问:
Showing that, if $|u_{n+1}|<\frac12|u_n|$ for $n>12$, the alternating series $\sum_{n>12}u_n$ converges to a non-zero number, and ...
This question was asked in my msters exam for which I am preparing and I was unable to solve two options and so asking for help here.

Let $(u_n)_{n\geq1}$ be a sequence real numbers satisfying following:

(i) $(-1)^n u_n \geq 0$ for all $n>0$.

(ii) $|u_{n+1}|<\frac12|u_n|$ for all $n>12$.

Prove that $\sum_{n>12} u_n$ converges to a non-zero real number, and that if $|u_{n-1}| < \frac12|u_n|$ for all $1<n<12$, then $\sum_{n\geq 1} u_n$ is a negative real number.

(I am writing only the options I couldn't solve.)

I am not able to prove these two assertions and would be really thankful for help.

答:
Why does no one do this problem? It is not that hard.

The first part is just the test for convergence. We just need to check if $|u_n|$ is decreasing to zero from a certain point $n$, which is obvious under your condition.

For more information on this test, you can take a look at this link :
 https://en.wikipedia.org/wiki/Alternating_series_test .

The idea is simple, you can easily imitate if you are not allowed to use it in your exam.

For the **second part** , it is not true. I know what your professor was trying to imply in this problem. However, there is not enough condition for the second part to hold.

That is mainly because this problem provides no control over the largeness of $u_{12}$ (also no connection amongst $(u_n,n<12)$, $u_{12}$, $(u_n,n>12)$ )

So we can choose $u_{12}$ large enough to make the limit positive.

We should expect more of something like:  $$ |u_{11}|> \dfrac{|u_{12}|}{2}> \frac{|u_{13}|}{4} $$

Full answer for part 1

For clarity, we define a new sequence $(a_n):= ( (-1)^{n}u_{n+12})$,clearly:

$(a_n)$ is a sequence of nonnegative numbers.

$a_n \ge \frac{1}{2} a_{n+1}$ hence it's a decreasing sequence. Furthermore, from that inequality, we get:

$0 \le a_{n+1} \le \dfrac{1}{2^n} a_1 \forall n>1$, thus:

$(a_n)$ is a sequence decreasing to zero.

Consider the following sequence of finite sums :
$$S_n := \sum_{k=1}^n (-1)^ka_k$$

So we see that:

$S_{2n+1}=S_{2n} -a_{2n+1} \le  S_{2n} $

$ S_{2n+1} -S_{2n-1}= a_{2n}-a_{2n+1} \ge 0 $( $(a_n)$ decreasing)

$ S_{2n+2}- S_{2n} = -a_{2n+1}+a_{2n+2 } \le 0$ (by the same reason)
So $(S_{2n+1})$ is an increasing sequence and $(S_{2n}) $ is a decreasing sequence.

Furthermore, $S_{2n+1}$ is bounded above by $S_2$ ( $S_{2n+1} \le S_{2n} \le S_2$ ) , $S_{2n}$ is bounded below by $S_1$ (by the same reasoning).

So they must have limit, we note $\lim S_{2n+1}= L_1$ and $\lim S_{2n}= L_2$ , respectively.

As $\lim a_n =0 $ , we see that:
$$ L_1-L_2= \lim S_{2n+1}- \lim S_{2n} =- \lim a_{2n+1}=0 $$
Thus $L_1=L_2=L$, from that we imply the convergence of $(S_n)$

By noticing that, $S_{n} = \sum_{k>12}^{12+n} u_n$ , we also derive the desired convergence.

On the other hand, as $S_{2n}$ decreasing, we have that:
$$ L = \lim S_{2n} \le S_2 = a_{2}-a_1 <0$$ ( as $ a_2< \frac{1}{2}a_1 $ )

Thus $L $ is non zero.
$\square $


问:
The relation between the supremum and the limit superior of a sequence
If $(a_n)_{n\ge 1} \subset \mathbb{R}$ is a bounded sequence, what is the relation between $\limsup\limits_{n\to \infty} a_n$ and $\sup\limits_{n\in \mathbb{N}}a_n$? My intuition tells me that they should be equal, but when I looked this up on Wikipedia I only found that $\limsup\limits_{n\to \infty}a_n\le \sup\limits_{n\in \mathbb{N}}a_n$. I can't produce an example where the inequality is strict and I believe that since $\sup\limits_{n\in \mathbb{N}}a_n$ is the limit of some subsequence of $(a_n)_{n\ge 1}$ and $\limsup\limits_{n\to \infty}a_n$ is the maximum of the limit points we should always have equality. So, please show me an example where this isn't true or prove that it is always true.

答:
Take any strictly decreasing and convergent sequence $(a_n)_{n\in\mathbb{N}}\subseteq\mathbb{R}$ and let $a$ be the limit of $(a_n)_{n\in\mathbb{N}}$. Then $a$ is the only accumulation point of $(a_n)_{n\in\mathbb{N}}$, hence $\limsup_{n\to \infty} a_n = \lim_{n\to\infty} a_n= a$. As $(a_n)_{n\in\mathbb{N}}$ is strictly decreasing by assumption, we have $a_1>a_2>...>a$, so that $\sup_{n\in \mathbb{N}} a_n =a_1>a=\limsup_{n\to \infty} a_n$.

For instance, the sequence $a_n:=\frac{1}{n}$, $n\in \mathbb{N}$ (with $a:=0$) has such properties.

ContentLicense: CC BY-SA 4.0


问:
If A is a n*n matrix with only 1 eigenvalue a, how can we show that A is an upper triangular matrix with all diagonal entries as a.
I have been trying for a long time now but not reaching anywhere. Can you please give me a hint on how to appraoch this question?
Will be very grateful.</p>

答:
This is not true. Take
$$
A=\pmatrix{ a & 0 \\1 & a}.
$$


问:
Existence of Cauchy Principal Value
Hey thanks for stopping by , I would really appreciate your help in proving this:

If $f$ is continuous on $(-\infty,\infty)$ and if $\int_{-\infty}^\infty {f(x)dx}$ converges to A , prove that
$$C.P.V. \int_{-\infty}^\infty {f(x)dx} = A$$
where C.P.V. stands for Cauchy Principal Value.

I thought of first taking the interval $(-s,s)$ which can later by varied to the original interval as $s$ approches $\infty$ and also thought of taking subdivisions in this interval so that i can discuss the integrability of $f$ in the component intervals of the subdivisions taken by using extreme value theoram in those component intervals , but all this is in my head and i don't think it is getting me anywhere . Please help me write the proof and also provide some pro-tip just to avoid rambling around in future. Thanks in Advance.

答:
Suppose $\int_{-\infty}^{\infty} f(x) \, dx$ converges. That means that $\int_{-\infty}^{0} f(x) \, dx = B$ and $\int_{0}^{\infty} f(x) \, dx = C$ for some real numbers $B$ and $C$ such that $A=B+C$. This again means that for every $\epsilon >0$ there exit constants $N$ and $M$ such that

$$\bigg|\int_{s}^{0} f(x) \, dx - B \bigg| < \frac{\epsilon}{2} \quad \text{and} \quad \bigg|\int_{0}^{r} f(x) \, dx - C \bigg| < \frac{\epsilon}{2}$$

when $s \leq N$ and $r \geq M$. Now set $L = \max(|N|, |M|)$. For all $\rho \geq L$, we have

$$\bigg|\int_{-\rho}^{0} f(x) \, dx - B \bigg| < \frac{\epsilon}{2} \quad \text{and} \quad \bigg|\int_{0}^{\rho} f(x) \, dx - C \bigg| < \frac{\epsilon}{2}.$$

Furthermore, observere that

\begin{align*}
\bigg| \int_{-\rho}^{\rho} f(x) \, dx -A \bigg| &= \bigg|\int_{-\rho}^{0} f(x) \, dx + \int_{0}^{\rho} f(x) \, dx - (B+C) \bigg| \\
&= \bigg|\int_{-\rho}^{0} f(x) \, dx - B + \int_{0}^{\rho} f(x) \, dx - C \bigg| \\
&\leq \bigg|\int_{-\rho}^{0} f(x) \, dx - B \bigg| + \bigg| \int_{0}^{\rho} f(x) \, dx - C \bigg| \\
&< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon,
\end{align*}

which shows that the Cauchy principal value converges to $A$.

However, as I pointed out in the comment section, this is easier to see by thinking about what convergence really means. The simple idea easily gets lost in this proof. The idea is that if the limit exits, you can take it in any way you want, i.e. you can approach $\infty$ and $-\infty$ at different rates. In particular, you can also approach them at the same rate, i.e. symmetrically, which is exactly what you do when you take the Cauchy principal value. Therefore the CVP has to converge to the integral.

Title: Existence of Cauchy Principal Value


问:
Proving a language is regular through the reverse language
Consider the language $L$ with symbols that can be written from the alphabet
$$ \Sigma = \left\{
\begin{bmatrix}
0 \\
0 \\
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
\end{bmatrix},
\begin{bmatrix}
1 \\
0 \\
\end{bmatrix},
\begin{bmatrix}
1 \\
1 \\
\end{bmatrix}
\right\}
$$
such that, when concatenated, the second line is three times plus one the first line (in binary). I have to show that $L$ is regular by providing a $DFA$.

What I know: The technique is to think of $L^R$, the reverse language, provide it's $DFA$ and apply an algorithm (that I already know) and get to ${L^R}^R=L$.

What I don't understand: I found the $DFA$ that recognizes $L^R$, ,

but I don't understand it's functioning.

I appreciate any help, and thanks in advance.

答:
The three nodes above represents the carry-over of current multiplication.
$q_0$ particularly stands for the operation on the least significant bit since we have to plus 1 additionally. If the last digit is $[0,0]$, then we knows it cannot be right. Thus no transition from $q_0$ marked with $[0,0]$ can be found. If it is $[1, 0]$, then this digit is verified and the carry-over is $\displaystyle\frac{1*3+1}2=2$, which means that we will have to add an additional 2 when we check the next digit.

The other nodes perform alike. I shall leave it to you. The accepting state is that we have checked all the digits and the residual carry-over is $0$.

One more thing, $q_0$ can be replaced by the middle node above which is marked by $1$. Think about it if you successfully figure out why the current DFA works for $L^R$.


问:
Using induction to prove $(ab)^n=a^n*b^n, a,b \neq 0 a,b \in \mathbb R, n \in \mathbb Z$
I am trying to prove that $(ab)^n=a^n*b^n, a,b \neq 0, a,b \in \mathbb R, n \in \mathbb Z$

I am not sure if we can use induction here.

If yes, we can see three options:

1) $n=0$, so $(ab)^0=1=1*1=a^0*b^0$

2) $n>0$, so for $n=1$ we have $(ab)^1=ab=a^1*b^1$

for $n+1$: $(ab)^{n+1}=(ab)^n*ab=a^n*a*b^n*b=a^{n+1} * b^{n+1}$

3) $n<0$, so here is the question. Am I allowed to use induction here?

If it is true for $-(n)$ and $-(n+1)$, does it mean that we prove whole thing using induction?

$(ab)^{-(n+1)}=(ab)^{-n}*(ab)^{-1}$=$(a^{-n}*b^{-n})/ab$=$a^{-n}*a^{-1}*b^{-n}*b^{-1}=a^{-(n+1)}*b^{-(n+1)}$

答:
$$(ab)^{n+1}=a(ba)^nb\overset{?}=a(ab)^nb\overset{*}=a(a^nb^n)b=a^{n+1}b^{n+1}$$

The step (?) works uses the commutativity of $\Bbb R$ under the operation of multiplication which suggests that we need an abelian group for the general result to hold.

The step (*) is applying the inductive hypothesis.


问:
Proving Morera's Theorem through the Cauchy-Goursat theorem for rectangles
In my book Morera's Theorem is stated as for if $f$ is a continuous function on an open set $U$, that has the property that around the boundary of any rectangle the closed integral is zero, then  $f$ is holomorphic.

I want to prove that this is equivalent to saying if $f$ is a continuous function on an open set $U$ that has the property $\oint f=0$ along any closed path in $U$, then $f$ his holomorphic.

A bit context: My book is Serge Lang's Introduction to Complex Analysis at an Undergraduate level. The book introduced the Cauchy-Goursat theorem for rectangles and then proved Cauchy's Integral Formula (for simply connected open sets)

Question:
Does anybody know how could I show that the conditions "around the boundary of any rectangle in $U$ the closed integral is zero" and "$\oint f$ along any closed path in $U$" are the same?

答:
I think I may have figured it out, if anybody sees any error in this please feel free to correct me. I will prove this as an iff statement.

First, we will consider the rectangle assumption.

If the closed integral along any rectangular path is zero then by the Gourast theorem we can define $U$ as a union of discs (defined by each rectangle) on which $f$ is holomorphic. Since $f$ is holomorphic it follows that $f's$ primitive along a path is a series of local primitives defined by the different discs (which are defined from the different rectangles). Since the primative exists it follows that $\oint f=0$ for any closed path.

We will now consider the assumption $\oint f =0$ first.

If $\oint f = 0$ for all closed paths in $U$. Then we can consider all rectangles inscribed in $U$ as such paths from which we get $\oint_{\partial R} f=0$.

Hence the iff statement is true.

If this is wrong please correct me.


问:
Terms of Stanley Sequences
Stanley Sequences (named after R. P. Stanley) are integer sequences beginning with 0, k where k is an integer bigger than 0, and where every following member is chosen to be the smallest integer bigger than the previous term so that no three terms (not necessarily consecutive) form an arithmetic sequence. Another way of saying this is that we want to lexicographically the first three-terms-arithmetic-sequence-free sequence beginning with 0, k. An even simpler way to put is that we search for the smallest integer bigger than the former term so that no three terms x, y, and z construct the equation x + z = 2y.

