问:
Is this $3 \times 3$ matrix with $2$ eigenvalues diagonalizable
$$
    \begin{bmatrix}
    1 & 1 & 0 \\
    3 & 0 & 1 \\
    1 & -1 & 2 \\
    \end{bmatrix}
$$

Is the $3$ by $3$ matrix above diagonalizable given the eigenvalues $-1$ and $2$? There are only $2$ eigenvalues so the one with the multiplicity two needs a full rank eigenspace for this to be diagonalizable. I think there should be a way to tell from the matrix alone whether that's possible without doing all the row operations, but I'm not sure what I should be looking for.

答:
A square matrix is diagonalizable if it has a complete set of eigenvectors. A complete set of eigenvalues guarantees you a complete set of eigenvectors, but having repeat eigenvalues means you need to check the eigenvectors for those eigenvalues.

The characteristic polynomial of your equation is:

$$\begin{split}0&=-x^3+3x^2-4&\\
&=-x^3+4x^2-4x-x^2+4x-4\text{ add and subtract 4x, also split the $3x^2$ in two parts}\\
&=-(x^3-4x^2+4x)-(x^2-4x+4)\\
&=-x(x^2-4x+4)-(x-2)^2\\
&=-(x+1)(x-2)^2\end{split}$$

Therefore the eigenvalue with multiplicity 2 is 2, and we need to check the eigenvectors.

$$\left(\begin{matrix}-1&1&0\\3&-2&1\\1&-1&0\end{matrix}\right)\left(\begin{matrix}v_1\\v_2\\v_3\end{matrix}\right)=0\\
\begin{cases}v_1=v_2\\
3v_1-2v_2+v_3=0\end{cases}\\
\left(\begin{matrix}v_1\\v_1\\-v_1\end{matrix}\right)\implies\left(\begin{matrix}1\\1\\-1\end{matrix}\right) \text{ is the only eigenvector for $\lambda=2$}$$

So it is not diagonalizable.

Sorry, I don't know a faster way.


问:
There are infinitely many choices of $(\alpha_1,\dots,\alpha_n)$ such that $f(\alpha_1,\dots,\alpha_n)\neq 0$
I'm trying to solve this exercise in the page 10 of this book

Maybe I'm forgetting something, but I couldn't solve this exercise, I need a hint or something to begin to solve this question.

Thanks in advance

答:
Deal  with the case $n = 1$ first. Here you use that a non-zero polynomial in one variable over a field has only finitely many roots.

When $n > 1$, argue by induction.

(Thanks to user78535 for suggesting some extra care here.)

If $X_1$ does not appear in $f$, then we have a polynomial in $n-1$ variables, and induction applies.

If $X_1$ does appear in $f$, write $f$ as a polynomial in $X_1$, with coefficients in $K[X_2, \dots, X_n]$. Clearly the coefficient of one of the powers $X_1^i$, for $i > 0$ has to be non-zero as a polynomial. Thus by induction there are infinitely many choices of $(\alpha_{2}, \dots, \alpha_{n}) \in (K \setminus \Lambda)^{n-1}$ for which this coefficient is non-zero as an element of $K$, when evaluated at $(\alpha_{2}, \dots, \alpha_{n})$.

Now appealing to the case $n = 1$, for each such choice of $(\alpha_{2}, \dots, \alpha_{n})$ there are infinitely many choices of $\alpha_{1}$ such that $f(\alpha_{1}, \dots, \alpha_{n}) \ne 0$.

ContentLicense: CC BY-SA 3.0


ContentLicense: CC BY-SA 4.0

Title: How can I prove that the results in the following sequence will always share the same factor?

Body:
The following sequence is built of $((A^z)^x -1)/6$ where $z$ increases by $1$ and the conditions for $A$ must be that $(A - 1) \mod 6 = 0$ and for $x$ must be that $x$ is odd and $x > 1$:

$((A^1)^x -1)/6 $ , $((A^2)^x-1)/6$ , $((A^3)^x-1)/6$ , $((A^4)^x-1)/6$ ...

How can i prove that the results in the sequence will always have the same factor?

Examples:

$(7^3-1)/6 =57$ , $(49^3-1)/6 =19608$ , $(343^3-1)/6 = 6725601$....

$57, 19608, 6725601$ all have a factor of $19$

$(7^5-1)/6 =2801$ , $(49^5-1)/6 = 47079208$ , $(343^5-1)/6 = 791260261657$....

$2801, 47079208, 791260261657$ all have a factor of $2801$

$(13^3-1)/6 =366$ , $(169^3-1)/6 =1 804468$ , $(2197^3-1)/6 = 1767416562$....

$366, 804468, 1767416562$ all have a factor of $61$

Body:
Hint:

Use $a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)$ with $a=A^x$.


问:
Need some assistance on the expression $\sum_{n=0}^\infty \frac{n!x^n}{\prod_{j=1}^n(1+jx)}$
In this thread, Markus Scheuer attempted to prove that:

$$\sum_{n=0}^\infty  \frac{n!x^n}{\prod_{j=1}^n(1+jx)}=\frac{1}{1-x}\tag{1}$$

He wrote as followed:

$$\sum_{n=0}^\infty  \frac{n!\color{red}{x^n}}{\color{green}{\prod_{j=1}^n(1+jx)}}=\sum_{n=0}^{\infty}\frac{n!}{\left(\color{blue}{1+\frac{1}{x}}\right)^{\overline{n}}}\tag{2}$$$$=\sum_{n=0}^{\infty}\frac{\color{purple}{1^{\overline{n}}1^{\overline{n}}}}{\left(1+\frac{1}{x}\right)}\,\color{ForestGreen}{\frac{1}{n!}}\tag{3}$$

where $q^{\overline{n}}=q(q+1)(q+2)...(q+n-1)$

I don't understand several points here:

First, on line $2$, why do $\color{red}{x^n}$ disspear? Did he multiply the denominator and numerator with $\dfrac{1}{x^n}$, if so, how can we multiply $\dfrac{1}{x^{n}}$ with the product $\prod_{j=1}^{n}(1+jx)$?, if so, isn't the product is going to be $\prod_{j=1}^{n}\left(\dfrac{1}{x^n}+\dfrac{j}{x^{1-n}}\right)$? I suspect that he factored the $x$ from the denominator, but then why $x^n$ disappear?

Second, how do you covert the product $\prod_{j=1}^{n}(1+jx)$ into $\left(1+\dfrac{1}{x}\right)^{\overline{n}}$? I try to write down to get a feel of it:

$\left(1+\dfrac{1}{x}\right)^{\overline{n}}=(1+\dfrac{1}{x})(2+\dfrac{1}{x})(3+\dfrac{1}{x})...$

The look similar to the product, but after factoring out $x$

Why can $\color{purple}{1^{\overline{n}}1^{\overline{n}}}$ replace $n!$ ? Also, why there is a new $\color{forestgreen}{\dfrac{1}{n!}}$

答:
Yes, he did do that but no, notice that you have $n$ copies of $x$ i.e., $x^n,$ if you put the $x^n$ inside the product like that, then you will have $x^{n^2}$ so $n^2$ copies. Instead, put one each per term in the product, so
$$\prod  _{j=1}^n\left (\frac{1+jx}{x}\right )=\prod  _{j=1}^n\left (\frac{1}{x}+j\right )$$

solves this mystery.

Cause $x^{\overline{n}}=x(x+1)\cdots (x+n-1),$so replacing $x=1,$ you get
$$1\cdot (1+1)\cdot (1+2)\cdots (1+n-1)=1\cdot 2\cdots (n-1)\cdot n=n!,$$
notice that there are two $1^{\overline{n}},$ so he multiplied and divide by $n!.$


问:
maximal bound for OU process
I would like to obtain the finiteness of the following expectation involving the OU process:

let $X_t = e^{-\alpha t} x + (1-e^{-\alpha t}) \mu + e^{-\alpha t} \int_{0}^{t} e^{\alpha u} \hspace{1mm} dB_u $ ,
where $ x $ and $ \mu $ are fixed constants;

for $\lambda > 0$, we would like to show

$E[\hspace{1mm} \sup_{t \in [0, \infty)} \hspace{1mm} e^{-\lambda t}\mid X_t \mid \hspace{1mm}] < \infty $.

A quick observation is that it is enough to show

$E[\hspace{1mm} \sup_{t \in [0, \infty)} \hspace{1mm} e^{-(\lambda + \alpha) t} \mid \int_{0}^{t} e^{\alpha u} \hspace{1mm} dB_u \mid \hspace{1mm}] < \infty $.

The above is a family of one dimensional Gaussian process; I am wondering if there is a direct argument available, or any existing inequality that would give us the desired bound.

答:
Allow me to answer my own question.

We can take $\lambda = 0$.

Note that $e^{−αt} \int_{0}^t e^{\alpha u} \hspace{1mm} dB_u$ defines a local martingale (since the integrand is deterministic, hence predictable).

Now we finish the proof by applying the BDG inequality with $p=1$ for $t \in [0,T]$ with some fixed $T$, to obtain a bound of the supremum of the stochastic integral by the square root of its quadratic variation on $[0,T]$; noting that the bound can be made uniform over $T$; we conclude by sending $T $ to infinity.

Title: maximal bound for OU process


问:
Choosing on the Number Line
Suppose a real number is chosen from the interval [-1, 1].

My homework asks to find the probability that the number is positive, and using geometric probability, I get 1/2.  Simple, right?

However, the next question asks what the probability of choosing a nonnegative real number is, and I do understand it is also 1/2

After finishing my homework, I'm wondering: Is it possible to choose 0, and what would the probability of choosing 0 be?

答: If the distribution is uniform, the probability of choosing $0$ from $[-1, 1]$ is zero. But you may say, if the probability of choosing any given number is zero, then the sum of all the probabilities would also be zero; but remember we are dealing with infinity.

Title: Choosing on the Number Line


问:
Show that $\frac12\|A\|_2\leq \sup_{\|u\|_2=1}|u^HAu|$.
Let $A\in\mathbb{C}^{n\times n}$, show that
$$\frac12\|A\|_2\leq \sup_{\|u\|_2=1}|u^HAu|,$$
where $\|A\|_2$ is the 2-norm of $A$.

答:
1.)  all vectors are constrained to have a length (2 norm) of one

2.) for notational simplicity, I assume WLOG that the Hermitian part of $A$  has a maximal singular value at least as large as the maximal singular value of the skew-hermitian part of $A$.  (If this wasn't true, the only alteration would be on the 3rd to last line, we'd change the plus sign to a minus sign)

$\Big \Vert A \Big \Vert_2$

$= \Big \Vert \frac{1}{2}\big(A+A^*\big) + \frac{1}{2}\big(A-A^*\big) \Big \Vert_2 $

$\leq \frac{1}{2}\Big \Vert \big(A+A^*\big)\Big\Vert_2 + \frac{1}{2}\Big\Vert \big(A-A^*\big) \Big \Vert_2 $

$= \max_{x} \frac{1}{2}\Big\{\Big \vert \mathbf x^*\big(A+A^*\big)\mathbf x\Big \vert\Big\}  + \max_{y} \frac{1}{2}\Big\{\Big \vert \mathbf y^*\big(A-A^*\big)\mathbf y\Big \vert\Big\} $

$\leq 2 \cdot \max_{x} \frac{1}{2}\Big\{\Big \vert \mathbf x^*\big(A+A^*\big)\mathbf x\Big \vert\Big\}   $

$\leq 2 \cdot \max_{x} \frac{1}{2}\Big\{\Big \vert \mathbf x^*\big(A+A^*\big)\mathbf x +\mathbf x^*\big(A-A^*\big)\mathbf x\Big \vert\Big\}   $

$= 2 \cdot \max_{x} \Big \vert \mathbf x^*A\mathbf x \Big \vert  $

justifications:

sub-additivity of operator 2 norm, normal matrices have maximal singular values given by modulus of a quadratic form.  Finally recall that a quadratic Hermitian form is purely real and a quadratic skew-hermitian form is purely imaginary


问:
Approximate standard error of the maximum likelihood estimate

Question: Suppose that $X$ is a discrete random variable with :

\begin{array}{|r|r|r|r|}
 \hline
         X &   0 &  1 &   2 &  3\\ \hline
         \mathbb P(X=x) &\frac{2}{3}\theta &\frac{1}{3}\theta &\frac{2}{3}(1-\theta) &\frac{1}{3}(1-\theta) \\ \hline
\end{array}
Where $0\leq\theta\leq1$ is a parameter. The following $10$ independent observations were taken from such a distribution: $\{3,0,2,1,3,2,1,0,2,1\}$

(a) Find the method of moments estimate of $\theta$.

(b) Find an approximate standard error for your estimate.

(c) What is the maximum likelihood estimate of $\theta$ ?

(d) What is an approximate standard error of the maximum likelihood estimate?

I figure out (a+b), $\theta$ using method of moment estimate and got, $\bar\theta=\frac{7}{3}-\frac{1}{2}\mathbb{E}(\bar{X})$ and $$Var(\bar\theta)=-\frac{1}{10}\theta^2+\frac{1}{10}\theta+\frac{1}{180}$$
Using the above information and with Log Likelihood function I also manage to find out (c), $MSE_{\bar{\theta}}=\frac{1}{2}$

for (d), I have no idea. What I only know that is,
$$\sqrt{n}(\bar{\theta}-\theta_0)⟶N\left(0,\frac{1}{I(\theta_0)}\right)\text{ Asymptotic normality of MLE}$$
But I really don't understand it truly, and how to solve (d) (Using it or other way)?

答:
(a)

Your MoM estimator doesn't look right. In fact, being

$$\mu=\frac{\theta}{3}+\frac{4}{3}(1-\theta)+ (1-\theta)=\frac{7}{3}-2\theta$$

You get

$$\hat{\theta}_{MM}=\Bigg[\frac{7}{3}-\overline{X}_{10}\Bigg]\frac{1}{2}$$

That is

$$\hat{\theta}_{\text{estimate}}=\frac{5}{12}$$

(b)

Variance of MoM estimator looks fine, anyway  the question is not asking MSE but just an estimation of standard error , that is standard deviation of the statistic sample distribution: $\approx 0.173$

(c) and (d): do the same procedure with ML estimation. Given that you know the asymptotic distribution of MLE I think it is not difficult for you to conclude.


