问:
Rudin's Definition of the limit of a function confusion
I was reading Baby Rudin, when I came across the definition he provides for the limit of a function, in terms of sequences. Namely, he says that
for function $f : E \rightarrow \mathbb{R}$,

$$\lim_{x \rightarrow q} f(x) = p$$

if and only if for every sequence $x_n$ in $E$, where $x_n \rightarrow q$, $x_n \neq q$, we have $f(x_n) \rightarrow p$.

I don't understand what would motivate the condition that $x_n \neq q$. Any insight as to why we need this condition would be appreciated.

答: For the same reason that $\lim_{x\to q}f(x)=p$ means$$(\forall\varepsilon>0)(\exists\delta>0):\color{red}{0<}|x-q|<\delta\implies\bigl|f(x)-p\bigr|<\varepsilon.$$Note that what happens when $x=q$ is irrelevant (since then $|x-q|\color{red}=0$). So, when we translate this definition into something equivalent using sequences, no term of those sequences can be equal to $q$.


答:
I don't understand what would motivate the condition that $x_n \neq q$. Any insight as to why we need this condition would be appreciated.

Say $f$ has a removable discontinuity at $q$, with $\lim\limits_{x\to q}f(x)=p,$ but $f(q)=r\ne p$.

Then if we allowed sequences $x_n$ with $x_n=q$, we could have $f(x_n)\to r$, not $p$.

ContentLicense: CC BY-SA 4.0

Title: Rudin's Definition of the limit of a function confusion

Body:
I was reading Baby Rudin, when I came across the definition he provides for the limit of a function, in terms of sequences. Namely, he says that
for function $f : E \rightarrow \mathbb{R}$,

$$\lim_{x \rightarrow q} f(x) = p$$

if and only if for every sequence $x_n$ in $E$, where $x_n \rightarrow q$, $x_n \neq q$, we have $f(x_n) \rightarrow p$.

I don't understand what would motivate the condition that $x_n \neq q$. Any insight as to why we need this condition would be appreciated.


问:
A $2021$ problem: $20\sim 21$ and $43\times 47$
Notice that $2021$ is a concatenation of consecutive integers:  $20\sim 21$

Also $2021$ is a product of consecutive primes: $43\times 47$.

What is the next number with both of these properties?

$24502451$ is close, $4943\times 4957$  but $4951$ is in between.

$2484224843$ is close, $49831\times 49853$ but $49843$ is in between.

$715353612\sim 715353607$ isn't quite there but is $845785793\times 845785799$

Are there any other numbers with the $2021$ property?

答:
If $p$ is prime and $q$ is the next prime and $pq$ has $2n$ digits, the probability that $pq$ is the concatenation of consecutive integers is heuristically about $10^{-n}$.  For $pq$ to have $2n$ digits (i.e. $10^{2n-1} \le pq < 10^{2n}$), $p$ must be between about $10^{n-1/2}$ and $10^n$, and there are approximately $$\frac{10^{n}}{n \ln 10} - \frac{10^{n-1/2}}{(n-1/2) \ln 10} \sim \frac{10 - \sqrt{10}}{10 \ln(10)} \frac{10^n}{n}$$
such primes.  Thus we expect about $0.297/n$ examples with $2n$ digits.  Since the harmonic series diverges, there should be infinitely many, but the next one could be quite large and hard to find.  I've checked by brute force that there are no further examples with up to $16$ digits.

EDIT: To expand on my comment above:

The concatenation of $y$ and $y+1$, where $y+1$ has $n$ digits,
is $(10^n+1) y + 1$, and thus $\equiv 1 \mod (10^n+1)$.

Given an even positive integer $d$ and positive integer $n$, you can solve the equation $x(x+d) \equiv 1 \mod (10^n+1)$, and then check in each solution
that $10^{2n} > x(x+d) \ge 10^{2n-1}$ and $x$ is prime and $x+d$ is the next prime.  For $n \le 36$ and $d \le 10000$ the only solutions found are
$n=2$, $d=4$, $x = 43$, $x(x+d) = 2021$ and that amazing
$n=36$, $d=2$, $x=891077215721081784886888257701070827$, $x(x+d)=794018604377235322848433897872605582794018604377235322848433897872605583$.


问:
Why is $\frac{987654321}{123456789} = 8.0000000729?!$
Many years ago,
I noticed that $987654321/123456789 = 8.0000000729\ldots$.

I sent it in to Martin Gardner at Scientific American
and he published it in his column!!!

My life has gone downhill since then:)

My questions are:

Why is this so?

What happens beyond the "$729$"?

What happens in bases other than $10$?

ContentLicense: CC BY-SA 3.0

Body:
Let
$$S_n(a)=1 +2a+\ldots +na^{n-1}=\frac{na^{n+1}-(n+1)a^n+1}{(a-1)^2},$$
$$T_n(a)=a^{n-1}+2a^{n-2}+\ldots +n=a^{n-1}S_n(a^{-1}).$$

Then
$$
\frac{S_n(a)}{T_n(a)}=\frac{na^{n+1}-(n+1)a^n+1}{a^{n+1}-(n+1)a+n}.$$
For $a=10,n=9$ we have
$$
\frac{S_n(a)}{T_n(a)}\approx\frac{8\cdot 10^{10}+1}{10^{10}}.
$$


问:
property about transient and recurrent states of a Markov chain
From here, we know that given a Markoc chain ans a state $i$, $i$ is transient if
$$ P(X_n = i, \text{ for some } n \geq 1 |X_0 =i) <1$$

Consider the following statements:

States in a class of a finite Markov chain is either all transient or all recurrent.

There is no finite Markov chain in which all states are transients.

All states of a finite irreducible Markov chain are recurrent.

So here is my attempt to prove the claims:

Suppose that state $i$ is transient and state $i$ and state $j$ belong to the same class in a Markov chain, then there exist $n \in \mathbb{N}$ such that $P^n(j,i)>0$ and there is a state $m$ where $m$ is accessible from $i$ but $i$ is not accessible from $m$. But then $m$ is accessible from $j$ via $i$ but $j$ is not accessible via $m$ or else $i$ would have been accessible from $m$.

Suppose there is a Markov chain where all of its states are transient, then given a initial distribution of being at any given states, the probability that we are at any given states after several time step tends to $0$. But this cannot be possible as the probability at any given state must sum up to one.(So here I think the claim is not rigorous, so any comment would be appreciated.)

A Markov chain is irreducible if it has only one class. Hence all states are reachable from all states. Therefore, the probability that a state comes back to itself is not null.

答:
I left some comments regarding my impressions of your proofs, so here's how I would do the proofs myself.

States in a class of a Markov Chain are either all recurrent or transient.

Suppose state $i$ and $j$ communicate, and state $i$ is recurrent. Let $p=P_n(i,j)>0$. Thus, the expected number of visits to $j$ is at least $p$ times the expected number of visits to $i$, which is infinite. Hence, the expected number of visits to $j$ is infinite, hence it is recurrent. If $i$ is transient, $j$ can't be recurrent, since by the previous argument, that would imply $i$ is recurrent.

There is no finite Markov chain in which all states are transient.

By the pigeonhole principle, there is at least one state that will be reached infinitely many times. In other words, if $M$ is a Markov chain with states $[k]=\{1,\ldots, k\}$, if $A_i$ for $i\in[k]$ represents the event that state $i$ is reached infinitely many times, then by inclusion exclusion $$1=P(A_1\cup\ldots\cup A_k)=P(A_1)+P(A_2)+\ldots+P(A_k)-P(A_1\cap A_2)-\ldots+P(A_1\cap A_2\cap A_3)+\ldots\pm\ldots\pm P(A_1\cap A_2\cap\ldots\cap A_k)$$But all of the terms on the right side of the equation are $0$ if each state is transient. This is a contradiction.

All states of a finite irreducible Markov chain are recurrent.

As irreducible Markov chains have one class, statement $1$ implies all states are either transient or recurrent. But statement $2$ implies that all states are not transient. Hence, they are all recurrent.


问:
Why the book used $c_x$ notation to denote a number in Taylor's remainder theorem?

(Taylor's remainder) Assume that $f(x)$ has $n+1$ continuous derivatives on an interval
$\alpha\le x \le\beta$, and let the point $a$ belong to the interval.
For nth degree Taylor polynomial $P_n(x)$ let $R_n(x)\equiv
 f(x)-P_n(x)$ denote the remainder in approximating $f(x)$ by $P_n(x)$.
Then

$$R_n(x)=\frac{(x-a)^{n+1}}{(n+1)!}f^{(n+1)}(c_x), \quad  \alpha\le
 x\le \beta$$with $c_x$ an unknown point between $a$ and $x$.

They denoted a number between $a$ and $x$ by $c_x$. so why not just write $c$ instead? and what does subscript $x$ exactly mean here and in general in mathematics?

答: It's just to remind you that the number depends on $x$.


问:
Give Lambda Calculus Term for Haskell Function
I am working on a larger project translating Haskell to Lambda calculus. I would like to give a lambda term to specific Haskell functions. I am not quite sure how to approach two of them. I went ahead and created the property each term should have, just to give you an idea of what I am trying to do.

What would the $\lambda$-terms for these Haskell functions be? If you could give a brief explanation that would be super helpful.

Function 1: $\mathsf{foldr}$: It should have the property

$$\mathsf{foldr} \, f \, u \, [N_1, \dots ,N_k] \to_\beta^* f N_1 (f N_2 (\dots (f N_k u)))$$

Function 2: $\mathsf{map}$: It should have the property

$$\mathsf{map} \, f \, [N_1, \dots,N_k] \to_\beta^* [f N_1,f N_2, \dots,f N_k]$$

答:
Given that a list $[N_1, \dots, N_k]$ of $\lambda$-terms is represented by the $\lambda$-term
$$\lambda c. \lambda n. c N_1 (c N_2( \dots(c N_k n) \dots ))$$
it is easy to define the $\lambda$-terms $\mathsf{foldr}$ and $\mathsf{map}$.

$\mathsf{foldr} \, f \, u \, [N_1, \dots, N_k] \to_\beta^* f N_1 (f N_2( \dots (f N_k u) \dots ))$.
Here, the intuition is to bring $\lambda c. \lambda n. c N_1 (c N_2( \dots(c N_k n) \dots ))$ (i.e. the list $[N_1, \dots, N_k]$) in "head position" and apply it to $f$ and then to $u$, so that, when the $\beta$-steps destroy $\lambda c$ and $\lambda n$, $f$ substitutes for $c$, and $u$ substitutes for $n$. Which $\lambda$-term $\mathsf{foldr}$ allows us to do that?

! Let $\mathsf{foldr} = \lambda f'. \lambda u'. \lambda l. lf'u'$. Then,
\begin{align}
\mathsf{foldr} \, f \, u \, [N_1, \dots, N_k] &= (\lambda f'. \lambda u'. \lambda l. lf'u')  f \, u \, [N_1, \dots, N_k]
\\
&\to_\beta  (\lambda u'. \lambda l. l f u') \, u \, [N_1, \dots, N_k]
\\
&\to_\beta (\lambda l. l f u) [N_1, \dots, N_k]
\\
&\to_\beta [N_1, \dots, N_k] \, f \, u
\\
&= \big( \lambda c. \lambda n. c N_1 (c N_2( \dots(c N_k n) \dots )) \big) f u
\\
&\to_\beta \big( \lambda n. f N_1 (f N_2( \dots(f N_k n) \dots )) \big) u
\\
&\to_\beta f N_1 (f N_2( \dots(f N_k u) \dots ))
 \end{align}

$\mathsf{map} \, f \, [N_1, \dots, N_k] \to_\beta^* [f N_1, \dots, f N_k]$.
Here the idea is a variant of the previous point: bring $\lambda c. \lambda n. c N_1 (c N_2( \dots(c N_k n) \dots ))$ (i.e. the list $[N_1, \dots, N_k]$) in "head position" and then replace each $N_i$ with $f N_i$, so as to obtain $\lambda c. \lambda n. c (fN_1) (c (fN_2) ( \dots(c (f N_k) n) \dots ))$, i.e. the list $[f N_1, \dots, f N_k]$.
How can each $N_i$ be replaced by $f N_i$? The $\lambda$-term $\lambda c. \lambda n. c N_1 (c N_2( \dots(c N_k n) \dots ))$ (i.e. the list $[N_1, \dots, N_k]$) should be applied to the $\lambda$-term $\lambda x . c (fx)$, so that, when the $\beta$-step destroys $\lambda c$ in the list, $\lambda x. c (f x)$ substitutes for $c$ and then $(\lambda x. c (f x))N_i \to_\beta c (f N_i)$.
Summing up, which $\lambda$-term $\mathsf{map}$ allows us to do all that?

! Let $\mathsf{map} = \lambda f'. \lambda l. \lambda c. l \, \lambda x. c (f'x)$. Then,
\begin{align}
 \mathsf{map} \, f \, [N_1, \dots, N_k] &= \big( \lambda f'. \lambda l. \lambda c. l \, \lambda x. c (f'x) \big) f \, [N_1, \dots, N_k]
 \\
 &\to_\beta \big( \lambda l. \lambda c. l \, \lambda x. c (f x) \big) [N_1, \dots, N_k]
\\
 &\to_\beta \lambda c. [N_1, \dots, N_k] \lambda x. c (f x)
 \\
 &=
 \lambda c. \big( \lambda c'. \lambda n. c' N_1 (c' N_2( \dots(c' N_k n) \dots )) \big) \lambda x.c (f x)
 \\
 &\to_\beta \lambda c.  \lambda n. (\lambda x.c (f x)) N_1 ((\lambda x.c (f x)) N_2( \dots( (\lambda x.c (f x)) N_k n) \dots ))
 \\
 &\to_\beta \lambda c.  \lambda n. c (f N_1) ((\lambda x.c (f x)) N_2( \dots( (\lambda x.c (f x)) N_k n) \dots ))
 \\
 &\to_\beta \lambda c.  \lambda n. c (f N_1) (c (f N_2) ( \dots( (\lambda x.c (f x)) N_k n) \dots ))
 \\
 &\to_\beta^* \lambda c.  \lambda n. c (f N_1) (c (f N_2) ( \dots( c (f N_k) n) \dots ))
 \\
 &= [f N_1, \dots, f N_k]
\end{align}


问:
Calculating the number of integers divisible by 8
Given the set [10000,70000], in order to calculate the number of terms divisible by 8 I know of two ways to calculate the answer

1st is
$$\frac{70000-10000}{8}+1=7501$$

2nd is
$$⌊\frac{60001}{8}⌋=7500$$

the fact that the results differ is very peculiar to me, can anyone explain what is going on, and which one is more accurate?

