问:
Solving Challenging Inequality $x^{\frac{p^2+p+1}{p^2}}-x^{\frac{p+1}{p}} < \log(p)$
This is a continuation of the problem asked here. I made some edits to the original problem, but enough changes were made to warrant a new question.

The goal is, given a value of $p$, find the maximum value of $x$ that satisfies
$$
x^{\frac{p^2+p+1}{p^2}}-x^{\frac{p+1}{p}} < \log(p)
$$
given $p\in\mathbb{Z}\geq 3$.

For purposes of usage, a lower bound of $x<x_{max}$ would be signifigantly more useful, although an upper bound may potentially be interesting.

In their answer, user Claude Leibovici showed that for sufficiently large $p$, $x_{max}\approx \frac{1}{2}p^2$.

Below is my work that I appended to the bottom of the original question. It provides a lower bound estimate that turns out to be signfigantly lower than the true value (but not too bad)
$$
e^{W(p^{k+1}-1} < x_{max}
$$

We can also think of this as the inequality of two continuous functions
$$
f(x;k,p) = x^{\frac{p^{k+1}-1}{p^k(p-1)}}
$$
$$
g(;a,k,p) = a + k\log(p).
$$

Note that $a$ is a constant derived from the problem that guarantees that $f(x; 1,p) < g(;a,1,p)$ for a given value of $x$. The initial question formulation was asking to find the largest value of $x$ that satisfied the following inequality
$$
f(x;2,p) - f(x;1,p) < g(;a,2,p) - g(;a,1,p)
$$
with the goal of if we know a priori that $f(x;1,p) < g(;a,1,p) $ we can then imply that $f(x;2,p) < g(;a,2,p).$

We can differentiate both sides of the second inequality with respect to $k$ to get
$$
x^{\frac{p^{k+1}-1}{p^k(p-1)}} \log(x)\left(\frac{\log(p)}{p^k(p-1)}\right) < \log(p)
$$
for which we want to find a value of $x$ max that makes this inequality true. Rearranging we get that
$$
x^{\frac{p^{k+1}-1}{p^k(p-1)}} \log(x) < p^k(p-1)
$$
setting $x = e^z$ and just using some temp placeholder variables we can say that
$$
e^{az}z < b
$$
which has solution
$$
z < W(ab) \implies \log(x) < W(p^{k+1}-1)
$$
so our max value is $x=e^{W(p^{k+1}-1)}$

Since the derivative of the left hand side seems to be decreasing on $[1,2]$, is it sufficient to say that if $x$ is less than our bound here it should also be less at the next point?

This may just be me rambling but maybe it is helpful to the task

答:
For this answer, I shall refer to what I wrote for this previous question.

Considering the problem of the zero of function
$$f(x)=x^a-x^b -\log(p) \quad \text{where} \quad a=\frac{p^2+p+1}{p^2}\quad \text{and} \quad b=\frac{p+1} p$$
$$f'(x)=a x^{a-1}-b x^{b-1} \quad \text{and} \quad f''(x)=a(a-1) x^{a-2}-b(b-1)  x^{b-2}$$ I suggested as an initial estimate
$$x_0=\frac{p^2\log (p)}{W\left(p^2 \log (p)\right)}$$ and Newton iterates will be
$$x_{n+1}=x_n-\frac{x_n^a-x_n^b -\log(p) }{a x_n^{a-1}-b x_n^{b-1} }$$ This converges very fast.

Using as before for a test, $p=123.456$ and working with unlimited precision we have
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 8150.36957189589 \\
 1 & 7636.04252044307 \\
 2 & 7634.23710338033 \\
 3 & 7634.23708015417 \\
 4 & 7634.23708015417
\end{array}
\right)$$ which shows that we know the analytical solution of the equation.

What has been checked is that, as a function of $p$, $f(x_n)$, $f'(x_n)$ and $f''(x_n)$ are always positive. So, by Darboux theorem, $x_n$ is an overestimate of the solution. So, to satisfy the inequality, $\color{red}{\text{a lower bound is } \lfloor x_n\rfloor ~~\forall n \ge 2}$

Now, let us consider the implicit function
$$g(x,p)=x^a-x^b -\log(p)=0$$
$$\frac{\partial g(x,p)}{\partial p}=\frac{x^b \log (x)}{p^2}-\frac{(p+2) x^a \log (x)}{p^3}-\frac{1}{p}$$
$$\frac{\partial g(x,p)}{\partial x}=a x^{a-1}-b x^{b-1}$$ and then, at solution,
$$\frac{dx}{dp}=-\frac{\frac{\partial g(x,p)}{\partial p} } {\frac{\partial g(x,p)}{\partial x} }$$

Numerical analysis shows that
$$\frac{dx_n}{dp}\sim p-2$$

For $n=3,4$ a quick and dirty linear regression $\frac{dx_n}{dp}=a+b\,p$ gives with $R^2=0.999977$
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & -1.86836 & 0.05913 & \{-1.98572,-1.75099\} \\
 b & +1.03026 & 0.00101 & \{+1.02825,+1.03227\}  \\
\end{array}$$

Edit

A much better estimate of the solution is given by the first iterate of Halley method
$$x_1=x_0+\frac{2 f(x_0) f'(x_0)}{f(x_0) f''(x_0)-2 f'(x_0)^2}$$
with the same $x_0$ as before.

For $p=123.456$, $x_1=7634.28$.

More spectacular would be the first iterate of Householder method
$$x_1=x_0+\frac{3 f(x_0) \left(f(x_0) f''(x_0)-2 f'(x_0)^2\right)}{f(x_0)^2 f'''(x_0)+6 f'(x_0)^3-6 f(x_0)
   f'(x_0) f''(x_0)}$$

Update

Just out of curiosity, I searched the maximum value of the ratio
$$R_n=\frac{x^a} {x^b +\log(p)}$$ for Newton successive iterates
$$\left(
\begin{array}{cc}
n & R_n \\
 0 & 1.11721588695 \\
 1 & 1.01728777375 \\
 2 & 1.00039248588 \\
 3 & 1.00000020331 \\
 4 & 1.00000000076
\end{array}
\right)$$


问:
Show that the Cauchy distribution, whose density is $f\left(x\right)=\frac{1}{\left[\pi\left(1+x^{2}\right)\right]}$ does not possess finite moments.
This from Hoel's Statistical Book:
Show that the Cauchy distribution, whose density is $f\left(x\right)=\frac{1}{\left[\pi\left(1+x^{2}\right)\right]}$ does not possess finite moments.

I've been reading the recommended section for this problem, but I have no clue of how to solve it.

答: $\int_R\frac{|x^n|}{1+x^2}dx$ does not exist for $n\ge 1$, since integrand ~$x^{n-2}$ as $x\to \infty$.

ContentLicense: CC BY-SA 4.0


问:
Is there a way to prove that $A \setminus C = (A \setminus (B \cup C)) \cup ((A \cap B) \setminus C)$?
Drawing out the Venn Diagrams I know this is true.

$(A \setminus (B \cup C))$ is just all the values that are solely in A.

$((A \cap B) \setminus C)$ is just all the values in both A and B but not in C.

And taking the union of these two leads us to the values that are either only in A or in only A and B (but not C).

I am unsure how to rigorously prove this, however. Would starting off by letting $x \in (A \setminus (B \cup C))$ be a good start?

Venn Diagram

答:
You pretty much have it.  It is a proof by cases that :

Take any $x$ where $x\in(A\smallsetminus(B\cup C))\cup((A\cap B)\smallsetminus C)$.

Thus we have that $x\in((A\smallsetminus(B\cup C))$ or $x\in ((A\cap B)\smallsetminus C)$

In the left case: $x\in A$ but $x\notin(B\cup C)$. The latter means $x\notin B$ and $x\notin C$. Thus in this case: $x\in A$ and $x\notin C$, which is to say: $x\in A\smallsetminus C$.

In the right case: $x\in(A\cap B)$ but $x\notin C$.  The former means $x\in A$ and $x\in B$. Thus in this case: $x\in A$ and $x\notin C$, which is to say: $x\in A\smallsetminus C$.

So in each case: $x\in A\smallsetminus C$.

Therefore $(A\smallsetminus(B\cup C))\cup((A\cap B)\smallsetminus C)\subseteq A\smallsetminus C$.

However, that is but half or the proof.

Demonstrating the converse uses the law of excluded middle: "Either $x\in B$ or $x\notin B$, so..."


问:
A problem on matrices : Powers of a matrix

If $ A=
        \begin{bmatrix}
        i & 0 \\
        0 & i \\
        \end{bmatrix}
, n \in \mathbb N$, then $A^{4n}$ equals?

I guessed the answer as $ A^{4n}=
        \begin{bmatrix}
        i^{4n} & 0 \\
        0 & i^{4n} \\
        \end{bmatrix}
=\begin{bmatrix}
        1 & 0 \\
        0 & 1 \\
        \end{bmatrix}$ which actually was the answer.

Can you provide me with the correct method of getting the answer?

 Thank you.

答: Isn't it enough to show $A^4=I$ ?</p>

ContentLicense: CC BY-SA 3.0

Body: Isn't it enough to show $A^4=I$ ?


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

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

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

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

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

答:
For the maximum, you can easily prove that $4$ of the $5$ product terms cannot be negative. Then, the maximum is achieved either all $5$ are positive or only $3$ are positive. In the former case, simple $AM-GM$ gives the maximum of $32.$ For the latter case, assume $a+b\leq 0$ and $b+c\leq 0$ while the other three are non-negative. In that case, $-a-b\leq 2$ and $-b-c\leq 2$ and $d+e = 5-a-b-c\leq 8.$

Then, $$S = (c+d)(d+e)(e+a)(-a-b)(-b-c)\leq 4(c+d)(e+a)(d+e)\leq(c+d+e+a)^2\cdot 8= 8(5-b)^2\leq 8\cdot 36 = 288.$$

Now the other possibility is when two terms involving $4$ different variables are negative - say $a+b\leq 0$ and $c+d\leq 0.$ In this case, $b+c\geq 0$ and so $a+d = (a+b+c+d) - (b+c)\leq 0.$ Finally, $e = 5-a-d -(b+c)\leq 5-a-d\leq 7.$ Now for the kill:
$$S = (-a-b)(-c-d)(d+e)(e+a)(b+c)\leq \dfrac{(e-5)^2}{4}\cdot \dfrac{(e+5)^3}{27}\leq \dfrac{12^3}{27} = 64.$$

You can do similar thing for the minimum.


问:
maximum absolute column sum (column norm) of a matrix and its largest eigenvalue
Given a matrix $A\in\mathbb{R}^{n\times n}$. How to prove the following statement?
$$||A||_{L_1}\leq \sqrt{n}\sigma_{\max}(A),$$
where $||A||_{L_1} = \max_{i\in\{1,...,m\}}||\mbox{col}_i(A)||_1$ and $\sigma_{\max}(A)$ is the largest singular value of $A$.

This conclusion is the Fact 9.8.12 xiv) on Page 573 from the book Matrix Mathematics Theory, Facts, and Formulas (Second Edition) by Dennis S. Bernstein. (I checked the reference therein but didn't get any clue.)

答:
The norm in question is conventionally denoted by $\|\cdot\|_1$. The inequality as it stands does not make sense, because the eigenvalues of $A$ can be non-real and there isn't any "maximum" eigenvalue in general, but I suppose that by $\lambda_\max(A)$ you mean the spectral radius $\rho(A)$ of $A$.

At any rate, the inequality is false. E.g.
$$
\left\|\pmatrix{1&1\\ 0&1}\right\|_1=2>\sqrt{2}=\sqrt{2}\rho(A).
$$
The inequality can be corrected by requiring that $A$ is normal, or more generally, by changing $\rho(A)$ to $\|A\|_2$, the induced $2$-norm (= the largest singular value) of $A$:
\begin{aligned}
\|A\|_1
&=\max_{j=1,2,\ldots,n}\|Ae_j\|_1\\
&\le\sup_{\|x\|_2=1}\|Ax\|_1\\
&=\sup_{\|x\|_2=1}\langle|Ax|,e\rangle\\
&\le\sup_{\|x\|_2=1}\|\,|Ax|\,\|_2\|e\|_2\\
&=\sqrt{n}\sup_{\|x\|_2=1}\|Ax\|_2\\
&=\sqrt{n}\|A\|_2.
\end{aligned}


问:
prove that $f_n$ conerges uniformly in $X$"
Please, help me to find a solution to this question:

"If a sequence of continuous functions $f_n:X \rightarrow \mathbb R$ converges uniformly in a dense set $D \subset X$, prove that $f_n$ conerges uniformly in $X$".

Obs:

Once $f_n$ converges uniformly in $D$, then $f_n$ is a Cauchy sequence, i.e., given $\varepsilon >0$, there exists $n_0 \in \mathbb N$, such that $m,n>n_0 \Rightarrow |f_m(d)-f_n(d)|<\varepsilon$, for every $d \in D$ Besides that, $f_n$ is continuous and every point of $X$ is the limit of a sequence of points pertaining to $D$. Then, $d_n \rightarrow x \Rightarrow |f(d_n) \rightarrow f(x)|$. How to conclude that there exists $n_0 \in \mathbb N$, such that $m,n>n_0 \Rightarrow |f_m(x)-f_n(x)|<\varepsilon$?

