答:
I'll rederive Cauchy-Schwarz inequality (specifically the equality condition) for continuous functions on an interval. Suppose $f,g : [a,b] \to \mathbb{R}$ are continuous. Throughout the following, all integrals are over $[a,b]$. Since $f$ and $g$ are continuous on a compact set, they are bounded, and so $\int f^2, \int g^2, \int fg$ all exist. I'll assume that $\int f^2 > 0$. I'll also assume that you can show the following result - if $h$ is continuous on an interval $[a,b]$, then $\int_{[a,b]} h^2 = 0$ if and only if $h$ is identically zero on the interval. [This is why I can assume $\int f^2 > 0$ - otherwise $f$ is the $0$ function and there's nothing to show]

Consider the function $$ J(t) := \int (t f - g)^2 = t^2 \int f^2 - 2 t \int fg + \int g^2.$$

We can derive Cauchy-Schwarz by observing that $J(t) \ge 0$ since it is the integral of a non-negative function, and then minimising the quadratic. Notice that as long as $\int f^2 > 0,$ the quadratic has a unique minima at $t_* = \int fg/ \int f^2$.  So we have that for all $t \in \mathbb{R},$ $$ J(t) \ge J(t_*) = \int g^2 -  \frac{(\int fg)^2}{\int f^2} \ge 0.$$

Now, we want to argue that if $(\int fg)^2 = \int f^2 \int g^2,$ then there must exist a $t \in \mathbb{R}$ such that $tf = g.$ But notice that $(\int fg)^2 = \int f^2 \int g^2$ implies that for $t_* = \int fg/ \int f^2,$  $J(t_*) = 0$. Due to the definition of $J$, this means that $\int (t_*f - g)^2 = 0.$ But $t_* f - g$ is continuous, so this means that $t_* f = g$ everywhere on the interval.

Now you can apply this equality condition to the continuous functions $xf$ and $f'$.

Title: Proving $\frac 1 4 < \left( \int_a^b [f']^2 \ dx\right)^{1/2}\left( \int_a^b x^2[f(x)]^2 \ dx\right)^{1/2}$

Body:
In chapter 6 of Rudin's Principles of Mathematical Analysis, problem 15, he writes

Suppose $f$ is a real, continuously differentiable function on $[a,b], f(a) = f(b) = 0$, and

$$\int_a^b f^2(x)\ dx = 1.$$

Prove that

$$ \int_a^b x f(x) f'(x) \ dx = -\frac 1 2 $$

and that

$$\int_a^b[f'(x)]^2 \ dx \cdot \int_a^b x^2f^2(x)\ dx > \frac 1 4. $$

I've shown everything up to demonstrating

$$\int_a^b[f'(x)]^2 \ dx \cdot \int_a^b x^2f^2(x)\ dx \ge \frac 1 4 $$

However, I can't figure out how to eliminate the case where equality holds.  I've seen several other resources say that equality would imply

$$ f'(x) = \lambda xf(x) $$

but I can't figure out why that would be implied.

Resources I've already looked at: How to show the inequality is strict?

Baby Rudin Chapter 6, Problem 15 : Strict inequality

Prob. 15, Chap. 6, in Baby Rudin: If $f$ is a real, continuously differentiable function on $[a, b]$, . . .

Proving a strict inequality (Application of Hölder's Inequality)

Show that if $f(a)=f(b)=0$ and $\int_a^b [f(x)]^2dx=1$, then $\int_a^b [f'(x)]^2dx\cdot\int_a^b [xf(x)]^2dx\gt \frac14$


问:
Solutions to $z^5-1=0$ using "root of unity"?
I have recently stumbled upon a problem which says:

Given that $z^5-1=0$, find all solutions for $z\in\mathbb{C}$.

I have researched this problem a bit, and I have discovered that this has something to do with a Root of Unity. What is a root of unity and how do I solve one? Also, is there a formula for finding all complex solutions of $z^n=1$, where n is a constant? If so, what does this have to do with a root of unity?

EDIT

Another question: I see the $\cos(x)+i\sin(x)$ formula for all $z^n=1$, but is there a formula for all $z^n=c$ for some $c \in \mathbb{R}$, or at least positive $c \in \mathbb{R}$?

答:
Definition (Root of unity). An element $\zeta$ of a ring $R$ with unity $1$ such that $\zeta^m = 1$ for some $m \ge 1$.  The least such $m$ is the order of $\zeta$.  -Encyclopedia of Mathematics

In the field of complex numbers, roots of unity of order $m$ take the form $$\cos\frac{2\pi k}{m} + i \sin\frac{2\pi k}{m},$$
where $k=0,1,\dots,m-1$.

We may write $1$ in polar form as $e^{2\pi i k}$ and therefore $\zeta^m=e^{2\pi i k}\Rightarrow\zeta=e^{\frac{2\pi i k}{m}}$. With Euler's formula, we can decompose the exponential into real and imaginary parts to get our final result, which is stated above. If we add the condition that $k$ must be relatively prime to $m$, we only get the so-called primitive roots.

The same can be done with $\zeta^m=c,$ where $c$ is a real number. Write $c$ as $ce^{2\pi i k}$ to get the slightly generalized result $\zeta=\sqrt[m]{c}\,(\cos\frac{2\pi k}{m} + i \sin\frac{2\pi k}{m})$.

Complex roots of unity 5 $\qquad\qquad\qquad\qquad\:\:$ The mapping $z\mapsto z^5-1$

ContentLicense: CC BY-SA 4.0


问:
Giving a 1-hour talk to highschool math club: any topic suggestion?
I've been invited (by my kid) to give a one hour talk to her highschool math club.  Last year (right before the pandemic hit) I did two such talks on probability, and they loved it.  I'm looking for topic suggestions.

Logistics: There will be about $10-15$ kids total, from grades $9-12$.  These kids are mostly "A" students, and some of them have skipped a math grade, but they're nowhere near Olympiad standard.  The talk will be on March $1$st, $2021$, and will be on zoom.  I will be making slides this weekend.

I don't want this to become an "educational" lecture, so I'm looking for fun examples to motivate certain math areas.  E.g. last year my talks on probability culminated with Polya Urn and Buffon's Noodle.  Also, I'd like everything (well, $95\%$) to be fully explainable within the one hour, i.e. I don't want to just spew facts and then tell them "just trust me".

My own math skill tends toward discrete: comp-sci, combinatorics, etc.  However I'm open to any idea - esp. if it has worked for you before! - as long as I can learn it during one weekend.  :)

Some ideas:

modulo arithmetic esp. finite fields: Perfect shuffle is fun.  Would cryptography (e.g. RSA) be too difficult?  What about codes e.g. Hamming?  Any other fun examples of modulo arithmetic?

non-Euclidean geometry: especially how a POINT in elliptic geometry is actually $2$ "ordinary points".  Can the alternates to the parallel postulate (and some consequence e.g. angle sum in a triangle) be explained sufficiently at highschool level in an hour?

finite geometry: Personally I find Finite projective planes very beautiful.  Do they (or finite affine planes) have any application?  (Besides the kiddie game Spot It?)

algorithm: This is actually my work area, but I don't know what I can cover in an hour.  Maybe the $O(N \log N)$ lower bound for sorting?  Some of these kids don't even have coding experience... :(

graph theory: Edge coloring of a complete graph is one of my favorite pictorial proofs.  Eulerian tour.  Euler characteristic (of a planar graph) is related.  What else?  Shortest path would require discussion of algorithm.  Hall's marriage theorem is surprising and neat but I don't think I can prove it from scratch in an hour.

combinatorics: Start with ${N \choose m}$ and then stars-and-bars for sure.  What are your favorite (elementary) examples?  Can Burnside's lemma be proved from scratch (at highschool level, without group theory) in an hour?

Comments on above and/or any other suggestion most welcome!

(Apologies if I'm tagging too widely...)

答:
When the numbers are not too bad, we can just do a one-step calculation over a few times:
Here I am using decimals; for worse numbers like $\sqrt{61}$

Lubin posted an algorithm (using whole numbers only) that appears to be what Fermat used; I then programmed that. There is a method due (largely) to Gauss and Lagrange that produces a chain or cycle of indefinite quadratic form, I programmed that long ago. Again, no decimal, all whole number, allows for any $\frac{A+\sqrt B}{C}.$  Finally, Conway's Topograph method. For that, I can recommend Weissman for self study or parent-child team study of some sort

Full treatment of $\sqrt{19}$

Method described  by Prof. Lubin  at Continued fraction of $\sqrt{67} - 4$

$$  \sqrt { 19} = 4 +     \frac{  \sqrt {19} - 4 }{ 1 }  $$
$$    \frac{ 1 }{   \sqrt {19} - 4 }  =  \frac{    \sqrt {19} + 4 }{3 } = 2 +  \frac{    \sqrt {19} - 2 }{3 } $$
$$    \frac{ 3 }{   \sqrt {19} - 2 }  =  \frac{    \sqrt {19} + 2 }{5 } = 1 +  \frac{    \sqrt {19} - 3 }{5 } $$
$$    \frac{ 5 }{   \sqrt {19} - 3 }  =  \frac{    \sqrt {19} + 3 }{2 } = 3 +  \frac{    \sqrt {19} - 3 }{2 } $$
$$    \frac{ 2 }{   \sqrt {19} - 3 }  =  \frac{    \sqrt {19} + 3 }{5 } = 1 +  \frac{    \sqrt {19} - 2 }{5 } $$
$$    \frac{ 5 }{   \sqrt {19} - 2 }  =  \frac{    \sqrt {19} + 2 }{3 } = 2 +  \frac{    \sqrt {19} - 4 }{3 } $$
$$    \frac{ 3 }{   \sqrt {19} - 4 }  =  \frac{    \sqrt {19} + 4 }{1 } = 8 +  \frac{    \sqrt {19} - 4 }{1 } $$

Simple continued fraction tableau:

$$
 \begin{array}{cccccccccccccccccc}
 & & 4 & & 2 & & 1 & & 3 & & 1 & & 2 & & 8 & \\
 \\
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 4 }{ 1 }   & &   \frac{ 9 }{ 2 }   & &   \frac{ 13 }{ 3 }   & &   \frac{ 48 }{ 11 }   & &   \frac{ 61 }{ 14 }   & &   \frac{ 170 }{ 39 }    \\
  \\
 & 1 & & -3 & & 5 & & -2 & & 5 & & -3 & & 1
 \end{array}
 $$

$$
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 19 \cdot 0^2 = 1 &     \mbox{digit}  &  4  \\
  \frac{ 4 }{ 1 }   & 4^2 - 19 \cdot 1^2 = -3 &     \mbox{digit}  &  2  \\
  \frac{ 9 }{ 2 }   & 9^2 - 19 \cdot 2^2 = 5 &     \mbox{digit}  &  1  \\
  \frac{ 13 }{ 3 }   & 13^2 - 19 \cdot 3^2 = -2 &     \mbox{digit}  &  3  \\
  \frac{ 48 }{ 11 }   & 48^2 - 19 \cdot 11^2 = 5 &     \mbox{digit}  &  1  \\
  \frac{ 61 }{ 14 }   & 61^2 - 19 \cdot 14^2 = -3 &     \mbox{digit}  &  2  \\
  \frac{ 170 }{ 39 }   & 170^2 - 19 \cdot 39^2 = 1 &     \mbox{digit}  &  8  \\
 \end{array}
 $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

$\sqrt{21}$ Simple continued fraction tableau:

4.58, 1.7, 1.39, 2.52, 1.89, 1.11,8.58

$$
 \begin{array}{cccccccccccccccccc}
 & & 4 & & 1 & & 1 & & 2 & & 1 & & 1 & & 8 & \\
 \\
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 4 }{ 1 }   & &   \frac{ 5 }{ 1 }   & &   \frac{ 9 }{ 2 }   & &   \frac{ 23 }{ 5 }   & &   \frac{ 32 }{ 7 }   & &   \frac{ 55 }{ 12 }    \\
  \\
 & 1 & & -5 & & 4 & & -3 & & 4 & & -5 & & 1
 \end{array}
 $$

$$
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 21 \cdot 0^2 = 1 &     \mbox{digit}  &  4  \\
  \frac{ 4 }{ 1 }   & 4^2 - 21 \cdot 1^2 = -5 &     \mbox{digit}  &  1  \\
  \frac{ 5 }{ 1 }   & 5^2 - 21 \cdot 1^2 = 4 &     \mbox{digit}  &  1  \\
  \frac{ 9 }{ 2 }   & 9^2 - 21 \cdot 2^2 = -3 &     \mbox{digit}  &  2  \\
  \frac{ 23 }{ 5 }   & 23^2 - 21 \cdot 5^2 = 4 &     \mbox{digit}  &  1  \\
  \frac{ 32 }{ 7 }   & 32^2 - 21 \cdot 7^2 = -5 &     \mbox{digit}  &  1  \\
  \frac{ 55 }{ 12 }   & 55^2 - 21 \cdot 12^2 = 1 &     \mbox{digit}  &  8  \\
 \end{array}
 $$

=======================================

$ \sqrt{41}$

6.403, 2.48, 2.08, 12.40,2.48, 2.08, 12.39

$$
 \begin{array}{cccccccccccccccc}
 & & 6 & & 2 & & 2 & & 12 & & 2 & & 2 & & 12 & \\
 \\
  \frac{ 0 }{ 1 }   &   \frac{ 1 }{ 0 }   & &   \frac{ 6 }{ 1 }   & &   \frac{ 13 }{ 2 }   & &   \frac{ 32 }{ 5 }   & &   \frac{ 397 }{ 62 }   & &   \frac{ 826 }{ 129 }   & &   \frac{ 2049 }{ 320 }    \\
  \\
 & 1 & & -5 & & 5 & & -1 & & 5 & & -5 & & 1
 \end{array}
 $$

$$
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 41 \cdot 0^2 = 1 &     \mbox{digit}  &  6  \\
  \frac{ 6 }{ 1 }   & 6^2 - 41 \cdot 1^2 = -5 &     \mbox{digit}  &  2  \\
  \frac{ 13 }{ 2 }   & 13^2 - 41 \cdot 2^2 = 5 &     \mbox{digit}  &  2  \\
  \frac{ 32 }{ 5 }   & 32^2 - 41 \cdot 5^2 = -1 &     \mbox{digit}  &  12  \\
  \frac{ 397 }{ 62 }   & 397^2 - 41 \cdot 62^2 = 5 &     \mbox{digit}  &  2  \\
  \frac{ 826 }{ 129 }   & 826^2 - 41 \cdot 129^2 = -5 &     \mbox{digit}  &  2  \\
  \frac{ 2049 }{ 320 }   & 2049^2 - 41 \cdot 320^2 = 1 &     \mbox{digit}  &  12  \\
 \end{array}
 $$

============================

$$
 \begin{array}{cccc}
  \frac{ 1 }{ 0 }   & 1^2 - 21 \cdot 0^2 = 1 &     \mbox{digit}  &  4  \\
  \frac{ 4 }{ 1 }   & 4^2 - 21 \cdot 1^2 = -5 &     \mbox{digit}  &  1  \\
  \frac{ 5 }{ 1 }   & 5^2 - 21 \cdot 1^2 = 4 &     \mbox{digit}  &  1  \\
  \frac{ 9 }{ 2 }   & 9^2 - 21 \cdot 2^2 = -3 &     \mbox{digit}  &  2  \\
  \frac{ 23 }{ 5 }   & 23^2 - 21 \cdot 5^2 = 4 &     \mbox{digit}  &  1  \\
  \frac{ 32 }{ 7 }   & 32^2 - 21 \cdot 7^2 = -5 &     \mbox{digit}  &  1  \\
  \frac{ 55 }{ 12 }   & 55^2 - 21 \cdot 12^2 = 1 &     \mbox{digit}  &  8  \\
 \end{array}
 $$

http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf
https://www.math.cornell.edu/~hatcher/TN/TNbook.pdf

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问:
Name for a pair of vertices connected by an edge
What is the proper name for a pair of vertices connected by an edge?

