问:
Contour integral of fraction (involving Gamma function) = Polynomial

Let $n$ be a natural number ( $0$ included ) and let
$x > 0$ be real.

Let $\gamma$ be a closed curve which encircles the points $0, 1, 2, \ldots, n$ in positive direction
( anti-clockwise ).
$$
\mbox{Define}\quad\operatorname{P}_{n}\left(x\right) = \frac{1}{2\pi\mathrm{i}}
\oint_{\gamma }\frac{\Gamma\left(t - n\right)}
{\Gamma^{2}\left(t + 1\right)}\,x^{t}\,\mathrm{d}t.
$$

Show that $P_{n}(x)$ is a polynomial of degree $n$.

My work so far: I tried to rewrite $\Gamma(t-n)$ by using the identity $\Gamma(z) = \frac{\Gamma(z+1)}{z}.$ Doing this I obtain that:

$P_{n}(x) = \frac{1}{2\pi i} \int_{\gamma}\frac{x^{t}}{\Gamma(t+1)(t-n)(t-n+1)...(t)}dt$.

From there, I can't see why that expression would be a polynomial. Any hints would be appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

The integrand numerator has poles at $\ds{1, 2, 3, \ldots, n}$: $\color{red}{The\ residue}$ at a pole $\ds{k \in \braces{1,2,3,\ldots,n}}$ is given by:
\begin{align}
&\lim_{t \to k}\bracks{\pars{t - k}
{\Gamma\left(t - n\right) \over
\Gamma^{2}\pars{t + 1}}\,x^{t}}
\\[5mm] = &\
{x^{k} \over
\Gamma^{2}\pars{k + 1}}
\lim_{t \to k}\bracks{\pars{t - k}
\Gamma\pars{t - n}}
\\[5mm] = &\
{x^{k} \over\pars{k!}^{2}}
\lim_{t \to k}\bracks{\pars{t - k}
{\pi \over \Gamma\pars{1 - t + n}
\sin\pars{\pi\bracks{t - n}}}}
\\[5mm] = &\
\pi\,{x^{k} \over\pars{k!}^{2}\pars{n - k}!}
\lim_{t \to k}\
{t - k \over \sin\pars{\pi\bracks{t - n}}}
\\[5mm] = &\
\pi\,{x^{k} \over\pars{k!}^{2}\pars{n - k}!}
\lim_{t \to k}\
{1 \over \cos\pars{\pi\bracks{t - n}}\pi}
\\[5mm] = &\
\pars{-1}^{k - n}\,
{x^{k} \over\pars{k!}^{2}\pars{n - k}!}
\end{align}

Then,
\begin{align}
\operatorname{P}_{n}\pars{x} & \equiv
\bbox[5px,#ffd]{{1 \over 2\pi\ic}
\oint_{\gamma }{\Gamma\pars{t - n} \over \Gamma^{2}\pars{t + 1}}\,x^{t}\,\dd t}
\\[5mm] = &
\bbx{\pars{-1}^{n}\sum_{k = 1}^{n}
{\pars{-1}^{k} \over\pars{k!}^{2}\pars{n - k}!}\,x^{k}}
\\ &
\end{align}


问:
How to find probability density of a transformed uniform RV?
Question: Let $X ∼ Unif[−3, 3]$, and $Y  =  (X + 1)^2$. Find  $f_{Y}(1)$ and  $f_{Y}(9)$, where  $f_{Y}(·)$ is the probability density function of Y.

My Approach:
The pdf of $X$ is the following: $\frac{1}{6}$ if $ -3 \le x \le 3 $ and 0 otherwise

$$F_{Y}(Y) = P(Y \le y) = P((X+1)^2 \le y) = P(-\sqrt{y}-1 \le X \le \sqrt{y}-1)$$
$$ = F_{x}(\sqrt{y}-1) - F_{x}(-\sqrt{y}-1)$$
$$ f_{Y}(Y) = f_{x}(\sqrt{y}-1)\frac{d}{dy}(\sqrt{y}-1) - f_{x}(-\sqrt{y}-1)\frac{d}{dy}(-\sqrt{y}-1)$$

The pdf $ f_{Y}(Y) $ is then:

$$ f_{Y}(Y) = f_{x}(\sqrt{y}-1)\frac{1}{2\sqrt{y}} - f_{x}(-\sqrt{y}-1)\frac{-1}{2\sqrt{y}}$$

I substitute Y=1 and Y=9 to get

$$ f_{Y}(1) = f_{x}(\sqrt{1}-1)\frac{1}{2\sqrt{1}} - f_{x}(-\sqrt{1}-1)\frac{-1}{2\sqrt{1}}$$

$$ f_{Y}(1) = (\frac{1}{6})\frac{1}{2} + (\frac{1}{6})\frac{1}{2}$$
$$ f_{Y}(1) = \frac{2}{12} $$

Which is just $\frac{1}{6}$. I repeated this same process with $f_{Y}(9)$ and got $\frac{1}{36}$, and both of those answers are incorrect. If anyone can help me understand where I went wrong I would greatly appreciate it. Thank you!

答:
$Y=(X+1)^2$ for $X\sim\mathcal{U}[-3..3]$ means the transformation will fold the support of $[-3..-1)\cup[-1..3]$ onto $[0..16]$.  Specifically it will map $[-3..-1)$ onto $(0..4]$ and $[-1..3]$ onto $[0..16]$.

Thus I obtain that:

$$\begin{align}f_{\small Y}(y)&=\left\lvert\dfrac{1}{2\surd y}\right\rvert\,(f_{\small X}(-\sqrt y-1)\,\mathbf 1_{y\in(0..4]}+f_{\small X}(\sqrt y-1)\,\mathbf 1_{y\in(0..16]})\\[1ex]&=\dfrac{\mathbf 1_{y\in [0..4]}}{6\surd y}+\dfrac{\mathbf 1_{y\in(4..16]}}{12\surd y}\end{align}$$

So $f_{\small Y}(1) = \tfrac 1{6}$ and $f_{\small Y}(9)=\tfrac 1{36}$ as you had.

ContentLicense: CC BY-SA 4.0


问:
Proving $\lim _{n\to\infty} \sqrt{n} \int_{1}^{\infty} e^{-n x^{2}} d x=0$
How to prove the following?
\begin{equation}
\lim _{n \rightarrow \infty} \sqrt{n} \int_{1}^{\infty} e^{-n x^{2}} d x=0
\end{equation}

答:
Some hints:

For $x \geq 1$ and $n \in \mathbb{N}$ you have $e^{-nx^2} \leq e^{-nx}$

$\int_1^{\infty}e^{-nx}dx = \frac{e^{-n}}{n}$

Now, you can bound $\sqrt{n} \int_{1}^{\infty} e^{-n x^{2}} dx$ easily from above.


问:
what is the meaning of this symbol $ ∧?$
I have some confusion in symbolics

what is the meaning of this symbol $\land?$

My attempt : I was googling and got the answer given below

But i didn't  understand the given above  answer

My confusion come from this, pic is given below

答:
The expression
$$  x^n=1\land \forall  k<n\colon x^k\ne 1 $$

is a logical conjunction of the two terms

$$ x^n = 1$$
and
$$ \forall  k<n\colon x^k\ne 1 .$$

You might write it like this to make it clearer how the symbols are grouped into expressions:
$$  (x^n=1)\land (\forall  k<n\colon x^k\ne 1). $$

This conjunction is true if and only if $x^n=1$ is true and
$\forall  k<n\colon x^k\ne 1$ is true.


问:
How are Sets different from Boxes?
I am reading the book
"Naive Set Thoery" by Sir Halmos. There, in

Section 2: Axiom of Specification, pg-5, first line, he writes
"The analogy between sets and boxes has many weak points"

What are these drawbacks? i.e Are there any properties which sets have and boxes do not, or vice-versa?

答:
One big issue is that an object can only be in one box at a time, but it can be in multiple sets.  For instance, the sets $\{1,2\}$ and $\{1,3\}$ cannot simultaneously be viewed as boxes, since the number $1$ would have to be in two different boxes at once.  You could get around this by allowing multiple "copies" of objects, so you have two different $1$s, one in each box.  This creates more problems, though: what if you put both $1$s in the same box?  A box with two $1$s in it is different from a box with one $1$, but $\{1,1\}$ is the same set as $\{1\}$.

Another issue is extensionality: two sets which have the same elements are equal, but is that true of boxes?  If you have two different boxes which each contain a $1$, are they actually the same box?  Well, maybe you say they are, as an extension of our policy of saying the two $1$s are "the same".  But what about two empty boxes?  Surely you can have two different boxes that are both empty, but extensionality means there can only be one empty set.

Title: How are Sets different from Boxes?


问:
Find the limit of such a sequence defined by recurrence (problematic induction because of $0$ in the denominator)
So I need to find the limit of such a sequence defined by recurrence:

$a_1 = \frac{1}{2}$

$a_{n+1}= \frac{\sqrt{1+a_n^2}-1}{a_n}$

I have calculated some elements of that seqence and those are:

$a_1 = 1$

$a_2= \sqrt{2}-1 \approx 0,41$

$a_3 = 1+ \frac{\sqrt{2}}{2} = \frac{\sqrt{4-\sqrt{2}}-1}{\sqrt{2}-1} \approx 0,22$

So I know that the values of next elements of the sequence are falling. Next I wanted to do prove by induction that the limit = $0$ , but I can't put $0$ in the denominator. So what do I do?

答: We can rearrange the given recurrence as $$a_{n+1}^2a_n+2a_{n+1}-a_n=0,$$ and take the positive solution of $a_{n+1}.$ I will prove that the sequence is bounded and monotonically decreasing by manipulating this new recurrence relation. Note that $$2a_{n+1}=a_n-a_{n+1}^2a_n= a_n(1-a_{n+1})(1+a_{n+1})\gt0$$ implies that $1-a_{n+1}\gt0,$ and hence $$1\gt a_{n+1}\gt 0$$ for all $n$ (including $0$). Also $$a_n-a_{n+1}=a_{n+1}+a_{n+1}^2a_n= a_{n+1}(1+a_{n+1}a_n)\gt 0$$ implies that $a_n\gt a_{n+1},$ and therefore the sequence is decreasing. Now by the (bounded) monotone convergent theorem, this sequence must have a limit $l$ in non-negative real numbers. Find it by solving $$l^3+2l-l=l(l^2+1)=0.$$


问:
How to verify the order of DOPRI Runge-Kutta method?
I've written code in Fortran based on the RK8(7)-13 method by Dormand and Prince to solve the system $\mathbf{y}'=\mathbf{f}(t,\mathbf{y})$. The method is Runge-Kutta 8$^\text{th}$ order with an embedded 7$^\text{th}$ order method; I want to verify that the error on the output agrees with this. For a method of order $p$ and timestep $h$, we expect the local truncation error to be $\mathcal{O}(h^{p+1})$. Therefore, if I advance the program by one time step and calculate $\text{error} = |y_{\text{RK}}-y_{\text{exact}}|$ (say I know $y_{\text{exact}}$) and plot $\log(\text{error})$ vs. $\log(h)$ for several values of $h$, the slope of the line should be 9.

Here's the problem: it isn't. I've tried simple polynomials like $f(t,y) = 1$ and $f(t,y) = 2t$, but the slope of the line always seems to be between 1 and 2; this is also the case for $f(t,y) = y$. However, for $y'' + y = 0$ (which translates into the coupled system $\mathbf{f}(t,y_{1}, y_{2}) = (y_{2}, -y_{1})$), I found that the slope is almost exact 9 for larger values of $h$ - for small values the slope goes back to ~1. Furthermore, the magnitude of the error is $~10^{-20}$ for this small $h$. I was wondering if anyone had any ideas as to what the issue could be.

My ideas:

I've been misinformed as to what order means, and the method I've described is not representative of the order of the RK scheme. This is entirely possible, but the fact that one of the slopes was 9 seems too coincidental.

Something to do with precision. This could explain why the small $h$ values have unreliable errors since I'm "only" carrying 30 digits of precision, but this wouldn't explain the disagreement for the polynomial $f$.

Something specific to RK methods that I'm unaware of. I'm hoping it's this.