答:
Hint: This is where those "flooring methods" came from:
$$10,000 \le 8k \le 70,000 \\[1mm] \frac{10,000}{8} \le k \le \frac{70,000}{8} \\[1mm] \\ 1250 \le k \le 8750 $$
Now, just count how many natural numbers there are between $1250$ and $8750$; that is, $8750 - 1249 = 7501$.

Since, the ratios above might not be integer in general, we can sharpen the bounds
$$\left\lceil\frac{10,000}{8}\right\rceil \le k \le \left\lfloor\frac{70,000}{8}\right\rfloor $$ which shows that your first approach is right.


答:
The problem is that both endpoints of the interval $[10000, 70000]$ are divisible by $8$, while neither endpoint of the interval $[1, 60001]$ is divisible by $8$.

Your first method is correct.  It works since both endpoints are divisible by $8$.

In your second method, your interval should start at $0$ and end at $60,000$ so that both endpoints are again divisible by $8$.  You then have to add $1$ to account for the fact that $0$ is divisible by $8$.  Observe that
$$\left\lfloor \frac{60000 - 0}{8} \right\rfloor + 1 = 7501$$

Title: Calculating the number of integers divisible by 8

Body:
Given the set [10000,70000], in order to calculate the number of terms divisible by 8 I know of two ways to calculate the answer

1st is
$$\frac{70000-10000}{8}+1=7501$$

2nd is
$$⌊\frac{60001}{8}⌋=7500$$

the fact that the results differ is very peculiar to me, can anyone explain what is going on, and which one is more accurate?


问:
Prove that raising a squared matrix to the power, keep its non linear rows
I need to prove  that raising a square matrix to any power, e.g. 3x3, will not change its non linear rows, meaning that in the end of gaussian
elimination, I have the same matrix. Condition: at least one of the diagonal elements is not zero

e.g.

A=
\begin{pmatrix}
1 & 2 &3 \\
 2& 4 & 6\\
0 & 1 & 2
\end{pmatrix}

after gaussian elimination:
\begin{pmatrix}
1 & 2 &3 \\
 0& 0 & 0\\
0 & 1 & 2
\end{pmatrix}
will still have same gaussian elimination result, even if I solve A^33 and then conduct gaussian elimination

答:
If I understand the statement, it isn't true.  Consider
$$A=\begin{pmatrix}0&0&0\\0&0&0\\0&1&0\end{pmatrix}\\
A^2=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}$$


问:
Statistics, Hypothesis and P-value - Check my answers
I am trying to solve a question but stuck with the steps. I can not find any similar questions. With help of some online resources to calculate some parts of the question but I can see that is not enough. I know my approach has lack of information but, this is the only thing I have reached, I was covid ill at the class hours and can not follow the class examples, I thought someone can help me to solve and learn the subject.

With help of the answers from here I try to give an answer. Still need some improvements but tried to do my best. I still do not have answer for question D and confused about CL(the part C) and Significance level(part B)

My answers:

$N\ =\ 9\ \ \ \ \ \ \ \ Sum\ of\ x\ =\ 3970\ \ \ \ \ \ \ \ Mean,\ µ = 441.1111  \      Variance,σ^2  = 161.1111$
$ \sigma\ =\ \sqrt{161.1111}  = 12.6929$

$t\ =\ \frac{m\ -\ \mu}{s\ /\ \sqrt n}$

$t\ =\ \frac{500\ -\ 441.1111}{12.6929\ /\ \sqrt9} = 13.918545$

We subtract 1 to get degrees free 9 - 1 = 8

Degrees of freedom = n – 1 = 8

$Probability: P( T ≤ 13.918545) = 0.00000069 $ So, this is the p-Value

$We\ will\ reject\ H_0\ at \ \alpha = 1% $ and also any > 1%

$$ (i) 0.10\
The\ information\ from\ the\ first\ question\, the\ critical\ t-value\ for\ α = 0.10\ and\ df = 8,\ t_c=\ 1.86
 \\ CI\ =\ (\bar{X}\ -\ \frac{t_c\ \times\ s}{\sqrt n},\ \bar{\ X}\ +\ \frac{t_c\ \times\ s}{\sqrt n})
 \\CI\ =\ (441.1111\ -\ \frac{1.86\ \times\ 12.6929}{\sqrt9},\ \ 441.1111+\frac{1.86\ \times\ 12.6929}{\sqrt9})
= (441.1111 – 7.868, 441.1111 + 7.868)
= (433.243, 448.979)
 $$

$For\ the\ other\ t_c\ values:$

(ii) 0.05  $t_c=\ 2.306$ CL = (431.354, 450.868) (iii) 0.01 $t_c=\
> 3.355$
= (426.915, 455.308)

Based on the answers in part 2 for (i) = 0.90, (ii) = 0.95, (iii) = 0.99, none of the confidence intervals contain 500.

The Question:

The worker says that the mean purchasing cost is 500 USD. We decide to test this.

For a random sample of 9 purchases drawn from a normally distributed population with unknown variance, the costs are:

```<code>430, 450, 450, 440, 460, 420, 430, 450, 440.
</code>```

A) Conduct a hypothesis test of whether the population mean purchasing equals 500 USD. Include all assumptions, the hypotheses, test statistic, and P-value and interpret the result in context.

B) For which significance levels can you reject $H_0?$ (i) 0.10, (ii) 0.05, or (iii)
0.01.

C) Based on the answers in part B), for which confidence levels would the confidence interval contain 500? (i) 0.90, (ii) 0.95, or (iii) 0.99.

D) Use part B) and part C) to illustrate the correspondence between results
of significance tests and results of confidence intervals.

答:
I will give some formulas, which may be proved by standard methods.

We have $H_0: X_i \sim N(a, \sigma^2)$,

$a= 500$, $n=9$, $\overline{X} = \frac{\sum_{i=1}^n X_i}{n} = 441.1111...$, $s^2 = \frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n} = \hat{\sigma^2}\frac{n-1}{n}$, $ \hat{\sigma^2} =\frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n-1} = 161.1111...$, $\hat{\sigma} = 12.69296...$.

We know that $\xi_n = \frac{\sqrt{n}(\overline{X} - a)}{\sigma} \sim N(0,1)$, $\eta_n = \frac{ns^2}{\sigma^2} \sim \chi_{n-1}^2$, $\xi_n$ and $\eta_n$ are independent,
$$t = \frac{\xi_n}{\sqrt{\frac{\eta_n}{n-1} }} = \sqrt{n-1}\frac{ \overline{X} - a}{s} =  \sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} \sim T_{n-1}.$$
and also $-t \sim T_{n-1}$, because $T_{n-1}$ is symmetric. Here $T_{n-1}$ is Student's distribution.

We have $t =\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} = -13.9185...$.

If $H_0$ is true then $|t| < u_{\frac{1+\gamma}2}$ with probability $\gamma$, where $u_c$ is a quantile function at $c$.

Put $\gamma = 0.99$. Hence $\frac{1+\gamma}2 = 0.995$. We know that $u_{0.995} = 5.04305...$ and hence $|t| < 5.04305...$ with probability $0.99$, if $H_0$ is true.

Hence we reject $H_0$ for all significance levels in $B$.

The confidence interval for $a$ we get form a condition $$|\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}.$$

It has the form:
$$ a \in (\overline{X}  - \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma} , \overline{X}  + \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma}).$$

Even in case $\gamma = 0.99$, when the  confidence interval is wider, than in cases $\gamma = 0.95$ and $\gamma = 0.90$, we saw that the condition $$|\sqrt{n}\frac{ \overline{X} - 500}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}$$ doesn't hold and hence the confinedce interval doesn't contain $a=500$.

So in $C)$ we get that $500$ is not contained is confinedce interval in any case.

D) We see that $H_0$ is accepted (significance level is fixed) if and only if $a$ is contained in a confidence interval of correspoding confidence level. The correspondence is illustrated.

Addition about D.

$H_0: X_i \sim N(a, \sigma^2)$.

Test has significanse level $1-\gamma$ (in other words, test has level of confidence $\gamma$). Test says that we shoud accept $H_0$ if and only if $$|\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}.$$

The confidence interval has the form:
$$ a \in (\overline{X}  - \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma} , \overline{X}  + \frac{ u_{\frac{1+\gamma}2}}{\sqrt{n}} \cdot \hat{\sigma}).$$

So the condition "$a$ belongs to the $\gamma$-confidence interval" is equivalent to condition $$|\sqrt{n}\frac{ \overline{X} - a}{\hat{\sigma}} | < u_{\frac{1+\gamma}2}$$

which is a necessary and sufficient condition to accept $H_0: X_i \sim N(a, \sigma^2)$ with  significanse level $1-\gamma$ .

Conclusion. "$a$ belongs to the confidence interval (with level of condidence $\gamma$)" if and only if we accept $H_0: X_i \sim N(a, \sigma^2)$, using the test with  significanse level $1-\gamma$ .


问:
The consistency of PA is falsifiable. Can the same be said of its soundness?
The statement that 'PA is consistent' is absolutely falsifiable (a term I just made up), in the sense that if it is false, then we can demonstrate its falseness independently of any metatheory (just deduce a contradiction within PA).

Can something similar be said of the statement that 'PA is sound'? I think not, because to even make sense of the claim that PA is sound, we need a metatheory (although see here).

Clarification. Let me attempt to clarify what I mean by "absolutely falsifiable." If we're given a formal system with a recursively ennumerable set of axioms, then if someone claims that "the string $\phi$ is not deducible in the system," well this is a falsifiable claim. In the sense that, if the claim is false, in other words if $\phi$ is deducible in the system, then we can verify this within the rules of the system, simply by deducing $\phi$.

For instance, if I claim that "ZFC cannot prove that the Ackermann function is defined for all inputs," well this is falsifiable: because, if it is a falsehood, then you can confirm that its a falsehood by using ZFC to prove that the Ackermann function is, in fact, defined for all inputs. In fact, since its already known that ZFC proves that the Ackermann function is total, you already know the claim is a falsehood.

On the other hand, if I simply claim that "the Ackermann function is not defined for all inputs" without reference to a foundational system, well this is a more problematic claim. If its true, we can never confirm this, because we have to check infinitely many inputs to make sure the program halts. If its false, then for some input, we'll have to wait an infinite amount of time before we can verify that, indeed, its not going to halt. So the claim that "the Ackermann function is not defined for all inputs" is not falsifiable (and neither is its negation).

答:
Suppose I say: "Here's my attempt to axiomatize the truths of arithmetic of successor, addition and multiplication over the natural numbers", and I start writing down some axioms. And for the sixth axiom I write down

(A6)$\quad\forall x\forall y(x \times Sy = x \times y + y)$

Then you could say "Hold on! That implies $2 \times 2 = 2 \times S1 = 2 \times 1 + 1 = 3$ so something has gone horribly wrong, for it's plain false that $2 \times 2 = 3$! Surely what you meant to write was $\forall x\forall y(x \times Sy = x \times y + x)$."

Now note, would you in this little exchange, be deploying a fancy metatheory?  Well, you would just be using some plain old arithmetic (hey, $2 \times 2 \neq 3$) in criticizing my attempt to axiomatize plain old arithmetic! (I suppose, if you really it is must, you can call at least some plain old arithmetic part of our metatheory, i.e. part of the background we have in play as we are trying to build a formalized arithmetic: but then don't build too much into the idea of "metatheory". Nothing fancy is going on.)

Well back to the drawing board. This time, I come up with a better axiomatization, our old friend PA.

Now suppose, just suppose, you could again deduce a sentence $\varphi$ in this new theory which is clearly false as a sentence of plain old arithmetic. Then you would have shown that PA is as bad as my first failed attempt (it isn't a sound axiomatization of arithmetic). And again, you wouldn't be appealing to a fancy metatheory in saying this -- just to arithmetic again. Sure in criticising PA you are appealing to something -- but to nothing other than what PA is trying to axiomatize in the first place.

Of course, few of us think that the supposition is going to get realized, and you get to find a falsehood in PA -- but that's another story!


答:
If you want a catch-all formula for the number of multiples of $n>0$ in $[a,b]$, you may use $$\left\lfloor \frac bn\right\rfloor-\left\lceil\frac an\right\rceil+1,$$

which essentially comes from the consideration that the map $x\mapsto nx$ defined on the integers in $\left[\left\lceil\frac an\right\rceil,\,\left\lfloor \frac bn\right\rfloor\right]$ is bijective onto the set of multiples of $n$ in $[a,b]$. In a number of instances the nuance of where to divide and where to take floor or ceiling is obscured by other facts.


问:
formula for relating number of successes to number of tries
Imagine we have a jar of marbles, where there are 10 different colors of marbles in the jar.
Let N be the number of marbles drawn in a sample and
K be the number of distinct colors in the sample.
Let C be the total number of colors in the jar (which is 10 in this example).

We know that:
K = F(C,N) + chance.

Intuitively, the more marbles we draw at a time, the more colors we are likely to obtain, up to the maximum.

This strikes me as a binomial. I am looking for a formula to express in more detail how
K = F(C,N).
An obvious feature is that C is the maximum value. So I figure this is a model where K is an increasing proportion of C as N increases.

I already know how to simulate the data, but wish to provide an expression of K depending on C and N.

Any suggestions will be appreciated.

Simulated data for the situation I intend will be as follows. I assume sampling with replacement. For this example, we have 10 colors in the jar (C=10). Note that we start by drawing one marble from the jar, which gives us a sample where N=1 and K=1. The number of colors found in a sample depends on the size of the sample. In this case it is seq(1:20). In this particular simulation, we have an equal representation for each color. The snip of R code I use to simulate the data is:

###snip of modified code for stack exchange exchange

T <- c(1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10)

for (i in 1:N){

for (s in 1:n.sims) {

plot.sample <- sample(T, size=N[i])

K[s,i] <- length(unique(plot.sample))

}

}

Note that the line is just a loess plot at this point and the proper function would level off at C, the maximum number of colors.

If I use the suggested formula
K=N*C/(N+C-1)
I get the figure on the right, which has the right shape, but finds colors very slowly.

If I estimate the function with a scaling coefficient, I get the following graph. Excellent fit: explained variance = 97%, though the visual suggests there could be a better function?

答:
Let $m_1,m_2,\dots,m_C$ be the numbers of murbles of the corresponding color, so that
$$\sum_{i=1}^C m_i=M,
$$
where $M $ is the total number of marbles in the jar (not given in the problem).

Then the probability that a sample of $N $ marbles will contain marbles of $K $ distinct colors is:
$$p_K(N,m_1,m_2,\dots,m_C)=\frac {[x^N]:\sum_{\mathbf a}^{ |\mathbf a|=K}\prod_{i=1}^C[(1+x)^{a_im_i}-a_i]}{\binom MN},
$$
where the operator $[x^n]: $ extracts the coefficient at $x^n $ in the subsequent expression and we have introduced binary vectors $\mathbf a=(a_1,a_2,\cdots, a_C) $ with $a_i=0,1$ equipped with the weight $|\mathbf a|\equiv \sum_i a_i $.

The expected value can then be computed as:
$$\bar K(N,m_1,m_2,\dots,m_C)=\sum_{K=0}^CKp_K(N,m_1,m_2,\dots,m_C).
$$

The expression can be drastically simplified if one assumes that the numbers of marbles of each color in the jar are equal, but this was not confirmed by the author.


问:
Orthogonal polynomials with respect to $e^{-|x|} \mathrm{d} x$ on the entire real line?
The Laguerre polynomials
https://en.wikipedia.org/wiki/Laguerre_polynomials
form a system of orthogonal polynomials with respect to the measure $e^{ -x} \mathrm{d} x$ on $(0,\infty)$.

Is anything known about the orthogonal polynomials with respect to the measure $e^{ -|x|} \mathrm{d} x$ on $(-\infty,\infty)$?

Thank you!

答:
It's a known result that symmetric weights lead to definite parity polynomials. For if $w(x)=w(-x)\ge0$ for $-a\lt x\lt a$ then if
$$\int_{-a}^aw(x)P_n(x)x^mdx=0$$
For all $0\le m\le n-1$, then
$$\begin{align}\int_{-a}^aw(x)P_n(x)x^mdx&=\int_a^{-a}w(-x)P_n(-x)(-x)^m(-dx)\\
&=(-1)^m\int_{-a}^aw(x)P_n(-x)x^mdx=0\end{align}$$
Also for $0\le m\le n-1$. And since orthogonal polynomials are unique up to a constant factor because if $\pi_n(x)$ and $\rho_n(x)$ are monic polynomials such that
$$\int_{-a}^aw(x)\pi_n(x)x^mdx=\int_{-a}^aw(x)\rho_n(x)x^mdx=0$$
for all $0\le m\le n-1$, then $\left(\pi_n(x)-\rho_n(x)\right)^*$ is a polynomial of degree at most $n-1$ so
$$\int_{-a}^aw(x)\left(\pi_n(x)-\rho_n(x)\right)^*\left(\pi_n(x)-\rho_n(x)\right)dx=0$$
So the integrand, being inherently nonnegative, must be identically zero. Thus $P_n(-x)$ is a constant times $P_n(x)$ and the polynomials $\frac12\left(P_n(x)+P_n(-x)\right)$ and
$\frac12\left(P_n(x)-P_n(-x)\right)$ are orthogonal polynomials with definite parity and exactly one of them is nonzero and of degree $n$. With this insight in hand we can crank out the first few orthogonal polynomials knowing that polynomials of one parity need only be orthogonal to polynomials of the same parity over the interval $[0,\infty)$:

