问:
Purely imaginary equation $p(x)=0$ with real coefficient
The quadratic equation $p(x)=0$ with real coefficient has purely imaginary roots. Then the equation $p(p(x))=0$ has

(A) only purely imaginary roots

(B) all real roots

(C) two real and two purely imaginary roots

(D) neither real nor purely imaginary roots

The official answer is (D)

My approach us as follow

Though the question has been solved in this website but dont consider it as a duplicate as I would like to put forward my method for perusal

$p\left( {p\left( x \right)} \right) = 0$

$p\left( x \right) = a{x^2} + bx + c = 0$

${b^2} - 4ac < 0$

$T = p\left( x \right)\& T > 0$

$a{x^2} + bx + c > 0$

$p\left( T \right) = 0 = a{T^2} + bT + c$

${b^2} - 4ac < 0$ but $T > 0$ contradiction so the equation is neither real nor imaginary

Is my approach of solving matches with the standard procedure

答:
As it has real coefficients. We can WLOG assume the roots are  $li,-li$ where $l$ is some real $l\neq 0$.Thus $$p(x)=a(x-li)(x+li)=a(x^2+l^2)$$ Here $a$ is some real $a\neq 0$

Now $$p(p(x))=0 \Rightarrow p(x)=li,-li$$

It is easy to  see $$a(x^2+l^2)=li,a(x^2+l^2)=-li$$ has neither real nor purely imaginary roots


答:
Let $r$ be a purely imaginary root of $p(x)$, then the other root is $\bar r=-r$ so we can write $p(x)=(x-r)(x-(-r))=x^2-r^2$.

Let $T=p(x)$, then we want to find roots of $p(p(x))=p(T)=0$. We know the roots to be $T=r,-r$. Thus, $x^2-r^2=\pm r\implies x^2=r^2\pm r$. Since $r^2\in\Bbb R-\{0\}$ and $r$ is purely imaginary non-zero complex number, $r^2\pm r$ is not real. So $x$ can't be purely imaginary or real.

ContentLicense: CC BY-SA 4.0

Title: Purely imaginary equation $p(x)=0$ with real coefficient

Body:
The quadratic equation $p(x)=0$ with real coefficient has purely imaginary roots. Then the equation $p(p(x))=0$ has

(A) only purely imaginary roots

(B) all real roots

(C) two real and two purely imaginary roots

(D) neither real nor purely imaginary roots

The official answer is (D)

My approach us as follow

Though the question has been solved in this website but dont consider it as a duplicate as I would like to put forward my method for perusal

$p\left( {p\left( x \right)} \right) = 0$

$p\left( x \right) = a{x^2} + bx + c = 0$

${b^2} - 4ac < 0$

$T = p\left( x \right)\& T > 0$

$a{x^2} + bx + c > 0$

$p\left( T \right) = 0 = a{T^2} + bT + c$

${b^2} - 4ac < 0$ but $T > 0$ contradiction so the equation is neither real nor imaginary

Is my approach of solving matches with the standard procedure


答: If it has purely imaginary roots, that basically means that $$p(x)=ax^2+b\,,$$ where $a,b$ are nonzero real numbers of the same sign. It follows that $$p(p(x))=a(ax^2+b)^2+b\,.$$ The solution to this is $$ax^2+b=\pm\sqrt{\frac{b}{a}}i\,,$$ which clearly shows that $x$ can neither be real nor purely imaginary because in that case the left-hand side would be a real number whereas the right-hand side is purely imaginary.


问:
Nick and Penny are independently performing independent Bernoulli trials. For con- creteness ...
Nick and Penny are independently performing independent Bernoulli trials. For con- creteness, assume that Nick is flipping a nickel with probability p1 of Heads and Penny is flipping a penny with probability p2 of Heads. Let X1, X2, · · · be Nick’s results and Y1, Y2, · · · be Penny’s results, with Xi ∼ Bern(p1), and Yj ∼ Bern(p2).

a) Find the distribution and expected value of the first time at which they are simul- taneously successful, i.e., the smallest n such that Xn = Yn = 1.

b) For p1 = p2, find the probability that their first successes are simultaneous, and use this to find the probability that Nick’s first success precedes Penny’s.

For part A, I made X = # of successful trials, got the geometric distribution to be X~Geom(p1p2), and got the expected value to be E(x) = 1/(p1p2).

For part B, I did X = # of trials for Nick's 1st success, Y = # of trials for Penny's 1st success, X,Y~Geom(p1) and using LOTP,

P(X=Y) = ($p1^2$)/($1-(1-p)^2$)

Then finally P(X < Y) = (1-P(X = Y))/(2) = [ 1 - (($p1^2$)/($1-(1-p1)^2$) ] / [2]

I feel like both of my answers are off but I do not know where they are off. Any help would be very much appreciated.

答:
Your answers are both right, but the results can be further simplified. Set $q=1-p$

$$\mathbb{P}[X=Y]=\sum_{k=1}^{\infty}p^2(q^2)^{k-1}=p^2\sum_{k=0}^{\infty}(q^2)^{k}=\frac{p^2}{1-q^2}=\dots=\frac{p}{2-p}$$

$$\mathbb{P}[X>Y]=\frac{1-\frac{p}{2-p}}{2}=\dots=\frac{1-p}{2-p}$$


问:
Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$
So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$

So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that the solution is $n^3$, do I have to prove $(n+1)^3$ is equal to that or any other method or I prove that $(n+1)^3-n = 2m^2+2m+2+n_m$ (the differience between $(n+1)$ and $n$

答:
Approaching by pattern,

We know that sum of first $k$ odd numbers is $k^2$.

We have $1,2,3,4\ldots$ terms which may be expressed as difference of triangular number of terms
$$ 1 = \underset{1}{\underbrace{1}} - \underset{0}{\underbrace{0}} $$ $$ 3+5 = \underset{\text{first 3 odd}}{\underbrace{(1+3+5)}} - \underset{\text{first odd}}{\underbrace{1}}$$ $$ 7+9+11 = \underset{\text{first 6 odd}}{\underbrace{(1+3+5+7+9+11)}}-\underset{\text{first 3 odd}}{\underbrace{(1+3+5)}}$$

In particular, $n^{th}$ sum is $$\text{sum of first $T_{n}$ odd numbers - sum of first $T_{n-1}$ odd numbers}$$

Thus $n^{th}$ sum is $$ 1 = 1^2 - 0^2 $$ $$ 3+5 = 3^2 - 1^2$$ $$ 7+9+11 = 6^2-3^2$$ $$ 13+15+17+19 = 10^2-6^2$$

In particular, $n^{th}$ sum is $$T_{n}^2-T_{n-1}^2$$
$$=\dfrac{n^2(n+1)^2}{4}-\dfrac{n^2(n-1)^2}{4}$$
$$=\dfrac{n^2}{4}(4n)$$
$$=n^3$$

Note : Once one makes these observations by naked eye, one can simply finish the problem by writing down step $3$ directly.


问:
Factorial Equality Problem
I'm stuck on this problem, any help would be appreciated.

Find all $n \in \mathbb{Z}$ which satisfy the following equation:

$${12 \choose n} = \binom{12}{n-2}$$

I have tried to put each of them into the factorial equation and then making them equal each other and manipulating a bit, although I get stuck at:

$$(12 - k)!\cdot k! - (14 - k)!\cdot (k - 2)! = 0$$

Maybe I'm on the completely wrong path? Maybe I just had a mathematical error?

Thanks in advance!

答:
The nonzero binomial coefficients with upper index $12$ are given by line $12$ in Pascal's triangle:
$$
  1,12,66,220,495,792,924,792,495,220,66,12,1.
$$
Among these equality only occurs for entries that are symmetric around the central number$~\binom{12}6$, which gives the solution $n=7$ to your equation. However all other numbers $\binom{12}k$ are zero, which gives infinitely many more solutions to you equation, namely for all integers $n<-1$ and $n>14$.

${}$

All in all the set of solutions is:
$
  \Bbb Z \setminus\{0,1,2,3,4,5,6,8,9,10,11,12,13,14\}.
$

ContentLicense: CC BY-SA 3.0

Title: Factorial Equality Problem


问:
How to simplify $\frac {\sin 3A - \cos 3A}{\sin A + \cos A} + 1$?
So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)

But after this I don't have any clue on how to proceed.

$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^{3} \theta-\sin \theta\right)}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4\sin^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta+\sin \theta+\cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+4 \cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta+\cos \theta-\sin ^{3} \theta-\cos ^{3} \theta}
{\sin \theta+\cos \theta}
$$

Original image

答:
$$\sin3A-\cos3A$$
$$=\sqrt2\sin\left(3A-\dfrac\pi4\right)$$

$$=\sqrt2\sin\left(3A+\dfrac{3\pi}4-\pi\right)$$

$$=-\sqrt2\sin3\left(A+\dfrac\pi4\right)$$

Similarly, $\sin A+\cos A=\sqrt2\sin(?)$

Finally for $\sin B\ne0,$ $$\dfrac{\sin3B}{\sin B}=\dfrac{\sin B(3-4\sin^2B)}{\sin B}=3-2(1-\cos2B)$$


问:
Which positive integers $a$ and $b$ make $(ab)^2-4(a+b) $ a square of an integer?
Which positive integers
$a$ and $b$ make
$(ab)^2-4(a+b)
$
a square of an integer?

I saw this in quora,
and found that
the only solutions with
$a \ge b > 0$
are
$(a, b, (ab)^2–4(a+b)) = (5, 1, 1)$
and $(3, 2, 16)$.
Another “solution” is
$a = b = 2, (ab)^2–4(a+b) = 0$.

My solution is
messy and computational,
and I wonder if
there is a more elegant solution.

Here is my solution.

Assume $a \ge b$
and write
$n^2 = (ab)^2–4(a+b)$ so $n < ab$.

Let $n = ab-k$
where $ab > k>0$ so
$(ab)^2–4(a+b) = (ab-k)^2 = (ab)^2–2kab+k^2$
or
$k^2–2kab+4(a+b) = 0$.

Then

$\begin{array}\\
k
&= \dfrac{2ab-\sqrt{4a^2b^2–16(a+b)}}{2}
\qquad \text{(use "-" since } k < ab)\\
&= ab-\sqrt{a^2b^2–4(a+b)}\\
&=(ab-\sqrt{a^2b^2–4(a+b)})\dfrac{ab+\sqrt{a^2b^2–4(a+b)}}{ab+\sqrt{a^2b^2–4(a+b)}}\\
&=\dfrac{4(a+b)}{ab+\sqrt{a^2b^2–4(a+b)}}\\
\end{array}
$

Therefore,
since $k \ge 1, 4(a+b) \ge ab$
so
$0 \ge ab-4(a+b) = ab-4(a+b)+16–16
 =(a-4)(b-4)-16$
or $16 \ge (a-4)(b-4)$.
This gives a finite number of possible $a, b$,
all at least $4$.

Computation shows that
none of these are solutions.

To get the possible values of
$a$ and $n$ in terms of $b$
for any fixed $b$,
do this:

Since
$n^2
= a^2b^2-4(a+b)$,

$\begin{array}\\
b^2n^2
&= a^2b^4-4b^2a-4b^3\\
&= a^2b^4-4b^2a+4-4b^3–4\\
&=(b^2a-2)^2–4(b^3+1)\\
\end{array}
$

so
$4(b^3+1)
= (b^2a-2)^2-b^2n^2
= (b^2a-2-bn)(b^2a+bn)
$.

For each factorization
$r*s = 4(b^3+1)$, try
$r=b^2a-2-bn, s=b^2a-2+bn$.

This gives
$s-r=2bn$,
so if $2b$ divides $s-r$,
then
$n=\dfrac{s-r}{2b}$.
Adding $s$ and $r$,
$2b^2a-4=s+r$ so if
$2b^2$ divides $s+r+4$,
then $a = \dfrac{s+r+4}{2b^2}$.

This allows us to compute all solutions for any fixed value of b.
Running this for $1 \le b \le 16$
gives the solutions above.

For $a \ge b \ge 5$,
the restriction
$16 \ge= (a-4)(b-4)$
gives a finite set of possibilities which computation shows yields no additional solutions.

I sure would like to see
a more elegant solution.
Also,
this messy algebra
provides lots of opportunities
for errors.

答:
We can find an upper bound for their "product" in the following way:
\begin{align}
{\left( {ab} \right)^2} - 4\left( {a + b} \right) &< {\left( {ab} \right)^2} \ \  (\because a,b>0) \\
 {\left( {ab} \right)^2} - 4\left( {a + b} \right) &\le {\left( {ab - 1} \right)^2} \\
(2ab-1)-4(a+b) &\le 0\\
ab - 2(a+b) - \frac 12 \color{blue}{+4} &\le 0 \color{blue}{+4}\\
 \left( {a - 2} \right)\left( {b - 2} \right) &\le \frac{9}{2} \\
 \left( {a - 2} \right)\left( {b - 2} \right) &\le 4 \\
\end{align}

Now, seeing some cases should finish the work.


答:
If $(ab)^2-4(a+b)$ is greater than $(ab-1)^2$ then it cannot be a square, since it is strictly between two consecutive squares. Hence

$$(ab)^2-4(a+b) \le (ab-1)^2=(ab)^2-2ab+1$$
$$2ab-4a-4b-1\le0$$
$$2(a-2)(b-2)=2ab-4a-4b+8\le 9$$

which, again, gives a finite set of possibilities to be checked.

WLOG suppose $a \ge b$. We need only consider the cases:

$b=1,2$

$b=3, a\le6$

$b\ge 4, a < 3$ (this case contradicts $a\ge b$)

For $b=1$, $(ab)^2-4(a+b) = a^2-4a-4$. For $a\ge7$, $a^2-4a-4> a^2-6a+9=(a-3)^2$. But $a^2 -4a-4 < a^2-4a+4=(a+2)^2$. So we only need to check $1 \le a \le 6$.

For $b=2$, $(ab)^2-4(a+b) = 4a^2-4a-8$, which is a square only if $a^2-a-2$ is. For $a\ge4$, $a^2-a-2> a^2-2a+1 =(a-1)^2$. But $a^2-a-2<a^2$, so we only need to check $a=2,3$.

For $b=3$, $(ab)^2-4(a+b) = 9a^2-4a-12$, and we only need to check $3 \le a \le 6$.

Body:
Which positive integers
$a$ and $b$ make
$(ab)^2-4(a+b)
$
a square of an integer?

I saw this in quora,
and found that
the only solutions with
$a \ge b > 0$
are
$(a, b, (ab)^2–4(a+b)) = (5, 1, 1)$
and $(3, 2, 16)$.
Another “solution” is
$a = b = 2, (ab)^2–4(a+b) = 0$.

My solution is
messy and computational,
and I wonder if
there is a more elegant solution.

Here is my solution.

Assume $a \ge b$
and write
$n^2 = (ab)^2–4(a+b)$ so $n < ab$.