Programming error. I'm hoping it isn't this.

Mistake entering in the Butcher table. I'm really hoping it isn't this.

Side note: I've also tried plotting the global error as a function of $h$, but the results were even stranger, so I think the problem may be more fundamental.

答: You're just hitting machine precision. Verify the slope of the log for error values larger than say $10^{-12}$ as for smaller errors, you'll see rounding errors due to the double precision arithmetic of your computer, and to the finite number of digits of the coefficients of the butcher table you have (quite often in textbooks, you don't even have the first 16 digits).</p>

ContentLicense: CC BY-SA 3.0


问:
Why does solving for the inverse function leads $x$ to be a constant?
I went down the wrong path in trying to show that; if $f(x)=ax+b$ then there is a $g(x)=dx+e$ such that $f(g(x)) = x$. This wrong path led me to a baffling result of x being a constant. Can anyone explain either my error or what it means?  Note that on my 2nd attempt I did derive $g(x)$, so that is not my question.  My question is what does the following mean?  How can x be a constant?

I did very simple substitution, just to see where it led me:

if $f(g(x)) = x$ then

$f(dx+e) = x$ and substituting;

$a(dx+e) +b = x$ and expanding

$adx +ae + b = x$ and grouping

$(ad-1)x = -ae - b$

$x = \frac{-ae-b}{ad-1}$ which is a constant

I know this is not how to get the inverse function.  I was able to figure that out.  But why do the above steps lead to an obviously impossible value for x?

答:
The problem with the reasoning is the same old division by zero. But this is very disguised.

Now, you start off with $f(g(x)) = x$, as an equality of functions : so  what this actually means, is that for every real number $r$ we have $f(g(r)) = r$.

Your manipulations led you to $(ad-1)x = -ae-b$, which continued to be an equality of functions.

At this point, you have divided both sides by $ad-1$. This has led to the functional equation $x = c$ for some constant $c$, which is a contradiction, and you feel you have done something wrong.

The truth is, you have obtained by contradiction that $ad-1 = 0$, because if indeed $ad-1 \neq0$ then the functional inequality $(6)$ would have held, which is not possible.

So no , $x$ cannot be a constant. Instead, the assumption that $ad-1\neq 0$ is false and has led you down this path, and once you tread it you know a contradiction awaits you. So that tells you $ad-1=0$.

Now, knowing that $ad-1 = 0$, from $(5)$ we get $ae+b = 0$. Thus, your objective of obtaining $d$ and $e$ in terms of $a$ and $b$ is achieved. Note that if $a= 0$ these constants can't be found.

Just want to add that if you want to derive $d$ and $e$ there's the easier method of substitution which you can use. As I mentioned earlier, a functional equality, upon substitution of appropriate numbers, becomes a numerical equality. So substitute numbers.

For example, suppose that $f(g(x)) = x$. Then substitute $x = 0$, to get $ae+b = 0$. Substitute $x = 1$ to get $a(d+e)+b = 1$, and solve these.


问:
Can the Infinite Product $\prod_{n=1}^\infty \frac{a_n}{a_n + 1}$ Converge or Diverge to Zero?
Can the infinite product $\prod_{n=1}^\infty \frac{a_n}{a_n + 1}$ converge or diverge to zero when the sequence $a_n$ is monotonically increasing? I conjecture that this is impossible given that $\lim_{n\to\infty}\frac{a_n}{a_n+1}$ by the monotonic increase of the $a_n$. However, I don't know of a rigorous proof or disproof of this, and analyzing convergence of infinite series and infinite products is not something I am very familiar with. If someone could provide a proof or disproof of this conjecture (preferably with a reference to a known theorem), or explain what additional restrictions need to be imposed on $a_n$ for the conjecture to hold, I would be very appreciative!

答:
Every term is less than $1$, so it converges to a constant less than 1.

Take $a_n=n$, and the product converge to $0$.

Take $a_n=2^n$, and the product converges, but not to $0$. This can be obtained by taking the logarithm and using the inequality $\ln(1+x)<x$.


问:
trying to understand a proof of menger's theorem

I'm trying to understand the short proof on Menger's Theorem from Wikipedia. I have several fairly (basic) questions that are preventing me from understanding the proof well, shown below.

This is probably a basic question, but in this proof, are they assuming a minimum AB-separator for G has size k?

I don’t understand how one can find an S to be an AB-separator of size less than k so that every AB-path in G has a vertex of S or the edge e (why does this S necessarily exist).

Also, why is it true that S together with either endpoint of e would be a better AB-separator of G? It clearly is larger, but why exactly would it be an AB-separator?

Why is S alone too small to be an AB-separator of G?

Why is it that because of the size of $C_1$, the endpoints of the paths in it must be exactly $S_1$?

Why is it true that every path in $C_1$ is internally disjoint from every path in $C_2$ because $S_1$ disconnects G? I was wondering if one could justify this by showing that if there were two paths $P_1$ in $C_1$ and $P_2$ in $C_2$ that weren’t internally disjoint, one would get a contradiction?

Even helping me to resolve one of these questions would be useful.

答:
If you go back and re-read the proof, you’ll see that an $AB$-separator of $G-e$ of cardinality less than $k$ doesn’t necessarily exist: as it says in the second paragraph of the proof, there might be one of cardinality $k$, in which case the induction hypothesis gives us an $AB$-connector of cardinality $k$ in $G-e$ and hence the desired one in $G$. It’s the rest of the proof that deals with the case in which there is no $AB$-separator of cardinality $k$ in $G-e$.

In that case $G-e$ must have an $AB$-separator $S$ of cardinality less than $k$. By definition of $AB$-separator every $AB$-path in $G-e$ goes through $S$, i.e., has a vertex in $S$. $S$ is too small to be an $AB$-separator in $G$, because by hypothesis the minimal cardinality of an $AB$-separator in $G$ is $k$. Thus, there must be $AB$-paths in $G$ that don’t go through $k$. The only edge that we removed was $e$, so those paths must all use the edge $e$: that’s the only way that they could exist in $G$ but not in $G-e$. And if they use $e$, they must pass through $v_1$ and $v_2$, the endpoints of $e$. That means that every $AB$-path in $G$ goes through $S\cup\{v_1\}$, and every $AB$-path in $G$ goes through $S\cup\{v_2\}$, i.e., that both $S\cup\{v_1\}$ and $S\cup\{v_2\}$ are $AB$-separators in $G$.

If $|S|<k-1$, then $|S\cup\{v_1\}|<k$, and $S\cup\{v_1\}$ would then be an $AB$-separator in $G$ of cardinality less than $k$. (The same holds for $S\cup\{v_2\}$.) By hypothesis, however, no such $AB$-separator exists. (The word better in the proof was poorly chosen: what is meant is smaller.) Thus, $|S|$ must be at least $k-1$. We were already assuming that $|S|\le k-1$, so in fact we must have $|S|=k-1$.

$C_1$ is an $AS_1$-connector of cardinality $k$: it consists of $k$ vertex-disjoint $AS_1$ paths. Since these $k$ paths have no vertices in common, their endpoints in $S_1$ must be $k$ different vertices. But $S_1$ contains only $k$ vertices, so each of these vertices must be the endpoint of one of the paths making up $C_1$.

Let $P_1$ be one of the paths making up $C_1$ and $P_2$ one of the paths making up $C_2$. Let $v$ be any internal vertex of $P_1$, and suppose that $v$ is also an internal vertex of $P_2$. Let $a$ be the endpoint of $P_1$ in $A$ and $b$ the endpoint of $P_2$ in $B$. Then the part of $P_1$ from $a$ to $v$ together with the part of $P_2$ from $v$ to $b$ would be an $AB$-path that did not go through $S_1$, which is impossible, since $S_1$ is an $AB$-separator. Thus, the paths making up $C_1$ must be internally vertex-disjoint from the paths making up $C_2$.

(The schematic diagram that accompanies the proof is actually quite helpful.)


问:
Does an isomorphism between quotient rings deduce an isomorphism between rings?
Exactly I mean if this proposition is true or in what case it stays true:

$I_1$ and $I_2$ are ideals of $R$, then $R/I_1\cong R/I_2$ if and only if $\exists\varphi\in Aut(R)$, such that $\varphi(I_1)=I_2$

The if part is naturally deduced by applying the First Isomorphism Theorem to $\pi\circ\varphi$, where $\pi$ is the canonical map from $R$ to $R/I_2$, but the only if part puzzles me.

If the only if part is true, how to construct an automorphism?

If it is false, what is the counterexample? Can I restrict $R$ to some kinds of rings so that the proposition stays true?(e.g. PIR, I tried to solve the problem in this case, because an automorphism maps $a$ to $b$ naturally maps $(a)$ to $(b)$, but I failed to get further)

答:
This is false, even for PIDs.  Let $R = \mathbf Z[i]$, $I_1 = (4+7i)$ and $I_2 = (1+8i)$.  Note $4^2 + 7^2 = 1^2 + 8^2 = 65$, $4+7i = (2+i)(3+2i)$, and $1+8i = (2+i)(2+3i)$.  The factors $2+i$, $3+2i$, and $2+3i = (i)(3-2i)$ are all prime in $R$ and by the Chinese remainder theorem
$$
R/I_1 \cong R/(2+i) \times R/(3+2i) \cong \mathbf Z/(5) \times \mathbf Z/(13) \cong \mathbf Z/(65)
$$ and

$$
R/I_2 \cong R/(2+i) \times R/(2+3i) \cong \mathbf Z/(5) \times \mathbf Z/(13) \cong \mathbf Z/(65),
$$
so $R/I_1$ and $R/I_2$ are isomorphic rings. The only ring automorphisms of $R$ are the identity and complex conjugation.  These automorphisms map the ideal $I_1 = (4+7i)$ to $(4+7i)$ and $(4-7i)$, neither of which is $(1+8i)$.

There is nothing that special about $\mathbf Z[i]$ here: you can build similar counterexamples with lots of quadratic rings.  The point is that you can have a lot of numbers in a quadratic ring with the same norm even though the numbers are not unit multiples of each other. In the case of $\mathbf Z[i]$ this amounts to saying there are positive integers that can be a sum of two squares $a^2 + b^2$ in lots of essentially different ways (not just from permuting the roles of $a$ and $b$ or making sign changes on $a$ or $b$).

The lesson here is that passing to quotient rings involves a lot of collapsing, so you should not really expect that the "only" way quotient rings of $R$ can be isomorphic in general should be from reducing some automorphism of $R$.


问:
Randomly Choosing a Number and Walking Along the Number Line
Let there be a bug that is on a number line. Let it originally start off at point 0. Each second, it chooses a number  in the interval [−2,2] randomly and uniformly and moves from position $x$ to $x+p$. Find the expected time it will take for the bug to leave the interval (−2,2).

I tried a couple of methods none of which worked. The first was using the Irwin-Hall Distribution, I thought to just use the cumulative density function to show the probability that the bug was outside of the interval $[-2,2]$ after $n$ seconds. The problem here was that this solution allows for the bug to leave the interval at say $2$ seconds but then it can come back in at $3$ seconds and then leave again at $5$ seconds and in the solution, it would count as a possibility.

One thing that I tried was that for each $n$, where $n$ represents the # of seconds passed, you could use the Irwin-Hall Distribution for $k=1$ to $n-1$, finding the overall probability that the bug never left the interval   $[-2,2]$ for all $n-1$ seconds and THEN you could use the probability density function to calculate the chance that it finally leaves the interval at $n$ seconds. The problem here was that the calculations get super messy and I don't see a clean way to do this.

The last possible solution idea that I have is setting up a function $E(x)$ which calculates the expected amount of time it will take the bug to leave the interval at position $x$. Then, we would have to calculate $E(0)$. The problem here is that I don't see a way to utilize this function well.

When I asked some other friends, they said that I have to use the function method to solve the problem (unfortunately they wouldn't explain to me HOW to use the function :( ).

Any hints or solutions for using this function method (or any method) would be GREATLY appreciated.

