问:
Metrics on $SO(n+1)/SO(n)$
I am having a question about finding the metric on $SO(n+1)/SO(n)$ such that
$\pi:SO(n+1)\rightarrow SO(n+1)/SO(n)$ is a Riemannian submersion, which $SO(n)$ are equipped with the standard bi-invariant metric. I try to use Schur's Lemma to show that it is isometric to the canonical metric on $S^n$ multiplied by a constant. I am confusing about how to find this constant directly.

答:
First, you are right that there is a unique (up to scaling) metric on $S^{n-1}$ that makes $\pi$ into a Riemannian submersion, as I explain in my answer here.  Note that the isotropy action in this case is transitive on the unit sphere, so is definitely irreducible.

Now that we know there is a constant we can scale by, let's figure it out.  I'm not exactly sure what you mean by the "standard bi-invariant metric" on $SO(n)$, but the bi-invariant metric I like to use is defined on $T_I SO(n)$ by $\langle X,Y\rangle =-Tr(XY)$.

The function $\pi:SO(n)\rightarrow S^{n-1}$ I'm going to use is $\pi(A) = A_n$ where $A_n$ denotes the last column of $A$.  This means that the preimage of the point $p=(0,...,0,1)\in S^n$ corresponds to matrices of the block form $diag(B,1)$ with $B\in SO(n-1)$.

Consider the tangent vector $\alpha'(0)\in T_p S^{n-1}$ with $\alpha(t) = (0,....,\sin(t),\cos(t))$.  Note that $\|\alpha'(0)\| = 1$ in the usual metric on $S^{n-1}$.

Now, the identity matrix $I\in SO(n)$ is an element of $\pi^{-1}(p)$, so let's find a tangent vector in$ (\ker \pi_\ast)^\bot\subseteq T_I SO(n) = \mathfrak{so}(n)$ which projects to $\alpha'(0)$.  (The notation $\pi_\ast$ refers to the differential $\pi_\ast: T_I SO(n)\rightarrow T_p S^{n-1}$.)  Then we can compute the length of this tangent vector to find out the scaling we need to have a Riemannian submersion.

To that end, first note that because $\pi$ is constant on the orbit $I \,\cdot SO(n-1)$, it follows that $\ker \pi_\ast$ contains $\mathfrak{so}(n-1)$, embedded in $\mathfrak{so(n)}$ as matrices with the block form $diag(B,0)$ with $B\in \mathfrak{so}(n-1)$.  Since $\pi$ is a subermsion, the kernel of $\pi_\ast$ cannot be any larger, so $\ker \pi_\ast = \mathfrak{so}(n-1)$.  A reasonable straightforward calculation now shows that $(\ker \pi_\ast)^\bot = \{M = (M)_{ij}\in \mathfrak{so}(n): M_{ij} = 0$ if both $i,j < n\}.$  In other words, $\ker \pi_\ast^\bot$ consists of matrices of the form $$M = \begin{bmatrix} 0 & \cdots & 0 & m_{1,n}\\ 0 & \cdots & 0 & m_{2,n}\\
\vdots &  & \ddots & \vdots\\ -m_{1,n} & -m_{2,n} & \cdots & 0\end{bmatrix}.$$

Now, consider $\gamma:\mathbb{R}\rightarrow SO(n)$ with $\gamma(t) = diag\left(1,...,1, \begin{bmatrix} \cos t & \sin t\\ -\sin t & \cos t\end{bmatrix}\right)$.  Then $\gamma(0) = I$ and $\gamma'(0)$ is a matrix whose only non-zero entries are $\gamma'(0)_{n-1,n} = -\gamma'(0)_{n,n-1} = 1$.  It follows that $\gamma'(0)\in (\ker\pi_\ast)^\bot.$

Finally, note that $\pi \circ \gamma = \alpha$, so $\pi_\ast(\gamma'(0)) = \alpha'(0)$.

Now, an easy calculation shows that $\langle \gamma'(0),\gamma'(0)\rangle  = 2$.  Since $\langle \alpha'(0), \alpha'(0)\rangle = 1$, we see that the submersion metric on $S^{n-1}$ is the usual metric scaled by a factor of $2$.

Title: Metrics on $SO(n+1)/SO(n)$


问:
How to make $f(x)$ differentiable.
How to solve \begin{equation*}
f(x) =  \left\{
        \begin{array}{ll}
            2x+b &\quad x < a \\
            x^{2} & \quad x \geq a\\
        \end{array}
    \right.
\end{equation*}

This only uses definition of a derivative, to make $f(x)$ differentiable at $a$.

答:
To make $f$ differentiable at $a$ you need $f'(a)$ to exist, so you need both limits $\lim\limits_{x\to a^+}\dfrac{f(x)-f(a)}{x-a}$ and $\lim\limits_{x\to a^-}\dfrac{f(x)-f(a)}{x-a}$ to exist and equal each other. Now, before you make $f$ differentiable you need to make sure it is continuous, since differentiability implies continuity.

To make $f$ continuous you need to make sure $\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a^-}f(x)=f(a)$. Once you've done it, try to see if you can get differentiability.

Note: Usually, to make $f$ differentiable at $a$, we compare $\lim_{x\to a^+}f'(x)$ and $\lim_{x\to a^-}f'(x)$, hoping they give the same value. This is actually stronger, since it says that $f$ is continuously differentiable. Since you said you have to use the definitions of differentiability, I wrote the exact thing, and not this unnecessarily (but common) stronger way to tackle the problem.

ContentLicense: CC BY-SA 4.0

Title: How to make $f(x)$ differentiable.


问:
Proof an Equation is Wrong

“For all numbers $j$ and $k$, $(j + k)^2 = j^2 + k^2$.”

How would you prove this is wrong??

When I times out the bracket with the number it makes $j^2 + k^2$ but that's wrong.

Should I put a number for $j$ and $k$ and then times it out?

答:
We develop this product term by term
$$(j+k)^2=(j+k)\times(j+k)=(j\times j)+\underbrace{(j\times k)+(k\times j)}_{=2\times j\times k}+(k\times k)$$
Therefore, we find this identity that you must learn by heart:
$$(j+k)^2=j^2+2jk+k^2$$
hence you can see that $$(j+k)^2=j^2+\underbrace{2jk}_{\text{a term more}}+k^2\neq j^2+k^2\quad\text{unless}\quad 2jk=0\quad\text{i.e unless}\quad j=0\,\text{or}\, k=0 $$</p>

ContentLicense: CC BY-SA 3.0

Title: Proof an Equation is Wrong


答:
First, we must have $\lim_{x\rightarrow a^{-}}f(x) = \lim_{x\rightarrow a^{+}}f(x)=f(a)$ for continuity. Then:

 $$2a + b = a^{2}$$

Next, we must have $\lim_{x\rightarrow a^{-}}f'(x) = \lim_{x\rightarrow a^{+}}f'(x)$ for differentiability. This gives:

$$\frac{d}{dx}(2x + b)\bigg\vert_{a} = \frac{d}{dx}(x^{2})\bigg\vert_{a}$$

$$2 = 2a\Rightarrow a = 1$$

Then,

 $2(1) + b = 1$, so $b = -1$, and $\boxed{(a, b) = (1, -1).}$

Title: How to make $f(x)$ differentiable.

Body:
How to solve \begin{equation*}
f(x) =  \left\{
        \begin{array}{ll}
            2x+b &\quad x < a \\
            x^{2} & \quad x \geq a\\
        \end{array}
    \right.
\end{equation*}

This only uses definition of a derivative, to make $f(x)$ differentiable at $a$.


问:
For any square matrix F, is it possible that $\operatorname{rank}(F F^T-F^T F) = 1?$
Given $F$ is any square matrix, let $F^T$ be the conjugate transpose.

Is it possible that $\operatorname{rank}(F F^T-F^T F) = 1?$?

It seems that it is impossible, but I don't know how to prove it.

On the other hand, is there specific name for the matrix in the form of $F F^T-F^T F$?

答:
since you mention conjugate transpose, I infer that the field is $\mathbb C$. The answer then is no.

$A:= FF^* - F^*F$

$A$ is Hermitian and hence diagonalizable.  This means $\text{rank}\big(A\big)=1$ can only occur because $\lambda_1 \neq0$ and $\lambda_k = 0$ for $k\in\big\{2,...,n\big\}$, but supposing this is true

$\lambda_1 $

$=  \lambda_1 + \big(\sum_{k=2}^n \lambda_k\big)$

$= \text{trace}\big(A\big) $

$= \text{trace}\big(FF^*-F^*F\big)$

$= \text{trace}\big(FF^*\big)-\text{trace}\big(F^*F\big)$

$=\text{trace}\big(FF^*\big)-\text{trace}\big(FF^*\big)$

$=0$

which is a contradiction


问:
Does a cube give the maximum volume for a fixed surface area?
If I am making a fish tank with an open lid and $1600 \,\rm{cm}^2$ of material, why is the maximum volume that the tank can occupy not given by a cube?

My work

Suppose the base of the fish tank is a square, with length $x$cm and height $h$cm.

Then the volume $V$ of the tank will be
\begin{align}
V = x^2h. \tag{1}
\end{align}
Given the constraint of the material,
\begin{align}
1600 = x^2 + 4xh. \tag{2}
\end{align}
Eliminating $h$: $$V = 400x - \frac{1}{4}x^3.$$
The maximum volume will be given by: $$\frac{dV}{dx} = 400 - \frac{3}{4}x^2 =0.$$
Hence $x = +\frac{40\sqrt{3}}{3}.$

Then using $(2)$, we get $h = \frac{20\sqrt{3}}{3}.$

Therefore (using $(1)$) the volume of the tank will be about $6158$cm$^3$.

But if the tank is a cube, then the volume will be $$\Bigg(\sqrt{\frac{1600}{5}}\, \Bigg)^3 = 5724.334022\text{cm}^3.$$

答:
You have already solved the problem, so you know the numerical reason.

The intuition is that if you put two of these open boxes together across the open face, you get a closed box with twice the surface area ($3200$) and twice the volume.

We know that the volume of the closed box is maximized when it's a cube (with side length $\sqrt{\frac{3200}{6}} = \frac{40\sqrt3}{3}$). So the optimal open box should be half of that cube.

That's exactly the solution you've found.


问:
Functions of infinite unions equality
There's a problem I can't really wrap my head around, and its this:

Let $f : A → B$ and $A_i \subset A$ for every $i \in \mathbb{N}$. Prove that

$f\left(\bigcup\limits_{i=1}^{\infty} A_{i}\right) = \bigcup\limits_{i=1}^{\infty} f(A_{i})$

How would I treat both finding $\subseteq$ and $\supseteq$? I would typically go about element chasing, and begin with $y \in f\left(\bigcup\limits_{i=1}^{\infty} A_{i}\right)$, then there $\exists i \in \mathbb{N}$ and $x \in A_i$ such that $y = f(x)$, and because of generality we get the RHS. I don't feel like this is sufficient reasoning, so could I be given a bit of direction of polishing my reasoning?

答:
I'm not clear why you "don't feel like this is sufficient reasoning", so I'll write out an answer in detail and maybe that will make it clearer what is the source of this feeling.

$\subseteq$: Suppose $y \in f \left( \bigcup_{i=1}^\infty A_i \right)$.  Then there is an $x \in \bigcup_{i=1}^\infty A_i$ such that $y = f(x)$.  So there is an $i \in [1,\infty)$ such that $x \in A_i$ and so $y \in f(A_i)$.  Since $f(A_i) \subseteq \bigcup_{i=1}^\infty f(A_i)$, we obtain $y \in \bigcup_{i=1}^\infty f(A_i)$.  Therefore, $f \left( \bigcup_{i=1}^\infty A_i \right) \subseteq \bigcup_{i=1}^\infty f(A_i)$.

$\supseteq$: Suppose $y \in \bigcup_{i=1}^\infty f(A_i)$.  Then there is an $i \in [1,\infty)$ such that $y \in f(A_i)$.  So there is an $x \in A_i$ such that $y = f(x)$.  Since $x \in A_i \subseteq \bigcup_{i=1}^\infty A_i$, we obtain $y \in f\left( \bigcup_{i=1}^\infty A_i \right)$.  Therefore, $\bigcup_{i=1}^\infty f(A_i) \subseteq f\left( \bigcup_{i=1}^\infty A_i \right)$.

Title: Functions of infinite unions equality


问:
Poisson or Exponential Distribution?
There are $10$ questions in a two-hour test. Suppose that the time taken for Jackson to answer each equations is independent of each other and has exponential distribution with $\lambda = 0.08$ questions per min.

What is the probability that he will take more than $30$ min to answer the first $3$ questions and less than $30$ min to answer the next $3$ questions?

My initial thought would be to use exponential distribution of $\lambda =\frac2{75}$ to solve the question, by finding $$P(Y < 30)\times(1 - P(Y < 30)),$$ where Y is the time taken for Jackson to finish 3 questions.

However, I have contesting views on using Poisson distribution as well. Perhaps the community could enlighten me. Cheers!

答:
\begin{align}
& \Pr( \,\overbrace{\text{time to answer the first three}}^{\begin{smallmatrix} \text{the sum of three exponentially} \\ \text{distributed random variables} \end{smallmatrix} }{} > 30 \text{ minutes}\, ) \\[8pt]
= {} & \Pr( \, \underbrace{ \text{number of questions answered in 30 minutes}}_\text{a Poisson-distributed random variable} {} <3 )
\end{align}

Numerically, finding the second of these two numbers is quick and easy, so my way of finding the first one would be to find the second one and rely on your understanding of why they must be equal.

Title: Poisson or Exponential Distribution?


问:
Let $g'(x)$ be constant. Find the fifth derivative of $f\circ g(x)$.
I have a question with some of my calculus homework. Here's the question:

Let $g'(x)$ be constant. Find the fifth derivative of $f\circ g(x)$.

First, is $f\circ g(x)$ just $fg(x)$? And how do I go about starting this question?
If $g'(x)$ is a constant, would that mean $g(x)$ equals a number?

Thanks in advance!