Let $n = ab-k$
where $ab > k>0$ so
$(ab)^2–4(a+b) = (ab-k)^2 = (ab)^2–2kab+k^2$
or
$k^2–2kab+4(a+b) = 0$.

Then

$\begin{array}\\
k
&= \dfrac{2ab-\sqrt{4a^2b^2–16(a+b)}}{2}
\qquad \text{(use "-" since } k < ab)\\
&= ab-\sqrt{a^2b^2–4(a+b)}\\
&=(ab-\sqrt{a^2b^2–4(a+b)})\dfrac{ab+\sqrt{a^2b^2–4(a+b)}}{ab+\sqrt{a^2b^2–4(a+b)}}\\
&=\dfrac{4(a+b)}{ab+\sqrt{a^2b^2–4(a+b)}}\\
\end{array}
$

Therefore,
since $k \ge 1, 4(a+b) \ge ab$
so
$0 \ge ab-4(a+b) = ab-4(a+b)+16–16
 =(a-4)(b-4)-16$
or $16 \ge (a-4)(b-4)$.
This gives a finite number of possible $a, b$,
all at least $4$.

Computation shows that
none of these are solutions.

To get the possible values of
$a$ and $n$ in terms of $b$
for any fixed $b$,
do this:

Since
$n^2
= a^2b^2-4(a+b)$,

$\begin{array}\\
b^2n^2
&= a^2b^4-4b^2a-4b^3\\
&= a^2b^4-4b^2a+4-4b^3–4\\
&=(b^2a-2)^2–4(b^3+1)\\
\end{array}
$

so
$4(b^3+1)
= (b^2a-2)^2-b^2n^2
= (b^2a-2-bn)(b^2a+bn)
$.

For each factorization
$r*s = 4(b^3+1)$, try
$r=b^2a-2-bn, s=b^2a-2+bn$.

This gives
$s-r=2bn$,
so if $2b$ divides $s-r$,
then
$n=\dfrac{s-r}{2b}$.
Adding $s$ and $r$,
$2b^2a-4=s+r$ so if
$2b^2$ divides $s+r+4$,
then $a = \dfrac{s+r+4}{2b^2}$.

This allows us to compute all solutions for any fixed value of b.
Running this for $1 \le b \le 16$
gives the solutions above.

For $a \ge b \ge 5$,
the restriction
$16 \ge= (a-4)(b-4)$
gives a finite set of possibilities which computation shows yields no additional solutions.

I sure would like to see
a more elegant solution.
Also,
this messy algebra
provides lots of opportunities
for errors.


问:
Is this approach to the limit solution kosher?$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$
Here is a non-rigourous approach to a limit:
$$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$$
I know the solution is correct. I'm suspicious of my methods.

Rewrite the expression:

$$ \lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$

I know the $\lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)=0$. So bear with me.

$$\frac{n!}{n^n}=\frac{n(n-1)(n-2)...[n-(n-2)][n-(n-1)]}{n^n}$$

Written this way, the numerator is a polynomial of degree $n$. For large $n$, the leading term will prevail. I will write that polynomial as a sum of $n^n$ and a polynomial of degree $n-1$.

$$\frac{n!}{n^n}\Rightarrow\frac{n^n+[\text{polynomial}]^{n-1}}{n^n}\Rightarrow 1+\frac{1}{n}$$

As noted in Step 2, this is not the expected result for this limit. But watch:

$$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{\tfrac{1}{n}}=\lim_{n\rightarrow\infty}\frac{1}{\left(1+\frac{1}{n}\right)^n}=\frac{1}{e}$$

This is the correct solution. But Step 2 is sketchy. Did I make another error which undid this error? Can this derivation be salvaged, or is my proof "not even wrong"?

Please be gentle with me.

答:
Too long for a comment.

Since you already received good answers, I shall try to go beyond the limit. I must precise that, as soon as I see a factorial, I automatically think about Stirling approximation.

$$a_n=\frac{(n!)^{\tfrac{1}{n}}}{n}\implies \log(a_n)=\frac 1n \log(n!)-\log(n)$$

$$\log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left({n}\right)\right)+\frac{1}{12
   n}+O\left(\frac{1}{n^3}\right)$$ which makes
$$\log(a_n)=-1+\frac{\log (2 \pi  n)}{2 n}+O\left(\frac{1}{n^2}\right)$$
$$a_n=e^{\log(a_n)}=\frac 1 e\left(1+\frac{\log (2 \pi  n)}{2 n} +O\left(\frac{1}{n^2}\right)\right)$$
which shows the limit and how it is approached.

Moreover, this gives you a shortcut eveluation of the $a_n$'s. Suppose $n=10$; the exact value is $0.453$ while the above truncated expression gives $0.444$ (that is to say a $2$% relative error).


问:
Asymptotic expansion of an integral with exponent times another function
According to this book page number 192

The asymptotic expansion of

$$\int_{0}^{\infty}[1-t \log t+(1-C-2 \log 2) t] e^{-2 t \log \left(\mu H / T_{1}\right)} d t$$

is

$$\frac{1}{2 \log \frac{\mu H}{T_{1}}}-\frac{\log 2}{\left(2 \log \frac{\mu H}{T_{1}}\right)^{2}}+\frac{\log \log \frac{\mu H}{T_{1}}}{\left(2 \log \frac{\mu H}{T_{1}}\right)^{2}}+\cdots$$

How is this computed?

Here is my try

The integrand can be written as

$$e^{{-2 t \log \left(\mu H / T_{1}\right)}+\log\left([1-t \log t+(1-C-2 \log 2) t] \right)}$$

Now, most of the contribution comes from the extremum of the function
which is given by

$$0=\frac{-C-\log (t)-2 \log (2)}{t (-C+1-2 \log (2))-t \log (t)+1}-2 \log \left(\frac{H \mu }{\text{T1}}\right)$$

Now, I do not know how to invert this relation to find an expression for $t_{ext}$.

Is there any other method to calculate the asymptotes other than the steepest decent method?

答:
I think splitting it would do the work. The formula is just the question of calculating something like:

$ \int_{0}^{\infty} (1- t\log t +at) e^{-\lambda t}dt$

After splitting, each term is .

i) $\int_{0}^{\infty} e^{-\lambda t} dt= \frac{1}{\lambda}$

ii)$\int_{0}^{\infty} te^{-\lambda t} dt= \frac{1}{\lambda^2}$

iii)$\int_{0}^{\infty}  t\log t .e^{-\lambda t}dt =\frac{1}{\lambda} \int_{0}^{\infty}( 1+\log t)e^{-\lambda t}dt $

For the last term, I don't know how to control but I guess as a physcist, you're provided with enough approximation formula to solve this term better than me.


问:
Show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$
Define the Fibonacci sequence via $f_1 = f_2 = 1$, and if $n > 2$, then $f_n =
f_{n−1} + f_{n−2}$ and show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$

I have been solving this problem using strong induction. This is my attemp.

For $n=1$.

$f_1=1$ and,

$\displaystyle \frac{\left ( \frac{1+\sqrt5}{2} \right )-\left ( \frac{1-\sqrt5}{2} \right )}{\sqrt5}=1$

For $n=2$

$f_1=2$ and,

$\displaystyle \frac{\left ( \frac{1+\sqrt5}{2} \right )^2-\left ( \frac{1-\sqrt5}{2} \right )^2}{\sqrt5}=1$ as well.

the statement is also true for $n = 3$

Now, assume that the statement is true for some $n-1$ and $n-2$. We get

$\displaystyle f_n =f_{n-1}+f_{n-2}=\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-1}-\left ( \frac{1-\sqrt5}{2} \right )^{n-1}}{\sqrt5}+\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-2}-\left ( \frac{1-\sqrt5}{2} \right )^{n-2}}{\sqrt5}$

From here, my intention is to show that the statement for $f_n$ is true. But I couldn't move on.

答:
Don't work so hard.  Define some notation to simplify the algebra.

Let $$\varphi = \frac{1 + \sqrt{5}}{2}, \quad \bar \varphi = \frac{1 - \sqrt{5}}{2}.$$  Then we can easily see that $$\varphi  + \bar \varphi = 1, \quad \varphi \bar \varphi = \frac{1^2 - (\sqrt{5})^2}{2^2} = -1, \quad \varphi - \bar \varphi = \sqrt{5}.$$  Consequently,
$$\varphi = \varphi(\varphi + \bar \varphi) = \varphi^2 - 1, \quad \bar \varphi = \bar \varphi (\varphi + \bar \varphi) = -1 + \bar \varphi^2,$$ or $$\varphi^2 = 1 + \varphi, \quad \bar \varphi^2 = 1 + \bar \varphi.$$
Now we have $$f_n = \frac{\varphi^n - \bar \varphi^n}{\varphi - \bar \varphi},$$ and $$\begin{align}
f_{n-1} + f_{n-2}
&= \frac{\varphi^{n-1} - \bar \varphi^{n-1} + \varphi^{n-2} - \bar \varphi^{n-2}}{\varphi - \bar \varphi} \\
&= \frac{\varphi^{n-2}(1 + \varphi) - \bar \varphi^{n-2}(1 + \bar \varphi)}{\varphi - \bar \varphi} \\
&= \frac{\varphi^{n-2} \varphi^2 - \bar \varphi^{n-2} \bar \varphi^2}{\varphi - \bar \varphi} \\
&= \frac{\varphi^n - \bar \varphi^n}{\varphi - \bar \varphi} \\
&= f_n.
\end{align}$$

This concludes the proof.


问:
Entire function defined by infinite product
I'm working on the following problem:

Suppose $|w|<1$. Prove that the function
$$ f(z)=\prod_{n=1}^\infty(1+w^{2n-1}e^z)(1+w^{2n-1}e^{-z}) $$
is entire.

I know of a theorem that states, if $(f_n)$ is a sequence of holomorphic functions defined on a domain $\Omega\subset\mathbb C$ and the series $\sum(1-f_n)$ converges uniformly and absolutely on compact subsets of $\Omega$, then the function $f=\prod f_n$ is holomorphic.

I don't know if this theorem can be applied here - any help on this problem is appreciated!

答:
Yes it applies here; an infinite product of the form $$\prod_n(1+h_n)$$ converges uniformly and absolutely when the infinite series $\sum_nh_n$ also converges uniformly and absolutely. Now consider that term in your product expression is $$(1+w^{2n-1}e^z)(1+w^{2n-1}e^{-z})=1+\left(2w^{2n-1}\cosh(z)+ w^{4n-2}\right)\,.$$ Clearly, because $|w|<1$, we have the easy evaluation $$\sum_n \left(2w^{2n-1}\cosh(z)+ w^{4n-2}\right)=2\frac{w}{1-w^2}\cosh(z)+\frac{w^2}{1-w^4}\,;$$
hence, your infinite product converges uniformly and absolutely (in compact subsets) for all $z$ and is therefore entire.


问:
Solving Systems by Gauss-Jordan Elimination
I'm going through my textbook solving the practice problems, I haven't had any trouble solving systems that are already in row-echelon form, or reduced row-echelon form. However, I'm struggling with using the Gaussian and Gauss-Jordan methods to get them to this point. One of the questions I have is:

$$x_1+x_2+2x_3=8$$
$$-x_1-2x_2+3x_3=1$$
$$3x_1-7x_2+4x_3=10$$

Which as an augmented matrix would be:

$$\begin{align}
\begin{bmatrix}
1 & 2 & 3 & 8\\
-1 & -2 & 3 & 1\\
3 & -7 & 4 & 10
\end{bmatrix}
\end{align}
$$

I understand the first step, which is to take a multiple of 1, that when added would set -1 to 0, which in this case would be 1. And this would give me:

$$\begin{align}
\begin{bmatrix}
1 & 2 & 3 & 8\\
0 & -1 & 4 & 2\\
3 & -7 & 4 & 10
\end{bmatrix}
\end{align}
$$

But after this I'm lost as to what I should do next, or if I even did the first step properly. Can any show me how to solve using this method?

答:
Follow the following thirteen steps,  Rx (like R1, R2 or R3) refers to the matrix row number:

Swap R1 with R3

Add R1/3 to R2

Multiply R2 by -3/13

Subtract R1/3 from R3

Multiply (3/2)R3

Swap R2 with R3

Subtract R2/5 from R3

Multiply (-5/6)R3

Subtract R3 from R2

Subtract (4) R3 from R1

Add (7/5) R2 to R1

Divide R1 by 3

Divide R2 by 5

You should arrive at a RREF of:

$$
\left[
\begin{array}{@{}ccc|c@{}}1 & 0 & 0 & 3\\0 & 1 & 0 & 1\\0 & 0 & 1 & 2\end{array}\right]
$$
Now, you read the solution from the bottom up, so we have:

$x_3 = 2$

$x_2 = 1$

$x_1 = 3$

You should of course check my work and verify that those value satisfy all three simultaneous equations!


问:
Is there any way to figure out if a system of two recurrence relations will converge?
Context: I'm trying to write code that generates 2D Iterated Function System (IFS) fractals based on some affine transformations. I want to generate the fractal till convergence when possible.

The system of two recurrence relations I'm looking at are affine transformations of the form:

$x_{n+1} = ax_n + by_n + e$

$y_{n+1} = cx_n + dy_n + f$

where a, b, c, d, e, f are real numbers and $x_i, y_i$ are integers.

I'm trying to figure out the conditions for which $x_i$ and $y_i$ would eventually converge but I have not made much progress yet. The only thing I can think of is that it will only converge if a, b, c and d are within the range (-1, 1). But I'm not sure if this is enough to ensure convergence.

答: You need the eigenvalues of the matrix $A=\begin {pmatrix} a&b\\c&d \end {pmatrix}$ to be less than $1$ in absolute value.  Unless there is something special about your problem, the matrix will have two eigenvalues, each with a corresponding eigenvector.  The requirement is the same as for a geometric series, which is the one dimensional equivalent of what you have here.  You can write your iteration in terms of $X=\begin {pmatrix} x\\y \end {pmatrix}$ and $E=\begin {pmatrix} e\\f \end {pmatrix}$ as $X_{n+1}=AX_n+E$ with the limit found by $X=AX+E, X=(1-A)^{-1}E$


问:
Construct the Kernel (Fundamental Set of Solutions) of a Differential Operator
How do I construct the kernel of the following differential operator in $[0,1]$:
$$
\mathcal{L} = \frac{d}{dx}(k(x)\frac{d}{dx})
$$
and $k(x) \in C([0,1])$, $k_{min} \leq k(x) \leq k_{max}, \forall x \in [0,1]$, for some $0<k_{min}<k_{max}$

In otherword, I am looking for the functions $\phi (x):[0,1] \to \mathbb{R} $ such that $\mathcal{L} (\phi (x))=0$

答:
Denote $D = d/dx$. Observe that the equation you want to solve is
$$D \cdot k(x) \cdot D \phi = k'(x) D \phi + k D^2 \phi = (k'(x) + k(x) \cdot D) (D \phi) = 0$$
Let us denote $\psi = D \phi$. Then we must solve the equation
$$k'(x) \psi + k(x) \cdot D \psi = 0 \implies D \psi = -\frac{k'(x)}{k(x)} \psi \implies \psi = \frac{C}{k(x)}$$
for some arbitrary constant $C$. This suggests a general solution
$$\phi(x) =  \int \frac{C}{k(x)} dx$$
Note that there will be another constant resulting from the indefinite integral above. (As expected, since your differential operator takes two derivatives.)

