问:
Formal proof for the limit of $\frac{\tanh(x)-1}{e^{-2x}}$ as $x \rightarrow \infty$
Formal proof for the limit of $\frac{\tanh(x)-1}{e^{-2x}}$ as $x \rightarrow \infty$.

So far

Keep in mind I have to use the definition for a limit. I.e for this would be a proof for the limit at $\infty$ with a finite value (I have checked this to be $-2$). I would have to use the definition that states $\lim_{x \ \to \  \infty} f(x) = L \Leftrightarrow
  \forall \ \epsilon>0\; (\exists \ \delta : (\;x>\delta\implies |f(x) - L|\leq\epsilon)).
$ i.e I start off by by finding that $$\left | \frac{\tanh{x}-1}{e^{-2x}}-(-2) \right | \iff \left | \frac{2}{1+e^{2x}} \right |$$ Or just
$\frac{2}{1+e^{2x}}$ since we will assume $\epsilon >0$. Now I would have to to start my proof and this is where I struggle.

I know that
"for alle $\epsilon>0$ let $\delta=\frac{2}{1+e^{2\epsilon}}>0$ and $x>\delta>0$....."

*Remark: I know we use $N,M,K$ and not $\delta$ normally.

答:
Based on feedback in comments I will add reasonable detail in the answer. Some of it may be skipped by those already well versed.

The definition of $\lim_{x\to\infty} f(x) =L$ requires us to show that for every $\epsilon >0$ we can find a corresponding $\delta>0$ such that the inequality $|f(x) - L|<\epsilon $ holds whenever $x>\delta$.

But this does not mean that we start with inequality $|f(x) - L|<\epsilon$ and manipulate it according to valid laws to transform it into "$x>\text{something} $" and then put $\delta=\text{something} $. This approach is what I call solving an inequality and I request readers never to do this while giving formal proofs of limits.

Instead what really needs to be done is to find an expression, say $g(x) $, which is very simple in form and satisfies $|f(x) - L|<g(x) $ and then try to solve $g(x) <\epsilon$. The simple form of $g(x) $ is needed to make this solving part trivial.

Here the desired limit $L=-2$ and we need to use the fact that $e^x>1+x$ if $x>0$ in order to find a suitable $g(x) $. By using it you get $$|f(x) - L|=\frac{2}{1+e^{2x}}<\frac{2}{2x+2}=\frac{1}{x+1}<\frac{1}{x}$$ Now the rightmost expression is less that $\epsilon$ if $x>1/\epsilon$. Thus we can take $\delta$ to be any number greater than or equal to $1/\epsilon$ (in particular $\delta=1/\epsilon$ works).

Essentially you have to do a very gross estimation of $|f(x) - L|$ by $g(x) $. The goal to have a very very simple form for $g(x)$ and you don't need to worry however gross the estimation by $g(x)$ is.


问:
Two statistics called equivalenet

Here is some step that I did. However, the hint from my professor is that drop the condition $\mathcal{P}-a.s$ and just try to establish the implication as(ie, "everywhere"). You can do this by constructing the functions $f$ and $g$. How could I construct the two functions for it? could some give ideas?

答:
$(\Leftarrow)$ Suppose that there exist functions f, g, such that T(X)=f(T*(X)) and T*(X)=g(T(X)) as. Then if you have $T(X)=T(Y)$, applying g to both sides gives you $T*(X)=T*(Y)$ as. If you have $T*(X)=T*(Y)$, applying f to both sides gives you $T(X)=T(Y)$ as.

$(\Rightarrow)$ Suppose $T(X)=T(Y) \Leftrightarrow T*(X)=T*(Y)$. This means $T(X)=T(Y)\Rightarrow T*(X)=T*(Y)$. Construct a function g such that g(T(X))=T*(X), e.g. maps T(X) to T*(X) where X is the observation. Also suppose $T*(X)=T*(Y)\Rightarrow T(X)=T(Y)$. Construct a function f such that f(T*(X))=T(X), e.g. maps T*(X) to T(X) for every observation X. So f and g have been constructed as.

ContentLicense: CC BY-SA 4.0

Title: Two statistics called equivalenet


问:
Why is "for all $x\in\varnothing$, $P(x)$" true, but "there exists $x\in\varnothing$ such that $P(x)$" false?
There exists an $X\in A$ such that $P(X)$. When $A$ is the empty set then this statement is false because there is nothing in $A$ that when plugged in for $X$, makes $P(X)$ come out True.

However, when the quantifier is the universal one, meaning "For all $X\in A$" (which is the empty set), then the statement is true !!

How is that ? .. Because by the reasoning in the first statement, there are not any values of $X$ that makes $P(X)$ come out True.

I'm missing something !!

(I'm studying Electrical Engineering and self studying velleman's how to prove it and the quantifiers are already kicking my butt so simplify the answers for me if possible :))

答:
Wittgenstein also had problems with it. I'm not sure how well I can explain this, but I'll give it a go.

Crudely, his claim was that you can't talk about things that don't exist.

He says, in typically abstruse form:

5.521

I dissociate the concept of all from truth-functions.

Frege and Russell introduced generality in association with logical product or sum. This made it difficult to understand the propositions '$(\exists x).fx$' [there exists $x$ and the proposition $f$ holds for it] and $'(x).fx'$ [all $x$ and the proposition $f$ holds for it].

His difficulties are associated with treating "$\exists x$" and "$\forall x$" as valid statements on their own (which may or not be held true these days). The [brackets] are my attempt to interpret the statements in the way he thought was silly. He goes on to say:

5.524

If we are given objects then we are given all objects.

By which he basically means that $\forall$ is redundant and adds nothing to a statement, though he puts it in a very theory laden way. You can always write a statement without it.

But, it seems he would also have a problem with doing this:

$$\forall x : P(x) \rightarrow \nexists x: ¬P(x) $$

or in words, replace "for all $x$, $P$ is true" with "there is no $x$ for which $P$ does not hold". The two statements are mathematically the same (in first order logic).  He would, I think, favor

$$P(x)$$

which one might read as "$P$ is true", with the subtext of "for all $x$ that exist, obviously! How could we know if it is true for those things that don't?"

ContentLicense: CC BY-SA 3.0


问:
Find the optimal weighting distribution for a die given a set prize values
Say you have $B$ boxes, each of which contains a prize of value $v_b$. You have a $B$-sided die, and you win the value of box $b$ by rolling $b$. You get $R$ rolls, and so can collect multiple prizes, but you can’t win the same box more than once.

Given you know $R$, $B$, and all the box values $v_b$,how do you choose to weight your die as to optimize your expected total reward?

My attempts so far
I've tried two approaches. Both share the definition of expected value:

$U = \sum_b \sum_r  u_{b,r} $

Where

$u_{b,r} = v_b (1 - w_b)^{r-1}w_b$

The difference in my two (failed) methods is in the expression of the constraints. In one method, I let

$w_b = W_b/\sum_i W_i$

and find the critical points by differentiating with respect to every $W_b$ and setting to zero.

In the other method, I use Lagrange multiplier to express the constraint. But in both, I keep ending up with contradictions.

Let's try a reduced version at the problem: I'm even having a problem with the very simple case where R=1 and B=2 is. In this case, the expected value is

$U = v_1w_1 + v_2w_2$

and the constraint is

$w_1 + w_2 = 1$

The answer I'd expect is that $w_1=1$ if $v_1 > v_2$ and $w_1=0$ if $v_1 < v_2$
But, using the Lagrange multiplier method I get

$L = v_1w_1 + v_2w_2 - \lambda (1 - w_1 - w_2)$

and thus the useless set of constraints

$0=v_1 - \lambda$

$0=v_2 - \lambda$

$0=1 - w_1 - w_2$

It's been a very long time since I've had to think about multivariable calculus, so I suspect I'm doing something silly. Since this simplified function is just a plane, I'm not sure the differential method of finding maxima should work, even with the constraint. If I understand my problem with this simple case, maybe I can figure it out for the general problem.

Context
I'm working on an analogous problem in the context of developing a resource distribution program.

答:
The attempt at a solution above is flawed, due to the fact that it doesn't take into account that each box can be only earned once. This creates a natural notion of stopping time within the problem that changes it's structure completely.

Let's keep this to $B=2$ now and generalize later. The weights of the die being $p_1, p_2$, we would like to calculate the expectation value of the winnings after $R$ attempts. After $R$ attempts it is still possible that the game has not ended yet; the strings of rolls (11...1), (22...2)  satisfy the condition. These are also the only strings in the game that earn you less than $v_1+v_2$ (all the others eventually stopped before the R-th attempt yielding the full amount). The winnings EV is

$$\langle W_2\rangle=\sum_i W_i P_{i}=v_1 p(11...1)+v_2p(22...2)+(v_1+v_2)(1-p(11..1)-p(22...2))$$

or more explicitly
$\langle W\rangle=v_1(1-p_2^R)+(1-p_1^R)v_2$

We can now use Lagrange multipliers on this to yield that the values that maximize the average winnings are

$$p_1=\frac{v_1^{\frac{1}{R-1}}}{v_1^{\frac{1}{R-1}}+v_2^{\frac{1}{R-1}}}~~, ~~p_2=\frac{v_2^{\frac{1}{R-1}}}{v_1^{\frac{1}{R-1}}+v_2^{\frac{1}{R-1}}}$$

For $B=3$ we consider the same setup and find the strings that do not terminate the game after $R$ attempts. These are comprised of the strings that contain only one or two of the 3 potential outcomes. Counting the probability of those strings is relatively simple and the expected winnings are

$$\langle W_3 \rangle=\sum_{i=1}^3 v_ip_i^R+(v_1+v_2)(p_1+p_2)^R+(v_2+v_3)(p_2+p_3)^R+(v_3+v_1)(p_3+p_1)^R+(v_1+v_2+v_3)(1-\sum_i p_i^R-(p_1+p_2)^R-(p_2+p_3)^R-(p_3+p_1)^R)$$

This already has become too difficult to maximize analytically. This can be generalized to a die of arbitrary number of sides $B$- just consider $S=$ all the $2^B-2$ subsets of the sets $\{1,2,..., B\}$ that are not empty and not the set itself and the winnings expected are

$$\langle W_B\rangle=\sum_{s\in S}\left[\left(\sum_{i\in s} v_i\right)\left(\sum_{j\in s}p_j\right)^R\right]+\left(1-\sum_{s\in S}\left(\sum_{i\in s}p_i\right)^R\right)\sum_{j=1}^{B}v_j$$

Still, other than the obvious situation when $v_1=v_2=...v_B$ where the maximization is achieved by $p_1=p_2=...=p_B=\frac{1}{B}$ for other winnings per box, the solution is not to my knowledge available analytically, and one has to resort to computational methods.


问:
For complex subspace $S$, there are vectors $u,w ∈ S$ such that $u\neq zw,\ \forall z\in \mathbb{C}$
The official question reads:

Suppose U and W are each 5-dimensional subspaces of $\mathbb{C}^8$.

Prove that there are two vectors  $u,w ∈ U∩W$ $\hspace{1mm}$  s.t. $\hspace{1mm}$  $u≠zw$  $\hspace{1mm}$  ∀ z ∈ $\mathbb{C}$

First I considered one subspace being in $\mathbb{R}^5$ and the other in $\mathbb{C}^5$, but I don't think that would even work here because it's asking for a generalized proof given two subspaces of $\mathbb{C}^8$.

But here's where I'm really stuck. The intersection of two subspaces is itself a subspace, right?
So then the three subspace properties hold (Additive Identity, Closure under Addition, Closure under Scalar Multiplication).

Since the question reads for any z ∈ $\mathbb{C}$, I'm thinking well if we let $z=a+bi$, then it's possible that $b=0$, at which point all we're dealing with is scalar multiplication, $\hspace{1mm}$ i.e. $\hspace{1mm}$ $u=aw$ $\hspace{2mm} ∀ a ∈ \mathbb{R}$.

But this is totally allowed since we're working within a subspace.

Perhaps I need $u$ & $w$ to both be linearly independent, which would mean I should look at the bases of $U$ and $W$? But then what would the intersection look like? Thank you for taking the time to assist.

答:
You're right with your second line of inquiry. Indeed, linear independence implies this condition (certainly if $u = zw$, then $u - zw = 0$ is a non-trivial linear combination of $u$ and $w$ that produces the zero vector). So, it suffices to show that $\dim(U \cap W) \ge 2$; if this is true, we can simply take the first two vectors of the basis to get two linearly independent vectors $u$ and $w$.

