问:
What is this notion in statistics?

Question and request

I'm sure the following idea is known, and must be basic, in statistics. I suspect it's related to the confidence interval (CI). However, CI is a statistical notion notorious of frequently being misinterpreted, so I cannot be sure. I hope you can point out what the following idea is, and also point out a rigorous account which supports your answer. The more rigorous the better.

The idea

Given a coin whose output set is $\{+,-\}$. Assume the distribution of the outcome is fixed. Assume moreover that after tossing it $10000$ times, exactly $7000$ times were $+$ and exactly $3000$ times were $-$.

Of course, we cannot assert that the underlying distribution is 70%:30%. However, given any possible distribution $p:(1-p)$ in terms of $p \in [0,1]$, we can calculated the probability, denoted by $\phi(p)$, of that "Exactly 7000 positive $+$ shows up in an 10000-time toss trial."

Define the subset $S:= \{p\in[0,1] \,|\, \phi(p) \geq 95\% \}$. What is this subset called?

答:
It is known as the 'empty set'.

Joking aside: the probability of getting EXACTLY 7000 out of 10000 is extremely small, even in the best case where the each coin independently has $p = .7$ chance of landing on $+$. Your function $\phi$ never exceeds 0.01 (in fact doesn't even get that high) so there are no values in its domain where it reaches $.95%$.

Related: suppose you want to test the hypothesis that the coin is fair (so that $p = .5$). Then of course tossing it 10000 times is a great way to settle this matter. However: the fact that $\phi(.5)$ is very small does not normally count as evidence against the Null hypothesis that $p = .5$. What would count as evidence is that the probability $\psi(p)$ of finding $70000$ or more $+$'s in 10000 tosses is still small for $p = .5$.

The set $S$ but with $\psi$ in the role of $\phi$ would be a (one-sided) confidence interval for the unknown true value of $p$ (though not the only possible choice of confidence interval!) after observing the data - the claim that we can reject our Null-hypothesis of $p$ being $0.5$ is equivalent to $0.5$ not lying in that interval.

By contrast: the fact that 0.5 does not lie in your interval $S$ proves nothing about the fairness of the coin since NO value of $p$ lies in your $S$.

I hope this is helpful. If you tell us some more about what intended use you had for $S$ I could perhaps give a more useful answer.

EDITED IN LATER: while your interval $S$ does not have its own name, it might be useful to note that your function $\phi$ does. It is called the likelihood of $p$. If you draw the graph of $\phi$ you see it reaches its highest value exactly where you expect it: at $p = 0.7$. Intuitively it is clear that the best estimate for $p$ you can give after seeing this data is indeed $\hat{p} = 0.7$. Now the fact that your function $\phi$ takes its largest value there helps you to justify this choice of estimator by using a pre-existing name for it: the maximum likelihood estimator.

Maximum likelihood estimators are widely used as estimators in statistics, also outside the context of coin tosses. They even have their own abbreviation: MLE.

You are looking for ways, it seem, to express how flat vs peaked the graph of $\phi$ is around its maximum. Confidence intervals do that, but not quite in the way you propose. I will think a bit about measures that are closer to your proposal. The conventional wisdom is of course that larger sample size (more coin tosses) give you more reliable estimates, so narrower intervals. This is true for confidence intervals. But in your case this might be tricky. It seems that picking the right value that may play the role of the .95 might itself depend on the sample size, making everything a bit messy. I will think a bit more about it.

Title: What is this notion in statistics?


问:
Why the proof of existence of Riemannian metric does not apply to pseudo Riemannian metric?
I just looked on the proof in Wikipedia Existence of Riemannian metric on smooth manifold, but I don’t see why this proof failed for pseudo-Riemannian metric, could anyone point out why this is not applicable for pseudo-Riemannian metric?</p>

答:
Just an example: a sum of two metrics of signature $(1,1)$ can be the zero metric:
$$\left(\begin{matrix}1 & 0 \\
0 & -1 \end{matrix}\right)+ \left(\begin{matrix}-1 & 0 \\
0 & 1 \end{matrix}\right)=\left(\begin{matrix}0 & 0 \\
0 & 0 \end{matrix}\right)$$

ContentLicense: CC BY-SA 4.0


问:
What is the radius of convergence of $\sum_{n=0}^\infty z^{n!}$?
Find the radius of convergence:
$$\sum_{n=0}^\infty z^{n!}$$

I suppose that this serie behaves in a similar way that $\sum_{n=0}^\infty z^{n}$ , so I have that it converges when $|z| < 1 $.

Could someone explain me if this approach is correct and give me a more rigorous proof?

Thanks in advance

答:
If $|z|<1$, then $|z|^{n!} \leq |z|^n$ so the series converges. If $z=1$, then the series obviously diverges.

So the radius is $1$.


问:
I know this solution but what happens when $ f(x) $ doesn't have derivative
Suppose a continuous function $ f \colon \mathbb{R} \rightarrow \mathbb{R} $ is not identically zero and satisfies the condition $$ f(x + y) = f(x)f(y) \ \ (x , y \in \mathbb{R}) $$
Prove that there exists a number $ a \in \mathbb{R} $ such that $ f(x) = a^x \ \ (x \in \mathbb{R})
$.

I wrote this :
Let's discus the case where $ y = 0 $; In this case the function will look like: $$ f(x + 0) = f(x) \cdotp f(0) $$ We see that: $$ f(x) = f(x) \cdotp f(0) \Rightarrow f(0) = 1 $$
Now let's find the derivative of $ f(x) $

$$ f^{'} (x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}   $$ using the initial formula we can change $ f(x + h) \text{ to } f(x) \cdotp f(h) $ $$  f^{'} (x) = \lim_{h \to 0} \frac{f(x) \cdotp f(h) - f(x)}{h}   $$
$$ f^{'} (x) = f(x) \cdotp \lim_{h \to 0} \frac{f(h) - 1}{h} $$
We can use the fact that: $ f^{'} (0) = \lim_{h \to 0} \frac{f(h) - 1}{h} $. It means that:
$$ f^{'} (x) = f(x) \cdotp f^{'} (0) $$
Let's suppose that $ f^{'} (0) = a $. After denoting $ f^{'} (0) \text{ by } a $, we get the following equation:
$$  f^{'} (x) = f(x) \cdotp a $$
Let's discus the derivative of: $$ f^{'} (x)^{\frac{1}{a}} = \frac{1}{a} \cdotp f (x)^{\frac{1}{a} - 1} \cdotp f^{'} (x) $$
If we plug in $  f^{'} (x) = f(x) \cdotp a $ in the main formula we will get this equation:
$$ f^{'} (x)^{\frac{1}{a}} = \frac{1}{a} \cdotp f (x)^{\frac{1}{a} - 1} \cdotp f(x) \cdotp a  $$
$$ f^{'} (x)^{\frac{1}{a}} = f (x)^{\frac{1}{a}} $$
If the derivative of $ f(x)^{\frac{1}{a}} $ equals to it's value, it means that this function is equal to $ e^x $. thus, we get the following equation: $$ a^{\frac{x}{a}} = e^x $$ $$ a^{\frac{1}{a}} = e $$

答:
Here's another idea: suppose $f(1) = a \neq 0$ and $f(x) = a^x g(x)$. Then $g(1) = 1$ and $g(x+y) = g(x)g(y)$. Note that $g$ is continuous and nonnegative, $g(0) = 1$ and $g(x+1) = g(x)$. We'd like to prove that $g \equiv 1$. Suppose that there exists an $r$ with $g(r) > 1$ (the other case is analogous). Then since $g$ is continuous, there is an interval $I$ on which $g > 1$. There thus exists a fraction $\frac{m}{n}$ such that $g(\frac{m}{n}) > 1$ since $\mathbb{Q}$ is dense in $\mathbb{R}$.

But then since $g(\frac{m}{n}) = g(\frac{1}{n})^m$, we have $g(\frac{1}{n}) > 1$ if $m\geq 0$ or $g(\frac{1}{n})\in (0,1)$ if $m<0$. Either way, $g(1) = g(\frac{1}{n})^n \neq 1$, contradiction.


问:
Proving that a group is closed under multiplication by scalars
I need to prove that the group $U = \{A \in M_n(\mathbb{R})\mid AB=BA\}$ is closed under multiplication by scalars.

So I let $\alpha \in \mathbb{R}$ and let $A \in U$, which means I need to prove that $\alpha A \in U$.

$(\alpha A)B = \alpha(AB) = \alpha(BA)$

But that does not prove what I need, I need to prove that $B(\alpha A) \in U$.

Where do I go from here?

答:
No, you are not asked to prove that $B(\alpha A)\in U$. What you should prove is the following:

if $A\in U$, then $\alpha A\in U$ for any $\alpha\in\mathbb{R}$,

or equivalently, if $AB=BA$, then $(\alpha A)B=B(\alpha A)$ for any scalar $\alpha$. This should be trivial because both sides of the last equality are equal to $\alpha(AB)$.

Alternatively, if we define $f:A\mapsto AB-BA$, then $f$ is a linear map defined on $M_n(\mathbb{R})$ and $U$ is its nullspace. Since every vector subspace is closed under scalar multiplication, the assertion follows.

ContentLicense: CC BY-SA 3.0


问:
How to prove that, if two numbers are equal $\mod2$ and$\mod3$ ,they are equal$\mod 6 $?
I was wondering if it could be proved without using the Chinese remainder theorem, arithmetically.

Thank you

答: $2,3$ are relatively prime so $2|(m-n)$, $3|(m-n)$ implies $6|(m-n)$ meaning $m\equiv n \mod 6$.


答:
If $c$ is a common multiple of $a$ and $b$, then $c$ is a multiple of their least common multiple $\operatorname{lcm}(a,b)$.

Apply this to $c=m-n$ and $a=2$ and $b=3$.

Title: How to prove that, if two numbers are equal $\mod2$ and$\mod3$ ,they are equal$\mod 6 $?

Body:
I was wondering if it could be proved without using the Chinese remainder theorem, arithmetically.

Thank you


问:
How to evaluate improper integral of odd function?
I recently was solving $\int_{0}^{\infty }\frac{x}{x^3+1}dx$. I wrote x as z to do contour integration and did the indentation for the pole at real axis but I don't know what to do with this odd function? How should I go further, write the z in terms of $\theta$ by euler formula? This is not homework question,I started learning complex analysis recently so I want to know the concepts used for such questions.

Body:
$$I=\int_{0}^{\infty} \frac{xdx}{1+x^3}=\int_{0}^{\infty}\left(\frac{-1}{3(1+x)}+\frac{(1+x)}{3(x^2-x+1)}\right)dx.$$
$$\implies I=\int_{0}^{\infty}\left(\frac{-1}{3(1+x)}+\frac{(2x-1)/2+3/2}{3(x^2-x+1)}\right)dx.$$
$$\implies I=\left .\left[\frac{-1}{3}\ln(1+x)+\frac{1}{6} \ln (x^2-x+1)\right]\right|_{0}^{\infty}+\frac{1}{2}\int_{0}^{\infty} \frac{dx}{(x-1/2)^2+3/4}$$
$$\implies I=\left .\left[\frac{1}{3}\ln\frac{\sqrt{x^2-x+1}}{1+x}+\frac{1}{\sqrt{3}} \tan^{-1}[(2x-1)/\sqrt{3}] \right] \right  |_{0}^{\infty}$$
$$\implies I=\frac{2\pi}{3\sqrt{3}}.$$
As $$\lim_{x\to \infty}\ln\frac{\sqrt{x^2-x+1}}{1+x}=0$$


问:
Examine the continuity and differentiability of $f(x)=\frac{\sin (\pi[x-\pi ])}{4+[x]^2}$

Prove that the function $$f(x)=\frac{\sin (\pi[x-\pi ])}{4+[x]^2}$$ where $[.]$ denotes the greatest integer function, is continuous as well as differentiable for all $x \in \mathbb R$

I tried to solve it like this:
$$\lim_{x\to c^+}f(x)=\lim_{x\to c^-}f(x)=f(c)$$
i.e.$$\lim_{x\to \frac{\pi}{2} ^+}f(x) =\lim_{h\to 0}f(\frac{\pi}{2}+h)=\lim_{h\to 0}\frac{\sin 0}{4+\pi^2}=0$$and similarly, $$\lim_{x\to \frac{\pi}{2} ^-}f(x) =\lim_{h\to 0}f(\frac{\pi}{2}-h)=\lim_{h\to 0}\frac{\sin -\pi}{4+(\pi-1)^2}=0$$

Therefore the function is continuous. Please check whether it is correct.
and help me check differentiability using

$$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}$$\

Thankyou

答: Your function $$f(x)=\frac{\sin (\pi[x-\pi ])}{4+[x]^2}$$ is equal to $0$ for all $x$ since sine of integer multiple of $\pi$ is $0$.</p>


问:
Intuitive Understanding of exponential distrbution

$\textbf{Problem}$: An athletic facility has $5$ tennis courts. Pairs of players arrive at the
courts and use a court for an exponentially distributed time with mean $40$ minutes.
Suppose a pair of players arrives and finds all courts busy and k other pairs waiting in
queue. What is the expected waiting time to get a court?

$\textbf{Answer}$: As long as the pair of players are waiting, all five courts are occupied by other players. When all five courts are occupied, the time until a court is freed up is exponentially distributed with mean $40/5=8 $ minutes. For our pair of players to get a court, a court must be freed up $k+1$ times. Thus, the expected waiting time is $8(k + 1)$.

The issue that I'm having in understanding is why does "The time until a court is freed up is exponentially distributed with mean $40/5=8$ minutes" rather than simply $40$ minutes. Since every court is independent and using the memoryless property.

On the contrary, if I look at $$Z=min(X_{1}, X_{2}, X_{3}, X_{4}, X_{5})$$. where $X_{i}$ is time taken by a pair to play in court $i$ and is exponentially distributed time with mean $40$ minutes.

Then $Z$ itself becomes and exponentially distributed with mean $40*5 = 200$ minutes.

I am unable to understand it intuitively. Any help is appreciated.

答: Each $X_i$ has rate $\lambda=\frac1{40}$ (with minutes as the time unit), so $Z$ has $\lambda=\frac5{40}=\frac18$ for a mean of $8$ minutes. You have applied the formula incorrectly: the rates add, not the means.


问:
Solving linear equations including semi-positive definite matrix
Suppose $A \in \mathbb{R}^{n \times n}$ is a positive semi-definite matrix. Can we solve the following for $x \in \mathbb{R}^n$?

\begin{equation}
\begin{split}
A(x-a) + \lambda I_{\bullet i}&=0\\
I_{\bullet i}^Tx&=0,\,\,\,\text{or }(x_i=0)
\end{split}
\end{equation}
where $\lambda \in \mathbb{R}$, $a \in \mathbb{R}^n$, $I_{\bullet i}$ is the i-th column of the identity matrix, and $i$ is a fixed and given index.
I did the following but cannot go further.

$x^TA(x-a) + \lambda x^TI_{\bullet i}=0$

$x^TA(x-a)=0$. Hence, $x- a \in N(A)$ or $x\in a+ N(A)$ and $\lambda=0$.

Also, $(x-a)^TAx=0$. Hence, $x \in N(A)$ and $Aa = \lambda I_{\bullet i}$ (I have no idea how to find $\lambda$ in this case).

Is there any other solutions for $x$ and consequently for $\lambda$?