答:
We have
$$(f\circ g)'(x)=f'(g(x))\cdot g'(x)$$ by the Chain Rule. Let $g'(x)=c$ for some constant $c.$ Thus, we have
$$(f\circ g)'(x)=f'(g(x))\cdot g'(x)=c\cdot f'(g(x)).$$
Claim that $$(f\circ g)^{(n)}=c^n\cdot f^{(n)}(g(x))$$ for all integers $n>0.$ Clearly, as we have shown above, this holds for $n=1.$
Now for the inductive step. Assume that $$(f\circ g)^{(n)}=c^n\cdot f^{(n)}(g(x)).$$ Then, we have
\begin{align*}
((f\circ g)^{(n)})'&=(c^n\cdot f^{(n)}(g(x)))'\\
&=(c^n)'\cdot f^{(n)}(g(x))+c^n\cdot (f^{(n)}(g(x)))'
\end{align*} by the Product Rule. Since $(c^n)'=0,$ we have
\begin{align*}
(c^n)'\cdot f^{(n)}(g(x))+c^n\cdot (f^{(n)}(g(x)))'&=0(f^{(n)}(g(x)))+c^n\cdot f^{(n+1)}(g(x))\cdot g'(n)\\
&= c^{n+1}\cdot f^{(n+1)}(g(x))
\end{align*} by the Chain Rule and that $g'$ is constant, which is want we want. Thus the inductive step is done. Putting in $n=5,$ the fifth derivative of $(f\circ g)$ is $c^5\cdot f^{(5)}(g(x))=\boxed{(g'(x))^5\cdot f^{(5)}(g(x))}.$
$\blacksquare$


问:
Why does every countable limit ordinal have cofinality $\omega$?
According to Wikipedia, if $\alpha$ is a countable limit ordinal, then $\mathrm{cf}(\alpha)=\omega$. It is intuitively clear to me that it should be so. Certainly the cofinality of such an ordinal must be $\geq\omega$. We can now look at what the ordinal ends with. If it has a single $\omega$ at the end, then that is a cofinal subset and we are done. It may, however, not end in $\omega$. It could, for example, end in $\omega$ $\omega$s or $\omega^2$. Then we can pick one element of each $\omega$ constituting the $\omega^2$, and the set of these elements is cofinal in $\alpha$, and its type is $\omega.$ If $\alpha$ ends in $\omega^3$, we can take one element of each $\omega^2$ constituting the $\omega^2$, and get a cofinal subset of type $\omega.$

It seems to me that this reasoning should work. It should probably be done inductively. How should I do it?

答: Hint:Let $\gamma$ be a countable ordinal. Enumarate $\gamma=\{\alpha_n:n<\omega\}$, construct $\beta_n$ as $\beta_n=\alpha_{m}$, where $m$ is the least $m<\omega$ with $\alpha_m>\beta_{n-1}$, then what is $\lim_{n\rightarrow \omega}\beta_n$?


问:
Finding a sequence of integrable functions on $[0,1]$ which is pointwise convergent to zero whose integrals increase without bound
As the title suggests I am looking for examples of sequence of integrable functions on $[0,1]$ which are pointwise convergent to zero. The sequence of integrals $\left\{\int_{0}^{1}f_n\right\}_{n=1}^{\infty}$ needs to increase without bound.

I've tried numerous integrable functions that are pointwise convergent to zero but I have yet to pinpoint one that satisfies the condition about integrals increasing without bound. Every single function I've found gives me smaller and smaller integrals. This is frustrating, to say the least.

Body: $$f_n(x)= \cases{n^2  & if $\ \ 0<x <1/n$ \\ 0 & otherwise}$$


问:
Combinatorics problem - $5$-digit code with $3,4,5$ present
I have a combinatorics problem that I'm struggling with. Here it is: "How many 5-digit codes have among their digits each of $3,4,5$?"

I do understand the general strategy, where we first see how we can permute $3,4,5$ with the other empty slots ($\frac{5!}{2} = 60$ ways) and then we multiply by $10^2$ (both of the other two slots can have any of the $10$ digits). However now we end up double counting, as some of those other two digits will be $3,4,5$ and will result in duplicating combinations between each other. I am struggling to come up with a formula for excluding these.

答:
There are a couple of strategies we could employ here.

Method 1:  We consider cases depending on how often each digit appears.

Case 1:  The digits $3$, $4$, $5$ each appear exactly once.

Choose one of the positions for the $3$, one of the remaining four positions for the $4$, and one of the remaining three positions for the $5$.  Then each of the remaining two positions may be filled with one of the remaining $10 - 3 = 7$ digits.  Hence, there are

  $$5 \cdot 4 \cdot 3 \cdot 7^2$$

such codes.

Case 2:  Exactly one of the digits $3$, $4$, $5$ appears twice and each of the others appears exactly once.

Choose which of the digits $3$, $4$, $5$ appears twice.  Choose two of the positions for that digit.  Choose one of the remaining three positions for the smaller of the two remaining digits from the set $\{3, 4, 5\}$ and one of the remaining two positions for the remaining digit from the set $\{3, 4, 5\}$.  Choose which of the remaining seven digits fills the remaining position.  There are

 $$\binom{3}{1}\binom{5}{2}\cdot 3 \cdot 2 \cdot 7$$

such codes.

Case 3:  Exactly two of the digits $3, 4, 5$ appear twice and the other one appears once.

Choose which of the three digits $3, 4, 5$ appears exactly once.  Choose which of the five positions that digit fills.  Choose two of the remaining four remaining positions for the smaller of the two remaining digits from the set $\{3, 4, 5\}$, then fill the remaining two positions with the remaining digit from the set $\{3, 4, 5\}$.  There are

 $$\binom{3}{1}\binom{5}{1}\binom{4}{2}$$

such codes.

Case 4:  Exactly one of the three digits $3, 4, 5$ appears three times and each of the others appears once.  Choose which of the three digits appears three times, then choose three of the five positions for that digit.  Choose one of the remaining two positions for the smaller of the remaining digits from the set $\{3, 4, 5\}$, then fill the final position with the other digit.  There are

 $$\binom{3}{1}\binom{5}{3}\binom{2}{1}$$

such codes.

Total:  Since these positions are mutually exclusive and exhaustive, add the above cases.

Method 2:  We use the Inclusion-Exclusion Principle.

There are $10^5$ codes.  We wish to exclude from these those in which at least one of the digits $3, 4, 5$ is missing.

We choose which of the three digits $3, 4, 5$ to exclude, which leaves nine ways to fill each of the five positions.  Thus, we subtract

 $$\binom{3}{1}9^5$$

from the total.

However, if we do so, we will have subtracted each case in which two of the digits $3, 4, 5$ are missing twice, once for each way we could have designated one of those digits as the missing digit.  We only want to subtract such cases once, so we must add them to the total.

We choose which two of the three digits $3, 4, 5$ to exclude, which leaves us eight ways to fill each of the five positions.  Thus, we add

 $$\binom{3}{2}8^5$$

to our running total.

However, if we first subtract those cases in which one of the digits $3, 4, 5$ is excluded and then add those cases in which two of the digits $3, 4, 5$ are excluded, we will not have excluded those cases in which all three digits $3, 4, 5$ are excluded at all.  This is because we first subtracted them three times, once for each way we could have designated one of those three digits as the excluded digit.  We then added these cases three times, once for each of the $\binom{3}{2}$ ways we could have designated two of those three digits as the excluded digits.  Thus, we must them from the total.

If all three digits $3, 4, 5$ are excluded, then we have seven choices for each of the five positions.  Thus, there are

 $$\binom{3}{3}7^5$$

cases in which all the digits $3, 4, 5$ are excluded.

By the Inclusion-Exclusion Principle, there are

 $$10^5 - \binom{3}{1}9^5 + \binom{3}{2}8^5 - \binom{3}{3}7^5$$

admissible codes.

As you can verify, the two methods yield the same answer.


Title: Finding a sequence of integrable functions on $[0,1]$ which is pointwise convergent to zero whose integrals increase without bound

Body:
As the title suggests I am looking for examples of sequence of integrable functions on $[0,1]$ which are pointwise convergent to zero. The sequence of integrals $\left\{\int_{0}^{1}f_n\right\}_{n=1}^{\infty}$ needs to increase without bound.

I've tried numerous integrable functions that are pointwise convergent to zero but I have yet to pinpoint one that satisfies the condition about integrals increasing without bound. Every single function I've found gives me smaller and smaller integrals. This is frustrating, to say the least.

Body:
Take
for $ n>0$, and $ x\in[0,1]$ :

$$f_n(x)=0 \;\;if\; \;x\ge \frac{1}{n^2},$$

$$f_n(x)=n^2e^n\; \; if\; \; 0<x<\frac{1}{n^2}$$

and $$f_n(0)=0$$

Check that
$$\int_0^1f_n(x)dx=e^n$$


问:
Exponential distribution Max(X,Y)
This problem appeared in a past Quiz. I'm still not able to solve it:

Note(λ = 1/B). Suppose X~exponential(λ = 1), Y ~ exp(λ = 1) and Z ~ exp ( λ = 3)

Find P(max(X,Y) > 10) and E(max(X,Z)).

I know that for exponential distribution

max(X,Y) = min(X,Y) + R. (Where R is the remaining lifetime).

However I don't know how to use this fact to solve a problem

答:
\begin{align}
& \frac d {dx} \Pr( \max\{X,Z\} \le x) = \frac d {dx} \Pr(X\le x\ \&\ Z\le x) \\[10pt]
= {} & \frac d {dx} \big( (1 - e^{-x})(1-e^{-3x}) \big) = \frac d{dx}( 1 -e^{-x} - e^{-3x} + e^{-4x}) \\[8pt]
= {} & e^{-x} + 3e^{-3x} - 4e^{-4x}.
\end{align}
So look for $\displaystyle\int_0^\infty x(e^{-x} + 3e^{-3x} - 4e^{-4x}) \, dx,$ remembering that $\displaystyle\int_0^\infty xe^{cx} \, dx = \frac 1 {c^2}. $

Title: Exponential distribution Max(X,Y)


问:
How to derive the axiom no. 15 from the Cantor's and Archimedean axiom?
How could one substitute the (15th) axiom of completeness with Archimedean and Cantor's axiom?

We discussed Cantor's axiom as well as Archimedean in analysis lectures and were told this question might appear on the exam. The first 14 axioms which also work for $\Bbb Q$ are left, and I need to derive, as mentioned in comments, the axiom of completeness of the field $\Bbb R.$
I am looking for proof that the supremum of a non-empty $S\subset(0,+\infty)$ is unique, therefore I have to rewrite the statement that, no matter how small positive real number $\varepsilon$ I subtract from the supremum, there will be some $x\in S$ such that $$\sup S-\varepsilon<x.$$
But, since I need to use the Archimedean axiom, I tried to use it here, but I'm not sure how.
Then I wanted to replace the term supremum when describing the properties of the non-empty intersection segments in Cantor's theorem using the statement:
$\forall x,y\in \Bbb R^+, \exists n\in\Bbb N,$ such that $ny\ge x$

Am I at least at the right starting trail?

答:
We can use the Cantor-intersection theorem and the Archimedean axiom to prove the Dedekind's theorem (source:Enrico Gregorio's answer).

Suppose $A,B\subseteq\Bbb R$ are non-empty s. t. $(\forall a\in A)(\forall b\in B)\quad a\le B$ and $A\cup B=\Bbb R$. Then $\exists ! c\in\Bbb R$ s. t. $a\le c\le b, \forall a\in A,\forall b\in B$.

Let's abbreviate the above with $A\le B$ and $A\le c\le B$.

Since $A\ne\emptyset\space\land\space B\ne\emptyset$, let's choose $a_0\in A$ and $b_0\in B$.

Now, we choose $c_0=\frac{a_0+b_0}2$.

If $A\le c_0\le B$, we're done. Otherwise, it is either
$$c_0b\text{ for some }b\in B.$$

If $c_0<a$, then let $a_1=a, b_1=b_0$.

If $c_0>b$, then let $a_1=a_0, b_1=b$.

In either case, $a_0\le a_1\space\land\space b_0\ge b_1\implies [a_0,b_0]\supseteq[a_1,b_1]$.

Let $d=b_0-a_0$. We have $b_0-c_0=c_0-a_0=\frac{b_0-a_0}2=\frac{d}2$.

If $c_0<a$, then $\frac{d}2=b_0-c_0>b_0-a=b_1-a_1$.

If $c_0>b$, then $\frac{d}2=c_0-a_0>b-a_0=b_1-a_1$.

We can repeat the procedure to build a sequence of nested closed intervals:

$$[a_0,b_0]\supseteq[a_1,b_1]\supseteq\cdots\supseteq[a_n,b_n]$$

If at some point $A\le c_n\le B$, we're done. Otherwise, we get a sequence of closed nested intervals $[a_n,b_n]$ with $a_n\in A,b_n\in B$ and $b_n-a_n\le\frac{d}{2^n}.$

By the Cantor-intersection theorem (now assumed as an axiom), there is $c\in\bigcap_n[a_n,b_n]$.

Now we want to prove $A\le c\le B$. Let's assume the opposite: there is either

$$a\in A\text{ with } cb.$$

Suppose $a>c$.

Let's take $\varepsilon=\frac{a-c}2$.
We can take $n\in\Bbb N$ s. t. $\frac{d}{2^n}<\varepsilon$. $(*)$

Then we obtain:
$b_n<a+\varepsilon\le c+\varepsilon=c+\frac{a-c}2=\frac{a+c}2\le a$ which contradicts $A\le B$.

Analogously in case of $c>b$.

$(*)$ As $2^n>n, n\in\Bbb N$, we  have $\frac1{2^n}<\frac1n\implies\frac{d}{2^n}<\frac{d}n$. In order to make $\frac{d}{2^n}<\varepsilon$, we could take $n$ s. t. $\frac{d}n<\varepsilon$, which is guaranteed to hold true by the Archimedean axiom.