Now, the key result here is to use the following formula:
$$\dim(U) + \dim(V) - \dim(U \cap V) = \dim(U + V).$$
Note that $U + V$ is a subspace of $\Bbb{C}^8$, and hence has dimension at most $8$. Also, $\dim(U) + \dim(V) = 5 + 5 = 10$ by assumption. Thus
$$\dim(U \cap V) = 10 - \dim(U + V) \ge 10 - 8 = 2,$$
as required.


问:
Prove that : $ \int_{0}^{2\ln{\varphi}}{\theta\ln{\left(2\sinh{\frac{\theta}{2}}\right)}\,\mathrm{d}\theta}=-\frac{1}{5}\zeta\left(3\right) $
Denoting $ \varphi=\frac{1+\sqrt{5}}{2}=\mathrm{Golden\ Ratio} $. How would you prove that : $$  \int_{0}^{2\ln{\varphi}}{\theta\ln{\left(2\sinh{\frac{\theta}{2}}\right)}\,\mathrm{d}\theta}=-\frac{1}{5}\zeta\left(3\right) $$

I came accross this strange identity while reading the following article : VALUES OF THE RIEMANN ZETA FUNCTION AND
INTEGRALS INVOLVING $\log{\left(2\sinh{\frac{\theta}{2}}\right)}$ AND $\log{\left(2\sin{\frac{\theta}{2}}\right)}$. (No complete proof was given in the article). I'm still trying to prove it. For now I did try to set $ x=2\ln{\left(\sinh{\frac{\theta}{2}}\right)} $ to get rid of the golden ratio, but it doesn't seem to get any better. I know I'll have to expand the integrand to get to $ \zeta $ using, for example, the fact that : $ \ln{\left(\sinh{x}\right)}=x-\ln{2}-\sum\limits_{n=1}^{+\infty}{\frac{\mathrm{e}^{-2nx}}{n}} $, but I'm sure I cannot do it now, because of that golden ratio in the upper bound. Any ideas, hints will be appreciated.

答:
A CAS gives the antiderivative (have a look here)
$$I(\theta)=\int\theta  \log \Bigg[2 \sinh \left(\frac{\theta }{2}\right)\Bigg]\,d\theta$$
$$I(\theta)=\frac{1}{12} \left(\theta ^3+6 \theta ^2 \left(-\log \left(1-e^{\theta }\right)+\log
   \left(\sinh \left(\frac{\theta }{2}\right)\right)+\log (2)\right)-12 \theta
   \text{Li}_2\left(e^{\theta }\right)+12 \text{Li}_3\left(e^{\theta }\right)-i \pi
   ^3\right)$$
$$\lim_{\theta \to 0} \, I(\theta )=\zeta (3)-\frac{i \pi ^3}{12}$$ and the surprise is that, after simplifications
$$I(2\log(\varphi))=\frac{4 }{5}\zeta (3)-\frac{i \pi ^3}{12}$$

What is also interesting is to see how fast converges the integral after a series expansion
$$\theta  \log \Bigg[2 \sinh \left(\frac{\theta }{2}\right)\Bigg]=\theta\log(\theta)+\sum_{n=1}^k a_k\,\theta^{2k+1}$$

$$\left(
\begin{array}{cc}
 k & \text{from series expansion}\\
 1 & -0.240365891808504 \\
 2 & -0.240411880810115 \\
 3 & -0.240411373694167 \\
 4 & -0.240411380739980 \\
 5 & -0.240411380630111 \\
 6 & -0.240411380631951 \\
 7 & -0.240411380631918 \\
 8 & -0.240411380631919
\end{array}
\right)$$


问:
State the approximate distribution of Y giving its Parameters and work out $ P(Y <22)$
This is the question
[
I tried solving it and the X-N (21,4.2^2/100) is correct but the probability is wrong its spouse to be 0.954. I have no teacher to ask as I am self-studying any help would be much appreciated. this is my work thank you!

This is similar example:

答:
When you correct your standard deviation for $\bar Y$, you will get with $n=50$ observations

$$Z = \frac{\bar Y - \mu}{\frac{\sigma}{\sqrt n}}\stackrel{\approx}{\sim}N(0,1)$$

and
$$P\left(\bar Y < 22\right)= P\left(Z < \frac{22-21}{\frac{4.2}{\sqrt{50}}}\right)=P\left(Z<\frac{\sqrt{50}}{4.2}\right)\approx 0.954$$


问:
Riemann - Prove that $ L(f, [a, b]) + L(g, [a, b]) \le L(f+g, [a, b]) $

Suppose $f, g: [a, b] \to \mathbb{R}$ are bounded functions. Prove that
\begin{equation}
    L(f, [a, b]) + L(g, [a, b]) \le L(f+g, [a, b])
\end{equation}

My attempt:

Suppose $f, g: [a, b] \to \mathbb{R}$ are bounded functions. Let $P$ be an arbitrarily fixed partition of $[a, b]$. Note that since $\inf (f+g)\ge\inf f+\inf g$, we have
\begin{align*}
L(f+g, P, [a, b]) &= \sum_{j=1}^n (x_j-x_{j-1}) \inf_{[x_{j-1}, x_j]} (f+g)\\
&\ge \sum_{j=1}^n (x_j-x_{j-1}) \inf_{[x_{j-1}, x_j]} f + \sum_{j=1}^n (x_j-x_{j-1}) \inf_{[x_{j-1}, x_j]} g \\
&= L(f, P, [a, b]) + L(g, P, [a, b])
\end{align*}

Now, taking suprema on both sides of the inequality above, we have
\begin{align*}
\sup_P L(f+g, P, [a, b]) &\ge \sup_P \{L(f, P, [a, b]) + L(g, P, [a, b])\}\\
&=\sup_P L(f, P, [a, b]) + \sup_PL(g, P, [a, b])
\end{align*}

where the suprema are taken over all partitions $P$ of $[a, b]$.

Can someone please read over my proof and let me know if there is something that needs to be revised? If revisions are needed, can someone please suggest how these revisions should be made?

答:
Like @Mason pointed out, the supremum part is incorrect. You need to show that for any partitions $P, P'$, there exists partition $P^*$ such that:

$$L(f,P,[a,b])+L(g,P',[a,b])≤L(f+g,P^*,[a,b])$$

An easy way to show that is to choose $P^* = P \cup P'$ and show that for any lower integrable function $h: [a,b] \to \mathbb{R}$, $L(h, P^*, [a,b]) \geq L(h, P, [a,b])$ and $L(h, P^*, [a,b]) \geq L(h, P', [a,b]) $.

Once you have that, if you still need help:

 $$L(f+g,P^*,[a,b]) \geq L(f,P^*,[a,b])+L(g,P^*,[a,b]) \geq L(f,P,[a,b])+L(g,P',[a,b])$$


问:
Representing affine transform of Legendre polynomials
I have a function defined as a set of weighted Legendre polynomials:
$f(x)=\alpha_0 P_0(x) + \alpha_1 P_1(x) + \alpha_2 P_2(x) +\ldots$.

I have another function similarly defined with Legendre basis weights, but it's over a different range. I want to compare the two functions over the range where they overlap. This means I need to transform one function by stretching and shifting it to match the other sequence's range.

This is not too hard. I basically want to express my function with an affine transform applied to the input, and simplify the resulting polynomial down to basic Legendre basis weights.  That is, finding the $\beta_i$ coefficients below.

$\begin{align*}
f(Ax+B) &= \alpha_0 P_0(Ax+B) + \alpha_1 P_1(Ax+B) +\alpha_2 P_2(Ax+B) +\ldots\\
&=  \beta_0 P_0(x) + \beta_1 P_1(x) + \beta_2 P_2(x) + \ldots
\end{align*}$

This is quite well defined. I could in theory do it by expanding the sum into a simple giant polynomial, then collect terms and find the Legendre weights again.

So for example, for the second order term only,

$\begin{align*}
P_2(Ax+B)&= {1\over2}(3(Ax+B)^2-1)\\
P_2(Ax+B)&= A^2(\frac12(3x^2-1)+ABx+{1\over2}(A^2+B^2)-\frac12\\
P_2(Ax+B)&= A^2 P_2(x) + ABP_1(x)+ {1\over 2}(A^2+B^2-1)P_0(x)
\end{align*}$

This gets more and more ugly with higher orders.

My question: is there a nice recurrence relation or simple way to generate these transformed basis weights given A, B, and the set of weights $\alpha_i$, producing the resultant set of weights $\beta_i$? It feels like there should be a method that does not drop down to expanding the polynomial in full.

答:
Well this is perhaps pretty much what you are doing, but it can be programmed quite simply. (Sorry about the lo-tech look, but I'm ASCII only).

