问:
Identity a subfield of ℝ that is isomorphic to fraction field of R={a+bsqrt(3)} where a, b are integers
It is easy to see that R is a subring of ℝ, but I don't know how to find an isomorphism between the fraction field of R and a subfield of ℝ .

My intuition is that it is isomorphic to R with rational coefficients instead of integer coefficients, is it correct?

答: Since $\Bbb Q(\sqrt 3)$ is a field and $\Bbb Z[\sqrt 3]\subset\Bbb Q(\sqrt 3)$ the fraction field $\widehat{\Bbb Z[\sqrt 3]}$ of $\Bbb Z[\sqrt 3]$ must be contained in $\Bbb Q(\sqrt 3)$. But $\sqrt 3\in\widehat{\Bbb Z[\sqrt 3]}$ where $\widehat{\Bbb Z[\sqrt 3]}$ is a field containing $\Bbb Q$ and thus $$\Bbb Q(\sqrt 3)\subseteq\widehat{\Bbb Z[\sqrt 3]}\subseteq\Bbb Q(\sqrt 3)\;\;\;\;\;\;\;\;\;\;\therefore\;\widehat{\Bbb Z[\sqrt 3]}=\Bbb Q(\sqrt 3)\;\;\;\;\;$$


问:
Closure and limit of a sequence
Let $E$ be a subset of a metric space $(S,d)$. I'm trying to show that an element is in $\overline{E}$ if and only if it is the limit of some sequence of points in $E$.

Suppose there is a sequence $(p_n) \subseteq E$. Then because $E \subseteq \overline{E}$, I know that $(p_n) \subseteq \overline{E}$. Since $\overline{E}$ is closed, $\overline{E}$ contains the limit of every convergent sequence of points in $\overline{E}$. So $p \in \overline{E}.$

This is how I approached the converse: I assumed $p \in \overline{E}$ and that there does not exist a sequence $(p_n) \subseteq E$ such that $p_n \to p$. Then I concluded that there exists  $\varepsilon > 0$ such that $d(p, x) \geq \varepsilon \ \forall x \in E$.

Then there exists an open ball centered at $p$ that does not contain a point of $E$ (I wasn't sure how to make this statement rigorous). So $p \notin \overline{E}$.

Could I get some feedback on my proof?

答:
You haven’t actually justified the conclusion that there is an $\epsilon>0$ such that $d(p,x)\ge\epsilon$ for each $x\in E$. Justifying it properly requires essentially the same work as a direct proof that if $p\in\operatorname{cl}E$, then there is a sequence $\langle p_n:n\in\Bbb Z^+\rangle$ in $E$ converging to $p$, so you might as well give the direct proof.

There are two cases.

If $p\in E$, there’s a trivial sequence in $E$ that converges to $p$; what is it?

If $p\in(\operatorname{cl}E)\setminus E$, you have to work just a little bit. For each $n\in\Bbb Z^+$ there’s a point $p_n\in E$ such that $d(p,p_n)<\frac1n$; why? What does this get you?

ContentLicense: CC BY-SA 3.0

Title: Closure and limit of a sequence


问:
logic question: How to say $A \cap (B \cup C)$ in english
$A$, the set of people who like to make bbq,

$B$, the set of people whose house is blue,

$C$, the set of people hosting a cat.

I've been asked to say $A \cap (B \cup C)$ in english but i'm not sure how.

I tried "People who like to bbq and people who live in a blue house or have a cat." but Im pretty sure my phrase translates to $(A \cap B) \cup C$ since $\cap$ is prioretary.

答:
The key is to translate the set notation into English.

Intersection: $\cap=$ "And"

Union: $\cup=$ "Or"

So the expression becomes:

$A \cap (B \cup C) \rightarrow A \ and \ (B \ or \ C)$

"People who like to make BBQ and either live in a blue house or host a cat, or both."

ContentLicense: CC BY-SA 4.0


问:
Showing that the set $S$ is countable
Here's the question:

Let $S$ be a set of positive real numbers. There is a constant $M>0$ such that for any $\{ a_1,a_2,\ldots, a_k \}\subset S$, we have that
$$\sum_{n=1}^{k}a_n\le M \tag{1}$$
Prove that $S$ is at most countable.

Here's my attempt:

If $S$ is finite then there is nothing to prove. So, we prove the following:

Claim. If $S$ is any infinite set of positive real numbers satisfying condition $(1)$ then $\max S$ exists.

Proof. Let $S$ be a set satisfying $(1)$. Then $M$ is an upperbound for the set $S$. By least upper bound axiom, $\alpha=\sup S$ exists. Since $S$ is a set of positive reals, $\alpha >0$. If the maximum existed then it must be equal to $\alpha$. So, suppose that it does not exist. Choose an $\varepsilon >0$ such that $a-\varepsilon >0$. Then by the Archimedean property, $M\le k(\alpha - \varepsilon)$ for some $k\in \mathbb{N}$.

Now, since $\alpha$ is the supremum and not the maximum, we can select distinct elements $a_1,a_2, \ldots , a_k$ from $S$ such that $\alpha - \varepsilon < a_i < \alpha$ for each $i$. Then $\sum_{n=1}^{k}a_n> k(\alpha - \varepsilon ) \ge M$ but this would contradict our assumption that $(1)$ holds. So, $\alpha$ must be the maximum. $\blacksquare$

Now, let $S$ be any set of positive real numbers satisyfing $(1)$. Then let $a_1= \max S$. Also $S \setminus \{ a_1 \}$ satisfies $(1)$ so, let $a_2=\max (S\setminus\{ a_1 \})$. We repeat this argument to obtain a sequence $\{ a_k \}$ such that $a_k=\max (S\setminus \{ a_1, a_2 , \ldots, a_{k-1} \})$.

Clearly this $\{ a_k \}$ is decreasing and there is no $a\in S$ such that $a_{k+1}<a<a_k$ for any $k$. Also, $\sum_{k=1}^{\infty}a_k$ converges, so, $a_n \to 0$.

If $a \in S\setminus \{a_k\}_{k\in \mathbb{N}}$ then $a<a_k$ for each $k$. So $a\le0$ (by taking limits) which is not possible. So $S=\{ a_k\}_{k\in \mathbb{N}}$.

Is this proof correct? Alternative proofs are also welcome!

答: It looks good to me, an alternative proof could be as follow. Suppose $S$ is not countable and let for $n\geqslant 1$, $S_n=\left\{s\in S,s\geqslant\frac{1}{n}\right\}$. Clearly $\displaystyle S=\bigcup_{n\geqslant 1}S_n$ and there is at least one $p\geqslant 1$ such that $S_p$ is not countable, otherwise $S$ would be countable as a countable union of countable sets. In particular $S_p$ is infinite so we can take $\{s_1,\ldots,s_k\}\subset S_p\subset S$ with $k>pM$. We thus have $\displaystyle\sum_{n=1}^k s_n\geqslant\frac{k}{p}>M$ which is not possible and therefore $S$ is countable.

Title: Showing that the set $S$ is countable


问:
If the mean of four of six given numbers is known, what is the mean of the other two?

Four of the six numbers
  $$1867,\quad 1993,\quad 2019,\quad 2025,\quad 2109,\quad \text{and}\quad 2121 $$
  have a mean of 2008. What is the mean of the other two numbers?

I would like to get help for this problem because i want to find a way to answer this problem without using guess and check. thank you :)

答:
The sum of the six given integers is $1867+1993+2019+2025+2109+2121=12134$.

The four of these integers that have a mean of 2008 must have a sum of $4(2008)=8032$. (We do not know which integers they are, but we do not actually need to know.)

Thus, the sum of the remaining two integers must be $12134-8032=4102$.

You know the next step! Just divide it by two.


问:
Collatz Conjecture, why an increment of $+6$ in the following?
Following my previous question at: Collatz Conjecture, why a rate of change of $*4$ in the following?

Following the rules of the Collatz Conjecture, in this experiment I have created a list of all odd numbers until $33333$. The list includes 3 columns, such as in the following sample:

A) Starting Odd $(X)$
B) $(X * 3) +1$
C) $X/2$ repeat until odd

1
4
2, (1)

3
10
(5) **

5
16
8, 4, 2, (1)

7
22
(11) **

9
28
14, (7)

11
34
(17) **

13
40
20, 10, (5)

15
46
(23) **

17
52
26 (13)

19
58
(29) **

...

You will notice that starting with $X=3$ in column A) and incrementing $X + 4$ in column A) will result in a consistent increment of $+6$ in column C) (marked by ** in the table). Why is that ?

Also the same doesn't apply when starting with $X=1$ in column A) and incrementing $X + 4$ in column A). Why is that?

答: The lines marked ** are those where the odd number is of the form $4k+3$  When you apply the Collatz iteration the first time you get $3(4k+3)+1=12k+10$.  This can only be divided by $2$ once, which is the second iteration.  Since the numbers of type $4k+3$ increment by $4$, after one iteration they increment by $12$ and after the second they iterate by $6$.  On the other hand, numbers of the form $4k+1$ will go to $12k+4$ and can be divided by $2$ at least twice on the second iteration.


答:
It wouldn’t hurt to say something to justify the assertion that $\sum_{n\ge 1}a_n$ converges, but the proof is correct. (There is a typo in that sum though: you have $\sum_{\color{red}k=1}^\infty a_{\color{red}n}$.) Here’s a slightly different approach.

Let $\alpha=\sup S>0$. For $n\in\Bbb N$ let $I_n=\left(2^{-(n+1)}\alpha,2^{-n}\alpha\right]$; then

$$S\subseteq(0,\alpha]=\bigcup_{n\ge 0}I_n\,,$$

and the intervals $I_n$ are pairwise disjoint. There are only countably many of these intervals, so if $S$ is uncountable, there must be an $n\in\Bbb N$ such that $S\cap I_n$ is uncountable. But then if $K$ is a $k$-element subset of $A\cap I_n$, we have

$$\sum K>2^{-(n+1)}\alpha k\,,$$

and this can be made arbitrarily large by choosing $k$ large enough, contradicting $(1)$.

Title: Showing that the set $S$ is countable

Body:
Here's the question:

Let $S$ be a set of positive real numbers. There is a constant $M>0$ such that for any $\{ a_1,a_2,\ldots, a_k \}\subset S$, we have that
$$\sum_{n=1}^{k}a_n\le M \tag{1}$$
Prove that $S$ is at most countable.

Here's my attempt:

If $S$ is finite then there is nothing to prove. So, we prove the following:

Claim. If $S$ is any infinite set of positive real numbers satisfying condition $(1)$ then $\max S$ exists.

Proof. Let $S$ be a set satisfying $(1)$. Then $M$ is an upperbound for the set $S$. By least upper bound axiom, $\alpha=\sup S$ exists. Since $S$ is a set of positive reals, $\alpha >0$. If the maximum existed then it must be equal to $\alpha$. So, suppose that it does not exist. Choose an $\varepsilon >0$ such that $a-\varepsilon >0$. Then by the Archimedean property, $M\le k(\alpha - \varepsilon)$ for some $k\in \mathbb{N}$.

Now, since $\alpha$ is the supremum and not the maximum, we can select distinct elements $a_1,a_2, \ldots , a_k$ from $S$ such that $\alpha - \varepsilon < a_i < \alpha$ for each $i$. Then $\sum_{n=1}^{k}a_n> k(\alpha - \varepsilon ) \ge M$ but this would contradict our assumption that $(1)$ holds. So, $\alpha$ must be the maximum. $\blacksquare$

Now, let $S$ be any set of positive real numbers satisyfing $(1)$. Then let $a_1= \max S$. Also $S \setminus \{ a_1 \}$ satisfies $(1)$ so, let $a_2=\max (S\setminus\{ a_1 \})$. We repeat this argument to obtain a sequence $\{ a_k \}$ such that $a_k=\max (S\setminus \{ a_1, a_2 , \ldots, a_{k-1} \})$.

Clearly this $\{ a_k \}$ is decreasing and there is no $a\in S$ such that $a_{k+1}<a<a_k$ for any $k$. Also, $\sum_{k=1}^{\infty}a_k$ converges, so, $a_n \to 0$.

If $a \in S\setminus \{a_k\}_{k\in \mathbb{N}}$ then $a<a_k$ for each $k$. So $a\le0$ (by taking limits) which is not possible. So $S=\{ a_k\}_{k\in \mathbb{N}}$.

Is this proof correct? Alternative proofs are also welcome!


问:
Why is it valid to use squared Euclidean distances in high dimensions in multiple regression?
Euclidean distance is not linear in high dimensions. However, in multiple regression the idea is to minimize square distances from data points to a hyperplane.

Other data analysis techniques have been considered problematic for their reliance on Euclidean distances (nearest neighbors), and dimensionality reduction techniques have been proposed.

Why is this not a problem in multiple regression?

答:
The nice thing about
squared Euclidean distance
is that is the simplest function
that is differentiable,
so you can set the
derivative
(or partial derivatives) to zero
to get an extreme value.

It also has other nice properties
such as easily giving
the minimizer of the distance
of a value to a set of values
is the mean
(i.e.,
minimizing
$\sum_{i=1}^n (x_i-a)^2$
with respect to $a$
gives
$a=(1/n)\sum_{i=1}^n x_i$).

Whether this is what you want
is another matter.


答: Another way to say ...${A \cap (B \cup C)= (A \cap B)\cup (A \cap C)}$ , i.e., A set of people who either live in blue house and like to make bbq  , or they have cat and also like to make bbq

Title: logic question: How to say $A \cap (B \cup C)$ in english

Body:
$A$, the set of people who like to make bbq,

$B$, the set of people whose house is blue,

$C$, the set of people hosting a cat.

I've been asked to say $A \cap (B \cup C)$ in english but i'm not sure how.

I tried "People who like to bbq and people who live in a blue house or have a cat." but Im pretty sure my phrase translates to $(A \cap B) \cup C$ since $\cap$ is prioretary.


答:
Your first part of your proof is fine, as long as you have proved that the limit of a sequence of points in a closed set is part of that set.

There's a bit of a gap in the second part your proof.  Notice that you write:

I assumed $p∈\bar{E}$ and that there does not exist a sequence $(p_n)⊆E$ such that $p_n→p$.

And then conclude:

Then I concluded that there exists $ε>0$ such that $d(p,x)≥ε ∀x∈E$.

But this statement isn't immediately obvious.  What this part requires you to do is actually find a sequence $p_n \to p$.

One way to do this would be to take a sequence of positive real numbers $r_n$ converging to $0$ (like $(\frac{1}{n} )$) and for each real number, find an $x_n \in E$ such that $d(p,x_n) < r_n$ (you know that at least one exists since $\bar{E}$ is closed; $x_n$ might be $p$ itself: it may be useful to divide up cases for when $p \in E$ and when $p$ is a limit point).