答:
For $x \in X$, pick a sequence $d_k \in D$ satisfying $d_k \to x$. Then
$$
|f_n(x) - f(x)| = \left|f_n\left(\lim_{k \to \infty} d_k\right) - f\left(\lim_{k \to \infty} d_k\right)\right| \\
 = \left|\lim_{k \to \infty} f_n(d_k) - \lim_{k \to \infty}f(d_k)\right| \\
= \lim_{k \to \infty} |f_n(d_k) - f(d_k)| \\
\le \sup_{d \in D} |f_n(d) - f(d)| \\
= \|(f_n|_D) - (f|_D) \|_{\infty} \underset{n \to \infty}{\longrightarrow} 0.
$$
The uniform convergence of $f_n|_D$ to $f|_D$ implies the last statement, and since $\|(f_n|_D)-(f|_D)\|_{\infty}$ is a quantity that depends only on $n$ and not on $x$, this proves uniform convergence of $f_n$ to $f$.

In fact, you can even show a bit more. Since $h(x) \overset{def}= |f_n(x) - f(x)|$ is a continuous function on $X$ and $D$ is dense in $X$, we have
$$
\sup_{d \in D} h(d) = \sup_{x \in X} h(x)
$$
or in other words, $\|(f_n|_D)-(f|_D)\|_{\infty} = \|f_n-f\|_{\infty}$. Since uniform convergence is the same as convergence under the supremum norm, the fact that the statement you are asking about is true is no surprise anymore, since now the proof simplifies to
$$
\|f_n-f\|_{\infty} = \|(f_n|_D)-(f|_D)\|_{\infty} \to 0.
$$

Hope that helps,


答:
Since $A=iI$ and scalars commute with matrices you get</p>

<p>$$A^{4n}=(i)^{4n}I^{4n}=I \,.$$ </p>

Title: A problem on matrices : Powers of a matrix

Body:
If $ A=
        \begin{bmatrix}
        i & 0 \\
        0 & i \\
        \end{bmatrix}
, n \in \mathbb N$, then $A^{4n}$ equals?

I guessed the answer as $ A^{4n}=
        \begin{bmatrix}
        i^{4n} & 0 \\
        0 & i^{4n} \\
        \end{bmatrix}
=\begin{bmatrix}
        1 & 0 \\
        0 & 1 \\
        \end{bmatrix}$ which actually was the answer.

Can you provide me with the correct method of getting the answer?

 Thank you.

Body:
Since $A=iI$ and scalars commute with matrices you get

$$A^{4n}=(i)^{4n}I^{4n}=I \,.$$


问:
Let $X\sim \pi(\lambda)$, find $E\Bigl[\frac{1}{(X+1)}\Bigr]$.
Let $X\sim\pi(\lambda)$, find $E\Bigl[\frac{1}{(X+1)}\Bigr]$.

It is a problem in my text book, here is my work:
$$\begin{align}
E\left[\frac{1}{(X+1)}\right]&=\sum_{x=0}^{\infty}\frac{1}{x+1}\cdot\frac{\lambda^xe^{-\lambda}}{x!}\\
&=\frac{1}{\lambda}\sum_{x=0}^{\infty}\frac{\lambda^{x+1}e^{-\lambda}}{(x+1)!}\\
&=\frac{1}{\lambda}\sum_{k=1}^{\infty}\frac{\lambda^ke^{-\lambda}}{k!}\space (let \space k=x+1)\\
&=\frac{1}{\lambda}\left[\sum_{k=0}^{\infty}\frac{\lambda^ke^{-\lambda}}{k!}-e^{-\lambda}\right]\\
&=\frac{1}{\lambda}\left[1-e^{-\lambda}\right]\\
\end{align}$$

Is it right?

答:
Although you have solved it, here's another way to solve this question.
Use MGF $E(e^{tx})=e^{\lambda(e^{t}-1)}$

Similarly $E(s^x)=e^{\lambda(s-1)}$

$\dfrac{1}{X+1}=\int_{0}^{1} s^Xds$

$E\left(\dfrac{1}{X+1} \right)=\int_{0}^{1} E(s^X)ds=\int_{0}^{1} e^{\lambda(s-1)}ds=\bigg(\dfrac{e^{\lambda s}e^{-\lambda}}{{\lambda}}\bigg)_{0}^{1}=\dfrac{e^{-\lambda}(e^{ \lambda  }-1)}{{\lambda}}=\dfrac{1}{\lambda}\left[1-e^{-\lambda}\right]$


问:
Question about surjective map $S^n \to S^n$ of degree zero
There exists surjective map $f:S^n \to S^n$ of degree zero.

By Hopf theorem, $f$ is homotopic to some constant map $c_{y}:S^n \to \{y\}$.

Let the homotopy be $H(x,t)$, $H(x,0) = f(x)$, $H(x,1) = c_y(x) = y$, so $H(-,t)$ gives a deformation retraction from $S^n$ to a point, but such deformation retraction shouldn't exist, so what's the problem here?

答: Let $f$ be the map $S^n\rightarrow S^n$ that maps an angle $\theta$ to $2\theta$ when $\theta\leq \pi$ or to $2\pi - 2\theta$. This is a map to the circle that winds around the circle once clockwise, and once counter clockwise. This is certainly homotopic to a constant, and there's no reason this function shouldn't exist, because it isn't a deformation retraction.


问:
Proving that for a CW pair $(X,A)$, $X \times I$ deformation retracts to $(X \times \left\{ 0 \right\}) \cup (A \times [0,1])$
I am trying to understand the proof of Allen Hatcher's Proposition 0.16 from "Algebraic Topology". If $X$ is a cell complex, he defines $(X,A)$ to be a CW-pair if $A \subseteq X$ is closed and is a union of cells i.e. images of open disks under extensions of their attachment maps via their boundaries. Let $D^n$ denote closed $n$-disks and $I=[0,1]$, as well as $X^n$ (resp. $A^n$) all the $n$-cells in $X$ (resp. $A$).

After proving that $D^n \times I$ deformation retracts to the base of the cylinder glued to its vertical boundary (that is $(D^n \times \left\{ 0 \right\}) \cup (\partial D^n \times I)$), which I get, he claims that $X^n \times I$ is obtained from $(X^n \times \left\{ 0 \right\}) \cup \left( (X^{n-1} \cup A^n) \times I \right)$ by attaching copies of $D^n \times I$ along
$(D^n \times \left\{ 0 \right\}) \cup (\partial D^n \times I)$. I don't understand why or how this is the case. By his definition, $X^n$ is obtained from $X^{n-1}$ by attaching the $D^n$ disks via maps $\varphi^n_\alpha: \partial D^n_\alpha \rightarrow X^{n-1}$ and taking $X^n := X^{n-1} \bigcup_{\varphi^n_\alpha} D^n_\alpha$. I see that we can extend these maps to $\tilde\varphi^n_\alpha: \partial D^n_\alpha \times I \rightarrow X^{n-1} \times I$ and hence see $X^n \times I$ as the disjoint union of $X^{n-1} \times I$ with $D^n_\alpha \times I$ under the equivalences given by $\tilde\varphi^n_\alpha$. But why is it that we can attach $D^n \times I$ precisely along $(D^n \times \left\{ 0 \right\}) \cup (\partial D^n \times I)$?

(I also looked at the answer here but it doesn't help alleviate my confusion.)

答:
I'm not sure of your exact dissatisfaction with Hatcher's discussion; I hope that the explanation of my interpretation (below) will help.

To make things slightly more transparent, let's write $(X^{n-1} \cup A^n) \times I$ as $(X^{n-1} \times I) \cup (A^n \times I)$ so that our goal is to form $X^n \times I$ from
$$
M_n := (X^{n} \times \{0\}) \cup (A^n \times I) \cup (X^{n-1} \times I)
$$
by attaching copies of $D^n \times I$. What we're missing in $M_n$ to get to $X^n \times I$ is, in some sense,
$$
X_n := \{n\text{-cells of } X \text{ not in } A\} \times (0,1].
$$
There are two "regions of attachment" along which $X_n$ will meet (after attaching) $M_n$, when we regard both as a subspace of $X^n \times I$:

along $X^n \times \{0\}$, and

along $X^{n-1} \times I$.

The first aligns the "missing" $n$-cells $X_n$ with the existing $n$-skeleton at $\{0\}$ while the second attaches the $n$-cells $X_n$ to the existing $(n-1)$-skeleton as in the inductive formation of a CW complex.

This brings us to why we attach $D^n \times I$ along $(D^n \times \{0\}) \cup (\partial D^n \times I)$. Attaching along $D^n \times \{0\}$ handles the first bullet while attaching along $\partial D^n \times I$ handles the second. Both attachments make sense because they naturally arise from the attaching maps in the inductive formation of the CW complex (in the precise way you mention in your question). Finally, we need both attachments -- as mentioned in the linked answer -- because otherwise we would wind up with "holes" in $D^n \times I$; we need the full cylinder attached along its entire base-union-sides in order to execute the deformation retraction discussed at the very beginning of the proof of proposition 0.16.


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

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

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

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

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

答:
Throw a dart at a square target, the probability of its landing in any subregion of the square being proportional to the area of the region.

The probability that it fails to land exactly on the diagonal from the upper left corner to the lower right corner is $1$, since the area of that diagonal is zero.

But that doesn't mean every point in the space of points where it could land fails to be on that diagonal.  Some points are on the diagonal.

That's the difference.

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


答:
We can solve the problem by substitution.

Step 1: Let
\begin{align}
  &A = a+b \\
  &B = b+c \\
  &C = c+d \\
  &D = d+e \\
  &E = e+a \\
\end{align}
Then the problem is equivalent to find the maximum and minimum of $ABCDE$, where $A,B,C,D,E\ge -2$ and $A+B+C+D+E = 10$.

\begin{align}
{}
\end{align}

Step 2: Consider each case separately, according to the sign of $A$, $B$, $C$, $D$, and $E$. Without loss of generality, we only need to consider cases as follows:
\begin{align}
 &(1):~(A,B,C,D,E) = (+,+,+,+,+)\\
 &(2):~(A,B,C,D,E) = (+,+,+,+,-)\\
 &(3):~(A,B,C,D,E) = (+,+,+,-,-)\\
 &(4):~(A,B,C,D,E) = (+,+,-,-,-)\\
 &(5):~(A,B,C,D,E) = (+,-,-,-,-)\\
\end{align}
where + means the number is non-negative, and - means the number is non-positive.

Step 3 (Minimum): Of course, the minimum can occur in cases (2) and (4). First, consider case (2). As your intuition suggested, the minimum occurs when $E=-2$ and $A=B=C=D=3$. We can show this as follows. First, observe that for all $(A,B,C,D,E)$, we have
\begin{align}
  -2ABCD \le ABCDE.
\end{align}
Next, observe that $4(ABCD)^{1/4}\le A+B+C+D = 10-E\le 12$, which implies that $ABCD\le 81$. Thus, in case (2), the minimum of $ABCD$ is $-162$. Similarly, you can find the minimum value under case (4), which is $-512$.

Step 4 (Maximum): In the same vein, the maximum can occur in cases (1), (3), and (5). And their maximum values are $32$, $784/9\approx 87.11$, and $288$.

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

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

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

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

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


答:
Examples of maps from $S^n$ to $S^n$ which are null-homotopic are just not deformation-retractions. Think of a sphere trying to wrap itself around a second sphere. Keep the first sphere on the surface of the second, but "unwrap it", and once it's unwrapped, collapse it to a point. I've formalized the idea below.

Write a point on the sphere as $(x_0, x_1,\cdots,x_n)$ where $\sum_{i=0}^n x_i^2 = 1$. Write such a point as $x = x_0 e_0 + v$ where $v = \sum_{i=1}^n x_i e_i$ and $\{e_0,\cdots,e_n\}$ is the standard basis of $\mathbb R^n$. We can re-write $x = (\cos \theta) x_0 + (\sin \theta) v_0$ where $v_0$ has norm $1$ and lies in the subspace spanned by $\{e_1,\cdots,e_n\}$; the parameter $\theta$ can take values from $0$ to $\pi$ while $v_0 \in S^{n-1}$ is arbitrary. This parametrization $\alpha$ of $S^n$ by $[0,\pi] \times S^{n-1}$ is bijective on the subset where $\theta \in ]0,\pi[$.

Consider the map $f : [0,\pi] \to [0,\pi]$ defined by $f(x) = 2x$ if $x \in [0,\pi/2]$ and $f(x) = 2x-\pi$ if $x \in [\pi/2, \pi]$. This map is homotopic to the map that sends everything to $0$ just by the homotopy $h(x,t) = tf(x)$.

Now consider $g = f \times \mathrm{id}_{S_n}$. By construction, this is homotopic to the map $(x \mapsto 0) \times \mathrm{id}_{S_n}$.

Note that $\alpha: [0,\pi] \times S^{n-1} \to S^n$ sends $\{0\} \times S^{n-1}$ to $\{e_0\}$ and $\{\pi\} \times S^{n-1}$ to $\{-e_0\}$. So there is indeed a well-defined null-homotopic map $\tilde g : S^n \to S^n$ defined by $\tilde g(\alpha(\theta, v_0)) = \alpha(f(\theta), v_0)$. (The reason why this is well-defined even though $\alpha$ is not bijective is because on the set of points where the fibers of $\alpha$ is not a singleton, namely $\{0\} \times S^{n-1}$ and $\{\pi\} \times S^{n-1}$, the above definition is independent of the choice of point in the fiber.)  The homotopy is basically given by $\tilde h(\alpha(\theta, v_0), t) = \alpha(tf(\theta), v_0)$.

Hope that helps,

Title: Question about surjective map $S^n \to S^n$ of degree zero

Body:
There exists surjective map $f:S^n \to S^n$ of degree zero.

By Hopf theorem, $f$ is homotopic to some constant map $c_{y}:S^n \to \{y\}$.

Let the homotopy be $H(x,t)$, $H(x,0) = f(x)$, $H(x,1) = c_y(x) = y$, so $H(-,t)$ gives a deformation retraction from $S^n$ to a point, but such deformation retraction shouldn't exist, so what's the problem here?