答:
We write $e_i$ for $I_{\bullet i}$. There are two possibilities:

$e_i\not\in\operatorname{range}(A)$, i.e. $AA^+e_i\ne e_i$. Since the LHS of $A(x-a)=-\lambda e_i$ lies inside $\operatorname{range}(A)$ but $e_i$ does not, we must have $\lambda=0$. The system of equations can then be rewritten as
$$
\pmatrix{A\\ e_i^T}x=\pmatrix{Aa\\ 0}.
$$
It is solvable if and only if $x=\pmatrix{A\\ e_i^T}^+\pmatrix{Aa\\ 0}$ is a solution.

$e_i\in\operatorname{range}(A)$, i.e. $AA^+e_i=e_i$. Then $e_i^TA^+e_i\ne0$ and one may verify that $(\lambda, x)=\left(\frac{a_i}{e_i^TA^+e_i},\ a-\lambda A^+e_i\right)$ is a solution.


问:
Is a closed set in $\ell^1$ closed in $\ell^2$?
Prove or disprove that the set $S= \{x=(x_1,x_2,...)\in \ell^2 :\displaystyle{\sum_{i=1}^{\infty}}x_i=1,x_i\geq0\}$ is a closed subset of $\ell^2$. Where $\ell^p$ is the sequence space equipped with the p- norm, $\|x\|_p=(\displaystyle{\sum_{i=1}^{\infty}}|x_i|^p)^{1/p}$, for $1\leq p<\infty$.

I proved that $S$ is a closed subset of $\ell^1$. But I don't know wether a closed set in $\ell^1$ is closed in $\ell^2$ or not.
Thanks in advance.

答: It is not closed in $l^{2}$. Consider the elements $(\frac 1 n,\frac 1 n,...,\frac 1 n,0,0,...)$ where $\frac 1 n$ is repeated $n$ times. These are elements of $S$ and they converge in $\ell^{2}$ norm to $(0,0,...)$ which is not in $S$.


问:
memorylessness of geometric distribution
We are given random variables  $X_k=I_k \cdot B_k$, where $I_k$ is Bernoulli with parameter $p$  and $B_k$ geometric with parameter $q$. So it holds: $$P(X_k=i)= p \cdot (1-q)^{i-1} \cdot q $$

Consider: $$Y_k =c+k-\sum_{j=1}^k X_k$$ and the stopping time: $$\tau=\inf\{ n\geq 1: Y_n \leq 0\} $$ for $k \in \mathbb{N}$ and $c\geq 1$

I want to prove: $$P(\tau=k, -Y_\tau=j)=P(\tau=k) \cdot (1-q)^j q$$

My approach: Rewrite

$$P( -Y_{\tau}=j \mid \tau=k) \cdot P(\tau=k) =\\
P( -Y_k=j) \cdot P(\tau=k)$$

Consider the first factor: The definition of $\tau$ tells us that:

$$Y_{k-1}>0$$  and therefore $$X_k > Y_{k-1}>0$$

This implies:

$$P(-Y_{k-1} + X_k-1=j \mid X_k > Y_{k-1}>0) \\= P(X_k =1+j+Y_{k-1}\mid X_k > Y_{k-1}>0, X_k>0) \\=P(X_k=j+1, X_k>0) =  (1-q)^j q$$

Can I argue like that?

Edit:
$$P( -Y_{k}=j \mid \tau=k) \overset{!}{=}P ( -Y_{k}=j )$$
The events are not independent, but if I plug in $\tau=k$, the event $\{\tau=k\}$ would not give further information?

答:
Here is the proof: Allthough not directly stated, I am going to assume $j\in\mathbb{N}$. First of all, as you have correctly followed: $$\textbf{P}(-Y_\tau=j,\tau=k)=\textbf{P}(-Y_k=j,\tau=k)$$ Now we proof the claim by induction over $k\in\mathbb{N}$

Let $k=1$. In this case, $$\textbf{P}(-Y_k=j,\tau=k)=\textbf{P}(Y_1=-j,\tau=1)=\textbf{P}(Y_1=-j),$$ since $Y_1=-j$ implies $\tau=1$. Since \begin{align*}\textbf{P}(\tau=1)&=\textbf{P}(Y_1\leq0)=\textbf{P}(c+1-X_1\leq0)=\textbf{P}(X_1\geq c+1)\\&=1-\textbf{P}(X_1\leq c)=1-\sum^c_{i=0}\textbf{P}(X_1=i)\\&=1-(1-p)-\sum^c_{i=1}\textbf{P}(X_1=i)\\&=p-pq(1-(1-q)^c)/(1-(1-q))=p(1-q)^c\end{align*} and \begin{align*}\textbf{P}(Y_1=-j)=\textbf{P}(c+1-X_1=-j)=\textbf{P}(X_1=c+1+j)=pq(1-q)^{c+j},\end{align*} the claim follows for $k=1$.

We are trying to prove that $k\Rightarrow k+1$. Since the structure of random walks is inherently recursive, conditioning on the first step should work (let the index of the probability measure denote the start of the random walk):

\begin{align*}\textbf{P}_c(\tau=k+1)&=\sum^{c}_{i=0}\textbf{P}_c(\tau=k+1|X_1=i)\textbf{P}_c(X_1=i)=\sum^{c}_{i=0}\textbf{P}_{c-i+1}(\tau=k)\textbf{P}_c(X_1=i)\\&=\sum^{c}_{i=0}\textbf{P}_{c-i+1}(\tau=k,Y_k=-j)\textbf{P}_c(X_1=i)(1-q)^{-j}/q\\&=\sum^{c}_{i=0}\textbf{P}_{c}(\tau=k+1,Y_{k+1}=-j|X_1=i)\textbf{P}_c(X_1=i)(1-q)^{-j}/q\\&=\textbf{P}_c(\tau=k+1,Y_{k+1}=-j)(1-q)^{-j}/q.\end{align*}

The first equation follows from the law of total probability, the second uses the fact that $(Y_n)_{n\in\mathbb{N}}$ is a homogenous markov chain, the third  uses the induction hypothesis. From there we repeat all of theses steps in reverse.


问:
Is this topology separable, first countable, second countable?
Consider a set $E$ with some point $p \in E$ and the topology:
$T = \left \{ \emptyset \right \}\cup \left \{ C \subset E: p \in E \right \}$.
Is this topology, separable, first countable or second countable?

I feel like it depends on the size of $E$, because $|P(E)| = 2^{|E|}$ but I'm struggling to figure out what a base for this topology would look like which makes it difficult to answer the questions about countability. Any solutions or hints would be appreciated!

答:
This is knwon as the
particular point tpology. You can find all you need in the wikipedia page. By the way $p$ is always dense so $T$ is always separable, for any $x$ the set $\{x,p\}$ is a local basis at $x$, so $T$ is first countable. As you mention, the second countability depends on the cardinality of $E$.


答: Okay, so I'm still not sure about countability but the set $\left \{ p \right \}$ is in fact dense because the only closed set containing $p$ is $E$ itself meaning that the closure $\bar{\left \{ p \right \}}$ is the intersection of only one set $E$ meaning that the set is dense and because the set is countable (finite in fact), the space is separable.

Title: Is this topology separable, first countable, second countable?

Body:
Consider a set $E$ with some point $p \in E$ and the topology:
$T = \left \{ \emptyset \right \}\cup \left \{ C \subset E: p \in E \right \}$.
Is this topology, separable, first countable or second countable?

I feel like it depends on the size of $E$, because $|P(E)| = 2^{|E|}$ but I'm struggling to figure out what a base for this topology would look like which makes it difficult to answer the questions about countability. Any solutions or hints would be appreciated!


问:
Reference request: overdetermined polynomial systems are almost surely inconsistent
If one has $m$ multi-variate polynomial equations with $n$ variables, and $m>n$ one obtains an overdetermined system.
According to wikipedia here "Most but not all overdetermined systems, when constructed with random coefficients, are inconsistent." Is this equivalent to say that among the space of coefficients for these polynomials, the set of overdetermined systems with solution is of measure zero? Or equivalently, that when choosing random coefficients, overdetermined systems are inconsistent almost surely?

If this is the case. Do you know a canonical reference (book/article) for this fact for me to cite? (The references in the wikipedia page didn't helped much).

Why I started a bounty? while the answer by @orangeskid (and the reference he provided) is awesome in explaining why this fact is true, I am limited in space in the manuscript I'm writing and I cannot add much more detail to it.
Note that the manuscript I'm writing lies in an area completely different from algebraic geometry and resultant theory, so introducing the resultant there might be confusing for the reader (And my advisor did not liked it). That is why I would like to keep as less details as I can regarding why the fact is true, while being sure that the sentence is true and correctly referenced.

Ideally, I would like to know if there is a reference one could use for a sentence like

Recall that the set of over-determined polynomial systems of equations are inconsistent for most coefficients used in them (in the sense of the Lebesgue measure for the coefficients space). [INSERT CITATION HERE]

Or something like that. Perhaps being still an amateur, I wasn't able to see if something in the reference @orangeskid provided can be cited directly. Maybe there is no other way but to add more detail (which I would like to avoid) as in the answer from @orangeskid but I would like to hear your suggestions!

答:
This involves the resultant of $N$ homogenous forms in $N$ variables.

For consider $n+1$ polynomials in $n$ variables $t_1$, $\ldots$, $t_n$. Homogenize and get $N$ forms in $N$ variables $x_0$, $\ldots$, $x_n$ ( $N=n+1$).

If the $n+1$ polynomials have a common solution then the $n+1$ homogenous forms have a  solution ( not at infinity). It follows that their resultant is $0$  (the converse : zero resultant implies there exists a solution, but it can be "at infinity"). Nevertheless, we see that if the polynomials have a common solution, then some non-zero polynomial in their coefficients is $0$.  But the set of zeroes of the resultant is small ( for instance, it has measure $0$).

A concrete example: consider a system of $n+1$ linear equations with $n$ unknowns $A t = b$. If it has a solution then the extended $(n+1)\times (n+1)$ matrix $(A,b)$ has $0$ determinant.

$\bf{Added:}$

Perhaps it is worth adding the proof of the following result.

If $P\in \mathbb{R}[X_1, \ldots, X_n]$ is a non-zero polynomial then the zero-set of $P$
$$Z(P) \colon = \{t=(t_1,\ldots, t_n)\in \mathbb{R}^n\ | \ P(t) = 0\}$$
has Lebesgue measure $0$.

Decompose $P$ into homogenous components:
$P=P_0+ \cdots + P_d$

$P_d$ homogenous of degree $d = \deg P$.  Assume that $P_d$ contains the monomial $x_n^d$ with a non-zero coefficient. Otherwise, apply a linear transformation of coordinates so this becomes the case. So we may assume
$$P(x) = x_n^d + Q_1(x_1,\ldots, x_{n-1}) x_n^{d-1} + \cdots + Q_d(x_1,\ldots, x_{n-1})$$

Now, the intersection of $Z(P)$ with any line vertical line $(x_1, \ldots, x_{n-1}) = (a_1, \ldots, a_{n-1})$ has at most $d$ points, and so measure $0$ in that line. Now apply Fubini.


问:
Find general solution of series.
I'm trying to find the series solution of:

$$2x^2y''+3xy'-(x^2+1)y=0$$

so far I found the indicial polynomial: $2k^2+k-1=0$ , so $k=1/2$ and $k=-1$.

Then I found the recurrence relation: $$a_n=\frac{a_{n-2}}{(2n+2k-1)(n+k)-1}$$

Now if I plug in $k=-1$: $$a_n=\frac{a_{n-2}}{2n^2+5n+2}$$

And if I plug in $k=1/2$: $$a_n=\frac{a_{n-2}}{2n^2+n-1}$$

I now $a_0$ and $a_1$ are free variables. But I'm confuse in how do I find the 2 independent series solutions? How do I find the coefficients?

Thanks!

答: The series for $k=-1$ and $k=\frac12$ are different objects, it is a little bit disingenuous to use the same symbols for their coefficients in close proximity. In both series the coefficient $a_0$ is free, this is the result of solving the indicial equation. The next equation only contains $a_1$ with a non-zero coefficient, so that $a_1=0$ in both cases, implying that all odd-index coefficients are zero. This means that you only get two independent solutions.

Title: Find general solution of series.


问:
Finding the power series representation of $\frac{x^2+1}{8x-1}$
So, I'm honestly at a loss of what I am doing wrong here. I know that the $x^2+1$ needs to be factored out of the summation, but apparently, this was not the answer to the first text box. Also, I'm not sure how to find the rate in this particular geometric series. I have tried to factor out $-1$ to make the rate $(8x)$, but I am not sure if this is even the right approach. Where did I go wrong in this problem?

答:
\begin{align*}
  & \frac{x^2+1}{8x-1} \\
    & \\
    & = \frac{\color{red}{x^2 - \frac{1}{8}x} + \frac{1}{8}x + 1}{\color{red}{8x-1}} = \color{red}{\frac{x}{8}} +  \frac{\frac{1}{8}x + 1}{8x-1}  \\
     & \\
    & = \frac{x}{8} + \frac{\color{blue}{\frac{1}{8}x - \frac{1}{64}} + \frac{65}{64}}{\color{blue}{8x-1}} = \frac{x}{8} + \color{blue}{\frac{1}{64}} + \frac{65}{64}\frac{1}{8x-1} \\
     & \\
    & = \frac{x}{8} + \frac{1}{64} - \frac{65}{64}\frac{1}{1-8x} = \frac{x}{8} + \frac{1}{64} - \frac{65}{64}(1-8x)^{-1}, \\
     & \\
  \end{align*}

and you can expand $(1-8x)^{-1} = (1+(-8x))^{-1}$ using the Binomial expansion, or Geometric series, as Sil pointed out in the comments.

If you're not comfortable with the method I used above, you can instead use polynomial long division:

$8x-1 \ \overline{)x^2 + 0x + 1}$

although my method is in fact polynomial long division in disguise.

$$$$

It looks like you went the quickest route and skipped the fraction/polynomial long division steps entirely:

$\frac{x^2+1}{8x-1} = \frac{-x^2-1}{1-8x} = (-x^2-1) \left(1 + (8x) + (8x)^2 + ...\right) = \ ...$

which is correct, but it isn't in the form required in your worksheet.