```<code>! plaguerre.f90
module plaguerre
   use ISO_FORTRAN_ENV, only: wp => REAL128, ik => INT64
   implicit none
   contains
      subroutine coeffs(n,det,num)
         integer, intent(in) :: n
         real(wp), intent(out) :: det
         real(wp), intent(out) :: num(n/2+1)
         real(wp) A(n/2,n/2)
         real(wp) b(n/2)
         integer i, k, m
         real(wp) mag
         integer(ik) d
         do i = 1, n/2
            A(i,1) = 1
            do k = 2, n/2
               A(i,k) = (2*i+2*k-4+2*mod(n,2))*(2*i+2*k-5+2*mod(n,2))*A(i,k-1)
            end do
            b(i) = -(2*i+2*k-4+2*mod(n,2))*(2*i+2*k-5+2*mod(n,2))*A(i,k-1)
         end do
         det = 1
         do k = 1, n/2
            do i = k, n/2
               mag = norm2(A(i,k:))
               A(i,k:) = A(i,k:)/mag
               det = det*mag
               b(i) = b(i)/mag
            end do
            m = k-1+maxloc(abs(A(k:,k)),1)
            if(k /= m) then
               A([k,m],k:) = A([m,k],k:)
               b([k,m]) = b([m,k])
               det = -det
            end if
            A(k,k+1:) = A(k,k+1:)/A(k,k)
            det = det*A(k,k)
            b(k) = b(k)/A(k,k)
            do i = k+1, n/2
               A(i,k+1:) = A(i,k+1:)-A(i,k)*A(k,k+1:)
               b(i) = b(i)-A(i,k)*b(k)
            end do
         end do
         do k = n/2, 1, -1
            b(k) = b(k)-dot_product(A(k,k+1:),b(k+1:))
         end do
         num = [det*b,det]
         if(any(abs(num) > huge(0_ik))) stop
         write(*,'(*(g0))',advance='no') '<span class="math-container">$$P_{',n,'}(x) ='
         if(n == 0) write(*,'(*(g0))',advance='no') 1
         do i = 1, n/2+1
            d = gcd(nint(abs(num(i)),ik),nint(abs(det),ik))
            if(i == 1) then
               if(num(i)/det < 0) write(*,'(*(g0))',advance='no') '-'
               call rational(nint(abs(num(i)/d),ik),nint(abs(det/d),ik))
               if(mod(n,2) == 1) write(*,'(*(g0))',advance='no') 'x'
            else
               write(*,'(*(g0))',advance='no') merge('-','+',num(i)/det < 0)
               call rational(nint(abs(num(i)/d),ik),nint(abs(det/d),ik))
               write(*,'(*(g0))',advance='no') 'x^{',2*i-2+mod(n,2),'}'
            end if
         end do
         write(*,'(*(g0))') '$$</span>'
      end subroutine coeffs
      recursive function gcd(a,b) result(c)
         integer(ik) c
         integer(ik), value :: a, b
         if(b == 0) then
            c = a
            return
         else
            c = gcd(b, modulo(a,b))
         end if
      end function gcd
      subroutine rational(n, d)
         integer(ik), value :: n, d
         if(d > 1) then
            write(*,'(*(g0))',advance='no') '\frac{',n,'}{',d,'}'
         else if( n > 1) then
            write(*,'(*(g0))',advance='no') n
         end if
      end subroutine rational
end module plaguerre

program test
   use plaguerre
   implicit none
   integer n
   real(wp) det
   integer, parameter :: MAXN = 12
   real(wp) num(MAXN/2+1)
   do n = 0, MAXN
      call coeffs(n,det,num)
   end do
end program test
</code>```

$$P_{0}(x) =1$$
$$P_{1}(x) =x$$
$$P_{2}(x) =-2+x^{2}$$
$$P_{3}(x) =-12x+x^{3}$$
$$P_{4}(x) =\frac{216}{5}-\frac{168}{5}x^{2}+x^{4}$$
$$P_{5}(x) =520x-\frac{220}{3}x^{3}+x^{5}$$
$$P_{6}(x) =-\frac{392400}{149}+\frac{382680}{149}x^{2}-\frac{20010}{149}x^{4}+x^{6}$$
$$P_{7}(x) =-\frac{5342400}{109}x+\frac{992880}{109}x^{3}-\frac{24360}{109}x^{5}+x^{7}$$
$$P_{8}(x) =\frac{339308928}{1073}-\frac{1917474048}{5365}x^{2}+\frac{4668384}{185}x^{4}-\frac{9216032}{26825}x^{6}+x^{8}$$
$$P_{9}(x) =\frac{48561690240}{6011}x-\frac{10820234880}{6011}x^{3}+\frac{364446432}{6011}x^{5}-\frac{3017160}{6011}x^{7}+x^{9}$$

Of course the coefficients of the $3$-term recurrence relations will depend on the normalization chosen, here as monic polynomials. I haven't checked these very thoroughly so some more verification might be in order.

EDIT:
$$P_{10}(x) =-\frac{1842353840275200}{29286409}+\frac{2318870138659200}{29286409}x^{2}-\frac{199736268379200}{29286409}x^{4}+\frac{3777252968160}{29286409}x^{6}-\frac{20526107490}{29286409}x^{8}+x^{10}$$
$$P_{11}(x) =-\frac{5535810553996800}{2686261}x+\frac{1408500020188800}{2686261}x^{3}-\frac{58597576329600}{2686261}x^{5}+\frac{679907186640}{2686261}x^{7}-\frac{2547041860}{2686261}x^{9}+x^{11}$$

Body:
It's a known result that symmetric weights lead to definite parity polynomials. For if $w(x)=w(-x)\ge0$ for $-a\lt x\lt a$ then if
$$\int_{-a}^aw(x)P_n(x)x^mdx=0$$
For all $0\le m\le n-1$, then
$$\begin{align}\int_{-a}^aw(x)P_n(x)x^mdx&=\int_a^{-a}w(-x)P_n(-x)(-x)^m(-dx)\\
&=(-1)^m\int_{-a}^aw(x)P_n(-x)x^mdx=0\end{align}$$
Also for $0\le m\le n-1$. And since orthogonal polynomials are unique up to a constant factor because if $\pi_n(x)$ and $\rho_n(x)$ are monic polynomials such that
$$\int_{-a}^aw(x)\pi_n(x)x^mdx=\int_{-a}^aw(x)\rho_n(x)x^mdx=0$$
for all $0\le m\le n-1$, then $\left(\pi_n(x)-\rho_n(x)\right)^*$ is a polynomial of degree at most $n-1$ so
$$\int_{-a}^aw(x)\left(\pi_n(x)-\rho_n(x)\right)^*\left(\pi_n(x)-\rho_n(x)\right)dx=0$$
So the integrand, being inherently nonnegative, must be identically zero. Thus $P_n(-x)$ is a constant times $P_n(x)$ and the polynomials $\frac12\left(P_n(x)+P_n(-x)\right)$ and
$\frac12\left(P_n(x)-P_n(-x)\right)$ are orthogonal polynomials with definite parity and exactly one of them is nonzero and of degree $n$. With this insight in hand we can crank out the first few orthogonal polynomials knowing that polynomials of one parity need only be orthogonal to polynomials of the same parity over the interval $[0,\infty)$:

! plaguerre.f90
module plaguerre
   use ISO_FORTRAN_ENV, only: wp => REAL128, ik => INT64
   implicit none
   contains
      subroutine coeffs(n,det,num)
         integer, intent(in) :: n
         real(wp), intent(out) :: det
         real(wp), intent(out) :: num(n/2+1)
         real(wp) A(n/2,n/2)
         real(wp) b(n/2)
         integer i, k, m
         real(wp) mag
         integer(ik) d
         do i = 1, n/2
            A(i,1) = 1
            do k = 2, n/2
               A(i,k) = (2*i+2*k-4+2*mod(n,2))*(2*i+2*k-5+2*mod(n,2))*A(i,k-1)
            end do
            b(i) = -(2*i+2*k-4+2*mod(n,2))*(2*i+2*k-5+2*mod(n,2))*A(i,k-1)
         end do
         det = 1
         do k = 1, n/2
            do i = k, n/2
               mag = norm2(A(i,k:))
               A(i,k:) = A(i,k:)/mag
               det = det*mag
               b(i) = b(i)/mag
            end do
            m = k-1+maxloc(abs(A(k:,k)),1)
            if(k /= m) then
               A([k,m],k:) = A([m,k],k:)
               b([k,m]) = b([m,k])
               det = -det
            end if
            A(k,k+1:) = A(k,k+1:)/A(k,k)
            det = det*A(k,k)
            b(k) = b(k)/A(k,k)
            do i = k+1, n/2
               A(i,k+1:) = A(i,k+1:)-A(i,k)*A(k,k+1:)
               b(i) = b(i)-A(i,k)*b(k)
            end do
         end do
         do k = n/2, 1, -1
            b(k) = b(k)-dot_product(A(k,k+1:),b(k+1:))
         end do
         num = [det*b,det]
         if(any(abs(num) > huge(0_ik))) stop
         write(*,'(*(g0))',advance='no') '$$P_{',n,'}(x) ='
         if(n == 0) write(*,'(*(g0))',advance='no') 1
         do i = 1, n/2+1
            d = gcd(nint(abs(num(i)),ik),nint(abs(det),ik))
            if(i == 1) then
               if(num(i)/det < 0) write(*,'(*(g0))',advance='no') '-'
               call rational(nint(abs(num(i)/d),ik),nint(abs(det/d),ik))
               if(mod(n,2) == 1) write(*,'(*(g0))',advance='no') 'x'
            else
               write(*,'(*(g0))',advance='no') merge('-','+',num(i)/det < 0)
               call rational(nint(abs(num(i)/d),ik),nint(abs(det/d),ik))
               write(*,'(*(g0))',advance='no') 'x^{',2*i-2+mod(n,2),'}'
            end if
         end do
         write(*,'(*(g0))') '$$'
      end subroutine coeffs
      recursive function gcd(a,b) result(c)
         integer(ik) c
         integer(ik), value :: a, b
         if(b == 0) then
            c = a
            return
         else
            c = gcd(b, modulo(a,b))
         end if
      end function gcd
      subroutine rational(n, d)
         integer(ik), value :: n, d
         if(d > 1) then
            write(*,'(*(g0))',advance='no') '\frac{',n,'}{',d,'}'
         else if( n > 1) then
            write(*,'(*(g0))',advance='no') n
         end if
      end subroutine rational
end module plaguerre

program test
   use plaguerre
   implicit none
   integer n
   real(wp) det
   integer, parameter :: MAXN = 12
   real(wp) num(MAXN/2+1)
   do n = 0, MAXN
      call coeffs(n,det,num)
   end do
end program test

$$P_{0}(x) =1$$
$$P_{1}(x) =x$$
$$P_{2}(x) =-2+x^{2}$$
$$P_{3}(x) =-12x+x^{3}$$
$$P_{4}(x) =\frac{216}{5}-\frac{168}{5}x^{2}+x^{4}$$
$$P_{5}(x) =520x-\frac{220}{3}x^{3}+x^{5}$$
$$P_{6}(x) =-\frac{392400}{149}+\frac{382680}{149}x^{2}-\frac{20010}{149}x^{4}+x^{6}$$
$$P_{7}(x) =-\frac{5342400}{109}x+\frac{992880}{109}x^{3}-\frac{24360}{109}x^{5}+x^{7}$$
$$P_{8}(x) =\frac{339308928}{1073}-\frac{1917474048}{5365}x^{2}+\frac{4668384}{185}x^{4}-\frac{9216032}{26825}x^{6}+x^{8}$$
$$P_{9}(x) =\frac{48561690240}{6011}x-\frac{10820234880}{6011}x^{3}+\frac{364446432}{6011}x^{5}-\frac{3017160}{6011}x^{7}+x^{9}$$

Of course the coefficients of the $3$-term recurrence relations will depend on the normalization chosen, here as monic polynomials. I haven't checked these very thoroughly so some more verification might be in order.