问:
Let $r \cdot u = s \cdot u$ exists in a vector space $V$. Show that if $r \ne s$, then $u = o$, the zero vector in $V$ and If $u \ne o$, then $r = s$.
Let $r \cdot u = s \cdot u$ exists in a vector space $V$. Show that:

If $r \ne s$, then $u = o$, the zero vector in $V$.

If $u \ne o$, then $r = s$.

EDIT:

Attempt:
I know that this question is a little bit easy because it is a basic concept. But, although so,
I got a little confused about my answer below.

Let $r,s \in \Bbb F$, a field, and $u \in V$. Now,
\begin{align*}
r \cdot u &= s \cdot u \\
r \cdot u - s \cdot u &= o \\
(r-s) \cdot u &= o \\
\end{align*}

What next?

Assume that $V$ is over field, say $F$. Then, if $r \ne s$,
\begin{align*}
(r-s) \cdot (r-s)^{-1} \cdot u &= o \cdot (r-s)^{-1} \\
u &= o,
\end{align*}
and if $u \ne o$, then we must have $r-s = 0$ i.e. $r=s$.

Does it true?

答:
EDIT: You are correct so far, but you must explain why $(r - s) \mathbf u = \mathbf 0$ implies that $r - s = 0$ or $\mathbf u = \mathbf 0.$ Equivalently, we can show that if $r - s \neq 0,$ then $\mathbf u = \mathbf 0.$ Given that $r - s \neq 0$ and $r$ and $s$ are elements of a field, what can you say about $r - s?$ Is there a way to "cancel" $r - s$ from the left-hand side? Once you study bases, you may use the below argument. (It can't hurt to practice.)

Consider a basis $\mathscr B$ of $V.$ Every vector of $V$ can be written uniquely as a finite $\mathbb F$-linear combination of basis vectors of $V.$ Particularly, there exist unique scalars $c_1, \dots, c_n$ and basis vectors $\mathbf v_1, \dots, \mathbf v_n$ such that $\mathbf u = c_1 \mathbf v_1 + \cdots + c_n \mathbf v_n.$ Given that $r \mathbf u = s \mathbf u,$ we have that $$(r c_1) \mathbf v_1 + \cdots + (r c_n) \mathbf v_n = (s c_1) \mathbf v_1 + \cdots + (s c_n) \mathbf v_n.$$ What can you say about the relationship between $r c_i$ and $s c_i$ for each integer $1 \leq i \leq n?$ From there, I believe that you can conclude the desired results.


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

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

答:
Too long for a comment.

SInce you already received good answer and comments, let me address the problem of the antiderivative about which you write

"this integral is a pain to solve, and online calculators aren’t able to solve it"

In fact, it is quite simple : let $\tan \left(\frac{\pi  x}{2}\right)=t$

$$I=\int \log (\tan \frac{\pi x}{2})\,dx=\frac 2\pi \int \frac{\log (t)}{t^2+1}\,dt$$
$$\frac{1}{t^2+1}=\frac{1}{(t+i)(t-i)}=\frac i 2\left(\frac{1}{t+i}-\frac{1}{t-i} \right)$$
$$I=\frac i \pi\left(\int \frac{\log(t)}{t+i}dt-\int \frac{\log(t)}{t-i}dt \right)$$
$$I=\frac i \pi\left(\Big[\text{Li}_2(i t)+\log (1-i t) \log (t)\Big]-\Big[\text{Li}_2(-i t)+\log (1+i t) \log (t)\Big]\right)$$
$$I=\frac 1 \pi\Big[2 \log (t) \tan ^{-1}(t)+i (\text{Li}_2(i t)-\text{Li}_2(-i t)) \Big]$$


问:
Partial cycles in projective resolutions of square-free algebra

Short version: Over a square-free algebra must every projective resolution of a simple module eventually terminate or contain a shift of itself as a direct summand?

I suspect not, but have not found a counterexample and have checked a few thousand non-isomorphic algebras.

Let $R$ be a finite-dimensional, associative, unital, connected, basic algebra over an algebraically closed field such no projective indecomposable module has two isomorphic composition factors. In other words, $R$ is a square-free algebra.

For an (single-sided) finitely generated $R$-module $M$, define $\Omega^0(M) = M$ and $\Omega^{n+1}(M)$ to be the kernel of the projective cover of $\Omega^n(M)$.

Since $\Omega^n(M)$ has finite-length, it has a unique Krull-Schmidt decomposition involving some isomorphism classes of indecomposable modules $M_{n,i}$.

There are two mutually exclusive possibilities: $\Omega^n(M)=0$ for some $n$, or $M \cong M_{n,i}$ for some $n \geq 1$ and some $i$. In the former case, $M$ has finite projective dimension, and in the latter the projective resolution “repeats” insofar as $\Omega^{n+k}(M)$ contains $\Omega^k(M)$ as a direct summand.

Are those the only two possibilities?

If $R$ is a square-free algebra, and $M$ is a simple module then must there be $n>0$ and $i$ such that either $\Omega^n(M) = 0$ or $M_{n,i} \cong M$?

Even more strongly,

If $R$ is a square-free algebra, and $M$ is a simple module, then is $\{ [ M_{n,i} ] \}$ a finite set of isomorphism classes of indecomposable modules?

If so, then $\Omega$ can be represented as a non-negative integer matrix in the Burnside ring, and we are asking if a certain diagonal entry in every proper power is 0, then must the matrix be nilpotent. I haven't found a quick way to verify this, but in the cases I've checked it works, which strikes me as crazy.

答:
Here is another more complicated example that shows Julian's ideas aren't just hiding the problem in a sneaky way, but that my dichotomy/theorem of the alternative is just completely wrong for simple modules:

Consider the (circular) quiver on 4 vertices with edges $1\to 2 \to 3 \to 4 \to 1$ and add the zero relations $1\to2\to3\to4$, $2\to3\to4\to1\to2$, $3\to4\to1\to2$, and $4\to1\to2\to3\to4$.

The set $\{M_{n,i}\}$ as $M$ ranges over the simple modules has 8 elements, all uniserial. The first four are the simple modules, and the rest are the radicals of the projective indecomposable modules, $J_1=“23”$, $J_2=“341”$, $J_3=“41”$, $J_4=“123”$.

$\Omega$ acts as the following transformation on modules: $$\begin{array}{cccc}
\Omega^0 & \Omega^1 & \Omega^{2k} & \Omega^{2k+1} \\ \hline
1 & 23 & 41 & 23 \\
2 & 341 & 0 & 0 \\
3 & 41 & 23 & 41 \\
4 & 123 & 0 & 0 \\
\end{array}$$

In particular, $\Omega^n(M)$ has no simple direct summands if $M$ is simple and $n \geq 1$, yet the 1st and 3rd simple modules all have infinite projective dimension.

It is true that the $8\times 8$ matrix of $\Omega$ is such that $\Omega^2$ has non-zeros on the diagonal, and it is true that the number of isomorphism classes of indecomposable modules required to write down $\Omega^n(M)$ is finite (8), so perhaps the matrix idea is not entirely crazy.


问:
Spectral Mapping Theorem in Continuous Functional Calculus
Let $A$ be a bounded self adjoint operator and $f$ be a continuous fucntion on $\sigma(A)$.
Let $\lambda \in \mathrm{Ran}(f)$. Prove that there are $\psi\in H,$ with $\|\psi\|=1$ and $\|(f(A)-\lambda)\psi\|$ arbitrarily small so that $\lambda \in \sigma(f(A))$.

This is an exercise from Reed & Simons' book chap 7, Q8.

I have managed to show the existence of $\psi$, but I am stuck at showing that leads to $\lambda \in \sigma(f(A))$. I am wondering if anyone could help me with that?

答:
I think I might have solved it by my own.
We will prove it by contradiction. Since we have shown the existence of $\psi.$ Then, we can find a sequence $\psi_n$ with $\|\psi_n\|=1$ and $\|(f(A)-\lambda) \psi_n\|\rightarrow 0.$
Assume $\lambda\in \rho(f(A)),$ then $(f(A)-\lambda)^{-1}$ is bounded.
$$
1 = \|\psi_n\|
= \|(f(A)-\lambda)^{-1}(f(A)-\lambda)\psi_n\| \\
\leq \|(f(A)-\lambda)^{-1}\|\|(f(A)-\lambda)\psi_n\|
\rightarrow 0.
$$
We derive the contradiction.


问:
Is there a positive integer $n$ such that the prime divisors of $n^3 - 1$ are $2$, $3$ and $7$?
For a positive integer $k$ write $\pi(k)$ for the set of all prime divisors of $k$.  For example, $\pi(24) = \{2,3\}$ and
$\pi(1) = \emptyset$.

Question.  Is there a positive integer $n$ for which $\pi(n^3 - 1) = \{2,3,7\}$?

An alternative point of view asks whether there are positive integers $a$, $b$ and $c$ such that $1 + 2^a3^b7^c$ is a perfect cube (of a positive integer).

To be clear, $n^3 - 1$ must be divisible by all three primes $2$, $3$ and $7$, and not divisible by any other prime.

I think the answer is "no", but have so far been unable to prove it.  I actually only need to find an answer for prime $n$, but it appears it may be true without assuming that $n$ is prime.  Using Maple, I have checked this for all $n \leq 10^9$, and for the first $106000000$-ish primes.

This is actually just one case ($r = 3$) of the problem I am working on which is to find all primes $p$ and integers $r>1$ for which $\pi(p^r - 1) = \{ 2, 3, 7 \}$.  I think I have managed to show that $r = 2$ and $p \in \{ 13, 97, 127 \}$, unless $r = 3$. (And, as you might guess, this whole thing is just one very special case of a more general problem I am looking at.)

答:
Given $n\in\mathbf N$ s.t. $n^3 - 1$ is only divisible by primes in the set $S = \{2,3,7\}$ we can write
$$n^3 - 1 = 2^{v_2}3^{v_3}7^{v_7}$$
if any of the $v_p \ge 2$ we can factor out a square to write
$$n^3 - 1 = \underbrace{2^{v'_2}3^{v'_3}7^{v'_7}}_{=:d} m ^2$$
where each $v_p'$ must be either $0$ or $1$.
So any solution of the original problem gives us an integral point on the curve
$$d m^2 = n^3 - 1$$
such curves (with one variable squared and the other cubed) are known as elliptic curves, normally the equation looks like
$$y^2 = x^3 + ax + b$$
but adding the term $d$ in front of the squared variable is an operation known as quadratic twisting. To "remove" this $d$ and get to a standard form we will rewrite our equation:

If $m,n,d$ are integers satisfying $dm^2= n^3- 1$ we have
$d^3 \cdot d m^2 = d^3 n^3 - d^3$
so
$$(d^2 m)^2 = (d n)^3 - d^3$$
and letting $y = d^2 m$ and $x = dn$ we have an integer solution of
$$y^2 = x^3 - d^3$$
(where in particular $y$ is only divisible by $2,3,7$)

The upshot of this is that we have a finite list of possible $d$s (as they are only divisible by $2,3,7$ and squarefree, $d$ must be one of $1, 2, 3, 7, 6, 14, 21, 42$).
Moreover given any elliptic curve, the set of integral points on that curve is finite (this is a theorem due to Siegel), and this set can be computed via Sage or other software. So we have a finite list of curves, and only finitely many points to check for each curve to see if they gave us a solution to the original problem (we have to check that the $(x,y)$ we found has $d|x$ and $d^2|y$ and all that the primes required divide $d m^2$).