Another way to see it: if $D [k(x) D] \phi(x) = 0$, then $k(x) D \phi(x) = C$ for some constant $C$, which suggests that $D \phi = C / k(x)$, giving the same exact result.


问:
Flaw in solving a PDE and Fourier Series

Consider the Laplacian equation in cylindrical coordinates with no theta $\left(~\theta~\right)$ dependence:
\begin{align}
&\mbox{So,}\quad\frac{1}{r} (r(u)_r)_r + (u)_{zz} = 0
\\[2mm]
&\mbox{subject to}\quad
\left\{\begin{array}{rcl}
u\left(r=R,z\right) & = & 1
\\
u\left(r=0,z\right) & = & \text{finite}
\\
u\left(r,z=\pm H\right) & = & 0
\end{array}\right.
\end{align}

Using separation of variables, $u(r,z) = A(r)B(z)$, we get
$$B''=-\lambda ^2 B \\
rA'' + A' - \lambda^2 r A = 0$$

Solving the $B(z)$ equation and imposing the boundary conditions, I get,
$$c_1 \cos(\lambda H) + c_2\sin(\lambda H) = 0 \\
c_1 \cos(\lambda H) - c_2\sin(\lambda H) = 0$$
I was told that this was incorrect, and I should be using $\sinh$ and $\cosh$ functions to this DE. My question is, why is that allowed? Aren't they non-periodic?

What am I doing wrong here, and how am I supposed to know when to use $\sin$ and $\cos$, and when to use $\sinh$ and $\cosh$ when solving PDEs?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$
{\Large ?}:\quad\left\{\begin{array}{l}
\ds{{1 \over r}\partiald{}{r}
\bracks{r\,\partiald{\on{u}\pars{r,z}}{r}} +
\partiald[2]{\on{u}\pars{r,z}}{z} = 0}
\\[3mm]
\ds{\on{u}\pars{r,\pm H} = 0}
\\[3mm]
\left.\begin{array}{rcl}
\ds{\on{u}\pars{0,z}} & \mbox{is} & \ds{finite}.
\\[1mm]
\ds{\on{u}\pars{R,z}} & \ds{=} & \ds{1}
\end{array}\right\}
\end{array}\right.
$

Lets $\ds{\on{u}\pars{r,z} \equiv
\sum_{n = 0}^{\infty}a_{n}\pars{r}
\cos\pars{k_{n}z}}$ where $\ds{k_{n} \equiv \pars{2n + 1}{\pi \over 2H}}$.

Note that
$\ds{\on{u}\pars{r,z}}$ already satisfies the boundary conditions at $\ds{z = \pm H}$ and, in addition, it exhibits the $\ds{z \mapsto -z}$ symmetry. Also,
\begin{equation}
\int_{-H}^{H}\cos\pars{k_{m}z}
\cos\pars{k_{n}z}\,\dd z = H\,\delta_{mn}
\label{1}\tag{1}
\end{equation}

$\ds{\on{u}\pars{r,z}}$ must satisfy the above differential equation:
\begin{align}
&\sum_{n = 1}^{\infty}\braces{%
{1 \over r}\totald{}{r}
\bracks{r\,\totald{a_{n}\pars{r}}{r}} -
k_{n}^{2}\,a_{n}\pars{r}}\cos\pars{k_{n}z} = 0
\end{align}
Use (\ref{1}) to obtain
$\ds{{1 \over r}\totald{}{r}\bracks{r\,\totald{a_{n}\pars{r}}{r}} -
k_{n}^{2}\,a_{n}\pars{r} = 0}$

The last equation solution is a linear combination of
$\ds{\on{J}_{0}\pars{\ic k_{n}r}\ \mbox{and}\ \on{Y}_{0}\pars{\ic k_{n}r}}$ which are
Bessel Functions.
Note that, as $\ds{r \to 0^{+}}$,
$$
\on{J}_{0}\pars{\ic k_{n}r} \to 1\quad
\mbox{and}\quad
\on{Y}_{0}\pars{\ic k_{n}r} \sim
{2 \over \pi}\ln\pars{\ic k_{n}r}.
$$
See this link.  The finite solution, when $\ds{r \to 0^{+}}$, is given by $\ds{b_{n}\on{J}_{0}\pars{\ic k_{n}r}}$ where $\ds{b_{n}}$ is a constant.

Then,
$$
\on{u}\pars{r,z} =
\sum_{n = 0}^{\infty}
b_{n}\on{J}_{0}\pars{\ic k_{n}r}
\cos\pars{k_{n}z}
$$

Moreover,
$$
1 = \sum_{n = 0}^{\infty}
b_{n}\on{J}_{0}\pars{\ic k_{n}\color{red}{R}}
\cos\pars{k_{n}z}
$$
Multiply both members by $\ds{\cos\pars{k_{n}z}/H}$ and use (\ref{1}) to integrate over $\ds{z \in \pars{-H,H}}$:
\begin{align}
&\overbrace{{1 \over H}
\int_{-H}^{H}\cos\pars{k_{n}z}\dd z}
^{\ds{{4 \over \pi}\,{\pars{-1}^{n} \over 2n + 1}}}
\\[5mm] = &\
\sum_{m = 0}^{\infty}
b_{m}\on{J}_{0}\pars{\ic k_{m}R}
\underbrace{\bracks{{1 \over H}\int_{-H}^{H}
\cos\pars{k_{n}z}\cos\pars{k_{m}z}\dd z}}
_{\ds{\delta_{nm}}}
\\[5mm] & \implies
b_{m} =
{4 \over \pi}\,{\pars{-1}^{n} \over \pars{2n + 1}\on{J}_{0}\pars{\ic k_{n}R}}
\end{align}

$\ds{\large\underline{Finally},}$
\begin{align}
&\on{u}\pars{r,z}
\\[2mm] = &\
{4 \over \pi}\sum_{n = 0}^{\infty}
{\pars{-1}^{n} \over \pars{2n + 1}
\on{J}_{0}\pars{\ic k_{n}R}}
\on{J}_{0}\pars{\ic k_{n}r}
\cos\pars{k_{n}z}
\end{align}


问:
An alternative definition for trace class.
While we know the definition for trace class is that $\sum_{n=0}^{\infty} \langle \mid A\mid x_n,x_n\rangle \leq +\infty$ holds for a orthogonal basis ${x_n}$, I wonder if it's true that  $\sum_{n=0}^{\infty} \mid\langle  A x_n,x_n\rangle\mid\leq +\infty$ holds for every orthogonal basis ${x_n}$ a sufficient condition for $A$ to be in the trace class.

Now I make some progress. We need prove the bounded self-disjoint operators with the property is compact.

Thanks in advance.

答:
Ok, I found one by google.

Define $A^+=\frac{|A|+A}{2}$ and $A^-=\frac{|A|-A}{2}$.

Since we have $2|\langle Bx,x\rangle|\leq|\langle Ax,x\rangle|+|\langle A^*x,x\rangle|=2|\langle Ax,x\rangle|$  , where $B=\frac{A+A^*}{2}$ and similarly for $C=\frac{A-A^*}{2i}$, we can only deal with the case when $A$ self-adjoint operator.

It can be easily checked that since $A$ is self-adjoint, which means $\left\| A \right\|=\sup_{\left\| \phi \right\|=1} |(A\phi,\phi)|$, $A$ is a bounded operator.
And it's a less obvious fact that $A|A|=|A|A$, where $|A|=\sqrt{A^*A}=\sqrt{A^2}$. The fact is often derived from $A|A|^2=A^3=|A|^2A$ and $|A|=\left\| A \right\|\sqrt{I-(I-\frac{A^2}{\left\| A \right\|^2})}$, a polynomial serie of $A$.

Then we have $A^+A^-=\frac{1}{4}(|A|A-A|A|)=0$, which implies $\mathrm{Im} A^-\subset \ker A^+$. The Hilbert space has a decomposition $H=\mathrm{Im} A^- \bigoplus \ker A^-$ . Choose a basis for $\ker A^-$ and , by Zorn's lemma or somehow, extend it to a basis of $H$. The basis chosen has the poperty that at least one of $ \langle A^+e,e\rangle$ and  $ \langle A^-e,e\rangle$ is $0$ and therefore:

$$\sum_k|\langle Ae_k,e_k\rangle|\leq\sum_k|\langle A^+e_k,e_k\rangle-\langle A^-e_k,e_k\rangle|=\sum_k|\langle A^+e_k,e_k\rangle+\langle A^-e_k,e_k\rangle|=\sum_k\langle |A|e_k,e_k\rangle$$

Then the conclusion can be obtained.


问:
Show that $\lim_{x\to 0}f(x) = 0$
Define the function $f(x) = x\cdot\chi_{\mathbb{Q}}(x)$ (where $\chi(x) = 1$ if $x \in \mathbb{Q}$ and $\chi(x) = 0$ if $x \in \mathbb{R}\setminus\mathbb{Q})$. Show that $\lim_{x\to 0}f(x) = 0$.

I tried solving this problem but I have no idea if I did it correctly. My professor mentioned something about functions like this oscillating too much to be continuous, so I figured it was okay to break this up into cases. Please let me know if this is wrong or if there are any errors, thanks!

Note that if $x\in\mathbb{Q}$, then $\chi(x) = 1$ and $2 > |\chi(x)|$. This implies,
\begin{equation*}
\frac{1}{2} < \frac{1}{|\chi(x)|}
\end{equation*}
\begin{equation*}
\frac{\epsilon}{2} < \frac{\epsilon}{|\chi(x)|}
\end{equation*}
So let $\delta \leq \frac{\epsilon}{2}$. Then $0 < |x-0| < \delta$, implies
\begin{equation*}
|x - 0| < \delta \leq \frac{\epsilon}{2} < \frac{\epsilon}{|\chi(x)|}
\end{equation*}
\begin{equation*}
|x||\chi(x)|  < \epsilon
\end{equation*}
\begin{equation*}
|x\chi(x)|  < \epsilon
\end{equation*}
\begin{equation*}
|x\chi(x) - 0|  < \epsilon
\end{equation*}

Note that if $x\in\mathbb{R}\backslash\mathbb{Q}$, then $\chi(x) = 0$ and $|x||\chi(x)| < |x|$, because $x$ is converging to (and never equals) $0$.

Suppose $\epsilon \in \mathbb{R}^+$, and choose $\delta \leq \epsilon$. Then $0 < |x - 0| < \delta$ implies,
\begin{equation*}
0 = |x||\chi(x)| < |x| < \delta \leq \epsilon
\end{equation*}
\begin{equation*}
|x\chi(x) - 0| < \epsilon
\end{equation*}

Therefore $\lim_{x\rightarrow0} f(x) = 0$.

答:
Aware of the definition of the limit, ie you have to check that for any sequence $(x_n)_n$ converging to 0, the sequence $(f(x_n))_n$ converges to 0. so to be formally correct use this.
You’re idea is okay, but you have to put both together in the proof, so for $\epsilon >0$ let $\delta= \epsilon/2$ will do the job.
Another hint for a fast solution: remark that for any x we have $0 \leq |f(x)|\leq |x|$.

Title: Show that $\lim_{x\to 0}f(x) = 0$


问:
extending bounded linear map
Let $H_0$ be a Hilbert subspace of a Hilbert space $H$. Suppose $T_0 \in B(H_0)$. Can I extend $T_0$ uniquely to $T \in B(H)$? I know extension exists. Can it be unique?

答: It cannot be extended uniquely, not even in the finite dimensional case. For example, take $\mathbb{R}^3$ and a basis $\{e_1,e_2,e_3\}$ of $\mathbb{R}^3$. Let $T_0(e_1)=e_1$ in $H_0=span(e_1)$. Then $T_1(e_2)=e_3,\ T_1(e_3)=e_2$ and $T_2=I$ ($I$ the identity). Then $T_1,\ T_2$ are two different extensions of $T$.

Title: extending bounded linear map


问:
Chain rule to find geodesics
I'm reading these notes but I don't know how is the chain rule being applied on page 92.

It says that we start with $\dfrac{dv}{du}=\dfrac{v'}{u'}$, having ruled out the case where $u'=0$. Then, he differentiates this wrt $u$ using the chain rule and goes this way

$$\dfrac{d^2v}{du^2}=\dfrac{d}{dv}\left(\dfrac{v'}{u'}\right)=\dfrac{u'v''-v'u''}{u'^2}\dfrac{1}{u'}$$

I don't really know why he got that. I understand $u$ and $v$ are functions of $t$, one does not depend on the other, so...

Thanks for any help.

答:
These are in fact my notes. You have a small typo: $\dfrac d{dv}$ should be $\dfrac d{du}$. But we're using the chain rule from single-variable calculus here:
$$\frac{dw}{du}=\frac{\frac{dw}{dt}}{\frac{du}{dt}}\,.$$</p>

Title: Chain rule to find geodesics


问:
Sequence of positives that is expressed as $3n$

This is a recurrence relation that my solution as a pic maybe the right one but im looking to confirm .

答: Looking for confirmation to my solution

Body: Looking for confirmation to my solution


问:
Prove Equivalency to Darboux Integrability
I've been working through a book on Introductory Real Analysis and I've been stumped by part of this problem.

Suppose we consider a partition that splits $[a,b]$ into $n$ partitions each with length $\frac{b-a}{n}$. Show that a function $f:[a,b]\rightarrow\mathbb R$ is Darboux integrable on just these partitions iff it is Darboux integrable over all partitions.

The if direction is very trivial, that one isn't too tricky.

However, the only if direction has me stumped. How can I show that integrability over this type of partition implies integrability over any partitions. I think I can show that it gives integrability over rational partitions (not sure if that's a formal term, basically a partition where all interval lengths are rational). However, I'm having trouble extending it to all partitions in general.

答:
Let $\mathcal{P}$ ($\mathcal{P}_U$) be the collection of all partitions (uniform partitions) of $[a,b]$.