EDIT:
$$P_{10}(x) =-\frac{1842353840275200}{29286409}+\frac{2318870138659200}{29286409}x^{2}-\frac{199736268379200}{29286409}x^{4}+\frac{3777252968160}{29286409}x^{6}-\frac{20526107490}{29286409}x^{8}+x^{10}$$
$$P_{11}(x) =-\frac{5535810553996800}{2686261}x+\frac{1408500020188800}{2686261}x^{3}-\frac{58597576329600}{2686261}x^{5}+\frac{679907186640}{2686261}x^{7}-\frac{2547041860}{2686261}x^{9}+x^{11}$$


问:
Let $V$ be a vector space and $T:V \to V$ is a linear transformation,show that the two following statements are equivalent:
Let $V$ be a vector space and $T:V \to V$ is a linear transformation,show that the two following statements are equivalent:

The intersection of $\text{ker(T)}$ and $\text{Im(T)}$ is the subspace $\left\{0\right\}$ of $V$.

If $T(T( \alpha))=0_V \implies T(\alpha)=0_V$

I don't know how to show $1 \implies 2$.

$2 \implies 1$

Let $x \in \text{ker(T)} \cap \text{Im(T)}$,then $T(x)=0_V$ since $x$ is in the kernel of $T$,and for some $\alpha \in V:T(\alpha)=x$ since $x$ is in the range of $T$,so $0_V=T(x)=T(T(\alpha))$ which by the assumption implies that $T(\alpha)=0_V$ and since $T(\alpha)=x$ so we have that $x=0_V$.

I would like to know how to prove the other direction.

答:
Suppose $\;\ker T\cap\text{Im}\,T=\{0\}\;$ , then

$$T(Tv)=0\implies Tv\in\ker T\,\; \text{. But trivially}\;\;Tv\in\text{Im}\,T\,,\,\,\text{so}$$

$$\;\;Tv\in\ker T\cap\text{Im}\,T=\{0\}\implies Tv=0$$


问:
Calculating $a^p+b^p$ given $a^2+b^2=c$
If one has $a^2+b^2=c$ where $a$,$b$ and $c$ are real numbers, is there any way to calculate $a^p+b^p$ where $p$ may be any real number? If there is, would you please explain with an example with $p=3$?</p>

答:
There are $a, b, A, B$ such that $a^2+b^2=A^2+B^2$ but $a^p+b^p \ne A^p+B^p$.
So the answer to your first question is NO.</p>

Title: Calculating $a^p+b^p$ given $a^2+b^2=c$


问:
Order of operations (BODMAS)
$$40-20/2+15\times1.5\\\hspace{.1in}\\40-20/2+15\times1.5=\\ 40-10+22.5=7.5$$

I'm studying and this is from an example.

In BODMAS, aren't addition and subtraction have same level?

So, in the 3rd line, it should be from left to right, correct?

答:
The example is incorrect under BODMAS, but typically, people use a different set of rules, to the effect of creating implied brackets.

Brackets unchanged

Of (ie functions) unchanged.

M  Multiplication of close-fit numbers.

D  If there is a horizontal fraction, then this is evaluated first.  For example, $/$ only is evaluated here.

MD  Multiplication and division are evaluated equally when set with equal-height signs.  This is a different division $\div$.  So $120/2 \div 60/2=2$ means 60/30, but $120 \div 2 \div 60 \div 2=0.5$.  This is a product of numbers, where $\div$ is read as a unitary inverse.  Rather like AS, really.

AS  addition and subtraction are equally evaluated, as if the sign were attached to the number, when set with equal-height signs.  eg 5-2+3 = 6

Here is an example of how this works.

$F = c/4\pi þ^2 * M^2/R^2$

In this equation, the multiplication represented by * is a lower operation (MD) then the multiplications represented by {4\pi þ^2}.  That is we see this as this, with the close-fit or inner multiplication, being developed done before the division /.  In turn, once these are evaluated, the product occurs from left to right in the manner of computers.

$F = (c/(4\pi þ^2)) * (M^2/R^2) = c/4/\pi \ þ^2 * M^2 / R^2$

Title: Order of operations (BODMAS)


问:
If the total number of $3$-element subsets of $(1,$ ... $, 23)$ with $S(A) < 36$ is $N$. Find $\frac{(N+ 45)}{25}$.

For any non-empty finite set $A$ of real numbers, let $S(A)$ be the sum of the elements in $A$. There are exactly $61$ $3$-element subsets $A$ of $(1,$ ... $,23)$ with $S(A) = 36$. The total number of $3$-element subsets of $(1,$ ... $, 23)$ with $S(A) < 36$ is $N$. Find $\frac{(N+ 45)}{25}$.

What I Tried: For some reason, the question did not really make that sense to me. For example, I did not fully understand the $2$nd sentence, did it mean that the set $A$ is actually the set $(1,$ ... $, 23) ?$ Probably not as then $S(A) \neq 36$. But then is $A$ another different set? If yes, how would we even start solving the question? Also if it meant that there are $61$ $3$-element subsets of $A$ , why is the set $(1,$ ... $, 23)$ even necessary?

As long as I am having problem understanding the question, I will have problems on solving it too.

Can anyone help?

答: We will use a bijection to solve this problem. For every 3-subset $(a,b,c)$ link this 3-subset to the 3-subset $(24-a,24-b,24-c)$. This will make a bijection between the 3-subsets with sum less than 36 to the 3-subsets with sum greater than 36, hence the quantity of them are equal. So we just have to decrease the number of 3-subsets with sum exactly equal to 36 -which is 61- from the number of all 3-subsets and divide it by 2 to find N. Because the number of all 3-subsets is $\binom{23}{3}=1771$ and so $N=\frac{1771-61}{2}=\frac{1710}{2}=855$ so the value required in the end of the problem is 36.


问:
Prove that there exists an integer $a$ such that $\frac{a^2-x}{p}\in \mathbb{Z}$
Let $x$ be an odd number, $y$ a positive integer, and $p$ an odd prime number such that $x^2+y^2=p$. Show that there exists an integer $a$ such that  $\dfrac{a^2-x}{p}\in \mathbb{Z}.$

This problem is one that my friend asked me, and I've considered it for some time, but I can't prove it. Thank you everyone.

答:
Hint: Because $p$ is odd and $p=x^2+y^2$ for some integers $x$ and $y$, we know that $p\equiv 1\bmod 4$. Let the prime factorization of $x$ be
$$x=q_1^{e_1}\cdots q_r^{e_r}$$
(which includes no $2$'s, since $x$ is odd). Use the fact that $x^2\equiv 0\bmod q_i$ for all $i$, together with quadratic reciprocity, to prove that $\big(\!\frac{x}{p}\!\big)=1$, so that there is an integer $a$ with $a^2\equiv x\bmod p$.

Full argument in spoiler:

 $$\begin{align*} \left(\frac{x}{p}\right)&=\left(\frac{q_1}{p}\right)^{e_1}\cdots\left(\frac{q_r}{p}\right)^{e_r}\\\\\\ &=\left(\frac{p}{q_1}\right)^{e_1}\cdots\left(\frac{p}{q_r}\right)^{e_r}\\\\\\ &=\left(\frac{x^2+y^2}{q_1}\right)^{e_1}\cdots\left(\frac{x^2+y^2}{q_r}\right)^{e_r}\\\\\\ &=\left(\frac{y^2}{q_1}\right)^{e_1}\cdots\left(\frac{y^2}{q_r}\right)^{e_r}\\\\\\ &=1^{e_1}\cdots 1^{e_r}\\\\\\ &=1 \end{align*}$$


问:
If $a,b$ have order $2^j$ mod $p$, then $a\cdot b$ has order $2^{j'}$ mod $p$ for some $j'<j$.
Let $a,b\in\mathbb{Z}^+$ and $p$ prime such that $(a,p)=(b,p)=1$. If $a,b$ have order $2^j$ mod $p$, then $a\cdot b$ has order $2^{j'}$ mod $p$ for some $j'<j$.

I have been trying this problem using that $ord_p(ab)\mid 2^j$. But I'm not sure.

答:
It simply suffices to show that $(ab)^m \equiv 1 \pmod{p}$ where $m=2^{j-1}$. We know that $(a^m)^2 \equiv 1 \pmod{p}$ but $a^m \not\equiv 1 \pmod{p}$. This means that $a^m \equiv -1 \pmod{p}$. Similarly, $b^m \equiv -1 \pmod{p}$. Hence:
$$(ab)^m \equiv -1 \cdot -1 \equiv 1 \pmod{p}$$
which proves the required.


问:
Solid angle of a pyramid
Suppose I have a rectangular pyramid. I partition the dihedral angle between a fixed pair of opposite faces into three parts and thereby obtain three sub-pyramids (within the original one). Consider the sum of the solid angles subtended at the vertex by these sub-pyramids. Is it equal or not to the solid angle subtended by the original pyramid at the vertex ? Please explain.</p>

答: The sum of the solid angles is equal to the total solid angle. The apparent contradiction with your formula  follows from the fact that it only works for a straight pyramid, and your subpyramides are not straight. Observe that an oblique pyramid with the same height and base subtends a smaller solid angle. The more oblique, the lesser the angle.</p>

Title: Solid angle of a pyramid


问:
Fourier transform in $\mathbb{L}^2(\mathbb{R})\setminus\mathbb{L}^1(\mathbb{R})$
In a book I've found  the following definition for the Fourier transform in  $\mathbb{L}^2(\mathbb{R})\\$:

$g(k)$=$\frac{1}{\sqrt{2\pi}}\frac{d}{dk}\int_{-\infty}^{+\infty}\frac{e^{-ikx}-1}{-ix}f(x)dx\\$
Now I don't understand why this definition is equivalent to the usual one that is:
$g(k)=\lim_{N \to \infty} \frac{1}{\sqrt{2\pi}}\int_{-N}^{+N}e^{-ikx}f(x)dx\\$  where the limit is in the sense of the norm $||\cdot||_2$. Where can I find a proof of the equivalence of the two definitions?

答:
This is not a proof, but a bit of unification of the two definitions.

Using the definition of derivative we have
$$
\frac{d}{dk} \int_{-\infty}^{\infty} \frac{e^{-ikx}-1}{-ix} \, f(x) \, dx
= \lim_{\delta\to 0} \frac{1}{\delta} \left( \int_{-\infty}^{\infty} \frac{e^{-i(k+\delta)x}-1}{-ix} \, f(x) \, dx - \int_{-\infty}^{\infty} \frac{e^{-ikx}-1}{-ix} \, f(x) \, dx \right)
\\= \lim_{\delta\to 0} \int_{-\infty}^{\infty} \frac{e^{-i(k+\delta)x}-e^{-ikx}}{-i\delta x} \, f(x) \, dx
= \lim_{\delta\to 0} \int_{-\infty}^{\infty} \frac{e^{-i\delta x}-1}{-i\delta x} \,e^{-ikx} \, f(x) \, dx
\\= \lim_{N\to\infty} \int_{-\infty}^{\infty} \frac{e^{-ix/N}-1}{-ix/N} \,e^{-ikx} \, f(x) \, dx
.
$$

The other definition is
$$
\lim_{N\to\infty} \int_{-N}^{N} e^{-ikx} \, f(x) \, dx
= \lim_{N\to\infty} \int_{-\infty}^{\infty} \chi_{[-N,N]}(x) \, e^{-ikx} \, f(x) \, dx
.
$$

Thus, both definitions are of the form
$$
\lim_{N\to\infty} \int_{-\infty}^{\infty} h_N(x) \, e^{-ikx} \, f(x) \, dx
,
$$
where $h_N \in L^2(\mathbb{R})$ and $h_N(x) \to 1$ pointwise (and on every compact set).


问:
System of multivariable equations with trig functions
I have the following system of equations:

\begin{cases} \frac{\cos (x)}{1+y^2}=0 \\ \frac{-2y\cdot  \sin(x)}{(1+y^2)^2}=0 \end{cases}

The first equation has the solution $x=k\pi-\pi/2$ for an integer $k$ and any real $y$. The second one has the solutions $x=k\pi$ for any real $y$ OR $y=0$ for any real $x$. My question is: how does one find the solution(s) of the system of equation?