By checking all these, we find there are no solutions, here is some Sage code to check these things (the last couple of conditions we can check by hand as there are only 4 candidates where $y$ is not divisible by other primes than those in $S$):

```<code>sage: S = [2,3,7]
</code>```
```<code>sage: list(Combinations(S).map(prod))
</code>```
```<code>[1, 2, 3, 7, 6, 14, 21, 42]
</code>```
```<code>sage: for d in Combinations(S).map(prod):
....:    for (x,y,_) in EllipticCurve([0,-d^3]).integral_points():
....:        if y.is_S_unit(S):
....:            print("d:",d," x:",x," y:",y)
....:            print("n:",x/d, " m:",y/(d^2))
....:            print("n^3 - 1 =",(x/d)^3 - 1, "factors as", ((x/d)^3 - 1).factor())
....:            print("--")
</code>```
```<code>d: 7  x: 14  y: 49
n: 2  m: 1
n^3 - 1 = 7 factors as 7
--
d: 7  x: 28  y: 147
n: 4  m: 3
n^3 - 1 = 63 factors as 3^2 * 7
--
d: 6  x: 10  y: 28
n: 5/3  m: 7/9
n^3 - 1 = 98/27 factors as 2 * 3^-3 * 7^2
--
d: 6  x: 33  y: 189
n: 11/2  m: 21/4
n^3 - 1 = 1323/8 factors as 2^-3 * 3^3 * 7^2
--
</code>```

Note that similar code if you include $-1$ in the set $S$ this does find Barry Cipra's example, so this should lend some confidence that it is doing the right thing. Even if we vary the set of primes here there doesn't seem to be solutions for many sets of 3 or more primes, so perhaps there is a deeper reason too, or maybe its just unlikely!

Body:
Given $n\in\mathbf N$ s.t. $n^3 - 1$ is only divisible by primes in the set $S = \{2,3,7\}$ we can write
$$n^3 - 1 = 2^{v_2}3^{v_3}7^{v_7}$$
if any of the $v_p \ge 2$ we can factor out a square to write
$$n^3 - 1 = \underbrace{2^{v'_2}3^{v'_3}7^{v'_7}}_{=:d} m ^2$$
where each $v_p'$ must be either $0$ or $1$.
So any solution of the original problem gives us an integral point on the curve
$$d m^2 = n^3 - 1$$
such curves (with one variable squared and the other cubed) are known as elliptic curves, normally the equation looks like
$$y^2 = x^3 + ax + b$$
but adding the term $d$ in front of the squared variable is an operation known as quadratic twisting. To "remove" this $d$ and get to a standard form we will rewrite our equation:

If $m,n,d$ are integers satisfying $dm^2= n^3- 1$ we have
$d^3 \cdot d m^2 = d^3 n^3 - d^3$
so
$$(d^2 m)^2 = (d n)^3 - d^3$$
and letting $y = d^2 m$ and $x = dn$ we have an integer solution of
$$y^2 = x^3 - d^3$$
(where in particular $y$ is only divisible by $2,3,7$)

The upshot of this is that we have a finite list of possible $d$s (as they are only divisible by $2,3,7$ and squarefree, $d$ must be one of $1, 2, 3, 7, 6, 14, 21, 42$).
Moreover given any elliptic curve, the set of integral points on that curve is finite (this is a theorem due to Siegel), and this set can be computed via Sage or other software. So we have a finite list of curves, and only finitely many points to check for each curve to see if they gave us a solution to the original problem (we have to check that the $(x,y)$ we found has $d|x$ and $d^2|y$ and all that the primes required divide $d m^2$).

By checking all these, we find there are no solutions, here is some Sage code to check these things (the last couple of conditions we can check by hand as there are only 4 candidates where $y$ is not divisible by other primes than those in $S$):

sage: S = [2,3,7]

sage: list(Combinations(S).map(prod))

[1, 2, 3, 7, 6, 14, 21, 42]

sage: for d in Combinations(S).map(prod):
....:    for (x,y,_) in EllipticCurve([0,-d^3]).integral_points():
....:        if y.is_S_unit(S):
....:            print("d:",d," x:",x," y:",y)
....:            print("n:",x/d, " m:",y/(d^2))
....:            print("n^3 - 1 =",(x/d)^3 - 1, "factors as", ((x/d)^3 - 1).factor())
....:            print("--")

d: 7  x: 14  y: 49
n: 2  m: 1
n^3 - 1 = 7 factors as 7
--
d: 7  x: 28  y: 147
n: 4  m: 3
n^3 - 1 = 63 factors as 3^2 * 7
--
d: 6  x: 10  y: 28
n: 5/3  m: 7/9
n^3 - 1 = 98/27 factors as 2 * 3^-3 * 7^2
--
d: 6  x: 33  y: 189
n: 11/2  m: 21/4
n^3 - 1 = 1323/8 factors as 2^-3 * 3^3 * 7^2
--

Note that similar code if you include $-1$ in the set $S$ this does find Barry Cipra's example, so this should lend some confidence that it is doing the right thing. Even if we vary the set of primes here there doesn't seem to be solutions for many sets of 3 or more primes, so perhaps there is a deeper reason too, or maybe its just unlikely!


问:
Let $f$ be a function defined by $f(x)= $\begin{cases} (x-r)^2&r-1\le x <r+1 \\ 1 & r+1\le x <r+2 \end{cases}$, where....

Let $f$ be a function defined by $f(x)= \begin{cases} (x-r)^2&r-1\le x <r+1 \\ 1 & r+1\le x <r+2 \end{cases}$, where $r=3k, k\in I$. Find $\int_0^{90} f(x).dx$

Here is how I am defining the function

$$f(x)=\begin{cases} x^2& -1\le x <1 \\ 1 & 1\le x \le 2 \\ (x-3)^2 & 2\le x\le 4 \\1& 4\le x \le 5 \\ .....\\.... \end {cases}$$

Now that the function is defined, I do not know how to use it in the integral. It has something to do with the periodicity of the function but I don’t know what the period is.

答:
Hints:

Break up the integral: $$I=\int_0^{90}f(x)dx= \int_0^1\phi_0(x)dx+\int_1^2g(x)dx+\int_2^4\phi_1(x)dx+\cdots+\int_{86}^{88}\phi_{29}(x)dx+\int_{88}^{89}g(x)dx+\int_{89}^{90}\phi_{30}(x)dx$$

where $g(x)=1$ and $\phi_{k}(x)=(x-3k)^2$ for $k \in \{0,\cdots,30\}$

Use periodicity:

$$\int_{-1}^1 x^2dx= \int_{-1+3k}^{1+3k} (x-3k)^2dx =\cfrac{2}{3} $$

Condense 1. using sigma notation:

$$I=\sum_{k=0}^{29}\int_{1+3k}^{3k+2} g(x)dx + \sum_{k=1}^{29}\int_{-1+3k}^{3k+1} \phi_{k}(x)dx +\int_0^1\phi_0(x)dx+\int_{89}^{90}\phi_{30}(x)dx$$

Evaluate the integrals and then the summations.

 Answer: $I=50$

Edit: I believe I owe an explanation to hint 2. You can justify it by using "transformations", just like in pre-calculus.

If $f(x)=x^2$ for $x\in[-1,1]$, then we can "shift" the graph from the interval $[-1,1]$ to $[2,4]$ by introducing the translation $g(x)=f(x-3)$, which translates $f(x)$ three units to the right.

Or, if you prefer, use a u-substitution:

Let $x=u-3$ (and $dx=du$) so then:

$$\int_{-1}^1 x^2dx= \int_{-1+3}^{1+3} (u-3)^2du =\int_{2}^{4} (u-3)^2du  $$


问:
Finding a Function That Approaches Another Function
One of my math professors gave me the following challenge. It isn't graded, it's just for fun.

Consider the function:

\begin{equation*}
f_n(x)=x+3^3x^3+5^3x^5+...+(2n-1)^3x^{2n-1},~x \in (0, 1).
\end{equation*}

I want to find which of the following functions $f_n$ is getting close to as $n$ gets larger:

$ \displaystyle a)\frac{x(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle b) \frac{x(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$

$\displaystyle c) \frac{x^2(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle d) \frac{x^2(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$

Based on some tests i ran in mathematica by giving $n$ and $x$ values, it looks like $b)$ is the answer, but I am not sure. Can anyone confirm or deny this, and show how one might find the right answer, either with pen and paper or by using mathematica or maple or some other software?

答:
Yes, b).

In Maple, if you do

```<code> map(factor, sum((2*i-1)^3*x^(2*i-1),i=1..n));
</code>```

The result will be 2 terms, one which is shape $(x^2)^{(n+1)}*r(n,x)$ with $r(n,x)$ polynomial in $n$ and rational in $x$, and the second term is (exactly) (b).

What this is is actually writing
$$\sum_{i=1}^{n} (2i-1)^3x^{2i-1} = \sum_{i=1}^{\infty} (2i-1)^3x^{2i-1} - \sum_{i=N-1}^{\infty} (2i-1)^3x^{2i-1}$$
(and the term order gets flipped in Maple's output).  Both the resulting sums are 'easy' to do by hand.

Body:
Yes, b).

In Maple, if you do

 map(factor, sum((2*i-1)^3*x^(2*i-1),i=1..n));

The result will be 2 terms, one which is shape $(x^2)^{(n+1)}*r(n,x)$ with $r(n,x)$ polynomial in $n$ and rational in $x$, and the second term is (exactly) (b).

What this is is actually writing
$$\sum_{i=1}^{n} (2i-1)^3x^{2i-1} = \sum_{i=1}^{\infty} (2i-1)^3x^{2i-1} - \sum_{i=N-1}^{\infty} (2i-1)^3x^{2i-1}$$
(and the term order gets flipped in Maple's output).  Both the resulting sums are 'easy' to do by hand.

ContentLicense: CC BY-SA 2.5


问:
What is the Type 1 error for the following test?
For testing the hypotheses H0: P = 0.8 versus H1: P ≠ 0.8 at 1% significance level, we
obtain a sample of n = 100 and p = 0.75.

Options:

A. 0.100

B. 0.140

C. 0.212  (Correct answer)

D. 0.010

E. 0.020

So far I have,

z= (P^-P)/std dev

= -1.25

za/2 = 2.575

Don't reject null

But I don't know how to get the answer

答: The probability of a type I error is the probability of rejecting the null when it's true, also known as a false positive. It's the alpha level of your test. So it's 1%, D is the answer. It is alpha because when the null is true, there's alpha percent chance that the sample is as extreme as the point that you're going to reject...which happens to be when you reject.</p>


问:
Determining independence of events
Given two events $A$ and $B$ such that $P(A)=P(B)=P(A \mid B^c)=\frac{1}{3}$, are $A$ and $B$ independent?

I'm having trouble figuring out how I can manipulate these equations to see if they're independent. Of course, I just need to check to see if $P(A \cap B)=\frac{1}{9}$ but most of the formulas I can think of involving the intersection of two events assumes independence. I was thinking about how $P(A)=P(A \mid B^c)$ implies that $A$ and $B^c$ are independent, but that isn't enough to make the leap.

Title: Determining independence of events

Body: $P(A\cap B)=P(A)-P(A\cap B^{c})=P(A)- P(A|B^{c}) (1-P(B))=\frac 1 3 -\frac 2 9=\frac 1  9=P(A)P(B)$


问:
Suppose we are given that $X$ is EXP($\lambda$) and $L$ is fixed. If $Y = X$, when $X<L$ and $Y = L$ when $X\geq L$. What is the CDF of $Y$
I am a bit lost on how to approach the solution to this problem.

I know that $\{Y \leq t\} = \{Y \leq t, X < L\}\cup\{Y\leq t, X\geq L\}$. But I do not know how to proceed from here. The reason being that I am not quite sure how to interpret the probability of the event $\{Y\leq t, X\geq L\}$. Any help will be greatly appreciated.

答:
$P(Y \leq t, X \geq L)=P(L \leq t, X \geq L)=0$ if $L >t$ and $P(Y \leq t, X \geq L)=P( X \geq L)= e^{-\lambda t}$ if $L \leq t$.

The event $L \leq t$ is not random. It is $\emptyset$ or $\Omega$ depending on whether $L >t$ or $L \leq t$.


问:
Find the radius of convergence of $\sum_{n=1}^\infty (\frac{5n^3+2n^2}{3n^3+2n})z^n$

Radius of convergence of $\sum_{n=1}^\infty (\frac{5n^3+2n^2}{3n^3+2n})z^n$

I know Hadamards lemma says that $\frac{1}{R}=\lim\sup \vert a_n\vert^{1/n}$

But I'm not sure how to compute $\lim\sup \vert a_n\vert^{1/n}$

I can see the $(a_n)\to 5/3$, so I believe $\lim\sup \vert a_n\vert$ should just be the limit of the sequence? But I'm not sure how to actually use the lemma to find $R$.

答:
Cauchy-Hadamard theorem is often a power hammer to kill a fly.

You can just notice that the series diverges for $z=1$ as its term doesn’t converge to zero in that case. And that the modulus of its term is less than $\frac{5}{2}\vert z\vert^k$. As the series $\sum z^k$ converges for $\vert z \vert \le 1$, the radius is equal to $1$.


答:
Rational functions don't matter
when finding the
radius of convergence
since
$\lim_{n \to \infty} (n^k)^{1/n}
=1
$
for any real $k$.

(This follows from
$\lim_{n \to \infty} n^{1/n}
=1
$.)

Also note that
$a^{1/n} \to 1$
for any $a > 0$.

So, in this case,
the radius is 1 -
this converges for $|z| < 1$.

Also,
since
$a_n \to 5/3,
a_n^{1/n} \to 1
$.

Title: Find the radius of convergence of $\sum_{n=1}^\infty (\frac{5n^3+2n^2}{3n^3+2n})z^n$

Body:
Radius of convergence of $\sum_{n=1}^\infty (\frac{5n^3+2n^2}{3n^3+2n})z^n$

I know Hadamards lemma says that $\frac{1}{R}=\lim\sup \vert a_n\vert^{1/n}$

But I'm not sure how to compute $\lim\sup \vert a_n\vert^{1/n}$

I can see the $(a_n)\to 5/3$, so I believe $\lim\sup \vert a_n\vert$ should just be the limit of the sequence? But I'm not sure how to actually use the lemma to find $R$.


答: At first glance I could tell that converges when $|z|<1$ ,maybe it will help you


问:
Using Mathematical Induction to Prove that $f^n (x) = 3^n e^{3x}$, where $f^n (x)$ is the $n$th derivative of $f$, and $n$ is a natural number.
Here is the question that I am not sure if I am doing right.

Let $f(x) = e^{3x}$ and suppose $f^n (x)$ represents the $n$th derivative of $f(x)$ with respect to $x$. Use mathematical induction to prove $f^n (x) = 3^n e^{3x}$, where $n$ is a natural number.

And here is my attempt:

First, we have to show that the base case is true. So for $n = 1$,

We know that $f'(x) = 3e^{3x}$. We want to see if $f^1 (x)$ is the same. So

\begin{align*}
f'(x) &= f^1 (x) \\
3e^{3x} &= 3^1 e^{3x} \\
3e^{3x} &= 3e^{3x}
\end{align*}

Since LS $\equiv$ RS for $n = 1$, the base case is true.

For the inductive step, we want to assume that $f^n(x) = 3^n e^{3x}$ is true for $n = k$, such that $f^k(x) = 3^k e^{3x}$. Then we want to show that this is also true for $n = k + 1$.

\begin{align*}
f^{k+1}(x) &= 3^{k + 1}e^{3x} \\
&= (3)^k(3)^1e^{3x} \\
&= (3)^1(3^k e^{3x}) \\
&= 3f^k(x)
\end{align*}

I don't know if this is correct or not, or if I am on the right track. Some help or some suggestions would be appreciated!

答:
We have$f^{(k)}=3^ke^{3x}$
And have to prove$f^{(k+1)}=3^{k+1}e^{3x}$.