问:
If $X$ is normally distributed, find a number $x_{0}$ such that $f\left(x\right)$
If $X$ is normally
distributed with $\mu=2$ , $\sigma=2$., find a number $x_{0}$ such that $f\left(x\right)$

a) $$P\left(X>x_{0}\right)=.10$$
My try:
$$Z=\frac{X-\mu}{\sigma}$$ $$P\left(Z>\frac{x_{0}-2}{2}\right)=1-X=.10$$
Where $X=\frac{x_{0}-2}{2}$, solving for $X$,$$X=0.9$$
$$\frac{x_{0}-2}{2}=0.9$$ Checking the table values for the normal distribution $$x_{0}=3.8$$
b) $$P\left(X>-x_{0}\right)=.20$$
My try:
$$\frac{-x_{0}-2}{2}=0.8$$
$$x_{0}=-3.6$$
$$P\left(Z>\frac{-3.6-2}{2}\right)=1-2.8=-1.85=0.08851$$

Can you please provide me some feedback, for both solutions, I tried to check if the values are correct but I don't get the same value for b)?

答:
You have to apply the normal distribution function. The equation is $P(X>x_0)=1-P(X<x_0)$ (converse probability).

This is the same as $P(X>x_0)=1-\Phi\left(\frac{x_0-2}{2} \right)=0.1$. Here $\Phi\left(z\right)$ is the cdf of the normal distribution. Solving the equation for $x$. Adding $\Phi\left(\frac{x_0-2}{2} \right)$

$1=0.1+\Phi\left(\frac{x_0-2}{2} \right) \quad |-0.1$

$1-0.1=\Phi\left(\frac{x_0-2}{2} \right) \quad $

$0.9=\Phi\left(\frac{x_0-2}{2} \right) \quad $

Now we have to use the inverse function in order to isolate $x_0$.

$\Phi^{-1}\left(0.9\right) =\frac{x_0-2}{2} \quad $

Now you look at a table and look where the function is approximately $0.9$. This is at $\Phi\left(1.28\right)= .89973\approx 0.9$. So $\Phi^{-1}\left(0.9\right)=1.28$. The equation becomes

$$1.28=\frac{x_0-2}{2} \quad $$

Is it comprehensible up to here? If yes, then I think you can go on by yourself.


问:
Lasker Nim, Sprague-Grundy Function Proof
It's been stated that the Sprague-Grundy function of Leskar's Nim is as follows:

$g (4k + 1) = 4k + 1\\ g (4k + 2) = 4k + 2\\ g (4k + 3) = 4k + 4\\ g (4k + 4) = 4k + 3$

The strategy to prove this claim is by induction, however I'm rather confused on how some claims are used in the proof of my textbook without much explination.

For the first form:

the followers of $4k + 1$ that have a single pile have Sprague-Grundy
values from $0$ to $4k$. Those that have two piles, $(4k, 1), (4k −1,2), . . . , (2k + 1, 2k)$, have even Sprague-Grundy values, and
therefore $g(4k + 1) = 4k + 1$.

I understand the values for moves that result in a single pile, but for moves that result in two piles, how do we know that the value of the function is even? We would have to know something about whether the the function evaulated at each of those two piles are both even or both odd, that makes the nim-sum even. But we don't know much about that.

Similar claims are given for the other proofs such as making claims on whether the function is odd or divisible by 4

How can we go by doing this?

答:
Suppose that the result is true for all $n<4k+1$; that’s the induction hypothesis. If we split $4k+1$ into two piles, the possibilities are of the forms $\langle 4\ell,4m+1\rangle$, $\langle 4\ell+1,4m\rangle$, $\langle 4\ell+2,4m+3\rangle$, and $\langle 4\ell+3,4m+2\rangle$, where $\ell+m=k$ in the first two, and $\ell+m=k-1$ in the last two. In each case both piles have fewer than $4k+1$ chips, so we can apply the induction hypothesis.

$\langle 4\ell,4m+1\rangle$: $g(4\ell)=g(4(\ell-1)+4)=4(\ell-1)+3$, and $g(4m+1)=4m+1$; both numbers are odd, so their Nim sum is even.

$\langle 4\ell+1,4m\rangle$: This is the same, but with the piles reversed.

$\langle 4\ell+2,4m+3\rangle$: $g(4\ell+2)=4\ell+2$, and $g(4m+3)=4m+4$; both numbers are even, so their Nim sum is even.

$\langle 4\ell+3,4m+2\rangle$: This is the same, but with the piles reversed.

Thus, every two-pile follower of $4k+1$ has an even Sprague-Grundy value. These values are at most $4k$, so $g(4k+1)=4k+1$.

The arguments for $4k+2,4k+3$, and $4k+4$ are similar. For $4k+2$, for instance, the possible divisions into two piles are of the forms $\langle 4\ell,4m+2\rangle$ and the same with the piles reversed, $\langle 4\ell+1,4m+1\rangle$, and $\langle 4\ell+3,4m+3\rangle$.

$\langle 4\ell,4m+2\rangle$: $g(4\ell)=g(4(\ell-1)+4)=4(\ell-1)+3$, and $g(4m+2)=4m+2$; the first is odd and the second, even, so their Nim sum is odd.

$\langle 4\ell+1,4m+1\rangle$: $g(4\ell+1)=4\ell+1$, and $g(4m+1)=4m+1$; both have binary expansions ending in $01$, so their Nim sum has a binary expansion ending in $00$ and is therefore a multiple of $4$.

$\langle 4\ell+3,4m+3\rangle$: $g(4\ell+3)=4\ell+4$, and $g(4m+3)=4m+4$; both are multiples of $4$, so their binary expansions end in $00$, and so does the binary expansion of their Nim sum, which is therefore a multiple of $4$.

Thus, every two-pile follower of $4k+2$ has a Sprague-Grundy value that is either odd or a multiple of $4$ and therefore is at most $4k+1$, so that $g(4k+2)=4k+2$.


问:
Notation for adding and taking away subsets in a family of subsets
Let $S:=\{A_1,A_2,\ldots,A_k\}$ for $A_i\subset\{1,\ldots,n\}$. I'm trying to figure out the correct notation to refer to $S_1:=\{A_1,A_2,\ldots,A_k,A_{k+1}\}$ and $S_2:=\{A_1,A_2,\ldots,A_{k-1}\}$. Is $S_1=S\cup{A_{k+1}}$ and $S_2=S\setminus{A_k}$ or rather $S_1=S\cup\{A_{k+1}\}$ and $S_2=S\setminus\{A_k\}$? I'm just unsure because the $A_i$ are subsets and not elements of $\{1,\ldots,n\}$. Also, if I wanted to denote the family $\{\{1,2\},\{1\}, \{2\},\ldots,\{n\}\}$ would it be $\{1,2\}\cup\left(\bigcup_{1\leq{i}\leq{n}}\{i\}\right)$ or $\{\{1,2\}\}\cup\left(\bigcup_{1\leq{i}\leq{n}}\{\{i\}\}\right)$ or something else? Thanks.

答: Your second assumption where we had $S_1 = S \cup \{A_{k+1}\}$ and $S_2 = S \setminus \{A_k\}$ is correct. Sure, the $A_i$ are sets themselves, but $S_1$ and $S_2$ are sets of sets. If we want to add or remove elements to (or from) the $S_i$, we also need to add or remove a set that contains sets (the $A_i$) as its elements. A similar logic would be true for your second question. You would want $$\{\{1,2\}\} \cup \left(\bigcup_{i=1}^n \{\{i\}\} \right)$$


答:
Your second choices are correct in all three cases:
$$
S_1 = S \cup \{A_{k+1}\} \\
S_2 = S \setminus \{A_k\} \\
\{\{1,2\},\{1\}, \{2\},\dots,\{n\}\} = \{\{1,2\}\}\cup\left(\bigcup_{i=1}^n\{\{i\}\}\right)
$$
But I think the versions with $\dots$ are the clearest.  Forget about $S,S_1,S_2$, and instead define
$$S_k := \{A_1,A_2,\dots,A_k\}$$

Title: Notation for adding and taking away subsets in a family of subsets

Body: Let $S:=\{A_1,A_2,\ldots,A_k\}$ for $A_i\subset\{1,\ldots,n\}$. I'm trying to figure out the correct notation to refer to $S_1:=\{A_1,A_2,\ldots,A_k,A_{k+1}\}$ and $S_2:=\{A_1,A_2,\ldots,A_{k-1}\}$. Is $S_1=S\cup{A_{k+1}}$ and $S_2=S\setminus{A_k}$ or rather $S_1=S\cup\{A_{k+1}\}$ and $S_2=S\setminus\{A_k\}$? I'm just unsure because the $A_i$ are subsets and not elements of $\{1,\ldots,n\}$. Also, if I wanted to denote the family $\{\{1,2\},\{1\}, \{2\},\ldots,\{n\}\}$ would it be $\{1,2\}\cup\left(\bigcup_{1\leq{i}\leq{n}}\{i\}\right)$ or $\{\{1,2\}\}\cup\left(\bigcup_{1\leq{i}\leq{n}}\{\{i\}\}\right)$ or something else? Thanks.


问:
Limiting Sum including Stirling numbers
I would like to simplify the following summation including $S(m,n)$, Stirling numbers of the Second kind.

$\begin{gather*}\\
&&
\frac{1}{n^{n+1}}\sum_{k=1}^{n-1}\binom{n}{n-k}S\left( n,n-k\right)
\left( n-k\right) !k\left( 1-\left( \frac{n-2}{n-1}\right) ^{k}\right)
&&
\end{gather*}$

I conjecture (in fact, computations show) that this sum approaches to $\frac{1}{e}(1-e^{-\frac{1}{e}})$ as $n$ approaches to $\infty$  . However, I do not have a proof.

It would be great if this can be simplified for any $n$ (I am not sure whether this is possible) or at least show the limiting result.

答:
We get for the sum term

$$\sum_{k=1}^{n-1} {n\choose k} {n\brace n-k} (n-k)!
\times k x^k
\\ = n \sum_{k=1}^n
{n-1\choose k-1} {n\brace n-k} (n-k)! \times x^k
\\ = n x \sum_{k=0}^{n-1}
{n-1\choose k} {n\brace n-1-k} (n-1-k)! \times x^k
\\ = n x \times n! [z^n] \sum_{k=0}^{n-1}
{n-1\choose k} (\exp(z)-1)^{n-1-k} x^k
\\ = n x \times n! [z^n] (\exp(z)-1+x)^{n-1}.$$

We find with $x=1$

$$n \times n! \times [z^n] \exp((n-1)z)
= n \times (n-1)^n.$$

We obtain for $x=(n-2)/(n-1)$

$$n \times  \frac{n-2}{n-1} \times
 n! [z^n] (\exp(z)-1/(n-1))^{n-1}.$$

Now the coefficient extractor is

$$\frac{1}{2\pi i}
\int_{|z|=1} \frac{\exp((n-1)z)}{z^{n+1}}
(1-\exp(-z)/(n-1))^{n-1} \; dz.$$

With $n$ large this becomes

$$\frac{1}{2\pi i}
\int_{|z|=1} \frac{\exp((n-1)z)}{z^{n+1}}
\exp(-\exp(-z)) \; dz.$$

We now apply the Saddle point Algorithm for Cauchy Coefficient Integrals
from pages 548 and 553 of Analytic Combinatorics by Flajolet and
Sedgewick.  (Theorem VIII.3) Our function $f(z)$ is

$$f(z) = (n-1)z-(n+1)\log z -\exp(-z)$$

and

$$f'(z) = (n-1) - (n+1)\frac{1}{z} + \exp(-z)$$

as well as

$$f''(z) = (n+1)\frac{1}{z^2} - \exp(-z).$$

Now let $\zeta_n$ be the positive real root of the saddle point equation
$f'(\zeta_n) = 0$ so that

$$\zeta_n (n-1) - (n+1) + \zeta_n \exp(-\zeta_n) = 0$$

and

$$\zeta_n = \frac{n+1}{n-1+\exp(-\zeta_n)}.$$

This implies

$$\frac{n+1}{n} \lt \zeta_n \lt \frac{n+1}{n-1}$$

so that with $n$ large $\zeta_n \approx 1.$ We get
for the asymptotics from equation 19 on the cited page

$$\frac{\exp(f(\zeta_n))}{\sqrt{2\pi f''(\zeta_n)}}$$

Now $f''(\zeta_n) = (n+1)/\zeta_n^2+(n-1)-(n+1)/\zeta_n$
and this evaluates to

$$\frac{\exp(n-1)\exp(-\exp(-1))}{\sqrt{2\pi (n-1)}}$$

Collecting everything from the two pieces we find
for the first piece

$$\frac{1}{n^{n+1}} \times n\times (n-1)^n
= \left(1-\frac{1}{n}\right)^n \sim \exp(-1)$$

and for the second piece

$$\frac{1}{n^{n+1}} \times n \frac{n-2}{n-1} n! \times
\exp(-1) \frac{\exp(n)\exp(-\exp(-1))}{\sqrt{2\pi (n-1)}}
\\ = \left(1-\frac{1}{n-1}\right)
\exp(-1) \exp(-\exp(-1))
\frac{n!}{n^n} \frac{\exp(n)}{\sqrt{2\pi (n-1)}}
\\ \sim \left(1-\frac{1}{n-1}\right)
\exp(-1) \exp(-\exp(-1))
\frac{n!}{n^n} \frac{\exp(n)}{\sqrt{2\pi n}}.$$

We apply Stirling's formula to conclude and get

$$\left(1-\frac{1}{n-1}\right)
\exp(-1) \exp(-\exp(-1))
\sim \exp(-1) \exp(-\exp(-1)).$$

We join the two pieces to obtain

$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{e}\left(1-e^{-1/e}\right)}$$

as proposed.

Title: Limiting Sum including Stirling numbers


问:
complex analysis -examples of complex functions that are bounded and the limits
What are some examples of complex functions that are bounded and the limits does not exists as $z\to 0$?</p>

答:
$f(z)=\frac{z}{|z|}$.

Also, let $g(z)$ be any analytic function with $g(0)\neq 0$ and let

$$h(z)=\frac{z}{|z|}g(z) \,.$$


问:
Insertion Sort Running Time Complexity
In the worst case, $j$ comparisons are required to insert the $jth$ element into the correct position. For the insertion sort algorithm, the running time complexity would be $\mathcal{\Theta}(n^2)$ for the number of comparisons as follows:

$$2+3+\textit{...}+n=\frac{n(n-1)}{2}-1$$

Question: Why the number of comparisons start from $2$ above although I think it should start from $1+2+3\textit{...}$?

答:
Cause it does it twice. First it asks: is $a_2>a_1,$ the answer is yes, then it does $i=i+1,$ so $i=2$ now and then it asks $a_2>a_2.$ is false and it breaks.

As you see there were two comparisions.