For a bounded function $f$, we have $L(P,f) \leqslant U(Q,f)$ for lower and upper Darboux sums corresponding to arbitrary partitions $P$ and $Q$.  It follows that

$$\sup_{P \in \mathcal{P}}L(P,f) \leqslant \inf_{P \in \mathcal{P}}U(P,f), \quad \sup_{P \in \mathcal{P}_U}L(P,f) \leqslant \inf_{P \in \mathcal{P}_U}U(P,f)$$

Since $\mathcal{P}_U \subset \mathcal{P}$ we have

$$\{U(P,f) \,|\, P \in \mathcal{P}_U\} \subset \{U(P,f) \,|\, P \in \mathcal{P}\}, \quad  \{L(P,f) \,|\, P \in \mathcal{P}_U\} \subset \{L(P,f) \,|\, P \in \mathcal{P}\}$$

Hence,

$$\tag{*}\sup_{P \in \mathcal{P}_U}L(P,f) \leqslant \sup_{P \in \mathcal{P}}L(P,f) \leqslant \inf_{P \in \mathcal{P}}U(P,f) \leqslant \inf_{P \in \mathcal{P}_U}U(P,f)$$

Darboux integrability with respect to uniform partitions means that

$$\sup_{P \in \mathcal{P}_U}L(P,f) =\inf_{P \in \mathcal{P}_U}U(P,f),$$

which, in view of (*), implies that $f$ is Darboux integrable with respect to all partitions since

$$0 \leqslant \inf_{P \in \mathcal{P}}U(P,f)- \sup_{P \in \mathcal{P}}L(P,f) \leqslant  \inf_{P \in \mathcal{P}_U}U(P,f)- \sup_{P \in \mathcal{P}_U}L(P,f) = 0,$$
and
$$ \inf_{P \in \mathcal{P}}U(P,f)= \sup_{P \in \mathcal{P}}L(P,f) $$


问:
If $F$ has characteristic $P>0$ prove that $A=\begin{pmatrix}1&\alpha\\0&1\end{pmatrix}$ satisfies $A^p=I$.
If $F$ has characteristic $P>0$ prove that $$A=\begin{pmatrix}1&\alpha\\0&1\end{pmatrix}$$ satisfies $A^p=I$.

I am trying to solve this problem but I could not proceed. This is from Herstein algebra book. Please give a hint.

答:
You can check that
$$
A^2=\left(\begin{matrix}
1&2\alpha\\ 0&1
\end{matrix}\right).
$$
By induction
$$
A^p=\left(\begin{matrix}
1&p\alpha\\ 0&1
\end{matrix}\right),
$$
but $p\alpha=0$ for $F$ has characteristic $p$, so $A^p=I$.


答:
You can find the $n^\text{th}$ power of your matrix:
$$
\begin{pmatrix}
1 & \alpha \\
0 & 1
\end{pmatrix}^2 =
\begin{pmatrix}
1 & 2\alpha \\
0 & 1
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & \alpha \\
0 & 1
\end{pmatrix}^3 =
\begin{pmatrix}
1 & 2\alpha \\
0 & 1
\end{pmatrix}\begin{pmatrix}
1 & \alpha \\
0 & 1
\end{pmatrix} =
\begin{pmatrix}
1 & 3\alpha \\
0 & 1
\end{pmatrix}
$$
$$...$$
$$\begin{pmatrix}
1 & \alpha \\
0 & 1
\end{pmatrix}^n = \begin{pmatrix}
1 & n\alpha \\
0 & 1
\end{pmatrix}$$

Title: If $F$ has characteristic $P>0$ prove that $A=\begin{pmatrix}1&\alpha\\0&1\end{pmatrix}$ satisfies $A^p=I$.

Body:
If $F$ has characteristic $P>0$ prove that $$A=\begin{pmatrix}1&\alpha\\0&1\end{pmatrix}$$ satisfies $A^p=I$.

I am trying to solve this problem but I could not proceed. This is from Herstein algebra book. Please give a hint.


问:
Inequalities $\pi(x^a+y^b)^\alpha\leq \pi(x^c)^\beta+\pi(y^d)^\gamma$ involving the prime-counting function, where the constants are very close to $1$
Let $\pi(x)$ be the prime-counting function, I'm curious about if a suitable variant of the second Hardy–Littlewood conjecture (this corresponding Wikipedia)

$$\pi(x^a+y^b)^{\alpha}\leq\pi(x^c)^{\beta}+\pi(y^d)^{\gamma}\tag{1}$$
can be proved, where the constants $0<a,b,c,d\leq 1$ and the constants $0<\alpha,\beta,\gamma\leq 1$ are very close to our upper limit $1$, for all real numbers $x<y$ with $L<x$ for a (yours) suitable choice of a constant $L$.

Question. Is it possible to prove any statement of the type $(1)$ under the cited requirements, for constants $0<a,b,c,d\leq 1$ and constants $0<\alpha,\beta,\gamma\leq 1$ all these (all together/ simultaneously) very close to $1$, for all real numbers $x$ and $y$ such that $x<y$ for a suitable $L<x$? Many thanks.

I don't know if this type of proposals $(1)$ are in the literature, or are essentially the same than the  original second Hardy–Littlewood conjecture, when we require that those constants are very close to $1$.

If there is relevant literature answer my question as a reference request and I try to search and read those statements from the literature.

References:

[1] G. H. Hardy and J. E. Littlewood, Some problems of ‘Partitio numerorum’ III: On the expression of a number as a sum of primes, Acta Math. (44): 1–70 (1923).

答:
I guess these constants lead to a wrong way to the second Hardy–Littlewood conjecture. Indeed, according to Wikipedia, for $x\ge L_0=355991$ we have
$$\frac x{\log x} \left( 1+\frac 1{\log x}\right)<\pi(x)< \frac x{\log x} \left( 1+\frac 1{\log x}+\frac {2.51}{(\log x)^2}\right).$$
Then for $x,y\ge L_0$ the difference between the left-hand side and the right-hand side of the conjecture experession is not so big. Namely,
$$\pi(x+y)-\pi(x)-\pi(y)\le$$ $$\frac {x+y}{\log (x+y)} \left( 1+\frac 1{\log (x+y)}+\frac {2.51}{(\log (x+y))^2}\right)- \frac x{\log x} \left( 1+\frac 1{\log x}\right)-\frac y{\log y} \left( 1+\frac 1{\log y}\right)\le$$
$$\frac {x+y}{\log (x+y)} \left( 1+\frac 1{\log (x+y)}+\frac {2.51}{(\log (x+y))^2}\right)- \frac x{\log (x+y)} \left( 1+\frac 1{\log (x+y)}\right)-\frac y{\log (x+y)} \left( 1+\frac 1{\log (x+y)}\right)=$$
$$\frac {2.51(x+y)}{(\log (x+y))^3}.$$

Thus I guess that if $\max\{a,b\}\alpha<\min\{c\beta, d\gamma\}$ then the right-hand side of the     expression from your question grows asymptotically faster than the left-hand side and so the inequality holds for sufficiently big $L$.

So a way to the conjecture is to obtain better upper bounds for $\pi(x+y)-\pi(x)-\pi(y)$ than $\frac {2.51(x+y)}{(\log (x+y))^3}$ and I guess advances in this direction can be found in papers related to the conjecture.

Remark that Trudgian’s upper bound for the difference between $\pi(x)$ and $\operatorname{li}(x)=\int_0^x \frac {dt}{\ln t}$,
$$|\pi(x)- \operatorname{li}(x)|\le f(x)= 0.2795\frac {x}{(\log x)^{3/4}}\exp\left(-\sqrt{\frac{\log x}{6.455}} \right)$$
for $x\ge 229$  implies that for $x,y\ge 229$ we have
$$\pi(x+y)-\pi(x)-\pi(y)\le$$ $$\operatorname{li}(x+y)- \operatorname{li}(x)- \operatorname{li}(y)+f(x)+f(y)+f(x+y)\le$$ $$
\frac 1{\log (x+y-1)}-2\operatorname{li}(1)+ f(x)+f(y)+f(x+y).$$


问:
Finding integrating factor for non-exact differential equation $(4y-10x)dx+(4x-6x^2y^{-1})dy=0$.
I am given this equation

$$(4y-10x)dx+(4x-6x^2y^{-1})dy=0$$

where I must find an integrating factor to turn this into an exact differential. The integrating factor is supposed to be in the form $\mu=x^ny^m$.

I have found $M_{y}=4$ and $N_{x}=(4-12xy^{-1})$. It is here where I get stuck. How do I go about finding the integrating factor in the form $\mu=x^ny^m$, and what would the $n$ and $m$ end up being? Any help would be greatly appreciated!

答:
The equation is of the form $M(x,y)dx + N(x,y)dy = 0$. Set $M(x,y) = 0$ to obtain $y_0(x) = \frac{5x}{2}$. Set $N(x,y) = 0$ to obtain $x_0(y) = \frac{2y}{3}$.

Let $g(y) = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}$, given $x = x_0(y)$. Similarly, let $f(x) = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{-N}$, given $y = y_0(x)$. We obtain $g(y) = \frac{3}{y}$ and $f(x) = \frac{3}{x}$

Note that $f(x)$ and $g(y)$ satisfy $$\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = M(x,y)g(y) - N(x,y)f(x)$$. Hence an integrating factor, $\mu$, can be found by $e^{\int f(x)dx + \int g(y)dy + c}$. Doing this gives us $\mu = kx^3y^3$, for arbitrary k.

The method used here can be found at https://www.ijesm.co.in/current_issue.php


问:
Probability Limits
What is the probability limit of $\frac{\frac{1}{n}{\sum}_{i = 1}^{n} X_i}{\frac{1}{n}{\sum}_{i = 1}^{n} Y_i}$ where $X_1$, ..., $X_n$, $Y_1$, ...$Y_n$ both come from a distribution with mean $\mu$ and variance $\sigma^2$

I know that the probability limit of $\frac{1}{n}{\sum}_{i = 1}^{n} X_i$ is just $\mu$ from the law of large numbers, but how do you solve for this when its these two fractions on top of each other? Does a probability limit even exist?

Thank you

答:
There I will assume, that $X_1, ... X_n, Y_1, ... Y_n$ are i.i.d.

In this case, your problem splits in two cases:

$\mu \neq 0$

Then $\frac{\sum_{i = 1}^n X_i}{\sum_{i = 1}^n Y_i}$ converges almost surely to $1$ by the Strong Law of Large Numbers.

$\mu = 0$

Then we have by CLT, that $\frac{\sum_{i = 1}^n X_i}{\sigma \sqrt{n}}$ converges in distribution to $N(0, 1)$. The same goes for $\frac{\sum_{i = 1}^n Y_i}{\sigma \sqrt{n}}$. And because they are independent, we have that $\frac{\sum_{i = 1}^n X_i}{\sum_{i = 1}^n Y_i}$ converges in distribution to $Cauchy(0, 1)$

Title: Probability Limits


答:
You are correct but you don't say what is your final choice of $\delta$ explicity. For $x\in\Bbb Q$, you chose $\delta\le\epsilon/2$ and for $x\in\Bbb R-\Bbb Q$, you chose $\delta\le\epsilon.$ So $\delta\le\epsilon/2$ works for both rational and irrational points in the $\delta$ neighbourhood of $0$.

Also note that for irrational points, we can infact choose any $\delta>0$ since $|f(x)|=0<\epsilon$ for all irrational points.

The last $3$ lines of your solution are sufficient proof if you make a small correction. For both rational and irrational $x, |f(x)|=|x\chi_{\Bbb Q}(x)|\le|x|$, so if we choose $\delta\le\epsilon$, then $0<|x|<\delta\implies |f(x)|\color{red}\le|x|<\delta\le\epsilon$.

Title: Show that $\lim_{x\to 0}f(x) = 0$

Body:
Define the function $f(x) = x\cdot\chi_{\mathbb{Q}}(x)$ (where $\chi(x) = 1$ if $x \in \mathbb{Q}$ and $\chi(x) = 0$ if $x \in \mathbb{R}\setminus\mathbb{Q})$. Show that $\lim_{x\to 0}f(x) = 0$.

I tried solving this problem but I have no idea if I did it correctly. My professor mentioned something about functions like this oscillating too much to be continuous, so I figured it was okay to break this up into cases. Please let me know if this is wrong or if there are any errors, thanks!

Note that if $x\in\mathbb{Q}$, then $\chi(x) = 1$ and $2 > |\chi(x)|$. This implies,
\begin{equation*}
\frac{1}{2} < \frac{1}{|\chi(x)|}
\end{equation*}
\begin{equation*}
\frac{\epsilon}{2} < \frac{\epsilon}{|\chi(x)|}
\end{equation*}
So let $\delta \leq \frac{\epsilon}{2}$. Then $0 < |x-0| < \delta$, implies
\begin{equation*}
|x - 0| < \delta \leq \frac{\epsilon}{2} < \frac{\epsilon}{|\chi(x)|}
\end{equation*}
\begin{equation*}
|x||\chi(x)|  < \epsilon
\end{equation*}
\begin{equation*}
|x\chi(x)|  < \epsilon
\end{equation*}
\begin{equation*}
|x\chi(x) - 0|  < \epsilon
\end{equation*}

Note that if $x\in\mathbb{R}\backslash\mathbb{Q}$, then $\chi(x) = 0$ and $|x||\chi(x)| < |x|$, because $x$ is converging to (and never equals) $0$.

Suppose $\epsilon \in \mathbb{R}^+$, and choose $\delta \leq \epsilon$. Then $0 < |x - 0| < \delta$ implies,
\begin{equation*}
0 = |x||\chi(x)| < |x| < \delta \leq \epsilon
\end{equation*}
\begin{equation*}
|x\chi(x) - 0| < \epsilon
\end{equation*}

Therefore $\lim_{x\rightarrow0} f(x) = 0$.


问:
A combinatorial sum

Any idea to evaluate the sum
$$
\sum_{j=m}^{k}\frac{\binom{m}{2m - j\,\,}}{\binom{k}{j}}
\quad\mbox{with}\quad m \leq k < 2m - 1.
$$

I have found the sum for $k=2m-1$. In fact, it is verified that
$$
\sum_{j = m}^{2m - 1}\ \frac{\binom{m}{2m-j\,\,}}
{\binom{2m - 1\,\,}{j}} = 2m\left(H_{2m} - H_{m}\right),
$$
where $H_{j}$ is the $j$-th harmonic number.

答:
For a short time, I hoped to be able to express the result in terms of hypergeometric functions but I failed.

If we let $k=m+n$, the problem reduces to
$$S_n=\sum_{j=m}^{m+n} \frac{\Gamma (j+1)\,\,\Gamma (m+n+1-j)}{\Gamma (m+n+1)} \,\binom{m}{2 m-j} $$ which can write
$$S_n=\frac{\Gamma(m+1)}{n!\ \Gamma(m+n+1)} P_{n}(m)$$ where $P_n$ is a  polynomial of degree $2n$ (it does not seem to be factorable for any $n$).

The very first are
$$\left(
\begin{array}{cc}
 n & P_n \\
 0 & 1 \\
 1 & m^2+m+1 \\
 2 & m^4+2 m^3+m^2+4 \\
 3 & m^6+3 m^5-2 m^4-9 m^3+13 m^2+18 m+36 \\
 4 & m^8+4 m^7-10 m^6-44 m^5+53 m^4+184 m^3+100 m^2+576
\end{array}
\right)$$

I have not been able to find any pattern for the coefficients [except that the constant term is $(n!)^2$ and that the coefficient of $m^{2n}$ is $1$ (!!) ].