To do that we just need to take derivative of $f^{(k)}(x)$:

$$f^{(k+1)}(x)=\frac{d}{dx}[f^{(k)}(x)]=(3^ke^{3x})'=3^k\times 3e^{3x}=3^{k+1}e^{3x}$$
It is done!


问:
Sequential approach to differentiability of complex valued function
Let $f$ be analytic function on an open set $G$. Let $z\in G$. Let $\{z_n\}$ and $\{w_n\}$ be two sequences in $G$ which converges to $z$. Then show that $lim_{n \to \infty} \frac{f(z_n)-f(w_n)}{z_n-w_n}=f'(z)$.
It feels like this should follow just by the definition of differentiability. But can we do the proof in $\epsilon$ and $n$ argument?
I also thought of keeping one n fixed say for $\{z_n\}$ and letting another n tend to infinity we may get $\frac{f(z_n)-f(z)}{z_n-z}$ which will converge to $f'(z)$. But I could not write proper argument for this. Any hint would be helpful.

答:
Let's use the Taylor expansion of $f$. Suppose $w_n,z_n \to z_0$ and let $\sum_{n \ge 0} a_n (z-z_0)^n$ be the Taylor expansion of $f$ around $z_0$. Then
$$
f(z_n) - f(w_n) = \sum_{n \ge 0} a_n \left( (z_n-z_0)^n - (w_n-z_0)^n \right) \\
= \sum_{n \ge 1} a_n \left( (z_n-w_n) \left(\sum_{i=0}^{n-1} (z_n-z_0)^i (w_n-z_0)^{n-1-i} \right) \right) \\
= a_1(z_n-w_n) + (z_n - w_n) \left( \sum_{n \ge 2} a_n \left(\sum_{i=0}^{n-1} (z_n-z_0)^i (w_n-z_0)^{n-1-i} \right) \right).
$$
Dividing by $z_n - w_n$ and letting $n \to \infty$ proves that
$$
\lim_{n \to \infty} \frac{f(z_n) - f(w_n)}{z_n-w_m} = a_1 = f'(z_0)
$$
since all the terms in the big sum go to zero (because $z_n,w_n \to z_0$). For details, if $|z_n-z_0|,|w_n-z_0| < \varepsilon$, then the big sum's absolute value is $\le \sum_{n \ge 2} |a_n| \varepsilon^{n-1}$ which goes to $0$ with $\varepsilon$ since this is equal to $g(\varepsilon)$ where
$$
g(w) = \frac{f(z_0+w) - a_0 - a_1w}{w}
$$
which is analytic when $f$ is analytic.

Hope that helps,


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

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

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

答:
Consider that you look for the minimum value of function$$ f(x)= \frac{2x-1}{x^2-x}-\ln\left(1+\frac{1}{x} \right)\qquad \text{with} \qquad x>1$$ Its first derivative is
$$f'(x)=\frac{-x^3-2 x^2+2 x-1}{(x-1)^2\, x^2\, (x+1)}$$ the denominator is positive $\forall x >1$. The numerator shows only one real root $(\Delta=-83)$ which is
$$x=-\frac{2}{3} \left(1+\sqrt{10} \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\frac{79}{20 \sqrt{10}}\right)\right)\right) \sim -2.83118$$ which is not in the acceptable range. So $f'(x)  <0 \quad \forall x >1$ and $f(x)$ is a decreasing function which starts from $+\infty$.

Around $x=1$, by Taylor
$$f(x)=\frac{1}{x-1}+(1-\log (2))-\frac{x-1}{2}+O\left((x-1)^2\right)$$

Now, for large values of $x$
$$f(x)=\frac 1x +\frac 3{2x^2}+\sum_{n=3}^\infty \frac{n+(-1)^n}{n\, x^n}$$ and all coefficients are positive.

Hence, $f(x) >0 \to 0^+$ and the inequality holds.


答:
I don't believe the question as posted is complete - unlike ordinary numbers, vector Dot Products do not have the cancellation property. From Wikipedia article for Dot Product

If $a ⋅ b = a ⋅ c$ and $a  ≠  0 $, then we can write: $ a  ⋅ ( b  −  c ) = 0$ by the Distributive Law; this just means that  $a$  is perpendicular to $( b  −  c )$, which still allows $( b  −  c ) ≠  0 $, and therefore allows $b  ≠  c $.

To put it another way, $r ⋅ u = s ⋅ u$ (with $r ≠ s$) can still be true if $u ⟂ (r - s)$ and $u ≠ 0$.

EDIT: Assuming a downvote means someone doesn't believe me, I'll work through a simple example with three perpendicular, non-zero unit vectors.

Let $r = (1,0,0)$, $s = (0,1,0)$, and $u = (0,0,1)$.

$$r ⋅ u = s ⋅ u$$
$$\left\lVert r \right\rVert \times  \left\lVert u \right\rVert \times \cos \theta_{ru} = \left\lVert s \right\rVert \times \left\lVert u \right\rVert \times \cos \theta_{su}$$
$$1 \times 1 \times \cos \theta_{ru} = 1 \times 1 \times \cos \theta_{su}$$
Obviously, being perpendicular unit vectors, $\theta_{ru} = \theta_{su} = 90^\circ$, so we get
$$1 \times 1 \times \cos 90 = 1 \times 1 \times \cos 90$$
$$1 \times 1 \times 0 = 1 \times 1 \times 0$$
$$0=0$$

Title: Let $r \cdot u = s \cdot u$ exists in a vector space $V$. Show that if $r \ne s$, then $u = o$, the zero vector in $V$ and If $u \ne o$, then $r = s$.

Body:
Let $r \cdot u = s \cdot u$ exists in a vector space $V$. Show that:

If $r \ne s$, then $u = o$, the zero vector in $V$.

If $u \ne o$, then $r = s$.

EDIT:

Attempt:
I know that this question is a little bit easy because it is a basic concept. But, although so,
I got a little confused about my answer below.

Let $r,s \in \Bbb F$, a field, and $u \in V$. Now,
\begin{align*}
r \cdot u &= s \cdot u \\
r \cdot u - s \cdot u &= o \\
(r-s) \cdot u &= o \\
\end{align*}

What next?

Assume that $V$ is over field, say $F$. Then, if $r \ne s$,
\begin{align*}
(r-s) \cdot (r-s)^{-1} \cdot u &= o \cdot (r-s)^{-1} \\
u &= o,
\end{align*}
and if $u \ne o$, then we must have $r-s = 0$ i.e. $r=s$.

Does it true?


问:
Expected number of tosses to get a head from a coin using integration formulae?
I recently started learning Expectation Probability , first of all Any Good resources to study it will be appreciated if any one can share

What I have learnt so far the expected  value of some Unknown Variable say $x$ be $ E(x)$

which boils down this equation  $ E(x) = \int_{-\infty}^\infty x*P(x) dx $ where P(x) is the Probability of some specific $x$ .

So I wanted to find the Expected Number of tosses to Get a  Heads from an Unbiased Coin So I used this equation to solve it

$E(Getting Heads) =>  \int_{0}^\infty\dfrac{x}{2^x}  dx$ , **IT IS A PRETTY STANDARD RESULT THAT E(GETTING  HEADS) = 2, But this integral is giving me another answer , which is $\dfrac{1}{\ln^2\left(2\right)}$
 Can Anyone tell me where am I going wrong with understanding stuff.

答: The integral formula you're using only applies to continuous random variables with probability density function $p(x)$. But the "number of coin tosses before heads appears" is a discrete random variable because it only takes integer values. The expected value of a discrete random variable is $E(X)=\sum_x x\cdot P(X=x)$, taking the sum over all possible values $x$.


问:
If A is a countable set, how can I show that A has countably many finite subsets?
Lets say that A is a countable set. How would I go about showing that A has countably many finite subsets?

I need to show that for every $n \in N$, the set $P^n(A)$ of finite subsets - with exactly $n$ elements - of A is countable. How would I then show how the union $U_{n \in N} P^n(A)$ of all these sets is countable?

答:
Hints:

By the definition of a countable set, there exists an injection $f: A \to \Bbb{N}$.

Let $\mathcal{G}$ denote the set of all finite subsets of $A$.

Define $\xi: \mathcal{G} \to \Bbb{N}$  by
$\xi(G) = \prod_{k \mathop \in f(G)} p_{k + 1}$

where $p_n$ denotes the $n$th prime number.


问:
Production Scheduling over Time
A company has a permit to operate for five seasons. It can manufacture only during the first four seasons and in the fifth period it is only allowed to sell any leftover products. It can manufacture two types of products. One unit of product 1 requires five man-hours in preparatory shop and three man-hours in finishing shop. Each unit of product 2 requires six man-hours in preparatory shop and one man-hour in finishing shop. During each season the company has at most 12,000 man-hours in preparatory and 15,000 man-hours in the finishing shop (only during the first four seasons). The product manufactured during some season can be sold anytime from the next season onward. However, selling requires some marketing effort and it is expected that 10 and 20 minutes of premium radio marketing time (premium time occur during high volume listening hours) are required to sell 10 units of products 1 and 2 per period, respectively. Correspondingly, 25 and 40 minutes of non-premium radio marketing minutes are required to sell 10 units of products 1 and 2 per period, respectively. The costs of both premimum and non-premium radio marketing minutes are given in the table below. If a unit of product is available for sale during a season, but is not sold in that season, the manufacturer has to pay carryover charges of $10 per unit to put it up for sale again in the next season. The selling prices in the various seasons are given in the table below. How should the company operate in order to maximize the total profit?

Edit: I answered my own question below. Please comment with any feedback you have. Thank you.

答:
I don't know what to do with the "10 and 20 minutes of premium radio marketing time (premium time occur during high volume listening hours) are required to sell 10 units of products 1 and 2 per period, respectively. Correspondingly, 25 and 40 minutes of non-premium radio marketing minutes are required to sell 10 units of products 1 and 2 per period, respectively" part.

You have to express it as part of the 'cost of radio minutes' that you have written in the expression to be maximized.

10 minutes of premium radio marketing time are required to sell 10 units of product 1, incurring a cost of $10P_i$. Therefore, for one product, the cost incurred for advertising product 1 through premium radio is $P_i$. Similarly, 25 minutes of non-premium radio marketing time is required to sell 10 units of product 1, i.e. 2.5 minutes per product. Thus the total cost of advertising product 1 is $P_i + 2.5N_i$ per unit.

We can do the same thing for product 2.

Title: Production Scheduling over Time


问:
Does a discrete (and closed) set have an accumulation point?
Let $(E,\tau)$ be a topological space, $\mathcal N_\tau(x):=\{N\subseteq E:N\text{ is a }\tau\text{-neighborhood of }B\}$ for $x\in E$ and $B\subseteq E$. Remember that $x\in E$ is called

$\tau$-accumulation point of $B$ if $$\forall N\in\mathcal N_\tau(x):B\cap N\setminus\{x\}\ne\emptyset;\tag1$$

$\tau$-isolated point of $B$ if $$\exists N\in\mathcal N_\tau(x):B\cap N=\{x\}\tag2.$$

Moreover, $B$ is called $\tau$-discrete if every point in $B$ is an $\tau$-isolated point of $B$; i.e. $$\forall x\in B:\exists N\in\mathcal N_\tau(x):B\cap N=\{x\}\tag3.$$

Assume $B$ is $\tau$-discrete. Can we show that $B$ has no $\tau$-accumulation point or do we need to additionally assume that $B$ is $\tau$-closed (i.e. $E\setminus B\in\tau$)?

Let $x\in E$. If $x\in B$, then $$B\cap N=\{x\}\tag4$$ and hence $$B\cap N\setminus\{x\}=\emptyset\tag5$$ for some $N\in\mathcal N_\tau(x)$; hence $x$ is not a $\tau$-accumulation point of $B$.

Now assume $x\in E\setminus B$. If $B$ would be $\tau$-closed, we would clearly find a $N\in\mathcal N_\tau(x)$ with $N\subseteq E\setminus B$ and hence $(5)$ would hold again and we we could conclude that $x$ is not a $\tau$-accumulation point of $B$.

But do we really need the $\tau$-closedness assumption? If so, can it be shown that the following equivalence holds: $B$ is $\tau$-closed and $\tau$-discrete if and only if $B$ has no $\tau$-accumulation point?

答:
Counterexample For Your First Question:

Take $E := \mathbb R$ with the standard topology. $B := \{1/n\}_{n = 1}^\infty$ is a discrete set w.r.t. the standard topology but clearly $0$ is an accumulation point of $B$. So discreteness by itself is not sufficient to guarantee no accumulation point.


答:
How will I solve this summation to get an answer 2 ?

From your phrasing of the question, I'm assuming that you threw the coin $n$ times and got tails each time, and then you got heads on the $(n+1)$-th time.  The probability of this occurring is $\frac{1}{2^{n+1}}$.  So you want to find the value of
$$ \frac{1}{2} \sum_{n=0}^\infty \frac{n+1}{2^n} $$

There is a trick for solving this kind of sum, assuming that you already know the sum of a geometric series and some basic calculus.  (If not, there are other ways of proving it, but the only ways I know are conceptually more difficult.)

$$ \begin{align}
\sum_{n=0}^\infty (n + 1) x^n &= \sum_{n=0}^\infty \frac{d}{dx} x^{n+1} \\
&= \frac{d}{dx} \sum_{n=0}^\infty x^{n+1} \\
&= \frac{d}{dx} \frac{x}{1-x} \\
&= \frac{1}{(1-x)^2} \\
\end{align} $$

The second line (moving the derivative outside the sum) is actually nontrivial, but it is okay in this instance because the sum converges uniformly.  (This is a technical detail that may or may not be interesting to you.)  The third line is just summing a geometric series, and the fourth line is basic calculus and some algebra.

