问:
Prove that $\mathbf{a}^H\mathbf{b}+\mathbf{b}^H\mathbf{a} = 2\operatorname{Re}(\mathbf{b}^H\mathbf{a})$?
Why would the following identity hold for any two complex vectors $\mathbf{a}, \mathbf{b} \in \mathbb{C}^N$
$$\mathbf{a}^H\mathbf{b}+\mathbf{b}^H\mathbf{a} = 2\operatorname{Re}(\mathbf{b}^H\mathbf{a})$$

This is taken from Steven M. Kay's Fundamentals of Statistical Signal Processing - Detection Theory page 475.

Does this identity have a name? Any help is appreciated, thanks.

答:
For any complex number $z=z_r+iz_i$ and its complex conjugate $\overline{z}=z_r-iz_i$ one can see that
$$z+\overline{z}=z_r+iz_i+z_r-iz_i=2z_r=2\operatorname{Re}(z)$$

The scalar product for any two complex vectors $\mathbf{a}, \mathbf{b} \in \mathbb{C}^N$ is
$$\mathbf{a} \cdot \mathbf{b} = \mathbf{a}^{\operatorname{H}}\mathbf{b} = \sum_{k=1}^{N} \overline{a}_kb_k$$ where $\mathbf{a}^{\operatorname{H}}=\overline{\mathbf{a}^{\operatorname{T}}}$ denotes the conjugate transpose.

Thus, one has
$$\mathbf{a}^H\mathbf{b}+\mathbf{b}^H\mathbf{a} \\
=\sum_{k=1}^{N} \overline{a}_kb_k +\sum_{k=1}^{N} \overline{b}_ka_k \\
=\sum_{k=1}^{N}\overline{a}_kb_k + \overline{b}_ka_k \\
=\sum_{k=1}^{N}\overline{\overline{b}_ka_k} + \overline{b}_ka_k \\
=\sum_{k=1}^{N}2\operatorname{Re}(\overline{b}_ka_k) \\
=2\operatorname{Re}\left(\sum_{k=1}^{N}\overline{b}_ka_k\right) \\
=2\operatorname{Re}(\mathbf{b}^H\mathbf{a})
$$

Thank you @Semiclassical for the comment that provided the answer.

References:
complex dot product,
complex identities


问:
Is $|(1 - k\epsilon)(2 - k\epsilon)...(n - k\epsilon)| \leq n!$
Let $\epsilon \in \mathbb{R}$, $0 \le \epsilon \le 1$. Let $n$ be a positive integer. Let $k$ be a non-negative integer, $0 \leq k \leq n+1$.

Is it true that $|(1 - k\epsilon)(2 - k\epsilon)...(n - k\epsilon)| \leq n!$

I tried some numbers and it seems to work but I am not sure if it's true in general. Any hints would be appreciated.

答:
Let's replace $n+1$ by $n$ and $k\epsilon$ by $x$. Then the question is:

For an integer $n>1$, and $x\in\mathbb{R}$, let $f_n(x)=\prod_{k=1}^{n-1}|k-x|$.
Show that $f_n(x)\leqslant f_n(0)$ for $0\leqslant x\leqslant n$.

We may assume that $x$ is not an integer. Since $f_n(x)=f_n(n-x)$, we may also assume that $x\leqslant n/2$. Now, if $x>1$ then $f_n(x)/f_n(x-1)=(x-1)/(n-x)<x/(n-x)\leqslant 1$. Thus, by induction (i.e. after repeated decrease of $x$ by $1$), we may assume that $0<x<1$. Finally, $f_n(x)$ is clearly decreasing on this interval, and we're done.

ContentLicense: CC BY-SA 4.0


问:
Locating roots of a complex function
Let $a,b,c$ be real numbers. $a>0$, $|c|>1$, and $b^2-2ac >0 $. Find the number of roots of the function $f(z) =az^2+bz+c-e^{-z}$ in the region $\{z : Re(z)\geq 0\}$ for every $a,b,c$. When will it have one real root on the real axis?

I have tried using Rouche's theorem in a semicircunference with center $0$ and radius R, but I didn't get any results. I also tried studying the roots of $g(z) =p(z)p(\overline{z})-e^{-z}$ where $p(z)=az^2+bz+c$ using again Rouche's theorem and I got to some results but I didn't really know how to link them with the roots of f. I would like to know how to aproach this problem.

答:
Let $p(z) = az^2+bz+c$. For $z=iy$ with $y \in \Bbb R$  we have
$$
 |p(z)|^2 = |(c-ay^2) + iby|^2 = (c-ay^2)^2 + b^2y^2 \\
= c^2 + (b^2-2ac)y^2 + a^2y^4 \ge c^2 > 1
$$
and therefore
$$
 |f(z) - p(z)| = |e^{-z}| = 1 < |p(z)| \, .
$$
Also
$$
 |f(z) - p(z)| = |e^{-z}| \le 1 < |p(z)| \, .
$$
if $|z| = R$ is sufficiently large. So Rouché's theorem shows that $f$ and $p$ have the same number of zeros in the right half-plane (which can be zero, one, or two).

Title: Locating roots of a complex function


问:
one limit divided into two limit factors, is it can one determine result directly but other don't?
there is a question:

$$
\lim_{x->0}{\frac{x-arcsinx}{x^3(sinx+1)}} = \lim_{x->0}{\frac{1}{sinx+1}} · \lim_{x->0}{\frac{x-arcsinx}{x^3}}
$$

[1]you know there have four place x.
we know in limit calculation, if numerator x tend to 0, the denominator x must tend to 0 too, so can get the correct result.

but in the given question, there divided into two lim's multiplication:$$\lim_{x->0}{\frac{1}{sinx+1}} · \lim_{x->0}{\frac{x-arcsinx}{x^3}}$$

there I named them as lim_A· lim_B.

in my opinion, the lim_A and lim_B comes from one entirety, if lim_A's x tends to 0, the lim_B's x should tends to 0 too.

but however, in the question's solution there only let lim_A's x ->0 ignore the lim_B's x,

$$\lim_{x->0}{\frac{1}{sinx+1}} · \lim_{x->0}{\frac{x-arcsinx}{x^3}} = \lim_{x->0}{\frac{x-arcsinx}{x^3}}$$

so, why ? is there any theorem state this step? (I don't understand because I think the lim_A and lim_B comes from one entity, if one x->0, the others(lim_B's x) should x->0 too.)

答:
Note:  your English or typing needs help!  The word is opinion,not option.However there is no reason why both limits (not limitations) should $\to 0$.

For lim_B, get the first two terms in the power series for $arcsin(x)=x+\frac{x^3}{6}+...$  The answer is lim_B$=-\frac{1}{6}$


问:
Proving that $P\left(\bigcup_{j=1}^n E_j^{(n)}\right)\sim\sum_{j=1}^n P(E_j^{(n)})$ for independent events $E_j^{(n)}$

For arbitrary events $\{E_j, 1\le j\le n\}$, we have

$$P\left(\bigcup_{j=1}^n E_j\right)\ge\sum_{j=1}^n P(E_j) - \sum_{1\le j < k \le n} P(E_jE_k)$$

If $\forall n: \{E_j^{(n)}, 1\le j\le n \}$ are independent events, and

$$P\left(\bigcup_{j=1}^n E_j^{(n)} \right)\to 0 \quad \text{as} \quad n\to\infty$$

then $P\left(\bigcup_{j=1}^n E_j^{(n)} \right) \sim \sum_{j=1}^n P(E_j^{(n)})$.

My attempt at showing this last claim:

I am going to guess that the ~ symbol in the problem means that

$$\lim_{n\to\infty}\frac{P\left(\bigcup_{j=1}^n E_j^{(n)} \right)}{\sum_{j=1}^n P(E_j^{(n)})}=1$$

From

$$\sum_{j=1}^n P(E_j^{(n)})\ge P\left(\bigcup_{j=1}^n E_j^{(n)}\right)\ge\sum_{j=1}^n P(E_j^{(n)}) - \sum_{1\le j < k \le n} P(E_j^{(n)}E_k^{(n)})$$

and dividing we have

$$1\ge \frac{P\left(\bigcup_{j=1}^n E_j^{(n)}\right)}{\sum_{j=1}^n P(E_j^{(n)})} \ge 1 - \frac{\sum_{1\le j < k \le n} P(E_j^{(n)}E_k^{(n)})}{\sum_{j=1}^n P(E_j^{(n)})}$$

so then the problem reduces to showing

$$\lim_{n\to\infty}\frac{\sum_{1\le j < k \le n} P(E_j^{(n)}E_k^{(n)})}{\sum_{j=1}^n P(E_j^{(n)})} = 0$$

But how to make progress after that?

答:
Notice that $$\sum_{1\leqslant j<k\leqslant n}\mu(E_jE_k)=\sum_{1\leqslant j<k\leqslant n}\mu(E_j)\mu(E_k)=\frac 12\left(\left(\sum_{j=1}^n\mu(E_j)\right)^2-\sum_{j=1}^n\mu(E_j)^2\right).$$
Dividing by $\sum \mu(E_j)$, the first term is not problematic. For the second one, write
$$\sum_{j=1}^n\mu(E_j)^2\leqslant \max_{1\leqslant k\leqslant n}\mu(E_k)\cdot \sum_{j=1}^n\mu(E_j)\leqslant \mu\left(\bigcup_{1\leqslant k\leqslant n}E_k\right)\cdot \sum_{j=1}^n\mu(E_j).$$</p>

ContentLicense: CC BY-SA 3.0


答:
We have $0 \le k\epsilon < n+1$.

For $k \epsilon \ge 1$, we take $m = \lfloor k \epsilon \rfloor$. For $0 \le k \epsilon < 1$, we take $m = 1$.

Either way we have $|m - k \epsilon| \le 1$ for some $1 \le m \le n$.

Now we have:
$$|1-k\epsilon|\le |m-k\epsilon| + |1-m| \le m$$
$$:$$
$$|(m-2)- k\epsilon| \le |m- k\epsilon|+|-2| \le 3$$
$$|(m-1)- k\epsilon| \le |m- k\epsilon|+|-1| \le 2$$
$$|m- k\epsilon| \le 1$$
$$|(m+1)- k\epsilon| \le |m- k\epsilon|+|1| \le 2\le m+1$$
$$|(m+2)- k\epsilon| \le |m- k\epsilon|+|2| \le 3 \le m+2$$
$$:$$
$$|n - k\epsilon| \le |m-k\epsilon| + |n-m| \le n-m+1\le n$$

Hence we see that $|(1-k\epsilon)(2-k\epsilon)\dots(n-k\epsilon)| \le m!(n-m+1)! \le n!$ by term-by-term comparison.

Title: Is $|(1 - k\epsilon)(2 - k\epsilon)...(n - k\epsilon)| \leq n!$

Body:
Let $\epsilon \in \mathbb{R}$, $0 \le \epsilon \le 1$. Let $n$ be a positive integer. Let $k$ be a non-negative integer, $0 \leq k \leq n+1$.

Is it true that $|(1 - k\epsilon)(2 - k\epsilon)...(n - k\epsilon)| \leq n!$

I tried some numbers and it seems to work but I am not sure if it's true in general. Any hints would be appreciated.


问:
If $|X|\leq c$ then $|\mathbb{E}(XY)|\leq c |\mathbb{E}(Y)|$
Assume that $X,Y$ are two $L^2$ random variables such that there exists a deterministic constant $c>0$ with $|X|\leq c$  almost surely. Do we have the following inequality ? $$|\mathbb{E}(XY)|\leq c |\mathbb{E}(Y)|$$

答: If $X \sim \text{Uniform}[-1,1]$ and $Y = X$ with probability $1$, then $|X| \le 1 =: c$ almost surely, but we have $|\mathbb{E}(XY)| = |\mathbb{E}(X^2)| = \tfrac{1}{3}$ and $c|\mathbb{E}(Y)| = 0$.


问:
Continuity of Functions with Vertical Tangents
I'm running into some confusion regarding properties of continuous functions. I'm comfortable with the epsilon-delta definition of limits and the basic definition of continuity at a point $a$ ($\lim_{x\rightarrow a}f(x)$ must exist, $f(a)$ must exist, and the two must equal one another), but I'm frequently encountering the statement that a function is continuous if and only if "a small change in $x$ produces a small change in $f(x)$". This seems reasonable enough, but continuous functions with vertical tangents seem to present a contradiction to this assertion. For example, the function $f(x)=\sqrt[3]{x}$ is continuous at $x=0$ by the definition of continuity, but its derivative at $x=0$ is undefined because the line tangent to the curve $y=f(x)$ is vertical. Shouldn't this mean, then, that at $x=0$ a small change in $x$ produces an infinite change in $y$? I'm not sure if $\left.\frac{dy}{dx}\right|_{x=0}=\infty$ constitutes an abuse of notation, but certainly it is true that $f^\prime(0)$ is undefined and $$\lim_{x\rightarrow 0}f^\prime(x)=\infty\text{.}$$ So I'm not sure how this "small change in $x$/small change in $f(x)$" description of continuity holds in this particular case.  Can anyone help resolve this confusion for me?

答:
This is exactly why we use algebraic proofs, not words and pictures, to establish what is fact in mathematics. Here is a proper proof that $x\mapsto x^{1/3}$ is continuous at $0$.

Definition: Continuity

A function $f:D\subseteq \Bbb{R}\to\Bbb{R}$ is continuous at a point $x_0$ if $\forall \epsilon>0$, $\exists \delta>0$ such that $|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$

In our example, let's test if this applies to $f:x\mapsto x^{1/3}$ at the point $x=0$.

$$|f(x)-f(x_0)|<\epsilon\text{ in our case,} ~~|x^{1/3}|<\epsilon$$
We'll check left and right continuity separately. For right continuity, we have
$$x^{1/3}<\epsilon\implies x=|x-0|<\epsilon^3$$
So we have found our delta - it is precisely $\epsilon^3$. For left continuity,
$$-x^{1/3}<\epsilon\implies x^{1/3}>-\epsilon\implies x>-\epsilon^3$$
I.e, as long as $x$ is within $\epsilon^3$ of zero the inequality holds.