 Then for any $\epsilon > 0$, simply find $N$ such that $\forall k \ge N$, $\epsilon > r_k > 0$ so that $d(p,x_k) < r_k < \epsilon$, proving convergence $x_n \to p$.

Title: Closure and limit of a sequence

Body:
Let $E$ be a subset of a metric space $(S,d)$. I'm trying to show that an element is in $\overline{E}$ if and only if it is the limit of some sequence of points in $E$.

Suppose there is a sequence $(p_n) \subseteq E$. Then because $E \subseteq \overline{E}$, I know that $(p_n) \subseteq \overline{E}$. Since $\overline{E}$ is closed, $\overline{E}$ contains the limit of every convergent sequence of points in $\overline{E}$. So $p \in \overline{E}.$

This is how I approached the converse: I assumed $p \in \overline{E}$ and that there does not exist a sequence $(p_n) \subseteq E$ such that $p_n \to p$. Then I concluded that there exists  $\varepsilon > 0$ such that $d(p, x) \geq \varepsilon \ \forall x \in E$.

Then there exists an open ball centered at $p$ that does not contain a point of $E$ (I wasn't sure how to make this statement rigorous). So $p \notin \overline{E}$.

Could I get some feedback on my proof?


问:
Some of approximations related to integrals and elliptic functions

I have just known about the elliptic functions and I saw three nice examples as following :

$$\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{4\ln^{2}\tan x+ \pi^{2}}{\rm d}x\cong\frac{1}{4}$$
$$\int_{0}^{1}\frac{{\rm d}x}{\sqrt{1- x^{3}}}\cong\frac{1}{\sqrt[4]{3}}\int_{0}^{\sqrt{4\sqrt{3}- 6}}\frac{{\rm d}x}{\sqrt{1- x^{2}}\sqrt{1- \frac{\sqrt{3}+ 2}{4}\cdot x^{2}}}$$
$$\int_{0}^{\frac{\pi}{3}}\int_{0}^{\frac{\pi}{3}}\frac{{\rm d}y{\rm d}x}{\sin^{2}x+ \cos^{2}y}\cong\frac{4}{\pi}$$
I'm looking forward to similar problems, any comments and solutions are welcome and appreciated

Thanks a real lot !

答:
Substitution
$$x=\arctan y$$
presents the first integral in the form of
$$\int\limits_0^\infty \dfrac{y\,\text d y}{(4\ln^2y+\pi^2)(1+y^2)} = \dfrac14$$
(see WA integration).


问:
Ratio of Sample Mean of Exponential Random Variables
Find the distribution of $\bar{X}/\bar{Y}$, i.e. the distribution of the ratio of the sample means of two independent random samples from the (same) exponential distribution with parameter $\lambda>0$.

So far I've played around with potential similarities between the exponential, chi-squared, and Fisher-Snedecors' probability density functions but have not yet found a concrete path for a solution. Any suggestions?

答:
First notice that the $\lambda$ in the numerator and the denominator cancels out, so we may as well assume the distribution of each observation is $e^{-x/2}(dx/2)$ for $x>0.$ This makes the expected value of each observation equal to $2.$ The reason we do that is that the exponential distribution with expected value $2$ is the same as the chi-square distribution with $2$ degrees of freedom. If the respective sample sizes are $m$ and $n,$ we then have a ratio of two independent random variables, each of which is the quotient of a chi-square random variable by half its degrees of freedom. And those two factors of one-half also cancel.

Therefore $\left.\overline X\right/\overline Y \sim F_{2m,2n}.$


问:
How many 5 digit numbers can be obtained using digits 0,2,9 at least once?
This was my approach:
Total numbers that can be formed from 0, 2, 9 will be=2×3×3×3×3=162

Out of these;
Numbers with only 0, 9= 1×2×2×2×2= 16

Numbers with only0, 2= 1×2×2×2×2= 16

Numbers with only 9, 2=
2×2×2×2×2= 32

Numbers with only 2=1
Numbers with only 9=1

Thus the answer I got=162-16-16-32-2= 96

However, in my textbook the answer given is 100.

Where I am going wrong?

答:
The main problem is while considering the cases, you haven't taken 'at least once'. That is, your approach would have been better if you counted all permutations with 0,2 at least once, rather than all together.

While counting numbers with 0,2 the case 22222 arises, similarly with 0,9 the case 99999 is counted. And also when you count 2,9 both cases arise once again. Here as you are counting them extra, they must be added.

Hence, the correct calculation will be  162-16-16-32+2=100


问:
Does this implication hold given this function?
Does $\lim_{x \to \infty} (f(x) - 2x) = 0$ imply that $\lim_{y \to \infty} (g(y) - \frac{y}{2}) = 0$?

My original function is

$f(x) = \sqrt{4x^2 + 1}$ for $x \in [0, \infty)$

with the inverse I found

$g(y) = \frac{\sqrt{x^2 - 1}}{2}$

答: Let $\epsilon >0$. Then there exists $T$ such that $x>T$ implies $|f(x)-2x| <\epsilon$. If $y >f(T)$ we can take  $x=g(y)$ to get $|y-2g(y)|<\epsilon$ or $|g(y)-\frac y 2|<\epsilon/2<\epsilon$. If you call $f(T)$ as $S$ it follows that $|g(y)-\frac y 2|<\epsilon$ whenever $y >S$.


问:
Solving a special rational equation on a very small interval
I need to solve the following equation (for $x$):
\begin{equation}
\mathcal f(x):=\sum_{i=1}^n b_i \left( \frac{a_i}{1+b_i x}\right)^2-\phi=0, \quad \text{with} \quad -1/b_1< x \le 0.
\end{equation}
Assume that the equation has at least one solution in $(-1/b_1,0]$ and $b_1\ge b_2\dots\ge b_n>0$. In practice, $b_1$ can be very large, so that the interval of interest $(-1/b_1,0]$ can be very small. Therefore, numerical issues can happen (I think because $1/(1+b_1x)$ can blow up, at least when a numerical method captures a solution that is very close to $-1/b_1$). What is the best practical idea to find a solution of $f(x)=0$ that falls into $(-1/b_1,0]$?

By the way, I am currently using the bisection method and $b_1$ is around 1000 for example. The quantity $n$ could be up to 50 in my case. Further, such an equation must be solved iteratively in my algorithm for different values of $ a_i$'s for $i\in [n]$.
Thank you very much in advance for your time and help.

答:
I prefer to add a second answer for some crucial points.

The solution you are looking for is not in the range $-\frac 1{b_1} < x <0$ but in the range $$\color{red}{-\frac 1{b_1}+\epsilon \leq x \leq-\frac 1{b_2}-\epsilon}$$ $\epsilon$ being at the level of machine accuracy. So, if your search method use bounds, these are the ones to be used.

Moreover, in this range, there are two possible solutions for $x$ since
$$\lim_{x\to {-\frac{1}{b_1}}+\epsilon}  \, f(x) = +\infty \qquad \text{and} \qquad \lim_{x\to {-\frac{1}{b_2}}-\epsilon}  \, f(x)= +\infty$$but, depending on the value of $\phi$, no root is also possible as well as a double root.

Reusing the example worked in my previous answer, if $\phi < 2.57\times 10^8$, there is no solution.

Using $\phi=10^{10}$, the roots are $-9.085\times 10^{-4}$ and $-1.170\times 10^{-4}$ each of them being close to the bounds.

So, considering before any rescaling
$$f(x)=\sum_{i=1}^n b_i \left( \frac{a_i}{1+b_i x}\right)^2- \phi$$ we should consider
$$g(x)=(1+b_1x)^2(1+b_2x)^2\sum_{i=1}^n b_i \left( \frac{a_i}{1+b_i x}\right)^2- \phi(1+b_1x)^2(1+b_2x)^2$$ that is to say
$$g(x)=a_1^2b_1(1+b_2x)^2+a_2^2b_2(1+b_1x)^2+(1+b_1x)^2(1+b_2x)^2\sum_{i=3}^n b_i \left( \frac{a_i}{1+b_i x}\right)^2-\phi(1+b_1x)^2(1+b_2x)^2$$

So, working around $x=-\frac{1}{b_1}$, we have
$$g\left(-\frac{1}{b_1}\right)=a_1^2 b_1 \left(1-\frac{b_2}{b_1}\right)^2$$
$$g'\left(-\frac{1}{b_1}\right)=2 a_1^2 b_1 b_2 \left(1-\frac{b_2}{b_1}\right)$$
$$g''\left(-\frac{1}{b_1}\right)=2 b_1 b_2 \left(a_1^2 b_2+a_2^2 b_1\right)-2 b_1^2 \left(1-\frac{b_2}{b_1}\right)^2 \phi+$$
$$2 b_1^2 \left(1-\frac{b_2}{b_1}\right)^2\sum_{i=3}^n \frac{a_i^2 b_i}{\left(1-\frac{b_i}{b_1}\right)^2}$$

Now, from these informations, build the parabola
$$h(x)=A+Bx+Cx^2$$
$$C=\frac 12 g''\left(-\frac{1}{b_1}\right)\qquad B=g'\left(-\frac{1}{b_1}\right)+\frac{2C}{b_1}\qquad A=g\left(-\frac{1}{b_1}\right)+\frac{B}{b_1}-\frac{C}{b_1^2}$$ Solve the quadratic $h(x)=0$ and keep the largest root as the approximate solution.

Applied to the worked example, this would give
$$x_0=-0.9998478974\times 10^{-4}$$ while the exact solution is
$$x=−0.9998479000\times 10^{-4}$$

Update

In the previous edit, there were $b_1$ and $b_2$ missing as mutliplying factors in the expansion of $g(x)$ but all the remaining was correct.

To allow test with the conditions $n=5$, $b_i=10^{5-i}$, $a_i=p_{i+5}^2$ and $\phi=10^k$, I produced some results
$$\left(
\begin{array}{ccc}
k & \text{estimation} & \text{solution} \\
 10 & -8.26703656651674 \times 10^{-5} & -8.30010445244925 \times 10^{-5} \\
 11 & -9.46205809895313 \times 10^{-5} & -9.46525763471424 \times 10^{-5} \\
 12 & -9.83067188617129 \times 10^{-5} & -9.83098989883711 \times 10^{-5} \\
 13 & -9.94652543341715 \times 10^{-5} & -9.94655718742864 \times 10^{-5} \\
 14 & -9.98309681582244 \times 10^{-5} & -9.98309998987002 \times 10^{-5} \\
 15 & -9.99465543307055 \times 10^{-5} & -9.99465575043391 \times 10^{-5} \\
 16 & -9.99830996825483 \times 10^{-5} & -9.99830999998987 \times 10^{-5} \\
 17 & -9.99946557190165 \times 10^{-5} & -9.99946557507511 \times 10^{-5} \\
 18 & -9.99983099968264 \times 10^{-5} & -9.99983099999999 \times 10^{-5} \\
 19 & -9.99994655747581 \times 10^{-5} & -9.99994655750754 \times 10^{-5} \\
 20 & -9.99998309999683 \times 10^{-5} & -9.99998310000000
\times 10^{-5}
\end{array}
\right)$$

Edit

If we are sure that $x$ will be very close to $-\frac 1{b_1}$, then, we can write $x= -\frac 1{b_1}+\epsilon$ and expand $f(x)$ as a series around $\epsilon=0$.

Defining
$$u_i=\frac{a_i^2 b(i)}{(1+ b_ix)^2}$$ we have
$$u_1=\frac{a_1^2}{b_1 \epsilon ^2}$$ and, for $i>1$,
$$u_i=b_1^2\sum_{k=0}^\infty (-1)^k (k+1)\frac{a_i^2\, b_1^{k}\, b_i^{k+1} } {(b_1-b_i)^{k+2} }\epsilon^k$$ Neglecting the positive powers of $\epsilon$ then leads to the approximate equation
$$\frac{a_1^2}{b_1 \epsilon ^2}+b_1^2 \sum _{i=2}^n \frac{a_i^2 b_i}{(b_1-b_i)^2}=\phi$$

This would  give as a first approximation
$$ x=-\frac 1{b_1}+\frac{a_1}{\sqrt{b_1}}\left(\phi -b_1^2 \sum _{i=2}^n \frac{a_i^2 b_i}{(b_1-b_i)^2}\right)^{-\frac 12} $$

Using it for the above conditions, for $\phi=10^{10}$ this gives  $x=--8.2997747\times 10^{-5}$ and for $\phi=10^{20}$ this gives $x=-9.9999831\times 10^{-5}$; this is much better than the parabolic approximation.


问:
$f(x)=e^{-x},x>0$ Find $\lim_{n\rightarrow\infty}P(X_n<2+\log_en)$
Let $X_1,X_2....X_n$ be i.i.d random variables with pdf

$f(x)=e^{-x},x>0$

if $X_n=\max(X_1,X_1....X_n)$ then $\lim_{n\rightarrow\infty}P(X_n<2+\log_en)$

I have calculated Pdf of $X_n$ Which is $n(1-e^{-x})^{n-1}e^{-x}$

As $n$ goes large our distribution is Normal but I am not able to find the expectation. How do I solve this problem?

Body: $P(X_n<2+log \, \, n)=[P(X_1<2+log \, \,  n)]^{n}=(1-e^{-(2+log n)})^{n}=(1-\frac 1 {e^{2}n})^{n} \to e^{-e^{-2}}$ since $(1+\frac x n)^{n} \to e^{x}$ for any real number $x$.


答:
I am assuming you meant $Y_n = \max(X_1,\dots,X_n)$, as $X_n = \max(X_1,\dots,X_n)$ does not really make sense.

You can compute the cdf exactly, without having to go through the pdf: note that for every $t$, $Y_n \leq t$ if, and only if, $X_i \leq t$ for every $1\leq i \leq n$ (by definition of the maximum). Then, using independence,
$$\begin{align}
\mathbb{P}\{ Y_n \leq 2+ \log n\}
&=
\mathbb{P}\{ \forall 1\leq i\leq n,\; X_i \leq 2+ \log n\}
= \prod_{i=1}^n \mathbb{P}\{  X_i \leq 2+ \log n\}\\
&= \mathbb{P}\{  X_1 \leq 2+ \log n\}^n
\end{align}$$
the last equality as all r.v.'s are identically distributed. But since $X_1$ has an exponential distribution, its CDF is $\mathbb{P}\{  X_1 \leq t\} = 1-e^{-t}$ for all $t\geq 0$, and so
$$
\mathbb{P}\{ Y_n \leq 2+ \log n\}
= \left(1-e^{-(2+ \log n)}\right)^n
= \left(1-\frac{e^{-2}}{n}\right)^n
$$
Can you conclude by computing $\lim_{n\to\infty}\left(1-\frac{e^{-2}}{n}\right)^n$?

Remark: the above outline is quite general, and useful when dealing with the maximum of independent random variables.

