightarrow\infty}\frac{x^{n}}{\exp(x^2)}=0$$

I am pretty sure that I have to use L'Hospital. I've tried induction:

$n=1$: $$\lim_{x\rightarrow\infty}\frac x{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac1{2x\exp(x^2)}=0$$

And for $n\rightarrow n+1$:
$$\lim_{x\rightarrow\infty}\frac{x^{n+1}}{\exp(x^2)}=\lim_{x\rightarrow\infty}\frac{(n+1)x^n}{2x\exp{x^2}}$$

And now I am stuck. The term $2x$ really annoys my for my induction hypothesis.

Any hints?

答:
From where you left it:

$$\lim_{x\to\infty}\frac{(n+1)\color{red}{x^n}}{2\color{red} xe^{x^2}}=\frac{n+1}2\lim_{x\to\infty}\frac{x^{n-1}}{e^{x^2}}\stackrel{\text{Inductive Hyp.}}=\frac{n+1}2\cdot 0=0$$


问:
Asymptotic complexity of $f(n) = f(\sqrt{n}) + n$
In this attached question, it is asked to find to time complexity of f(n). We can simplify this equation to $f(n) = f(\sqrt{n}) + n$ which can be further modified as $s(m) = s(m/2) + 2^m$ by taking $ n = 2^m $ and $s(m) = f(2^m)$. How can I find the complexity of $f(n)$ from here?

答:
By repeated application of the recurrence on powers $n=2^{2^m}$,
$$n+n^{1/2}+n^{1/4}+\cdots4+f(2)\sim n$$ because there are $\log_2(\log_2(n))$ terms and $\sqrt n\log_2(\log_2(n))=o(n)$.

Hence

$$f(n)=\Theta(n).$$


问:
Neighbourhoods of a point with respect to the intersection of topologies
Let $X$ be a set and $\{\mathcal{T}_i \;;\;i\in I\}$ a family of topologies on $X$. Then $\mathcal{T}:=\bigcap_i\mathcal{T}_i$ is again a topology on $X$.

Do the neighbourhoods of a point $x\in X$ with respect to $\mathcal{T}$ admit a simple description in terms of the neighbourhoods with respect to the $\mathcal{T}_i$? Clearly we have
$$\mathcal{N}_{(X,\mathcal{T})}(x)\subseteq \bigcap_i \mathcal{N}_{(X,\mathcal{T}_i)}(x)\,,$$
but I don‘t see why the converse should hold.

答: Given a set $X$ and two $x\in A \subsetneq B\subsetneq X$. Take two topologies $\tau_1$ and $\tau_2$ that are generate only by $A$ and $B$ respectively. The intersection of the two topologies is the trivial one $\{X,\emptyset\}$ so the only neighborhood of $x$ is $X$. But $\bigcap_i \mathcal{N}_{(X,\tau_i)}(x)$ is all the sets that have $B$ as a subset. So clearly the converse does not hold since $B$ is in the intersection but being different than $X$ it can't be a neighborhood of $x$ in the interseciton topology.


问:
How to prove this inequality $ x + \frac{1}{x} \geq 2 $
I was asked to prove that:

$$x + \frac{1}{x}\geqslant 2$$

for all values of $ x > 0 $

I tried substituting random numbers into $x$ and I did get the answer greater than $2$. But I have a feeling that this is an unprofessional way of proving this. So how do I prove this inequality?

答: Let $y= \frac{1}{x}$ then $xy= 1,$ we have ${xy}'= y+ x{y}'= 0$ with $x+ y$ reaches its minimum value when $1+ {y}'= 0.$ Therefore, we have $x= y= 1$ so that $x+ y= 2.$


问:
Little mistake in proof mean value theorem inequality?
I have to proof the following inequality: $\cos x>1-\frac{x^2}{2}$ for all $0<x\leq \frac{\pi}{2}$ given the following function: $g(x)=1-\frac{x^2}{2}-\cos x$.