答:
If $x < -2$ or $x > 2$, then $E(x) = 0$. Otherwise, $$E(x) = 1+\frac{1}{4}\int_{-2}^2 E(x+p) dp \tag 1$$
By symmetry, $E(x) = E(-x)$, so assuming that $x > 0$, this simplifies to $$1+\frac{1}{2}\int_{0}^{2-x}E(p)dp + \frac{1}{4}\int_{2-x}^{2}E(p)dp \tag 2$$

Then, differentiate both sides to get $$E'(x) = -\frac{1}{4}E(2-x) \tag 3$$
Differentiate again to get $$E''(x) = -\frac{1}{16}E(x) \tag 4$$

The solution to this is $E(x) = c_1\cos\left(\frac{x}{4}\right) + c_2\sin\left(\frac{x}{4}\right)$, but since $(2)$ must be satisfied, this means that (skipping the tedious calculus steps) $$c_{1}\cos\left(\frac{x}{4}\right)+c_{2}\sin\left(\frac{x}{4}\right) = 1+c_{1}\sin\left(\frac{1}{2}\right)+\left(2-\cos\left(\frac{1}{2}\right)\right)c_{2}+\left(c_{1}\sin\left(\frac{1}{2}\right)-c_{2}\cos\left(\frac{1}{2}\right)\right)\cos\left(\frac{x}{4}\right)+\left(-c_{1}\cos\left(\frac{1}{2}\right)-c_{2}\sin\left(\frac{1}{2}\right)\right)\sin\left(\frac{x}{4}\right)$$

Skipping the messy algebra again, this means that $c_{1}=\frac{\cos\left(\frac{1}{2}\right)}{2-\cos\left(\frac{1}{2}\right)-2\sin\left(\frac{1}{2}\right)}, c_2 = -\frac{2-\cos\left(\frac{1}{2}\right)-2\sin\left(\frac{1}{2}\right)+\cos\left(\frac{1}{2}\right)\sin\left(\frac{1}{2}\right)}{\left(2-\cos\left(\frac{1}{2}\right)-2\sin\left(\frac{1}{2}\right)\right)\left(2-\cos\left(\frac{1}{2}\right)\right)}$. Plugging in $x = 0$ yields that $$E(0) = c_1 = \frac{\cos\left(\frac{1}{2}\right)}{2-\cos\left(\frac{1}{2}\right)-2\sin\left(\frac{1}{2}\right)} \approx 5.37$$ which matches up numerically with simulations.


问:
Solve quartic equation $a^4-6a^2b-8ac-3b^2=0$
Please help me to find roots of this quartic equation for a:
$$a^4-6a^2b-8ac-3b^2=0$$

Wolfram Alpha gave this result.

But may be there is simple way to get all a?

答:
Rewrite the equation by completing the squares
\begin{align}
a^4-6a^2b-8ac-3b^2
=(a^2+s)^2 -( 2s+6b)\left( a +\frac{4c}{2s+6b} \right)^2\tag1\\
\end{align}
where $s$ happens to satisfy $(s+b)^3=8(c^2-b^3)$, or
$s=2\sqrt[3]{c^2-b^3}-b$.
For notational convenience, denote
$$p=\frac12\sqrt{2s+6b}= \sqrt{\sqrt[3]{c^2-b^3}+b }\tag2$$
and then $s=2p^2-3b$.
Substitute them into (1) and then factorize to get
$$\left(a^2+2pa+2p^2-3b-\frac{2c}p\right) \left(a^2-2pa+2p^2-3b+\frac{2c}p\right)=0
$$
which yields the solutions
$$a=p\pm \sqrt{ 3b-p^2+\frac{2c}p },\>\>\> -p\pm \sqrt{ 3b-p^2-\frac{2c}p } $$
with $p$ given by (2).


问:
Factorise $x^{9}-1$ in $\Bbb{F}_7[x]$
I don't know much about how to solve it since we are just learning the topic, but I tried to answer it, so far I got
$$(x-1)(x-2)(x-4)({x}^{6}+{x}^{3}+1)$$
but the answer is
$$(x-1)(x-2)(x-4)(x^{3}-2)(x^{3}-4).$$

Title: Factorise $x^{9}-1$ in $\Bbb{F}_7[x]$

Body: Mod $7$, $x^6+x^3+1\equiv x^6+x^3-6= (x^3-2)(x^3+3)\equiv (x^3-2)(x^3-4)$.


问:
Evaluate the sum by writing it as a definite integral
I am asked to evaluate the following sum by writing it as a definite integral. I was hoping if my methods we correct for the following sums.

$\displaystyle{\lim_{n \to \infty} \frac{3}{n}\sum_{i=1}^{n}\frac{2+\frac{3i}{n}}{\sqrt{1+\frac{3i}{n}}}}$

We know $\Delta x=\frac{3}{n}$. We know know that $x_i^*=a+i\Delta x.$ Is $a$ starting at $0$ or $1$? I am tempted to say it's starting at $1$ but I am not sure how to justify this. How would I turn this into an integral after?

My second sum is the following:

$\displaystyle{\lim_{n \to \infty} \sum_{i=1}^{n}\frac{1}{\sqrt{n^2+i^2}}}$

I could rewrite the sum as follows, which gives:

$\displaystyle{\lim_{n \to \infty} \sum_{i=1}^{n}\frac{1}{\sqrt{1+\frac{i^2}{n^2}}}}$

$\displaystyle{\lim_{n \to \infty} \sum_{i=1}^{n}\frac{1}{\sqrt{1+\left(\frac{i}{n}\right)^2}}}$

Here $\Delta x=\frac{1}{n}$. I am pretty sure here, $a$ starts from $0$ since $x_i^*=a+i\Delta x.$ If we only have $\frac{1}{n}$ then no terms were added to $x_i^*$ so this means $a$ starts at $0$. Of course, this means $b=1$.

So as an integral, I can write this as $\displaystyle{\int_{0}^{1}} \frac{1}{\sqrt{1+x^2}}dx=\sinh^{-1}(x)|_{0}^{1}=\sinh^{-1}(1)$

Is this correct? It would be great if someone could guide me in the first question and verify my 2nd sum.

答:
The first sum

Whether the integral starts at $0$ or $1$ actually depends on you. Here's what I mean:

$$\displaystyle{\lim_{n \to \infty} \frac{3}{n}\sum_{i=1}^{n}\frac{2+\frac{3i}{n}}{\sqrt{1+\frac{3i}{n}}}}=\lim_{n\to\infty}\frac{3}n\sum_{i=1}^n\frac{1+\left(1+\frac{3i}n\right)}{\sqrt{1+\frac{3i}n}}=\int_1^4\frac{1+x}{\sqrt x}\,dx$$

but you can also write $$\displaystyle{\lim_{n \to \infty} \frac{3}{n}\sum_{i=1}^{n}\frac{2+\frac{3i}{n}}{\sqrt{1+\frac{3i}{n}}}}=\lim_{n\to\infty}\frac{3}n\sum_{i=1}^n\frac{2+\left(\frac{3i}n\right)}{\sqrt{1+\left(\frac{3i}n\right)}}=\int_0^3 \frac{2+x}{\sqrt{1+x}}\,dx$$

You can see that it's just about shifting, and it's arbitrary. For the sake of easier computing, I will pick the first one.

The second sum

There is a typo in your post. It should be $$\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{\sqrt{n^2+i^2}}=\lim_{n\to\infty}\color{red}{\frac{1}{n}}\sum_{i=1}^n\frac{1}{\sqrt{1+\left(\frac{i}n\right)^2}}$$

and the rest of your work seems fine to me.


问:
power of a number in a factorial
what is the largest power of 24 in 150! ?

HINT : answer is 48

I need to know the method for solving such questions when the highest power of the number to be found is non-prime ..

答:
I assume from what you wrote that you know how to find the highest power of a prime $p$ that divides $n!$.

Compute $a$, the largest number such that $2^a$ divides $150!$.

Calculate $b$, the largest number such that $3^b$ divides $150!$.

Which is smaller, $a/3$ or $b$? We are comparing these because we need three $2$'s for every $3$.

That tells you what the limiting resource is, $2$'s or $3$'s.

Title: power of a number in a factorial


答:
Your second sum is evaluated correctly.

General trick is to replace the sum by an integral,$\frac{1}{n}$ by $dx$ and $\frac{i}{n}$ by $x$.

About the bounds suppose we calculate $$\lim_{n \to \infty}\frac{1}{n} \sum_{f(n)}^{g(n)} f(i/n)$$
The lower bound of integral is taken $$\lim_{n\to \infty} \frac{f(n)}{n}$$
and upper bound replace $f(n)$ by $g(n)$

Title: Evaluate the sum by writing it as a definite integral

Body:
I am asked to evaluate the following sum by writing it as a definite integral. I was hoping if my methods we correct for the following sums.

$\displaystyle{\lim_{n \to \infty} \frac{3}{n}\sum_{i=1}^{n}\frac{2+\frac{3i}{n}}{\sqrt{1+\frac{3i}{n}}}}$

We know $\Delta x=\frac{3}{n}$. We know know that $x_i^*=a+i\Delta x.$ Is $a$ starting at $0$ or $1$? I am tempted to say it's starting at $1$ but I am not sure how to justify this. How would I turn this into an integral after?

My second sum is the following:

$\displaystyle{\lim_{n \to \infty} \sum_{i=1}^{n}\frac{1}{\sqrt{n^2+i^2}}}$

I could rewrite the sum as follows, which gives:

$\displaystyle{\lim_{n \to \infty} \sum_{i=1}^{n}\frac{1}{\sqrt{1+\frac{i^2}{n^2}}}}$

$\displaystyle{\lim_{n \to \infty} \sum_{i=1}^{n}\frac{1}{\sqrt{1+\left(\frac{i}{n}\right)^2}}}$

Here $\Delta x=\frac{1}{n}$. I am pretty sure here, $a$ starts from $0$ since $x_i^*=a+i\Delta x.$ If we only have $\frac{1}{n}$ then no terms were added to $x_i^*$ so this means $a$ starts at $0$. Of course, this means $b=1$.

So as an integral, I can write this as $\displaystyle{\int_{0}^{1}} \frac{1}{\sqrt{1+x^2}}dx=\sinh^{-1}(x)|_{0}^{1}=\sinh^{-1}(1)$

Is this correct? It would be great if someone could guide me in the first question and verify my 2nd sum.


答:
So, it appears the only issue was that you didn't fully factorise the sextic polynomial. You needed to either factorise it further or prove that it was irreducible (as it turns out, it was reducible).

Note that substituting elements of $\Bbb{F}_7$ into the polynomial will find all linear factors (i.e. factors of the form $x - \alpha$ for $\alpha \in \Bbb{F}_7$), but it's possible for a reducible polynomial of degree $4$ or more to not have any linear factors. In this case, the sextic polynomial has no roots, but it still factorises into two irreducible cubic roots.

So, in this case, I would note that $x^6 + x^3 + 1$ is a quadratic in $x^3$, and use this fact as inspiration to see if we can factorise it into the form $(x^3 - \alpha)(x^3 - \beta)$. There's a few ways we can do this, but a simple way is to simply look at the corresponding quadratic $x^2 + x + 1$, and find its roots in the same way you did for the original polynomial.

Substitution into $x^2 + x + 1$ should reveal that
$$x^2 + x + 1 = (x - 2)(x - 4),$$
and hence
$$x^6 + x^3 + 1 = (x^3)^2 + (x^3) + 1 = (x^3 - 2)(x^3 - 4).$$
Now, are we done? Can we reduce these cubics any further? If we could, then they would have to factorise into a linear factor and a quadratic factor. The linear factor would mean that the cubic polynomial has a root. But, substituting all of $\Bbb{F}_7$ into these polynomials will yield no roots at all, which means these cubics are irreducible, and we are indeed done.

(Again, remember that the fact that these polynomials are cubics is very important in this step! If we had a quartic, it'd be possible to factorise into two irreducible quadratics, and there'd still be no roots.)

Title: Factorise $x^{9}-1$ in $\Bbb{F}_7[x]$

Body:
I don't know much about how to solve it since we are just learning the topic, but I tried to answer it, so far I got
$$(x-1)(x-2)(x-4)({x}^{6}+{x}^{3}+1)$$
but the answer is
$$(x-1)(x-2)(x-4)(x^{3}-2)(x^{3}-4).$$


问:
Painleve's theorem on removable singularities for compact subsets of zero length
I have this problem:

The exercise mentioned in the above problem is here:

How do I prove this using the given hint? The question I have is if I consider a portion of S and cover it with connected disks, they are no longer disjoint and I cant use the given exercise. I am confused about what limit I need to take.