问:
How can I evaluate this integral using the delta function?
I am unsure how to answer this question using the delta function. I am unfamiliar with the summation inside of the integral

答:
The sum of delta functions is sometimes called a Dirac comb. Here, this sum is $1$ when $x$ is a nonnegative even integer ($0,2,4,\dots$) and $0$ otherwise. The integral is therefore turned into an infinite sum, since $e^{-x}$ is only evaluated where the comb returns $1$:
$$\int_{-\infty}^\infty e^{-x}\sum_{n\ge0}\delta(x-2n)=\sum_{n=0}^\infty e^{-2n}=\frac1{1-e^{-2}}$$


答: Considering that the function $z\mapsto z^3 + 1$ is invariant under rotation by $\frac{2\pi}{3}$, it will be useful to start with the contour integral $$\oint_{\Gamma(R)} \frac{z}{z^3 + 1}\, dz$$ where $\Gamma(R)$ is the positively oriented sector of radius $R$ in the first and second quadrants subtended by angle $\frac{2\pi}{3}$. Namely, $\Gamma(R) = [0,R] \cup \gamma_R \cup L_R$ where $\gamma_R$ is the arc of radius $R$ and $L_R$ is the line segment from $Re^{2\pi i/3}$ to $0$. For large enough $R$ the sector contains the simple pole at $z = e^{i\pi/3}$. By the residue theorem the contour integral is $2\pi i$ times the residue of $\frac{z}{z^3 + 1}$ at $z = e^{\pi i/3}$, which evaluates to $$2\pi i \frac{e^{\pi i/3}}{3(e^{\pi i/3})^2} = \frac{2\pi i}{3 e^{\pi i/3}}$$ The integrand is $O(R^{-2})$ as $R\to \infty$ so by the ML-inequality its integral over the arc $\gamma_R$ is $O(R^{-1})$ as $R\to \infty$. Now using parametrization $z = xe^{2\pi i /3}$, $0 \le x \le R$, we write $$\int_{L_R} \frac{z}{z^3 + 1}\, dz = -\int_0^R \frac{xe^{2\pi i/3}}{x^3 + 1}\, e^{2\pi i/3}\, dx = -e^{4\pi i/3}\int_0^R \frac{x}{x^3 + 1}\, dx$$ Therefore $$\frac{2\pi i}{3e^{\pi i/3}} = (1 - e^{4\pi i/3})\int_0^R \frac{x}{x^3 + 1}\, dx + O(R^{-1})\quad \text{as $R\to \infty$}$$From this we deduce $$\int_0^\infty \frac{x}{x^3 + 1}\, dx = \frac{2\pi}{3\sqrt{3}}$$

Title: How to evaluate improper integral of odd function?

Body: I recently was solving $\int_{0}^{\infty }\frac{x}{x^3+1}dx$. I wrote x as z to do contour integration and did the indentation for the pole at real axis but I don't know what to do with this odd function? How should I go further, write the z in terms of $\theta$ by euler formula? This is not homework question,I started learning complex analysis recently so I want to know the concepts used for such questions.


问:
Measures associated to polynomial complex dynamics
This is probably easy questions but I could not find a good explanation for myself.

When I'm reading about polynomial complex dynamics, I found a technical term measure-valued Laplacian for a Green's function associated to a polynomial.

Many of books/authors say that it's Laplacian $\color{blue}{in\,\, the\,\, sense \,\,of\,\, distribution}$.

My question is what does it means to be Laplacian in the sense of distribution?

Plus, this defines a measure so-called $\color{blue}{harmonic\,\,measure}$. Actually, I have seen these terms used interchangeably with $\color{blue}{equilibrium\,\,measure}$.

Can someone explain me these notion of measures? I think I understand that the equilibrium measure means in the sense of potential theory which minimizes the energy integral. But, the harmonic measure I'm not quite get it (both in terms of definition and usage).

答:
Distributions are roughly speaking generalized functions. Measures are particular examples of distributions. This is a technical subject, and I won't try to give here a precise definition of distributions.

One of the greatest advantage of distributions is that you can always differentiate a distribution and get another distribution. Here, Laplacian in the sense of distribution means: take a function $u$, think of it as a distribution, apply the the Laplacian, and you get a distribution $\Delta u$. It turns out that if $u$ is a subharmonic function, this procedure gives you in fact a measure $\Delta u$.

The equilibrium measure, as you say, is defined as maximizing the energy. The harmonic measure of a compact $K \subset \mathbb C$ is defined as $\Delta u$, where $u$ is the Green function associated to $K$. It is subharmonic, and so $\Delta u$ is indeed a measure. Again, it turns out that these two things are the same.

I don't know what you've been reading to prompt these questions, by Ransford's book "Potential theory in the complex plane" is a very good place to find more detailed answers.


问:
How to determine if Standard Deviation is high/low
I have derived the following response time data for a performance test I am running:

Min - 8sec
Max - 284sec
Average - 28sec
Standard Deviation - 27sec

What does the standard deviation say about the response time data distribution? When you say low/high standard deviation, what does this actually mean? Is this in comparison to the Average/Min/Max?

I know what standard deviation is and how it's computed. I'm just not sure how to tell if it is high or low.

答:
When trying to figure this out myself I opted for using the Std deviation as a percentage of the range. Looking at this graph:

from that image I would I would say that the SD of 5 was clustered, and the SD of 20 was definitionally not, the SD of 10 is borderline.

More mathematically,

The SD of 5 has 68% of the values within 10% of the range

The SD of 10 has 68% of the values within 20% of the range

The SD of 20 has 68% of the values within 40% of the range

So as a purely internal measure of High / Low Std deviation I chose to say if the SD was less than 10% of the range then its low, greater than 10% of the range then high.

But you could of course choose different percentages based on your own data sets.

https://www.scribbr.com/statistics/standard-deviation/


问:
Showing that, if a conic represents an empty set, then $\Delta \neq 0$ and $h^2<ab$.
In one of the books I referred for conic sections, it gave a statement

If a conic represents an empty set, then $\Delta \neq 0$ and $h^2<ab$.

Can someone please explain this statement..? What is the meaning of a conic representing an empty set, actually..?

答:
If $a,b,c,f,g,h$ are chosen randomly, then the chance that the quadratic (1) will represent a null set is the highest. There is no unique parametric condition. See three cases below

The equation of conic/ pair of lines is given by $$ax^2+by^2+2hxy+2gx+2fy+c=0~~~~(1)$$
See it as a quadratic of $x$ and treat $y$ as constant,
then
$$x=\frac{-(hy+g)\pm \sqrt{(h^2-ab)y^2+2(gh-af)y+g^2-ac}}{a}~~~~(2)$$
Eq. (1) will represent null set if the discriminant is negative drfinite (for all real values of $y$). $f(x)=Ax^2+Bx+C \le 0 ~\forall~ y\in R ~~if~~
A<0~~and ~~ \Delta=B^2-4AC<0$. This implies
$$h^2<ab~~~(3)$$
and
$$\Delta=(gh-af)^2-(h^2-ab)(g^2-ac)=abc+2fgh-af^2-bg^2-ch^2 <0~~~~(4).$$
(3) and (4) note $\Delta <0$ not ($\Delta \ne 0).$ (Case 1).

When $h^2=ab, gh=af$ but $g^2<ac$, we will have $\Delta=0$ so two parallel (imaginary)lines (2). This also corresponds to a null set. (Case 2)

Let $g^2-ac=0, gh-af=0, h^2<ab$, then you get $\Delta=2fgh-af^2-ch^2$ may be positive or negative. One again gets two (imaginary) lines (1). (Case 3)

Case 3, justifies OP's obervation in a book.


问:
How to find PDF and expected value of $\max(x,0)$ for random variable $x$
Let $a,b,c, \tau$ be positive constants and $x$ is an exponentially distributed variable with parameter $\lambda = 1$, i.e. $x\sim\exp(1)$.

$$E = \tau\Big[a\frac{1+a}{1+e^{-bx+c}} - 1 \Big]^+$$

where $[z]^+ = \max(z,0)$

How can I find

The PDF for $E$

The expectation of $E$.

答:
first calculate the interval where $Z=\frac{a(a+1)}{1+e^{-bX+c}}-1\leq0$

The probability of this interval is a mass probability assigned to $Z=0$. The remainig probability is distributed over $Z$ according to the specified random law. The resulting "PDF" is a mixed density, not absolutely continuous.

Finally calculate  the density of $Y=\tau[Z]^+$ by a monotonic transformation and its expectation in the way you are used to do

Do not forget that Random Variables need Capital Letters...


问:
$\frac{a^{2}-1}{b+1}+\frac{b^{2}-1}{a+1}$ an integer $\Rightarrow \frac{a^{2}-1}{b+1}$ and $\frac{b^{2}-1}{a+1}$ are integers.
If $\frac{a^{2}-1}{b+1}+\frac{b^{2}-1}{a+1}$ is an integer, prove that also $\frac{a^{2}-1}{b+1}$ and $\frac{b^{2}-1}{a+1}$ are integers.

By doing the math, I get $\frac{(a-1)(a+1)^{2}+(b-1)(b+1)^{2}}{(a+1)(b+1)}$ is an integer which means that $(a+1)(b+1)$ divides $(a-1)(a+1)^{2}+(b-1)(b+1)^{2}$.

but I don't know how to continue.

By the way, this is not homework or anything. I just found it in a book.

答:
Call the fractions $r,s.\,$ Note $\,rs = (a\!-\!1)(b\!-\!1) := m\in \Bbb Z,\,$ so if $\,r\!+\!s = n\in\Bbb Z\,$ then $\,r,s\,$ are roots of $\, (x-r)(x-s)\, =\ \color{#c00}1\cdot x^2 - n\, x + m\,$
so $\,r,s\in \Bbb Z\,$ by Rational Root Test (i.e. if $\,c/d\,$ is a reduced rational root then $\,d\mid\color{#c00}1,\,$ so $\,d\!=\!1,\,$ so $\,c/d\in\Bbb Z)$.

Generally this shows $\,\bbox[5px,border:1px solid #c00]{\text{two rationals with integer sum & product must be integers}}$


问:
A problematic integral: $\int_0^{2\pi} e^{-2\pi i\lambda\cos(t)}\,dt$
Is there a special trick to calculate this integral?
$$\int_0^{2\pi} e^{-2\pi i\lambda\cos(t)}\,dt$$
for $\lambda>0$.</p>

答:
The Bessel function of the zeroth order can be represented by the integral:
$$J_0 (z) = \frac{1}{\pi} \int_{0}^\pi e^{i z \cos(t)} dt$$
Since the $i\sin(z\cos x)$ part of the exponent is zero over the integral until $2\pi$, only the $\cos(z \cos x))$ remains. This term remains identical through $x\to x+\pi$, so that:
$$\int_{0}^{2\pi} e^{-2\pi i \lambda \cos(t)} dt = 2\pi J_0(2\pi\lambda)$$
As for the original integral, for a proof that this is indeed a solution of Bessel's ODE, see here.</p>


问:
Divisibility of cyclic sums
Lately I have been studying the divisibility of some cyclic sums, and I was wondering about the following

Conjecture

Let it be a set of distinct positive integers $S=\{x_1,x_2,...,x_n\}$ such that $2\leq{x_1}<x_2<...<x_n$. Then,

$$\prod_{k=1}^{n}x_{k}\nmid\left(\sum_{cyc}\left(\prod_{k=1}^{n-1}x_{k}\right)+\sum_{cyc}\left(\prod_{k=1}^{n-2}x_{k}\right)+...+\sum_{k=1}^nx_k\right)$$

I would appreciate any help regarding this conjecture proof or refutation.

I already noted that there exist sets of distinct integers $S=\{x_1,x_2,...,x_n\}$ such that $$\prod_{k=1}^{n}x_{k}\mid\sum_{cyc}\left(\prod_{k=1}^{n-1}x_{k}\right)$$

For example, for $S=\{2,3,6\}$, $$2*3*6\mid (2*3)+(3*6)+(6*2)$$

Thanks in advance!

EDIT

I found that the conjecture is true if $\min\{x_1,x_2,...,x_n\}>n$, as it follows from the following

Proof

Assuming $x_{1}=x_{2}=...=x_{n}=n+1$, we get that

$$\prod_{k=1}^{n}x_{k}=\left(n+1\right)^{n}$$

$$\left(\sum_{cyc}\left(\prod_{k=1}^{n-1}x_{k}\right)+\sum_{cyc}\left(\prod_{k=1}^{n-2}x_{k}\right)+...+\sum_{k=1}^{n}x_{k}\right)=n\left(\left(n+1\right)^{n-1}+\left(n+1\right)^{n-2}+...+\left(n+1\right)\right)$$

$$n\left(\left(n+1\right)^{n-1}+\left(n+1\right)^{n-2}+...+\left(n+1\right)\right)=\left(\left(n+1\right)-1\right)\left(\left(n+1\right)^{n-1}+\left(n+1\right)^{n-2}+...+\left(n+1\right)\right)$$

$$\left(\left(n+1\right)-1\right)\left(\left(n+1\right)^{n-1}+\left(n+1\right)^{n-2}+...+\left(n+1\right)\right)=\left(n+1\right)^{n}-\left(n+1\right)$$

Subsequently,

$$\left(\sum_{cyc}\left(\prod_{k=1}^{n-1}x_{k}\right)+\sum_{cyc}\left(\prod_{k=1}^{n-2}x_{k}\right)+...+\sum_{k=1}^{n}x_{k}\right)=\left(n+1\right)^{n}-\left(n+1\right)$$

Thus,

$$\prod_{k=1}^{n}x_{k}>\left(\sum_{cyc}\left(\prod_{k=1}^{n-1}x_{k}\right)+\sum_{cyc}\left(\prod_{k=1}^{n-2}x_{k}\right)+...+\sum_{k=1}^{n}x_{k}\right)$$

And therefore,

$$\prod_{k=1}^{n}x_{k}\nmid\left(\sum_{cyc}\left(\prod_{k=1}^{n-1}x_{k}\right)+\sum_{cyc}\left(\prod_{k=1}^{n-2}x_{k}\right)+...+\sum_{k=1}^{n}x_{k}\right)$$

Subsequently, the conjecture is true if $\min\{x_1,x_2,...,x_n\}>n$.

Title: Divisibility of cyclic sums

Body:
The conjecture is true for $n=2$ since
$$\small x_1+x_2=kx_1x_2\implies (kx_1-1)(kx_2-1)=1\implies kx_1-1=kx_2-1\implies x_1=x_2$$

The conjecture is false for $n=3$ since for $(x_1,x_2,x_3)=(2,4,14)$, we have$$\frac{x_1x_2+x_2x_3+x_3x_1+x_1+x_2+x_3}{x_1x_2x_3}=1$$

The conjecture is false for every $n\ge 3$.

Proof :

Let $$S(x_1,x_2,\cdots,x_n):=\sum_{cyc}\left(\prod_{k=1}^{n-1}x_{k}\right)+\sum_{cyc}\left(\prod_{k=1}^{n-2}x_{k}\right)+...+\sum_{k=1}^nx_k$$

We have
$$S(x_1,x_2,\cdots,x_n)=S(x_1,x_2,\cdots,x_{n-1})+x_nS(x_1,x_2,\cdots,x_{n-1})+x_n+\prod^{n-1}_{k=1}x_k$$

So, taking
$$x_n=S(x_1,x_2,\cdots,x_{n-1})+\prod^{n-1}_{k=1}x_k$$
we get
$$\frac{S(x_1,x_2,\cdots,x_n)}{\displaystyle\prod_{k=1}^{n}x_k}=\frac{2+S(x_1,x_2,\cdots,x_{n-1})}{\displaystyle\prod_{k=1}^{n-1}x_k}=\frac{2+S(x_1,x_2,\cdots,x_{n-2})+x_{n-1}S(x_1,x_2,\cdots,x_{n-2})+x_{n-1}+\displaystyle\prod^{n-2}_{k=1}x_k}{\displaystyle\prod_{k=1}^{n-1}x_k}$$

Now, taking
$$x_{n-1}=2+S(x_1,x_2,\cdots,x_{n-2})+\prod^{n-2}_{k=1}x_k$$
we get
$$\frac{S(x_1,x_2,\cdots,x_n)}{\displaystyle\prod_{k=1}^{n}x_k}=\frac{2+S(x_1,x_2,\cdots,x_{n-1})}{\displaystyle\prod_{k=1}^{n-1}x_k}=\frac{2+S(x_1,x_2,\cdots,x_{n-2})}{\displaystyle\prod_{k=1}^{n-2}x_k}$$

Similarly, taking
$$x_{i}=2+S(x_1,x_2,\cdots,x_{i-1})+\prod^{i-1}_{k=1}x_k$$
for $i=3,4,\cdots,n-2$, we get
$$\frac{S(x_1,x_2,\cdots,x_n)}{\displaystyle\prod_{k=1}^{n}x_k}=\frac{2+S(x_1,x_2)}{\displaystyle\prod_{k=1}^{2}x_k}=\frac{2+x_1+x_2}{x_1x_2}$$
which equals $1$ when $(x_1,x_2)=(2,4)$.