Let's look at the sequence where k = 1.
The first few terms of this sequence are: 0, 1, 3, 4, 9, 10, 12, 13, 27, 28.

We notice an interesting thing if we write the indexes of the terms of S(1) (the sequence where k=1), where a a_0=0, a_1=1, and so on in binary.

We get 0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001. Now let's write the values for these indexes in ternary => 0, 1, 10, 11, 100, 101, 111, 1000, 1001. This goes on forever. Another way of putting it is that each term of S(1) is obtained by writing the index in binary and reading it in ternary. Another way to look at this is to collect all the numbers in ternary without a '2' in their representation.

What is an easy way to prove this? Also, can we suggest other numerical systems we can use that give us a similar relationship when k is different than 1?

What I've tried doing is proving that since the only way to obtain a '2' when we multiply by 2 in ternary is by starting with a '1' (without carrying over, I mean) there are only a few cases which we should aim to avoid and then form a case by case analysis to prove that these are in fact the smallest remaining options but that is both unclear and tedious. Do you have any suggestions?

Thank you in advance, and please excuse any language mistakes I may have made...

(If someone wonders I'm researching the topic for fun. It's not for school work or sth like this)

答:
The $k=1$ case is not hard.

a) The sequence of ternary-2-free numbers is free of solutions to $x+z=2y$ with $x\ne z$: Since $x,z$ are $2$-free, adding produces no carries, hence $x+z$ has at least one digit $1$, namely at a place where $x$ and $z$ differ. The addition in $y+y$ alo produces no carries, hence $y+y$ consists only of digits $0$ and $2$. Therefore $x+y\ne 2z$

b) Any lexically smaller sequence has a solution to $x+z=2y$: Let $b_0, b_1, b_2,\ldots $ be lexically smaller than our ternary-2-free sequence $a_0,a_2,a_2,\ldots$. Let $n$ be minimal with $b_n\ne a_n$ (and hence with $b_n<a_n$). Since the ternary-2-free numbers $<a_n$ are precisely $a_0,\ldots, a_{n-1}$, we conclude that $b_n$ is not 2-free.
Let $x$ be the number that has ternary digit $0$ where $b_n$ has digit $0$ and has digit $1$ where $b_n$ has digit $1$ or $2$.
Let $y$ be the number that has ternary digit $1$ where $b_n$ has digit $1$ and has digit $0$ where $b_n$ has digit $0$ or $2$.
Then $x,y$ are $<b_n$ and 2-free, hence occur among $b_0,\ldots, b_{n-1}$.
Also, we verify that $b_n+y=2x$ because adding $y$ turns every $1$ of $b_n$ into a $2$.

Title: Terms of Stanley Sequences


问:
Prove that iterated $x_{i + 1} = x_i - \log x_i$ is sublinear.
Define a recurrence as: $$x_0 = n, x_{i + 1} = x_i - \log x_i, \forall i \geq 0.$$ Let $N_n$ be such that $x_{N_n} < 1$. Is it possible to show that $N_n \in o(n)$?

Some thoughts: if $x_{i + 1} = x_i - f(x_i)$ and $f$ is constant then $N_n = \frac{n}{f(1)}$ and is thus not sublinear. If $f$ is $\Omega(n)$ then $N_n$ is constant and thus trivially sublinear.  Is $f(x) = \log x$ sufficient to make $N_n$ sublinear?

Edit: As pointed out in the comments, the criteria $x_{N_n} < 1$ can never be met for this question. The correct question should be

Let $N_n$ be the smallest integer such that $x_{N_n} < 10$. Is it possible to show that $N_n \in o(n)$?

答:
The other answers adress the speed of convergence to $1$, which I think is not your question.

I want to reverse the problem. Write $f(x) = x-\log(x)$. Let $M>1$ and $(u_n)$ be the sequence defined by:

$$u_0 = M \text{ and } u_{n+1} = g(u_n),$$

where $g : [1, +\infty) \to [1, +\infty)$ is the inverse bijection to $f$. Then $x_i = g(x_{i+1})$, so that $n = x_0 = g^{(N_n)} (x_{N_n}) < g^{(N_n)} (M) = u_{N_n}$ and $n = x_0 = g^{(N_n-1)} (x_{N_n-1}) \geq g^{(N_n-1)} (M) = u_{N_n-1}$. Note also that

$$\lim_{n \to + \infty} \frac{f(n)}{n} = 1 \text{ so } \lim_{n \to + \infty} \frac{g(n)}{n} = \lim_{n \to + \infty} \frac{f(g(n))}{g(n)} \frac{g(n)}{n} = 1.$$

Since $u$ is increasing, we get that $N_n = \inf \{k \geq 0 : \ u_k > n\}$. This is convenient, since we have translated the initial problem to the study of the growth rate of a single sequence.

Let $L>0$. For $x \geq M$, we have $f(x) = x-\log(x) \leq x-\log(M)$, so that $x \leq g(x-\log(M))$, whence $g(x) \geq x+\log(M)$. It follows that $u_n \geq M + n \log(M)$, so that $\lim_{n \to + \infty} u_n = +\infty$.

Now, let me prove that $u_n$ grows super-linearly. Let $L>0$. Let $n_0$ be such that $u_n \geq 10^L$ for all $n \geq n_0$. By the same argument as above, $g(x) \geq x+L$ for all $x \geq 10^L$. Hence, for all $n \geq n_0$,

$$u_n = u_{n_0} + (u_n-u_{n_0}) \geq u_{n_0} + (n-n_0)L.$$

It follows that $\liminf_{n \to +\infty} u_n/n \geq L$. Since this holds for all $L> 0$, we finally get

$$\lim_{n \to + \infty} \frac{u_n}{n} = +\infty.$$

Specializing to the subsequence $(N_n)$, which also converges to $+\infty$,

$$\lim_{n \to + \infty} \frac{u_{N_n}}{N_n} = +\infty.$$

But $n < u_{N_n} \leq g(n)$, so

$$\lim_{n \to + \infty} \frac{g(n)}{N_n} = +\infty.$$

Finally, since $g(n) \sim n$, we get

$$\lim_{n \to + \infty} \frac{N_n}{n} = 0,$$

that is, $N_n = o(n)$.

The reasoning above can be adapted to many functions instead of $\log$, although many details have to be checked. I think that the result should hold for the recursion $x_{i+1} = x_i-h(x_i)$, where $\inf h>0$ and $\lim_{x \to +\infty} h(x) = +\infty$ as well as $\lim_{x \to +\infty} h(x)/x = 0$, but that is not completely trivial.


问:
Show that $\int_{1}^{x} \frac{\sin t}{t}dt -x +1 <0,\: x>1$?
I made the following attempt but it did not feel convincing. I also learned afterward that differentiation is not allowed across inequalities.

The original expression on the LHS is equivalent to

$\frac{d}{dx}\int_{1}^{x}\left(\frac{\sin(t)}{t}-1\right)\:dt = \frac{\sin(x)}{x}-1$
So the original inequality now says $\sin(x)<x, \: x>1$

But this is a trivial inequality as $\sin(x)$ is bounded by $[-1,1]$.
Therefore, the inequality holds for $x>1$.

答:
Let $f(x)=\int_{1}^{x} \frac{\sin t}{t} dt -x+1 \implies f(1)=0$$

D.w.r.t. $x$ using Lebnitz:
$$f'(x)=\frac{\sin x}{x}-1 <0, x\in R.$$
So $f(x)$ is decreasing in $x \in R$,  hence
$$ x>1 \implies f(x)<f(1)=0.$$


问:
Conjugacy classes, irreducible representations, center of the group
I am trying to prove that the number of irreducible representations of a finite group $G$ equals the number of conjugacy classes in it.

It seeems the action of the group on itself by conjugation $g(x)=gxg^{-1}$ is related to this. Let us call this the conjugation representation. It is not irreducible, so we consider its decomposition into irreducibles.

If $x$ is such that $gx=xg$ for every $g$, than this leads to a trivial representation. So the multiplicity of the trivial representation, $m_1$, equals the number of such $x$'s. This is the order of the center of the group, $|Z_G|$.

The character $\chi(g)$, in the conjugation representation, equals the order of the centralizer of $g$. This is $|G|/|C_g|$ where $C_g$ is the conjugacy class of $g$. Then the multiplicity of representation $R$ is given by $m_R=\frac{1}{|G|}\sum_{C}|C|\chi^R(C)\chi(g)=\sum_{C}\chi^R(C)$. When $R$ is the trivial representation, we get that $m_1$ equals the number of conjugacy classes.

Fine. We are almost there. It remains to prove that the order of the center of the group equals the number of irreducible representations. What is the simplest way of proving this?

PS: Wait a second, this cannot be. The center of the permutation group $S_n$ is trivial, $|Z_{S_n}|=1$, but the number of conjugacy classes/irreps is not 1. I am confused.

答:
(1) Is wrong. You are failing to take into account there can be linear combinations of non-fixed-points which are fixed points of the conjugation action. Thus, $|Z(G)|$ is just a lower bound for the multiplicitiy of the trivial rep in the conjugation rep. Indeed, the trivial-rep isotypic component (i.e. sum of all the trivial subreps) of the conjugation rep admits a basis of the form $\{\sum_{g\in C}g\mid C\}$ as $C$ runs over the conjugacy classes of $G$. Since central elements form singleton conjugacy classes, these are the single-summand elements of this basis. But all you get is that the number of conjugacy classes equals itself.

You will basically have to learn Artin-Wedderburn if you want to understand this fact. You should already know there are algebra homomorphisms $\mathbb{C}[G]\to\mathrm{End}(V)$ for each irrep $V$. This means you can put them together into a homomorphism $\mathbb{C}[G]\to\bigoplus\mathrm{End}(V)$ (the sum is over irreps). This must have trivial kernel, since any element of the kernel would act as $0$ on any irrep, hence on any rep (by semisimplity), hence on the regular rep $\mathbb{C}G$ hence it times $1$ would be $0$ within $\mathbb{C}[G]$. To see that the map is also onto (and hence an algebra isomorphism), we can show the dimensions match.

To do this, let's find the multiplicity of any irrep $V$ within $\mathbb{C}G$. That equals the dimension of the hom space $\hom_G(\Bbb CG,V)$ (which you can show using semisimplicity and the distributivity of $\mathrm{hom}$). But any equivariant map $\mathbb{C}G\to V$ is determined by where $1$ is sent to, which is arbitrary, hence this hom-space is naturally isomorphic (as a vector space) to $V$ itself, so the dimension (hence the multiplicity) is $\dim V$ itself. As a result, we get $\dim\mathbb{C}[G]=\sum(\dim V)^2$.

On the one hand, the multiplicity of the trivial rep in the conjugaction action is the number of conjugacy classes. (More generally, if $\Omega$ is a $G$-set, the multiplicity of the trivial rep in the permutation rep $\mathbb{C}\Omega$ equals the number of orbits, by using a basis like the one I described for the conjugation rep. Can you show this? It is a nice exercise, and relevant to Burnside's counting lemma.) On the other hand, the trivial representation under the conjugation action corresponds to the center $Z(\mathbb{C}[G])$ of the group algebra, which has the same dimension as $\bigoplus Z(\mathrm{End}(V))$, which equals the number of irreps.


答:
The Mean Value Theorem for integrals states that exists a $c \in [1,x]$ such that

$$\frac{1}{x-1}\int_{1}^{x} \frac{\sin t}{t}dt= \frac{\sin c}{c}<1$$

Title: Show that $\int_{1}^{x} \frac{\sin t}{t}dt -x +1 <0,\: x>1$?

Body:
I made the following attempt but it did not feel convincing. I also learned afterward that differentiation is not allowed across inequalities.

The original expression on the LHS is equivalent to

$\frac{d}{dx}\int_{1}^{x}\left(\frac{\sin(t)}{t}-1\right)\:dt = \frac{\sin(x)}{x}-1$
So the original inequality now says $\sin(x)<x, \: x>1$

But this is a trivial inequality as $\sin(x)$ is bounded by $[-1,1]$.
Therefore, the inequality holds for $x>1$.


问:
Figuring out notation of random variables

Hi, I'm currently working on probability and statistics question and have become a little stuck on understanding this question.

Could someone explain to me what an N(3,4) variable is as I haven't come across it in my notes at all.

Thanks!

答:
Could someone explain to me what an N(3,4) variable is ?

It is a Gaussian random variable with mean 3 and variance 4

Assuming independence between $X$ and $Y$, using properties of

Gaussian distribution

linearity of expectation

quadratics property of variance,

You get that

$$Y\sim N(11;100)$$

Then you can solve the request by standardizing and using Z-table

Further explanation:

Expectation is a linear operator thus

$$E(5X-4)=5\times 3-4=11$$

Variance is a quadratic operator thus

$$V(5X-4)=5^2\times 4=100$$

Linear combination of gaussian is still gaussian thus

$$P(Y>18.5)=P(Z>\frac{18.5-11}{\sqrt{100}})=1-\Phi(0.75)=1-0.7734=0.2266$$

$$\Phi(0.75)=\int_{-\infty}^{0.75} \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt=0.7734$$

is tabulated in Z-table. If you are not familiar with Z-table you cannot solve the problem.


问:
What wrong with pointwise convergent
It is true that the uniform limit of continuous functions is continuous as it has proved as a theorem, but what is wrong with pointwise limit? I mean why this theorem it doesn't work for if the convergence is pointwise?

Can any one clarify this to me may be using epsilon definition if possible?

Thank you.

答: The standard example is the sequence of function $f_n:[0,1]\to \mathbb R$, given by $f_n (x)=x^n$. Each function is continuous but the point-wise limit is (easily seen to be) the function $f(x)=0$ if $0\le x<1$ and $f(1)=1$, a discontinuous function. </p>

ContentLicense: CC BY-SA 3.0

Title: What wrong with pointwise convergent


问:
Let $f$ have a pole of order $m$ at $z_0$ and let $g$ have a pole of order $n$ at $z_0$. Classify the isolated singularity of $f/g$ at $z_0$.
Please help me understand and how to start. Let $f$ have a pole of order $m$ at $z_0$ and let $g$ have a pole of order $n$ at $z_0$.  Classify the isolated singularity of $\displaystyle{f/g}$ at $z_0$.