The only thing I observed is that, if $n$ is even, the term is $m$ is systematically missing.

Without any proof of it, I do not think that a closed form could exist for a general $n$.

Title: A combinatorial sum


问:
Eisenstein Criterion with Polynomials of Rational Coefficients?
I'm wondering if it's possible to apply the Eisenstein criterion to polynomials that have rational, non-integer coefficients? I am thinking about using a shift to go to integers and apply the Eisenstein criterion there. Then, since f(x) is irreducible iff f(x+c) is irreducible, that should show that the original polynomial is irreducible. For instance, with the polynomial $x^5-(3/5)x^4+9x-3$, I wonder if there's some way to shift this so it gets integer coefficients and we can do the procedure I said?

答:
That polynomial is Eisenstein at $3$.  It is actually irrelevant what other prime factors are doing in the coefficients, whether they are in numerators or denominators.  A way to deal with this is to give up on trying to work in $\mathbf Z[x]$ and instead allow yourself to invert all primes other than $3$: let the coefficients be in the ring $\mathbf Z_{(3)}$, which denotes fractions $a/b$ where $(3,b) = 1$. In this ring (a "localization" at the prime 3), which has fraction field $\mathbf Q$, the only prime element is $3$ and the Eisenstein criterion applies with the prime $3$ (not with other primes!).

Think about it this way: if $A$ is a UFD with a prime element $p$, the notion of Eisenstein polynomials at $p$ in $A[x]$ makes sense and such polynomials are irreducible in $A[x]$ and in $K[x]$ where $K$ is the fraction field of $A$.  Can you prove that? In your example, take $A$ to be not $\mathbf Z$, but $\mathbf Z_{(3)}$ (a UFD with just one prime element up to unit multiple, namely $3$) and $K = \mathbf Q$.


问:
Range of a multivariable function $f(x,y) = \frac{4x^2+(y+2)^2}{x^2+y^2+1}$
A function is defined in $\mathbb{R}^2$
\begin{align*} f(x,y) = \frac{4x^2+(y+2)^2}{x^2+y^2+1}\end{align*}
Find the range of possible values for the function.

Here we can see that when $x=0$ and $y=-2$ the minimum value of the range is $0$. To find the maximum value of the range I have found the partial derivatives and equate them to zero, then I got $y=\frac{4}{3}$ and $x^2= \frac{-25}{9}$, but $x$ cannot be complex ($x \in \mathbb{R}$). So how do we find the max values of this function? Could anyone please give me a hint, is this the way of finding the range of a multivariable function?

答:
You've already noticed that the minimum of $f$ is $0$.

The $x^2$ term in the numerator can be easily removed by considering $$g(x,y):=f(x,y)-4=\frac{\frac{4}{3}-3(y-\frac{2}{3})^2}{x^2+y^2+1}.$$

It is easy to show that $$\max_{x,y}{g(x,y)}=\max_y{g(0,y)=\max_y{\frac{4y-3y^2}{y^2+1}}}=1.$$

Therefore, the range of $f$ is $[0,5]$. The maximum is obtained when $x=0,y=\frac{1}{2}$.


问:
Limiting Value Sum $c^X$ for Uniform $RV$
I have $n$ draws $X_i$ from a Uniform r.v. on $[0,1]$ and want to find the limit of $\frac{1}{n} \sum_{i=1}^n c^{X_i}$ where $c$ is a known constant in $[0,1]$. My thought is to transform this summation into a function of $\frac{1}{n} \sum_{i=1}^n X_i$ and use the continuous mapping theorem, but have not had success finding the right transformation.

答: If $X_i$ are i.i.d., then by Law of Large Numbers the limit of $\frac{\sum_{i=1}^n c^{X_i}}{n}$ is just $E[c^{X_1}] = M_{X_1}(\ln(c))$ almost surely, where $M_{X_1}(t) = E[e^{t X_1}]$ is the moment generating function of $X_1$. It is known, that if $X_1 \sim U[0;1]$, that $M_{X_1}(t) = \frac{e^t - 1}{t}$. Thus $\frac{\sum_{i=1}^n c^{X_i}}{n}$ converges to $\frac{c-1}{\ln(c)}$ almost surely.


答: Just multiply  the polynomial by the common denominator of the coefficients. Eisenstein criterion works not only for monic polynomials, but also polynomials with leading coefficient not divisible by $p$.

Title: Eisenstein Criterion with Polynomials of Rational Coefficients?

Body: I'm wondering if it's possible to apply the Eisenstein criterion to polynomials that have rational, non-integer coefficients? I am thinking about using a shift to go to integers and apply the Eisenstein criterion there. Then, since f(x) is irreducible iff f(x+c) is irreducible, that should show that the original polynomial is irreducible. For instance, with the polynomial $x^5-(3/5)x^4+9x-3$, I wonder if there's some way to shift this so it gets integer coefficients and we can do the procedure I said?


问:
Check my answer for harmonic conjugate of $e^x Sin(y)$
I’m new to the topic
Can someone please check my answer for harmonic conjugate of $U(x,y)=e^x Sin(y)$?

My answer is $V(x,y)=-e^x Cos(y)+C$

答:
Yes you are correct.

First let's check that $u(x,y)$ is a harmonic function, that is it satisfies the  two-dimensional Laplace equation
\begin{equation}
\frac{\partial^{2 }u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0
\end{equation}
Since $\frac{\partial^{2}u}{\partial x^{2}}=e^x\sin(y)$ and $\frac{\partial^{2} u}{\partial y^{2}}=-e^{x}\sin(y)$, we conclude that $u(x,y)$ is harmonic.

Next, we find a harmonic conjugate to $u(x,y)$. We have \begin{equation*}
u_{x}=v_{y} \iff e^{x}\sin(y)=v_{y}
\end{equation*}
So integrating with respect to $y$
\begin{equation}
v(x,y)=-e^{x}\cos(y)+f(x) \tag{2}
\end{equation}
where $f(x)$ is a function depending only on $x$. We can determine this function from the Cauchy-Riemann equation $-u_{y}=v_{x}$ , which gives
\begin{align*}
-u_{y}=v_{x} &\iff -e^{x}\cos(y)=-e^{x}\cos(y)+f'(x) \\
&\iff f'(x)=0
\end{align*}
Then integrating with respect to $x$ we obtain
\begin{equation}
f(x)=C
\end{equation}
where $C$ is an arbitrary constant. Therefore the harmonic conjugate f $u(x,y)$ is given by $$v(x,y)=-e^{x}\cos(y)+C$$


问:
A mathematical model for rotations of a die
I have a normal 6-sided die and I would like to find a model for how it transforms under rotatins (north, south, east, west), that can help me determine which side is up. Just to make sure we are using the same die, here's how mine is numbered:

To give a few examples, when the die is in that position (1 facing up, 2 facing at me), if I rotate it south it will end up with 5 facing up and 1 facing me. Similarly, a rotation to the west gives 3 facing up and 2 facing at me.

I tried representing the die and its rotations as a graph where each node is one of the states the die can be in and there are four transitions for every node (corresponding to each of the four rotations). The problem is that there is a total of $24$ states and $4 * 24 = 96$ transitions and writing down all of them is laborious and very error prone.

The reason there are $24$ states is because when 1 is facing up and 2 is facing at me is quite different from when 1 is facing up and 3 is facing me. So a state of the die is a 2-tuple of the form $(\text{number facing up},\ \text{number facing me})$. (There are $6$ numbers that can face up and for each of them there are are $4$ numbers that can face me, so $24$.)

I need some help in finding a nicer mathematical model for determining what face is up, given a starting position and a sequence of rotations.

答:
See the expanded view of the cube.

I added an imaginary square (the half one) to the expanded view.
Let's define two sets such that;

$$\left( { y }_{ 0 },{ y }_{ 1 },{ y }_{ 2 },{ y }_{ 3 } \right) =\left( 1,2,6,5 \right) \\ \left( { x }_{ 0 },{ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 } \right) =\left( 1,3,6,4 \right) $$

The values in the first and third indices in both series are equal whenever you rotate, not ${ x }_{ 0 }={ y }_{ 0 }$ and ${ x }_{ 2 }={ y }_{ 2}$, i.e everytime you rotate the values comes to that indices will be equal.

Let's n be number of south or north. If it is south say $n>0$, if it is north say $n<0$. Every four rotation in that directions the series $\left( { y }_{ 0 },{ y }_{ 1 },{ y }_{ 2 },{ y }_{ 3 } \right) =\left( 1,2,6,5 \right)$ will repeat. Before completing four rotations if you pass east-west rotations you have to work with $ \left( { x }_{ 0 },{ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 } \right) =\left( 1,3,6,4 \right) $ by changing the first and third indices. So I will just explain for south-north rotation. By the way the first indices in the series is what you see on the upper face of the cube.

For any $n$ rotations in south-north direction your series will be:

$$\\ \left( { y }_{ 3n\; { mod }\; 4 },{ y }_{ 3n+1\; { mod }\; 4 },{ y }_{ 3n+2\; { mod }\; 4 },{ y }_{ 3n+3\; { mod }\; 4 } \right) $$

So what you see on the upper face will be $ { y }_{ 3n\; { mod }\; 4 }$. If you past to east-west rotations don't forget, the first and third indices of series $y$ should pass series $x$.


问:
Homomorphism from $S_3$ to ($\mathbb{Q},+)$
I am solving exercise in abstract algebra and could not solve this 1 correctly.

Does there exists a homomorphism from $S_3$ to the additive group ($\mathbb{Q},+)$ of rational numbers?

I think it exists. Map $A_3$ to $1$ and remaining elements to $-1$. But answer is no.

So, what mistake I am making? Please tell.

答: $\{0\}$ is the only finite subgroup of $(\Bbb Q,+)$, and hence there is no nontrivial homomorphism from any finite group $G$ to $(\Bbb Q,+)$.


问:
Determine a group's order by Universal Property (Mapping Property)
For a group represented as $\langle x,y\mid x^4,y^5,xyx^{-1}y\rangle$, how to determine its precise order? I guess I may need to use the universal property, but how to construct functions to determine size?

The question does have some steps or hints like the show it has at most 20 elements first, then construct some subjective to get its minimal order, etc.

I wonder what is the standard procedure to solve such questions

答:
For instance, in this particular case, first the third "anticommutativity relation" allows you to write every element in the form $a^ib^j$.  That together with the first two relations tells us that the order is at most $20$.

Next note that the elements of a certain semi-direct product $\Bbb Z_5\rtimes_\varphi\Bbb Z_4$ satisfy the given relations.  Namely the one with $\varphi:\Bbb Z_4\to\rm{Aut}\Bbb Z_5\cong\Bbb Z_4$ given by $\varphi(1)=2$.

This implies that there is a surjective homomorphism from $G$ to this group.  Thus we have a lower bound of $20$ for the order as well.


问:
Can I show that this sequence is bounded?
Let $(a_n)$,$(b_n)$ be real sequences with the following properties:

$$ a_n \to 0 \hspace{0.2cm} \text{as} \hspace{0.2cm} n\to \infty$$

$$ \frac{1}{n} \sum_{k=1}^n b_k  \hspace{0.2cm} \text{is a bounded sequence}$$

Question: Can I show that  $\frac{1}{n} \sum_{k=1}^n a_k b_k$ is a bounded sequence?

答:
No. One counterexample is
$$
a_k = \begin{cases}
1/\sqrt k, &\text{if $k$ is even}, \\
0, &\text{if $k$ is odd};
\end{cases}
\qquad
b_k = (-1)^kk.
$$


问:
Intuitive understanding of why the sum of nth roots of unity is $0$
Wikipedia says that it is intuitively obvious that the sum of $n$th roots of unity is $0$.  To me it seems more obvious when considering the fact that $\displaystyle 1+x+x^2+...+x^{n-1}=\frac{x^n-1}{x-1}$ and that $\{\alpha \in \mathbb{C}: \alpha^n=1\}$ is a cyclic group.

For me, it is hard to, intuitively, see this in the case that $n$ is an odd composite natural number.

Here is my reasoning thus far:

Let $n=5$. (I understand this is simple when considering the minimal polynomial).

Since the conjugate of an $n$th root of unity is also an $n$th root of unity, it is easy to see that, when written in trigonometric form, the sum (possible reordering here)

$1+\alpha^1+\alpha^2+\alpha^3+\alpha^4$

$=1+( \cos (72)+i \sin (72))+( \cos (144) + i \sin (144))+( \cos (72)-i \sin (72))+( \cos (144)-i \sin (144))$

$=1+2 \cos (72)+2 \cos (144)$.

Somehow $2 \cos (72)+2 \cos (144)=-1$, but I'm not sure this is obvious, or perhaps I'm being silly here.

What do you think?  Silly?

答:
If $\rm\:n>1\:$ then $\rm\:x^n - b\ x^{n-1} + \cdots\:$ has root sum $\rm\:= b\:,\:$ hence $\rm\:x^n - 1\:$ has root sum $= 0\:.$

At the heart of the matter is the following addition law:

$$\rm\ \ \ (x^k-{\color{red} r}\ x^{k-1}+\cdots\:)\ (x^n - {\color{red} s}\ x^{n-1}+\cdots\:)\:\ =\ \ x^{k\:+\:n} - ({\color{red}{r+s}})\ x^{k\:+\:n-1} + \cdots$$


问:
How do we find a lower bound on (P) without actually solving it?
So, if we have $(P)$ min $\{ 4x_1 - x_2 : 3x_1 - x_2 - x_3 = -2, x_1 + x_2 - x_4 =2, x \ge 0 \}$ how to find the lower bound without actually doing it ?

Body:
Substitute $x_{1,2}$ in terms of $x_{3,4}$.

$4x_1=x_3+x_4\\x_2=2-x_3/4+3x_4/4$

So $4x_1-x_2=5x_3/4+x_4/4-2$ which has minimum value $-2$ when $x_{3,4}=0$. Now check whether for $x_{3,4}=0$, we get $x_{1,2}\ge0$.


问:
Probability question - money notes into jars

Charlie and Mr. Z. are playing the following game:

Mr. Z. brings $10$ notes of 100£ and $10$ notes of 20£.

Charlie distributes the $20$ notes into $2$ jars, in a way he chooses.

Mr. Z. chooses one of the jars randomly.

Charlie chooses one note from the chosen jar randomly.

Charlie receives the chosen note and pays Mr. Z. 60£. (In other words, Charlie can either gain 40£ or lose 40£).

How should Charlie distribute the notes into the jars in order to maximize his chances of winning?

My first intuitive thought was that he should distribute the notes evenly, but I found out that it isn't the answer and didn't know how to prove that.