答:
Well, we are trying to solve the following system of equations:

$$
\begin{cases}
\cos\left(x\right)=0\\
\\
\text{n}\sin\left(x\right)=0
\end{cases}\tag1
$$

It is not hard to see, from the first equation, that:

$$x=\pi\text{k}\pm\frac{\pi}{2}\tag2$$

Where $\text{k}\in\mathbb{Z}$.

Substituting that in the second equation, gives:

$$\text{n}\sin\left(\pi\text{k}\pm\frac{\pi}{2}\right)=0\tag3$$

We can simplify the LHS:

$$\text{n}\left(\mp1\right)=0\space\Longleftrightarrow\space\text{n}=0\tag4$$


问:
In how many ways $n$ dishes around a round table can be colored with 3 colors with adjacent dishes with different colors

Consider a round table with $n$ places. In each place there is a colored dish. In how many ways the dishes in this table can be colored with 3 colors, in such a way that adjacent dishes have different colors?

Solution: If $P(n)$ is the number of possibilities for a table with $n$ places, it is true that $$P(n)=P(n-1)+2P(n-2).$$By inspection it is easy to show that $P(2)=6$ and $P(3)=6$, and we can find the solution for any $P(n)$ by substitution using the recursion, or by the solution for the previous difference equation given by:
$$P(n)=2^n+2(-1)^n$$.

My problem: I don't understand why $$P(n)=P(n-1)+2P(n-2)$$ is true, or why, directly, $$P(n)=2^n+2(-1)^n$$ is true. Yes, I know how last equation is obtained from the previous equation, but cannot find a reasonable argument why it makes sense from a combinatoric argument. Any help?

答:
To find the number of possibilities we label the three colors to be blue green and red. Now we assign positions $a_1,a_2...a_n$ around the circle(cyclic).

Case 1:when $a_2,a_3...a_n$ are such that when arranged in cyclic order no two o adjacent positions have same colour($P(n-1)$ ways).In that case $a_2,a_n$ are of different colours say blue and green and we have to  assign $a_1$ to be red .Hence number of possibilities is $P(n-1)$ .

Case 2: suppose $a_2,a_3...a_{n-1}$  when arranged in cyclic order no two o adjacent positions have same colour-($P(n-2)$ ways).Also inorder to make it disjoint with case 1 we take $a_2,a_n$ to be of same colours say blue,blue.In that case $a_1$ can be assigned any of the two other colours  (here green/red) Hence number of possibility is $2(P(n-2))$


问:
How do you solve 9k ≡ 3 (mod 21)?
I know how to solve it when the GCD is 1, as in "solving 9k≡2 (mod 29). You just use the Euclidean algorithm backward and find what the multiplicative inverse of 9 is. But what is the general solution for the questions like this one where there's no multiplicative inverse?</p>

答:
$$9k\equiv3\pmod{21}\iff3k\equiv1\pmod7$$

Now $3\cdot5\equiv1\pmod7\iff3^{-1}\equiv5$

Title: How do you solve 9k ≡ 3 (mod 21)?


问:
How to find the following limit?

Can anyone explain how to find that limit?
$$\underset{x\to 0}{\text{lim}}\frac{\int_0^x \cos \left(t^2\right) \, dt}{x} $$

answer suppose to equal $1$

I tried to solve integral by using the define integral formula and I got $\cos x^2$, however it seems to be wrong, because limit of $\frac{\cos x^2}{x}$ does not exist.

答:
Hint:

Since $\;\cos^2 t\;$ , or even $\;\cos t^2\;$  (whatever you meant) is a continuous function, the function $\;g(x):=\int\limits_0^x\cos^2t\,dt\;$ is a differentiable one, and $\;g(0)=0\;$. Thus, you can apply L'Hospital's rule to calculate the limit, using the Fundamental Theorem of Calculus:

$$g'(x)=\cos^2x\; ...$$

whereas the derivative of the denominator is...

Title: How to find the following limit?


问:
Find the $\sigma$-algebra that makes the following map a measure($X=\mathbb{R}$)
$v(A)=\begin{cases}
      0 & \text{if $A$ is finite} \\
      1 & \text{if $A^c$ is finite} \\
   \end{cases} $

This is a basic question and I don't think $P(\mathbb{Q})$ is a desired $\sigma$-algebra, nor are $P(\mathbb{R})$ and $\{\emptyset, \mathbb{R}\}$. Any idea?

答:
Considering $X=\mathbb{R}$, you want to find the $\sigma$-algebras that make the following map a measure:
$\nu(A)=\begin{cases}
      0 & \text{if $A$ is finite} \\
      1 & \text{if $A^c$ is finite} \\
   \end{cases} $

Note that there are more than one such $\sigma$-algebra.

$1$. Let $\Sigma$ be any $\sigma$-algebra generated by a finite collection of finite subsets of $\Bbb R$. That means there is $\{F_i\}_{i \in \{1, \dots, n\}}$ where each $F_i$ is finite and
$\Sigma = \sigma(\{F_i\}_{i \in \{1, \dots, n\}})$.

It is straight forward to check that $\nu$ is a measure defined on $\Sigma$.

$2$. Suppose $\Sigma$ is $\sigma$-algebra such that $\nu$ is measure defined on $\Sigma$.

First, if $E \in \Sigma$ is any infinite set, then $E^c$ is finite. In fact, if $E$ is a infinite set and $E^c$ is not finite, $\nu$ is not defined for $E$, which is a contradiction to the fact that $\nu$ is defined on $\Sigma$.

Let $\mathcal{F}=\{F \in \Sigma : F \textrm { is finite }\}$. From the previous paragraph, it follows immediately that $\Sigma = \sigma(\mathcal{F})$.

Second, let us prove that the set $\mathcal{F}$ is a finite collection.

Proof: Suppose  $\mathcal{F}$ is infinite and let $\mathcal{F}_1 \subseteq \mathcal{F}$ be a countably infinite subset of $\mathcal{F}$. Then $E=\bigcup_{F\in \mathcal{F}_1}F\in \Sigma$ is infinite. So, we know that $E^c$ is finite, and then we have
$$1= \nu(E) = \nu \left (\bigcup_{F\in \mathcal{F}_1}F \right) \leqslant \sum_{F\in \mathcal{F}_1} \nu(F)=0 $$
Contradiction.  $\square$

Combining $1$ and $2$ above we have that $\nu$ is a measure defined on the  $\sigma$-algebra $\Sigma$ if and only if $\Sigma$ is a $\sigma$-algebra generate by a finite collection of finite subsets of $\Bbb R$.


问:
Level Sets Questions
1) In the following link, question 1:
http://mathquest.carroll.edu/libraries/MVC.student.14.01.pdf

Is it true that both partial derivatives are negative ? If so, can someone help me find an example of a contour plot with $f_x<0 , f_y>0$ in one of the points ? I just want to verify

2) In the following (question 4, with $t=1$ instead of 2 ):
http://mathquest.carroll.edu/libraries/MVC.student.14.06.pdf

At $t=1$ i think that the derivative is positive... this is because the $x'(1)=0 ,y'(1)>0$ and $  z_y(3,0)  > 0 $ .. Is it correct?

Thanks in advance

答:
Honestly, I could just make a guess of the function in part one. I could just imagine and so the following function may help us to find out what exactly happens in that point.

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

Title: Level Sets Questions


问:
Geodesics on a cylinder: straight line or helix?
It is known that geodesics on a cylinder are helical lines (helix). As a special case of a helix--- straight line. But one can take two points so cleverly that you can draw two helixes between them (see picture), one of which is a straight line. Both satisfy geodesics differential equation. The question: is, can they both be considered as geodesics or only the shortest one is "true" geodesic?

答:
I think the only reasonable definition of a geodesic in such situations involves locally optimal (shorter) paths.  After all, consider the case where the two end points are on precisely opposite positions on the cylinder.  There are two equivalent arcs (geodesics).  (There are an infinite number of geodesics on a sphere linking anti-podes, such as the north and south poles.)  For a cylinder there are, then, an infinite number of geodesics in the arbitrary case, indexed by the number of rotations of the helix around the cylinder (like a winding number).

"A geodesic is a locally length-minimizing curve."


问:
What is the simplest way to do this combinatorics problem?
Question: Bob is about to hang his eight shirts in the wardrobe. He has 4 different styles of shirt, two identical ones of each particular style. How many different arrangements are possible if no two identical shirts are next to each other?

Here are the solutions:

In this case, we take the first case of all possible combinations: $\frac{8!}{2^4}$
But we see that we counted the possibilities with at least one pair together
so we have to subtract $\frac{7!}{2^3}\cdot \binom{4}{1}$ if we tie the two offending shirts together.

^This I don't get - why and what is being double-counted?

However, we see now that we double subtracted ones that have two pairs thus we must add the chance of two pairs so we must add $\frac{6!}{2^2}\cdot \binom{4}{2}$ and so on

Thus the final answer is equal to $864$

Is there a better way that doesn't involve BOTH adding AND subtracting?

答:
We denote with $a,b,c,d$ the four different types of shirts. We are looking for words of length $8$ built with all letters from the multiset $\mathcal{V}=\{a,a,b,b,c,c,d,d\}$ which do not contain the bad words in $\mathcal{B}=\{aa,bb,cc,dd\}$.

Here we use PIE the inclusion-exclusion principle to count the number of valid words.

In order to do the job some kind of bookkeeping is helpful. We consider
\begin{align*}
&\left(.\ .\ .\ .\ .\ .\ .\right)\tag{1}\\
&\quad-\left(aa\ .\ .\ .\ .\ .\ .|bb\ .\ .\ .\ .\ .\ .\ .|cc\ .\ .\ .\ .\ .\ .|dd\ .\ .\ .\ .\ .\ .\right)\tag{2}\\
&\quad+\left(aa\ bb\ .\ .\ .\ .|aa\ cc\ .\ .\ .\ .|aa\ dd\ .\ .\ .\ .\right.\\
&\quad\qquad \left.|bb\ cc\ .\ .\ .\ .|bb\ dd\ .\ .\ .\ .|cc\ dd\ .\ .\ .\ .\right)\tag{3}\\
&\quad-\left(aa\ bb\ cc\ .\ .|aa\ bb\ dd\ .\ .|aa\ cc\ dd\ .\ .|bb\ cc\ dd\ .\ .\right)\\
&\quad+(aa\ bb\ cc\ dd)
\end{align*}

Comment:

In (1) we count all $8$-letter words from $\mathcal{V}$ indicated by eight dots which gives $\frac{8!}{2!2!2!2!}$ words.

In (2) we subtract all words which contain at least one bad word. Note that the first pattern
\begin{align*}
aa\ .\ .\ .\ .\ .\ .
\end{align*}
represents all words which have two adjacent $a$'s and which gives $\frac{7!}{1!2!2!2!}$ words. In fact it counts all words which contain at least $aa$ besides possibly, occurrences of other adjacent letters e.g. $bb$ represented by two of the other dot's. Such words are counted more than once, because they are also counted with $(bb\ .\ .\ .\ .\ .\ .)$.

In oder to compensate this we add the words which contain at least twice occurrences of adjacent letters in (3).

In (3) we add words containing two bad words as compensation for those which we've subtracted twice in (2), which gives $\frac{6!}{1!1!2!2!}$ words. We  continue in the following lines accordingly.

No more cases are left to consider, since words containing five or more bad words have length greater than $8$.

We obtain according to (1) to (3):
\begin{align*}
&\frac{8!}{2!2!2!2!}-\binom{4}{1}\frac{7!}{1!2!2!2!}+\binom{4}{2}\frac{6!}{1!1!2!2!}\\
&\qquad-\binom{4}{3}\frac{5!}{1!1!1!2!}+\binom{4}{4}\frac{4!}{1!1!1!1!}\\
&\quad=2\,520-4\cdot 630+6\cdot 180-4\cdot 60+1\cdot 24\\
&\quad\,\,\color{blue}{=864}
\end{align*}
in accordance with OPs statement.


问:
Find the least order differential equation that accepts $y=a\cos{(\ln{x})}+b\sin{(\ln{x})}$ general solution.
There are two coefficients but I found the least order differential equation as first order one. When I put the given general solution to the differential equation that I found, It provides the equation.

Here's how I found the differential equation:

$$y=a\cos{(\ln{x})}+b\sin{(\ln{x})} \Rightarrow y\prime=-\frac{a}{x}\sin{(\ln{x})}+\frac{b}{x}\cos{(\ln{x})}$$

Multiply both sides with $\frac{ax}{b}$,

$$\frac{ax}{b}y\prime=-\frac{a^2}{b}\sin{(\ln{x})}+a\cos{(\ln{x})}$$

$$a\cos{(\ln{x})}=y-b\sin{(\ln{x})}\Rightarrow\frac{ax}{b}y\prime=-\frac{a^2}{b}\sin{(\ln{x})}+y-b\sin{(\ln{x})}$$

$$\frac{ax}{b}y\prime=y-\left(\frac{a^2+b^2}{b}\right)\sin{(\ln{x})}$$

Multiply both sides with $\frac{b}{ax}$ to simplify,

$$y\prime=\frac{b}{ax}y-\frac{a^2+b^2}{ax}\sin{(\ln{x})}$$

$$y\prime-\frac{b}{ax}y=-\frac{a^2+b^2}{ax}\sin{(\ln{x})}$$

The last formula is the first order differential equation that I found. When I put the given solution and it's derivative, it provides the eqution. Is my solution correct? Or should have I found second order differential equation for some reason, because there are 2 coefficients, a and b?

答:
You have to eliminate both constants $a$ and $b$. Better to rewrite the solution as :
$$y=a\cos{(\ln{x})}+b\sin{(\ln{x})}$$
$$y=a\cos{(t)}+b\sin{(t)}$$
With $x=e^t$.

Then the characteristic polynomial is
$$r^2+1=0$$
And the DE is:
$$y''(t)+y(t)=0$$
And
$$\dfrac {dy}{dt}=\dfrac {dy}{dx}\dfrac {dx}{dt}=x\dfrac {dy}{dx}$$
$$\dfrac {d^2y}{dt^2}=x\dfrac {d}{dx} \left (x\dfrac {dy}{dx} \right)=......$$
The differential equation becomes:
$$x^2y''+xy'+y=0$$


问:
Proof that $N/1*(N-2)/2*(N-3)/3...$ is always integral? (Algorithm for Combinations)
I am trying to understand the algorithm for calculating combination N Choose R. It basically relies on the fact that $(N)/1 * (N-1)/2 * (N-2)/3 * ... * (N-R+1)/R$ is always an integer for any $R$.