My proof: Note that $g'(x)=\sin x-x$. It follows that for every $x>0$ there exists a $c\in ]0,x[$ such that
\begin{equation}
\frac{g(x)-g(0)}{x-0} =g'(c)=\sin c-c.
\end{equation}
Now, noting that $g(0)=0$ and $\sin(c)<1 \implies \sin(c)-c<1$ gives us:
\begin{equation}
\frac{g(x)}{x} =\frac{1-\frac{x^2}{2}-\cos x}{x}<1.
\end{equation}
But now, the inequality that follows is $\cos x>1-\frac{x^2}{2}-x$ and not the desired one. Did i make a little mistake somewhere?

答: Use the inequality $\sin x<x$ for all $0<x\leq \pi/2$ and all your troubles are gone.


问:
For any integer $a$, there is an integer $k$ such that $a^2=3k$ or $a^2=3k+1$

Let $a$ be an integer. Prove that there exists an integer $k$ such that $a^2=3k$ or $a^2=3k+1$.

Here is what I did:

I said: $a\in\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$. When I put 2 into $a^2$, I get $4$ but when I put $2=k$, $3k$, I get the number $4$ on the LHS and the number $6$ on the RHS. But $4$ is not equal to $6$. So how can I prove that $a^2=3k$ equivalent?

I've learned what the division algorithm is. However, when I tried to solve this problem, I got confused.

答:
So, if you divide $a$ by 3, there are only three possible remainders: 0, 1, or 2. So $a$ can be expressed in one of the forms:

$a = 3m + 0$,

$a = 3m + 1$, or

$a = 3m + 2$,

where $m$ is some integer that can be found using the Euclidean Algorithm.

If we just accept for the moment that we know $m$, we can try to see the form of $a^2$ for each of these cases:

$a^2 = (3m)^2 = 9m^2$, which if we consider your two cases, this falls under $a^2 = 3k$ if we set $k = 3m^2$.

$a^2 = (3m+1)^2 = 9m^2 + 6m + 1$, which can be expanded as $3(3m^2 + 2m) + 1$, and if we set $k = 3m^2 + 2m$, this falls under the $a^2 = 3k + 1$ case.

$a^2 = (3m+2)^2 = 9m^2 + 12m + 4$, which can be expanded as $3(3m^2 + 6m + 1) + 1$, and if we set $k = (3m^2 + 6m + 1)$, this falls under the $a^2 = 3k + 1$ case.

So you can see that your two cases cover ALL the possible forms of $a^2$, and we're done.


问:
How to provide a counter example to sequence limit question?
Suppose $(a_n)^∞_{a=1}$is a convergent sequence and $a_n \in [a, \infty)$ for all n.  Is it necessarily true that the limit of  $(a_n)^∞_{a=1}$  lies in
$(a,∞)$ if $a_n \in (a, \infty)$ for all $n$? Prove or provide a counterexample.

So if the limit doesn't lie in $(a,∞)$, then the limit must lie in $(-∞, a]$? I guessing it would be $a$?  I don't understand how can all the terms lie in a certain interval but the limit somehow is in the other interval?

答:
$$(\forall n) \: a_n \geq a \Rightarrow \lim_{n \rightarrow \infty} a_n \geq a$$
and,
$$(\forall n) \: a_n > a \Rightarrow \lim_{n \rightarrow \infty} a_n \geq a$$

An example where $(\forall n) \: a_n > a$ and $\lim_{n \rightarrow \infty} a_n = a$ is the sequence $(\frac{1}{n})$. Clearly $\frac{1}{n} > 0$ but $\frac{1}{n} \rightarrow 0$.

"I don't understand how can all the terms lie in a certain interval but the limit somehow is in the other interval?" ---- Limits of sequences from a closed set will always lie in that closed set, but limits of sequences from an open set will lie in the closure of that set.


问:
norm of the derivative for multivariable mapping
Let $m,n,d\in \mathbb{N}^*$. We define the following $C^1$ mapping: $f\colon\mathbb{R}^n \times\mathbb{R}^m \rightarrow\mathbb{R}^d$. Do we have the following :

$$ \|Df(\cdot,\cdot)\|_{\mathbb{R}^{(n+m)\times d}} = \|D_xf(\cdot,\cdot) \|_{\mathbb{R}^{n\times d}} + \|D_y f(\cdot,\cdot)\|_{\mathbb{R}^{m\times d}},$$
or we only have the inequality
$$\|Df(\cdot,\cdot)\|_{\mathbb{R}^{(n+m)\times d}} \leq\|D_xf(\cdot,\cdot) \|_{\mathbb{R}^{n\times d}} + \|D_y f(\cdot,\cdot) \|_{\mathbb{R}^{m\times d}},$$

答: Equality holds for the $\ell^1$ norm on matrix entries, but is not generally true with other norms.