So, a counterexample is
$$x_i=\begin{cases}S(x_1,x_2,\cdots,x_{n-1})+\displaystyle\prod^{n-1}_{k=1}x_k&\ \ \text{if $\ \ i=n$}
\\\\2+S(x_1,x_2,\cdots,x_{i-1})+\displaystyle\prod^{i-1}_{k=1}x_k&\ \ \text{if $\ \ 3\le i\le n-1$}
\\\\4&\ \ \text{if $\ \ i=2$}\\\\2&\ \ \text{if $\ \ i=1$}\end{cases}$$
where
$$S(x_1,x_2,\cdots,x_n)=\sum_{cyc}\left(\prod_{k=1}^{n-1}x_{k}\right)+\sum_{cyc}\left(\prod_{k=1}^{n-2}x_{k}\right)+...+\sum_{k=1}^nx_k$$

For example,

$(x_1,x_2,x_3,x_4)=(2,4,16,254)$ is a counterexample for $n=4$.

$(x_1,x_2,x_3,x_4,x_5)=(2,4,16,256,65534)$ is a counterexample for $n=5$.$\quad\blacksquare$

Added :

Let $$S(x_1,x_2,\cdots,x_n):=\sum_{cyc}\left(\prod_{k=1}^{n-1}x_{k}\right)+\sum_{cyc}\left(\prod_{k=1}^{n-2}x_{k}\right)+...+\sum_{k=1}^nx_k$$

Let us call your conjecture Conjecture 1.

Conjecture 1 : If $x_1,x_2,\cdots,x_n$ are positive integers satisfying $2\le x_1\lt x_2\lt\cdots\lt x_n$, then $\displaystyle\prod_{k=1}^{n}x_k\not\mid S(x_1,x_2,\cdots,x_n)$.

We've already seen that Conjecture 1 is true for $n=2$, and is false for $n\ge 3$.

Now, let us consider the following conjecture :

Conjecture 2 : If $x_1,x_2,\cdots,x_n$ are positive integers satisfying $2\le x_1\lt x_2\lt\cdots\lt x_n$ and $\color{red}{\gcd(x_1,x_2,\cdots,x_n)=1}$, then $\displaystyle\prod_{k=1}^{n}x_k\not\mid S(x_1,x_2,\cdots,x_n)$.

Conjecture 2 is true for $n=2$ since Conjecture 1 is true for $n=2$.

In the following, I'm going to prove the following claim :

Claim : Conjecture 2 is true for $n=3,4$.

Proof :

Let $A:=S(x_1,x_2,x_3)$.

Suppose that there are positive integers $x_1,x_2,x_3$ satisfying $2\le x_1\lt x_2\lt x_3, \gcd(x_1,x_2,x_3)=1$ and $x_1x_2x_3\mid A$.

If exactly one of $x_1,x_2,x_3$ is even, then $A$ is odd, so $x_1x_2x_3\not\mid A$.

If exactly one of $x_1,x_2,x_3$ is odd, then $A$ is odd, so $x_1x_2x_3\not\mid A$.

So, we see that $x_1,x_2,x_3$ have to be odd to have
$$x_1x_2x_3\mid A\implies x_1x_2x_3\mid \frac A2\implies \frac A2\ge x_1x_2x_3\implies 4A\ge 8x_1x_2x_3$$

$$\implies X_1X_2X_3\le 3(X_1+X_2+X_3)+10$$
$$\implies X_1X_2\le 3\bigg(\frac{X_1}{X_3}+\frac{X_2}{X_3}+1\bigg)+\frac{10}{X_3}\lt 3\times 3+\frac{10}{13}\lt 10\tag1$$
where $X_i:=2x_i-1$ for $i=1,2,3$.

Also, since $x_1\ge 3$ and $x_2\ge 5$, we have
$$X_1X_2=(2x_1-1)(2x_2-1)\ge 5\times 9=45\tag2$$

There is no $(x_1,x_2)$ satisfying both $(1)$ and $(2)$.

So, Conjecture 2 is true for $n=3$.

Next, let $B:=S(x_1,x_2,x_3,x_4)$.

Suppose that there are positive integers $x_1,x_2,x_3,x_4$ satisfying $2\le x_1\lt x_2\lt x_3\lt x_4, \gcd(x_1,x_2,x_3,x_4)=1$ and $x_1x_2x_3x_4\mid B$.

If exactly one of $x_1,x_2,x_3,x_4$ is even, then $B$ is odd, so $x_1x_2x_3x_4\not\mid B$.

If exactly two of $x_1,x_2,x_3,x_4$ are even, then $B$ is odd, so $x_1x_2x_3x_4\not\mid B$.

If exactly three of $x_1,x_2,x_3,x_4$ is odd, then $B$ is odd, so $x_1x_2x_3x_4\not\mid B$.

So, we see that $x_1,x_2,x_3,x_4$ have to be odd to have
$$x_1x_2x_3x_4\mid B\implies x_1x_2x_3x_4\mid \frac B2\implies \frac B2\ge x_1x_2x_3x_4\implies 8B\ge 16x_1x_2x_3x_4$$

$$\implies X_1X_2X_3X_4\le 3(X_1X_2+X_1X_3+X_1X_4+X_2X_3+X_2X_4+X_3X_4)+12(X_1+X_2+X_3+X_4)+31$$
$$\implies X_1X_2\le 3\bigg(\frac{X_1X_2}{X_3X_4}+\frac{X_1}{X_4}+\frac{X_1}{X_3}+\frac{X_2}{X_4}+\frac{X_2}{X_3}+1\bigg)+12\bigg(\frac{X_1}{X_3X_4}+\frac{X_2}{X_3X_4}+\frac{1}{X_3}+\frac{1}{X_4}\bigg)+\frac{31}{X_3X_4}$$
$$\lt 3\times 6+12\bigg(2+\frac{1}{13}+\frac{1}{17}\bigg)+\frac{31}{13\times 17}=\frac{569}{13}\lt 44\tag3$$

where $X_i:=2x_i-1$ for $i=1,2,3,4$.

Also, since $x_1\ge 3$ and $x_2\ge 5$, we have
$$X_1X_2=(2x_1-1)(2x_2-1)\ge 5\times 9=45\tag4$$

There is no $(x_1,x_2)$ satisfying both $(3)$ and $(4)$.

So, Conjecture 2 is true for $n=3,4$.$\quad\blacksquare$


问:
How to evaluate improper integral of odd function?
I recently was solving $\int_{0}^{\infty }\frac{x}{x^3+1}dx$. I wrote x as z to do contour integration and did the indentation for the pole at real axis but I don't know what to do with this odd function? How should I go further, write the z in terms of $\theta$ by euler formula? This is not homework question,I started learning complex analysis recently so I want to know the concepts used for such questions.

答:
Let's assume $I=\int_0^\infty\frac{x}{x^3+1}dx$. Now we will make a substitution which is $x=\frac{1}{t}$ & $dx=\frac{-dt}{t^2}$.

So, now $I=\int_0^\infty\frac{1}{t^3+1}dt$. Since limits are same for both integrals, we can add both integrals.

Now, $2I=\int_0^\infty\frac{x+1}{x^3+1}dx=\int_0^\infty\frac{1}{x^2-x+1}dx=\int_0^\infty\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}dx=[\frac{2}{\sqrt{3}}tan^{-1}(\frac{(x-\frac{1}{2})}{\frac{\sqrt{3}}{2}})]_0^\infty=\frac{2}{\sqrt{3}}(\frac{\pi}{2}-(-\frac{\pi}{6}))$

So, $2I=\frac{2}{\sqrt{3}}(\frac{4\pi}{6})$ which equates to $I=\frac{2\pi}{3\sqrt{3}}$

Hope it helps!


问:
Finding solution of the PDE : An Initial value problem
I've been given a PDE of the form $$xu_x+(x^2+y)u_y=1-\left(\frac{y}{x}-x\right)u~~; ~~u(1,y)=0$$
Attempt : Firstly it's a first order linear PDE which has the general form $$a(x,y)u_x+b(x,y)u_y=c(x,y)u+d(x,y)$$Where $a,b,c,d \in \mathcal{C}^1(\Omega)$, where $\Omega \subseteq \mathbb{R}^2$ (open and connected) and we also have $a^2+b^2 \neq 0$ on $\Omega_2$. Then first of all the largest $\Omega_2$ on which $F:=a^2+b^2=x^2+(x^2+y)\neq 0$ is : $(x,y) \in \Omega_2\setminus \{(0,0)\}$ right?

We have $s \mapsto \Gamma_s:=(f(s_0):=1,g(s_0):=s,h(s_0):=0))$. Now, from transversality conditions we get :$$J:=\mathrm{det}\, \begin{pmatrix}a&f'\\b&g'\end{pmatrix}\Bigg{|}_{f(s_0),g(s_0),h(s_0)}=\mathrm{det}\begin{pmatrix}1&0\\1+s&1\end{pmatrix}=1$$
So we atleast expect solutions to exist $\textit{locally}$ for all $s \in (-\delta,\delta),\delta>0$.
Now to find the solutions explicitly i'll employ the method of characteristics: \begin{align}\frac{\mathrm{d}x}{\mathrm{d}t}&=x~~;~x(0,s):=1 \implies x(t,s):=e^t\\\frac{\mathrm{d}y}{\mathrm{d}t}&=x^2+y~~;~ y(0,s):=s \\&=y+e^{2t} \end{align}
Solving this we get
\begin{align}   y(t,s)&=e^{2t}+c_2e^{t}~~;~y(0,s):=s \implies
y(t,s)=e^{t}\left(e^t+s-1\right)\end{align} Lastly \begin{align}\frac{\mathrm{d}z}{\mathrm{d}t}&=1-\left(\frac{y}{x}-x\right)z~~;~z(0,s):=0 \\&=1-\left(\frac{e^t(e^t+s-1)}{e^t}-e^t\right)z=1+(1-s)z\end{align}
Which gives me $$z(t,s)=\frac{1}{s-1}\left(1-e^{t(1-s)}\right)$$
As the Jacobian was non-zero by inverse function theorem i can expect that i can invert the map $$(t,s)\mapsto \left(x(t,s),y(t,s)\right)$$ and i can solve $x=x(t,s),y=y(t,s)$ uniquely for $t$ and $s$ and the map $$(x,y)\mapsto \left(T(x,y);S(x,y)\right)$$ should be of class $\mathcal{C}^1$.

So i have $x(t,s)=e^t \implies t=\ln\,x=:T(x,y)$ and $y(t,s)=e^{\ln\,x}\left(e^{\ln \,x}+s-1\right)=x(x+s-1)$ i.e $s=(y/x)-x+1=:S(x,y)$. From here i can write $$u=z(T(x,y),S(x,y))=\frac{x}{y-x^2}\left(1-x^{\frac{x^2-y}{x}}\right)$$
Which is valid when $y>x^2$. I'm not sure about my workings here, appreciate any hints and help. Thanks.

答:
I agree with your result as shown below.
$$xu_x+(x^2+y)u_y=1-(\frac{y}{x}-x)u$$
Charpit-Lagrange system of characteristic ODEs :
$$\frac{dx}{x}=\frac{dy}{x^2+y}=\frac{du}{1-(\frac{y}{x}-x)u}$$
A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{x^2+y}$ which is a first order linear ODE :
$$\frac{y}{x}-x=c_1$$
A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{1-(\frac{y}{x}-x)u}=\frac{du}{1-c_1u}$ which is a first order linear ODE :
$$x^{c_1}(u-\frac{1}{c_1})=c_2$$
The general solution of the PDE (any boundary condition) expressed on the form of implicit equation $c_2=F(c_1)$ with arbitrary function $F$ is :
$$x^{c_1}(u-\frac{1}{c_1})=F(c_1)$$
$$x^{\frac{y}{x}-x}(u-\frac{1}{\frac{y}{x}-x})=F\left(\frac{y}{x}-x\right)$$
$$\boxed{u(x,y)=\frac{1}{\frac{y}{x}-x}+x^{-\frac{y}{x}+x}F\left(\frac{y}{x}-x\right)}$$
Determination of the function $F$ in order to satisfy the condition $u(1,y)=0$ :
$$u(1,y)=0=\frac{1}{y-1}+F\left(y-1\right)$$
Let $X=y-1 \quad;\quad y=X+1$
$$0=\frac{1}{X}+F\left(X\right)$$
$$F(X)=-\frac{1}{X}$$
Now the function $F(X)$ is known. we put it into the above general solution where $X=\frac{y}{x}-x$

$F\left(\frac{y}{x}-x\right)=-\frac{1}{\frac{y}{x}-x}$

The particular solution which satisfies both the PDE and the condition is :

$${u(x,y)=\frac{1}{\frac{y}{x}-x}-x^{-\frac{y}{x}+x}}\frac{1}{\frac{y}{x}-x}$$
$$\boxed{{u(x,y)=\frac{x}{y-x^2}}\left(1 -x^{\frac{x^2-y}{x}}\right)}$$


问:
Cases in which $Pr(X|C) = 1$
My conjecture is that there are only two cases in which $Pr(X|C) = 1$ (assume that $Pr(C) > 0$): if $Pr(X) = 1$ or if the probability of the intersection of $C$ and NOT $X$ is $0$. What do you think? Can you prove your answer? Does a theorem already exist somewhere?

答:
Regarding the comments you actually meant: $$Pr(X|C) = 1 \iff Pr(C\cap X^c)=0$$

But this is pretty obvious:
$\Rightarrow$ follows by:
$$Pr(X|C) = 1 \iff Pr(X \cap C) = Pr(C)$$ as well as
$$Pr(C) = Pr(C \cap X^c) + Pr(C \cap X)$$
So in total we get:
$$  Pr(X \cap C) = Pr(C \cap X^c) + Pr(C \cap X)$$

Hence: $$Pr(C \cap X^C) = 0$$

The other direction can be seen by reading the proof upside down.

Title: Cases in which $Pr(X|C) = 1$


问:
Cost minimization problem
The problem is as follows:

A firm uses $k$ units of capital and $l$ units of labor to produce
$(k^{\alpha}l^{1-\alpha})^{1/\beta}$
units of output, where $\alpha$ and $\beta$ satisfying $0 < \alpha < 1$ and $\beta  > 1$, are parameters.

If the firm obtains these inputs at the competitive rental rate $r$ for capital and the competitive wage rate $w$ for labor, and seeks to minimize the cost of producing at least $y$ units of output, it solves the problem

$$\min rk + wl \text{ subject to  } (k^{\alpha}l^{1-\alpha})^{1/\beta}\geq y$$

where, in this cost-minimization problem, the output requirement y is another parameter.

(1) Using Kuhn-Tucker theorem, show that the minimum cost function $c(r,w,y)$ is

$$c(r,w,y) = \frac{1}{\alpha^{\alpha}\times(1-\alpha)^{(1-\alpha)}} \times r^{\alpha}w^{(1-\alpha)}y^\beta $$

(2) Using the Envelop theorem, and the equation (1) derive the optimal solutions $k*(r,w,y)$ and $l*(r,w,y)$.