It is obvious that the numerator is divisible by any number in the denominator, but I don't understand how the numerator is divisible by all numbers in the denominator. For two co-prime numbers, it makes sense, but what about numbers that aren't coprime to each other?

For example, consider this: $(N)/1 * (N-1)/2 * (N-2)/3 * (N-3)/4 * (N-4)/5 * (N-5)/6$. What if 4 needs $N-3=12$ and 6 needs $N-3=12$. Individually, $N-3$ is divisble by 4 or 6, but $N-3$ is not divisible by 24.

https://stackoverflow.com/questions/9330915/number-of-combinations-n-choose-r-in-c
https://stackoverflow.com/questions/1838368/calculating-the-amount-of-combinations

答:
The individual number-to-number fraction product does not have all integer values. It is the total denominator product that divides the numerator product. Another way to phrase this is by saying that the product of any $r$ consecutive positive integers is divisible by $r!$ (i.e. the product of the first $r$ consecutive positive integers). Combinatorially speaking, the fact that this value is the number of ways of choosing objects is proof of divisibility, but I shall try to reason this in a intuitive number-theoretic manner.

Consider the prime factorization of $r!$. Let $p^k$ be the largest power of $p$ that divides $r!$. It suffices to show that the product of any $r$ consecutive positive integers is divisible by $p^k$. For any product of numbers, we can see that the power of $p$ dividing this product is:
$$n_1+n_2+n_3+\cdots$$
where $n_i$ is the number of values in the product divisible by $p^i$. This is because if the maximum power of $p$ dividing a certain term in the product is $p^i$, then it is counted $i$ times in the above sum (in $n_1,n_2,\cdots,n_i$).

Now, we show that this sum is least when our $r$ consecutive numbers are the first $r$ positive integers, and this would show that since $r!$ is the case where power of $p$ is least, we would also have $p^k$ dividing any product of $r$ consecutive positive integers.

To minimize $n_1$, what would are first number in our $r$ consecutive numbers have to satisfy? Well, it is minimized when the first number is $1 \pmod{p}$ since that way, it delays the occurrence of the first number divisible by $p$ as long as possible. For example, if $r=9$ and $p=5$, if the first number was $1 \pmod{5}$, there would only be $1$ number in the product that is divisible by $5$, but in any other case, there would be $2$ numbers divisible by $5$.

It similarly works that for any $n_i$, the way to minimize $n_i$ is by making the first number $1 \pmod{p^i}$. Now, it becomes evident why $1,2,3,\cdots,r$ minimizes the power of $p$. The first term, $1$, is $1 \pmod{p^i}$ for all integers $i$! It is in some sense, the farthest away from all the numbers divisible by $p$! Hope this provides a nice intuitive perspective to the divisibility aspect!


答:
Suppose you have your circle full of dishes. Locate yourself in one of the dishes, say the one most at north. Take it out and close the circle. There are two options: Either the dishes still form a circle with the desired property $P(n-1)$ or they are not, meaning, when you form the circle they connect in the same color. There are two options for this because they could not have been the same color as the first dish you took out. Take out one of them. Now they have to be different colors and so $P(n-2)$ in two ways so $2\cdot P(n-2).$

The second equation goes as this. Put one dish, you have $3$ options. Then you have $2$ options on the right, then two options and so on. So you will end up having $3\cdot 2^{n-1}=2^n+2^{n-1}.$ Notice that if you close the circle, perhaps the last dish is equal to the first one so if they are equal eliminate them so you will have $2^n+2^{n-1}-(2^{n-1}+2^{n-2}-(2^{n-2}+2^{n-3}-\cdots )).$  If you keep doing this is clear that most of it cancels out except the last term of the last one ($P(2)=2^2+\color{red}{2^1}=6$) which mimics, combinatorially, the telescopic sum in the difference equation. Notice that the sign alternates when you distribute it.

Title: In how many ways $n$ dishes around a round table can be colored with 3 colors with adjacent dishes with different colors

Body:
Consider a round table with $n$ places. In each place there is a colored dish. In how many ways the dishes in this table can be colored with 3 colors, in such a way that adjacent dishes have different colors?

Solution: If $P(n)$ is the number of possibilities for a table with $n$ places, it is true that $$P(n)=P(n-1)+2P(n-2).$$By inspection it is easy to show that $P(2)=6$ and $P(3)=6$, and we can find the solution for any $P(n)$ by substitution using the recursion, or by the solution for the previous difference equation given by:
$$P(n)=2^n+2(-1)^n$$.

My problem: I don't understand why $$P(n)=P(n-1)+2P(n-2)$$ is true, or why, directly, $$P(n)=2^n+2(-1)^n$$ is true. Yes, I know how last equation is obtained from the previous equation, but cannot find a reasonable argument why it makes sense from a combinatoric argument. Any help?


答:
$$\underset{x\to 0}{\text{lim}}\frac{\int_0^x \cos \left(t^2\right) \, dt}{x}=^*\underset{x\to 0}{\text{lim}}\frac{\frac{d}{dx} \left(\int_0^x \cos \left(t^2\right) \, dt\right)}{1}=\underset{x\to 0}{\text{lim}}\cos x^2=1$$
$*$ applied L'Hopital rule

Title: How to find the following limit?

Body:
Can anyone explain how to find that limit?
$$\underset{x\to 0}{\text{lim}}\frac{\int_0^x \cos \left(t^2\right) \, dt}{x} $$

answer suppose to equal $1$

I tried to solve integral by using the define integral formula and I got $\cos x^2$, however it seems to be wrong, because limit of $\frac{\cos x^2}{x}$ does not exist.


问:
When is this sum equal to $0$?
When, for arbitrary positive integers $m$ and $n$, is the following sum equal to $0$?

$$
\sum_{i=0}^{mn-1} (-1)^{\lfloor i/m \rfloor +\lfloor i/n\rfloor}
$$

答:
The sum is $0$ if and only if $m$ and $n$ contain different numbers of factors of $2$.

If neither contains a factor of $2$, the number of summands is odd, so the sum can't be $0$.

If both contain a factor of $2$, let $m=2k$ and $n=2l$, and $i=2kla+2b+c$ with $a,c\in\{0,1\}$ and $0\le b\lt kl$. Then

$$
\left\lfloor\frac im\right\rfloor=\left\lfloor\frac{2kla+2b+c}{2k}\right\rfloor=2la+\left\lfloor\frac bk\right\rfloor
$$

and likewise

$$
\left\lfloor\frac in\right\rfloor=\left\lfloor\frac{2kla+2b+c}{2l}\right\rfloor=2ka+\left\lfloor\frac bl\right\rfloor\;,
$$

so we just get $4$ times the contributions for the case $k,l$ for the $4$ different pairs of values of $a$ and $c$.

Thus, by induction we can divide out common factors of $2$. That leaves the case where one of $m$ and $n$ is even and the other is odd. In this case $i\to mn-1-i$ flips the parity of the exponent, and thus the sign of the summand, so the sum is $0$.

Title: When is this sum equal to $0$?


问:
Why can points on the frontier of a convex set be identified through a minimization program.
Think, for instance, of a radius $1$ circle centered at $(1,1)$. If we want to identify the points on the southwestern quadrant of its circunference, one can solve the program

\begin{align}
\min &\quad w x +(1-w)y\\
\text{s.t.}&\quad\begin{cases}y\geq 1-\sqrt{1-(x-1)^2}\\x\geq 0\\y\geq 0\end{cases}
\end{align}

for $w\in(0,1)$, i.e, minimizing a convex combination of the components of the points belonging to the circle. For instance, $w=0$ results in $(1,0)$, $w=1$ in $(0,1)$ and $w=\frac{1}{4}$ in $(1-\frac{1}{\sqrt{10}},1-\frac{3}{\sqrt{10}})$. It's easy to check that all these lie on the circumference.

My question is why.

答: The slopes of the tangent lines of a decreasing function are in one-to-one correspondence with the interval $(0,1)$. When you minimize the linear function $wx+(1-w)y$ for a fixed $w$, you are finding the point on the corresponding line that is tangent to the curve. For the northeastern quadrant of the circle, you would instead maximize (equivalently, minimize the negative). For the other two quadrants, you would negate the coefficient of either $x$ or $y$ in the linear objective function.


问:
Show that $a_1a_2 \cdots a_k$ has order $1$ or $2$.

Let $G = \{a_1, \cdots , a_k \}$ be a finite abelian group.  Show that $a_1a_2 \cdots a_k$ has order $1$ or $2$.

So far I have done this:

I think that the product $a_1a_2 \cdots a_k$ will have order $1$ when none of the elements of $G$ is inverse of itself. since by reorganizing the elements in an adequate way we arrive at the identity. Otherwise, if for some $a_i$ in $G$ $a_i$ is inverse of itself, then the product will have order $2$. If so, how could you formalize this?

答:
For any finite abelian group $G=:\{g_1,\dots,g_k\}$, there are $n\in \Bbb N$ and positive integers $l_1,\dots,l_n$ such that $G\cong \Bbb Z_{l_1}\times\dots\times \Bbb Z_{l_n}=:\{a_1,\dots,a_k\}$, with $k=\prod_{i=1}^nl_i$ (fundamental theorem of finite abelian groups). In additive notation, we get:

\begin{alignat}{1}
\sum_{i=1}^{k}a_i &\stackrel{\dagger}{=} \biggl(l_2\dots l_n\sum_{i=0}^{l_1-1}i,\space\space\dots,\space\space l_1\dots l_{n-1}\sum_{i=0}^{l_n-1}i\biggr) \\
&= \biggl(l_2\dots l_n\frac{(l_1-1)l_1}{2},\space\space\dots,\space\space l_1\dots l_{n-1}\frac{(l_n-1)l_n}{2}\biggr) \\
\tag 1
\end{alignat}

whence:

\begin{alignat}{1}
2\sum_{i=1}^{k}a_i &= \bigl(l_2\dots l_n(l_1-1)\color{red}{l_1},\space\space\dots,\space\space l_1\dots l_{n-1}(l_n-1)\color{red}{l_n}\bigr) \\
&= (0,\dots,0)
\tag 2
\end{alignat}

Therefore, the element $\sum_{i=1}^{k}a_i$ has order at most $2$, and so does its "isomorphic image" $\sum_{i=1}^{k}g_i$.

$\dagger$ See here.


问:
How to calculate the perimeter of a polygon with n sides using chords.
This question comes from the first problem set of the MIT OCW calc course. It asks for the formula of the perimeter of a regular polygon with n sides with a radius of 1. The answer sheet has this for the answer

$$n\sin(2\pi/n)$$

If each wedge of the polygon creates a chord length of $2\sin(\theta/2)$ (the half-angle formula) and each wedge is $2\pi/n$ radians why is the answer not $2n\sin(\pi/n)$? I get the $(\pi/n)$ by multiplying $2\pi/n$ by $1/2$.

I know I must be missing something, but I'm not sure where I'm making the mistake.

答:
Assuming regular polygon,

$$ AB=u = R \cdot \sin \dfrac{\pi}{n} $$

perimeter

$$ 2u n = 2nR \sin \dfrac{\pi}{n} $$

when $n\to \infty $ perimeter $\to 2 \pi R$

Take $R=1$ here if so desired in this example.

It may be a typo error in the student's answer sheet.

Here the polygon drawn is for $n=8$ sides.


问:
Find the closed-form solution to a integral with the floor function
I have the following integral:

$$\int_0^1k^{\left\lfloor\frac{1}{x}\right\rfloor}dx$$

My question is, is there a nice closed form of this integral?

I have not any idea where to start.

Maybe I could use the substitution $u=\frac{1}{x}$, so I get:

$$\int_0^1k^{\left\lfloor\frac{1}{x}\right\rfloor}dx=-\int_\infty^1\frac{k^{\left\lfloor u\right\rfloor}}{u^2}du$$


问:
Uniform c.r.v and volume of slice (probability). two variable
I have this homework I'm struggling with.

Given Uniform
$$f_{X,Y}(x,y) = \begin{cases} \frac{1}{4} ,  \  -1\leq x\leq   1, \  -1\leq y\leq   1 \\ 0, \ \textrm{otherwise}  \\ \end{cases}$$

I need to find the probability of
$$ \mathbb{P}(X^{2}+Y^{2} \leq 1) $$

this what I'm thinking

The volume of this slice of a sphere with centrum (0,0,0) and r= 1 is the probability.

Is this assumption correct and if so how do I find the volume?
2

答:
A brief simulation in R provides useful graphs and an approximate answer.

```<code>set.seed(2020)
x = runif(10^5, -1,1)
y = runif(10^5, -1,1)
w = x^2 + y^2
mean(w < 1)
[1] 0.78491     # aprx P(W < 1)
pi/4
[1] 0.7853982   # exact
</code>```

The left panel below illistrates the uniform distribution of $(X,Y)$ in the square of area $4$ with vertices at $(-1,-1)$ and $(1,1).$ Orange points correspond to $P(W < 1).$

The right panel shows a histogram of 100,000 simulated values
of $W.$ The desired probability $P(W < 1)$ is the area of the
histogram to the left of the vertical orange line.