Now, we want to prove the Dedekind's theorem is equivalent to the axiom of completeness (like Henno Brandsma did here):

Every non-empty subset of $\Bbb R$ that is bounded above has a supremum in $\Bbb R$.

Direction $\boxed{\Rightarrow}$

Suppose Dedekind's theorem holds.

Let $S\subset\Bbb R$ with some upper bound $u$ and let's define:
$$A=\{x\in\Bbb R\mid\exists s\in S, x<s\}\\\text{and}\\ B=\{x\in\Bbb R\mid\forall s\in S, x\ge s\}$$

$B\ne\emptyset$ because $u\in B$.

Then our $c=\sup S$.

Direction $\boxed{\Leftarrow}$

Assume the axiom of completeness holds.

Then we take $c=\sup A$.

Body:
We can use the Cantor-intersection theorem and the Archimedean axiom to prove the Dedekind's theorem (source:Enrico Gregorio's answer).

Suppose $A,B\subseteq\Bbb R$ are non-empty s. t. $(\forall a\in A)(\forall b\in B)\quad a\le B$ and $A\cup B=\Bbb R$. Then $\exists ! c\in\Bbb R$ s. t. $a\le c\le b, \forall a\in A,\forall b\in B$.

Let's abbreviate the above with $A\le B$ and $A\le c\le B$.

Since $A\ne\emptyset\space\land\space B\ne\emptyset$, let's choose $a_0\in A$ and $b_0\in B$.

Now, we choose $c_0=\frac{a_0+b_0}2$.

If $A\le c_0\le B$, we're done. Otherwise, it is either
$$c_0<a,\text{for some }a\in A\space\text {or}\space c_0>b\text{ for some }b\in B.$$

If $c_0<a$, then let $a_1=a, b_1=b_0$.

If $c_0>b$, then let $a_1=a_0, b_1=b$.

In either case, $a_0\le a_1\space\land\space b_0\ge b_1\implies [a_0,b_0]\supseteq[a_1,b_1]$.

Let $d=b_0-a_0$. We have $b_0-c_0=c_0-a_0=\frac{b_0-a_0}2=\frac{d}2$.

If $c_0<a$, then $\frac{d}2=b_0-c_0>b_0-a=b_1-a_1$.

If $c_0>b$, then $\frac{d}2=c_0-a_0>b-a_0=b_1-a_1$.

We can repeat the procedure to build a sequence of nested closed intervals:

$$[a_0,b_0]\supseteq[a_1,b_1]\supseteq\cdots\supseteq[a_n,b_n]$$

If at some point $A\le c_n\le B$, we're done. Otherwise, we get a sequence of closed nested intervals $[a_n,b_n]$ with $a_n\in A,b_n\in B$ and $b_n-a_n\le\frac{d}{2^n}.$

By the Cantor-intersection theorem (now assumed as an axiom), there is $c\in\bigcap_n[a_n,b_n]$.

Now we want to prove $A\le c\le B$. Let's assume the opposite: there is either

$$a\in A\text{ with } c<a\space\text{ or }\space b\in B\text{ with } c>b.$$

Suppose $a>c$.

Let's take $\varepsilon=\frac{a-c}2$.
We can take $n\in\Bbb N$ s. t. $\frac{d}{2^n}<\varepsilon$. $(*)$

Then we obtain:
$b_n<a+\varepsilon\le c+\varepsilon=c+\frac{a-c}2=\frac{a+c}2\le a$ which contradicts $A\le B$.

Analogously in case of $c>b$.

$(*)$ As $2^n>n, n\in\Bbb N$, we  have $\frac1{2^n}<\frac1n\implies\frac{d}{2^n}<\frac{d}n$. In order to make $\frac{d}{2^n}<\varepsilon$, we could take $n$ s. t. $\frac{d}n<\varepsilon$, which is guaranteed to hold true by the Archimedean axiom.

Now, we want to prove the Dedekind's theorem is equivalent to the axiom of completeness (like Henno Brandsma did here):

Every non-empty subset of $\Bbb R$ that is bounded above has a supremum in $\Bbb R$.

Direction $\boxed{\Rightarrow}$

Suppose Dedekind's theorem holds.

Let $S\subset\Bbb R$ with some upper bound $u$ and let's define:
$$A=\{x\in\Bbb R\mid\exists s\in S, x<s\}\\\text{and}\\ B=\{x\in\Bbb R\mid\forall s\in S, x\ge s\}$$

$B\ne\emptyset$ because $u\in B$.

Then our $c=\sup S$.

Direction $\boxed{\Leftarrow}$

Assume the axiom of completeness holds.

Then we take $c=\sup A$.


问:
Abstract Proof that Exponential Map is Surjective onto $\mathrm{GL}_n(\mathbb{C})$
It is well-known that the exponential map associated to any compact connected Lie group is surjective (the proof is a simple application of the Lefschetz fixed point theorem). As it happens, the exponential map associated to $\mathrm{GL}_n(\mathbb{C})$ is also surjective, although $\mathrm{GL}_n(\mathbb{C})$ fails to be compact. The only way I know how to prove this latter claim is by expressing each element of $\mathrm{GL}_n(\mathbb{C})$ in its Jordan canonical form and then showing that any Jordan block is the exponential of some matrix.

My question is this: is there a wider class of Lie groups (more general than just compact and connected, and perhaps including examples like $\mathrm{GL}_n(\mathbb{C})$) for which we can say that the exponential map is surjective?

答:
Not an answer, but another proof for surjectivity of $\exp$ for $\operatorname{GL}_n(\mathbb{C})$ is by the Cauchy integral formula for Banach space valued holomorphic functions:

If $x\in A$, a complex Banach algebra, and the spectrum $\sigma(x)$ of x does not separate $0$ from $\infty$, then $x$ has a logarithm (and also nth roots) in $A$. That is, there is a $y\in A$ with $x=\sum\frac{y^n}{n!}$. I seen this on page 264 of 'functional analysis' by Rudin.


问:
Formula for lines that can be drawn using $n$ points
Please help me!
How many lines can be drawn using $6$ points? Each line is made by connecting $2$ points.</p>

答: Depends on where the points are. If no three are on a line, then any two determine a line, so the answer is the number of ways to choose two things from six. Do you know how to do that?</p>


问:
$(n + a)^{b} = \Theta(n^{b})$ using limits?
At university we were told that $(n + a)^{b} = \Theta(n^{b})$ can be proven by finding $c_1$, $c_2$ and $n_0$.

$a, b \in \mathbb{R} $ with $b > 0$

The professor said it is not possible to solve this by using limits. But I do not quite understand why this is the case. Could somebody please tell me why limits cannot be applied here?

Thanks.

答:
It depends on what you are saying by "using limits". Here is a proof using the definition of a limit:

As $\dfrac{(n+a)^b}{n^b} \longrightarrow_{n\to \infty} 1$, then there exists $n_0 \geqslant 1$ such that for every $n \geqslant n_0$ :
\begin{align}
1-\dfrac{1}{2} \leqslant \dfrac{(n+a)^b}{n^b} \leqslant 1+\frac{1}{2}
\end{align}
(I applied the definition of the limit with $\varepsilon = \frac{1}{2}$) Thus, thus, for all $n \geqslant n_0$:
\begin{align}
\frac{1}{2}n^b \leqslant (n+a)^b \leqslant \frac{3}{2}n^b
\end{align}
and $(n+a)^b = \Theta(n^b)$.

But you cannot say that if $u_n$ and $v_n$ both have the same limit at infinity, then $u_n=\Theta(v_n)$. A counter-example is $u_n = n$ and $v_n = n^2$. They both go to infinity but $u_n \neq \Theta(v_n)$.


问:
Are there general tips for determining whether a graph is planar or nonplanar? (And, if planar, an approach for finding a planar embedding?)

Are there general tips for determining whether a graph is planar or nonplanar?

If a graph is planar, is there a general approach for finding a planar embedding? I usually try to draw out the vertices one by one, connecting them so that they don't intersect, and if there is an obstruction, I try to find a Kuratowski subgraph. However, I think this method doesn't usually work very well and is very tedious (especially for large graphs).

There are also quite a few theorems on Wikipedia on planarity testing and oftentimes stronger conditions than the basic test using Euler's formula (i.e. that $|E| \leq 3|V|-6$ or $|E|\leq 2|V|-4$ if the girth is greater than $3$) are required.  As an example, I was wondering if one could demonstrate the use of a general technique for the following graph, which I think is nonplanar:

答:
Here's what I did.  First, I redrew the graph with fewer edge crossings, to make it easier to look for a Hamilton cycle:

I did this by drawing the graph, more or less as pictured, in Geogebra. Then you can drag the points around, and Geogebra adjusts the connecting segments for you.  In this form, it's easy to see the Hamilton cycle EGFJDHIBKCAE.

So then I followed the procedure described in the answer I linked in my comment:

and found the graph to be non-planar.


问:
Does the cartesian product have a neutral element?
Let $A$ be any set. Is there a set $E$ such that $A \times E = E \times A = A$?

I thought of the empty set, but Wikipedia says otherwise. This operation changes dimension, so an isomorphism might be needed for such element to exist.

答: In some sense, the whole reason we have these things called addition and multiplication and the ring axioms is because of certain properties satisfied by the Cartesian product and disjoint union. Both are associative and commutative (up to natural isomorphism). One distributes over the other (up to natural isomorphism). Both have identity elements (up to natural isomorphism). Decategorify, restricting to finite sets, and you get the non-negative integers. Take the Grothendieck group, and you get the integers, and then at some point you are led to write down the ring axioms in general. But it's good to keep in mind where it all comes from. </p>


问:
Proving the given inequality
Let $a_1,...,a_n$ be positive numbers. Show that $(a_1+...+a_n)(\frac{1}{a_1}+...+\frac{1}{a_n})\ge n^2$. Just not really sure how to approach this problem, any tips/solutions would be greatly appreciated.

答: You should probablly use either the C-S inequality or you can devide the set of all a_i into bigger than one and smaller than one and then neglacte them from the opposite sum (the a's that are smaller than one neglect from the sum of a's to the power of one and vice versa)</p>

Title: Proving the given inequality


问:
Quotient of wreath group by commutator
[Self studying Robinson, ex 1.6.20]

Robinson asks us to prove that $G/[B,K] \cong(H/H') \times K$ (where $G=H \wr K$, $K\neq 1$ and $B$ is the base group.) As a hint, we are told to show first that $B'\leq [B,K]$.

I can prove the hint.  First, the generators of $B'=[B,B]$ are elements of the form $g^{-1}h^{-1}gh$.  Elements in different components of B commute, so w.l.o.g. g and h can be taken as elements of the same, single component of B, say the y component $H_y$ (where $y\in K$).  It is therefore enough to show that this $[g,h]\in [B,K]$

Since K is not 1 we can choose a non-trivial k in K.  Then $[g,k]$ is the element of B with $g$ as the $yk$ component and $g^{-1}$ as the $k$ component.

Then $[g,h] = [g,k][h,k][(gh)^{-1},k]$ and so lies in $[B,K]$.  $\square$

But now I have no idea how to proceed to the conclusion.  I need another hint, please.

答:
Hint: Consider the following map $f:G\to (H/H')\times K$:
$$((h_y)_{y\in K},\,k)\mapsto\left(\prod_y\overline{h_y},\,k\right)$$
where the elements $\overline{h_y}\in H/H'$ already commute.

Prove that this map is a surjective homomorphism and show that its kernel is just $[B,K]$, using your lemma for one direction.

Title: Quotient of wreath group by commutator


问:
Matrix Vector Product, as a Derivative?
Let $p\in\mathbb{R}^d$ and $A\in \mathbb{R}^{d\times d}$, can the product $Ap$ be the derivative of some function?

In particular given a matrix $A$, how can I construct $H:\mathbb{R}^d\to \mathbb{R}$,  such that $\nabla H(p)=Ap$. Note below I can do this for specific $A$ but I want it for general $A$ (say positive semi-definite, but possibly singular).

$\textbf{My Work :}$

If $A$ is the identity matrix then we have $\nabla H(p)=\nabla \frac{1}{2}\|p\|^2=p=Ap$.

If $A$ can be written as $B^TB$ for some matrix $B\in\mathbb{R}^{d\times d}$, then take $\nabla H(p)=\nabla \frac{1}{2}\|Bp\|^2=B^TBp=Ap$.

What about more general $A$?

答:
If $A$ is symetric you can set $H(p) = \frac{1}{2}p^TAp $.

If you then calculate the derivative using the chain rule you get $\nabla H(p) = Ap$.

If $A$ is not symetric you can't find such a function $H$.

For example let $a_{12} \neq a_{21}$.
You can write down the product of $Ap$ and start integrating the first element of the vector with respect to $p_1$:

Then

$H(p) = \frac{1}{2}a_{11}p_1^2 + p_1a_{12}p_2 + p_1a_{13}p_3 +...$

Similarly when you integrate the second element with respect to $p_2$ you get:

$H(p) = p_1a_{21}p_2 + \frac{1}{2}a_{22}p_1^2 +...$

So we get a contradiction. You can do a similar argument for any non symetrical matrix.

Title: Matrix Vector Product, as a Derivative?


问:
$\left| \int_Q \phi dx \right| \leq M \frac{m(Q)}{(1 + m(Q))} $ for every cube, then $\lim_{k \to \infty} \int \phi(kx)f(x) dx = 0$ for $f \in L^1$
Assume $\phi: \mathbb{R}^d \to \mathbb{R}$ and has the property that $\left| \int_Q \phi dx \right|  \leq M \frac{m(Q)}{(1 + m(Q))} $ holds for every rectangle $Q$ for some constant $M$ that is independent of $Q$, then $\lim_{k \to \infty} \int_{\mathbb{R}^d} \phi(kx)f(x) dx = 0$ for ever $f \in L^1(\mathbb{R}^d)$

I know that you can just verify the claim for an increasing sequence of cubes. If $Q_n = [-n,n] \times[-n,n] \times [-n,n]$ ($d$-times), then we can do a change of variables so that

\begin{equation}
\int_{Q_n} \phi(kx)f(x) dx = \frac{1}{k^d} \int_{Q_{nk}} \phi(y)f(y/k) dy
\end{equation}

This looks closer to what I want. If I can get some control of $f$ I could use the fact that $\frac{1}{k^d}\left| \int_Q \phi dx \right|  \leq \frac{M}{k^d} \frac{m(Q)}{(1 + m(Q))} $ and as $k \to \infty$ we would get the answer. But I do not know what I can do with the function $f$.