NOTE: Lots of googling didn't yield an answer.

DISCLAIMER: I am no math expert.

答: Adjacent vertices</p>

ContentLicense: CC BY-SA 3.0

Body: Adjacent vertices


问:
Why is $ \|T(x) \|\le \| T\| \|x \|$?
Let $X$ and $Y$ be normed linear spaces and let $T$ be a bounded linear operator from $X$ to $Y$. The norm of $T$ is defined as
$$\|T \|=\sup\{\|T(x) \|\;:\;\|x \|\le 1\}. $$

From the definition of the norm, we can say that $\|T(x) \|\le \|T \|\;\forall \;x,$ and that $\|T \|\|x \|\le \|T \|$ as $\|x\|\le 1$.

However I don't understand why we write $\|T(x)\|\le \|T\|\|x\|$?

答:
Clearly $\Vert T(x) \Vert \le \Vert T \Vert$ if $\Vert x \Vert = 1$.

Now, just remember that we're dealing with linear operators.

So if $\Vert x \Vert = c$, then $\Vert \frac{x}{c} \Vert = 1$.

So $$\left\Vert T(x) \right\Vert = c \left\Vert T\left(\frac{x}{c} \right) \right\Vert \le c \Vert T \Vert$$

Title: Why is $ \|T(x) \|\le \| T\| \|x \|$?


问:
$f(z)$ is analytic on $\{ |z| \leq 1\}$, and $f(z)$ is real on $\{|z| = 1\}$, show $f(z)$ is a constant.
Let $f(z)$ be analytic on $\{|z| < 1\}$ that extends to $\{|z| \leq 1\}$. Assume $f(z)$ is real on $|z| = 1$. Prove that $f(z)$ is a constant.

Here is my attempt,
define $g(z) = e^{if(z)}$, for $|z| = 1$, we have $|g(z)| = 1$. Then $|\frac{1}{g(z)}| = 1$ as well. So that $|g(z)| = 1$ for $|z| = 1$.

(1) Is there any theorem I can use to show that this leads to $g(z)$ is constant on $|z| \leq 1$?

(2) If $g(z)$ is constant, can I show that $f(z)$ is constant?

答: The imaginary part of your function is a harmonic function equal zero on the boundary. By uniqueness for the Dirichlet problem, it is identically zero. The real part is the harmonic conjugate which, using Cauchy-Riemann conditions, has zero partial derivatives and hence is a constant. Your function is a real constant.</p>


问:
Prove that $a ≤ x ≤ b ⇒ |x| ≤ |a|+|b|$
Can you please help me with this proof $$a ≤ x ≤ b ⇒ |x| ≤ |a|+|b|$$ ? I am literally stuck for hours.

This is what i thought, but i don't know if it counts as a proof.
First of all, if a ≤ x then  $x ≤ -a$ so $a ≤ x ≤ -a$ if $x ≤ -a$ then $|x| ≤ |-a| = |a|$
Now we look at b:
if x ≤ b then $-b ≤ x ≤ b$ so  $$|-b| = |b| ≤ |x|$$ OR  $$|x| ≤ |-b| = |b|$$
let's choose the "worst" option which is |b| ≤ |x|
so $$|b| ≤ |x| ≤ |a| ⇒  |x| ≤ |a| - |b|  ≤ |a| + |b| ⇒  |x| ≤ |a| +|b|$$

Is it correct? Please show me other ways to prove it.

答:
Hint

$x$ can not be further away from $0$ than both $a$ and $b$.

Consider that $x$ must have the same sign as at least one of $a,b$.

Title: Prove that $a ≤ x ≤ b ⇒ |x| ≤ |a|+|b|$


问:
About convergence in Schwartz space

The Question

Let $f,\varphi\in\mathcal S(\mathbb R^d)$, $\varepsilon>0$ and denote $\varphi_\varepsilon(x)=\frac{1}{\varepsilon^d}\varphi\left(\frac
x\varepsilon\right)$. Prove that $\varphi_\varepsilon\ast f\to f\int\varphi$ as $\varepsilon\to 0$ in $\mathcal S(\mathbb R^d)$.

Some Ideas!

I've been working around this question in a course in Fourier Analysis. Some of the ideas that I've worked around are as such.

First we've got to see what $\varphi_\varepsilon\ast f$, so we expand it in terms of the integral:

$$(\varphi_\varepsilon\ast f)(x)=\int\limits_{\mathbb R^d}\frac{1}{\varepsilon^d}\varphi\left(\frac
y\varepsilon\right)f(x-y)\ dy=\int\limits_{\mathbb R^d}\varphi\left(y\right)f(x-\varepsilon y)\ dy,$$

where the last equality is true after a change of variables $y\mapsto \varepsilon y$.  So, if we were searching for pointwise convergence we would be done. Applying Lebesgue's DCT we obtain that the last term converges to $f\int\varphi$ as $\varepsilon\to 0$.

However, reality sets in and then I realized that this was not the question I was supposed to answer. So I go back to Grafakos' book to check the definitions and now I've got to prove that
$$\rho_{\alpha,\beta}\left(\varphi_\varepsilon\ast f- f\int\varphi\right),$$
where $\rho_{\alpha,\beta}$ is a seminorm in $\mathcal S(\mathbb R^d)$, is a really small quantity as $\varepsilon\to 0$. Writing this out without a supremum we obtain that its absolute value is bounded above by

$$C\|x\|^{|\alpha|}\left|\partial^\beta\int\limits_{\mathbb R^d}\varphi(y)(f(x-\varepsilon y)-f(x))\ dy\right|.$$

Now, this quantity converges to zero pointwise when $x$ is fixed. It doesn't converge uniformly because of the norm outside the integral and I require that because of the supremum in the seminorm.

One idea could be to use the Fourier transform in $\mathcal S$, since I know that if $f_n\to f$ in $\mathcal S$, then $\hat f_n\to \hat f$ in $\mathcal S$ as well (the same result holds for the inverse transform). However, I'm not fully in control of that because I still don't know how to handle the Fourier transform very well.

There's also the fact that applying the Fourier transform four times in a row is the identity operator. But I can't see how to use that.

I also thought for a moment that I could bound the norm using a compact support, but then I recalled that $\mathcal S\subseteq \mathcal C_0$ and not $\mathcal C_c$. So my argument went down.

Any hints, suggestion or comments are welcome. Kind thanks for reading the whole thing.

答:
Let $C=\int_{\Bbb{R}^d} \varphi(x)dx$.

$f \ast \varphi_\epsilon \to f$ in the $\sup$ norm, this follows from $$(\varphi_\varepsilon\ast f)(x)-C f(x)=\int_{\mathbb R^d}\varphi\left(y\right)(f(x-\varepsilon y)-f(x))\ dy$$ that $\varphi\in L^1$ and $f$ is bounded uniformly continuous so that $\sup_{x,y}|f(x-\varepsilon y)-f(x)|\to 0$ as $\epsilon\to 0$.

$(1+|x|^k) (f \ast \varphi_\epsilon) \to (1+|x|^k) f$ in the $\sup$ norm:
$$(1+|x|^k)((\varphi_\varepsilon\ast f)(x)-C f(x))=\int_{\mathbb R^d}\varphi\left(y\right)(1+|x|^k)(f(x-\varepsilon y)-f(x))\ dy$$ $$\sup_{x,y}\frac{1+|x|^k}{1+|y|^k}|f(x-\varepsilon y)-f(x)|\to 0, \qquad (1+|y|^k) \varphi(y) \in L^1$$
($f$ is rapidly decreasing)

Finally use that $$\partial^\alpha(f \ast \varphi_\epsilon)=(\partial^\alpha f) \ast \varphi_\epsilon \to C \partial^\alpha f$$

Title: About convergence in Schwartz space


问:
How many function $f\colon A\to B$ with the following properties: $f (x + y) = f (x) + f (y)$ for every $x, y\in A$
Given $A = \{1, 2, 3,\dots , 111\}$ and $B = \{1, 2, 3,\dots , 2021\}$.

How many function $f\colon A\to B$ with the following properties:

$$f (x + y) = f (x) + f (y)$$

for every $x, y \in A$.

All I can think of is substituting $x = y$

$$f(2x)=2f(x)$$

After that I was confused what should I do.

答:
Think of A on the x-axis  and  B on the y-axis. Then consider the (integer) lattice points. Note that the function is completely determined by $f(1)$. (Can you argue this rigorously?)

Now note that $18<2021/111<19$. So as not to “overshoot” the range, you have to have $1\le f(1)\le 18$. So there  are $18$ such functions.


答:
Given one solution of $z^n=c$, the other $n-1$ can be found by multiplying by the $n-1$ respective $n$-th roots of unity other than $1$.
In general those are the first $n-1$ powers of a primitive $n$-th root.

So take an example:  $z^5=3$.

Then take one solution, say $z=\sqrt[5]3$.  Then if $\zeta_1,\dots,\zeta_4$ are the primitive fifth roots of unity, in this case the ones other than $1$, we have that the five roots are $\sqrt[5]3\zeta_1,\dots,\sqrt[5]3\zeta_4, \sqrt[5]3$.

And note that we have $\zeta_k=\zeta_1^k$.

(All of this follows rather easily from the fact that for any $n$-th root of unity, the $n$-th power is $1$.  And using Euler's formula: $e^{i\theta}=\cos\theta+i\sin\theta$, we also get that $e^{2\pi i k/n}$ is primitive iff $(n,k)=1$.)

Title: Solutions to $z^5-1=0$ using "root of unity"?

Body:
I have recently stumbled upon a problem which says:

Given that $z^5-1=0$, find all solutions for $z\in\mathbb{C}$.

I have researched this problem a bit, and I have discovered that this has something to do with a Root of Unity. What is a root of unity and how do I solve one? Also, is there a formula for finding all complex solutions of $z^n=1$, where n is a constant? If so, what does this have to do with a root of unity?

EDIT

Another question: I see the $\cos(x)+i\sin(x)$ formula for all $z^n=1$, but is there a formula for all $z^n=c$ for some $c \in \mathbb{R}$, or at least positive $c \in \mathbb{R}$?


问:
Representation by Complex Surfaces
Consider the topological manifold $M=\mathbb CP^2\#\overline{CP^2}$.  This has $H_2(M,\mathbb Z)=\mathbb Z H+\mathbb Z E$ where $H^2=1$ and $E^2=-1$ are the standard generators of the homology of $\mathbb CP^2$ and $\overline{\mathbb CP^2}$ respectively, each with self-intersection as given.

The class $5H-E$ is representable by a complex curve with respect to the standard complex structure on $M$ induced by blow-up from $\mathbb CP^2$.  This curve has genus 6 by adjunction.

Now consider the class $H-5E$.  As $M$ admits an orientation-reversing diffeomorphism, this class is also smoothly represented by an embedded surface of genus 6.

Does $M$ admit a complex structure, inducing the opposite orientation of the standard complex structure, in which this class has a complex representative?  If so, what is the corresponding first Chern class of this structure?

答:
The answer is yes. It is a somewhat tautological construction.

Suppose $f : M \to M$ is a diffeomorphism and $\mathcal{A} = \{(U_i, \varphi_i)\}_{i\in I}$ is a complex atlas. Then $\mathcal{A}' = \{(f^{-1}(U_i), \varphi_i\circ f)\}_{i \in I}$ is also a complex atlas on $M$ and $f : (M, \mathcal{A}') \to (M, \mathcal{A})$ is a biholomorphism, so $c_1(M, \mathcal{A}') = f^*c_1(M, \mathcal{A})$. Moreover, if $i : \Sigma \to (M, \mathcal{A})$ is an embedding of a complex submanifold of $(M, \mathcal{A})$, then $f^{-1}\circ i : \Sigma \to (M, \mathcal{A}')$ is also an embedding of a complex submanifold of $(M, \mathcal{A}')$.