To sum up in words, the phrase "a small change in $x$ produces a small change in $f(x)$" is kind of correct, but needs to be interpreted correctly. Your remark

but its derivative at $x=0$ is undefined because the line tangent to the curve $y=f(x)$ is vertical. Shouldn't this mean, then, that at $x=0$ a small change in $x$ produces an infinite change in $y$?

Is not correct. Yes, it is true that the derivative of $x^{1/3}$ at zero is undefined-
$$\lim_{x\to 0}\frac{x^{1/3}}{x}=\infty$$
But for any nonzero value, say $t$, the change is definitely finite.
$$\frac{t^{1/3}}{t}=t^{-2/3}$$
Which is well defined $\forall t\neq 0$.


问:
Proof if $f$ is continuous then the preimage is closed.
I want to show that if $f: \mathbb{R}\to\mathbb{R}$ is continuous then $f^{-1}(l) = \{x\in\mathbb{R}|f(x) = l\}$ is closed.

I am not sure of my proof as I feel like I am missing a step:

I decided to use sequential continuity and consider a convergent sequence $(x_n)\subset f^{-1}(l)$.

By sequential continuity if $x_n \to L \implies f(x_n) \to f(L) = L'$

I want to say that I am now done - but I think I need to prove that $L \in f^{-1}(l)$, or $L' \in Imf$.

答:
The following shows that $L\in f^{-1}(l)$, which completes the OP's solution.

Since $x_n \in f^{-1}(l)$, we have $f(x_n)=l$, for all $n$. By sequential continuity of $f$,

$$x_n\to L \implies f(x_n)\to f(L)$$

But $f(x_n)$ is the constant sequence $l$, which converges (trivially) to $f(L)$, i.e. $f(L)=l$, so that $f^{-1}(l)=L$, as required.

If you're familiar with the equivalent definition of continuity that the inverse image of a closed set is closed, then you will appreciate this super quick "proof":

Proof: $\{l\}$ is closed in $\mathbb{R}$. QED


问:
Prove $\frac{d}{dt} \Big|_{t=0}\mbox{tr}(e^{X+tY})=\mbox{tr}(e^XY)$
I’m asked to prove $\frac{d}{dt}\Big|_{t=0}\mbox{tr}(e^{X+tY})=\mbox{tr}(e^XY)$ for any $X,Y$ in $M_n(\mathbb{C})$. My attempt is to assume both $X$ and $Y$ are diagonalizable, and since the set of all diagonalizable matrices is dense in $M_n(\mathbb{C})$, if we can show this is true for diagonalizable matrices, then we are done. I expected this will somehow simplify the proof, but seems it does not work well unless I further assume $X,Y$ can be diagonalizable at the same time. Any suggestions on this?

答:
Truly, definitely, only a side-comment, irrelevant to anybody's immediate needs, but a comment worth making - I think! maybe? I hope??? - and one truly, definitely, too much of a pain to enter in the comment section.

From the Lie theoretic point of view,
$${d\over dt}\Bigg\vert_{t=0}\exp (X+tY)=  d\exp_X(Y),$$
where $d\exp_X$ denotes the derivative of the (Lie theoretic!) exponential function
$$\exp\colon {\mathfrak g} \to G,$$
at $X\in {\mathfrak g} = M_n({\mathbb C})$,
with $G = GL_n({\mathbb C})$, the invertible matrices,  and $Y\in T_X{\mathfrak g}$.

It turns out that, suitably interpreted,
$$ d\exp_X (Y)= g(\text{ad }X)Y \cdot \exp X, $$
where ad is the adjoint operator, and
$$g(z) = {e^z -1 \over z  }.$$
("Suitably interpreted": for instance, above, the multiplication on the right by $\exp X$ on the tangent space is the 'functorial' operation  arising from multiplication on the right on the Lie group.)

All of this turns up - for instance - in the (or 'a') proof of the Hausdorff-Campbell-Baker formula.  For a non-power series proof (sort of), albeit one arguably a bit heavy on notation (but certainly more explicit than mine), see 4.26, and the following few sections (this answer/comment is basically lemma 4.27) of  Natural Operations in Differential Geometry (Kolar, Michor, Slovak)

Taking traces 'kills' ad, and one ends up with your formula, almost, except that $\exp X$ is on the right - but, that doesn't matter. (I did it that way in keeping with this reference.)


问:
What operation is $\circ$?
Given the following exercise:

"Consider the operation $\circ$ defined on $G_4 =\{ 0, 1, 2, 3\}$ by Table 3. Check whether it satisfies each of the three axioms from the definition of a group, and decide whether it is a group."

Table 3:

The properties of a group:

$1$. Asssoiativity: for any $a, b, c \in G_4$, we have $a\circ (b\circ c)= a\circ (b\circ c)$

$2$. Identity: for any $a \in G_4$, we have $a\circ e = e\circ a=a$

$3$. Inverse: for every $a \in (G_4,\circ ) $, theres is an inverse $a^{-1}$, such that $a\circ a^{-1}=a^{-1}\circ a=e$

Previosly I have concluded whether the sets $(G_4,+)$ and $(G_4,*)$ are groups by doing the following:

But I dont know what the $\circ$ operation for Table 3 is doing. Since we are working with numbers and not functions I am stuck trying to figure this out. I know that to find the inverses I can check if there's a value in every row that's equal to the identity value $e$, but I cannot find the identity value because I'm stuck wondering what the operation is doing. How do I read from Table 3 if it is a group or not?

答: There's no $1$ in the last two rows.  But left translation by any element of the group should be a permutation of the group elements.

Title: What operation is $\circ$?


答:
Consider the integer lattice consisting of the points $1,\dots,n$. We shift this through the reals until the point $1$ is on the point $-n$.

Each point in the lattice that is positive is less than or equal to the nearest integer above or equal to it.

Each point in the lattice that is negative, add $n$ to it. Then they are less than or equal to the remaining integers from the original lattice.


问:
Divisibility by n for 3- digit number
We have 3 numbers:
$\; \overline{xyz}=100x+10y+z$, $\; \overline{yzx}$ and $\; \overline{zxy}$, all 3 divisible by the same number $n$.

Prove that the number $P=x^3+y^3+z^3-3xyz$ is also divisible by n.

$n \mid 100x+10y+z$

$n \mid 100y+10z+x$

$n \mid 100z+10x+y$

$n \mid 111x+111y+111z \Leftrightarrow n \mid 111(x+y+z)$ but since 111 is not a prime, we can't say for sure if $n \mid (x+y+z)$.

Also $P = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$.

So if I managed to prove that $n \mid (x+y+z)$ I think it would be sufficient, right?

Can you help me?

答:
Introduction

$\newcommand{\d}[1]{\overline{#1}}$As Greg mentions in the comments, we can make significant improvements. Let $P = x^3+y^3+z^3 - 3xyz$ for $x,y,z$ in context. For $\d{xyz}$ a number, we refer to both $\d{yzx},\d{zxy}$ as its cyclic counterparts or cyclic permutations. I am sure you can generalize this to $m$ digits, which I require only for the last extra section.

$\gcd(n,x,y,z) = 1$ WLOG

Indeed, suppose that we proved the statement for the case $\gcd(n,x,y,z) = 1$. Then, for $\gcd(n,x,y,z)=d>1$, we may consider the statement for $\frac{n}d, \frac xd,\frac yd,\frac zd$ : note that $\frac nd$ will still divide each of the cyclic counterparts of $\d{\frac xd\frac yd\frac zd}$, so by the statement , $\frac{x^3+y^3+z^3-3xyz}{d^3}$ an integer multiple of $\frac nd$, which implies that $x^3+y^3+z^3-3xyz$ is an integer multiple of $n$.

$n$ divides $999$

Now, suppose that $\gcd(n,x,y,z)  =\gcd(n,\gcd(x,y,z)) = 1$. Note that $n$ divides $10\d{xyz} - \d{yzx} = 999x$, and similarly $n$ divides $999y$ and $999z$.

Now, by Bezout's theorem, there is an integer linear combination of $x,y,z$ given by  $ax+by+cz$ which equals $\gcd(x,y,z)$. Then $n$ divides $999(ax+by+cz) = 999\gcd(x,y,z)$ implying that $n$ divides $999$ by our assumption. Thus, we may restrict ourselves to divisors of $999$.

The obvious cases

The divisors of $999$ are : $1,3,9,27,37,111,333,999$.

For $1$ the statement is obvious. For each of $999,333,111$, if $\d{xyz}$ is a multiple then $x=y=z$ so $P=0$.

For $3,9$ the statement is clear , because $\d{xyz}$ is a multiple of $3$ or $9$ if and only if $x+y+z$ is , respectively (follows from a divisibility test for $3$ and $9$ : take the sum of digits etc.), and $P$ is an integer multiple of $x+y+z$. Thus, we are left with just $27$ and $37$. The techniques for these two are somewhat different.

The case $n=37$

Let $n=37$.

We know that $\d{xyz}$ and all its cyclic permutations are multiples of $37$. Let us pick the cyclic permutation that has the lowest unit digit i.e. with $z\leq x$ and $z \leq y$. For example, in the family $592,925,259$ we choose $592$, so that $x=5,y=9,z=2$. So we assume $z\leq x , z \leq y$.

Since $37 \times 3 = 111$, we can subtract $1$ from each digit of a ($3$-digit) multiple of $37$ and we will still get a multiple of $37$. For example, from $592$ we get $481$ which is still a multiple of $37$. In algebraic terms , we get that $\d{(x-1)(y-1)(z-1)}$ is also a multiple of $37$ (this is the number with digits $(x-1)$,$(y-1)$,$(z-1)$ in that order). Going like this $z$ times, we get that $\d{(x-z)(y-z)0}$ is a multiple of $37$ i.e. $\d{(x-z)(y-z)}$ is an at most (since $x-z=0$ is possible) two digit multiple of $37$.

For example, starting from $592$, we get to $481$ then to $370$. Since $592$ is a multiple of $37$, so is $370$ and therefore so is $37$. To take another example, take $962 = 37 \times 26$, subtracting one from each digit gives $851$ and then $740$. So $740$ is a multiple of $37$, hence $74$ is a multiple of $37$.

The only (at most) two digit multiples of $37$ are $0,37,74$.  So, we must have that $\d{(x-z)(y-z)} = 0,37,74$. This gives three cases : $x=y=z$, $x-z=3,y-z=7$ and $x-z=7,y-z=4$.

Now if $x=y=z$ then $P = 0$.

If $x-z=3,y-z = 7$ then $x=z+3,y=z+7$ so we have : $P = (z+3)^3+(z+7)^3+ z^3 - 3z(z+3)(z+7) = 37(3z+10)$ is a multiple of $37$.

If $x-z=7,y-z = 4$ then $x=z+7,y=z+4$ so we have : $P = (z+7)^3+(z+4)^3+z^3 - 3z(z+4)(z+7) = 37(3z+11)$ is a multiple of $37$.

Thus, $n=37$ works out.

The case $n=27$

For $27$ the task can be accomplished from the factorization $$P = \frac 12(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$$ Note that the $\frac 12$ is redundant, we only need to prove that $27$ divides $2P$.

Now, $9$ divides $x+y+z$ from the divisibility rules for $9$, so we only need to prove that $3$ divides $x^2+y^2+z^2 -xy-yz-xz$ since then $27 = 3 \times 9$ will divide the product of these. Recall that $3$ divides $x+y+z$, but then:
$$
x^2+y^2+z^2 -xy-yz-xz = (x^2+y^2+z^2 +2xy+2yz+2xz) - 3(xy+yz+zx) \\
= (x+y+z)^2 - 3(xy+yz+zx)
$$

which is a multiple of $3$. Hence, we are done.

A related property of divisors of $999$

All the divisors of $999$ have a property, one that I interestingly had a crush on some fourteen years ago, because I'd discovered some generalization to $n$ digits as well :

For $n$ a divisor of $999$, let $\d{xyz}$ be a multiple of $n$. Then $\d{yzx},\d{zxy}$ are also multiples of $n$.

That is, if one of these cyclic permutations is a multiple of $n$ then the rest of them all are. This generalizes :

For $n$ a divisor of $10^m-1$ and any $m$ digit multiple of $n$ , any cyclic permutation of the digits of $m$ (while retaining leading zeros) results in a number which is also a multiple of $n$.

For example, $271 | 99999$, so qualifies. To generate free multiples of $271$ (this was pastime in school, if you like) we have $271|27100$ and we would get $271|10027$ and $271|71002$ for free.

I leave the above as an exercise.

Title: Divisibility by n for 3- digit number


问:
Big cap notation in probability
Let $X_i$ be a random variable. Is the following true?

$$P\left\{\bigcap^k_{i=1}(X_i < 0)\right\} \equiv P\left\{(X_1 < 0) \land (X_2 < 0) \land \dots \land (X_k < 0) \right\}$$

or does the "big cap" notation mean something different in a probability context?

答:
For two "events" $E$ and $F$, set-theoretic intersection $E\cap F$ and boolean algebra conjunction $E\wedge F$ means the same thing.

Although, in the foundation of probability theory, people may want to distinguish events from models of events.

Title: Big cap notation in probability


问:
Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$
Here is my attempt:

Let $(x, y)$ be any such point.

Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we have the equalities
$$
\lvert x \rvert = \lvert y \rvert = \sqrt{ (x-4)^2+(y-2)^2}.
$$
From $\lvert x \rvert = \lvert y \rvert$, we obtain $y = \pm x$.