Any thoughts or suggestions would be appreciated.

答:
Define the family $\mathcal{D} \subset L^1(\mathbb{R}^d)$ by

$$ \mathcal{D} = \Biggl\{ \sum_{i=1}^{n} c_i \mathbf{1}_{Q_i} : \text{$c_1, \cdots, c_n$ constants, $R_1, \cdots, R_n$ rectangles} \Biggr\}.$$

Then $\mathcal{D}$ is a dense subset of $L^1(\mathbb{R}^d)$. Moreover, by the assumption,

$$ T_k(f) = \int_{Q} \phi(kx)f(x) \, \mathrm{d}x \tag{*} $$

defines a linear functional on $\mathcal{D}$. Now let us make some observations:

For each rectangle $Q$,
$$ \left| T_k(\mathbf{1}_Q) \right|
= k^{-d}\left| \int_{kQ} \phi(x) \, \mathrm{d}x \right|
\leq \frac{M}{k^d} \frac{m(kQ)}{1+m(kQ)}
\leq M m(Q). $$
In particular, the inequality in the second step tells that $T_k(\mathbf{1}_Q)$ converges to $0$ as $k \to \infty$. Then by linearity, $T_k(f) \to 0$ for any $f \in \mathcal{D}$.

For $f = \sum_{i=1}^{n} c_i \mathbf{1}_{Q_i} \in \mathcal{D}$, we may assume that $Q_i$'s are non-overlapping. Then
$$ \left| T_k f \right|
\leq \sum_{i=1}^{n} \left| a_i \right| \left| T(\mathbf{1}_{Q_i}) \right|
\leq M \sum_{i=1}^{n} \left| a_i \right| m(Q_i)
= M \| f \|_{L^1}. $$  This tells that each $T_k$ extends to a continuous linear functional on $L^1(\mathbb{R}^d)$, which we also write as $T_k$. Then the extension $T_k$ is again given by $\text{(*)}$ and the inequality $\left| T_k(f) \right| \leq M\|f\|_{L^1}$ continues to hold for $f \in L^1(\mathbb{R}^d)$.

Combining altogether, for any $f \in L^1(\mathbb{R}^d)$ and $g \in \mathcal{D}$,

$$ \limsup_{k\to\infty} \left| T_k(f) \right|
\leq \limsup_{k\to\infty} \left| T_k(g) \right| + M \| f - g \|_{L^1}
= M \| f - g \|_{L^1}. $$

So by letting $g \to f$ in $L^1$, the desired claim follows.


问:
Branch and Bound algorithm on an integer problem
When performing the branch and bound algorithm on an integer problem, multiple integer feasible solutions may be found.

Generally, is the second best integer feasible solution found during the branch and bound process
necessarily the second best integer solution of the Integer Problem??

答: No, a second-best solution might be pruned away and never encountered.  A simple modification to find the best two solutions is to keep a pair of solutions (instead of just one) as the incumbent, updating the incumbent whenever a solution is found that is better than the worse one.  This idea extends to the $k$ best solutions.


问:
Finding the limit of a pseudo geometric recurring sequence
$$
a_{n+1}=\frac{1}{2^{a_n}}
$$
Whilst practising to find the n-th term of a sequence, I haven't been able to successfully generalize this recurring sequence. I've called it pseudo geometric but honestly I got no idea what it's called, I would love to learn!

Things I've tried:

In my calculator, this sequence converges to a value around $0.641185744$.

We can assume the sequence to be geometric, so $k^{n+1}=\frac{1}{2^{k^n}}$, then $\log_2(k^{n+1})=\frac{1}{k^n}$ but I believe this is a dead end.

答:
The $n$-th term depends on the first term of the sequence. However this sequence do converge to the value you found. This value satisfies:

$$L = \frac1{2^L}, \text{ or } L\times 2^L = 1$$

A sequence of its decimal expansion is at https://oeis.org/A104748.

Its closed form involves the Lambert W Function, which is non-elementary:

$$L\times 2^L = 1 \iff (L\ln 2)e^{L\ln 2} = \ln 2 \iff L \ln 2= W(\ln2)\iff L = \frac {W(\ln 2)}{\ln 2}$$

so the only way to obtain its value is by numerical approximation.

EDIT: The recurrence relation given is equivalent to:

$$a_{n+1} = \left(\frac12\right)^{a_n}$$

Given that $a_1=1$, we have the sequence:

$$\left\{1,\frac12, \left(\frac12\right)^\frac12, \left(\frac12\right)^{\left(\frac12\right)^\frac12},\left(\frac12\right)^{\left(\frac12\right)^{\left(\frac12\right)^\frac12}}, \dots\right\}$$

or, in Tetration notation:

$$\left\{^0\left(\frac12\right),^1\left(\frac12\right),^2\left(\frac12\right),^3\left(\frac12\right)\dots\right\}$$

giving the "closed form" $a_n = ^{n-1}\left(\frac12\right)$.


问:
How to solve an infinite series given some terms
Given that $a_0 = -3, a_1 = 5, a_2 = -4, a_3 = 2$, and $\sum_{k=2}^{\infty}a_k = 7$, how can $\sum_{k=1}^{\infty} a_k$ be found?

It doesn't make much sense to me that there is enough information here to solve the problem.

Body:
For example, if $ a_4=9$ and

$$a_5=a_6=a_7=...=0$$

then

$$\sum_{k=2}^\infty a_k=$$
$$a_2+a_3+a_4+a_5+...=7$$
and

$$\sum_{k=1}^\infty a_k=$$
$$a_1+a_2+...=5+7=12$$


问:
Why does every countable limit ordinal have cofinality $\omega$?
According to Wikipedia, if $\alpha$ is a countable limit ordinal, then $\mathrm{cf}(\alpha)=\omega$. It is intuitively clear to me that it should be so. Certainly the cofinality of such an ordinal must be $\geq\omega$. We can now look at what the ordinal ends with. If it has a single $\omega$ at the end, then that is a cofinal subset and we are done. It may, however, not end in $\omega$. It could, for example, end in $\omega$ $\omega$s or $\omega^2$. Then we can pick one element of each $\omega$ constituting the $\omega^2$, and the set of these elements is cofinal in $\alpha$, and its type is $\omega.$ If $\alpha$ ends in $\omega^3$, we can take one element of each $\omega^2$ constituting the $\omega^2$, and get a cofinal subset of type $\omega.$

It seems to me that this reasoning should work. It should probably be done inductively. How should I do it?

答:
There are two approaches here:

$\omega\leq\operatorname{cf}(\alpha)\leq|\alpha|\leq\alpha$, for limit ordinals $\alpha$. If you show this inequality then you are done, since countable ordinals have cardinality $\omega$.

You can actually construct a cofinal $\omega$-sequence. Let $f\colon\omega\to\alpha$ be a bijection. Let $\alpha_0=f(0)$, and let $\alpha_{n+1}=f(k)$ such that $k$ is the least for which $f(k)>\alpha_n$. Because ordinals are well-ordered there are no decreasing sequences, and therefore there exists such $k$. It is not hard to show that the sequence $\alpha_n$ is cofinal.


问:
Multiplication module which is not cyclic
By the definition from Barnard (1980), a multiplication module is an $R$-module $M$ in which for all submodules $N$, there exists an ideal $I$ of $R$ such that $N=IM$. Note that a cyclic $R$-module (a module generated by one element), say $\langle m\rangle$ is a multiplication module since all of its submodules, say $N$, can be written as $Im=I(Rm)$ where $I=\langle r_i\ |\ r_im\in N\rangle$.

My question is: is there any example for a non-cyclic multiplication module? I've tried some examples such as $M_{2\times1}(\mathbb Z)$ over $M_{2\times2}(\mathbb Z)$, since I thought it is non-cyclic over $\mathbb Z$, but it turns out that it is cyclic over $M_{2\times2}(\mathbb Z)$. Is there any basic example of a non-cylic multiplication module? Thanks in advance!

答:
Two easy examples:

(1) An invertible ideal is a multiplication module.  Indeed, if $I$ is invertible  and $J \subseteq I$ then $K = I^{-1}J \subseteq R$ is an ideal and $J = IK$.  This points to any Dedekind domain which is not a PID as a source of multiplicative ideals that are not principal.

(2) Every ideal generated by idempotents is a multiplication module.  This is also easy to check.  So for example, in any infinite product of rings $\prod R_\alpha$, the ideal generated by the standard basis vectors is a multiplicative module which is not principal.


问:
Does an extraneous root of a polynomial in trigonometric functions have meaning? (Eg, the root $\cos\theta=-2$ of $\cos^2\theta+3\cos\theta+2=0$.)

Does an extraneous root of a polynomial in trigonometric functions have meaning?

E.g.,
$$\cos^2{\theta} + 3\cos{\theta} + 2 = 0$$ has two roots, $\cos{\theta} = -1$ and $\cos{\theta} = -2$. The second root is usually discarded as it is beyond the range of the cosine function. But does it have any geometric, analytical, etc, meaning?

答:
Geometrically, discarding $\cos\theta =-2$ is to discard those roots lying outside the the real axis of the complex plane, i.e. restricting the domain of $\theta$ to real values.

If, instead, the domain of $\theta$ is the full complex space, you need to solve for the complex roots, as follows

$$2=-\cos\theta= \cos(2\pi n + \pi-\theta) = \cosh[-i (2\pi n + \pi-\theta)]$$

which yields

$$\theta = (2n+1)\pi \pm i \cosh^{-1}(2)$$


问:
Finding relation of two sets
If A = {1, 2, 3} and B = {1, 2, 3, 4}, would R = {(a, b) ∈ A × B | b = a^2} be {(1,1), (2,4)} since the only time that b = a^2 is true is when (A,B) = (1,1) and (A,B) = (2,4)?</p>

答:
Short answer: Yes and you also explained correctly why.

Maybe in other words to help your understanding:

The set $R$ contains all pairs of numbers $(a,b)$ for which $a$ is an element of $A$ and $b$ is an element of $B$ and furthermore the pair must also satisfy the condition  $b = a^2$. The pairs you mentioned are the only ones that satisfy these three conditions.

Title: Finding relation of two sets


答:
The Cauchy-Schwarz Inequality can be stated as follows:

$$(x_1^2+x_2^2+\dots+x_n^2)(y_1^2+y_2^2+\dots+y_n^2) \ge(x_1y_1+x_2y_2+\dots+x_ny_n)^2$$

Now taking $x_i = \sqrt {a_i}$ and $y_i = \frac 1 {\sqrt {a_i}}$ (which is valid since each $a_i > 0$):

$$(\sqrt {a_1}^2+\sqrt {a_2}^2+\dots+\sqrt {a_n}^2)\left(\frac 1 {\sqrt {a_1}^2}+\frac 1 {\sqrt {a_2}^2}+\dots+\frac 1 {\sqrt {a_n}^2}\right)\ge\left(\frac {\sqrt{a_1}}{\sqrt{a_1}}+\dots+\frac {\sqrt{a_n}}{\sqrt{a_n}}\right)^2$$

which yields your inequality after simplification.

Title: Proving the given inequality

Body: Let $a_1,...,a_n$ be positive numbers. Show that $(a_1+...+a_n)(\frac{1}{a_1}+...+\frac{1}{a_n})\ge n^2$. Just not really sure how to approach this problem, any tips/solutions would be greatly appreciated.


问:
Question regarding the domain of an inverse function
My calculus professor mentioned the following lemma:

Lemma: Let $f$ be a continuous function on $[a,b]$. If $f$ is a strictly increasing, then $\operatorname{Dom}f^{-1}=[f(a),f(b)]$. If $f$ is a strictly decreasing, then $\operatorname{Dom}f^{-1}=[f(b),f(a)]$.

After mentioning it, he proceed to say that the case where $f$ is strictly increasing and continuous on $(a,b)$ is more difficult to analyze. But that's all he said. So I'm wondering, what can we say about the domain of $f^{-1}$ when $f$ is strictly increasing and continuous on $(a,b)$? Is there any property that the domain of $f^{-1}$ holds whenever $f$ is strictly increasing and continuous on $(a,b)$

答: Not much can be said about $\text{dom}(f^{-1})$ when the domain of $f$ is an open interval. This is because the image of $f$ under such circumstances can become unbounded, as is the case with $f(x)=\tanh^{-1}(x)$. The domain of $f$ in this case is $(-1,1)$, but the image is $(-\infty,\infty)$.


问:
power series and sequence
Let $\{ $a_n$ : n\geq 1\}$ be a sequence of real numbers such that radius of convergence of the power series : P(t) = $ \sum\limits_{n=0}^\infty a_n t^n $ satisfies $R > 0$.Then $ a_n \rightarrow 0$???

What i was able to do is :

To find R, I applied root test of convergence:

$$ \lim_{n\to\infty} |a_n t^n|^{1/n} < 1 $$

on further solving, we will get:

$$  |t| < [1/{\lim_{n\to\infty} |a_n|^{1/n}}] $$

therefore we get the value of R as :

$$\frac{1}{\lim_{n\to\infty} |a_n|^{1/n}} [1]$$

Now, $R>0$ (given)

Thus, by [1]

$$\frac{1}{(\lim_{n\to\infty}) |a_n|^{1/n}}> 0$$

or we can also write it as :

$$ \lim_{n\to\infty} |a_n|^{1/n}  > 0 $$

or

$$ \lim_{n\to\infty} |a_n| > 0 $$

on removing the modulus sign,this will give us:

$ \lim_{n\to\infty} a_n > 0$    or  $ \lim_{n\to\infty} a_n  < 0 $

As a result we see that $a_n$ never converges to $0$.it either converges to some number greater than $0$ or less than $0$.