The Legendre polynomials satisfy $P_{n+1} = d_n \cdot X\cdot P_n + f_n \cdot P_{n-1}$. If we define polynomials $p_n (x) = P_n(ax+b)$ then we have $p_n = \sum\limits_{j \leq n} \mathbf A_{n,j} P_j$ for a ( lower triangular) matrix $\mathbf A$, and the coefficient vector you seek will be $\mathbf A^T\dot  \alpha$. Since the $p$'s also satisfy a recurrence relation, the rows of $\mathbf A$ can be computed recursively. I tried this and got:

```<code>A(0,0) = 1
A(1,0) = b    A(1,1) = a
n= 1..N-1
    A(n+1, 0) = p[n] * b * A( n , 0)
              - p[n] * a * (q[1]/p[1]) * A( n , 1)
              + q[n] * A( n-1, 0)

    j=1..n-1
        A(n+1, j) = p[n] * b * A( n, j)
                  + p[n] * (a/p[j-1]) * A( n, j-1)
                  - p[n] * (a * q[j+1]/p[j+1]) * A( n, j+1)
                  + q[n] * A( n-1, j)

    A(n+1, n) = p[n] * b * A( n  , n)
              + p[n] * (a/p[n-1]) * A( n , n-1)

    A(n+1, n+1)= a*A( n , n)
</code>```

which appears to work, though I cannot vouch for stability and suspect there has to be a better way.

Body:
Well this is perhaps pretty much what you are doing, but it can be programmed quite simply. (Sorry about the lo-tech look, but I'm ASCII only).

The Legendre polynomials satisfy $P_{n+1} = d_n \cdot X\cdot P_n + f_n \cdot P_{n-1}$. If we define polynomials $p_n (x) = P_n(ax+b)$ then we have $p_n = \sum\limits_{j \leq n} \mathbf A_{n,j} P_j$ for a ( lower triangular) matrix $\mathbf A$, and the coefficient vector you seek will be $\mathbf A^T\dot  \alpha$. Since the $p$'s also satisfy a recurrence relation, the rows of $\mathbf A$ can be computed recursively. I tried this and got:

A(0,0) = 1
A(1,0) = b    A(1,1) = a
n= 1..N-1
    A(n+1, 0) = p[n] * b * A( n , 0)
              - p[n] * a * (q[1]/p[1]) * A( n , 1)
              + q[n] * A( n-1, 0)

    j=1..n-1
        A(n+1, j) = p[n] * b * A( n, j)
                  + p[n] * (a/p[j-1]) * A( n, j-1)
                  - p[n] * (a * q[j+1]/p[j+1]) * A( n, j+1)
                  + q[n] * A( n-1, j)

    A(n+1, n) = p[n] * b * A( n  , n)
              + p[n] * (a/p[n-1]) * A( n , n-1)

    A(n+1, n+1)= a*A( n , n)

which appears to work, though I cannot vouch for stability and suspect there has to be a better way.


问:
Response of a specific LTI system to an exponential
I am taking a Signals & Systems course in my Telecommunications Engineering degree, and now I am trying to put together everything I have seen.
I came across this basic problem of determining the output of a system whose transfer function is $H(s)=\frac{1}{s+2}$ when it is fed with a signal $x(t)=e^{2jt}$ (where $j^2=-1$).
As every other student of S&S I know that exponentials are eigenfunctions to LTI systems, hence I tried this
$$y(t)=H(s_0)e^{s_0t}=H(2j)e^{2jt}=\frac{1}{2j+2}e^{2jt}=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}e^{2jt}$$
"Nice!" I said. But then I wanted to test the convolution theorem, and so I happily did
$$Y(s) = H(s)X(s) = \frac{1}{s+2}\frac{1}{s-2j}=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}\left(\frac{1}{s-2j}-\frac{1}{s+2}\right)$$
Seeing this, and that its time-domain counterpart would be $y(t)=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}\left(e^{2jt}-e^{-2t}\right)$ I thought I made a mistake, but Wolfram Alpha spat the same. Where does the $e^{-2t}$ term come from?. My first idea was that maybe I was messing it up with the ROC's or something, but all the present signals are right-sided.

Then I tried the convolution with the impulse response, which is precisely $h(t)=\mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\}=e^{-2t}$.
$$(x\ast h)(t)=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau=\int_{0}^{\infty}e^{-2\tau}e^{2j(t-\tau)}d\tau=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}e^{2jt}$$
I calmed down a bit when I saw that something matched what was predicted by the eigenfunction property.
After this I attempted the same by applying the commutativity of the convolution:
$$(h\ast x)(t)=\int_{0}^{\infty}e^{2j\tau}e^{-2(t-\tau)}d\tau$$
And it diverges, but I found out that
$$(h\ast x)(t)=\int_{0}^{t}e^{2j\tau}e^{-2(t-\tau)}d\tau=\int_{0}^{t}e^{-2\tau}e^{2j(t-\tau)}d\tau=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}\left(e^{2jt}-e^{-2t}\right)$$

I must admit that I do not quite understand the difference between integrating from $0$ to $\infty$ and from $0$ to $t$ in the convolution, and therefore I am unable to explain why with the integration interval $[0,\infty)$ the convolution gives the same as the inverse Laplace transform. And I definitely cannot understand why all this methods do not give the same answer.

On the other hand, I tried feeding the system with $e^{-2t}$. The convolution with the interval of integration $[0,\infty)$ is divergent, but the following is certainly true:
$$\int_{0}^{t}e^{-2\tau}e^{-2(t-\tau)}d\tau = te^{-2t} = \mathcal{L}^{-1}\left\{\frac{1}{(s+2)^2}\right\}$$
Beside, a (very simple) "tool" I created in geogebra which shows graphically the convolution between two signals showed the same function (and it has worked very well for every pair of real signals I have tried).

But I don't understand how to apply the property of eigenfunction here since $H(-2)=\frac{1}{0}$.

May someone help me understand this?

答:
Your computations are correct. Your assumption that the "Laplace transform of the convolution is the product of the Laplace transforms is incorrect" for the definition of convolution you are considering. The convolution you are considering has integral limit infinity whereas the theorem applies to the convolution with integral limit $t$. There is an important difference between these two formulas: Note that restricting the integral limit to $t$ is equivalent to having the integral limit to infinity but with $x(t)$ defined only for $t>0$. That's why it is the Fourier (and not the Laplace) transform that dominates signal & systems analysis (including telecommunications)

EDIT (on the convolution integral limits):

In linear time invariant (LTI) systems theory you have a system with an impulse response $h(t)$ and an input signal $x(t)$. In general, these two functions (signals) are defined for all $t\in \mathbb{R}$ and the output of the system is given by the convolution integral

$$
y(t) = \int_{-\infty}^{\infty} h(\tau) x(t-\tau)d\tau, t\in \mathbb{R}.
$$
Note that the output $y(t)$ is in general non-zero for both positive and negative values of $t$.

The above is the most general formula for the output of an LTI system and applies to arbitrary $h(t)$ and $x(t)$ (as long as the integral converges).

Now, if the support of $h(t)$ and/or $x(t)$ is finite, you can change the integral limits to reflect that.

Let's start with the impulse response defined only for $t\geq 0$, i.e., $h(t)=0,t<0$, corresponding to a, so called, causal system. Then, it is easy to see that the integrad in the above expression is zero for all $\tau<0$ and thus we can write

$$
y(t) = \int_{0}^{\infty} h(\tau) x(t-\tau)d\tau, t\in \mathbb{R}.
$$

Let, in addition, the input $x(t)$ only defined for positive $t$, i.e., $x(t)=0,t<0$. Notice that the term $x(t-\tau)$ appearing in the integrad will be non-zero only whenever it holds $t-\tau \geq 0$ or, equivalently, $\tau \leq t$. But this condition effectively imposes a restriction to the upper integral limit leading to
$$
y(t) = \int_{0}^{t} h(\tau) x(t-\tau)d\tau, t\in \mathbb{R}.
$$
Finally, note that the above integral will only be non-zero for $t\geq 0$, ultimately leading to
$$
y(t) = \int_{0}^{t} h(\tau) x(t-\tau)d\tau, t\geq 0,
$$
with the understanding that $y(t)=0,t<0$.

The last convolution integral is effectively the one assumed when working the Laplace transform, i.e., all systems and input/output signal are causal.


问:
Point within the interior of a given angle
The point $M$ is within the interior of given angle $\alpha$. Find the distance between $M$ and the vertex of the angle ($OM=?$) if $a$ and $b$ are the distances from $M$ to the sides of the angle.

We can see that $$OM^2=OP^2+PM^2=OK^2+KM^2$$
Let $OP=x;PK=y$. Then $$2OM^2=OP^2+PM^2+OK^2+KM^2\\=x^2+b^2+y^2+a^2.$$ I can't approach the problem further. How can we use the given angle $\alpha$? Thank you in advance!

答:
A two-liner :

Since $OPMK$ is cyclic, $\angle PMK = \pi - \angle POK = \pi - \alpha$; by cosine-rule in $\triangle PMK$, $PK^2=a^2+b^2+2ab\cos \alpha$.

Now $OM$ is simply the diameter of the circle which can be found by sine-rule in either of $\triangle POK$ or $\triangle PMK$,
$$OM=2r=\frac{PK}{\sin \angle POK}=\frac{PK}{\sin \alpha}=\frac{\sqrt{a^2+b^2+2ab\cos \alpha}}{\sin \alpha}$$


问:
What is the difference between "singular value" and "eigenvalue"?
I am trying to prove some statements about singular value decomposition, but I am not sure what the difference between singular value and eigenvalue is.

Is "singular value" just another name for eigenvalue?

答:
Consider the comparison between largest singular value and largest eigen value of a matrix with real entries, $A\in \mathbb{R}^{m \times n}$:

\begin{align*}
 \sigma_{\max}(A)&= \sup_{x \in S^{n-1}} ||Ax||_2 =\sup_{x \in S^{n-1}} \sqrt{||Ax||_2^2}= \sup_{x \in S^{n-1}} \sqrt{x^\top A^\top A x} = \sqrt{\lambda_{\max}(A^\top A)}
\end{align*}

For $A^\top =A$, we have, $\sigma_{\max}(A)=\sqrt{\lambda_{\max}(A^2)}=\sqrt{\lambda^2_{\max}(A)}=|\lambda_{\max}(A)|=\sup\limits_{x \in S^{n-1}}| x^\top A x|$


问:
Intuitively understanding why the volume of a cylinder of height $h$ and radius $R$ is $\pi R^2h$
It's well known that a cylinder with height $h$ and the base radius $R$ will have a volume of

$$V = \pi R^2h$$

I'm trying to derive an intuitive understanding of it. I have two ways, one seems to be "working" (as in - it yields the correct result) and another one is failing, see below.

Method 1

If we agree that the area of a circle is known as $\pi R^2$ then we can think of a volume as a sum of all "areas of circles along the heights". So if we imagine that there is such an "object" that has a height of "one point" then its volume is equal to the area of the circle it has as a base. We then see that there will be exactly $h$ points in the height of the cylinder, so there will be exactly $h$ such "objects". Hence we can just multiply a "volume" of one "object" by $h$ which will yield the correct cylinder volume. I believe this method is just a "naive" version of a more rigorous calculus approach with defining the limit for $dx$ where $x$ will be along the height of the cylinder

Method 2

Now let's look at the cylinder as a rectangle rotated along one of its sides. If we have $h$ as the cylinder height and $R$ as a radius, then it's an $h \times R$ rectangle rotated by $2\pi$ around the side with length $h$. So the volume is the "sum" of all the "objects" with the height of "one point" and the base a rectangle. There are exactly $2\pi R$ points along the circumference of the circle that the rectangle will follow. Therefore we will have $2\pi R$ of these "objects". So we just need to multiple the "volume" of one such "object" by the number of these "objects" to get the cylinder volume. The "volume" of this "object" will be just the area of a rectangle, since it has a height of "one point". To conclude, we have $2\pi$ of these "objects", thus the volume must be $V = 2\pi R \times V_{\text{object}} = 2\pi R \times h R = 2\pi R^2h$.

But method 2 yields incorrect result, it's exactly $2$ times more than it should be. Why is that? I.e. why is that the point-based construction worked in the first case but failed in the second case? And how to make it work properly? What about rotation in general (i.e. not only a cylinder)?

答:
I will try and keep the answer simple and certain parts may be more simplified than the real situation.

First reduce the problem to a $2D$ problem as the height $h$ is not your concern. It really boils down to area of a circle being $\pi R^2$ vs. $2\pi R^2$.

a) Draw a rectangle with length of $\pi R$ and width of $2 R$. Now its area is $2 \pi R^2$. If we added height $h$, its volume would be $2\pi R^2h$. Now take a pin of length $R$ (if we added height $h$ to it, we would have a rectangular blade), place it in the rectangle with one end of the pin at the center of the rectangle.

Then rotate the pin around that end. Can you see that as we rotate, it will touch the rectangle at exactly two points (as the rectangle width is $2R$) and at all other points, it leaves space around. This clearly shows that the area of rotation of a pin of length $R$ is less than $2\pi R^2$. Now add height $h$ to it and we see that the solid generated by revolution of a rectangular blade of dimension $R \times h$ has lower volume than $2 \pi R^2 h$.

b) Now if you rotate a pin of length $R$ around one of its end by a very small angle $d\theta$, the arc length is $R d\theta$. Considering the angle is really small, we can approximate the area to a triangle of base $R d\theta$ and height $R$. So, $dA = \frac{1}{2} R \times R d\theta$. Over $2\pi$, this translates to $\pi R^2$.


问:
Finding pmf by using law of total probability
A television quiz game operates as follows. In the first part of the game, a contestant is asked a series of difficult questions; the probability of answering any question correctly is $p$, independently of all other questions. This first part of the game ends as soon as the contestant answers a question incorrectly. Denote by $N$ the total number of questions answered correctly in this first part of the game (excluding the final incorrect one). In the second part of the game, the contestant is asked a series of easy questions; here, the probability of answering any question correctly is $r$, independently of all other questions. The total number of questions asked in this second part of the game is $N$ (i.e. if $N = n$ in the first part of the game, the contestant answers $n$ questions in the second part). The contestant's final score, $y$ say, is equal to the number of correct answers in the second part of the game.

(a) Find the probability mass function of $N$. Also state the conditional distribution of $Y$ was given $N = n$, including the value(s) of its parameter(s).

(b) Use the Law of Total Probability to find the probability mass function of $Y$. Show that this pmf has the same form as that of $N$, but with $p$ replaced by another quantity which you should define.

My answer:

(a) $N$ is obviously a geometric distribution, and $Y$ binomial.

$P(N=n)=p^n(1-p)$, $P(Y=k|N=n)={n \choose k}r^k(1-r)^{n-k}$

(b) Use Law of total probability,

$P(Y=k)=\sum_{n=k}^\infty {n \choose k}r^k(1-r)^{n-k}p^n(1-p)$

Could you please check if my answer is correct. And then help me with the last step. I have no idea how to simplify this series into a geometric distribution-like form. It gives me a hint to use binomial expansion, but I don't know how to use it with $p^n$ presents.

答:
You are completely right. The final proof is the following

$$\mathbb{P}[Y=k]=(pr)^k(1-p)\sum_{n-k=0}^{\infty}\binom{n}{n-k}[p(1-r)]^{n-k}=$$

$$=(pr)^k(1-p)\sum_{n-k=0}^{\infty}\binom{(n-k)+k}{n-k}[p(1-r)]^{n-k}=$$

$$=(pr)^k(1-p)\underbrace{\sum_{i=0}^{\infty}\binom{i+k}{i}[p(1-r)]^i}_{\text{It's a known series}}=$$

$$=(pr)^k(1-p)\frac{1}{[1-p(1-r)]^{k+1}}=$$

$$=\left[\frac{pr}{1-p(1-r)}\right]^k\cdot\frac{1-p}{1-p(1-r)}$$

Thus

$$Y\sim Geo\left[\frac{1-p}{1-p(1-r)}\right]$$