One straight-forward approach came to my mind suggests to search for the maxima of the following function, which represents the probability of Charlie's winning:

for $a$=number of 100£ notes in the first jar,  $b$=number of 20£ notes in the first jar, we get:

$$\mathbb{P}\left(Winning\right)=\frac{1}{2}\left(\frac{a}{a+b}+\frac{\left(10-a\right)}{\left(10-a\right)+\left(10-b\right)}\right)$$

but it doesn't involve much probability considerations (and honestly I'm not sure I recall the technique for that).

I would be glad to hear your suggestions to the problem. (It should be related to the subject of conditional probability.)

答:
Charlie should put one $100$ pound note in one jar and all the other notes in the other jar.  By doing so, he guarantees that he will win half the time, while still winning nearly half the time if Mr. Z picks the other jar.  Using your formula shows that the probability of Charlie winning with this choice is
$$\Pr(\text{winning}) = \frac{1}{2}\left(\frac{1}{1 + 0} + \frac{9}{9 + 10}\right) = \frac{1}{2}\left(1 + \frac{9}{19}\right) = \frac{1}{2} \cdot \frac{28}{19} = \frac{14}{19}$$


问:
How to transform quadratic terms to something like $u^2+v^2$?
I am told that by the substitution of $x = (a\cos \theta)u - (b\sin \theta) v$ and $y = (a\sin \theta)u + (b\cos \theta)v$, we can reduce $3x^2 + 2xy + 3y^2$ to $u^2 + v^2$.

I tried to plug in the substitutions, but I did not see how the quadratic term is reduce to $u^2 + v^2$. What I got was something complicated containing $a$, $b$, $\cos \theta$ and $\sin \theta$.

I am asked to reduce $3x^2 + 2xy + 3y^2 - x - 2y$ to $u^2 + v^2$ by similar substitutions. I do not know how to construct this substitution. Could you explain how do find this kind of substitutions in general?

答:
I think you are using a substitution more complicated than necessary. You also asked as to how such substitutions work in the general case, so I'll address that in this answer.

Consider the general bivariate quadratic polynomial,
$$f(x,y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c$$

There's a really neat way of writing this using matrices,
$$
f(x,y) =
\begin{pmatrix}x & y & 1\end{pmatrix}
\begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c\end{pmatrix}
\begin{pmatrix} x \\ y \\ 1\end{pmatrix} $$

Or like this,
$$
f(x,y) =
\begin{pmatrix}x & y\end{pmatrix}
\begin{pmatrix}a & h\\ h & b\end{pmatrix}
\begin{pmatrix}x \\ y\end{pmatrix} +
2\begin{pmatrix}g & f\end{pmatrix}
\begin{pmatrix}x \\ y\end{pmatrix}
+c
$$

which is even better as now the terms of degree $0, 1, 2$ are all separate. What we want to do given such a quadratic polynomial, is substitute the variables $(u, v)$ which will make the cross term $xy$ or $uv$ vanish. Note that we cannot always eliminate both the cross and linear terms.

Such a substitution is essentially a rotation of the axes of the plane from $x, y$ to $u, v$. So our problem reduces to finding the appropriate angle by which the axes should be rotated.

Consider the rotation matrix,
$$
R(\theta) =
\begin{pmatrix}
\cos \theta & \sin \theta\\
-\sin \theta & \cos \theta
\end{pmatrix}
$$

$R(\theta)$ is an orthonormal matrix with determinant 1, and hence, its inverse is simply its transpose. This can also be seen as the inverse operator must be a rotation by $-\theta$. So we get,

$$
R^{-1}(\theta) = R^{T}(\theta) = R(-\theta) =
\begin{pmatrix}
\cos \theta & -\sin \theta\\
\sin \theta & \cos \theta
\end{pmatrix}
$$

Now do the substitution
$$
\begin{pmatrix}u \\ v\end{pmatrix} =
R(\alpha)\begin{pmatrix}x \\ y\end{pmatrix}
$$
where $\alpha$ is the appropriate angle.

We get,
$$
f(u,v) =
\begin{pmatrix}u & v\end{pmatrix}
\underbrace{
R(\alpha)
\begin{pmatrix}a & h\\ h & b\end{pmatrix}
R^{-1}(\alpha)}_{M(\text{Let})}
\begin{pmatrix}u \\ v\end{pmatrix}
+
2\begin{pmatrix}g & f\end{pmatrix}
R^{-1}(\alpha)
\begin{pmatrix}u \\ v\end{pmatrix}
+c
$$

Doing the calculations, we find that,
$$
M =
\begin{pmatrix}
a\cos^2 \alpha + b\sin^2 \alpha + h\sin 2\alpha & h\cos 2\alpha - \dfrac{a-b}{2}\sin 2\alpha\\
h\cos 2\alpha - \dfrac{a-b}{2}\sin 2\alpha & a\sin^2 \alpha + b\cos^2 \alpha - h\sin 2\alpha
\end{pmatrix}
$$
where I've used the double angle identities $\cos 2t = \cos^2 t - \sin^2 t$ and $\sin 2t = 2\sin t \cos t$.

Therefore, to make the cross term vanish, the angle $\alpha$ must satisfy
$$h\cos 2\alpha - \dfrac{a-b}{2}\sin 2\alpha = 0$$

There are two cases:

$\alpha = \dfrac{\pi}{4}$ when $a = b$

$\tan 2\alpha = \dfrac{2h}{a-b}$

In your particular case, $a = b = 3$, so $\alpha = \dfrac{\pi}{4}$. The substitution is just $$(x, y) = \left(\dfrac{1}{\sqrt{2}}u - \dfrac{1}{\sqrt{2}}v, \dfrac{1}{\sqrt{2}}u + \dfrac{1}{\sqrt{2}}v \right)$$


问:
Convergence of a series of integrals: $S=\sum_{n\geq 1}(-1)^{n}\int_{n}^{n+1}\frac{1}{t e^t}\,dt$
Let

$$S = -\int_{1}^{2}\frac{1}{te^t}\,dt + \int_{2}^{3}\frac{1}{te^t}\,dt-\int_{3}^{4}\frac{1}{te^t}\,dt + \cdots +\text{ad inf}$$

Does the series $S$ converge? Clearly,

$$S=\sum_{n\geq 1}(-1)^{n}\int_{n}^{n+1}\frac{1}{t e^t}\,dt$$

So I thought of using alternating series test as there is a $(-1)^{n}$ term, but I am not sure how to estimate the integral term.

答:
Notice that $1/te^t$ is a monotone decreasing function as $t$ grows. Thus (readily evident from the interpretation of an integral as signed area),

$$\int_{n}^{n+1} \frac{1}{te^t} dt \le \frac{1}{ne^n}$$

I believe this should be enough to get you to the desired conclusion.


问:
Show that there is no pair s.t. $a^1+a^2+a^3+...+a^n=b^m$
Let $a,b,m$ and $n$ belong to integer. Is there any pair exist such that