Now,this is what i have been able to do.But i am not sure if i am right on this one.Can anyone let me know if i am right and if i am wrong somewhere,plz do tell where i have made mistake.

One more question..the power series can start at any number "n" right?? i know this fact.just wanted to confirm.

Thanks a lot for all the help.

答: The radius of convergence of $\sum_n n^{100000000000000}x^n$ is $1$, so $(a_n)$ doesn't converge necessarily to $0$. </p>

Title: power series and sequence


问:
Efficiently find at least one lattice point $(x,y)$ on hyperbola of equation $axy+bx+cy+d=0$
Even after a lot of search on internet, I didn't come up with a solution.
I need an algorithm to efficiently find at least one lattice point on hyperbola of equation $axy+bx+cy+d=0$.

Lattice point means integer coordinates and equation with integer means diophantine equations. The most closed resources I've found are:

This Link and this other post.

But the former look for solutions of equations and the latter is asking something linked but different.

So, there is an algorithm (better if efficient of course) or this is an unsolved/open problem?

答:
Given integers $a$, $b$, $c$ and $d$ we want to find a pairs of integer $(x,y)$ such that
$$axy+bx+cy+d=0.$$
Note that if $(x,y)$ is such a pair, then
\begin{eqnarray*}
(ax+c)(ay+b)&=&
a^2xy+abx+acy+bc\\
&=&a(axy+bx+cy+d)-ad+bc\\
&=&bc-ad.
\end{eqnarray*}
Let $N:=bc-ad$. This shows that lattice points on the hyperbola correspond bijectively to factorizations $N=uv$ with $u\equiv b\pmod{a}$ and $v\equiv c\pmod{a}$. Of course factorization methods have been studied plenty, and this particular factorization problem may allow for some slightly more effective methods than for general factorization of integers.

Note that such factorizations need not exist; for example
$$5xy+3x+2y+1=0,$$
has no integral solutions, because $N=bc-ad=1$ has no factors congruent to $2$ or $3$ mod $5$.


问:
Differentiate this trigonometric function
Let $h$ be defined as

$$h(x) = \begin{cases}\sin(x^2)\cos\Big(\frac{1}{x}\Big) \quad& x \neq 0\\ 0\quad&x=0\end{cases}$$

The question is:

Prove that $h$ is differentiable at $0$ and find $h'(0)$.

I showed that the left and right limits exist and equal to each other. I also showed that the function $h$ is continuous at $x=0$ but I could not show the right and left derivatives at $x=0$ are equal.

Thanks for your help.

答:
when $x>0, 0< |f(x)| < x^2$

The function is continous at $0,$ and $f(0) = 0$

The right-hand derivative of $f(0)$

$|f'(0)| = \lim_\limits{h\to 0^+}  |\frac {f(h)- 0}{h}| < |h|$

And the left-hand derivative is 0.

As the function is continuous, and the left-hand limit equals the right-hand limit, the function is differentiable and the derivative is 0.


问:
Show that the ring $\mathbb Z/6\mathbb Z[x]$ has infinitely many prime ideals
I’ve shown the case for $\mathbb F_3[x]$ but this is a lot simpler since this a field and hence a UFD and follows the same proof for prime numbers. I’ve tried using that maximal ideals are prime and that an ideal is maximal if and only if $R/I$ is a field but struggling to construct any maximal ideals from this

答: Well you are in luck because $(\mathbb Z/6\mathbb Z)[x]\cong F_2[x]\times F_3[x]$, and using what you just said about $F_3[x]$ having infinitely maximal ideals, you should see why this has infinitely many maximal ideals (hence infinitely many prime ideals.)


问:
convergence of an integral ( with an inner integral)
I need to figure out for which values of $p \in R $ does the following integral converge?

$$\int_0^{1} \frac{x^p}{\int_0^{x}\ln(1 + \sin(t) + t)dt} {dx} $$

Please note that I don't have to calculate anything, just to prove convergence. In addition, I may use Riemann or Dirichlet tests to prove it.

Thank you for your help!

In addition, I need to figure out whether the following converges, where the limits of integration have changed but the integrand is the same:

$$\int_1^\infty \frac{x^p}{\int_0^{x}\ln(1 + \sin(t) + t)dt} {dx} $$

答:
For the first question, just note that $\log(1 + t + \sin t) = 2t + O(t^2)$, so that

$$F(x) = \int_{0}^{x} \log (1 + t + \sin t) \, dt = x^2 + O(x^3) $$

near $x = 0$. Also, since the integrand is positive for $t > 0$, we have $F(x) > 0$ for $x > 0$. Thus

$$ \int_{0}^{1} \frac{x^p}{F(x)} \, dx \text{ converges} \quad \Longleftrightarrow \quad \int_{0}^{1} x^{p-2} \, dx \text{ converges}. $$

Therefore we must have $p-2 > -1$, or equivalently $p > 1$.

For the second question, note that $\log t \leq \log(1 + t + \sin t) \leq \log (3t)$ for $t > 1$. That is, we have the asymptotic relation

$$F(x) = x \log x + O(x) \quad \text{as } x \to \infty. $$

So we find that

$$ \int_{1}^{\infty} \frac{x^p}{F(x)} \, dx \text{ converges} \quad \Longleftrightarrow \quad \int_{2}^{\infty} \frac{x^{p-1}}{\log x} \, dx \text{ converges}. $$

Now it is not hard to show that the latter converges if and only if $p-1 < -1$, or equivalently, $p < 0$.


问:
Is the set of sequences with values in a finite set separable?
Let $\mathcal{A}$ be a finite set and consider the set of all sequences $\mathcal{A}^{\mathbb{Z}}$ on $\mathbb{Z}$ with values in $\mathcal{A}$. This set has a cardinality of $\mathcal{A}^{\mathbb{Z}}$ which is not countable, right? (is there actually a simple explanation?)

My actual question however is if there is at least a countable and dense subset (w.r.t. to the product topology when endowing each $\mathcal{A}$ with the discrete topology), i.e. is $\mathcal{A}^{\mathbb{Z}}$ separable?

Sadly, I couldn't come up with anything myself. I ask this to answer another question from ergodic theory, namely if the two-sided full shift over a finite alphabet is transitive.

答: The set of sequences $a_i$ that are constant for all but finitely many $i$ is countable and dense in $A^\Bbb{Z}$.


问:
Find an elementary matrix such that $EA=B$
I am given two matrices, and I have to find an elementary matrix $A$ such that $EA=B$.

$$E = \begin{bmatrix}2&4\\2&-6\end{bmatrix}$$

$$B = \begin{bmatrix}10&4\\-10&-6\end{bmatrix}$$

I tried "transposing" the equation, meaning $(EA)^T = B^T$. The equation given would then be $(A^T)(E^T) = B^T$.

I, however, can't manage to end up with the matrix $B$.

答:
You don't need to go through transposing.

Just find the inverse matrix of $E$

Then $A = E^{-1} B$

How do you find the inverse of $E$? See e.g. here:

https://www.mathsisfun.com/algebra/matrix-inverse.html

See in particular the section 2x2 Matrix

It says that:

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1} = \frac{1}{(ad-bc)} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

OK, so you have

$$E = \begin{bmatrix}2&4\\2&-6\end{bmatrix}$$

Therefore

$$E^{-1} = \frac{1}{(-20)} \begin{bmatrix}-6&-4\\-2&2\end{bmatrix} $$

and then

$$A = \frac{-1}{20} \begin{bmatrix}-6&-4\\-2&2\end{bmatrix} \begin{bmatrix}10&4\\-10&-6\end{bmatrix} =  \begin{bmatrix}1&0\\2&1\end{bmatrix}$$


答:
You have to determine the limit of $\dfrac{h(x)}x$ when $x\to 0$.

Now, using equivalence of functions, we get
$$\:\dfrac{h(x)}x\sim_0\dfrac{x^2}x\cos\dfrac1x=x\cos \dfrac1x=O(x),$$
so $\;\lim_{x\to 0}\dfrac{h(x)}x =0$.

Title: Differentiate this trigonometric function

Body:
Let $h$ be defined as

$$h(x) = \begin{cases}\sin(x^2)\cos\Big(\frac{1}{x}\Big) \quad& x \neq 0\\ 0\quad&x=0\end{cases}$$

The question is:

Prove that $h$ is differentiable at $0$ and find $h'(0)$.

I showed that the left and right limits exist and equal to each other. I also showed that the function $h$ is continuous at $x=0$ but I could not show the right and left derivatives at $x=0$ are equal.

Thanks for your help.


问:
Can you factor a matrix out of a summation?
Just checking, is this valid?

$\sum_{i=1}^n AV = A\sum_{i=1}^nV$

Where $A$ is a matrix, and $V$ is a vector? And more generally?

答:
It depends. If $A$ is a fixed matrix which does not depend on $i$, then yes you can factor out $A$ of the summation. In the same vein, if $V$ does not depend on $i$ in any way then you can factor out $V$ although it would factor out to the right

$$\sum_{i=1}^n AV=\left(\sum_{i=1}^n A\right)V$$

In general, $AV_1+AV_2=A(V_1+V_2)$ and $A_1V+A_2V=(A_1+A_2)V$.


答:
Hint what about the power series
$$\sum_{n=0}^\infty t^n$$

Title: power series and sequence

Body:
Let $\{ $a_n$ : n\geq 1\}$ be a sequence of real numbers such that radius of convergence of the power series : P(t) = $ \sum\limits_{n=0}^\infty a_n t^n $ satisfies $R > 0$.Then $ a_n \rightarrow 0$???

What i was able to do is :

To find R, I applied root test of convergence:

$$ \lim_{n\to\infty} |a_n t^n|^{1/n} < 1 $$

on further solving, we will get:

$$  |t| < [1/{\lim_{n\to\infty} |a_n|^{1/n}}] $$

therefore we get the value of R as :

$$\frac{1}{\lim_{n\to\infty} |a_n|^{1/n}} [1]$$

Now, $R>0$ (given)

Thus, by [1]

$$\frac{1}{(\lim_{n\to\infty}) |a_n|^{1/n}}> 0$$

or we can also write it as :

$$ \lim_{n\to\infty} |a_n|^{1/n}  > 0 $$

or

$$ \lim_{n\to\infty} |a_n| > 0 $$

on removing the modulus sign,this will give us:

$ \lim_{n\to\infty} a_n > 0$    or  $ \lim_{n\to\infty} a_n  < 0 $

As a result we see that $a_n$ never converges to $0$.it either converges to some number greater than $0$ or less than $0$.

Now,this is what i have been able to do.But i am not sure if i am right on this one.Can anyone let me know if i am right and if i am wrong somewhere,plz do tell where i have made mistake.

One more question..the power series can start at any number "n" right?? i know this fact.just wanted to confirm.

Thanks a lot for all the help.


答:
For the side question, as long as $|\mathcal{A}|\ge 2$ we have $2^{\aleph_0}\le|\mathcal{A}^{\mathbb{Z}}|$ so that it's uncountable.

For the actual question, I'll consider $\mathcal{A}^{\mathbb{N}}$, but the same idea works for $\mathbb{Z}$.  We can see that the product topology on such a set has a basis of cones $\{x\in\mathcal{A}^{\mathbb{N}}:\sigma\text{ is contained in }x\}$ for $\sigma\in\mathcal{A}^{<\mathbb{N}}$.  As a result, if we consider (say) the sequences which are almost always some $a\in\mathcal{A}$ (in the sense that there are only finitely many entries not $a$), then this set will be dense of size $\mathcal{A}^{<\mathbb{N}}=\aleph_0$.

Title: Is the set of sequences with values in a finite set separable?

Body:
Let $\mathcal{A}$ be a finite set and consider the set of all sequences $\mathcal{A}^{\mathbb{Z}}$ on $\mathbb{Z}$ with values in $\mathcal{A}$. This set has a cardinality of $\mathcal{A}^{\mathbb{Z}}$ which is not countable, right? (is there actually a simple explanation?)

My actual question however is if there is at least a countable and dense subset (w.r.t. to the product topology when endowing each $\mathcal{A}$ with the discrete topology), i.e. is $\mathcal{A}^{\mathbb{Z}}$ separable?

Sadly, I couldn't come up with anything myself. I ask this to answer another question from ergodic theory, namely if the two-sided full shift over a finite alphabet is transitive.


答:
Hint: Can you find a linear combination of the columns of $E$ that would produce the first column of $B$? In other words, can you find constants $k_1$ and $k_2$ such that

$$k_1\begin{bmatrix}2\\2 \end{bmatrix} + k_2\begin{bmatrix}4\\-6 \end{bmatrix} = \begin{bmatrix}10\\-10 \end{bmatrix}?$$

Once you've done that, can you do the same for the second column of $B$? (This will need two more constants, $l_1$ and $l_2$.)

Finally, arrange the four constants you found into a $2 \times 2$ matrix.

Title: Find an elementary matrix such that $EA=B$

Body:
I am given two matrices, and I have to find an elementary matrix $A$ such that $EA=B$.

$$E = \begin{bmatrix}2&4\\2&-6\end{bmatrix}$$

$$B = \begin{bmatrix}10&4\\-10&-6\end{bmatrix}$$

I tried "transposing" the equation, meaning $(EA)^T = B^T$. The equation given would then be $(A^T)(E^T) = B^T$.

I, however, can't manage to end up with the matrix $B$.


问:
Number of distinct paths in directed graph visiting all vertices?
The following child's homework problem states:

Jo is going on an 8-day activity holiday. Each day she can choose one of the water sports: kayaking or sailing, or land-based sports. She never does different water sports on consecutive days. She also wants to try all three options on at least one day of her holiday.
How many different schedules are possible?

This is effectively "How many distinct walks of length 7 visiting each node at least once are there on the following directed graph?":

Or alternatively, "How many sequences of length 8 on the alphabet $\{L,K,S\}$ exist not containing the subsequence $SK$, or $KS$, and which feature each letter at least once?".

How would one solve such a problem, without exhaustively enumerating all paths?