Plugging in $x=\frac{1}{2}$ recovers your original problem.  The answer is
$$ \frac{1}{2} \sum_{n=0}^\infty \frac{n + 1}{2^n} = \frac{1}{2} \frac{1}{\left( 1 - \frac{1}{2} \right)^2} = 2 $$

Title: Expected number of tosses to get a head from a coin using integration formulae?

Body:
I recently started learning Expectation Probability , first of all Any Good resources to study it will be appreciated if any one can share

What I have learnt so far the expected  value of some Unknown Variable say $x$ be $ E(x)$

which boils down this equation  $ E(x) = \int_{-\infty}^\infty x*P(x) dx $ where P(x) is the Probability of some specific $x$ .

So I wanted to find the Expected Number of tosses to Get a  Heads from an Unbiased Coin So I used this equation to solve it

$E(Getting Heads) =>  \int_{0}^\infty\dfrac{x}{2^x}  dx$ , **IT IS A PRETTY STANDARD RESULT THAT E(GETTING  HEADS) = 2, But this integral is giving me another answer , which is $\dfrac{1}{\ln^2\left(2\right)}$
 Can Anyone tell me where am I going wrong with understanding stuff.


问:
Let $r \cdot u = s \cdot u$ exists in a vector space $V$. Show that if $r \ne s$, then $u = o$, the zero vector in $V$ and If $u \ne o$, then $r = s$.
Let $r \cdot u = s \cdot u$ exists in a vector space $V$. Show that:

If $r \ne s$, then $u = o$, the zero vector in $V$.

If $u \ne o$, then $r = s$.

EDIT:

Attempt:
I know that this question is a little bit easy because it is a basic concept. But, although so,
I got a little confused about my answer below.

Let $r,s \in \Bbb F$, a field, and $u \in V$. Now,
\begin{align*}
r \cdot u &= s \cdot u \\
r \cdot u - s \cdot u &= o \\
(r-s) \cdot u &= o \\
\end{align*}

What next?

Assume that $V$ is over field, say $F$. Then, if $r \ne s$,
\begin{align*}
(r-s) \cdot (r-s)^{-1} \cdot u &= o \cdot (r-s)^{-1} \\
u &= o,
\end{align*}
and if $u \ne o$, then we must have $r-s = 0$ i.e. $r=s$.

Does it true?

答:
I suppose the complete problem statement is:

Let $V$ be a vector space over the field $F$. Suppose we have $ru=su$ for some $r,s\in F$ and $u\in V$. Show that $r=s$ or $u=0$.

Transform the equality to $(r-s)u=0$. Assume $r\ne s$. Then $r-s\ne 0$ and has a multiplicative inverse $\frac1{r-s}$ in $F$. It follows that $$u=1\cdot u=\frac1{r-s}\cdot(r-s)\cdot u=\frac1{r-s}\cdot 0=0.$$


问:
Calculus derivatives
What is the second derivative of $h(x)=g(f(x))$
I was able to find the first derivative which is $h'(x)=g'(f(x))\cdot f'(x)$
I know for the second derivative we are going to need to use both the chain rule and product rule but I am unsure how

Title: Calculus derivatives

Body:
$$h(x)=g(f(x))$$
$$h'(x)=g'(f(x))\times f'(x)$$
$$h''(x)=g''(f(x))\times f'(x)\times f'(x) + g'(f(x))\times f''(x)$$
$$h''(x)=g''(f(x))\times (f'(x))^2 + g'(f(x))\times f''(x)$$


问:
Finding total number of vertices of a polygon
A polygon has 20 sides .

Then , find total number of triangles using vertices of polygon but not the sides of polygon. Ans is 800.

I used Pnc concept here .

Out of 20 vertices , I select 3 vertices to make a triangle.

$20C_3$ , then I subtract 16 *20.

Why 16? Since every 4 vertices of a polygon can make triangles with 2 sides. That gives us triangle which have one side of a polygon .

I am not getting how to go further than this .

答: You count all triangles and subtract those with exactly one polygon edge. You also need to subtract those with two such edges.


问:
Random list lengths
I am trying to analyze analytically what would happen if I build a random simulator as follow:

I begin with $ N $ lists of length 1, at each iteration, I will pick a random (non-empty) list, reduce its length by 1. And then, with probably $ p $, I create a new list of length 1, otherwise, I pick a random (non-empty) list and increase its length by 1. Every iteration keeps the total lengths of the lists constant, and I am interested in the distribution of list lengths.

A simple Monte Carlos simulation yield some interesting findings. With $ N = 100 $ and $ p = 0.5 $, we have, approximately $ P(L = 1) \approx 2 P(L = 2) $ and $ P(L = 2) \approx 2 P(L = 3) $, and so on. As it is just a simulation, the number is not exact of course. I run the simulator 1,000,000 times to be sure we reach stationary.

I really wonder there could be some time-reversible Markov Chain that could explain this, but my skill there is really rusty and some help would be greatly appreciated.

Attached below is the code I used to perform the Monte Carlos simulation, just for reproducibility.

```<code>from random import *

Size = 100
D = {}
N = [1] * Size
L = Size
R = Size
p = 50

for i in range(0, Size + 1):
    D[i] = 0

for i in range(0, 1000000):
    toremove = randint(0, R - 1)
    N[toremove] = N[toremove] - 1
    if N[toremove] == 0:
        N[toremove] = N[R - 1]
        N[R - 1] = 0
        R = R - 1
    tocreate = randint(0, 100)
    if tocreate < p:
        N[R] = N[R] + 1
        R = R + 1
    else:
        if R == 1:
            toappend = 0
        else:
            toappend = randint(0, R - 1)
        N[toappend] = N[toappend] + 1
    for n in N:
        D[n] = D[n] + 1
print(D)
</code>```

EDIT:

To make the problem more tractable for a manual calculation. I reduced the list size to 3. In this case, we have only 3 meaningful states. We have 3 lists of size 1, 1 list of size 2 together with 1 size of size 1, and 1 list of size 3.

The transition matrix for these states is as follow (the probability values can be reasoned by a case by case analysis).

$\left(\begin{array}{ccc}
p & 1-p & 0 \\
\frac{p}{2} & \frac{1}{2} & \frac{1-p}{2} \\
0 & p & 1-p
\end{array}\right)$

For $ p = \frac{1}{2} $, the stationary probability can be found as the eigen vector of $ T' $ corresponding to eigenvalue 1 to be

$\left(\begin{array}{c}
\frac{1}{4} \\
\frac{1}{2} \\
\frac{1}{4}
\end{array}\right)
$

That explains the list length distribution because the expected number of lists of length 1 will be $ N \times (\frac{1}{4} \times 3 + \frac{1}{2} \times 1) = \frac{5N}{4} $. The number of list of length 2 would be $ N \times \frac{N}{2} =  $ and the number of lists of length 1 would be $ N \times \frac{1}{4} = \frac{N}{4} $, which is roughly the doubling that we are seeing. In fact $ \frac{5N}{4} $ matches better to the experiment data than $ N $.

Attached is the octave code for computing the Markov chain stationary state probability as well as the expected list lengths. The 10000 power is used as an easy way to approximate. With this code, now I can easily tune $ p $ and figure $ p $ can be used effectively to change list length distributions.

```<code>p = 0.5
T = [p, (1-p), 0;p/2,1/2,(1-p)/2;0,p,(1-p)];
([1,0,0] * T^10000) * [3,0,0;1,1,0;0,0,1]
</code>```

All the above shown Markov chain could be used to analyze the situation. However, it requires a lot of effort to analyze the states in the case by case manner and it certainly won't scale for big $ N $. I am planning to run this simulator with $ N = 1000000 $ or above.

答:
I decided to write out the details assuming $4$ initial lists of length $1$. The five partitions are

$$4,3+1,2+2,2+1+1,1+1+1+1.$$
The transition probabilities are listed, with all others being zero.

\begin{align}
&P(4\to 4)=p\\
&P(4\to 3+1)=1-p\\
&P(3+1\to 4)=\frac{1}{4}(1-p)\\
&P(3+1\to 3+1)=\frac12\\
&P(3+1\to 2+2)=\frac14(1-p)\\
&P(3+1\to 2+1+1)=\frac12p\\
&P(2+2\to 3+1)=\frac12(1-p)\\
&P(2+2\to 2+2)=\frac12(1-p)\\
&P(2+2\to 2+1+1)=p\\
&P(2+1+1\to 3+1)=\frac13(1-p)\\
&P(2+1+1\to 2+2)=\frac13(1-p)\\
&P(2+1+1\to 2+1+1)=\frac13(1+p)\\
&P(2+1+1\to 1+1+1+1)=\frac13 p\\
&P(1+1+1+1\to 2+1+1)=1-p\\
&P(1+1+1+1\to 1+1+1+1)=p
\end{align}

Hence the transition matrix is

$$
\begin{pmatrix}
p & 1-p & 0 & 0 & 0 \\
\frac14(1-p) & \frac12 & \frac14(1-p) & \frac12 p & 0\\
0 & \frac12(1-p) & \frac12 (1-p) & p & 0 \\
0 & \frac13(1-p) & \frac13 (1-p) & \frac13(1+p) & \frac13 p\\
0 & 0 & 0 & 1-p & p
\end{pmatrix}
$$

Specializing to $p=1/2$, the left-eigenvector corresponding to eigenvalue $1$ is $\frac{1}{15}(1,4,2,6,2)$.

$$4,3+1,2+2,2+1+1,1+1+1+1.$$

Hence the expected number of lists of lengths $1$, $2$, $3$, $4$ are respectively
\begin{align}
N\cdot 1\cdot \frac{4}{15}+N\cdot 2\cdot \frac{6}{15}+N\cdot 4\cdot \frac{2}{15}&=\frac{24}{15}N,\\
N\cdot 2\cdot \frac{2}{15}+N\cdot 1\cdot \frac{6}{15}&=\frac{10}{15}N,\\
N\cdot 1\cdot \frac{4}{15}&=\frac{4}{15}N,\\
N\cdot 1\cdot \frac{1}{15}&=\frac{1}{15}N.
\end{align}
These gives ratios of $2.4,2.5,4$ which are roughly in agreement with the conjectured 2-fold ratio for $N\to \infty$.

In comments I noted a possible way to visualize the states and transitions for various list sizes. To display this I'll need to introduce a few concepts. The first is that of using a Young diagram to visualize integer partitions: For each term of the partition, we draw a row of that many boxes. We then join these stacks horizontally, yielding a quantity of boxes equal to the integer being partitioned. For instance, the partition $(5,4,1)$ of $10$ would be represented by the Young diagram (Wikipedia source)

One convenient feature of a Young diagram is that it's easy to visualize adding or removing boxes. For instance, if we remove $1$ from $3$ to get the partition $(5,3,1)$ of $9$, and then add it to $1$ to get the partition $(5,3,2)$ of $10$, the sequence of Young diagrams is

What this idea in turn leads to is the concept of Young's lattice: We partially order the integer partitions according to whether one Young diagram can be obtained by removing some number of boxes. This is presented graphically as a Hasse diagram:

Note that said diagram is organized so that each level consists of those partitions for a given integer $(N=1,2,3,4,\ldots)$. By removing one box and adding another, we move down one level and then back up to our original level.

With all that built up, how do we use this to visualize the transitions in the problem of interest? If we take $N=4$, then the possible states are the diagrams in the highest level shown in the above diagram. Taking one of these Young diagrams as our state, we randomly choose one of the rows and remove the last box. This moves us down one level, with probability depending on multiplicity of identical rows (e.g., there are two ways to remove one box from the partition $(2,2)$ to reach $(2,1)$.) At this point, we choose with probability $p$ whether to place that box at the bottom as a new row. If not, then we choose one of the remaining rows at random and place it there. In either case, we end up moving back to the original highest level of the lattice.

If we consider various ways of moving in this lattice, we can work out the transition probabilities to go from one diagram in a given row to another. (It's still tedious but at least it's organized.) The real hope, though, would be to use this presentation to find a more systematic way of determining the probabilities. It does give one point of visual intuition, however: If $p=1/2$, then the most probable move is always to add one row. This amounts to picking the right-most upward move after moving down. As such, we expect our random walk on the Young lattice to pile up on the right. That, in turn, means we expect to see partitions with many short rows more often than a few long ones. This is at least qualitatively in agreement with the proposed $1/2^L$ scaling. Hopefully more quantitative insights are possible as well.

Title: Random list lengths


答:
Number the vertices 1 through 20.

For the moment, assume that the order in which the vertices are selected matters.

Suppose the first vertex is #1. Then, if the second vertex is #3, there are 15 possible triangles, as the remaining vertex cannot be vertex 20, 1, 2, 3, or 4. Similarly, if the second vertex is #19, there are 15 possible triangles. For any other second vertex (#4 through #18), there are 14 possible triangles, because #20, #2, and the two vertices on either side of the second vertex are ineligible. So, all in all 15 + 15 + 15 * 14 = 240.

Since the first vertex can be any one of the 20, there are 20 * 240 = 4800 possible permutations.

But, order does not matter, so we must divide by $3!=6$ and $4800 / 6 =800$.

Title: Finding total number of vertices of a polygon

Body:
A polygon has 20 sides .

Then , find total number of triangles using vertices of polygon but not the sides of polygon. Ans is 800.

I used Pnc concept here .

Out of 20 vertices , I select 3 vertices to make a triangle.

$20C_3$ , then I subtract 16 *20.

Why 16? Since every 4 vertices of a polygon can make triangles with 2 sides. That gives us triangle which have one side of a polygon .

I am not getting how to go further than this .