答:
Since $f$ has a pole of order $m$ at $z_0$, $f$ can be written on the form

$$f(z)=\frac{\phi(z)}{(z-z_0)^m},$$

for an analytic function $\phi$ in a neighborhood around $z_0$. Correspondingly, $g$ may be written as

$$g(z)=\frac{\psi(z)}{(z-z_0)^n},$$

for an analytic function $\psi$ in a neighborhood around $z_0$. Thus,

$$\frac{f(z)}{g(z)}=\frac{\phi(z)}{(z-z_0)^m} \cdot \frac{(z-z_0)^n}{\psi(z)}= \frac{\phi(z)}{\psi(z)} (z-z_0)^{n-m}.$$

Now you see that if $n \geq m$, $\frac{f}{g}$ has a removable singularity at $z_0$. If $m > n$ however, we have a pole of order $m-n$.


问:
Is ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$?
This is a follow-up question of a previous one.

Is it true ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$ ?

It seems to me it's true because for any $x \in{}^{*}\mathbb{R}$, there is a $y \in {}^{*}\mathbb{Q}$ such that $y < x$ and vice versa, and because for any $x' \in \mathbb{R}$, there is a $y' \in {}^{*}\mathbb{Q}$ such that $y' = x'$.

答:
The simplest argument is to transfer the formula $\mathbb{Q}\not=\mathbb{R}$.

In more detail, the symbols "Q" and "R" in the language are interpreted as respectively the rationals $\Bbb Q$ and the reals $\Bbb R$ in the standard model, and as $^\star\Bbb Q$ and $^\star\Bbb R$ in the non-standard model.  The fact that $\Bbb Q$ is PROPERLY contained in $\Bbb R$, and therefore distinct from it, is encoded by the formula Q$\not=$R.  By the transfer principle, the same formula remains true when interpreted over the nonstandard model.  Hence $^\star\Bbb Q\not= {}^\star\Bbb R$.

Title: Is ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$?


问:
About intersection of sets
Consider sets $A_1, A_2,  \ldots, A_n$. Let $A_i$ be a set of functions missing the element $i$ in their image. Turns out $\bigcap A_i$ is a set of functions missing $n$ elements in their image. Having some difficulty seeing that. For example, $\{(1, a), (2, a)\} \cap \{(1, b), (2, b)\} = \emptyset$ where the intersection of the given sets are not missing two elements. Can, someone, please, elaborate on this? Thanks.

答:
A function is a triple $(X,Y,E)$ where $X$ is the domain, $Y$ the codomain and $E\subseteq X\times Y$ the association. When someone says $i$ is missed in the image, means that your function can be represented by the triple $(X,Y\setminus \{i\},E)$ because $E$ does not contain $(x,i)\in E.$

In this way, you seemed to be taking the intersection of the association $E$ of two different functions (i.e., $X$ and $Y$). Instead $A_i$ is the collection of sets you are describing with the common property that the codomain is $Y\setminus \{i\}$, so $X\in A_b$ and $Y\in A_a$ in your notation. $A_a\cap A_b=\emptyset$ if you are considering the codomains to be $\{a,b\}$ but $A_a = \{Y\}$ and $A_b=\{X\}$ so you have to be doing $\{X\}\cap \{Y\}$ which is different from taking $X\cap Y$ as you are suggesting.

Edit/Example:
Consider the functions from $f:\{1,2,3\}\longrightarrow \{a,b,c\}.$ I am going to drop the set notation and the tuples notation and denote, for example, the function $\{(1,a),(2,a)\}$ by $aa$ where the index is the preimage and the letter is the image at that index. So, for example
$$A_a=\{bbb,ccb,bcb,cbb,bbc,\color{red}{ccc},bcc,cbc\}$$ and $$A_b=\{aaa,cca,aca,caa,aac,\color{red}{ccc},acc,cac\},$$
so $A_a\cap A_b=\{\color{red}{ccc}\}=\{(1,c),(2,c),(3,c)\}.$

Notice that then $|A_a|=|A_b|=(3-1)^3$ and $|A_a\cap A_b|=(3-2)^3=1$

Title: About intersection of sets


问:
How to take the derivative of linear algebra inside a norm
If I have an equation like this...

$$
L(X_s, y_s) = \frac{1}{2}|| y_t - K_{ts}(K_{ss} + \lambda I)^{-1}y_s ||^2
$$

and I want to find the minimum with respect to $y_s$, how could I go about differentiating what is inside of the norm?

答:
$\def\p{\partial}$
To reduce the vistual clutter, define the following variables
$$\eqalign{
A &= K_{ts}(K_{ss} + \lambda I)^{-1} \\
x &= y_s \\
b &= y_t \\
}$$
Then the objective function can be written in the standard form
$$\eqalign{
L &= \tfrac 12 \|Ax-b\|^2 \\
}$$
whose gradient and optimum
$$\eqalign{
\frac{\p L}{\p x}
 &= A^T(Ax-b) \quad\xrightarrow{\frac{\p L}{\p x}=0}\quad
x = (A^TA)^{-1}A^Tb \\
}$$
have been derived many times on this site
(aka the Normal Equation of the system).


问:
Closed form for zeros.
Let $t \in [0,1]$ and $f(x) = (x+t)e^{1/(1+x^2)}-\frac{3}{2}t -x$. Is it true that this function has exactly one root on $[0,\infty)$ and is there a closed form for it in terms of $t$? I'm not quite sure how to show that this is true or false, and in particular finding the formula. If there does not exist a formula for the root, is it true that as $t \to 0$, the roots diverge? Is there any hint or suggestion?

答:
I am a bit skeptical about a closed form for the root. There is a simple formula for $t$
$$t(x)=\frac{2 \left(e^{\frac{1}{x^2+1}}-1\right) x}{3-2 e^{\frac{1}{x^2+1}}}$$ but its inversion does not seem possible even using special functions.

If we consider the problem for small values of $t$ (that is to say for large values of $x$), by Taylor we have
$$t=\frac{2}{x}+\frac{3}{x^3}+\frac{13}{3 x^5}+\frac{77}{12 x^7}+\frac{187}{20
   x^9}+O\left(\frac{1}{x^{11}}\right)$$ and using series reversion
$$x=\frac{2}{t}+\frac{3 t}{4}-\frac{7 t^3}{24}+\frac{11 t^5}{48}-\frac{163
   t^7}{720}+O\left(t^9\right)$$ Trying for $t=\frac 1 {1000}$, this truncated series gives
$$x=\frac{1440000539999790000164999837}{720000000000000000000000}$$  which is $$x\approx 2000.00074999970833356249977361111$$ while the "exact" solution is    $$x=2000.00074999970833356249977361136$$

For $t=\frac 12$, the approximation gives
$$x=\frac{400337}{92160}\approx 4.34393$$ while the exact solution is $x=4.344314$.

Edit

There is another way to get more and more accurate closed form approximations when $x$ is large.

Let $x=\frac 1y$ and consider the function $f(y)=0$. Now, build the $[1,n]$ Padé approximant of it built around $y=0$. It will write
$$f(y) \sim \frac {a^{(m)}y -t } {1+\sum_{i=1}^m b_i^{(m)} y)^i}\implies y=\frac t {a^{(m)}} \implies x=\frac  {a^{(m)}} t$$ All coefficients $a^{(m)}$ and $b_i^{(m)}$ are defined by the values of function and its derivatives at $y=0$.

For the fun of it, I give you the one for $m=9$. It write
$$x_{(9)}=\frac{2 \left(8521 t^8+94080 t^6+269280 t^4+276480 t^2+92160\right)}{3t
   \left(561 t^8+15600 t^6+64000 t^4+80640 t^2+30720 \right)}$$

For $t=\frac 1 {1000}$ it gives a relative error of $7.73 \times 10^{-33}$%.

Title: Closed form for zeros.


问:
Spherical Geometry Distance Between 2 Points
I try to calculate distance between 2 points on Earth. I have an a car and this car goes with speed 100 km/h and I know the start point latirude and longitude and car goes 5 hours. I want to calculate "How far has the vehicle traveled?". Actually answer is so simple I know distance = speed * time . However does this rule apply to spherical geometry? I am not sure.</p>

答: If the vehicle is traveling along the surface of the sphere, then yes: $distance = speed * time$.


问:
Smooth convex functions
Let $E\subseteq\mathbb R^d$ be a convex set, $\beta\geq 0$ be a given real number and $f:E\to\mathbb R$ be a convex and differentiable function satisfying:
$$f(y)\leq f(x)+\nabla f(x)^\top (y-x) +\frac{\beta}{2}\|x-y\|_2^2, \quad \forall x,y\in E.$$
Show that $\nabla f$ is $\beta$-Lipschitz, i.e.,
$$\|\nabla f(x)-\nabla f(y)\|_2\leq\beta\|x-y\|_2, \quad \forall x,y\in E.$$

Edit: I am interested in the case when $E$ can be any convex subset of $\mathbb R^d$. When $E=\mathbb R^d$, the claim can be proven as proposed by the answer below. However, in many convex optimisation textbooks, the claim is stated for general domains $E$ without proof.

答:
Let $x,y,z\in \mathbb R^d$. Then by the smoothness condition
$$
f(x+z) \le f(x) + \nabla f(x)^Tz + \frac\beta2 \|z\|_2^2
$$
and by convexity
$$
\nabla f(y)^T (x+z-y) \le f(x+z)-f(y).
$$
Adding both inequalities gives
$$
(\nabla f(y) - \nabla f(x))^Tz - \frac\beta2 \|z\|_2^2 \le f(x) - f(y) -\nabla f(y)^T(x-y).
$$
The left-hand side is maximal for $z= \frac1{\beta}(\nabla f(y) - \nabla f(x))$, which gives
$$
\frac1{2\beta} \|\nabla f(y) - \nabla f(x)\|_2^2 \le f(x) - f(y) -\nabla f(y)^T(x-y).
$$
The inequality is true if we exchange $x$ and $y$. Adding the resulting both inequalities yields to
$$
\frac1{\beta} \|\nabla f(y) - \nabla f(x)\|_2^2 \le (\nabla f(x)-\nabla f(y))^T(x-y).
$$

Title: Smooth convex functions


问:
Why do each of these cylindrical triple integrals evaluate differently?
The problem in question is thus:
Find the volume cut out of the sphere of a radius $a$ centered at the origin by the polar curve $r = a\cos\theta$. I attempted to solve the problem using this cylindrical triple integral, taking advantage of $x$-axis symmetry.
$$
2\int_0^{\frac{\pi}{2}}\int_0^{a\cos\theta}\int_{-\sqrt{a^2 - r^2}}^{\sqrt{a^2 - r^2}}r\,dz\,dr\,d\theta = \frac{2\pi a^3}{3} - \frac{8a^3}{9}
$$
However, when this, similar integral is evaluated (which in my mind should yield the same results as the first one), the answer is different as the last term cancels out.
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{a\cos\theta}\int_{-\sqrt{a^2 - r^2}}^{\sqrt{a^2 - r^2}} r \, dz\,dr\,d\theta = \frac{2\pi a^3}{3}
$$
Which one of these answers is correct and why is the other one wrong? I understand that it has something to do with the theta integral, but I have no idea where my error is in assuming that these two triple integrals should produce identical results.

Thanks! This is my first time on the Math Stack Exchange so let me know if there is anything else I can do to improve the nature of my question.

THANK YOU ALL!!! Many wonderful explanations below, have a great day everyone!

答:
$2\pi a^3/3$ is the volume of the hemisphere so it is clear that the second answer is
wrong. But the integral itself seems to be correct since $2\int_0^{\pi/2}f(\cos\theta)\,d\theta=\int_{-\pi/2}^{\pi/2} f(\cos\theta)\,d\theta$ due to the evenness of $\cos\theta$ so it is likely that you made some calculation error.

In fact it seems symbolab is making the same error. The second integral is$$2\int_{-\pi/2}^{\pi/2}\int_0^{a\cos\theta}r\sqrt{a^2-r^2} \, dr~d\theta=-\int_{-\pi/2}^{\pi/2} \int_{a^2}^{a^2\sin^2\theta}\sqrt m\,dm~d\theta$$where $m=a^2-r^2$. This gives$$V=-\frac23\int_{-\pi/2}^{\pi/2}(\color{red}|a\sin\theta\color{red}|^3-|a|^3)\,d\theta$$So you have probability forgotten to take the modulus and landed at the answer of $2\pi a^3/3$.


问:
Proving that $\frac{\sigma_{n-1}}{\omega_n} = n$ in $\mathbb{R}^n$
If $\sigma_{n-1}$ was the surface area of the unit sphere in $\mathbb{R}^n$ and $w_{n}$ was the area of the unit ball in $\mathbb{R}^n$, my lecture notes prove that $$\frac{\sigma_{n-1}}{\omega_n} = n$$ by the proof $$\omega_n = \int_{B_1(0)} dx$$$$=\int_0^1\sigma_{n-1} z^{n-1} dz$$$$=\sigma_{n-1}\int_0^1z^{n-1}dz$$$$=\frac{\sigma_{n-1}}{n}$$ but I don't understand how we got from the 1st line to the second line, namely that $$ \int_{B_1(0)} dx =\int_0^1\sigma_{n-1} z^{n-1} dz$$ could someone please explain this to me? Thanks!</p>

答: It just has a polar  transformation </p>

Body: It just has a polar  transformation


问:
Parametrizing fractals and space filling curves
I am a 2nd year undergraduate doing a BS-MS in an Indian Institute and am taking a course in Elementary Differential Geometry (studied till the basics of surfaces). Our instructor has assigned a term-project on fractals and space filling curves. One of the things he has asked us is, to prove that the Hilbert's curve is parametric and write down it's parametric equations.