Title: Insertion Sort Running Time Complexity


问:
A set has the property that $A\subset B$ holds for any set $B$. Prove that $A=\emptyset$

A set has the property that $A\subset B$ holds for any set $B$. Prove that $A=\emptyset$.

In the question, do we know that empty set is  subset of for all set? If yes, then the question is clear. But, if no, then how can I prove? Can you help, hint?

答:
For any set B, we are given that A ⊂ B. Assume B = ∅ to show a contradiction.

A ⊂ ∅ means x ∈ A ⟹ x ∈ ∅, which is false.

If you reference the truth table for implication, the only way this would imply true is if x ∈ A is False. So if x is not an element of A then x = ∅.


问:
Showing that a number is a perfect square
How do I show that A=999...982000...081 is a square number when the oly information i have is that there are just as many 9s as there are 0s (with the number being greater than or equal to 4) ?</p>

答:
Actually the mathematical representation of your number is $$a_k=(10^k-1)\times10^{k+4}+82\times10^{k+2}+81$$ To show that this is a perfect square, let's get some hints:

If $a_k$ is to be equal to some $x^2$, then $\big((10^{k+2}-9)-9\big)\big((10^{k+2}-9)+9\big)=10^{k+2}(10^{k+2}-18)=(10^k-1)\times10^{k+4}+82\times10^{k+2}=x^2-81=(x-9)(x+9)$

So if this number is to be a perfect square, it is the perfect square of $10^{k+2}-9$.

Now use induction (base of the induction is $a_0=8281=(10^2-9)^2=91^2$) and go through the exact same procedure to prove that $a_k=(10^{k+2}-9)^2$


问:
An example for a complex-valued test function
Does anyone have an easy example for a complex-valued test function $f \in C_c^\infty(\mathbb{R},\mathbb{C})$? By test function I mean a $C^\infty$-function with compact support. </p>

答: Norbert gives a nice answer. I would like to remark, and this is what might confuse you, is that a (non-zero) test functions can never be analytic. This is because the only analytic functions that are constant in some interval, are constant everywhere.</p>


问:
Fibonacci sequence proof, how should I proceed?
Given the Fibonacci sequence   $F_{0}=0 \;,\; F_{1}=1 \;,\;  F_{n+2}= F_{n+1}+ F_{n+1} ; \;\; n\geq 0$

Prove that $ \sum _{i=1}^{2n}F_{i}F_{i-1} = F_{2n}^{2}\\$

Base Cases:   $ F_{1}F_{0} = F_{0}^{2}\\ 0 = 0 \\    F_{2}F_{1} = F_{2}^{2}\\ 1 = 1   $

Which are true.

Now, with  $\;n=k\;$  we have the inductive step,

$ \sum _{i=1}^{2k}F_{i}F_{i-1} = F_{2k}^{2}\\$

We have to prove then that for $n = k+1$

$ \sum _{i=1}^{2k+2}F_{i}F_{i-1} = F_{2k+2}^{2}\\$

We can do the following:

$ F_{2k+2}F_{2k+1} +\sum _{i=1}^{2k}F_{i}F_{i-1} = F_{2k+2}^{2}\\$

and from the inductive step we can do,
$ F_{2k+2}F_{2k+1} +  F_{2k}^{2}= F_{2k+2}^{2}\\$

I get stuck here, please help me.

答:
You are missing a term in the sum for $n=k+1$. It should be
$$
\sum_{i=1}^{2k+2}F_{i}F_{i-1} = F_{2k+1}F_{2k} + F_{2k+2}F_{2k+1} + \sum_{i=1}^{2k}F_{i}F_{i-1}\,.
$$

Now try see if you can finish the proof.


问:
Identity map between two Banach spaces always bounded
Let $X$ be a Banach spaces, $\| \cdot \|_1$ and $\| \cdot \|_2$ be two different norms.

Suppose that $\| x\|_1 \leq M\| x\|_2$ for all $x\in X$.

I've seen many places that say that the identity map $I: (X, \| \cdot \|_1) \rightarrow (X, \|\cdot \|_2)$ is always bounded, i.e. continuous.

But I'm not sure why it's true. And where does it use the completeness?

This is how I have attempted to prove it.

Let $x\in X$.

Then $\|I(x) \|_1= \|x \| _2 \leq \| I\|_1 \| x\|_1 \leq M\| x\|_2$.

Not sure how to proceed.

Thank you.

答:
Edit: this answer only makes sense if the $X$ is different; see Robert Israel's answer instead.

Given your assumption $\lVert x \rVert_1 \leq M \lVert x \rVert_2$ for some $M > 0$, we can only say that the identity map $I: (X, \lVert \cdot \rVert_2) \to (X, \lVert \cdot \rVert_1)$ is continuous.

To see this, note that if $\lVert x \rVert_2 < \epsilon / M$, then $\lVert I(x) \rVert_1 = \lVert x \rVert_1 \leq M \lVert x \rVert_2 < \epsilon$.

The fact that $X$ associated with either metric being a Banach space doesn't really matter here; this is a more general result from metric/topological spaces. We see that the topology induced by the norm $\lVert \cdot \rVert_1$ is finer than the one induced by the norm $\lVert \cdot \rVert_2$.

If the two norms were equivalent, i.e. $m \lVert x \rVert_2 \leq \lVert x \rVert_1 \leq M \lVert x \rVert_2$, then their topologies would be the same, and so the identity map would be continuous.

For a counterexample to the original direction, we can consider the Banach spaces $\ell^1$ and $\ell^\infty$ (where $\ell^1 \subseteq \ell^\infty$ or equivalently $\Vert \cdot \rVert_\infty \leq \lVert \cdot \rVert_1$). Consider the identity map $I: \ell^\infty \to \ell^1$, and the element $x = (1, 1, 1, \dots)$ which has $\Vert x \rVert_\infty = 1$ but $\lVert x \rVert_1 = \infty$.


Title: Fibonacci sequence proof, how should I proceed?

Body:
Given the Fibonacci sequence   $F_{0}=0 \;,\; F_{1}=1 \;,\;  F_{n+2}= F_{n+1}+ F_{n+1} ; \;\; n\geq 0$

Prove that $ \sum _{i=1}^{2n}F_{i}F_{i-1} = F_{2n}^{2}\\$

Base Cases:   $ F_{1}F_{0} = F_{0}^{2}\\ 0 = 0 \\    F_{2}F_{1} = F_{2}^{2}\\ 1 = 1   $

Which are true.

Now, with  $\;n=k\;$  we have the inductive step,

$ \sum _{i=1}^{2k}F_{i}F_{i-1} = F_{2k}^{2}\\$

We have to prove then that for $n = k+1$

$ \sum _{i=1}^{2k+2}F_{i}F_{i-1} = F_{2k+2}^{2}\\$

We can do the following:

$ F_{2k+2}F_{2k+1} +\sum _{i=1}^{2k}F_{i}F_{i-1} = F_{2k+2}^{2}\\$

and from the inductive step we can do,
$ F_{2k+2}F_{2k+1} +  F_{2k}^{2}= F_{2k+2}^{2}\\$

I get stuck here, please help me.

Body:
$$
\begin{align*}
\sum _{i=1}^{2(k+1)}F_{i}F_{i-1} &= F_{2k+2}F_{2k+1} + F_{2k+1}F_{2k} + \sum_{i=1}^{2k}F_{i}F_{i-1}\\
&=F_{2k+2}F_{2k+1} + F_{2k+1}F_{2k}+F_{2k}^2\\
&=F_{2k+2}F_{2k+1} + F_{2k}(F_{2k+1}+F_{2k})\\
&=F_{2k+2}F_{2k+1} + F_{2k}(F_{2k+2})\\
&=F_{2k+2}(F_{2k+1} + F_{2k})\\
&=F_{2k+2}(F_{2k+2})\\
&=F_{2k+2}^2
\end{align*}
$$


问:
An equivalence relation on regions of the plane.
Let $R\subseteq\mathbb{R}^2$. Consider the set of all "horizontal sections" $H_R =${$Rb|b\in\mathbb{R}$}, where $Rb=${$a\in\mathbb{R} | (a,b)\in R$}. Similarly consider the set of "vertical sections" of $R$, $V_R =${$ aR|a\in\mathbb{R}$} where $aR=${$ b\in\mathbb{R} | (a,b)\in R$}.  Now define the equivalence relation on $\wp (\mathbb{R^2})$ such that $R \sim S$ if, and only if, $H_R=H_S$ and $V_R=V_S$.

QUESTION: What is the equivalence class of a disk?

答:
Denote with $\pi_1,\pi_2: \Bbb R^2 \to \Bbb R$ the coordinate projections.

First, let me make explicit what exactly the elusive sets $H_R$ and $V_R$ are (since I tripped into a pitfall about that earlier on).

We have that $H_R = \{Rb: b \in \Bbb R\}$, where $Rb = \{a \in \Bbb R: (a,b)\in\Bbb R\}$. In terms of the projections, this comes down to $H_R = \{\pi_1[R \cap \pi_2^{-1}(b)]: b \in \Bbb R\}$. We can imagine this as the sets we see when moving a horizontal slit over the set $R$.

Similarly, $V_R = \{\pi_2[R \cap \pi_1^{-1}(a)]: a \in \Bbb R\}$, and it can be imagined as the subsets of $\Bbb R$ obtained from sliding over $R$ with a vertical slit.

For finding the equivalence class, let us reduce to the entirely generic case of the (closed) unit disk $B$.

Then we observe that $H_B = V_B = \{[-a,a]: a \in [0,1]\}$. The apparent symmetry allows us to restrict attention to the upper right quadrant, since $R$ will have to be extended in a unique way, by said symmetry. We will hereafter drop distinction between $R$ and the intersection of $R$ with the upper right quadrant.

The conditions $H_R = H_B$ and $V_R = V_B$ imply that for each $r,s \in [0,1]$, there are unique $a, b$ with $\pi_1[R \cap \pi_2^{-1}(s)] = [0,a]$ and $\pi_2[R \cap\pi_1^{-1}(r)] = [0,b]$. We define mappings $i_H, i_V$ by $i_H(s) =a$ and $i_V(r)=b$.

Now suppose that for given $a$, we find an $s \in [0,1]$ such that $i_H(s) = a$. If for some $s' > s$ we would have that $i_H(s') = a' > a$, then for $i_V(a')$ to be an interval, we would need $(a',s) \in R$ since $s \le s'$, contradicting that $a' > a$ and $i_H(s)=a$.

This contradiction provides us with the information that $s \le s'$ implies $i_H(s) \supseteq i_H(s')$. A similar reasoning can be applied to obtain that $r \le r'$ implies $i_V(r) \supseteq i_V(r')$.

Now what happens when $s < s'$ and $i_H(s) = i_H(s')$? Then $i_V(i_H(s)) = s'$, while for $a' > i_H(s)$ we have necessarily that $i_V(a') < s$; the above analysis hence renders it impossible that there exists an $a''$ such that $s < i_V(a'') < s'$, contradicting our assumptions.

Therefore, both $r \mapsto i_V(r)$ and $s \mapsto i_H(s)$ are strictly decreasing surjections, and by the above argument they are also injective; A cursory remark is that $i_H(i_V(r))=r$ and $i_V(i_H(s)) = s$ for all $r,s$.

In conclusion, $B \sim R$ iff $R$ is the area below a strictly decreasing bijection $f:[0,1]\to[0,1]$, reflected first the horizontal, then the vertical coordinate.

Real analysis tells us that on intervals, "strictly decreasing bijection" is synonymous with "continuous injection mapping endpoints to opposite endpoints". Thus this answer matches Brian M. Scott's comment.


问:
Calculation of conditional probability for an expression
I have this question on conditional probability

A biased four-sided die is rolled, and the down face is a random variable $X$, described  by the following probability mass function :

$$p_{\small X}(x)=\begin{cases}x/10&:& x\in\{1,2,3,4\}\\0&:&\textsf{otherwise}\end{cases}$$

Given the random variable $X$ a biased coin is flipped, and the random variable $Y$ is one or zero according to the result. The conditional mpf is

$$p_{\small Y\mid X}(y\mid x) = \left(\tfrac{x+1}{2x}\right)^y\left(1-\tfrac {x+1}{2x}\right)^{1-y}\quad[y\in\{0,1\} $$

The conditional probability formula is

$$p_{\small X\mid Y}({x}\mid{y})=\frac{p_{\small Y\mid X}({y}\mid{x})\cdot p_{\small X}(x)}{p_{\small Y}(y)}$$

Now, $p_{\small Y\mid X}({y}\mid{x})=p_{\small Y\mid X}({0}\mid{x})+p_{\small Y\mid X}({1}\mid{x})$ where $x\in\{1,2,3,4\}$ and $p_{\small X}(x)=\frac{x}{10}$  at $x\in\{1,2,3,4\}$

My question is how to calculate P(Y) as Y takes {0,1}

答:
My question is how to calculate P(Y) as Y takes {0,1}

$$p(Y=y)=\sum_{x=1}^4p(x, y)$$

Thus $Y$ takes the values $\{0;1\}$ with probabilities $\{6/20;14/20\}$ respectively


问:
categorical interpretation of quantification
Many constructions in intuitionistic and classical logic have relatively simple counterparts in category theory.  For instance, conjunctions, disjunctions, and conditionals have analogues in products, coproducts, and exponential objects.

More esoteric concepts still permit reasonably simple categorical interpretations.  For instance, the modal operator $\Box$ (necessity) can be described as a monoidal endofunctor (see, for instance, Paiva & Ritter's Basic Constructive Modality (PDF)).

What are the similar constructions that are suitable interpretations for universal and existential quantification in first-order logic?

答:
The simplest, and most direct, way to include quantifiers that stays within the bounds of first-order logic (as you wish for), while avoiding the better-known constructions seen in higher-order type theory, is to expand the treatment used for finite conjunctions to handle conjunctions indexed by sets $T$. Taking $T$ to be finite, this subsumes both the ‟true” predicate $⊤$ (by taking $T = ∅$), and the ‟and” connective ∧ (by taking $T = 2$), as well as the universal quantifier $∀$ for a single-sorted logic (taking $T$ to be the set of all terms) or for a multi-sorted logic, (where we may have different instances of $T$, one for each type of term). In the last case, we get an algebra for bounded quantifiers that avoids the need for any excursion into dependent types or higher-order type theory / logic and stays strictly first-order.

The key observation is this: if all of the inferences hold $f(t): C ⊢ A(t)$, for every $t∈T$ - in particular, if we have any inference $f(c): C ⊢ A(c)$, where $c$ is a free variable not present in either $C$ or $(∀x)A(x)$ - then we may apply generalization to $f(c)$ to produce the inference $‹x¦f(x)› ≡ {Gen}_c^x f(c): C ⊢ (∀x)A(x)$. We can make this independent of $c$ by imposing an identity ${Gen}_c'^x f(c') = {Gen}_c^x f(c)$, where $c'$ is any other free variable satisfying the same requirement.