There is an orientation-reversing diffeomorphism $f : \mathbb{CP}^2\#\overline{\mathbb{CP}^2} \to \mathbb{CP}^2\#\overline{\mathbb{CP}^2}$ which 'switches summands', so $f_*$ swaps $H$ and $E$. If $\mathcal{A}$ is the complex atlas from the standard complex structure on $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$ (i.e. the blowup of $\mathbb{CP}^2$ at a point), then $c_1(\mathbb{CP}^2\#\overline{\mathbb{CP}^2}, \mathcal{A})$ is the Poincaré dual of $H - E$. On the other hand, $\mathcal{A}'$ defines a complex structure on $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$ which induces the opposite complex structure, and $c_1(\mathbb{CP}^2\#\overline{\mathbb{CP}^2}, \mathcal{A}')$ is the Poincaré dual of $E - H$.

Suppose $i : \Sigma \to (\mathbb{CP}^2\#\overline{\mathbb{CP}^2}, \mathcal{A})$ is an embedding of a genus $6$ curve with $i_*[\Sigma] = 5H - E$. Then $f^{-1}\circ i : \Sigma \to (\mathbb{CP}^2\#\overline{\mathbb{CP}^2}, \mathcal{A}')$ is an embedding of a genus $6$ curve with $$(f^{-1}\circ i)_*[\Sigma] = (f^{-1})_*\circ i_*[\Sigma] = (f_*)^{-1}(i_*[\Sigma]) = (f_*)^{-1}(5H - E) = -5E + H = H - 5E$$ as desired.

Title: Representation by Complex Surfaces


问:
$\forall a,b,c,d\in \mathbb{R}$ prove that $(a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c) \geq 0$ and $a + d = b + c$
$\forall a,b,c,d\in \mathbb{R}$ prove that $(a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c)$ and
$a + d = b + c$
Attempt

consider $(a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c)=2(a-b)(c-d)$ since $2 >0$ we also want show that
$(a-b)(c-d)\geq 0$

When $a=c$ or $c=d$ we get that the expression can be $0$. Otherwise

Suppose that
$(a-b)(c-d)<0$ for $a,b,c,d$ it implies that the expression is  upper bounded by $0$
but for $a=1,b=-1,c=3,d=1$ it not occur therefore must be $(a-b)(c-d) \geq 0$.

I don´t sure of the part above, is mi proof right?
or I should consider an argue of say that
for real numbers $a,b,c,d$ we have that if $(a-b)<0$ then $(c-d)$ cannot be $(c-d)>0$ and similary in the other cases. Argue that we cannot have two factors with differnt sign.

Thanks in advice, I appreciate any hint or correction.

答:
Let $a,b,c,d \in \mathbb{R}$ such that $a+d=b+c \Rightarrow a-b=c-d$
$$(a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c)=(a-b)^2+(a-c)(b-d)+(d-a)(b-c)$$
Notice that $$(a-c)(b-d)+(d-a)(b-c)=ab-ad-cb+dc+db-dc-ab+ac$$
$$=ac-ad+db-cb$$
$$=a(c-d)+b(d-c)=a(c-d)-b(c-d)=(a-b)^2$$
Remplazing

$$(a-b)^2+(a-b)^2=2(a-b)^2 \geq 0$$ which is true


问:
Finding a particular solution in Matlab
So I have a question about finding the particular solution to a non-homogeneous system of equations in Matlab. This is building on an assignment where I had to find the spanning set (I think this is the same as the column space?) of a system.
The snippet of code shows my solution to the problem and here is my reasoning:

Begin by finding the ordered set of indexes of the pivot and the non-pivot positions in the reduced coefficient matrix A (with dimensions m x n).

Next, reduce the augmented matrix and call it R.

The first for loop sets all of the free variables (non-pivot positions) in R to zero. All I have left are the basic variables and the solution column. Since all pivot positions are equal to 1, I can say that variable is equal to the solution column plus some combination of free variables. (Right now, I am only interested in the solution column - for example, if x1 = 3 + x2, I only want the 3)

The second for loop assigns entries in the last column to a row in matrix p for the particular solution.

My code has worked for all the test cases I have tried so far, but I am wondering whether my solution is actually correct or if I just got lucky on a few examples? If I'm wrong (or if I'm right) is there a better way to solve this?

```<code>%finds the ordered set of indexes of the pivot columns
[~,pivot_c]=rref(A);

%finds the ordered set of indexes of the non-pivot columns
S=1:n;
nonpivot_c=setdiff(S,pivot_c);

%set R equal to the reduced echelon form of the augmented matrix
R = rref([A,b]);

%pre-allocate space for the particular solution p
p = zeros(n,1);

%set all free variables in R equal to zero
for i = 1:n
    if i == nonpivot_c
    R(:,i) = zeros(m,1);
    end
end

%now the only values remaining are the basic variables
%the last column is the values of the particular solution
%when the free variables are equal to zero
for i = 1:m
    for j = 1:n
        if R(i,j) == 1
            p(j,1) = R(i, (n+1));
        end
    end
end

%display the solution
p

</code>```

答:
What you have done usually works (as long as it is known a priori that the equation $Ax = b$ has a solution). Unfortunately, if i == nonpivot_c does not do what you think it does: Matlab will never treat i == nonpivot_c as a true statement (and will therefore never zero out the free columns) because it is an array of Booleans. If any of your free columns happens to contain $1$ as an entry, this will lead to trouble in your final for loop.

To correctly check whether i is an element of nonpivot_c, use if ismember(i,nonpivot_c).

In any case, the method you outlined is not the most efficient way to do what you're trying to do. Consider the following approach:

```<code>A = [0,0,2,3,4;...
    1,2,3,4,5;...
    -1,-2,-5,-7,-9];
b = [2;0;-2];        % Has a solution
% b = [1;0;0];         % Has no solution

[m,n] = size(A);
[R,p_cols] = rref([A,b]);
if p_cols(end)==n+1
    disp('---NO SOLUTION---')
else
    p=zeros(n,1);
    for i = 1:length(p_cols)
        p(p_cols(i)) = R(i,end);
    end
    disp('SOLUTION:')
    disp(p)
end
</code>```