$a^1+a^2+a^3+...+a^n=b^m$ where $a,b,m,n\ge 2$

Source code Pari/GP

```<code>for(m=2,10,for(n=2,10,for(a=2,100,for(b=2,100,if(sum(i=1,n,a^i)==b^m,print([a,n,b,m]))))))
</code>```

Edit: more on observation

There always have at least one solution for $\sum_{i=1}^{n}a^i=b^m+k$ for every integer $k\ne 0$

Apologies for the short question and shows no effort towards solution. Thank you.

答:
Your identity can be expressed as $$a\frac{a^n-1}{a-1}=b^m\,.$$
Observing that $\gcd(a, \frac{a^n-1}{a-1})=1$, this reduces to $$\frac{x^{mn}-1}{x^m-1}=y^m$$ assuming $a=x^m$ and $b=xy$ with $\gcd(x,y)=1$.
Now, when $n=2$, we obtain $$y^m-x^m=1\,,$$ which cannot have any solution for nonzero $x,y$. Thus $n\ge 3$; this essentially becomes part of the equation $$\frac{a^n-1}{a-1}=y^m$$ with $n\ge 3, |a|\ge 2,|y|\ge 2,m\ge 2$ that was investigated by Nagell and Ljunggren in the first half of the last century, which is conjectured to have the only known solutions $$ (a^n,y^m)\in\left\lbrace (3^5,11^2), (7^4,20^2),(18^3,7^3),((-19)^3,7^3)\right\rbrace\,,$$ none of which has the values of $a\in\{3,7,18,-19\}$ being an $m$th power. So your problem has no solutions conditional on the Nagell-Ljunggren conjecture. In fact, thanks to the $abc$-conjecture (if Mochizuki’s proof is\has been confirmed by experts), then Shorey (1986) showed that there are only finitely many solutions. (It may well be that your problem is easier to solve in general because the $a$ variable is an $m$th power; however, I have not considered this possibility carefully in its own right).


问:
Number of draws to see every object at least once
I woke up this morning with the following question on my mind:

What is (the distribution of) the number of draws you need to sample (with replacement) from an urn with $n$ distinct objects until you have seen every object at least once.

Calling this number $N$, it is quite easy to see that $\mathbb{E}[N]$ is of order $n\log n$. Moreover, simulations suggest that
$$
\frac{N-n \log n}{n}
$$
converges in distribution to some non-trivial, non-normal distribution $\mu$.

So my more precise question is: Has this observation been proven, and, if so, what is the distribution $\mu$?

Thanks.

答:
This problem is known as the Coupon Collector's Problem. For instance, it is known that $E(N) = n \log n + \gamma n + \frac{1}{2} + o(1)$, where $\gamma$ is the Euler-Mascheroni constant $\gamma \approx 0.577$, and $\mathrm{Var}(N) \leq 2n^2$. Hence, the quantity $\frac{N - n \log n}{n}$ you are interested in, satisfies:

\begin{align}
E\left(\frac{N - n \log n}{n}\right) &= \gamma + \frac{1}{2n} + o\left(\frac{1}{n}\right), \\
\mathrm{Var}\left(\frac{N - n \log n}{n}\right) &\leq 2.
\end{align}

Several other results to bound the tails of the distribution are known; see the Wikipedia-page for more on that.


答:
The function
\begin{equation}
f(t)=\frac{1}{te^t}
\end{equation}
is continuous, strictly decreasing on $[1,+\infty)$ and its limit as $t\to+\infty$ is $0$. This is enough to conclude that the terms
\begin{equation}
a_n=\int_n^{n+1}\frac{1}{te^t}\,\text{d}t
\end{equation}
are decreasing and converge to $0$ as $n\to+\infty$. Indeed, by the mean value theorem, there are $t_n\in[n,n+1]$ and $t_{n+1}\in[n+1,n+2]$ such that $a_n=f(t_n)$ and $a_{n+1}=f(t_{n+1})$, thus $a_n=f(t_n)>f(t_{n+1})=a_{n+1}$ and $a_n=f(t_n)\to 0$ as $n\to+\infty$. All the requirements of the alternating series test are therefore fulfilled and allow to conclude that the series converges.

Title: Convergence of a series of integrals: $S=\sum_{n\geq 1}(-1)^{n}\int_{n}^{n+1}\frac{1}{t e^t}\,dt$

Body:
Let

$$S = -\int_{1}^{2}\frac{1}{te^t}\,dt + \int_{2}^{3}\frac{1}{te^t}\,dt-\int_{3}^{4}\frac{1}{te^t}\,dt + \cdots +\text{ad inf}$$

Does the series $S$ converge? Clearly,

$$S=\sum_{n\geq 1}(-1)^{n}\int_{n}^{n+1}\frac{1}{t e^t}\,dt$$

So I thought of using alternating series test as there is a $(-1)^{n}$ term, but I am not sure how to estimate the integral term.


问:
How do I solve $32x \equiv 12 \pmod {82}$?
I am able to solve simpler linear congruences, for example $3x \equiv 2 \pmod 5$. What I would do in this case is use that $0 \equiv 10 \pmod 5$ and then utilising a theorem: $3x \equiv 12 \pmod 5$. Then I can divide by $3$ leaving me $x \equiv 4 \: \left( \mathrm{mod} \: {\frac{5}{\mathrm{GCD}(5,3)}} \right) \quad \Longleftrightarrow x \equiv 4 \pmod{5}$ which means the solution is $x = 4k + 5$ where $k \in \mathbb{Z}$.

But I cannot apply the same method to this congruence:
$$ 32x \equiv 12 \pmod {82} $$
This is how far I got:
$$ 8x \equiv 3 \: \left( \mathrm{mod} \: \frac{82}{\mathrm{GCD}(82, 4)} \right) $$
$$ \Updownarrow $$
$$ 8x \equiv 3 \pmod {41} $$

What could I do next? Please provide solutions without the Euclidean algorithm.

EDIT:

What I found later is that I can say that
$$ 0 \equiv 205 \pmod {41} $$
And then I can add it to the congruence in question and divide by $8$.
So I guess my question is essentially 'How can I find a number that is a multiple of $41$ (the modulus) and which, if added to $3$ gives a number that is divisible by $8$?'

I reckon the Euclidean algorithm is something which gives an answer to these kinds of questions?!

答:
$32x \equiv 12 \pmod{82} \; \text{ iff }$

$\quad (\exists \, x, k \in \Bbb N)\, [ 0 \lt x \lt 82 ; \land \; k \gt 0 \; \land \; 32x = 12 + 82k\,]$

Employing Euclidean division,

$\quad 82 = 32 \cdot 2 + 18$

So,

$\quad 32x = 12 + 32 \cdot (2k) + 18k$

Using this 'reduction pattern' we employ the corresponding algorithm:

$\alpha-$Solve:

$\;32x \equiv 12 \pmod{82}, \text{ and } \; 82 = 32 \cdot 2 + 18,\quad -12 + 32 = 20$

$\alpha-$Solve:

$\;18x \equiv 20 \pmod{32}, \text{ and } \; 32 = 18 \cdot 1 + 14,\quad -20 + 2\cdot 18 = 16$

$\alpha-$Solve:

$\;14x \equiv 16 \pmod{18}, \text{ and } \; 18 = 14 \cdot 1 + 4,\quad -16 + 2\cdot 14 = 12$

$\alpha-$Solve:

$\;4x \equiv 12 \pmod{14}, \text{ ANS: } \; x = 3 \text{ is the least residue solution}.$

Propagating backward,

Solve $14x = 16 + 3 \cdot 18, \text{ ANS: } \; x = 5$

Solve $18x = 20 + 5 \cdot 32, \text{ ANS: } \; x = 10$

Solve $32x = 12 + 10 \cdot 82, \text{ ANS: } \; x = 26$

$\text{ANSWER: }  x = 26 \text{ is the least natural number satisfying }$

$\quad 32x \equiv 12 \pmod{82}$


问:
The Simplicial Flabby Resolution of a Sheaf
I study sheaf cohomology by Demailly's book and I have a trouble. Is the inductive formula at the end of page 198
$$\mathcal{A}^{[q]}=(\mathcal{A}^{[q-1]})^{[0]}$$
right? I think that $(\mathcal{A}^{[0]})^{[0]} = \mathcal{A}^{[0]}$, is it true?

May be

$$\mathcal{A}^{[q]}=(\mathcal{A}^{[q-1]}/\mathcal{A})^{[0]}$$
or
$$\mathcal{A}^{[q]}=(\mathcal{A}^{[q-1]}/\mathcal{A}^{[q-2]})^{[0]}?$$
Is this resolution have any other name?

Thanks a lot!

答:
The explicit expression is indeed
$$\mathcal{A}^{[q]}=(\mathcal{A}^{[q-1]}/\mathcal{A}^{[q-2]})^{[0]}.$$

And the flabby sheaves $\mathcal A^{[q]}$ are also called the Godement sheaves of $\mathcal A$, while the long exact sequence

$$0\longrightarrow \mathcal A\longrightarrow \mathcal A^{[0]}\longrightarrow\cdots\longrightarrow \mathcal A^{[q]}\longrightarrow \mathcal A^{[q+1]}\longrightarrow \cdots$$
is also called the Godement canonical flabby resolution.


答:
Set $$b_n=\int_{n}^{n+1}\frac{1}{t e^t}\,dt,$$
then $$\frac{1}{(n+1)e^{n+1}}\leq b_n\leq\frac{1}{ne^{n}},$$
and using squeeze theorem,
$$\lim_{n\to +\infty}{b_n}=0.$$

Obviously, $\{b_n\}$ is decreasing, therefore $S=\sum_n{(-1)^nb_n}$ is convergent.


问:
How to prove that arithmetic progression contains non coprime pairs
There is a progression $a_n=1+bn$. Prove that there is at least one pair $(a_i,a_j)$ such as $\gcd(a_i, a_j) \neq 1$.

When the $a_0$ is not |1|, it quite easy to prove:
$$a_{|a|} = a_0 + |a_0|b$$
$$\gcd(a_0, a_0 + |a_0|b) \geq |a_0|$$
But when the $a_0$ is |1|, I don't know how to play with it.

答: If $a_n\ne \pm 1$ then $a_n$ is divisible by a prime $p,$  so $a_{(n+p)}=a_n+bp$ is also divisible by $p.$


问:
Pell's type equation in sum
I have an another observation on concatenation in sum. For instance, take $12^2 + 33^2 = 1233$. Find as many such pairs (x, y) with $x^2 + y^2 = xy\,$(here xy is concatenation)is possible. Also, discuss how this is happening and is there any connection between this kind of problem and pell's equations?
I got $10^2 + 1^2 = 101$;
$0^2 + 1^2 = 01$;
$88^2 + 33^2 = 8833$;
$12^2 + 33^2 = 1233$;
such pairs.</p>

答:
This is equivalent to $x^2 + y^2 = 10^n x + y$ with size constraints on $y$ (preferably $10^{n-1} \le y < 10^n$, though we will shortly see that the right-hand inequality is superfluous).

Completing the square gives $(2x - 10^n)^2 + (2y - 1)^2 = 10^{2n} + 1$, so we are looking for a way to write $10^{2n}+1$ as the sum of two squares $A^2+B^2$, where $A = 2x-10^n$ is even and $B=2y-1$ is odd.  Since clearly $B \le 10^n$, we needn't worry about $y$ being too large, but we'd like $B \ge \tfrac15\cdot10^n - 1$ so that $y$ doesn't need to be zero-padded before concatenation (otherwise we could use the trivial solution $A=10^n$, $B=1$ giving $x=10^n$, $y=1$).

A simple way to construct solutions is to choose $n$ in certain congruence classes so that $10^{2n}+1$ has a particular prime factor that yields a non-trivial decomposition as the sum of two squares.  For instance if $n = 16k+12$ we have $10^n \equiv -4 \pmod{17}$ and $17 \mid 10^{2n}+1$, which gives the decomposition

$$10^{2n}+1 = [\tfrac1{17}(15\cdot 10^{n}-8)]^2 + [\tfrac1{17}(8\cdot 10^n +15 )]^2.$$

Here, $B = \tfrac1{17}(8\cdot 10^n + 15)$ so we may take $y = \tfrac4{17}(10^n+4)$.  Similarly, $x = \tfrac4{17}(4\cdot 10^n-1)$.  Taking $k=0$ this gives

$$ \color{red}{941176470588}^2 + \color{blue}{235294117648}^2 = \color{red}{941176470588}\color{blue}{235294117648}.$$

I don't see a strong connection to Pell equations but it certainly is possible when working with conics like these.

Title: Pell's type equation in sum


问:
Lebesgue Differentiation Theorem-Type Problem
I just read over a proof of the Lebesgue differentiation theorem in $\mathbb{R}^n$ and attempted a related exercise in the book that I am using (the Tao Measure Theory book). I am unable to figure this out and was hoping to get some tips or hints:

For each $h > 0$, let $E_h \subset B(0, h)$ such that $m(E_h) \geq c \cdot m(B(0, h))$ for some $c > 0$ which is independent of $h$. If $f \colon \mathbb{R}^n \to \mathbb{C}$ is locally integrable, with $x$ a Lebesgue point of $f$, then prove that
$$\lim_{h \to 0} \frac{1}{m(E_h)} \int_{E_h + x} f(y) \, dy = f(x).$$

I know that this is something that follows from the Lebesgue differentiation theorem (LDT), but after a few steps I'm not really sure where to go. I started by noting that by translation invariance we have that $m(E_h) \geq c \cdot m(B(x, h))$ so since we have
$$\lim_{h \to 0} \frac{1}{m(B(x, h))} \int_{B(x, h)} f(y) \, dy = f(x),$$
by the LDT it follows that
$$\lim_{h \to 0} \frac{c}{m(E_h)} \int_{B(x, h)} f(y) \, dy \leq f(x),$$
but I have zero idea where to go from there. I'm not sure how to change the domain of integration without complicating things further, and I don't know how to get the other inequality either. Any tips or suggestions would be great!

答:
Wow, I feel stupid for not getting this sooner but never mind. By translation invariance of Lebesgue measure, we have that $m(B(0, h)) = m(B(x, h))$ for all $x \in \mathbb{R}^n$ since $B(x, h) = x + B(0, h)$. Note that if $E_h \subset B(0, h)$ then it follows that $x + E_h \subset B(x, h)$. Thus putting these two facts together along with the assumption that $m(E_h) \geq c \cdot m(B(0, x))$ for some $c > 0$, we see that $m(E_h) \geq c \cdot m(B(x, h))$ for all $x \in \mathbb{R}^n$. Then by the Lebesgue differentiation theorem, we have that that

$$\lim_{h \to 0} \frac{c}{m(E_h)} \int_{B(x, h)} |f(y) - f(x)| \, dy \leq \lim_{h \to 0} \frac{1}{m(B(x, h)} \int_{B(x, h)} |f(y) - f(x)| \, dy = 0$$

And since (E_h + x \subset B(x, h),$ since out integrand is nonnegative we have by monotonicity of domain that

$$\lim_{h \to 0} \frac{c}{m(E_h)} \int_{x + E_h} |f(y) - f(x)| \, dy \leq \lim_{h \to 0} \frac{c}{m(E_h)} \int_{B(x, h)} |f(y) - f(x)| \, dy \leq 0$$

And then obviously since our integrand is nonnegative this implies that

$$\lim_{h \to 0} \frac{c}{m(E_h)} \int_{x + E_h} |f(y) = f(x)| \, dy = 0 \Rightarrow \lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} |f(y) - f(x)| \, dy = 0,$$

meaning that

$$\left|\lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} f(y)- f(x) \, dy \right| \leq \lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} |f(y) - f(x)| \, dy = 0,$$

and thus

$$\left|\lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} f(y)- f(x) \, dy \right| = 0 \Rightarrow \lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} f(y)- f(x) \, dy = 0,$$

therefore by linearity of the integral we have that

$$\lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} f(y) \, dy = \lim_{h \to 0} \frac{f(x)}{m(E_h)} \int_{x + E_h} \, dy = \lim_{h \to 0} \frac{f(x)}{m(E_h)} \int_{\mathbb{R}^d} \mathbb{1}_{x + E_h}(y) \, dy$$

the right integral obviously equalling $m(x + E_h) = m(E_h)$ by translation invariance, and thus we have that

$$\lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} f(y) \, dy = f(x).$$


问:
If not associative, then what?
Consider a binary operation $*$ acting from a set $X$ to itself. It's useful and standard to work with operations which are associative, such that $(a*b)*c = a*(b*c)$. What about operations which are not associative?

Is there any way to characterize all different possible types of such binary operations $*$ which are not associative? Eg. Can we say that if $*$ is not associative,  then it must instead satisfy one of set of other possible properties, depending on any other additional operations that we have on our set $X$?

If we also add some additional structure to our set $X$ so that we can add elements together and multiply by scalars, it's standard to quantify the amount that two elements of $X$ commute with each other under $*$ by calculating the commutator $[a,b] = a*b - b*a$.
Is it ever useful to consider an 'associative commutator' $[abc] = (a*b)*c - a*(b*c)$, for a given non-associative $*$?

Finally, I know from Lie algebras that if $*$ anticommutes then it can be natural to consider a Jacobi identity

$(a*b)*c = a*(b*c) - b*(a*c)$

Are there other natural extensions of associativity in different settings?
Why do Lie algebras use this Jacobi identity and not for example

$(a*b)*c = a*(b*c) + k b*(a*c)$

Where k is a scalar?

答:
"What about operations which are not associative?" In many areas we encounter non-associative algebra structures, e.g., in operad theory, homology of partition sets, deformation theory, geometric structures on Lie groups, renormalisation theory in physics and many more.

In a certain sense one can answer your question what else can happen. One way is, to classify all nonassociative algebras defined by the action of invariant subspaces of the symmetric group $S_3$ on the associator of the considered laws, see for example here. But of course these are not all possibilities.

A well known example of a non-associative algebra structure related to Lie algebras are pre-Lie algebras (also called left-symmetric algebras).
They satisfy the identity
$$
(x,y,z)=(y,x,z)
$$
for all $x,y,z\in A$, where $(x,y,z)$ is the associator. In particular, associative algebras are a trivial example where both sides are zero, i.e., with $0=0$. Then the commutator
$$
[x,y]=xy-yx
$$
is a Lie bracket, see Is there a relationship between associators and commutators?

Pre-Lie algebra arise in algebra, geometry and and physics, see my survey article here. They play an important role for crystallographic groups, fundamental groups of affinity flat manifolds (Milnor), Gerstenhaber deformation theory, Rota-Bater operators and Yang-Baxter equations, just to name a few key words.

Title: If not associative, then what?


问:
Solving $\lim_{x\to 0}\left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)$ without series expansion or L'Hopital's rule
I want to find the following limit without using series expansion or L'Hopital's rule. I tried replacing $x$ with $2x$.

$$\begin{aligned}L_{1}&=\lim_{x\to 0}\left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)\end{aligned}$$

$$\begin{aligned}L_{2}&=\lim_{x\to 0}\left(\frac{4x+2x(\cos^2x-\sin^2x)-6\sin x\cos x}{32\sin x\cos x}\right)\end{aligned}$$