This is my homework problem. Humble request to anyone for the solution. Thanks.

答:
Set up the Lagrangian function:

$$\min_{k,l} rk+wl-\lambda\left((k^{\alpha}l^{1-\alpha})^{1/\beta}-y\right)$$

Note that the function $(k^{\alpha}l^{1-\alpha})^{1/\beta}$ is quasi-concave and homogeneous and of degree less than 1 (since $\beta>1$). A quasi-concave function of degree between 0 and 1 is concave. The negative of a concave function is convex ($\lambda$ is positive). Further, the two linear functions $rk$ and $wl$ are convex. The sum of convex functions is convex. Thus, the entire Lagrangian function is convex. The first order conditions will be sufficient for a minimum. It is also noted that the production function has infinite marginal product for either capital or labor when one is set to $0$. Thus, we will not have to worry about corner solutions where only one of the two inputs is used. Further, prices are strictly positive and so no more than y
  will be produced. The first order conditions are:

$$r=\lambda\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\alpha\left(\frac{l}{k}\right)^{1-\alpha}$$

$$w=\lambda\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\left(1-\alpha\right)\left(\frac{k}{l}\right)^{\alpha}$$

$$(k^{\alpha}l^{1-\alpha})^{1/\beta}=y$$

Solving these gives:

$$\frac{r}{\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\alpha\left(\frac{l}{k}\right)^{1-\alpha}}=\frac{w}{\frac{1}{\beta}\left(k^{\alpha}l^{\left(1-\alpha\right)}\right)^{\frac{1}{\beta}-1}\left(1-\alpha\right)\left(\frac{k}{l}\right)^{\alpha}}$$

$$\frac{\left(1-\alpha\right)\left(\frac{k}{l}\right)^{\alpha}}{\alpha\left(\frac{l}{k}\right)^{1-\alpha}}=\frac{w}{r}$$

Gives one key relationship:

$$\frac{k}{l}=\frac{\alpha}{1-\alpha}\frac{w}{r}$$

Now, we use this along with the production constraint to solve for the inputs:

$$(k^{\alpha}l^{1-\alpha})=y^{\beta}$$

$$\left(\frac{\alpha}{1-\alpha}\frac{w}{r}\right)^{\alpha}l=y^{\beta}$$

$$l^{*}=\left(\frac{1-\alpha}{\alpha}\frac{r}{w}\right)^{\alpha}y^{\beta}$$

$$k^{*}=\left(\frac{\alpha}{1-\alpha}\frac{w}{r}\right)^{1-\alpha}y^{\beta}$$

Plugging these back into the cost function gives the desired result:

$$y^{\beta}w^{1-\alpha}r^{\alpha}\left[\left(\frac{\alpha}{1-\alpha}\right)^{1-\alpha}+\left(\frac{1-\alpha}{\alpha}\right)^{\alpha}\right]$$

$$y^{\beta}w^{1-\alpha}r^{\alpha}\left[\left(1+\left(\frac{\alpha}{1-\alpha}\right)\right)\left(\frac{1-\alpha}{\alpha}\right)^{\alpha}\right]$$

$$y^{\beta}w^{1-\alpha}r^{\alpha}\left[\left(\frac{1}{1-\alpha}\right)\left(\frac{1-\alpha}{\alpha}\right)^{\alpha}\right]$$

$$y^{\beta}w^{1-\alpha}r^{\alpha}\frac{1}{\alpha^{\alpha}\left(1-\alpha\right)^{1-\alpha}}$$

By the envelope theorem, the derivative of an optimized function with respect to one of its parameters can be taken treating the choice variables as fixed. At the optimum, a very small change in these choice variables has no effect on the optimized value. This means, when we take the derivative of the optimized cost function with respect to $r$ and $w$ we will get back $k^*$ and $l^*$ respectively. I will leave this to you to check.

Title: Cost minimization problem


问:
indefinite integral of $\int \frac{dx}{\sqrt{(x)(x+1)(x+2)}}$
I tried integration by parts considering $dx=du$

$\frac{1}{\sqrt{(x)(x+1)(x+2)}}=v$ but im not getting the answer.

My attempt....

$uv=\sqrt{\frac{x}{(x+1)(x+2)}}$ and-$\int{dv u}$=$\frac{I}{2}$+$\int{\frac{\sqrt{x}}{2\sqrt{(x+1)}(x+2)^{\frac{3}{2}}}}$+$\int{\frac{\sqrt{x}}{2\sqrt{(x+2)}(x+1)^{\frac{3}{2}}}}$

How do we calculate $\int{\frac{\sqrt{x}}{2\sqrt{(x+1)}(x+2)^{\frac{3}{2}}}}$ and
$\int{\frac{\sqrt{x}}{2\sqrt{(x+2)}(x+1)^{\frac{3}{2}}}}$

How do we do it?

Please help!

答:
In general, an integral with square-root of a polynomial of degree $3$ or $4$ is an elliptic integral.

Maple evaluates this in terms of the elliptic integral of the first kind $F$:
$$
\int \!{\frac {1}{\sqrt {x \left( x+1 \right)  \left( x+2 \right) }}}
\,{\rm d}x=2\;{ F} \left( \frac{\sqrt {x+2}}{\sqrt{2}},\sqrt {2} \right)
$$


问:
What is the p.d.f of $Z$?
Let $f(x)=\frac{1}{2}\sin x \;  (0<x<\pi )$ be the p.d.f of $X$ and $g(x)=\begin{cases}
x & \text{ if } x<\frac{\pi }{3} \\
\frac{\pi }{9} & \text{ if } x\geqslant \frac{\pi }{3}
\end{cases}$
and $Z=g(X)$. What is the p.d.f. of $Z$?  Since $g$ is not injective on $x\geq \frac{\pi }{3}$, I struggle with this problem. please help me..

答:
In the range $(0<X<\frac{\pi}{3})$ the density is the same $f_X$. When $X\geq \frac{\pi}{3}$ the rv is discrete, and in the point $Z=\frac{\pi}{9}$ the rv cumulates a positive probability mass...

Thus the requested "mixed" pdf is

$$ f_Z(z) =
\begin{cases}
\frac{sin (z)}{2},  & \text{if $0<z<\frac{\pi}{9} \cup \frac{\pi}{9}<z<\frac{\pi}{3}  $ } \\
\frac{3}{4}, & \text{if $z=\frac{\pi}{9}$}\\
0, & \text{elsewhere}
\end{cases}$$

Title: What is the p.d.f of $Z$?


问:
Parabolic structure on direct sum of parabolic vector bundles
Let $X$ be an irreducible smooth projective curve defined over the field of complex numbers $\mathbb{C}$. Let $E$ be an algebraic vector bundle over $X$. Let $P\in X$.

Defn : A quasi-parabolic structure on $E$ over the point of $P$ is given by a strictly decreasing filtration of subspaces (such a filtration is called a flag) of the fibre $E_{P}$ over $P$
$$ E_{P}=F^1(E_{P})\supsetneq F^2(E_{P})\supsetneq  \cdots F^{n_i}(E_{P})\supsetneq F^{n_i+1}(E_{P})=\{0\}$$

Defn : A parabolic structure on $E$ over a point $P$ is given by a quasi-parabolic structure with the following additional data: rational numbers $\{\alpha^j_{P}\}$ associated to each subspace $F^j(E_{P})$ satisfying the condition, $$0\leq \alpha^1_{P}\lneq \alpha^2_{P}\cdots \lneq \alpha^{n_i}_{P}\lneq 1$$

Defn :  Let $\mathcal{E}:=\big(E,P, \{F^j(E_{P})\},\{\alpha^j_{P}\}\big)$ and $\mathcal{F}:=\big(F,P, \{F^j(F_{P})\},\{\beta^j_{P}\}\big)$ be two parabolic vector bundles over $X$ with quasi-parabolic structures at the same point $P\in X$. A parabolic morphism from $\mathcal{E}$ to $\mathcal{F}$ is given by a vector bundle morphism $\psi$  such that given a weight $\alpha^s_{P}$ if  $\beta^t_{P}$ is the smallest weight such that $\alpha^s_{P}\leq\beta^t_{P}$ (i.e., $\beta ^{t-1}_{P}< \alpha^s_{P}\leq\beta ^{t}_{P}$) then we demand $$ \psi_{P}(F^{s}(E_{P})) \subset F^{t}(F_{P}) $$
If there is no such $\beta^t_{P}$ then we demand $$ \psi_{P}(F^{s}(E_{P})) =0 $$

Given two rank $n$ parabolic vector bundles $\mathcal E=\big(E,P,\{F^j(E_P)\}_{j=1}^{n+1},\{\alpha_P^j\}\big)$ and $\mathcal F=\big(F,P,\{F^j(F_P)\}_{j=1}^{n+1},\{\beta_P^j\}\big)$ over $X$ with parabolic structure at same point and full flags. We can define their direct sum and even give it a natural quasi parabolic structure at $P\in X$. (Just give the flag as $\{F^j((E\oplus F)_P):=F^j(E_P)\oplus F^j(F_P)\}$).

My question : How do you give $\mathcal{E\oplus F}$ parabolic structure? That is what weights do you give each member of the flag?

The weights need to be given so that the natural maps $\mathcal{E\to E\oplus F}$ and $\mathcal{F\to E\oplus F}$ are parabolic morphisms. So assigning the new weights $\gamma_P^j$ as $\max\text{ or }\min\{\alpha_P^j,\beta_P^j\}$ won't work. Is there some natural way to do it?

Thank you.

答:
Maybe it's a little bit late to answer after three years but I'll try the same...

I think that the problem is that the definition of the flag for $E \oplus F$ over $P$ should not be the natural one. If I got it correctly it should depend also on the weights for $E,F$ in the following manner.

You start putting $F^1(E \oplus F)=E \oplus F$. We can then consider the following situations:

$\beta_1 > \alpha_1$. We then put $F^2(E\oplus F)=F^2E \oplus F$.

$\beta_1=\alpha_1$. We then put $F^2(E \oplus F)=E \oplus F$.

$\beta_1 < \alpha_1 $. This is symmetric to the first one.

One then proceeds inductively in the same way to complete the flag.
One should then assign the following weights $\gamma_i$ to this quasiparabolic structure. I work out just the case $\gamma_1$, the others are defined in the same way.

So for each of the case above we put:

$\gamma_1=\alpha_1$

$\gamma_1=\alpha_1$

$\gamma_1=\beta_1$


问:
Number of Total orders of a dependency graph
 Define a dependency graph to be a graph $G=(V,E)$ such that an edge between vertices $v$ and $u$ in $V$ is present if $v<u$ i.e. $v$ comes before $u$ in our ordering (I'm not very concise here, I just want to give you my idea).

 My quesion now is how many total orders do we have on such a graph ? Is there a closed formula for this?

 What I thought of doing is make something like a decision tree that has as root the first element in the ordering and as leaves the final one and do topological sorting on the graph to produce paths from the root to the leaves

 As a concrete example lets look at the graph below. I know that it has 30 total orders.

答:
It appears in your example that your dependency graph consists of a collection of total orders, each independent from each other.  With this restriction, the solution you seek is $$\frac{n!}{n_1!n_2!\cdots n_k!}$$

Where there are $k$ total orders, and the $i^\textrm{th}$ contains $n_i$ elements, and there are $n=n_1+n_2+\cdots+n_k$ total elements in your graph (not counting the max and min).

In the example shown $n_1=2, n_2=1, n_3=2, n=5$.

The explanation is as follows: we begin with $n!$ orders on the $n$ elements.  However, within each of the $k$ columns, we must choose a linear order that respects the original order of that column; exactly $\frac{1}{n_i!}$ of the linear orders respects the order of the $i^\textrm{th}$ column.


问:
Right invertible elements in a monoid.
Prove that if every element in the monoid is right invertible, then every element has exactly one right inverse.

That is in the monoid $(M,\circ )$ with the identity element e,
$$\forall a\in M \; \exists a^\prime \in M\; : \; a\circ a^\prime=e \quad \Rightarrow \quad \forall a\in M \; \exists! a^\prime \in M\; : \; a\circ a^\prime=e $$

答:
Ok, so there is a standard argument for this kind of questions (unit of a group is unique; the unit element of a ring is unique etc). The proof is basically the same as below:

For $\forall a \in M$, assume we have two right inverses $b_1$ and $b_2$, then we have $ab_1=ab_2=e$. By hypothesis, we knew there are right inverses for both $b_1$ and $b_2$, say they are $c_1$ and $c_2$ respectively. Then multiply $c_1$ to the equation $ab_1=e$ from right, we have $ab_1c_1=c_1$, then since $b_1c_1=e$, we have $a=c_1$. Similarly, multiplying $c_2$ to the equation $ab_2=e$ from right, we get $a=c_2$. Therefore we showed that $b_1$ and $b_2$ have the same right inverse, namely just $a$. So $b_1a=b_2a=e$. Multiplying $b_1$ to the equation $b_2a=e$ from right, we have $b_2=b_2ab_1=b_1$. Therefore there exists exactly one right inverse.

Actually, this proof gets you a stronger that, this monoid is actually a group.

Title: Right invertible elements in a monoid.


问:
Finding the power series representation of $\frac{x^2+1}{8x-1}$
So, I'm honestly at a loss of what I am doing wrong here. I know that the $x^2+1$ needs to be factored out of the summation, but apparently, this was not the answer to the first text box. Also, I'm not sure how to find the rate in this particular geometric series. I have tried to factor out $-1$ to make the rate $(8x)$, but I am not sure if this is even the right approach. Where did I go wrong in this problem?

答:
Using the method as suggested in the comments and by Adam Rubinson:

You just have $(x^2+1) \left( \frac{-1}{1 - 8x} \right)$. Now expand $\left( \frac{-1}{1 - 8x} \right)$ using geometric series, where $a = -1$, $r = 8x$ to get:

$$x^2 \left(- 1-8x-64x^2 - \cdots \right) + \left(- 1-8x-64x^2 - \cdots \right)$$
$$= -1-8x+(-1-64)x^2+(-8-512)x^3$$

Notice that for the $x^n$ term, the coefficient of the second summation is $8^2$ times that of the coefficient in the first summation, because the first summation is shifted two places to the right (which means it is a factor of $8^2$ behind).

Hence, after $-1-8x$, the general term is $(-8^{n-2} - 8^n)x^n$, or $-8^{n-2} (1+8^2)x^n = -1 \cdot 8^{n-2} \cdot 65x^n$ starting from $n = 2$. In summation notation, this is just:

$$f(x) = -1-8x - 65 \sum_{k=2}^\infty 8^{k-2} \cdot x^k, |x| < \frac{1}{8}.$$

and factoring the $8^{-2}$ outside the summation gives the nicer form:

$$f(x) = -1-8x - \frac{65}{64} \sum_{k=2}^\infty (8x)^k, |x| < \frac{1}{8}.$$

which is equivalent to the form stated in the comments.


问:
Prove inequality so I can use Banach fixed point thrm
I have the function $\varphi(x)=\sqrt{2-\sqrt{4-x^2}}$.
I want to prove that $d(\varphi(x),\varphi(y)\leq k\cdot d(x,y)$ for some $k,\ \forall x,y$.
I thoight it would be simple, just by saying that

$\left|\sqrt{2-\sqrt{4-x^2}}-\sqrt{2-\sqrt{4-y^2}}\right|\leq k\cdot |x-y|$ and for x>y I started to compare only $\sqrt{2-\sqrt{4-x^2}}-\sqrt{2-\sqrt{4-y^2}}=x-y$. But I tried everything, rationalizing, multiplying by the compose, squaring, and I don't come any further. I would be really glad if someone could tell me if there is a trick to it or something.

