问:
Failure of a Subring to be an Ideal
Is there any measure of the failure of a subring to be an ideal, using for example homological methods? I just remembered that for $\phi : R\rightarrow S$, where $R,S $ commutative rings with unity, that $\phi^{-1}(\phi(I))=I+Ker\phi$, while $\phi(\phi^{-1}(J))=J\cap Im\phi$, where $I$ an ideal of $R$ and $J$ an ideal of $S$, and it's not as symmetrical as I would want. Of course $Im\phi$ is a subring, and $J\cap Im\phi$ is an ideal of it, but there's something "off-putting" about this.

答:
Question: "Is there any measure of the failure of a subring to be an ideal, using for example homological methods?"

Answer: The following Theorem seems to hold:

Theorem. If $R \subseteq S$ is a unital subring which is an ideal then $R=S$.

Proof: If $R \subseteq S$ is a unital subring where $R$ is an ideal, it follows for any element $s \in S$ that $s=s 1\in R$ hence $R=S$. Hence the only unital subring $R$ which is an ideal is the trivial sub ring $R=S$. QED

Example 1. Let $S:=\mathbb{Q}$ be the field of rational numbers. Let $R \subseteq S$ be a unital sub ring which is an ideal. This means that for any element $x\in R$ and any element $s\in S$ it follws $sx \in R$. Since $1\in R$ this implies that $s1 =s \in R$ hence $R=S$.

Example 2. If $S$ is any unital associative ring and $R \subseteq S$ is an ideal then $R \subseteq S$ is a (non-unital) sub ring.

Proof: Since $R$ is an ideal, it follows $R$ is closed under addition and for any $s\in S, x\in R$ it follows $sx\in R$. In particular for any $s,x\in R$ it follows $sx\in R$ hence $R$ has a multiplication. It follows $R$ is a (non-unital) associative sub ring of $S$.

Definition. Let $R^{id}$ be the set of elements (finite sums) $\sum_i s_ix_i$ with $s_i \in S, x_i \in R$.

It follows $R  \subseteq R^{id}$ and $R^{id}\subseteq S$ is an ideal in $S$.

Lemma. $R$ is an ideal in $S$ iff $R=R^{id}$.

Proof: If $R$ is an ideal is tollows $R^{id} \subseteq R$ by definition. If $R=R^{id}$ it follows $R$ is an ideal. QED

Hence the ideal $R^{id}$ measures when the ring $R$ is an ideal. You may define the abelian group $R^{id}/R$. By the above argument it has the property that $R^{id}/R=0$ iff $R$ is an ideal, hence the quotient $R^{id}/R$ "detects" when $R$ is an ideal.

Title: Failure of a Subring to be an Ideal


问:
Does $\sum_{n=1}^\infty \frac{\cos{(\sqrt{n})}}{n}$ converge?
The series is:
$$\sum_{n=1}^\infty \frac{\cos(\sqrt{n})}{n}$$

Considering it isn't always positive, I replace $\frac{\cos{\sqrt{n}}}{n}$ with its absolute value and I find that:
$$\vert \frac{\cos{\sqrt{n}}}{n}\vert\gt \frac{\cos^2{\sqrt{n}}}{n}=\frac{\ 1+\cos{2\sqrt{n}}}{2n}=\frac{1}{2n}+\frac{\cos{2\sqrt{n}}}{2n}$$
if $\sum_{n=1}^\infty \vert\frac{\cos{\sqrt{n}}}{n}\vert $   converges, then
$\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$ converges.Using Comparison test,we can draw the conclusion that  $\sum_{n=1}^\infty\frac{1}{2n}$  converges , which is impossible.
So I get that $\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$  absolutely diverges. But I can't figure out whether  $\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$ converges or not.

I have tried Dirichlet's test, but I can't figure out whether $$S_{n}=\sum_{k=1}^n \cos{\sqrt{k}}$$ is bounded.

(This is my first time to ask question.Maybe there exist some mistakes in my conclusion.Thanks. :)

答:
Outline

I haven't fully thought out this answer but leaving it here as a response to a comment on a duplicate question.

At the cost of introducing a constant $C_N$ we can start taking the sum at a large index $N$. $\sum_{n=1} \cos (\sqrt n) = C_N +\sum_{n=N} \cos (\sqrt n)$

Define $x_n = \sqrt n$ so have $\sum_{n=N} \cos (\sqrt n) / n = \sum_{n=N} \cos (x_n) / x_n^2$.

Defining $\Delta x_n = x_{n+1} - x_{n}$ we have $\Delta x_n \approx \frac 1 {2 \sqrt n} = \frac 1 {2 x_n}$ via first order taylor expansion of $x(n+1) \approx x(n) + x'(n)\cdot 1$

Substituting into the sum, we have $ \sum_{n=N} \cos (\sqrt n) / n \approx 2 \sum_{n=N}\cos (x_n) / x_n \Delta x_n$. The right hand side is a Riemann / Darboux integral for $2 \int_{\sqrt N} \frac {\cos x} x dx$ and the integral is finite.

ContentLicense: CC BY-SA 4.0


问:
Interchanging from a function defined by an integral to indefinite integral
This question might be trivial, but I have the following:

$$\operatorname{E}_n(x) = \int_{0}^{x} e^{-t^2}t^n \, dt$$

If I have $\operatorname{E}_n(x)$ as a function defined as an integral, it is correct to say

$$\operatorname{E}_n = \int e^{-x^2}x^n \, dx ?$$

答:
$\int e^{-x^2}x^n \, dx$ denotes the set of all antiderivatives of $e^{-x^2}x^n$.

$E_n(x) = \int_{0}^{x} e^{-t^2}t^n \, dt$ is a special antiderivative of $e^{-x^2}x^n$ because of $E_n(0)=0.$


答:
As the name indicates, an indefinite integral is not completely defined, it represents a family of functions. So, no it is not really correct.

To some extent one might admit the equivalence

$$\int f(t)\,dt=\int_a^x f(t) dt$$ where $a$ is unspecified.

Title: Interchanging from a function defined by an integral to indefinite integral

Body:
This question might be trivial, but I have the following:

$$\operatorname{E}_n(x) = \int_{0}^{x} e^{-t^2}t^n \, dt$$

If I have $\operatorname{E}_n(x)$ as a function defined as an integral, it is correct to say

$$\operatorname{E}_n = \int e^{-x^2}x^n \, dx ?$$


问:
What is wrong in this choice of $\delta$?
Suppose I have a wrongly evaluated limit$-$
$$\lim_{x \to2} 3x+3=6$$Then we get $-$ $$3x+3-6<\varepsilon
$$
$$3x-3<\varepsilon
$$$$|x-1|<\frac\varepsilon {3}$$$$|x-2|<\frac\varepsilon 3 -1$$
$$|x-2|<\frac{\varepsilon-3} 3=\delta $$ What is wrong with this choice of $\delta$? (Also, please try to provide a diagram, which will help beginners like me!)

P.S I had posted another question, based on this question (which was closed recently). I wrote this post making the question explicit, so would expect a decent answer.Thanks!

答:
The third to fourth line, $|x-1| < \frac \epsilon 3 \implies |x-2| < \frac \epsilon 3 -1$ is dead wrong.

$|a| < M$ most certainly does NOT imply $|a-k| < M-k$.

If $k > M$ then $M -k< 0$ which would make this impossible.  And as we are taking $\epsilon$ to be arbitrarily small, it will almost certainly be the case that

$|x-1| < \frac \epsilon 3 < 1$ so

$|x-1| - 1 < \frac \epsilon 3 -1 < 0$.

You seem to assume that $|x-1| - 1 = |x-1| - 1 = |x-2|$ but that would only be true if $x-1 \ge 1$.  But as we are assumimg $|x-1| < \frac \epsilon 3$ that will not be true at all (even if we assume $x-1 > 0$ we don't have $x-1 \ge 1$).

(As well if $a < 0; k > 0$ then $a-k < a < 0$ and $|a-k| > |a|$.  In fact $|a-k| = |a| + k$.  So $|a| < M$ will not imply $|a-k| = |a| +k < M -k$.)

....

Actually the triangle inequality that says $|a| + |k| \ge |a + k|$ can be manipulated to show $|a - k|\ge |a| -|k|$

so we have $|x-2| = |x-1 -1| \ge |x-1| -1 < \frac \epsilon 3 -1$ just .... kills this.


问:
Calculus book recommendation
I'm a high school student who has just finished learning single variable calculus(or calculus-2). I want to study some elementary(to not overburden myself) multivariable calculus which would be very useful in understanding the heart of electrodynamics and magnetism. I do not want to learn this directly from books like Griffiths where only the necessary stuff for physics is taught(I want to enjoy the math obviously!).

I went through some standard books like Hubbard's and Spivak's but they're just way too advanced.

Is there any such elementary multivariable calculus book that you think is apt for me?

答:
First thing to understand is that there exists vectors calculus and multivariable calculus, both are related but not the same. The vector calculus is what you use in EM, in griffith for the first pages or so, he explains the basic vector calculus stuff but without really detailing the mathematics. If you just want physics, then just do griffith cover to cover.

There is another book for E&M by purcell, which is 'simpler' mathematically i.e: not requiring you to 'really' understand most of vector calculus results to approach E&M. However, this book's difficulty is that in the magnetism chapter, it uses concepts of relativity to bring in the connection between electricity and magnetism... which was sort of difficult for me to understand (And...also where I stopped reading this book )

It's not something too bad, because some of the techniques used in the book have no rigorous explanation like the method of separation of variable for solving partial differential equation.

In my profile bio, I've linked an animated video textbook which covers both multivariable and vector calculus. It is where I learned it, and I think it suffices for most part, but in that course too, a good knowledge of the basics is assumed. There is also a good YouTube channel called mu prime math which has playlists of short videos where you can understand the MV concepts.

It has it's pitfalls in that it doesn't give you problems to solve, but I guess you could search up individually search up problems to do or study some questions of integration on multiple integrations in MSE(There are many many solved examples on this site, you just have to look)

A great book is the one by Strang here, it has very nice pictures to aid in understanding the multivariable concepts.

Title: Calculus book recommendation


问:
Closed form of $\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}$?
I considered the following function:
$$f:\, \mathbb{C}\mapsto\mathbb{C}_{\infty},\, z\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}\right)^{-1}.$$

It can be seen that $f(z)=0$ at every integer (division of a non-zero complex number by zero is assumed to be complex infinity, so its reciprocal is $0$). It also seems that the function is periodic on the real axis with period $2$, i.e.
$$\forall x\in\mathbb{R}:\, f(x+2)=f(x),$$
though I was not able to prove that rigorously. I tried to use the $\sinh$ addition formula, but that's probably not useful.

On the whole real axis, $f$ is very close to
$$g:\, \mathbb{C}\mapsto \mathbb{C},\, z\mapsto 0.31837572\sin \pi z,$$
but on the imaginary axis, there's a big difference between $f$ and $g$, for instance:
$$f(2i)\approx 5.3796i,\,\text{but}\, g(2i)\approx 85.2499i.$$

Why is that? Can a closed form of $f$ in terms of Weierstrass/Jacobi elliptic functions be found?

Note:
Curiously, any
$$h:\,\mathbb{R}\mapsto\mathbb{R},\, x\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{a^{x+n}-a^{-x-n}}\right)^{-1}$$
where $a\gt 0$ and $a\ne 1$ seems to be $2$-periodic...

Edit:
The function
$$f:\, \mathbb{C}\mapsto\mathbb{C}_{\infty},\, z\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}\right)^{-1}$$
is probably a doubly-periodic elliptic function with periods $2$ and $2\pi i$, so I added an appropriate tag.

答:
We can evaluate it in terms of Jacobi theta functions.

This identity is occasionally useful:
$$
\sum _{n=-\infty }^{\infty }{\frac {b{u}^{n}}{1-a{q}^{2\,n}}}=
\frac{b}{2i}\;
{\frac {
{\theta_1} \left( \frac{i}{2}\ln  \left( au \right) ,q \right) {
\theta_2} \left( 0,q \right) {\theta_3} \left( 0,q
 \right) {\theta_4} \left( 0,q \right) }{{\theta_1}
 \left( \frac{i}{2}\ln  \left( u \right) ,q \right) {\theta_1} \left(
\frac{i}{2}\ln  \left( a \right) ,q \right) }}
$$
provided $1 > |u| > |q^2|$ so that the series converges.

For this problem, write $\sinh$ in terms of $\exp$ to get the left side of this,
with
$$
a=e^{-2 z},\quad b=2e^{-z},\quad q=e^{-1},\quad u=-e^{-1}
$$
to get the answer
$$
\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}
=
{\frac {i{{\rm e}^{-z}}{\theta_2} \left( iz+i/2,{{\rm e}^{-1}}
 \right) {\theta_2} \left( 0,{{\rm e}^{-1}} \right) {
\theta_3} \left( 0,{{\rm e}^{-1}} \right) {\theta_4}
 \left( 0,{{\rm e}^{-1}} \right) }{{\theta_2} \left( i/2,{
{\rm e}^{-1}} \right) {\theta_1} \left( iz,{{\rm e}^{-1}}
 \right) }}
$$
Note, despite all those $i$s in there, this is real when $z$ is real.