答:
Instead of having an extended discussion in the comments, let me provide an answer.

As in the hint, we work locally in some disk which we shall also call $U$ with boundary given by the contour $\gamma_0$, oriented counterclockwise.  We want to show that for all $z_0 \in U \setminus S$, $$f(z_0) = \frac{1}{2\pi i} \int_{\gamma_0} \frac{f(z)}{z-z_0} \, dz,$$ so that $$h(z_0) = \frac{1}{2\pi i} \int_{\gamma_0} \frac{f(z)}{z-z_0} \, dz, \quad \forall z_0 \in U$$ is an analytic continuation of $f$ from $U \setminus S$ to $U$.

Fix $z_0 \in U \setminus S$.  Since $S$ is compact, we have $d := d(z_0,S) > 0$.  Let $0 < \epsilon < \frac{d}{3}$.  By hypothesis $S$ can be covered by balls $\{B_1, \ldots, B_n\}$ whose radii sum to $< \epsilon$, and without loss of generality we suppose that each ball intersects $S$ in a nonempty set.  Therefore, we have $d(z_0, \bigcup_n B_n) > \frac{d}{3}$.

Group the balls by connected component, and let $\gamma_1, \ldots, \gamma_m$ be the boundary of each connected component, oriented counterclockwise.  In other words, $\gamma_1, \ldots, \gamma_m$ are disjoint contours with disjoint interiors such that $\partial(\bigcup_n B_n) = \bigcup_j \gamma_j$.  (Some minor modifications may be needed to achieve this.)

By Exercise 2, we have $$f(z_0) = \frac{1}{2\pi i} \int_{\gamma_0} \frac{f(z)}{z-z_0} \, dz - \sum_j \frac{1}{2\pi i} \int_{\gamma_j} \frac{f(z)}{z-z_0} \, dz,$$ so
$$\begin{align}
\left\lvert f(z_0) - \frac{1}{2\pi i} \int_{\gamma_0} \frac{f(z)}{z-z_0} \, dz \right\rvert &= \left\lvert \sum_j \frac{1}{2\pi i} \int_{\gamma_j} \frac{f(z)}{z-z_0} \, dz \right\rvert \\
&\leq \sum_j \frac{1}{2\pi} (\operatorname{length}(\gamma_j) \cdot \sup_{z \in \gamma_j} \left\lvert \frac{f(z)}{z-z_0}\right\rvert)
\end{align}$$

We can see that $$\frac{1}{2\pi} \sum_j \operatorname{length}(\gamma_j) \leq \frac{1}{2\pi} \sum_n \operatorname{length}(\partial B_n) < \epsilon.$$  Moreover, since $f$ is bounded on $U \setminus S$, say $\left\lvert f(z) \right\rvert \leq M$, we have $$\left\lvert \frac{f(z)}{z-z_0} \right\rvert \leq \frac{M}{d(z_0, \bigcup_j \gamma_j)} \leq \frac{M}{d/3}$$ whenever $z \in \bigcup_j \gamma_j$.  Hence, $$\left\lvert f(z_0) - \frac{1}{2\pi i} \int_{\gamma_0} \frac{f(z)}{z-z_0} \, dz \right\rvert < \frac{3M\epsilon}{d}.$$

Taking the limit as $\epsilon \to 0$, we get $$f(z_0) = \frac{1}{2\pi i} \int_{\gamma_0} \frac{f(z)}{z-z_0} \, dz$$ as desired, and this holds for all $z_0 \in U \setminus S$.


问:
Show that $\Gamma\left(3^{-3}+1\right)<\Gamma \left(\operatorname{W}(3)^{-\operatorname{W}(3)}+1\right)$
Hi it's one of my old problem that I cannot solve let me propose it  :

Prove that
$$\Gamma\left(3^{-3}+1\right)<\Gamma \left(\operatorname{W}(3)^{-\operatorname{W}(3)}+1\right)\quad \tag{1}$$
where $W(\cdot)$ is the Lambert W function and $\Gamma(\cdot)$ is the gamma function.

The central function is :

$$f(x)=\Gamma\left(x^{-x}+1\right)$$

The derivative is :

$$f'(x)= -x^{-x}(\ln(x)+1)\Gamma\left(x^{-x}+1\right)\psi(x^{-x}+1)$$

Unfortunately it doesn't helps since $f(x)$ is increasing around $x=3$ while $f\left(\operatorname{W}(x)\right)$ is decreasing around $x=3$ .

Maybe I think we can do the same as the Claude Leibovici's answer  here Show that $\Gamma(\Omega)\leq \Gamma\Big(\operatorname{W}\Big(x^{x}\Big)\Big)<2$ on $(0,1]$ .

My question :

How to show $(1)$ ?

Thanks in advance !

答:
Mking the problem more general, considering
$$\Delta(x)=\Gamma \left(\operatorname{W}(x)^{-\operatorname{W}(x)}+1\right)-\Gamma\left(x^{-x}+1\right)$$ around $x=e$, use Taylor expansions to order $O\left((x-e)^{n+1}\right)$

For example
$$x^{-x}+1=\left(1+e^{-e}\right)-2 e^{-e} (x-e)+\frac{1}{2} e^{-1-e} (4 e-1)
   (x-e)^2+O\left((x-e)^3\right)$$ gives
$$\Gamma\left(x^{-x}+1\right)=\Gamma \left(1+e^{-e}\right)-2 (x-e) \left(e^{-e} \Gamma \left(1+e^{-e}\right)
   \psi ^{(0)}\left(1+e^{-e}\right)\right)+\frac{1}{2} e^{-1-2 e} (x-e)^2 \Gamma
   \left(1+e^{-e}\right) \left(-e^e \psi ^{(0)}\left(1+e^{-e}\right)+4 e^{1+e}
   \psi ^{(0)}\left(1+e^{-e}\right)+4 e \psi ^{(0)}\left(1+e^{-e}\right)^2+4 e
   \psi ^{(1)}\left(1+e^{-e}\right)\right)+O\left((x-e)^3\right)$$

Similarly,
$$W(x)=1+\frac{x-e}{2 e}-\frac{3 (x-e)^2}{16 e^2}+O\left((x-e)^3\right)$$
$$\log(W(x))=\frac{x-e}{2 e}-\frac{5 (x-e)^2}{16 e^2}+O\left((x-e)^3\right)$$
$$-W(x)\log(W(x))=-\frac{x-e}{2 e}+\frac{(x-e)^2}{16 e^2}+O\left((x-e)^3\right)$$
$$\operatorname{W}(x)^{-\operatorname{W}(x)}=e^{-W(x)\log(W(x))}=1-\frac{x-e}{2 e}+\frac{3 (x-e)^2}{16 e^2}+O\left((x-e)^3\right)$$

The formulae are already messy but workable. Computing the numerical value of $\Delta_n(3)$ as a function of the order $n$ of Taylor expansions, we have

$$\left(
\begin{array}{cc}
 n & 10^6 \times \Delta_n(3) \\
 1 & -4892.218349 \\
 2 & +265.5544022 \\
 3 & +68.15191015 \\
 4 & -13.15976659 \\
 5 & +6.138334905 \\
 6 & +4.440578904 \\
 7 & +4.374006576 \\
 8 & +4.408250996 \\
 9 & +4.405368867 \\
 10 & +4.405111061 \\
 11 & +4.405194838 \\
 12 & +4.405189100 \\
\cdots & \cdots \\
\infty & +4.405188465
\end{array}
\right)$$

In fact $\Delta(x)=0$ for $x=3.000042164$.

But, there is another to this equation close to $x=\frac 1e$. A first order Taylor expansion gives it equal to $0.44121$ while the exact solution is $0.42655$.


问:
First and second derivative of a summation
Consider the function $f(\mu) = \sum_{i = 1}^{n} (x_i - \mu)^2$,  where $x_i = i,\,i=1, 2,\dots, n$.

What is the first and second derivative of $f(\mu)$?

答:
Adding an answer here to further clarify the other ones which are simply answers without steps. To get the first derivative, this can be re-written as:

$$\frac{d}{d\mu}\sum(x-\mu)^{2}
\\
= \sum\frac{d}{d\mu}(x-\mu)^{2}
$$

After that it's standard fare chain rule

$$
= \sum-1 \cdot 2(x-\mu)
\\
= -2\sum(x-\mu)
$$

Second derivative: you can observe the same property of linear summation:
$$
\frac{d}{d\mu}-2\sum(x-\mu)
\\
=-2\sum\frac{d}{d\mu}(x-\mu)
\\
=-2\sum(-1)
\\
=2n
$$


问:
Doubt in exercise 8.2 in Linear Algebra by Hoffman and Kunze
Here is the problem from the book.

Consider $R^4$ with the standard inner product. Let $W$ be the subspace of $R^4$ consisting of all vectors which are orthogonal to both $\alpha = (1, 0, -1, 1)$ and $\beta = (2, 3, -1, 2)$. Find a basis for $W$.

I am trying to solve this problem using Gram-Schmidt orthogonalization process but it isn't working out. Is it possible to solve this using Gram-Schmidt process or there is some other way? In either case, please help me as I am reading the book on my own.

答: Hint: If you just let a general vector $r=(w,x,y,z)$ be any such vector, then you have 2 equations expressing that $r$ is orthogonal to each of $\alpha,\ \beta$. Work with those equations using say row reduced matrix.


问:
Determinant of matrix with zero diagonal and some structure
$$|A|={\begin{vmatrix} 0& a_1&a_2&\cdots&a_n\\
a_1&0&a_2&\cdots&a_n\\
a_2&a_1&0&\ddots&\vdots\\
\vdots&\vdots&\ddots&\ddots&a_n\\
a_n&a_{n-1}&\cdots&a_1&0
\end{vmatrix}}_{n+1}$$

Please give me some hints or something that relates this to me . Thank in advance! I have tried it many times on this exercise. So now please help me about this.

答:
The conclusion is true, but step 4 isn't. I'm not sure how to fix this as yet.

Observe the following (fill in gaps as needed).

Clearly $ \sum a_i$ is an eigenvalue with the all 1's eigenvector.

The determinant is a degree $n$ polynomial in the variables $a_i$.

Taking the difference of the first and second rows, $ a_1$ is a factor of the determinant.

(This is true only for $k=1, 2$. I thought it generalizes but doesn't.) Taking the difference of the first and $k+1$th row, $a_k$ is a factor of the determinant.

Hence, the determinant is of the form $k \times (\sum a_i ) \prod (a_i)$ for some constant $k$.

Find the term $ a_n \prod a_i$ to show that $k = (-1)^n$.

Conclusion: The determinant is $ (\sum a_i ) \prod (-a_i)$


问:
Decreasing sequence $(f_n)_{n\geq 1}$ of positive measurable functions $f_n \downarrow f$ in $(0,1]$ or $(0,1)$.
Is there an example where the Lebesgue integrals have the property
\begin{align*}
\int_0^1 f_n \not \to \int_0^1 f?
\end{align*}

My attempt so far is to disprove the statement by showing
\begin{align*}
\int_0^1 f_n \to \int_0^1 f?
\end{align*}
but this requires Dominated Convergence Theorem which would require $f_1$ to be integrable and this is not a given property. Any hints or counterexamples would be appreciated.

Body: $f_n(x)=\frac1  {nx}, f(x)=0$ in $(0,1)$.


问:
$X \cong Y$, $X$ complete $\implies Y$ complete?

Let $\cong$ denote the homeomorphic notation.

Let $X,Y$ be metric spaces, and let $X \cong Y$. If $X$ is a complete metric space does it imply $Y$ is also complete.

答: Hint: $(0,1)$ in the standard metric is incomplete, and $\Bbb R$ is complete.