As a bonus, we can eliminate the for loop:

```<code>[m,n] = size(A);
[R,p_cols] = rref([A,b]);
if p_cols(end)==n+1
    disp('---NO SOLUTION---')
else
    p=zeros(n,1);
    p(p_cols) = R(1:length(p_cols),end);
    disp('SOLUTION:')
    disp(p)
end
</code>```

Title: Finding a particular solution in Matlab

Body:
What you have done usually works (as long as it is known a priori that the equation $Ax = b$ has a solution). Unfortunately, if i == nonpivot_c does not do what you think it does: Matlab will never treat i == nonpivot_c as a true statement (and will therefore never zero out the free columns) because it is an array of Booleans. If any of your free columns happens to contain $1$ as an entry, this will lead to trouble in your final for loop.

To correctly check whether i is an element of nonpivot_c, use if ismember(i,nonpivot_c).

In any case, the method you outlined is not the most efficient way to do what you're trying to do. Consider the following approach:

A = [0,0,2,3,4;...
    1,2,3,4,5;...
    -1,-2,-5,-7,-9];
b = [2;0;-2];        % Has a solution
% b = [1;0;0];         % Has no solution

[m,n] = size(A);
[R,p_cols] = rref([A,b]);
if p_cols(end)==n+1
    disp('---NO SOLUTION---')
else
    p=zeros(n,1);
    for i = 1:length(p_cols)
        p(p_cols(i)) = R(i,end);
    end
    disp('SOLUTION:')
    disp(p)
end

As a bonus, we can eliminate the for loop:

[m,n] = size(A);
[R,p_cols] = rref([A,b]);
if p_cols(end)==n+1
    disp('---NO SOLUTION---')
else
    p=zeros(n,1);
    p(p_cols) = R(1:length(p_cols),end);
    disp('SOLUTION:')
    disp(p)
end


问:
Proving $\text{vol}(A) \le \sum_{i=1}^N \text{vol}(A_i)$ where $A \subset \bigcup_{i=1}^N A_i$
I am trying to prove that
$\displaystyle\text {vol}(A) \le \sum_{i=1}^N \text {vol}(A_i)$ where $\displaystyle A \subset \bigcup_{i=1}^N A_i$. This is intuitively very clear but I cannot think of another way to prove it other than saying
$$
\text {vol}(A) = \int_A \chi_A \le \text {vol}(B) = \int_B \chi_B \le \sum_{i = 1}^N \text {vol}(A_i) = \sum_{i = 1}^N \int_{A_i} \chi_{A_i}
$$
where $B = A_1 \cup A_2 \cup \cdots \cup A_N$. Does what I wrote down here complete the proof? Are there better ways of showing what I am trying to prove here? Thanks!

答:
Note that

$$\chi_{A_1\cup A_2} = \chi_{A_1}+ \chi_{A_2}- \chi_{A_1\cap A_2}\leqslant \chi_{A_1}+ \chi_{A_2}$$

By an easy inductive argument,

$$\chi_{\cup_{j=1}^NA_j} = \chi_{\cup_{j=1}^{N-1}A_j\cup A_N}\leqslant  \chi_{\cup_{j=1}^{N-1}A_j}+ \chi_{A_N} \leqslant \sum_{j=1}^N \chi_{A_j}$$

By monotonicity and additivity of the integral with ambient space $X$,

$$\int_X\chi_{\cup_{j=1}^NA_j}\leqslant \int_X \sum_{j=1}^N \chi_{A_j} = \sum_{j=1}^N \int_X\chi_{A_j}$$

Since $A \subset \cup_{j=1}^NA_j$, we have $\chi_A \leqslant \chi_{\cup_{j=1}^NA_j}$, and

$$vol(A) = \int_X \chi_A \leqslant \int_X\chi_{\cup_{j=1}^NA_j}\leqslant \sum_{j=1}^N \int_X\chi_{A_j} = \sum_{j=1}^N vol(A_j)$$


问:
Inequality with a floor function
I'm trying to prove that $8x^2-5\text {floor} (x)^2 \ge 10$ for all $x \ge 4$? I understand that I can use $0 \le  x - \text {floor} (x) < 1$ to assist in solving the equation but I'm having trouble fitting it in.

答:
we know $8x^2 \geq 10 + 5x^2$ for all $x \geq 4$... Just solve the inequality or draw graph///

and $x^2 \geq \text{floor(x)}^2$ for all $x \geq 0$... Obviously floor of x will be less or equal to x when x is postive.

Hence $8x^2 \geq 10 + 5x^2\geq 10 + 5\text{floor(x)}^2$ for all $x \geq 4$

Title: Inequality with a floor function


问:
How are the open sets of $[0,1]^\omega$ endowed with the uniform topology?
My atempt: Obviously, the basic sets are open balls $B(x,\epsilon)$. It is easy to see that $B(x,\epsilon) \subset \Pi_{\alpha \in \omega} B(x_{\alpha}, \epsilon)$, but the 'reverse inclusion' is not necessarely true. For instance, consider $ \phantom {.}x=(\frac{1}{2})$, $ \phantom{.}\epsilon = \frac{1}{4}$ and $ y=(\frac{1}{4}+\frac{1}{n+3})_{n \in \mathbb{N}}$. $\phantom{.} \bar{\rho} (x,y) = 1/4$, which implies that $y \notin B(x,\epsilon)$.

答:
Perhaps the simplest way to describe the open balls:

$$B(x,r)=\bigcup\left\{\prod_{n\in\omega}B(x_n,s):0<s<r\right\}\,;$$

I’ll leave it to you to prove that it’s correct.


问:
Simplification - Pre Algebra
I have $$\frac{x^4+2x}{x^4}$$ which becomes $$\frac{x^3+2}{x^3}$$ after pulling out the common factor.  I wanted to know why you you don't simplify $\frac{x^4+2x}{x^4}$ to $2x$ and the two $x^4$ cancel each other out, or is this because you must apply the denominator fraction to all numerator parts?  So $\frac{x^4}{x^4}$ and $\frac{2x}{x^4}$ which doesn't work.

答:
You can either do

$$\frac{x^4+2x}{x^4} = \frac{x(x^3+2)}{x(x^3)} = \frac{x^3+2}{x^3}$$

and then split the fraction as

$$\frac{x^3+2}{x^3} = \frac{x^3}{x^3} + \frac{2}{x^3} = 1 + \frac{2}{x^3}$$

Alternatively, split up first, and cancel common factors:

$$\frac{x^4+2x}{x^4} = \frac{x^4}{x^4} + \frac{2x}{x^4} = 1 + \frac{2}{x^3}$$

Either way works, though the second is quicker and a bit more straightforward.

Title: Simplification - Pre Algebra


问:
Finding an integral submanifold of an involutive distribution
I've been given the following vector field on the manifold $M = \mathbb{R}^3 \setminus \{\text{$z$-axis}\}$ $$X = (x-2y)\,\frac{\partial}{\partial x} + (2x+y)\, \frac{\partial}{\partial y} + z\, \frac{\partial}{\partial z},\quad Y = -y\,\frac{\partial}{\partial x} + x\, \frac{\partial}{\partial y}$$
It was straightforward to show that they're linearly independent and commute, therefore they define an involutive rank $2$ distribution, say $D$. The question is find and describe its integral submanifolds. I tried the following methods:

I was able to figure out that the form $\omega = - xz\,dx - yz\,dy + (x^2 + y^2)\,dz$ annihilates $X$ and $Y$ and $d\omega = 3x\,dx\,dz + 3y\,dy\,dz$. Since our distribution is involutive, naturally $d\omega\wedge \omega = 0$. But I've been unable to figure out an $\alpha$ such that $d\omega = \omega\wedge\alpha$. This could then help me figure out the integral submanifolds

Alternatively, we can the flows of $X$ and $Y$, which I did after a painstaking computations but I'm not sure how to proceed after that.

Any help would be appreciated.

答:
Just for the record, here's a way to do it with minimal guessing: The key observations needed are seeing that $\omega$ looks simpler written in cylindrical coordinates and recognizing the formula for the exterior derivative of a quotient.
\begin{align*}
  \omega &= -xz\,dx - yz\,dy + (x^2+ y^2)\,dz\\
         &= -z(x\,dx + y\,dy) + (x^2+y^2)\,dz\\
         &= -\frac{z}{2}(2x\,dx + 2y\,dy) + (x^2+y^2)\,dz\\
         &= -\frac{z}{2}d(x^2+y^2) + (x^2+ y^2)\,dz\\
         &= -\frac{z}{2}d(r^2) + r^2\,dz\\
         &= -z(r\,dr) + r^2\,dz\\
         &= r(-z\,dr + r\,dz)
\end{align*}
where $x^2 + y^2 = r^2$. Now observe that the second factor looks like the numerator of the exterior derivative of a quotient, namely
$$
  d\left(\frac{z}{r}\right) = \frac{r\,dz - z\,dr}{r^2}.
$$
Therefore,
\begin{align*}
  \omega &= r^3\,d\left(\frac{z}{r}\right)
\end{align*}
It follows that the integral manifolds are the level sets given by
$$
  \frac{z}{r} = c
$$
outside the $z$ axis. These are of course just the cones given by
$$
  z = cr
$$


问:
proving (inf,min,max)
I need to prove for this function:

that

and in addition if this function got min/max

i found the derivative but got stuck

答: the exponential function is always positive so it will add value in total but sin is periodic and minimum value is -1 so if we take limit to exponential to infinity so it goes to zero and take sin some periodic value of -1 so min value of function is -1.</p>

Title: proving (inf,min,max)


问:
The definition of an interior chart
I'm confused about the definition of an interior chart adopted by John M. Lee. The following is a snapshot of the text, and the part that bothers me is underlined with red. Please tell me why an open subset of $\mathbb{H}^n$ that does not intersect $\partial\mathbb{H}^n$ can be seen as an open set in the Euclidean space. I have a hard time using the subspace topology to conclude this result.

答: If $U$ is an open set of $\mathbb H^n$, then there exists an open set $V$ of $\mathbb R^n$ such that $U = V \cap \mathbb H^n$. Now, clearly $\mathbb H^n = \partial \mathbb H^n \cup \operatorname{Int} \mathbb H^n$, which means that $$U = (V \cap \partial \mathbb H^n) \cup (V \cap \operatorname{Int} \mathbb H^n).$$ If $U \cap \partial \mathbb H^n = \varnothing$, then $V \cap \partial \mathbb H^n = \varnothing$ and that implies that $U = V \cap \operatorname{Int} \mathbb H^n$, that is, $U$ is an open set in $\mathbb R^n$ since it is a intersection of two open sets in $\mathbb R^n$.

Title: The definition of an interior chart


问:
Which operators commute with curl?
Let $X$ be the space of infinitely differentiable maps from $\mathbb{R}^3$ to $\mathbb{R}^3$. Let $C:X\rightarrow X$ denote the curl map. What are all the linear maps from $X$ to $X$ that commute with $C$? For example, rotations and translations commute with $C$. Also of course $C$ commutes with itself. Other than products of these three types of linear operators (translation, rotation, and curl) what other operators commute with $C$?

答:
I don't claim to produce an exhaustive list, but some notable ones that are missing:

'scalar derivative operators' (I don't know a better way to say it), i.e. differential operators that don't see the vector field structure. For example $\partial_1Cf=C\partial_1f$. Of course this extends to any $P(\nabla)$ where $\nabla = (\partial_1,\partial_2,\partial_3)$ and $P$ is any scalar valued polynomial of three variables.

Convolutions. If $f$ is integrable, scalar valued, and compactly supported, then for any $g\in X$, the function $f*g$ defined by
$$ f*g(x):=\int_{\operatorname{supp} f} f(y) g(x-y)dy$$
is (1) well-defined (2) an element of $X$, and (3) commutes with $C$. Point (3) follows because of the standard fact that convolutions commute with any derivative:
$$ \partial_i (f*g) =  f*( \partial_i g) $$
(this fact is of course used to prove (2).)

If you include enough decay at infinity,  you also get lots of others like (scalar) Fourier multipliers and the projection to the gradient or curl parts of the Helmholtz decomposition.

Title: Which operators commute with curl?


问:
Figure eight knot is not a torus knot
After seeing this picture of the figure eight knot:

Why isn't the figure eight knot considered a $(2,3)$-torus knot?

答:
The $(3,2)$ torus knot is also known as the "trefoil" knot.  It admits a tricoloring, while the figure eight knot (that you drew) does not.  Here is a reference: http://en.wikipedia.org/wiki/Tricolorability

To answer your question directly: The diagram you drew certainly looks like it lies on the standard two-torus in $\mathbb{R}^3$.  But stare at one of four arcs that connect an outer crossing with an inner crossing.  The crossing pattern at the ends of the arc prevent it from lying in the two-torus. At one of the ends, it will have to go into (or out of, respectively) the two-torus, to avoid crashing into another arc.

Title: Figure eight knot is not a torus knot


问:
Let $t_n$ be the number of strings of length $n$ that do not contain $11$ as a substring over the alphabet $\{ 1,2,3 \}$, prove $t_n =$ (cont.)
If $t_n$ denotes the number of strings of length $n$ that do not contain $11$ as a substring over the alphabet $\{ 1,2,3 \}$, then prove $t_n = \sum_{i=0}^{\lfloor{\frac{n+1}{2}} \rfloor}2^{n-i} C(n-i+1,i)$.

Here's my attempt:

My attempt was by induction, but I'm stuck at the induction step. We verify the base case and first show $n=1$ holds. $t_1 = 3$, and the desired sum with $n=1$ yields $3$.

For our induction step, suppose $t_n = \sum_{i=0}^{\lfloor{\frac{n+1}{2}} \rfloor} 2^{n-i} C(n-i+1,i)$. Given this, we wish to show $t_{n+1} = \sum_{i=0}^{\lfloor{\frac{n+2}{2}} \rfloor} 2^{n-i+1} C(n-i+2,i)$. Here's where I'm stuck. Is induction even the way to go? If it is, what's the step I should make? If it isn't, where should I look to?

Thanks.

答: Forget about induction and instead condition on the number $i$ of times that $1$ appears.  The remaining $n-i$ digits are $2$ or $3$.


问:
Is this twisted cube like graph planar?

I couldn't find any planar representation of this graph neither could I find a subdivision or minor of $K_{5}$ or $K_{3,3}$. So I don't know how to go about it. Please help.

答:
If you contract the two long vertical edges, you are left with $K_{3,3}$
So $K_{3,3}$ is a minor of your graph, which is in turn not planar.

Title: Is this twisted cube like graph planar?


问:
Generalize Oddtown
Suppose we have a city with $n$ people. In this city there are $m$ clubs. The number of members in each club is not divisible by $s$ (for a given $s$), but the number of people in the intersection of each two clubs is. Show that $m<cn$, where $c$ is the number of (distinct) primes that divide $s$.

My ideas so far: I can solve this when $s=p^k$ for $p$ prime. In this case, consider the n-coordinate vectors $v_1,v_2..v_m$ such that the $j$ entry of $v_i$ is $1$ if the person number $j$ belongs to the club number $i$ and else it's $0$. If these vectors are LD over the rationals, they are also LD over the integers, ie, we can find integers $r_i$ such that $\sum_i r_i v_i=0$. We can assume that there is a $j$ such that $r_j$ is not a multiple of $p$, otherwise we divide everyone for $p$ until that happens. Taking inner product with $v_j$ we have $\sum_i r_i v_iv_j=0$. We have $p^k \mid r_iv_iv_j$ for all $i$ different from $j$, therefore $p^k\mid r_jv_jv_j$ which implies $p \mid r_j$, contradiction. So, the vectors are LI and $m\leq n$.