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

答: $$f(z)={\Big\{}\begin{array}[ll]{ll}0 & \Re z\in\mathbb{Q}\\ 1& \Re z\in\mathbb{Q}\end{array}$$</p>


问:
An example of minimizing Mean Square Error (MSE)
I've established that the function $f(x)$ that minimizes $$ MSE = \mathbb{E} \left[ (Y - f(x))^2 \right] $$

is $$f(x) = \mathbb{E} \left[ Y \mid X \right]$$

The next part of the question (from Shumway and Stoffer, 3.14)  is to "apply the above result to the model:"

$$ Y = X^2 + W $$

where $X$ and $W$ are independent, and both $N(0,1)$.  Additionally, I'm to "show that $MSE = 1$".

It's been a while since I've done anything related to "mathematical statistics" so I'm like a fish out of water right now.  I'm not even sure where to begin.  I can tell right away that the sum of a chi-squared and normal is not going to be pretty.  Is there some trick with $\mathbb{E} \left[ Y \mid X \right]$ here that I'm not seeing?  I'm not sure that integral calculation will be a pleasant experience...

Many thanks for any solution/first steps/pointers.

答:
MSE is also known as risk, and is the expected value of the loss function.

$$\begin{split}MSE&=E[(Y-f(x))^2]\\
&=E[(X^2+W-E(X^2+W|X))^2] \text{ plug in Y in both places}\\
&=E([X^2+W-E(X^2|X)-E(W|X)]^2)\text { linearity of expectation}\\
&=E[(X^2+W-X^2-E(W))^2]\text { $E(g(X)|X)=g(X)$ property, and W is independent of X}\\
&=E(W-E(W))^2 \text { I believe you can see}\\
&=var(W) \text{ definition of variance}\\
&=1\end{split}$$


问:
Infinite sum $\frac{e^{-a n^2}}{n^2+b^2}$
I was wondering if the function
\begin{equation}
f(a,b) = \sum_{n=1}^\infty \frac{e^{-a n^2}}{n^2 + b^2},   \qquad\qquad (a,b>0)
\end{equation}
has a representation in terms of known elementary/special functions (like theta functions etc)?

答:
It is not too difficult to see that we have the differential equation
$$\frac{\partial}{\partial a}f(a,b)-b^2f(a,b)=\frac12(1-\vartheta_3(e^{-a})).$$
Using the integrating factor $M(a)=e^{-ab^2}$, we have
$$\frac{\partial}{\partial t}\left(e^{-tb^2}f(t,b)\right)=\frac12e^{-tb^2}(1-\vartheta_3(e^{-t})),$$
then integrating from $t=0$ to $t=a$, we have
$$e^{-ab^2}f(a,b)-f(0,b)=\frac12\int_0^a e^{-tb^2}(1-\vartheta_3(e^{-t}))dt.$$
Then from here, we have
$$f(0,b)=\frac{\pi}{2b}\coth\pi b-\frac1{2b^2},$$
so that
$$f(a,b)=\frac{e^{ab^2}}{2b^2}\left\{\pi b\coth\pi b-1+b^2\int_0^a e^{-tb^2}(1-\vartheta_3(e^{-t}))dt\right\}.$$


答:
A set $B$ being $\tau$-discrete does not imply it has no accumulation points, as the standard example $B=\{\frac1n\mid n \in \Bbb N^+\}$ in $\Bbb R$ in the standard topology shows. Here $0 \in B'$.

If $B$ is closed and $\tau$-discrete then it indeed does not have an accumulation point: if $p$ were one, it must lie in $B$ by closedness and this would contradict the discreteness of $B$.

And the condition is necessary of course: a set $B$ has no acculation points iff it is closed and $\tau$-discrete. We just saw the left to right implication, and if a set has no accumulation points, it is closed (voidly, as a set is closed iff it contains all its accumulation points, which is trvial when there are none) and $\tau$-discrete (as no $b \in B$ is an accumulation point it must be an isolated point (it's one or the other).

Title: Does a discrete (and closed) set have an accumulation point?

Body:
Let $(E,\tau)$ be a topological space, $\mathcal N_\tau(x):=\{N\subseteq E:N\text{ is a }\tau\text{-neighborhood of }B\}$ for $x\in E$ and $B\subseteq E$. Remember that $x\in E$ is called

$\tau$-accumulation point of $B$ if $$\forall N\in\mathcal N_\tau(x):B\cap N\setminus\{x\}\ne\emptyset;\tag1$$

$\tau$-isolated point of $B$ if $$\exists N\in\mathcal N_\tau(x):B\cap N=\{x\}\tag2.$$

Moreover, $B$ is called $\tau$-discrete if every point in $B$ is an $\tau$-isolated point of $B$; i.e. $$\forall x\in B:\exists N\in\mathcal N_\tau(x):B\cap N=\{x\}\tag3.$$

Assume $B$ is $\tau$-discrete. Can we show that $B$ has no $\tau$-accumulation point or do we need to additionally assume that $B$ is $\tau$-closed (i.e. $E\setminus B\in\tau$)?

Let $x\in E$. If $x\in B$, then $$B\cap N=\{x\}\tag4$$ and hence $$B\cap N\setminus\{x\}=\emptyset\tag5$$ for some $N\in\mathcal N_\tau(x)$; hence $x$ is not a $\tau$-accumulation point of $B$.

Now assume $x\in E\setminus B$. If $B$ would be $\tau$-closed, we would clearly find a $N\in\mathcal N_\tau(x)$ with $N\subseteq E\setminus B$ and hence $(5)$ would hold again and we we could conclude that $x$ is not a $\tau$-accumulation point of $B$.

But do we really need the $\tau$-closedness assumption? If so, can it be shown that the following equivalence holds: $B$ is $\tau$-closed and $\tau$-discrete if and only if $B$ has no $\tau$-accumulation point?


问:
If grad f is in (H^(-1)(U))^3 is f in L^2(U)?
I know that if $f \in L^2(U)$ then $\nabla f \in H^{-1}(U)$. However, is the converse true? I was thinking that maybe I could use the characterization of $H^{-1}(U)$ that says that if $f \in H^{-1}(U)$ then there exists $f_0 \in L^2(U)$, $f_1=(L^2(U))^3$ such that
$$\langle f, v \rangle=(f_0, v)_{L^2(U)}+(f_1, \nabla v)_{L^2(U)}.$$ However, I'm not sure exactly how. Does anyone know of any books that would have details related to this?

答: This is a highly nontrivial result. It is stated in Temam, Navier-Stokes equations, Proposition I.1.2. For the proof it refers to Necas, Equations aux derivees partielles, which I was not able to verify. Basically, every book that treats Stokes or Navier-Stokes problems should deal with this result. But most of them do not provide a proof.</p>


问:
Contour Integrals and Residues
I'm trying to figure out what it is all about, but my mind is blowing up. First of all, I have turned back and looked at the general definitions of integrals. Then I have looked to line integrals. Basically, the integral is helping us to find the area between two points along the curve. For example when force $F$ is constant but height of the object varies point to point, we can use the integral to find the total work done by $F$.

The line integral is helping us to find the total work done by any vector or scalar fields along any path. Arch length formula gives us the length of curve. So if all these definitions are Ok, please help me to understand about Contour and Residues? Why do we need to use them? Is there any physical meaning or are we just expanding our number set to use The Euler formula and obtain the result easily?

答:
In complex analysis, you might want to think of contour integrals as anti-derivatives. This helps to explain why the path can be moved (as long as it does not pass over a singularity) without altering the integral; it is just the difference of the anti-derivative at the endpoints.

Another way to express this is using Green's Theorem applied to a complex differentiable function:
$$
\oint_{\partial\Omega} f(z)\,\mathrm{d}z
=\iint_\Omega(if_x-f_y)\,\mathrm{d}x\,\mathrm{d}y\tag{1}
$$
The Cauchy-Riemann Equations are simply
$$
if_x=f_y\tag{2}
$$
Applying $(2)$ to $(1)$ yields Cauchy's Integral Theorem:
$$
\oint_{\partial\Omega} f(z)\,\mathrm{d}z=0\tag{3}
$$
when $f$ is holomorphic (i.e. satisfies the Cauchy-Riemann Equations) in $\Omega$.

$\hspace{2.5cm}$

Thus, the difference of the integral along the green path and the integral along the red path is the integral along the blue path. According to $(3)$, that is $0$.

There are points at which otherwise holomorphic functions go bad. These are singularities. The residue is $\frac1{2\pi i}$ times the integral counter-clockwise around such a point.

$\hspace{2.5cm}$

Note that the difference between the integrals along the green and red paths is the integral along the blue path, taking the integral in along the bottom of the blue line and out along the top (the integrals along the lines cancel, being in opposite directions). The blue contour is the boundary of the "C"-shaped area which avoids the singularity, so the integral along the blue contour is $0$. Therefore, the integral along the green path equals the integral along the red path. Thus, it does not matter which contour around the singularity we take, the residue will be the same.

Title: Contour Integrals and Residues


问:
Common false/unproven "popular mathematics" claims
Motivated by the common unproven claim in memes that every finite sequence of digits appears in the digits of $\pi$ (see Does $\pi$ contain all possible number combinations?), I started wondering what other similar common claims are out there?

Criteria

A claim about math that's "commonly" stated as true by math enthusiasts who are non-mathematicians. For instance, something that one might find posted somewhere as a meme.

The claim must be either false or unproven.

答:
A lot of things in cryptography are like this. When people say that something like Blum Blum Shub is a provably secure pseudorandom number generator, what is really meant is that cracking the generator is provably equivalent to factoring. Since $P\neq NP$ is unproven, it's often necessary to make that kind of hardness assumptions to "prove" that something is secure. I imagine this confuses a lot of laypeople. (Of course, it's not always necessary. One time pads actually are provably secure, in a completely $P=NP$-proof way.)

Other examples:

"There is no such thing as a formula for primes."

"$|\mathbb{R}|=\aleph_1$"

"$P(X\text{ and } Y) = P(X)P(Y)$" (You need independence for this to be true. Easy to pick up the error by watching someone do this, and not realizing they were using independence.)

"If the random variables $X, Y, Z$ are pairwise independent, then $X, Y, Z$ are independent of each other."

"The Godel sentence for Peano Arithmetic is necessarily true." (There exist very strange models of PA where it's false.)


问:
Does the solution to $x^i = 2$ have infinitely many solutions, or only one?
So in this video from bprp: https://www.youtube.com/watch?v=chcwYQBOcRw (the second question so skip ~5 mins), he writes only one solution. However, in the comments, there is debate over whether there are infinitely many solutions or not. Can someone clarify this?

Haziq provides an interesting argument but Stranger 01 answers it. However, plugging the solution into wolfram alpha gives multiple solutions:https://www.wolframalpha.com/input/?i=x%5Ei+%3D+2

答:
What is "$x^i$" in the first place? The most common meaning for that is $\exp(i\ln(x))$, since $\exp$ has a clear definition, and $\ln$ has a clear definition as the logarithm with a branch cut on the negative reals, and range in the strip between $\Im{z}=-\pi$ and $\Im{z}=\pi$.

If that is accepted, then
$$\begin{align}x^i&=2\\
\exp(i\ln(x))&=2\\
\ln\left(\exp(i\ln(x))\right)&=\ln2\\
\end{align}$$

Now be careful. For real $y$, $\ln(\exp(y))=y$. But for general complex $z$, $\ln(\exp(z))\neq z$. In general, $\ln(\exp(z))=z-2\pi i k$ for the right integer value of $k$ that lands $z-2\pi i k$ in that strip range of $\ln$.

$$\begin{align}
i\ln(x)-2\pi i k&=\ln2\\
i\ln(x)&=\ln2+2\pi i k\\
\ln(x)&=-i\ln2+2\pi k\\
\exp(\ln(x))&=\exp(-i\ln2+2\pi k)\\
x&=\exp(-i\ln2+2\pi k)=e^{2\pi k}\left(\cos\ln2-i\sin\ln2\right)
\end{align}$$

The logic shows that the only candidate solutions are these numbers $e^{2\pi k}\left(\cos\ln2-i\sin\ln2\right)$ as $k$ ranges over the solutions. But all these numbers are solutions, as you can verify:

$$\begin{align}
\exp(-i\ln2+2\pi k)^i&=\exp(i\ln(\exp(-i\ln2+2\pi k)))\\
&=\exp(i(-i\ln2+2\pi k-2\pi \ell i))
\end{align}$$

This is only for the right integer value of $\ell$. But the right value of $\ell$ is $0$, since $-i\ln2+2\pi k$ is already in the strip range of $\ln$.

$$\begin{align}
&=\exp(i(-i\ln2+2\pi k))\\
&=\exp(\ln2+2\pi ik)\\
&=\exp(\ln2)\exp(2\pi ik)\\
&=2
\end{align}$$

This was all based on one definition of "$x^i$" with one option for a logarithm function. Different things might happen with different choices.


Title: Production Scheduling over Time

Body:
A company has a permit to operate for five seasons. It can manufacture only during the first four seasons and in the fifth period it is only allowed to sell any leftover products. It can manufacture two types of products. One unit of product 1 requires five man-hours in preparatory shop and three man-hours in finishing shop. Each unit of product 2 requires six man-hours in preparatory shop and one man-hour in finishing shop. During each season the company has at most 12,000 man-hours in preparatory and 15,000 man-hours in the finishing shop (only during the first four seasons). The product manufactured during some season can be sold anytime from the next season onward. However, selling requires some marketing effort and it is expected that 10 and 20 minutes of premium radio marketing time (premium time occur during high volume listening hours) are required to sell 10 units of products 1 and 2 per period, respectively. Correspondingly, 25 and 40 minutes of non-premium radio marketing minutes are required to sell 10 units of products 1 and 2 per period, respectively. The costs of both premimum and non-premium radio marketing minutes are given in the table below. If a unit of product is available for sale during a season, but is not sold in that season, the manufacturer has to pay carryover charges of $10 per unit to put it up for sale again in the next season. The selling prices in the various seasons are given in the table below. How should the company operate in order to maximize the total profit?