问:
Geometric Maximization Exercise
What is the height of a circular cone with surface line ("Mantellinie") s, which has the maximal volumina?
My problem here is, that I am not really sure what surface line is..is this the same thing as just to say surface?</p>

答:
The "Mantellinie" $s$ is what you think it is: It's the distance from the tip of the cone to the rim at the bottom. Given $s$, you are free to choose the height $h$ in the interval $[0,s]$, and the radius of the bottom circle will then be $r=\sqrt{s^2-h^2}$. Therefore we have to maximize the function
$$V(h)={\pi\over3} r^2 h={\pi\over3}(s^2 h-h^3)\qquad(0\leq h\leq s)\ .$$
As $V(0)=V(s)=0$ this maximum is taken for some $h\in\ ]0,s[\ $, which will come to the fore by solving $V'(h)={\pi\over3}(s^2-3h^2)=0$ for $h$. There is only one  solution in the interval $\ ]0,s[\ $, namely $$h={s\over\sqrt{3}}\ ,$$
and this is the height you wanted to know.</p>

ContentLicense: CC BY-SA 3.0

Title: Geometric Maximization Exercise


问:
Path difference on a sphere due to geodesic deviation
Between two points of same latitude but different longitudes [spherical coordinates $(\theta,\phi)$] we measure two lengths of arcs  $(s_1,s_2)$ along (1) the geodesic or red great circle (a graphic of numerical integration) and (2) along the constant $ \phi$ non-geodesic thicker purple arc  or latitude.

There is no difference of these distances at the equator or poles but is a maximum at some intermediate latitude value.

How to find the $\phi$ at which the difference of these arc distances $ (s_2-s_1) $  is a maximum?

Thanking you in advance.

答:
For convenience of the application we will use earth-like spherical coordinates with $\phi=0$ referring to the equator (so that the poles are at $\phi=\pm\frac\pi2$).
Further we let the azimuthal angle between $s_1$ and $s_2$ be $2\theta\,\, (0<2\theta\le\pi)$.

We seek to maximize over $\phi$ the expression:
$$
D(\theta,\phi)=2\theta\cos\phi-\arccos(\cos^2\phi\cos2\theta+\sin^2\phi)=
2\theta\cos\phi-\arccos(1-A_\theta\cos^2\phi),
$$
with $A_\theta=1-\cos2\theta=2\sin^2\theta$.

Equating the derivative of $D(\theta,\phi)$ over $\phi$ to $0$ we find (besides the trivial case $\sin\phi=0$) the equation
$$
\boxed{\cos^2\phi=\frac{2\theta^2-A_\theta}{\theta^2A_\theta}=\frac1{\sin^2\theta}-\frac1{\theta^2}},
$$
which being resolved gives the maximizing latitude $\phi_\text{max}$.

For small azimuthal separations ($\theta\to0$) $\phi_\text{max}=\arccos\frac1{\sqrt3}\approx 54.74^\circ.$


问:
Help with the proof of Zorn's Lemma: if $A$ and $B$ are conforming subsets, why must it be that $A\subseteq B$ or $B\subseteq A$?
Here we can find a simple proof of Zorn's Lemma, but I find something that I can't really understand. There is an statement that say: If A and B are conforming subsets of $X$ and $A\neq B$, then one of these two sets is an initial segment of the other.

If I take, by example, the set $X=\{0,1,2,3,4,5,6\}$, I can create two conforming subsets which satisfy the definition but aren't initial segment of each other, like $B=\{0,2,4,5\}$ and $C=\{0,2,3,4\}$.

This proof is similar to another used by Gorodentsev in his book Algebra I, so there is a thing that I am missing here. Sorry for my bad English and thanks in advantage.

Definition

We shall say that a subset $A$ of $X$ is conforming if the following two conditions hold:

The order $\le$ is a well order of the set A.

For every element $x \in A$, we have $x = f(P(A, x))$, where $f$ maps the initial segment to its strict upper bound x.

Definition

If $C$ is a chain in $X$ and $x\in C$, then we define
$P(C, x) = \{ y \in C: y < x\}$. A subset of a chain $C$ that has the form $P(C, x)$ is called an initial segment in $C$.

答:
No. You can't.

If $A$ is conforming, then its minimum element is $f(\varnothing)$. So that's unique, say $x_0$; then the next element is the one chosen by $f$ as an upper bound of $\{x_0\}$. So again, that's unique, call that $x_1$, etc. etc.

By transfinite induction we get that if $A$ and $B$ are both conforming, they must have been built by essentially iterating the choice function to a certain length. Since a choice function is a function its output are well-defined, that means that $A\subseteq B$ or $B\subseteq A$.

The reason you think you have two conforming subsets is that you didn't specify the choice function which maps a chain to an upper bound. Once you do, you have to ask yourself, is $f(\{x\mid x>2\})$ going to be equal to $3$ or to $4$?


问:
Counting permutation of duplicate items
I am not really sure I fully understand the formula for finding the number of permutation of duplicate items.

The formula is:

$$\frac{n!}{p!q!r!\cdots}$$

Where do the factorials in the denominator exactly come from?

If we asssume the string "ANNA" and we want the count of the permutation of duplicate items.

We have $4$ characters so since we have $4$ options for the first character, $3$ for the second, $2$ for the third and $1$ for the last we have $4!$ different permutations.

But some of the characters are duplicates.

We have "A" and "N" repeated, meaning it adds more permutations to the outcome.

If $p=2$ is the number of occurences of "A" and if $q = 2$ is the number of occurences for "N"  then would the $p!$ and $q!$ essentially be selecting $1$ item at a time from $2$ (or $p$) duplicates? So essentially are the denominators the binomial $\binom{2}{1}$ (or generally $\binom{p}{1}$ where $p$ are the number of repeated characters) where order does not matter?

Is my understanding correct?

答:
A different way of viewing this formula which avoids "division by symmetry" arguments:

Given $n$ items, $p$ of which are identical of one type, $q$ of which are identical of another type, $r$ of which are identical of a third type, etc... with $p+q+r+\dots = n$, approach the count by following the steps:

Pick which of the spaces are used by the $p$ objects of the first type.  There are $\binom{n}{p}$ ways to make this choice

Pick which of the remaining spaces are used by the $q$ objects of the second type.  There are $\binom{n-p}{q}$ ways to make this choice given the earlier choice

Pick which of the remaining spaces are used by the $r$ objects of the third type.  There are $\binom{n-p-q}{r}$ ways to make this choice given the earlier choices

$\vdots$

This gives a final total of:

$$\binom{n}{p}\binom{n-p}{q}\binom{n-p-q}{r}\cdots$$

Note, this explanation was able to completely avoid division as an operation as binomial coefficients could have been defined recursively purely using addition and multiplication.

Now... if you insist on rewriting this using fractions, you should know that $\binom{a}{b}=\dfrac{a!}{b!(a-b)!}$ in which case you have

$$\binom{n}{p}\binom{n-p}{q}\binom{n-p-q}{r}\cdots = \dfrac{n!}{p!\color{red}{(n-p)!}}\cdot \dfrac{\color{red}{(n-p)!}}{q!\color{blue}{(n-p-q)!}}\cdot\dfrac{\color{blue}{(n-p-q)!}}{r!(n-p-q-r)!}\cdots$$

Cancelling the numerators of each term after the first in the product with a portion of the denominator from the previous term gives the well known $$\dfrac{n!}{p!q!r!\cdots}$$

Title: Counting permutation of duplicate items


问:
Why is $f_1(n)$ not computable but $f_2(n)$ is?
I have the following two functions, where the first one is not computable and the second one is.

$$f_1(n)= \begin{cases}
      1 & ,\text{if in the decimal representation of n appears in the decimal fraction expansion of} \ \pi \\
      0 & ,\text{else}
   \end{cases}
$$

For example, if we have $\pi = 3.1415926 ...$. Then $f_1(14195) = 1$ but we don't know if $f_1(333)$ has a solution.

$$f_2(n)= \begin{cases}
      1 & ,\text{if in the decimal representation of pi, there are n many consecutive 7's } \ \\
      0 & ,\text{else}
   \end{cases}
$$

I am having a nightmare trying to understand why one is computable and the other one is not. If we take that $n =3$, then for $f_2$, we are trying to determine if $777$ appears in the decimal representation of $\pi$. But in $f_1$ we already saw that determining if $f(333)$ has a solution is not computable.

How are these two any different?

PS - the $n$ used in $f_1(n)$ is not the same as the one used in $f_2(n)$.

答:
We start with $f_2(n)$. We have two cases:

Case 1: There are arbitrarily long sequences of consecutive $7$'s in $\pi$, then $f(n)=1, \forall n$, and this is clearly computable.

Case 2: There is a longest sequence of length $k$ of consecutive $7$'s somewhere in $\pi$ and $k$ is fixed. Hence, $$
     f_2(n)=\left\{\begin{array}{ll} 1, & \text{if } n \leq k \\
         0, & k < n\end{array}\right.
  $$
This is also clearly computable.

The function can be generalized (e.g. the apperance of the sequence $333$ is also computable).

$f_1(n)$ on the other hand claims to compute whether or not an arbitrary long sequence of numbers appears in $\pi$ (e.g. $1415926\cdots$). This is a dramatic difference! Now, recapitulate the following definition:

Def.: A (total) function $f: \Sigma^* \to \Sigma^*$ is computable if there is a Turing machine $M$ that halts on every input $\omega$ with $f(\omega)$ as its output on the tape.

For $f_2(n)$, this is clear from the provided cases. Now, for $f_1(n)$ we assume that we have a sequence $\omega$ which is not in $\pi$ (and I do not want to discuss whether or not $\pi$ contains every possible number or not), then there is no way to build a Turing machine which halts on $\omega$ (because in order for our Turing machine to halt, we need to go through all of $\pi$), and therefore $f_1$ is not computable.

Important note for $f_2$: We only proved that $f_2$ is computable. Which function does the actual job is unknown (but also not important for this problem).


问:
Find best approximation in Hilbert space $H = L_2[0; 1]$ for element $a(t)$
I try to solve this problem all day, but can't reach any progress in it. Can you give me some hints?

Find best approximation in Hilbert space $H = L_2[0; 1]$ for element $a(t) = \frac{1}{\sqrt[4]t}$ by elements from subset $M = \{ x \in H: \int\limits_0^1 x(t) dt = 1 \}$

I know that best approximation for this function is projector $P_M (a) = \sum\limits_{k=1}^{n}(x_k, a) x_k$, where $(x_k, a) = \int\limits_{0}^{1} x_k(t) a(t)dt$ is scalar product. But I can't get form of functions $x_k(t)$

答:
You want to solve the problem
\begin{align*}
\text{Minimize} \quad & \int_0^1 [x(t) - a(t)]^2 \, \mathrm{d}t \\
\text{w.r.t.} \quad & x \in L^2(0,1) \\
\text{such that}\quad & \int_0^1 x(t) \, \mathrm{d}t = 1
\end{align*}
The Lagrangian $L : L^2(0,1) \times \mathbb R \to \mathbb R$ is given by
$$L(x, \lambda) = \int_0^1 [x(t) - a(t)]^2 \, \mathrm{d}t + \lambda \, \left( \int_0^1 x(t) \, \mathrm{d}t - 1\right).$$
Now, you have to solve the following system (the left-hand sides are the derivatives of $L$ w.r.t. $x$ and $\lambda$) for $x$ and $\lambda$:
\begin{align*}
2 \, (x(t) - a(t) ) + \lambda &= 0 \quad\text{f.a.a. } t \in (0,1)\\
\int_0^1 x(t) \, \mathrm dt &= 0\end{align*}
The solution of this system should be straight-forward.</p>


问:
Linear Independence of $v_1+x, v_2+x, v_3+x$
Let $v_1, v_2, v_3$ be linear independent vectors in vector space V. Suppose $V \neq span\{v_1, v_2, v_3\}$ and let $x\notin span\{v_1, v_2, v_3\}$ Prove that $v_1+x, v_2+x, v_3+x$ is linearly independent.

My attempt goes like this:

Need to show that $a(v_1+x)+b(v_2+x)+c(v_3+x)=0$ only when $a=b=c=0$.

This statement rearranges to : $a(v_1)+b(v_2)+c(v_3)+(a+b+c)x=0$

Assume $a+b+c=0$ when $a,b,c$ are not all $0$. Then $a=-b-c$. To satisfy the equation above, this would mean that $v_1,v_2,v_3$ is not linearly independent, which is a contradiction.

Therefore one of $a,b,c$ must be zero. WLOG assume  $a=0$ then $b=-c$ where they are not both zero. This would mean that $a(v_1) + (-c)v_2+c(v_3) = 0$, which can not be since it is L.I.

Hence, $a=b=c=0$, showing linear independence.

Is there a flaw with this approach?

答:
I think you're close. It's good to try the $a + b + c = 0$ case, but you haven't talked about what happens in the other case. Also, I think the logic could in this case could be made a little more clear.

When $a + b + c = 0$, then the equation simplifies to
$$av_1 + bv_2 + cv_3 = 0,$$
which implies $a = b = c = 0$ by the linear independence of $v_1, v_2, v_3$. On the other hand, if $a + b + c \neq 0$, then rearranging the equation, we get
$$x = \frac{-a}{a + b + c}v_1 + \frac{-b}{a + b + c}v_2 + \frac{-c}{a + b + c}v_3$$
(note that $a + b + c \neq 0$ is necessary to perform the division), and hence $x \in \operatorname{span}\{v_1, v_2, v_3\}$ against assumption. This takes care of the missing case, which turned out not to be possible.


问:
$\iint_A\nabla\times\textbf{u}\cdot \textbf{n}\ dS$ with $\nabla\times\textbf{u}$ known.

Let $$\textbf{v}=\operatorname{curl}\textbf{u}= (-2+43x-86y+43z, 3-2x+4y-2z, 2-47x+94y-47z)$$ and let $A$ be the part of $z(x,y)=9-x^2-9y^2$ that is above the plane $z(x,y)=x+45y+\frac{113}{2}$. Now calculate $$\iint_A\nabla\times\textbf{u}\cdot \textbf{n}\ dS$$

A hint is given to use Stoke's theorem twice.