问:
Showing that the graph of a function lies entirely above the x-axis
I have the following question here.

a) Let $f(x)=x^4-x^3+1$. Show that the graph of the function $f$ lies above the $x$-axis

Is there a way to approach this in a "nice" way? I could just sketch the graph using derivatives, concavity and such but is there a more nicer way of doing this? I feel like I could do this using integration in some way but I am not really sure.

b) Consider the region bounded by the graph of $f$, the $x$-axis, the line $x = a$, and the line $x = a + 1$. What is the value of a for which the area of this region reaches its minimum? What is the value of this minimum?

Would I simply take the bounds as $a$ and $a+1$ and then integrate the function in terms of $a$, and then maximize the function by taking the derivative with respect to $a$?

Any help would be very much appreciated!

答:
Hint: take cases when $x\le 0$ ,
$x\in[0,1]$ and $x\ge 1$

$x\le 0$:$x^4+1+\underbrace{(-x^3)}_{\ge 0}> 0$

$x\in [0,1]$:$x^4+\underbrace{(1-x^3)}_{\ge 0}>0$

for $x\ge 1$ its obvious


答: You can write your space as $$W=\{(x,y,z,w)\in\mathbb{R}^4: \langle(x,y,z,w),(1,0,-1,1)\rangle=0,\,\langle(x,y,z,w),(2,3,-1,2)\rangle=0\}.$$ The two inner product conditions give you a (homogeneous) system of linear equations with two free variables, so you should get a 2-dimensional space.

Title: Doubt in exercise 8.2 in Linear Algebra by Hoffman and Kunze

Body:
Here is the problem from the book.

Consider $R^4$ with the standard inner product. Let $W$ be the subspace of $R^4$ consisting of all vectors which are orthogonal to both $\alpha = (1, 0, -1, 1)$ and $\beta = (2, 3, -1, 2)$. Find a basis for $W$.

I am trying to solve this problem using Gram-Schmidt orthogonalization process but it isn't working out. Is it possible to solve this using Gram-Schmidt process or there is some other way? In either case, please help me as I am reading the book on my own.


问:
Number of real roots of $\sum_{k=0}^{n}\frac{x^{k}}{k!}=0$
Prove the following, without induction. Is it possible?

The equation $\sum_{k=0}^{n}\frac{x^{k}}{k!}=0$ has no real root if $n$ is even.

  And if $n$ is odd, it has only one real root.

I also tried searching the proofs many times with search key words "number of real root" or "exponential function" or "$\sum_{k=0}^{n}\frac{x^{k}}{k!}$".
But failed. What is the way to search the related topics?

答: Note that,  $f'(x)+\frac{x^k}{k!}=f(x).$ If $k$ is even and $x_0$ is the smallest root of $f,$  then $f'(x_0)<0.$  So immediately after $x_0$ $f(x)<0$ and the function decreases. Thus, in order for the function $f$ to have its second zero which it must have as a polynomial of even degree (the multiplicity of each root is clearly one) it has to increase. However, if the function $f$  increases at a point then $f(x)>0$ (follows from our identity). This shows that our assumption is wrong. So, $f$ does not have a root. Once even case is established, odd case follows immediately.</p>


问:
If $A$, $B$ and $C$ are subsets of a finite set $S$, how do we show how many elements of $S$ are contained in at least two of the subsets?
Obviously I know that I have to use the inclusion-exclusion principle but I'm not sure how to start the proof. I have been told that the answer I have to get to is
$$|A ∩ B| + |A ∩ C| + |B ∩ C| − 2|A ∩ B ∩ C|$$
Is this correct? or should I be aiming for another answer?

**EDIT **

so far I have that we want to find
$|(A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) - (A ∩ B ∩ C)|$
Is this a correct first step to take?

答:
Expanding on Cosmo5's comment, and against the backdrop of Inclusion-Exclusion, the foundation of the principle is that when each region is added, subtracted, added, subtracted, ... the net effect is that each region is counted once.

From the constraints of the problem, you can consider the following 4 mutually exclusive regions:

Region 1 : in A, in B, not in C.

Region 2 : in A, not in B, in C.

Region 3 : not in A, in B, in C.

Region 4 : in A, in B, in C.

As stated, your formula mishandles Region 4, because the effect of the first three terms is that Region 4 is included three times.

To offset this, the final term must deduct Region 4 twice rather than once.


问:
Simultaneous algebra question
Let $x$, $y$ be real numbers such that
$$x+3y=4$$$$x^2+3y^2=8$$

Find $$x^3+3y^3$$

Also, let $$ax+by=v_1$$$$ax^2+by^2=v_2$$
Find the value of $ax^3+by^3$ in terms of $v_1$, $v_2$

I've tried $(x+3y)(x^2+9y^2−3xy)−6y^3$ and was left with $48+20y^2−6y^3$

Title: Simultaneous algebra question

Body:
$$x+3y=4 \implies x=4-3y \implies (4-3y)^2+3y^2=8 \implies 3y^2-6y+2=0 \implies y=1\pm \frac{1}{\sqrt{3}}$$
We get $x=1-\sqrt{3}$ for $y=1+\frac{1}{\sqrt{3}}$, then $F=x^3+3y^3=(1-\sqrt{3})^3+3(1+\frac{1}{\sqrt{3}})^3=16-\frac{8}{\sqrt{3}}.$

or $x=1+\sqrt{3}$ for $y=1-\frac{1}{\sqrt{3}}$
Finally, $F=x^3+3y^3=(1+\sqrt{3})^3+3(1-\frac{1}{\sqrt{3}})^3=16+\frac{8}{\sqrt{3}}$


问:
Random variables, Correlation and expectation
We know that if $E(X|Y) = E(X)$, then $X$ and $Y$ are uncorrelated. I found out that the reverse statement is not always true, ie. $X$ and $Y$ are uncorrelated, but $E(X|Y)$ is not equal to $E(X)$. But I don't get why... If I am right, can you give an example where this happens??

Body: $Y\sim N(0,1)$ and $X=Y^{2}$. Note that $E(X|Y)=Y^{2}$ and $EX=1$. Now $E(XY)=EY^{3}=0=(EX) (EY)$.


问:
Can the Infinite Product $\prod_{n=1}^\infty \frac{a_n}{a_n + 1}$ Converge or Diverge to Zero?
Can the infinite product $\prod_{n=1}^\infty \frac{a_n}{a_n + 1}$ converge or diverge to zero when the sequence $a_n$ is monotonically increasing? I conjecture that this is impossible given that $\lim_{n\to\infty}\frac{a_n}{a_n+1}$ by the monotonic increase of the $a_n$. However, I don't know of a rigorous proof or disproof of this, and analyzing convergence of infinite series and infinite products is not something I am very familiar with. If someone could provide a proof or disproof of this conjecture (preferably with a reference to a known theorem), or explain what additional restrictions need to be imposed on $a_n$ for the conjecture to hold, I would be very appreciative!

答:
Let $q_n=\dfrac{1}{1+a_n}$ be such that $0\lt q_n\lt 1,$ then by the elementary theory of infinite products $\prod_{n=1}^\infty (1-q_n)$ converges to a non-zero number if and only if $\sum_{n=1}^{\infty}q_n$ converges.

This clearly gives you a criteria to find the convergence to a non-zero number.

For example: if you set $a_n$ to be the $n$th prime (as mentioned in your comment), then $\sum_{n=1}^{\infty}q_n$ diverges and therefore ....


问:
Easy way of finding the residue of $\int^\infty_0 \frac{dx}{x^4+x^2+1}dx$
In the lecture note I am reading, it finds that the pole of $\int^\infty_0 \frac{dx}{x^4+x^2+1}dx$ is at $w_1=e^{i\pi/3}$ and $w_2=e^{2i\pi/3}$ in the upper half plane. It is obvious that those poles are simple, and it asserts that

$\text{Res}_{w_j}(f)=\frac{1}{Q'(w_j)}=\frac{1}{4w_j^3+2w_j}$ where $Q(z)=z^4+z^2+1$.

I only know the formula for finding a simple pole $\text{res}_{z_0}f=\lim_{z\to z_0}(z-z_0)f(z)$.

How does the above equation hold?

答: More generally, if $f(z)=\frac{p(z)}{q(z)}$, where $p,q$ are analytic at $z_0$, with $p(z_0)\neq 0,\,q(z_0)=0,\,q'(z_0)\neq 0$, then $f$ has a simple pole at $z_0$, and you can write $$\frac{p(z)}{q(z)}=\frac{p(z_0)+p'(z_0)(z-z_0)+\cdots}{q'(z_0)(z-z_0)+\cdots}.$$ Now you can multiply by $z-z_0$, take the limit for $z\rightarrow z_0$ and you will get the formula $$Res(f,z_0)=\frac{p(z_0)}{q'(z_0)}.$$


问:
Good book for self study of a First Course in Real Analysis
Does anyone have a recommendation for a book to use for the self study of real analysis? Several years ago when I completed about half a semester of Real Analysis I, the instructor used "Introduction to Analysis" by Gaughan.

While it's a good book, I'm not sure it's suited for self study by itself.  I know it's a rigorous subject, but I'd like to try and find something that "dumbs down" the material a bit, then between the two books I might be able to make some headway.

答: Mathematical Analysis a straightforward approach by K.G Binmore is good for self study since it contains solutions to the exercises......The best book ever written is probably  Introduction to Real Analysis by mapa.....</p>


问:
finding a probability density function of a random variable given a relation between another random variable
I am trying to find a pdf for a random variable $X$ where $X=-2Y+1$ and $Y$ is given by $N(4,9)$

Here is my attempt:

we know $\mu=4$ and $\sigma=3$. so that the normal distribution of $Y$ is given by $\frac{1}{3\sqrt{2\pi}}e^\frac{-(y-4)^2}{18}$

We can differentiate the cumulative function of $X$ to get the pdf for $X$.

cdf of $X = P(X<x)$ = $P(-2Y+1<x)=P(Y<\frac{-(x-1)}{2})=\int_{-\infty}^{\frac{-(x-1)}{2}}\frac{1}{3\sqrt{2\pi}}e^\frac{-(y-4)^2}{18}dy$

so $\frac{d}{dx}(\int_{-\infty}^{\frac{-(x-1)}{2}}\frac{1}{3\sqrt{2\pi}}e^\frac{-(y-4)^2}{18}dy)=f(x)$, which is the density function for $X$

$f(x)=-\frac{1}{6\sqrt{2\pi}}e^\frac{-(\frac{-(x-1)}{2}-4)^2}{18}$

Is this a correct way to approach the problem? I feel like my answer is very funky.

答:
$-2Y+1 <x$ is not equivalent to $Y <-\frac {x-1} 2$.

$P(X\leq x)=P(-2Y+1 \leq x)=P(Y \geq \frac {1-x} 2)=\int_{\frac  {1 -x} 2}^{\infty} \phi (t)dt$ where $\phi$ is the standard normal density. Hence the density of $X$ is $\frac 1 2\phi (\frac {1-x} 2)$


答: Yes, you can use the Gram-Schmidt procedure. Extend the set of vectors $\ \{\alpha, \beta\}\ $ to a basis of $\ R^4\ $ by adding another two appropriate vectors to it. The vectors $\ \gamma=(0,0,1,0)\ $ and $\ \delta=(0,0,0,1)\ $ will do for this purpose because $\ \alpha, \beta, \gamma, \delta\ $ are linearly independent.  Now use the Gram-Schmidt procedure to orthonormalise this basis in the order $\ \alpha, \beta, \gamma, \delta\ $.  If the resulting orthonormal basis is $\ u_1,u_2,u_3,u_4\ $, then $\ u_1=\frac{\alpha}{\|\alpha\|}\ $ and $\ u_2=\frac{\|\alpha\|^2\beta-\langle\beta,\alpha\rangle\alpha}{\left\| \|\alpha\|^2\beta-\langle\beta,\alpha\rangle\alpha\right\|}\ $ will span the same space as $\ \alpha\ $ and $\ \beta\ $, and $\ u_3, u_4\ $ will be a basis for $\ W\ $.


问:
power of a number in a factorial
what is the largest power of 24 in 150! ?

HINT : answer is 48

I need to know the method for solving such questions when the highest power of the number to be found is non-prime ..