Conversely, if we have the inference $h: C ⊢ (∀x)A(x)$, then we may combine it with the instantiation rule $π_At: (∀x)A(x) ⊢ A(t)$ to produce the result $(Πt)h ≡ π_At∘h: C ⊢ A(t)$. It is understood that the usual convention is adopted to take everything modulo α-equivalence to avoid any issues with free-bound capture.

If the categorical algebra is suitably-defined, then these two processes should be inverses, i.e. $‹x¦(Πx)h› = h$ and $(Πt)‹x¦f(x)› = f(t)$ for every $t∈T$, and certain naturality conditions should also hold true. Natural bijections, alone, entail an underlying infrastructure which provides a bundle of rules/operators and related proof algebra sufficient to capture all of what we need.

In the following, I will use the conventions that for a category $$:

$||$ = the objects of $$,

for $X, X' ∈ ||$, $(X,X')$ = the set of arrows/morphisms $X → X'$,

along with the definitions

$^+$ is the dual category:

$|^+| = ||$,

for $X, X' ∈ ||$, $^+(X,X') = (X',X)$,

identities in $^+$ are the same as those in $$,

compositions in $^+$ are the reverse of those in $$.

$×$ is the direct product category:

$|×| = ||×||$,

for $(X,Y),(X',Y') ∈ |×|$, $(×)((X,Y),(X',Y')) = (X,X')×(Y,Y')$.

$1$ is the unit category:

$|1| = \left\{0\right\}$,

$1(0,0) = \left\{1_0\right\}$,

$^T$ is the indexed product category, with indexing set $T$:

$|^T| = ||^T$,

for $A,B ∈ ||^T$, $^T(A,B) = ⨅_{t∈T} (A(t),B(t))$

for $A ∈ ||^T$, $1_A = [t∈T ↦ A(t)] ∈ ^T(A,A)$,

for $A,B,C ∈ ||^T$, $f: B → C$, $g: A → B$, $f∘g ≡ [t∈T ↦ f(t)∘g(t)] ∈ ^T(A,C)$

In a category $$ for logic, objects are predicates, arrows $A → B$ are inference relations $A ⊢ B$ between predicates $A$ and $B$, while morphisms $f: A → B$ are proofs that bear witness to the respective inference, the relation which we may also write $f: A ⊢ B$.

Within this framework, the distinguished object $⊤$ (for ‟true”), and connective $∧$ (for ‟and”) may be directly implemented by functors

True (‟terminal object”) ― $⊤: 1 → $,

And (‟product”) ― $∧: × → $,

and proof rules *all* implemented via *natural bijections*, as follows:
$$(C,⊤) ⇔ 1(0,0) = \{1_0\},$$
$$(C,A∧B) ⇔ (C,A)×(C,B),$$

To extend this to the universal quantifier - treating it as an indexed product - we add in another functor

All (‟indexed product” indexed by a set $T$) ― $∀: ^T → $,

and another natural bijection:
$$(C,∀A) ⇔ ^T(C,A).$$

For True, the bijection is trivial and is given by
$$h: C ⊢ ⊤ ⇔ 1_0: 0 → 0.$$
For And (conjunctions), the bijection is given by
$$h: C ⊢ A∧B ⇔ f: C ⊢ A, g: C ⊢ B.$$
For All (universal quantifiers), it is given by
$$h: C ⊢ (∀x)A(x) ⇔ [t∈T ↦ f(t): C ⊢ A(t)].$$

So, let's see how this works for conjunctions, first, and then more generally for indexed conjunctions - subsuming $⊤$, $∧$ and $∀$ within this treatment as special cases.

The Natural Bijection for Conjunctions

For conjunctions, the bijection is given by
$$f: C ⊢ A, g: C ⊢ B ⇔ h: C ⊢ A∧B$$
this works as follows:

There is a functor
$$∧:  ×  → $$
such that

Objects: $A,B∈|| ↦ A∧B∈||$,

Morphisms: for $A,A',B,B'∈||$, $f: A ⊢ A', g: B ⊢ B' ↦ f∧g: A∧A' ⊢ B∧B'$.

The defining properties of a functor, when applied here, entail the identities:

for $A,B ∈ ||$, $1_A∧1_B = 1_{A∧B}$,

for $A,A',A'',B,B',B'' ∈ ||$, and for $f:A⊢A', f':A'⊢A'', g:B⊢B', g':B'⊢B''$, $(f'∘f)∧(g'∘g) = (f'∧g')∘(f∧g)$.

There is a bijective natural transform

for $A,B,C∈||$, $(f: C ⊢ A, g: C ⊢ B) ↦ ‹f,g›: C ⊢ A∧B$,

for $A,B,C∈||$, $(h: C ⊢ A∧B) ↦ (⊤h: C ⊢ A, ⊥h: C ⊢ B)$,

with inverse relations:
$$⊤‹f,g› = f, ⊥‹f,g› = g, ‹⊤h,⊥h› = h,$$
and naturality conditions described equivalently either as:
$$‹k∘f∘m,l∘g∘m› = k∧l∘‹f,g›∘m$$
or
$$⊤(k∧l∘h∘m) = k∘⊤h∘m, ⊥(k∧l∘h∘m) = l∘⊥h∘m$$

More precisely:

$‹⋯,⋯›:  → $ and $(⊤⋯,⊥⋯):  → $ are natural transforms between the functors $$ and $$ given by

$: ××^+ → $, with

Objects: $: A,B,C∈|| ↦ (C,A) × (C,B)$,

Morphisms: $: (k:A⊢A',l:B⊢B',m:C'⊢C) ↦ ((k,l,m): (f,g) ∈ (C,A) × (C,B) ↦ (k∘f∘m, l∘g∘m) ∈ (C',A') × (C',B'))$,

$: ××^+ → $, with

Objects: $: A,B,C∈|| ↦ (C,A∧B)$,

Morphisms: $: (k:A⊢A',l:B⊢B',m:C'⊢C) ↦ ((k,l,m): h ∈ (C,A∧B) ↦ k∧l∘h∘m ∈ (C',A'∧B'))$,

iven by

For $A,B,C∈||$, $‹⋯,⋯›_{A,B,C}: (f,g) ∈ (A,B,C) ↦ ‹f,g› ∈ (A,B,C),$

For $A,B,C∈||$, $(⊤⋯,⊥⋯)_{A,B,C}: h ∈ (A,B,C) ↦ (⊤h,⊥h) ∈ (A,B,C),$

which satisfies the requisite properties for natural transforms that for
$$(k,l,m) ∈ ××^+((A,B,C),(A',B',C'))$$
(i.e. for $k: A⊢A'$, $l: B⊢B'$, $m: C'⊢C$),
$$‹⋯,⋯›_{A',B',C'}∘(k,l,m) = (k,l,m)∘‹⋯,⋯›_{A,B,C}$$
(i.e. $‹k∘f∘m,l∘g∘m› = k∧l∘‹f,g›∘m$, for $(f,g) ∈ (A,B,C)$), and
$$(⊤⋯,⊥⋯)_{A',B',C'}∘(k,l,m) = (k,l,m)∘(⊤⋯,⊥⋯)_{A,B,C}$$
(i.e. $(⊤(k∧l∘h∘m),⊥(k∧l∘h∘m))$ = $(k∘⊤h∘m,l∘⊥h∘m)$ for $h ∈ (A,B,C)$)

The operators $‹⋯,⋯›$, $⊤(⋯)$, $⊥(⋯)$ are equivalently described as ‟polymorphic operators” with polymorphism parametrized by $A,B,C∈||$. On account of these identities, each of the operators, as well as the functor $∧$ can be reduced to a small set of primitives, given by
$$L_{A,B} ≡ ⊤ 1_{A∧B}: A∧B ⊢ A,$$
$$R_{A,B} ≡ ⊥ 1_{A∧B}: A∧B ⊢ B,$$
as follows:

for $A,B,C∈||$ and $h: C⊢A∧B$,
$$⊤h = ⊤(1_{A∧B}∘h) = ⊤1_{A∧B}∘h = L_{A,B}∘h,$$
$$⊥h = ⊥(1_{A∧B}∘h) = ⊥1_{A∧B}∘h = R_{A,B}∘h,$$

for $A,A',B,B'∈||$, and $k:A⊢A'$, $l:B⊢B'$,
$$k∧l = k∧l∘‹⊤1_{A∧B},⊥1_{A∧B}›∘1_{A∧B} = ‹k∘L_{A,B},l∘R_{A,B}›$$

with the smaller set of identities:

for $A,B,C∈||$ and $f:C⊢A, g:C⊢B$, $L_{A,B}∘‹f,g› = f$ and $R_{A,B}∘‹f,g› = g$,

for $A,B,C.C'∈||$ and $f:C⊢A, g:C⊢B, m:C'⊢C$, $‹f∘m,g∘m› = ‹f,g›∘m$,

for $A,B∈||$, $‹L_{A,B},R_{A,B}› = 1_{A∧B}$

from which *all* of the previously stated identities follow (which are all well-known, by the way, seen for instance in Girard's formalism).

These, too, are polymorphic operators, and it is generally the case that all of the above relations can be stated more succinctly without the extra parametrization made explicit, without any loss of relevant context, like so:
$$⊤h = L∘h, ⊥h = R∘h,$$
$$L = ⊤1, R = ⊥1,$$
$$L∘‹f,g› = f, R∘‹f,g› = g,$$
$$‹L,R› = 1, k∧l = ‹k∘L,l∘R›.$$

The Natural Bijections for Indexed Conjunctions and Universal Quantifiers

For the universal quantifier, this works as follows:

There is a functor
$$∀: ^T → $$
such that

Objects: $A∈|^T| ↦ (∀x)A(x) ≡ ∀A∈||$,

Morphisms: for $A,A'∈|^T|$, $(∀x)f(x) ≡ ∀f: ∀A ⊢ ∀A'$ in $$

The key defining properties, for $∀$ to be a functor, are

for $A∈|^T|$, $(∀x)1_{A(x)} = 1_{(∀x)A(x)}$,

for $A,A',A''∈|^T|$ and for $f:A⊢A'$, $f':A'⊢A''$, $∀(f'∘f) = ∀f'∘∀f$, i.e. $(∀x)(f'(x)∘f(x)) = (∀x)f'(x)∘(∀x)f(x)$

By taking $T$, to be finite sets or natural numbers, we can use the same formalism described here to formulate the natural bijections for finitely-indexed conjunctions. That subsumes the treatments for both the conjunction $A∧B = ⋀(A,B)$ for $T = 2$ and $(A,B) ∈ ||²$ and $⊤ = ⋀()$ for $T = 0$, for $() ∈ ||⁰ ≅ 1$, as special cases.

A bijective natural transform is given by

for $A∈|^T|$, $C∈||$, $[t∈T ↦ f(t): C ⊢ A(t)] ↦ ‹x¦f(x)›: C ⊢ ∀A$,

for $A∈|^T|$, $C∈||$, $h: C ⊢ ∀A ↦ [t∈T ↦ (Πt)h: C ⊢ A(t)]$,

with inverse relations $(Πt)‹x¦f(x)› = f(t)$, for $t∈T$ and $‹x¦(Πx)h› = h$, and naturality conditions described equivalently as:
$$‹x¦k(x)∘f(x)∘m› = (∀x)k(x)∘‹x¦f(x)›∘m$$
or
$$(Πt)((∀x)k(x)∘h∘m) = k(t)∘(Πt)h∘m$$
for $t∈T$.

More precisely:

$‹⋯¦⋯›:  → $ and $((Πt)⋯¦t∈T):  → $ are natural transforms between the functors $$ and $$ given by

$: ^T×^+ → $

Objects: $: A∈|^T|, C∈|| ↦ ^T(C,A)$,

Morphisms: for $A,A'∈|^T|, C,C'∈||$, $: (k:A⊢A',m:C'⊢C) ↦ ((k,m): f ∈ ^T(C,A) ↦ (t∈T ↦ k(t)∘f(t)∘m) ∈ ^T(C',A'))$

$: ^T×^+ → $

Objects: $: A∈|^T|, C∈|| ↦ (C,∀A)$,

Morphisms: for $A,A'∈|^T|, C,C'∈||$, $: (k:A⊢A',m:C'⊢C) ↦ ((k,m): h ∈ (C,∀A) ↦ ∀k∘h∘m ∈ (C',∀A'))$,

iven by

for $A∈|^T|, C∈||$, $‹⋯¦⋯›_{A,C}: f ∈ (A,C) ↦ ‹x¦f(x)› ∈ (A,C)$,

for $A∈|^T|, C∈||$, $((Πt)⋯¦t∈T)_{A,C}: h ∈ (A,C) ↦ ((Πt)h:t∈T) ∈ (A,C),$

which satisfies the requisite properties for natural transforms that for
$$(k,m) ∈ ^T×^+((A,C),(A',C')),$$
i.e. $k: A⊢A', m: C'⊢C$,
$$‹⋯¦⋯›_{A',C'}∘(k,m) = (k,m)∘‹⋯,⋯›_{A,C}$$
i.e. $‹x¦k(x)∘f(x)∘m› = (∀x)k(x)∘‹x¦f(x)›∘m$, for $f ∈ (A,C)$, and
$$((Πt)⋯:t∈T)_{A',C'}∘(k,m) = (k,m)∘((Πt)⋯:t∈T)_{A,C}$$
i.e. $(Πt)((∀x)k(x)∘h∘m) = (k(t)∘(Πt)h∘m)$ for $h ∈ (A,C)$ and $t∈T$.

The operators $‹⋯¦⋯›$, $(Πt)(⋯)$ are, likewise, described equivalently as ‟polymorphic operators” with polymorphism parametrized by $A∈|^T|$ and $C∈||$.

A reduction similar to that for conjunctions yields, equivalently, the following reduced set of operators and identities:
$$π_A(t) ≡ (Πt)1_{∀A}: (∀x)A(x) ⊢ A(t)$$
for $t∈T$ as follows:

for $A∈|^T|$, $C∈||$ and $h: C⊢∀A$, $t∈T$, $(Πt)h = π_A(t)∘h$,

for $A,A'∈|^T|$, and $k:A⊢A'$, $∀k = ‹x¦k(x)∘π_A(x)›$

with the smaller set of identities:

for $A∈|^T|$, $C∈||$, $[t∈T ↦ f(t):C⊢A(t)]$, $π_A(t)∘‹x¦f(x)› = f(t)$,

for $A∈|^T|$, $C,C'∈||$, $[t∈T ↦ f(t):C⊢A(t)]$, $m:C'⊢C$, $‹x¦f(x)∘m› = ‹x¦f(x)›∘m$,

for $A∈|^T|$, $‹x¦π_A(x)› = 1_{∀A}$

from which *all* of the previously stated identities follow.