This is as far as I go...

$\frac{\sqrt{4-y^2}-\sqrt{4-x^2}}{\sqrt{2-\sqrt{4-x^2}}+\sqrt{2-\sqrt{4-y^2}}}=x-y
$

$\frac{x^2-y^2}{(\sqrt{2-\sqrt{4-x^2}}+\sqrt{2-\sqrt{4-y^2}})(\sqrt{4-y^2}+\sqrt{4-x^2})}=x-y \Leftrightarrow \frac{x+1}{(\sqrt{2-\sqrt{4-x^2}}+\sqrt{2-\sqrt{4-y^2}})(\sqrt{4-y^2}+\sqrt{4-x^2})}=1
$

答:
[edited]

The domain of $\varphi$ is $[-2,2]$ and its only fixed point is $z=0$. If I understood, your  question concerns the convergence of $x_{n+1} = \varphi(x_n), \quad x_0 \in [-2,2]$, to the fixed point $z=0$.

You can start by noting that $\varphi([-2,2]) = [0,\sqrt{2}]$ and so, in terms of convergence, we can suppose right form the start that this is all about the interval $I=[0, \sqrt{2}]$.

The sequence $(x_n)$ is bounded (all terms belong to $I$) and decreasing (assuming, without loss of generality that $x_0 \in I$) and so it is convergent. There is no need for the fixed point theorem.

if you really want to use the fixed point theorem, you can still use this idea that the original domain can be restricted and proof the boundness of $\varphi'$ in $I$. This way you can conclude that $\varphi$ is Lipschitz on $I$, with $k = \frac{1}{2 \sqrt{2-\sqrt{2}}}<1$.


问:
An integral which is derived from Jacobi's Sum of Two Squares Theorem by Complex Analysis
Here is an integral that I'm having trouble with. I have got to this integral by trying to prove Jacobi's sum of two squares by complex analysis. And here is it:
$$I:= \int_{-\infty}^{\infty} \frac{\operatorname{csch}(x) \sin (2x)}{\cos (2x) - \cosh (\pi)}\mathrm{d}x=\pi \coth\left(\frac{\pi}{2}\right) - \frac{1}{2}B \left(\frac{1}{4}, \frac{1}{4}\right)$$
My step so far:
It is well-known that:
$$\sum_{n=1}^\infty 2 \pi (-1)^{n-1} \mathrm{e}^{-\omega \pi n} \sin(\pi \kappa n) =
    \frac{2 \pi  e^{\pi  \omega } \sin (\pi  \kappa )}{2 e^{\pi  \omega } \cos (\pi
   \kappa )+e^{2 \pi  \omega }+1}$$$$=\frac{\pi   \sin (\pi  \kappa )}{\cos (\pi
   \kappa )+\cosh\left( \pi  \omega \right)}$$
By substitution $k = t+1$ cause we know that $\cos\pi (t+1) = - \cos \pi t$:
$$2\cdot\sum_{n=1}^\infty  \mathrm{e}^{-\omega \pi n} \sin(\pi t n) =  \frac{   \sin (\pi t )}{\cosh\left( \pi  \omega \right)-\cos (\pi  t)}$$
Now let $\omega = 1$ and $ t = \frac{2x}{\pi}$. We obtain:
$$2\cdot\sum_{n=1}^\infty  \mathrm{e}^{-\pi n} \sin\left( 2nx\right) =  \frac{   \sin (2x )}{\cos (2x)-\cosh\left( \pi  \right)}$$
Plug this infinite representation in our integral:
$$ I = -2\cdot\sum_{n=1}^\infty  \mathrm{e}^{-\pi n} \int_{-\infty}^{\infty} \frac{\sin (2nx)}{\sinh(x)} \mathrm{d}x$$
Each integral inside is fairly easy to compute. Indeed, one can use Laplace transform:
$$\int_{-\infty}^{\infty} \frac{\sin (2nx)}{\sinh(x)} \mathrm{d}x = 2 \int_{0}^{\infty} \frac{\sin (2nx)}{\sinh(x)} \mathrm{d}x=4  \int_{0}^{\infty} \frac{e^{-x}\sin (2nx)}{1- e^{-2x}} \mathrm{d}x$$$$=4\sum_{i=0}^{\infty} \int_{0}^{\infty} \sin(2nx) e^{-(2n+1)x} \mathrm{d}x= \pi \tanh\left(n\pi \right) $$
The last inequality is obtained by the fact that: $\tanh \left(\frac{\pi x}{2}\right) = \frac{4x}{\pi} \sum_{k\geq 1} \frac{1}{(2k-1)^2 + x^2}$. Therefore:
$$I: = 2\pi \sum_{n=1}^{\infty} e^{-\pi n} \tanh(\pi n)$$
And I'm stuck here because I don't know how to connect this sum with the result above. Indeed, I can derive a further result by using $\tanh(x)$ generating summation which is:
$$\tanh (x) = 1+ 2\sum_{n=1}^{\infty}\frac{(-1)^n}{e^{2nx}}$$
However, I can't still be able to get the Beta Function to appear. Hope anyone can help me to derive the result above. Thank you so much.

UPDATE: So I am trying to prove the relation without using Jacobi's two square theorem or elliptical function:
$$\left(\sum_{n=-\infty}^{\infty}e^{-\pi n^2}\right)^2 = \sum_{n = - \infty}^{\infty} \frac{1}{\cosh \left(\pi n\right)}$$
Actually, I intend proving this by constructing an complex integral and make use of Residue Theorem of the function $\displaystyle f(z) = \frac{1}{2i} \cot (\pi z) \mathrm{sech} (\pi z) $ over the contour which has a square shape from $-N - \frac{1}{4}$ to $N + \frac{1}{4}$ on the real axis and likewise on the imaginary axis. Since $N \to \infty$, $f(z) \to 0$, by calculating the residue, I obtained:
$$\sum_{n=-\infty}^{\infty} \mathrm{sech}(\pi n) = \sum_{n=-\infty}^{\infty} (-1)^n \coth \left(\frac{\pi}{2}(2n+1)\right)$$
Then, in order to evaluate the LHS of $(1)$, I used Abel–Plana formula for the divergent series on the RHS which is:
$$\sum_{n=0}^{\infty} (-1)^n f(n) = \frac{f(0)}{2} + i\int_0^{\infty} \frac{f(it) - f(-it)}{2\mathrm{sinh}(\pi t)} \mathrm{d}t$$
Rewrite the RHS of $(1)$ and then apply this formula since $\coth (z)$ is holomorphic in the region $\Re{(z)} \geq 0$. Finally, I use the result of:
$$\sum_{n=-\infty}^{\infty}e^{-\pi n^2}=\frac{\sqrt[4]{\pi}}{\Gamma\left(\frac{3}{4}\right)}$$
By Euler's reflection formula, I got the integral above. Therefore, if one can prove the result above indenpendently, we can have another way proving Jacobi's two square theorem via complex analysis.

答:
This is a long comment (exposition of my comment to the question). Please do respond by updating your question.

The Jacobi two square theorem is equivalent to the following identity $$\left(\sum_{n\in\mathbb {Z}} q^{n^2}\right)^2=1+4\sum_{n\geq 1}\frac{q^n}{1+q^{2n}}\tag{1}$$ The final sum expression for your integral $I$ can be written as $$I=-2\pi\sum_{n\geq 1}\frac{q^n(1-q^{2n})}{1+q^{2n}}=2\pi\sum_{n\geq 1}\left(q^n-\frac{2q^n}{1+q^{2n}}\right)=\frac{2\pi q} {1-q}-4\pi\sum_{n\geq 1}\frac{q^n}{1+q^{2n}}$$ where $q=e^{-\pi} $ (you have a minor sign typo in your question).

Now an explicit evaluation of the above sum is itself based on the equation $(1)$ and the integral becomes $$I=\pi(1-\vartheta_{3}^2(q))+\frac{2\pi q} {1-q}$$ where $\vartheta_3(q)=\sum_{n\in\mathbb {Z}} q^{n^2}$ is a theta function defined by Jacobi.

Explicit evaluation of $\vartheta_{3}(q)$ is possible if $q=\exp(-\pi\sqrt{r}), r\in\mathbb {Q} ^{+} $ via a complicated theorem of Selberg and Chowla. The evaluation is easy and famous for $q=e^{-\pi} $.

However I don't see an explicit evaluation without using the Jacobi two square theorem $(1)$. Can you give some more details about your version of Jacobi two square theorem and some indication of how your arrive at your integral while trying to prove it. A direct algebraic proof of $(1)$ is available on this website.


问:
Functions which satisfy $f(m^2+n^2)=f(m)^2+f(n)^2$
Find all $f$: $f(m^2+n^2)$=$f(m)^2+f(n)^2$ for all $m,n\in \mathbb{N_0}$ where

$f(1)>0$,

$f\in\mathbb{N_0}$ for all inputs

I did $f(0)=2f(0)^2\implies f(0)=0$

$f(1)=f(1)^2\implies f(1)=1$

From this we can get $f(2)=2, f(4)=4$ etc. easily and also use some tricks like

$f(2^2+1^2)=f(5)=5, f(4)=4$

so $f(3)=\sqrt{5^2-4^2}=3$

I tried using the parametrization of Pythagorean triples but am unable to prove that $f(7)=7$

Please help

Body:
We see that $f(2n^2) = 2f^2(n)$. So $f(50)=2f^2(5)=2(25)=50$
So $f(50)=50$. But $50 = 1+49$. So $f(50) = 1 + f^2(7)$.

So $f(7)=7$


问:
Intuitive Understanding of exponential distrbution

$\textbf{Problem}$: An athletic facility has $5$ tennis courts. Pairs of players arrive at the
courts and use a court for an exponentially distributed time with mean $40$ minutes.
Suppose a pair of players arrives and finds all courts busy and k other pairs waiting in
queue. What is the expected waiting time to get a court?

$\textbf{Answer}$: As long as the pair of players are waiting, all five courts are occupied by other players. When all five courts are occupied, the time until a court is freed up is exponentially distributed with mean $40/5=8 $ minutes. For our pair of players to get a court, a court must be freed up $k+1$ times. Thus, the expected waiting time is $8(k + 1)$.

The issue that I'm having in understanding is why does "The time until a court is freed up is exponentially distributed with mean $40/5=8$ minutes" rather than simply $40$ minutes. Since every court is independent and using the memoryless property.

On the contrary, if I look at $$Z=min(X_{1}, X_{2}, X_{3}, X_{4}, X_{5})$$. where $X_{i}$ is time taken by a pair to play in court $i$ and is exponentially distributed time with mean $40$ minutes.

Then $Z$ itself becomes and exponentially distributed with mean $40*5 = 200$ minutes.

I am unable to understand it intuitively. Any help is appreciated.

答:
Let $Z=\min(X_1,...,X_5)$.

We have $X_1=Z$ with probability ${\large{\frac{1}{5}}}$, and by the memoryless property we have
$$
E[X_1-Z{\,\mid\,}X_1 > Z]=40
$$
hence
\begin{align*}
&
E[X_1-Z]
=
\left({\small{\frac{1}{5}}}\right)\!(0)
+
\left({\small{\frac{4}{5}}}\right)\!(40)
\\[4pt]
\implies\;&
E[X_1-Z]=32
\\[4pt]
\implies\;&
E[X_1]-E[Z]=32
\\[4pt]
\implies\;&
40-E[Z]=32
\\[4pt]
\implies\;&
E[Z]=8
\\[4pt]
\end{align*}


问:
formula or method for intermediate values within simple sequence?
I have a series of data points and I am trying to find out the formula to calculate what intermediate values might be, for example if A is 5, what value would B have? This seems like it's pretty easy, but I don't know the terminology for this type of curve so searching for an answer has not been very fruitful. I am looking to use this date in a simple program, where I can give any value of (A) and get the corresponding (B) value.

```<code>1000   3.2
100    1.6
10     0.8
1      0.4
0.1    0.2
(A)    (B)
</code>```

Examples ...

```<code>var valueB = someFormula(valueA: 1000) // would return 3.2
var valueB = someFormula(valueA: 10)   // would return 0.8
var valueB = someFormula(valueA: 0.1)  // would return 0.2
var valueB = someFormula(valueA: 5)    // would return ?
</code>```

Much appreciated in advance.

答:
Do you notice a pattern between the two columns? When $A$ is multiplied by $10$, then $B$ is always multiplied by $2$. Therefore, it's not difficult to see that
$$\tag{1}
\log_{10} (A) +1  =\log_2 (5B)
$$
from which we can solve for $B$ ...