答:
You have to compute the highest power of each prime factor
of the number you want
(for 24, these are 2 and 3)
and then for each of these,
compute how many times
that power of the primes divide the factorial
(for 24 it is $2^3$ and $3^1$),
and choose the smaller.

For your case,
2 goes 75+37+18+9+4+2+1=146,
so $2^3$ goes $[146/3] = 48$.
3 goes 50+16+5+1 = 73.
The smaller of these is 48.

Title: power of a number in a factorial


问:
Solve $\lim\limits_{z\to 2i} 2x + y^2 = 4$ using the formal definition of limit
First of all we need to remember that $z=x+iy$. Well, I'm confusing about a tip from the author where he says to consider $|z-z_{0}| < 1$. Then we need to prove $|x| < 1$ and $|y|<3$ . This is the part I'm not understanding! How can I prove that? Then he says to use this affirmation to conclude:

$$\begin{align}|f(z) - f(z_0)| &= |2x+y^2 - 4| \\&= |2x + (y-2)(y+2)| \\&\leq 2|x| + |y-2|(|y|+2)\\& \leq 5|x| + 5|y-2|\end{align}$$
and then $5|x| + 5|y-2| \leq 10|z-2i|$

答:
$|a+ib|=\sqrt {a^{2}+b^{2}}$ if $a$ and $b$ are real numbers. Hence $|a+ib| \geq \sqrt {a^{2}}=|a|$ and $|a+ib| \geq \sqrt {b^{2}}=|b|$. Thus $|(x+iy)- (0+2i)| =|x+i(y-2)| <1$ implies that $|x| <1$ and $|y-2| <1$ so $|y| <1+2=3$.

[To finish the proof take $\delta =\frac {\epsilon} {10}$]


问:
Use of the concept of subgroup vs field extension
Why is it popular to use the idea of subgroups in cases of groups and field extensions in case of fields? In both case one set is the subset of the other along with the restriction of some additional properties.

Edit: As per suggestion of @Lubin, I would rephrase the question as - why do we emphasize on the substructures in case of groups, while for fields, we emphasize the overlying structures.

答: I think it's due to history. Groups were discovered/invented as a complete unit and sub groups were found later(Starting with Lagrange).First field existed intuitively as rational numbers,Later Galois extended them with roots looking for 5th degree equation solutions.</p>


Title: Showing that the graph of a function lies entirely above the x-axis

Body:
I have the following question here.

a) Let $f(x)=x^4-x^3+1$. Show that the graph of the function $f$ lies above the $x$-axis

Is there a way to approach this in a "nice" way? I could just sketch the graph using derivatives, concavity and such but is there a more nicer way of doing this? I feel like I could do this using integration in some way but I am not really sure.

b) Consider the region bounded by the graph of $f$, the $x$-axis, the line $x = a$, and the line $x = a + 1$. What is the value of a for which the area of this region reaches its minimum? What is the value of this minimum?

Would I simply take the bounds as $a$ and $a+1$ and then integrate the function in terms of $a$, and then maximize the function by taking the derivative with respect to $a$?

Any help would be very much appreciated!

Body:
Another way: By AM-GM $$x^4+x^4+x^4+1\ge 4|x|^3$$
thus $$3(x^4-x^3+1)=\underbrace{(3x^4+1-3x^3)}_{\ge 0}+2>0$$

Here we used the fact that $4|x|^3-3x^3\ge 0$


问:
Solve quartic equation $a^4-6a^2b-8ac-3b^2=0$
Please help me to find roots of this quartic equation for a:
$$a^4-6a^2b-8ac-3b^2=0$$

Wolfram Alpha gave this result.

But may be there is simple way to get all a?

答:
$$\text{Given}\quad a^4-6a^2b-8ac-3b^2=0\quad\implies 3b^2+6a^2b -(a^4-8ac)=0$$

The solution for  $b$ using the quadratic equation is
$$b = \pm\frac{2 \sqrt{a^4 - 2 a c}}{\sqrt{3}} - a^2\quad\text{for}\quad a\in\mathbb{R}\land c\in\mathbb{R}\land b\in\mathbb{R}\iff a^4-2ac\ge0$$
Technically, this defines all values of $a$ but perhaps you seek integers.

Regular algebras solves for $c$ as $$c = \frac{a^4 - 6 a^2 b - 3 b^2}{8 a}$$
The second equation looks easier to work with (though you can try the first if you want) and experimentation in a spreadsheet offers these sample integer triples.
\begin{equation}
(1,-11,-37)\quad
(1,-7,-13)\quad
(1,-3,-1)\quad
(1,1,-1)\quad
(1,5,-13)\quad
(1,9,-37)\quad\\
(2,-8,1)\quad
(2,-4,4)\quad
(2,0,1)\quad
(2,4,-8)\quad
(2,8,-23)\quad
(2,12,-44)\quad\\
(3,-11,13)\quad
(3,-7,13)\quad
(3,-3,9)\quad
(3,1,1)\quad
(3,5,-11)\quad
(3,9,-27)\quad
(3,13,-47)\quad\\
(4,-8,26)\quad
(4,0,8)\quad
(4,8,-22)\quad\\
(5,5,-5)\quad\\
(6,-8,59)\quad
(6,-4,44)\quad
(6,0,27)\quad
(6,4,8)\quad
(6,8,-13)\quad
(6,12,-36)\quad\\
(7,-7,77)\quad\\
(8,-8,109)\quad
(8,0,64)\quad
(8,8,13)\quad
\end{equation}

WolframAlpha offer only the roots $a=0,b=0$ or
$a\ne0\land c = \frac{a^4 - 6 a^2 b - 3 b^2}{8 a}$ as shown here. This solution for $c$ (as I pointed out above) appears to have infinite integer solutions and, when $a=1$, it appears that these have a symmetry for values of $b$ around zero. Let me know if you have other questions about this.


Body:
Try to express it as a sum of squares. One way that I did it was

$$ x^4 - x^3 + 1 = (x^2 - \frac{1}{2} x - \frac{1}{2} ) ^2 + \frac{1}{12} (3x-1)^2 + \frac{2}{3}.$$

Another way is

$$ x^4 - x^3 + 1 = (x^2 - \frac{1}{2} x - \frac{1}{4} ) ^2 + \frac{1}{16} (2x-1)^2 + \frac{7}{8}.$$


问:
Why $\int_{-\infty-\frac{i\omega}{2k^2}}^{{\infty-\frac{i\omega}{2k^2}}} e^{-k^2x^2}dx=\int_{-\infty}^{{\infty}} e^{-k^2x^2}dx$?
In complex analysis lecture, my professor said the following

$$\int_{-\infty-\frac{i\omega}{2k^2}}^{{\infty-\frac{i\omega}{2k^2}}} e^{-k^2x^2}dx=\int_{-\infty}^{{\infty}} e^{-k^2x^2}dx$$

the equality holds because of Cauchy Theorem and this is because the function is analytic.

My question is that how do I use Cauchy Theorem to see this?   Both end points are still not the same.

答: Take the rectangle with corners at $R,-r,-r-\frac{i\omega}{2k^2},R-\frac{i\omega}{2k^2}$. The integrand is analytic, so by Cauchy-Goursat theorem, the integral around the rectangle is zero. Let $R \to \infty$ and $-r \to -\infty$. The integral along the two vertical line segments will vanish, which can be shown using the ML-inequality. The remaining terms will give you the expression above.


问:
If a group has order $2p^a$ where p is an odd prime and a $\geq$ 1, then G has a proper, nontrivial normal subgroup
I have this problem and I'm stuck. I only know that $Z(G)$ is a normal subgroup of $G$ but don't know how to continue from there.

I would really appreciate any help.

答: The Sylow $p$-group has index 2, so it has to be normal.


答:
Let us denote distribution functions by $F$, and density functions by $f$. Then,
\begin{align*}
F_X(x) &= P(X \leqslant x)\\
&= P\left(Y \geqslant \frac{1-x}{2}\right)\\
&= 1 - F_Y\left(\frac{1-x}{2}\right)\\
&= 1 - \int_{-\infty}^{\frac{1-x}{2}} f_Y(t) dt\\
&= 1 - \int_{0}^{\frac{1-x}{2}} f_Y(t) dt - \int_{-\infty}^{0} f_Y(t) dt\\
&= \frac{1}{2} - \int_{0}^{\frac{1-x}{2}} f_Y(t) dt
\end{align*}
Therefore, by Fundamental Theorem of Calculus, we have : $$ f_X(x) = \frac{1}{2} \cdot f_Y\left(\frac{1-x}{2}\right) = \frac{1}{6\sqrt{2\pi}} \exp\left[-\frac{(x+7)^2}{72}\right]$$
So, $~X \sim N(-7,36)$ . Hope it helps.

Title: finding a probability density function of a random variable given a relation between another random variable

Body:
I am trying to find a pdf for a random variable $X$ where $X=-2Y+1$ and $Y$ is given by $N(4,9)$

Here is my attempt:

we know $\mu=4$ and $\sigma=3$. so that the normal distribution of $Y$ is given by $\frac{1}{3\sqrt{2\pi}}e^\frac{-(y-4)^2}{18}$

We can differentiate the cumulative function of $X$ to get the pdf for $X$.

cdf of $X = P(X<x)$ = $P(-2Y+1<x)=P(Y<\frac{-(x-1)}{2})=\int_{-\infty}^{\frac{-(x-1)}{2}}\frac{1}{3\sqrt{2\pi}}e^\frac{-(y-4)^2}{18}dy$

so $\frac{d}{dx}(\int_{-\infty}^{\frac{-(x-1)}{2}}\frac{1}{3\sqrt{2\pi}}e^\frac{-(y-4)^2}{18}dy)=f(x)$, which is the density function for $X$

$f(x)=-\frac{1}{6\sqrt{2\pi}}e^\frac{-(\frac{-(x-1)}{2}-4)^2}{18}$

Is this a correct way to approach the problem? I feel like my answer is very funky.


问:
Method of finding "closest" common multiple?
I am trying to program a script to do this, but I can't find any existing methods, so I hope you have any ideas.
The problem is similar to LCM, but the numbers are inaccurate, so I have to find the closest solution.

Example: I have the numbers 1296, 306, 8928, 2466, 1881.
They should be multiples of the number 180, but since the numbers are inaccurate,
I need to find a number that is close to 180, that gives the least amount of error as multiple for the set of numbers.

Any ideas on how I can solve this?

答:
I am confused by what you intend by "I need to find a number that is close to 180, that gives the least amount of error as multiple for the set of numbers."

I am guessing that you want each number $n$ converted to $m$, where $m$ is the closest multiple of $180$ to $n$.  Assuming so, after the conversion, you could then (for example) compute the lcm of $\{m_1, m_2, m_3, m_4, m_5\}$ using standard methods.

If my guess and subsequent analysis is accurate, then your query reduces to how to convert a positive integer $n$ into $m$.

Apply the Euclidean Algorithm so that

$n = (90\times P) + r$,

where $P \,\in \{0,1,2,\cdots\}$ and $r \in \{0,1,2,\cdots, 89\}.$

Then, if $P$ is even, then the closest multiple
$$m = 180 \times \left(\frac{P}{2}\right).$$

Alternatively, if $P$ is odd, then the closest multiple
$$m = 180 \times \left(\frac{P+1}{2}\right).$$

The only wrinkle is that if $P$ is odd and $r = 0$, then $n$ is equidistant between two multiples of 180 and my algorithm is (then) automatically "rounding up".


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:
@to Mods: Please kindly understand that we are discussing.

Almost Sure Convergence

(This is the continuation of my discussion in the above post)

Let $(\sigma_n ; n \ge 0)$ and $(X^{\sigma_n} ; n \ge 0 )$ be the sequence of volatilities and their respective stochastic processes as defined above.

As we had argued, we are able to show the almost sure convergence of $X^{\sigma_n}$ to $\tilde{X}$ at a countable dense subset, say $\mathbb{Q}^+$, of our time horizon $\mathbb{R}^+$, that is: Almost surely, for all $t \in \mathbb{Q}^+$, we have:
$$\lim_n X^{\sigma_n}_t = \tilde{X}_t$$
Now if we fix a real positive number $T$. We see that, for almost all $\omega \in \Omega$, the sequence of continuous function $(X^{\sigma_n}(\omega) ; n \ge 1) \subset \mathcal{C}([0,T]) $ is equicontinous.