I tried to google this but got no information on this topic. So could anybody give some idea as to how to do this not only for the Hilbert's curve but for any fractal or space filling curve?

Your help will be highly appreciated.

答:
Hint Construct the Hilbert $H(t)$ curve piecewise out of smaller copies of the Hilbert curve:

$$
H(t) = \begin{cases}
f_1(H(4t)) & 0 \le t \le \frac{1}{4} \\
f_2(H(4t-1)) & \frac{1}{4} \le t \le \frac{1}{2} \\
f_3(H(4t-2)) & \frac{1}{2} \le t \le \frac{3}{4} \\
f_4(H(4t-3)) & \frac{3}{4} \le t \le 1
\end{cases}
$$

Geometric transformations $f_k$ are left as an exercise.  You may need to prove continuity at the transition points.


问:
$x+y+z=a+b+c$ and $xyz=abc$. Suppose that $a≤x<y<z≤c$ and $a<b<c$. Prove that $a=x$, $b=y$ and $c=z$.
Let $a,b,c,x,y,z$ be six positive real numbers satisfying
$$x+y+z=a+b+c$$ and $$xyz=abc$$ Further, suppose that $a≤x<y<z≤c$ and $a<b<c$. Prove that $a=x$, $b=y$ and $c=z$.

I have tried to convert this problem into two cubic polynomials using Vieta's Theorem.

But I don't have idea to proceed.
I am adding the two graphs for a visualization of the idea.

答:
Let $f(X) = (X-a)(X-b)(X-c), g(X) = (X-x)(X-y)(X-z)$.

By the equality condition $ f(X) - g(X) = k X$, where $ k = ab+bc+ca - (xy+yz+za)$.

Hint: Now, prove that $k = 0$, from which we can conclude that $ f(X) = g(X)$ so $ a=x, b =y, c = z$.

 Hint: Which condition have we not used as yet?

 By the condition of the roots, $f(a) = 0, g(a) \leq 0$ so $f(a) - g(a) = k a \geq 0 \Rightarrow k \geq 0$.

 By the condition of the roots, $ f(c) = 0, g(c) \geq 0 $ so $ f(c) - g(c) = k c \leq 0 \Rightarrow k \leq 0$.

Note: Technically, with $ f(X) = g(X)$, all that we can conclude is$ \{ a, b, c \} = \{ x, y, z \}$. Adding in $ a < b < c$ and $ x < y < z$, we can then conclude that $ a=x, b =y , c = z$.


问:
Conditional probability with consonants and vowels
A group is made up of ten people: four from Modena and six from Parma. A person from the group, chosen at random, writes the name of their city. A letter of the word written in this way is chosen at random which turns out to be a vowel. What is the probability that the person who wrote is from Parma?

I think I had to calculate $ P(P|V) = \frac{ P(V|P)P(P) }{P(V) } $ where $ P(P) $ is the probability that a person is from Parma, so  $ \frac{6}{10} $. $ P(P) $ is the probability that a letter is a consonants or a vowel so $ \frac{1}{2} $ and $ P(P|V) = \frac{3}{6} $ because this is the probability to have a vowel if we choosed a person from Modena. So I obtained that $ P(P|V) = 0.72 $ but I don’t know if is correct

答:
If a person is from PARMA the probability to get a vowel is $\frac{2}{5}$, not $\frac{1}{2}$

Asssuming that a person doesn't lie when writing down the name of his City, the requested probability is

$$\mathbb{P}[\text{PARMA}|V]=\frac{\frac{6}{10}\cdot\frac{2}{5}}{\frac{6}{10}\cdot\frac{2}{5}+\frac{4}{10}\cdot\frac{1}{2}}=\frac{6}{11}\approx0.545$$


问:
Limit of $a_{n+1}/a_n$
So it's given that
$$\lim_{n\to\infty}a_n = a$$ also $a\neq 0$.
I need to prove, using the definition of the limit $$\forall\epsilon\geqslant0.\exists n_\epsilon\in \mathbb{N}.\forall n>n_\epsilon.|a_n-L|<\epsilon$$
that $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=1$$
I tried saying that if $\forall\epsilon\geqslant0.\exists n_\epsilon\in \mathbb{N}.\forall n>n_\epsilon.|a_n-a|<\epsilon$ then, $a-\epsilon\leqslant a_n \leqslant a+\epsilon$ and the same for $a_{n+1}$: $a-\epsilon\leqslant a_{n+1}\leqslant a+\epsilon$, then
$$\left|\frac{a_{n+1}}{a_n}-1\right|=\left|\frac{a+\epsilon}{a-\epsilon}-1\right|=\left|\frac{a+\epsilon-a+\epsilon}{a-\epsilon}\right|=\left|\frac{2\epsilon}{a-\epsilon}\right|$$
I don't know if what I did was right, but I also don't know where to go from here, I would appreciate hints or guidance.

答: Write $\left|\frac{a_{n+1}}{a_n}-1 \right|=\left|\frac{a_{n+1}-a_n}{a_n}\right|$. Assume that $a_n \to a \neq 0$ so $|a_n| \to |a|>0$. Then for sufficiently large $n$, it follows that $|a_n|>\frac{|a|}{2}$ (why?). For the numerator, consider that the sequence $\{a_n\}$ must be Cauchy since it converges.

Title: Limit of $a_{n+1}/a_n$


问:
Finding a suitable function $f$ not to be Riemann integrable.
Given, $f:[\frac{1}{2},2] \to \mathbb{R} $ a strictly increasing function.

Let, $g(x)=f(x)+f(\frac{1}{x})$ ,$\forall x \in [1,2]$

(1) there is a suitable $f$ such that $U(P,g)=L(P,g)$, where $U(P,g)$ is upper Riemann sum and $L(P,g)$ is lower Riemann sum.

(2) there is suitable $f$ such that $U(P,g)\neq L(P,g)$

Option (1) is clearly true by taking $f=x$.

But what about option (2)?

As, $f$ is strictly increasing, it can have atmost countable number of discontinies.

So, then, $g(x)$ also have atmost countable number of discontinies $\implies$ $g$ is Riemann integrable.

But, I don't understand,how  option (2) is given as true in answer?

Edit : in comment, I have seen option (1) is true only if $f$ is constant. But, $f$ is strictly increasing, so, $f$ can't be constant. But option  (1) is also given as true in answer.

Please help me by giving an example so that option (1) is also true.

答: Actually, if $f(x)=x$, then $f(x)=x+\frac1x$, which is strictly increasing on $[1,2]$. Therefore, for any partition $P$ of $[1,2]$, $L(P,g)<U(P,g)$. This shows that the first option is false and that the second option is true.


答:
I did the case $a=1$.  The inner integral is
$$
\int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} r\;dz = 2r\sqrt{1-r^2}
$$
The next integral is
$$
\int_0^{\cos\theta} 2r\sqrt{1-r^2}\;dr = \frac{2}{3}\left(
1+(\cos^2\theta-1)\sqrt{1-\cos^2\theta}\right)
$$
In case $\theta > 0$, this simplifies to
$$
\frac{2}{3}\left(1+(\cos^2\theta-1) \sin \theta\right)
$$
But in case $\theta < 0$, it simplifies to
$$
\frac{2}{3}\left(1-(\cos^2\theta-1) \sin \theta\right)
$$
Now it is clear that for the outer integral, $\int_{-\pi/2}^{\pi/2}\cdots \;d\theta$
is not equal to $2 \int_{0}^{\pi/2}\cdots\;d\theta$ .

This is a good thing to warn calculus students about.  Do not (without thinking) simplify $\sqrt{1-\cos^2\theta}$ to $\sin \theta$.

Title: Why do each of these cylindrical triple integrals evaluate differently?

Body:
The problem in question is thus:
Find the volume cut out of the sphere of a radius $a$ centered at the origin by the polar curve $r = a\cos\theta$. I attempted to solve the problem using this cylindrical triple integral, taking advantage of $x$-axis symmetry.
$$
2\int_0^{\frac{\pi}{2}}\int_0^{a\cos\theta}\int_{-\sqrt{a^2 - r^2}}^{\sqrt{a^2 - r^2}}r\,dz\,dr\,d\theta = \frac{2\pi a^3}{3} - \frac{8a^3}{9}
$$
However, when this, similar integral is evaluated (which in my mind should yield the same results as the first one), the answer is different as the last term cancels out.
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{a\cos\theta}\int_{-\sqrt{a^2 - r^2}}^{\sqrt{a^2 - r^2}} r \, dz\,dr\,d\theta = \frac{2\pi a^3}{3}
$$
Which one of these answers is correct and why is the other one wrong? I understand that it has something to do with the theta integral, but I have no idea where my error is in assuming that these two triple integrals should produce identical results.

Thanks! This is my first time on the Math Stack Exchange so let me know if there is anything else I can do to improve the nature of my question.

THANK YOU ALL!!! Many wonderful explanations below, have a great day everyone!


问:
Does a limit of a function in a limit point equals only lateral limit on one side?
I have a function that goes like this:

$$f(x) = \begin{cases}\log(x)^{\log(x)}&x >  1\\
      a&x = 1
\end{cases}$$
I need to find a so that this function is continuous for $x = 1$.
Therefore, i need to calculate the lateral limits of $f(x)$ as $x$ approaches $1$.
How do i calculate the limit as $x$ approaches $1$ from left if the function is not defined for those values?

答:
The function is not defined for $x<1$, therefore we need to evaluate

$$\lim_{x\to 1^+} \log(x)^{\log(x)} =\lim_{x\to 1^+} e^{\log x \log(\log(x))}$$

form which we can find the value for $a$ which assures continuity.


答:
The working is not shown, so the following is conjecture on how the first and second integrals were worked out.

$$I(\theta)=\int_0^{a\cos\theta}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}}r\, dz\,dr=\int_0^{a\cos\theta}2r\sqrt{a^2-r^2}\,dr=\frac{4}{3}\Big[(a^2-r^2)^{3/2}\Big]_0^{a\cos\theta}=\frac{4a^3}{3}(\sin^3\theta-1)$$

Then $2\int_0^{\pi/2}\sin^3\theta\,d\theta=\frac{4}{3},\qquad\int_{-\pi/2}^{\pi/2}\sin^3\theta\,d\theta=0$ give out different answers.

The problem occurs because the last expression $\sin^3\theta-1$ is not correct. Substituting $\cos\theta$ into $(1-r^2)^{3/2}$ gives $(\sin^2\theta)^{3/2}=|\sin\theta|^3$ which is always non-negative.


问:
show that holomorpic function f such that f(1/2n) = f(1/2n+1) are constant
this is from my homework assignment:

let $f$ be holomorphic on the unit ball such that $f(\frac{1}{2n}) = f(\frac{1}{2n+1})$ for all $n \in \mathbb{N}$
show that f is constant.

答:
Uniqueness Theorem. Let $G$ be a connect domain where $\{x_n\}\subseteq G$ and $x_n\to x_0\in G$. Suppose $f,g$ are analytic and $f(x_n)=g(x_n)$, then $f=g$ on all of $G$.

Proof.

We will assume that $f$ is entire and $f(x_n)=0$ where $x_n\to 0$. We wish to show that $f$ is the constant zero function. This special case will establish the full theorem.

Since $f$ is continuous, $\lim f(x_n) = 0$ and so $f(0)=0$. We now have

$\displaystyle \left|f'(0)\right|=\lim_{R\to 0}\left|\frac{1}{2\pi i} \cdot \int_{B(0,R)} \frac{f(z)}{z} dz\right|\le \lim_{R\to 0} \frac{1}{2\pi} \cdot \frac{max_{B(0,R)}|f(z)|}{R} \cdot 2\pi R=0$

So, $f'(0)=0$. $f'$ is continuous at zero, so we now have

$\displaystyle \left|f''(0)\right|=\lim_{R\to 0}\left|\frac{1}{2\pi i} \cdot \int_{B(0,R)} \frac{f'(z)}{z} dz\right|\le \lim_{R\to 0} \frac{1}{2\pi} \cdot \frac{max_{B(0,R)}|f'(z)|}{R} \cdot 2\pi R=0$

So, $f''(0)=0$. In general if $f^{(k)}(0)=0$, then

$\displaystyle \left|f^{(k+1)}(0)\right|=\lim_{R\to 0}\left|\frac{1}{2\pi i} \cdot \int_{B(0,R)} \frac{f^{(k)}(z)}{z} dz\right|\le \lim_{R\to 0} \frac{1}{2\pi} \cdot \frac{max_{B(0,R)}|f^{(k)}(z)|}{R} \cdot 2\pi R=0$

Thus, $f^{(k)}(0)=0$ for every $k$. Thus, the taylor series coefficients of $f$ are zero and so $f$ is the zero function.


问:
If a sentence doesn't belong to any complete and closed theory...
the question is:

If $\varphi$ doesn't belong to any $T$ complete and closed under $\vdash$ then $\left \{ \varphi \right \}\vdash \neg\varphi$.

I initially thought about solving it through a reciprocal proof ($A$ implies $B$ iff not $B$ implies not $A$), but because the general statement has the form "not $A$ and $B$ implies $C$", then I got into problem while attempting to find the real "not $A$", then I proceeded with a contradiction argument, and I proved that for a formula $\varphi$ that would belong to a complete theory and with this deduction closeness cannot satisfy such condition '$C$', but Im not sure if this is right.

Thank you in advance.

答:
Lemma: If $\Sigma$ is a consistent set of $\mathcal{L}$-sentences, then there is a complete $\mathcal{L}$-theory $T$ containing $\Sigma$. (Proof below, but try proving it yourself first! The main idea is to use Zorn's lemma and the compactness theorem.)