So my take is: $$\iint_A\nabla\times\textbf{u}\cdot \textbf{n}\ dS=\iint_A\textbf{v}\cdot \textbf{n}\ dS=\int_{\partial A} \nabla\times \textbf{v} \cdot d\textbf{r}$$

So I get $d\textbf{r}$ as the boundary to an ellipse that satisfies $1=(\frac{x+\frac{1}{2}}{3})^2+(y+\frac{5}{2})^2$, so we have $\textbf{r}=(x,y,(\frac{x+\frac{1}{2}}{3})^2+(y+\frac{5}{2})^2)$ and
$\nabla\times\textbf{v}= (96,90,84)$

I am not sure  how to proceed from here. Should I parametrize $\textbf{r}$ in terms of trig functions?c

答:
The hint essentially means that you should translate this to another surface (with the same boundary curve) than the one you currently have. In other words,

$\displaystyle \iint_{S} (\nabla\times\textbf{u})\cdot \textbf{n}\ dS = \int_{C} \textbf{u}\cdot d\textbf{r} = \iint_{S_1} (\nabla\times\textbf{u})\cdot \textbf{n}\ dS$.

The boundary curve of the elliptic paraboloid surface, as you correctly found, is -

$\displaystyle (\frac{x+0.5}{3})^2+(y+2.5)^2 = 1, z = 9 - x^2 - 9y^2$.

Now using Stokes' Theorem a second time, we can equate line integral of the vector field along the boundary curve to the surface integral of the curl of $\textbf{u}$ over elliptic disk given by,

$\displaystyle (\frac{x+0.5}{3})^2+(y+2.5)^2 \leq 1, z = x + 45y + \frac{113}{2}$.

$\nabla \times \textbf{u}= (-2+43x-86y+43z, 3-2x+4y-2z, 2-47x+94y-47z)$

For the dot product, use the fact that normal vector to the plane is $(-1, -45, 1)$ and the dot product is simply $-131$.

So your task is just finding the area of the ellipse. You can, Either use the formula $\pi ab$ for area of the ellipse and multiply by $131$ which gives you answer of $-393 \pi$. Or you can do the integral as below.

Use substitution, $x = - 0.5 + 3r \cos\theta, y = -2.5 + r \sin\theta$ for the surface integral, $0 \leq r \leq 1, 0 \leq \theta \leq 2\pi$ and plug in Jacobian $3r$ instead of usual $r$. That is $3$ times the area of a unit circle( = $3 \pi$).


问:
apply Paley-Wiener to $\frac{\sin z}{z}$
is it true that
$$ \left| \frac{\sin z}{z} \right| \leq C e^{A|z|}$$
for every complex number $z$ ?
is $A=2$ minimal?</p>

答:
it was trivial indeed: using
$$ \sin z = \sum_{n=0}^\infty (-1)^n \frac{z^{2n+1}}{(2n+1)!},$$
dividing and taking abs, one arrives at
$$ \left| \frac{\sin z}{z} \right| \leq \exp |z|.$$</p>


问:
How do I show that $\ell^{\infty}$ is a normed linear space?
How do I show that $\ell^{ \infty}$ is a normed linear space,
where $\ell^{ \infty}$ is define as $$\|\{a_n\}_{n=1}^{\infty}\|_{\ell^\infty}=\sup_{1 \leq k \leq \infty} |a_k|?$$
There are three properties that I need to check in order for this to be a normed linear space. Nonnegativity, positive homogeneity, and the triangle inequality.

I am having trouble working with the supremum. Any idea will be greatly appreciated thanks

答:
Let $(a_n),(b_n) \in \ell^\infty$ Then we have for each $k$:

$$|a_k+b_k| \le |a_k|+|b_k| \le ||(a_n)||_\infty+||(b_n)||_\infty.$$

This shows that $(a_n+b_n) \in \ell^\infty$ and that

$$||(a_n+b_n)||_\infty \le ||(a_n)||_\infty+||(b_n)||_\infty.$$


答:
Well begin by noticing that for $a,b\in \mathbb{R}$ and $(x_n),(y_n)\in \ell^\infty$ you have that $(ax_n+by_n)$ is bounded, therefore it belongs in $\ell^\infty$ as well.

For $\|\cdot\|_{\infty}$ to be a norm, check that it is non-negative and $\|x_n\|_{\infty}$ becomes zero only when $(x_n)=0$.

$\|\cdot\|_{\infty}$ also satisfies $\|ax_n\|_{\infty}=|a|\|x_n\|_{\infty}$ and use the triangle inequality for the absolute value to show that  $\|x_n+y_n\|_{\infty}\leq \|x_n\|_{\infty}+\|y_n\|_{\infty}$. These prove $\|\cdot\|_{\infty}$ is a norm on $\ell^\infty$.

Title: How do I show that $\ell^{\infty}$ is a normed linear space?

Body:
How do I show that $\ell^{ \infty}$ is a normed linear space,
where $\ell^{ \infty}$ is define as $$\|\{a_n\}_{n=1}^{\infty}\|_{\ell^\infty}=\sup_{1 \leq k \leq \infty} |a_k|?$$
There are three properties that I need to check in order for this to be a normed linear space. Nonnegativity, positive homogeneity, and the triangle inequality.

I am having trouble working with the supremum. Any idea will be greatly appreciated thanks


问:
Is: $\sum_{k=-15}^{5}{(e^{\frac{i}{10}\pi })^{k}}$ equal to i or -i
This is what i did:

I hope it is readable, but to write everything in Latex is really painful.

My answer is i, and  it makes total sense to me, but some people disagree and say the answer is -i.

So if that is really the case can someone explain to me what I did wrong.

答: It is $i$. Think about how the complex numbers you are adding are going around the unit circle in the complex plane. First consider the Real parts, for every point with a nonzero real part on the left there is a point added that has an equally long nonzero real part on the right, cancelling out all nonzero real parts in the sum. Now consider the Imaginary parts, for $k$ from $-15$ to $4$ you have a similar argument, where the imaginary part of each point above zero is cancelled out by the equivalently long imaginary part of a point below zero. Thus the only point in the sum that has not been cancelled out is the last point when $k=5$, which is the point $i$.


问:
Conics Section: Equation of a circle
I have a question that states

Find the equation of a circle with center $(2,-3)$ and passing through $(3, -5)$.

I arrived at $x^2 + y^2 -4x + 6y - 5 = 0 $. It was marked wrong. The answer is said to be $x^2 + y^2 -4x + 6y +8 = 0$.

My question is isn't the approach of solving this question same as $(x -a)^2 + (y-b)^2 = (x₁-a)^2 + (y₁-b)^2$ where $(a,b)$ is the coordinate of the center of the circle and $(x_1, y_1)$ is the coordinate of the point.

Title: Conics Section: Equation of a circle

Body:
The circle has the equation

$$(x-2)^2+(y+3)^2=r^2,$$

where $r$ is unknown. The point $(3,-5)$ is on the circle, hence

$$(3-2)^2+(-5+3)^2=r^2.$$

This gives $r^2=5.$

Hence

$$(x-2)^2+(y+3)^2=5$$

or

$$x^2 + y^2 -4x +6y +8 =0.$$


问:
Does a dense subspace of test functions induce the (weak-$\ast$) topology on the space of distributions?
Suppose $E$ is a dense subspace of $\mathcal{S}(\mathbb{R}^n)$. Is it true that the topology on $\mathcal{S}’(\mathbb{R}^n)$ induced by the linear functionals $u\mapsto u(\phi)$, $\phi\in \mathcal{S}(\mathbb{R}^n)$ coincides with the usual weak-$\ast$ topology on $\mathcal{S}’(\mathbb{R}^n)$? What about $\mathcal{D}(\mathbb{R}^n)$ and $\mathcal{D}’(\mathbb{R}^n)$?

The specific problem that I have in mind is this. Let $(u_j)_j$ be a sequence in $\mathcal{S}’(\mathbb{R}^n)$. Suppose we can show that $u_j(\phi)\rightarrow 0 $ as $j\rightarrow\infty$ for all $\phi$ that can be factorized as $$\phi(x_1, x_2, \cdots, x_n) = \phi_1(x_1)\phi_2(x_2)\cdots\phi_n(x_n)$$ where $\phi_1, \phi_2, \cdots, \phi_n\in\mathcal{S}(\mathbb{R})$. Does it then follow that $u_j\rightarrow 0 $ in $\mathcal{S}’(\mathbb{R}^n)$?

答:
The answer is no. By (IV.1.2) in [1], there is a one-to-one correspondence between "weak-topologies"  and "dual spaces".

In other words, if $\phi$ lies in $\mathcal{S}(\mathbb{R}^n)\setminus E$,  then the evaluation at $\phi$, namely the
functional
$$
  u\in \mathcal{S}'(\mathbb{R}^n)\mapsto  u(\varphi )\in \mathbb R
  $$
is not continuous with respect to the topology described in the question, and hence that topology is not the same
as the weak-$*$ topology.

[1] Schaefer, Helmut H.; Wolff, M. P., Topological vector spaces., Graduate Texts in Mathematics. 3. New York, NY: Springer. vi, 346 p. (1999). ZBL0983.46002.


Title: Conics Section: Equation of a circle

Body:
I have a question that states

Find the equation of a circle with center $(2,-3)$ and passing through $(3, -5)$.

I arrived at $x^2 + y^2 -4x + 6y - 5 = 0 $. It was marked wrong. The answer is said to be $x^2 + y^2 -4x + 6y +8 = 0$.

My question is isn't the approach of solving this question same as $(x -a)^2 + (y-b)^2 = (x₁-a)^2 + (y₁-b)^2$ where $(a,b)$ is the coordinate of the center of the circle and $(x_1, y_1)$ is the coordinate of the point.

Body:
For a quick check,

$$3^2+(-5)^2-4\cdot3+6(-5)-5\ne 0$$

and

$$3^2+(-5)^2-4\cdot3+6(-5)+8=0.$$

The constant term in the equation must be

$$a^2+b^2-(x_1-a)^2-(y_1-b)^2=4+9-1-4=8.$$

As noticed by others,

$$-1-4=-5.$$


问:
How do I prove that finite Stars are not defined in First Order Logic
Consider $ D $ as Set and $v \notin D$. A Star is a graph G with a set of nodes $B \cup \{v\}$ with a set of edges defined as $\{\{v,d\}| d\in D\}$

Show that the class of finite stars is not defined in First Order Logic.

I have two questions regarding that, why would that not be defined in First Order Logic?
Because something like

$$
\exists v \forall b  E(v,b)
$$

Wouldn't that be a class that is defined in first order Logic? And has the same attributes as a class?

答:
I will assume, alhough you should probably make this clearer in your question, that you are quantifying over nodes and you have a binary predicate $E$ stating that two nodes are connected by an edge.

Your formula $\exists v\forall b E(v,b)$ just says that there exists an element (let us call it center) which connects to every other element, but it doesn't state that two elements that are not the center can not be connected, what is necessary in a star graph. Of course, you could describe star graphs by the more complicated
$$\exists v\forall a\forall b E(v,a)\wedge \Big(E(a,b)\rightarrow (a=v\vee b=v)\Big),$$
but nothing is said about their cardinalities... And that is where you find your problem. Assume $T$ is an axiomatization of the theory of finite star graphs: if you consider the formula
$$\varphi_{n}=\exists a_{1}\cdots\exists a_{n} \bigwedge_{i=1}^{n}\bigwedge_{j\neq i}\neg(a_{i}=a_{j}),$$
any model satisfying $\varphi_{n}$ has at least $n$ elements. Consider then the theory $T\cup\{\varphi_{n} : n\in\mathbb{N}\}$. It has a model from the compactness theorem, since, for any finite subset $T_{0}\cup F_{0}$ of $T\cup\{\varphi_{n}:n\in\mathbb{N}\}$, with $T_{0}\subseteq T$ and $F_{0}\subseteq \{\varphi_{n} : n\in\mathbb{N}\}$, there exists $N=\max\{n\in\mathbb{N} : \varphi_{n}\in F_{0}\}$, and so a star graph with $N+1$ elements is a model of $T_{0}\cup F_{0}$: since any finite subset of $T\cup\{\varphi_{n} : n\in\mathbb{N}\}$ has a model, the whole thing must have a model as well.

But what is a model of $T\cup\{\varphi_{n}:n\in\mathbb{N}\}$? Is a model of $T$, that is, a finite star graph, which is also a model of $\{\varphi_{n} : n\in\mathbb{N}\}$, meaning that it is infinite. A contradiction, hence there is no axiomatization of finite star graphs.


答:
Yes, you can use the formula
$$
(x-a)^2+(y-b)^2 = (x_1-a)^2+(y_1-b)^2
$$
where $(a,b)$ is the centre of the circle and $(x_1,y_1)$ is a point that lies on the circle. This formula works because the RHS tells you the length of the $\text{radius}^2$, which is what you need. Hence, the equation of the circle is
$$
(x-2)^2+(y+3)^2=(3-2)^2+(-5+3)^2=5 \, .
$$
If we expand the brackets, we get
$$
x^2-4x+4+y^2+6y+9=5
$$
which simplifies to the textbook answer.

Personally, I prefer methods that require less memorisation. Hence, if I was solving this question, I would find the length of the radius using the distance formula, and then write the equation of the circle in the form
$$
(x-a)^2 + (y-b)^2 = r^2 \, ,
$$
before expanding the brackets if necessary.