Title: $f(x)=e^{-x},x>0$ Find $\lim_{n\rightarrow\infty}P(X_n<2+\log_en)$

Body:
Let $X_1,X_2....X_n$ be i.i.d random variables with pdf

$f(x)=e^{-x},x>0$

if $X_n=\max(X_1,X_1....X_n)$ then $\lim_{n\rightarrow\infty}P(X_n<2+\log_en)$

I have calculated Pdf of $X_n$ Which is $n(1-e^{-x})^{n-1}e^{-x}$

As $n$ goes large our distribution is Normal but I am not able to find the expectation. How do I solve this problem?


问:
Confusion in circular permutations

From point 2 I have marked ,

If the 4 persons are sitting in a round table, then they can be shifted four times. You can see by looking at figure a,b,c and d.

Now , I understand that by making ABCD sit in a row and changing their positions . It looks like their arrangements are different . Like CDAB , BCDA but also they have written only 4 orders for this. In real , isn’t it that there will be total 4! Arrangements for it and not 4.

Also , from the round table perspective. If you keeping Changing the angle you’re looking at the ABCD. You can see different arrangements but they are also same right. So their total arrangements should be equal to 1.

I didn’t understand for 2 ,

No of linear arrangements is n? I think it should be n!.

Also , how did we get 1/n in 4 and n! For 3.

答:
You're thinking about it the right way - the explanation in the book is just not very clear. The point is: if we counted circular arrangements the same way we count linear arrangements, we'd get $n!$, but this would count each arrangement exactly $n$ times (once in each of its equivalent rotated forms, as illustrated by the figure on top), so the true count is $n!/n$.

For example, with 4 elements, there are 24 linear arrangements:

ABCD BCDA CDAB DABC

ABDC BDCA DCAB CABD

ACBD CBDA BDAC DACB

ACDB CDBA DBAC BACD

ADBC DBCA BCDA CADB

ADCB DCBA CBAD BADC

but only 6 circular arrangements. Each row above represents one of the circular arrangements (depicted in four different ways).

Title: Confusion in circular permutations


问:
Understanding the proof of product of completely regular space is completely regular under product topology
Below is the proof:

Proof. Let $a=\prod_{i\in I}a_i\in\prod_{i\in I}X_i$ and $U$ be any open set containing $a$. Then there exists a finite subset $J$ of $I$ and sets $U_i\in\tau_i$ such that
$$a\in\prod_{i\in I}U_i\subseteq U$$
where $U_i=X_i$ for all $i\in I\setminus J$. As $(X_j,\tau_j)$ is completely regular for each $j\in J$, there exists a continuous mapping $f_j:(X_j,\tau_j)\longrightarrow[0,1]$ such that $f_j(a_j)=0$ and $f_j(y)=1$, for all $y\in X_j\setminus U_j$. Then $f_j\circ p_j:\prod_{i\in I}(X_i,\tau_i)\longrightarrow[0,1]$, where $p_j$ denotes the projection of $\prod_{i\in I}(X_i,\tau_i)$ onto $(X_j,\tau_j)$.

If we put $f(x)=\max\{f_j\circ p_j(x):\ j\in J\}$, for all $x\in\prod_{i\in I}X_i$ then $f:\prod_{i\in I}(X_i,\tau_i)\longrightarrow[0,1]$ is continuous (as $J$ is finite). Further, $f(a)=0$ while $f(y)=1$ for all $y\in X\setminus U$. So $\prod_{i\in I}(X_i,\tau_i)$ is completely regular. $\qquad\square$

I am trying to understand the line $f(x)=\max\{f_j(p_j(x)):\ j\in J\}$. What does this function mean and what is the point of it? What exactly are we taking the max of? We have not assumed that any of the $X_i$ are ordered or are sets that even contains a max.

答:
For each $j\in J$ you have a continuous function $f_j:X_j\to[0,1]$ such that $f_j(a_j)=0$, and $f_j(y)=1$ for each $y\in X_j\setminus U_j$. Now let $x\in X=\prod_{i\in I}X_i$. For each $i\in I$ we have the projection map $p_i:X\to X_i$. In particular, $p_j(x)$ is a point in $X_j$ for each $j\in J$, so we can apply $f_j$ to it to get

$$(f_j\circ p_j)(x)=f_j\big(p_j(x)\big)\in[0,1]\,.$$

That is, for each $j\in J$ we have a number $(f_j\circ p_j)(x)$ in $[0,1]$, and

$$\{(f_j\circ p_j)(x):j\in J\}$$

is therefore a finite subset of $[0,1]$. Since this is a finite set of real numbers it has a maximum element;

$$\max\{(f_j\circ p_j)(x):j\in J\}$$

is that maximum element. We now define a function $f$ from $X$ to $[0,1]$ by setting $f(x)$ equal to this maximum element:

$$f:X\to[0,1]:x\mapsto\max\{(f_j\circ p_j)(x):j\in J\}\,.$$

Thus,

$$\begin{align*}
f(a)&=\max\{(f_j\circ p_j)(a):j\in J\}\\
&=\max\left\{f_j\big(p_j(a)\big):j\in J\right\}\\
&=\max\{f_j(a_i):j\in J\}\\
&=0\,,
\end{align*}$$

since $f_j(a_j)=0$ for every $j\in J$. And if $x\in X\setminus U$, there must be a $j\in J$ such that $x_j\in X_j\setminus U_j$. so $(f_j\circ p_j)(x)=f_j(x_j)=1$, and therefore $f(x)=1$. Thus, $f$ will be the desired function if it is continuous, and continuity follows (with a little work) from the fact that $J$ is finite. (If you have trouble proving this, try it first on the assumption that $|J|=2$.)


问:
Transform quadratic to perfect square
If one wanted to transform $y(x) = ax^2 + bx + c$ to the form $(ux + v)^2,$ is there a transformation you can make in terms of $y(\text{subsitution}) = (ux + v)^2$?

答:
A polynomial with three parameters cannot be transformed into a polynomial with two parameters, as this would lose information.  Instead, we can transform $y(x)$ into an equation of the form $(ux+v)^2+w$. To do this, we first try expanding the new equation:
$$(ux+v)^2+w=(u^2)x^2+(2uv)x+(v^2+w)$$
Then equate each term of the two polynomials:
$$a=u^2$$
$$b=2uv$$
$$c=v^2+w$$

Now you can solve this set of three equations with substitution.
$$u=\sqrt{a}$$
$$v=\frac{b}{2\sqrt{a}}$$
$$w=c-\frac{b^2}{4a}$$

Note that this is called "Completing the Square".

Title: Transform quadratic to perfect square


问:
Application of implicit theorem
Find conditions on the function $f$ and $g$ which permit you to solve the equations $$f(xy)+g(yz)=0\ \ \textrm{and} \ \ g(xy)+f(yz)=0 $$ for $y$ and $z$ as functions of $x$, near the point $x=y=z=1$ and $f(1)=g(1)=0$.

Attempt: This problem seems to be an application of the implicit function theorem. Usually on this kind of problems we define special transformations, but I don't now how to define it in this case since I have several variables $x,y,z$ and two functions $f$ and $g$. Any idea on how to start?

答:
Let $F((x,y,z)) = \begin{bmatrix} f(xy) + g(yz) \\ f(yz)+g(xy)\end{bmatrix}$ and show that
$Df((1,1,1)) = \begin{bmatrix} f'(1) & f'(1)+g'(1) & g'(1) \\
g'(1) & f'(1)+g'(1) & f'(1)\end{bmatrix}$. Pick any $2 \times 2$ submatrix and find a condition that ensures that the submatrix is invertible.

Title: Application of implicit theorem


问:
How to demonstrate that there is no all-prime generating polynomial with rational cofficents?
It seems like there is no polynomial with finite variables known, which could generate all prime numbers, by integer assignments. Is there a proof that such polynomial can not exist and does anyone have one in his/her stack?</p>

答:
Here, it is clear that by hand-picking values, it is clear that there is no all-prime generating polynomial. It is even possible to demonstrate that there are infinite composite values to the function as well.

First of all, we state that a polynomial with integer input, and integer coefficients will only give integer output. This is because any positive power (the exponent in the function) of an integer is also an integer, and an integer multiplied by an integer (coefficient) is also an integer.

We write a general polynomial function in the form: $$f(x)=a_mx^m+a_{m-1}x^{m-1}+\cdots+a_2x^2+a_1x^1+a_0x^0.$$
Note that it is allowed in this proof for individual values of $a$ to be $0$, even the $a_m$, but just that there must be at least one non-constant term!

Now, we let the arbitrary integer $n$ be of the form $ka_0$ where $k$ can be any integer. We substitute $x=ka_0$ into the polynomial form, which gives us: $$f(n)=f(ka_0)=a_m(ka_0)^m+a_{m-1}(ka_0)^{m-1}+\cdots+a_2(ka_0)^2+a_1(ka_0)^1+a_0(ka_0)^0.$$
Let's expand that:
$$f(ka_0)=a_mk^ma_0^m+a_{m-1}k^{m-1}a_0^{m-1}+\cdots+a^2k^2a_0^2+a^1k^1a_0^1+a_0k^0a_0^0.$$
We notice that every term in the new expression has a factor of $a_0$! This means that $a_0$ is a factor of the value of the polynomial output.

Here, we notice that we defined $k$ as any arbitrary integer, so we can set it to any value we want. This gives us infinite outputs for the function as composite numbers, since $k$ can be any desired integer. $\square$


问:
The number of partial orders of finite set.
My question is about the number of partial orders of finite set. For example, if $S=\{a, b\}$, then there are 4 partial orders $\{ (a, a), (b, b) \}$, $\{ (a, a), (a, b), (b, b) \}$, $\{ (a, a), (b, a), (b, b) \}$, $\{ (a, a), (a, b), (b, a), (b, b) \}$. Then, how many are there partial orders if $|S|=n$?

I tried to count for the case $|S|=3$, $\{ (a, a), (b, b), (c, c) \}$, $\{ (a, a), (b, b), (c, c), (a, b) \}$, $\{ (a, a), (b, b), (c, c), (a, b), (b, c), (a, c) \}$, and so on. But it is too inefficient and does not give any insight.

If this is a difficult question, then I want to just know the case of $|S|=3$.

答:
Each finite pos (partially ordered set) has to have at least one minimal element.

Consider 3-set $\ X:=\{a\ b\ c\}.\ $ First, let $a$ be the unique minimal element. There are exactly three different partial orders in $\{b\ c\}.\ $ The same when $\ b\ $ or $\ c\ $ is the only minimal element in $\ X.\ $ Each of those partial orders in a 2-element set extends uniquely over the third element that is minimal. Every two of these partial orders in $\ X\ $ are different. Thus, there are exactly $9=3\cdot 3\ $
partial orders in $\ X\ $ that have exactly one minimal element.

Now, let $\ a\ $ and $\ b\ $ be the only minimal elements in $\ X\ $ (so that $\ c\ $ is not minimal). We have exactly three partial orders like this in $\ X.\ $ If the lonely non-minimal element is $\ b\ $ or $\ a\ $, each time we get 3 different partial orders, and all of such $\ 3\cdot 3=9\ $ partial orders are different. Thus, so far, we have 9+9=18 different partial orders.

Finally, there is exactly one partial order such that all elements of $\ X\ $ are minimal. This gives us finally the answer 1+18=19.

THEOREM There are exactly $\ 19\ $ different partial orders in any fixed 3-set.


问:
How can we be sure that the tensor identity $\partial_{\mu}\phi^*\partial^{\mu}\phi -\partial_{\mu}\phi\partial^{\mu}\phi^* = 0 $ is true?
Consider the complex scaler field $\phi$.

Why is it that the tensor identity $$\partial_{\mu}\phi^*\partial^{\mu}\phi -\partial_{\mu}\phi\partial^{\mu}\phi^*  = 0 $$
is true.

I thought the fact that we specify up and down indices is important and we cannot just assume that swapping them will allow for such an identity to hold.

Furthermore how can we tell that $\partial_{\mu}\phi\partial^{\mu}\phi^* = -\phi^*\partial_{\mu}\partial^{\mu}\phi$

This is probably quite obvious I am just a little confused.

答:
Assuming $\phi$ is a classical field. Recall that $\partial_\mu$ denotes the covariant derivative, meaning $\partial_\mu = (\partial_0, \partial_1,\partial_2, \partial_3)$ and $\partial^\mu = g^{\mu \nu}\partial_\nu$ (contract in the $\nu$ variable) where $g^{\mu\nu}$ is the inverse metric. For simplicity, take $g = \eta$ to be the Minkowski metric with signature $(-, +, +, +)$ then it follows that
$\partial^\mu = (-\partial_0, \partial_1, \partial_2, \partial_3).$ Finally, we see that
\begin{align}
\partial^\mu\phi^\ast \cdot \partial_\mu \phi =&\ (-\partial_0\phi^\ast, \partial_1\phi^\ast, \partial_2\phi^\ast, \partial_3\phi^\ast)\cdot (\partial_0\phi, \partial_1\phi, \partial_2\phi, \partial_3\phi) \\
=&\  (\partial_0\phi, \partial_1\phi, \partial_2\phi, \partial_3\phi)\cdot (-\partial_0\phi^\ast, \partial_1\phi^\ast, \partial_2\phi^\ast, \partial_3\phi^\ast)\\
=&\ \partial_\mu\phi \cdot \partial^\mu\phi^\ast.
\end{align}

Also, it is not true that
\begin{align}
\partial^\mu\phi^\ast \cdot \partial_\mu \phi= -\phi^\ast \partial^\mu\partial_\mu \phi
\end{align}
since the left hand side only involves first order derivatives whereas the right hand side involves second order derivatives. However, it is true that
\begin{align}
\int\ \partial^\mu\phi^\ast \cdot \partial_\mu \phi= -\int\ \phi^\ast \partial^\mu\partial_\mu \phi
\end{align}
because of integration by parts.


问:
A equivalent condition of nonzero linear functional?

** problem.** Let $X$ be an infinite-dimensional normed vector space and let $φ$ be a nonzero linear functional defined in $\,X.\,$ Then the following are equivalent.
(i) $φ$ is bounded.
(ii) The kernel of $φ$ is a closed linear subspace of X.
(iii) The kernel of $φ$ is not dense in $X$.

My attempt.:(i)$\Rightarrow$(ii) because $φ$ is bounded,then it is continuous so The kernel of $φ$ is a closed linear subspace of $X$.

(ii)$\Rightarrow$(iii): because $φ$ is nonzero amd from (iii), (iii) is ture.

(iii)$\Rightarrow$(i): I don't know how to do it, i try to assmue $φ$ is not bounded,then we know $φ$ is not continuous at every point of $X$, i want use it to prove the kernel is dense in $X$,but I failed. I also try to use riesz lemma because from (iii) the closure of kernel is proper closed subspace of $X$,I still failed.