答:
For every $n\geqslant 1$, $\cos \dfrac{1}{n}$ gets closer and closer to $1$.
That is,

$$\cos \dfrac{1}{n}\approx 1-\dfrac{1}{2n^2}+\dfrac{1}{24n^4}-\dfrac{1}{720n^6}$$ and for large enough $n$, the larger terms go to zero, making the approximation more accurate; as $n$ approaches $\infty$, our polynomial, and the cosine, go to $1$. Thus in the sum

$$\sum_{n\geqslant 1}\ln\left(\cos\dfrac{1}{n}\right)$$ the terms

$$\ln\left(\cos\dfrac{1}{n}\right)\longrightarrow 0,$$ since $\ln 1 = 0$. The approximate value of $$\ln\cos \dfrac{1}{n}\approx -\dfrac{1}{2n^2} - \dfrac{1}{12n^4} - \dfrac{1}{45n^6} - \dfrac{17 }{2520n^8}$$ and the terms go to zero for sufficiently large $n$. Therefore your series converges and is (for fun) approximately
$$-\left[\frac{1}{2}\zeta(2)+\dfrac{1}{12}\zeta (4)+\dfrac{1}{45}\zeta(6)+\dfrac{17}{2520}\zeta(8)\right]$$ though this agrees only to two decimal places with the correct figure $-0.945369$.

Body:
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.

Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
\lim\limits_{n \to \infty} \frac{\ln\left(\cos \frac{1}{n} \right)}{\frac{1}{n^2}} &= -\frac{1}{2} \cdot \lim\limits_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\cos \frac{1}{n}} \\
&= -\frac{1}{2}.
\end{align*}
Hence, the limit is equal to $-\frac{1}{2} \ne 0$. Since the $p-$ series is convergent with $p=2$, by the limit comparison test, the series is convergent.

Am I true?


问:
Expression of $\cos^{-1}\left(4x^3-3x\right)$
The Original question : (translated from French)

Let $f$ be the function
$$
f\left(x\right)=\cos^{-1}\left(4x^3-3x\right)
$$

Find the definition domain of $f$.

Compute $f'\left(x\right)$. Find an expression of $f\left(x\right)$ in terms of $\cos^{-1}\left(x\right)$.

My attempt :
$f$ is defined on $\left[-1;1\right]$ and
$$
f'\left(x\right)=\frac{3-12x^2}{\sqrt{1-\left(4x^3-3x\right)^2}}
$$
To answer the question I thought i could express it as
$$
f'\left(x\right)=\frac{3-12x^2}{\sqrt{-1+3x-4x^3}\sqrt{1-3x+4x^3}}
$$
and try a kind of decomposition to find it as
$$
f'\left(x\right)=a\frac{3-12x^2}{\sqrt{-1+3x-4x^3}}+b\frac{3-12x^2}{\sqrt{1-3x+4x^3}}
$$
What am I missing ?

答:
The problem with your approach is that $f$ is not differentiable at all points of $[-1,1]$.

Since domain is $[-1,1]$, let $x=\cos \theta$, where $\theta \in [0,\pi]$ so that $\cos $ function becomes a bijection and hence $\cos^{-1}x=\theta$. Note that $0\le 3\theta \le 3\pi$.

Now using trig. identities, it follows that:

$f(x)=\cos^{-1} (\cos 3\theta)$

Case 1: $0\le 3\theta \le \pi\implies 0\le \theta\le \pi/3$

$f(x)=3\theta =3 \cos^{-1}x$ for $\frac 12\le x \le 1$

Case 2: $\pi\lt 3\theta \le 2\pi \implies \pi/3\lt \theta \le 2\pi/3$

$\cos 3\theta=\cos (2\pi-(2\pi-3\theta))=\cos(-3\theta+2\pi)\implies f(x)=\cos^{-1}\cos 3\theta=2\pi-3\theta=2\pi-3\cos^{-1}x$ for $-\frac 12\le x\lt \frac 12$.