For the list of math series see here; the one I used is in the list of Binomial Coefficients [3]


问:
What is the range of the following function?
I have difficulties in understanding the concept of range. Let $f:\mathbb Z_{12}\to \mathbb Z_3$, $f(x)=x$. What is the range of it? Here is what i think: Range of $f$ is the set $\{a \mid a\equiv x \bmod 3 \text{ and } x=0,1,2\}$. Is that right?

Thanks

答:
The range is the set of values which you can get as result of applying the function. That is, for a given function $f:X\to Y$, the range is $\{f(x):x\in X\}$.

If I understand your function correctly, its range is $Z_3$.


问:
Calculate polygon using some integrals
I have a polygon below. The length of it is $a$, base is $b$, and height is $h$. I want to have an integral to calculate volume of it. This is so different from what I know before, and it's not graph volume with disks. But, I have a big hint that the volume is
$$
\frac{\sqrt{3} h}{12} \bigl( a^2 + ab + b^2 \bigr).
$$
I want to ask what to make to set up the calculation. And how to show that the volume is the answer here. I have ideas, like using the method of calculating volume of disk, $\pi r_{\rm out} - \pi r_{\rm in}$. But where do I find radius? Thank you for helping..

答:
Assuming that the base of this frustum of a pyramid has an equilateral triangle cross section, the first thing you have to work out is the area of the triangle in terms of its side length:
$$
A = \frac{\sqrt{3}}{4} s^2
$$

Next you imagine slicing the solid region perpendicular to the height, so that each slice is an equilateral triangle. These triangles vary in size, with their side lengths varying linearly from $s = b$ to $s = a$ over the course of a height change of $h$. Let's parametrize this side length in terms of this vertical coordinate $z$:
$$
s = b + \frac{a-b}{h} z.
$$
Notice that this parametrization gives $s=b$ when $z=0$ and $s=a$ when $z=h$, as it must. Substituting this expression for $s$ into the triangle area formula yields:
$$
\begin{align*}
A(z) &= \frac{\sqrt{3}}{4} s^2
= \frac{\sqrt{3}}{4} \biggl( b + \frac{a-b}{h} z \biggr)^{\!2}
= \frac{\sqrt{3}}{4} \cdot \frac{1}{h^2} \Bigl( bh + (a-b) z \Bigr)^2 \\
&= \frac{\sqrt{3}}{4h^2} \bigl( b^2h^2 + 2bh(a-b)z + (a-b)^2 z^2 \bigr)
\end{align*}
$$

Now, we assemble all these cross-sectional areas in an integral to find the volume:
$$
\begin{align*}
V = \int_0^h A(z) \, dz
&= \frac{\sqrt{3}}{4h^2} \int_0^h \bigl( b^2h^2 + 2bh(a-b)z + (a-b)^2 z^2 \bigr) \, dz \\
&= \frac{\sqrt{3}}{4h^2} \biggl. \Bigl( b^2h^2z + bh(a-b)z^2 + (a-b)^2 \cdot \frac{1}{3} z^3 \Bigr) \biggr\rvert_0^h \\
&= \frac{\sqrt{3}}{4h^2} \Bigl( b^2h^2(h) + bh(a-b)(h)^2 + (a-b)^2 \cdot \frac{1}{3}(h)^3 \Bigr) \\
&= \frac{\sqrt{3}}{4h^2} \cdot \frac{h^3}{3} \Bigl( 3b^2 + 3b(a-b) + (a-b)^2 \Bigr) \\
&= \frac{\sqrt{3}h}{12} \Bigl( a^2 + ab + b^2 \Bigr)
\end{align*}
$$

Title: Calculate polygon using some integrals


问:
Question in metric spaces.
Let $(X,d)$ be a metric space and $A\subset X$. Let
$d(x,A) = \inf \{ d(x,a) \mid a \in A\}$.
If $d(x,A) = 0$ for all $x$ then which of the following must be true?

$A$ is compact

$A$ is closed

$A$ is dense

$A = X$

My attempt: Since if $A$ is dense in $X$, we can always find elements from $A$ that approximate a particular element in $X$. So I chose 3. But the book says 4. Can somebody tell me which is correct.

答:
Given $\alpha \in \mathbb R$, and a non-empty bounded subset $B$ of $\mathbb R$, we have \begin{align}
\alpha = \inf B & \iff \textrm{for any $\varepsilon>0$, $\alpha+\varepsilon$ is not a lower bound of $B$} \\ & \iff \textrm{for any $\varepsilon>0$ there is $b_\varepsilon \in B$ such that $b_\varepsilon<\alpha+\varepsilon$.}
\end{align}
In your case, $d(x,A)=0$ if and only if for any $\varepsilon>0$ there is $a_\varepsilon \in A$ such that $d(x,a_\varepsilon)<\varepsilon$, which means, $d(x,A)=0$ if and only if for any $\varepsilon>0$ we have $B(x,\varepsilon) \cap A \neq \varnothing$. Thus, $d(x,A)=0$ is equivalent to $x \in \overline A$.

Title: Question in metric spaces.


问:
$M\subset N$ for $R$-modules $M,N$ if $S_{\mathfrak m}^{-1}M\subset S_{\mathfrak m}^{-1}N$ for all maximal ideals $\mathfrak m\subset R$?
Consider the following proposition (with proof) taken from S. Lang's "Algebraic Number Theory":

Proposition $\mathbf{18}$. Let $A$ be a Dedekind domain and $M,N$ two modules over $A$. If $\mathfrak p$ is a prime of $A$, denote by $S_{\mathfrak p}$ the multiplicative set $A-\mathfrak p$. Assume that $S_{\mathfrak p}^{-1}M\subset S_{\mathfrak p}^{-1}N$ for all $\mathfrak p$. Then $M\subset N$.

Proof. Let $a\in M$. For each $\mathfrak p$ we can find $x_{\mathfrak p}\in N$ and $s_{\mathfrak p}\in S_{\mathfrak p}$ such that $a=x_{\mathfrak p}/s_{\mathfrak p}$. Let $\mathfrak b$ be the ideal generated by the $s_{\mathfrak p}$. Then $\mathfrak b$ is the unit ideal, and we can write

$$1=\sum y_{\mathfrak p}s_{\mathfrak p}$$

with elements $y_{\mathfrak p}\in A$ all but a finite number of which are $0$. This yields

$$a=\sum y_{\mathfrak p}s_{\mathfrak p}a=\sum y_{\mathfrak p}x_{\mathfrak p}$$

and shows that $a$ lies in $N$, as desired.

My first question is mainly concerned with the emphasised part and my second, more generally, with which of the assumptions of the proposition are necessary.

So, first regarding the emphasised part. From my understanding we consider the ideal

$$I=\sum_{\mathfrak p} Rs_{\mathfrak p}$$

and claim that this ideal is the unit ideal. As far as I can tell this is a consequence of the following observation:

Maximal ideals are in particular prime and hence occur among the $\mathfrak p$. Pick any maximal ideal $\mathfrak m$ and note that $s_{\mathfrak m}\in I$ but $s_{\mathfrak m}\notin\mathfrak m$ by definition. Hence $I$ is not contained in any maximal ideal and so has to be the unit ideal (as every proper ideal is contained in some maximal ideal).

Is this the correct way to think about it? I have found similar claims at some other places but never the full argument spelled out (which I have problems to grasp).

On to the second question: what is really needed for this to work? I think Dedekind domain can be drastically weakened to commutative (from what I understand this is not really needed but a subtle detail I do not deem too important for my question) unital ring. The Dedekind assumption is likely due to the book being on Algebraic Number Theory. However, I am also not sure about the assumption on all primes. The argument I gave above only uses the assumption on all maximal ideals (which makes no difference in a Dedekind domain but in general). I know that for a similar theorem equivalence in the cases of all primes and all maximal ideals holds (see here for instance).

So I claim the following from the aforegoing analysis:

Claim. Let $R$ be a commutative unital ring and $M,N$  two modules over $R$. If $\mathfrak m$ is a maximal ideal of $R$, denote by $S_{\mathfrak m}$ the multiplicative set $R\setminus\mathfrak m$. Assume that $S_{\mathfrak m}^{-1}M\subset S_{\mathfrak m}^{-1}N$ for all $\mathfrak m$. Then $M\subset N$.

It might be possible to reduce this statement to the one I linked earlier (this one) but I am not sure if and how. Also, I could not find a more general reference for Prop. $18$ which makes me wonder. The books on Commutative Algebra and Abstract Algebra I checked did not seem to contain something similar either.

Thanks in advance!

答:
For your first question: yes to everything you said.

For your second question: yes, that's fine, though if you're not working in a domain you need to be careful to not accidentally assume $M \to S_{\mathfrak{p}}^{-1} M$ is injective, and the "subset" relations you wrote need to be treated with additional care since we can't just define localizations as taking place in the field of fractions. For instance, ponder $A=M=\mathbb{Z}/6\mathbb{Z}$ and $\mathfrak{p} = (3) = \{0, 3\}$. Then $S_{\mathfrak{p}}^{-1} M \cong \mathbb{Z}/3\mathbb{Z}$, essentially since in that ring we have $3 = \frac{6}{2} = \frac{0}{2} = 0$. It is nonetheless a standard fact that localization is exact and hence preserves injections, so there's no important funny business going on. In particular you'd need to use $[a/1] = [x_{\mathfrak{p}}/s_{\mathfrak{p}}]$ and slightly tweak the argument to write $a \in N$.


答:
The side length of the equilateral triangle at the base is $a$ and at height $h$ is $b$. So at height $x$, the side length will be $a - \frac{(a-b)x}{h}, 0 \leq x \leq h$.

The area of equilateral triangle at height $x$, $A(x) = \frac{\sqrt3}{4}\big(a - \frac{(a-b)x}{h}\big)^2$.

Now at height $x$, if we consider the volume of a very thin wedge of this solid, of thickness $dx$,

$dV = A(x) \cdot dx$

So $V = \displaystyle \int_0^h \frac{\sqrt3}{4h^2}\big(a(h-x) + bx\big)^2 \ dx$

$ = \frac{\sqrt3}{4h^2}\displaystyle \int_0^h \big(a^2(h-x)^2 + b^2x^2 + 2 abx(h-x)\big) \ dx$

$ = \displaystyle \frac{\sqrt3}{12} (a^2 + ab + h^2) h$

Title: Calculate polygon using some integrals

Body:
I have a polygon below. The length of it is $a$, base is $b$, and height is $h$. I want to have an integral to calculate volume of it. This is so different from what I know before, and it's not graph volume with disks. But, I have a big hint that the volume is
$$
\frac{\sqrt{3} h}{12} \bigl( a^2 + ab + b^2 \bigr).
$$
I want to ask what to make to set up the calculation. And how to show that the volume is the answer here. I have ideas, like using the method of calculating volume of disk, $\pi r_{\rm out} - \pi r_{\rm in}$. But where do I find radius? Thank you for helping..


问:
Partition Properties
I'm currently trying to understand the proof of Erdős–Rado in Jech's Set Theory book.

But since I've already problems in the first steps, I went back to the proof of:
$$2^\kappa \not\rightarrow (\omega)_{\kappa}^{2}$$

He wants to find a partition that has no homogeneous set of size 3. $S=\{0,1\}^{\kappa}$ and $F:[S]^2\rightarrow \kappa$ by $F(\{f,g\})=$ the least $\alpha < \kappa$ such that $f(\alpha)\not=g(\alpha)$.

Now comes the part I don't understand:

If $f,g,h$ distinct elements of $S$, then $F(\{f,g\})=F(\{f,h\})=F(\{g,h\})$ is impossible.

Why is it impossible? why can't there be an $\alpha$ s.t. $f(\alpha)\not= g(\alpha)\not=h(\alpha)$?

If someone also knows a nice proof of the Erdős–Rado theorem I'd be happy to hear about it!

Best, Luca

答:
Assume that $F(\{f,g\})=F(\{g,h\})=F(\{f,h\})=\alpha$. Then for every $\beta<\alpha$ we have $f(\beta)=g(\beta)=h(\beta)$.

Suppose that $f(\alpha)=0$ then $g(\alpha)=1$, but then $h(\alpha)\neq g(\alpha)$ so $h(\alpha)=0$, in which case $h(\alpha)=f(\alpha)$.

This is a contradiction, since we assumed that $\alpha$ was such that $f$ and $h$ disagreed on its value (and agreed on its predecessors).

Title: Partition Properties


问:
Inequality containing logarithms
I've been reading an algorithm book lately.
There is this one question that I've been struggling with and can't solve it:

Consider a computer that can do 10^10 operations per second.
With an algorithm that has a time complexity of $n *log(n)$.
What is the largest input size $n$ that this computer can complete within an hour?

Perhaps this question is pretty simple, but none of the online calculators that I know of (things like symbolab.com) have been able to solve it.

Any help would be greatly appreciated.

答:
$$n \log(n) = x\implies n=\frac{x}{W(x)}$$ where $W(x)$ is Lambert function.

If $x$ is large, you will have a very good approximation using

$$W(x) = L_1 - L_2 + \frac{L_2}{L_1} + \frac{L_2\left(-2 + L_2\right)}{2L_1^2} + \frac{L_2\left(6 - 9L_2 + 2L_2^2\right)}{6L_1^3} +$$ $$\frac{L_2\left(-12 + 36L_2 - 22L_2^2 + 3L_2^3\right)}{12L_1^4} + \cdots $$ where $L_1=\log(x)$ and $L_2=\log(L_1)$.

Title: Inequality containing logarithms


问:
Sylvester's sequence: is there an exact closed form?
I'm afraid this is one of those "amateur mathematician with no journal access" questions. Anyhow, Wikipedia (here) and OEIS give this closed form for the terms Sylvester's sequence:

$S_n = \lfloor E^{2^{n+1}} + \frac12\rfloor$ (where $E \approx 1.264$).

Is this the best we can do for a closed form, or is there an exact formula known? Is it known that there is no such formula?

答:
To quote that Wikipedia article (emphasis mine):

This would only be a practical algorithm if we had a better way of calculating $E$ to the requisite number of places than calculating $s_n$ and taking its repeated square root.


问:
Limit of a rational expression involving sinus
The task is to find
$$L= \lim_{n\rightarrow \infty} n^2 - \frac{n}{\sin(\frac{1}{n})}.$$
For sufficiently large $n$ we have $\sin(\frac{1}{n}) \approx \frac{1}{n}$ so that
$$ L \approx n^2 - \frac{n}{\frac{1}{n}} = n^2 - n^2 = 0, $$
so the limit $L$ is zero. Am I correct?

答:
Set $f(x):=\frac{1}{x^2}-\frac{1}{x\sin x}=\frac{\sin x-x}{x^2\sin x}$.

By applying L-Hospital's Rule three times, you can see $\lim_{x\rightarrow 0}f(x)=-\frac{1}{6}$.

Hence in your case, $\lim_{n \rightarrow \infty}f(\frac{1}{n})=-1/6$.


问:
Finding All Combinations
Question

Consider the following prices on balls:

\begin{alignat}{2}
  &\text{Red (R):}   &       15&\\
  &\text{Green (B):} &        6&\\
  &\text{Blue (G):}  &\qquad  8&
\end{alignat}

Question

Is there an "elegant" way of finding all the possible ways of bying balls for a total sum of $90$?

My own approach

By a brute force search, I have found the following possibilities, but can I somehow find this in a more "elegant" way?

\begin{alignat}{3}
  (0\,&\text{R},{} & 3\,&\text{G},{} &9\,&\text{B})\\
  (0\,&\text{R},   & 7\,&\text{G},   &6\,&\text{B})\\
  (0\,&\text{R},   &11\,&\text{G},   &3\,&\text{B})\\
  (0\,&\text{R},   &15\,&\text{G},   &0\,&\text{B})\\
  (2\,&\text{R},   &10\,&\text{G},   &0\,&\text{B})\\
  (2\,&\text{R},   & 2\,&\text{G},   &6\,&\text{B})\\
  (2\,&\text{R},   & 6\,&\text{G},   &4\,&\text{B})\\
  (4\,&\text{R},   & 1\,&\text{G},   &3\,&\text{B})\\
  (4\,&\text{R},   & 5\,&\text{G},   &0\,&\text{B})\\
  (6\,&\text{R},   & 0\,&\text{G},   &0\,&\text{B})
\end{alignat}

Update

$(2\,\text{R},{} 6\,\text{G},{} 4\,\text{B}) \to (2\,\text{R},{} 6\,\text{G},{} 3\,\text{B})$

答:
Your answer $(2,6,4)$ is incorrect.

Idk what method you have adopted, here is one systematic approach that you could using simple Diophantine equations.

Firstly reduce $15r + 6b + 8g = 90$ to $5r +2b + {\frac8 3}g = 30$.

So $g$ can only be from $0, 3,6,9$ with the corresponding reduced values of the totals being $30$, $22$, $14$, and $6$.

For each of these totals, we can form Diaphontine equations in two variables, eg for the total of $30$.

$5r + 2b = 30,\, b = 15 - \frac 5 2 r$

which immediately yields $(r,b,g)$ values of $(0,15,0), (2,10,0), (4,5,0)$ and $(6,0,0)$.

Similarly for the other three cases, you get the equations

$b = 11 - \frac5 2r,\quad b = 7 - \frac5 2r,\quad b = 3 - \frac5 2 r$

which you can enumerate similarly.

Title: Finding All Combinations


问:
Polar Coordinate function of a Straight Line
I was having some problem when trying to come out a polar coordinate function with straight line equation.

I know it is not good to post images here, but please bear with me as the question requires us to solve the equation from the straight line in the image.

What I have done is I tried to come out with an implicit function for the straight line L.