Proof of Lemma: We will apply Zorn's lemma to the set of all consistent $\mathcal{L}$-theories containing $\Sigma$, partially ordered by inclusion. Since $\Sigma$ is consistent, it is in particular a member of this set, which is hence non-empty.

Now, suppose $(T_i)_{i\in I}$ is a chain of consistent theories, each of which contains $\Sigma$. Then let $T=\bigcup_{i\in I} T_i$. Certainly $T$ is a set of $\mathcal{L}$-sentences containing $\Sigma$, so we just need to show that it is consistent. By the compactness theorem, it suffices to show that $T$ is finitely consistent. However, if $\Delta\subseteq T$ is a finite subset of $T$, then there is some $i\in I$ such that $\Delta\subseteq T_i$, and hence (as a subset of a consistent set of sentences) $\Delta$ is consistent.

So $T$ is finitely consistent, hence consistent, and we can apply Zorn's lemma to obtain a maximal (with respect to inclusion) consistent $\mathcal{L}$-theory $T$ such that $\Sigma\subseteq T$. Then we just need to show that $T$ is complete; but indeed, if $\phi$ is an $\mathcal{L}$-sentence such that $T$ is consistent with $\phi$ (resp. $\neg\phi$), then by maximality of $T$ we have $\phi\in T$ (resp. $\neg\phi\in T$), as desired.

Given this lemma we may prove the result you want. Indeed, if $\varphi$ doesn't belong to any complete theory $T$, then applying the above lemma to the set $\Sigma=\{\varphi\}$ shows that $\varphi$ must be inconsistent. This means by definition that $\emptyset\models\neg\varphi$, and hence, by the completeness theorem, that $\emptyset\vdash\neg\varphi$, so in particular we have $\{\varphi\}\vdash\neg\varphi$, as desired.


问:
About the cardinality of a set
I'm trying to understand the following exercise about set theory:

We have:     $\Omega= \{a,b,c,d\}$

and we are supposed to figure out the cardinality of the following set:

$\textbf{{Y$\subseteq$ $\Omega$ | Y $\cap$ {a,b} = Y \ {c}}}$

According to my tutor the cardinality is $8$, because this set is a power set using the formula of the cardinality for a power set, $2^k$, $2^3$ we get $8$.

I think that I more or less understand what a power set is, in general, and how to compute the cardinality of a given power set.

But I still do not know how I can come to that conclusion in this exercise, or the answer for the cardinality of the set above. Or why the set above has anything to do with a power set.

答:
First of all, let's answer this part: "I don't understand how we get the cardinality of a power set".

So, consider a set $\mathrm{A}$ whose cardinality is $n$. Now, the power set $\mathcal{P}(\mathrm{A})$, will contain all sets which have any or no number of elements from $\mathrm{A}$. If you know combinatorics, then the number of ways, is you know,
$$\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n$$

Now, as for your question, let the considered set be $\mathrm{A}$.
$$\therefore A = \{\mathrm{Y}\subset\Omega :\ \mathrm{Y}\cap \{a, b\}=\mathrm{Y}\setminus \{c\}\}$$
Now, note that in any case $\mathrm{Y}\cap \{a,b\}\subset \{a,b\}$. Hence, the possible values of $\mathrm{Y}$ as of now are $\emptyset$, $\{a\}$, $\{b\}$ and $\{a,b\}$. Now, for every set $\mathrm{B}$ that does not include $c$, and the same set including $c$, the set $\mathrm{B}\setminus \{c\}$ remains the same. We had got four values for $\mathrm{Y}$, there exist two copies making a valid value of $\mathrm{A}$.

Therefore, the cardinality of $\mathrm{A}$ is $8$, it contains the sets $\emptyset$, $\{c\}$, $\{a\}$, $\{a,c\}$, $\{b\}$, $\{b,c\}$, $\{a,b\}$ and $\{a,b,c\}$.

Hope this helps :)

Title: About the cardinality of a set


问:
Is softmax "omni-periodic"?
A function is said to be "periodic" if there's a finite $T$ such that $f(t) = f(t \pm T), \forall$. Usually there's either one such $T$, or none at all. How about infinitely many $T$, that can also be complex?:

$$
f(t)=f(t \pm T),\ \forall t, T \in \mathbb{C} \tag{1}
$$

In some sense, Softmax qualifies; take any set of values, $(x_0, x_1, ..., x_{N-1})$, and add or subtract any number, real or complex, and the output's the same ($e^T$ factors from numerator and denominator).

It's easier to conceptualize if compared to a sine; with frequency $f=2$, we'd need to add (or subtract) $T=1/2$ from every input array value for every output array value to remain the same.

Is this interpretation sensible? Is there an entire class of such functions (with a formal name)?

To be clear, softmax cannot be directly expressed as $(1)$ since it can't directly be represented as $f(t)$, but if we assume all inputs are unique and sorted (and uniformly-spaced?), we obtain a discretization whose continuous variant may qualify for said interpretations.