Case 3: $2\pi\lt 3\theta \le 3\pi\implies 2\pi/3\lt \theta \le\pi$

$f(x)=\cos^{-1}\cos 3\theta=\cos^{-1}\cos (3\theta-2\pi)=3\theta -2\pi=3\cos^{-1}x-2\pi$ for $-1\le x\lt -1/2$.

Can you find the derivative now?


问:
Finding a clever solution to a game of chance
This question may smell like my older question A probability game, but this time my intentions are different.

The above problem came from my friend. Or so I thought, it really came from me incorrectly remembering his problem. The problem he actually gave me was:

Suppose you start with 1 dollar and flip a weighted coin. With probability 3/4 (call this heads) you win a dollar and with probability 1/4 (call this tails) you lose a dollar. What is the probability you eventually run out of money? (Suppose the house has as many dollar bills as needed).

I know how to solve this problem in the following way:
First observe the number of tails must equal the number of heads plus one. If a string of Hs and Ts satisfy this property, call it Good. Now we want to count the number of Good strings that don't have Good proper prefixes. For a string of length 2n-1 this is just the nth Catalan number. Now we just take the appropriate summation and we arrive at the probability we eventually run out of money which turns out to be 1/3.

Now when my friend told me this problem, he said he was able to apply a clever trick and solve it almost instantly. The thing is, he doesn't remember how he did it. He says he vaguely remembers using a recurrence relation and then telescoping a sum. (Which sounds very plausible)

So can you come up with a short and sweet solution?

On a side note, I am aware my title is poor at best. If you can think of a better one, please let me know.

答:
Here is a more formal way to derive the answer: Define $q_n$ to be the probability that you eventually go broke when you start with $n$ dollars. Then $q_0 = 1$, $\lim_{n \to \infty} q_n = 0$, and $q_n = \frac{1}{4} q_{n-1} + \frac{3}{4} q_{n+1}$.

The last equation defines a recursive sequence with constant coefficients, whose general solution is
$q_n = c_1 + c_2 3^{- n}$ (the factors $1$ and $\frac{1}{3}$ come from solving the quadratic equation of Moron's answer, which is the fundamental equation of the recursion). To satisfy the other two conditions on $q_n$, we see that $c_1 = 0$ and $c_2 = 1$ i.e.
$q_n = 3^{-n}$.

In particular, the probability to go broke starting with one dollar is $q_1 = \frac{1}{3}$.

ContentLicense: CC BY-SA 2.5


答:
I'll try stripping down some of the terminology and keep it as simple as possible. Any integer can be written in the form of $3l$, $3l+1$ or $3l+2$ for some $k\in\mathbb{Z}$. (This follows from the division algorithm.) We don't know which of the three our $a$ happens to be so we need to check what happens when we square it for each case.

Case 1: $a = 3l$.

$a^2 = 9l^2 = 3(3l^2)$. By setting $k = 3l^2$, we have $a^2 = 3k$. This is one of the two possibilities given to us in the problem.

Case 2: $a = 3l+1$.

$a^2 = (3l+1)^2 = 9l^2+6l+1 = 3(3l^2+2l)+1$. By choosing $k=3l^2+2l$, we have $a^2 = 3k+1$, again this is one of the two possibilities given to us in the problem.

Can you see how to do the third one? Do you see how this solves the problem at hand?


问:
Vector parallel and distinct
Equation 1:
$$x=3-t\ ,\ y=2t\ ,\ z=2-3t$$
Equation 2:
$$x=3+s\ ,\ y=-3-2s\ ,\ z=6+3s$$

Now I proved that they are distinct by showing that the system does not have a solution.
But how am I supposed to prove that they are parallel when I can't get them to be multiples of each other since they are different variables.

Also, how can I find the equation of a unique plane that contains both of them?