I think the general case will follow in a similar fashion, but for that we will need $cn$ coordinate vectors, and it's not obvious how to construct that.

答:
Let $p_1,p_2,\dots,p_c$ be the distinct primes dividing into $s$, where $p_i$ occurs with multiplicity $m_i$.

For each $i\in \{1,\dots,p\}$, let $S_i$ be the set of clubs whose member sets are not a multiple of $p_i^{m_i}$. Your logic proves that $|S_i|\le n$, because $S_i$ is a collection of clubs where each has have a size which is not a multiple of $p_i^{m_i}$, but any two have an intersection whose size is a multiple of $s$, and therefore is a multiple of $p_i^{m_i}$.

Furthermore, every club must belong to one of the sets $S_i$; if a club were in none of the sets, then its size would be a multiple of $\mathrm{lcm}(p_1^{m_1},\dots,p_c^{m_n})=s$. Therefore, the total number of clubs is at most $\sum_i |S_i|=cn$.

Title: Generalize Oddtown


问:
Inverse Laplace Transform properities
If you have the inverse Laplace transform of a function of the form $F(\omega)/\omega$ , how to find the inverse Laplace transform of the function $F(\omega)/w$?
I have attached the formula for the inverse laplace transform of the $((1/\omega)\cdot F(\omega))$ and mu function $F(\omega)$.

general formula

My function

答:
Hello I hope I am not too late. In this answer I'll give you an example how to calculate $\mathcal{L}^{-1} \{F(s)\}(t)$ as you requested in the comments.

Definition

An integral formula for the inverse Laplace transform, is given by the line-integral$$f(t) = \mathcal{L}^{-1} \{F(s)\}(t) =  \frac{1}{2\pi i}\lim_{T\to\infty}\int\limits_{\gamma-iT}^{\gamma+iT}e^{st}F(s)\,\mathrm{d}s,$$ where the integration is done along the vertical line $Re(s) = γ$ in the complex plane such that γ is greater than the real part of all singularities of $F(s)$ and $F(s)$ is bounded on the line, for example if contour path is in the region of convergence.

Example Let $F(\omega)=O(\lvert\omega\rvert^{-1})$ be entire. Evaluate $\mathcal{L}^{-1} \left\{\frac{F(\omega)}{\omega}\right\}(t).$

Proof. We will evaluate the integral by expressing it as a limit of contour integrals along the contour $\mathcal{C}$ that goes along the line from $\gamma−iT$ to $\gamma+iT$ and then counterclockwise along a semicircle $\Gamma$ centered at the origin from $\gamma+iT$ to $\gamma−iT$. Take $T$ to be greater than $\gamma$, so that $0$ is enclosed within the curve. The contour integral is
$$\oint\limits_{\mathcal{C}}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}\omega$$
Since $F(\omega)e^{\omega t}$ is an entire function (having no singularities at any point in the complex plane), the integrand has singularities only where the denominator $\omega$ is zero. According to the residue theorem, then, we have
$$\oint\limits_{\mathcal{C}}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}\omega=2\pi i\operatorname{Res}_{\omega=0}\left[\frac{F(\omega)}{\omega}e^{\omega t}\right]=2\pi i\lim_{\omega\to0}\left[\omega\cdot\frac{F(\omega)}{\omega}e^{\omega t}\right]=2\pi i F(0).$$
The contour $\mathcal{C}$ may be split into a "straight" part and a curved arc, so that $$\int\limits_{\gamma-iT}^{\gamma+iT}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}s+\int\limits_{\Gamma}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}\omega=2\pi iF(0) \Rightarrow \int\limits_{\gamma-iT}^{\gamma+iT}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}s=2\pi iF(0)-\int\limits_{\Gamma}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}\omega.$$
Now we can show that the integral over $\Gamma$ equals zero:
$$\left\lvert\int\limits_{\Gamma}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}\omega\right\rvert = \int\limits_{\pi/2}^{3\pi/2}\frac{\lvert F(Te^{i\varphi})\rvert}{T}e^{T\cos(\varphi)t}\,\mathrm{d}\varphi\to0\qquad (T\to\infty).$$
This is because $\cos\varphi<0, \,\varphi\in(\pi/2,3\pi/2)$. In conclusion we have
$$\lim_{T\to\infty}\frac{1}{2\pi i}\int\limits_{\gamma-iT}^{\gamma+iT}\frac{F(\omega)}{\omega}e^{\omega t}\,\mathrm{d}s=\mathcal{L}^{-1} \left\{\frac{F(\omega)}{\omega}\right\}(t)=F(0).$$

Title: Inverse Laplace Transform properities


问:
Dominant and Injective Morphisms reverse implication
As mentioned before, I am self-studying some commutative algebra out of "A Course in Commutative Algebra" by Kemper. In the text, Kemper has the following problem:

Let $X$ and $Y$ be affine varieties over a field $ K$, and let $f: X → Y$ be a morphism with induced
homomorphism $ϕ: K[Y ] → K[X].$ We say that $f$ is dominant if the image
$f(X)$ is dense in $Y$ , i.e., $\overline{f(X)} = Y$ .

Show that $f$ is dominant if and only if ϕ is injective.

He provides a proof of the forward implication and the reverse implication. I have come terms in understanding the forwards implication, but I have two questions regarding the reverse implication. Here is the proof given:

Assume that $ϕ$ is injective. Write $K[X] = K[x_1, . . . , x_m]/I,
K[Y ] = K[y_1, . . . , y_n]/J,$ and let $f$ be given by $f_1, . . . , f_n ∈ K[x_1, . . . , x_m].$
Take $g ∈ \mathcal{I}_{K[y_1,...,y_n]} (f(X)).$ Then the polynomial $g(f_1, . . . , f_n) ∈
K[x_1, . . . , x_m]$ vanishes on $X$, so $ϕ(g + J) = 0.$ This implies $g ∈ J.$
We have shown that $\mathcal{I}_{K[y_1,...,y_n]} (f(X)) ⊆ J$, so

$$Y ⊆ \mathcal{V}_{K^n} (J) ⊆ \mathcal{V}_{K^n} (\mathcal{I}_{K[y_1,...,y_n]}(f(X)))
= \overline{f(X)} ⊆ Y$$$ \square$

Now form here, I can see why $f(X)$ is dominant.

Before I ask my questions let me clarify some of the author's notations:

$\mathcal{I}_{K[y_1,...,y_n]} (f(X))$ is the vanishing ideal of $f(X)$.

$\mathcal{K^n}(J)$ is the affine variety given by $J$.

For the most part, I am okay with the proof, my questions are as follows:

What justifies the containment $Y ⊆ \mathcal{V}_{K^n} (J)$?

What justifies the equality $\mathcal{V}_{K^n} (\mathcal{I}_{K[y_1,...,y_n]}(f(X)))
= \overline{f(X)}$?

Any help would be appreciated.

答:
We have $k[Y]=K[y_1,\ldots,y_n]/J$, this means that $Y\subset K^n$ is exactly the vanishing set of $J$. Or we can say that the class of any function $a\in J$ is equal to zero in $K[y_1,\ldots,y_n]/J$, hence for any maximal (or prime) ideal $\mathfrak m\subset K[y_1,\ldots,y_n]/J$ the element $a$ is equal to zero in $(K[y_1,\ldots,y_n]/J)/\mathfrak m$, i.e. it vanishes on the corresponding point.

$\mathcal V(\mathcal I_{K[y_1,\ldots,y_n]}(f(X)))$ is a closed set containing $f(X)$, hence $\overline{f(X)}\subset \mathcal V(\mathcal I_{K[y_1,\ldots,y_n]}(f(X)))$. Conversely, if $p\in K^n$ does not lie in the closure, then we can choose a distinguished open set $D(b)$ containing $p$ and not intersecting with $\overline{f(X)}$. This $b$ vanishes on $f(X)$, and does not vanish on $p$, hence $p\notin\mathcal V(\mathcal I_{K[y_1,\ldots,y_n]}(f(X)))$.


问:
Math extrema problem
I have a math extremum problem,  but I am unable to solve it.

"Car gasoline use per hour is: $w = 6 - 0.15v + 0.0025v^2$.  At what speed does the car have to travel,  so that it would be cheapest. "

I used derivative to find a global minimum and got $30$km/h,  but my textbook says the correct answer is $49$km/h.

Can anyone help?

Thank you in advance!

答:
Your are right

$$w=6+\frac{v^2-60v}{400}=6+\frac{(v-30)^2-900}{400}\ge 6-\frac94$$, the minimum value occurs if $v=30$

Title: Math extrema problem


问:
What is the relationship between $M(f)$, $I(f)$ and $T(f)$ where M is the midpoint rule and T is the trapezoid rule?

It is necessary to approximate

$$I(f) = \int\limits_a^bf(x)\,\mathrm{d}x$$ where f is a function that has $2$ continuous
derivatives. If $M(f)$ and $T(f)$ are the approximations obtained
through the Midpoint Rule and the Trapezoid Rule, respectively, and
$f''$ is constant, what is the relationship between $M(f),T(f)$ and
$I(f)$?

I did this

$$|E_T|= \frac{k(b-a)^3}{12h^2} \\
|E_M| = \frac{k(b-a)^3}{24h^2} \\
\frac{|E_T|}{|E_M|} = 2 \\
|E_I| = 0$$

The relationship is $|E_I| < |E_M| < |E_T|$.

Is this correct?

答:
You appear to be beginning to analyze estimates of the potential errors in the uses of $M(f)$ and $T(f)$ as opposed to the estimates themselves.

You may be expected to notice that "$f''(x)$ is constant" (quite a hypothesis!) means that $f(x)$ is a quadratic function, for which the "Simpsons Rule" approximation, although it is not expressly mentioned here but you may have heard of, will be exact (i.e. "$S(f)$," the Simpsons approximation, will be $I(f)$).

And there is a well known linear relationship between the midpoint, trapezoid, and Simpsons approximation methods that is described below:

Relation between Simpson's Rule, Trapezoid Rule and Midpoint Rule

Note that the problem does not provide what number of partitions (or which partitions) are being used to evaluate the rules. The answer linked above (and simplicity of $f$) may shed light on that.


问:
$\operatorname{Aut}(\mathbb{H})$ acts on circles
Suppose $\mathbb{H}=\{z \in \mathbb{C}\mid \operatorname{Im}(z)>0\}$.
Is it true that $\operatorname{Aut}(\mathbb{H})$ can map any circle in $\mathbb{H}$ to any circle in $\mathbb{H}$ (i.e. for any two circles $C_1,C_2$ with all their points in $\mathbb{H}$ there's some element of $\operatorname{Aut}(\mathbb{H})$ that carries $C_1$ to $C_2$)?

The only useful fact I know is that
$f\in \operatorname{Aut}(\mathbb{H}) \Rightarrow f=\frac{az+b}{cz+d}, a,b,c,d \in \mathbb{R}, ad-bc \neq 0 $.

Thank you for any help!

答:
The automorphism group here is $G=\mathrm{SL}_2\mathbb{R}$ acting by Mobius transformations (kernel $\{\pm I_2\}$).

Consider the stabilizer $\mathrm{Stab}(i)=\mathrm{SO}(2)$. Its orbits are circles symmetric about the imaginary axis intersecting it at points $\lambda i$ and $\lambda^{-1}i$ with $\lambda>1$. Thus, $K=\mathrm{SO}(2)$ fixes each of these circles. In the upper-half plane model of hyperbolic geometry (which isn't necessary for this answer, just background) $G$ acts by isometries, and these are the hyperbolic circles centered at $i$.

The Iwasawa decomposition $G=KAN$ is relevant here. That is, $K=\mathrm{SO}(2)$ has a complementary subgroup $AN$ consisting of upper triangular matrices. More precisely, $A$ consists of diagonal matrices and $N$ consists of unitriangular matrices - the letters stand for kompact, abelian, nilpotent (the explanation for these terms involves Lie theory). In other words, $G$ is a "knit product" of these subgroups, and $AN$ forms a transversal (set of coset representatives) for $G/K$.

The subgroup $AN$ is enough to turn any circle into an aforementioned one with hyperbolic center $i$. For convenience, let's call these "central" circles. You can use homotheties from $A$ to ensure a circle's max and min imaginary parts are reciprocal, then use $N$'s elements to translate it until it is symmetric about the imaginary axis. Moreover, this element of $AN$ that maps a circle to a central one is unique, and no nontrivial element of $AN$ can map one central circle to another.

Let $\Gamma$ be the set of all "central" circles. Putting the above information together, we may conclude $\Gamma$ is a complete transversal for the $G$-orbits (with $G$ acting on circles). That is, each $G$-orbit of circles may be represented by exactly one central circle from $\Gamma$. In the context of hyperbolic geometry, this means $G$ preserves hyperbolic radii and any two circles of the same hyperbolic radius are related by an element of $G$.


问:
Prove that $n\leq a_1+a_2+...+a_n \leq n+1$
If $a_1=2,a_{n+1}=\sqrt{a_n+8}-\sqrt{a_n+3}$. Prove that $n\leq a_1+a_2+...+a_n \leq n+1$ for every $n\ge1$ and $\lim a_n=1$.

I have showed that by squaring and inequality techniques:

$a_i<\sqrt{3}$ for every $i>1$.

If $a_i>1$ then $a_{i+1}<1$ for every $i\ge 1$

I think that $\sqrt{3}$ can be improved, but I am not sure if it's useful.

Body:
We have

$
a_{n+1}-1 = (a_n-1) \left( \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right)
$

and

$(a_{n+1}-1)+(a_n-1) = (a_n-1) \left( 1+ \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right)$,

a) $ \left| \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right| < \left| \cfrac{1}{3+\sqrt{8+a_n}} + \cfrac{1}{2+\sqrt{3+a_n}} \right| < \left| \cfrac{1}{3+\sqrt{8+0}} + \cfrac{1}{2+\sqrt{3+0}} \right| = 5-(\sqrt 3 + \sqrt8) < 1 $

This way we have $|a_{n+1}-1| < q |a_n-1| < q^n |a_1-1|$ by induction. This way, $|a_{n+1}-1|$ tends to 0.

b) We have to show some results:

$ \left( \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right) < 0$, therefore $a_{n+1}-1$ and $a_n-1$ have opposite signals. But $a_1-1=1>0$, then the even guys are negative, the odd ones are positive: $a_{2k} < 1 < a_{2k+1}$.

$ \left( 1+ \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right) > 0$, therefore $a_{n+1}+a_n-2$ and $a_n-1$ have the same signals: $a_{2k}+a_{2k+1} < 2 < a_{2k+1}+a_{2k+2}$

Now we have two cases to consider:

$ (a_1+a_2)+\ldots+(a_{2k+1}+a_{2k+2}) > 2+\ldots+2=2k+2$ and $ a_1+(a_2+a_3)+\ldots+(a_{2k}+a_{2k+1})+a_{2k+2} < 2+\ldots+2+1=2k+3$

$ (a_1+a_2)+\ldots+(a_{2k+1}) > 2+\ldots+1=2k+1$ and $ a_1+(a_2+a_3)+\ldots+(a_{2k}+a_{2k+1}) < 2+\ldots+2+2=2k+2$


问:
What's the meaning of $\sigma^3$ and $\sigma^4$ and how to find them
I know you can use $\sigma^3$ and $\sigma^4$ with $\mu$3 and $\mu$4 respectively to find the skewness and the kurtosis, but by themselves what do they mean + what's the formula to find them?

Thank you in advance!

答:
skewness=$\frac{\mu_3}{\sigma^3}$, kurtosis=$\frac{\mu_4}{\sigma^4}$, where $\mu_k=E((X-\mu)^k)$ and $\sigma^2=E((X-\mu)^2)=E(X^2)-\mu^2$

Using Google:

https://brownmath.com/stat/shape.htm


问:
Wondering how these two sets are countably infinite