Thus we have the equations
$$
\sqrt{ (x-4)^2 + (\pm x -2)^2} = \lvert x \rvert,
$$
which implies
$$
(x-4)^2 + ( \pm x - 2)^2 = x^2,
$$
which simplifies to
$$
x^2 - 2(4 \pm 2)x + 20 = 0.
$$

Thus we have the following two quadratic equations
$$
x^2 - 12x + 20 = 0 \qquad \mbox{ and } \qquad x^2 -4x + 20 = 0,
$$
and the solutions of these quadratic equations are
$$
x = \frac{12 \pm 8 }{ 2 } \qquad \mbox{ and } \qquad x = \frac{ 4 \pm 8 \iota }{ 2 },
$$
that is,
$$
x = 10, 2 \qquad \mbox{ and } \qquad x = 2 \pm 4 \iota.
$$
We will of course only need the real values for our $x$.

Thus there are eight possible points satisfying the condition given in the problem, namely
$$
(10, 10), (10, -10), (-10, 10), (-10, -10), (2, 2), (2, -2), (-2, 2), (-2, -2).
$$

Is my solution correct in terms of the technique employed as well as the answers obtained? Or, are there any mistakes?

答:
(-4,2) (-4,-2) (4,-2)

The original point is 4 units from the y axis and two units from the x axis. i.e distance from each axis. The question is asking to find every point which is 4 units from the y axis AND 2 units from the x axis. There is one such point in each quadrant, which are the three additional points I listed. If you graph it, it should make sense. Those three points are the only ones which will satisfy the condition of distance of 4 from the y axis and distance of 2 from the x axis. It is basically a question about symmetry around each axis.

These represent the remaining three corners of an 8x4 rectangle, centered on the origin.  One point in each quadrant.  I believe most people are overthinking what the problem actually asks for.


问:
How to Show that an $n$-finned Hyperplane is not Homeomorphic to $\mathbb{R}^m$
This question came up again in an algebraic topology course, so I'm placing a bounty on it.  The answers below use algebraic topology - I am asking for a proof that does not use algebraic topology, e.g. homology, higher homotopy groups, or the Jordan Curve/Separation Theorem.  No smooth structures, just point-set stuff.

Original question:

Suppose that $X_n$ is the union of $n$ half closed, straight line segments, all joined at a common point (their boundary point).  So $X_2$ is just an open interval, and otherwise is an $n$-pointed asterisk with open ends.  How can one show that for $n \geq 2$ all the spaces $X_n \times \mathbb{R}$ are topologically distinct? These spaces are like $n$ sheets attached along a copy of $\mathbb{R}$.

More generally if we take $n$ copies of the $k$-dimensional closed half-space $\mathbb{H}^k$ and identify them along a $(k-1)$-plane, how do we show that this space is not homeomorphic to $\mathbb{R}^k$ for $n > 2$?

Discussion:

In the more general form, the $n=1$ case is known to be equivalent to the Invariance of Dimension Theorem, but for $n > 2$ it seems more geometrical that the spaces aren't homeomorphic.

Be careful trying to use a fact like "a homeomorphic image of a hyperplane disconnects Euclidean space into two pieces."  This may be the key ingredient, but would need a point-set proof; I only know of algebraic ones.

There is a theorem of Moore, simplified by Bing, that there is no uncountable collection of copies of a simple triod (three arcs attached at a common vertix, i.e. a three-legged asterisk) contained in the plane.  There is a Cantor Set of them along the spine of our three-finned 'plane.'  However, the proofs of this theorem depend crucially on the Jordan Curve Theorem and its corollaries.  It has higher-dimensional analogues, but they also depend on the higher-dimensional form of the Jordan Separation Theorem.

Similarly, evaluating the simplicial complex structures could differentiate them, but that is a lot of theory.  If we look at neighborhoods on, say, $X_3 \times \mathbb{R}$ of the points along the 'spine' we will get connected neighborhoods whose boundaries are the theta curve.  It can be shown that no connected set has boundary of a theta curve in $\mathbb{R}^2 = X_2 \times \mathbb{R}$ and this method also generalizes inductively, but again the details are not trivial and we are depending on at least the polygonal version of the Jordan Curve Theorem.  Putting a circle into one of the legit-planes doesn't separate an $n$-finned plane; similarly (equivalently, as an aside), the complete bipartite graph $K_{3,3}$ embeds in it.

Yet another method would be counting the number of components that removing the spine creates, which could distinguish $X_2$ from the others but seems harder to generalize.  This is one of the classic separation theorems, that a hyperplane cuts Euclidean space into two components, but we can't assume it.

And generalizing any of these methods (except for analyzing as complexes) to higher dimensions seem to have even more difficulties, for example by attaching three copies of threespace to a plane.  There is also the possibility of using symmetry groups and relating those to the bundle structure, but again it would use a lot of theory.

What is the point-set way to prove this?  Or is it equivalent to Jordan Curve Theorem, which would be really interesting?

答:
You can use the idea that the continuous image of a connected set is connected.

We will take a look at the sets $X_{n}$ that you first described

Remove a small ball around the point where you have connected the half open intervals in U, let's call that point the origin in both U and V where a homeomorphism f:U -> V. With U consisting of m half open interval gluings and v consisting of k half open interval gluings are two such examples of the line segment gluings with $m,k>2$ described in your question and $m\neq k$. Take a Neighborhood around the origin and its image and look at U\N and V\f(N). For a small enogh neighborhood U\N has m connected components, and V\f(N) has either k connected components or 2 connected components depending on whether the origin in U maps to the origin in V or not. We can provide a contradiction in either case. If V\f(N) has fewer connected components then consider the map $f^{-1}$. Since the map is a bijection there must be a connected component in V\f(N) whose image is in two or more connected components in U\N. This contradicts the idea that the continuous image of a connected set is connected.

If U\N has fewer connected components then consider the map $f$. It is the same idea with the map $f$. Since the map is a bijection there must be a connected component in U\N whose image is in two or more connected components in V\f(N). This contradicts the idea that the continuous image of a connected set is connected.

With the hyperplane version you can use a more sophisticated version of the same idea.


问:
How may random unique combinations a 4x4 matrix can make
Let's say we have a matrix (grid) of 4x4 that has values from 0 to 15 in it.

```<code>    0  1  2  3
a = 4  5  6  7
    8  9  10 11
    12 13 14 15
</code>```

If this array is randomized or shuffled how many unique combination can it make?

For example:

By unique I mean let's say 10 in the matrix a above is at 3x3 and in an attempt of a shuffle of the whole matrix, only 10 position shuffled with let's say 15 (10 is at the place 15 is and vice versa) so the new position fo 10 is now 4x4 and the matrix will look like this:

```<code>    0  1  2  3
b = 4  5  6  7
    8  9  15 11
    12 13 14 10
</code>```

Now, b is a unique combination of a 4x4 matrix a. So, like that how many unique combinations can array a can have?

答:
If you reduce from $4 \times 4$ to $16 \times 1$ with index from $1\ -> 16$, we can make a calculation like this.

With set of number can choose from $0$ to $15$

Choose a number for index $1$, we have $16$ choices

Choose a number for index $2$, we have $15$ choices (cause it has to be different with index $1$)

Choose a number for index $3$, we have $14$ choices (cause it has to be different with index $1$ and $2$)

And so on.

So, the number of choices will be $16 \times 15 \times 14 \times ... \times 1$ or it can be simplified is $16!$ according to the counting theory.

Note that, every time you select a number to fill in an index, it will be different from other index, so when you complete all $16$ choices, you will have an unique result.

Another way to approach this problem is, $4 \times 4 $ matrix has $16$ places to fill.

Let start with $0$, how many places you can put $0$ into the matrix? It's clearly $16$ cause our matrix is empty now.

Continue with $1$, our matrix right now is containing $0$, which means, you can not put $1$ at the same place as $0$. So how many places does the matrix left? It's $15$

You keep do it with $2, 3, 4, ..., 15$ and with the counting theory, you will have same result is $16!$.