Test code:

```<code>import numpy as np

def is_omniperiodic(fn):
    x = np.random.randn(10)
    ref = fn(x)
    for T in 100 * np.random.randn(100):
        assert np.allclose(fn(x + T), ref)

def softmax(x):
    return np.exp(x) / np.sum(np.exp(x))
</code>```

答:
For a continuous function of a real variable, you can have one (minimal) period.  If there are two or more periods (not rational multiples of each other) your function is constant.  I do not know of any reason to study real discontinuous functions with two or more periods.

For a continuous function of a complex variable, you can have two (minimal) periods, as long as one is not a real multiple of the other.  The study known as "elliptic functions" deals with such "doubly periodic" cases.  If they are non-constant, they cannot be entire functions, but they can be meromorphic.

See https://en.wikipedia.org/wiki/Doubly_periodic_function

I looked at the link on "softmax".  It seems to be a function on $K$-tuples to $K$-tuples like this:
$$
S(x_1,x_2,\dots, x_K) :=
\left(\frac{\exp(x_1)}{\sum_{k=1}^K \exp(x_k)},
\frac{\exp(x_2)}{\sum_{k=1}^K \exp(x_k)},
\dots,
\frac{\exp(x_K)}{\sum_{k=1}^K \exp(x_k)}\right)
$$
Now it is true that if you add the same thing $T$ to all the arguments, then the result is unchanged.
$S(x_1+T,x_2+T,\dots,x_K+T) = S(x_1,x_2,\dots,x_K)$.  But that is not what we would call "periodic".

Title: Is softmax "omni-periodic"?


问:
When is 2 a quadratic residue mod p
I'm trying to prove that 2 is a QR when $p = \pm \ 1 \text{ mod } 8.$ I know how to prove via a "factorial" proof, butI'm given the hint to use the fact that if $ \zeta $ is a primitive 8-root of unity, then $(\zeta + \bar{\zeta})^2 = 2$, and would like to solve the exercise using the hint. This is where I got so far: $$ \text{ I can use the fact that } 2 \text{ is a quadratic residue if and only if } 2^{\frac{p-1}{2}}=1 \text{ mod p}$$
$$ 2^{\frac{p-1}{2}} = (\zeta + \bar{\zeta})^{p-1} = \sum_{k = 0} ^{p-1} {p-1 \choose k} \bar\zeta^k \zeta^{p-1-k} = \sum_{k = 0} ^{p-1} {p-1 \choose k}\zeta^{p-1+6k}$$ Now I guess I should use the fact that $ \zeta^4 = -1$, and thus the symmetry of the sum cancels many factors, but I'm stuck at this point, and even computing the sum for small p didn't help much. Can someone help me out? Thanks!

答: Since $\zeta^p + \bar{\zeta}^p \equiv (\zeta + \bar{\zeta})^{p} \equiv (\zeta + \bar{\zeta})\left(\frac{2}{p}\right) \pmod{p}$, you can do casework on if $p \equiv \pm 1 \pmod{8}$ or $p \equiv \pm 3 \pmod{8}$ and the fact that $\zeta^4 \equiv -1 \pmod{p}$.

Title: When is 2 a quadratic residue mod p


问:
About the solutions of the diophantine equation $ x×(2a-x)=b$
Let us consider a real dynamical system $s′=g(s)$. In order to study the stability of the central manifold, we reformulate the problem as follows: For given fixed positive integers $a,b$, I am asking if this equation
$$ x(2a-x)=b$$
has positive integer solutions $1≤x<a$.

We can find the solution:

$$x=a\pm\sqrt{a^2-b}$$

however, we need $x$ to be an integer for a given $a,b$. So, my question is about ideas that permit us to conclude that a positive integer solution $x$ exists. I notice that we do not need the concepts from mapping theory and this is just an algebraic equation.

答:
Let $b=b_0b_1$ a factorization of $b$. If we have

$$x=b_0,\\2a-x=b_1$$ then $$2a=b_0+b_1.$$

So $a$ must be the arithmetic mean of two factors of $b$.


问:
Permutation of sets with fixed points problem

I asked some confusions regarding this in a previous post and now want to see if I am doing the problems correctly.

My solutions:
a) $\Omega$ sample space contains $5!=120$ elements.

b) by the inclusion exclusion principle $|A_1 \cup A_2 \cup A_3|= 4!+4!+4!-(3!+3!+3!)+(2!)= 56$

d) $\frac{56}{120}$

I am not sure if this is correct could anyone help me see if I am on the right track?

答:
Yes this is all correct. $A_i$ is the set of fixed permutations $n_i=i$ i.e., in which $i^{th}$ student collects his own gift.

So $A_1 \cup A_2 \cup A_3 $ describes when at least one of students $1,2,3$ collect their own gift.


问:
Why are the central angles of a regular polygon equal to the exterior angles?
I am currently in Geometry, doing proof, when me, my teacher, and my classmates realized that the exterior angles of a polygon are equal to the central angles of a polygon (for a regular polygon). What I want to know it why. Why is this true for any regular polygon? What proof gives this explanation? I will put a picture to help you visualize. [1. So can anyone explain why this is, and pictures to show markings would be appreciated.

答:
Hint. Notice that the sum of the interior angles of an $n$-gon is $180^\circ\cdot (n-2)$. Triangulate the $n$-gon and use induction in order to prove it.

Then, the sum of the exterior angles is $$180^\circ\cdot n-180^\circ\cdot (n-2)=360^\circ$$

Notice that this works for non-regular polygons too.

If you want to prove it for regular polygons without using this well-known formula, you may also approach it this way: consider you regular decagon (even though this is valid for any regular $n$-gon)

Due to symmetry, both of the smaller green angles are not only equal, but also add up to an "entire" green angle, which you see at the bottom. Since the sum of angles in a triangle is $180^\circ$, which is the same as one flat angle, we find that the two blue angles are the same. Can you take it from here?


问:
Relation between "harmonic form" and fourier series?
I am currently prepping for uni having been a few years out of the studying loop (programming as it happens). Anyway, I've been re-reading my A-level notes/exercises and looking through OpenCourseWare stuff and I noticed this assertation in my notes:

Given $a \cos\theta \pm b \sin\theta = c$ then:

$a \sin\theta \pm b\cos\theta = r\sin(\theta\pm\alpha)+c$

$a \cos\theta \pm b\sin\theta = r\cos(\theta\pm\alpha)+c$

Given $r \in \mathbb{N}$, $0 \leq \theta < 2\pi$.

Now, I've heard of fourier series which have a very similar form to these equestions.

So, my question is, is there a relation between the two?

Please bear in mind I know I'm stepping off a cliff and into "unknown unknowns" territory if they are - I have absolutely no idea what harmonic analysis is and I don't (yet) understand fourier series fully, although I grasp roughly how they work. I'm just interested to know if they are related and of course any further information / direction / reading I should follow up. I ask because this was one of those "odd topics" at A-level that we derived in class from a geometric argument then I've never seen again.

Thanks.

答:
I can't really make sense of those equations, but it looks something like the statement that for any $ a, b \in \mathbb{R} $, then for $ r = \sqrt{a^2 + b^2 }$ and $ \phi = \arctan(b/a) $ we have

$ a \sin(\theta) + b \cos(\theta) = r \sin(\theta + \phi).$

If you visualize x and y axes with $ cos(\theta) $ along the x-axis and $ sin(\theta) $ along the y-axis, then this says we can identify any sinusoidal, i.e. any function like $ r \sin(\theta + \phi) $, by its cartesian coordinates $(a,b)$ in this plane, or its polar coordinate $(r, \phi )$.  $r$ is called the magnitude of the sinusodial, and $ \phi $ the phase.

This is directly related to Fourier Analysis.  The main idea in Fourier Analysis is to decompose a function into its sinusodial components.  For instance, if $f$ is a real-valued function on the interval $ [0, 2\pi] $ that is suitably regular, it turns out you can write $f$ uniquely as an infinite sum of sine and cosines, called the Fourier Series:

$ f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)). $

The numbers $ a_n $ and $ b_n $ are called the Fourier coefficients of your function.

I think this representation of Fourier series is sometimes difficult to grasp.  But using the relationship above, we see we could have also wrote:

$ f(x) = a_0 + \sum_{n=1}^{\infty} r_n \sin(nx + \phi_n) $

In this form I think you can see more clearly what's going on: For each $n$, there is only one sinusoidal component of $f$ of frequency $n$; $r_n = \sqrt{a_n^2 +b_n^2}$ tells you its magnitude and $\phi_n$ tells you its phase.

ContentLicense: CC BY-SA 2.5


问:
Is there a usual method for finding the minimal polynomial of trigonometric values?
I've been thinking a bit about finding the minimal polynomials of side lengths of regular $n$-gons inscribed in the unit circle. For example, I recently wanted to find the minimal polynomial of the side length of an inscribed regular nonagon. Using the law of cosines, I was able to find that the side length is $a=\sqrt{2-2\cos(\frac{2\pi}{9})}$.

So $\displaystyle \frac{2-a^2}{2}=\cos \left(\frac{2\pi}{9} \right)$, but since I can't express $\displaystyle\cos \left(\frac{2\pi}{9} \right)$ in terms of rational numbers or their square roots, I'm unsure of how to proceed exactly. Is there a usual method to attack values like this? Possibly for say $\displaystyle \cos \left(\frac{m\pi}{n} \right)$?

Thanks.

答:
EDIT : Following Marty Green's approach, I managed to actually find the minimal polynomial explicitly. I think both of you (Marty and Dom) will appreciate. Write
$$
2 \cos \left( \frac{2\pi}9 \right) = \zeta_2 + \zeta_7
$$
where $\zeta_2$ and $\zeta_7$ are the second and seventh rooth of unity, respectively. Now we try to find the minimal polynomial for $\zeta_2 + \zeta_7$ (if we found it then substituting $\zeta_2 + \zeta_7 = x = 2y$ in the polynomial gives us our wished polynomial). Notice that
$$
x^9 -1 = \prod_{d \mid n} \Phi_d(x) = (x-1)(x^2+x+1)(x^6 + x^3 + 1)
$$
where $\Phi_d(x)$ is the cyclotomic polynomial of the $d^{\text{th}}$ roots of unity. Now since
$$
\phi_9(x) = x^6 + x^3 + 1 = \prod_{(i,9)=1}(x-\zeta_i)
$$
and developping this product to obtain the symmetric polynomials in the $\zeta_i$ as coefficients of $\phi_9(x)$, we obtain the two useful identities :
$$
\sum_{(i,9)=1} \zeta_i = 0, \quad \underset{i \neq j}{\sum_{(i,9)=(j,9)=1}} \zeta_i \zeta_j = 0.
$$
Now we also have the identity $\zeta_6 + \zeta_3 + 1 = 0$ since $\zeta_3$ is a root of $x^2 + x + 1$. Hence we have
$$
(x-(\zeta_1 + \zeta_8))(x-(\zeta_2 + \zeta_7))(x-(\zeta_4 + \zeta_5)) =
$$
$$
x^3 - ((\zeta_1 + \zeta_8) + (\zeta_2 + \zeta_7) + (\zeta_4 + \zeta_5))x^2 +
$$
$$
((\zeta_1 + \zeta_8)(\zeta_2 + \zeta_7) + (\zeta_1 + \zeta_8)(\zeta_4 + \zeta_5) + (\zeta_2 + \zeta_7)(\zeta_4 + \zeta_5))x
$$
$$
 - (\zeta_1 + \zeta_8)(\zeta_2 + \zeta_7)(\zeta_4 + \zeta_5).
$$
Since the sum of the roots of $\Phi_9$ is $0$, the quadratic term vanishes. For the linear term, notice that all pairs of roots appear as products of two roots beside the ones that are together in the sum, i.e. $\zeta_1 \zeta_8$, $\zeta_2 \zeta_7$ and $\zeta_4 \zeta_5$. Hence if we call the linear term $\alpha$,
$$
\alpha = \left( \underset{i \neq j}{\sum_{(i,9)=(j,9)=1}} \zeta_i \zeta_j \right) - \zeta_1\zeta_8 - \zeta_2 \zeta_7 - \zeta_4 \zeta_5 = 0 - \zeta_9 - \zeta_9 - \zeta_9 = -3.
$$
Now for the constant in the end, develop it manually to have something like this :
$$
 - (\zeta_1 + \zeta_8)(\zeta_2 + \zeta_7)(\zeta_4 + \zeta_5) =
$$
$$
 - (\zeta_1 \zeta_2 \zeta_4 + \zeta_1 \zeta_2 \zeta_5 + \zeta_1 \zeta_7 \zeta_4 + \zeta_1 \zeta_7 \zeta_5 +
$$
$$
\zeta_8 \zeta_2 \zeta_4 + \zeta_8 \zeta_2 \zeta_5 + \zeta_8 \zeta_7 \zeta_4 + \zeta_8 \zeta_7 \zeta_5) =
$$
$$
- (\zeta_7 + \zeta_8 + \zeta_3 + \zeta_4 + \zeta_5 + \zeta_6 + \zeta_1 + \zeta_2) = -(\zeta_3 + \zeta_6) = 1
$$
The last line is because the sum of the roots is $0$. Hence a polynomial for which $\zeta_2 + \zeta_7$ is a root is $x^3 - 3x + 1$. Making the variable change to get the polynomial for $\cos(2\pi / 9)$ gives us $8x^3 - 6x + 1 = 0$, which is irreducible (there are many possible answer to this irreducibility question non-rationality of the roots, rational root theorem, etc.) which I leave up to you.

P.S. I will never do that again, it's so painful... but it was fun. XD Hope you liked it.


问:
What is the best way to solve a limit of $\log(x)^{\log(x)}$?
I have this function which I need to calculate the limit of $\log(x)^{\log(x)}$ as $x$ approaches 1.
While my mind tells me that with logic, $0+^0+$ equals to number to the power of 0, therefore = 1,
but I don't know how to prove it.

Body:
Let $y = \lim_{x\to1} (\ln x)^{\ln x}$. Then we have
$$\ln y = \lim_{x\to1}\frac{\ln(\ln x)}{1/\ln x} = \lim_{x\to1} \frac{\frac{1}{x \ln x}}{-\frac{1}{x(\ln x)^2}} = - \lim_{x\to 1} \ln x = 0$$
So $y = 1$.


问:
Steps to turn on $2^n$ lamps on a roulette wheel

There are $2^n$ lamps on a roulette wheel. Each time player A decides
on a list of lamps to flip (so on becomes off and off becomes on).
Player B sees player A's list, and rotates the roulette wheel to a
different location. Player A still flip the lights as if they were on
the original location. For example, say there are four lights
$1,2,3,4$. Player A decides to flip the switch of lamp $1$. Player B
rotates the wheel to $3,4,1,2$. Then after this round, lamp $3$ is
flipped.

At the beginning, some lamps are on and some are off. Prove that regardless of the initial state, for no more than
$2^{2n}$ rounds, player A can always turn on all the lights.

There is a hint to this problem. Prove by mathematical induction. Suppose there is a way for $2^{k}$ lamps. Consider the case when $2^{k+1}$. Pair up the lamps that are opposite of each other...

I am thinking after we pair up the lamps that are opposite of each other, We define a new state for this $2^{k}$  pairs. If the pair has different states, we treat them as an "off" lamp. Otherwise we treat them as an "on" light. But there is no way to turn on this "off" light in this new, paired up roulette. My idea could be completely off.

Certainly, a solution without using this hint is welcome too.

答:
It looks to me that you've made significant progress. Before reading further, I encourage you to push through from what you have, with the knowledge that your path does indeed lead to a solution. In fact, my solution is just what you have and what's in the comments.

If you want to read further, scan through the hints as a suggestion, before reading in more detail.

I also provided some motivational steps that hint we're proceeding in the right direction. These obviously could be skipped when writing up an actual solution.

Hint 1: Consider the last step: What possible state could A be in which guarantees no matter what B choses, A will win?

 The only state is where all lamps are off, and A chooses all lamps to be turned on. Then, no matter what B does, A will win.

 Conversely, given any state where some lamp is on, B can rotate so that it is turned off, so A will not win (yet).

 Observation: This step is unique. It suggests that we can push this train of thought further.

Hint 2: Consider the second last step: How can A get to the last step.

 One possible approach is that every other lamp is on, and A decides to turn on every other lamp. Then, if B lands on an on lamp, A turns all of them off (getting us to the last step). If B lands on an off lamp, A turns all of then on (and is done).

 Note that this approach isn't unique, for example the pattern could be "on on off off, ..... " so we might not be able to backtrack further.

 However, what do these approaches have in common? It seems like the diametrically opposite lamps have to be the same state.

 In fact, we can show that if a state has diametrically opposite lamps that are on/off, then we cannot reach an "all on" or "all off" state. (How?)

So, this line of reasoning indicates that we likely really want to get to "diametrically opposite lamps are both on or both off".

Hint 3: Setting up the induction.

 Given $2^{k+1}$ lights, we'd call it scenario X.

 We construct scenario Y with $2^k$ lights as follows:

 - If lights $i, 2^k + i$ of X are in the same on/off, then light $i$ in Y is on.

 - If lights $i, 2^k + i$ of X are in different on/off, then light $i$ in Y is off.

 Fill in this gap: If A wants to turn on lamps 1 and 3 in Y, what is the equivalent sequence in X?

Hint 4: Run the induction, see what we get.

 No matter what the $2^k$ lamps in Y are, after at most $2^{2k}$ steps we can get them to all on.

 Running the equivalent sequence on X, we get that lamps $ i, 2^k + i$ are all both on/off. Call this X'.

 Obviously this isn't sufficient, or is it?

Hint 5: Setup the induction again.

 Now, after $2^{2k}$ steps, X' has $2^{k+1}$ lights that are matched up with their opposite partners. We construct scenario Z with $2^k$ lamp in the following manner:

 - Lamp $i$ in X' has the same on/off as lamp $i$ scenario Z for $1 \leq i \leq 2^k$.

 Fill in this gap: If A wants to turn on lamps 1 and 3 in Z, what is the equivalent sequence in X'?

Hint 6: Run the induction again.

 No matter what the $2^k$ lamps in Z are, after $2^{2k}$ steps we can get them to all on.

 This means that lamps $ i$ and $ 2^k + i$ for $ 1 \leq i \leq 2^k$ in X' are all on.

Hence, we are done.

Bonus: Note that in the induction, if $2^k$ lamps takes $f(k)$ steps, then $2^{k+1}$ lamps takes $2f(k)$ steps. Since we can easily verify that $2^0$ lamps takes (at most) $1$ step, we can conclude that $f(k) = 2^k$ steps, which is much better than the $2^{2k}$ steps that they gave us.


问:
Example of a subspace of the p-adic numbers which is not locally compact
Let $p$ be a prime number, and let $\mathbf{Q}_p$ be the field of $p$-adic numbers together with its topology derived from the $p$-adic valuation. Can someone give me an example (together with an argument) of a subspace of $\mathbf{Q}_p$ which is not locally compact? Thank you in advance.

P.S. My motivation arises from the fact that given the field of real numbers $\mathbb{R}$ together with its euclidean topology, the subspace of rational numbers $\mathbb{Q}$ is not locally compact. Maybe the same happens if we complete $\mathbb{Q}$ with respect to the metric arising from the $p-$adic valuation?

答:
A subspace $A\subseteq X$ of a locally compact Hausdorff space $X$ is locally compact if and only if it is locally closed. That is if and only if $A$ is open in its closure $\overline A$.

Since $\mathbb{Q}_p$ is Hausdorff and locally compact the statement applies. The subspace $\mathbb{Q}$ is dense in $\mathbb{Q}_p$, but not open. Hence it cannot be locally compact in the subspace topology.

Note that the same arguement shows that $\mathbb{Q}$ is not locally compact in the subspace topology inherited from $\mathbb{R}$.


问:
Struggling to find solution: question 2
$$xy'-2y=\frac{x^6}{(y+x^2)}$$

Can someone please help with this one?

It looks like a bernoulli equation, I have tried to multiply both sides by denominator and gather them in parantheses to leave the $y'$ alone but to no avail.

答:
$xy'-2y=\dfrac{x^6}{y+x^2} \, $ i.e, $\displaystyle y'- \frac{2y}{x}=\dfrac{x^5}{y+x^2}$

LHS form is $y' + f(x)y = g(x)$.

Integrating factor $I = \displaystyle e^{\int f(x)dx} = e^{\int \frac{-2}{x}dx} = e^{-2 \, lnx} =  \frac{1}{x^2}$.

So, $\displaystyle \frac{y'}{x^2} - \frac{2y}{x^3}= \dfrac{x^3}{y+x^2}$

$\displaystyle \frac{d}{dx}\left({\frac{y}{x^2}}\right) = \dfrac{x}{(y/x^2)+1}$

It is straightforward from here.

Title: Struggling to find solution: question 2


问:
Find two functions $ f,g: \mathbb{R}\rightarrow \mathbb{R}$
Given a function $\sigma: \mathbb{R}\rightarrow\mathbb{R}$ which satisfies the following properties$(x\in\mathbb{R}, n\in \mathbb{N}):$

$$\sigma(n)=0, \sigma(x+1)=\sigma(x), \sigma(n+\frac{1}{2})=1$$