These, too, can be written more succinctly, without loss of context, by omitting the extra parametrization of the polymorphic operators as:
$$(Πt)h = π(t)∘h, π(t) = (Πt)1, π(t)∘‹x¦f(x)› = f(t)$$
for $t∈T$ and
$$‹x¦π(x)› = 1, ∀k = ‹x¦k(x)∘π(x)›.$$

By dualizing everything, the treatments for $⊤$, $∧$ and $∀$ can be transformed into treatments respectively for their duals False ($⊥$), Or ($∨$) and Some ($∃$), the existential quantifier, as well as indexed disjunctions.


问:
When would standard topology "equal" to lower limit topology?
I know that the lower limit topology is strictly finer than the standard topology. Therefore, for any set $A$, $T_{std}|_A\subset T_{ll}|_A$. But is there any condition that $T_{std}|_A= T_{ll}|_A$, I feel like countability of $A$ would do the job, but I cannot prove it yet.

答:
Countability of $A$ is necessary but not sufficient: $\Bbb Q$ is countable, but $\Bbb Q\cap[0,1)$ is open in the lower limit topology but not not in the standard topology.

You need to make sure that for each $a\in A$, $[a,\to)\cap A$ is a relatively open subset of $A$ in the usual topology. This means that for each $a\in A$ there must be an $x\in\Bbb R$ such that $(x,\to)\cap A=[a,\to)\cap A$, which amounts to saying that $x<a$ and $(x,a)\cap A=\varnothing$. In other words, no point of $A$ can be a limit from the left of other points of $A$.


答: The identity map from $(X, \|\cdot\|_1)$ to $(X, \|\cdot\|_2)$ has a closed graph (because its inverse is continuous), so the Closed Graph Theorem says it is continuous.

Title: Identity map between two Banach spaces always bounded

Body:
Let $X$ be a Banach spaces, $\| \cdot \|_1$ and $\| \cdot \|_2$ be two different norms.

Suppose that $\| x\|_1 \leq M\| x\|_2$ for all $x\in X$.

I've seen many places that say that the identity map $I: (X, \| \cdot \|_1) \rightarrow (X, \|\cdot \|_2)$ is always bounded, i.e. continuous.

But I'm not sure why it's true. And where does it use the completeness?

This is how I have attempted to prove it.

Let $x\in X$.

Then $\|I(x) \|_1= \|x \| _2 \leq \| I\|_1 \| x\|_1 \leq M\| x\|_2$.

Not sure how to proceed.

Thank you.


问:
Can the orbits of a semigroup touch without one being included in the other?

Question. Let $(S(t))_{t \ge 0}$ be a continuous semigroup of linear operators on some Banach space $X$. Might there exist $f, g\in X$ and $0<t_0<t_1$ such that
  \begin{equation}S(t_0)f=S(t_1)g\end{equation}
  but \begin{equation} S(t_0-\varepsilon_0)f\ne S(t_1-\varepsilon_1)g \end{equation}
  for all $0<\varepsilon_0\le t_0$ and $0<\varepsilon_1\le t_1$ (in particular, $f\ne g$)?

Pictorially, I am wondering if the following configuration is possible:

Of course we know that, when the evolution is given by a group, this is not possible: orbits either coincide or are disjoint. This is the case of autonomous ODE systems or of the Schrödinger equation. But here we have a semi group, such as the heat one, which only goes forward in time, not backwards. So the only obvious thing that we can say is that, as soon as they touch, orbits merge into one. But in principle I don't see why they should coincide in the past.

Added: After some searches, I have found that for the special case of the heat equation, the answer is negative. This is commonly referred to as backward uniqueness property. Here is a simplified version, which takes into account classical solutions on bounded domains:

Theorem (Taken from Evans's book on PDE, 2nd ed., pag.64) Let $U\subset \mathbb{R}^n$ be an open and bounded domain. Suppose $u, \bar{u}$ are classical solutions of
\begin{equation}
\begin{cases} u_t=\Delta u & \text{in }U\times(0, T) \\  u=0 &\text{on }\partial U \times [0, T] \end{cases}
\end{equation}
If at time $T$ we have
\begin{equation}
u(x, T)=\bar{u}(x, T),\quad \forall x\in U,
\end{equation}
then $u\equiv \bar{u}$ on the whole parabolic cylinder $U\times (0, T]$.

The property holds in much more general functional settings, as I read here (look for the keyword backward uniqueness).

All of this leaves the general question open. Is the backward uniqueness property true for all continuous linear semigroups? I guess that the answer should be negative, otherwise this would not be regarded as a special feature of the heat equation. However, I cannot find an explicit example.

答:
I think that it is a good question. To simplify, let us study the equation
$$
u_t=-\Lambda u,
$$
where $\Lambda=\sqrt{-\Delta}$. This equation is locally well-posed (both, forward and backward in time) for analytic initial data in a complex strip around the real axis. Then, at some time $T>0$, two solutions with a different initial data (at time $t=0$) can not coincide. Otherwise the backward problem posed at time $T$ would be ill-posed.

I think that for the heat equation the situation is similar.


答:
Now, $p_{\small Y\mid X}({y}\mid{x})=p_{\small Y\mid X}({0}\mid{x})+p_{\small Y\mid X}({1}\mid{x})$ where $x\in\{1,2,3,4\}$ and $p_{\small X}(x)=\frac{x}{10}$  at $x\in\{1,2,3,4\}$

No.   $p_{\small Y\mid X\!}(y\mid x) = \left(\tfrac{x+1}{2x}\right)^y\left(1-\tfrac {x+1}{2x}\right)^{1-y}\quad[\textsf{where }y\in\{0,1\}, x\in\{1,2,3,4\}]$

Note: this means that $p_{\small Y\mid X}(0\mid x)=(\tfrac{x-1}{2x})$ and $p_{\small Y\mid X}(1\mid x)=(\tfrac{x+1}{2x})$ , for all $x\in\{1,2,3,4\}$

These do not sum to give $p_{\small Y\mid X\!}(y\mid x)$.   They sum to give $1$, as a well behaved probability mass function should.

My question is how to calculate P(Y) as Y takes {0,1}

The law of total probability says: when $y\in\{0,1\}$ then:

$$\begin{align}p_{\small Y\!}(y) &= \sum_{x\in\{1,2,3,4\}} p_{\small Y\mid X\!}(y\mid x)\,p_{\small X\!}(x)\\[2ex]&= {p_{\small Y\mid X\!}(y\mid 1)\,p_{\small X\!}(1)}+{p_{\small Y\mid X\!}(y\mid 2)\,p_{\small X\!}(2)}+{p_{\small Y\mid X\!}(y\mid 3)\,p_{\small X\!}(3)}+{p_{\small Y\mid X\!}(y\mid 4)\,p_{\small X\!}(4)}\\[2ex]&=\sum_{x\in\{1,2,3,4\}}\left(\tfrac{x+1}{2x}\right)^y\left(\tfrac {x-1}{2x}\right)^{1-y}\tfrac x{10}\\&=\tfrac 1{20}\sum_{x\in\{1,2,3,4\}}(x+1)^y(x-1)^{1-y}\qquad[\textsf{where }y\in\{0,1\}]\\[2ex]&~~\vdots \end{align}$$

Title: Calculation of conditional probability for an expression

Body:
I have this question on conditional probability

A biased four-sided die is rolled, and the down face is a random variable $X$, described  by the following probability mass function :

$$p_{\small X}(x)=\begin{cases}x/10&:& x\in\{1,2,3,4\}\\0&:&\textsf{otherwise}\end{cases}$$

Given the random variable $X$ a biased coin is flipped, and the random variable $Y$ is one or zero according to the result. The conditional mpf is

$$p_{\small Y\mid X}(y\mid x) = \left(\tfrac{x+1}{2x}\right)^y\left(1-\tfrac {x+1}{2x}\right)^{1-y}\quad[y\in\{0,1\} $$

The conditional probability formula is

$$p_{\small X\mid Y}({x}\mid{y})=\frac{p_{\small Y\mid X}({y}\mid{x})\cdot p_{\small X}(x)}{p_{\small Y}(y)}$$

Now, $p_{\small Y\mid X}({y}\mid{x})=p_{\small Y\mid X}({0}\mid{x})+p_{\small Y\mid X}({1}\mid{x})$ where $x\in\{1,2,3,4\}$ and $p_{\small X}(x)=\frac{x}{10}$  at $x\in\{1,2,3,4\}$

My question is how to calculate P(Y) as Y takes {0,1}


问:
How to write proofs using set theory
I want to learn how to write proofs. I have read about different proof techniques, like direct proofs and contradiction, etc. But I can't seem to get past basic examples.

I have seen how naive set theory and ZFC is the standard for writing proofs now.

But I have no idea how to write, say, a direct proof using set theory.

I read this page and the first example uses set builder notation to create sets (that consist of algebra), that then use set theory to write the proof. Is this how you write proofs with set theory? Or am I missing something? How do I incorporate algebraic expressions into the proofs while also using sets.

Basically, I am asking for help on how to write proofs, like a direct proof, using set theory.

答: Welcome to MSE. The methods in the pdf you linked are rather standard, if most students were to prove those same things, they would probably prove them in more or less the same way (having seen similar proofs enough times, the answer just sort of jumps out, but this takes some practice). For example, if you want to prove that two sets are equal, you show they are subsets of each other often times (this is not always how it is done, there are other ways depending on the context, sometimes $A\subseteq B$ and there is some property A has, which guarantees A to be all of B, but for very basic set theory, often times yes that is how it's done), like they did in that pdf you linked. And yes, for that first example, the way they proved that is really the standard way you would prove this fact. Now as you get farther in mathematics, you have more tools at your disposal (like say: topology, abstract algebra, and real analysis) so the proofs get more complicated, but the methods you see in this pdf are very standard and I would make sure you understand the proofs. There are some very good books on introductory set theory I would suggest reading. "A TeXas Style Introduction to Proof" for example. To build these skills it is really up to you to just practice basic proofs, that's not really something someone else can do for you (for the most part). If you post a solution to a problem on MSE and ask for verification, we can also check your proof, and comment on it (/make sure it's correct).

Title: How to write proofs using set theory


问:
Basis for smooth vector fields given a Lie Group structure
Given a Lie Group $G$ with $\dim G = n$, I can take a basis of $T_e G$ and left translate it to obtain a basis at any point $g$ in $G$. This way I can  construct $n$ left translated vector fields
$\underline{E}_i, i=1,...,n$ that span $L(G)$.

My question is, is it possible to express any smooth vector field as a linear combination:

$\underline{X} = f^i\underline{E}_i$, where $f^i$ smooth functions ?

More abstract, if a smooth manifold $M$ can be turned into a Lie Group, is the basis of the Lie algebra of vector fields also a basis for the module of all smooth vector fields ?

Thanks

答:
Yes. Pick any left invariant metric $\langle , \rangle$ on $G$ then given an orthonormal frame $X_1, \ldots, X_n$ of left invariant vector fields you can write any vector field $Y$ as
$$
Y = \sum_{i=1}^n\langle Y, X_i \rangle X_i\,.
$$


问:
Why do you use triangle inequality in this proof?
I just read this question (The non-empty intersection of two open discs contains an open disc.) and I don't understand how the triangle inequality lines complete the proof. Please be nice - I've just started learning topology but hopefully I'll get there!

Thanks in advance for your help.

答:
Say we have three points $x, y$, and $z$ in a metric space $X$ (a metric space is simply a set where we have a notion of a distance - i.e., I can tell you the distance between any two points in the space $X$). Say I know the distance $d(x,y)$ between points $x$ and $y$, and I know the distance $d(x,z)$ between $x$ and $z$. The question is, can I say anything about the distance between $y$ and $z$? It might seem like the answer is no, at first, but the triangle inequality says something very important: It tells us the largest possible value the distance $d(y,z)$ between $y$ and $z$ can be. More specifically, the triangle inequality tells us that $$d(y,z) \leq d(y,x) + d(x,z)$$

The triangle inequality gives us information about $d(y,z)$ that we wouldn't otherwise have. But now, how does this concept relate back to the problem at hand? Say that I have a ball $B_r(x)$ of radius $r>0$ centered at a point $x$. That is, $$B_r(x) = \{m \in X: d(m,x) < r\}$$. Similarly, say I have a ball $B_{r'}(y)$ of radius $r' > 0$ centered at a point $y \in X$. Say $z$ is a point in both balls. I want to show that there is a ball centered at $z$ that is contained in both of the previously mentioned balls (which are centered at $x$ and $y$).

Since $z \in B_r(x)$, we know that $d(z,x) < r$. But because $z \in B_{r'}(y)$, we also know $d(z,y) < r'$. Can we find a ball centered at the point $z$ contained in the intersection $B_{r}(x) \cap B_{r'}(y)$? My answer is yes! Consider the ball $B_{\epsilon}(z)$ centered at $z$ of radius $\epsilon = \min \{r - d(z,x),r' - d(z,y)\}$. Let's show that $B_{\epsilon}(z)$ is a subset of $B_r(x)$. Let $m \in B_{\epsilon}(z)$. I want to show that $m \in B_r(x)$. That is, I need to show $d(m,x) < r$. How am I going to do this?

Well, since $m \in B_{\epsilon}(z)$, I know that $d(m,z) < \epsilon$. I know something about the distance between points $m$ and $z$ (namely $d(m,z) < \epsilon$), and I also know something about the distance between points $z$ and $x$ (namely, $d(z,x) < r$). But, how in the world would I find something that tells us about the distance between $m$ and $x$?? Well, this is where the triangle inequality comes in!! The triangle inequality directly tells me that $$d(m,x) \leq d(m, z) + d(z, x)$$ Because I know $d(m,z) < \epsilon$, I can say that the right-hand side of the above inequality is less than $\epsilon + d(z,x)$. In other words, $$d(m,x) \leq d(m,z) + d(z,x) < \epsilon + d(z,x)$$ But by the way we defined $\epsilon$, I know that $\epsilon \leq r-d(z,x)$. Hence, I can say that $$d(m,x) < \epsilon + d(z,x) \leq (r- d(z,x)) + d(z,x) = r$$

Hence, $d(m,x) < r$ as we wanted to show (showing this  wouldn't have been possible without the triangle inequality!)