答:
Calculate $x-x_0$ and so on, where the $0$ subscript means it's the constant term:
$$x-3=-t, y=2t, z-2=-3t\\x-3=s, y+3=-2s, z-6=3s$$
Now plug in $s=-t$

Write equation of a plane given by a line and a point. Take line given by equation 1, and then take any point on line 2. You can choose $s=0$, but you don't have to.

Title: Vector parallel and distinct


问:
If $S\subseteq \{1,2,...,15\}$ has more than $5$ elements, then there are two distinct subsets of $S$ with the same element sum.

For any finite set $T$ of real numbers, $s(T)$ is the sum of its elements. Let $S\subseteq \{1,2,...,15\}$.  If $s(B)\ne s(C)$ for any disjoint subsets $B,C$ of $S$, then show that $|S|\le 5$.

Source : Principles and Techniques in Combinatorics

I know that $|S|\le 6$.  Because for any $X\subseteq S$, $0\le s(X) \le 1+2+...+15=120$.  If $|S|\ge 7$, then $S$ has $2^7>121$ subsets, so there are two subsets $B,C$ of $S$ with equal element sum.  We can assume that $B$ and $C$ are disjoint (otherwise, remove the intersection from $B,C$).

I also know that there is an example with $|S|=5$.  My example is $\{8,11,13,14,15\}$.  But how to exclude the case $|S|=6$?

Oh, if you know which contest this problem came from, that'd be great.  Apparently, the book cited a wrong competition because it says USAMO 1986, but this problem is not in USAMO 1986.  Thank you very much.

BTW here is what I know about the case $|S|=6$.  Let $S=\{x_1,x_2,...,x_6\}$ where $x_1<x_2<...<x_6$.  If $x_6<14$, then $s(S)\le 13+12+...+8=63$.  Since $S=\{8,9,...,12,13\}$ doesn't work we must have $s(S)<63$.  So $s(X)$ for $X\subseteq S$ can have at most $63$ possible values but $63<2^6$, so this is a contradiction.  Therefore, $x_6\ge 14$.  Then there are too many cases to consider for $x_5$.

答: Suppose $|S|=6$. Note that the maximum sum of any $4$-element (or smaller) subset is $12+13+14+15=54$, so at most $55$ different sum values are taken by subsets of $S$ of size at most $4$. But at most $7$ different values are taken by subsets of $S$ of size $5$ or $6$ (since it only has this many subsets of size $5$ or $6$). So in total the number of sum values for subsets of $S$ is at most $62<2^6$.


问:
solve $|\sin(a)|=\cos(3a)$
$|\sin(a)|=\cos(3a)$ is an alternative version of an equation $\sqrt{1-x^2}=4x^3-3x$, where I made a substitution $x=\cos(a)$ for $x \in [-1, 1]$. Unfortunately, I have no idea how to solve trigonometric equations.

答:
Consider

Case 1:  $0 \le a \le \frac \pi 6$ then $0 \le 3a \le \frac \pi 2$.  Than $\sin a \ge 0$ and $\cos 3a \ge 0$.

Then $a$ and $3a$ are both in the first quandrant.  And the identity that will apply is $\sin a = \cos (\frac \pi 2 - a)$.

So we must have $\frac\pi 2 - a= 3a$ or $a =\frac {\pi} 8$.

Case 2:  $\frac \pi 6 < a < \frac \pi 2$.  Then $\frac \pi 2 < 3a < \frac {3\pi}2$.  But then $\cos 3a < 0$ and that's not the case.

Case 3: $\frac \pi 2 \le a \le \frac {2\pi}3$.  And $\frac {3\pi}2 \le 3a \le 2\pi$.  Then $\sin a \ge 0$ and $\cos 3a \ge 0$.

Now $a$ is in the second quadrant and $a$ is in the fourth and the indentity that is relevent is.

$\sin (\frac \pi 2 + \alpha) = \cos (2\pi - \alpha)$.

So $a = \frac \pi 2 + \alpha$ and $3a = 2\pi - \alpha$ so $a+ 3a = 2\pi + \frac \pi 2$ so $4a = \frac {5\pi }2$ and $a = \frac {5\pi} 8$.