We have $A = \mathbb{R}$ and $B = \{x|x \in \mathbb{R} \land \exists y (y\in \mathbb{Z} \land |x-y| < \frac{1}{2})\}$

How is $A-B$ countably infinite?

I know the definition of set minus is $A \cap \overline{B}$, but I don't know how to translate $B$ and to see if this is a countable set.

答: $|x-y|<1/2\iff x\in(y-1/2,y+1/2)$ and thus $B$ contains all intervals of the form $(y-1/2,y+1/2),y\in\Bbb Z$. The only real numbers not in $B$ are of the form $y+1/2,y\in\Bbb Z$ and thus $A-B=\{y+1/2:y\in\Bbb Z\}$ is countably infinite since $\Bbb Z$ is countably infinite.


问:
Hodge Star Operator and definition of divergence operator on riemannian manifold
From my lecture notes, on a general Riemannian manifold $(M, g)$, the divergence operator $\operatorname{div}: T^\infty(TM)\rightarrow C^\infty(M)$ is defined as $\operatorname{div}X := \star^{-1}d(x\neg\operatorname{Vol})$ , where $\star$ is the Hodge star operator $C^\infty\rightarrow T^\infty(\bigwedge^nT^\ast M)$ defined as $f\rightarrow f\operatorname{Vol}_g$.

Now I've read this from other place that there is another characterized definition of divergence operator given by:

For $\vec{F}\in C^\infty(U)$ with an open subset $U\subset X$, the function $\operatorname{div}(\vec{F})\in C^\infty$, $$d(\star w(\vec{F})) = \operatorname{div}(\vec{F})dV_X|_U$$

The detail of this definition can be found here.

Now my question. How does the definition given in my lecture imply the characterization property of divergence operator?

答: If $ \overline{F} = (f_1,f_2,...,f_n) $ then you can identify it locally with the vector field  $ X = \sum_i f_i\partial_i $ and thus you have definitions for both of them identical via this. There are several equivalent definitions of divergence, like defining $ div(X)dV = d(i_X dV) $, trace of Levi-Civita connection, etc.</p>


问:
Complement of a Vitali set
We know that it holds for the Vitali set that $V\subset [0,1]$ and that Vitali set is unmeasurable. I know that the complement $\mathbb{R}\setminus V$ is also not measurable. However, I was wondering if we can say that $[0,1]\setminus V$ is also not measurable?

答:
If $U = [0, 1] \setminus V$ is measurable, then so is $U^c$ and in turn, $[0, 1] \cap U^c = [0, 1] \setminus U = V.$

Therefore, $U$ cannot be measurable.

In general, if $B$ is a measurable set and $A \subset B$, then $A$ is measurable iff $B \setminus A$ is.

Title: Complement of a Vitali set


问:
Is $x2^x = O(2^x)$?
Is $x2^x = O(2^x)$? Intuitively it seems like the answer is no. But if we write $x2^x=2^{\log x + x}$ then the answer (to me) is unclear. Thanks in advance!

答:
Answer is no. Assume that

$$x2^x=O(2^x)$$

Then there exists $M>0$ and $X>0$ such that $x\geq X$ implies

$$x2^x\leq M2^x$$

But then for $x=\max\{2M,X\}$ we have that

$$x2^x\geq 2M 2^x>M2^x$$

which is a contradiction since $x\geq X$. We conclude $x2^x\neq O(2^x)$.

Title: Is $x2^x = O(2^x)$?


问:
Prove $a^n \rightarrow 0$ as $n \rightarrow \infty$ for $\left|a\right| < 1$ without use of log properties
$a^n \rightarrow 0$ as $n \rightarrow \infty$ for $\left|a\right|  < 1 $

Hint $u_{2n}$ = $u_{n}^2$

I have totally no idea how to prove this, this looks obvious but i found out proof is really hard...
I am doing a real analysis course and there's a lot of proving and I stuck there.
Any advices? Practice makes perfect?

答:
Proof sketch:  Maybe try showing that $|\frac{1}{a^n}| \to \infty$.

Fix $p = |\frac{1}{a}| > 1$, and let $p = ( 1 + b )$.  Show by induction that $p^n \ge 1 + nb$, and conclude the statement above using the Archimedean property of the reals.


问:
Smooth hypersurfaces of the blow-up.
Let's assume $X$ is the blow up of $\mathbb{P}^n$ along a smooth subvariety $Z$. Especially $X$ is smooth. I was wondering what the hypersurfaces of $X$ look like? The hypersurfaces should give an ample divisor, so it should have positive intersection number with any curve. Hence I think this implies that it should contain the exceptional divisor which is a projective bundle over $Z$. So it seems like every hypersurface section has two components one should be the exceptional divisor and possibly another divisor (which is blow of some divisor in the projective space). But this seems to imply that hypersurfaces are not smooth (they have at least two components) and contradicts Bertini's theorem. (So a better question is what smooth hypersurfaces look like?)

答: I think you are asking a very difficult question. If you want to know the very ample divisors on $X$ then you are asking, at minimum, "what is the structure of the cone of ample/nef divisors on $X$?", and that is ignoring the ample/very ample distinction. If you Google the words "ample" and "blowup" together I think you will find some papers that will quickly demonstrate that it can take quite a bit of effort to say relatively little. Fixing $d > 1$, if you are blowing up a point $p$, then you are asking for something like a bound on the multiplicity of the divisor $D \subset \mathbb P^n$ at $p$ such that it can still be moved in a linear system $L \subset |D|$ which is large enough that strict transforms of members of $L$ cover $X$. And while this is straightforward enough for a single reduced point, blowing up at multiple and/or fat points will quickly get confusing, and blowing up a positive dimensional subvariety is even worse (in particular, the geometric characterization of a given linear system downstairs will become trickier and tricker). If you haven't already, I would advise that you start reading Lazarsfeld's Positivity in Algebraic Geometry to learn more about this kind of thing.

Title: Smooth hypersurfaces of the blow-up.


问:
$f(z)$ is analytic on $\{ |z| \leq 1\}$, and $f(z)$ is real on $\{|z| = 1\}$, show $f(z)$ is a constant.
Let $f(z)$ be analytic on $\{|z| < 1\}$ that extends to $\{|z| \leq 1\}$. Assume $f(z)$ is real on $|z| = 1$. Prove that $f(z)$ is a constant.

Here is my attempt,
define $g(z) = e^{if(z)}$, for $|z| = 1$, we have $|g(z)| = 1$. Then $|\frac{1}{g(z)}| = 1$ as well. So that $|g(z)| = 1$ for $|z| = 1$.

(1) Is there any theorem I can use to show that this leads to $g(z)$ is constant on $|z| \leq 1$?

(2) If $g(z)$ is constant, can I show that $f(z)$ is constant?

答:
Note that your function $g$ is analytic on $U = \{z : |z| < 1\}$ as well. By the Maximum Modulus Theorem, we know that $|g| \le 1$ on $U$.

Consider the Cayley transform $$\phi(z) = \frac{z - i}{z + i}$$
defined on the closed upper half plane $\{z : \Im z \ge 0\}.$ This maps the open upper half plane $\Bbb H$ onto the open unit circle bihilomorphically and the real line onto the boundary.

Thus, we may consider the function $\tilde g = g \circ \phi$ on $\overline{\Bbb H}$. Note that $\tilde g$ is real on the real line.

Now, by the Schwarz reflection principle, $\tilde g$ extends to $G$ on $\Bbb C$ defined by $$G(z) = \begin{cases}\tilde g(z) & z \in \overline{\Bbb H},\\ \overline{\tilde g(\bar z)} & z \notin \overline{\Bbb H}, \end{cases}$$
holomorphically.

However, $|\tilde g| \le 1$ on $\overline{\Bbb H}$ and thus, the above formula shows that $|G| \le 1$ on $\Bbb C$ as well. Since $G$ is entire, $G$ is constant, by Liouville's theorem.

Thus, $\tilde g$ is constant and in turn, $g$ is constant. ($\phi$ is bijective.)

Thus, $z \mapsto e^{i f(z)}$ is constant. This means that $e^{i f(z)} = e^{i f(0)}$ for all $z \in U$. This means that $f(z) - f(0) = 2\pi k_z$ for some integer $k_z \in \Bbb Z$. A priori, it is possible that $z \mapsto k_z$ is not constant. However, $k_z$ is a continuous function of $z$ and takes value in the discrete set $\Bbb Z$. Since $U$ is connected, it follows that $z \mapsto k_z$ is constant.

Since $k_0 = 0$, it follows that $$f(z) = f(0),$$
for all $z \in U$.


问:
In a group of 6 people either we have 3 mutual friends or 3 mutual enemies. In a room of n people?
A group of 6 people each pair is either a friend (acquaintance) or an enemy (stranger). It is to be proven that there are either 3 mutual friends or 3 mutual enemies in this group. I have an ad-hoc reasoning for this,  but it is not yielding a big picture.

I would appreciate an approach or way of thinking about this which would make it easier to generalize it for n friends. i.e find the minimal number of mutual friends or enemies that can exist for a group of n people. My approach is getting cumbersome when I try for larger n.

(My reasoning)
Any person either has (atleast 3 friends) OR (atleast 3 enemies). This is true because each person can be friend\enemy with five others and by the pigeonhole principle atleast one of the two holes (friend , enemy ) must contain 3 or more people. Consider one of the friends in the former case (or enemies in the latter). He has a relationship with the other two. There are three possibilities. FF, EE, EF. The first and last case results in 3 mutual friends. In the second case, we must consider the relationship between the remaining two. If they are friends, then there are three mutual friends. If they are enemies, then there are 3 mutual enemies.

答:
Prove that in a group of six people, atleast three must be
mutual friends or at least three must be mutual strangers</p>


问:
Hitting time for a random walk where step probabilities linearly depend on the distance to an absorbing wall
Consider the case where I have a discrete random walk on the integers over an interval $[0, M]$, where I start at some position $0 < k < M$, and both endpoints (i.e. $0$ and $M$) are fully absorbing.  The caveat, to this otherwise simple random walk problem, is that the probability of taking a step depends on the distance between the walker and the absorbing boundaries.  Letting $x_i$ represent the position of the walker, we have $P[+1] = \frac{x_i}{M}$, and $P[-1] = 1 - P[+1]$.

In other words, the probability of the walker taking a $+1$ step is equal to the probability of sampling from the integers over the interval $[1, M]$, with uniform probability, and seleting an integer $j$ such that $j \leq x_i$.

Can we state the probability that the walker reaches the absorbing target $M$?

答:
The usual way: write $u(x)$ for the probability to reach $M$ before $0$ for the random walk starting from $x$, and note that $(u(x))_{0\leqslant x\leqslant M}$ is the unique solution of the linear system $u(0)=0$, $u(M)=1$, and, for every $1\leqslant x\leqslant M-1$,
$$
Mu(x)=xu(x+1)+(M-x)u(x-1).
$$
Hence, for every $0\leqslant x\leqslant M$,
$$
u(x)=\frac1{2^{M-1}}\sum_{k=0}^{x-1}{M-1\choose k}.
$$</p>


问:
Computing the cohomology ring of the orientable genus $g$ surface by considering a quotient map.
I'm attempting to do Hatcher Chapter 3.2, Exercise 1.

Let $M_g$ be the genus $g$ orientable surface, and let $q$ be the quotient map from $M_g$ to the wedge product of $g$ tori obtained by collapsing a subspace of $M_g$ homeomorphic to a $g$-times punctured sphere to a point.

We know that $H^*(T^2) = \wedge[\alpha,\beta]$, so $H^*(\vee_i T^2) = \oplus_{i = 1}^g \wedge[\alpha_i,\beta_i]$. The map $q$ induces a ring homomorphism $q^*: H^*(\vee_i T^2) \to H^*(M_g)$. Choosing appropriate representatives for the cohomology classes $\alpha_i$ and $\beta_i$ (which I do not denote any differently), we can see that $q^*(\alpha_i)$ and $q^*(\beta_i)$ are represented by $\alpha_i \circ q$ and $\beta_i \circ q$ respectively, and the ring homomorphism preserves the exterior algebra relations.

My only problem is determining if there are other relations in $H^*(M_g)$ because I don't believe $q^*$ is an isomorphism. Is there some way to determine these without directly computing the cup product structure of $M_g$ via a gluing diagram?

答:
The comments on the previous answer express a desire for additional details, so I'll add some here for any future students looking at this question. Robin nicely explains how the only interesting multiplicative structure inside $H^*(M_g)$ comes from the cup product $H^1(M_g) \times H^1(M_g) \to H^2(M_g)$ so we focus on that. Fix the following notation: let $X := \bigvee_{i = 1}^g T_i$ with each $T_i$ a torus, let $A$ denote the subspace of $M_g$ equal to a $g$-punctured sphere, let $i : A \to M_g$ denote inclusion, and let $q: M_g \to M_g/A = X$ denote the quotient map. As Robin mentions, the strategy is use the fact (Hatcher proposition 3.10) that the square
$$
\require{AMScd}
\begin{CD}
H^1(X) \times H^1(X) @>{\cup}>> H^2(X) \\
@V{q^* \times q^*} VV @V{q^*}VV \\
H^1(M_g) \times H_1(M_g) @>{\cup}>> H^2(M_g)
\end{CD}
$$
commutes to get at the cup product structure on $M_g$ from (i) the (known) cup product structure on $X$ and (ii) the action of $q$ on cohomology. As (i) is addressed in the question and in Robin's answer, we focus on the details of (ii).

Disclaimer: I am still very much a student of algebraic topology, so I welcome any comments for improving my argument!

We consider a piece
$$
\cdots \rightarrow H_1(A) \xrightarrow{i_*} H_1(M_g) \xrightarrow{q_*} H_1(X) \xrightarrow{\delta} H_0(A) \xrightarrow{i_*} H_0(M_g) \to \cdots
$$
of the long exact sequence of homology induced by the pair $(M_g, A)$. Regard $H_1$ as the abelianisation of $\pi_1$ to see that the map $i_* : H_1(A) \to H_1(M_g)$ is zero (because loops in the subspace $A$ lie in the commutator subgroup of $\pi_1(M_g)$). Note also that $i_* : H_0(A) \to H_0(M_g)$ injects (by the definition of the zero-th homology) so that $\delta = 0$. Thus, $q_*$ is encased by zero maps and so induces an isomorphism on first homology. By the universal coefficient theorem for cohomology (noticing that the Ext terms vanish here), it follows that $q^*$ induces an isomorphism on first cohomology.