Showing that $B_{\epsilon}(z) \subset B_{r'}(y)$ is almost exactly the same. Hence, $B_{\epsilon}(z) \subset B_r(z) \cap B_{r'}(y)$

As a side note, I've been using the term "Balls" to refer to subsets of a metric space $X$. If you're working in the metric space $\mathbb{R}^2$, as the question you linked certainly is, a ball is also called a disc.


问:
How do I evaluate the limit?
I am trying to evaluate limit$$\lim_{x\to 0}\frac{\sqrt{x+2}-\sqrt{2}}{x}$$ For my answer I got $\frac{1}{2\sqrt{2}}$. My professor answer sheet says $\frac{\sqrt2}{4}$. I am not sure why that is. My work and problem is attached below. Thank you.

答:
$$\frac{\sqrt{2}}{4}=\frac{\sqrt{2}}{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2}}=\frac{1}{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2}}=\frac{1}{2\sqrt{2}}$$

Please remember to check your work thoroughly before submitting it, anywhere for that matter!!

Title: How do I evaluate the limit?


问:
How do I solve the convergence $\lim_{n \to \infty}(3^n+1)^{\frac{1}{n}}$
$$\lim_{n \to \infty}(3^n+1)^{\frac{1}{n}}$$

I'm fairly sure I can't bring the limit inside the 1/n and I don't think I can use l'Hôpital's rule. I'm pretty sure I'm meant to use the sandwich theorem but I'm not quite sure how to do that in this circumstance.

答:
Since $\ln\big(3^n+1\big)^{\frac{1}{n}}=\frac{\ln(3^n+1)}{n}$ , one can observe $\displaystyle\lim_{n\to\infty}{(3^n+1)^{\frac{1}{n}}}=\displaystyle\lim_{n\to\infty}e^{\frac{\ln(3^n+1)}{n}}=e^{\tiny{{\displaystyle\lim_{n\to\infty}{\frac{\ln(3^n+1)}{n}}}}}$

Now by L'hopital's rule, $$\displaystyle\lim_{n\to\infty}{\frac{\ln(3^n+1)}{n}}=\displaystyle\lim_{n\to\infty}{\frac{\frac{3^n\cdot\ln(3)}{3^n+1}}{1}}=\displaystyle\lim_{n\to\infty}{\frac{3^n\cdot\ln(3)}{3^n+1}}=\ln(3)$$

So finally $\displaystyle\lim_{n\to\infty}{\frac{\ln(3^n+1)}{n}}=\ln(3)$ and therefore $\displaystyle\lim_{n\to\infty}{(3^n+1)^{\frac{1}{n}}}=e^{\tiny{{\displaystyle\lim_{n\to\infty}{\frac{\ln(3^n+1)}{n}}}}}=e^{ln(3)}=3$

P.S. The result was also clear of the fact that $1$ does not have a significant impact on the limit as $n\to\infty$ and hence the result could've been achieved with ignoring the $1$ in the parentheses.


答:
Your answers are equivalent, as noted in the comments.

$$\frac{1}{2\sqrt 2} = \frac{1}{2 \sqrt 2} \cdot \underbrace{\frac{\sqrt 2}{\sqrt 2}}_{=1} = \frac{\sqrt 2}{2 \sqrt 2 \sqrt 2} = \frac{\sqrt 2}{2 \cdot 2} = \frac{\sqrt 2}{4}$$

Title: How do I evaluate the limit?

Body: I am trying to evaluate limit$$\lim_{x\to 0}\frac{\sqrt{x+2}-\sqrt{2}}{x}$$ For my answer I got $\frac{1}{2\sqrt{2}}$. My professor answer sheet says $\frac{\sqrt2}{4}$. I am not sure why that is. My work and problem is attached below. Thank you.


Title: How do I solve the convergence $\lim_{n \to \infty}(3^n+1)^{\frac{1}{n}}$

Body:
$$\lim_{n \to \infty}(3^n+1)^{\frac{1}{n}}$$

I'm fairly sure I can't bring the limit inside the 1/n and I don't think I can use l'Hôpital's rule. I'm pretty sure I'm meant to use the sandwich theorem but I'm not quite sure how to do that in this circumstance.

Body: $$\lim_{n \to \infty}(3^n+1)^{\frac{1}{n}}=\lim_{n \to \infty}3(1+\frac{1}{3^n})^{\frac{1}{n}}=\lim_{n \to \infty}3e^{\frac{1}{3^nn}}=3$$


问:
What is the meaning of the PI function.
I am solving for a configuration problem and i have seen a function π

This is a function not 3.14 which is the value of pi. While accessing some lectures i found out that they also call this symbol as product of something. I don't have much to show for it but here's the symbol .

Please lead me to some resource.

答:
As most of us know, Pi is defined as the ratio of a circle’s diameter to its circumference. As a retired engineer, l still have it memorized to 7 decimal points, ie 3.1415926. It is called a transcendental number, or function, because using finite math it Never divides out. In (about) 1990, MIT used a Cray supercomputer to see if the machine could finish in a day. It could not... even after 10e670,000,000,000+/- permutations. (l honestly don’t remember how many orders of magnitude it ran but it certainly convinced me.) The most recent and accurate attempt was In 2014: the world record was a computer that calculated Pi (π) correctly to 13,300,000,000,000 decimal places.

Many mathematicians now call it an “indefinite finite” function, or an “irrational” number, but to me it is asymptotic. Ask your pc to find sq rt 145 and you will see similar results. There are lots of them out there.  Fortran (or formula translation) language run on a mainframe will generally throw you out after 10e7 with an error code saying Excceeds Maximum Allowed Permutations.

For the record, the formula used for calculating pi is:

Gregory-Leibniz Series:
\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\ldots.  Written non-code: pi/4=1-1/3+1/5-1/7+1/9-...  Notice the denominators are prime numbers.

Where frac = fraction

And ldots   = ellipsis (continue series - ...)

If one uses this method he or she will quickly substitute ldots with lDIOTS for it’s tedious nature is... Tedious    :-)

There are about five accepted formulas but l consider the G-L series the most fundamentally beautiful.  Hope this helped.

Title: What is the meaning of the PI function.


问:
Find $\lim_{n\to \infty} \left [\tan \frac{\pi}{2n}\tan \frac{2\pi}{2n}..\tan \frac{n\pi}{2n}\right]^{1/n}$
$$\ln y = \lim _{n\to \infty} \frac 1n \left [\ln (\tan \frac{\pi}{2n})+....\ln (\tan \frac{n\pi}{2n}) \right]$$
$$\ln y = \int _0^1 \ln (\tan \frac{\pi x}{2}).dx$$

Now this integral is a pain to solve, and online calculators aren’t able to solve it, so is this method even correct? If yes, then how do I solve the integral, and if not, what is the correct way?

答: Hint: Assuming the final $\tan(\frac{n\pi}{2n})$ isn't there (that being $\tan\frac{\pi}{2}=\infty$), just use $\tan(\frac{\pi}{2}-\theta)=\cot\theta$


问:
$\begin{align}\color{#c00}a&\mid b\\\color{#c00}c&\mid d\end{align}\Rightarrow\ \color{#c00}{ac}\mid bd$ $\ \ \bf\small [Divisibility\ Product\ Rule]$
I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?

Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.

Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$.

答:
Your reasoning goes in the wrong direction; you want to demonstrate that $ac\mid bd$, i.e. that there exists an $x$ such that $acx=bd$, using the assumption that $a\mid b$ and $c\mid d$. This is done simply by showing that $x=jk$ works.

In other words, the sentence beginning with "Then $ac\mid bd$ implies..." cannot be part of an argument that $ac\mid bd$ is true. If $P$, $Q$, and $R$ are statements, then showing that $P\Rightarrow R$ and $Q\Rightarrow R$ does not show that $P\Rightarrow Q$.


问:
Given commuting, positive definite A and B, why are the eigenvalues of AB strictly positive?
I am asked to prove that if A and B are positive definite, and AB = BA, then AB is positive definite.

I would first need to prove that AB has positive eigenvalues.

From there, the rest of the proof is easy:
(AB)* = B* A* = BA = AB. Therefor AB is Hermitian. Therefor, having only positive eigenvalues, it is positive definite.

I have taken a look at this proof, but I don't know what congruent matrices are.
I was wondering if there is another proof, or, alternatively, if anyone can explain the above proof in greater detail.

I am new to Stack Exchange, apologies for not using appropriate formatting.

I am also new to Linear Algebra so go easy on me:

I am somewhat familiar with the square root of matrices.

I understand eigenvalues and eigenvectors basic theory.Namely, I understand that
if two matrices are similar, they have the same eigenvalues.

I have just been introduced to positive definite matrices, but understand the very basic theory.

Thanks,
Miguel

答:
I have since found a pretty cool proof (not made by me).

$Proof$ that if $A$ and $B$ are positive definite, $AB$ has only positive eigenvalues:

($A^*$ denotes the conjugate transpose of $A$. According to this definition, a positive definite matrix is necessarily hermitian. $ \langle ·,· \rangle $ denotes the standard inner product, dot product.)

Because $A$ is positive definite, there exists invertible $\sqrt A$ such that $A = \sqrt A\sqrt A$.

Because $B$ is positive definite, there exists invertible $\sqrt B$ such that $B = \sqrt B\sqrt B$.

Note that $(\sqrt A)^* = \sqrt A.$ (The same is true for $B$)

Let $\lambda$ be an eigenvalue of $AB$, and $v$ a corresponding eigenvector.
Let $u = \sqrt A^{-1}v$. (Then, $v = \sqrt A u$)

$ABv = \lambda v \iff \sqrt A \sqrt A B \sqrt A u = \lambda \sqrt A u \iff \sqrt A^{-1} \sqrt A \sqrt A B \sqrt A u = \sqrt A^{-1} \sqrt A \lambda u \iff  \sqrt A B \sqrt A u = \lambda u$

Therefor, $AB$ and $\sqrt A B \sqrt A$ have the same eigenvalues.

I'll now show that all eigenvalues of $\sqrt A B \sqrt A$ are positive. Let $\lambda$ be an eigenvalue of $\sqrt A B \sqrt A$, and $u$ a corresponding eigenvector.

$\lambda \langle u,u\rangle = \langle\lambda u,u\rangle = \langle\sqrt A B \sqrt A u,u\rangle = \langle (\sqrt A \sqrt B)(\sqrt B \sqrt A) u, u\rangle = \langle (\sqrt B \sqrt A) u, (\sqrt A \sqrt B)^* u\rangle = \langle \sqrt B \sqrt A u, \sqrt B^* \sqrt A^* u\rangle = \langle \sqrt B \sqrt A u, \sqrt B \sqrt A u\rangle > 0$

, by the positive-definite property of the inner prooduct.

(Note that $\sqrt B \sqrt A u \neq  0$, as that would mean $\sqrt B \sqrt A$ is not invertible.)

Therefor, $\lambda \langle u,u\rangle > 0 \Rightarrow \lambda > 0$, which concludes this proof.


答:
Yes, because the frame $(\underline{E}_i)$ spans the tangent space $T_g M$ at each $g \in G$; this follows immediately from the left-invariance of the frame.

Explicitly, let $(\underline{e}^i)$ denote the (again, left-invariant and global) coframe dual to $(\underline{E}_i)$. For any smooth vector field $X$ on $G$ we can recover the coefficients $f^j$ explicitly:
$$X = e^j (\underline{X} ) \underline{E}_j = e^j \left(f^i \underline{E}_i \right) \underline{E}_j = f^i e^j(\underline{E}_i) \underline{E}_j = f^i \delta^j{}_i \underline{E}_j= f^j \underline{E}_j .$$
So, $f^j = e^j(X)$; in particular, $f^j$ is the contraction of smooth tensors and so is itself smooth.

Title: Basis for smooth vector fields given a Lie Group structure

Body:
Given a Lie Group $G$ with $\dim G = n$, I can take a basis of $T_e G$ and left translate it to obtain a basis at any point $g$ in $G$. This way I can  construct $n$ left translated vector fields
$\underline{E}_i, i=1,...,n$ that span $L(G)$.

My question is, is it possible to express any smooth vector field as a linear combination:

$\underline{X} = f^i\underline{E}_i$, where $f^i$ smooth functions ?

More abstract, if a smooth manifold $M$ can be turned into a Lie Group, is the basis of the Lie algebra of vector fields also a basis for the module of all smooth vector fields ?

Thanks


Body:
We have

$$3= (3^n)^{1/n}\leq (3^n+1)^{1/n}\leq (3^n+3^n)^{1/n}=(2\cdot 3^n)^{1/n}=3\sqrt[n]{2}$$

Thus

$$3\leq \lim_{n\to\infty}(3^n+1)^{1/n}\leq \lim_{n\to\infty}3\sqrt[n]{2}=3$$

We conclude the limit is $3$.


问:
How is $ 2n + 10 \leq cn $ manipulated to $ (c - 2) n \geq 10 $?
I have the inequality $ 2n + 10 \leq cn $, for all $n \geq n_0 $

I don't understand how this inequality is manipulated to $ (c - 2) n \geq 10 $

What are the steps involved?

Has ", for all $n \geq n_0 $" been incorrectly omitted from the derived inequality?

Body: $$2n+10\le cn\iff 10\le cn-2n\iff 10\le (c-2)n\iff (c-2)n\ge10$$


问:
In a triangle is there an angle splitter formula like angle bisector theorem but generalized?
A segment starting from a vertex of a triangle splits that angle in 2 arbitrary parts.
It also splits the opposite triangle side in 2 parts.

Can a relationship be established between the angle parts and the side parts, like the proportional relationship in the case of bisector?

Here's a picture:

For simplicity I drew the triangle as right, with one leg being twice as long as the other. The blue segment is bisector, magenta is a median and red splits the right angle in two of 30 and 60 degrees respectively.

I tried to reason something from the picture but I couldn't.

答:
Let's say we have a figure as the one below. Stewart's Theorem states the following relationship between the sides of a triangle and a cevian:

$$ d^2=\frac{c^2n+b^2m}{m+n}-m\cdot n$$

$\hspace{3cm}$

For a relationship between sides and angles, you should use the law of sines and the law of cosines. Alternatively, you can try to form special triangles ($30°-60°-90°$, $45°-45°-90°$, $15°-75°-90°$, etc.) to solve specific problems.


问:
A problem on matrices: Find the value of $k$

If $
        \begin{bmatrix}
        \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\
        \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7}  \\
        \end{bmatrix}^k
 =
        \begin{bmatrix}
        1 & 0 \\
        0 & 1 \\
        \end{bmatrix}
$, then the least positive integral value of $k$ is?

Actually, I got no idea how to solve this.