Indeed, we can prove this fact either by Komolgorov's continuity theorem( a bit overkill here) or just by obversing that:
$$ X^{\sigma_n}_t=x_0 +\underbrace{ \int_{0}^t \left( f(s,X_s) -\dfrac{\sigma_n^2}{2} \right) ds }_{A^{(n)}_t} +\underbrace{\sigma_n W_t}_{B^{(n)}_t}$$

The part $A$ gives us a Lipschitz continuous function with the Lipschitz coefficient clearly bounded for all $\sigma_n$. ( by our choice of $\sigma_n$ and the boundedness of $f$). Hence, $(A^{(n)}(\omega) )$ is a family of equicontinuous functions for every $\omega$

The part $B$ clearly gives a family of equicontinuous functions as it is formed by the scalar multiplication between a bounded real number and a continuous function shared by all $n$.

Note: Two arguments above are constructed in the perspective that we are only considering all the processes restricted to the compact subset $[0,T]$ of time horizon)

Thus for almost every $\omega$, the family of functions $$( X^{\sigma_n}(\omega); n \ge 1 )$$ is a relative compact subset of $\mathcal{C}([0,T])$. Thus for every subsequence of that sequence, we can always extract a subsubsequence that converges uniformly to a limit function.

However, our very first result on the almost sure convergence of this sequence tells that if that limit exists, that limit function must be identical to $\tilde{X}$.

So a.s, $X^{\sigma_n}(\omega)$ converges uniformly to $\tilde{X}(\omega)$

Thus the conclusion $\square$.

Comments:

In the end, we even have the a.s convergence of $X^{\sigma}$ to $\tilde{X}$ under the usual metric of $\mathcal{C}( \mathbb{R_+})$

It seems that I haved blah-blah a lot in my arguments. By all means, the central idea is the equicontinuity of paths when restricted to $[0,T]$.

Some relaxation on the boundedness of $f$ can be done (if needed). While a replacement by an upper bound for $|f|$ only dependent of $t$ is easy to make, a bound depending also on $x$ is more not that straigthforward.


问:
sequence of functions, limits and integral
For the following two sequences of functions $f_n:[0,1]\to \mathbb{R}$, find $f(x)=\lim_{n\to\infty}f_n(x)$, and compare $\lim_{n\to\infty}\int_{0}^{1}f_n$ (if it exists) to $\int_{0}^{1}f$ (if it exists).

(a) $f_n(x)=\begin{cases}
          n^2x, \quad  &  0\leq x \leq\frac{1}{2n} \\
          n-n^2x, \quad &\frac{1}{2n}\leq x \leq\frac{1}{n} \\
          0, \quad &\frac{1}{n}\leq x \leq 1 \\
     \end{cases}$

(b) $f_n(x)=\begin{cases}
          1, \quad &\text{if} \, x=\frac{a}{b} \in \mathbb{Q}  \text{ with b}\leq n, \\
          0, \quad &\text{otherwise.}  \\
     \end{cases}$

For (a), I tried to take limits of $f_n$ piecewise. In the first interval, limits of $f_n(x)=+\infty$. In the second one, I'm not sure whether limits of $f_n(x)=-\infty$. Since I graphed several $f_n$ on the second interval, their function value are always positive. But if I simply calculate $n-n^2x$ as $n\to\infty$, it would be $-\infty$. Besides, I'm also stuck on how to get $\lim_{n\to\infty}\int_{0}^{1}f_n$ and $\int_{0}^{1}f$ . If it's infinity, can I get  its integral on piecewise interval?

For(b), I think $f_n(x)$ is always jumping back and forth between 1 and 0, so its integral would be zero and the limit of its integral would be 1 on rational numbers and 0 on irrational numbers. But I'm not sure about $f(x)$ here, and get stuck again on how to get $\lim_{n\to\infty}\int_{0}^{1}f_n$ and $\int_{0}^{1}f$ .

答:
(a): For any $x>0$ the sequence $(f_n(x))$ is $0$ after some stage: $f_n(x)=0$ if $x > \frac 1 n$ or $n >\frac  1x$. Hence the limit of this sequence is $0$ and $f(x)=0$ for all $x$. (Note that $f_n(0)=0$ for all $n$). I will let you compute the integral of $f_n$ to see that $\int f_n$ does not tend to $0=\int f$.

(b): For any rational number $x=\frac p q$ we have $f_n(x)=1$ whenever $ n >q $ so $f(x)=\lim f_n(x)=1$. For $x$ irrational  $f(x)=\lim f_n(x)=0$. $\int f(x) dx$ does not exist as  a Riemann integral but its Lebesgue integral is $0$.

$\int f_n(x)dx=0$ for all $n$: If $x=\frac a b$ is a rational number(where $a$ and $b$ are integers) with $b \leq n$ and $0 \leq x\leq 1$ then there are only $n$ possible values for $b$ and only $b+1$ values of $a$ for  a given $b$. Hence $f_n(x)=0$ except for  a finite number of points $x$.


问:
How to check for range condition when a equation containing other function is converted into quadratic?
I came across a question in my textbook which is something like this

The least positive value of a for which $4^x-a\times2^x-a+3 \le 0$ is satisfied by at least one real value of $x$ is ______

So to solve this question I used this approach:

Converted it into Quadratic by assuming "$2^x$" as '$t$'

Since the "leading coefficient in the inequality != zero". Hence the equation is quadratic.

Now the leading coefficient is greater than $0$. And the inequality condition is $\le 0$.Hence,
discriminant must be greater than or equal to zero.

I got range of a which comes out to be $a \in (-\infty, -6] \cup [2,\infty)$

I think the answer should be $2$ because it is the least positive value. But, I have some kind of confusion in my mind. As I have assumed "$2^x$' as '$t$'.. I'm not sure about the answer.. Please help me after this

答:
The interval where the discriminant is nonnegative is $a \in (-\infty, -6] \cup [2, \infty)$.

For $a =2$, we have $t^2-2t+1 = (t-1)^2$ as the quadratic equation, and it is equal to zero when $t = 2^x = 1$, which is certainly satisfied by some real value of $x$.

For $0 < a < 2$, by your discriminant and leading coefficient argument, $t^2 - at - a+3 > 0$ for all $t$.

Hence the least positive value of $a$ must be $2$.


问:
Find the finite power series of $\frac{u^3+3u^2+3u+1}{u^2+2u+3}$
I tried to do the partial fraction of $\frac{x^3}{(x-1)^4(x^2+2)}$. By letting $u=x-1$ I get $\frac{1}{u^4}\frac{u^3+3u^2+3u+1}{u^2+2u+3}$. I was able to get the answer by doing the long division
of the $\frac{u^3+3u^2+3u+1}{u^2+2u+3}$ part.
So, I wonder is it possible to turn $\frac{u^3+3u^2+3u+1}{u^2+2u+3}$ into a finite power series  $a_{n}(x-c)^n $ with $c=0$ or turns $\frac{x^3}{(x^2+2)}$ into a finite power series $a_{n}(x-c)^n $ with $c=1$ so it can be divided by $(x-1)^4$ ?

答:
$$\frac{u^3+3u^2+3u+1}{u^2+2u+3}=1+u-2\frac{u+1 }{u^2+2 u+3}$$
Now, let $$u^2+2u+3=(u-a)(u-b)\qquad\qquad a=-1-i \sqrt{2}\quad b=-1+i \sqrt{2}$$
$$\frac{u+1 }{u^2+2 u+3}=\frac{a+1}{(a-b) (u-a)}-\frac{b+1}{(a-b) (u-b)}$$ composing the series and playing with the complex numbers
$$\frac{u^3+3u^2+3u+1}{u^2+2u+3}=\frac{1}{3}+\frac{7 }{9}u+\sum_{n=2}^\infty c_n\, u^n$$ with
$$c_n=-\frac{\left(\sqrt{2}-2 i\right) \left(\frac{1}{3} i
   \left(\sqrt{2}+i\right)\right)^n+\left(\sqrt{2}+2 i\right) \left(-\frac{1}{3} i
   \left(\sqrt{2}-i\right)\right)^n}{3 \sqrt{2}}$$ which generate the sequence
$$\left\{\frac{10}{27},-\frac{14}{81},-\frac{2}{243},\frac{46}{729},-\frac{86}{2187}
   ,\frac{34}{6561},\frac{190}{19683},-\frac{482}{59049},\frac{394}{177147}\right\}$$

Now, as your properly did, let $u=x-1$ to get
$$\frac{x^3}{(x^2+2)}=$$ but this one is much simpler. Let $x^2=2t^2$ to make
$$\sqrt{2} \frac{t^3}{1+t^2}$$ Use the series expansion of $\frac 1{1+y}$, replace $y$ by $t^2$ and multiply by $\sqrt{2} {t^3}$ and go backward from $t$ to $x$ and from $x$ to $u$.


问:
Is $\sum^\infty_{n=0}(\frac z{1+z})^n$ holomorphic on the right half plane?
Since $\sum^k_{n=0}(\frac z{1+z})^n$ is holomorphic for every $k$ in the right half plane, it suffices to show that the series is uniformly convergent in every compact subset of right half plane. But how do we show this?

答:
The geometric series $\sum_{n=0}^\infty w^n$ converges to $(1-w)^{-1}$ for $|w|<1$ and diverges otherwise.

So $\sum_{n=0}^{\infty} (\frac{z}{1+z})^n=(1-\frac{z}{1+z})^{-1}=1+z$ for $|\frac{z}{1+z}|<1$ and diverges otherwise. Of course, $1+z$ is holomorphic, so this series is holomorphic everywhere it exists (where $|\frac{z}{1+z}|<1$).

Note $|\frac{z}{1+z}|<1 \iff |z|^2<|1+z|^2$ and $|1+z|^2=1+2\mathrm{Re}(z)+|z|^2$; can you simplifty?


问:
Improper Double integral to Proper by change of variables.
Let $D$ be the region bounded by $y = x^2, y = \frac12 x^2$ and $y = 6x$. I want to find the following
$$\iint_D \frac1y dy\ dx.$$

Note that $\frac 1y$ is not defined at $(0,0)$, so I apply a change of variables to convert into a proper integral. Let $x = uv$ and $y = u^2$.

Calculating the Jacobian we have $dx\ dy = 2u^2 du\ dv$. Next, I am getting confused with the limits of the integral. Need some help!

答: You can compute the integral directly as $$\int_0^6\int_{\frac{1}{2}x^2}^{x^2}\frac{1}{y}dydx+\int_6^{12}\int_{\frac{1}{2}x^2}^{6x}\frac{1}{y}dydx=6\ln(2)+\int_6^{12}\left(\ln(6x)-\ln\left(\frac{1}{2}x^2\right)\right)dx.$$


问:
What is the determinant of an inversed Matrix where the matrix is an upper triangular matrix?
Given that

$
U = \begin{bmatrix}1&4&6\\0&2&5\\0&0&3\end{bmatrix}
$

find

$
det(U^{-1})
$

I just have one question.. would I get the answer if I simply inversed U and found its determinant? or do i need to first inverse the U matrix, get the determinant, and finally divide it by det(U)?

In any case, I looked up two different methods of inverting a 3-by-3 matrix..

creating a matrix of cofactors and finding its determinants and dividing it by det(U) = 1/6

the second method:

Row reduction but not dividing by det(U) = 1/6...

So in other words, I get 1/6 as the determinant of inverse U when I divide it by det(U) on the one hand, but also, I get 1/6 as determinant of inverse U when I do not divide it by det(U)...

答:
$U = \begin{bmatrix}1&4&6\\0&2&5\\0&0&3\end{bmatrix}. \tag 1$

The determinant is a multiplicative function of its argument; thus

$\det(U) \det(U^{-1}) = \det(UU^{-1}) = \det(I) = 1; \tag 2$

since $U$ is upper triangular, its eigenvalues are $1$, $2$, and $3$, and we have

$\det(U) = 1 \cdot 2 \cdot 3 = 6; \tag3$

therefore in accord with (2)

$\det(U^{-1}) = (\det(U))^{-1} = \dfrac{1}{6}. \tag 4$

This is the same answer as is obtained by first computing $U^{-1}$ and taking the determinant of the result.