Case 4: $\frac {2\pi}3 < a \le  \frac {5\pi}6$ and $2\pi < 3a \le 2\pi + \frac {\pi}2$.

Then $a$ is in second quadrant but $3a$ is in the first but over $2\pi$.

The identity that applies here is $\cos (2\pi + \alpha) = \sin (\frac \pi 2 + \alpha)$.

So $a = \frac \pi 2 + \alpha$ and $3a = 2\pi + \alpha$ ans so $3a - a = \frac {3\pi }2$ and $a= \frac {3\pi}4$.

Case 5: $\frac {5\pi}6 < a < \frac {7\pi}6$ and $\frac {5\pi}2 < 3a < \frac {7\pi}2$.  But then $\cos 3a < 0$ which isn't possible.

Case 6: $\frac {7\pi}6 \le a \le \frac {4\pi}3$ so $\frac {7\pi}2 \le 3a \le 4\pi$ then $\sin a \le 0$ but $\cos 3a = -\sin a \ge 0$.

The identity here is $-\sin (\pi + \alpha) = \sin (\pi - \alpha)=\cos(\frac {7\pi}2 + \alpha)$.

So $a = \pi + \alpha$ and $3a = \frac {7\pi}2 + \alpha $ so $2a = \frac {5\pi}2$ and $a =\frac {5\pi}4$.

Case 7: $\frac {4\pi}3 \le a \le \frac {3\pi}2$ so $4\pi \le 3a \le \frac {9\pi}2$.

Then identity to use here is $-\sin (\frac {3\pi} 2 - \alpha)= \cos (4\pi + \alpha)$

So $a = \frac {3\pi}2  -\alpha$ and $3a =4\pi +\alpha$ so $4a = 4\pi + \frac {3\pi}2$ so $a =\pi +\frac {3\pi}8 = \frac {11\pi}8$.

Case 8: $\frac {3\pi}2 < a < \frac {11\pi } 6$ so $\frac {9\pi}2< 3a < \frac {11\pi}2$ so $\cos 3a < 0$.

And final final case

9:  $\frac {11\pi}6 \le a < 2\pi$ then $\frac {11\pi}2 \le 3a < 6\pi$.

The identity we need is $-\sin (\frac {3\pi}2 + \alpha) = \cos(6\pi - \alpha)$.

So $a + 3a = 6\pi + \frac {3\pi }2$ and so $a = \frac {15\pi}8$.

Title: solve $|\sin(a)|=\cos(3a)$


问:
How to find the maximum value of this expression?
expression:$$\sum_{i=1}^{n-1} \frac{\sqrt{1+\cos(\theta_i)}}{\sqrt{2}} +\frac{\sqrt{1-\cos(\sum_{i=1}^{n-1}\theta_i)}}{\sqrt{2}}$$ I guess its maximum value is obtained when all $\theta_i=\pi/n$ are the same and the the max value is $n\cos(\frac{\pi}{2n})$.The partial derivatives is so complex that I can't analytically  solve it.

答:
We seek the maximum of  $\displaystyle f(\theta_1,\ldots,\theta_{n-1})=\sum_{i=1}^{n-1} \frac{\sqrt{1+\cos(\theta_i)}}{\sqrt{2}} +\frac{\sqrt{1-\cos(\sum_{i=1}^{n-1}\theta_i)}}{\sqrt{2}}$ on $\mathbb R^{n-1}$. By $2\pi$-periodicity and parity, we can seek this maximum on $[0,\pi]^{n-1}$. Then, for $(\theta_1,\ldots,\theta_{n-1}) \in[0,\pi]^{n-1}$:

$f(\theta_1,\ldots,\theta_{n-1})=\displaystyle \sum_{i=1}^{n-1} \cos\dfrac{\theta_i}{2} + \left|\sin \dfrac{\sum_{i=1}^{n-1} \theta_i}{2}\right|$

Let $k\in\mathbb Z$ such that: $\ \ \displaystyle 2k\pi \leqslant \sum_{i=1}^{n-1} \theta_i \leqslant 2(k+1)\pi\ \ $

Let  $\ \ \displaystyle \theta_n= (2k+1)\pi - \sum_{i=1}^{n-1}\theta_i \ \  $.