 I did trial and error method and got 7 as the answer.

how do i solve this?
 Can you please offer your assistance? Thank you

答:
We start with polar coordinates of a point $P=(r,\theta)$ vs its Cartesian form $(x,y)$. You know $x=r\cos\theta, y=r\sin\theta$. Now rotate the point $P$ by an angle of $\alpha$ with respect to origin so its angle with respect to $x$ axis becomes $\theta+\alpha$. Its new polar coordinates is $P'=(r,\theta+\alpha)$ and its new Cartesian coordinates $(x',y')$, where $x'=r\cos(\theta+\alpha), y'=r\sin(\theta+\alpha)$. Use trig identities to expand these

$x'=r(\cos \theta \cos \alpha -\sin \theta \sin \alpha)=x\cos \alpha -y \sin\alpha$, and

$ y'=r(\sin\theta\cos\alpha+\cos\theta\sin\alpha)=y\cos\alpha+x\sin\alpha$.

If you write these as a matrix equation you can finish your problem.
$
\left[ \begin{array}{c} x' \\ y' \end{array} \right] = \begin{bmatrix} \cos \alpha &  -\sin \alpha \\
        \sin \alpha  & \cos \alpha \\  \end{bmatrix} \times \left[ \begin{array}{c} x \\ y \end{array} \right]
$

If you have several rotation $\alpha_1, \cdots, \alpha_n$ then the effect of multiplying all the related matrices is the same as adding the given angles. In particular if you have $n$ rotations by the same angle $\alpha$ the result will be same as one rotation of $n\alpha$. In short
$
 \begin{bmatrix} \cos \alpha &  -\sin \alpha \\
        \sin \alpha  & \cos \alpha \\  \end{bmatrix}^n=  \begin{bmatrix} \cos n\alpha &  -\sin n\alpha \\
        \sin n\alpha  & \cos n\alpha \\  \end{bmatrix}$


Title: Find $\lim_{n\to \infty} \left [\tan \frac{\pi}{2n}\tan \frac{2\pi}{2n}..\tan \frac{n\pi}{2n}\right]^{1/n}$

Body:
$$\ln y = \lim _{n\to \infty} \frac 1n \left [\ln (\tan \frac{\pi}{2n})+....\ln (\tan \frac{n\pi}{2n}) \right]$$
$$\ln y = \int _0^1 \ln (\tan \frac{\pi x}{2}).dx$$

Now this integral is a pain to solve, and online calculators aren’t able to solve it, so is this method even correct? If yes, then how do I solve the integral, and if not, what is the correct way?

Body:
$$I=\int_0^1\ln (\tan \frac{\pi (1-x)}{2})dx=\int_0^1\ln (\cot \frac{\pi x}{2})dx$$

$$I+I=\int_0^1\ln (\cot \frac{\pi x}{2})dx+\int_0^1\ln (\tan \frac{\pi x}{2})dx=\int_0^1\ln (1)dx=0$$


问:
$\det( J_f^T J_f)$
$f:R^n \rightarrow R^{n+1}, f(x) =\frac{(x_1,\dots,x_n,1) }{(1+x_1^2+\dots+x_n^2)^{\frac{1}{2}}}  $

$J_f$ is the Jacobi matrix of f

I want to calculate $\det( J_f^T J_f)$ but:

$J_f^t J_f =  \left( \begin{array}{cc} |(x,1)|^2 - x_1^2 & -x_1 x_2 & \dots & -x_1 x_n \\ \vdots & & & \vdots \\- x_1 x^n & -x_2 x^n & \dots & |(x,1)|^2 -x_n^2 \end{array} \right) \cdot \frac{1}{|(x,1)|^{4}}$

I don't know how it's possible to find a general solution for this determinant.
It seems impossible to me.

答:
Let me use $n = 3$ to make this easier to write and picture

The trick here is that the matrix

$$\begin{pmatrix} x^2 & xy & xz \\ xy & y^2 & yz \\ xz & yz & z^2 \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \begin{pmatrix} x & y & z \end{pmatrix}$$

is rank 1.

So we have

$$J_f^\top J_f = \frac{1}{c^2}\left( cI - u u^\top \right)$$

where $c = (1 + x^2 + y^2 + z^2)$ and $u^\top = \begin{pmatrix} x & y & z \end{pmatrix}$.

And now we use a common formula for the determinant of a rank-1 update:

$$\det(I + uv^\top) = 1 + u^\top v.$$

See e.g. Determinant of rank-one perturbations of (invertible) matrices or https://en.wikipedia.org/wiki/Matrix_determinant_lemma

So we obtain

\begin{align}
\det(J_f^\top J_f) &= \frac{1}{c^{2n}}\det(cI - u u^\top) \\
&= \frac{1}{c^n} \det\left(I - \frac{1}{c} u u^\top \right) \\
&= \frac{1}{c^n} \left( 1 - \frac{1}{c} u^\top u \right) \\
&= \frac{1}{c^{n+1}} \left[ (1 + x_1 + \dots + x_n) - (x_1 + \dots + x_n) \right] \\
&= \frac{1}{c^{n+1}}.
\end{align}

So the final answer is

$$\det(J_f^\top J_f) = \frac{1}{(1 + x_1 + \dots + x_n)^{n+1}}.$$

Title: $\det( J_f^T J_f)$


问:
How can I prove that $A \setminus (B \setminus C) \subseteq (A \setminus B) \cup C$?
I found the Venn Diagram of the relation using Wolfram so I know this is a true statement. I'm lost on how to prove it though.

Venn Diagram

I'm thinking of going with the $x \in A \setminus (B \setminus C)$ method but I don't know what to do after this.

答:
Let $x\in A\setminus(B\setminus C)$. By definition, we have that $x\in A$ and $x\notin B\setminus C$.

Suppose $x\notin C$. Then, $x\notin C\cup (B\setminus C)$, so since $B\subseteq C\cup(B\setminus C)$, we have $x\notin B$. Thus, $x\in A\setminus B$.

Thus, $A\setminus (B\setminus C)\subseteq (A\setminus B)\cup C$.


问:
Calculating the integral $\int_{z}^{\infty} e^{-ax^4+bx^2}dx$
I am looking for a way to calculate the integral
$$\int_{z}^{\infty} e^{-ax^4+bx^2}dx$$
with $z>0$. I have only found this so far, which is not quite what I need. I realize that there may not be a simple expression for this integral, in which case can a function of $z$ approximating the integral be found?

答:
First, note that by rescaling $x$, we can eliminate one of $a$ or $b$, so it will suffice to consider an integral of the form

$$ \int\limits_z^\infty dx \ \exp\left(-ax^4+x^2 \right)$$

For $z\rightarrow 0$

Relabeling the constant $a$ to keep future expressions simpler, let

$$ I(z)=\int\limits_z^\infty dx \ \exp\left(-\frac{x^4}{8a}+x^2 \right)$$

Fortunately $I(0)$ is expressible in terms of Bessel functions

$$ I(0)\stackrel{\text{M}}{=} \frac{1 }{2}\pi \sqrt{a} e^a \left(I_{-1/4}(a)+I_{1/4}(a) \right)$$

Where $I_n$ are modified Bessel functions of the first kind, and the symbol $\stackrel{\text{M}}{=}$ denotes Mathematica. Split up the integral

$$ I(z)=I(0)-\int\limits_0^z dx \ \exp\left(-\frac{x^4}{8a}+x^2 \right)$$

For the second term on the right we may expand the exponential around $x=0$

$$ \exp\left(-\frac{x^4}{8a}+x^2 \right)=1+x^2+\left(\frac{1}{2}-\frac{1}{8a} \right)x^4+\cdots$$

Term by term integration yields

$$ I(z)\sim I(0)-\left( z + \frac{z^3}{3}\right) \ \ , \ \ z\rightarrow 0$$

Of course, you may continue the series for higher powers of $z$. Here is plot of the integral for small $z$ and $a=1.2$.

For $z \rightarrow \infty$

Let

$$I(z)= \int\limits_z^\infty dx \ \exp\left(-ax^4+x^2 \right)=\int\limits_z^\infty dx \ \frac{e^{x^2}}{-4ax^3} \frac{d}{dx}\left( e^{-ax^4}\right)$$

We now integrate by parts

$$ I(z)=\frac{e^{-ax^4+x^2}}{-4ax^3} \Bigg\vert_z^\infty-\int\limits_z^\infty dx \ \frac{2x^2 -3}{4ax^4} e^{-ax^4+x^2}$$

For a fixed $a$, we have

$$ I(z) >> \int\limits_z^\infty dx \ \frac{2x^2 -3}{4ax^4} e^{-ax^4+x^2} \ \ , \ \ z \rightarrow \infty$$

Neglecting the small term above, we are left with

$$ I(z) \sim \frac{e^{-az^4+z^2}}{4az^3}  \ \ , \ \ z \rightarrow \infty$$

In principle, you could continue to get more terms in the asymptotic series by repeated integration by parts.


问:
Prove that $\displaystyle\ln\left(1+\frac{1}{x} \right) < \frac{2x-1}{x^2-x}$ when $x > 1$
We want to prove that

$$\forall x>1, \,\,\,\displaystyle\ln\left(1+\frac{1}{x} \right) < \frac{2x-1}{x^2-x}.$$

I tried to look up inequalities involving $\log$ that might be useful, but I couldn't finagle any of them to fit this particular situation.

答:
By strict concavity of $f(x) = \ln(x)$, we have that
$$
f(y) < f(x) + f'(x)(y-x) \implies \ln(y) < \ln(x) + \frac{y-x}{x},\forall x,y\in\mathbb{R}_{>0}.
$$
Setting $y=1+x$ and letting $x$ be itself, we then have for $x>1$:
$$
\ln(1 + \frac{1}{x}) = \ln(\frac{x+1}{x}) = \ln(1+x) - \ln(x) < \frac{1+x-x}{x} = \frac{1}{x} < \frac{1}{x} + \frac{1}{x-1} = \frac{2x-1}{x(x-1)}
$$
where the latter equality follows from partial fraction expansion.


问:
Group action definition
The Wikipedia for group action and several posts here have what seems to me to be two, slightly different definitions for a group action.$\ $   Let G be a group acting on space X.$\ $  These two definitions are:

A group homomorphism $\varphi$ from the group to the automorphism group of the space.
$$\varphi : G\ \longrightarrow \text{Aut}(X)$$

A map that takes an element of the group and an element of the space, and returns an element of the space.
$$\alpha : G\, \times X\ \longrightarrow X$$

Which of these is the the group action?$\ $   Is it a function of one argument or two?$\ $  What is the return type?  Does it return a function (automorphism) or an element of the space?$\ $ The first definition is in line with what I think of as a representation, ex. from a symmetry group C$_\text{2v} \longrightarrow $ GL(V) that returns a matrix, rather than a vector.

I (think) I understand the basic idea of what is going on, but I'd like to know what the actual mathematical definition is.

Thanks in advance.

答:
The definitions end up being more or less the same in the sense that if you have a map $f: G \times X \to X$ satisfying the second definiton, there exists a group homomorphism (induced by $f$) satisfying the first definition (and vice versa).

To show you how this might work, suppose you have $f: G \times X \to X$. For each $g \in G$, you can define a map $\phi_g: X \to X$ by letting $\phi_g(x) = f(g,x)$. This will be a bijection since $\phi_g^{-1}$ is precisely $\phi_{g^{-1}}$. Then, the homomorphism $G \to \text{Aut}(X)$ is the one sending $g \mapsto \phi_g$. If instead you start with a homomorphism $f: G \to \text{Aut}(X)$ and let $f_g := f(g)$, we can define a map $\alpha: G \times X \to X$ by letting $\alpha(g,x) = f_g(x)$. Hence, there is a one-to-one correspondence between maps $G \times X \to X$ (which satisfy other conditions as mentioned in the comments) and homomorphisms $G \to \text{Aut}(X)$

Though, in my experience, the 2nd definition you mentioned is a more conventional starting place for defining group actions. If you look in many classic Algebra books like the one by Dummit and Foote, they will define a group action as a map $G \times X \to X$ satisfying various conditions. However, as you learn more about Abstract Algebra, the difference between these two notions becomes miniscule, and you will start to use both notions of a group action interchangeably.

Title: Group action definition


问:
What does a probability of $1$ mean?
From a textbook on probability on the Law of Large Numbers:

Theorem 3-19 (Law of Large Numbers): Let $X_1,X_2, \ldots , X_n$ be mutually independent random variables (discrete or continuous), each having finite mean  and variance. Then
  if $S_n = X_1 + X_2 +\dots+ X_n$,

$$ \lim_{n \to\infty} P\left(\left|\frac{S_n}{n} - \mu\right| \geq \varepsilon\right) = 0 $$

Since $S_n$ is the arithmetic mean of $X_1,X_2, \ldots , X_n$, this theorem states that the probability of the arithmetic mean $\frac{S_n}{n}$ differing from its expected value $\mu$ by more than $\varepsilon$ approaches zero as $n \to \infty$. A stronger result, which we might expect to be true, is that $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $  but this is actually false. However, we can prove that $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $
with probability one.

The only difference between the last sentence and the one before that is the phrase 'with probability one'. What does probability one mean here ? The usual definition is that the event occurs with 100% certainty. If that is the case, why is the original assertion $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $ false ?

答:
Probability zero does not mean "never happens", and probability one does not mean "always happens".  For example, if you choose a real number uniformly in the interval $[0,100]$, you will choose an integer with probability zero, and a transcendental number with probability one.  That doesn't mean there aren't any integers though.

To understand what "probability zero" means, you need to know some measure theory; it means that the measure of that event is zero.  Similarly, "probability one" means that the measure of that event is one.  This idea is used quite frequently in probability theory, because measure-zero events are often annoying and we want to not think about them.

Title: What does a probability of $1$ mean?


问:
Why does mapping to unit disk and close to infinity imply non-normalcy of family of iterates?
I am reading Devaney's "Introduction to Chaotic Dynamical Systems", the chapter on the exponential family Prop. 9.1 I don't really understand how he arrives at the conclusion that if for $x\in\mathbb R$ and $U$ any neighborhood of $x$ you can find points $z_1,z_2\in U$ such that there is an $n\in \mathbb N$ for which $|E^n(z_1)|<1$ and $|E^n(z_2)|$ is as large as you wish, then $\{E^n\mid n\in\mathbb N\}$ is not a normal family. If someone could tell me what property or equivalence of normal families he is using I would be grateful.

答:
By definition of normality, every subsequence of $\{E^n \ | \ n \in \mathbb{N}\}$ must have a, possibly finer, subsequence that either converges uniformly on compact subsets or converges uniformly to $\infty$.