Edit: I answered my own question below. Please comment with any feedback you have. Thank you.

Body:
$x_i=$ units of product 1 sold in season $i$

$y_i=$ units of product 2 sold in season $i$

$a_i=$ units of product 1 built in season $i$

$b_i=$ units of product 2 built in season $i$

$N_i=2.5x_i+4y_i$ (non-premium radio minutes in season $i$)

$P_i=x_i+2y_i$ (premium radio minutes in season $i$)

maximize $200x_2+450y_2+250x_3+400y_3+300x_4+400y_4+150x_5+300y_5-80P_2-20N_2-100P_3-40N_3-55P_4-10N_4-75P_5-25N_5-10(r_1+r_2+r_3+r_4+s_1+s_2+s_3+s_4-x_2+x_3+x_4+x_5+y_2+y_3+y_4+y_5)$

subject to

$5r_i+6s_i\leq 12000$

$3r_i+s_i\leq15000$

$N_2\leq 1000$

$P_2\leq200$

$N_3\leq 800$

$P_3\leq 400$

$N_4\leq 1000$

$P_4\leq200$

$N_5\leq 900$

$P_5\leq 300$

$x_i,y_i,r_i,s_i\geq 0$


问:
How to find the required coefficients of the expansion of the determinant?
I have a questionw where it's given that the determinant $\Delta(x)$ such that,
\begin{align*}
\Delta(x)
&=
\begin{vmatrix}
\cos(2x) & \sin^2(x) & \cos(4x) \\
\sin^2(x) & cos(2x) & \cos^2(x) \\
cos(4x) & \cos^2(x) & \cos(2x) \\
\end{vmatrix}\\
&=a_0+a_1\sin x +a_2\sin^2x+....
\end{align*}

In this question, I need to find the values of $a_0,a_1,a_2$.

Here's how I approached it, first, I put $x=0$ for both the LHS and RHS, this leaves me with only $a_0$ in the RHS and in the LHS, all the terms containing $\sin$ become $0$ and $\cos$ terms come out to be $1$, which yields the value $a_0=-1$

Then, I differentiated both sides of the equation and ended up with $3$ determinants (I differentiated column wise) and on the RHS, I got $\cos x$ as the coefficient of $a_1$, and now putting $x=0$, I also got that $a_1=0$.

But, I cannot get how to find out $a_2$, one way I thought was to divide the 1st derivative by $\cos x$ and then again differentiating but that complicates stuff a lot. Any help would be appreciated. The answer says that $a_2=18$

答:
Taking into account what you are asked for, let $t=\sin^{-1}(x)$ to make first
$$\Delta(t)=\left(
\begin{array}{ccc}
 1-2 t^2 & t^2 & 8 t^4-8 t^2+1 \\
 t^2 & 1-2 t^2 & 1-t^2 \\
 8 t^4-8 t^2+1 & 1-t^2 & 1-2 t^2
\end{array}
\right)$$ and now $u=t^2$. It is simple (done by hand).


问:
$\int_{\mathbb{R}}|f(t)|^2dt=\int_{\mathbb{R}}|f'(t)|^2dt$ implies $f(t)=\mathbb{x}_{i}|f(t)|$
Let $f \in C^{1}(\mathbb{R},\mathbb{R}^m)$ be such that $f$ and $f'$ are square integrable and
$$\{t:f(t)=0\} \subset \{t:f'(t)=0\}$$
$$ |\{t:f(t)=0\}|=n\in \mathbb{N}$$
Prove that if $\int_{\mathbb{R}}|f'(t)|^2\, dt=\int_{\mathbb{R}}|f|'^2(t)\,dt$, then there are $n+1$ points
$$\mathbb{x}_{1},\mathbb{x}_{2},...,\mathbb{x}_{n+1}\in S^{m-1}=\{\mathbb{x}\in \mathbb{R}^m:|\mathbb{x}|=1\} $$
such that for each $t \in \mathbb{R}$, $f(t)=\mathbb{x}_{i}|f(t)|$ for some $i$.

I got this problem from Postech Math Competition.

Thanks in advance.

答:
We have $ a_1<a_2<...<a_n \in \mathbb{R} $ such that $ f(a_i) = f'(a_i) = 0 $ and $f$ is non zero everywhere else. Let $I_1 = (-\infty,a_1), I_{j+1} = (a_j,a_{j+1}) $ for $ 1\leq j\leq n-1 $ and $I_{n+1} = (a_n,\infty) $.  Now if $ f = (f_1,....,f_m) $ then $ |f| = \sqrt{f_1^2+f_2^2+...f_m^2} $, hence $f$ is differentiable in $\cup_j I_j $ and
$$ |f|' = \frac{2f_1f_1' + ...+2f_mf_m'}{2\sqrt{f_1^2+...+f_m^2}} = \frac{f.f'}{|f|} $$
Thus Cauchy-Swartz Inequality gives that $ |f|'^2\leq |f'|^2 $. So $ |f'|^2-|f|'^2 \geq 0 $ for $ f\neq 0 $, is continuous and non-negative for $ t \neq a_i $. But if we have as given
$$ \int_\mathbb{R} (|f'(t)|^2-|f|'(t)^2 )dt = 0 $$ We can conclude from above that $ |f|'^2 = |f'|^2 $ in $\cup_jI_j $ hence equality for the C-S inequality holds except at those $n $ points. Thus we have $ \lambda_j \in \mathbb{R} $ such that $ f' = \lambda_j f $ in $ I_j $(from linear independence) and hence $|f|' = \lambda_j|f| $. Now we have a piecewise differentiable function $ g : \mathbb{R} \rightarrow S^{m-1}\cup \{0\} $ defined as
$$ g = \begin{cases} f/|f| &\mbox{if } f \neq 0 \\
0 & \mbox{if } f = 0 \end{cases} $$
So in $I_j$ we have
$$ g' = \frac{|f|f'-f|f|'}{|f|^2} = \frac{\lambda_j|f|f-\lambda_j|f|f}{|f|^2} = 0 $$
So $ g $ is piecewise constant,i.e there exists $ x_1,x_2,...,x_{n+1} $ such that $ g(t) = x_j $ for $ t \in I_j$. Thus $ f(t) = x_j|f(t)| $ for some $j$ for all $ t \in \mathbb{R} $.</p>


问:
What is the Type 1 error for the following test?
For testing the hypotheses H0: P = 0.8 versus H1: P ≠ 0.8 at 1% significance level, we
obtain a sample of n = 100 and p = 0.75.

Options:

A. 0.100

B. 0.140

C. 0.212  (Correct answer)

D. 0.010

E. 0.020

So far I have,

z= (P^-P)/std dev

= -1.25

za/2 = 2.575

Don't reject null

But I don't know how to get the answer

答:
The test statistic is $$Z \mid H_0 = \frac{\hat p - p_0}{\sqrt{p_0 (1 - p_0)/n}} = \frac{0.75 - 0.8}{\sqrt{0.8 (0.2)/100}} = -1.25.$$  This is approximately standard normally distributed, hence the two-sided $p$-value is $$\Pr[|Z| > 1.25 \mid H_0] \approx 0.2113.$$  The discrepancy may be due to the answer choice (C) being derived from a normal distribution table with limited decimal precision, whereas my value is generated from a computer.

That said, the Type I error of the test remains $\alpha = 0.01$.  This is because the $p$-value that was obtained would lead us to conclude not to reject the null.  The $p$-value is not the same thing as the Type I error of the test.  To understand the difference, the $p$-value would be the Type I error probability if the conclusion were to always reject $H_0$.  So, taking the $p$-value we calculated above, what that means is if we decided to conclude $p \ne 0.8$ based on the observed data, the probability that this is an error is about $0.2113$.  But because $\alpha = 0.01$ is already set, the conclusion is that you would not reject and that the sample furnishes insufficient evidence to suggest the true proportion is not $0.8$.  The Type I error is controlled at $0.01$ and in fact there is no point in computing the $p$-value of the test in order to know that the Type I error is $0.01$, because you don't need to actually perform the test in order to control this error probability through setting the significance level of the test.

In another vein, the test statistic we used above is only an approximation; specifically, it is the normal approximation to the binomial distribution.  The exact $p$-value corresponds to a binomial probability.  If $n = 100$ and the observed proportion is $\hat p = 0.75$, this means there were $X = 75$ "successes" observed.  Under the null hypothesis, $$X \mid H_0 \sim \operatorname{Binomial}(n = 100, p_0 = 0.8).$$  Due to the discrete nature of the test statistic, the actual Type I error will be less than the nominal level of the test $\alpha$, and  is computed by first finding a rejection region $\mathcal R_\alpha = [0, l] \cup [u, n]$ with $l$ as large as possible and $u$ as small as possible such that $\Pr[X \ge u \mid H_0] \le 0.005$ and $\Pr[X \le l \mid H_0] \le 0.005$.  The bounds $l, u$ could be found empirically by computer, but a good approach is to use the normal approximation to find initial guesses.  Since the critical value for a standard normal distribution at the $99.5^{\rm th}$ percentile is $z^*_{\alpha/2} = 2.57583$, we compute $$np_0 \pm z^*_{\alpha/2} \sqrt{np_0(1-p_0)} \approx (69.6859, 90.3141).$$  This suggests that we should guess $l \approx 70$ and $u \approx 90$.  Now using a computer, we find $$\sum_{x=0}^{70} \Pr[X = x \mid H_0] \approx 0.011249, \quad \sum_{x=90}^{100} \Pr[X = x \mid H_0] \approx 0.00569638.$$  The lower limit is too big and the upper limit is just a hair too small, so we decrement our guess for $l$ until we obtain $l = 68$, and increment our guess for $u$ until we get $u = 91$:  $$\sum_{x=0}^{68} \Pr[X = x \mid H_0] \approx 0.0031297, \quad \sum_{x=91}^{100} \Pr[X = x \mid H_0] \approx 0.00233356.$$  Hence our rejection region is $$\mathcal R_\alpha = [0, 68] \cup [91, 100].$$  Then the actual Type I error is the sum of these probabilities, $0.00546326 < 0.01$.  Notice how we did not actually need to perform the hypothesis test or compute a $p$-value.

That said, the $p$-value corresponds to computing $$\Pr[X \le 75 \mid H_0] + \Pr[X \ge x' \mid H_0],$$ where $x'$ is the smallest integer above $80$ such that $\Pr[X = x'] \le \Pr[X = 75] \approx 0.0438778$.  Using a computer, this gives us $x' = 86$, hence the two-sided $p$-value is $$\sum_{x = 0}^{75} \Pr[X = x \mid H_0] + \sum_{x=86}^{100} \Pr[X = x \mid H_0] \approx 0.211797.$$  This agrees very closely with our normal approximation.


问:
Calculus Derivatives In
What is the derivative of the function h(x)=In(f(x))?
I know it would end up being a fraction but I am unsure</p>

答:
$h(x)=\ln (f(x))$

Applying Chain Rule:$$h'(x)=\frac{f'(x)}{f(x)}$$

Title: Calculus Derivatives In


问:
Why can we assume these inequalities without loss of generality? (Measure Theory, the Lebesgue Integral)
I am reading "Introduction to the Lebesgue Integral" (in Japanese) by Yoiti Yosida.

There is the following proposition in this book:

Let $I := [a, b), I_i := [a_i, b_i)$ for $i \in \{1, \dots, n\}$.

Let $|I| := b-a, |I_i| := b_i-a_i$ for $i \in \{1, \dots, n\}$.

If $I \subset \bigcup_{i=1}^{n} I_i$, then $|I|\leq\sum_{i=1}^{n} |I_i|$.

And the proof is the following.

Proof:

Without loss of generality, we can assume $$a_1\leq a <b_1,$$$$a_n<b\leq b_n,$$$$a_{i+1}<b_i\leq b_{i+1} (i=1,\dots,n-1).$$
$$b-a\leq b_n-a_1=(b_1-a_1)+\sum_{i=1}^{n-1}(b_{i+1}-b_i)\leq \sum_{i=1}^{n} (b_i-a_i).$$

Why can we assume the above inequalities without loss of generality?

答: They are just ordering the intervals from left to right. There is one interval that contains $a$ and we may reaname that as $(a_1,b_1]$. Next $b_1$ must be in some interval and we may call that $(a_2,b_2]$ [unless the first interval already convers the whole of $[a,b]$]. And so on until you reach $b$. Note that $b_1 \in (a_2,b_2]$ gives $a_2 < b_1\leq b_2$ etc.


问:
Why $\frac{1}{2i}(e^{i\omega t} - e^{-i\omega t}) = \frac{i}{2} (e^{-i \omega t} - e^{i\omega t})$
Let $i := \sqrt{-1}$, $f$ be the frequency ($\frac1p$), and  $\omega := 2 \pi f$.

From page 3 here, why does $\frac{1}{2i}(e^{i\omega t} - e^{-i\omega t}) = \frac{i}{2} (e^{-i \omega t} - e^{i\omega t})$?

I understand that the LHS's $\frac{1}{2i}$ is multiplied by $\frac ii$ to get the RHS's $\frac i2$, but I don't understand how the contents of the parentheses changes?

答:
Just that $\frac{1}{i} = - i$, so by distributing the minus sign,
$$\frac{1}{2i}(e^{i\omega t} - e^{-i\omega t}) = -\frac{i}{2}(e^{i\omega t} - e^{-i\omega t}) = \frac{i}{2}(e^{-i\omega t} - e^{i\omega t}). $$</p>