EDIT: I found it, in Python, you can check the answer with some line of code like this:

```<code>import itertools

a = set(itertools.permutations([*your set here*]))

print(len(a))
</code>```

But as I've tried, Python can not execute the answer if the set is big.

Body:
If you reduce from $4 \times 4$ to $16 \times 1$ with index from $1\ -> 16$, we can make a calculation like this.

With set of number can choose from $0$ to $15$

Choose a number for index $1$, we have $16$ choices

Choose a number for index $2$, we have $15$ choices (cause it has to be different with index $1$)

Choose a number for index $3$, we have $14$ choices (cause it has to be different with index $1$ and $2$)

And so on.

So, the number of choices will be $16 \times 15 \times 14 \times ... \times 1$ or it can be simplified is $16!$ according to the counting theory.

Note that, every time you select a number to fill in an index, it will be different from other index, so when you complete all $16$ choices, you will have an unique result.

Another way to approach this problem is, $4 \times 4 $ matrix has $16$ places to fill.

Let start with $0$, how many places you can put $0$ into the matrix? It's clearly $16$ cause our matrix is empty now.

Continue with $1$, our matrix right now is containing $0$, which means, you can not put $1$ at the same place as $0$. So how many places does the matrix left? It's $15$

You keep do it with $2, 3, 4, ..., 15$ and with the counting theory, you will have same result is $16!$.

EDIT: I found it, in Python, you can check the answer with some line of code like this:

import itertools

a = set(itertools.permutations([*your set here*]))

print(len(a))

But as I've tried, Python can not execute the answer if the set is big.


问:
Geometric meaning of surface revolution
It's known that the volume of revolution of the function $f(x)$ (assuming it's real, continuous...) is
$$V=\pi\int_a^b f(x)^2dx$$

This can be modelized as if we add together all the infinitesimal cylinders of radius $|f(x)|$ integrating over all the domain $x \in [a,b]$

However, what's the geometric interpretation of the surface

$$S=2\pi\int_a^b f(x) \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}dx$$

And this isn't the sum of the perimeter of all the small circumference with radius $|f(x)|$ (at least I can figure out why) ?

It will be $$S=2\pi\int_a^b ydx$$
Can anyone explain the geometric meaning of $\sqrt{1+\bigg(\dfrac{dy}{dx}\bigg)^2}$

Thanks in advance.

答:
I moved this answer to the more active:
Why is surface area not simply $2 \pi \int_{a}^{b} (y) dx$ instead of $2 \pi \int_a^b (y \cdot \sqrt{1 + y'^2}) dx$?</p>

Title: Geometric meaning of surface revolution


问:
Is there an asymptotic for following recurrences?
The recurrence I am interested is

$$T(2^k)=2T(2^{k-1})+aT(2^{k-2})+c_1\cdot 2^k$$
at an $a\in\mathbb R_{>0}$ and at a $c_1>0$.

Is there a closed form or a sharp asymptotic?

答: Set $A_k = T(2^k, 2^k)$ and explore solutions of the form $A_k = a + 2^kb$.


问:
How to prove that $\lim_{n\to\infty}\frac{n^k}{3^{\sqrt{n}}}=0$
I know we can apply the L'Hospital's rule repeatedly $k$ times but $3^{\sqrt{n}}$ becomes unwieldy while taking derivative $k$ times. Is there another way to solve this?

答:
the $\sqrt n$ throws a monkey wrench in taking derivatives of the denominator.  The chain rule will lead to products and the derivatives get longer and more complex.  But we replace  $m = \sqrt n$ then

$\lim\limits_{n\to \infty} \frac {n^k}{3^{\sqrt n}} = \lim\limits_{\sqrt n\to \infty} \frac {m^{2k}}{3^m}=\lim\limits_{m\to \infty} \frac {m^{2k}}{3^m}$.

That stops the denominator's derivative from "spreading".  $(3^m)' = (\ln 3) 3^m$ and so the $2k$ derivative of $3^m$ is $(\ln 3)^{2k} 3^m$.

And the $2k$th derivative of $n^{2k} = (2k)!$, a constant. (Clear? Yes?  first derivative is $2k m^{2k-1}$ and second is $2k(2k-1)m^{2k-2}$ and induction....?)

So

$\lim\limits_{n\to \infty} \frac {n^k}{3^{\sqrt n}} = \lim\limits_{m\to \infty} \frac {m^{2k}}{3^m}=\lim\limits_{m\to \infty} \frac {(2k)!}{(\ln 3)^{2k} 3^m} = 0$

......

I suppose it is mildly interesting that if $\lim \frac {f(x)}{g(x)}$ is in indeterminate form that l'hopitals rule not only states $\lim \frac {f(x)}{x} = \frac {\frac {df(x)}{dx}}{\frac {dg(x)}{dx}}$ but $\lim \frac {f(x)}{x} = \frac {\frac {df(x)}{dy}}{\frac {dg(x)}{dy}}$ for any $y$ dependent on $x$.....  which is really obvious and unsurprising if we care to think about it... which I'm not sure I ever had.....


答:
Let us assume

$$f(x) = \frac{x^k}{3^\sqrt{x}}$$

$$\implies f'(x) = \frac{3^\sqrt{x}kx^{k-1} - 3^\sqrt{x} \ln 3 \frac{1}{2\sqrt x} x^k}{3^{2\sqrt x}}$$

Examining critical points

$$kx^{k-1} = x^k \ln 3 \frac{1}{2\sqrt x} \implies x^* = \left(\frac{2k}{\ln 3}\right)^2$$

Now, we can verify that for this function, and for $k>0$

$$f'(x) \begin{cases}>0 & x<x^* \\ =0  & x = x^* \\ <0  & x> x^*\end{cases}$$

Using the monotone convergence theorem - this means that the limit exists. and must be zero

Title: How to prove that $\lim_{n\to\infty}\frac{n^k}{3^{\sqrt{n}}}=0$

Body: I know we can apply the L'Hospital's rule repeatedly $k$ times but $3^{\sqrt{n}}$ becomes unwieldy while taking derivative $k$ times. Is there another way to solve this?


问:
Are elliptic curve irreducible?
Suppose we have a typical elliptic curve (smooth, projective curve with a distinguished point $\mathcal{O} = [0,1,0]$)  over some algebraically closed field $\overline{K}$
$$E: F(X,Y,Z) = Y^2Z + a_1XYZ + a_3YZ^2 - X^3 - a_2X^2Z - a_4XZ^2 - a_6Z^3 =0$$
is $F$ irreducible in $\overline{K}[X,Y,Z]$? I cant find anything definitive online.

答:
Just to get this off the unanswered list.

If $F$ is reducible (e.g. $F(x,y,z)=(x-y)(z-w)(y-z)$) then it cannot be smooth. The reason is the following nice observation:

Observation 1: Let $X$ be a Noetherian regular connected scheme. Then, $X$ is integral.

Proof: We need to show that $X$ is irreducible and reduced. Since $\mathcal{O}_{X,x}$ is regular for all $x$, and regular rings are integral (e.g see [1, Tag00NP]) we see that $X$ is clearly reduced. To see that it’s irreducible note that we can decompose $X$ into finitely many irreducible components $C_i$. Note then that if $X$ is not irreducible we must have that $C_i\cap C_j\ne \varnothing$ for some $i\ne j$ else then $\displaystyle X=\bigsqcup_i C_i$ which contradicts that $X$ is connected. But, let $x$ be a closed point of $C_i\cap C_j$ (for the $i,j$ where they intersect). Note then that $\mathcal{O}_{X,x}$ cannot be integral since $\mathrm{Spec}(\mathcal{O}_{X,x})$ contains $\eta_{C_i}$ which are distinct minimal primes. Thus, we arrive at a contradiction. Thus, $X$ is irreducible as desired. $\blacksquare$

This seems eminently useful if you know, a priori, that your hypersurface $E:=V(F)\subseteq \mathbb{P}^2_k$ is connected. But, again, this is actually automatic!

Observation 2: Let $F\in k[x,y,z]$ be a non-zero homogenous polynomial. Then, $C:=V(F)\subseteq \mathbb{P}^2_k$ is connected.

Proof: There are several proofs of this fact. But, one is to note that if $n=\deg(F)$ then we have a short exact sequence

$$0\to \mathcal{O}(-n)\to \mathcal{O}_{\mathbb{P}^2_k}\to i_\ast \mathcal{O}_C\to 0$$

where $i:C\hookrightarrow \mathbb{P}^2_k$ is the inclusion. From the LES in cohomology we see that we have an exact sequence

$$0\to \mathcal{O}_{\mathbb{P}^2_k}(\mathbb{P}^2_k)\to (i_\ast \mathcal{O}_{\mathbb{P}^2_k})(\mathbb{P}^2_k)\to H^1(\mathbb{P}^2_k,\mathcal{O}(-n))$$

But, this last group is zero (e.g. see [1, Tag01XT]) and so we see that

$$k\cong \mathcal{O}_{\mathbb{P}^2_k}(\mathbb{P}^2_k)\cong (i_\ast\mathcal{O}_C)(\mathbb{P}^2_k)=\mathcal{O}_C(C)$$

and so $\mathcal{O}_C(C)$ has no non-trivial idempotents, and so is connected. $\blacksquare$

EDIT: This is somewhat tangential, but alternatively to Observation 1 if you want to assume that your $E$ is a group, then it’s connected if and only if it’s irreducible (e.g. see Lemma 8 of this blog post of mine).

References:

[1] Stacks project authors. The Stacks Project. https://stacks.math.columbia.edu

Title: Are elliptic curve irreducible?


问:
In $\mathcal{P}(\mathbb{Z})$, $A \equiv B \iff \exists_{r \in \mathbb{Z}} ( \forall_{x \in A} (x + r \in B) ∧ \forall_{y \in B} (y - r \in A))$...
We define an equivalence relation $≡$ in the set $\mathcal{P}(\mathbb{Z})$ as:

$$A \equiv B \iff \exists_{r \in \mathbb{Z}} ( \forall_{x \in A} (x + r \in B) ∧ \forall_{y \in B} (y - r \in A))$$

Determine $|\mathcal{P}(\mathbb{Z})/_≡|$ and $∣[\mathbb{N}]_≡∣ $.

I know that for $r = 0$: $A = B$, for $r = 1$: $A$ and $B$ lay next to each other, have $1$ common element and both cover a part of $X$ axis that is $1$ wide. Similar facts are true for $r = 2$, $r =3$...

That would mean that: $|\mathcal{P}(\mathbb{Z})/≡| = |\mathbb{N}|$.

However, I don't know what to do with $∣[\mathbb{N}]_≡∣ $.

答:
Since the quotient map $\mathcal{P}(\mathbb{Z}) \to \mathcal{P}(\mathbb{Z})/\equiv$ given by $A \mapsto [A]_\equiv$ is surjective,  $$\left|\mathcal{P}(\mathbb{Z})/\equiv\right| \leq |\mathcal{P}(\mathbb{Z})| = |\mathbb{R}|.$$

To show $|\mathcal{P}(\mathbb{Z})/\equiv| \geq |\mathbb{R}|$, consider the map $\mathcal{P}(\mathbb{Z} \cap [1,\infty)) \to \mathcal{P}(\mathbb{Z})/\equiv$ given by $A \mapsto [\{0\} \cup A]_\equiv$.  Show that this map is injective.

For example, $[\mathbb{N}]_\equiv = \{\mathbb{Z} \cap [r,\infty) : r \in \mathbb{Z}\}$ has cardinality $|\mathbb{Z}|$ and the only representative with minimal element $0$ is $\mathbb{N}$ itself.


问:
Riemannian connection is flat iff there exists a Euclidean coordinate system
I read the following statement in Methods of Information Geometry by S. Amari, chapter 1 (Screenshot below)

Riemannian connection is flat iff there exists a Euclidean coordinate
system

While I am able to intuitively understand the 'if' statement; I cannot quite figure out the 'only if' part.

Here is my approach for the 'if' part:

Let $X,Y,Z$ be vector fields on a manifold $S$. Since Riemannian connection is a metric connection, we have
$$Z\langle X, Y\rangle=\left\langle\nabla_{Z} X, Y\right\rangle+\left\langle X, \nabla_{Z} Y\right\rangle$$
If the manifold is flat with respect to the Riemannian connection, we can find an affine coordinate system $\left[\xi^{i}\right]$ such that

$$Z\langle \partial_i, \partial_j\rangle=\left\langle\nabla_{Z} \partial_i, \partial_j\right\rangle+\left\langle \partial_i, \nabla_{Z} \partial_j\right\rangle = 0 + 0 = 0$$

In other words, we have that $\langle \partial_i, \partial_j\rangle$ is constant along any direction. Further, because any affine transformation of the affine coordinate system is also parallel, we can choose $\langle \partial_i, \partial_j\rangle = \delta_{ij}$. Hence, there exists an affine coordinate system which is Euclidean.

My attempt for the 'only if' part:

Given that there is a coordinate system which is Euclidean, we have that $\langle \partial_i, \partial_j\rangle = \delta_{ij}$. We again have, by the metric connection property,

$$\left\langle\nabla_{Z} \partial_i, \partial_j\right\rangle+\left\langle \partial_i, \nabla_{Z} \partial_j\right\rangle = Z\langle \partial_i, \partial_j\rangle = 0$$

This implies that $\nabla_{Z} \partial_i = \nabla_{Z} \partial_j = 0$ as $Z$ is an arbitrary vector field. Further, this implies that the coordinate vectors $(\partial_i,\partial_j)$ are parallel with respect to $\nabla$. Hence, the Riemannian connection $\nabla$ is flat because we are able to find a coordinate system which is parallel.

My doubts:

Is my approach for the 'only if' part correct?

I feel that 'only if' part violates common sense - e.g. on a sphere $S^2$, the unit vectors along the latitudes and longitudes are orthogonal. But the Levi-Civita connection with respect to the induced metric from $\Bbb R^3$ is not flat - which contradicts the statement.

答:
I think there is a mistake in my premise that the spherical coordinate system is Euclidean. By definition, an Euclidean coordinate system should satisfy $<\partial_i,\partial_j> = \delta_{ij}$. In case of the Spherical coordinates, this does not hold true as   $ds^2=d\theta^2+sin^2\theta\,d\phi^2$, which shows that $<\partial_\phi,\partial_\phi>\ne1$.

So, spherical coordinates on a sphere is not a counterexample to the statement that "A Riemannian manifold is flat iff it has an Euclidean coordinate system".


问:
Reference request: Auslander-Reiten quivers of which types of algebras have been found?
I would like to know Auslander-Reiten quivers of which types of algebras have been found.

For quantum groups, there is a paper http://arxiv.org/abs/1202.1714 by Julian Külshammer. I search the Internet, but was not able to find other papers about Auslander-Reiten theory for quantum groups.

We know that the irreducible representations of quantum groups are parameterized by Drinfeld polynomials. I think that the the vertices of Auslander-Reiten quivers can be given by Drinfeld polynomials (or correspond to Drinfeld polynomials). Are there some papers describe Auslander-Reiten quivers for quantum groups whose vertices correspond to Drinfeld polynomials explicitly?

For group algebras, there are some papers by Karin Erdmann.

There are also many papers about Auslander-Reiten quivers for projective algebras by Christof Geiss, Bernard Leclerc, and Jan Schroer.

Are there some other papers about Auslander-Reiten quivers of finite dimensional algebras? Thank you very much.

答:
This is not an answer, but the comments are getting too long, so I just write it here:

This question is difficult to answer, in the way that there are so many classes of algebras where people have worked on AR-quiver, and I dare not to say I have the complete list. But I can try to list some, and the list would be even more messy if you want AR-quiver of the related categories (derived/homotopy/functorially finite subcat).

Finite/tame-type hereditary algebras - well-known and in many literature.

Finite-type self-injective algebras - work of (mainly) Riedtmann and others.

Tame-type group algebras - work of (mainly) Erdmann, as you said.

Many classes of tame-type self-injective algebras, done by too many people and I couldn't keep track, but probably most are related to works of Skowronski.  He also has a survey on tame-type self-injective algebras, which should contain some material on this.

I think there is also a theorem stating that AR component of some of the wild-type self-inejctive algebras (such as group algebras) with infinitely many indecomposable modules must be of type $\mathbb{Z}A_\infty$ or $\mathbb{Z}A_\infty/\langle \tau^m\rangle$.  And there are probably more work done in studying the types of AR components which are tubes.

Skorwonski's series of books (Elements of Rep. Thy and Assoc. Algebras) also has quite detail treatmeant to AR-quiver components of tame/wild-type algebras, arising from tilting (tilted algebras) and such like (quasi-tilted algebras).  This kind of extends the list from (1).

Preprojective algebras, as you said there. (you missed the "pre-" prefix there, which is quite important, because it is originally about preprojective components of AR-quiver).

Certain quasi-hereditary algebras arising from Lie theory/type A representation theory.  The AR-quviers are known to some people, such as Ringel, Reiten, Miemietz, Madsen, and probably Koenig and Xi, and some other Chinese scholars influenced by the work of Xi whose name I forgot.  But I doubt if any of these are written down explicitly anywhere.

I think Julian Külshammer also has some works on the AR-quiver of Frobenius kernel of Lie algebra, but not entirely sure (see comments).

I should also point out that the term 'quantum group' is misleading, I think you mean small quantum group whenever you said quantum group in your question?  There is probably some hope in working in this direction, because small quantum groups are finite dimensional algebras.  Apart from Kulshammer, Mazorchuk and H.H.Anderson probably also know something on this issue.  In the case of 'other quantum group', I think studying AR-quiver would be incredibly difficult.  I may be wrong though.  However, there is something done in this direction by concerning the BGG category $\mathcal{O}$ instead, which is what I mentioned in point 8 above.


问:
Discrete approximation to continuous normal
Im reading Gelman's "Regression and other stories" book and I'm finding it difficult to wrap my head around one idea: in chapter 3 (p. 46) he approximates a discrete density using a normal density function. The question he's asking is what is the probability of tied election if we assume that the election outcome is forested to be N(0.49, 0.04). He than goes on and, assuming a population of voters of n=200000, calculated that P(tied election) = 1/n * f, where f is the PDF of this forecast. Where does the 1/n comes from in this formula? He uses the same rationale in his " Estimating the Probability of Events That Have Never Occurred: When Is Your Vote Decisive?" Gelamn et al. (1998) paper (J of the American Statistical Association):

Pr(your vote is decisive in your state)
= Pr(np = .5n) = Pr(p = .5), which is approx.  f(.5)/n

答:
This is from the approximating Binomial distribution to a normal when $n$ is large enough, as the skew of the distribution is not too great. In this case a reasonable approximation to $B(n, p)$ is given by the
$ \mathcal{N}(np,\,np(1-p))$


问:
How to simplify my if conditions using Boolean Logic?
My code:

```<code>def save_model(self, request, obj, form, change):
    if "_continue" in request.POST:
        if "PAN_ID" in request.POST and not change:
            try:
                data = Dematad.objects.filter(
                    Q(PAN1=obj.PAN_ID) | Q(PAN2=obj.PAN_ID) | Q(PAN3=obj.PAN_ID)
                ).get()

                obj.email= data.EMAIL1.strip() or data.EMAIL2.strip() or data.EMAIL3.strip() or None

                obj.full_name =  data.NAME

                obj.address =  ", ".join(
                        [data.AD1, data.AD2, data.AD3, data.AD4, data.PIN]
                    )
            except Dematad.DoesNotExist:
                    self.message_user(
                        request, "User created and saved. No data available in database."
                    )

    if not change:
        obj.is_superuser = False
        obj.is_staff = True
        obj.set_password(User.objects.make_random_password(20))
</code>```

return super().save_model(request, obj, form, change)

I want to simplify the code and thus tried to model the if conditions through logic. Consider the try, except block together as D

S1:  $A \implies ((B  \land \tilde C) \implies D)$

S2:  $ \tilde C \implies E $

How do I connect these statements? How do I simplify my if conditions?

EDIT:
By simplification I' m meaning to remove the inner block, or avoid nested if's. I want to know how can expressions be simplified.

答:
I personally don't think you can combine or simplify because:

You don't have repeated expressions, so playing with different variables is not a good idea IMHO

We don't know if combining the if conditions would lead into a different program, so better a programmer should answer that question

Edit. WolframAlpha cannot reduce the combined expressions.