```<code>par(mfrow=c(1,2))  # enables two panels per plot
 plot(x, y, pch=".")
  points(x[w < 1],y[w < 1], col="orange")
 hist(w, prob=T, col="skyblue2")
  abline(v=1, col="orange", lwd=2)
par(mfrow=c(1,1))
</code>```

Can you find the density function of $W?.$

Body:
A brief simulation in R provides useful graphs and an approximate answer.

set.seed(2020)
x = runif(10^5, -1,1)
y = runif(10^5, -1,1)
w = x^2 + y^2
mean(w < 1)
[1] 0.78491     # aprx P(W < 1)
pi/4
[1] 0.7853982   # exact

The left panel below illistrates the uniform distribution of $(X,Y)$ in the square of area $4$ with vertices at $(-1,-1)$ and $(1,1).$ Orange points correspond to $P(W < 1).$

The right panel shows a histogram of 100,000 simulated values
of $W.$ The desired probability $P(W < 1)$ is the area of the
histogram to the left of the vertical orange line.

par(mfrow=c(1,2))  # enables two panels per plot
 plot(x, y, pch=".")
  points(x[w < 1],y[w < 1], col="orange")
 hist(w, prob=T, col="skyblue2")
  abline(v=1, col="orange", lwd=2)
par(mfrow=c(1,1))

Can you find the density function of $W?.$


答:
You may also use the Maclaurin expansion of $\cos x$ to avoid L’Hôpital’s:

$$\lim_{x\to 0} \frac{\int_0^x \left(1-\frac{t^4}{2!} +\dots\right) dt }{x} \\ =\lim_{x\to 0} \frac{x-\frac{x^5}{5\cdot 2!} +\dots}{x} \\ =\lim_{x\to 0} 1-\frac{x^4}{5\cdot 2!} +\dots \\ =1$$

Title: How to find the following limit?


问:
Is it true that $a \wedge b \wedge c \wedge d = (a \wedge b) \wedge (c \wedge d)$ for the exterior product?
Let us suppose we have a (finite-dimensional) vector space $X$ and let $\bigwedge^k X$ be its exterior product of order $k$. Then $a_1 \wedge a_2 \wedge \dots \wedge a_k \in \bigwedge^k X$ denotes the exterior product of $a_1, a_2, \dots, a_k \in X$.

Question: What is the relation between $\bigwedge^4 X$ and $\bigwedge^2 ( \bigwedge^2 X )$?

For example, suppose we have two exterior products $a \wedge b$ and $c \wedge d$. What is the relation between $(a \wedge b) \wedge (c \wedge d)$ and $a \wedge b \wedge c \wedge d$?

It seems that in $(a \wedge b) \wedge (c \wedge d)$ obeys the (anti)symmetries

$(a \wedge b) \wedge (c \wedge d) = -(b \wedge a) \wedge (c \wedge d)$

$(a \wedge b) \wedge (c \wedge d) = -(a \wedge b) \wedge (d \wedge c)$

$(a \wedge b) \wedge (c \wedge d) =  (c \wedge d) \wedge (a \wedge b)$

but I do not see any antisymmetry such as

$(a \wedge b) \wedge (c \wedge d) =  -(a \wedge c) \wedge (b \wedge d)$

I suspect I am getting confused about the notation. Are there two different types of wedge products in the play?

答:
The wedge product is usually defined to be associative. The fact that the expression $a\wedge c\wedge b\wedge d$ is well-defined already requires associativity, since otherwise we would need to specify the order of the wedge products.

That said, it is not the case that $\bigwedge^2(\bigwedge^2 X)\cong\bigwedge^4 X$. One easy way to see this is by dimension counting: Since $\dim(\bigwedge^kX)=\binom{\dim X}{k}$, we have $\dim(\bigwedge^2(\bigwedge^2 X))=3\binom{\dim X+1}{4}$ and $\dim(\bigwedge^4 X)=\binom{\dim X}{4}$, and these two dimensions are never equal unless both are zero. The issue is exactly as you suspect, there are two different wedge products, one introduced by each application of the $\bigwedge^2$ functor, and the "outer" one only acts on elements of $\bigwedge^2 X$.


问:
Bounding $P(|X-\mu|\geq \alpha)$ using $E[|X|]$
Given $ X $ is a random variable with $E[X] = \mu < \infty$, show that for every $\alpha \in \mathbb{R} $ $$ P(|X-\mu|\geq \alpha) \leq \frac{2E[|X|]}{\alpha} $$ I tried using Chebyshev's inequality, but I don't know that $\text{Var}(X)$ exists, and even if it does exist, it only gives me $$ P(|X-\mu|\geq\alpha) \leq \frac{\text{Var}(X)}{\alpha^{2}}=\frac{E[X^2]-\mu^2}{\alpha^{2}} $$Which doesn't seem to help me very much.

答:
Hint:

$$
E\big(|X-\mu|\big)\le E\big(|X|+|\mu|\big)
$$
by the triangle inequality.

(Note that you need to specify $\ \alpha>0\ $ for the inequality to be true, not merely $\ \alpha\in\mathbb{R}\ $, as you currently have)


问:
In how many ways $n$ dishes around a round table can be colored with 3 colors with adjacent dishes with different colors

Consider a round table with $n$ places. In each place there is a colored dish. In how many ways the dishes in this table can be colored with 3 colors, in such a way that adjacent dishes have different colors?

Solution: If $P(n)$ is the number of possibilities for a table with $n$ places, it is true that $$P(n)=P(n-1)+2P(n-2).$$By inspection it is easy to show that $P(2)=6$ and $P(3)=6$, and we can find the solution for any $P(n)$ by substitution using the recursion, or by the solution for the previous difference equation given by:
$$P(n)=2^n+2(-1)^n$$.

My problem: I don't understand why $$P(n)=P(n-1)+2P(n-2)$$ is true, or why, directly, $$P(n)=2^n+2(-1)^n$$ is true. Yes, I know how last equation is obtained from the previous equation, but cannot find a reasonable argument why it makes sense from a combinatoric argument. Any help?

答:
This will be a different approach and probably not the one you are looking for but still may be interesting so I am posting:

This problem can be considered as finding chromatic polynomial $\chi_{C_n}(k)$ of $C_n$ computed at $k=3$ where $C_n$ is an $n$-cycle graph. And we know an algorithm for finding chromatic polynomial of a graph, called Deletion-Contraction Formula. According to this formula, we have
$$\chi_{C_n}(k) = \chi_{C_n\cdot e}(k) + \chi_{C_n \backslash e}(k), \text{for any edge }e \in E(C_n)$$

And here, you can find chromatic polynomial of a cycle graph $C_n$ found by Deletion-Contraction Formula, which yields:
$$\chi_{C_n}(k) = (k-1)^{n} + (k-1)(-1)^{n}$$
So, for $k = 3$, we have
$$P(n) = \chi_{C_n}(3) = 2^n+2(-1)^n$$


问:
the existence of a maximun for $f(X)$ don't depend on *all* topologies of X
Let $X$ be a compact topological space, and $Y$ a topological space under the order topology. Let $f:X \rightarrow Y$ a continuous function under these topologies. Then, $f(X)$ has largest and smallest elements (*), by the extreme value theorem.

If I endow $X$ with other topology for which $f$ is not continuous or $X$ is not compact, it seems not to influence the fact (*), about $f(X)$, which I can assert for the first topology of X. So, is it sufficient to find one topology that fulfills compactness and continuity to assert the fact (*), or is there some detail that I'm forgetting?

答:
If I understand the question correctly, you have a set $X$, a linearly ordered space $Y$ with the order topology, and a function $f:X\to Y$, and you want to know whether the following statement is true:

If there is a topology on $X$ that makes $X$ compact and $f$ continuous, then $f[X]$ has a largest and a smallest element in the order on $Y$.

Yes, this is true. If there is such a topology on $X$, $f[X]$ is a compact subset of $Y$ and therefore must have a maximum and a minimum element.

This is not a necessary condition, however. Let $Y=[0,1)\cup(1,2]$ with the order topology, let $X=Y$, and let $f$ be the identity map. Then $f[X]$ has both a maximum and a minimum element, but there is no topology on $X$ that makes $X$ compact and $f$ continuous: if there were such a topology, $Y$ would be compact, and it isn’t.


问:
Does the relation $\exp(\log T)=T(\|T-I\|<1)$ hold in a Banach space?
In real analysis of one variable, we have , $e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$ for $x\in \mathbb{R}$ and $\log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}(|x|<1).$

In case of Banach space $E$ and for a bounded linear operator $T$ we can similarly define,
$\log(I+T)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}T^n}{n}(||T||<1)$ and $\exp(T)=\sum_{n=0}^{\infty} \frac{T^n}{n!}.$

Now my question is that, does the relation $\exp(\log T)=T(||T-I||<1)$ hold in this case?

I know it's true, when $T\in M_n(\mathbb{C})$ but my question is for any banach space $E$

答:
This can be done right away using Holomorphic Functional Calculus. On a disk of radius less than $1$, centered at $0$, your function $z\longmapsto \log(1+z)$ is analytic. From $1+z=\exp(\log(1+z))$ on such a disk, using that the functional calculus is a homomorphism we get
$$
I+T=\exp(\log(I+T)),\qquad \|T\|<1.
$$


问:
What is the interval of convergence of the Taylor series $f(x) = \ln(1-x^2)$?
What is the interval of convergence of the Taylor series $f(x) = \ln(1-x^2)$ centered at $x = 0$?

For $f(x) = \ln(1-x^2) = \ln(1-x) + \ln(1+x) $    I've found that the nth derivative for $x=0$ is $-(n-1)! + (-1)^{(n+1)}*(n-1)!$

Thus, the Taylor series centered at $x = 0$ is $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}(x-0)^n$ =  $\sum_{n=0}^\infty \frac{-(n-1)! + (-1)^{(n+1)}*(n-1)!}{n!}(x)^n$ = $\sum_{n=1}^\infty \frac{-1+ (-1)^{(n+1)}}{n}(x)^n$ = $0-x^2 +0 - \frac{x^4}{4}+\ldots$

Which is equal to $\sum_{n=1}^\infty \frac{-1}{2^{2n-1}}(x)^{2n}$, when $n$ is a even number and $=0$ when $n$ is an odd number.

Because of this, we can't apply the ratio test.

Then I tried to see if the series is absolutely convergent by putting it inside modulo. The resulting series is $\sum_{n=0}^\infty \frac{1}{2^{2n-1}}(x)^{2n}$

I applied the ratio test here:

$\lim_{n\to\infty} \frac{x_n}{x_{n+1}} $ = $\lim_{n\to\infty} \frac{x^{2n}}{2^{2n-1}} \frac{2^{2n-3}}{x^{2n+2}} = \lim_{n\to\infty} \frac{1}{4x^2} <1 $ for any value of  $x$. Thus the series is never absolutely convergent.

Because of this, I said that the interval of convergence is $\emptyset$.

Is this alright?

答: Since$$\log(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\cdots$$when $|x|<1$ and since $|x|<1\iff|x^2|<1$, you have$$\log(1-x^2)=-x^2-\frac{x^4}2-\frac{x^6}3-\cdots\tag1$$when $|x|<1$. It's easy to see that the series $(1)$ diverges when $|x|>1$ and therefore the radius of convergence is $1$.


问:
Why does the short exact sequence for projective module split?
I want to understand the definition of surjective module in terms of splitting sequence. The definition says for a projective $R$-module $P$, the following short exact sequence $$0 \to A \xrightarrow{f} B \xrightarrow{g} P \to 0$$ splits, where $A,B$ are also $R$-modules.

I want to see why $B \cong A+P$ ?

The Wikipedia, says, then there is a section map $h:P \to B$ such that $gh=1_P$.

$(1)$ Why is so ?

This then says $B=\text{Im}(h)\oplus \text{ker}(g)$.

$(2)$ why is so ?

I am trying in the following way:

Since the given sequence is short exact sequence, the map $g$ is surjective. This means that $\text{Im}(g)=P$.

Now as $P$ is surjective module any module homomorphism factors through an epimorphism to $B$   i.e.,  for any $R$-module $C$  there exists an epimorphism (surjective module homomorphism) $i: C \twoheadrightarrow  B$ and a module homomorphism $j: P \to C$  such that $$ij=g.$$
I can't go further. What is the section map $h$ here ?

Any explanation of above two questions ?

答:
Consider the diagram:
\begin{align}
&P \\
&\Big\|\operatorname{id_P} \\[-3ex]
0\longrightarrow A\xrightarrow{\enspace f\enspace} B \xrightarrow{\enspace g\enspace}&P\longrightarrow 0
\end{align}
As $P$ is projective, and $g$ is onto, the map $\operatorname{id}_P$ factors through $B$, i.e. there exists a map $s:P\longrightarrow B$ such that  $g\circ s=\operatorname{id}_P$.

Some more details:
Any element $b\in B$ can be written as the sum of an element in $f(A)$ and an element in $s(P)$:

Indeed you easily check that $g\bigl(b-s(g(b))\bigr)=g(b)-(g\circ s)\bigl(g(b)\bigr)=g(b)-g(b)=0$, so there exists a (unique because $f$ is injective) $a$ such that
$$b-s\bigl(g(b)\bigr)=f(a)\iff b= f(a)+s\bigl(g(b)\bigr).$$


问:
Calculate Expected Value in dice game.
We play a game in 2 stages:

In stage one, we throw a dice until we get number 6, Let N represent the number of times played until we got 6 for the first time.

In stage two, we throw N dices (each one only once).

Question: Let $X$ represent the sum of results we got in stage 2, calculate $E(X|N=n)$:

What I know?
I know that $N$ is $\operatorname{Geo}(1/6)$ and this $E(N)=1/(1/6)=6$ to continue I need to know the distribution of $X|N=n$, can I get help?

答:
If we throw $n$ dice, then the expected value of their sum is $3.5n$. This follows directly from the fact that the average score on one die is $3.5$ (and expectation is linear).

Let $A_i$ equal the outcome of the $i$th roll of the die. $E(A_i)$ can be calculated in the following way:$$\frac{1+2+3+4+5+6}{6}=3.5 \, .$$ Let $B$ equal the sum of $n$ rolls.
\begin{align}
E(B)&=E(A_1)+E(A_2)+\ldots+E(A_n) \\
&= \underbrace{3.5 + 3.5 + \ldots + 3.5}_{\text{$n$ times}} \\
&= 3.5n \, .
\end{align}

Title: Calculate Expected Value in dice game.


问:
Field extension with infinite degree
I'm currently working on the proof of the following statement:

Let $L$ be a field and $K$ and $F$ be two subfields such that $F\subseteq K\subseteq L$. Then the following holds for the field extensions $L/F, L/K$ and $K/F$:

$$L/F \ \ \text{is infinite iff.} \ \ L/K \ \ \text{is infinite or} \ \ K/F \ \ \text{is infinite}$$

I've managed to prove the implication from left to right, and, from right to left in the case of $K/F$ is infinite.