How to proceed. Any hints are appreciated. Thanks.

答:
Sorry for my previous misunderstanding on the question. I will add some detail for the path from @user's Hint.

Above all, for first order we have
$$
A_1=\lim_{x\to0} \frac{x-\sin x}{x^3}=\frac1{6}, \quad B_1=\lim_{x\to0} \frac{1-\cos x}{x^2} =\frac1{2}
$$
which is proved in that post before.

To pursue higher order result exampled by $B_2$ denoted as
$$
B_2=\lim_{x\to0} \frac{\cos x-(1-\frac1{2}x^2)}{x^4}=\lim_{x\to0} \frac1{x^2}\left(\frac1{2}-\frac{1-\cos x}{x^2}\right)
$$
so we have
$$
B_2=\lim_{x\to0} \frac1{4x^2}\left(\frac1{2}-\frac{1-\cos2x}{4x^2}\right)
$$
hence
$$
4B_2=\lim_{x\to0} \frac1{x^2}\left(\frac1{2}-\frac{1-\cos2x}{4x^2}\right)
$$
$$
\frac{B_2}{4}=\lim_{x\to0} \frac1{x^2}\left(\frac1{8}-\frac{1-\cos x}{4x^2}\right)
$$
whose subtraction gives
$$
\begin{aligned}
\frac{15}{4}B_2
&=\lim_{x\to0} \frac1{x^2}\left(\frac3{8}-\frac{\cos x-\cos2x}{4x^2}\right)=\lim_{x\to0} \frac1{x^2}\left(\frac3{8}-\frac{\sin^2 x-\sin^2\frac{x}{2}}{2x^2}\right)\\
&=\lim_{x\to0} \frac1{x^2}\left(\frac3{8}-\frac{4\sin^2\frac{x}{2}(1-\sin^2\frac{x}{2})-\sin^2\frac{x}{2}}{2x^2}\right)\\
&=\lim_{x\to0} \frac1{x^2}\left(\frac3{8}-\frac{3\sin^2\frac{x}{2}}{2x^2}\right) + \lim_{x\to0} \frac{2\sin^4\frac{x}{2}}{x^4}\\
&=\lim_{x\to0} \frac3{8}\left(\frac{x^2-4\sin^2\frac{x}{2}}{x^4}\right) + \lim_{x\to0} \frac{2\sin^4\frac{x}{2}}{x^4}\\
&=\lim_{x\to0} \frac3{8}\left(\frac{x+2\sin\frac{x}{2}}{x}\right)\left(\frac{x-2\sin\frac{x}{2}}{x^3}\right) + \lim_{x\to0} \frac{2\sin^4\frac{x}{2}}{x^4}\\
&=\lim_{x\to0} \frac3{8}\left(\frac{\frac{x}{2}+\sin\frac{x}{2}}{\frac{x}{2}}\right)\left(\frac{\frac{x}{2}-\sin\frac{x}{2}}{4\cdot(\frac{x}{2})^3}\right) + \lim_{x\to0} \frac{\sin^4\frac{x}{2}}{8\cdot(\frac{x}{2})^4}
=\frac5{32}
\end{aligned}
$$
where we need to recall the value of $A_1$, and obtain
$$
B_2=\frac1{24}
$$
I think you can as well get $A_2$, which is
$$
A_2=\lim_{x\to0} \frac{\sin x-(x-\frac1{6}x^2)}{x^5}=\frac1{120}
$$
by almost the same approach, and that is the resources to solve $L_1$.

However $L_2$ is trivial
$$
L_2=\lim_{x\to0} \frac{2x}{16\sin 2x}\left(2+\cos2x-\frac{3\sin2x}{2x}\right)=\lim_{x\to0} \frac{x}{16\sin x}\left(2+\cos x-\frac{3\sin x}{x}\right)=0
$$
for we do not need to calculate the infinitesimal with high order, or if you want, you can get the proper result from $L_1$, which is
$$
2+\cos x-\frac{3\sin x}{x} \sim \frac{x^4}{60} \quad (x\to0)
$$
From another view, this is a hint how to make up this expression, just taking these lower order items back we easily have
$$
2+\cos x-\frac{3\sin x}{x} = \left(\cos x-1+\frac{x^2}{2}\right)-3\left(\frac{\sin x}{x}-1+\frac{x^2}{6}\right) \sim (B_2-3A_2)x^4 = \frac{x^4}{60}
$$
which is very similar to the path of series expansion.


问:
$f^{-1}f(\Sigma)$ has measure zero for analytic map from manifolds $M$ to $N$
Differential Topology Hirsch Chapter 3 Section 1 Problem 4: (a) Let $M$ be a connected manifold and $f: M \rightarrow N$ an analytic map. Let $\Sigma \subset M$ be the set of critical points. If $\Sigma \neq M$ then $f^{-1}f(\Sigma)$ has measure zero.

(b) If $f$ is merely $C^{\infty}$ The conclusion (a) can be false.

For part (a) A function is analytic iff its Taylor series about $x_0$ converges to the function in some neighborhood for every $x_0$ in the domain. The definition of analytic function is the same as $C^{\infty}$ function except for the notion of convergence in a neighborhood of nonzero radius in the former. Therefore, we can apply the Morse-Sard Theorem and conclude that the set of regular values is dense. A critical point cannot be a regular value so $f^{-1}f(\Sigma)$, which is the set of critical points must have measure 0.

For part (b) I didn't use anything about the function being analytic in part (a) so I'm not sure how I would approach this part.

Thank you!

答:
I am trying to do this question myself, and I have not been able to do it yet . Now for the counterexample part of a smooth function $f: M\rightarrow N$ such that $f^{-1}(f(\Sigma))$ doesn't have null measure, I think an idea would be to consider $M=N=\mathbb{R}$ and take a bump function. We would have that $\Sigma\neq \mathbb{R}$ since there are points where the derivative is non-zero and so the differential is surjective, and we would have that $f^{-1}(f(\Sigma))$ wouldn't have measure zero.  So I don't think your argument works, but I haven't been able to prove this myself. I am guessing it will use some local property of real analytic functions , but I am not very aware of these so I will try to look things up.

Here's the attempt at proving the statement , the idea is similar to what Hirsch did to prove the Morse-Sard theorem, and notice we can assume that $\dim M\geq \dim N$:

Notice that we can do this locally and assume we have a function $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$ analytic.
Let $\Sigma_f$ denote the set of critical points of $f$.We will see that $m(\Sigma_f)=0$.  We have that $f_i(x_1,...,x_m)=\sum a^i_{k_1}...a{k_m}(z)(x_1-z_1)^{k_1}...(x_m-z_m)^{k_m}$. We can divide $\Sigma_f$ into two sets, let $\Sigma_1$ denote the set of points $x\in \Sigma$ such that $\nabla f_i(x)\neq 0$ for some differential operator of order $\geq 2$, and $\Sigma_2$ the set of $x\in \Sigma$ such that $\frac{\partial f_i}{\partial x_j}(x)\neq 0$ for some $i\in \{1,..,n\}$ and $j\in \{1,..,m\}$. We notice that $\Sigma\subset \Sigma_1 \cup \Sigma_2$ since if we have that $\frac{\partial f _i}{\partial x_j}(z)=0$ and $\nabla f_i(z)=0$ for every $i,j$ and differential operator of order $\geq 2$ then we will have that $a^i_{k_1}...a_{k_m}(z)=0$ which would contradict the fact that $f\neq 0$ since $M$ is connected.

Now Let's see that $\Sigma_1- \Sigma_2$ has null measure. We can consider the set of points $x\in M$ such that $\nabla f(x)= 0$ and $\frac{\partial f_i}{\partial x_j}\neq 0$, and this set will form a submanifold of dimension $m-1$ which we will call $M_{\nabla },i,j$. We have that $\Sigma_1-\Sigma_2 \subset \cup_{\nabla,i,j}X_{\nabla,i,j}$ where this union is over a countable set , and since $M$ is a $n-1$\submanifold it will have null measure so that $\Sigma_1-\Sigma_2$ has null measure.

If $x\in \Sigma_2$ we will have that there exists $f_i$ such that $\frac{\partial f_i}{\partial x_j}(x)\neq 0$ and so there exists an open set $U$ such that $\frac{\partial f_i}{\partial x_j}(x)\neq 0$ for $x\in U$.Now using the Implicit function theorem for analytic functions we have that there exists a function $h:A\times B\rightarrow U$ such that $f_i(h(x,t))=t$, for $(x,t)\in A\times B$. And so $f(h(x,t))=(u_t(x),t)$ and so being a critical point of $f$ is equivalent to being a critical point for $u_t:A\subset \mathbb{R}^{n-1}\rightarrow \mathbb{R}^n$, and so by induction hypothesis we have that $m(\Sigma_{u_t})=0$ and then using the fubini theorem we get that $m(\Sigma_f\cap U)=0$. Since we can cover $\Sigma_f$ by a coutable number of these open sets we get the desired result.

Now we already know that $\Sigma_f\subset f^{-1}(f(\Sigma))$ has zero measure , we just need to see that $f^{-1}(f(\Sigma))-\Sigma_f$ also has null measure. Let $x\in f^{-1}(f(\Sigma_f))-\Sigma_f$ then we will have that $d_x f$ is surjective. And so there will exist an open set $x\in U=A\times B\subset M$ and parametrizations such that $\bar f=\psi^\circ f\circ \phi^{-1} : U\rightarrow \mathbb{R}^n$ looks like the canonical projection. Since $K:=\hat f(\phi(\Sigma_f))\subset A$ has null measure by Sard's theorem, we will have that $\hat f^{-1}(K)=K\times B$ will have null measure by Fubini's theorem, and so $m(\hat f^{-1}(f(\Sigma_f)))=0$. Now since $\phi^{-1}$ takes null measure sets into null measure sets we have that $m(\phi^{-1}(\hat f^{-1}(f(\Sigma_f))))=m( f^{-1}(f(\Sigma))\cap U)=0$.


问:
Is Socle of evey locally Noetherian module a direct summand
Definition: An $R$-module $M$ is called locally Noetherian if, any finitely generated submodule of $M$ is Noetherian.

Question: Let $R$ be a ring with identity and $M$ be a  locally Noetherian $R$-module. Is $soc(M)$   a direct summand of $M$.

My attempt: Let $S=soc(M)$=sum of  simple submodules of $M$. Since each simple submodule is finitely generated $R$-module, by hypothesis, $S$ is a sum of Noetherian simple $R$-modules. Now, how should I proceed?

答:
You can't prove it, because it's false. An obvious counterexample is $\mathbb{Z}/4\mathbb{Z}$ which is Noetherian, hence locally Noetherian, but indecomposable.

If you want a counterexample with a locally Noetherian module which is not Noetherian, take the Prüfer group $\mathbb{Z}(p^\infty)$ for any prime $p$. This is indecomposable as well.

Both examples are Artinian modules and the socle of an Artinian module is nonzero and never a direct summand except in the trivial case when the module is semisimple and finitely generated.


问:
Banach Vector Subspace of Banach Space if and only if closed
Question: Verify that if $E$ is a Banach Space, then the Vector subspace $V \in E$ is Banach (with restriction of the norm) if and only if it's closed.

I managed to prove that if, in the question conditions, $V$ is closed, then it's Banach.

But if $V$ is banach, i'm not able to prove that it is closed.

I tried to make the following work:
Suppose that $V$ isn't closed. So, there is a sequence $\{x_n\} \in V$ with ${x_n} \rightarrow x_0$ and $x_0 \notin V$. From that, i tried to prove that i can construct $x_0$ with elements of the sequence and get a contradiction, but i didn't manage to do it.

Also, i cannot fit the fact that $E$ and $V$ are Banach into solving the question. Maybe this info just doesn't matter (don't know).

So, if you have any hints, would be really helpful.
Thanks!

答: If $\{x_n\}\subset V$ converges to $z\in E$, then $\{x_n\}$ is a Cauchy sequence. Since $V$ is complete, there must be a limit $y\in V$. By the uniqueness of the limit, $y=z$ and so $z\in V$. As pointed out in the comments, this proof is valid in any complete metric space.


问:
$V / K \times K \cong V$ for vector spaces
Claim. If $G$ is a group with a normal subgroup $K$, then $G/K \times K \cong G$.

Proof. $G/K$ is quotient group consisting of equivalence classes. Pick a representative $[g]$ for each equivalence class. Then consider the map $\varphi : G / K \times K \longrightarrow G$ defined by $([g], k) \mapsto gk$. The map is well-defined since we picked a single representative for each equivalence class. If $\varphi([g], k) = gk = g'k' = \varphi([g'], k')$, then $g = g' k' k^{-1} \in g'K$, so $[g] = gK = g'K = [g']$ and thus $g = g'$ since we have chosen one specific representative. It follows from $g = g'$ and $gk = g'k'$ that $k = k'$, so $([g], k) = [g'], k')$; therefore $\varphi$ is injective. Clearly $\varphi$ is surjective: if $x \in G$, then $x \in [g]$ for some representative $g \in G$, so $x = g\tilde{k}$ for some $\tilde{k} \in K$, so $\varphi([g], \tilde{k}) = g \tilde{k} = x$. $\tag*{$\square$}$
The proof is fake and the claim is false. All we have is a bijection between $G/K \times K$ and $G$. There are counterexamples: Is $ G \cong G/N \times N$?. Also relevant: When does the isomorphism $G\simeq ker(\phi)\times im(\phi)$? hold?.

Suppose $V, W$ are vector spaces over a field $K$. Let $\phi : V \to W$ be a linear transformation. If $V$ is finite-dimensional, then $V \cong \ker \phi \times \text{im} \, \phi$ via some isomorphism $\tau$. Then we can decompose $\phi$ as $\phi = i \circ p \circ \tau$, where $p: \ker \phi \times \text{im} \, \phi \to \text{im} \, \phi$ is projection $(a, b) \mapsto b$ and $i : \text{im} \, \phi \to W$ is inclusion $w \mapsto w$. So when $V$ is finite-dimensional, any linear transformation $\phi : V \to W$ is a composition of isomorphisms, inclusions, and projections. I want to prove that this fact is still true when $V$ is infinite-dimensional. It suffices to prove that if $\phi : V \to W$ is linear, then $V \cong \ker \phi \times \text{im} \, \phi$.

My first thought was to use $\text{im} \, \phi \cong V / \ker \phi$ and conclude $\text{im} \, \phi \times \ker \phi \cong V / \ker \phi \times \ker \, \phi \cong V$. But based on the above fake fact, I don't immediately expect the last equality to be true. It turns out that if $V/K$ is finite-dimensional, then $V/K \times K \cong V$ (see here). So I don't know how to handle the case that both $V$ and $V/\ker \phi$ are infinite-dimensional.

I thought I would find an answer at Is a vector space isomorphic to the kernel $\oplus$ image of a map out of it?. But that question seems to assume that the bijection in the fake proof above is a vector space isomorphism. Just before posting the question, I found an answer in the "similar questions" suggestion: if $\phi : V \to W$ is a linear transformation between vector spaces, then $V \cong \ker \phi \times \text{im} \, \phi$. But I am still curious if the original proof idea will work. Specifically:

If $V$ is an infinite-dimensional vector space and $K$ is a subspace of $V$ such that $V/K$ is infinite-dimensional, then is $V/K \times K \cong V$?

答:
Yes, if you assume the Axiom of Choice. Assuming choice, any family of linearly independent vectors of a vector space may be extended to a basis  (and in particular any vector space has a basis).

Now pick a basis $B'$ of $K$, extend it to a basis $B$ of $V$. Set
$L=Span(B\setminus B')$. Then $B\setminus B'$ is a basis of $L$, and it is not difficult to prove that $V=L\oplus K$. Then projection on $L$ induces an isomorphism $V/K\simeq L$. But then $V=L\oplus K\simeq L\times K\simeq V/K\times K.$


问:
Probability puzzle with cards
Given one hundred cards with numbers 1-100 on each card and two baskets, you throw each of these cards with 50% of probability falling in the left basket and 50% to the right. What is the expected value of the minimum number in the left basket?

My thoughts are:

Let S be the r.v. corresponding to the minimum number in the left basket.
Then $E(S) = \sum_{s=0}^{100} P(S > s)$. For s > 0, I think the probability that S > s is $\frac{1}{2^s}$, because this event happens when the numbers 1, 2, ..., s all landed in the right basket. I think $P(S > 0) = 1 - \frac{1}{2^{100}}$ because the minimum number is greater than 0 as long as any card landed in the left bin. Does this seem correct?

Extension

What changes if we ask what is the expected value of the minimum in the basket that contains the number 100?

答:
Suppose there are $n$ cards with numbers from $1$ to $n$ and the probability of any of them falling into the left basket is $p$.

Suppose, that when you toss cards you start with the smallest numbers, so that the number of the toss coincides with the number f the card tossed (we can do this, because all cards are tossed independently and their actual order does not determine anything). Suppose $X_i$ is $1$ if card with number $i$ is in the left basket and $0$ otherwise.

Then the minimal number in the left basket will be $\min\{i| X_i = 1\}$ if it is non empty, and $0$ otherwise (the probability of which is $(1-p)^n$)

Then for any $k < n$ the probability that $\min\{i| X_i = 1\} = k$ is $(1-p)^{k-1}p$.

Thus, your expectation is $\sum_{k = 1}^n k(1-p)^{k-1}p$.

EDIT 1:

There is also a way to calculate the same number differently (with the method, that OP was using).

The probability, that $\min\{i| X_i = 1\} \geq k$ for $k \in 1 ... n$ is the probability, that $X_1 = ... = X_{k-1} = 0$, but $\max{X_k, ... , x_n} = 1$. And that is $(1-p)^{k-1}(1 - (1-p)^{n-k + 1}) = (1-p)^{k-1} - (1-p)^n$.

Thus our expectation can be alternatively expressed as $$\sum_{k = 1}^n P(\min\{i| X_i = 1\} \geq k) = \sum_{k = 1}^n ((1-p)^{k-1} - (1-p)^n) = \frac{1 - (1-p)^n}{p} - n(1-p)^n$$

Thus we get an alternative solution, as well as prove algebraic identity

$$\sum_{k = 1}^n k(1-p)^{k-1}p = \frac{1 - (1-p)^n}{p} - n(1-p)^n$$

EDIT 2:

If there is already have a card with number $n+1$ in the basked before we started tossing the cards, then the only value different will be the value in the case when $X_1 = ... = X_n = 0$ (it will change from $0$ t $n+1$). Because the probability of this event is $(1-p)^n$ the expectation will increase by $(n+1)(1 - p)^n$ and become

$$\frac{1 - (1-p)^n}{p} - n(1-p)^n + (n+1)(1-p)^n = \frac{1 - (1-p)^n}{p} + (1-p)^n = \frac{1 - (1-p)^{n+1}}{p}$$

Title: Probability puzzle with cards