问:
Cubic formula confusion/clarification
So the cubic formula yields ultimately a degree 6 equation which is easily solved as it just involves solving a quadratic in $z^3$ but this gives 6 solutions. My question is at what step in the process can we say that 3 of the solutions are equal to each other (as in 6 solutions boils down to 3)?

答:
For a reduced cubic $$y^3=3py+2q$$

the resolvents are symmetric in roles:

$$u^3+v^3=2q$$

$$uv=p$$

they are conjugate to each other,

$$u=\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}} \implies
v=\frac{p}{\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}}
 =\sqrt[3]{q\mp \sqrt{q^{2}-p^{3}}}$$

One of the roots $y=u+v$ is just a matter of writing order.

Also one list of roots namely

$$y=u\omega_{3}^{k}+v\omega_{3}^{2k}$$

is equivalent to

$$y=u\omega_{3}^{2k}+v\omega_{3}^{k}$$

where $\omega_n=e^{2\pi i/n}$.

There're cyclic properties for polynomial equations.  The Galois theory is a group theory about polynomial zeros.  For your interests, see also other posts of mine here, here, here and here.

Title: Cubic formula confusion/clarification


问:
Homology of the unit ball $D^3$ with identification of boundary points by $180^\circ$ degree rotation around the vertical axis
This problem comes from Topology and Geometry of Bredon :

Let $X$ result from $D^3$ by identifying points on its boundary $S^2$ taken into one another by the $180^\circ$ rotation about the vertical axis. Give $X$ a CW-complex structure and compute its homology.

Here is my concern : For two different CW-complex structures, I get different homology groups.

The first one consists of one $0$-cell, $x$, one $1$-cell, two $2$-cells $\sigma_1, \sigma_2$, and one $3$-cell $\rho$. See the figure below:

Then the $2$-cell are attached to the $1$-cell by the word $a^{\pm 2}$ depending on the orientation, which are degree $2$ maps, and the $3$ cell is attached to each $2$-cell by a degree $2$ map since there are two identified preimages which are orientation preserving. is this correct? This gives $H_1(X) \cong \mathbb{Z_2}$, $H_2(X) = 0$, and $H_3(X) = 0$.

The second one consists of two $0$-cells, $x_0, x_1$, one $1$-cell $a$, one $2$-cell $\sigma$, and one $3$-cell $\rho$. See the figure below :

Is the $2$-cell $\sigma$ -which is the left part of the sphere in the figure above- attached by the word $aa^{-1}$, and the $3$-cell a degree $2$ map since two preimage differ by a rotation, which is orientation preserving?
I get then $H_1(X) = 0$ which seems weird because this space looks like a lens space except that there is no reflection.

答:
First, I think you need to convince yourself that $X\approx S(\Bbb{R}P^2)$, the unreduced suspension of $\Bbb{R}P^2$ defined as the quotient space of $\Bbb{R}P^2\times I$ by collapsing $\Bbb{R}P^2\times\{0\}$ to a point and $\Bbb{R}P^2\times\{1\}$ to another point.

If you take $\Bbb{R}P^2\approx D^2/{(x\sim-x)},\forall x\in S^1$, then you will see that $\Bbb{R}P^2\times I$ is just the solid cylindrical space. Identifying the two copies indicated above, it forms the spherical objects that you drew.

Suppowe we're working with the standard cell structure, i.e., $\Bbb{R}P^2=e_1^0\cup e_1^1\cup e_1^2$ and $I=[0,1]=e_2^0\cup e_3^0\cup e_2^1$, then we can deduce the cell structure on $S(\Bbb{R}P^2)$:
\begin{align}
\text{0-cells: }e_2^0,\text{ }e_3^0\\
\text{1-cell: }e_1^0\times e_2^1\\
\text{2-cell: }e_1^1\times e_2^1\\
\text{3-cell: }e_1^2\times e_2^1\\
\end{align}
because any $e_1^\alpha\times e_\beta^0$ gets quotient out by the definition of unreduced suspension.

At this point, you might realize that the second cell structure in your post agrees with this one, which yields a cellular chain complex
$$0\to\Bbb{Z}\overset{\partial_3}{\to}\Bbb{Z}\overset{\partial_2}{\to}\Bbb{Z}\overset{\partial_1}{\to}\Bbb{Z}\oplus\Bbb{Z}\to 0$$
By direct computation $\operatorname{im}(\partial_1)=\langle e_3^0-e_2^0 \rangle=\langle x_1-x_0\rangle$, which kills one copy of $\Bbb{Z}$ in the 0-th cellular chain group. $\ker(\partial_1)=0$, which implies the triviality of $H_1$. Next, $\operatorname{im}(\partial_2)=0$ as argued by you in the post. $\partial_3$ is a multiplication by $2$. Now, just apply the definition of cellular homology to get the answer.

The first "cell structure" is not even a cell structure, notice that points in $\sigma_1$ need to be identified according to the rotation about $z$-axis, so it's not a "nice" cell attached to the $1$-skeleton. So number $1$ is invalid.

There is a significant difference between $X$ and the lens space. A Mayer-Vietoris sequence argument shows that for a complex $K$
$$H_{k+1}(S(K))\cong \tilde{H}_k(K)$$
In particular, $H_1(S(\Bbb{R}P^2))\cong\tilde{H}_0(\Bbb{R}P^2)\cong 0$, so there is nothing weird.


问:
Derivative of the exponential map
I'm new to Riemannian geometry and always confused by the derivatives there. Consider on a Riemannian manifold $M$, define the following map

$$u(r,\theta)=\text{exp}_{x} (rf(\theta))$$
where $x \in M$, $r \in \mathbb{R}$ and $f(\theta)\in T_xM$.

What is $\partial_{r}u$? I know it is equal to $f(\theta)$ when$ r=0$, what about other $r$? Also, how to compute $\partial_{\theta}u$ ?

答:
If $x \in M$ is fixed, and you want to compute the partial derivative $(\partial_ru)(r,\theta)$, you compute the derivative of the map $$r \mapsto rf(\theta) \mapsto \exp_x(rf(\theta)),$$which is the composition $\exp_x \circ (r \mapsto f(\theta))$, where $\theta$ is fixed. Thus $$(\partial_ru)(r,\theta) = {\rm d}(\exp_x)_{rf(\theta)}(f(\theta)),$$where one thinks of $f(\theta)$ as an element of $T_{rf(\theta)}(T_xM) \cong T_xM$. At $r = 0$, of course this boils down to $(\partial_ru)(0,\theta) = f(\theta)$.

Similarly, for $(\partial_\theta u)(r,\theta)$, you compute the derivative of the map $$\theta \mapsto rf(\theta) \mapsto \exp_x(rf(\theta)),$$which is the composition $\exp_x \circ (\theta \mapsto rf(\theta))$, where $r$ is fixed, and this yields $(\partial_\theta u)(r,\theta) = {\rm d}(\exp_x)_{rf(\theta)}(rf'(\theta))$. This time you see that $(\partial_\theta u)(0,\theta) = 0$, which makes sense geometrically. So essentially, the question boils down to what you know about the derivative ${\rm d}(\exp_x)_v$ for $v \neq 0$. Even if you don't know much, you may be able to combine the above computations with results such as the Gauss lemma to achieve whatever is it that you're trying to do here.

Title: Derivative of the exponential map


问:
Critical value example where partial derivative does not exist
Each of the following functions has a critical value where the partial derivatives do not exist.

$f(x,y)=(x^2+y^2)^{1/3}$

$f(x,y)=1-\sqrt{x^2+y^2}$

$f(x,y)=3-[(x-1)(y-2)]^{2/3}$

Does anyone have any cuter examples?

答: You can also consider $f(x,y) = |x|$, which has infinitely many critical points along the line $x=0$, where its partial derivative does not exist.


问:
The differential of the exponential map on a Riemannian manifold
Let $M$ be a Riemannian manifold and $exp_{x}: T_{x}M \to M$ the usual exponential map. It is an important fact that $D_{0}exp_{x}(v) = v$ for any $v \in T_{0}(T_{x}M)$.

The proof goes like this:

$D_{0} exp_{x}(v) = \frac{d}{dt} exp_{x}(tv)|_{t = 0} =  \gamma(t)'|_{t = 0} = v$, where $\gamma(t)$ is the geodesic starting at $v$. I am struggling to understand this proof because I find it hard to interpret $\frac{d}{dt} exp_{x}(tv)|_{t = 0}$. Is it supposed to be understood as $\lim_{t \to 0} \frac{exp_{x}(tv) - exp_{x}(0)}{t}?$ This difference expression does not make sense since we cannot take difference on a manifold without extra structure.

I guess my problem is most of the texts I encountered tend to treat calculations like this like they are in the Euclidean space rather than on a manifold. I suppose it should be correct to do so, but is it trivially true without any justification?

The way I would prove this is to take $f \in C^{\infty}(M)$, and try to figure out what is the action of $D_{0}exp_{x}(v)$ on $f$, but none of the reference I find take this approach.

答:
To compute ${\rm d}(\exp_x)_0(v)$, you compute $$\frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \exp_x(\alpha(t))$$where $\alpha \colon (-\epsilon,\epsilon) \to T_xM$ is any curve with $\alpha(0) = 0$ and $\alpha'(0) = v$.  How to understand the above derivative? Read it as the composition it is: $\exp_x\circ \alpha \colon (-\epsilon, \epsilon) \to M$. In other words, this is just a particular curve in the manifold $M$, and you should be comfortable taking velocities of curves. The result will not depend on the choice of $\alpha$, so one takes $\alpha(t) = tv$, meaning that we look at the curve $t \mapsto \exp_x(tv)$. This is precisely the geodesic starting at $0$ with intitial velocity $v$. But the derivative $$\frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \exp_x(tv)$$is the initial velocity of this curve! Which you know to be $v$. Thus $${\rm d}(\exp_x)_0(v) = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0}\exp_x(tv) = v.$$

In more detail, what is the derivative of a curve $\gamma\colon I \to M$? For each $t \in I$, you expect to have $\gamma'(t) \in T_{\gamma(t)}M$, right? And what is a tangent vector at $\gamma(t)$? It is a derivation acting on germs of smooth functions at $\gamma(t)$. How is the derivation $\gamma'(t)$ defined? Simple: take $f \in \mathcal{C}^\infty(M)$ and set $\gamma'(t)f = (f\circ \gamma)'(t)$. The derivative on the right side here is a usual derivative you learnt in single-variable calculus, because the composition $f\circ \gamma$ is a function $I \to \Bbb R$, i. e., a real-valued function defined on some interval. Once you understand things like the chain rule, one deals with $\gamma'(t)$ itself without going back every time to what $\gamma'(t)f$ is.


问:
Logarithm to trigonometry conversion
I have an expression like
$$\frac{1}{2} i \log \left(\frac{a-b+i c}{a+b-i c}\right)$$

I was wondering if it is possible to write it in terms of trig functions. I guess it is possible to write all logs in some form of trig due to Euler's relation.

As an attempt to solve the problem, I was trying to play around the relation

$$\frac{1}{2} i \log \left(\frac{a+i c}{-a+i c}\right)=\tan^{-1}\left(\frac{a}{c}\right)$$ to change into arctan function but does not look like this alone is going to help in this case.

Any suggestions?

答:
Assuming $a,c>0$, if you write $a+ic = re^{i\theta}$ where $r=\sqrt{a^2+c^2}$ and $\theta=\tan^{-1}(\frac{c}{a})$, then
$$-a+ic = re^{i(\pi - \theta)}$$
Since both numbers are positive the argument will be $\pi - \theta$ for $-a+ic$.

Now put this in your expression to get:
$$\frac{1}{2}i\log{\frac{a+ic}{a-ic}}=\frac{1}{2}i\log{\frac{re^{i\theta}}{re^{i(\pi -\theta)}}} = \frac{1}{2}i\log e^{i(2\theta-\pi)}=\frac{1}{2}i\times i(2\theta - \pi)$$
which becomes
$$-\frac{1}{2}(2\theta - \pi)=\frac{\pi}{2}-\theta=\frac{\pi}{2}-\tan^{-1}\frac{c}{a}=\tan^{-1}\frac{a}{c}$$
Use the fact $$\tan^{-1}x+\tan^{-1}\frac{1}{x}=\frac{\pi}{2}$$ for all $x>0$
.The expression might lead to different expressions depending on the values of $a,c$ I took the case where both are positive. It can be shown for rest of the cases as well. Hope this helps....

Title: Logarithm to trigonometry conversion


问:
Check my proof that $(ab)^{-1} = b^{-1} a^{-1}$
The following question is problem Pinter's Abstract Algebra. And to put things in context: $G$ is a group and $a, b$ are elements of $G$.

I want to show $(ab)^{-1}$ = $b^{-1}a^{-1}$.

I originally thought of proving the fact in the following manner:
\begin{align*}
(ab)^{-1}(ab) &= e \newline
(ab)^{-1}(ab)(b^{-1}) &= (e)(b^{-1}) \newline
(ab)^{-1}(a)(bb^{-1}) &= (b^{-1}) \newline
(ab)^{-1}(a)(e) &= (b^{-1}) \newline
(ab)^{-1}(a) &= (b^{-1}) \newline
(ab)^{-1}(a)(a^{-1}) &= (b^{-1})(a^{-1}) \newline
(ab)^{-1}(e) &= (b^{-1})(a^{-1}) \newline
(ab)^{-1} &= (b^{-1})(a^{-1}) \newline
\end{align*}

I know this may seem extremely inefficient to most, and I know there is a shorter way. But would this be considered a legitimate proof?

Thanks in advance!

答:
The definition of  inverse is
a*a-1 = I       (ie a operated with inverse should give identity element)
a-1*a = I      (ie a inverse operated with a should also give identity element)

so here

$(AB) B^{-1} A^{-1} = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$

$B^{-1} A^{-1} (AB) = B^{-1}(A^{-1}A)B = B^{-1} I B = B^{-1}B = I$

Here when $B^{-1}A^{-1}$ Operated to AB on both sides and in both case it given I (Identity Matrix ).