Understanding $q^*$ on second cohomology requires a bit more topology and in this case we'll think about $H_2$ in terms of simplicial homology. Let $q_i : X \to T_i$ denote the quotient map which collapses the wedge sum $X$ to the $i$-th torus in the sum. Then we have a commutative diagram
$$
\require{AMScd}
\begin{CD}
H_2(M_g) @>{q_*}>> H^2(X) @>{(q_i)_*}>> H^2(T_i) \\
@V{\cong} VV @V{\cong}VV @V{\cong}VV \\
\mathbb{Z} @>>> \mathbb{Z}^g @>>> \mathbb{Z}.
\end{CD}
$$
Letting $\sigma : \Delta^2 \to M_g$ represent the generating class in $H_2(M_g) \cong \mathbb{Z}$, that $\sigma$ is not a boundary in $H_2(M_g)$ implies that $q_i \circ q \circ \sigma$ is itself not a boundary in $H_2(T_i) \cong \mathbb{Z}$. In particular, $q_i \circ q \circ \sigma$ generates $H_2(T_i)$ so we see that $(q_i)_* \circ q_*$ maps 1 to 1 along the bottom row of the diagram. Now, the isomorphism $H_2(X) \cong \mathbb{Z}^g$ comes from the isomorphism $H_2(X) \cong \bigoplus_{i = 1}^g H_2(T_i)$ which is induced by the inclusions $T_i \hookrightarrow X$ (Hatcher corollary 2.25). So if $\sigma_i : \Delta^2 \to T_i$ represents a generator of $H_2(T_i)$ then the map $(q_i)_*$ sends $(0, \ldots, 0, \sigma_i, 0, \ldots, 0) \in H_2(X)$ to $\sigma_i$. Along the bottom row of the diagram, if we regard $\mathbb{Z}^g$ as generated by the standard basis vectors $\{e_1, \ldots, e_g\}$, then $(q_i)_*$ acts as $\varepsilon_i$ for $\varepsilon_i(e_j) := \delta_{ij}$. We conclude that for each $i$ we have $\varepsilon_i(q_*(1)) = 1$ from which it follows that $q_*(1) = (1, \ldots, 1) \in \mathbb{Z}^g \cong H_2(X)$. Again apply the universal coefficient theorem for cohomology (noting that the Ext terms still vanish) to see that $q^* : H^2(X) \to H^2(M_g)$ maps each of the $g$ generators of $H^2(X)$ to the same generator of $H^2(M_g)$. In terms of the isomorphisms $H^2(X) \cong \mathbb{Z}^g$ and $H^2(M_g) \cong \mathbb{Z}$, we have concluded that $q^*(e_i) = 1$ for all $i$.

With a precise understanding of the action of $q$ on cohomology, we may return to the original commuting square. We adopt the notation of the OP here: let $H^1(X)$ have generators $\{\alpha_i, \beta_i\}_{i = 1}^g$ and $H^2(X)$ have generators $\{\gamma_i\}_{i = 1}^g$ with $\alpha_i \beta_i = \gamma_i$ and all other products of generators zero (note that we use here the known cup product structure on the torus and that we adopt the convention that multiplication denotes a cup product). Then that $q^*$ induces an isomorphism on first cohomology implies that $H_1(M_g)$ has generators $\alpha_i' := q^*(\alpha_i)$ and $\beta_i' := q^*(\beta_i)$. And the action of $q^*$ on $H^2$ means that $H^2(M_g)$ has generator $\gamma' := q^*(\gamma_1) = \cdots = q^*(\gamma_g)$. By our very first commuting square, we conclude that

Theorem: The cohomology ring $\tilde{H}^*(M_g)$ is generated by $\{\alpha_i', \beta_i'\}_{i=1}^g \cup \{\gamma'\}$ with $\alpha_i'$ and $\beta_i'$ of weight one, $\gamma'$ of weight two, and nonzero cup product relations $\alpha_i' \beta_i' = \gamma'$ for all $i$ (all other cup products of generators are zero).


问:
Giving a 1-hour talk to highschool math club: any topic suggestion?
I've been invited (by my kid) to give a one hour talk to her highschool math club.  Last year (right before the pandemic hit) I did two such talks on probability, and they loved it.  I'm looking for topic suggestions.

Logistics: There will be about $10-15$ kids total, from grades $9-12$.  These kids are mostly "A" students, and some of them have skipped a math grade, but they're nowhere near Olympiad standard.  The talk will be on March $1$st, $2021$, and will be on zoom.  I will be making slides this weekend.

I don't want this to become an "educational" lecture, so I'm looking for fun examples to motivate certain math areas.  E.g. last year my talks on probability culminated with Polya Urn and Buffon's Noodle.  Also, I'd like everything (well, $95\%$) to be fully explainable within the one hour, i.e. I don't want to just spew facts and then tell them "just trust me".

My own math skill tends toward discrete: comp-sci, combinatorics, etc.  However I'm open to any idea - esp. if it has worked for you before! - as long as I can learn it during one weekend.  :)

Some ideas:

modulo arithmetic esp. finite fields: Perfect shuffle is fun.  Would cryptography (e.g. RSA) be too difficult?  What about codes e.g. Hamming?  Any other fun examples of modulo arithmetic?

non-Euclidean geometry: especially how a POINT in elliptic geometry is actually $2$ "ordinary points".  Can the alternates to the parallel postulate (and some consequence e.g. angle sum in a triangle) be explained sufficiently at highschool level in an hour?

finite geometry: Personally I find Finite projective planes very beautiful.  Do they (or finite affine planes) have any application?  (Besides the kiddie game Spot It?)

algorithm: This is actually my work area, but I don't know what I can cover in an hour.  Maybe the $O(N \log N)$ lower bound for sorting?  Some of these kids don't even have coding experience... :(

graph theory: Edge coloring of a complete graph is one of my favorite pictorial proofs.  Eulerian tour.  Euler characteristic (of a planar graph) is related.  What else?  Shortest path would require discussion of algorithm.  Hall's marriage theorem is surprising and neat but I don't think I can prove it from scratch in an hour.

combinatorics: Start with ${N \choose m}$ and then stars-and-bars for sure.  What are your favorite (elementary) examples?  Can Burnside's lemma be proved from scratch (at highschool level, without group theory) in an hour?

Comments on above and/or any other suggestion most welcome!

(Apologies if I'm tagging too widely...)

答:
The $2$-dimensional case of Sperner's lemma is a good introduction to graph theory. The proof is self-contained aside from using the Handshaking lemma, which is also easy to show. Take a triangle and subdivide it into several smaller triangles. Call a triangle minimal if it contains no other triangles. If whenever two minimal triangles touch, they share a side, then we call this subdivision a triangulation. (This is to avoid the possibility that the side of one minimal triangle is part of a side of another triangle.)

Color the nodes of the original triangle three colors, say red, blue, and green. Then color all of the other nodes with those colors, under the condition that if a node lies on a side of the original triangle, then it has to be given one of the colors used for the original two nodes of that side. Nodes inside the triangle can be given any of the three colors. See the picture below. (Apologies for the poor quality- this was made in mspaint and then hastily resized.) Observe that there's a red-blue side, a red-green side, and a blue-green side.

There are five minimal triangles that have a vertex of each color. Sperner's lemma states that there must always be a minimal triangle with a node of each color, and in fact, there must be an odd number of such triangles. To see this, we will construct a graph from this triangulation. We do so by associating a vertex to each face, including the outside face, and draw an edge between two vertices if the side shared by their corresponding triangles has two different colors. See the picture below.

The internal vertices can have degree either $0$, $2$, or $3$. The degree is $0$ if the nodes of the corresponding triangle all have the same color, $2$ if the nodes use two different colors, and $3$ if the nodes all use different colors. We want to show that there are an odd number of internal vertices of odd degree. The Handshaking Lemma tells us that a graph must have an even number of vertices with odd degree, so if we can show that the outside vertex has odd degree, then the lemma follows. For example, the outside vertex in our picture has degree $7$.

To see this, look at any side of the original graph, say the red-green side. Moving from the red node to the green node, that side contributes $1$ whenever the colors switch from red to green or green to red. Since we start with a red node and end with a green node, there must be an odd number of switches. Hence, each side of the original triangle makes an odd contribution to the degree of the outside vertex so the outside vertex has odd degree. By the Handshaking Lemma, Sperner's Lemma follows.


问:
Primitive element of $\mathbb{F}_{16}^{\times}$
I have to find a primitive element of $\mathbb{F}_{16}^{\times}$. I defined $\mathbb{F}_{16} = \frac{\mathbb{F}_{2}[a]}{\langle a^4+a+1 \rangle}$.

I tried to prove that $a^3$ and $a^5 \neq 1$.

I got $a^3=a^3 \neq 1$ and $a^5 = a^2 + a \neq 1$.

But, when I try to check if $a^{15} = 1$, I get $a^{15} = a^3 + a^2 \neq 1$. I checked several times and can not get it, it may be something I am missing.

Are my calculations right?

How can I find the primitive element? Which is the primitive element?

答:
in ${GF(2^4)}$ The polynomial ${1+x+x^4}$ has 8 primitive elements.

Condition: ${x^i \in GF(2^r)}$ is primitive iff ${gcd(i, 2^{n}-1)=1)}$, Here, ${2^4-1=15}$ ,so we've possible candidates are ${\{1,2,4,7,8,11,13,14\}}$.

you can check, an element  ${\alpha \in GF(2^r)}$ is primitive iff ${\alpha ^m \ne 1}$ for ${1 \leq m < 2^r -1}$

Let's take ${x^4=\alpha}$ then ${\alpha^2=x^8=1+x^2, \alpha^{14}=1+x^3,\alpha^{15}=x+x^4=x+1+x=1}$, so we don't have any exponent m less than 15 which gives 1. So, 4 is a primitive element.

Now, lets check ${x^5=\alpha}$, now ${\alpha^2=1+x+x^2, \alpha^3= x^{15}=1}$, here m=3 (<15), hence 5 in not a primitive element.


问:
Prove $ (\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1$
Let $\triangle ABC$ be a acute triangle. Prove that: $$(\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1 $$
My attempt:
$$\Leftrightarrow (1-cosA)(1-cosB)(1-cosC)\ge cosA.cosB.cosC$$
$$\Leftrightarrow 1-2cosA.cosB.cosC + cosA.cosB + cosB.cosC+cosA.cosC \ge cosA+cosB+cosC $$
$$\Leftrightarrow cos^2A+cos^2B+cos^2C + cosA.cosB + cosB.cosC+cosA.cosC \ge cosA+cosB+cosC $$
$$ 0<cos A,cos B,cosC<1$$
$$cos^2A+cos^2B+cos^2C=1-2cosA.cosB.cosC\ge 3\sqrt[3]{cosA.cosB.cosC}$$
$$\Rightarrow cosA.cosB.cosC \le \frac{1}{8}$$
And I was stuck here. Could you help me ?

答: Simple. Use AM-GM and Jensen's inequalities: $\cos A\cos B\cos C \le \dfrac{(\cos A+\cos B+\cos C)^3}{27}\le \dfrac{1}{27}\cdot (3\cos (\frac{A+B+C}{3}))^3=\left(\frac{1}{2}\right)^3 =\dfrac{1}{8}.$


问:
Show that $f(z) = z^n$ is an entire function
Show that $f(z) = z^n$ is an entire function where $n = 0,1,2...$

I can show that $z$, $z^2$ and $z^3$ are entire functions but what about $z^n$ how do you crack that one.

Body:
Hint. For $z \neq z_0$ and $n \ge 2$, we have

$$\frac{z^n - z_0^n}{z - z_0} = \sum_{k = 0}^{n-1}z^{n-1-k}z_0^k.$$


答: Hint: As you noticed, $|g|=1 =|1/g|$ on the unit circle. What does the maximum modulus theorem tell you about $g,1/g$ in the open unit disc?

Title: $f(z)$ is analytic on $\{ |z| \leq 1\}$, and $f(z)$ is real on $\{|z| = 1\}$, show $f(z)$ is a constant.

Body:
Let $f(z)$ be analytic on $\{|z| < 1\}$ that extends to $\{|z| \leq 1\}$. Assume $f(z)$ is real on $|z| = 1$. Prove that $f(z)$ is a constant.

Here is my attempt,
define $g(z) = e^{if(z)}$, for $|z| = 1$, we have $|g(z)| = 1$. Then $|\frac{1}{g(z)}| = 1$ as well. So that $|g(z)| = 1$ for $|z| = 1$.

(1) Is there any theorem I can use to show that this leads to $g(z)$ is constant on $|z| \leq 1$?

(2) If $g(z)$ is constant, can I show that $f(z)$ is constant?


答:
Replacing $a$ by $|a|$, one can assume without loss of generality that $a$ is a nonnegative real number. If $a=0$, the result is direct. If $0\lt a\lt1$, the sequence defined by $u_n=a^n$ is decreasing and positive hence it converges to some finite nonnegative limit $\ell$. Since $u_{n+1}=au_n$, $\ell=a\ell$. Since $a\ne1$, the only possible limit is $\ell=0$, QED.

The hint that $u_{2n}=u_n^2$ can probably be used as follows, once one knows that the limit $\ell$ exists and is finite: $\ell=\ell^2$ hence $\ell=0$ or $1$ and, since $u_n\leqslant u_1=a\lt1$ for every $n\geqslant1$, $\ell\ne1$ hence $\ell=0$.

Title: Prove $a^n \rightarrow 0$ as $n \rightarrow \infty$ for $\left|a\right| < 1$ without use of log properties

Body:
$a^n \rightarrow 0$ as $n \rightarrow \infty$ for $\left|a\right|  < 1 $

Hint $u_{2n}$ = $u_{n}^2$

I have totally no idea how to prove this, this looks obvious but i found out proof is really hard...
I am doing a real analysis course and there's a lot of proving and I stuck there.
Any advices? Practice makes perfect?


问:
Is the inverse image function injective?
It seems that the inverse image function(see my previous question) is injective. Where a proof of that can be found?</p>

答:
Still waiting for an answer, but here is my proof:

It is clear that the set

$part(f) := \{f^{-1}(\{f(x)\}) : x \in dom(f)\}$ is a partition of $dom(f)$.

It is also clear that there exists a bijection between $im(f)$ and $part(f)$;

Then exists the induced bijection between $P(im(f))$ and $P(part(f))$;

From the other side, because $f^{-1}(X \cup Y) = f^{-1}(X) \cup f^{-1}(Y)$ where $X,Y \subset im(f)$

$f^{-1}(B) = \{x \in dom(f) : f(x) \in B\} = U[b \in B, f^{-1}(\{f(b)\}) = U[b \in B, \pi(b)] $.

where $ \pi(b) := f^{-1}(\{f(b)\})$ is an element of $part(f)$.

If the inverse image function is not injective, then for some
$B_1,B_2 \subset im(f), B_1 \neq B_2$ holds:

$U[b \in B_1, \pi(b)] = U[b \in B_2, \pi(b)]$.

But this is impossible, because  all $\pi(b)$ are disjoint and in the left and right parts
of the above equation the collections of $\pi(b)$ are different ( because  $B1 \neq B2$   and (3)).


答:
Hint:

Set $t = (x^2 - 2020x + 2020),$ express the LHS in terms of $t$, see what the value of $t$ is when $x=1$, and then see what happens to both the LHS and RHS when $x = 1.$