```<code>2x + 3y - 6 = 0
</code>```

But then I not sure how can I continue from there to come out with a polar coordinate function. Any ideas?

Thanks!

答:
If  $\alpha $ is CCW angle made to the normal then in Cartesian form ( not exactly as shown in your question, it should be normal to the line L.
$$ x \cos\alpha + y \sin  \alpha = p $$

and in polar coordinates after transformation

$$ r = p \sec \left( \theta - \tan^{-1} \frac{a}{b} \right ) $$


问:
Find whether the set $S$ in $\mathbb{R}^3$ is open or closed.
My set $S=\{(x, y, z) :-1 \le x \le 1,-1 < y <1, -1<z<1\}$.i am trying to prove that the set $S$ is open .

attempt  a:

So I pick up a point $(x_0,y_0,z_0)$ then I would try to show that it is an interior point. I pick up the space $(x_0-r,y_0-r,z_0-r) \times (x_0+r,y_0+r,z_0+r)$

.From here I should get that $x_0-r \ge -1$ and $x_0+r \le 1$ from here it is easy to conclude that $r \le x_0+1$ and $r \le 1-x_0$ from here we can conclude that $r \le 1$ and from the other equationsi am also able to prove that $r$ should be $0 <r < 1$.

attempt b:

I am trying to show that  closure($S$) $\ne S$.Then I am done. Now I want to show that the closure of $S$ is essentially $[-1,-1,-1] \times [1,1,1]$.We pick up a sequence $(-1,-1-\frac{1}{n},-1-\frac{1}{n})$then it is a sequence in $S$ converging to $(-1,-1,-1)$.Sothe point $(-1,-1,-1)$ must be in the set $S$. Since cl($S$) $\ne S$ we are done.

attempt c:

Now I wanted to find a continuous map $f$ from $S$ such that the map $f(S)$ is open. How do I find such an $f$?

答:
The set $S$ is not open. Any ball centered at $(1,0,0)$ contains points $(x,y,z)$ with $x>1$, and no such point belongs to $S$.

But the set $S$ is not closed either, since, say $(1,1,1)\in\overline S$, but $(1,1,1)\notin S$.


问:
Chinese Remainder Theorem with 0 mod n
I'm trying to get the least x from a system of congruences by applying the Chinese Remainder Theorem. Keep running into issues.

System of congruences:
$$
x \equiv 0 (_{mod} 7) \\
x \equiv 5 (_{mod} 6) \\
x \equiv 4 (_{mod} 5) \\
x \equiv 3 (_{mod} 4) \\
x \equiv 2 (_{mod} 3) \\
x \equiv 1 (_{mod} 2)
$$

I made the moduli relatively prime by removing 2nd and 6th congruence:

$1(_{mod} 2)$ is just a special case of $3 (_{mod} 4)$

and the 6th congruence because

$5 (_{mod} 6)$ splits into
$2 (_{mod} 3)$ and $1 (_{mod} 2)$, both of which are already represented in the system.

Product of moduli $m = 7 . 5 . 4 . 3 = 420$

Each respective $M_n$
$$
M_1 = 420/7 = 60 \\
M_2 = 420/5 = 84 \\
M_3 = 420/4 = 105 \\
M_3 = 420/3 = 140 \\
$$

Each respective modular inverse $y_n$
$$
y_1 = 0 \\
y_2 = 4 \\
y_3 = 3 \\
y_4 = 2 \\
$$

Trying to find solutions via CRT, I get

$x \equiv a_1  M_1 . y_1 + a_2  M_2 . y_2 + a_3  M_3 . y_3 + a_4  M_4 . y_4$

Plugging in the values:

$$
x \equiv 0 + 4 . 84 . 4 + 3 . 105 . 3 + 2 . 140 . 2 = 2849 \\
\equiv 329 (mod 42)
$$

And the "least" value being #329$ .

However, 329 doesn't satisfy the equation $329 (_{mod} 4) \equiv 3$.

What / where am I messing up?

答:
This is very similar to aid78's solution.
\begin{align}
x &\equiv 0 \pmod 7 \\
x &\equiv 5 \pmod 6 \\
x &\equiv 4 \pmod 5 \\
x &\equiv 3 \pmod 4 \\
x &\equiv 2 \pmod 3 \\
x &\equiv 1 \pmod 2
\end{align}

can be rewritten as

\begin{align}
x &\equiv \phantom{-}0 \pmod 7 \\
x &\equiv -1 \pmod 6 \\
x &\equiv -1 \pmod 5 \\
x &\equiv -1 \pmod 4 \\
x &\equiv -1 \pmod 3 \\
x &\equiv -1 \pmod 2
\end{align}

Since $\operatorname{lcm}(2,3,4,5,6)=60$, then the last five congruences are equivalent to $x \equiv -1 \pmod{60}$ So now you have to solve

\begin{align}
x &\equiv \phantom{-}0 \pmod 7 \\
x &\equiv -1 \pmod{60}
\end{align}

$x \equiv -1 \pmod{60}$ is equivalent to saying $x = 60n -1$ for some integer $n$.

This is one way to finish the problem.

\begin{align}
   x &\equiv 0 \pmod{7} \\
   60n - 1 &\equiv 0 \pmod{7} &\text{(Note $60\equiv 4 \pmod 7$)} \\
   4n &\equiv 1 \pmod 7 \\
   8n &\equiv 2 \pmod 7 \\
   n &\equiv 2 \pmod 7 \\
   n &= 7m + 2 \\
   x &= 60n - 1 \\
   x &= 60(7m + 2) - 1 \\
   x &= 420m + 119 \\
   x &\equiv 119 \pmod{420}
\end{align}

Title: Chinese Remainder Theorem with 0 mod n


问:
Show that if either $X$ or $Y$ is disconnected then $X \times Y$ is disconnected as well
Here is the statement: Let $X$ and $Y$ be two non-empty metric spaces. Show that if either $X$ or $Y$ is disconnected, then the Cartesian product $X\times Y$ is also disconnected

My idea(s): The placement of this exercise in my book is around a section where the connection between connectedness and continuous functions is established, leading me to believe that one could also prove the claim by constructing some function. However, I think that it is also possible to prove it by using the definition of disconnectednessn and choosing an appropriate metric for a product space.

Namely, let $X$ be disconnected and $Z = X \times Y$. Then there exists two non-empty open subsets $S_1, S_2 \subset X$ such that $S_1 \cup S_2 = X$ and $S_1 \cap S_2 = \varnothing$. Since for any $z \in Z: z = (x, y), x \in X = S_1 \cup S_2, y \in Y$ it follows that $\left(S_1 \times Y\right) \cup \left(S_2 \times Y\right) = Z$. Also $\left(S_1 \times Y\right) \cap \left(S_2 \times Y\right) = \varnothing$, as $S_1$ and $S_2$ are disjoint. $\left(S_1 \times Y\right)$ and $\left(S_2 \times Y\right)$.

Now, if we can show that $\left(S_1 \times Y\right)$ and $ \left(S_2 \times Y\right)$ are both open w.r.t. the metric in $Z$, we are done. Thus for $(a, b), (c, d) \in Z$, let $d_{Z}((a, b), (c, d)) = d_{X}(a, c)$. Since $S_1$ and $S_2$ are open w.r.t. $d_{X}$, and the metric does not depend on $Y$, it follows that $\left(S_1 \times Y\right)$ and $\left(S_2 \times Y\right)$ are also open w.r.t. $d_{X}$. Hence there exists two non-empty, disjoint and open subsets of $Z$ whose union is $Z$. Thus $Z$ is disjoint.

Edit: It is a shame that I cannot accept multiple answers, since so far all answers given here have been brilliant!

答:
The problem with this idea is that $d_Z$ is not a metric unless $Y$ consists of a single point: if $y_0$ and $y_1$ are distinct points of $Y$, and $x\in X$, then

$$d_Z(\langle x,y_0\rangle,\langle x,y_1\rangle)=0$$

even though $\langle x,y_0\rangle\ne\langle x,y_1\rangle$.

However, it is true that $S_1\times Y$ and $S_2\times Y$ are open in $Z$. This is immediate from the definition of the product topology: both are actually basic open sets in the product.