$X Z  = I$ means that $Z$ is the inverse of $X$
Similarly
If $ABB^{-1}A^{-1}$ is giving identity element then $B^{-1} A^{-1}$ is the inverse of $AB$.


问:
Does the following lower bound improve on $I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)}$, where $q^k n^2$ is an odd perfect number?
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Define the abundancy index
$$I(x)=\frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the classical sum of divisors of $x$.

Since $q$ is prime, we have the bounds
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1},$$
which implies, since $N$ is perfect, that
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$

By considering the negative product
$$\bigg(I(q^k) - \frac{2(q-1)}{q}\bigg)\bigg(I(n^2) - \frac{2(q-1)}{q}\bigg) < 0,$$
since we obviously have
$$\frac{q}{q-1} < \frac{2(q-1)}{q},$$
then after some routine algebraic manipulations, we arrive at the lower bound
$$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$

Now, a recent MO post improves on the lower bound for $I(n^2)$, as follows:
$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg)$$

Added February 3, 2021 - 8:10 PM (Manila time) Here is a quick way to show the improved lower bound for $I(n^2)$.  We have
$$I(n^2)=\frac{2}{I(q^k)}=\frac{2q^k (q - 1)}{q^{k+1} - 1}=\frac{2q^{k+1} (q - 1)}{q(q^{k+1} - 1)}=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} - 1}\bigg)$$
$$=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{(q^{k+1} - 1) + 1}{q^{k+1} - 1}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1} - 1}\bigg)$$
$$>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)$$
since $q^{k+1} > q^{k+1} - 1$.  But, of course, we obtain
$$I(n^2)>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg).$$
QED.

Repeating the same procedure as above, we have the negative product
$$\Bigg(I(q^k) - \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg)\Bigg(I(n^2) - \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg) < 0.$$

This implies, after some algebraic manipulations, that
$$I(q^k) + I(n^2) > \frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}.$$

But WolframAlpha says that the partial fraction decomposition of the new lower bound is given by
$$\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}} = \frac{3q^2 - 4q + 2}{q(q - 1)} + \frac{2(q - 1)}{q^{k+2}} - \frac{q}{(q - 1)(q^{k+1} + 1)}.$$

So essentially, my question boils down to:

QUESTION: Is it possible to produce an unconditional proof (that is, for all $k \geq 1$ and for all special primes $q \geq 5$) for the following inequality?
$$\frac{2(q - 1)}{q^{k+2}} > \frac{q}{(q - 1)(q^{k+1} + 1)}$$

MY ATTEMPT

I tried to ask WolframAlpha for a plot of the above inequality, it gave me the following GIF image:

So it does appear that the inequality is unconditionally true, which would mean that the new lower bound for $I(q^k) + I(n^2)$ improves on the old.  Is it possible to prove this analytically?

And lastly:  Based on this answer to a closely related question, since we appear to have obtained an improved lower bound for $I(q^k) + I(n^2)$, can we then say that there is indeed an integer $a$ such that $k \leq a$?

答:
And lastly:  Based on this answer to a closely related question, since we appear to have obtained an improved lower bound for $I(q^k) + I(n^2)$, can we then say that there is indeed an integer $a$ such that $k \leq a$?

I think that we cannot.

In the following, I'll explain the reason why I think that we cannot.

In the answer, I wrote

"If you get an improved lower bound for $f(k)$, then you can say that there is an integer $a$ such that $k\le a$ "

where $f(k)=I(q^k) + I(n^2)$.

Also, in a comment below the answer, I wrote

"if you get an improved lower bound $g(q)$, then from the above fact, we see that there is only one $k=k_0$ such that $f(k)=g(q)$. Then, we have $k\le \lfloor k_0\rfloor$ "

where "the above fact" is the fact that $f(k)$ is decreasing.

In short, we can say that there is an integer $a$ such that $k\le a$ if we get a function $g(q)$ (which is a fuction only on $q$) such that
$$I(q^k)+I(n^2)\ge g(q)\gt 3 - \frac{q-2}{q(q - 1)}$$

Now, let us see what you've got.

Let $$G(k,q)=\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}$$

You have
$$I(q^k) + I(n^2) > G(k,q)\gt 3 - \frac{q-2}{q(q - 1)}$$
which is correct since

$$I(q^k) + I(n^2)-G(k,q)=\frac{  (q- 4)q^{k+2}+2 q^{k + 1}  + (3 q - 4 )q + 2}{q^{k+2}(q - 1) (q^{k + 1} - 1) (q^{k + 1} + 1)}\gt 0$$

$$G(k,q)-\bigg(3 - \frac{q-2}{q(q - 1)}\bigg)=\frac{  (q - 4) q^{k + 2} +2 q^{k + 1} + 2( q - 2 )q + 2}{q^{k+2}(q - 1) (q^{k + 1} + 1)}\gt 0$$

Also, we have
$$\frac{\partial G(k,q)}{\partial k}=-\frac{  (q- 4) q^{2 k + 3}+ 2 q^{2 k + 2}+ 4( q- 2)q^{k + 2}  +4 q^{k + 1}    + 2( q - 2) q + 2}{q^{k+2}(q - 1) (q^{k + 1} + 1)^2}\ln q\lt 0$$
$$\lim_{k\to\infty}\bigg(I(q^k) + I(n^2) \bigg)=\lim_{k\to\infty}G(k,q)= 3 - \frac{q-2}{q(q - 1)}$$

These imply that you don't get a function $g(q)$ (which is a fuction only on $q$) such that
$$I(q^k)+I(n^2)\ge g(q)\gt 3 - \frac{q-2}{q(q - 1)}$$

So, I think that you cannot say that there is an integer $a$ such that $k\le a$.

Added :

I'm going to add another explanation with graphs.

Let $$f(k)=I(q^k) + I(n^2),\qquad G(k)=\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}},$$
$$Q=3 - \dfrac{q-2}{q(q - 1)}$$
where $q,Q$ are seen as constants.

Since we see that $f(k)\gt G(k)\gt Q$ holds for all $k,q$ and that both $y=f(k)$ and $y=G(k)$ are decreasing with $\displaystyle\lim_{k\to\infty}f(k)=\displaystyle\lim_{k\to\infty}G(k)=Q$, we see that their graphs are as follows :

Now, if you take a fixed positive integer $k=k'$, then we have the following :

This explains why you cannot say that
$$f(k)\gt G(k′)$$
holds for all $k$.

Added 2 :

I'm going to add more explanations with steps :

(1) $f(k)\gt G(k)\gt Q$ holds for all $k,q$.

(2) Both $y=f(k)$ and $y=G(k)$ are strictly decreasing.

(3) $\displaystyle\lim_{k\to\infty}f(k)=\displaystyle\lim_{k\to\infty}G(k)=Q$.

(4) If we let $k=k'$ where $k'$ is a fixed positive number larger than $1$, then $G(k')$ is entirely in terms of $q$ alone.

(5) $Q\lt G(k')\lt f(1)$

(6) It follows from $(2)(3)$ that, for any $Y$ satisfying $Q\lt Y\lt f(1)$, there exists only one $k$ such that $f(k)=Y$.

(7) It follows from $(5)(6)$ that there exists only one $k$ such that $f(k)=G(k')$. Let $k_0$ be this $k$.

(8) For $k\lt k_0$, we have $f(k)\gt G(k')$.

(9) For $k\ge k_0$, we have $f(k)\le G(k')$.

(10) It follows from $(8)(9)$ that $f(k)\gt G(k')$ does not hold for all $k$.

Added 3 :

You seem to think that the following claim is true :

Claim : If there are two functions $f(x),g(x)$ such that
$$f(x)\gt g(x)\gt 1$$
holds for all $x\gt 0$, then there is a fixed positive integer $m$ such that $f(x)\gt g(m)$ holds for all $x\gt 0$.

This claim is false. Take $f(x)=\dfrac 1x+1$ and $g(x)=\dfrac{1}{x+1}+1$ for which $f(x)\gt g(x)\gt 1$ holds for all $x\gt 0$, but $f(x)\gt g(m)$ does not hold for $x\ge m+1$.


问:
Show that the sequence $x_n$ converges.
$$x_n = 1 + \frac{2}{4} + \frac{3}{4^2} + \dotsb + \frac{n}{4^{n-1}}$$

So I need to see if the sequence is monotone and bounded. By doing $x_{n+1} - x_n$ I get $\frac{n+1}{4^n} > 0$ so its increasing. How do I show that it is bounded? Also I have a hint to use the $2^n > n$ inequality.

答:
Consider
$$x_n=\sum_{k=1}^n k y^{n-1}=\frac d {dy}\left(\sum_{k=1}^n  y^{n}\right)=\frac d {dy}\left(\frac{y \left(y^n-1\right)}{y-1}\right)$$
$$x_n=\frac{(n (y-1)-1) y^n+1}{(y-1)^2}$$Now, let $y=\frac 14$ to obtain after expansion
$$x_n=\frac{16}{9}-\frac{1}{3} 4^{1-n} n-\frac{4^{2-n}}{9}$$

Title: Show that the sequence $x_n$ converges.


问:
You believe that there's a fly somewhere less then 6 feet away from you.
You believe that there is a fly somewhere less than $6$ feet away from you. If you believe that he is located Uniformly in a circle of radius $6$ feet away from you, what is the probability that he is more than $2$ feet away from you?

My work:
This is clearly a continuous uniform distribution question, and I know that we need to account for area here for the density function since the fly is within a circular radius. So I believe we use the joint pdf $f(x,y) = \cfrac{1}{36\pi}$ to help us calculate the probability the fly is more than $2$ feet away from me since if I integrate over the area of the circle I get $1$. But I don't know where to go from here and how I can use this info to help me calculate the probability. Also what would the bounds of $f(x,y) = \cfrac{1}{36\pi}$ be? I know that $(x,y)$ would be whatever's in the circle, but what would that be?

Any help with finding the bounds or probability would be greatly appreciated!
Thank You!

答:
Just for fun, you could do an integral in polar coordinates, where $dA=rdrd\theta$. That is how you would do it using the "uniform density".

$\displaystyle \int_0^{2\pi}\int_2^6 \frac r {36\pi} dr d\theta=\frac 1 {36\pi}\int_0^{2\pi}16d\theta=\frac{32\pi}{36\pi}=\frac 8 9$


问:
How to prove that $\lim_{n\to\infty}\frac{n^k}{3^{\sqrt{n}}}=0$
I know we can apply the L'Hospital's rule repeatedly $k$ times but $3^{\sqrt{n}}$ becomes unwieldy while taking derivative $k$ times. Is there another way to solve this?

答:
Here is a "non-L'Hospital" way using $e^t =\sum_{m=0}^{\infty}\frac{t^m}{m!} > \frac{t^m}{m!}$ for $t>0$.

Hence,

\begin{eqnarray*} \frac{n^k}{3^{\sqrt{n}}}
& = & \frac{n^k}{e^{\sqrt n \ln 3}} \\
& < & \frac{n^k}{\frac{\left(\sqrt n \ln 3\right)^{2k+2}}{(2k+2)!}} \\
& = & \underbrace{\frac{(2k+2)!}{\ln^{2k+2}3}}_{=C_k}\cdot \frac{n^k}{n^{k+1}}\\
& = & C_k\cdot\frac 1n \stackrel{n\to\infty}{\longrightarrow} 0
\end{eqnarray*}


问:
Three points $x,y,z$ are chosen at random on the point interval $(0,1)$. What is the probability that $x \le y \le z$?
I have been trying this problem for a while. But, couldn't find any solution. How do I solve this?</p>

答:
$\displaystyle \int_0^1\int_0^z\int_0^ydxdydz=\int_0^1\int_0^zydydz=\int_0^1\frac{z^2}{2}=\frac 1 6$

or... read the comment

You can also choose another of the six possible integration orders:

$\displaystyle \int_0^1\int_x^1\int_x^zdydzdx=\int_0^x\int_x^1(z-x)dzdx=\int_0^1\left[\frac{z^2}2-xz\right]_{z=x}^1dx=\int_0^1\left(\left(\frac 1 2 -x\right)-\left(\frac{x^2}2-x^2\right)\right)dx=\int_0^1\left(\frac 1 2 -x+\frac {x^2}2\right)dx=\left[\frac x 2 - \frac {x^2} 2+\frac{x^3}6\right]_0^1=\frac 1 6$


答:
First do a drawing

Your probability is simply the "favourable area" (purple area) divided by the "total area" (big circle area). Integration is not needed because the distribution is Uniform. Anyway if you want to integrate you will find the same result. Try it!

Simply do the ratio of the 2 areas to get the probability. It is easier to calculate the probability that the fly is LESS than 2 feet from you.... thus you requested probability is the complement

$$P[X>2]=1-\frac{\pi 2^2}{\pi 6^2}=1-\frac{4}{36}=\frac{8}{9}$$

Title: You believe that there's a fly somewhere less then 6 feet away from you.

Body:
You believe that there is a fly somewhere less than $6$ feet away from you. If you believe that he is located Uniformly in a circle of radius $6$ feet away from you, what is the probability that he is more than $2$ feet away from you?

My work:
This is clearly a continuous uniform distribution question, and I know that we need to account for area here for the density function since the fly is within a circular radius. So I believe we use the joint pdf $f(x,y) = \cfrac{1}{36\pi}$ to help us calculate the probability the fly is more than $2$ feet away from me since if I integrate over the area of the circle I get $1$. But I don't know where to go from here and how I can use this info to help me calculate the probability. Also what would the bounds of $f(x,y) = \cfrac{1}{36\pi}$ be? I know that $(x,y)$ would be whatever's in the circle, but what would that be?

Any help with finding the bounds or probability would be greatly appreciated!
Thank You!