Note: If the 'must visit all vertices' condition were removed it would just be $|A^7|$, the sum of the entries of the 7th power of the adjacency matrix of the graph: British Olympiad; Combinatorics Recursion

Note: this is not the number of Hamiltonian paths since vertices can (and at least one must) be visited more than once.

答:
If we temporarily forget about the condition that requires Jo to try all three options once, then this is indeed the number of walks of length seven (not eight!) in the directed graph, which can be counted by summing the entries of
$$
   \begin{bmatrix}1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{bmatrix}^7 = \begin{bmatrix}
 120 & 169 & 119 \\
 169 & 239 & 169 \\
 119 & 169 & 120 \end{bmatrix}.
$$
(It's a bit faster to compute this  if instead we sum the middle row of $A^8 = ((A^2)^2)^2$.)

Then, it is easiest to eliminate the cases we don't like by inclusion-exclusion:

Subtract the $2^8$ walks which only visit Land or Kayak;

Subtract the $2^8$ walks which only visit Land or Sail;

Add back the $1$ walk which only visits Land.

This gives us $1393 - 2 \cdot 256 + 1 = 882$ as the final answer.


问:
Proving that a morphism of schemes is monic if the underlying continuous map is injective and the map on sheaves is epic.
I am trying to understand Tennison / Hitchin's Sheaf Theory proof of the following result.

Let $f:X\to Y$ be a morphism of ringed spaces such that the underlying map of topological spaces is a monomorphism and the morphism $f^\sharp:\mathscr{O}_Y\to f_\ast \mathscr{O}_X$ is an epimorphism. Then $f$ is a monomorphism in the category of ringed spaces.

The proof begins like this: (this is not a citation, it's just what I understood)

Let $(g_1,g_1^\sharp),(g_2,g_2^\sharp):(Z,\mathscr{O}_Z)\to (X,\mathscr{O}_X)$ be morphisms of ringed spaces such that the diagram

commutes. Passing to the diagram of underlying continuous maps, we get that $g_1=g_2$. Henceforth, we'll denote both $g_1$ and $g_2$ by $g$. By the adjunction between direct and inverse images, the diagram of sheaves over $Z$

is also commutative. Finally, passing to the stalks we obtain the commutative diagram

The authors then affirm that, by hypothesis $f^\flat_{g(p)}$ is an epimorphism. I can't see why that is true. The answer Relation between $\mathcal{O}_Y\rightarrow f_*\mathcal{O}_{X}$ and $f^{-1}\mathcal{O}_Y\rightarrow \mathcal{O}_X$ being epimorphism/monomorphism. even shows that this exact adjunction does not preserves epimorphisms. What am I missing?

答:
If $f : X \to Y$ is an injective continuous map, then $f_* : \textbf{Sh} (X) \to \textbf{Sh} (Y)$ is fully faithful, so the adjunction counit $f^{-1} f_* A \to A$ is an isomorphism for every sheaf $A$ on $X$. Hence, if $f : X \to Y$ is an injective continuous map and $f^\sharp : O_Y \to f_* O_X$ is an epimorphism, then $f^\flat : f^{-1} O_Y \to O_X$ is also an epimorphism.

That said, in my view it would have been better to state the condition in terms of $f^\flat$ in the first place, since $f^\sharp$ may not be an epimorphism even if $f^\flat$ is.


问:
Notation $\Delta_a$ to show absolute error
I have numerical methods course and studying the subject "errors". Our teacher showed Absolute Error with the notation $\Delta_a$(where $a$ is subscript). For example he wrote $\Delta_a=|a-a_0|$ and later wrote the error of functions like $\Delta_{\sin x}$ or $\Delta_{\cos x}$.

But after I studied some books I saw they write for example $\Delta a$(Where $a $ is not a subscript) and I think it is the correct notation. Am I right about that? If so, then should I use the same notation (with subscript) in the homework problems or exams?

答:
Notation is to ease the communication of mathematics to other people. Thus there is no absolute "right" and "wrong" notation as long as its clear to the reader what you mean by it. If you are writing an article or in a forum like this then you should strive to use the notation that is most commonly used.

In a course - where the homework is you communicating with your lecturer - then its a good idea to use the same notation as used in the lectures or in you textbook. This will make it easier for that person to understand you. If you want to use a different notation then make sure you clearly define it.

For this particular notation if you wrote $\Delta_a$ without defining it then I don't think I would have understood it right away so I would agree that $\Delta a$ is probably more understandable. But then again if by $\Delta a$ you also mean $|a-a_0|$ then this could also be confusing. $\Delta a$ is very often used to denote just $a-a_0$ (no absolute value), e.g. in the definition of the derivative which is sometimes written as $\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$. This might be the reason for using this notation with a subscript - to separate these two meanings.


问:
How can I show a the Cartan subalgebra being abelian implies its adjoint representation on the original Lie algebra completely commutes?
I am reading a set of notes on the classification of simple Lie algebras, in which the claim is made that since $\mathfrak h\subset g$, a Cartan subalgebra is abelian it then follows that the adjoint action of $\mathfrak h$ on $\mathfrak g$ completely commutes. My attempt at showing this is the following:

$$(\text{ad}(h_1)\circ \text{ad}(h_2))(g)=[h_1,[h_2,g]]=[[h_1,h_2],g]=[[h_2,h_1],g]=[h_2,[h_1,g]]=(\text{ad}(h_2)\circ \text{ad}(h_1))(g).$$ However, the second equality relies on the fact that the algebra is associative, which isn't necessarily true for a Lie algebra. I'm also concerned that since $[h_1,h_2]=0$ this doesn't make sense. Is there a simple way to see this?

答: You have to use a root system: write ${\cal g}={\cal h}\oplus_{\alpha\in A}g_a$ where $A$ is a root system, if $x\in {\cal g}_a, u,u'\in {\cal h}, [u,[u',x]]=a(u)a(u')x$, if $x\in {\cal h}, [u,[u',x]]=0$.


问:
How many terms are there in a cyclic summation?
I was wondering whether there is any formula that can calculate the number of terms in a cyclic sum:

If $i = 1, \ldots, n$, then for $\sum_{\text{cyc}}x_ix_j$, there are $\frac{n(n-1)}{2}$ terms in the summation.

For example

If $i = 4$, then there are $6$ terms, like,  $x_1x_2, x_1x_3, x_1x_4, x_2x_3, x_2x_4, x_3x_4.$  which equals to $ \frac{4\times3}{2} = 6$.

However, is there any formula that can calculate the cases like,

$i = 1, \ldots, n$,  $\sum_{\text{cyc}}x_ix_jx_k$

$i = 1, \ldots, n$,  $\sum_{\text{cyc}}x_ix_jx_kx_l$

答:
The following are just some thoughts that may help:

For your first summation, we need $i \neq j$ and that the summation cycles through in the manner you described. Notice that since we are counting terms we see that clearly terms of the form $x_{1}x_{j}$, where $j$ is such that $1 \leq j \leq i$.we count $j-1$ distinct terms. Clearly the number possibilities decreases by $1$ as the index of the first term increases by $1$. Recall that
$\sum_{k=1}^nk=\frac{n(n+1)}2$ and then set $j-1 = n$ to get the desired formula.
For your second summation, we can look at the problem the same way where we fix the first symbol $x_{1}$ and look at the distinct terms possible of $x_{1}x_{j}x_{k}$. By the previous work there are $\frac{j(j-1)}2$ terms of this form. Now if we fix the terms to be of the form $x_{2}x_{j}x_{k}$ we see that there are $\frac{j(j-1)}2 - 4$ if $i = 4$....

I am not sure yet of a general form for how much to subtract off. I am sure there is a combinatorial number to make your problem easy, but I do not know one.


问:
Expected number of trials for either A or B to occur
Suppose we are playing a game where the player gets 3 points for a win, 1 point for a draw and 0 points for a loss. The probabilities are as follows

P(Win) = 0.2

P(Draw) = 0.5

P(Loss) = 0.3

What is the expected number of games before a player has 3 points or more?

I am unsure quite how to approach the problem, the expected number of games for a win would be 5 and the expected number of games for 3 draws would be 6 games so we would expect it to take less than 5 games to reach the 3 point threshold. My instinct would be to add together the probabilities in some way but I think that needing at least 3 games to get the points via draws is confusing me.

答:
By the definition of the expectation we get:

$$ E[X] = \overset{\text{win}}{0.2} \cdot \overset{\text{pts.}}{3} \cdot \overset{\text{no. of games}}{X} + 0.5 \cdot 1 \cdot X + 0.3 \cdot 0 \cdot X = 1.1X$$

You basically need to check when $$ 1.1X \geq 3$$
This is not a whole number, but use the fact that $X$ is a natural number (you can't have half games / negative games right? ). So you will need to choose the closest whole number to $\frac{3}{1.1}$


答:
If $f$ is strictly increasing and continuous on $(a,b)$, then $\lim_{x\to a^+}f(x)$ and  $\lim_{x\to b^-}f(x)$ exist (possibly $-\infty$ and $+\infty$ respectively), and
$$\operatorname{Dom} f^{-1}=\bigl(\lim_{x\to a^+}f(x),\lim_{x\to b^-}f(x)\bigr).$$

Title: Question regarding the domain of an inverse function

Body:
My calculus professor mentioned the following lemma:

Lemma: Let $f$ be a continuous function on $[a,b]$. If $f$ is a strictly increasing, then $\operatorname{Dom}f^{-1}=[f(a),f(b)]$. If $f$ is a strictly decreasing, then $\operatorname{Dom}f^{-1}=[f(b),f(a)]$.

After mentioning it, he proceed to say that the case where $f$ is strictly increasing and continuous on $(a,b)$ is more difficult to analyze. But that's all he said. So I'm wondering, what can we say about the domain of $f^{-1}$ when $f$ is strictly increasing and continuous on $(a,b)$? Is there any property that the domain of $f^{-1}$ holds whenever $f$ is strictly increasing and continuous on $(a,b)$


问:
Extension of unbounded symmetric operator ranges equal implies trivial extension
This is a problem I have encountered in my work and studies in operator theory and functional analysis.

We take a Hilbert space $H$. We take a symmetric (possibly unbounded) operator $C$ which extends $A$, $A \subseteq C$. We are given that $\text{Range}(A+i)=\text{Range}(C+i)$. I need to prove $C=A$, or basically that the domains are equal $D(A)=D(C)$.

Here are the definitions I have used.
If $T$ is a densely defined linear operator on a Hilbert space $H$, the domain $D(T^*)$ is the set of $\phi \in H$ for which there is a $\eta \in H$ with
$$
\langle T\psi,\phi \rangle = \langle \psi,\eta \rangle
$$
for all $\psi \in D(T)$. For each such $\phi \in D(T^*)$ we define $T^* \phi = \eta$, and $T^*$ is called the adjoint of $T$. A densely-defined operator is said to be symmetric if $\langle T\phi,\psi \rangle = \langle \phi,T\psi \rangle$ for all $\phi,\psi \in D(T)$, and in this case $D(T) \subseteq D(T^*)$ and $T=T^*$ on $D(T)$ and $T^*$ is said to extend $T$. A symmetric operator is self-adjoint iff $D(T)=D(T^*)$ and thus $T=T^*$.

So $C$ is densely defined but $A$ may not be. In fact, I have no idea how to do this. I do not know how to use the fact that $C$ is symmetric. I would appreciate any help with this. I thank all helpers.

答:
Let $\gamma \in \mathscr D(C)$ but not  $\in \mathscr D(A)$.

Then there is an $\alpha \in \mathscr D(A)$ such that $(A-i) \alpha = (C-i)\gamma$.

Then $(C-i)(\alpha - \gamma) =0$, which implies

$$((\alpha - \gamma),C(\alpha - \gamma))=i((\alpha - \gamma),(\alpha - \gamma))$$
which cannot hold for symmetric $C$.


答:
You call the sums cyclic sums, but in fact they are symmetric sums. Not all textbooks distinguish clearly between the two, which is unfortunate, because the distinction is useful. I'll call them symmetric sums here. See below for an explanation of the difference.

The number of terms in the symmetric sum
$$\sum_{\text{sym}}x_1x_2\cdots x_d,$$
over the products of $d$ distinct variables out of $n$, is precisely $\binom{n}{d}$. After all, each term is obtained by choosing $d$ variables from the $n$ variables, and every choice of $d$ variables yields a term in the symmetric sum.

For more general symmetric sums of the form
$$\sum_{\text{sym}}x_1^{n_1}x_2^{n_2}\cdots x_d^{n_d},$$
the number of terms is a multinomial coefficient. If $d_k$ denotes the number of $k$-th powers in the product, then the number of terms is
$$\binom{n}{d_0,\ldots,d_n}=\frac{n!}{d_0!d_1!\cdots d_n!}.$$
Note that $d_0+d_1+\ldots+d_n=n$.

The difference between symmetric and cyclic sums is clearest with an example; if $n=4$ then
$$\sum_{\text{sym}}x_1x_2=\sum_{\sigma\in S_4}x_{\sigma(1)}x_{\sigma(2)}=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4,$$
which is the sum over all permutations of the indices of $x_1x_2$. On the other hand
$$\sum_{\text{cyc}}x_1x_2=\sum_{\sigma\in C_4}x_{\sigma(1)}x_{\sigma(2)}=x_1x_2+x_2x_3+x_3x_4+x_4x_1,$$
which is the sum over all cyclic shifts of the indices.

Title: How many terms are there in a cyclic summation?

Body:
I was wondering whether there is any formula that can calculate the number of terms in a cyclic sum:

If $i = 1, \ldots, n$, then for $\sum_{\text{cyc}}x_ix_j$, there are $\frac{n(n-1)}{2}$ terms in the summation.

For example

If $i = 4$, then there are $6$ terms, like,  $x_1x_2, x_1x_3, x_1x_4, x_2x_3, x_2x_4, x_3x_4.$  which equals to $ \frac{4\times3}{2} = 6$.