Title: Conics Section: Equation of a circle


答:
You can also show it using the Paley-Wiener theorem:

$$\frac{\sin z}{z}=\frac{1}{2}\int_{-1}^{1}e^{ixz}dx$$

From this it follows - according to the Paley-Wiener theorem - that

$$\left| \frac{\sin z}{z} \right| \leq Ce^{|z|}$$

Title: apply Paley-Wiener to $\frac{\sin z}{z}$

Body:
is it true that
$$ \left| \frac{\sin z}{z} \right| \leq C e^{A|z|}$$
for every complex number $z$ ?
is $A=2$ minimal?</p>


问:
Regularity of heat equation with inhomogeneous Dirichlet boundary
Does anyone know a reference for the regularity theory of the Heat equation with inhomogeneous Dirichlet boundary condition:
$$\begin{cases} u_t(t,y) = u_{yy} (t,y) & (t,y)\in [0,T]\times \mathbb{R}_+\\
u(t,0)=h(t) & t \in [0,T]\\
u(0,y)=0 & y \in \mathbb{R}_+,
 \end{cases}\qquad (\star) $$
where $h\in C([0,T])$, $h(0)=0$, and $u_0$ continuous and rapidly decaying at infinity.

Specifically, I'm looking (if it exists) for an estimate of the type
$$\sup_{t\in [0,T]}|\partial^k_y u(t,y)| \leq C_k\sup_{t\in [0,T]}|h(t)|\Phi(y),\qquad \forall k\in\mathbb{N},\qquad (\spadesuit)$$
where $\Phi$ is rapidly decaying in $y$ (think of the heat kernel or one of its spatial derivatives, for instance).

I know that a solution of  $(\star)$ is given by the convolution
$$u(t,y)=-2\int_0^t\partial_y K(\tau,y)h(t-\tau)\,\mathrm{d}\tau= \int_{0}^t\frac{y}{\sqrt{4\pi \tau^3}}e^{-\frac{y^2}{4\tau}}h(t-\tau)\,\mathrm{d}\tau, $$
where $K(\tau,y)$ is the standard heat kernel.  Since $y\mapsto K(\tau,y)$ is smooth for all $t,y>0$, the derivatives in ($\spadesuit$) exist for these values. But do they exist at $y=0$ (it seems this would depend on the fact that $h(0)=0$)? And, can we bound them uniformly in time?  I'm not sure how to manipulate the kernel to get what I need.

答:
The estimate doesn't work as stated because we need higher order derivatives of the boundary function $h$ in the right-hand side.

Since $\lim_{y\to 0}-2\partial_yK(\tau,y)=\delta_\tau$ as distributions in $[0,t]$, we have
$$\lim_{y\to 0}\partial^{2k}_yu(t,y)= \lim_{y\to 0}\partial_t^ku(t,0)=\lim_{y\to 0}\int_{0}^t-2\partial_y K(\tau,y)h^{(k)}(t-\tau)\,\mathrm{d}\tau=h^{(k)}(t).$$
Because $y\mapsto \partial_yK(\tau,y)$ is smooth and rapidly decaying, we deduce that there exists a rapidly decaying smooth function $\Phi_k$ such that
$$|\partial_y^{2k}u(t,y)|\leq |h^{(k)}(t)|\Phi_k(y),\qquad \forall t\in [0,T],y\geq 0.$$
For odd derivatives we can just interpolate using the inequality
$$\|\partial_y^{2k+1}u(t)\|_{L^{\infty}(y,\infty)}\leq C_k\left( \|u(t)\|_{L^{\infty}(y,\infty)}+\|\partial_y^{2k+2}u(t)\|_{L^{\infty}(y,\infty)}\right), \quad \forall y \geq 0,$$
where $C_k$ is independent from $t$ and $y$. Hence we deduce
$$|\partial_y^ku(t,y)| \leq \left(\left|h^{(\left\lceil k/2\right\rceil)}(t)\right|+\left|h(t)\right|\right)\Phi_k(y),\qquad \forall k\in \mathbb{N},t\in [0,T],y\geq 0.$$

I believe, this is the correct version of the estimate.


问:
Check that a surface over $\mathbb Z$ is regular
Let $X$ be an arithmetic surface over $\mathbb Z$ with smooth generic fibre (which is a curve over $\mathbb Q$). Moreover assume that $X$ is given by one equation $F=0$.

added later: In other words, the generic fibre is expressed as a plane curve over $\mathbb Q$.

Now let $x\in X$ be a closed point. How do I check if $X$ is regular at $X$? Ok, formally I must show that $\mathcal O_{X,x}$ is a regular local ring (of Krull dimension 2). But in more concrete terms what should I do with my $F$? I mean, if $X$ was a variety over a field, I could use the Jacobian criterion, but what about this particular case where the base is $\operatorname{Spec} \mathbb Z$?

Remark: With the term arithmetic surface I mean an integral projective scheme of dimension $2$ which is flat over $\mathbb Z$.

答:
$\def\FF{\mathbb{F}}\def\ZZ{\mathbb{Z}}\def\fm{\mathfrak{m}}$You can approach this problem by a variant of the Jacobian criterion. If $A$ is a regular ring of dimension $n$, $\fm$ a maximal ideal, and $f_1$, $f_2$, ..., $f_m \in \fm$, then $A/\langle f_1,\ldots,f_m \rangle$ is regular of dimension $n-m$ at $\fm$ if and only if images of the $f_j$ in $\fm/\fm^2$ are linearly independent. I'll abbreviate $\fm/\fm^2$ to $V$ from now on; it is called the Zariski cotangent space. For $g \in \fm$, we'll write $\overline{g}$ for the class of $g$ in $V$

When $A = k[x_1, \ldots, x_n]$ and $\fm = \langle x_1, \ldots, x_n \rangle$, then a basis for $V$ is the $\overline{x_j}$. For $f \in \fm$, the expansion of $\overline{f}$ in the $\overline{x_j}$ basis is  $\overline{f} = \sum  \tfrac{\partial f}{\partial x_j}(0) \cdot \overline{x_j}$. This is where the Jacobian criterion comes from.

Let's see how to adopt this to $A = \ZZ[x_1, \ldots, x_n]$ and $\fm = \langle x_1, x_2, \ldots, x_n, p \rangle$. Now a basis for $V$ is $\overline{x_1}$, ..., $\overline{x_n}$, $\overline{p}$. For $f \in \fm$, we have
$$\overline{f} = \sum \frac{\partial f}{\partial x_j} \cdot \overline{x_j}(0) + \frac{f(0)}{p} \cdot \overline{p}.$$

Thus, given $f_1$, $f_2$, ..., $f_m$ in $\langle x_1, x_2, \ldots, x_n, p \rangle$, the quotient $\ZZ[x_1, \ldots, x_n]/\langle f_1, \ldots, f_m \rangle$ is regular of dimension $n+1-m$ if and only if the $m \times (n+1)$ matrix

$$\begin{bmatrix}
\tfrac{\partial f_1}{\partial x_1}(0) & \tfrac{\partial f_1}{\partial x_2}(0) & \cdots & \tfrac{\partial f_1}{\partial x_n}(0) & \tfrac{f_1(0)}{p} \\
\tfrac{\partial f_2}{\partial x_1}(0) & \tfrac{\partial f_2}{\partial x_2}(0) & \cdots & \tfrac{\partial f_2}{\partial x_n}(0) & \tfrac{f_2(0)}{p} \\
\vdots&\vdots&&\vdots&\vdots \\
\tfrac{\partial f_m}{\partial x_1}(0) & \tfrac{\partial f_m}{\partial x_2}(0) & \cdots & \tfrac{\partial f_m}{\partial x_n}(0) & \tfrac{f_m(0)}{p} \\
\end{bmatrix}$$
has rank $m$ over $\FF_p$.

This doesn't fully answer your question because a maximal ideal of $\ZZ[x_1, \ldots, x_n]$ could be the kernel of a map $\ZZ[x_1, \ldots, x_n] \to \FF_q$ for $\FF_q$ an extension of $\FF_p$. (EG the maximal ideal $\langle x^2+1, 3 \rangle$ in $\ZZ[x]$.) I started to write out how to deal with that case, but it started getting messy; I'm going to think if I can think of a simpler way to present it.


问:
How to solve this Confidence Level Question?

A company has conducted a test on $400$ people. Out of $400$ people, $224$ people like chocolate. Find the percentage of people who like chocolate in the $99\%$ confidence interval.

The problem is the question doesn't give the standard deviation for the calculation.
I believe the mean is
$$\begin{align*}
 \frac{224}{400} &= 0.56\quad\text{(the mean)}\\[5pt]
N &= 400\\[5pt]
 Z &= \frac{2.576}{(99\% CI)} \\[5pt]
  4^n &= 262,144 \\[5pt]
CI &= \text{mean} \mp \frac{s \cdot Z}{\sqrt{N}} \\[5pt]
\end{align*}$$

How do I find the $s$ value?

答:
your distribution is a bernulli $Bern(0.56)$

Its standard deviation is $\sigma=\sqrt{p(1-p)}$

A good estimation of $\sigma$ is

$$\hat{\sigma}=\sqrt{\overline{X}(1-\overline{X})}$$

This is your

$$s=\sqrt{0.56\cdot\left(1-0.56\right)}\approx0.4964$$


问:
Calculus book recommendation
I'm a high school student who has just finished learning single variable calculus(or calculus-2). I want to study some elementary(to not overburden myself) multivariable calculus which would be very useful in understanding the heart of electrodynamics and magnetism. I do not want to learn this directly from books like Griffiths where only the necessary stuff for physics is taught(I want to enjoy the math obviously!).

I went through some standard books like Hubbard's and Spivak's but they're just way too advanced.

Is there any such elementary multivariable calculus book that you think is apt for me?

答:
Try Calculus by James Stewart.

An amazing book, with easy-to-understand explanations, and some really good intuitive exercises. Don't forget to solve the Problems Plus:)

But if you go a little short on exercise problems, you can opt for Demidovich.

Title: Calculus book recommendation


问:
Interesting integral $\int_0^{2\pi}\tan(\cos(x))dx=0$
Consider the integral  $$I=\int_0^{2\pi}\tan(\cos(x))dx$$

I would like to show this integral is $0$ via elementary methods (excluding complex analysis, special functions, series representations).

The bounds of integration suggest some kind of symmetry argument to show that the integral vanishes.

I tried $x=\pi/2-u\implies dx=-du\implies$ $$I=-\int^{-\frac{3\pi}{2}}_{\frac{\pi}{2}}\tan(\sin(u))du$$ From here I don’t see a good route.

I also tried $$I=\int_0^{2\pi}\tan(\cos(x))dx=\int_0^{2\pi}\frac{\sin(\cos(x))}{\cos(\cos(x))}dx$$ Then let $t=\cos(\cos(x))\implies dt=-\sin(\cos(x))\cdot-\sin(x)=\sin(x)\sin(\cos(x)) \space dx$

Now $$I=\int_0^{2\pi}\frac{\csc(x)}{t}dt$$

The question now would be how to invert $t=\cos(\cos(x))$? But this obviously would be tough. Again, I think there’s a simple symmetry argument I’m missing. Can anyone help?

答:
We can deal with this kind of integral question by switching to the variable: $y=\pi+x$.$$I=\int_{0}^{2\pi} \tan(\cos(x))dx=\int_{-\pi}^{\pi} \tan(\cos(\pi+x))dx=-\int_{-\pi}^{\pi}\tan(\cos(x))dx$$
Antiderivatives of odd functions are even and vice versa. Since tangent function is odd, its antiderivative
is even. This boils down to the fact that: $$I=\int_{0}^{2\pi}\tan(\cos(x))dx=-\int_{-\pi}^{\pi} \tan(\cos(x))dx=0$$


问:
Fatou’s lemma equality

I found a sequence of functions that makes Fatou’s lemma be a strict unequality, but I’m not able to find one that makes it be an equality.

Can you help me to find one?

Thank you

答: $S=[0,1]$ , $\mu=$ Lebesgue measure and $f_n=1$ on $S$ for all $n$

Title: Fatou’s lemma equality


问:
How to rotate any plane in 3d to (XY) plane?
I want to rotate a given plane in 3d to (XY) so i can work like in normal 2d. Because if they ask like what is the equation of a circle on a given plane, i'll just rotate it to (XY) plane and apply the normal 2d formula. But i'm in final year of highschool and my teacher said that you don't need to make that complicated, but i want to know even if i didn't use it in a test. so if can anyone help me about it.
P.S: can you wrap this topic up in terms of cross product, dot product or some basic stuff not like rotation matrix or so, if you have to do it can you explain really well because like i said i'm in final year in highschool not college. thanks in advance</p>

答: A generic plane $\prod\in \mathbb{R}^3$ has a normal unit vector $\hat n$ so we need a rotation matrix $R(\theta,\hat u)$ which transforms $\hat n$ into $\hat e_z$. Taking $L$ as the intersection line between $\prod$ and the $XY$ plane, we can use Rodrigues formula around the line $L$ with direction $\hat u$ to perform a rotation of $\theta=\arccos\left(\hat n\cdot \hat e_z\right)$ Rodrigues formula


答:
The standard deviation of a sample proportion is
$$\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}.$$

The confidence interval should then be

$$\left(\hat{p} - Z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \, \hat{p} + Z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right).$$

Title: How to solve this Confidence Level Question?

Body:
A company has conducted a test on $400$ people. Out of $400$ people, $224$ people like chocolate. Find the percentage of people who like chocolate in the $99\%$ confidence interval.

The problem is the question doesn't give the standard deviation for the calculation.
I believe the mean is
$$\begin{align*}
 \frac{224}{400} &= 0.56\quad\text{(the mean)}\\[5pt]
N &= 400\\[5pt]
 Z &= \frac{2.576}{(99\% CI)} \\[5pt]
  4^n &= 262,144 \\[5pt]
CI &= \text{mean} \mp \frac{s \cdot Z}{\sqrt{N}} \\[5pt]
\end{align*}$$

How do I find the $s$ value?