问:
Grothendieck topology on pre/sheaves
Given a Grothendieck topology $T$, say subcanonical, on a category $C$, we are able to talk about sheaves in $(C, T)$. Since pre/sheaves can be viewed as generalized objects of $C$ (via Yoneda functor), one natural question arises, is $Shv(C, T)$ or $Psh(C)$ a site extending the site $(C, T)$? If it is, then we take pre/sheaves on that new site, can we get more pre/sheaves?</p>

答:
Yes. Every topos has the so-called canonical topology, in which a sieve is covering if and only if it is generated by a small jointly epimorphic family. (In the Grothendieck topos case, this is the same as being jointly epimorphic, because the small-generation condition is automatic.) Moreover, the category of sheaves in the canonical topology of a Grothendieck topos is equivalent to the original topos, so there are no new sheaves. See Proposition 2.2.7 in [Johnstone, Sketches of an elephant, Part C].

That said, it is not legitimate to form presheaves on a large site: for one thing, the resulting category is not a topos.

Title: Grothendieck topology on pre/sheaves


问:
one limit divided into two limit factors, is it can one determine result directly but other don't?
there is a question:

$$
\lim_{x->0}{\frac{x-arcsinx}{x^3(sinx+1)}} = \lim_{x->0}{\frac{1}{sinx+1}} · \lim_{x->0}{\frac{x-arcsinx}{x^3}}
$$

[1]you know there have four place x.
we know in limit calculation, if numerator x tend to 0, the denominator x must tend to 0 too, so can get the correct result.

but in the given question, there divided into two lim's multiplication:$$\lim_{x->0}{\frac{1}{sinx+1}} · \lim_{x->0}{\frac{x-arcsinx}{x^3}}$$

there I named them as lim_A· lim_B.

in my opinion, the lim_A and lim_B comes from one entirety, if lim_A's x tends to 0, the lim_B's x should tends to 0 too.

but however, in the question's solution there only let lim_A's x ->0 ignore the lim_B's x,

$$\lim_{x->0}{\frac{1}{sinx+1}} · \lim_{x->0}{\frac{x-arcsinx}{x^3}} = \lim_{x->0}{\frac{x-arcsinx}{x^3}}$$

so, why ? is there any theorem state this step? (I don't understand because I think the lim_A and lim_B comes from one entity, if one x->0, the others(lim_B's x) should x->0 too.)

Body:
You may also do it as $$L=\lim_{x \to 0} \frac{x-\sin^{-1}x}{x^3}$$
by letting $x=\sin t$ as
$$L=\lim_{t\to 0} \frac{\sin t- t}{\sin^3 t}= \lim_{t \to 0} \frac{1-t^3/3!+t^4/5!-...-t}{(t-t^3/3!+...)^3}=-\frac{1}{6}.$$


Title: Logarithm to trigonometry conversion

Body:
I have an expression like
$$\frac{1}{2} i \log \left(\frac{a-b+i c}{a+b-i c}\right)$$

I was wondering if it is possible to write it in terms of trig functions. I guess it is possible to write all logs in some form of trig due to Euler's relation.

As an attempt to solve the problem, I was trying to play around the relation

$$\frac{1}{2} i \log \left(\frac{a+i c}{-a+i c}\right)=\tan^{-1}\left(\frac{a}{c}\right)$$ to change into arctan function but does not look like this alone is going to help in this case.

Any suggestions?

Body:
$\frac {i}{2}\ln \frac {a+b+ic}{a-b-ic} = x\\
\ln \frac {a+b+ic}{a-b-ic} = -2ix\\
\frac {a+b+ic}{a-b-ic} = e^{-2ix}\\
a+b+ic = e^{-2ix}(a-b-ic)\\
a+b+ic = e^{-2ix}(a-b-ic)\\
(ic-b)(1+e^{-2ix}) = a(e^{-2ix} - 1)\\
\frac {ic - b}{a} = \frac {(e^{-2ix} - 1)}{e^{-2ix} + 1}$

At this point the right hand side equals very nearly $\tan x.$  That is:

$i\frac {c}{a} - \frac {b}{a} = \frac {(e^{-2ix} - 1)}{e^{-2ix} + 1}\\
\frac {ic - b}{a} = \frac {(e^{-ix} - e^{ix})}{e^{-ix} + e^{ix}}\\
\frac {-c - bi}{a} = \frac {i(e^{-ix} - e^{ix})}{e^{-ix} + e^{ix}}\\
\frac {-c - bi}{a} = \frac {e^{ix} - e^{-ix}}{i(e^{-ix} + e^{ix})}\\
\frac {-c - bi}{a} = \tan x\\
x = \tan^{-1} (-\frac {c}{a} - i\frac {b}{a})$


问:
Permutation and combination #probability
I am currently studying Extension 1 Mathematics. I missed two classes and I figured out that tomorrow I will have a quiz. Can you help me to solve this permutation and combination question:

In how many ways 3 cards be selected from a pack of 52 playing cards
if:

(i) at least one of them is an ace;

(ii) not more than one is an ace.

In how many ways can 9 books be distributed amongst a man, a woman
and a child, if the man receives 4, the woman 3, and the child 2?

I know the solution (below), but I don't understand how to get there.

1 (i) 4804 (ii) 21808

2 1260

Thanks in advance!

答:
As a starting point, consider the rule of products and the rule of sums. To calculate the number of ways there can be at least one ace, you have to calculate how many ways there can be $1$ ace, $2$ aces, and $3$ aces, and add them together.

To calculate the number of $n$ card hands with $k$ aces, you must first consider that there are $4$ aces in a deck. The number of ways to draw $k$ of them is:

$$\binom{4}{k}$$

Then you must draw $n-k$ non-aces. There are $48$ non-aces, so the number of ways to draw $n-k$ of them is

$$\binom{48}{n-k}$$

Use the rule of products on these two factors to find the total number of $n$ card hands with $k$ aces.


问:
Solutions in a field
Suppose $F$ is a field. Suppose further that $F$ is an ordered field. Consider the following equation $x^3 = b$, where $b \in F$. Prove that the equation above has, at most, one solution.

My attempt:

Suppose we have $x = b^{1/3} \in F$. We have that

$$x^3 = (b^{1/3})^3) = b^{\frac{1}{3} \cdot 3} = b^1 = b$$

Thus, there is one solution.

Assume there exists another solution and denote this by $c^{1/3}$. Since $b^{1/3}$ is a solution, $x = b^{1/3} \implies (b^{1/3})^3 = b \implies (b^{1/3})^3 - b = 0$. In a similar fashion, since $c^{1/3}$ is another solution, $x = c^{1/3} \implies (c^{1/3})^3 = b \implies (c^{1/3})^3 - b = 0$. Together we see that

$$(b^{1/3})^3 - b = 0 = (c^{1/3})^3 - b \implies (b^{1/3})^3 =(c^{1/3})^3 \implies b^{1/3} = c^{1/3} \implies b = c$$

Therefore there is at most one solution.

Is this correct for a field, $F$? Or did I make a mistake? Any assistance is much appreciated.

答: I think this only means that "one solution is unique", you still need to show $a_1^3>a_2^3$ if $a_1,a_2 \in F$

Title: Solutions in a field


答:
Just because you have $(b^{1/3})^3$ equals to $(c^{1/3})^3$, that doesn't mean that $(b^{1/3})$ equals $(c^{1/3})$. In fact, that's the very thing you're being asked to prove.

The theorem isn't true for all fields: for example, if $F=\mathbf{Z}_7$, then $x^3-1$ has three solutions, namely 1, 2 and 4.

However, you only have to prove it for ordered fields. $\mathbf{Z}_7$ is not an ordered field. Your proof will likely use the fact that the field is ordered.

Here's one such proof, but maybe you can find a simpler one:

Let $p$ and $q$ be two solutions to $x^3-b=0$, and let $p<q$ (note I'm using the fact that it's an ordered field) and let $q=p+d$.

Since $p^3-b=(p+d)^3-b=0$, it follows that $d(3p^2+3pd+d^2)=0$, but this equals $d((3p/2+d)^2+3p^2/4)$ (valid because $F$, being ordered, does not have characteristic $2$).

The only way a positive weighted sum of two squares can equal $0$ is if (because $F$ is ordered) both squares are $0$, so that means either $d=0$ or $3p/2+d=0$ and $p=0$, in which case $d=0$ anyway.

Thus $p^3=q^3$ if and only if $p=q$, hence $x^3-b=0$ has at most one solution for any $b$ (it may have no solutions).

There are probably simpler ways to prove this, but you will have to rely on the fact that the field $F$ is ordered. It's sufficient to prove that $f(x)=x^3$ is strictly monotonic.

Title: Solutions in a field

Body:
Suppose $F$ is a field. Suppose further that $F$ is an ordered field. Consider the following equation $x^3 = b$, where $b \in F$. Prove that the equation above has, at most, one solution.

My attempt:

Suppose we have $x = b^{1/3} \in F$. We have that

$$x^3 = (b^{1/3})^3) = b^{\frac{1}{3} \cdot 3} = b^1 = b$$

Thus, there is one solution.

Assume there exists another solution and denote this by $c^{1/3}$. Since $b^{1/3}$ is a solution, $x = b^{1/3} \implies (b^{1/3})^3 = b \implies (b^{1/3})^3 - b = 0$. In a similar fashion, since $c^{1/3}$ is another solution, $x = c^{1/3} \implies (c^{1/3})^3 = b \implies (c^{1/3})^3 - b = 0$. Together we see that

$$(b^{1/3})^3 - b = 0 = (c^{1/3})^3 - b \implies (b^{1/3})^3 =(c^{1/3})^3 \implies b^{1/3} = c^{1/3} \implies b = c$$

Therefore there is at most one solution.

Is this correct for a field, $F$? Or did I make a mistake? Any assistance is much appreciated.


问:
Which number is greater? $2\sqrt{2}$ or $e$
I have to determine which number is greater, $2\sqrt{2}$ or $e$.

I had a similar question as well, it was $2^\sqrt{2}$ compared to $e$.

For that one I managed to prove the inequality by using the increasing sequence converging to $e$: $(1+\frac1n )^n $

So I just searched for a value to assign to n such that $(1+\frac1n )^n \gt 2^\sqrt2$

I tried to proceed in a similar way with $2\sqrt{2} \gt \lt e$ , but it seems I can't get nowhere (I used the sequence decreasing and converging to $e$:  $(1+\frac1n )^{n+1}$ )

Is there another way to prove the inequality without the use of the calculator and maybe using derivatives? The question was in a derivatives file, so I'm wondering is there's a way to get to the end using them.

Any hint would be much appreciated, thanks.

答:
You aleady received good and simple answers to the question.

Now just for your curiosity: during your steps, your tried to find $n$ such that
$$\left(1+\frac{1}{n}\right)^n > k$$
The only explicit solution for the equality is given in terms of Lambert funnction and it is
$$n=\frac{\log (k)}{W_{-1}\left(-\frac{\log (k)}{k}\right)+\log (k)}$$ For $k=2^\sqrt2$, the exact solution would be, as a real, $n=24.6624$, then $\lceil n \rceil=25$.

Computing
$$\left(\frac{26}{25}\right)^{25}=\frac{236773830007967588876795164938469376}{88817841970012523233890533447265625}\sim 2.66584$$ while $2^\sqrt2\sim 2.66514$.


问:
does the following dynamic system converge to a steady state?
This is an economics problem, but I'm pretty sure this kind of thing comes up elsewhere. I've used dynamic programming to find the optimal path of a system (law of motion), which is:

$k_{t+1}=\beta\alpha k_t^\alpha$,

where $\alpha,\beta \in R(0,1)$

now I want to find out if this expression converges to a steady state, defined as

$\lim_{t\to\infty} k_{t+1}=k_t=k^*.$

I thought about trying to tackle it as an eigenvalue problem, of the form

$\left[ \begin{matrix} \log k(t+1)\\ \log k(t+2) \end{matrix} \right] = \left[ \begin{matrix} 0 &1\\ 0 &\alpha \end{matrix}\right] \left[ \begin{matrix} \log k(t)\\ \log k(t+1)\end{matrix} \right] + \left[ \begin{matrix} 0\\ \log\beta + \log\alpha \end{matrix} \right]$

or

$U_{t+1} = AU_t +B$

My question is this: will the system converge as long as the eigenvalues of A are inside the unit circle? I'm concerned about how the contribution of B behaves over time. Am I on the right track with this thing at all?

答:
This dynamical system is explicitly solvable.

Let $x_t=\log k_t$, then $x_{t+1}=\alpha x_t+b$ with $b=\log(\alpha\beta)$. Since $\alpha\ne1$, this is equivalent to $x_{t+1}-x^*=\alpha (x_t-x^*)$ with $x^*=b/(1-\alpha)$, hence $x_t=x^*+\alpha^t(x_0-x^*)$ is the explicit solution of your dynamical system. Since  $|\alpha|<1$, $x_t\to x^*$ for every starting point $x_0=\log k_0$, which means that $k_t\to k^*$ with $k^*=\exp(x^*)=(\alpha\beta)^{1/(1-\alpha)}$.


问:
Is there an asymptotic for following recurrences?
The recurrence I am interested is

$$T(2^k)=2T(2^{k-1})+aT(2^{k-2})+c_1\cdot 2^k$$
at an $a\in\mathbb R_{>0}$ and at a $c_1>0$.

Is there a closed form or a sharp asymptotic?

答:
Set

$$S(k) \triangleq \frac{T(2^k)}{2^k}$$

Then after dividing everything in sight by $2^k$, your recurrence becomes

$$S(k) = S(k-1) + a S(k-2) + b$$

Here $a$ and $b$ are still arbitrary positive constants, but they are not the same constants we started with. This is a minor piece of bookkeeping at the end, though, which I will leave to you.

Now we have a linear recurrence, which is easily solved. According to wolframalpha:

$$S(k) = \frac{a c_1 (1 - \sqrt{4a+1})^k + a c_2 (1+\sqrt{4a+1})^k - b 2^k}{2^ka}$$
though you can use generating functions, linear algebra, etc. if you prefer. Here $c_1$ and $c_2$ are arbitrary parameters (related to the two base cases of the recursion).