By his remarks, given $U$ a neighborhood we may construct a subsequence $\{E^{n_j} \ | \ j \in \mathbb{N}\}$ such that for each $j$ there exists $z_{1_j}, z_{2_j} \in U$ such that $|E^{n_j}(z_{1_j})| < 1$ and $|E^{n_j}(z_{2_j})| >j$. We can actually take $z_{2_j} = x \in \mathbb{R}$ for all $j \in \mathbb{N}$, since this point is in particular compact and converges to $\infty$, then the subsequence we constructed cannot have a subsequence that converges uniformly on compact sets. This means that the only chance of it being normal is that it has a subsequence that uniformly converges to $\infty$, but there's always a point such that its image is in the unit disk, so that can't happen either.


答:
Note that, for $x>1$,

\begin{align}
f(x)&=\frac{2x-1}{x^2-x}-\ln\left(1+\frac{1}{x} \right)\\
&=-\int_x^\infty f’(t)dt=\int_x^\infty \frac{(t-1)^3+5(t-\frac12)^2+\frac34}{(t^2-t)^2(t+1)}dt>0
\end{align}

Title: Prove that $\displaystyle\ln\left(1+\frac{1}{x} \right) < \frac{2x-1}{x^2-x}$ when $x > 1$

Body:
We want to prove that

$$\forall x>1, \,\,\,\displaystyle\ln\left(1+\frac{1}{x} \right) < \frac{2x-1}{x^2-x}.$$

I tried to look up inequalities involving $\log$ that might be useful, but I couldn't finagle any of them to fit this particular situation.


问:
Existence of a specific continuous linear operator
Let $B_1$ and $B_2$ be Hilbert spaces, and $\mathbb{L}$ be the collection of continuous linear operators from $B_1$ to $B_2$. Then, I wonder whether this statement is true: Given $x\in B_1,$ $y\in B_2,$ there always exists some $l \in \mathbb{L}$ such that $l(x)=y.$

答: Let $f(x)=1$ defined on $Vect(x)$ you can extend it to a continuous linear function $g$ on $B_1$ (Hahn Banach). Write $h(z)=g(z)y$, ut is bounded since $\|h(z)\|\leq |g(z)|\|y\|\leq \|g\|\|y\|\|z\|$.


答:
Consider the natural logarithm of the limit, $$\lim_{n\to\infty}\frac1n\ln\left(3^n+1\right)$$

Note that $\lim_{n\to\infty}\ln\left(3^n\right)-\ln\left(3^n+1\right)=\ln\left(\lim_{n\to\infty}\frac{3^n}{3^n+1}\right)=\ln1=0$. Hence we have:

$$\lim_{n\to\infty}\frac1n\ln\left(3^n+1\right)$$

$$=\lim_{n\to\infty}\frac1n\ln\left(3^n+1\right)+\lim_{n\to\infty}\frac1n\left[\ln\left(3^n\right)-\ln\left(3^n+1\right)\right]$$

$$=\lim_{n\to\infty}\frac1n\left[\ln\left(3^n+1\right)+\ln\left(3^n\right)-\ln\left(3^n+1\right)\right]$$

$$=\lim_{n\to\infty}\frac1nn\ln3$$

$$=\ln3$$

Now $$\lim_{n \to \infty}(3^n+1)^{\frac1n}$$

$$=\exp\left(\lim_{n\to\infty}\frac1n\ln\left(3^n+1\right)\right)$$

$$=\exp(\ln3)=3$$


问:
Frobenius coin problem
Suppose that you only have coins worth, say 3 and 5 euros. According to Sylvester result we can find the Frobenius nr $g(3,5)=15-3-5=7$ so 7 is the largest integer that cannot be written as $a_{1}k_{1}+a_{2}k_{2}$ for $k_{1},k_{2}\in\mathbb{N}$ and $a_{1},a_{2}$ are the values of these coins.

a) how do you pay 8€,9€ and 10€ with these coins?

b)use a) to show that it is possible to pay all amounts that are greater than 10€ with the coins 3€ and 5€.

c) show that it is impossible to pay the amount of 7€ with these coins.

I am afraid I do not understand 100% the whole idea behind the Frobenius numbers.

a) can we just take 3€+5€=8€ and 3€+3€+3€=9€ and 5€+5€=10€ this seems suspicious of how easy it is....

b)do I have to use both coins? or just 3€ or 5€?
11€=3€+5€+3€

12€=3€+3€+3€+3€

13€=3€+5€+5€

14€=5€+3€+3€+3€

.

.

c) if we could pay 7€ with these coins we could have written

$7€=k_{1}5€+k_{2}3€$ but this is impossible as $k_{1},k_{2}\in\mathbb{N}$

can someone please explain to me what should be done in this exercise and how?

答: Your answer to (a) is correct.  Your answer to (b) needs to have induction, three dots is insufficient.  Your answer to (c) needs to go into a lot of detail about the word "impossible"; why is this so?</p>

Title: Frobenius coin problem


问:
Split output from one conveyor belt equally to five other conveyor belts.
I was asked a question about the game "Satisfactory" and how to split an output into $5$ equal parts using "splitters" with one input and 2-3 outputs and mergers that accept 2-3 inputs.

A splitter accepts input from one conveyor belt and its output feeds two or three other conveyor belts. A merger accepts input from two or three conveyor belts and its output feeds a single conveyor belt.

$$\frac{a}{2}+\frac{b}{3}=\frac{c}{5}
\implies c = \frac{5 a}{2} + \frac{5 b}{3}$$

I'm out of ideas.  I'm beginning to think it's not possible. How do I split something into $5$ equal parts when I can only split/merge things by $2$s or $3$s?

答:
There's a simple way to solve this problem. As noted in the comments, there are no integer solutions to

$$2^x3^y=5^z$$

due to the fundamental theorem of arithmetic, so it's impossible to represent $\frac15$ using finite sums of multiples of $\frac12$ and $\frac13$.

However, we can "cheat" by using the sum of the geometric progression:
$$\frac15 = \frac16 + \frac1{36} + \frac1{216} + \frac1{1296} + \dots$$

We can implement this with the mergers and splitters by using one 3-splitter and three 2-splitters to do a 6-way split,
then feed one of those outputs back into the input stream. Here's a diagram:

src: the source

m2: 2-way merger

s3: 3-way splitter

s2a, s2b, s2c: 2-way splitters

o1-o5: the outputs


问:
Error with Differential Forms
I'm learning a little bit about differential forms in my Complex Analysis course, and I'm not sure if I'm making an error in what follows. I wonder if anyone could offer any insight into this. Thank you.

So, the question is to show that $\alpha = \frac{xdy - ydx}{x^2 + y^2}$ is closed. To show this, I need to show that $d\alpha = 0$, which I can show as follows (according to my understanding).

Let $f = \frac{1}{x^2 + y^2}, g = xdy - y dx$. Then $d \alpha = f \wedge dg + df \wedge g$.

So, calculate $df = \frac{-1}{(x^2+y^2)^2}2x dx + \frac{-1}{(x^2 + y^2)^2}2y dy = \frac{-2}{(x^2+y^2)^2} (x dx + y dy)$ and $dg = 1 dx \wedge dy - 1 dy \wedge dx = 2 dx \wedge dy$.

Then, substituting these in the right places, we get

\begin{align*}
d \alpha &= \left(\frac{1}{x^2 + y^2}\right) \wedge (2 dx \wedge dy) + \left(\frac{-2}{(x^2+y^2)^2} (x dx + y dy)\right) \wedge (xdy - y dx)\\
&= \frac{2}{x^2+y^2} dx \wedge dy - \frac{2}{(x^2+y^2)^2}(x^2 dx \wedge dy - y^2 dy \wedge dx)\\
&= \frac{2}{x^2 + y^2} dx \wedge dy - \frac{2}{x^2 + y^2} dx \wedge dy\\
&= 0.
\end{align*}

This seems to work.

However, when I interchange $f$ and $g$, I seem to run into a problem, and I'm not sure why. Here's what happens. I keep $d \alpha = f \wedge dg + df \wedge g$. However, now I let $f = xdy - ydx$ and $g = \frac{1}{x^2 + y^2}$.

In this case, $df$ and $dg$ are no different (except of course for interchanging them), so we have $df = 2 dx \wedge dy$ and $dg = \frac{-2}{(x^2+y^2)^2} (x dx + y dy)$.

Now, substituting in, we get

\begin{align*}
d \alpha &= (xdy - ydx) \wedge \left(\frac{-2}{(x^2+y^2)^2} (x dx + y dy)\right) + (2 dx \wedge dy) \wedge \left(\frac{1}{x^2 + y^2}\right)\\
&= \frac{-2}{(x^2+y^2)^2} (xdy - ydx) \wedge (xdx + ydy) + \frac{2}{x^2 + y^2} dx \wedge dy\\
&= \frac{-2}{(x^2+y^2)^2}(x^2 dy \wedge dx - y^2 dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\
&= \frac{-2}{(x^2+y^2)^2}(-x^2 dx \wedge dy - y^2 dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\
&= \frac{-2}{(x^2+y^2)^2}((-x^2 -y^2)dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\
&= \frac{-2}{(x^2+y^2)^2}(-(x^2 +y^2)dx \wedge dy)+ \frac{2}{x^2 + y^2} dx \wedge dy\\
&= \frac{2}{x^2+y^2} dx \wedge dy + \frac{2}{x^2+y^2} dx \wedge dy\\
&= \frac{4}{x^2 + y^2}\\
&\neq 0.
\end{align*}

I've done this several times by hand and keep having this issue, so I assume the issue is not a simple calculation error on my part but a real conceptual misunderstanding about how to do these calculations.

It seems like just interchanging which function is $f$ and which is $g$ might change the sign I wind up with, but I don't see how it could change the result from 0 (which is what it's supposed to be) to something nonzero.

Any insights would be appreciated! Please remember that I have very little experience with differential forms.

答: If $\phi$ is a $p$-form and $\psi$ is a $q$-form, then $d(\phi\wedge\psi) = d\phi\wedge\psi + (-1)^p\phi\wedge d\psi$. In the second case, $f$ is a one-form and $g$ is a $0$-form, so $d\alpha = d(f\wedge g) = df\wedge g + (-1)^1f\wedge dg = df\wedge g \color{red}{-} f\wedge dg$.

Title: Error with Differential Forms


答:
Doing it below with random variables — same idea than in your case (where it s the event `"something happens to that sequence of random variables")

The difference between $X=a$ and $X=a$ a.s. (almost surely, i.e. with probability one) is better understood when you actually see a random variable as what is it: a function from some set $\Omega$ equipped with a probability distribution to some set of values $\mathcal{X}$.

"$X=a$" means "$\forall\omega\in\Omega,\ X(\omega)=a$", i.e "$\{\omega\in \Omega: X(\omega)=a\}=\Omega$".

However, "$X=a$ a.s" (or, equivalently, $\mathbb P\{X=a\}$) only means the measure of the set $\{\omega\in \Omega: X(\omega)=a\}$ is 1. There might be some $\omega$'s for which $X(\omega)\neq a$; but the measure of the set of such $\omega$'s is $0$ (that is, the set of "bad" $\omega$'s is negligible).

Title: What does a probability of $1$ mean?

Body:
From a textbook on probability on the Law of Large Numbers:

Theorem 3-19 (Law of Large Numbers): Let $X_1,X_2, \ldots , X_n$ be mutually independent random variables (discrete or continuous), each having finite mean  and variance. Then
  if $S_n = X_1 + X_2 +\dots+ X_n$,

$$ \lim_{n \to\infty} P\left(\left|\frac{S_n}{n} - \mu\right| \geq \varepsilon\right) = 0 $$

Since $S_n$ is the arithmetic mean of $X_1,X_2, \ldots , X_n$, this theorem states that the probability of the arithmetic mean $\frac{S_n}{n}$ differing from its expected value $\mu$ by more than $\varepsilon$ approaches zero as $n \to \infty$. A stronger result, which we might expect to be true, is that $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $  but this is actually false. However, we can prove that $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $
with probability one.

The only difference between the last sentence and the one before that is the phrase 'with probability one'. What does probability one mean here ? The usual definition is that the event occurs with 100% certainty. If that is the case, why is the original assertion $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $ false ?


问:
maximum and minimum value of $(a+b)(b+c)(c+d)(d+e)(e+a).$

Let $a,b,c,d,e\geq  -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$

I couldn't proceed much, however I think I got the minimum and maximum case.

For minimum, we get $-512$ with equality on $(-1,-1,-1,-1,9).$

For maximum, we get $288$ with equality on   $(-1,-1,-1,4,4).$

答:
This is the problem of China national in math olympiad here is my proof:
.
.
.
Wlog let $|e|=Max(|a|,|b|,|c|,|d|,|e|)$ so we have $(e+d)(e+a)>0$ (take note we are to find the min at first ) for it to be negative we have $2$ cases one is that all $3$ of $(b+c),(c+d),(a+b)$ are negative which we have from AM_GM that :
$$(-a-b)(-b-c)(-c-d)(d+e)(e+a)\le (\frac{-a-2b-2c-d}{3})^3(\frac{a+d+2e}{2})^2=(\frac{a+d+2e-10}{3})^3(\frac{a+d+2e}{2})^2 \le 512$$
so its done the second case is that only one of them is negative. Let $a+b$ be negative (the way we use AM_GM gives us this freedom to assume this) now we have
$$(-a-b)(b+c)(c+d)(d+e)(e+a)\le (\frac{2(c+d+e)}{5})^5=(\frac{2(5-a-b)}{5})^5\le(\frac{14}{5})^5<  512$$
so we are done here. now we want to find the Max which we do in the ofllowing way.
if all of $a+b,b+c,c+d$ are possitive we have from AM_GM that
$$S\le (\frac{2(a+b+c+d+e)}{5})^5=32<288.$$
the next case is the following . we have $a+b\le0$, $b+c\le0$, $c+d\ge0$,  then by AM_GM
$$(-a-b)(-b-c)(c+d)(d+e)(e+a)\le (\frac{-a-2b-c}{2})^2(\frac{c+d+e+a}{2})^2(d+e)=(\frac{d+e-b-5}{2})^2(\frac{5-b}{2})^2(d+e)\le 288$$
the last case is :
$a+b\le0$, $b+c\ge0$, $c+d\le0$, then by AM_GM we have
$$S=(-a-b)(b+c)(-c-d)(d+e)(e+a)\le (\frac{-a-b-c-d}{2})^2(\frac{b+c+d+e+e+a}{3})^3=(\frac{e-5}{2})^2(\frac{5+e}{3})^3< 288$$
and done done.