How do I prove the other case, i.e. that if $L/K$ is infinite then so is $L/F$?

答:
Suppose that $L/K$ is infinite but $L/F$ is finite.

Then $L$ admits a finite set of generators as a $F$-vector space.
These generarators would be also a set of generators as a $K$-vector space, since any linear combination with coefficients in $F$ is also a linear combination with coefficients in $K$. This contradicts the assumption.

Title: Field extension with infinite degree


问:
For commutative rings, if the ring of polynomials is a PID then the ring is an integral domain.
I am reading Dummit and Foote's proof that if $R$ is a commutative ring, and if $R[x]$ is a PID, then $R$ is a field.  At the start of the proof they write

Proof: Assume $R[x]$ is a Principal Ideal Domain. Since $R$ is a subring of $R[x]$ then
$R$ must be an integral domain (recall that $R[x]$ has an identity if and only if $R$ does). ...

I understand how $R$ has an identity if and only if $R[x]$ does, but I don't see how this tells us that $R$ is an integral domain.

To try to work it out myself, if $a,b\in R$ and $ab=0$ then we need to relate this to the other assumptions, so perhaps we consider the ideal $(a,b)$ in $R[x]$.  Then there is some $c \in R$ such that $(c) = (a,b)$ and therefore $c = am+bn$ for some $m,n\in R$.  Then

$$ abn = 0 $$

$$ a(c-am) = 0 $$

$$ ac = a^2m $$

This seems to me a dead-end.  We are not guaranteed by anything that I can see, that there is a unitary element or a unit in any of these rings.

答:
Remember that PID means Principal Ideal Domain.

If $A\subset B$ is an inclusion of rings and $B$ is a domain, $A$ must be as well because any relation
$$
xy=0
$$
in $A$ holds in $B$ too.


答:
Let the colours be red, green, blue. To have different colours, you can either go forward on th colour cycle (red$\to$ green$\to$blue$\to$red) or backwards. So you can pick any of three colours for the first field and then make $n$ binary choices (forward or backwards). This suggests that the answer is somehow related to $2^n$, though details are left out in this simplified argument.

For the recursion, if you remove the $n$th Place, you either obtain a valid colouring for $n-1$ (namely when the neighbours had different colours, and in this case you can reconstruct the removed colour), or you need to remove a duplicate colour and thereby obtain a valid colouring for $b-2$ places, but cannot reconstruct the original colour of the $n$th place (namely, there are two possibilities).

Convince yourself, that you can obtain every colouring of $n-1$ / $n-3$ places. This we conclude $$P(n)=P(n-1)+2P(n-2).$$


问:
explicit value of the integral.
If $I= \int_1^4(1+x)^{1-m}x^{-1} \, dx$ where $0<m<1.$ Is it possible to compute the explicit value of the convergent integral $I$.

答:
Mathematica gives:

$$\frac{\, _2F_1(m,m;m+1;-1)}{m}+4^{-m} \left(\frac{\left(\frac{2}{5}\right)^m \left(2\
   5^m-5\ 2^m\right)}{m-1}-\frac{\, _2F_1\left(m,m;m+1;-\frac{1}{4}\right)}{m}\right),$$

where $\mbox{}_2F_1$ is a hypergeometric function.

Title: explicit value of the integral.


问:
Composing continuous functions on abs.cont R.V
I'm wondering what are sufficient conditions I cas ask for a function $f:\mathbb{R} \mapsto \mathbb{R} $ s.t given ab abs.continous R.V $X:\Omega \mapsto \mathbb{R}$, the composition $ f \circ \Omega$ will remain an abs continous R.V .

Is it enough that f will be continous?

答:
If $A$ is a Borel set, then $\mathbb{P}\{F(X) \in A\}$ is given by
\begin{equation*}
\mathbb{P}\{f(X) \in A\} = \int_{F^{-1}(A)} f_{X}(x) \, dx.
\end{equation*}
Hence the question boils down to: when is it possible to change variables in the right-hand side of the expression above?

There is a somewhat well-known sufficient condition, though I haven't seen it written down formally anywhere.  Roughly speaking, if $F^{-1}(\{x\})$ is finite for every $x \in \mathbb{R}$, then $F(X)$ has the following PDF:
\begin{equation} \label{E: perron frobenius density}
f_{F(X)}(y) = \sum_{x \in F^{-1}(\{y\})} \frac{f_{X}(x)}{|F'(x)|}.
\end{equation}

When $f$ is a $C^{1}$ diffeomorphism of $\mathbb{R}$, this follows directly from the change-of-variables formula:
\begin{equation*}
\int_{F^{-1}(A)} f_{X}(x) \, dx = \int_{A} f_{X}(F^{-1}(y)) |(F^{-1})'(y)| \, dy = \int_{A} \frac{f_{X}(F^{-1}(y))}{|F'(F^{-1}(y))|} \, dy.
\end{equation*}

More generally, one can prove the following (e.g. using Sard's Theorem):

If $F : \mathbb{R} \to \mathbb{R}$ is a piecewise $C^{1}$ map (i.e.
$C^{1}$ on $\mathbb{R} \setminus D$, where $D$ is a closed set with no
accumulation points) with the property that $F^{-1}(N)$ is a Lebesgue
null set whenever $N$ is a Lebesgue null set, then, for any random
variable $X$ with absolutely continuous law, the random variable
$F(X)$ also has absolutely continuous law given by \begin{equation*}
 f_{F(X)}(y) = \sum_{x \in F^{-1}(\{y\})} \frac{f_{X}(x)}{|F'(x)|}.
 \end{equation*}

The case where $F$ is piecewise constant shows that the assumption concerning null sets is necessary to get absolute continuity of the law of $F(X)$.  (If $F$ is piecewise constant with countable range, then $F(X)$ has a singular law.)

In ergodic theory, the operator $f \mapsto \sum_{x \in F^{-1}(\{y\})} \frac{f_{X}(x)}{|F'(x)|}$ is called the Perron-Frobenius transfer operator; that's where I encountered it first.  I've also seen this formula (which is clearly of interest elsewhere) in a probability book for electrical engineers.


问:
Find the least order differential equation that accepts $y=a\cos{(\ln{x})}+b\sin{(\ln{x})}$ general solution.
There are two coefficients but I found the least order differential equation as first order one. When I put the given general solution to the differential equation that I found, It provides the equation.

Here's how I found the differential equation:

$$y=a\cos{(\ln{x})}+b\sin{(\ln{x})} \Rightarrow y\prime=-\frac{a}{x}\sin{(\ln{x})}+\frac{b}{x}\cos{(\ln{x})}$$

Multiply both sides with $\frac{ax}{b}$,

$$\frac{ax}{b}y\prime=-\frac{a^2}{b}\sin{(\ln{x})}+a\cos{(\ln{x})}$$

$$a\cos{(\ln{x})}=y-b\sin{(\ln{x})}\Rightarrow\frac{ax}{b}y\prime=-\frac{a^2}{b}\sin{(\ln{x})}+y-b\sin{(\ln{x})}$$

$$\frac{ax}{b}y\prime=y-\left(\frac{a^2+b^2}{b}\right)\sin{(\ln{x})}$$

Multiply both sides with $\frac{b}{ax}$ to simplify,

$$y\prime=\frac{b}{ax}y-\frac{a^2+b^2}{ax}\sin{(\ln{x})}$$

$$y\prime-\frac{b}{ax}y=-\frac{a^2+b^2}{ax}\sin{(\ln{x})}$$

The last formula is the first order differential equation that I found. When I put the given solution and it's derivative, it provides the eqution. Is my solution correct? Or should have I found second order differential equation for some reason, because there are 2 coefficients, a and b?

答:
In order to answer the question you address as a comment:

Let us work somehow backwards. If, in your first degree equation

$$y^\prime-\frac{b}{ax}y=-\frac{a^2+b^2}{ax}\sin{(\ln{x})}\tag{1}$$

you plug

$$y=A\cos{(\ln{x})}+B\sin{(\ln{x})}\tag{2}$$

(please note that I have taken variables capital $A,B$):

$$-A\sin{(\ln{x})}+B\cos{(\ln{x})}\dfrac{1}{x}-(A\cos{(\ln{x})}+B\sin{(\ln{x})})\frac{b}{ax}=-\frac{a^2+b^2}{ax}\sin{(\ln{x})}$$

otherwise said:

$$-aA\sin{(\ln{x})}+aB\cos{(\ln{x})}-bA\cos{(\ln{x})}-bB\sin{(\ln{x})}=-(a^2+b^2)\sin{(\ln{x})}$$

You see that, identifying the LHS and the RHS:

$$aA+bB=-(a^2+b^2)  \ \ \text{and} \ \ aB-bA=0$$

a linear system whose solutions are $A=a$ and $B=b$.

It is in this particular case

Therefore a possible answer to your question is that the set of functions (2) is the solution of the whole family of differential equations (1), for all possible $a,b$.


问:
The Galois type:a realization
What does it mean in terms of basic set-theoretic operations in the definition 1.0.25 that $M$ realizes $S(N)$ ?
All the relevant definitions are in the snippets, I believe.

As $N\preceq M$ it seems to me that $M$ always relaizes $S(N)$ ?
I.e. with $(N,N,a)\in K^3$, say. Here $a$ is any element in $N$.

答:
It looks like the context here is Abstract Elementary Classes, but actually the question already makes sense even in first-order logic.

We should think of $\operatorname{tp}(a, M, N)$ as the type of $a$ as over the parameters $M$ as computed in the model $N$.  Then the definition of $S(M)$ is that it is the collection of all types over $M$ in all possible elementary extensions of $M$.  The definition of $M$ being full over $N$ is then that every type over $N$ that can be realized in some elementary extension of $N$ is already realized in $M$.

I think the problem here is that the notation is a bit unfortunate - in Definition 1.0.24(2) the $N$ is being universally quantified, so when we later look at $S(N)$ the role of $N$ has changed.  $S(N)$ contains not only types of the form $\operatorname{tp}(a, N, N)$ (that is, types realized in $N$) but also types $\operatorname{tp}(a, N, N')$ for any $N'$ with $N \preceq N'$.

For a concrete example of something not full, in first-order logic it's never true that $N$ is full over $N$ because $N$ never realizes the type $\{x \neq n : n \in N\}$.

Title: The Galois type:a realization


问:
Prove that the matrix representing the tangent map $f_{*q_0} $ is the jacobian matrix of f

Proposition: Let $\Omega \subseteq \mathbb{R}^n$ an open set, $f: \Omega \rightarrow \mathbb{R}^m$ with $f \in C^1(\Omega)$, now $$(f_{*q_0}(\omega))(h) =\omega(h \circ f) \tag{1}$$
where $\omega \in T_{q_0} \Omega$ , $h:\mathbb{R}^m \rightarrow \mathbb{R}$ and $ f_{*q_0}: T_{q_0}\Omega \rightarrow T_{f(q_0)} \mathbb{R}^m$ is the tangent map of $f$ at $q_0$.

Problem: Using the previous proposition , prove that the matrix representing the tangent map $f_{*q_0} $ is the jacobian matrix.
Suggestion: Compute $f_{*q_0}(\frac{\partial}{\partial x_i})(y_j)$, where $(x_1,\dots,x_n)$ is a coordinate system over $\Omega$ and $(y_1,\dots,y_m)$ is a coordinate system over $\mathbb{R}^m$.

I think that the suggestion is wrong for two reasons:

The "$|_{q_0}$"is missing: It  should be $f_{*q_0}(\frac{\partial}{\partial x_i}|_{q_0})(y_j)$ because a basis of the space $T_{q_0}\Omega$ is given by $\big\{ \frac{\partial}{\partial x_i}|_{q_0}: i=1,\dots,n \big\}$ as far as I know.

In order to apply the proposition I need a function $h:\mathbb{R}^m \rightarrow \mathbb{R}$ and I don't think $y_j$ has that form because it is just a coordinate, how can a coordinate be a  function of $m$ variables? what would  its arguments be? Therefore the composition $y_j\circ f$ would not make sense

Can someone shed some light?

Somehow I have to find that $f_{*q_0}(\frac{\partial}{\partial x_i}|_{q_0})= \frac{\partial f_k}{\partial x_i}|_{q_0}\frac{\partial}{\partial y_k}|_{f(q_0)}$ and from that conclude that the matrix $A$ that represents the tangent mapping is $A_{ki}=\frac{\partial f_k}{\partial x_i}|_{q_0}$.

答:
For your two questions:

The "$|_{q_0}$"is missing: It  should be $f_{*q_0}(\frac{\partial}{\partial x_i}|_{q_0})(y_j)$ because a basis of the space $T_{q_0}\Omega$ is given by $\big\{ \frac{\partial}{\partial x_i}|_{q_0}: i=1,\dots,n \big\}$ as far as I know.

Usually this is omitted from notation, and it is understood from the context. So yes, technically speaking it should be $\left. \frac{\partial}{\partial x_i} \right|_{q_0}$, but you should understand that writing $\frac{\partial}{\partial x_i}$ can often mean either the vector field or at a specific point, depending on context.

In order to apply the proposition I need a function $h:\mathbb{R}^m \rightarrow \mathbb{R}$ and I don't think $y_j$ has that form because it is just a coordinate, how can a coordinate be a  function of $m$ variables? what would  its arguments be? Therefore the composition $y_j\circ f$ would not make sense

Coordinates make perfect sense as functions. The coordinate "$y_j$" really means the function
$$ h(y_1,y_2,\dots,y_m) = y_j $$
That is a perfectly valid function of $m$ variables: it just spits out the $j^\mathrm{th}$ input. A function $f \colon \Bbb{R}^n \to \Bbb{R}^m$ is given by $m$ functions of $n$ variables:
$$ f(x_1,\dots,x_n) = \Big( f_1(x_1,\dots,x_n),\dots,f_m(x_1,\dots,x_n) \Big) $$
The composition is then just $y_j \circ f = f_j$.