答:
(iii) im plies that the kernel is closed as shown here: $\ker f$ is either dense or closed when $f$ is a linear functional on a normed linear space

Now if $f$ is not bounded there exsits $(x_n)$ such that $|f(x_n)| >n\|x_n\|$ . Fix any $x_0$ with $f(x_0) \neq 0$ and consider $\frac {x_n}{f(x_n)} -\frac {x_0} {f(x_0)}$. This sequence lies in the kernel and its limit is $-\frac {x_0} {f(x_0)}$. The limit must be in the kernel but this is a contradiction to the choice of $x_0$.


问:
Suppose $\lim_{n\to \infty} \frac{a_n}{n}=0 $
Suppose $$\lim_{n \to \infty} \frac{a_n}{n}=0. $$
Prove $$\lim_{n\to \infty} \frac{\max\{a_1,a_2,\ldots, a_n \}}{n}=0. $$

Below is what I tried , but I am not sure about my proof.

Denotes that
$$a_{n_1}=\max\{a_1\}$$
$$a_{n_2}=\max\{a_1,a_2\}$$
$$\vdots$$
$$a_{n_k}=\max\{a_1,a_2,\ldots, a_k \}$$
Then what we want proof can write as
  $$\lim_{k\to \infty} \frac{a_{n_k}}{k}=0 $$
It's easy to see $n_k\ge n_{k-1} $, hence $\{a_{n_k}\}$ is a subsequence of $\{a_n\}$. Also $\{\frac{a_{n_k}}{k}\} $ is a subsequence of $\{\frac{a_n}{n} \} $. Then we have
$$\lim_{k\to \infty} \frac{a_{n_k}}{n_k}=0 $$
Since $k\ge n_k $, then
$$\frac{|a_{n_k}|}{k}\le \frac{|a_{n_k}|}{n_k} $$
Here we can have $$\lim_{k\to \infty} \frac{a_{n_k}}{k}=0 $$
So any problem with my proof, since I am not so satisfied with this kind of notation, any more convenient ways to do it ?

答:
Let $\epsilon >0$ be fixed and arbitrary. Find $N_1$ so that $n>N_1$ implies $\Big|\frac{a_n}{n}\Big|<\epsilon$. Next find $N_2$ so that $n>N_2$ implies $\frac{\max\{|a_1| , \dots , |a_{N_1}|\}}{n}<\frac{\epsilon}{2}$. Put $N=\max\{N_1,N_2\}$. So we get for $n=N+1,N+2,\ldots$ that
\begin{eqnarray*}
\Bigg|\frac{\max\{a_1, \dots , a_n\}}{n}\Bigg| & \leq & \frac{\max\{|a_1|, \dots , |a_n|\}}{n}
\\ & = & \max\Big\{\Big|\frac{a_1}{n}\Big|, \dots ,\Big|\frac{a_n}{n}\Big|\Big\} \\
 & \leq & \max\bigg\{\frac{\max\{|a_1| , \dots , |a_{N_1}|\}}{n},\epsilon\bigg\} \\ & \leq & \epsilon
\end{eqnarray*}


问:
Identity for projection and inclusion of the canonical one-form on the cotangent bundle
I'm currently reading through Introduction to Mechanics and Symmetry by Marsden and Ratiu, specifically the section on Cotangent bundles. I'm trying to do the following exercise:

Let $N$ be a submanifold of $M$ and denote by $\Theta_N$ and
$\Theta_M$ the canonical one-forms on the cotangent bundles
$\pi_N: T^*N\to N$ and $\pi_M: T^*M\to M$, respectively. Let $\pi: (T^*M)|N\to T^*N$ be the projection defined by
$\pi(\alpha_n)=\alpha_n|T_nN$, where $n\in N$ and $\alpha\in T^*_nM$.
Show that $\pi^*\Theta_N=i^*\Theta_M$ where $i:(T^*M)|N\to T^*M$ is
the inclusion.

I believe we have $\pi^*\Theta_N(v)=\Theta_N(\pi_* v)$. I believe that $\pi_* v$ just turns $v\in T_\alpha(T^*M)$ into an element of $T_\alpha(T^*N)$ by removing whatever part of it was in $T_nM$. Is it true that $i_* v$ does the same thing, i.e. it takes a vector in $T_\alpha(T^*M)$ and turns it into a vector in $T_\alpha(T^*N)$? If not, how would I go about showing the identity in question?

答:
Firstly, to answer your questions:

I believe that $\pi_*v$ just turns $v\in T_\alpha(T^*M)$ into an element of $T_\alpha(T^*N)$ by removing whatever part of it was in $T_nM$.

I don't really know what this means - there's no `part' of $\pi_*v$ in $T_nM$.

Is it true that $i_*v$ does the same thing, i.e. it takes a vector in $T_\alpha(T^*M)$  and turns it into a vector in $T_\alpha(T^*N)$?

Not quite. In Marsden and Ratiu's notation, the vector would be $T_\alpha i(v)$ ($i_*$ is reserved for vector fields). Then we have
$$
i:(T^*M)\vert N\to T^*M \implies T_\alpha i :T_\alpha ((T^*M)\vert N)\to T_{i(\alpha)}(T^*M).
$$

The way to prove this result is to use the definitions of the canonical one-forms, along with the identity
$$
\pi_N\circ \pi = \pi_M\circ i
$$
(both sides map $\alpha_n\in (T^*M)\vert N$ to $n\in N$). Then for any $\alpha \in (T^*M)\vert N$ and $v_\alpha \in T_\alpha((T^*M)\vert N)$
\begin{align*}
(\pi^*\Theta_N)_\alpha(v_\alpha) &= (\Theta_N)_{\pi(\alpha)}(T_\alpha\pi(v_\alpha)) \\
&= \pi(\alpha)(T_{\pi(\alpha)}\pi_N(T_\alpha\pi(v_\alpha))) \qquad\text{by the definition of $\Theta_N$}\\
&= (\alpha\vert TN)(T_\alpha(\pi_N\circ \pi)(v_\alpha)) \qquad\text{by the chain rule}\\
&= i(\alpha)(T_\alpha(\pi_M\circ i)(v_\alpha)) \qquad\text{by the above identity}\\
&= i(\alpha)(T_{i(\alpha)}\pi_M(T_\alpha i(v_\alpha))) \qquad \text{by the chain rule}\\
&= (\Theta_M)_{i(\alpha)}(T_\alpha i(v_\alpha)) \qquad\text{by the definition of $\Theta_M$}\\
&= (i^*\Theta_M)_\alpha(v_\alpha).
\end{align*}
Hence $\pi^*\Theta_N = i^*\Theta_M$.


问:
Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$
Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$

答:
Writing
$$\frac 1{x^3-1}=\frac{1}{(a-1) (a-b) (x-a)}+\frac{1}{(b-1) (b-a) (x-b)}+\frac{1}{(a-1) (b-1) (x-1)}$$ where
$$a=-\frac{1}{2}-\frac{i \sqrt{3}}{2} \qquad \text{and} \qquad b=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$$ we face three integrals
$$I(c)=\int \frac{\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{x}{c}\right)+\log (x) \log \left(1-\frac{x}{c}\right)$$
$$J(c)=\int_0^t \frac{\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{t}{c}\right)+\log (t) \log \left(1-\frac{t}{c}\right)$$
Recombining the three terms, computing at the bounds and using the values of the polylogarithms leads to
$$\int_0^\infty\frac{\ln(x)}{x^3-1}dx=\frac{5 \pi ^2}{54}-\left(-\frac{\pi ^2}{18} \right)=\frac{4 \pi ^2}{27}$$


问:
How do I show that the following set is a vector space?
Question

Determine if the following set is a vector space:
$$ \left\{ \begin{bmatrix} a \\ b \\ c \\ d\end{bmatrix} : \begin{matrix} a - 2b & = & 4c \\ 2a & = & c + 3d\end{matrix}\right\}$$

I have a clue that I'm suppose to use the axioms to prove that it is a vector space, like that there exists the zero vector, its closed under addition and multiplication but I have no clue how to go about it

答: Presumably you are using the usual operations of addition in the vector space and multiplication by real scalars.  You now need to verify the axioms of a vector space with this subset of $\Bbb R^4$ and these operations.  The same $0$ vector works.  You need to show that it is in your subset.  Once you do that, the fact that it is the additive identity is inherited from $\Bbb R^4$.  Now you need to show that it is closed under multiplication by scalars, so show that any vector in your set can be multiplied by a real number and stay in the set.  Addition works the same way:  you are to take two vectors in your set and show their sum is still in the set.  Just write it out and see where it takes you.


问:
Prove that $C:\mathbb{R}_x^n \to T_xU$ defined as $v \to D_v$ is a bijection

Prove that the map $C:\mathbb{R}_x^n \to T_xU$ defined as $C(v)=D_v$ is bijection. Where

$D_v(f)=\frac{d}{dt} |_{t=0} f(x+tv)=\sum_{i=1}^{n} \frac{\partial}{\partial x} (x)v^i$

for $f \in C^\infty (U)$.

Show that $C$ is a linear map.

I know how to prove that.

Show that $C$ is injective.

If for all $f\in C^\infty (U)$, $D_v(f)=0$ that means

$\frac{d}{dt} |_{t=0} f(x+tv)=\sum_{i=1}^{n} \frac{\partial}{\partial x} (x)v^i=0$

Since this is a derivative that equal zero, does that mean $f(x+tv)$ is constant?

Show that $C$ is onto. Hint: use Taylor's Theorem.

Not sure either.

Any help would be appreciated. Thanks

答:
First off, $f(x+tv)$ having derivative 0 at $t=0$ does not mean that $f$ is constant. For instance, take $f(t)=t^2$, then $f'(0)=0$ but $f$ is not constant. You need the derivative to vanish on the whole domain to conclude that $f$ is constant.

Anyways, let's turn to injectivity of this map. Suppose $D_v = 0$. Then for every $f \in C^\infty (U)$, $D_v(f) = 0$. So let's define $f(w) = \langle w - x, v\rangle$. The brackets here denote the usual inner product, and I'll leave checking smoothness up to you. Essentially, $f$ comes from projecting $w$ to the $v$ direction. We compute $f(x + tv) = \langle tv, v\rangle = t ||v||^2$. Hence, the derivative of this function at $t=0$ is $||v||^2$. As we assumed that $D_v = 0$, we have that $||v||^2 = D_v(f) = 0$. Hence, $v = 0$ so the kernel of this map is trivial.

For surjectivity, could you specify what exactly your definition of $T_x U$ is? The hint is correct that you have to apply Taylor's theorem, but the specifics of how this argument will be written will depend on this definition.

EDIT: Let's let $Y \in T_x U$. Let's also define by $p_i \in C^\infty (U)$, by $p_i(w) = \pi_i(w - x)$ where $\pi_i$ is projection onto the $i^{th}$ coordinate. We can therefore consider the values $\lambda_i = Y(p_i)$. Let $X = \sum \lambda_i D_{e_i}$, which is in the image of $C$. We claim $X = Y$. Indeed, take some $f \in C^\infty (U)$. By Taylor's theorem, we will write $f(w) = f(x) + \sum a_i p_i(w) + \sum (a_{ij} p_i p_j)(w)$. Here, the $a_i$ are constants and the $a_{ij}$ are smooth functions. Let $Z = X - Y$. Then $Z(f) = Z(f(x)) + \sum a_i Z(p_i) + \sum Z(a_{ij} p_i p_j)$. By the Leibniz rule, $Z(f(x)) = 0$. Furthermore, $Z(p_i) = X(p_i) - Y(p_i) = \lambda_i - Y(p_i) = 0$. Now, we are left to consider $Z(a_{ij} p_i p_j)$. Again, applying the Leibniz rule, this is $Z(a_{ij} p_i) p_j(x) + (a_{ij}p_i)(x) Z(p_j) = 0$. Hence, we have shown that $Z = 0$ so $X = Y$. We chose $X$ to be in the image of $C$, and $Y \in T_x U$ was arbitrary. Hence, $C$ is onto.


问:
Is this math game always winnable?
Kent Haines describes the game of Integer Solitaire, which I find to be excellent for young kids learning arithmetic. I'm sure they will be motivated by this game to get a lot of practice.

Kent asks a question about his game, which I find very interesting, and so I am asking here, in the hopes that Math.SE might be able to answer.

The child draws 18 cards from an ordinary deck of cards, and then regards the cards to have values Ace = 1, 2, 3, ..., Jack = 11, Queen = 12, King = 13, except that Black means a positive value and Red means a negative value.

Using 14 of the 18, the child seeks to find solutions of four equations:

For example, a successful solution would look like:

Question. Does every set of 18 cards admit a solution?

Kent Haines says, "I have no idea whether all combinations of 18 cards are solvable in this game. But I have played this game for five years with dozens of students, and I have yet to see a combination of 18 cards that is unsolvable."

Follow up Question. In the event that the answer is negative, what is the probability of having a winning set?

For the follow up question, it may be that an exact answer is out of reach, but bounds on the probability would be welcome.

答:
28 card counterexample:

$$ black: K, K, J, J, 9, 9, 7, 7, 5, 5, 3, 3, A, A $$
$$ red: K, K, J, J, 9, 9, 7, 7, 5, 5, 3, 3, A, A $$

cannot satisfy __ + __ = __ because 2 odds make an even (whether added or subtracted), and there are no evens in the set.

Edit: it's 28 cards, not 26.

The 29th counterexample is easy: with only one additional even, it isn't enough to satisfy both of the top two equations. So, 2 evens are needed to be added.

Title: Is this math game always winnable?


问:
Find non-zero vectors $x, y, b $belonging to $R^3 $such that $Ax=b, By=b$
where $A$ and $4$ are as follows
\begin{equation*}
A =
\begin{pmatrix}
0 & 1 & 2\\
1 & 1 & 1\\
2 & 1 & 0
\end{pmatrix}
\end{equation*}
\begin{equation*}
B =
\begin{pmatrix}
1 & -1 & 0\\
0 & 0 & 1\\
1 & 1 & 0
\end{pmatrix}
\end{equation*}

This problem throws me off in the sense that I have to find 3 vectors. I suspect I have to use the LU factorization first in order to determine the b vector.
So I have found a decomposition for both A and B. $PA=LU; PB=LU$. I'm not sure if I started correctly or I am taking the wrong approach. Any observations are much appreciated.

PA=LU

PB=LU

答: Note that the matrix $B$ is invertible (non-singular). Therefore, the equation $B\mathbf{y}=\mathbf{b}$ is equivalent to $\mathbf{y}=B^{-1}\mathbf{b}$. So first you can pick vectors $\mathbf{x}$ and $\mathbf{b}$ satisfying $A\mathbf{x}=\mathbf{b}$, which you can do in lots of different ways, and then calculate $\mathbf{y}=B^{-1}\mathbf{b}$.