If you’re required to argue in terms of a metric on $X\times Y$, you have to work a little harder, and the argument will depend somewhat on the specific product metric that you use. The most common product metrics define $d_Z(\langle x,y_0\rangle,\langle x,y_1\rangle)$ to be

$$\sqrt{d_X(x_0,x_1)^2+d_Y(y_0,y_1)^2}\,,$$

$$d_X(x_0,x_1)+d_Y(y_0,y_1)\,,$$

or

$$\max\{d_X(x_0,x_1),d_Y(y_0,y_1)\}\,.$$

All of these generate the same topology, so you can use any of them. Let $p=\langle x,y\rangle\in S_1\times Y$. There is an $\epsilon>0$ such that $B_{d_X}(x,\epsilon)\subseteq S_1$, and for each of these three product metrics it’s straightforward to show that $B_{d_Z}(p,\epsilon)\subseteq S_1\times Y$.


答:
Your proof is almost correct, except for the justification of openness of $(S_1\times Y)$-- the space $X\times Y$ comes equipped with a very specific topology (the product topology) and your task is to show $(S_1\times Y)$ is open in that particular topology. (Otherwise, you could just say "let $d_Z$ be the discrete metric, then every set is open hence we are done." There is an obvious problem with the argument.)

The topology on $X\times Y$ can be defined in two ways:

Either by the definition of product topology, where the open sets in $X\times Y$ are generated by (i.e. unions of) the sets in $\{U\times V\;|\; U\overset{\text{open}}\subset X, V\overset{\text{open}}\subset Y\}$. Using this definition, it is immediate that $S_1\times Y$ is open.

Or by using one of the product metrics:

$d_{1}((a,b),(c,d))=d_X(a,c) + d_Y(b,d)$

$d_{2}((a,b),(c,d))=\sqrt{(d_X(a,c))^2 + (d_Y(b,d))^2}$

$d_{\infty}((a,b),(c,d))=\max\{d_X(a,c),d_Y(b,d)\}$

It is a non-trivial fact that all the metrics above induce exactly the product topology on $X\times Y$ so it does not matter which one you choose to prove the openness of $S_1\times Y$-- using $d_\infty$ should be the easiest.

Title: Show that if either $X$ or $Y$ is disconnected then $X \times Y$ is disconnected as well

Body:
Here is the statement: Let $X$ and $Y$ be two non-empty metric spaces. Show that if either $X$ or $Y$ is disconnected, then the Cartesian product $X\times Y$ is also disconnected

My idea(s): The placement of this exercise in my book is around a section where the connection between connectedness and continuous functions is established, leading me to believe that one could also prove the claim by constructing some function. However, I think that it is also possible to prove it by using the definition of disconnectednessn and choosing an appropriate metric for a product space.

Namely, let $X$ be disconnected and $Z = X \times Y$. Then there exists two non-empty open subsets $S_1, S_2 \subset X$ such that $S_1 \cup S_2 = X$ and $S_1 \cap S_2 = \varnothing$. Since for any $z \in Z: z = (x, y), x \in X = S_1 \cup S_2, y \in Y$ it follows that $\left(S_1 \times Y\right) \cup \left(S_2 \times Y\right) = Z$. Also $\left(S_1 \times Y\right) \cap \left(S_2 \times Y\right) = \varnothing$, as $S_1$ and $S_2$ are disjoint. $\left(S_1 \times Y\right)$ and $\left(S_2 \times Y\right)$.

Now, if we can show that $\left(S_1 \times Y\right)$ and $ \left(S_2 \times Y\right)$ are both open w.r.t. the metric in $Z$, we are done. Thus for $(a, b), (c, d) \in Z$, let $d_{Z}((a, b), (c, d)) = d_{X}(a, c)$. Since $S_1$ and $S_2$ are open w.r.t. $d_{X}$, and the metric does not depend on $Y$, it follows that $\left(S_1 \times Y\right)$ and $\left(S_2 \times Y\right)$ are also open w.r.t. $d_{X}$. Hence there exists two non-empty, disjoint and open subsets of $Z$ whose union is $Z$. Thus $Z$ is disjoint.

Edit: It is a shame that I cannot accept multiple answers, since so far all answers given here have been brilliant!


问:
Proof without using truth tables
I need some further explanation, as I feel I am just missing some critical piece of information. I have shown my work thus far (I excluded the truth tables as they take up so much space but included the pertinent values). Can someone shed some light on what I'm overlooking here?

Which of the following formulas are equivalent? Without showing truth tables, explain why which of the following five statements are equivalent. Hint – analyze the possible values of P, Q, and R that make each of the statements false.

a. P → (Q → R).

b. Q → (P → R).

c. (P → Q) ∧ (P → R).

d. (P ∧ Q) → R.

e. P → (Q ∧ R).

My work:

So using truth tables I was able to determine that a,b,d are all equivalent to each other AND c and e are equivalent to each other.
A, B, D are all false ONLY when p and q are true and r is false, as that is the  only time that the conditional statement becomes true implies false.
C and E are false for several instances, when p=T q=T r=F; p=T q=F r=T; p=T q=F r=F.

What has me stumped is how to write a proof from that information. I have tried using the logical equivalence identities, but am unable to  come up with the correct answer, so I feel like that is not the proper way to do it. So I'm wondering, is there somethin I'm missing? Is this supposed to be a proof by implication, or and indirect proof, and if so, how am I supposed to start that? I understand that these are conditional statements, but again, I feel as if I am missing something.

答:
(a), (b), and (d) can be proven equivalent by using (five times in total) the fact that $A\implies B$ is equivalent to $(\lnot A)\lor B$, together with de Morgan's law for negating an "and" statement and the associativity and commutativity of "or" statements.

Similarly, (c) and (e) can be proven equivalent by using that $A\implies B$ is equivalent to $(\lnot A)\lor B$ together with the distributive laws for "and"s and "or"s.

(I agree that the given hint is basically the same as using truth tables.)

Title: Proof without using truth tables


答: Here is a proof that uses continuous functions: since $X$ is disconnected, there is a continuous surjection $f:X\rightarrow \{0,1\}$ where the latter is equipped with the discrete topology. Then $g:X\times Y \rightarrow \{0,1\}$ given by $g(x,y)=f(x)$ is a continuous surjection as well (it is equal to $f$ composed with the projection map onto the first coordinate), proving disconnectedness of $X\times Y$.


问:
Recommendations for Numerical Analysis texts?
I'm in a numerical analysis course right now and it's pretty rigorous but I'm enjoying it a lot. I took a lower level course before that was more oriented towards implementation of numerical methods, so it's not my first time with the material.

Anyway, my professor wrote his own course notes and they're fantastic but I wanted some reference material since I'm currently outlining my own notes. Just wondering what the big ones out there are

edit: I would say I'm at an advanced undergraduate level

答:
I feel worth recommending the book [1] of Solomon Mikhlin. The first thing to say is that it is by no means an introductory level text: however, it offers a very accurate analysis of the error causes in numerical processes, along with the description of the "methods" which should be adopted to minimize/mitigate their effect. Definitely a worth second reading in numerical analysis.

[1] Mikhlin, Solomon G. Error analysis in numerical processes, Translated and revised from the German. Translated by Reinhard Lehmann, (English) Pure and Applied Mathematics, 1237. A Wiley-Interscience Series of Texts, Monographs and Tracts. Chichester: John Wiley & Sons Ltd., pp. 283 (1991), ISBN 0-471-92133-5, MR1129889, Zbl 0786.65038.


问:
What is the probability that water will arrive to its destination?
I have the following picture:

A, B, C, D are all INDEPENDENT stations pumping water so that water
arrives at L, the probability of each one of them working is written
under them. Water arrives if we can get an entire "line" of stations
working. What is the probability that water arrives to L?

Now in the solution given the author does the following:

P(System works)=$P(A\cap B \cap (C \cup D))$

then proceeds by saying $P(C\cup D)=P(C)+P(D)-P(C\cap D)$

Now my question is: how is that correct? namely, this doesn't take into account all stations working (we remove $P(C\cap D)$) which is one of the possible cases the system works.

For the system to work we need one of these 3 cases to happen:

1.$A\cap B\cap C \cap D'$

2.$A\cap B\cap C' \cap D$

3.$A\cap B\cap C \cap D$

It looks to me like the author didn't take into account the third option, any help would be appreciated.

答:
A and B are connected in Series, while C and D are connected in Parallel

Thus the system will work only if

A and B both work, $p_1=0.9\times 0.9$

AND

At leas one of C and D works, $p_2=1-0.2\times 0.2$

Concluding

$$p=p_1\cdot p_2=0.7776$$

This is exactly the same solution of your textbook


答:
$$\sin \left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{6 n^3}+\frac{1}{120
   n^5}+O\left(\frac{1}{n^7}\right)$$ Now long division
$$\frac 1{\sin \left(\frac{1}{n}\right)}=n+\frac{1}{6 n}+\frac{7}{360 n^3}+O\left(\frac{1}{n^5}\right)$$
$$n^2-\frac n{\sin \left(\frac{1}{n}\right)}=-\frac{1}{6}-\frac{7}{360 n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached.

Title: Limit of a rational expression involving sinus

Body:
The task is to find
$$L= \lim_{n\rightarrow \infty} n^2 - \frac{n}{\sin(\frac{1}{n})}.$$
For sufficiently large $n$ we have $\sin(\frac{1}{n}) \approx \frac{1}{n}$ so that
$$ L \approx n^2 - \frac{n}{\frac{1}{n}} = n^2 - n^2 = 0, $$
so the limit $L$ is zero. Am I correct?


问:
Proof of the derivative of an elementary function is also elementary
I heard of the result that the derivative of an elementary function is also elementary long ago. Now I want to prove it rigorously. I found this answer(I didn't comment because it was an old post):https://math.stackexchange.com/q/2195559. It actually proved that: (1). the derivative of $\exp,\log,\operatorname{id}$ and constant is elementary. (2). If we assume that f and g have elementary derivatives, then $f+g,f-g,fg,f/g,f\circ g$ also have elementary derivatives. So far so good. But what confuses me is how does this 2 facts imply the original proposition(the derivative of any elementary function is also elementary)? I mean isn't the fact (2) seems too weak?

I think the keypoint lies in the construction of the elementary functions. I use the definition of the elementary functions to be:

$\exp,\log,\operatorname{id}$ and constant is elementary.

the sum, difference, product, quotient, composition of 2 elementary functions is elementary.

Need help!

答:
The closure under differentiation can be proven by induction.

The result is clear for « base functions » ($\ln x, e^x)...$

If a map is a sum, the product, the composition... of two simple functions, then it follows that its derivative is a simple function based on the induction hypothesis and the differentiation rules of the sum, the product, the composition... of two functions.

As any simple function can be obtained by induction using previous rules according to the very definition of a simple function, we’re done.


问:
Show that $θ=2sinθ$
This is probably an easy question for most people but for me I am stuck I don't get how this works. And I have no teacher as I am self studying . Any help is much appreciated.

It says this is the working out put I don't get it.

答:
Hint

The area of a triangle is given by, $$\frac{1}{2}ab\sin(\theta)$$
where a and b are the sides and $\theta$ is the included angle.

The area of a sector is given by, $$\frac{1}{2}\;\theta \ r^2$$
where r is the radius.

Title: Show that $θ=2sinθ$


问:
A Combinatorics Question Choosing Paths on a Grid

This is for a Year 11 HSC Maths Course.

I'm currently stuck at the first part. At each point there should be 2 ways of select where you go (except the bottom and right edges). Since there are $5*7=35$ points on the gird, and there are 11 points on the bottom and right edges, there are $2^{35-11}=2^{24}$ possible routes without restrictions.

But this approach seems to be flawed as it does not match the answers. Could someone suggest a way of thinking about this?

答:
To get from $A$ to $B$ you must take a total of $4$ steps to the right and $6$ steps down, and you can take those $10$ steps in any order. Thus, there is one path for each possible ordering of the $10$ steps. The order is completely determined if you know which $4$ of the $10$ are steps to the right, and they can be any $4$, so there are $\binom{10}4$ possible paths.

If you must pass through $C$, you can combine any path from $A$ to $C$ with any path from $C$ to $B$.

If you cannot move along the top line, your first move must be down, and then you have another problem like the first one.

And if you cannot move along the second row from the top, your first two steps down cannot be split by a step to the right. You can think of this as a combination of several problems like the second one, depending on how far you go along the top line before going down, or you can think of it as a problem just like the first one, but on a slightly smaller grid.


问:
Intuitively understanding why the volume of a cylinder of height $h$ and radius $R$ is $\pi R^2h$
It's well known that a cylinder with height $h$ and the base radius $R$ will have a volume of

$$V = \pi R^2h$$

I'm trying to derive an intuitive understanding of it. I have two ways, one seems to be "working" (as in - it yields the correct result) and another one is failing, see below.