To me, given the query's tags, the real challenge of the problem is to find a pre-calculus approach to determining whether any solution other than $x=1$ is possible.

Frankly, this seems to exclude taking derivatives, since that would require a knowledge of Calculus.  The only pre-calculus approach that I can think of is experimenting with what happens when $x < 0, ~x=0, ~0 < x < 1,$ and $~x > 1.$

Note that it is presumed that the problem is based on the Real Numbers only, rather than the Complex Numbers.  Therefore, since the LHS terms of $\sqrt{t+1}$ and $\sqrt{t-1}$ are involved, you can not allow $(t+1) < 0$ or $(t-1) < 0$.

Body:
Find the sum of all real numbers x that satisfy

$(\sqrt{x^2-2020x+2021})+\sqrt{x^2-2020x+2019})^{\frac{3x}{2}}+(\sqrt{x^2-2020x+2021})-\sqrt{x^2-2020x+2019})^{\frac{3x}{2}}=2^{1+\frac{3x}{4}}$

My Answer :

Square the two sides

$(\sqrt{x^2-2020x+2021})+\sqrt{x^2-2020x+2019})^{3x}+(\sqrt{x^2-2020x+2021})-\sqrt{x^2-2020x+2019})^{3x}=2^{1+\frac{3x}{2}}$

Let $a=(\sqrt{x^2-2020x+2021})+\sqrt{x^2-2020x+2019})^x, b=(\sqrt{x^2-2020x+2021})-\sqrt{x^2-2020x+2019})^x$

So $a^3+b^3=(a+b)((a+b)^2-6)=2^{1+\frac{3x}{2}}$

i'm stuck:)


问:
Solving recurrence using characteristic equation
I am recently learning the how to use the various methods to solve recurrences. So far I have acquainted myself with the Master's Theorem and Substitution method. One method I just can't seem to understand is the following question which needs to be solved by characteristic equation:

$$ T(n) = 2T(n/3) + 1,T(1) = 1$$

I watched certain tutorials and readings and I imagine I needed to derive some sort of degree and simultaneous equation out of this?

How do I do it in this case?

Sorry for being amateur at this.

答:
Write $n=3^m$ so
$T(3^m)=2T(3^{m-1})+1$.

Then let $T(3^m)=u(m)$ so
$u(m)=2u(m-1)+1$.

You should be able to solve this.


问:
Does $[0,+ \infty)$ counts for a closed interval in extended real number system?
In Topology by Munkres, page 84, chapter 2, section 14 the order topology:

So, we are discussing the situation when $X=\mathbb{R}\cup \{+\infty, -\infty\}$. Then $[0,+\infty):= \{t\in X: 0 \leq t < +\infty \}$, it is not of the form $[a,b]:= \{t \in X: a\leq t \leq b \}$, hence $[0, +\infty)$ is not a closed interval. And this example disprove the statement: "a closed interval is an interval which is also a closed set". Also, above reasoning shows that for any $a\in X, [a,+\infty), (-\infty, a]$ and $\mathbb{R} = (-\infty,+\infty)$ are not closed intervals.

Is there any flaw in my understanding?

答:
$[0,+\infty)$ is a semi-open interval in the extended real line with respect to the order topology. For some $M>0$, consider the neighborhood $N=\{x \in X : x>M\}$ of $+\infty$, then $N$ intersects $[0,+\infty)$ and thus $+\infty$ is limit point of $[0,+\infty)$.

Same reason applies to the rest of the sets.


问:
Minimizing an expression (over the integers)
In the context of Hurwitz groups and manifolds, one comes by what wikipedia defines as a "remarkable" fact, that $1-\dfrac1 a -\dfrac 1 b - \dfrac1 c > 0$ has a minimal value of $1/42$ if $a,b,c \in \mathbb{Z}$ and $a<b<c$ (I'm not sure this last part is needed, but let's consider it anyway).

I was wondering how one would go about solving a problem like this. I am totally at loss when I see problems "over the integers": I never actually learned any techniques to approach such problems. So the question is: is there a suitable general method to minimize expressions over the integers? I of course feel that there is not standard way, but I'm interested in some examples as to how to proceed, maybe taking this expression as a base for the examples.

答:
For this problem you just need some intelligent case analysis. Increasing the value of $a, b$ or $c$ makes the sum of reciprocals smaller and thus increases the expression on the right hand side. So you need the smallest values you can get.

$a=1$ leaves the left hand side negative, and the best we could do with $a\ge 3$ would be to set $a=3, b=4, c=5$ giving us $\frac {13}{60}$. So we need to test $a=2$ to see if we can do better than this.

The expression then becomes $\frac12-\frac1b-\frac1c$ with $2\lt b \lt c$

If $b\ge 4$ the best we can do is $b=4, c=5$, giving us a value of $\frac 1{20}$.

So we are left testing $b=3$, when our expression becomes $\frac16-\frac1c$. The lowest value of $c$ for which this is positive is $c=7$ and this gives the minimum value $\frac1{42}$.


问:
Contraction function on $\mathbb{Q}$ that does not have a fixed point
We know the contraction principle holds on the complete metric space.

I am thinking about cases where the space is not complete, such as $\mathbb{Q}$. Could anyone help giving an example of a contraction function $f: \mathbb{Q}\rightarrow\mathbb{Q}$ that is ONTO and does not have a fixed point? How about if a bijection is required?

答:
Let $t$ be an irrational real number.  Let rational approximations $(a_n)_{n\geq 1}$ strictly increasing to $t$ and $(b_n)_{n\geq 1}$ strictly decreasing to $t$, satisfying $a_{n+1}-a_n\leq\frac12(a_n-a_{n-1})$ and $b_n-b_{n+1}\leq\frac12(b_{n-1}-b_n)$.  Then define $T\colon\mathbb{Q}\to\mathbb{Q}$ to be continuous, piecewise-linear map sending $a_n\mapsto a_{n+1}$ and $b_n\mapsto b_{n+1}$, and scales distance by $\frac12$ outside $[a_1,b_1]$, i.e.,
$$
T(q)=
\begin{cases}
\frac12 (q-a_1)+a_2 & q\leq a_1\\
\frac12 (q-b_1)+b_2 & q\geq b_1\\
\dfrac{(a_n-q)a_n+(q-a_{n-1})a_{n+1}}{a_n-a_{n-1}} & q\in[a_{n-1},a_n]\\
\dfrac{(b_{n-1}-q)b_{n+1}+(q-b_n)b_n}{b_{n-1}-b_n} & q\in[b_n, b_{n-1}].
\end{cases}
$$
By construction, $T\colon\mathbb{Q}\to\mathbb{Q}$ is bijective with $\operatorname{Lip}(T)=\frac12<1$ but has no fixed point (since $t\notin\mathbb{Q}$).


问:
Binomial distribution greater than
If $n = 6$ and $p = 0.50$, what is the probability that $x ≥ 1$?

$P(x ≥ 1 | n = 6 \,\text{and}\, p = 0.50) = ?$

Attempt:

$p=0.5$

$n=6$

$x=1 \,\text{or}\, 2$ etc

$(\frac{n!}{x!(n-x)!})(p^x(1-p)^{n-x})$

So I calculated 0.0938 for 1 and 2,3 etc etc but none of the answers were right, what am i missing and need to do to arive at the answer?

Title: Binomial distribution greater than

Body:
$P(x\geq 1)=\sum_{x=1}^6 \frac{n!}{(n-x)!x!}(\frac{1}{2})^x (\frac{1}{2})^{6-x}  = \sum_{x=1}^6 \frac{n!}{(n-x)!x!}(\frac{1}{2})^6=(\frac{1}{2})^6 (2^6-1)
=0.984375$ (by identity).

(Since, $(1+y)^n=^nC_0+^nC_1 y +^nC_2 y^2+...+^nC_n y^n$, take $y=1$ to get the above identity.)

Let me know if this is correct.


问:
Prove that for every natural number $n$, there exists a natural number $m$ such that $\frac{1}{m}-\frac{1}{m+1} < \frac{1}{n}$
I am not sure how to proceed with my proof. I can easily explain in words why this is so (just by maximizing the values of the LHS and RHS then noting the left will always decrease faster than the RHS due to denominator getting smaller), but I do not think this is mathematically valid and doesn't necessarily prove anything.

This is how I did it.

$\frac{1}{m} - \frac{1}{m+1} < \frac{1}{n}$

$=\frac{1}{m(m+1)} < \frac{1}{n}$

Now, I am sadly stuck and have no way to proceed. I am thinking to expand this inequality like $\frac{1}{m(m+1)}  < \text{something} < \frac{1}{n}$

or

$\text{something}  < \frac{1}{m(m+1)}  < \frac{1}{n}$, but I am not sure how to go about this or if this is even the right approach.

答:
You have $\frac{1}{m(m+1)} < \frac{1}{n}$, which is equivalent to
$m(m+1) > n$. Surely you can find a value of $m$ that will make this true.


问:
Prove $a^n \rightarrow 0$ as $n \rightarrow \infty$ for $\left|a\right| < 1$ without use of log properties
$a^n \rightarrow 0$ as $n \rightarrow \infty$ for $\left|a\right|  < 1 $

Hint $u_{2n}$ = $u_{n}^2$

I have totally no idea how to prove this, this looks obvious but i found out proof is really hard...
I am doing a real analysis course and there's a lot of proving and I stuck there.
Any advices? Practice makes perfect?

答:
Since $0\le|a|\lt1$, we have $0\le|a|^{n+1}\le|a|^n$. Since $|a|^n$ is a non-increasing sequence, bounded below, $A=\lim\limits_{n\to\infty}|a|^n$ exists. Then,
$$
\begin{align}
|a|A
&=|a|\lim_{n\to\infty}|a|^n\\
&=\lim_{n\to\infty}|a|^n\\
&=A
\end{align}
$$
Thus, $(|a|-1)A=0\implies A=0$. Therefore,
$$
\left|\lim_{n\to\infty}a^n\right|=\lim_{n\to\infty}|a|^n=A=0
$$</p>


答:
In order that the inverse $f^{-1}$ to exist, $f$ must be injective (1-1).
Because, if not, $\exists x_1, x_2: f(x_1) = f(x_2) \land  x_1 \neq x_2
   \Rightarrow \exists y_1, y_2: y_1 = y_2 \land  f^{-1}(y_1) \neq
   f^{-1}(y_2)$, violating the definition of a function.

Now, $f^{-1} = F[y \in im(f)), \{x \in dom(f): f(x) = y \}]$. We have $y_1 \neq y_2 \Rightarrow f(x_1) \neq f(x_2) \Rightarrow x_1 \neq x_2 \Rightarrow f^{-1}(y_1) \neq f^{-1}(y_2)$ (by injectivity of $f$). Which implies $f^{-1}$ is injective.

Title: Is the inverse image function injective?

Body: It seems that the inverse image function(see my previous question) is injective. Where a proof of that can be found?</p>


问:
page 74-75 Problem 2 Hungerford algebra
Consider the following problem from Hungerford algebra:

A subset $X$ of an abelian group $F$ is said to be linearly independent if $n_1x_1 + \ldots +
n_kx_k = 0 $ always implies $n_i = 0$ for all $i$ (where $n\in \mathbb{Z}$ and $x_1, \ldots , x_k$ are distinct elements of $X$).

Prove that

(a) If $F$ is free abelian, it is not true that every linearly independent subset of $F$ may be extended to a basis of $F$.

(b) If $F$ is free abelian, it is not true that every generating set of $F$ contains a basis of $F$. However, if $F$ is also finitely generated by $n$ elements, $F$ has rank $m\leq n $.

I think a is false, as let a linearly independent subset $\{x_1 , \ldots , x_k \}$, then it can be extended to basis by simply adding the basis elements which are not in span of $\{x_1, \ldots ,x_k \}$.

For b, I am clear with the part that which is after however but couldn't find reasoning of before however part.

答:
(a)  Let $F= \Bbb Z, X= \{ 2 \}$.

(b)  Let $F=\Bbb Z, X=\{ 2, 3 \}$.

Adding elements that are not in the span of $X$ may destroy linear independence.

Title: page 74-75 Problem 2 Hungerford algebra


问:
Vector perpendicular to 3d vector?
Given a unit vector $v_1 \in \mathbb{R}^3 : |v_1| = 1$, let $v_1 = (x_1,y_1,z_1)$

What's the simplest way to find another unit vector $v_2 = (x_2, y_2, z_2)$ such that they are perpendicular (ie $v_1 \cdotp v_2 = 0$) ?

Notice the constraints are:

\begin{align}
    x_1x_2 + y_1y_2 + z_1z_2 &= 0,\\
    {x_1}^2 + {y_1}^2 + {z_1}^2 &= 1,\\
    {x_2}^2 + {y_2}^2 + {z_2}^2 &= 1.
\end{align}

What is a solution for $x_2$, $y_2$, $z_2$ in terms of $x_1$, $y_1$, $z_1$ that is easy to express and calculate?

(Yes there is an infinite family of such vectors $v_2$ for a given $v_1$, I just want one, any one)

答:
It looks like OP is looking for an algorithm to find a perpendicular vector.

Let $\mathbf{p}_1 = (x, y, z)$ be the original unit vector, $x^2 + y^2 + z^2 = 1$.  I shall use notation
$$\mathbf{p}_1 = (x, y, z) = \left[\begin{matrix}x\\y\\z\\ \end{matrix}\right]$$
where the parenthesised form $(x, y, z)$ is just shorthand for the proper vector/matrix form.

Construct two helper vectors by rotating $\mathbf{p}$ 90° around two different axes (the axes being perpendicular to each other), say around the $z$ axis and the $y$ axis,
$$\begin{aligned}
\mathbf{q}_1 &= (y ,\, -x ,\, z) \\
\mathbf{q}_2 &= (z ,\,  y ,\, -x) \\
\end{aligned}$$
and calculate their vector cross products wrt. the original vector:
$$\begin{aligned}
\mathbf{d}_1 = \mathbf{p} \times \mathbf{q}_1 &= ( y z + x z ,\, y z - x z ,\, - y^2 - x^2 ) \\
\mathbf{d}_2 = \mathbf{p} \times \mathbf{q}_2 &= ( - y z - x y ,\, z^2 + x^2 ,\, x y - y z ) \\
\end{aligned}$$
One of these may be a zero vector (or very small), depending on how close the original $\mathbf{p}$ was to the respective rotation axis, but if nonzero, they are perpendicular to the original vector.  So, pick the larger one in magnitude:
$$\mathbf{p}_2 = \begin{cases}
\displaystyle \frac{\mathbf{d}_1}{\left\lVert \mathbf{d}_1 \right\rVert}, & \left\lVert \mathbf{d}_1 \right\rVert \ge \left\lVert \mathbf{d}_2 \right\rVert \\
\displaystyle \frac{\mathbf{d}_2}{\left\lVert \mathbf{d}_2 \right\rVert}, & \left\lVert \mathbf{d}_1 \right\rVert \lt \left\lVert \mathbf{d}_2 \right\rVert \\
\end{cases}$$

Title: Vector perpendicular to 3d vector?


Title: Prove $ (\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1$

Body:
Let $\triangle ABC$ be a acute triangle. Prove that: $$(\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1 $$
My attempt:
$$\Leftrightarrow (1-cosA)(1-cosB)(1-cosC)\ge cosA.cosB.cosC$$
$$\Leftrightarrow 1-2cosA.cosB.cosC + cosA.cosB + cosB.cosC+cosA.cosC \ge cosA+cosB+cosC $$
$$\Leftrightarrow cos^2A+cos^2B+cos^2C + cosA.cosB + cosB.cosC+cosA.cosC \ge cosA+cosB+cosC $$
$$ 0<cos A,cos B,cosC<1$$
$$cos^2A+cos^2B+cos^2C=1-2cosA.cosB.cosC\ge 3\sqrt[3]{cosA.cosB.cosC}$$
$$\Rightarrow cosA.cosB.cosC \le \frac{1}{8}$$
And I was stuck here. Could you help me ?

Body:
We have
$$1- \cos A = 1-  \frac{b^2+c^2-a^2}{2bc} = \frac{(a+b-c)(c+a-b)}{2bc},$$
so
$$(1-\cos A)(1-\cos B)(1-\cos C) = \frac{(a+b-c)^2(b+c-a)^2(c+a-b)^2}{8a^2b^2c^2},$$
and
$$\cos A \cos B \cos C = \frac{(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)}{8a^2b^2c^2}.$$
Thefore, the original inequality become
$$(a+b-c)^2(b+c-a)^2(c+a-b)^2 \geqslant (a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2).$$
It's remain to prove that
$$(a+b-c)^2(c+a-b)^2 \geqslant (a^2+b^2-c^2)(c^2+a^2-b^2).$$
But, this is true because
$$(a+b-c)^2(c+a-b)^2 - (a^2+b^2-c^2)(c^2+a^2-b^2) = 2(b-c)^2(b^2+c^2-a^2) \geqslant 0.$$
The proof is completed.