问:
$3^n+4^n<5^n$ for all $n>2$
I'm doing the following induction proof and wanted to know if this was valid. I think it is, but I'm seeing more complicated solutions than what I did. What I did seems much easier.

Prove that $3^n+4^n<5^n$ for all $n>2$.

When $n=3$ we get $91<125$. No problem, now assume the result is true from $k<n$, $(5^k>3^k+4^k)$ and consider $5^{k+1}=5 \times 5^k>5(3^k+4^k)=5\times 3^k + 5\times 4^k>3\times 3^k+4\times 4^k=3^{k+1}+4^{k+1}$ since $5\times 3^k>3\times 3^k$ and $5\times 4^k> 4\times 4^k$.

答:
Another proof, NO INDUCTION! -- just for fun.

Let $\ n\ge 3.\ $ then:

$$ 3^n + 4^n <\, 3^2\cdot 5^{n-2} + 4^2\cdot 5^{n-2}
 \,=\, (3^2+4^2)\cdot 5^{n-2}\, =\ 5^n $$

REMARK   If you look at the above proof under a magnifying glass then you may see induction after all -- just a trace.

Title: $3^n+4^n<5^n$ for all $n>2$


问:
Expressing $\sin(2x)$ as a polynomial of $\sin{x}$
Using trignometric identities ( double angle forumlas) one can see that $\cos{2x} = 2 \: \cos^{2}{x} - 1$ can be expressed as a polynomial of $\cos{x}$, where $p(\cos{x})=2 \: \cos^{2}{x}-1$. Then its natural to ask the same question for the sine function.

Can we express $\sin{2x}$ as a polynomial of $\sin{x}$?

答:
Here's a slightly different way of proving that $\sin nx$ is not
a polynomial in $\sin x$ when $n$ is even. Indeed $\sin nx$
cannot be written as $f(\sin x)$ for any function $f$.

To prove this all one has to do is to write down two numbers $a$
and $b$ such that $\sin a=\sin b$ but $\sin na\ne\sin nb$.
Let $a=\pi/(2n)$ and $b=\pi-a$. Then $\sin a=\sin b$. But
$na=\pi/2$ and $nb=n\pi-\pi/2$ so that $\sin na=1$ and $\sin nb=-1$
(recalling that $n$ is even).

ContentLicense: CC BY-SA 2.5


问:
How to use the explicit formula for divergence in spherical coordinates
I'm not sure how to use correctly
$$\nabla \cdot \vec{F} = \frac{1}{r^2} \partial_r (r^2 F^r) + \frac{1}{r \sin \theta} \partial_\theta (\sin \theta F^\theta) + \frac{1}{r \sin \theta} \partial_\phi F^\phi$$

For exemple, I have $\vec{F}$ = $r^3\phi sin(\theta)(\hat{u_r} + \hat{u_\phi} + \hat{u_\theta})$

Do I just replace $F^r$ with $r^3$ and so on then find the derivative?

Is this just a plug and go formula?

答:
Yes you simply take partial derivative as stated in the formula. In this case,

$\vec F = r^3 \phi \sin \theta \ (\hat{u_r} + \hat{u_\phi} + \hat{u_\theta}) = r^3 \phi \sin \theta \ \hat{u_r} + r^3 \phi \sin \theta \ \hat{u_\phi} + r^3 \phi \sin \theta \ \hat{u_\theta}$

Now simply take derivative for each component.

$\frac{1}{r^2}\partial_r (r^2 F^r) = \frac{1}{r^2}\partial_r (r^5 \phi \sin \theta) = 5r^2 \phi \sin\theta$

$\frac{1}{r \sin\theta}\partial_{\phi} (F^{\phi}) = \frac{1}{r \sin\theta}\partial_{\phi} (r^3 \phi \sin\theta) = ?$

$\frac{1}{r \sin\theta}\partial_{\theta} (\sin \theta F^{\theta}) = \frac{1}{r \sin\theta}\partial_{\theta} (r^3 \phi \sin^2 \theta) = ?$


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答:
Asserting you are a mathematician without immediately volunteering further details is just a bad thing. It's very closed.

The principle is simple: As soon as you mark yourself as part of a world that people are not part of and don't understand, you alienate them. Unless you give them something that relates to their own experience, you are going to have a hard time.

Consider this conversation:

— Yeah, I'm a school teacher.

  — Cool, do you teach little ones or big ones?

  — Little ones, they're constantly surprising...

  — Yeah, I've got two kids about to go to school.

  [Conversation ensues]

It's OK for a teacher, because everyone knows what a teacher does, roughly. Even if they don't really understand what a teacher does day to day, they know why they do what they do.

— Yeah, I'm a mathematician.

  — ... (blank look as they realize they know nothing about what a professional mathematician does) ... Yeah, I was bad at maths at school. (nothing else to say)

  [Awkward silence ensues]

In this case they have nothing to work with. With something like maths, you have to give them something extra to relate to, or they will just jump to their only experience of mathematics: school, where more than likely they were not working at the standard of a profession mathematician.

It is easy to find yourself in the Ivory Tower. People responding in this way should be taken as an indication that your work has become very detached from social reality — although some mathematicians think this is a virtue, it really is not, it's more like laziness.

Things you could say instead of "I'm a mathematician":

"I investigate the properties of prime numbers."

"I look at how fluids move."

"I'm interested in symmetry."

"Do you know X? Well, I do something like X."

"I try to find out why Y."

Just give a more open answer that give people a chance to grasp what you do. Do not require them to be part of your world to understand it. If you do this you will find people will suddenly become quite good at maths.


问:
Prove $\int_{2}^{\infty}\ln\sec\frac{\pi}{x}\,dx$ is convergent
I have proved that  $\int_{2}^{3}\ln\sec\frac{\pi}{x}\,dx$ is convergent by first proving that $$\sec x\leq\frac{1}{1-\frac{2}{\pi}x}$$ and therefore proving that if $\int_{2}^{3}\frac{1}{1-\frac{2}{x}}\,dx$ converges then $\int_{2}^{3}\ln\sec\frac{\pi}{x}\,dx$ converges. I am stuck now on how to prove that $\int_{3}^{\infty}\ln\sec\frac{\pi}{x}\,dx$ converges. I have tried using limit comparison test with $\sec\frac{\pi}{x}-1$ but I can't prove that $\int_{3}^{\infty}\left(\sec\frac{\pi}{x}-1\right)dx$ converges either.

答:
$\cos y \geq 1-\frac {y^{2}} 2$ for all $y \geq 0$. [Proof: $(\cos y - 1+\frac {y^{2}} 2)'=-\sin y +y\geq 0$ and $\cos y - 1+\frac {y^{2}} 2$ vanishes when $y=0$].

Hence, on $(\pi, \infty)$ we have $ln (\sec \frac {\pi} x)=-\ln (\cos  \frac {\pi} x)\leq -ln (1- (\frac {\pi} x)^{2}))$. Now I  will let you finish using the inequality $-\ln (1-t) =\ln (\frac 1 {1-t})=\ln [1+\frac t {1-t}] \leq \frac t {1-t}$.


答:
I am going to show a part of it just to give you the idea. You can do the rest. If $V_{1}$ and $V_{2}$ belongs to the vector space (S), we want to show that $V = V_{1}+V_{2}$ also belongs to that vector space. In other words, we want to show $V$ has the properties defined in the set.

$$ V_{1} = \begin{bmatrix} a_{1} \\ b_{1} \\ c_{1} \\ d_{1}\end{bmatrix}, \space  V_{2} = \begin{bmatrix} a_{2} \\ b_{2} \\ c_{2} \\ d_{2}\end{bmatrix} \longrightarrow V = \begin{bmatrix} a=a_{1}+a_{2} \\ b=b_{1}+b_{2} \\ c=c_{1}+c_{2} \\ d=d_{1}+d_{2}\end{bmatrix} $$

Since the vectors $V_{1}$ and $V_{2}$ belong to S, we can say:

$$ \text{Because $V_{1} \in S$}: \space a_{1} - 2b_{1} = 4c_{1} $$
$$ \text{Because $V_{2} \in S$}: \space a_{2} - 2b_{2} = 4c_{2} $$

Now, let's add Left-hand side together, and Right-hand side together:

$$ (a_{1}+a_{2}) - 2(b_{1}+b_{2}) = 4(c_{1}+c_{2}) $$

$$ a - 2b = 4c $$

Similarly, you can show $2a=c+3d$ can be satisfied as well.

Good Luck.

Title: How do I show that the following set is a vector space?

Body:
Question

Determine if the following set is a vector space:
$$ \left\{ \begin{bmatrix} a \\ b \\ c \\ d\end{bmatrix} : \begin{matrix} a - 2b & = & 4c \\ 2a & = & c + 3d\end{matrix}\right\}$$

I have a clue that I'm suppose to use the axioms to prove that it is a vector space, like that there exists the zero vector, its closed under addition and multiplication but I have no clue how to go about it


问:
If $D=\{(x,y):x^2+y^2\leq 1\}$. Show there is $p_0\in D$ such that $T(p_0)=(0,0)$
Let $D=\{(x,y):x^2+y^2\leq 1\}$, $T$ transformation of class $C'$ on an open containing $D$, $$T:\left\lbrace \begin{array}{rcl} u &=& f(x,y) \\ v&=&g(x,y)  \end{array}\right.  $$  whose Jacobian is never $0$ in $D$, and $|T(p)-p|\leq \frac{1}{3}$ for all $p\in D$. Show there is $p_0\in D$ such that $T(p_0)=(0,0)$

Idea: Seems like fixed point theorem can help, for example if  I define  $H(p)=T(p)-p$, then $H(p)=p$ iff $T(p)=0$, so it's enough to find such point for $H$, by hypothesis $|H(p)|\leq \frac{1}{3}$ . Usually the fixed theorem is stated as: if $f:D\to D$ is continuous then there is a fixed point, or also there is no retraction of class $C'$ from $D$ onto the unit circle (only the boundary of $D$). Here I know that the jacobian of $T$, $J(p)\neq 0$ for all $p\in D$ so there is a neighborhood of any $p\in D$ where $T$ is $1$-to-$1$, i.e., in where $T^{-1}$ exists. But at least for now I can't have a precise idea. Any hints are welcome.

For context, this is a problem #17 section 7.5 from Buck's Advance Calculus. In this section we have the results:

For $T:D\subset\mathbb{R}^n \to \mathbb{R}^n$ of class $C'$, $D$ open, with $J(p)\neq 0$ for all $p\in D$ then $T$ is locally $1$-to-$1$ in $D$.

If $J(p_0)\neq 0$ only for a specific value, then $T$ is $1$-to-$1$ only in a small neighborhood of $p_0$, where $T^{-1}$ exists.

If conditions as in 1.,then $T(D)$ is an open set.

答:
It took me a bit to figure out how to do this without homotopy or winding numbers, which is how it's always proved.  This is an interesting exercise; to see how this normally arises, look up the "dog on a leash" or "walking the dog" lemma of complex analysis.

Define the boundary $\partial(A)$ of a set $A \subset \mathbb{R}^n$ to be the intersection $\overline{A} \cap \overline{\mathbb{R}^n \setminus A}$.  Thus it's the points which are arbitrarily close to both $A$ and its complement; for example, the unit circle is the boundary of $D$.

Note by definition that if $x \in A$ and $x \notin \partial(A)$, then $x$ is in the interior of $A$ - because if every open neighborhood of $x$ intersected the complement of $A$, then $x$ would be in the boundary.  Thus a set is composed of its interior points and its boundary points.

Lemma: If $D \subset \mathbb{R}^n$ and $T$ is $C'$ with non-vanishing Jacobian on (a neighborhood of) $D$ then $T$ doesn't send an interior point of $D$ to the boundary of $T(D)$.

Proof: If $x \in U \subset D$ is a sufficiently small open neighborhood of $x$ in $D$, then $T(U) = V$ is open by the open mapping theorem, and $V = T(U) \subset T(D)$ so that $x$ is an interior point of $T(D)$.  QED

Now returning to your problem, if $B$ is the closed ball centered around $0$ with radius $\frac{1}{3}$ then $T(0) \in B$ since $|T(0) - 0| = |T(0)| \leq \frac{1}{3}$.  However, $T(\mathbb{S}^1) \cap B = \varnothing$ since $|T(p)-p| \leq \frac{1}{3}$ and thus restricting to $\mathbb{S}^1 = \partial(D)$ gives us points at least $\frac{2}{3}$rds away from $0$.  Since $\mathbb{S}^1$ is the boundary of $D$, by the above lemma $T(D)$ has no boundary points in $B$.

Note that $T(D) \cap B$ is non-empty by virtue of containing at least the point $T(0)$.  Thus $T(D)$ must contain all of $B$ in order to not have any boundary points there: Otherwise, there would be no point contained in both $\overline{T(D)}$ and $\overline{B \setminus T(D)}$.  But then these two sets would be a separation of $B$, which is connected - impossible.

Thus $0 \in B \subset T(D)$, completing the proof.


问:
Can we prove a set of a set by the Pairing Axiom?
According to the Pairing Axiom, suppose we have two sets $a$ and $b$, we have a set $\{a, b\}$.

But can we have a set $\{k\}$ from a set $k$ by the Pairing Axiom?

答:
Yes. Just take $a=b=k$. In the pairing axiom, $a,b$ are not required to be different.

Also have in mind that for $a=b$, $\{a,b\}=\{a,a\}=\{a\}$.


问:
Prove that $e^a e^b = e^{a+b}$
I've read the argument in Rudin, but I think I need a little clarification

\begin{align}
e^a e^b &= \sum_{k=0}^{\infty} \frac{a^k}{k!} \sum_{m=0}^{\infty} \frac{b^m}{m!}\\ &= \sum_{n=0}^{\infty} \frac{ n!}{n!}  \sum_{k=0}^{n} \frac{a^k}{k!} \frac{b^{n-k}}{(n-k)!} \\
&= \sum_{n=0}^{\infty} \frac{(a+b)^n}{n!}\\
&=e^{a+b}\\
\end{align}

I'd like to understand how we get from the first line to the second. Of course $m =n-k$, but how do we ensure that we haven't missed any addends in the limit changing process?

Changing limits: $m=0 \rightarrow n-k=0 \rightarrow k=n$ and $m=\infty \rightarrow n-k=\infty \rightarrow k=0$, assuming $n \to \infty$.

\begin{align}
\frac{b^m}{m!} \sum_{k=0}^{\infty} \frac{a^k}{k!}
&= \sum_{k=0}^{\infty} \frac{a^k}{k!} \frac{b^{m}}{m!} \\
&=\sum_{k=0}^{?} \frac{a^k}{k!} \frac{b^{n-k}}{(n-k)!} \\
\end{align}

how to justify that "?"$=n$ ?