问:
A Probability Question of Infinite situation
Consider this situation:

The computer will randomly output 1 or 0, the probability of each one is 1/2.

The computer will continuously output until the number of 1 reaches 100.

Now,the number of 0 will be 1 or 2 or 3 or ...

My question is the expected number of 0s.

(

There is another explanation:

Let N be the number of zeros before there are a hundred ones

I just want to know the expected value of N
)

Maybe it does not meet the definition, I just want to say----which number of 0 is the most possible?

I do this question with the binormial distribution,I think it's may be

$$\displaystyle\sum_{n=0}^{\infty} C_{100+n}^{n}\displaystyle\frac{n}{2^{100+n}}$$

However,I don't know how to deal with it.

On the other hand, I'm not sure , I think my method is strange...

答:
So basically, let’s formalize and generalize the problem.

Let $(X_i)$ be the bits the computer outputs. $(X_i) \sim B(1/2)$ iid.

We can modify the problem a bit and say (equivalently) $(X_i)$ is infinite, and we are looking for the number $N$ of zeroes before we see a hundred 1s

Actually let us generalize the problem a bit and consider the number

$N_n$ of zeroes before we reach n ones.
We are looking for $N_{100}$

let us use a recursive strategy on $N_n$ :

$N_1$ is pretty easy : it is the number of zeroes before the first 1, it follows a geometric law $\mathcal{G}(1/2)$

let us now look at $N_{k + 1}$ knowing $N_{k}$ :
fun thing is that $$N_{k + 1} - N_{k} | N_{k}$$ also follows a geometric law (it is the number of zeroes between the kth one and the (k+1)th one

afterward
$$\mathbb{E}(N_{k+1}) = \mathbb{E}(\mathbb{E}(N_{k+1} | N_{k})) \\
= \mathbb{E}(\mathbb{E}(N_{k+1} - N_k + N_k | N_{k}))\\
= \mathbb{E}(\mathbb{E}(N_{k+1} - N_k | N_{k}) + N_k) \\
= \mathbb{E}(N_k) + \mathbb{E}(\mathcal{G}(1/2))$$

From this, I think you can infer the end of the induction


问:
Find the inradius of a triangle situated within an ellipse

Question: $∆ABC$ is situated within an ellipse whose major axis is of length $10$ and whose minor
axis is of length $8$. Point $A$ is a focus of the ellipse, point $B$ is an endpoint of the minor axis
and point $C$ is on the ellipse such that the other focus lies on $BC$. Compute the inradius
of ∆ABC.

I am given the hint of the area formula involving the inradius $A = rs$.

First I call the other focus $D$. I know that $P=20$ and therefore $s=10$ because $AB+BD=10$ and $AC+CD=10$. So I have $A = rs = 10r$. If I can get $A$, I immediately have $r$ but I'm not sure how to get there.

I've figured out that the focal length ($AD$) is 3 and that $AB = BD = 5$, but it hasn't helped. Can someone point me in the right direction?

答:
Line $BC$ has equation $$\frac{x}{3}+\frac{y}{4}=1$$
Ellipse has equation
$$\frac{x^2}{25}+\frac{y^2}{16}=1$$
Intersection points are
$$B(0,4);\;C\left( \frac{75}{17},-\frac{32}{17}\right)$$
Perimeter of $\triangle ABC=20$. Half perimeter $p=10$

Area of $\triangle ABC;\;\mathcal{A}=\frac{300}{17}$

radius of inscribed circle is
$$r=\frac{A}{p}=\frac{30}{17}$$


答:
I'll be using coordinate geometry to arrive at my hint.
First off, we now that A is situated right at the focus of the ellipse. For our convenience, let us assume that it is the positive focus, (5,0).
Also, we know that B is situated at the endpoint of the minor axis. Again, for our convenience, let us assume that it is the positive endpoint, and so we fix B as (4,0).

Well, first of all, we know that the other focus lies on the line connecting B and C.
BUT, we also know that the other focus is (-5,0).

Find the equation of the line joining these points, and then substitute y into the equation of the ellipse.
You'll get a quadratic equation in x. Solve it, however tedious it is, get the value of (x,y).

Henceforth, use the distance formula to evaluate s,a,b,c.

Substitute them into your formula and solve the rest.

Title: Find the inradius of a triangle situated within an ellipse

Body:
Question: $∆ABC$ is situated within an ellipse whose major axis is of length $10$ and whose minor
axis is of length $8$. Point $A$ is a focus of the ellipse, point $B$ is an endpoint of the minor axis
and point $C$ is on the ellipse such that the other focus lies on $BC$. Compute the inradius
of ∆ABC.

I am given the hint of the area formula involving the inradius $A = rs$.

First I call the other focus $D$. I know that $P=20$ and therefore $s=10$ because $AB+BD=10$ and $AC+CD=10$. So I have $A = rs = 10r$. If I can get $A$, I immediately have $r$ but I'm not sure how to get there.

I've figured out that the focal length ($AD$) is 3 and that $AB = BD = 5$, but it hasn't helped. Can someone point me in the right direction?


问:
Show that $f$ is Riemann-Integrable
Let $f$ be a function defined on $[0,1]$ by: $f(x)=1$ if $x=\frac{1}{n}, n \in \mathbf{N}^*$ and $f(x)=0$ if not. Show that $f$ is Riemann integrable on $[0,1]$.

I know that there is already a post on this function and this problem might not be difficult, but I didn't really find an answer to my question. I'm not really good at proving if a function is integrable, so I would like to understand how it works correctly and become good x)

To show that $f$ is Riemann intergable on $[a,b]$, we consider the upper and lower Darboux sum. By definition we have to show the following: $\exists$ a subdivision $\sigma$: $\overline{S}_{\sigma}(f)\le \underline{S}_{\sigma}(f)+\epsilon \ \forall \epsilon >0$. By densition of irrational numbers, we have that $\underline{S}_{\sigma}(f)=0 \ \forall \sigma$. So, we have to show that $\overline{S}_{\sigma}(f)\le\epsilon$.

I remark that there exists $\epsilon$ such that there is a finite number of discontinuities on the interval $[\epsilon,1]$, so probably I can bound the things in the sum quite "easily". My problem is how to choose my subdibision correctly (and to be sure that it's correct) and how to work correctly on the interval $[0,\epsilon]$ as more we approach to 0, more there are of discontinuities (because of density of irrational numbers). I ask this question not only for this case, but in general how it works. If someone could explain these things in details, I would really appreciate it. Thank you very much in advance!

答:
Hint

For $0 \lt \epsilon \lt 1$, consider the subdivision

$$\begin{aligned}\sigma \equiv x_ 0 &= 0 \lt x_1 = \frac{\epsilon}{2}\\&\lt x_2 = \frac{1}{N} -\frac{\epsilon}{2 2^{N+1}} \lt x_3 = \frac{1}{N} +\frac{\epsilon}{2 2^{N+1}}\\
&\dots\\
&\lt x_{N} = \frac{1}{2} -\frac{\epsilon}{2 2^{2}} \lt x_3 = \frac{1}{N} +\frac{\epsilon}{2 2^{2}}\\
&\lt x_{N+2} = \frac{1}{2^0} -\frac{\epsilon}{2 2^{1}} \lt x_{N+3} = 1
\end{aligned}$$

Where $N$ is the largest integer such that $\frac{1}{N}\gt \epsilon / 2$.

To understand why that works, I suggest that you take an example, i.e. $\epsilon = 1/10$, and make a drawing of the subdivision vs. the function. From there, you can compute the Darboux lower and upper sum to complete the proof. You'll essentially see that Darboux lower sum is equal to zero while Darboux upper sum is less than

$$\frac{\epsilon}{2} + \frac{\epsilon}{2}\left(\frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{N+1}}\right) \lt \epsilon.$$

The idea, as you have seen is to isolate zero on one side from the other discontinuities of the function.

Title: Show that $f$ is Riemann-Integrable


答:
From $a+b+c=0$, we can replace every occurrence of $a$ with $-b-c$, turning this into a two variable problem.  Then, after simplifying the various expressions, we find
\begin{align*}
\frac{(-b-c)^2+b^2+c^2}{2} &= b^2+bc+c^2  \\
\frac{(-b-c)^3+b^3+c^3}{3} &= -b^2c-bc^2  \\
\frac{(-b-c)^5+b^5+c^5}{5} &= -b^4c -2b^3c^2 -2b^2c^3 -bc^4
\end{align*}
It is straightforward to show the product of the first two is the third.

Title: $a + b + c = 0$ Prove that $\frac{a^2 + b^2 + c^2}{2} * \frac{a^3 + b^3 + c^3}{3} = \frac{a^5 + b^5 + c^5}{5}$

Body:
The question says: $a + b + c = 0$

Prove that $$\frac{a^2 + b^2 + c^2}{2} * \frac{a^3 + b^3 + c^3}{3} =
\frac{a^5 + b^5 + c^5}{5}$$

So I started with simplifying each part of expression

$$ \frac{a^2 + b^2 + c^2}{2} = \frac{(a+b+c)^2 -2ab-2ac-2bc}{2} =
-(ab + ac + bc)$$
Because $a+b+c=0$

$$ \frac{a^3+b^3+c^3}{3} = \frac{(a+b+c)^3 - 3a^2b-3ab^2-3a^2c-3ac^2-3b^2c-3bc^2-6abc}{3}
 = -(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2 -2abc) $$

So, my question is, what can I do with the second part? Like $\frac{(a+b+c)^2(a+b+c)^3 - ...}{5}$


问:
Showing that a sequence $\left(x_n\right)_{n \in \mathbb{N}}$ in a metric space is bounded iff the set of its elements is bounded
Let $(X, d)$ be a metric space, $\left(x_n\right)_{n \in \mathbb{N}}$ a sequence in $X$ and $E = \{x_n \mid n \in \mathbb{N}\}$. I want to show that $\left(x_n\right)_{n \in \mathbb{N}}$ is bounded iff $E$ is bounded. So far I can show the "if" part, but I am stuck at the "only if".

To show "if", suppose $\left(x_n\right)_{n \in \mathbb{N}}$ is bounded, i.e. $\exists z \in X, r > 0: \forall n \in \mathbb{N}: x_n \in \mathcal{B}_r(z)$. Let $x_k, x_l \in E$. Then,
$d(x_k, x_l) \leq d(x_k, z) + d(z, x_l) \leq r + r \Longleftrightarrow \sup \{d(x_k, x_l)\mid k, l \in \mathbb{N}\} \leq 2r < \infty$. Hence $\mathrm{diam}(E) \leq 2r < \infty$, that is $E$ is bounded.

To show the "only if", suppose  that $\mathrm{diam}(E) = C < \infty$. Let $z \in X$ and set $r = \mathrm{dist}(z, E) + \mathrm{diam}(E)$, so that the ball centered at $z$ can reach every element of $E$. Again, let $x_k, x_l \in E$, so that $d(x_k, z) \leq d(x_k, x_l) + d(x_l, z)$. (And from here on one must show that $d(x_k, z) \leq r$).

So, currently I am stuck at finishing the last inequality. How should I proceed?

答:
You are almost done:
Observe that
$$d(x_k,x_l)\leq \mathrm{diam}(E)$$
holds (by using the definitions of $\mathrm{diam}$).
Thus, we have
$$
d(x_k,z) \leq \mathrm{diam}(E)+ d(x_l,z).
$$
By taking the infimum over $x_l\in E$, we get
$$
d(x_k,z)= \inf_{x_l\in E} d(x_k,z) \leq \mathrm{diam}(E)+ \inf_{x_l\in E} d(x_l,z)
= \mathrm{diam}(E)+ \mathrm{dist}(z,E) = r.
$$


问:
Prove that the following statements are equivalent (isometric)
Let $\phi: \mathbb{R}^2 \to \mathbb{R}^2$ be a lineair mapping.
Prove that the following statements are equivalent:

$\phi$ is isometric.

For all $x \in \mathbb{R}^2$ the following holds: $|\phi(x)| = |x|$ (distance to (0,0))

For all $x,y \in \mathbb{R}^2$ the following holds: $\langle \phi(x),\phi(y) \rangle = \langle x,y\rangle$ (scalar product)

To prove the equivalence I want to prove the following implications:

$1 \Rightarrow 2$, $2 \Rightarrow 3$ and $3 \Rightarrow 1$.

I have no difficulties proving the first implication. Because $\phi$ is a linear mapping $(0,0)$ stays at its place and because $\phi$ is isometric the distance from $(0,0)$ to $x$ and $\phi(x)$ must be the same.

The latter two I don't quite understand.

Is anybody willing to give me some tips to solve this problem?

答:
If you always have $\langle\phi(x),\phi(x)\rangle=\langle x,x\rangle$, then, if $x,y\in\Bbb R^2$,$$\langle\phi(x+y),\phi(x+y)\rangle=\langle x+y,x+y\rangle.\tag1$$But\begin{align}(1)&\iff\langle\phi(x)+\phi(y),\phi(x)+\phi(y)\rangle=\langle x,x\rangle+2\langle x,y\rangle+\langle y,y\rangle\\&\iff\|\phi(x)\|^2+2\langle\phi(x),\phi(y)\rangle+\|\phi(y)\|^2=\|x\|^2+2\langle x,y\rangle+\|y\|^2\\&\iff\langle\phi(x),\phi(y)\rangle=\langle x,y\rangle.\end{align}

On the other hand if you always have $\langle\phi(x),\phi(y)\rangle=\langle x,y\rangle$, then, in particular, $\|\phi(x)\|^2=\|x\|^2$, and therefore $\|\phi(x)\|=\|x\|$. And so$$\|\phi(x)-\phi(y)\|=\|\phi(x-y)\|=\|x-y\|,$$and so $\phi$ is an isometry.