Then unsubstituting gives

$$T(2^k) = \frac{a c_1 (1 - \sqrt{4a+1})^k + a c_2 (1+\sqrt{4a+1})^k - b 2^k}{a}$$

Of course, it is easy to get the asymptotics from this, if that's what you're interested in.

I hope this helps ^_^


问:
Prove: If $f(x)$ is a local maximum, then $f(x)$ is a global maximum.
Let $C \subseteq \mathbb{R}^n$ be a convex set. A function $f:C \to \mathbb{R}$ is convex if for all $x,y\in C$ and all $t \in [0,1]$,
$$f(tx+(1-t)y) \le tf(x)+(1-t)f(y).$$
Prove or disprove the following. Assume that $C \subseteq \mathbb{R}^n$ is a convex set and $f: C \to \mathbb{R}$ is a convex function.

If $f(x)$ is a local maximum, then $f(x)$ is a global maximum.

If $f(x)$ is a local minimum, then $f(x)$ is a global minimum.

My instincts tell me that 1 is false and 2 is true. I am not sure how to find a counter-example for number 1 but I think I can prove the second one using a contradiction.

答:
I would agree with your instincts.

To show 1, you just need an example. The function $f(x) = x^2$ on $[-1,2]$ will do.

Hint: what are the local maxima? And don't forget to prove $f$ is convex.


答:
You offer no justification for your claim that $b^{1/3} = c^{1/3} \implies b = c$.

In essence, that's what the problem requires you to prove.

As an indication that your proof is not valid, note that you never used the hypothesis that $F$ is an ordered field.

Thus, let $F$ be an ordered field, and suppose $x,y\in F$ are such that $x^3=y^3$.

We want to show $x=y$.

If $x=0$, then $x^3=0$, so $y^3=0$, hence $y=0$, so $x=y$.

Similarly, if $y=0$, then $y^3=0$, so $x^3=0$, hence $x=0$, so $x=y$.

Thus suppose $x,y\ne 0$, and assume $x\ne y$.

Our goal is to derive a contradiction.

\begin{align*}
\text{Then}\;\;&
x^3=y^3
\\[4pt]
\implies\;&
x^3-y^3=0
\\[4pt]\implies\;&
(x-y)(x^2+xy+y^2)=0
\\[4pt]
\implies\;&
x^2+xy+y^2=0
\\[4pt]
\implies\;&
4x^2+4xy+4y^2=0
\\[4pt]
\implies\;&
(4x^2+4xy+y^2)+3y^2=0
\\[4pt]
\implies\;&
(2x+y)^2+3y^2=0
\\[4pt]
\end{align*}
conradiction, since $(2x+y)^2 \ge 0$ and $3y^2 > 0$.

Title: Solutions in a field


问:
Uniform Continuity: Is this a good proof?
Show that $f(x)=1/x^2$ is not uniformly continuous one the set $(0,1]$

Using the Sequential Criterion for Nonuniform Continuity - which states that

a function $f:A \rightarrow  $ R fails to be uniformly continuous on A iff there exists a particular $\epsilon_0$>0 and two sequences ($x_n$) and ($y_n$) in A, satisfying $|x_n -y_n| \rightarrow 0$, but $|f(x_n) - f(y_n)|\ge \epsilon_0$

I would say:

Take ($x_n$) = $\frac{1}{n}$ and ($y_n$)=$\frac{1}{n^2}$. Obviously $|\frac{1}{n} - \frac{1}{n^2}| \rightarrow 0$, but $|f(x_n) - f(y_n)| = |n^2 - n^4| \ge 12 $, for example for n $\ge$ 2

答:
This CW answer is intended to remove this question from the Unanswered queue.

There is nothing wrong with your argument involving the Sequential Criterion.

As pointed out in the comments, you could also have approached this by:

exhibiting a Cauchy sequence $(x_n)_n$ such that $(f(x_n))_n$ is not Cauchy;

showing that the range of $f$ is not bounded, while its domain is.


问:
show that the hypothesis $\mu(X) < \infty$ can not be omitted
Let $(X,\mathfrak{M},\mu)$ be a measure space and $\mu(X)<\infty$ and $\{f_n\}$ be a sequence of bounded complex measurable functions on $X$, and $f_n\rightarrow f$ uniformly on $X$  then  $\lim_{n\to\infty}\int_X f_n d\mu =\int_X f d\mu$.

show that  the  hypothesis $\mu(X)  < \infty$  can not be omitted

My attempt : I was  thinking about constant function  $f_n= c $

Is there any other example ?

答: A constant function does not work, since $\int_X f_n \mathrm{d}\mu = \int_X f \mathrm{d}\mu$ in this case. However, you can choose $f_n = 1/n$.


答:
A different method:

$$\lim_{n\to\infty}\frac{n^k}{3^\sqrt n}=\lim_{n\to\infty}\exp\left(\ln\left(\frac{n^k}{3^\sqrt n}\right)\right)=\exp\left(\lim_{n\to\infty}\ln\left(\frac{n^k}{3^\frac n2}\right)\right)$$

We can try to find $\lim_{n\to\infty}\ln\left(\frac{n^k}{3^\frac n2}\right)$ and plug the result back in.

$$\lim_{n\to\infty}\ln\left(\frac{n^k}{3^\frac n2}\right)=\lim_{n\to\infty}k\ln n-\frac n2\ln3=\lim_{n\to\infty}k\ln n-\frac{\ln3}2 n$$

Note that the derivative of $k\ln n-\frac{\ln3}2 n$ with respect to $n$ is $\frac kn-\frac{\ln 3}2$. As the derivative approaches $-\frac{\ln 3}2$ as $n$ tends to infinity, this implies the function will decrease without bound, i.e. $\lim_{n\to\infty}k\ln n-\frac{\ln3}2 n=-\infty$.

Plugging this back:

$$\exp\left(\lim_{n\to\infty}\ln\left(\frac{n^k}{3^\frac n2}\right)\right)=\exp(-\infty)=0$$

Q.E.D.


问:
Show that automorphism Aut(N/$\mathbb{Q}$) is isomorphic to symmetric group $S_3$
Let the extension $N/\mathbb{Q}$  be the splitting field of the of the polynomial $f (X) = X^3 - 2 ∈ Q [X]$, where $N ⊂ \mathbb{C}$. Show that:
The automorphism group Aut$(N / \mathbb{Q})$ is isomorphic to the symmetric group $S_3$.

Im currently studying for my algebra exam and this one exercise of an old algebra exam is rather hard. I really don't know where to begin, so maybe someone could give me hint? Thanks in advance!

Edit: Thanks for the quick answer! Ok, so the splitting field should be:
The polynom has the  zero points $\sqrt[3]{2}$, $\sqrt[3]{2}\zeta_3$ and $\sqrt[3]{2}\zeta_3^2$ with $\zeta_3=\frac{-1+i\sqrt{3}}{2}$. The splitting field of $f$ over $\mathbb{Q}$ should be $L=\mathbb{Q}(\sqrt[3]{2},\zeta_3)$

But now I am not sure how to determine $\mathbb{Q}$-conjugates of $N$

答:
Your edit is very much on the right direction. The splitting field is the smallest field containing $\sqrt[3]{2}\zeta_3^j$ for $j = 0,1,2$. This is certainly contained in $\Bbb Q(\sqrt[3]{2}, \zeta_3)$ but since $\zeta_3 = (\sqrt[3]{2}\zeta_3^2) / (\sqrt[3]{2}\zeta_3^2)$ we have the other inclusion, hence the equality.

Now, this extension is Galois and it sits in a tower $N = \Bbb Q(\sqrt[3]{2})(\zeta_3) - \Bbb Q(\sqrt[3]{2}) - \Bbb Q$.

Hint: prove that

$$
[N:\Bbb Q] = [N : \Bbb Q(\sqrt[3]{2})][\Bbb Q(\sqrt[3]{2}) : \Bbb Q] = 6.
$$

Indeed,

 Since $X^2+X+1 \in \Bbb Q[X] \subset \Bbb Q(\sqrt[3]{2})[X]$ vanishes at $\zeta_3$ and $\zeta_3 \in \Bbb C\setminus \Bbb R$, we have $[N : \Bbb Q(\sqrt[3]{2})] = 2$. On the other hand since $X^3-2$ is irreducible over $\Bbb Q$ by Eisenstein's criterion, we have $m(\sqrt[3]{2},\Bbb Q) = X^3-2$ and $[\Bbb Q(\sqrt[3]{2}) : \Bbb Q] = 3$.

Hint': This gives $[N:\Bbb Q] = 6$ and so $G = \mathbf{Gal}(N/\Bbb Q)$ has order $6$.
If you already know that a splitting field of a polynomial of degree $n$ embeds as a subgroup of $S_n$, then you are done. Otherwise, since groups of order $6$ are characterized, analize them and describe some explicit automorphisms to conclude.

 There are only two such groups up to isomorphism, namely, $S_3$ and $\Bbb Z_6$. It suffices to show that $G$ is not $\Bbb Z_6$. One possible argument is to note that the maps defined by $\tau(\zeta_3) = \zeta_3^2, \tau(\sqrt[3]{2}) = \sqrt[3]{2}$ and $\sigma(\zeta_3) = \zeta_3, \sigma(\sqrt[3]{2}) = \sqrt[3]{2}\zeta_3$ are a transposition and $3$-cycle respectively who do not commute.


问:
Convex hull of points in $\mathbb{S}^n$ not contained in a hemisphere
Suppose I have $n+2$ points in $\mathbb{S}^n$ in the induced metric. In this question, the asker states that the convex hull of any four points in $\mathbb{S}^2$ which are not contained in a hemisphere must be all of $\mathbb{S}^2$ (they give a citation, but I don't have access to it). I have tried to prove it myself, but I haven't made much progress. How is this result proved? Also, I suspect that that a similar result holds in higher dimensions. Is this the case?

答: Partial answer: the result continues to hold in higher dimensions. Several authors (Danzer, Grünbaum and Klee, Helly's theorem and its relatives, citing e.g. Robinson, Spherical theorems of Helly type and congruence indices of spherical caps and Blumenthal, Metric methods in linear inequalities) assert without proof that convex sets (in the sense that every pair of non-antipodal points are connected by a minimising geodesic, the first reference also uses some other definitions) are intersections of hemispheres.</p>


问:
Writing a sum as a fraction
Express

$$\sum^{20}_{i=2}f(x)^i$$where $$f(x)=\sum_{i\geq 1}2^{i-1}x^{3i}$$ as a fraction of polynomials $p(x)/q(x)$ and simplify as much as possible.

Hmm. How to do it? Wolfram is really stupid on this one, just expanding the sum. Plugging into geometric series gives intractable mess.

答:
The sum of any geometric series is:

$$\sum_{k=0}^{n}r^k=\frac{1-r^n}{1-r}$$

Since your index starts at 2, you'll have to subtract the first two terms:

$$\sum_{k=2}^nr^k=\frac{1-r^n}{1-r}-r-1$$

Your $f(x)$ is an infinite series, so use the same equation to calculate the sum of the series. As long as $x^3<\frac{1}{2}$:
$$f(x)=\frac{1}{2}\sum_{i=1}^\infty{(2x^3)}^i=\frac{x^3}{1-2x^3}$$

The final result will be the geometric series from 1 to 20 with $r=f(x)$.

Title: Writing a sum as a fraction


问:
Limit problem. Do not understand how do I evaluate it.
What do I do with it?

I cannot use L'Hopital's rule to equaliate limits right now, thats what the teacher said.

Body: $$\frac{\sqrt{n+3} - \sqrt[3]{8n^3+3}}{\sqrt[4]{n+4}-\sqrt[5]{n^5+5}} = \frac{\dfrac{\sqrt{n+3} - \sqrt[3]{8n^3+3}}n}{\dfrac{\sqrt[4]{n+4}-\sqrt[5]{n^5+5}}n}$$ and note that $$\frac{\sqrt[p]{\textrm{something}}}{n} = \sqrt[p]{\frac{\textrm{something}}{n^p}}$$


问:
Prove n is always a multiple of 4 for this sum to be 0.
Each of the numbers $x_1,x_2,x_3,...,x_n$ equals 1 or -1, and
$$x_1x_2x_3x_4 + x_2x_3x_4x_5 + ... + x_nx_1x_2x_3 = 0.$$
Prove that n is divisible by 4.

I attempted to solve this problem first by letting $x_1x_2x_3...x_n$ = $c$ where $c=-1$ or $c=1$ and rewriting but it didn't provide much use. It's pretty clear that n must be even, so from there we must prove that if n is even with only 1 factor of 2 this fails. Another idea I had was that I noticed we can write the term i+1 in terms of term i, by multiplying by $x_{i+5}/{xi}$. In the case of n=6 this provides some nice cancellations, but overall, nothing that I think helps me solve the problem. Any insights would be appreciated.

答: There are $n$ terms with $4$ variable each, and each variable occurs $4$ times in the sum.  Since the sum is $0$, there must be the same number of terms each to $1$ and $-1$, so that $n$ is even, say $n=2m$.  There are $m$ terms with an odd number of variables equal to $-1$ and $m$ terms with an even number of variables equal to $-1$.  Thus, if $m$ is odd, the total number of negative variables, counting each occurrence of each variable separately, is odd.  This is impossible, since the total number of negative occurrences is $4$ times the number of negative variables.  Then $m$ is even, and $m$ is divisible by $4$.


问:
Why is $d(x^2)= 2xdx$?
This seems a bit obvious, but I'm not clear on one part. My teacher had told us $d(x^2) =2xdx$.
However on asking why, we were told $d(x^2)= (x+dx)^2-x^2= 2xdx+(dx)^2$. We were told that we can ignore the $(dx)^2$ because its small.
For obvious reason I feel this step is wrong, why can we ignore $(dx)^2$, is their any other way of getting the result?

答: By definition, $df$ is the linear part of $\Delta f$ when you give an increment $dx$ to $x$. You first compute $\Delta x^2$ as you did. Then you only keep the linear part in $dx$: $dx^2=2xdx$.

Title: Why is $d(x^2)= 2xdx$?