Method 1

If we agree that the area of a circle is known as $\pi R^2$ then we can think of a volume as a sum of all "areas of circles along the heights". So if we imagine that there is such an "object" that has a height of "one point" then its volume is equal to the area of the circle it has as a base. We then see that there will be exactly $h$ points in the height of the cylinder, so there will be exactly $h$ such "objects". Hence we can just multiply a "volume" of one "object" by $h$ which will yield the correct cylinder volume. I believe this method is just a "naive" version of a more rigorous calculus approach with defining the limit for $dx$ where $x$ will be along the height of the cylinder

Method 2

Now let's look at the cylinder as a rectangle rotated along one of its sides. If we have $h$ as the cylinder height and $R$ as a radius, then it's an $h \times R$ rectangle rotated by $2\pi$ around the side with length $h$. So the volume is the "sum" of all the "objects" with the height of "one point" and the base a rectangle. There are exactly $2\pi R$ points along the circumference of the circle that the rectangle will follow. Therefore we will have $2\pi R$ of these "objects". So we just need to multiple the "volume" of one such "object" by the number of these "objects" to get the cylinder volume. The "volume" of this "object" will be just the area of a rectangle, since it has a height of "one point". To conclude, we have $2\pi$ of these "objects", thus the volume must be $V = 2\pi R \times V_{\text{object}} = 2\pi R \times h R = 2\pi R^2h$.

But method 2 yields incorrect result, it's exactly $2$ times more than it should be. Why is that? I.e. why is that the point-based construction worked in the first case but failed in the second case? And how to make it work properly? What about rotation in general (i.e. not only a cylinder)?

答:
Method 2 fails because rectangular slices overlap each other, especially near the axis of the cylinder. This results in the volume being over-counted.

Instead we need triangular slices (which have half the volume of the rectangular slices).

This is best understood with a diagram:

Source code (Mathematica):

```<code>(* Cylinder dimensions *)
cylinderHeight = 2;
cylinderRadius = 1;
(* Slice dimensions *)
numSlices = 16;
sliceAngle = 2 Pi / numSlices;
sliceHalfThickness = cylinderRadius * sliceAngle / 2;
(* Slices *)
rectangularSlice =
  Cuboid[
    {0, -sliceHalfThickness, 0},
    {cylinderRadius, +sliceHalfThickness, cylinderHeight}
  ];
triangularSlice =
  Prism @ {
    {0, 0, 0},
    {cylinderRadius, -sliceHalfThickness, 0},
    {cylinderRadius, +sliceHalfThickness, 0},
    {0, 0, cylinderHeight},
    {cylinderRadius, -sliceHalfThickness, cylinderHeight},
    {cylinderRadius, +sliceHalfThickness, cylinderHeight}
  };
(* Make diagrams *)
showSlices[slice_] :=
  Graphics3D[
    Table[
      {
        FaceForm[Black],
        EdgeForm @ Directive[Thick, Hue[n / numSlices]],
        Rotate[slice, n * sliceAngle, {0, 0, 1}]
      }
      , {n, numSlices}
    ]
    , Boxed -> False
    , ImageSize -> 240
    , Lighting -> {"Ambient", White}
  ];
Row[
  showSlices /@ {rectangularSlice, triangularSlice}
]
</code>```

Body:
Method 2 fails because rectangular slices overlap each other, especially near the axis of the cylinder. This results in the volume being over-counted.

Instead we need triangular slices (which have half the volume of the rectangular slices).

This is best understood with a diagram:

Source code (Mathematica):

(* Cylinder dimensions *)
cylinderHeight = 2;
cylinderRadius = 1;
(* Slice dimensions *)
numSlices = 16;
sliceAngle = 2 Pi / numSlices;
sliceHalfThickness = cylinderRadius * sliceAngle / 2;
(* Slices *)
rectangularSlice =
  Cuboid[
    {0, -sliceHalfThickness, 0},
    {cylinderRadius, +sliceHalfThickness, cylinderHeight}
  ];
triangularSlice =
  Prism @ {
    {0, 0, 0},
    {cylinderRadius, -sliceHalfThickness, 0},
    {cylinderRadius, +sliceHalfThickness, 0},
    {0, 0, cylinderHeight},
    {cylinderRadius, -sliceHalfThickness, cylinderHeight},
    {cylinderRadius, +sliceHalfThickness, cylinderHeight}
  };
(* Make diagrams *)
showSlices[slice_] :=
  Graphics3D[
    Table[
      {
        FaceForm[Black],
        EdgeForm @ Directive[Thick, Hue[n / numSlices]],
        Rotate[slice, n * sliceAngle, {0, 0, 1}]
      }
      , {n, numSlices}
    ]
    , Boxed -> False
    , ImageSize -> 240
    , Lighting -> {"Ambient", White}
  ];
Row[
  showSlices /@ {rectangularSlice, triangularSlice}
]


答:
Let me try to organize what you have a little more neatly, maybe this will clear things up.

First, your definition of elementary function:

$
\bullet \exp, \log, \text{id}, \text{constant}\\
\bullet \text{The sum of two elementary functions, i.e. }f+g \\
\bullet \text{The difference of two elementary functions, i.e. }f-g \\
\bullet \text{The product of two elementary functions, i.e. }fg \\
\bullet \text{The quotient of two elementary functions, i.e. }\frac{f}{g} \\
\bullet \text{The composition of two elementary functions, i.e. }f \circ g
$

Now, the proof you posted proves first that $\exp, \log$, id and constant functions have elementary derivatives, so the claim holds for the first bullet point in the list above. Then it goes on to prove that it holds for sums (2nd bullet point), differences (3rd bullet), products (4th bullet), quotients (5th bullet) and compositions (6th bullet).

So it really has shown that all the functions that fall under the definition of elementary function have elementary derivatives! And as mathcounterexamples.net pointed out in his answer, this can be expanded to the sum, product, etc. of more than two functions by induction.

Title: Proof of the derivative of an elementary function is also elementary

Body:
I heard of the result that the derivative of an elementary function is also elementary long ago. Now I want to prove it rigorously. I found this answer(I didn't comment because it was an old post):https://math.stackexchange.com/q/2195559. It actually proved that: (1). the derivative of $\exp,\log,\operatorname{id}$ and constant is elementary. (2). If we assume that f and g have elementary derivatives, then $f+g,f-g,fg,f/g,f\circ g$ also have elementary derivatives. So far so good. But what confuses me is how does this 2 facts imply the original proposition(the derivative of any elementary function is also elementary)? I mean isn't the fact (2) seems too weak?

I think the keypoint lies in the construction of the elementary functions. I use the definition of the elementary functions to be:

$\exp,\log,\operatorname{id}$ and constant is elementary.

the sum, difference, product, quotient, composition of 2 elementary functions is elementary.

Need help!


问:
Integral with differential is purely imaginary
Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $\gamma$, then $\int_\gamma\overline{f(z)}f'(z)dz$ is purely imaginary.

I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $\int_a^b\overline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?

答:
Let us write $f = u + iv$. Then from analyticity of $f$ we have:
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{ and }\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}  $$

Now,
$$\int_{\gamma} \bar{f}f'dz = (u-iv)(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}) (dx + idy)\\
=\int_{\gamma}[(u\frac{\partial u}{\partial x} + v\frac{\partial v}{\partial x}) + i(u\frac{\partial v}{\partial x} - v\frac{\partial u}{\partial x})][dx + idy]$$
After multiplication we find it's real part to be:
$$\int_{\gamma} (u\frac{\partial u}{\partial x}dx -u\frac{\partial v}{\partial x}dy) + (v\frac{\partial v}{\partial x}dx + v\frac{\partial u}{\partial x}dy)$$

After replacing $ - \frac{\partial v}{\partial x}$ to $\frac{\partial u}{\partial y}$ and $\frac{\partial u}{\partial x}$ to $\frac{\partial v}{\partial y}$ we get:
$$\int_{\gamma} (u\frac{\partial u}{\partial x}dx + u\frac{\partial u}{\partial y}dy) + (v\frac{\partial v}{\partial x}dx + v\frac{\partial v}{\partial y}dy)$$

The integrand is easily seen to be an exact differential of $\frac{1}{2} (u^2 + v^2)$ and hence the integral over any closed curve $\gamma$ is $0$.


问:
Prove that : $ \int_{0}^{2\ln{\varphi}}{\theta\ln{\left(2\sinh{\frac{\theta}{2}}\right)}\,\mathrm{d}\theta}=-\frac{1}{5}\zeta\left(3\right) $
Denoting $ \varphi=\frac{1+\sqrt{5}}{2}=\mathrm{Golden\ Ratio} $. How would you prove that : $$  \int_{0}^{2\ln{\varphi}}{\theta\ln{\left(2\sinh{\frac{\theta}{2}}\right)}\,\mathrm{d}\theta}=-\frac{1}{5}\zeta\left(3\right) $$

I came accross this strange identity while reading the following article : VALUES OF THE RIEMANN ZETA FUNCTION AND
INTEGRALS INVOLVING $\log{\left(2\sinh{\frac{\theta}{2}}\right)}$ AND $\log{\left(2\sin{\frac{\theta}{2}}\right)}$. (No complete proof was given in the article). I'm still trying to prove it. For now I did try to set $ x=2\ln{\left(\sinh{\frac{\theta}{2}}\right)} $ to get rid of the golden ratio, but it doesn't seem to get any better. I know I'll have to expand the integrand to get to $ \zeta $ using, for example, the fact that : $ \ln{\left(\sinh{x}\right)}=x-\ln{2}-\sum\limits_{n=1}^{+\infty}{\frac{\mathrm{e}^{-2nx}}{n}} $, but I'm sure I cannot do it now, because of that golden ratio in the upper bound. Any ideas, hints will be appreciated.

答:
This may not be the exact answer you're looking for since it requires two polylogarithmic identities that must be calculated beforehand. And given how most of the terms in the penultimate step seem to cancel each other out, it makes me believe that there is a better solution. Nevertheless, I will present this solution in hopes of possibly inspiring someone else to come up with a more concise and efficient answer!

Calling the integral $\mathfrak{I}$, enforce the transformation $x\mapsto\tfrac 12x$ to get

$$\mathfrak{I}=4\int\limits_0^{\log\varphi}\mathrm dx\, x\log(2\sinh x)$$

Now, recall that $\sinh x$ can be rewritten in terms of $e$ as such

$$\log(2\sinh x)=\log\left(e^{x}-e^{-x}\right)=x+\log\left(1-e^{-2x}\right)$$

After splitting up the integral and evaluating each component, one arrives at

$$\mathfrak{I}=\frac 43\log^3\varphi+4\int\limits_0^{\log\varphi}\mathrm dx\, x\log\left(1-e^{-2x}\right)$$

For the natural logarithm, use its Taylor Expansion to rewrite the integral with an infinite sum.

$$\mathfrak{I}=\frac 43\log^3\varphi-4\sum\limits_{n\geq1}\frac 1n\int\limits_0^{\log\varphi}\mathrm dx\, xe^{-2nx}$$

Integrate the resulting expression using integration by parts on $u=x$ and $\mathrm dv=e^{-2nx}\mathrm dx$ to see that

$$\int\limits_0^{\log\varphi}\mathrm dx\, xe^{-2nx}=-\frac 1{2n}e^{-2n\log\varphi}\log\varphi-\frac 1{4n^2}e^{-2n\log\varphi}+\frac 1{4n^2}$$

Hence

$$\begin{align*}\mathfrak{I} & =\frac 43\log^3\varphi-4\sum\limits_{n\geq1}\frac 1{n}\left(\frac 1{4n^2}-\frac 1{4n^2}e^{-2n\log\varphi}-\frac 1{2n}e^{-2n\log\varphi}\log\varphi\right)\\ & =\frac 43\log^3\varphi-\zeta(3)+\sum\limits_{n\geq1}\frac 1{n^3}\varphi^{-2n}+2\sum\limits_{n\geq1}\frac 1{n^2}\varphi^{-2n}\log\varphi\\ & =\frac 43\log^3\varphi-\zeta(3)+\operatorname{Li}_3(\varphi^{-2})+2\operatorname{Li}_2(\varphi^{-2})\log\varphi\end{align*}$$

The complex polylogarithmic results can be simplified with the following two identities

$$\begin{align*}\operatorname{Li}_2\left(\varphi^{-2}\right)& =\frac {\pi^2}{15}-\log^2\varphi\\\operatorname{Li}_3\left(\varphi^{-2}\right) & =\frac 45\zeta(3)-\frac {2\pi^2}{15}\log\varphi+\frac 23\log^3\varphi\end{align*}$$

To give

$$\int\limits_0^{2\log\varphi}\mathrm dx\, x\log\left(2\sinh\frac x2\right)\color{blue}{=-\frac 15\zeta(3)}$$


问:
Solving $y'' + 2xy' - y = -1$ where $y = y(x)$
I want to solve $y'' + 2xy' - y = -1$ where $y = y(x)$. Since this is a second-order non-homogenous ode I must obtain the solution to the homogenous equation first: $$y'' + 2xy' - y = 0$$ However, the non-constant term has me stuck on how to actually approach this.

Can anyone please suggest a method for me to try?

答:
We need to solve the second order linear ordinary differential equation: $$\color{red}{y''+2xy'-y=-1, \quad y=y(x)}.$$

Since this is a second-order non-homogenous ode I must obtain the solution to the homogenous equation first:
$$y''+2xy'−y=0.$$

Yes, that's correct.

Now, for solve $$\color{red}{y''+2xy'-y=0}$$ we can use the Yuval's hint, for that note that we don't have singular points, so we can find solutions in power series centered in $0$, convergent to $|x|<\infty$.

Let, $$\color{blue}{(\operatorname{V.C}):\begin{cases} \displaystyle y=\sum_{n=0}^{+\infty} a_{n}x^{n}\\ \displaystyle y'=\sum_{n=1}^{+\infty} a_{n}\cdot n x^{n-1}\\
\displaystyle y''=\sum_{n=2}^{+\infty}a_{n}\cdot n \cdot (n-1) x^{n-1} .\end{cases}}$$
Then, we can re-write $$y''+2xy'-y=0, $$as $$\left( \sum_{n=2}^{+\infty}a_{n}\cdot n \cdot (n-1) x^{n-1} \right)+2x\left( \sum_{n=1}^{+\infty} a_{n}\cdot n x^{n-1} \right)-\left( \sum_{n=1}^{+\infty} a_{n}\cdot n x^{n-1} \right)=0.$$

Now, you must find the recurrence relation of $c_{k}$ such that you can choose from a certain subset of the set of nonzero coefficients. Then, you can find $y(x)$.

Finally, you need to solve $$y''+2xy'-y=-1$$ for that you need find the particular solution, for that you can use many methods: indeterminate coefficients, the null method or the parameter variation method.

On the other hand, it seems to me that the solution to the problem cannot be found in a closed form.


问:
Integral over a contour in complex plane
I was trying to calculate $\int_\gamma \frac{z^2}{z-1}dz$ where $\gamma$ is a circle of radius 2, center at origin, oriented counterclockwise.
I think it has a hole at z=1? I do not know how it affects the integral, what I did was using the formula $\int_\gamma f(z)dz=\int^b_af(\gamma(t))\gamma'(t)dt,$ so I got $\int_0^{2\pi}\frac{(2e^{it})^2}{2e^{it}-1}2ie^{it} dt,$ but since its starting and end points are the same, shouldn't it be zero? However the answer is not zero.

The next question is over the same contour, but a different integral $\int_\gamma \frac{e^{2z}}{(z-1)^3}dz$, and its answer is zero, I am so confused now, please help and thank you in advance

答:
You can use Cauchy's Integral Formula:
\begin{equation}
f(z) = \frac{1}{2 \pi i} \int_{C_{(z_{0},r)}}\frac{f(w)}{w-z}dw
\end{equation}

To use this formula $f$ must be holomorphic at an open set $\ U$ containing $\overline{D}(z_{0}, r)$ and $z \in D(z_{0}, r)$

In your case, $f(w) = w^{2}, \ z_{0} = 0, \ r = 2, \ z = 1$ so you can use the formula.

Onto your second question, the same formula can be generalized to
\begin{equation}
\frac{f^{n)}(z_{0})}{n!} = \frac{1}{2 \pi i} \int_{C(z_{0},r)}\frac{f(w)}{(w-z)^{n+1}}dw
\end{equation}
where $f^{n)}(z_{0})$ refers to the $n^{th}$ derivative.

With this formula you can solve easily both questions.


问:
Compute determinant
I have to compute the following determinant of an $(n+1)\times(n+1)$ matrix

$$\Delta_{n+1} :=\begin{vmatrix}
1 & 1 &  \dots & 1 & 1 & 1 \\
1 & 1 &  \dots & 1 & x_1 & 1 \\
1 & 1 &  \dots & x_2 & 1 & 1 \\
\vdots\ & \vdots\ & \ddots & \vdots & \vdots & \vdots\\
x_n & 1 &  \dots & 1 & 1 & 1 \\
\end{vmatrix}$$

If I subtract the first row from all the other rows, I easily get that

$$\Delta_{n+1} = (-1)^{n+2}(x_1-1)(x_2-1)\cdots(x_n-1)$$

by using Laplace expansion on the last column. However, the answer given in my book is:

$$\Delta_{n+1} = (-1)^{\frac{n(n+1)}{2}}(x_1-1)(x_2-1)\cdots(x_n-1)$$

I wonder which one is correct. I tried computing

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

to prove that this number is always even but that doesn't seem true. So, what is the real answer here?

答: We have\begin{align}\begin{vmatrix}1&1&1\\1&x_1&1\\x_2&1&1\end{vmatrix}&=\begin{vmatrix}1&1&1\\0&x_1-1&0\\x_2-1&0&0\end{vmatrix}\\&=\begin{vmatrix}0&x_1-1\\x_2-1&0\end{vmatrix}\\&=-(x_1-1)(x_2-1).\end{align}As you can see, the sign is negative here. So, in this case, your book is correct, but you are not.

Title: Compute determinant


问:
Rubik's Cube: Number of Permutations of the Corner Position Orientations
The Rubik's Cube has 8 corners, and each corner has 3 stickers.  A corner can be in 1 of 3 orientations, i.e. any of the three stickers can point up, giving $3^8$ possible permutations of the corner orientations.

However, 1/3 of those permutations are unreachable (more detail here).  It's not possible, for example, to disorient one corner.  So, there are $3^7$ reachable permutations of the corner orientations.

As an aside, there are 8! arrangements of the corners, so there are $8!*3^7$ possible permutations of the positions and orientations of the corners.

What I can't figure out is how many ways the pieces in the corner positions can be oriented.  Instead of looking at the ways the corner pieces--the red-blue-yellow, red-green-yellow, red-green-white, etc.--can be oriented, I want to know how many ways the pieces in the up-left-back, up-right-back, up-right-front, etc. positions can be oriented.

As an example, I'll arbitrarily define a corner cubie as "oriented" if it has a red or orange sticker facing up or down, rotated once if it has a yellow or white sticker pointing up or down, or rotated twice if it has an blue or green sticker pointing up or down.  Thus, rotating the UP or DOWN face does not change the orientation of any corner cubies, whereas twisting any of the other four sides changes the orientation of four corner cubies.

I'll also arbitrarily index the corners cubies and positions as follows:

```<code>0   1   2   3   4   5   6   7
ULB URB URF ULF DLF DLB DRB DRF
RBY RGY RGW RBW OBW OBY OGY OGW
</code>```

(In the solved state, the up-left-back position is occupied by the red-blue-yellow cubie, ..., and the down-right-front position is occupied by the orange-green-white cubie.)

So, twisting the FRONT face 90 degrees changes the orientation of four corners, 2, 3, 4, and 7.  Subsequently twisting the TOP face 90 degrees, note that the permutation of the orientations of the corner cubies remains the same; however, the orientations of the positions change.  E.g. the RBY (0) cubie is still oriented, but the cubie in position ULB (0) is now disoriented.

By position, then, how many ways can the orientations be permuted?  For what it's worth, I've empirically found the number to be 6,513.  I would like to verify mathematically that that number is correct.

Edit: Here's a snippet of code that illustrates what I'm looking for.

```<code>    // Permutation of the orientations of the pieces occupying the
    // 8 corners.
    array<uint8_t, 8> cornerOrientations =
    {
      iCube.getCornerOrientation(CORNER_POSITION::ULB),
      iCube.getCornerOrientation(CORNER_POSITION::URB),
      iCube.getCornerOrientation(CORNER_POSITION::URF),
      iCube.getCornerOrientation(CORNER_POSITION::ULF),
      iCube.getCornerOrientation(CORNER_POSITION::DLF),
      iCube.getCornerOrientation(CORNER_POSITION::DLB),
      iCube.getCornerOrientation(CORNER_POSITION::DRB),
      iCube.getCornerOrientation(CORNER_POSITION::DRF)
    };

    // Treated as a base-3 number, converted to base-10.
    uint32_t permInd =
      cornerOrientations[0] * 2187 +
      cornerOrientations[1] * 729 +
      cornerOrientations[2] * 243 +
      cornerOrientations[3] * 81 +
      cornerOrientations[4] * 27 +
      cornerOrientations[5] * 9 +
      cornerOrientations[6] * 3 +
      cornerOrientations[7];

    return permInd;
</code>```

Versus:

```<code>    // Permutation of the orientations of the 8 corner pieces.
    array<uint8_t, 8> cornerOrientations =
    {
      iCube.getCornerOrientation(CORNER_PIECE::RBY),
      iCube.getCornerOrientation(CORNER_PIECE::RGY),
      iCube.getCornerOrientation(CORNER_PIECE::RGW),
      iCube.getCornerOrientation(CORNER_PIECE::RBW),
      iCube.getCornerOrientation(CORNER_PIECE::OBW),
      iCube.getCornerOrientation(CORNER_PIECE::OBY),
      iCube.getCornerOrientation(CORNER_PIECE::OGY),
      iCube.getCornerOrientation(CORNER_PIECE::OGW)
    };

    // Treated as a base-3 number, converted to base-10.
    uint32_t permInd =
      cornerOrientations[0] * 2187 +
      cornerOrientations[1] * 729 +
      cornerOrientations[2] * 243 +
      cornerOrientations[3] * 81 +
      cornerOrientations[4] * 27 +
      cornerOrientations[5] * 9 +
      cornerOrientations[6] * 3 +
      cornerOrientations[7];

    return permInd;
</code>```

Note the CORNER_PIECES and CORNER_POSITIONS.  For CORNER_PIECES I calculate (and empirically find) $3^7$ permutations.  For CORNER_POSITIONS I empirically find 6513 permutations with a breadth-first search to depth 9, but don't understand why.

答:
That's a strange definition of orientation you are using. Half the corner pieces turn clockwise if you change it from oriented to twisted once, and the other half turn anti-clockwise.

With the normal definition of orientation, only $3^7=2187$ orientations are possible instead of $3^8=6561$ since the total twist of the pieces must be zero modulo 3. With your definition, you negate the twist of any four of the locations. They can indeed be any four, since all permutations are possible making any arrangement of the two types of corner pieces possible. This original twist constraint no longer holds, but instead you have the constraint that you can split them into two sets of four locations with the same total amount of twist mod 3.

It turns out that with this new definition, only $48$ of the $6561$ orientations cannot be reached. These are the configurations where there are 7 locations with one twist value and the 8th location with a different value. Whichever way you split this into two sets of four, the two sets will have different total twist values. There are 8 possibilities for the odd-one-out location, and 3 twist values it can be, and 2 twist values for the rest, giving a total of $8*3*2=48$ impossible configurations.

Proving that all other arrangements are reachable is a bit of case work, but it can be reduced by realising that

the order of the locations does not matter,

twisting all the locations by the same amount is possible (since four pieces are twisted in opposite directions to the other four), and

we can negate all the twists.

From this we can assume without loss of generality that there are at least as many 0s as 1s, and at least as many 1s as 2s. That leaves only the following cases to check:

00000000 = 0000+0000

00000001 = (impossible)

00000011 = 0001+0001

00000012 = 0000+0012

00000111 = 0000+0111

00000112 = 0011+0002

00001111 = 0011+0011

00001112 = 0001+0112

00001122 = 0000+1122

00011122 = 0011+0122


答:
The continuous image of a connected space is connected, so from this the "positive" version of your statement follows:

If $X \times Y$ is connected then $X$ and $Y$ are connected,

which is logically equivalent to

$X$ disconnected or $Y$ disconnected implies $X \times Y$ disconnected.

(let $p="X \text{ connected}"$, $q = "Y \text{ connected}"$, $r = "X \times Y\text{ connected}"$ then the first is $$r \implies (p \land q)$$

while the latter is $$(\lnot p \lor \lnot q) \implies \lnot r$$

and simple FOL tells us those are equivalent statements.)

And the first can be seen immediately by noting that $X= \pi_1[X \times Y], Y = \pi_2[X \times Y]$ and applying the "continuous image of connected spaces" fact twice. All we need are that both projections are continuous.