答:
Suppose we have two series

$$p(x)=\sum_{k=1}^\infty a_kx^k$$

$$q(x)=\sum_{k=1}^\infty b_kx^k$$

Then multiplying and collecting terms gives $$(p\cdot q)(x)=a_0b_0+(a_1b_0+b_1a_0)x+(a_2b_0+a_1b_1+a_0b_2)x^2+\dots \\=\sum_{k=0}^\infty \left(\sum_{j+i=k}a_jb_i\right)x^k\\=\sum_{k=0}^\infty \left(\sum_{j=0}^k a_jb_{k-j}\right)x^k$$

This informal approach actually works whenever either one of the series converges and the other converges absolutely. Try to see what happens when we choose the series to be those of $e^a,e^b$.

ADD The case when the series are not power series, but just plain series gives
$$\sum\limits_{k = 0}^\infty  {\left( {\sum\limits_{j = 0}^k {{a_j}} {b_{k - j}}} \right)}  = \sum\limits_{k = 0}^\infty  {{a_k}}  \cdot \sum\limits_{k = 0}^\infty  {{b_k}} $$

This gives when $$\eqalign{
  & {a_k} = \frac{{{a^k}}}{{k!}}  \cr
  & {b_k} = \frac{{{b^k}}}{{k!}} \cr} $$ that

$$\displaylines{
  {e^a}{e^b} = \sum\limits_{k = 0}^\infty  {\left( {\sum\limits_{j = 0}^k {\frac{{{a^j}}}{{j!}}} \frac{{{b^{k - j}}}}{{\left( {k - j} \right)!}}} \right)}  \cr
   = \sum\limits_{k = 0}^\infty  {\frac{1}{{k!}}\left( {\sum\limits_{j = 0}^k {\frac{{k!}}{{j!\left( {k - j} \right)!}}{a^j}} {b^{k - j}}} \right)}  \cr
   = \sum\limits_{k = 0}^\infty  {\frac{1}{{k!}}{{\left( {a + b} \right)}^k}}  \cr
   = {e^{a + b}} \cr} $$

Title: Prove that $e^a e^b = e^{a+b}$


问:
$T:V\rightarrow W$ is surjective $\iff \exists S:W\rightarrow V, TS=I_W$
My textbook has this problem, and I saw some proofs in the net, but this idea seems wrong to me. Can anyone point out where is the loophole in my understanding?

$T:V\rightarrow W$ is surjective $\iff \exists S:W\rightarrow V, TS=I_W$

My counter example:

Let $V $ be $R^3$, $W$ be $R^2$, and $T$ be the transformation $z=0$. So $T$ maps $3$ dimensional space to $2$ dimensional space $xy$-plane. This mapping is surjective, because for each point in $xy$-plane there is a point in $R^3$. Specifically, for each $(x,y)\in R^2$ all the points in the line parallel to $z$-axis and passing though $(x,y,0)$ in $R^3$ map to $(x,y)$ in $R^2$. So $T$ is surjective here, because range(T) = $R^2$. But since mapping T is not injective, no mapping  $S$ can convert $(x,y)$ in $R^2$ back to $(x,y,z)$ in $R^3$, because we can not decide, to which point of the line, parallel to z-axis, should $(x,y)$ be mapped to.

答:
Your mistake is in

But since mapping $T$ is not injective, no mapping $S$ can convert $(x,y)$ in $\mathbb{R}^2$ back to $(x,y,z)$ in $\mathbb{R}^3$, because we can not decide to which point of the line parallel to $z$-axis, should $(x,y)$ be mapped to.

In a sense, that's precisely what we want to do here — decide which point on this line to map to. Yes, we have infinitely many choices. But it doesn't mean that we can't decide and choose one of them. Of course, this choice can't be random, since we have to come up with a linear transformation. How about we decide that $S$ sends each $(x,y)\in\mathbb{R}^2$ to $(x,y,0)\in\mathbb{R}^3$? You can verify that this $S$ is indeed a linear transformation and that it satisfies the desired requirements.

I think you may be thinking about finding an inverse to the transformation $t$, which indeed is impossible when $T$ is only surjective but not injective (and thus not bijective). But note that the definition of an inverse requires inverting $T$ from both sides. Viz., $S:W\to V$ is an inverse to $T:V\to W$ iff $ST=\operatorname{Id}_V$ and $TS=\operatorname{Id}_W$. Of course, that's impossible when $T$ is not bijective. But here we're inverting from the right side only.


问:
Find a matrix $A$ with no zero entries such that $A^3=A$
I took a standard $2 × 2$ matrix with entries $a, b, c, d$ and multiplied it out three times and tried to algebraically make it work, but that quickly turned into a algebraic mess. Is there an easier method to solve this?

答: One way to simplify the problem is to impose some further assumptions on $A$. For example, suppose we further assume that $A$ is singular. Then $A=uv^T$ for some entrywise vectors $u$ and $v$. The requirements that $A^3=A$ and $A\ne0$ entrywise are now equivalent to $(v^Tu)^2=1$ and $u,v\ne0$ entrywise, which are utterly easy to fulfil (when the characteristic of the underlying field is not $2$).


答:
Consider the function $f(a,b,c,d)=(a-2b-4c,\,2a-c-3d)$ from $\Bbb R^4$ to $\Bbb R^2.$

Confirm that if $v_1,v_2\in \Bbb R^4$ and $r\in \Bbb R$

then $(**)\;f(v_1+rv_2)=f(v_1)+rf(v_2).$

Let $V=\{v\in \Bbb R^4: f(v)=(0,0)\}.$ Then

(i). $V\ne \emptyset$ (e.g.$(0,0,0,0)\in V$).

(ii). If  $v_1,v_2\in V$ and $r\in \Bbb R$ then $f(v_1+rv_2)=f(v_1)+rf(v_2)=(0,0)$ by $(** )$ so $v_1+rv_2\in V.$

And you're done.


问:
How to solve for x in modular arithmetic when the equation has exponent?
Say I have the following equation-->

$x^2=180345$ and modulo is $8$. I am using an online calculator to solve this equation, and the answer comes out to be $x=1$ or $3$ or $5$ or $7$.

I am not getting the maths behind this. It will be very helpful if someone can refer to some materials where these kind of equations are solved or atleast show me the algorithm for solving this equaions. Thank you.

答:
$$x^2 = 180345 \bmod 8 = 1 \bmod 8$$

Now, $x$ can have only the following set of values $\bmod 8$:

$\{0, 1, 2, 3, 4, 5, 6, 7\}$

You can verify that only $x = 1, 3, 5, 7$ satisfy $x^2 = 1 \bmod 8$ and the other values do not.


答:
First thing get that $180345$ to something reasonable.  $180345 \equiv 1$ so we have

$x^2\equiv 1 \pmod 8$ so $x^2 -1 = \equiv 1 \pmod 8$.  Now we can just test them one after another as there are just $7$ of them....

But let's be clever.  $x^2 \equiv 1 \pmod 8$

$x^2 -1 \equiv \pmod 8$

$(x+1)(x-1) \equiv 0 \pmod 8$.

So $x\equiv \pm 1\pmod 8$ are two immediate answers.

Any other solutions must relate to the zero divisors of $8$.

$(x+1)(x-1)= 8k$ for some integer $k$. and $x+1, x-2$ are both even.  So we have $x-1, x+1 = 2,4$ or $4,6$ or $6,8\equiv 0$ or $0,2$.

So $x = 3,5, 7\equiv -1, $ or $1$.

=======

Also as player3236 stated in the comments, for any odd number $n = 2k+1$ we have $n^2 \equiv 4k^2 + 4k + 1\equiv 4(k^2+k)+ 1\pmod 8$.  And as $k^2 + k$ is always even....

$n^2 \equiv 4(k^2+k) + 1 \equiv 8\frac {k^2 + k}2 + 1\equiv 1\pmod 8$.

So every odd number is a solution.  $x\equiv 1,3,5,7$

And clearly $x$ even is not a solution so... that's it.

Title: How to solve for x in modular arithmetic when the equation has exponent?

Body:
Say I have the following equation-->

$x^2=180345$ and modulo is $8$. I am using an online calculator to solve this equation, and the answer comes out to be $x=1$ or $3$ or $5$ or $7$.

I am not getting the maths behind this. It will be very helpful if someone can refer to some materials where these kind of equations are solved or atleast show me the algorithm for solving this equaions. Thank you.


问:
span equality $span\{v_1,v_2\}=span\{av_1,bv_2\}$
Prove for nonzero scalars a and b such that $$span\{v_1,v_2\}=span\{av_1,bv_2\}$$

My try:

$$span\{v_1,v_2\}=c_1v_1+c_2v_2=c'_1(av_1)+c'_2(bv_2)=span\{av_1,bv_2\}$$ by splitting contants. Is it correct to show ? then what is the role of non zero scalar word here ? i feel it is okay for 0 also

Body: $span\{v_1,v_2\}=\{xv_1+yv_2:x,y\in\mathbb{K}\}=\{\frac{x}{a}(ax)+\frac{y}{b}(bv_2):x,y\in\mathbb{K}\}=\{x'(av_1)+y'(bv_2):x',y'\in\mathbb{K}\}=span\{av_1,bv_2\}$


答:
Writing

$$span\{v_1,v_2\}=c_1v_1+c_2v_2$$ is not correct. It should be

$$span\{v_1,v_2\}=\{c_1v_1+c_2v_2 \mid (c_1,c_2) \in \mathbb F^2\}.$$

Then you could notice that for any $(c_1,c_2) \in \mathbb F^2$ you have

$$c_1v_1+c_2v_2= \frac{c_1}{a}(a v_1)+ \frac{c_2}{b}(b v_2)$$ as $a,b$ are not vanishing by hypothesis.

Title: span equality $span\{v_1,v_2\}=span\{av_1,bv_2\}$

Body:
Prove for nonzero scalars a and b such that $$span\{v_1,v_2\}=span\{av_1,bv_2\}$$

My try:

$$span\{v_1,v_2\}=c_1v_1+c_2v_2=c'_1(av_1)+c'_2(bv_2)=span\{av_1,bv_2\}$$ by splitting contants. Is it correct to show ? then what is the role of non zero scalar word here ? i feel it is okay for 0 also


问:
Solving a special rational equation on a very small interval
I need to solve the following equation (for $x$):
\begin{equation}
\mathcal f(x):=\sum_{i=1}^n b_i \left( \frac{a_i}{1+b_i x}\right)^2-\phi=0, \quad \text{with} \quad -1/b_1< x \le 0.
\end{equation}
Assume that the equation has at least one solution in $(-1/b_1,0]$ and $b_1\ge b_2\dots\ge b_n>0$. In practice, $b_1$ can be very large, so that the interval of interest $(-1/b_1,0]$ can be very small. Therefore, numerical issues can happen (I think because $1/(1+b_1x)$ can blow up, at least when a numerical method captures a solution that is very close to $-1/b_1$). What is the best practical idea to find a solution of $f(x)=0$ that falls into $(-1/b_1,0]$?

By the way, I am currently using the bisection method and $b_1$ is around 1000 for example. The quantity $n$ could be up to 50 in my case. Further, such an equation must be solved iteratively in my algorithm for different values of $ a_i$'s for $i\in [n]$.
Thank you very much in advance for your time and help.

答:
The current framing of the Question suggests that the $b_i$'s are fixed but the problem (root-finding) will need to be solved repeatedly for varying coefficient's $a_i$.  My remarks are intended to help with the selection of an initial approximation of the root, to be improved iteratively.

Note that the problem specifies only that the $b_i$'s are ordered by weak inequalities:

$$ b_1 \ge b_2 \ge b_3 \ge \ldots \ge b_n $$

Unless the intention is vary those $b_i$'s, we can eliminate a priori consecutive terms with equal $b_i$'s (by lumping them together) in the summation that defines $f(x)$:

$$ f(x) := \sum_{i=1}^n b_i \left( \frac{a_i}{1+b_i x} \right)^2  - \phi $$

So let's assume the $b_i$'s are distinct, although to be sure there is no requirement that they are widely spaced.

Then all the terms in the summation are continuous and strictly decreasing on the interval $(-1/b_1,0]$ in which the root is sought, and only the first term fails to be continuous at left hand endpoint $x=-1/b_1$ because the first term tends to $+\infty$ there.

It follows that any root $f(r) = 0$ with $r\in (-1/b_1,0]$ is unique, and such a root exists if and only if $f(0) \le 0$.  The case $f(0)=0$ is easily checked as it amounts to $\sum a_i^2 b_i = \phi$, so we assume hereafter that the check shows $f(0) \lt 0$.

Locating $r$ is striking the right balance between the contributions of the first term, which are "potentially" unbounded, and the contributions of the rest of the summands, which are at most a finite value we can directly evaluate at $x=-1/b_1$.

My suggestion is that we evaluate those terms:

$$ g(x) := \sum_{i=2}^n \frac{a_i^2 b_i}{(1 + b_i x)^2} $$

at two points and use linear interpolation to approximate their sum in between.  Since those terms are concave up (a convex function), that linear interpolation is consistently an overestimate of their true contribution.

Initially we would evaluate $g(x)$ at the two points $x=-1/b_1$ and $x=0$.  Then we would solve the simpler rational equation (as equivalent to a cubic):

$$ \frac{a_1^2 b_1}{(1 + b_1 x)^2} + b_1(g(0) - g(-1/b_1)) x + g(0) = \phi $$

The solution $r_1$ of this equation will yield $f(r_1) \lt 0$ because of the error introduced by linear interpolation.  Therefore $r_1 \gt r$ overestimates the actual root.

We can then repeat this procedure, replacing evaluation of $g$ at the upper endpoint $x=0$ with evaluation at new upper endpoint $x=r_1$.  The attraction of this method is that it conservatively approaches the actual root from above, and therefore does not risk passing the singularity at $x = -1/b_1$.


问:
Three sigma confidence interval for rayleigh distributed quantity
Please consider a one dimensional measurement device, which can measure a position in $x$ direction and is specified to have a precision of $3\sigma(x) < thr$, where $thr$ is a threshold value, for example a few nanometer. That means 99,73% of the measurements have an error $|\Delta x|$ of less than $thr$.

Now I combine two of these devices to measure an $(x,y)$ position. How can I write the specification for $r=\sqrt{\Delta x^2+\Delta y^2}$ to indicate that 99,73% of the measurement errors $r$ are lower than $thr$?

I understand that when $\Delta x$ is normal distributed, then $r$ is Rayleigh distributed. I tried a spec like $\bar{r}+3\sigma(r) <  thr$, but the deviation is quite high.