问:
Let $p:\tilde{G}\to G$ be a universal cover. Then who are the outer automorphisms $\operatorname{out}(G)$ in terms of $\operatorname{out}(\tilde{G})$?
For example, let $p:\operatorname{SU}(3)\to G$ be a universal cover. May I say that the outer automorphisms of $G$ are of the form $p(g)\mapsto p(\overline{g})$, since the outer automorphisms of $\operatorname{Out}\operatorname{SU}(3)=\langle \sigma\rangle$, sigma being the complex conjugation?

In general, is there any simple way to write $\operatorname{Out}(G)$ in terms of $\operatorname{Out}(\tilde{G})$?

We have that if $\varphi\in\operatorname{Out}(G)$, as $\tilde{G}$ is simply connected, $p\circ\varphi$ lifts to a homomorphism $\tilde{\varphi}:\tilde{G}\to\tilde{G}$. But $\ker \tilde{\varphi}=p^{-1}(1)$. Is there any way to turn this $\tilde{\varphi}$ into an automorphism? In my specific case I may assume that $G=\tilde{G}/Z(\tilde{G})$.

答:
Let me assume $G$ is connected so that all the lifts make sense and are uniquely defined and group morphisms.

$\ker\tilde\varphi $ is not $p^{-1}(1)$ ! In fact $\tilde\varphi$ is always an automorphism.

Indeed, you can also lift $\psi = \varphi^{-1}$ to $\tilde \psi$ and $\tilde \psi\tilde\varphi$ is a lift of $id_G$ which sends $1$ to $1$ so it must be $id_{\tilde G}$, and similarly for $\tilde\varphi\tilde \psi$.

In fact, this clearly provides a group monomorphism $Aut(G)\to Aut(\tilde G)$. Now if $\varphi$ is inner, say $x\mapsto gxg^{-1}$, and if $g_0$ is a lift of $g$, then $y\mapsto g_0yg_0^{-1}$ is a lift of of $\varphi$ (which in fact doesn't depend on the choice of $g_0$), so it factors as a morphism $Out(G)\to Out(\tilde G)$.

In fact, because $\ker p$ is central in $\tilde G$, $Inn(G)\to Inn(\tilde G)$ is an isomorphism, so this map on outer automorphism groups is also injective.

In the specific case you describe, $G = \tilde G/Z(\tilde G)$, since $Z(\tilde G)$ is not only normal but characteristic, that is, it's preserved by arbitrary automorphisms, it follows that in fact, $Aut(G)\to Aut(\tilde G)$ is an isomorphism.

More generally, I'm not sure what can be said; e.g. if we look at $\mathbb R\to S^1$, then  $Aut(\mathbb R) = \mathbb R^\times$ (I'm thinking in terms of continuous automorphisms for everything to make sense of course), whereas $Aut(S^1) = C_2$, the injection being given by $-1\mapsto -1$ (if we view $C_2$ as $\{\pm 1\}$)


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答:
Skolem’s Paradox:

From the Wikipedia article: “Skolem’s paradox is that every countable axiomatisatin of set theory in first-order logic, if it is consistent, has a model that is countable.”

Here’s the link:

http://en.wikipedia.org/wiki/Skolem%27s_paradox

LastEditorDisplayName: anon


问:
Exact short sequence

Given a subspace $V_1$ of $V$, then
$$
\{0\} \xrightarrow{} V_1 \xrightarrow{i}  V \xrightarrow{\pi_{V/ V_1}}  V/V_1
\xrightarrow{} \{0\}
$$
is a exact short sequence.

Q: Is this something that I have to prove? What exactly do I have to prove?

答: For a sequence $A \overset{f}{\rightarrow} B \overset{g}{\rightarrow}C$, being exact in B means that $\operatorname{Ker}(g)=\operatorname{Im}(f)$. So you have to prove that your sequence is exact in $V_1$, $V$ and $V/V_1$, i.e. that the map $i$ is injective, the projection $\pi_{V/V_1}$ is surjective and that the image of the latter is equal to the kernel of the former. It is quite obvious but if you have any doubt, just prove it!

Title: Exact short sequence


问:
Matsubara sum arising from QFT and contour integral
In the lecture of E. Fradkin on quantum field theory, an example of Matsubara sum is performed using contour integration (see eq. 5.214 in the lecture). It reads
$$
\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2}
= \oint_{C_{+}\cup C_{-}} \frac{dz}{2\pi i} \frac{e^{iz\tau}}{z^2+x^2} \pi\cot(\pi z)\quad(\tau, x>0),
$$
where $C_{\pm}$ denotes the lines $z\pm i\epsilon$ followed counter clockwise (see this figure). I see that above equation can be deduced applying residue theorem to the finite box, obtained by truncating $C_{+}\cup C_{-}$, and taking a limit.

To evaluate the right hand side, the author deforms the contour into $C^{+}\cup C^{-}$ and calculate instead residues at $z=\pm ix$. I am curious about the mathematical justification of this process.

To me, equating $\oint_{C_{-}}(\text{integrand})$ with residue at $-ix$ seems illegitimate since $e^{iz\tau}$ blows up as $z\rightarrow -i\infty$. I don't get what 'deforming $C_{-}$ to $C^{-}$' even means.

One approach that I tried is the change of variable $z\rightarrow 1/z$, which amounts to passing to the other coordinate chart of Riemann sphere. It goes like
$$
(RHS)=\oint_{\Gamma} \frac{dz}{2\pi i} \frac{e^{i\tau/z}}{1+z^2x^2} \pi\cot(\pi/z),
$$
where $\Gamma$ is the image of $C_{+}\cup C_{-}$ followed counter clockwise. The problem here is twofold; first, $\Gamma$ passes through the essential singularity at $z=0$. Second, if $C_{+}\cup C_{-}$ is taken to be the limit of boxes $B_n$ and $\Gamma$ the limit of their images $B'_n$, each contour $B'_n$ encircles an infinite subset of the poles $\{\pm 1/n: n\in\mathbb{Z}\}$. I have no idea if I can apply the residue theorem and take a limit in this case.

edit

Following Svyatoslav's suggestion, I obtained (for $0<\tau<2\pi$)
$$
\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2}
= \oint_{C_{+}\cup C_{-}} \frac{dz}{2\pi i} \frac{e^{iz\tau}}{z^2+x^2} \frac{\pi e^{-i\pi z}}{\sin\pi z}
= \frac{\pi}{x} \frac{\cosh(\pi-\tau)x}{\sinh(\pi x)} \\
= \frac{\pi}{x} \left[ \cosh(\tau x) \coth(\pi x) - \sinh(\tau x) \right].
$$
However, it is claimed in (5.215) of the lecture that
$$
\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2}
\approx \frac{\pi}{2x}\coth(\pi x) e^{-|\tau|x} \quad \text{for small } \tau.
$$
Did the author make a mistake?

答:
This is a duplicate of what I wrote on Physics Stack exchange. I hope this is OK because there is a separate readership here.

You can also derive @Svyatoslav's correct expression   by Poisson Summation:
$$
 \frac 1{2\pi} \sum_{n=-\infty}^\infty \frac{e^{in\tau}}{n^2+M^2}=
   \sum_{n=-\infty}^\infty   \frac 1{2|M |} e^{-|M||\tau+2\pi n|}, \quad \hbox{(Poisson Summation)}\\
   =  \frac 1 {2M} \frac{\cosh(\pi -\tau)M}{\sinh \pi M}, \quad 0<\tau<2\pi,\nonumber
$$
The first line come from applying Poisson summation to the zero temperature expression

$$
\int_{-\infty}^{\infty} \frac{dk}{2\pi}\frac{e^{ik\tau}} {k^2+M^2}=\frac 1 {2|M|}e^{-|\tau||M|}
$$
and has the  physical interpretation as  the method-of-images sum over the  $n$-fold winding of the particle trajectory around the periodic imaginary time direction.
The passage from the first  to second  lines is just summing  the two  geometric series from  $n=0$ to $ \infty$ and $n=-\infty$ to $-1$.


问:
Solve $x^2+2=y^3$ using infinite descent?
Just so this doesn't get deleted, I want to make it clear that I already know how to solve this using the UFD $\mathbb{Z}[\sqrt{-2}]$, and am in search for the infinite descent proof that Fermat claimed to have found.

I've alaways been fascinated by this Diophantine equation $x^2+2=y^3$ in particular ever since I saw it, and I still have no clue how to attack it without $\mathbb{Z}[\sqrt{-2}]$. What's disappointing is that no one else seems interested in the hunt (an elementary proof using infinite descent). I know it's been studied extensively, and there have even been generalizations, such as Mordell's equation. However, I've never seen Fermat's original proof that $(x,y)=(\pm 5, 3)$ is the only integer solution. Obviously, Fermat probably knew nothing of UFD's, which is why I believe there has to be an infinite descent proof like he claimed. Has anyone apart from him actually seen this proof? People mention it all the time, yet I can't find anything about it. As I said, I know that it involved infinite descent, but I've never seen it anywhere and no one seems to have any idea about it.

Does anyone have ideas for this approach? I mean, infinite descent seems more effective for showing a contradiction, e.g. showing there are no solutions. But how could it work here? Also, why isn't it published anywhere in all this time? Could it really be that only Fermat knew his method of descent well-enough to make this problem submit to it?

Thanks!

答:
A definitive (?) answer to your question can be found on page 561 of the Nov 2012 edition of The Mathematical Gazette, where a totally elementary descent mechanism first used by Stan Dolan in the March 2012 edition is adapted (as per his challenge to the “interested reader”) to solve Fermat’s two “elliptic curve” theorems. The method uses math which was clearly available in Fermat’s time, and in particular to Fermat himself.

I personally believe this finally puts to rest any questions of whether Fermat could have had a proof of these two claims.


问:
Prove that a polynomial of degree $d$ has at most $d$ roots (without induction)

Let $p(x)$ be a non-zero polynomial in $F[x]$, $F$ a field, of degree $d$. Then $p(x)$ has at most $d$ distinct roots in $F$.

Is it possible to prove this without using induction on degree? If so, how can I do this?

答:
If $p(x)=0$ has a root $x=a$, then $p(a)=0$ and $p(x)=p(x)-p(a)$.

$p(x)-p(a)$ is the sum of terms like $c_rx^r-c_ra^r=c_r(x-a)(x^{r-1}+ax^{r-2}+a^2x^{r-3}+ \dots +a^{r-1})$

And since $x-a$ is a factor of every term, it is a factor of $p(x)$. So every root gives us a linear factor.

Suppose $p(x)=k(x-a_1)(x-a_2) \dots (x-a_d)$ of degree $d$ has the $d$ roots $a_1 \dots a_d$ and $b$ is distinct from all of these, then $p(b)$ is a product of non-zero factors, so cannot be equal to zero.

Is this a proof without induction? Difficult to say. For example, how do we prove that if $r$ divides each of $e_1, e_2 \dots e_n$ then it divides their sum?

But the division step, with $p(x)=(x-a)q(x)$ and $q(x)$ having strictly lower degree than $p(x)$ leads to a descending sequence of integers (the orders of the polynomials obtained by successive divisions) - and what we need to know for that is that (i) any strictly descending sequence of non-negative integers is finite; and (ii) we can bound the length of the sequence by $d$ - and we can do this by observing that there are $d$ non-negative integers less than $d$.

So it all depends on the properties of integers that we are allowed to assume - and that is because we need to say things about the degree of $p(x)$ - an integer, and we also need to do things like indexing the coefficients.


问:
The Diophantine equation $x^2 + 2 = y^3$
How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y>0$ ?
($x,y$ are integers.)</p>

答:
A completely elementary proof can be found on page 561 of the Nov 2012 edition of The Mathematical Gazette, where a descent mechanism first used by Stan Dolan in the March 2012 edition is adapted (as per his challenge to the “interested reader”) to solve Fermat’s two “elliptic curve” theorems. The method uses math which was clearly available in Fermat’s time, and in particular to Fermat himself.

I personally believe this finally puts to rest any questions of whether Fermat could have had a proof of these two claims.


问:
Weird condition implies isomorphism between vector spaces?
Let $U$ and $V$ be vector spaces. It can be shown that the existence of atleast one space $W$ such that $\mathscr L(U,W) \cong \mathscr L(V,W)$ is not sufficient for $U$ and $V$ to be isomorphic as, for example, if we consider $\mathbb R[X]$ as the space of all real polynomials, it is true that $End(\mathbb R[x]) \cong \mathscr L(\mathbb R,\mathbb R[X])$ although it is obviously not isomorphic to $\mathbb R$. My question is: If we strengthen the condition and require that for ANY vector space $W$ we have that $\mathscr L(U,W) \cong \mathscr L(V,W)$, does that imply that $U \cong V$? This is a way more powerful restriction and so I can imagine the proposition would hold, specially since it seems almost impossible to think of any possible counter-examples to this. What seems like a plausible thing to do is to try to extract some argument about the cardinality of the basis of $U$ and $V$, but it's not entirely clear to me  how to do that.

答:
this is an example of "Yoneda lemma". it is a general principle wich says that $HOM(X,-)$(or $Hom(-,X)$) is always an injection from any category to the category of sets.so if there are functorial isomorphism between $Hom(U,W)=Hom(V,W)$ for each $W$ then $U,V$ are isomorphism.(functorial means that the isomorphism are compatible with maps $W\to W'$)

for the proof just take $W=V,U$ you get a map $U\to V$ and a map in the inverse direction(by taking inverse image of identity) and functoriality says that this morphism are inverse to each other


问:
Prove that a nonempty hypersurface in $\Bbb A^n_k$ has at least one non-singular point.