However, is there any formula that can calculate the cases like,

$i = 1, \ldots, n$,  $\sum_{\text{cyc}}x_ix_jx_k$

$i = 1, \ldots, n$,  $\sum_{\text{cyc}}x_ix_jx_kx_l$


答:
As you point out yourself, the Lie bracket is in general not associative. Instead one has the Jacobi identity which tells us that

$$[h_1, [h_2, g]] + [h_2, [g, h_1]] + [g, [h_1, h_2]]].$$

Now, that $h_1, h_2$ commute means by definition that $[h_1, h_2]=0$, hence the last term is $=0$, hence

$$[h_1, [h_2, g]] = - [h_2, [g, h_1]].$$

Now use anticommutativity and bilinearity of the Lie bracket to write this as

$$= -[h_2, -[h_1, g]] = [h_2, [h_1,g]].$$

Or, since you write "adjoint action", maybe you know already (as follows from the Jacobi identity) that the adjoint action defines a Lie algebra representation, i.e.

$$ad([x,y]) =ad(x) \circ ad(y) -ad(y) \circ ad(x) \qquad \text{for all } x,y \in g,$$

and now you apply this to $h_1, h_2$ with $[h_1, h_2] = 0$.In short, if two Lie algebra elements commute, their adjoints commute.

Title: How can I show a the Cartan subalgebra being abelian implies its adjoint representation on the original Lie algebra completely commutes?

Body:
I am reading a set of notes on the classification of simple Lie algebras, in which the claim is made that since $\mathfrak h\subset g$, a Cartan subalgebra is abelian it then follows that the adjoint action of $\mathfrak h$ on $\mathfrak g$ completely commutes. My attempt at showing this is the following:

$$(\text{ad}(h_1)\circ \text{ad}(h_2))(g)=[h_1,[h_2,g]]=[[h_1,h_2],g]=[[h_2,h_1],g]=[h_2,[h_1,g]]=(\text{ad}(h_2)\circ \text{ad}(h_1))(g).$$ However, the second equality relies on the fact that the algebra is associative, which isn't necessarily true for a Lie algebra. I'm also concerned that since $[h_1,h_2]=0$ this doesn't make sense. Is there a simple way to see this?


问:
Let $f:c_0\to \mathbb{F} $ be given by $f(x_1, x_2, . . .) =x_1$. Determine all the Hahn-Banach extensions of $f$ to the space $c$.
This question had to parts.

$a)$ Calculate $‖f‖$.

$b)$ Determine all the Hahn-Banach extensions of $f$ to the space $c$.

My work:

$a)$ Let $x=(x_1,x_2,\ldots)$, then $||f(x)||=||(x_1)||\leq||x||_{\infty}$ . This means that $||f||\leq1$. Choose $x=(1,0,\ldots)$ then $||x||_{\infty}=1$ and $||f(x)||=1$. Which means that $||f||=1$

$b)$ is where I get stuck. I know that I can write $c=c_0\oplus \lambda(1,1\ldots)$. However I am unsure how to continue from here

答:
You're off to a good start.  By noting the direct sum decomposition of $ c $, you know that any extension is completely determined by $ f $ along with its value at $ \mathbf{w} = (1, 1, \ldots) $.  Call this value $ f(\mathbf{w}) = k \in \mathbb{F} $.  Then you can calculate
$$
\begin{align}
\bar{f}(\mathbf{x}) &= f(\pi_{c_0} \mathbf{x} + \pi_{c_0^\perp}\mathbf{x})\\
&= f(\pi_{c_0} \mathbf{x}) + \bar f(\pi_{c_0^\perp} \mathbf{x})\\
&= f(\mathbf{x} - (\lim{\mathbf{x}}) \mathbf{w}) +  \bar f((\lim{\mathbf{x}}) \mathbf{w})\\
&=f(\mathbf{x} - (\lim{\mathbf{x}})\mathbf{w}) + \lim \mathbf{x}\ \cdot \bar f(\mathbf{w})\\
&=f(\mathbf{x} - (\lim{\mathbf{x}})\mathbf{w}) + k \lim \mathbf{x}
\end{align}
$$
Edit: I noticed from our calculations in the comments that there is only one value of $ k $ which yields an extension $ \bar f $ of norm no larger than that of $ f $.  So this problem is more interesting than I at first realized.


问:
Definite integral is an antiderivative of?
Is the following definite integral $\int_a^x w^3dw$ an antiderivative of $w^3$?

答:
If we assume $a$ is a constant and $x$ is a variable then $$F(x) = \int_a^x w^3dw$$ is antiderivative of $f(x)=x^3$. The $w$ is just an integration variable.

The expression $\int_a^x w^3dw$ is a function of $x$, not of $w$. You can also write it as   $\int_a^x t^3dt$, it's still a function of $x$ (no matter which variable you use under the integral).


问:
Show that $I=\langle x+1, x^2+1\rangle$ is maximal in $\mathbb Z[x]$.
I’m trying to show this by showing that $\mathbb Z[x]/I$ is a field. Since in this ring $x=-1$ and so $x^2=1$, but also $x^2=-1$ from which it follows that $1=-1$ and so it is probably isomorphic to $\mathbb F_2$? But I can’t find a homomorphism from $\mathbb Z[x]\ \longrightarrow\ \mathbb F_2$ with kernel $I$.

答:
Note that $I$ is maximal iff $\mathbb Z[x]/I$ is a field.

But $\mathbb Z[x]/(x+1,x^2+1)\cong \mathbb Z[-1]/((-1)^2+1)\cong\mathbb Z/2\mathbb Z=\mathbb F_2$ which is a field.


问:
help integrating
I'm really struggling to integrate this.

$$\int_0^t \sqrt{\cos^2\left(1\right)\sinh^2\left(u\right)+4\sin^2\left(\dfrac{1}{2}\right)\cosh^2\left(\dfrac{u}{2}\right)} \,\mathrm{d}u$$

This is to find the arc length of a plane curve. An online integral calculator gives me such a complicated result with $\ln$ and I don't know how they got to that result.

I also then need to use the result to find the inverse.

答:
Hint:

Set $a={{\cos }^{2}}\left( 1 \right),b={{\sin }^{2}}\left( \frac{1}{2} \right)$
and using   $\sinh \left( u \right)=2\sinh \left( \frac{u}{2} \right)\cosh \left( \frac{u}{2} \right)$ the integral becomes:

$$\begin{align}
  & =\int{\sqrt{4a{{\sinh }^{2}}\left( \frac{u}{2} \right){{\cosh }^{2}}\left( \frac{u}{2} \right)+4b{{\cosh }^{2}}\left( \frac{u}{2} \right)}du} \\
 & \ =\int{\sqrt{4{{\cosh }^{2}}\left( \frac{u}{2} \right)\left( a{{\sinh }^{2}}\left( \frac{u}{2} \right)+b \right)}du} \\
 & =\int{2\cosh \left( \frac{u}{2} \right)\sqrt{a{{\sinh }^{2}}\left( \frac{u}{2} \right)+b}du}\ \quad ,x=\sinh \left( \frac{u}{2} \right) \\
 & =4\int{\sqrt{a{{x}^{2}}+b}\ dx} \\
\end{align}$$

Title: help integrating


问:
Convergence Newton's Method for this system
I'd like to prove that the following non-linear system

$$
 F(x) =
 \begin{pmatrix}
  x_1^3 + x_2^3 - 4 \\
  x_1^3 - x_2^3 \\
 \end{pmatrix}
=
 \begin{pmatrix}
 0 \\
 0 \\
 \end{pmatrix}
$$
will converge when using the Newton's Method for the start vectors $x^{(0)} = (x_1^{(0)}, x_2^{(0)})^T \in [1,2] \times [1, 2] \subset \mathbb R^2$.

I'm trying to use this lemma I learned on class

Let $f: D \subset \mathbb R^n \rightarrow \mathbb R^n$ be continuously
  differentiable and let $D$ be convex as well open, $f'$ is the jacobian matrix. If $$q = \sup_{x\in
 D}||f'(x)|| < 1$$ then $f$ is a contraction mapping using the Norm $||
 \cdot ||$ with a contraction rate $q$.

I have calculated the Jacobian matrix, and it is as follows
$$
\begin{pmatrix}
  3x_1^2 & 3x_2^2 \\
  3x_1^2 & - 3x_2^2 \\
 \end{pmatrix}
$$
Using the maximum norm (sum of lines) for matrixes, I get:
$||F' (x_1, x_2)||_\infty = \max \{3x_1^2 + 3x_2^2, 3x_1^2 - 3x_2^2\} = 3x_1^2 + 3x_2^2$ which is for sure not $< 1$ in the desired interval. Where have I gone wrong? Thanks a lot guys!

答:
For a more general approach on the convergence of Newton method ( and different from his attempt ) see the this article here. In Theorem 1.1 make $E = F = \mathbb{R}^2$ both with norm
$$
\| (x_1,x_2)\|= |x_1|+|x_2|.
$$
and make your initial point in $B(x_0,r)$ whit $x_0=(\frac{3}{2},\frac{3}{2})$ and $r=\frac{1}{2}$, i.e.
\begin{align}
B(x_0,r)= &
\left\{ (x_1,x_2)\in \mathbb{R}^2 : \bigg\| (x_1,x_2)-\left(\frac{3}{2},\frac{3}{2}\right)\bigg\|= |x_1-\frac{3}{2}|+|x_2-\frac{3}{2}| < \frac{1}{2} \right\}
\\
=
&
(1,2)\times(1,2)
\end{align}
Nothe that $\overline{B(x_0,r)}=[1,2]\times[1,2]$.


问:
Finding a sequence of integrable functions on $[0,1]$ which is pointwise convergent to zero whose integrals increase without bound
As the title suggests I am looking for examples of sequence of integrable functions on $[0,1]$ which are pointwise convergent to zero. The sequence of integrals $\left\{\int_{0}^{1}f_n\right\}_{n=1}^{\infty}$ needs to increase without bound.

I've tried numerous integrable functions that are pointwise convergent to zero but I have yet to pinpoint one that satisfies the condition about integrals increasing without bound. Every single function I've found gives me smaller and smaller integrals. This is frustrating, to say the least.

答:
We can do this without piecewise defintions. Let $f_n(x)= n^3x^n(1-x).$ Then $f_n(x)\to 0$ at $x=0,x=1.$ On $(0,1),$ we have $f_n(x)\to 0$ pointwise, simply because exponential decay always beats out polynomial growth.

Does $\int_0^1 f_n$ increase to $\infty?$ Yes, because

$$\int_0^1 n^3x^n(1-x)\,dx = \int_0^1 n^3(x^n-x^{n+1})\,dx$$ $$ = n^3\left (\frac{1}{n+1}- \frac{1}{n+2}\right ),$$

which increases to $\infty.$


问:
Probability of success in four free throws
I have the following problem:

There is a basketball player with a mean of success of $0.8$ per free throw. In a game, he has four free throws.

a) What is the probability of him succeeding in at least 3 of the free throws?

b) What is the probability of him succeeding in exactly two of the free throws?

答:
I will show answers computed in R where dbinom is a binomial PDF and pbinom is a binomial CDF. I guess
you are supposed to use the formula for the binomial PDF, under the assumption that free
throws are independent of one another:

$$P(X = x)= {4\choose x}(.8)^x(.2)^{4-x},$$
for $x = 0,1,2,3,4.$

First, find the individual probabilities
by hand computation and [in (a)] sum the relevant
probabilities. You can check your results against
the probabilities from R. [Because $n = 4$ is so small,
using a normal approximation is not feasible.]

Let $X \sim\mathsf{Binom}(n = 4, p=.8).$ You seek
(a) $P(X \ge 3) = 1 - P(X\le 2)$ and (b) $P(X = 3).$

```<code># a
1 - pbinom(2, 4, .8)
[1] 0.8192
sum(dbinom(3:4, 4, .8))
[1] 0.8192

# b
dbinom(2, 4, .8)
[1] 0.1536
</code>```

Here is a graph of the PDF of $\mathsf{Binom}(5, .8).$

```<code>hdr = "PDF of BINOM(4, .8)"
x = 0:4;  PDF = dbinom(x, 4, .8)
plot(x, PDF, type="h",lwd=3, col="blue", main=hdr)
abline(v = 0, col="green2")
abline(h = 0, col="green2")
</code>```

Body:
I will show answers computed in R where dbinom is a binomial PDF and pbinom is a binomial CDF. I guess
you are supposed to use the formula for the binomial PDF, under the assumption that free
throws are independent of one another:

$$P(X = x)= {4\choose x}(.8)^x(.2)^{4-x},$$
for $x = 0,1,2,3,4.$

First, find the individual probabilities
by hand computation and [in (a)] sum the relevant
probabilities. You can check your results against
the probabilities from R. [Because $n = 4$ is so small,
using a normal approximation is not feasible.]

Let $X \sim\mathsf{Binom}(n = 4, p=.8).$ You seek
(a) $P(X \ge 3) = 1 - P(X\le 2)$ and (b) $P(X = 3).$

# a
1 - pbinom(2, 4, .8)
[1] 0.8192
sum(dbinom(3:4, 4, .8))
[1] 0.8192

# b
dbinom(2, 4, .8)
[1] 0.1536

Here is a graph of the PDF of $\mathsf{Binom}(5, .8).$

hdr = "PDF of BINOM(4, .8)"
x = 0:4;  PDF = dbinom(x, 4, .8)
plot(x, PDF, type="h",lwd=3, col="blue", main=hdr)
abline(v = 0, col="green2")
abline(h = 0, col="green2")


问:
Simplify Fraction with a Square Root?
I'm trying to simplify a problem that I have the answer to but when I check it with Wolfram it seems a bit more simplified and I can't think of what steps are taking place to get that result.

Somehow this: $\frac{1}{2\sqrt{1-\frac{x^2}{4}}}$

Becomes this: $\frac{1}{\sqrt{4-x^2}}$

And I'm not sure how. I'm guessing its some simple Algebra operation that I'm totally forgetting right now. Any help breaking it down is much appreciated.

Title: Simplify Fraction with a Square Root?