答:
To formalize this problem, let the following be true:
$$\Delta x \sim \mathcal{N}(0,\sigma^2)$$
$$\Delta y \sim \mathcal{N}(0,\sigma^2)$$
$$thr=3\sigma \ \therefore \ \Biggl\{ \begin{matrix} \mathbb P(-thr \le \Delta x \le thr)\approx 99.73\% \\
\mathbb P(-thr \le \Delta y \le thr)\approx 99.73\%\end{matrix}$$
$$r=\sqrt{\Delta x^2 + \Delta y^2}$$

Then, as you said, random variable $r$ draws from a Rayleigh Distribution with a scale parameter of $\sigma$.  Now if you want to find the $F_r=99.73\%$ quantile of $r$, then you can just use the equation:

$$Q(F_r,\sigma)=\sigma \sqrt{-2 \ \mathrm{ln}(1-F_r)}$$
$$Q(0.9973,\sigma)=\sigma \sqrt{-2 \ \mathrm{ln}(1-0.9973)}=\frac{thr}{3} \sqrt{-2 \ \mathrm{ln}(1-0.9973)}$$


问:
How to determine each point of a circle?
I have a circle with a predefined radius (e.g r=5), so X(+/-)max and Y(+/-)max equals (+/-)5.
How do I determine all other points of the circle with a function?
For example: How do I get the +Y/-Y Value for X=2,5?

Thank you :)

答:
Well the general equation of that circle is $$x^2+y^2=5^2\tag 1$$
If the centre is not the origin but instead $(a, b)$, the equation would be $$(x-a) ^2+(y-b)^2=25$$

In this case the centre is the origin, so we rearrange $(1)$ to:
$$y^2=25-x^2=25-\frac52^2=\frac{75}{4}\implies y=\pm\frac{\sqrt{75}}{2}=\pm \frac52 \sqrt 3$$


问:
find $f \in L^2(0,1)$ such that $f = g + K(f)$
I'm preparing for my qualifying exam and the question bank had the following question under functional analysis section and I'm stuck with this problem.

Let $L^2(0,1)$ be the Hilbert space of square integrable functions on the interval $[0,1]$ with Lebesgue measure. Define $K : L^ 2 (0,1) \rightarrow L^ 2 (0,1)$ by $K(f)(t) =\int_0^t (t − s)f(s)\,ds$. Show that
$||K|| < 1$. Given $g \in L^ 2 (0,1)$ find $f \in L^ 2 (0,1)$ such that $f = g + K(f)$.

My idea for the second part of the problem: I feel like it is the kernel of some map and that the question is asking to find an element in the coset of $g$ (the coset that you get after you quotient out by the kernel).

I really have no idea how to proceed. Any help would be appreciated.

答:
There is another way to prove the second statement. Since $K: L^2\to L^2$ for a given $g\in L^2(0, 1)$ you can define a map $T_g: L^2(0, 1)\to L^2(0, 1)$ by
$$T_g(f)=g+K(f).$$
If you show that $T_g$ is a contraction mapping, since $L^2$ is a complete metric space then by the Banach Fixed Point Theorem there exists a unique fixed point $f\in L^2(0, 1)$ of $T_g$, that is
$$f=T_g(f)=g+K(f)$$
which is the solution you are looking for. This is indeed the case, since
$$\|T_g(f_1)-T_g(f_2)\|_{L^2}=\|K(f_1)-K(f_2)\|_{L^2}\leq \|K\|\|f_1-f_2\|_{L^2}$$
and you already proved that $\|K\|<1$.

EDIT: If you want to compute $f$ explicitly, you can go as follows (i'll skip any technical detail). First we write the equation as
\begin{align}f&=g+t\int_0^1 f(s)ds-\int_0^1sf(s)ds\\&=g+C_1t-C_2\end{align}
where
\begin{align}C_1&=\int_0^1f(s)ds,\\ C_2&=\int_0^1 sf(s)ds.\end{align}
Then by replacing $f$ in $C_1$ we get
\begin{align} C_1=\int_0^1 (g(s)+C_1s-C_2)ds=\int_0^1g(s)ds+\dfrac{C_1}{2}-C_2\end{align}
and then
$$\dfrac{C_1}{2}+C_2=\int_0^1 g(s)ds.$$
Similarly by replacing $f$ in $C_2$ we get
$$\dfrac{C_1}{3}-\dfrac{3C_2}{2}=-\int_0^1 sg(s)ds.$$
With these two equations we can obtain $C_1$ and $C_2$, hence by replacing it in the first equation for $f$ you get the full expression only in terms of $g$.


问:
Find the posterior density of $\log \theta$ and Bayes estimator under 0-1 loss

Let $X_1,...,X_n$ be iid from the uniform distribution on $(0,\theta)$
where $\theta>0$ is unknown. Suppose the prior distribution of
$\theta$ is log-normal with parameters $\mu, \sigma^2$, which are
known constants.

Find the posterior density of $\log \theta$.

Suppose the loss function is $L(\delta, \theta)=\begin{cases}0 & if \delta=0\\1&otherwise\end{cases}$. Find the Bayes estimator of
$\theta$ under this loss function. Hint: Part (1) is related. With 0-1
loss the Bayes estimator would be the Maximum A Posteriori (MAP)
estimator...in this case is the mode of the posterior.

We have $$\begin{split}\xi(\log\theta|x)&\propto f(x|\log\theta)\xi(\log\theta)\\
&=\frac 1 {\theta^n}\frac 1 {\sigma\sqrt{2\pi}}e^{-\frac 1 2\left(\frac{\theta-\mu}{\sigma}\right)^2}\end{split}$$

Is this correct; what distribution is this?

答:
reading  you posts I think you are a clever guy thus with a sketch I think you can do the rest by yourself.... I am just an amateur, but this is what I would do in this case...

Find the posterior density of $log \theta$.

The model is uniform thus

$$p(\mathbf{x}|\theta)=\frac{1}{\theta^n}\cdot\mathbb{1}_{(0;\theta)}(x_{(n)})=\frac{1}{\theta^n}\cdot\mathbb{1}_{(x_{(n)};+\infty)}(\theta)$$

The Prior is lognormal

$$\pi(\theta)\propto \frac{1}{\theta}e^{-(\log\theta-\mu)^2/(2\sigma^2)}$$

First I find the posterior of $\theta$

$$\pi(\theta|\mathbf{x})\propto \frac{1}{\theta^{n+1}}e^{-(\log\theta-\mu)^2/(2\sigma^2)}\cdot\mathbb{1}_{(x_{(n)};+\infty)}(\theta)$$

Now I transform the posterior getting the density of its log with the standard transformation theorem

$$\lambda=\log\theta$$

$$\theta=e^{\lambda}$$

$$\theta'=e^{\lambda}$$

Thus

$$\pi(\lambda|\mathbf{x})\propto e^{-\lambda n}e^{-(\lambda-\mu)^2/(2\sigma^2)}$$

Let's focus on the exponent

$$-\frac{1}{2\sigma^2}[2\sigma^2 n \lambda+\lambda^2+\mu^2-2\lambda\mu]$$

Waste the element $\mu^2$ because not depending on $\lambda$ (it will be part of the normalizing constant) do some algebraic manipulation completing the square inside the bracket and you will find the kernel of a certain Gaussian density... truncated in the support

$$(\lambda|\mathbf{x}) >\log x_{(n)}$$


问:
An interesting stopping time problem using PGF's
This is a question I thought of the other day:

Suppose $U_1, U_2, U_3, \cdots \overset{\text{iid}}{\sim} \text{Uniform}(0, 1)$ and let $N = \min\{n: \sum_{i=1}^{n}U_i \ge 1\}$. Compute $\mathbb{P}(N \in 2\mathbb{Z})$ (i.e. probability $N$ is even). How about $\mathbb{P}(N \in 3 \mathbb{Z})$? In general, $\mathbb{P}(N \in k\mathbb{Z})$. Similarly, what is $\mathbb{E}[N|N \in k\mathbb{Z}]$?

I'll post the answer below, but challenge yourself before looking!

答:
We are going to need a few results to answer this question.

Definition The probability generating function (PGF) of a random variable $X$ with support $\mathbb{N}_0 = \{0, 1, 2, \cdots\}$ is defined as $g_X(s) = \mathbb{E}[s^X]$.

Fact 1 By definition,
\begin{align*}
g_X(s) = \sum_{k=0}^{\infty} \mathbb{P}(X = k)s^k
\end{align*}

Fact 2 We have
\begin{align*}
H_X(s) \overset{\text{def}}{=}\frac{1-g_X(s)}{1-s} = \sum_{k=0}^\infty\mathbb{P}(X > k)s^k
\end{align*}

Fact 3 We have
\begin{align*}
\mathbb{P}(X \in k\mathbb{Z}) = \frac{1}{k}\sum_{j=0}^{k-1}g_X(\omega_{k,j})
\end{align*}
where $\omega_{k,j} = \exp\left(\frac{2j\pi \mathbf{i}}{k}\right)$ are the $k$th roots of unity, with $\mathbf{i}$ being the imaginary unit.

Now we can attack this problem. First, note that
\begin{align*}
\mathbb{P}(N > n) = \mathbb{P}\left(\sum_{k=1}^{n}X_k < 1\right) = \frac{1}{n!}
\end{align*}
Since the probability corresponds to finding the volume of an $n$-dimensional right tetrahedron with leg side lengths 1. So we have
\begin{align*}
H_N(s) = \sum_{k=0}^{\infty}\mathbb{P}(N > k)s^k = e^s
\end{align*}
And therefore $g_N(s) = 1 - (1-s)H_N(s) = 1 - (1-s)e^s$. We can now see that
\begin{align*}
\mathbb{P}(N \in 2\mathbb{Z}) &= \frac{1 + g_N(-1)}{2} = 1 - e^{-1} \approx 0.6321 \\
\mathbb{P}(N \in 3\mathbb{Z}) &= \frac{1 + g_N(\exp(2\pi \mathbf{i}/3)) + g_N(\exp(4\pi \mathbf{i}/3))}{3} = 1 - \frac{\sin\left(\frac{\sqrt{3}}{2}\right)}{\sqrt{3e}} - \frac{\cos\left(\frac{\sqrt{3}}{2}\right)}{\sqrt{e}} \approx 0.3403 \\
\mathbb{P}(N \in 4\mathbb{Z}) &= \frac{1 + g_N(\mathbf{i}) + g_N(-1) + g_N(-\mathbf{i})}{4} = 1 - \frac{1}{2}e^{-1} - \frac{\sin(1)}{2} - \frac{\cos(1)}{2}\approx 0.1252 \\
&\vdots
\end{align*}
and so on. For expected values, we use the formula that
\begin{align*}
\mathbb{E}[N|N \in k\mathbb{Z}] = \frac{\sum_{n \in k\mathbb{N}_0}n \mathbb{P}(N = n)}{\mathbb{P}(N \in k\mathbb{Z})}
\end{align*}
Define
\begin{align*}
L_N(s) = \sum_{n = 0}^\infty n \mathbb{P}(N = n) s^n
\end{align*}
Note that $L_N(s) = s \frac{d}{ds} g_N(s) = s^2 e^s$ and
\begin{align*}
\sum_{n \in k\mathbb{N}_0}n \mathbb{P}(N = n) = \frac{1}{k} \sum_{j=0}^{k-1} L_N(\omega_{k,j})
\end{align*}
Plugging in these values, we have
\begin{align*}
\mathbb{E}[N|N \in 2\mathbb{Z}] &= \frac{\frac{1}{2}(e^1 + e^{-1})}{1 - e^{-1}} \approx 2.44 \\
\mathbb{E}[N|N \in 3\mathbb{Z}] &= \frac{\frac{e}{3} + \frac{2 \sin(\frac{\sqrt{3}}{2} - \frac{\pi}{6})}{3\sqrt{e}}}{1  - \frac{\sin\left(\frac{\sqrt{3}}{2}\right)}{\sqrt{3e}} - \frac{\cos\left(\frac{\sqrt{3}}{2}\right)}{\sqrt{e}}} \approx 3.06 \\
\mathbb{E}[N|N \in 4\mathbb{Z}] &= \frac{\frac{1}{4}e^{-1} + \frac{1}{4}e - \frac{\cos(1)}{2}}{1 - \frac{1}{2}e^{-1} - \frac{\sin(1)}{2} - \frac{\cos(1)}{2}} \approx 4.0055
\end{align*}
and so on.

Challenge Do the same thing to compute $\text{Var}(N|N \in k \mathbb{Z})$.


问:
help faulhaber's formula summation problem
I'm given that $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

and told to prove $\sum_{k=1}^n (2k-1)=n^2$

$\sum_{k=1}^n (2k-1)=2\sum_{k=1}^n k- \sum_{k=1}^n 1=2\frac{n(n+1)}{2}-n$

$=2\frac{n(n+1)}{2}-n$

$=\frac{n\left(n+1\right)\cdot \:2}{2}-n$

$=n\left(n+1\right)-n$

$=n^2+n-n$

$=n^2$

$\sum_{k=1}^n (2k-1)=n^2$

next I need to write the summation that gives the sum
of the odd numbers found in the first $n$ rows of this sequence.

$1 = a_1$

$3 + 5 = a_2$

$7 + 9 + 11 = a_3$

$13 + 15 + 17 + 19 = a_4$

$21 + 23 + 25 + 27 + 29 = a_5$

$\vdots \vdots \vdots$

I'm givin that $n=3$ should yield the same result as $\sum_{k=1}^6 (2k-1)=36$

if you sub in $n^2$ to the formula for $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ you get $\frac{n^{2}\left(n+1\right)^{2}}{2^{2}}$

at $n=3$; $\frac{3^{2}\left(3+1\right)^{2}}{2^{2}}=36$

I'm having trouble getting that formula into the correct form I believe this is the end result but I'm unsure how to prove the equality $\frac{n^{2}\left(n+1\right)^{2}}{2^{2}}=\sum_{k=1}^{n}k^{3}$

Then I need to replace that sum with an explicit formula of n

答:
Hint:  Write the rows of your sequence this way:
$$
\begin{aligned}
a_1 &= (2\cdot1 -1)\\
a_2 &= (2\cdot2-1) + (2\cdot3-1)\\
a_3 &= (2\cdot4-1) + (2\cdot5-1) + (2\cdot6-1)\\
a_4 &= (2\cdot7-1) + (2\cdot8-1) + (2\cdot9-1) + (2\cdot10-1)
\end{aligned}
$$
When you the sum the first $n$ rows of your sequence, you are summing $2k-1$ from $k=1$ up to a final value for $k$.

For $n=1$, the final value of $k$ is $1$.

For $n=2$, the final value of $k$ is $3$.

For $n=3$, the final value of $k$ is $6$.

For $n=4$, the final value of $k$ is $10$.

Find an expression for the final value of $k$, as a function of $n$. Then plug this final value into your formula
$$
\sum_{k=1}^{\text {final}}(2k-1) = ({\text {final}})^2.
$$


答:
Note that $a_i$ is a sum of $i$ terms. The first term is $(i^2 - i + 1)$ and the common difference of the terms is $2$.