Let $k$ be an algebraically closed field. Prove that a nonempty hypersurface in $\Bbb A^n_k$ has at least one non-singular point

My work is that i'm using the Hilbert’s Nullstellensatz , but i have no idea what to do next.

答:
This is proved in Hartshorne's book Algebraic Geometry GTM-52, Theorem I.5.3. If $Y \subseteq \mathbb{A}^n_k$ is an affine variety it follows the set of singular points $Y_{sing} \subseteq Y$ is a proper closed subset of $Y$. In partucular it holds for a hypersurface $H:=V(f)$.

If the ideal $I$ defining $Y$ is generated by the polynomials $f_1,..,f_l$,
the proof in HH.I.5.3 shows that the ideal of $Y_{sing}$ is defined using the
jacobian matrix of the polynomials $f_j$. This is used to prove that $Y_{sing} \subsetneq Y$ is a proper subset. Let me cite the proof for hypersurfaces: If $Y:=V(f) \subseteq \mathbb{A}^n_k$ and $char(k)=0$ and $Y_{sing}=Y$ it follows

$$\partial f/\partial_{x_i} \in (f)$$

but $deg(\partial f/\partial_{x_i}) =deg(f)-1$ hence $\partial f/\partial_{x_i}=0$ - a contradiction since $deg(f) \geq 1$ and $char(k)=0$.


问:
Upper bound for dependent variables
If i have a set of $n$ dependent variables $\{n_1,...n_n\}$

The probability of success for every one of those variables is equal to $k$ but if there is one variable that succeeded the probability that another variable will succeed is $k'<k$.

Is there any problem if i assume that the variables are indenpendent and follow a binomial distribution to derive an upper bound on the probability of $x$ variables succeed?

答:
So the idea is (roughly) to use stochastic domination.
Consider the sequence $(X_1,\dots,X_n)$ of random variables that you are considering. Consider also a sequence $(Y_1,\dots,Y_n)$ of random variables of parameter $(m,k)$ ($m$ tries, $k$ probability of success for each try). You have
$$ \mathbb{P}(\sum_{i=1}^n X_i \ge x) \le \mathbb{P}(\sum_{i=1}^n Y_i \ge x) = \mathbb{P}(Z_n \ge x), $$
where $Z_n = \sum_{i=1}^n Y_i$.

The last quantity in the RHS can be computed explicitely: $Z_n$ is a binomial r.v. of parameter (mn, k). Now my inequality above is not fully obvious, see below if you need the details.

Bonus track : to justify this with a coupling, let $U$ be a uniform random variable on $(0,1)$ and $k' < k$. Consider
$$ X = \mathbf{1}_{U \le k'}, \qquad Y = \mathbf{1}_{U \le k}. $$
With this construction, $X \sim \mathcal{B}(k')$, $Y \sim \mathcal{B}(k)$ and $X \le Y$ for all $\omega$ in your probability space (i.e. for all realization of $U$). Now you can build

First of whole sequence $A^1, \dots, A^m$ and $C^1,\dots, C^m$ that you sum to get your binomial $X_1 \sim \mathcal{Bin}(m,k')$ and $Y_1 \sim \mathcal{Bin}(m,k)$. As $A^i \le C^i$ for all $i$, $X_1 \le Y_1$.

Once this is clear, you get your whole sequence of $(X_i)_{i \ge 1}$ and $(Y_i)_{i \ge 1}$ using this and obtain exactly the sequence $(X_1,\dots,X_n)$ and $(Y_1,\dots,Y_n)$ desired, with $X_j \le Y_j$ for all $j \ge 1$, which fully justifies my computation.

Title: Upper bound for dependent variables


问:
Bayes theorem tree diagram

A bag I contain $4$ white and $6$ black balls while another Bag II contains $4$ white and $3$ black balls. One ball is drawn at random from one of the bags, and it is found to be black. Find the probability that it was drawn from Bag $I$.

By using NLP, I can detect spam e-mails in my inbox. Assume that the word ‘offer’ occurs in $80$% of the spam messages in my account. Also, let’s assume ‘offer’ occurs in $10$% of my desired e-mails. If $30$% of the received e-mails are considered as a scam, and I will receive a new message which contains ‘offer’, what is the probability that it is spam?

I have drawn tree digram for both and it works for the first question but not for the second question

Can you explain it?

Thank you!

答:
My tree diagram looks like the following:

Then it is asked for $P(S|O)=\frac{P(O\cap S)}{P(O)}=\frac{0.24}{0.24+0.07}$. It is all comprehensible? If not, feel free to ask.

Title: Bayes theorem tree diagram


答:
The surprising thing is not that the second calculation is wrong, but that the first calculation is right.

You shouldn't be using probabilities in the diagram.  Instead, you should be counting events.  Let's look at the second diagram first. The idea is that we receive $100$ typical emails, so that $30$ are spam and $70$ are desired.  Of the $30$ spam emails, $80\%$, or $24$ contain the word "offer" and $6$ do not.  Of the $70$ desired emails, $7$ contain the word "offer" and $63$ do not.  We have received $31$ emails containing the word "offer" and $24$ of them are spam.  We conclude that the probability that an email is spam, given that it contains the word "offer" is $\frac{24}{31}$.

The first problem is peculiar.  You are interpreting the phrase "A ball is drawn at random from one of the bags," to me that each of the $17$ balls has an equal probability of being selected.  So, if we do this $17$ times, we'll draw $9$ black balls, $3$ from bag II and $6$ from bag $I$ and $$\Pr(I|B)=\frac23$$ if we ignore the denominators in the diagram.  How come the book says the answer is $\frac7{12}$?  I believe they intend that one of the bags be chosen at random, and then a ball be chosen at random from that bag.  Bayes' rule rule gives $$\begin{align}
\Pr(I|B)&=\Pr(B|I)\frac{\Pr(I)}{\Pr(B)}\\
&=\frac{\Pr(B|I)\Pr(I)}{\Pr(I)\Pr(B|I)+\Pr(II)\Pr(B|II)}\\
&=\frac{\frac{6}{10}\cdot\frac12}{\frac12\cdot\frac6{10}+\frac12\cdot\frac37}\\
&=\frac{\frac{6}{10}}{\frac6{10}+\frac37}=\frac{42}{42+30}=\frac7{12}\\
\end{align}$$

Title: Bayes theorem tree diagram

Body:
A bag I contain $4$ white and $6$ black balls while another Bag II contains $4$ white and $3$ black balls. One ball is drawn at random from one of the bags, and it is found to be black. Find the probability that it was drawn from Bag $I$.

By using NLP, I can detect spam e-mails in my inbox. Assume that the word ‘offer’ occurs in $80$% of the spam messages in my account. Also, let’s assume ‘offer’ occurs in $10$% of my desired e-mails. If $30$% of the received e-mails are considered as a scam, and I will receive a new message which contains ‘offer’, what is the probability that it is spam?

I have drawn tree digram for both and it works for the first question but not for the second question

Can you explain it?

Thank you!


问:
How to "explain in words" this expression involving liminf
Let $\Omega\subset\mathbb{R}^N$ and $F:\Omega\times\mathbb{R}\to\mathbb{R}$ be a function and consider the expression
$$ \liminf_{| u|\to+\infty} \frac{F(x, u(x))}{|u|^p +| u|^q} <0,$$
for some $p, q >1$. I am trying to "explain in words" this expression in a easy-to-understand way.

Could anyone please suggest me something?

I was thinking something like

"No matter how far you look toward infinity, there will be $|u|$ such that $F(x, u)$ is a negative quantity"

but I am not totally convinced.

Could anyone please help? Thank you in advance!

答:
Your sentence is weaker than what the limit is saying.  As an illustration, $-1/x$ on $x \in (0,\infty)$ has all its values negative, but its limit infimum is zero.

It is generally useful to speak in terms of "tail"s when talking about limits infimum and supremum.  Here, a "tail" is: for each choice of $u_0$, restrict the limit as $|u| \rightarrow \infty$ to $|u| > u_0$.  Taking the limit on the "complement of a bounded ball" might be more descriptive than "tail", depending on your setting.

For the limit infimum to be negative...

There is a quantity, $m < 0$ such that for each tail of the limit there are infinitely many choices of $(x, u(x))$ in that tail having $F(x, u(x)) < m$.

Note that the limit requires you to sort by magnitude(s) of $u$, but there is no restriction on the choice of $x$ among those $u(x)$ having the same magnitude.  So $x$ is part of your choice for points giving the various infima.

We can probably take this a step further...

All tails of the limit have infinitely many $(x,u(x))$ where $F$ is bounded away from and below zero.

"Bounded away from" captures the existence of $m$ in the prior version.  It means there is a gap between zero and all the values of $F$ from the called out $(x,u(x))$ points.


问:
Linearize equation
I have the following equation. I need to write it in a linear form $y=mx$.

$$M(x) = N \left(3-\exp\left(-\frac{x}{c}\right)\right)$$

Here, $M$ is the dependent variable and $x$ is the independent variable.  $N$ is a constant. My attempt so far:

$$\ln(M) = \ln (N ) + \ln(3 - \exp(-\frac{x}{c}) )$$

Thank you.

答:
you can rewrite your function as

$$\frac{M-3N}{N}=e^{-x/c}$$

that is

$$L=e^{-x/c}$$

Than take the log on both sides getting

$$\log L=-\frac{x}{c}$$

that is

$$y=-\frac{1}{c}\cdot x$$

...as you desired

Title: Linearize equation


答: Suppose that $\textrm{diam}(E) = C < +\infty$, so $E$ is bounded in the diameter sense, let $z=x_1$ and $r= C+1 >0$. Then I claim that all points of $(x_n)_{n \in \Bbb N}$ are in $B_r(z)$: Let $x_m$ be arbitrary, then $d(x_1, x_m) \le \textrm{diam}(E)$ as the diameter is an upper bound for all the distances between members of $E$. In particular $d(z,x_m) < r$ and so $x_m \in B_r(z)$ and we're done.

Title: Showing that a sequence $\left(x_n\right)_{n \in \mathbb{N}}$ in a metric space is bounded iff the set of its elements is bounded

Body:
Let $(X, d)$ be a metric space, $\left(x_n\right)_{n \in \mathbb{N}}$ a sequence in $X$ and $E = \{x_n \mid n \in \mathbb{N}\}$. I want to show that $\left(x_n\right)_{n \in \mathbb{N}}$ is bounded iff $E$ is bounded. So far I can show the "if" part, but I am stuck at the "only if".

To show "if", suppose $\left(x_n\right)_{n \in \mathbb{N}}$ is bounded, i.e. $\exists z \in X, r > 0: \forall n \in \mathbb{N}: x_n \in \mathcal{B}_r(z)$. Let $x_k, x_l \in E$. Then,
$d(x_k, x_l) \leq d(x_k, z) + d(z, x_l) \leq r + r \Longleftrightarrow \sup \{d(x_k, x_l)\mid k, l \in \mathbb{N}\} \leq 2r < \infty$. Hence $\mathrm{diam}(E) \leq 2r < \infty$, that is $E$ is bounded.

To show the "only if", suppose  that $\mathrm{diam}(E) = C < \infty$. Let $z \in X$ and set $r = \mathrm{dist}(z, E) + \mathrm{diam}(E)$, so that the ball centered at $z$ can reach every element of $E$. Again, let $x_k, x_l \in E$, so that $d(x_k, z) \leq d(x_k, x_l) + d(x_l, z)$. (And from here on one must show that $d(x_k, z) \leq r$).

So, currently I am stuck at finishing the last inequality. How should I proceed?


问:
How to interpret that a distribution does not have defined entropy (or has infinite entropy)?
An entropy (is Shannon sense) can be interpreted as uncertainty or missing knowledge. When the knowledge is added, the entropy decreases. Hence it can also be interpreted as information content.

However, there are discrete distributions which do not have an entropy because $-\sum_{i=1}^{\infty}p_i\log p_i$ tends to infinity (see here).

How to interpret this situation? Does non-existent or infinite entropy means that it is not possible to remove uncertainty and get complete knowledge about values of variable described by such distribution?

答:
A discrete distribution (with infinite outcomes) can have an infinite entropy. I don't see much to interpret here. Actually, one might argue that, on the contrary, the surprising thing is that some (most) discrete distributions with infinite outcomes have finite entropy, i.e. they produce  a finite amount of information, so they can be described (in average) with a finite number of bits.

I think that the most natural example of infinite entropy is a continuous distribution (say, uniform on $[0,1]$), which can be regarded as the limit of a discrete uniform distribution for increasingly large number of values. It's not surprising that a real number in $[0,1]$ has infinite entropy, because to describe it you need an infinite number of bits (think of its binary fractional representation). Put in other way, in a real number in $[0,1]$ you can code an arbitrarily big amount of information.