$\theta_n$ is in $[-\pi,\pi]$  and $\ \ f(\theta_1,\ldots,\theta_{n-1})=\displaystyle \sum_{i=1}^n \cos\dfrac{\theta_i}{2}$

By concavity of $\cos$ on $[-\frac{\pi}{2},\frac{\pi}{2}]$, we can write that:

$f(\theta_1,\ldots,\theta_{n-1}) \leqslant n \displaystyle  \cos \dfrac{\sum_{i=1}^n \theta_i}{2} = n \cos \dfrac{(2k+1)\pi}{2n} \leqslant n \cos\dfrac{\pi}{2n}$

And:   $\fbox{$\displaystyle\max_{(\theta_1,\ldots,\theta_{n-1})\in\mathbb R^{n-1}} f(\theta_1,\ldots,\theta_{n-1}) = f\left( \frac{\pi}{2n},\ldots,\frac{\pi}{2n}\right) = n\cos \dfrac{\pi}{2n}$}$


问:
coordinates of points minimizing the infinity norm distance between two non-intersecting circles
We are given two non-intersecting circles centered at $O_{1}$ and $O_{2}$. For simplicity sake, let $O_{1} = (0,0)$ and $r_{1} = r_{2} = 10$.  Let $t$ be the distance between the two origins minus $2r$. In the picture, I call the distance between origins $deltaOrig$.

The circles do not overlap in either the x or y-dimensions.

In the attached image, $O_{2} = (27,23)$. Thus, as shown in the picture, $$t_{x} = (27-0-(2*10)) = 7 \\t_{y} = (23-0-(2*10) = 3$$.

To minimize the infinity norm or Chebyshev distance between the two circles, I need to find the point on each circle such that the quadrilateral between those two points is square with side length k.

How do I know this?  I have this formulated as a QCP or quadratically constrained program optimization problem where I minimize the distance of the arc connecting the circles subject to each endpoint lying on a circle. Using this, I also find that the coordinates that minimize this distance form the angle $\theta$ as shown in the picture (pardon my imperfect drawing).
For example in this instance, I find that $k = 11$ and the minimizing coordinates are $$(8,6)\\(19,17)$$.

Question: how do I find the geometric solution to this problem (what is $\theta$ as $O_2$ changes? I suspect there's a simple geometric argument but I am stumped except when the circles are distance enough that the square becomes a rectangle.

Here are some results from my CPLEX solution where $O_1$ remains $(0,0)$ and $r=10$:
$O_2 = (50,30)$ then $k=30.0$ and $\theta = 0^{\circ}$

$O_2 = (49,30)$ then $k=29.024$ and $\theta = 2.798^{\circ}$

...

$O_2 = (30,30)$ then $k=15.857$ and $\theta = 45^{\circ}$

$O_2 = (29,30)$ then $k=15.367$ and $\theta = 47.026^{\circ}$

...

$O_2 = (11,30)$ then $k=10.024$ and $\theta = 87.202^{\circ}$

$O_2 = (10,30)$ then $k=10.0$ and $\theta = 90^{\circ}$

Thank you!

Body:
Calling

$$
\cases{
p_1 = (0,0)\\
p_2 = (27,23)\\
r_1 = 10\\
r_2 = 10\\
p=(x_1,y_1)\\
q=(x_2,y_2)\\
p_c=(y_1,x_2)
}
$$

we have the conditions

$$
\mathcal{R(x_1,y_1,x_2,y_2)}=\cases{
\|p-p_1\|^2=r_1^2\\
\|q-p_2\|^2=r_2^2\\
(p-p_c)\cdot(q-p_c) = 0\\
\|p-p_c\|^2=\|q-p_c\|^2
}
$$

so the problem can be enunciated as

$$
\min_{x_1,x_2,y_1,y_2}\left(\|p-p_c\|^2+\|q-p_c\|^2\right),\ \ \text{s. t.}\ \ \mathcal{R}(x_1,y_1,x_2,y_2)
$$

In our case we have

$$
\{
x_1 = 8, y_1 = 6, x_2 = 19, y_2 = 17
\}
$$