Body:
We have that $2=\sqrt{4}$. Therefore
$$2\sqrt{1-\frac{x^2}{4}} = \sqrt{4}\sqrt{1-\frac{x^2}{4}} = \sqrt{4\left(1-\frac{x^2}{4}\right)}=\sqrt{4-x^2}.$$


问:
Find this limit, When n to infinity
I don´t understand this problem, I need help with this

Find the next limit. Justify your procedure
$$\displaystyle \lim_{n \to \infty} \sqrt[n]{n + c} $$

I Tried do this, but I think it's wrong

$$\displaystyle \lim_{n \to \infty} \sqrt[n]{n + c} = \displaystyle \lim_{n \to \infty} {n + c}^n $$

After that, I tried use a stolz criterium $$\displaystyle \lim_{n \to \infty} \sqrt[b_{n}]{a_{n}} = \displaystyle \lim_{n \to \infty} \sqrt[b_{n+1}-b_{n}]{\dfrac{a_{n+1}}{a_{n}}}$$
where $$a_{n} = n+c , b_{n} =n , a_{n+1} = n+c+1 , b_{n+1} = n+1 , \displaystyle \lim_{n \to \infty} b_{n} = \displaystyle \lim_{n \to \infty} n = \infty$$
so $$\displaystyle \lim_{n \to \infty} \sqrt[n]{n+c} = \displaystyle \lim_{n \to \infty} \sqrt[n+1-n]{\dfrac{n+c+1}{n+c}} = \displaystyle \lim_{n \to \infty} \sqrt[1]{\dfrac{n+c+1}{n+c}} = \displaystyle \lim_{n \to \infty} \dfrac{n+c+1}{n+c} = \dfrac{\displaystyle \lim_{n \to \infty} n+c+1 }{\displaystyle \lim_{n \to \infty} n+c }$$

I got this but I don't know if this is right or not,

答:
Using the formula $e^{ln(x)}=x$:
$$\displaystyle \lim_{n \to \infty} \sqrt[n]{n + c}=\lim_{n \to \infty} (n + c)^{\frac{1}{n}}= e^{\lim_{n \to \infty}\frac{ln(n + c)}{n}}= e^{\lim_{n \to \infty}\frac{1}{n+c}}=e^0=1$$

If you had persisted with what you were trying:
$$\dfrac{\displaystyle \lim_{n \to \infty} n+c+1 }{\displaystyle \lim_{n \to \infty} n+c } = \displaystyle \lim_{n \to \infty}\dfrac{ n+c+1 }{ n+c } = \displaystyle \lim_{n \to \infty}(1+\frac{1}{n+c})=1+0=1$$

Title: Find this limit, When n to infinity


答:
$a\sqrt{b} = \sqrt{a^2b}$.

This moves the 2 inside the $\sqrt{}$ where it becomes $2^2 = 4$.

Title: Simplify Fraction with a Square Root?

Body:
I'm trying to simplify a problem that I have the answer to but when I check it with Wolfram it seems a bit more simplified and I can't think of what steps are taking place to get that result.

Somehow this: $\frac{1}{2\sqrt{1-\frac{x^2}{4}}}$

Becomes this: $\frac{1}{\sqrt{4-x^2}}$

And I'm not sure how. I'm guessing its some simple Algebra operation that I'm totally forgetting right now. Any help breaking it down is much appreciated.


问:
Interesting properties of Green's function
I am trying to improve my understanding of the Green's Function and its use to solve second-order linear ODEs.

Is it correct to assume that the Green's functions can be used to solve all second order linear ODEs?

Is there any examples when using the Green’s functions to solve second-order linear ODEs that gives an interesting or special result?

Is there any properties of Green's function that has a special case or is special?

答:
It would not print my just prior answer and i cannot see it nor can I delete it and seems the interface here is just 'hung up' on that 'muffed' answer so I try the same again.

With respect to (1) would say 'No'; Greens function method is  not a general universal choice for solving the most general second order diff eq. It is usually for inhomogenous eq's and more often for those such as wave equation, Laplacian and the like as opposed to your particular example which I guess you deleted because I don't see it now. For now will give a couple of rather elementary examples and afterward example a little less trivial so you may be able to have more of an idea of under what circumstances it may be applied and depending upon what you call interesting etc. it may qualify for your note (2) and (3). Green's fct is also quite often used in partial diff eq's and more than one dimension.

Briefly or simply put in it's most basic and familiar form the greens fct is often put in terms of primed and unprimed coordinate(s) in which it is generally symmetric and such that applying differential operator the result is zero everywhere except where $x\!=\!x'\;$ or it gives $\delta(x\!-\!x')$ and in short it's formal definition is the solution{written in a rather sloppy and non-rigorous form} to $Dop\; G(x\!-\!x')\!=\!\delta(x\!-\!x')$ usually also satisfying some boundary conditions depending upon the case considered where have written $Dop$ to mean a general(or partial) differential operator usually linear.
Though it can be and often is considered in finite regions of various boundary conditions, for now just consider open infinite space in one dimension. and the examples to follow should make it more clear.

For example to solve for $g(x)$ in
$\;\left(1-\frac{d^2}{d\:x^2}\right)g(x)=f(x)\;$ we form $G(x\!-\!x')$ such that $\left(1-\frac{d^2}{d\:x^2}\right)G(x\!-\!x')=\delta(x\!-\!x')$ so then $\left(1-\frac{d^2}{d\:x^2}\right)\int G(x\!-\!x')f(x')\:dx'=\int\delta(x\!-\!x')f(x')\:dx'=f(x)$
and $$g(x)\!=\!\int G(x\!-\!x')f(x')\:dx'$$

Depending upon the form of $f(x)$ using
$\int_{-\infty}^{\infty}e^{ik(x-x')}dk\!=\!2\pi\delta(x\!-\!x')$ obtain

$$\left(1-\frac{d^2}{d\:x^2}\right)\int_{-\infty}^{\infty}\left \{\frac1{2\pi}\!\int_{-\infty}^{\infty}\frac1{(1+k^2)}e^{ik(x-x')}dk\right\}f(x')dx'\;$$

$$=\int_{-\infty}^{\infty}\left \{\frac1{2\pi}\!\int_{-\infty}^{\infty}e^{ik(x-x')}dk\right\}f(x')dx'\!=\!f(x)$$
So $$G(x)\!=\!\int_{-\infty}^{\infty}\frac1{2\pi(1+k^2)}e^{ikx}dk$$

and $$\;g(x)\!=\!\int_{-\infty}^{\infty}\left \{\frac1{2\pi}\!\int_{-\infty}^{\infty}\frac1{(1+k^2)}e^{ik(x-x')}dk\right\}f(x')dx'\!=\!\int G(x\!-\!x')f(x')\:dx'\;$$ So for this particular example we need $f(x)$ such that the double integral may be evaluated perhaps by interchanging order of integration or whatever. Also it should be clear that we could also solve $\;C\!-\!\frac{d^2}{d\:x^2}\;$ where $\;C\;$ is any positive constant replacing the prior $\;1$.

Now the above is not the only way to form a delta function and perhaps related is that often in other than rectangular coordinates, solutions to the homogenous eq. are not valid over all space but consider {such as perhaps in spherical coord's for example} one solution goes to 0 at inf but becomes infinite at 0 and another solution vice-versa so a solution is to put them together sort of back to back such that there is a discontinuity in first derivative so that taking 2nd derivative yields proportional to delta fct. In fact can demonstrate this same idea in open infinite one dim rectangular. Suppose we require a solution, $g(x)$ say, to $\left(1\!-\!\frac{d^2}{d\:x^2}\right)g(x)\!=\!f(x)$ and require the Green's fct to vanish at infinity. A solution may be written in terms of a Green's fct $$G(x)\!=\!\frac1{2}e^{-|x|}\;$$ because $\;\left\{1\!-\!\frac{d^2}{d\:x^2}\right\}G(x)\!=\!\delta(x)\;$ and $$g(x)\!=\!\int_{-\infty}^{\infty}G(x-x')f(x')\:dx'\;$$ and this green's fct is same as just prior expressed in a fourier integral type form.

For a less trivial example I had occasion to analyze the kinetic energy or Laplacian plus a constant, which took as unity  , on a group of particles in quantum theory of total $3n$ rectangular dim's.
in hyperspherical coordinates. Without going into detail, after certain angular solutions we have at last the solution to the hyperradius coord. with parameter(s) depending upon
particular discrete quantum numbers according angular eigensolutions, such as $\mathit{lk}_2$ which may be considered as total generalized
angular momentum of which detail we may not concern here
$$\left\{-{\frac{d^2}{{\mathit{dx}_1}^2}}-\frac{(3n\!-\!1)\dfrac{d}{\mathit{dx}_1}}{x_{1}}+\mathit{lk}_2
\frac{(\mathit{lk}_2+3n-2)}{{x_{1}}^2}+1\right\}\mathit{fr}(x_1,y_1)=0$$ except at $x_1\!=\!y_1$ and the variables
$x_1,y_1$ in which it is symmetric are analogous to the prime and unprimed coord's prior. A delta fct singularity is effected by a discontinuity in the first derivative
at $x_1\!=\!y_1\;.$ We set the magnitude of the Wronskian between two solutions of which one is regular at zero and the other at infinity equal to unity after multiplied by
a volume normalization  $x_1^{3n\!-\!1}$ integrating factor. More specifically the correct solution for n even is of form
$$\mathit{fr}(x_1,y_1)=\frac{\pi\,i\,J_{(3n\!/2+\mathit{lk}_2\!-\!1)}(i\,\mathit{rl})\left(J_{(3n\!/2+\mathit{lk}
_2\!-\!1)}(i\,\mathit{rg})+iY_{(3n\!/2+\mathit{lk}_2\!-\!1)}(i\,\mathit{rg})\right)}{2x_1^{(3n\!/2\!-\!1)}y_1^{(3n\!/2\!-\!1)}}$$ where $\mathit{rl},\mathit{rg}$ is the lesser,greater of $x_1,y_1.\,$ For n odd $\;\mathit{fr}\;$ can be written in terms of elementary functions of finite power series times an exponential which is actually simpler but we don't write that detail here because it is somewhat involved and we can demonstrate this property for odd n more directly on a similar but simpler case.{ Note in the following $r$ means the magnitude of a hyperradius of $3n$ rectangular dim's and not just one dim $r$ nor in general only 3 dim's.}
Instead to demonstrate we attempt a solution  $G_1$ to $\{1-\nabla^2\}G_1=0,r\ne 0$, for a similar but different green's fct which has a simple terminating solution just as the prior that did not write only for odd n but not even and easier to show than the prior, and has singularity at $r=0$ in form ${\rm{G1}}\!=\!{\rm{G}}\,e^{-r}(1+\mathit{Ps})$ where $\mathit{Ps}$ is a power series in $r$ to be determined with the
constant term removed,\begin{gather*}-\left (\frac{d^{2}{\rm{G1}}}{dr^{2}}+\frac{(3n-1)}{r}\frac{d{\rm{\,G1}}}{dr} \right)+{\rm{G1}}=0,r\ne 0.\end{gather*}
For n odd the series terminates after (3n-1)$/$2 terms. For example for n$=$3\begin{gather*} {\rm{G1}}=\frac{(15+15r+6r^{2}+r^{3})e^{-r}}
{32\pi ^{4}r^{7}}\text{  and for n$=$5  }\\[8pt] {\rm{G1}}=\frac{(10395+10395r+4725{r}^{2}+1260{r}^{3}+210{r}^{4}+21{r}^{5}+r^{6})e^{-r}}{256\pi^{7}r^{13}}\end{gather*}  and in general for n odd $>$ 2  \begin{gather*}{\rm{G1}}=\Gamma{\left (\frac{3n}{2}\right)}\frac{\left(\sum
_{i=1}^{\frac{3(n-1)}{2}}\prod_{j=1}^{i}{\frac{(3n-2j-1)}{j(3n-j-2)}}r^{i}+1\right)e^{-r}}{2\pi^{3n/2}(3n-2)r^{3n-2}}.\end{gather*}


问:
Expectation of $X^2$ where $X = x_1 + x_2 + x_ 3+ \dots +x_k$ with $P(x_i = 0) = 1-p$ and $P(x_i = x) = p$
We are told that $X = x_1 + x_2 + x_ 3+ \dots +x_k$ and that the $x$'s are independent. The probability of $x_i = x$ is $p$, while the probability that $x_i = 0$ is $1-p$. Use this to find $E(X^2)$.

The first thing I noticed was that if $P(X = x)$ that's just like saying what's the probability that only one of the $x_i$ is equal to $x$. So this makes me think it's binomial distribution. If that is the case $E(X)$ is just the sum of binomial distributions. However, I am not sure how you can handle $E(X^2)$ as $E(X^2) = E((x_1 + x_2 + x_ 3+ \dots +x_k)^2)$.

Any suggestion?

答:
Here's my understanding of what you've asked, using slightly different notation to make things a bit clearer. Let $X_1, ..., X_k \sim \text{iid}$ with $P(X_i = 0) = 1-p$ and $P(X_i = c) = p$ for some constant $c$. We are asked to calculate $E[(X_1 + ... + X_k)^2]$.

This problem becomes simpler if we define $Z_1, ..., Z_k \sim \text{iid Bernoulli}(p)$. Note that $X_i$ is equal in distribution to $cZ_i$. Thus, we have:
$$E[(X_1 + ... + X_k)^2] = E[(cZ_1 + ... + cZ_k)^2] = c^2 E[(Z_1 + ... + Z_k)^2].$$
Now, since the $Z_i$ are iid Bernoulli$(p)$, it follows that $Z_1 + ... + Z_k \sim \text{Binomial}(k, p)$. The variance of a Binomial$(k,p)$ is $kp(1-p)$ while the mean is $kp$. Thus:
$$E[(Z_1 + ... + Z_k)^2]= Var[(Z_1 + ... + Z_k)] + (E[Z_1 + ... + Z_k])^2 = kp(1-p) + (kp)^2$$
and it follows that
$$E[(X_1 + ... + X_k)^2] = c^2[kp(1-p) + (kp)^2]$$