I'll show how I obtained these values. I immediately saw that there is a ratio of 10 between the consecutive values of A and a ratio of 2 between the consecutive values of B, prompting me to calculate $\log_{10}A$ and $\log_2 B$. In addition, because the values of $B$ are not whole numbers, I saw that I should multiply them by $5$ to get integer values. Therefore, I should compute the value of $\log_2 (5B)$. I put the values into Excel and got a table that looks like this:

```<code> A      Log_10(A)   B     Log_2 (5B)
 1000   3           3.2   4
 100    2           1.6   3
 10     1           0.8   2
 1      0           0.4   1
 0.1    -1          0.2   0
</code>```

Now, it's pretty easy to see that you just need to add 1 to get from the second column to the last column. Thus, Equation ($1$).

When we solve Equation (1) for $A$, we get
$$\tag{2}
A  =10^{\log_2 (B/2)}
$$

On the other hand, solving for $B$ results in
$$\tag{3}
B = \frac{2}{5} 2^{\log_{10} (A)}
$$

Body:
Do you notice a pattern between the two columns? When $A$ is multiplied by $10$, then $B$ is always multiplied by $2$. Therefore, it's not difficult to see that
$$\tag{1}
\log_{10} (A) +1  =\log_2 (5B)
$$
from which we can solve for $B$ ...

I'll show how I obtained these values. I immediately saw that there is a ratio of 10 between the consecutive values of A and a ratio of 2 between the consecutive values of B, prompting me to calculate $\log_{10}A$ and $\log_2 B$. In addition, because the values of $B$ are not whole numbers, I saw that I should multiply them by $5$ to get integer values. Therefore, I should compute the value of $\log_2 (5B)$. I put the values into Excel and got a table that looks like this:

 A      Log_10(A)   B     Log_2 (5B)
 1000   3           3.2   4
 100    2           1.6   3
 10     1           0.8   2
 1      0           0.4   1
 0.1    -1          0.2   0

Now, it's pretty easy to see that you just need to add 1 to get from the second column to the last column. Thus, Equation ($1$).

When we solve Equation (1) for $A$, we get
$$\tag{2}
A  =10^{\log_2 (B/2)}
$$

On the other hand, solving for $B$ results in
$$\tag{3}
B = \frac{2}{5} 2^{\log_{10} (A)}
$$


问:
How to obtain the Lyapunov candidate of this system
I have been trying to find a suitable Lyapunov function for this system

$\dot{x_{1}} = \sin(x_{2})$

$\dot{x_{2}} = -x_{1}-x_{2}$

As can be noted it has multiple equilibrium points:

Equilibrium Points: $(-n\pi,n\pi)$ where $n = ... -2,-1, 0,+1,+2 ...$

When n is even the equilibrium point is stable focus, but when odd the equilibrium point is a saddle.
I tried this:

$V = x^{T}Px + (1-\cos(x_{2}))$

For V to be positive definite we need $P_{11}+P_{12} >0$ and $P_{11}P_{22}-(P_{12})^{2}>0$, where $P$ is a $2\times 2$ symmetric matrix ($P_{12} = P_{21}$). The time derivative of Lyapunov candidate is:

$$ \dot{V} = \dot{x_{1}}x_{1}P_{11} + \dot{x_{2}}x_{1}P_{12} + \dot{x_{1}}x_{2}P_{12} + \dot{x_{2}}x_{2}P_{22} + \dot{x_{2}} \sin(x_{2})  $$

By solving I get to the point:

$ \dot{V} = (-P_{12} - P_{22})x_{1}x_{2} - P_{12}x_{1}^{2} - P_{22}x_{2}^{2} + (P_{12} -1)x_{2}sin(x_{2}) +(P_{11} -1)x_{1}\sin(x_{2}) $

In this point I tried to make:
$$-P_{12}-P_{22} = 0\longrightarrow P_{12} = -P_{22}\tag1$$
$$P_{11} -1 =0 \tag2$$

And we also assume that:

$$P_{12}>0\tag3$$

$$P_{22}>0 \tag4$$

$$P_{12}-1 <0\longrightarrow P_{12}<1\tag5$$

So we can guarantee that $\dot{V}$ is negative definite in the domain $-\pi<x_{2}<\pi$. But the problem manifests here. If we are suppose to satisfy $(5)$ and $(3)$ we obtain that:

$$0<P_{12}<1$$

If I assign $P_{12} = 0.5$ from $(1)$ we can say that $P_{22} = -0.5$ but this value does satisfy $(4)$ and this is my MAIN issue with this. I don't know I my math went wrong, or if I just went trough the wrong process, the process I used for this exercise was similar to the process one can follow to obtain a Lyapunov function that guarantee asymptotic stability for a pendulum with friction problem, but it is not working for me.

答:
Your first constraint from $(1)$, that $P_{12} = -P_{22}$, it not necessary. If we still set $P_{11}=1$ and $P_{12}=1/2$ yields

$$
\dot{V} = -\frac{1}{2} x^\top
\underbrace{
\begin{bmatrix}
1 & \frac{1}{2}+P_{22} \\
\frac{1}{2}+P_{22} & 2\,P_{22}
\end{bmatrix}
}_Q x - \frac{1}{2}x_2\,\sin(x_2). \tag{1}
$$

The matrix $Q$ can only be made positive semi-definite when $P_{22} = 1/2$, which simplifies $(1)$ to

$$
\dot{V} = -\frac{1}{2} (x_1 + x_2)^2 - \frac{1}{2}x_2\,\sin(x_2), \tag{2}
$$

which is negative definite for $-\pi < x_2 < \pi$.

Another option is to make all non-polynomial terms of the time derivative of the Lyapunov function zero, which can be done using $P_{11}=P_{12}=1$. This only enables you to get a negative semi-definite result at best when using $P_{22}=1$, yielding

$$
\dot{V} = -(x_1 + x_2)^2. \tag{3}
$$

Using LaSalle's invariance principle you can still show that all initial conditions converge to one of the equilibria (though, this does not show that most initial conditions converge to the stable equilibria with even $n$).


问:
Matrix with eigenvalues as $0$ and rank as $n-1$
How do I find a matrix $A \in \mathbb{R}^{n * n}$ with all eigenvalues as zero and rank being $n-1$.

Is this possible? Because the number of non-zero eigenvalues is the rank. But this question has cropped up in my assignment.

Any example will do. Thanks!

答:
Consider a $3\times3$ matrix as below:

$A=\begin{bmatrix} 0 & 3 & 0 \\ 0 & 0 & 3\\ 0 & 0 & 0  \end{bmatrix}.$

Here $rank(A)=2$ and $\lambda_1=\lambda_2=\lambda_3=0$.

The above matrix is a nilpotent matrix.


问:
Relationship determinant symmetric matrix and skew-symmetric counterpart
Say we have a square matrix A and we write it as the sum of its symmetric and skew-symmetric counterpart. Is there any formula which relates the determinants of A, its symmetric and skew-symmetric parts?</p>

答:
Let $A=B+C$, where $B$ is symmetric and $C$ is (hollow and) skew-symmetric.

When $n\le2$, one may directly verify that $\det(A)=\det(B)+\det(C)$.

When $n\ge3$, $\det(A)$ is not a function of $\det(B)$ and $\det(C)$. It suffices to construct two symmetric matrices $B_0,B_1$ and two skew-symmetric matrices $C_0,C_1$ such that $\det(B_0)=\det(B_1)$ and $\det(C_0)=\det(C_1)$ but $\det(B_0+C_0)\ne\det(B_1+C_1)$.

Here are our constructions (that work over all fields, including those of characteristic is $2$). When $n\ge3$ is odd, let
$$
B_k=\pmatrix{1&1&0\\ 1&1&0\\ 0&0&k}\oplus I_{n-3},
\quad C_0=C_1=\pmatrix{0&-1&0\\ 1&0&0\\ 0&0&0}\oplus0.
$$
Then $\det(B_0)=\det(B_1)=\det(C_0)=\det(C_1)=0$ but $\det(B_0+C_0)=0\ne1=\det(B_1+C_1)$.

When $n\ge4$ is even, let
\begin{aligned}
B_0&=I_2\oplus\pmatrix{1&1\\ 1&1}\oplus I_{n-4},\\
B_1&=I_2\oplus\pmatrix{0&0\\ 0&0}\oplus I_{n-4},\\
C_0&=\pmatrix{0&-1\\ 1&0}\oplus\pmatrix{0&0\\ 0&0}\oplus0,\\
C_1&=\pmatrix{0&0\\ 0&0}\oplus\pmatrix{0&-1\\ 1&0}\oplus0,\\
\end{aligned}
Again, $\det(B_0)=\det(B_1)=\det(C_0)=\det(C_1)=0$ but $\det(B_0+C_0)=0\ne1=\det(B_1+C_1)$.


问:
Is this Stochastic Process bounded from above?
I'm dealing with a problem in my thesis that involves proving the boundedness of a Stochastic Process. This question is related to another one I did recently, but the results found in the other one sadly didn't help me sufficiently.

Let $\text{d}X_t= f(t,X_t) \text{d}t +\text{d}W_t$ be a stochastic process defined for the times $t\ge0,$ such that $X_0=0$ and $f(t,x)<a<0$ for all $t\ge0,x>k$ for a certain constant $k$.

Prove that $$\mathbb{P}[\exists C>0: X_t<C \ \forall t\ge 0]=1.$$

In other words, I wish to show that if the function $f$ is negative when $X_t$ goes above a certain value $k$, then the stochastic process is bounded from above, because the drift part is going to be negative.

I know that $Y_t=at+W_t$ is bounded from above if $a<0$, but what I can't manage to deal with is the fact that we cannot control the moments when $X_t$ goes above $k$: every time it goes above $k$ it might reach a higher peak since the bound on $Y_t$ is not $\omega$-wise uniform, and the sequence of those peaks might be unbounded.

Here's a simulation of the SDE $\text{d}Y_t=-Y_t\text{d}t+\text{d}W_t$, which seems to fit these hypotheses, as the time taken is quite long.

EDIT: my SDE, in particular, is $$\text{d}X_t=X_t\frac{D(X_t)-S}{D(X_t)+S}\text{d}t+X_t\sigma\text{d}t,$$ with $D$ being $D(x)=D_0\exp(-\alpha x),$ for some $D_0>S>0$, and $X_0>0$. This SDE is equivalent to $$\text{d}Y_t=\left(\frac{D(\exp(Y_t))-S}{D(\exp(Y_t))+S}-\frac{\sigma^2}2\right)\text{d}t+\sigma\text{d}t,$$ thanks to the substitution $Y_t=\log(X_t)$: this second equation is the one I was referring to. Actually, $S$ is not a constant but a bounded function, which breaks the Markovianity, but I think that if the result is not true for constant $S$ the it's not true also for bounded $S$.

Any help would be immensely appreciated.

答:
This post is meant to explain why your toy example also fails and why I think your simulation did not show a good illustration here

Your toy example is Ornstein-Uhlenbeck process ( me too, I was having a little trouble solving until I realize how trivial it is when writing this answer), with solution:
$$ Y_t= e^{-t} \underbrace{ \int_{0}^t e^sdW_s}_{=: A_t}$$

Moreover, $A_t$ is just a time change Brownian, that is, there is a Brownian process $B$ such that:
$$ (A_t)= \left( B_{e^{2t}-1}\right)$$
and clearly, $(Y_t)$ is a.s unbounded.

Remark: We do have the same manipulation for a suitable SDE of form :
$$dY_t= -g(t)Y_tdt+dW_t$$
or even
$$dY_t= -g(t,Y_t)Y_tdt+dW_t$$
under some conditions on $g$

@to Mods: Since this topic is now more like a discussion. I would like to break my long comment in multiple posts for readability. Please kindly understand.


答:
Consider  $A\in \mathbb{R}^{n\times n} $

$$a_{ij} = \begin{cases} 1&  j=i+1,\\
0& \text{otherwise} \end{cases}$$

Title: Matrix with eigenvalues as $0$ and rank as $n-1$

Body:
How do I find a matrix $A \in \mathbb{R}^{n * n}$ with all eigenvalues as zero and rank being $n-1$.

Is this possible? Because the number of non-zero eigenvalues is the rank. But this question has cropped up in my assignment.

Any example will do. Thanks!


问:
Is it possible to do an integration over a Bézier curve "area"?
I have a shape defined by a Bézier spline that has a width,
and I want to give it an attractive force.

Is such a thing even doable without approximating it?

Edit:

To integrate the area, I think I will first have to express the answer for a line (crossection) of the path, then integrate it through the path.

Edit:

Moving into the teritory of actually answering the question, but I think I'll keep editing the question until I actually have a final answer.

So, now I want to rename the spline, it will new be defined by $$Q(t) = (1-t)^3\ Q_0+(1-t)^2t\ Q_1+(1-t)t^2\ Q_2+t^3\ Q_3$$ rearranging gives $$Q_3\ t^3+Q_2\ (t^2-t^3)+Q_1(t-2t^2+t^3)+Q_0(1-3t+3t^2-t^3)\\=\\(\underbrace{Q_3-Q_2+Q_1-Q_0}_{\mathcal{A}})\ t^3+(\underbrace{Q_2-2Q_1+3Q_0}_{\mathcal{B}})\ t^2+(\underbrace{Q_1-3Q_0}_{\mathcal{C}})\ t+Q_0$$

Finding the speed vector and its square $Q′(t)$:$$Q′(t)=3\mathcal{A}t^2+2\mathcal{B}t+\mathcal{C}$$
$$(Q′(t))^2=9\mathcal{A}^2t^4+4\mathcal{B}^2t^2+\mathcal{C}^2+12\mathcal{A}\mathcal{B}t^3+6\mathcal{A}\mathcal{C}t^2+4\mathcal{B}\mathcal{C}t$$

Now I'm going to find $P_0(t)$ and $P_1(t)$.

Then, I want to rotate the speed vector left and right, putting $P_0$ on the left side and $P_1$ on the right. Rotating a vector $(x,\ y)$ gives $(-y,\ x)$. Rotating right gives $(y,\ -x)$.

We also devide by the length of the speed vector to get a unit vector. Then multiplying with the spline width $W$.

We name the length of the speed vector $$\Gamma(t)=\sqrt{Q′_x(t)^2+Q′_y(t)^2}$$

We get $$P_0(t) = Q(t) + W{(-Q′_y(t),\ Q′_x(t))\over\Gamma(t)}\\[1.5em]P_1(t) = Q(t) + W{(Q′_y(t),\ -Q′_x(t))\over\Gamma(t)}$$

Can I now just do this? $$\iint{1 \over \lVert P(u,t)\rVert ^2}\Gamma(t)\ du\ dt$$

I'm trying to express: the force of attraction by a spline cross section times how far the cross section moves at time $t$.

答:
Reposting an answer that I got on paper for posterity.

Part 1

$$C(t)=P_0B_{0,3}+P_1B_{1,3}+P_2B_{2,3}+P_3B_{3,3}$$ where B is a Bernstein polynomial

$$t\in[0,1]$$
Force on the line:
$$F_{line}=\int_{t=0}^1{1\over(Q(t))^2}\ dt\\[2em]
=\int_{t=0}^1{1\over((1-t)^3\ Q_0+(1-t)^2t\ Q_1+(1-t)t^2\ Q_2+t^3\ Q_3))^2}\ dt$$

Assume: Roots $\alpha$, $\beta$, $\gamma$ of $Q(t)$ exist
$${1\over((t-\alpha)(t-\beta)(t-\gamma))^2}\\[2.5em]=\overbrace{{A\over(t-\alpha)}+{B\over(t-\beta)}+{C\over(t-\gamma)}+{D\over(t-\alpha)^2}+{E\over(t-\beta)^2}+{F\over(t-\gamma)^2}}^{constants}\\[3.5em]
\begin{align}\int_0^1{A\over t-\alpha}\ dt &= A(log(1-\alpha)-log(-\alpha))\\[1.5em]
\int_0^1{D\over (t-\alpha)^2}\ dt &= \left[\begin{array}{}u=t-\alpha\\du=dt\\t=0\implies u=-\alpha\\t=1\implies u=1-\alpha\end{array}\right]\\[1.5em]&=D\int_{-\alpha}^{1-\alpha}{1\over u^2}\ du=D\left[{-1\over u}\right]_{-\alpha}^{1-\alpha}\\[1.5em]&=D\left({-1\over 1-\alpha}+{1\over\alpha}\right)\end{align}$$

So $\int_0^1{1\over(Q(t))^2}\ dt$ turns into six solvable integrals.

As an approximation of your desired force over the area, you could do $F_{area}\approx wF_{line}$. $w$ is width of band.

Part 2

More serious attempt. No approximations.

Say we have two equations $Q_l(t)$ and $Q_r(t)$ that fulfill $${Q_l(t)+Q_r(t)\over2}=Q(t)$$

$$Q_l(t) = Q(t) + \delta\ \vec{v}(t)$$

$$Q_r(t) = Q(t) - \delta\ \vec{v}(t)$$

$\vec{v}(t)$ is always orthogonal to $Q(t)$ and has length 1

$$\vec{v}(t)={\vec{T}′(t)\over\lVert\vec{T}′(t)\rVert},\ \text{where}\ \vec{T}(t)={\vec{Q}′(t)\over\lVert\vec{Q}′(t)\rVert}$$

So $Q_l(t)$ and $Q_r(t)$ exist and should be not too hard to find. We could also construct them by moving $Q$'s control points.

Now introduce a new variable $\omega$ that goes from 0 to 1, and swap out $Q(t)$ in the original  problem with $(1-\omega)\ Q_l(t)+\omega\ Q_r(t)$. We can set up the integral $$\int_{\omega=0}^1\int_{t=0}^1{1\over((1-\omega)\ Q_l(t)+\omega\ Q_r(t))^2}\ dt\ d\omega$$

Solve for the $\omega$ integral first
$$\begin{align}\int_{\omega=0}^1{1\over((Q_r-Q_l)\ \omega+Q_l)^2}\ d\omega&=\int_0^1{1\over((Q_r-Q_l)(\omega+{Q_l\over{Q_r-Q_l}}))^2}\ d\omega\\[2.5em]&={1\over(Q_r-Q_l)^2}\int_0^1{1\over(\omega+{Q_l\over Q_r-Q_l})^2}\ d\omega\\[2.5em]&= {1\over(Q_r-Q_l)^2}\left[{-1\over(1+{Q_l\over Q_r-Q_l}}+{Q_r-Q_l\over Q_l}\right]\\[2.5em]&={-1\over((Q_r-Q_l)^2+Q_l(Q_r-Q_l))}+{1\over Q_l(Q_r-Q_l)}\\[2.5em]&={-1\over Q_r^2-Q_rQ_l}+{1\over Q_lQ_r-Q_l^2}\\[2.5em]&={1\over Q_rQ_l-Q_r^2}+{1\over Q_lQ_r-Q_l^2}\end{align}$$

Two very similar terms. Let's solve the integral for one of them, the other will be analogous. $$\require{cancel}\int_{t=0}^1{1\over Q_lQ_r-Q_l^2}\ dt\\[2.5em]
{1\over Q_lQ_r-Q_l^2}={1\over(Q(t)+\delta\ v(t))\ (Q(t)-\delta\ v(t))-(Q(t)+\delta\ v(t))^2}\\[2.5em]{1\over\cancel{Q^2}-(\delta\ v)^2-\cancel{Q^2}-2Q\delta\ v- (\delta\ v)^2}\\[2.5em]={-1\over2(Q\delta\ v+(\delta\ v)^2)}$$

So you get $$-{1\over2\delta}\int_0^1{1\over Q(t)v(t)+\delta\ v(t)^2}\ dt$$

$v(t)$ should be 1st order polynomial.
Denominator is then 4th order polynomial.
Its roots can be found with some closed form equation.
$$-{1\over2\delta}\int_0^1{1\over(t-\epsilon)(t-\zeta)(t-\eta)(t-\theta)}\ dt$$

The integrand can be split into partial fractions, and we get
$$-{1\over2\delta}\left(\int_0^1\overbrace{{G\over(t-\epsilon)}\ dt+\int_0^1{H\over(t-\zeta)}\ dt+\int_0^1{I\over(t-\eta)}\ dt+\int_0^1{J\over(t-\theta)}}^{constants}\ dt\right)$$

Each integral is solvable with variable substitution like we did before, we get
$$F_{Area}={-1\over2\delta}\left(G(log(\ldots)-log(\ldots))+H(log(\ldots)\cdots\right)$$


问:
some questions about complex analysis
1a) Find the largest open disc on which the series $\sum_{n = 0}^\infty n \cdot (2z+1)^n $ converges.

1b) Find the function represented by the derivative of this power series.

I already found when it converges but my question is if the function which I have to differentiate is $n\cdot(2z+1)^n$ or an equivalent sum.

2)I have to sketch the curve defined by this equation:$|z + 1| + |z - 1| = r $ for $ r, z \in [2, \infty) \times \mathbb{R}^+ $.

The only thing that cames to my mind is to simplify the expression using $ |z| \ge z $, but I don't know if this correctly produces the sketch.

答:
For the second part of #1, you are correct in that you shouldn't just differentiate your given sum and then try to arrive at an equivalent sum. Rather, note that $$\begin{align*} \sum_{n = 0}^\infty n \cdot (2z + 1)^{n} &= {2z + 1} \sum_{n = 0}^\infty n \cdot (2z + 1)^{n-1}\\& = \frac{2z + 1} {2} \sum_{n = 0}^\infty \frac{d}{dz} (2z + 1)^n \\&= \frac{2z + 1}{2} \cdot\left[ \frac{d}{dz}\sum_{n = 0}^{\infty} (2z + 1)^n \right] \\&= \frac{2z + 1}{2}\cdot \left[\frac{d}{dz} \frac{1}{1 - (2z + 1)}\right]\\& = \frac{2z + 1}{4z^2} \end{align*}$$.

Hence, its derivative is simply $ \frac{d}{dz} \frac{2z+1}{4z^2} = - \frac{z+1}{2z^3} $.

For #2, you are going to have a piece-wise function with 3 different line segments. Let $r = f(z) $ denote the value of $ r $ as a function of $ z $. Using the basic definition of the absolute value function, $$ f(z) = \begin{cases}-2z &\ z \ < -1 \\\ \ \ 2 & |z| \le\ \ \ 1 \\ \ \ \ 2z & \ z \ > \ \ \ 1 \end{cases} $$

Title: some questions about complex analysis


问:
How to prove this inequality using Taylor approximation?
I'd like to prove the following via Taylor approximation :

$$x > 0 \implies \sin x > x-\frac{x^3}{3!}$$

I tried to estimate the error but I found $E_3(x) \leq |\frac{x^4}{4!}| $, which doesn't give information about the sign of the error. Then, I tried to compare the sign of each couple of terms in the error but it gives the right result just when $0<x \leq1 $ because in this case

$$(\frac{x^5}{5!} - \frac{x^7}{7!} )+(\frac{x^9}{9!} -\frac{x^{11}}{11!} ) + \dots$$

all the couples are positive. But from here I don't know how to continue to include the case $x>1$.

How can I find a way using Taylor approximation ?  (I know other methods like derivatives can be used, but I want just to use Taylor if possible).

Besides, Is there a general method to find the intervals where the sign of the error is positive/negative?

Body:
Hint

If $x\in [0,\pi]$, there is $b\in [0,x]$ s.t. $$\sin(x)=x-\frac{x^3}{3!}+\frac{\sin(b)}{4!}x^4.$$

If $x\geq \pi$ then $x-\frac{x^3}{3!}< -1$.


问:
Proving if monotonic or strictly monotonic
I have a function and its derivative in the given interval is always greater than or equal to 0, f'(x)>=0. I am asked to speak about the monotonicity of this function. I know it is monotonic and increasing in the interval from its graph and also the derivative however, how do i know if it is strictly monotonic or not?</p>

答:
Since the derivative $f'(x)\geq 0$, there is a point $t$ where $f'(t)=0$, at which point $f(x)$ will be flat, not increasing.  Hence, the function is just monotonically increasing; it would be strictly increasing if $f'(x)>0$.

In this sense, "increasing" really means "non-decreasing".


问:
How do I finish this probability question?
A fair 4-sided is rolled 6 times. What is the probability that the total number of 1’s and
2’s in the first three rolls equals the total number of 3’s and 4’s in the last three rolls?

Here's what I've done:
$P\{X=i\}=1/4$ for all $i\in\{1,2,3,4\}$ hence $E[X]=(1+2+3+4)/4=5/2$, $E[X^2]=(1+4+9+16)/4=15/2$, $Var(X)=15/2-(5/2)^2=5/4$ and hence the standard deviation=$5^(1/2)/2$. How can I follow up on this?

答: Don't think too complicated on this one. This can be solved only using 'basic' probability: You might want to simplify this to a 2 sided coin in either head or tails, as all the chances are equivalent to a half. You need to find the sum of the square of the probability of having zero, one, two or three heads, which is $\frac{1^2+3^2+3^2+1^2}{8^2} = \frac{5}{16}$


问:
Poisson Distribution in Nature
The Poisson distribution is utilized for the number of events occurred in a specific interval. There are a lot of events that could be modeled with this distribution such as the number of meteorites or the patients arriving at the hospital at a given time. The formula of Poisson Distribution probability mass function is
$$
f(k;\lambda) = \frac{\lambda^ke^{-\lambda}}{k!}
$$
where $\lambda$ is called Poisson parameter and $k$ is the number of occurrences. We had this distribution as an approximation of Binomial when $np \leq 10$. However, I would like to know if there is a explanation or Theorem (like Central Limit Theorem For Gaussian Distribution) why some specific events follow this distribution and which relevant characteristics they share?

答:
The main theorem you can use is the one that links Poisson and Negative Exponential (it is very easy to be proved).

the interarrival time (time between an arrival and another) in a Poisson process with parameter $\theta$ is Exponentially distributed with mean $\frac{1}{\theta}$

Thus any event represented with a negative exponential (i.e. lifetime of some devices) but viewed as # of arrival can be represented by a Poisson

Title: Poisson Distribution in Nature


问:
Tricky Expectation
Let $X$ denote a Poisson distributed random variable and let $Z = \max(X-\min(X,c),0)$, where $c$ is a constant. How do I compute $\mathbb E[Z]$?

Any help is greatly appreciated!

答: Hint: write $Z=(X-c)\mathbb{1}_{\{X\geq c\}}$.</p>

Title: Tricky Expectation


问:
On the convergence of the Cauchy's product of two convergent alternating series
I want to show the following result:

If $(a_n)$ and $(b_n)$ are monotone sequences with limit $0$, then the Cauchy's product series of $\sum(-1)^na_n$ and $\sum(-1)^nb_n$ is convergent if, and only if, the numbers $\sigma_n = a_n(b_0+b_1+\ldots+b_n)$ and $\tau_n = b_n(a_0+a_1+\ldots+a_n)$ also form a null sequence.

I think that this result is quite interesting: for example we can infer that the Cauchy's product series of two alternating harmonic series is convergent.
We can assume both $a_n, b_n \ge 0$ monotone descending. It's easy to show that those conditions are necessary, by just looking at the general term $c_n$ of the product series: $$c_n = (-1)^n(a_0b_n+a_1b_{n-1}+\ldots + a_nb_0)$$ for which the absolute value can be lower bounded by both $\sigma_n$ and $\tau_n$ (and hence cannot tend to $0$ if $\limsup \sigma_n > 0$ or $\limsup \tau_n > 0$). I don't really know how to prove that these conditions are also sufficient. Can you just give me a hint? Thanks in advance!

答:
I'll assume that $a_n$ and $b_n$ are nonnegative, otherwise we just have to multiply everything with $-1$ once or twice. Further I'll write $A_n = a_0 + \ldots + a_n$ and similarly for $B_n$, and deviating from your notation I'll write
$$c_n = a_0b_n + \ldots + a_nb_0\,.$$
Thus the Cauchy product of the two series $\sum (-1)^na_n$ and $\sum (-1)^nb_n$ is $\sum (-1)^nc_n$.

Now trivially a necessary condition for the convergence of the Cauchy product is $c_n \to 0$. Since
$$\sigma_n = a_nB_n \leqslant c_n \qquad\text{and}\qquad \tau_n = A_nb_n \leqslant c_n$$
it is immediate that convergence of the Cauchy product implies $\sigma_n \to 0$ and $\tau_n \to 0$. In the other direction, we first observe that splitting the sum defining $c_n$ in the middle yields the inequality
$$c_n \leqslant a_{\lfloor n/2\rfloor} B_{\lfloor n/2\rfloor} + A_{\lfloor n/2\rfloor} b_{\lfloor n/2\rfloor}\,.$$
Hence $\sigma_n \to 0$ and $\tau_n \to 0$ together imply $c_n \to 0$. If we could show that $c_n$ is eventually decreasing, convergence of the Cauchy product would then follow by the Leibniz criterion. But I don't see how that could be shown, so I'll take a less direct route.

It is clear that the sequence of products of the partial sums,
$$P_n = \Biggl(\sum_{k = 0}^{n} (-1)^k a_k\Biggr)\Biggl(\sum_{k = 0}^{n} (-1)^k b_k\Biggr)\,,$$
converges, and therefore it is sufficient to show that
$$P_n - \sum_{k = 0}^{n} (-1)^k c_k$$
tends to $0$. We can write this difference as
$$(-1)^{n+1} \sum_{j = 1}^{n} (-1)^{j-1} s_{n,j}\,,$$
where
$$s_{n,j} = \sum_{m = j}^{n} a_mb_{n+j-m}\,.$$
For $1 \leqslant j < n$ we have
$$s_{n,j+1} = \sum_{m = j+1}^{n} a_mb_{n+j+1-m} \leqslant \sum_{m = j+1}^{n} a_m b_{n+j-m} \leqslant s_{n,j}$$
by monotonicity of $(b_n)$ and nonnegativity, whence
$$\Biggl\lvert\, P_n - \sum_{k = 0}^{n} (-1)^k c_k\Biggr\rvert \leqslant s_{n,1} \leqslant c_{n+1} \to 0\,.$$


问:
Optimal way to do linear approximation of function on an interval.
Assume that I want to find the line which in squared integral sense ends up being closest to another function on an interval:

$$\min_{k,m}\left\{\int_{a}^{b} |f(x)-(kx+m)|^2dx\right\}$$

Expanding the square:
$$f(x)^2 - 2 f(x)(kx+m) + (kx+m)^2$$

Let's assume we know $$\int_a^b f(x) dx = [F(x)]_a^b\\\int_a^b f(x)^2 dx = [F_2(x)]_a^b$$

How can we derive optimal $k$ and $m$? Is this sufficient information or do we need to know something more?

答:
As Damien writes, we will need the expression $$\int_a^b xf(x) dx$$

We can try integration by parts:

$$\int_a^b u(x)v'(x)dx = [u(x)v(x)]_a^b - \int_a^b u'(x)v(x)dx$$
If we set $u(x)=x,v'(x)=f(x)$

$$\int_a^b xf(x)dx = [xF(x)]_a^b - \int_a^b F(x)dx$$

Here we will also need $\int_a^b F(x)dx$.

And since we don't know $F(a),F(b)$ separately we won't be able to calculate $bF(b)-aF(a)$ either.

Another thing we can do is to approximate $f$ by straight line between $f(a)$ and $f(b)$ on $[a,b]$ for only the purpose of approximating the above integral.

$$f_{appr}(x) = f(a) + \frac{(x-a)(f(b)-f(a))}{b-a}$$

Now $\int xf_{appr}(x)dx $ can be easily calculated as a polynomial.

How good this approximation will be, I don't know. Experiments will probably need to be conducted.


问:
If 550 represents 0 and 1280 represents 15, a value between 550 and 1280, what value does it represent according to the above?
I am trying to improve my math skills, and I am trying to solve this problem:

I have values that can be in the range between 550 and 1280.