问:
Is the proof to a recursive sequence $a_{n+1}=1+\frac{1}{5}a_n^2$ correct?
We have a recursive sequence:
$$a_1 = 1$$
$$a_{n+1}=1+\frac{1}{5}a_n^2$$
We need to show that the sequence is convergent. We also need to analyse what would happen if we would pick a different first term $a_n$

What I did:

$$
a_{n+1} \geq a_n\\
1+\frac{1}{5}a_n^2 \geq a_n\\
a_n^2-5a_n+5\geq 0\\
(a_n-\frac{5-\sqrt5}{2})(a_n -\frac{5+\sqrt5}{2})\geq 0\\
\implies$$
$\text{so this implies that if all the terms are bellow}$ $\frac{5-\sqrt5}{2}$ or above $\frac{5+\sqrt5}{2}$ then the sequence is increasing

We see that $a_1 = 1$ is bellow $\frac{5-\sqrt5}{2}$, so let's prove that $a_n \leq \frac{5-\sqrt5}{2}$ by induction:

$n = 1:$
$$a_1 = 1 \leq \frac{5-\sqrt5}{2}$$
$$\text{holds}$$
$n \rightarrow n+1:$
$$
\frac{5-\sqrt5}{2} \geq a_{n+1} \\
\frac{5-\sqrt5}{2} \geq 1+\frac{1}{5}a_n^2\\
\frac{15-5\sqrt5}{2} \geq a_n^2\\
\text{for this to hold $a_n$ should not be between the positive and the negative root.}\\
\text{But we see that $\frac{5-\sqrt5}{2}$ is more than the $-\sqrt{\frac{15-5\sqrt5}{2}}$ sp this means that the sequence isn't increasing.}
$$

I do not know how to prove this, or if my process is correct and I made a mistake somewhere.

答:
$a_1=1<\frac{5-\sqrt{5}}{2}$

suppose now that $a_k\le \frac{5-\sqrt{5}}{2}$

$$a_{k+1}=1+\frac15 a_k^2\le 1+\frac15 \cdot\left( \frac{5-\sqrt{5}}{2}\right)^2=\frac{5-\sqrt{5}}{2}$$

therefore $a_n\le \frac{5-\sqrt{5}}{2}$ for any $n$.

Edit
As it is monotone increasing and bounded the sequence converges to $a$.

To find $a$ we use the definition of the sequence
$$a=\frac{1}{5}a^2+1$$
which gives $a=\frac{5-\sqrt{5}}{2}$.


问:
Integers of a series
The first $n$ consecutive positive integers in a series are $1, 2, 3, 4...n-1, n$. The integers $a, b, c, d$ are removed from the list. $100<a<b<c<d$ and at least $3$ of $a, b, c, d$ are consecutive the average of the remaining integers is $89.5625$ and i need help finding all the possible values of $d$.

$86.5625 = 89 \dot 9/16$. I made the sum of $a+b+c+d = s$. And I know that the average of the numbers before removed is $89\cdot 9/16 + 0.5\cdot2 \approx 91 9/16$. And since the average is equal to around the middle number then so $n$ is around $183$. After this point I don't know what to do.

答:
After removing, you have $n-4$ different positive integers. Their average is $89.5625$, and since this is at least the average of the first $n-4$ positive integers, you must have $n-4<179$.

Also, the total of these numbers is $(n-4)\times 89.5625$. For this to be an integer, $n-4$ must be a multiple of $16$. Also, you know $(n-4)\times 89.5625$ is less than the sum of the first $n$ positive integers. These two facts should allow you to determine the value of $n$ (only one $n$ is possible).

Title: Integers of a series


问:
Is $\gcd(a,bc)=\gcd(a,b)\gcd({a\over\gcd(a,b)}, c)$?
Is it true that $\gcd(a,bc)=\gcd(a,b)\gcd({a\over\gcd(a,b)}, c)$?

It is true in quite a few examples that I came up with, e.g.

$a = 18, b = 21, c = 33$

$\gcd(18,21)\gcd({18\over\gcd(18,21)}, 33) = 3 \gcd({18 \over 3}, 33) = 3 \times 3 = 9$

$\gcd(18,693) = 9$

Here is my (probably not very good) attempt at a proof:

For a $b$ where $\gcd(a, b) = 1$

For a $c$ where $\gcd(a, c) = 1$

$\gcd(k_1a, k_1b) = k_1$

$\gcd(k_2a, k_2c) = k_2$

$\gcd(k_1k_2a, k_1k_2bc) = k_1k_2$

Hence, $\gcd(k_1k_2a, k_1b) \gcd({k_1k_2a \over \gcd(k_1k_2a, k_1b)}, k_2b) = k_1 \gcd({k_1k_2a \over k_1}, k_2c) = k_1k_2 = \gcd(k_1k_2a, k_1k_2bc)$

Is this statement true, or did I just stumble across several lucky coincidences?

答:
It is generally true (for positive arguments) that $\gcd(dx,dy)=d\gcd(x,y)$. Now  $d=\gcd(a,b)$ divides both $a$ and $b$, so you get from this that
$$
  \gcd(a,bc)=d\gcd(a/d,(b/d)c).
$$
Moreover $a/d$ and $b/d$ are relatively prime (a common divisor, multiplied by $d$, would divide both $a$ and $b$, and this can only happen for $1$). So what you are asking boils down to: if $a'$ and $b'$ are relatively prime, does it hold for all $c$ that $\gcd(a',b'c)=\gcd(a',c)$? The answer is yes, this is a well known fact attributed to Gauss.</p>


问:
Problems with the complex square root
I have a problem understanding the following procedure. (
It's from a script)

Consider the domain C[0,$\infty$) and the branch of the logarithm given by

$ log(z)=ln(|z|)+i \cdot arg(z)$ ,with $arg(z) \in(0,2\pi)$.

Then we find: $z^{\frac{1}{2}} = \sqrt{|z|} e^{i arg(z)/2}$, where $arg(z^{\frac{1}{2}}) \in (0,\pi)$
The square root of any complex number that is not a positive real then lies in the upper halfplane and in this case we find $(-1)^{\frac{1}{2}}=i$

I don't understand why the last one follows. Why applies $(-1)^{\frac{1}{2}}=i$ ?

I know I can write: $-1^{\frac{1}{2}}= e^{\frac{1}{2}ln(|-1|)+arg(-1)}=e^{\frac{1}{2}arg(-1)}$ and arg(-1) ist $\pi$ and with that the upper equation would be correct....BUT that's actually wrong, since $\pi$ is not in the domain of arg. Can someone explain that to me?

答: The only argument of $-1$ in $(0,2\pi)$ is $\pi$. Therefore, by that definition we have$$z^{1/2}=\sqrt1e^{i\pi/2}=i.$$

Title: Problems with the complex square root


问:
Set theory difficult question on proving that set $\mathbf {H}$ always exists
I have been puzzling over this question for literally hours:

Let H be a 1011-element subset of the set {0, 1, 2, ..., 2021}. Prove that it has two (not necessarily distinct) elements a and b such that a+b is a power of 2.

After watching a heap of Youtube videos about set theory, I have finally come up with this:
$$\mathbf {H} \subset \mathbb {N} | \mathbf {H} _ z = m, m \in \mathbb {N}, 0 \leq m \leq 2021, |\mathbf {H}| = 1011 \\
\mathbf {J} = \{(a, b) | a + b = 2 ^ {n} | n \in \mathbb {N}\} \\
(\mathbf {H} _ x, \mathbf {H} _ y) \in \mathbf {J} | x \in \mathbb {N}, y \in \mathbb {N}, x \neq y$$

However, since I am very inexperienced, I can't solve this...

I would really appreciate a solution and an explanation on how you have got to that answer :)

答:
I claim that we can pair up the numbers $0,\ldots,2021$ into $1010$ pairs so that the sum of each pair is a power of $2$, or at least one element of the pair is a power of $2$.

Note that we can start by pairing $27-2021,28-2020,\ldots,1023-1025$. So the only numbers remaining to be paired are $0,\ldots,26,1024$.

Now, we pair $0-1024,6-26,7-25,\ldots,15-17,3-5$. So all that is left is $1,2,4,16$. Pair $1-2,4-16$.

Now we are done, since at least one pair has $2$ elements inside the set. And, either the sum of those elements is a power of $2$, or one of the elements is a power of two, in which case the sum of it with itself is a power of $2$.


问:
If $x\equiv a\pmod{n}$, prove that either $x\equiv a\pmod{2n}$ or $x\equiv a+n\pmod{2n}.$

If $x\equiv  a\pmod{n}$, prove that either $x\equiv a\pmod{2n}$ or $x\equiv a+n\pmod{2n}.$

Well I am a bit stuck in this congruence problem. I tried out :-

$n\vert x-a$ so for $2n|x-a$, $x-a$ must be even but I am not able to prove what if $x-a$ is odd, how can I prove that $x-(a+n)$ will be even and divisible by $2n$. Is this the right approach? Though it seems a very vague one, I can't think of any other.

答: Hint: The hypothesis means that $x = a + kn$ for some integer $k$. What can you conclude about $x$ modulo $2n$ when $k$ is even? How about when $k$ is odd?


问:
Rotation composition in $R^3$ and linear mappings
I'm trying to understand why the result of a composition of linear transformations (two rotations) it is different from mapping the base vectors from $R^3$ to $R^3$ twice.
I'll explain.
I have a mapping: ϕ that maps vectors in $R^3$ to $R^3$ as a linear combinations of the canonical base vectors (e1,e2,e3).
The transformation matrix associated with ϕ is the following:

\begin{bmatrix}\cos(a)&0&\sin(a)\\0&1&0\\-\sin(a)&0&\cos(a)\end{bmatrix}
That is the classic counterclockwise rotation with respect to the e2 axis.

Now I want to map the canonical base vectors (e1,e2,e3) using ϕ.

$ϕ(\mathbf{e1}) = \cos(a) \bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) - \sin(a) \bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr)$

$ϕ(\mathbf{e2}) = 0 \bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr) + 1 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$

$ϕ(\mathbf{e3}) = \sin(a) \bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + \cos(a) \bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr)$

After that, I want to perform a counterclockwise rotation around the rotated base vector e3 (e.g. ϕ(e3)).

To do this, i'm using this transformation matrix:

\begin{bmatrix}\cos(b)&-\sin(b)&0\\\sin(b)&\cos(b)&0\\0&0&1\end{bmatrix}

And I map the "old" rotated basis onto the new basis using Ψ:

$Ψ(ϕ(\mathbf{e1})) = \cos(b) \bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr) + \sin(b) \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} \cos(a)\cos(b) \\ \sin(b) \\ -\sin(a)\cos(b) \end{smallmatrix} \bigr)$

$Ψ(ϕ(\mathbf{e2})) = -\sin(b) \bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr) + \cos(b) \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} -\cos(a)\sin(b) \\ \cos(b) \\ \sin(a)\sin(b) \end{smallmatrix} \bigr)$

$Ψ(ϕ(\mathbf{e3})) = 0 \bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 1 \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr)$

Assuming that what is written is true, now I have tried to do the verification, checking if the composition of the two rotations actually gives me the same result.

Therefore:

\begin{equation}
\begin{bmatrix}\cos(b)&-\sin(b)&0\\\sin(b)&\cos(b)&0\\0&0&1\end{bmatrix}*\begin{bmatrix}\cos(a)&0&\sin(a)\\0&1&0\\-\sin(a)&0&\cos(a)\end{bmatrix}=\begin{bmatrix}\cos(a)\cos(b)&-\sin(b)&\sin(a)\cos(b)\\\sin(b)\cos(a)&\cos(b)&\sin(a)\sin(b)\\-\sin(a)&0&\cos(a)\end{bmatrix}
\end{equation}

If now I tried to make a linear mapping transforming the canonical basic vectors with this composition of rotations, I would find a different result from the one found before.

Therefore (calling τ this linear mapping):

τ(e1) = \cos(a)\cos(b) $\bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr)$ + \sin(b)\cos(a) $\bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$ - \sin(a) $\bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} \cos(a)\cos(b) \\ \sin(b)\cos(a) \\ -\sin(a) \end{smallmatrix} \bigr)$

τ(e2) = -\sin(b) $\bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr)$ + \cos(b) $\bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$ + 0 $\bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} -\sin(b) \\ \cos(b) \\ 0 \end{smallmatrix} \bigr)$

τ(e3) = \sin(a)\cos(b) $\bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr)$ + \sin(a)\sin(b) $\bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$ + \cos(a) $\bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} \sin(a)\cos(b) \\ \sin(a)\sin(b) \\ \cos(a) \end{smallmatrix} \bigr)$

Clearly:
Ψ(ϕ(e1)) ≠ τ(e1), Ψ(ϕ(e2)) ≠ τ(e2), Ψ(ϕ(e3)) ≠ τ(e3)

When instead I should have expected the same result...

I noticed that by reversing the order of the matrix product, I get the same result instead... I can't understand why.

答: You are not applying a matrix correctly to find the image of a column vector. When you are computing $\Psi(\phi(\mathbf{e1}))$, you can only use the entries of $\Psi$ and of $\phi(\mathbf{e1})$; in stead you are also using $\phi(\mathbf{e2})$ and $\phi(\mathbf{e3})$, which cannot be right. Use the entries of $\phi(\mathbf{e1})$ as coefficients in forming a linear combination of the columns of $\Psi$, not the entries of (just) one column of $\Psi$ as linear combination of things that make no sense.


问:
Golden ratio rectangles
I'm designing a layout and I would like to use four golden ratio rectangles. The total width of the layout is 960px. How do I find the height (x)? Below is a diagram of the layout.

答:
Finding the required height amounts to expressing $a$ and $c$ in terms of the height $h = a+b$. I will do this below; denote with $\phi$ the golden ratio.

We have that $\dfrac h a =\phi$, i.e. $a = \dfrac h \phi$. Now $d = \dfrac h2$, and $\dfrac c d = \phi$.

Using some trivial algebra, we obtain:

$$c = \phi d = \frac\phi2 h$$

Thus we have reduced to solving the equation:

$$960 = 2a + c = \left(\frac2\phi+\frac\phi 2\right)h$$

In conclusion, $h = \dfrac{960}{\frac2\phi+\frac\phi 2} \approx 469.42$.

Title: Golden ratio rectangles