Find two functions $f,g:\mathbb{R}\rightarrow \mathbb{R}$ with $g(x)\neq0$ for all $x$ such that $\sigma(x)=g(x)(f(x)+1)$.

I have tried:

We have $\sigma(n)=0, \forall n\in\mathbb{N}$
$$\Rightarrow \sigma(1)=\sigma(2)=\cdots=0$$
And $\sigma(x+1)=\sigma(x)\Rightarrow \sigma(x)$ is periodic function with period $=1$.

$\sigma(n+\frac{1}{2})\Rightarrow \sigma(\frac{3}{2})=\sigma(\frac{5}{3})=\cdots=1 $

$\sigma(x)=g(x)(f(x)+1)$ with $g(x)\neq 0$
Since, we have periodic function, so we will use trigonometric function for $f(x)$ and $g(x)$

…………………………

Please kindly help me on this.

答:
This diagram shows the set of points that are defined by the description you have given
: zero for integer values and 1 at the mid-points.

You need a curve that will pass through these points, and you are right to think that something trigonometric will do the task.

You should be aware that one obvious candidates is $\sigma(x)=\sin^2 (\pi x)$

Let $g(x)=1$ as suggested in the comments and let $f(x)=\sin^2 (\pi x)-1$ and then you have $\sigma(x)=g(x)(f(x)+1)$

Pick any periodic function. $g(x)=0.4 \sin^2 (\pi x) + 0.1$ has the characteristics you require since $0.1 \le g(x) \le 0.5$. You will then need to change $f(x)$ to $f(x)=2 \sin^2 (\pi x)-1$ to keep your peaks at $1$.


问:
Implication: F $\implies$ T
Why is F $\implies$ T taken as true? Why is this the "convention"?</p>

答:
It's a convention in classical logic. Not all logical systems make this convention.

To see why this convention makes sense, consider a statement of the form $P\implies Q$, which I make, swearing that I'm telling the truth. When will you be able to sue me for lying? Well, if $P$ is true, but $Q$ is false then certainly you can sue me. But if $P$ is not true and $Q$ is, and you try to sue me, the judge will wonder what the hell you want. I only said that if $P$ is true then $Q$ is true. So, if $P$ is false, I did not commit to anything and thus your case is dismissed (i.e., I was telling the truth).

The point is that the implication is material implication, not that of causal implication. You may disagree with that interpretation and develop a logic where it is interpreted differently (for instance, in paraconsistent logic a statement of the form $F\implies T$ is dismissed as meaningless, not having any truth value (simply saying, don't bother about such nonsense as pondering what results from a false assumption)). Still, using classical logic in mathematics is very powerful, and so we make this convention.

Title: Implication: F $\implies$ T


答:
The figure below shows a decagon.

Let us call the central angle $\angle AKB = \angle BKC = \theta$. It has measure $\dfrac{360^\circ}n$ by rotational symmetry, but this value is not needed here.

Since both $\triangle AKB$, $\triangle BKC$ are isosceles, we have

$$\angle KAB = \angle KBA = \angle KBC = \angle KCB = \frac12(180^\circ - \theta)$$

So we have, by adjacent angles on straight lines:

$$\angle LBC = 180^\circ - \angle KBA - \angle KBC = 180^\circ - (180^\circ - \theta) = \theta$$

proving our assertion.

If a rigorous proof is not needed, try to rotate the regular polygon. When $AB$ is rotated with respect to the center $K$ to $BC$, the angle of rotation is the center angle $\theta$. At the same time, the angle rotated is also the angle formed by $AB$ and $BC$, which is the exterior angle $\angle LBC$. (This argument can also be extended to a proof of the sum of exterior angles of a convex polygon theorem)

Title: Why are the central angles of a regular polygon equal to the exterior angles?

Body: I am currently in Geometry, doing proof, when me, my teacher, and my classmates realized that the exterior angles of a polygon are equal to the central angles of a polygon (for a regular polygon). What I want to know it why. Why is this true for any regular polygon? What proof gives this explanation? I will put a picture to help you visualize. [1. So can anyone explain why this is, and pictures to show markings would be appreciated.


问:
Proof of Gaussian elimination/Why does it work
I have just had a class on linear algebra and the professor explained how to solve matrixes. While he could explain how to solve them by using Gaussian's elimination, he failed to explain how does that work.

Why does matrix before doing any operations have the same solutions as the matrix after "changing" a row with Gaussian elimination?

Where can I read the proof?

答:
This is the best I can find online: https://www.youtube.com/watch?v=bnC848ie16Q&ab_channel=MyWhyU

Essentially, the elimination creates a new plane with the same liner solution as the old plane (before the subtraction) - i.e.: the two planes still intersect, but when you eliminate the "x", "y" or "z" dimension, that plane reorients (twists) to maintain the solution parallel to the axis on which the dimension was collapsed.

This makes sense if you think about subtracting two linear equations that intersect - the pivot point will remain the same.

Enjoy !

OwnerDisplayName: khajvah


问:
Are all minimum cuts on a graph equivalent?
In my discrete math class, our professor is insistent on using a labeling algorithm to find the max flow of a graph. This is fine, although the Ford-Fulkerson algorithm makes more sense to me. When the labeling algorithm terminates, it has the side-effect of giving us sets $P$ and $\bar{P}$ separated by a minimum cut based on which vertices have been labelled when the algorithm terminates. My understanding is that the Ford-Fulkerson algorithm won't provide a specific minimum cut, but will give you the min-cut value.

Are all minimum cuts equivalent, at least in these small introductory problems?

Extra question: I can't even begin to think of how I'd make this rigorous, but would it be reasonable to say that the final residual graph given by the Ford-Fulkerson algorithm could give us minimum cut sets $P$ and $\bar{P}$ based on where the path from $s$ to $t$ breaks down on the saturated edges?

答: I'm not sure what you mean by "equivalent", but there may be many different minimum cuts. As an extreme example, if your graph consists of a source $s$, sink $t$, and $n$ other vertices each having an edge from $s$ of weight $1$ and an edge to $t$ of weight $1$, then all of the $2^n$ possible cuts are minimum cuts of value $n$. But all minimum cuts have the same value, by definition.


问:
quasi-component in nested rectangles counterexample
Example 115 of Steen & Seebach's "Counterexamples in Topology" is of a pair of vertical lines in the Euclidean plane at x = -1 & x = 1, called $L_1$ & $L_2$ along with an infinite collection of nested rectangles, $R_n$, which lay between the 2 vertical lines, and whose vertical sides get arbitrary close to $L_1$ & $L_2$ (each $R_n$ has height $2n$ and width $\frac{2n}{(n+1)}$ for $n=0,1,...$). The set of points in these lines and rectangles, $X$, inherit the topology of the Euclidean plane.

They state that $L_1\cup L_2 = X-\cup_n R_n$ is a quasi-component of X, i.e. the intersection of all sets containing a point in $L_1\cup L_2$ which are both open & closed (according to the definition on page 29). However, this doesn't make sense to me, since surely $L_1$ is both open and closed (it's a union of open sets, hence open, and it's the complement of $L_2\cup \cup_n R_n$ which is a union of open sets, hence $L_1$ is also closed).

Can anyone help me understand this example?

答:
$L_1$ and $L_2$ are not open in $X$. Due to the way the sides of the rectangles $R_n$ become increasingly close to $L_1$ and $L_2$ (and increasingly tall), any neighbourhood of any point in $L_1$ must intersect infinitely many of the rectangles $R_n$.

Fix $(-1, y_0) \in L_1$. Suppose $C \subseteq X$ contains $(-1, y_0)$ and is clopen. Since $C$ is open, there exists an $\varepsilon > 0$ such that, for any $(x, y) \in X$,
$$d((x, y), (-1, y_0)) \le \varepsilon \implies (x, y) \in C.$$
Let $A_n$ be the left line segment of $R_n$, i.e.
$$A_n = \operatorname{conv}\left\{\left(-\frac{n}{n + 1}, -n\right), \left(-\frac{n}{n + 1}, n\right)\right\}.$$
Let $N \in \Bbb{N}$ such that
$$N \ge \max\left\{\frac{1}{\varepsilon} - 1, |y_0|\right\}.$$
Suppose $n \ge N$. Then $|y_0| \le n$, hence
$$\left(-\frac{n}{n+1},y_0\right) \in A_n.$$
Additionally,
$$d\left(\left(-\frac{n}{n+1},y_0\right), (-1, y_0)\right) = \left|-\frac{n}{n+1} + 1\right| = \frac{1}{n+1} \le \varepsilon \implies \left(-\frac{n}{n+1},y_0\right) \in C.$$
Note that $C \cap R_n$ is clopen. Since $R_n$ is connected, and $C \cap R_n$ is non-trivial by the above argument, $R_n \subseteq C$ for all $n \ge N$.

On the right side, we similarly see
$$n \ge N \implies \left(\frac{n}{n+1},y_0\right) \in R_n \subseteq C,$$
and this sequence converges to $(1, y_0)$. As $C$ is closed, $(1, y_0) \in C$ too. That is, a point in $L_1$ implies a point in $L_2$ too. As $L_1$ and $L_2$ are connected subsets of $X$, we get $L_1 \cup L_2 \subseteq C$.

Conversely, $L_1 \cup L_2 \cup \bigcup_{n \ge N} R_n$ is a family of clopen subsets of $X$ (as $N$ ranges over $\Bbb{N}$) whose intersection is $L_1 \cup L_2$, and which contain $(-1, y_0)$. Hence $L_1 \cup L_2$ is indeed the quasicomponent of $(-1, y_0)$.


问:
An integral of the function $f(x^2)$ when $x^2$ is set equal to $t$
Question

Given that $F^{\prime}(x)=f(x)$,find the value of the integral below:
$$
\int_{a}^{b} x f\left(x^{2}\right) d x
$$

Work: (wondering if the work and the answer is correct. Thank you very much):
Original image

Given:
$$
F^{\prime}(x)=f(x)
$$
Let $$I=\int_{a}^{b} x \cdot f\left(x^{2}\right) d x\tag1\label{eq1}$$
Let $$x^{2}=t \quad \Rightarrow \quad x d x=\frac{d t}{2}$$

So from Equation $\ref{eq1}$

$$I=\frac{1}{2} \int_{a}^{b} f(t) d t$$
$$I=\frac{1}{2}[F(t)]_{a}^{b}$$
$$I=L_{2}^{L}\left[F\left(x^{2}\right)\right]_{a}^{b}$$
$$\Rightarrow \quad I=\frac{1}{2}\left(F\left(b^{2}\right)-F\left(a^{2}\right)\right)$$
So
$$I=\frac{F\left(b^{2}\right)-F\left(a^{2}\right)}{2}$$

答: Yes, it's right. It can be made more concise viz.$$t=x^2\implies xdx=\tfrac12dt\implies\int_a^bxf(x)dx=\int_{a^2}^{b^2}\tfrac12f(t)dt=\tfrac12(F(b^2)-F(a^2)).$$


答:
The metric space $\Bbb{Z}$ (with the $p$-adic metric) is covered by the collection of open sets $$U_n = n+p^{|n|+3} \Bbb{Z},n\in\Bbb{Z}$$ but no finite subcollection $$\bigcup_{n=-N}^N U_n$$
covers it, because $U_n$ covers $p^{N-|n|}$ residue classes modulo $p^{N+3}$, so $\bigcup_{n=-N}^N U_n$ covers at most $\sum_{n=-N}^N p^{N-|n|}$ residue classes modulo $p^{N+3}$, ie. not them all.

Title: Example of a subspace of the p-adic numbers which is not locally compact

Body:
Let $p$ be a prime number, and let $\mathbf{Q}_p$ be the field of $p$-adic numbers together with its topology derived from the $p$-adic valuation. Can someone give me an example (together with an argument) of a subspace of $\mathbf{Q}_p$ which is not locally compact? Thank you in advance.

P.S. My motivation arises from the fact that given the field of real numbers $\mathbb{R}$ together with its euclidean topology, the subspace of rational numbers $\mathbb{Q}$ is not locally compact. Maybe the same happens if we complete $\mathbb{Q}$ with respect to the metric arising from the $p-$adic valuation?


问:
Algebraic structure $(Z^{3},\star)$
I need help solving this task, if anyone had a similar problem it would help me.

The task is:

Prove the associativity of an algebraic structure
$(Z^{3},\star) $
if operation $\star$ is defined with
$(a,b,c)\star(x,y,z)=(a+(-1)^{b}x,b+(-1)^{c}y,c+(-1)^{a}z)$

I tried this:

$((a,b,c)\star(x,y,z))\star(d,e,f) =  (a,b,c)\star(((x,y,z)\star(d,e,f)))$

$(a+(-1)^{b}x,b+(-1)^{c}y,c+(-1)^{a}z)\star (d,e,f)=(a,b,c)\star(x+(-1)^{y}d,y+(-1)^{z}e,z+(-1)^{x}f)$

And now, I don t know what to do next?

Thanks in advance !

答:
If we expand out the left hand side, we obtain \begin{align*}((a,b,c)\star(x,y,z))\star(d,e,f) &= (a + (-1)^bx,b+(-1)^cy,c+(-1)^az)\star(d,e,f)\\&= (a + (-1)^bx + (-1)^{b+(-1)^cy}d ,b + (-1)^cy + (-1)^{c + (-1)^az}e,c+(-1)^az+(-1)^{a+(-1)^bx}f\end{align*}

The right hand side similar expands as

\begin{align*}(a,b,c)\star((x,y,z)\star(d,e,f))&= (a,b,c)\star(x + (-1)^yd,y+(-1)^ze,z+(-1)^xf)\\&=(a + (-1)^b(x + (-1)^yd), b + (-1)^c(y + (-1)^ze),c+(-1)^a(z + (-1)^xf)\end{align*}

So now, all we need to do is show that these two are equal. To keep track of that, we'll subtract them, and see what we end up with:

\begin{align*}&(a,b,c)\star((x,y,z)\star(d,e,f))-((a,b,c)\star(x,y,z))\star(d,e,f)\\&=((-1)^{b + (-1)^cy}d-(-1)^{b+y}d,(-1)^{c+(-1)^az}e-(-1)^{c+z}e,(-1)^{a+(-1)^bz}f-(-1)^{c+z}f)
\\&=\left((-1)^bd\left(\left((-1)^{(-1)^c}\right)^y-(-1)^{y}\right),(-1)^ce\left(\left((-1)^{(-1)^a}\right)^z-(-1)^z\right),(-1)^af\left(\left((-1)^{(-1)^b}\right)^x-(-1)^x\right)\right).\end{align*}

Now, note that $(-1)^{(-1)^n} = -1$ for all $n$ (since both $1$ and $-1$ are odd), so that last line simplifies down to

$$\left((-1)^bd\left((-1)^y-(-1)^y\right),(-1)^ce\left((-1)^z-(-1)^z\right),(-1)^af\left((-1)^x-(-1)^x\right)\right) = 0.$$

Thus, $\star$ is associative.

Title: Algebraic structure $(Z^{3},\star)$


问:
Properties of compact operators
I'm working on the following problem, regarding compact operators.

Let $T$ and $S$ be bounded linear operators on an infinite-dimensional Hilbert space. Prove or disprove the following statements:

If $T^n=I$ for some $n$, then $T$ is not compact.

If $T^2=0$ then $T$ is compact.

If $TS$ is compact then either $T$ or $S$ is compact.

The definition of compactness that I'm familiar with is: a linear operator $T$ is compact if and only if whenever $(x_n)$ is a bounded sequence, $(Tx_n)$ contains a convergent subsequence.

What I've done: For (1), I have seen the proof that an idempotent operator is compact if and only if it has finite rank. For (2), I am not sure how to start. For (3), I have only been able to prove the converse (if either $T$ or $S$ is compact than $TS$ is compact).

Any help on this problem? Thanks!

答:
Here are some hints to help you along the way.

Use the converse of (3) to show that the negation of (1) is false (i.e., (1) is true).

For (2), consider weighted shift operators, with sequence of weights $(1,0,1,0,1,0,\ldots)$.  Show this operator is non-compact but squares to zero.

For (3), the answer to (2) shows that this is false.

Title: Properties of compact operators


答:
If $T^n=I$ and $T$ is compact, then $T^n$ would be compact and so would be the identity, which is not possible in an infinite dimensional Hilbert space.

Let $H$ be the space of square summable sequences and $T\colon (x_k)_{k\geq 1}\mapsto (\lambda_k x_{k+1})_{k\geqslant 1}$, where $\lambda_k=1$ if $k$ is even and $\lambda_k=0$ if $k$ is odd.

Title: Properties of compact operators

Body:
I'm working on the following problem, regarding compact operators.

Let $T$ and $S$ be bounded linear operators on an infinite-dimensional Hilbert space. Prove or disprove the following statements:

If $T^n=I$ for some $n$, then $T$ is not compact.

If $T^2=0$ then $T$ is compact.

If $TS$ is compact then either $T$ or $S$ is compact.

The definition of compactness that I'm familiar with is: a linear operator $T$ is compact if and only if whenever $(x_n)$ is a bounded sequence, $(Tx_n)$ contains a convergent subsequence.

What I've done: For (1), I have seen the proof that an idempotent operator is compact if and only if it has finite rank. For (2), I am not sure how to start. For (3), I have only been able to prove the converse (if either $T$ or $S$ is compact than $TS$ is compact).

Any help on this problem? Thanks!


问:
Approximation in series
How to approximate $[1 - 1/(2x)]$ to $[1/(1 + (1/2x))]$ when $x$ very large. What power series should be used?. This problem is used to find the precession velocity with the Yukawa potential.

答: $1/(1-x)$ approximate $1+x$ when $x$ very small.

OwnerDisplayName: Cesar Antonio Hoyos Pelaez

Title: Approximation in series


答:
I assume you mean how is

\begin{equation}
\frac{1}{1+\frac{1}{2x}}
\end{equation}
approximated by

\begin{equation}
1-\frac{1}{2x}.
\end{equation}
Consider Taylor expanding the function

\begin{equation}
(1+x)^n
\end{equation}
around zero and only keeping terms up to first order. The first derivative is

\begin{equation}
\frac{d}{dx}(1+x)^n = n (1+x)^{n-1},
\end{equation}
so the expansion becomes

\begin{align}
(1+x)^n \approx & (1+x)^n |_{x=0} + n (1+x)^{n-1}|_{x=0} (x-0)\\
= & 1 + n x.
\end{align}
In your case, the function is

\begin{equation}
\left( 1+\frac{1}{2x} \right)^{-1}.
\end{equation}
The second term becomes small for $x \rightarrow \infty$ and $n=-1$, so

\begin{equation}
\frac{1}{1+\frac{1}{2x}} \approx 1 - \frac{1}{2x}.
\end{equation}

Title: Approximation in series

Body: How to approximate $[1 - 1/(2x)]$ to $[1/(1 + (1/2x))]$ when $x$ very large. What power series should be used?. This problem is used to find the precession velocity with the Yukawa potential.

OwnerDisplayName: Nicolai


问:
explain $df(tx).x = \sum_{i=1}^n {\partial f\over \partial x_i}(tx)x_i \hspace{1cm} x\in \mathbb R^n$
the question is :

let $U$ be a Neighbourhood of the origine of $R^n$ and :

$x\in U \Rightarrow tx \in U , \forall t\in U $

let f be a numeric function defined in U , and $f(0)= 0$

if we have $f\in C^1$ , prove the existence of $n$ numeric functions $f_i$ defined in U /
$f_i(0)= {\partial f\over \partial x_i}(0) \, \, \forall i$ and :

$$ f(x) = \sum_{i=1}^n f_i(x) x_i  \hspace{1cm} \forall x=(x_1,x_2,....,x_n)\in U $$

and the answer is :

let $x\in U$ , the function $\phi : [0,1]\to \mathbb R$ , $t \to f(xt)$ . we have $\phi\in C^1$ with :
$\phi(0)=0 \hspace{1cm} \phi(1)=f(x) , \phi'(t)= df(tx).x $

and then :

$df(tx).x = \sum_{i=1}^n {\partial f\over \partial x_i}(tx)x_i $  , i dont understand this !, first of all , x is a vector and $\phi'(t)$ is not ! (cuz i saw that $\phi'(t)=df.vector$ wich make it a vector) , and how they came up with $=\sum_{i=1}^n {\partial f\over \partial x_i}(tx)x_i$ next ?

the rest of the solution is :

$f(x)=\int_0^1\phi(t)dt = \sum_{i=1}^n f_i(x).x_i \Rightarrow f_i(x)=\int_0^1 {\partial f\over \partial x_i}(tx)dt $

im having this type of problem everywhere , can someone explain to me what should happen to the vector when we took it outside after we devided  ?

答:
This is just chain rule, $ \phi(t) = f(tx) $ implies $$ \phi'(t) = \nabla f(tx).x = \sum_{i=1}^n\frac{\partial f}{\partial x_i}(tx)x_i $$
You have gradient defined as $$ \nabla f = \Big(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},....,\frac{\partial f}{\partial x_n}\Bigr) $$
and for any curve $\gamma $ chain rule gives you that $ (f\circ \gamma)'(t) = \nabla f(\gamma(t)).\gamma'(t) $. In this case $ \gamma(t) = tx $.</p>


问:
Benford's law and voting in Georgia
With Benford's law "the leading digits of the number found in real-world data sets" should make sense.  But this video

https://www.youtube.com/watch?v=DoF3WS42w3M&ab_channel=RobertA.Bonavito%2CCPA

claims that by applying Benford's law they can prove electoral fraud.

He claims that since Benford's law is "broken" in these graphs, it appears that there has been fraud committed, but I wonder if the artificiality of breaking things into counties and only having 156 of them violates the "naturally occuring" idea that is the foundation of Benford's Law.

I am curious why Benford's law does not seem to apply here.

答:
Using his data for first digits, neither the Biden nor the Trump first digits are significantly different from what would be predicted by Benford's law.  Using R:

```<code>countyBiden <- c(47,40,16,21,7,9,9,5,5)
countyTrump <- c(43,36,12,15,9,17,9,9,9)
benford <- log10(2:10) - log10(1:9)
chisq.test(countyBiden, p=benford)
#
#         Chi-squared test for given probabilities
#
# data:  countyBiden
# X-squared = 12.601, df = 8, p-value = 0.1263
#
chisq.test(countyTrump, p=benford)
#
#         Chi-squared test for given probabilities
#
# data:  countyTrump
# X-squared = 11.231, df = 8, p-value = 0.189
</code>```

That is even before considering whether Benford's law should apply here.

Title: Benford's law and voting in Georgia

Body:
Using his data for first digits, neither the Biden nor the Trump first digits are significantly different from what would be predicted by Benford's law.  Using R:

countyBiden <- c(47,40,16,21,7,9,9,5,5)
countyTrump <- c(43,36,12,15,9,17,9,9,9)
benford <- log10(2:10) - log10(1:9)
chisq.test(countyBiden, p=benford)
#
#         Chi-squared test for given probabilities
#
# data:  countyBiden
# X-squared = 12.601, df = 8, p-value = 0.1263
#
chisq.test(countyTrump, p=benford)
#
#         Chi-squared test for given probabilities
#
# data:  countyTrump
# X-squared = 11.231, df = 8, p-value = 0.189

That is even before considering whether Benford's law should apply here.


问:
How is convolving Haar measure with itself over subsets enough to define subgroups?
In this post, Terry Tao says that Gaussians are "a subgroup of $\Bbb R$".

When you convolve Haar probability measure on a (compact) subgroup with itself, you get back the same measure, and this can in fact be used as a definition of such subgroups. If you convolve a Gaussian probability measure with itself, you almost get back the same Gaussian measure, but it has spread out by a factor of $\sqrt 2$. So Gaussians are in some sense a "$\sqrt 2$-approximate group".

I don't understand this comment, or how it means that Gaussians can be viewed as a subgroup of $\Bbb R$.

How can you define subgroups through this? Perhaps this?

A subset, $H$ of a group $G$ with Haar measure $\mu$ is called a subgroup if $\mu |_H=\mu\ast\mu|_H$.

How does this show that a Gaussian is a subgroup of $\Bbb R$?

答:
Obviously, I am not Terry Tao and this answer has some speculation on my part. The interesting bit of theory is that subgroups (of a compact group) can be identified by algebraic properties in the semigroup of probability measure. This post only aims to answer the question, "How can you define a subgroup through this?".

Let $G$ be a compact group and $\mathcal{P}(G)$ be the collection of regular Borel probability measures on $G$. This is a semigroup under the convolution operation. There is a classical theorem of Wendel which connects idempotent elements in the semigroup $\mathcal{P}(G)$ (i.e. measures such that $\mu * \mu = \mu$) with compact subgroups of $G$. In particular, each idempotent measure is actually the Haar measure of a closed subgroup. So, from an idempotent measure, taking the support gives a compact subgroup, and from a compact subgroup, taking the Haar measure results in an idempotent measure.

So, we can say that a subset $A \subset G$ is a (compact) subgroup if there exists an idempotent measure $\mu \in \mathcal{P}(G)$ such that $\sup(\mu) = A$.

Rudin extended this theorem to the locally compact abelian case. In particular, any idempotent probability measure here concentrates on a compact subgroup.


问:
understanding relation of matrix and linear transformation
I  was reading Sheldon Axler's book question 3.c.3 which is answered here:
3.c.3
but I can't understand how it works.
can anyone give me explanation and some intuition about relation of them please.
proof</p>

答:
One way to relate a transformation to its matrix is to note that the $j$th column of a matrix specifies the image of the $j$th basis element of the domain.

In our case, the bases are $\{v_1,\dots,v_n,u_1,\dots,u_m\}$ and $\{w_1,\dots,w_p\}$.

For $1 \leq j \leq n$, the image of the $i$th basis element is given by
$$
T(v_j) = w_j = 0 \cdot w_1 + \cdots + 0 \cdot w_{j-1} + 1 \cdot w_j + 0 \cdot w_{j+1} + \cdots + 0 \cdot w_n.
$$
Correspondingly, the $j$th column of $\mathcal M(T)$ will be $(0,\dots,0,1,0,\dots,0)$. Thus, the $j$th entry of the $j$th column is $1$, and all other entries of the column are $0$. This gives us the expected result for the first $n$ columns.

For $j>n$, we find that the image of the $j$th basis vector is given by
$$
T(u_{j-n}) = 0 = 0 \cdot w_1 + \cdots + 0 \cdot w_p,
$$
which means that the $j$th column will be $(0,\dots,0)$. In other words, all entries outside the first $n$ columns are $0$.


问:
Finding differential of inverse?
Let $f(x)=6x^3-11x^2-2$, $x\geqslant 1.
5$. Find $\frac{\text{d}f^{-1}}{\text{d}x}$ at the point $x=1517=f(7)$.

I thought you should first find the inverse and then differentiate but from looking at the inverse, that's clearly not what I should do. Any ideas? Also, why did they give me x>=1.5?

答:
Hint: Indeed, you don't need to know the inverse function of $f$.
Note that by inverse function theorem, you can find $$\frac{df^{-1}}{dx}(b)=(f^{-1})'(b)=\frac{1}{f'(a)}$$
where $b=f(a)$ and only you need  show that $f'(a)\not=0$ and then you need to find $\frac{1}{f'(a)}$.

Title: Finding differential of inverse?


问:
How to find integers $p$ and $q$ such that $(p\sqrt{2}+q)^2=34-24\sqrt{2}$
Find integers $p$ and $q$ such that $(p\sqrt{2}+q)^2=34-24\sqrt{2}$.

I approached this question first by expanding the the left-hand side to get:

$$2p^2 +2\sqrt{2}pq+q^2 = 34-24\sqrt{2}$$

The problem becomes intractable for me at this point

Body:
Note

$$\sqrt{34-24\sqrt2}
=\sqrt{2(17-2\sqrt{72})}
= \sqrt{2(\sqrt9-\sqrt{8})^2}= 3\sqrt2-4
$$
Thus, $(p,q)=(3,-4),(-3,4)$.


答: To apply Benford's law, it's required that the numbers in question are distributed over several orders of magnitude. I'm not going to waste my time watching a propaganda / disinfo video; I'll just note that voting precincts are of relatively uniform size and so Benford's law is inappropriate in this setting.</p>

Title: Benford's law and voting in Georgia

Body:
With Benford's law "the leading digits of the number found in real-world data sets" should make sense.  But this video

https://www.youtube.com/watch?v=DoF3WS42w3M&ab_channel=RobertA.Bonavito%2CCPA

claims that by applying Benford's law they can prove electoral fraud.

He claims that since Benford's law is "broken" in these graphs, it appears that there has been fraud committed, but I wonder if the artificiality of breaking things into counties and only having 156 of them violates the "naturally occuring" idea that is the foundation of Benford's Law.

I am curious why Benford's law does not seem to apply here.