Hence $$a_i = (i^2 - i + 1) + (i^2 - i + 3) + \ldots + (i^2 - i + 1 + 2(i-1))$$

Or, $$a_i = \sum_{j=1}^i \left(i^2 - i + (2j-1) \right)$$

Or, $$a_i = i^3 - i^2 + i^2 = i^3$$

Finally $$\sum_{i=1}^n a_i = \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$

Title: help faulhaber's formula summation problem

Body:
I'm given that $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

and told to prove $\sum_{k=1}^n (2k-1)=n^2$

$\sum_{k=1}^n (2k-1)=2\sum_{k=1}^n k- \sum_{k=1}^n 1=2\frac{n(n+1)}{2}-n$

$=2\frac{n(n+1)}{2}-n$

$=\frac{n\left(n+1\right)\cdot \:2}{2}-n$

$=n\left(n+1\right)-n$

$=n^2+n-n$

$=n^2$

$\sum_{k=1}^n (2k-1)=n^2$

next I need to write the summation that gives the sum
of the odd numbers found in the first $n$ rows of this sequence.

$1 = a_1$

$3 + 5 = a_2$

$7 + 9 + 11 = a_3$

$13 + 15 + 17 + 19 = a_4$

$21 + 23 + 25 + 27 + 29 = a_5$

$\vdots \vdots \vdots$

I'm givin that $n=3$ should yield the same result as $\sum_{k=1}^6 (2k-1)=36$

if you sub in $n^2$ to the formula for $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ you get $\frac{n^{2}\left(n+1\right)^{2}}{2^{2}}$

at $n=3$; $\frac{3^{2}\left(3+1\right)^{2}}{2^{2}}=36$

I'm having trouble getting that formula into the correct form I believe this is the end result but I'm unsure how to prove the equality $\frac{n^{2}\left(n+1\right)^{2}}{2^{2}}=\sum_{k=1}^{n}k^{3}$

Then I need to replace that sum with an explicit formula of n


答:
At the point where you say that you're a mathematician, that is then the most recent topic thrown into the "conversation pot", and the only connection most people have to that topic is that they were bad at it once upon a time.  So that is all they can say, if they say anything at all.

That person you are speaking with was bad at math at a time when they had to work with math on a daily basis, and of whatever they did learn, they probably remember a quantity that, in the vocabulary of those who were bad at math, is expressed as "jack squat".

So, that person having revealed everything they know about math, that topic is thereby exhausted.  At that point, someone's conversational skills will have to kick in and move things in a different direction. One principle that works very well in conversation is "Me too, I'm just like you". Well, not said literally, of course.

"I'm a mathematician"
"Oh, man, I was bad at math in school"
"Believe it or not, so was I! I ended up in this remedial class, even. That's where things turned around for me because there was a great teacher who really helped us to "get it". Still, I had no idea at that time I'd end up an academic, let alone in math. Life is funny that way."

Just be glad the response isn't this: "Oh, you're a mathematician? Like Euler and Gauss and those guys? What have you discovered in math that people are using to solve problems?"

Now you're screwed in the conversation.

Title: How do you respond to "I was always bad at math"?

Body:
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.


问:
Matrix, cofactor, minor, Determinant proof
I'm looking at Apostol's Calculus and I found something that might be trivial but I can't see it directly.

$det A_{12}'=\begin{vmatrix}
0 & 1 & 0 & ...& 0\\
a_{21} & 0 & a_{23} & ...& a_{2n}\\
. & . & . & ...& .\\
. & . & . & ...&.\\
.&.&.&.&.&\\
a_{n1}&0&a_{n3}&...&a_{nn}
\end{vmatrix}=detA_{12}\begin{vmatrix}
0 & 1 & 0 & ...& 0\\
1 & 0 & 0 & ...& 0\\
0 & 0 & 1 & ...& 0\\
0 & 0 & 0 & 1...& 0\\
.&.&.&.&.&\\
0&0&0&...&1
\end{vmatrix}
$

$detA_{12}'$ here is the cofactor and $A_{12}$ is the minor. It said the equation above is from uniqueness theorem. I can see this function satisfies Axiom 1-3, but I don't know why the $d(A_1,...,A_n)=detA_{12}$ and why $f(I)$ would be the right-hand side determinant of this non-identity matrix. Can anyone explain this? I'm not sure if I misunderstood some basic concepts here...

The uniqueness theorem I have is: Let $d$ and $f$ be functions satisfying AXIOMS 1-4 and 1-3 respectively. Then,
$f(A_1, . . . , A_n) = d(A_1, · · · , A_n)f(I_1, . . . , I_n).$
In particular, if $f$ satisfies, in addition, AXIOM 4, then $f = d$.

Besides, Apostol explained $J$ is the identity matrix of order $n-1$ but I don't know why $f(J)$ would look like the matrix below. I thought if it's identity matrix then $f(J)$ would be 1. Thank you for helping me clarify on this! I'm really confused.

$f(J)$ is by definition of the detriment of the matrix:
\begin{matrix}
0 & 1 & 0 & ...& 0\\
1 & 0 & 0 & ...& 0\\
0 & 0 & 1 & ...& 0\\
0 & 0 & 0 & 1...& 0\\
.&.&.&.&.&\\
0&0&0&...&1
\end{matrix}

答:
I haven't the book at hand and I don't know which pages or section you are referring to, but the idea is to view
$$
\det A_{12}'=\begin{vmatrix}
0 & 1 & 0 &\cdots& 0\\
a_{21} & 0 & a_{23} &\cdots& a_{2n}\\
\vdots&\vdots&\vdots&\cdots&\vdots\\
\vdots&\vdots&\vdots&\cdots&\vdots\\
a_{n1}&0&a_{n3}&\cdots&a_{nn}
\end{vmatrix}\tag{1}
$$
as a function $f(B_1,B_3,\ldots,B_n)$ of the rows/columns (depending on the convention adopted by Apostol) of $(n-1)\times(n-1)$ submatrix
$$
B=\,\begin{bmatrix}
a_{21} & a_{23} &\cdots& a_{2n}\\
\vdots&\vdots&\cdots&\vdots\\
\vdots&\vdots&\cdots&\vdots\\
a_{n1}&a_{n3}&\cdots&a_{nn}
\end{bmatrix}.
$$
It should be straightforward to verify that $f$ is an alternating multilinear function (i.e. a function satisfying the axioms you mentioned) of $B_1,B_3,\ldots,B_n$. It follows from the uniqueness theorem that
$$
f(B_1,B_3,\ldots,B_n)=d(B_1,B_3,\ldots,B_n)f(J_1,J_2,\ldots,J_{n-1}),
$$
where $J_1,J_2,\ldots,J_{n-1}$ are the rows/columns of the $(n-1)\times(n-1)$ identity matrix $J$. However, $d(B_1,B_3,\ldots,B_n)$ is precisely the minor $\det A_{12}$ and
$f(J_1,J_2,\ldots,J_{n-1})$ is precisely the RHS of $(1)$ when the $a_{ij}$s are replaced by the entries of $J$. Hence
$$
f(J_1,J_2,\ldots,J_{n-1})=
\begin{vmatrix}
0&1&0&0&\cdots&0\\
1&0&0&\ddots&\ddots&\vdots\\
0&0&1&0&\ddots&\vdots\\
0&\ddots&0&1&\ddots&0\\
\ddots&\ddots&\ddots&\ddots&\ddots&0\\
0&\cdots&0&0&0&1
\end{vmatrix}
$$
and the result follows.


问:
$f(x+n)$ converges for any $x$ when $n\to +\infty$, then $f(x)$ converges when $x\to +\infty$?
Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function. I have two questions as follows.

Suppose that $f(x+n)\to L$ for all $x\in \mathbb{R}$ when $n\to \infty$. Is it guaranteed that $f(x) \to L$ when $x\to +\infty$?

<s>2. Suppose that $f(x+n)$ converges for all $x\in \mathbb{R}$ when $n \to \infty$. Is it guaranteed that $f(x)$ converges when $x \to +\infty$?</s> (As pointed out by @Ninad Munshi, the answer is obviously negative by considering a simple example with $f(x) = \sin(2\pi x)$. Thank @Ninad Munshi.)

Remarks:

Here we may require that the limit of any convergence must be a finite number. However, what if we consider a generalized situation that the limit is allowed to be infinity?

The question is an additive analogs of $f(nx)\to 0$ as $n\to+\infty$ , If $\,\lim_{n\to\infty}f(nx)\,$ exists, for all $x\in\mathbb R$, then so does $\,\lim_{x\to\infty}f(x)\,$ , and If $\lim_{n \to +\infty} \ f(\alpha n)$ exists , does this imply $\lim_{x \to +\infty} f(x)$ exists? , which can be solved using Baire's category theorem.

Any comments or criticism will be appreciated. Thank you.

答:
Let $f_n(x)$ denote the function whose graph is an isosceles triangle with base $[0,1/n]$ and height $1$. Specifically:
$$f_n(x) = \begin{cases}
2nx & \text{if }0 \leq x \leq \frac{1}{2n} \\
2 - 2nx & \text{if }\frac{1}{2n} \leq x \leq \frac{1}{n} \\
0 & \text{otherwise}
\end{cases}$$
and let
$$f(x) = \sum_{n=1}^{\infty}f_n(x-n)$$
So, the graph of $f$ consists of a series of increasingly narrow triangles, where the $n$'th triangle has base $[n, n+1/n]$ and height $1$. It's straightforward to verify that
$$\lim_{n \to \infty}f(x+n) = 0$$
for all $x$, but $\lim_{x \to \infty}f(x)$ does not exist.


问:
If $M$ is a monomial matrix with entries only from ${0,-1,1}$, then $C$ is self dual if and only if $CM$ is self dual.
I am interested to prove the theorem here but I am stuck. Can you please help me?  Here it is

Let $\mathcal{C}$ be a code over $\mathbb{F}_q$.  Then

If $M$ is a monomial matrix with entries only from $\{0,-1,1\}$, then $\mathcal{C}$ is self-dual if and only if $\mathcal{C}M$ is self-dual.

If $q=3$ and $\mathcal{C}$ is equivalent to $\mathcal{C}_1$, then $\mathcal{C}$ is self-dual if and only if $\mathcal{C}_1$ is self-dual.

If $q=4$ and $\mathcal{C}$ is equivalent to $\mathcal{C}_1$, then $\mathcal{C}$ is Hermitian self-dual if and only if $\mathcal{C}_1$ is Hermitian self-dual.

Now, when a code $\mathcal{C}$ is self-dual if and only if $\mathcal{C}$ is equal to its dual $\mathcal{C}^\perp$. A monomial matrix is a square matrix with exactly one nonzero entry each row and column.

So for number 1, what I did is by assuming the $\mathcal{C}$ is self-dual, then I would $\mathcal{C} = \mathcal{C}^\perp$.  Then $\mathcal{C}M = \mathcal{C}^\perp M$.  How can I be able to attain the equality that $\mathcal{C}M = (\mathcal{C}M)^{\perp}$ so that I can say that $\mathcal{C}M$ is self-dual?

For number 2, I am also confused that what is it something to do with the dimension of the field $\mathbb{F}$?  If $\mathcal{C}$ is equivalent to $\mathcal{C}_1$, then there exists  a monomial matrix $M$ and an automorphism $\gamma$ of the field such that $\displaystyle{\mathcal{C}_1 = CM\gamma}$.  Then if $\mathcal{C}$ is self-dual, then I would $\mathcal{C} = \mathcal{C}^\perp$, but what it is something to do with arriving $\mathcal{C}_1 = \mathcal{C}_1^\perp$.  What is in the dimension of the field?

For number 3, same case and confusion of number 2.

Can you please help me with that? I would surely appreciate the help.

答:
Hints:

The condition implies that $MM^T=I_n$. If $\mathcal{C}$ is self-dual then its generator matrix $G$ satisfies $GG^T=0$. What can you say about $GM$?

$\mathcal{C}$ and $\mathcal{C}'$ are equivalent if and only if there exists a monomial matrix $M$ such that $\mathcal{C}'=\mathcal{C} M$. When $q=3$, $\Bbb{F}_q=\{0,1,-1\}$. Therefore the claim follows from $1$.

With $q=4$ we have $\Bbb{F}_4=\{0,1,\omega,\overline{\omega}\}$ with $\omega^2=\overline{\omega}=\omega+1=\omega^{-1}$. This time you need to show that $M\overline{M}^T=I_n$.


问:
Trouble solving $\lim _{x\to -\infty }\frac{3x^7-25x^5+e^{-x^3/(1-x)^3}}{-5x^7+\left(111!\right)\cdot \:x}$
$$\lim _{x\to -\infty }\frac{3x^7-25x^5+e^{-\frac{x^3}{\left(1-x\right)^3}}}{-5x^7+\left(111!\right)\cdot \:x}$$

not sure how to approach this limit. I have tried simplifying the function but I can't get it into a usable form. am I missing something when it comes to limit rules on how to approach this?

答:
First deal with the one complicated part, the term

$$\lim_{x\to-\infty}e^{-\frac{x^3}{(1-x)^3}}\tag{1}$$

in the numerator. Suppose that $x=-u$, where $u$ is positive. Then $1-x=u+1$, and $-x^3=-(-u)^3=u^3$, so

$$e^{-\frac{x^3}{(1-x)^3}}=e^{\frac{u^3}{(u+1)^3}}=e^{\left(\frac{u}{u+1}\right)^3}\,.$$

And $u\to\infty$ as $x\to-\infty$, so you should be able to calculate the limit $(1)$ now.

Once you’ve done that, divide numerator and denominator by $x^7$, and then take the limit as $x\to-\infty$; you should find that many of the terms approach $0$.


问:
how would you find the solutions for $x$ where the quadratic $-27x^2+234x+169$ is a perfect square
how would you find the solutions for $x$ where the quadratic $-27x^2+234x+169$ is a perfect square of an integer and $x$ is positive integer

I know that $x$ is from $1$ to $11$ because anything greater than $11$ makes the polynomial negative ,but how would compute the correct $x$ values without testing each of the 11 values?

Body:
$$ y = -27x^2 + 234x + 169 $$

This is a parabola with a vertex (in this case, the highest point)  at $x = -\frac{b}{2a}=-\frac{234}{2(-27)}=4\frac{1}{3}$ and $y=676$.

The roots are at $\dfrac{13}{9}(3 \pm 2\sqrt 3) \approx \{-0.67, 9.34\}$

So we only need to test $x = 0,1,2,\dots,9$

\begin{array}{r,|,c}
   x & y=-27 x^2 + 234 x + 169 \\
\hline
   0 & \color{red}{169=13^2} \\
   1 & 376 \\
   2 & \color{red}{529 = 23^2} \\
   3 & 628 \\
   4 & 673 \\
   5 & 664 \\
   6 & 601 \\
   7 & \color{red}{484 = 22^2}\\
   8 & 313 \\
   9 & 88 \\
\hline
\end{array}