```<code>550 -> 0  (minimum value represents zero)
1280 -> 15 (maximum value represents 15)

780 -> ??
</code>```

What is the mathematical formula to calculate the corresponding value if, for example, the input value for this formula is 780?

excuse my ignorance

答:
Hint:

Let $y=ax+b$

If $x=550,y=0\implies 0=550a+b$

If $x=1280,y=15\implies ?$

Can you solve the two simultaneous equations in $a,b?$


问:
Can I multiply both sides with a polynomial to eliminate the denominator?
I have this equation $\frac{x-1}{x+3} = 0$. Solving this one through cross multiplication would give me $x = 1$.

But, If I'm correct, we can also express this one as $\frac{x}{x+3} = \frac{1}{x+3}$. If I cross-multiply, I'd get $x = 1$ or $x = -3$. But if I multiply both sides with $(x+3)$ to remove the denominator, I'd arrive with the same answer from the first approach.

So, how should I tackle this one? Can I multiply both side of an equation with a polynomial to cancel a denominator?

答:
As a general rule of thumb:

$$\bigg\{x\ \bigg|\ \frac{f(x)}{g(x)}=0\bigg\}=\bigg\{x\ \bigg|\ f(x)=0; g(x)\neq 0\bigg\}$$

So yes, multiplying or dividing by an expression is fine, but you must discard all solutions to $f(x)=0$ when $g(x)$ is zero too.