To proceed the "long way", and actually find $U^{-1}$ explicitly, one might follow the method outlined here.


答: It can be shown that $det(AB)=det(A)det(B)$. In particular, if $A\cdot A^{-1}=I$, you get that $det(A\cdot A^{-1})=det(A)det(A^{-1})=det(I)=1$. Hence $det(A^{-1})=det(A)^{-1}$.

Title: What is the determinant of an inversed Matrix where the matrix is an upper triangular matrix?

Body:
Given that

$
U = \begin{bmatrix}1&4&6\\0&2&5\\0&0&3\end{bmatrix}
$

find

$
det(U^{-1})
$

I just have one question.. would I get the answer if I simply inversed U and found its determinant? or do i need to first inverse the U matrix, get the determinant, and finally divide it by det(U)?

In any case, I looked up two different methods of inverting a 3-by-3 matrix..

creating a matrix of cofactors and finding its determinants and dividing it by det(U) = 1/6

the second method:

Row reduction but not dividing by det(U) = 1/6...

So in other words, I get 1/6 as the determinant of inverse U when I divide it by det(U) on the one hand, but also, I get 1/6 as determinant of inverse U when I do not divide it by det(U)...


问:
A binomial identity from Mathematical Reflections
Here is the problem:

Let $m,n$ be positive integers with $n>m$. Prove that

$\displaystyle\sum_{k=0}^{\lfloor\frac{n+m}2\rfloor} (-1)^{k}\binom{n}{k}\binom{m+n-2k}{n-1}=\binom{n}{m+1}$

This problem is O243 of Mathematical Reflections. A solution had been published using complex integration (https://www.awesomemath.org/assets/PDFs/MR5sol(1).pdf). However, I would like to see a solution using the difference operator, if any, since the form of the summand brings this to mind.

答:
First of all, it is worth stating explicitly that the problem assumes that $\binom{n}{k}$ is zero when $n < 0$ or $n>k$, even for $k\geqslant 0$. Indeed, otherwise

```<code>In[61]:= Table[
 Sum[(-1)^k Binomial[n, k] Binomial[n + m - 2 k, n - 1], {k, 0,
   n}], {n, 1, 5}, {m, 0, n - 1}]

Out[61]= {{0}, {0, 0}, {0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0, 0}}
</code>```

With those restrictions in place the claimed result is indeed reproduced:

```<code>In[65]:= Table[
 Sum[(-1)^k Binomial[n, k] Binomial[n + m - 2 k, n - 1] Boole[
    0 <= n - 1 <= n + m - 2 k], {k, 0, n}], {n, 1, 5}, {m, 0, n - 1}]

Out[65]= {{1}, {2, 1}, {3, 3, 1}, {4, 6, 4, 1}, {5, 10, 10, 5, 1}}

In[66]:= Table[Binomial[n, m + 1], {n, 1, 5}, {m, 0, n - 1}]

Out[66]= {{1}, {2, 1}, {3, 3, 1}, {4, 6, 4, 1}, {5, 10, 10, 5, 1}}
</code>```

With this said, the upper bound of summation over $k$ is $m_\ast = \left\lfloor \frac{m+1}{2} \right\rfloor$:
$$
   \mathcal{S}(n,m)= \sum_{k=0}^{m_\ast} (-1)^k \binom{n}{k} \binom{n+m-2k}{n-1}
$$
The summand is the hypergeometric term, meaning that
$$
 r(k) = \frac{c_{k+1}}{c_k} = \frac{(-1)^{k+1} \binom{n}{k+1} \binom{n+m-2(k+1)}{n-1} }{(-1)^k \binom{n}{k} \binom{n+m-2k}{n-1} } = \frac{-n+k}{k+1} \frac{-\frac{m+1}{2} + k}{-\frac{m+n-1}{2} + k} \frac{-\frac{m}{2} + k}{-\frac{m+n}{2}+k}
$$
and therefore
$$
   c_k = c_0 \prod_{q=1}^{k} r(q) = \binom{m+n}{n-1} \frac{(-n)_k}{k!} \frac{\left(-\frac{m}{2}\right)_k \cdot \left(-\frac{m+1}{2}\right)_k}{\left(-\frac{m+n}{2}\right)_k \cdot \left(-\frac{m+n-1}{2}\right)_k}
$$
And thus, since $r(m_\ast) = 0$ we have
$$
   \mathcal{S}(n,m) = \binom{n+m}{n-1} \cdot {}_3F_2\left(\left.\begin{array}{cll} -n & -m/2 & -(m+1)/2 \\ & -(m+n)/2 & -(m+n-1)/2 \end{array} \right| 1 \right)
$$
Now, per this identity:
$$
  {}_3F_2\left(\left.\begin{array}{cll} -n & a & b \\ & d & 1+a+b-d-n \end{array} \right| 1 \right) = \frac{(d-a)_n \cdot (d-b)_n}{(d)_n \cdot (d-a-b)_n}
$$
Using this identity for $a = -m/2$, $b = -(m+1)/2$ and $d=\epsilon-(m+n)/2$ with the intent to consider the limit of $\epsilon \to 0$ we have
$$
   \mathcal{S}(n,m) = \binom{n+m}{n-1} \lim_{\epsilon \to 0} \frac{ \left(-\frac{n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n}{2} + \epsilon \right)_n}{ \left(-\frac{m+n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n-m}{2} + \epsilon \right)_n }
$$
Using
$$
   \left(-\frac{n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n}{2} + \epsilon \right)_n = \frac{1}{2^{2n}} \frac{\Gamma(n+2 \epsilon)}{\Gamma(-n+2\epsilon)} = (-1)^n \frac{\Gamma(1+n-2\epsilon) \Gamma(n+2\epsilon)}{2^{2n} \pi} \sin(2 \pi \epsilon)
$$
$$
 \left(-\frac{m+n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n-m}{2} + \epsilon \right)_n = \frac{\Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) }{\Gamma\left(\frac{-n-m}{2} + \epsilon\right)\Gamma\left(\frac{1+m-n}{2} + \epsilon\right)} = \frac{\Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) }{-\frac{2\pi^2}{\sin(\pi m) + \sin(\pi n - 2 \pi \epsilon)}} \Gamma\left(1 + \frac{m+n}{2} - \epsilon\right) \Gamma\left( \frac{1+n-m}{2} - \epsilon\right) = (-1)^n \frac{\sin(2\pi \epsilon)}{2 \pi^2} \Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) \Gamma\left(1 + \frac{m+n}{2} - \epsilon\right) \Gamma\left( \frac{1+n-m}{2} - \epsilon\right)
$$
Combining, and using the duplication formula we get
$$
    \mathcal{S}(n,m) = \binom{n+m}{n-1} \frac{n! (n-1)!}{(n+m)! (n-m-1)!} = \binom{n}{m+1}
$$

Body:
First of all, it is worth stating explicitly that the problem assumes that $\binom{n}{k}$ is zero when $n < 0$ or $n>k$, even for $k\geqslant 0$. Indeed, otherwise

In[61]:= Table[
 Sum[(-1)^k Binomial[n, k] Binomial[n + m - 2 k, n - 1], {k, 0,
   n}], {n, 1, 5}, {m, 0, n - 1}]

Out[61]= {{0}, {0, 0}, {0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0, 0}}

With those restrictions in place the claimed result is indeed reproduced:

In[65]:= Table[
 Sum[(-1)^k Binomial[n, k] Binomial[n + m - 2 k, n - 1] Boole[
    0 <= n - 1 <= n + m - 2 k], {k, 0, n}], {n, 1, 5}, {m, 0, n - 1}]

Out[65]= {{1}, {2, 1}, {3, 3, 1}, {4, 6, 4, 1}, {5, 10, 10, 5, 1}}

In[66]:= Table[Binomial[n, m + 1], {n, 1, 5}, {m, 0, n - 1}]

Out[66]= {{1}, {2, 1}, {3, 3, 1}, {4, 6, 4, 1}, {5, 10, 10, 5, 1}}

With this said, the upper bound of summation over $k$ is $m_\ast = \left\lfloor \frac{m+1}{2} \right\rfloor$:
$$
   \mathcal{S}(n,m)= \sum_{k=0}^{m_\ast} (-1)^k \binom{n}{k} \binom{n+m-2k}{n-1}
$$
The summand is the hypergeometric term, meaning that
$$
 r(k) = \frac{c_{k+1}}{c_k} = \frac{(-1)^{k+1} \binom{n}{k+1} \binom{n+m-2(k+1)}{n-1} }{(-1)^k \binom{n}{k} \binom{n+m-2k}{n-1} } = \frac{-n+k}{k+1} \frac{-\frac{m+1}{2} + k}{-\frac{m+n-1}{2} + k} \frac{-\frac{m}{2} + k}{-\frac{m+n}{2}+k}
$$
and therefore
$$
   c_k = c_0 \prod_{q=1}^{k} r(q) = \binom{m+n}{n-1} \frac{(-n)_k}{k!} \frac{\left(-\frac{m}{2}\right)_k \cdot \left(-\frac{m+1}{2}\right)_k}{\left(-\frac{m+n}{2}\right)_k \cdot \left(-\frac{m+n-1}{2}\right)_k}
$$
And thus, since $r(m_\ast) = 0$ we have
$$
   \mathcal{S}(n,m) = \binom{n+m}{n-1} \cdot {}_3F_2\left(\left.\begin{array}{cll} -n & -m/2 & -(m+1)/2 \\ & -(m+n)/2 & -(m+n-1)/2 \end{array} \right| 1 \right)
$$
Now, per this identity:
$$
  {}_3F_2\left(\left.\begin{array}{cll} -n & a & b \\ & d & 1+a+b-d-n \end{array} \right| 1 \right) = \frac{(d-a)_n \cdot (d-b)_n}{(d)_n \cdot (d-a-b)_n}
$$
Using this identity for $a = -m/2$, $b = -(m+1)/2$ and $d=\epsilon-(m+n)/2$ with the intent to consider the limit of $\epsilon \to 0$ we have
$$
   \mathcal{S}(n,m) = \binom{n+m}{n-1} \lim_{\epsilon \to 0} \frac{ \left(-\frac{n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n}{2} + \epsilon \right)_n}{ \left(-\frac{m+n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n-m}{2} + \epsilon \right)_n }
$$
Using
$$
   \left(-\frac{n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n}{2} + \epsilon \right)_n = \frac{1}{2^{2n}} \frac{\Gamma(n+2 \epsilon)}{\Gamma(-n+2\epsilon)} = (-1)^n \frac{\Gamma(1+n-2\epsilon) \Gamma(n+2\epsilon)}{2^{2n} \pi} \sin(2 \pi \epsilon)
$$
$$
 \left(-\frac{m+n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n-m}{2} + \epsilon \right)_n = \frac{\Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) }{\Gamma\left(\frac{-n-m}{2} + \epsilon\right)\Gamma\left(\frac{1+m-n}{2} + \epsilon\right)} = \frac{\Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) }{-\frac{2\pi^2}{\sin(\pi m) + \sin(\pi n - 2 \pi \epsilon)}} \Gamma\left(1 + \frac{m+n}{2} - \epsilon\right) \Gamma\left( \frac{1+n-m}{2} - \epsilon\right) = (-1)^n \frac{\sin(2\pi \epsilon)}{2 \pi^2} \Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) \Gamma\left(1 + \frac{m+n}{2} - \epsilon\right) \Gamma\left( \frac{1+n-m}{2} - \epsilon\right)
$$
Combining, and using the duplication formula we get
$$
    \mathcal{S}(n,m) = \binom{n+m}{n-1} \frac{n! (n-1)!}{(n+m)! (n-m-1)!} = \binom{n}{m+1}
$$


问:
A geometry problem with the reflection of the incenter
Consider $\triangle ABC$ and its incenter $I$. Let $M$ be the midpoint of arc $BC$ (not containing $A$) and $P$ be the second intersection point of the circumcircle of $\triangle ABC$ and the circle with center $I$ and radius $IA$. Prove that if $I'$ is the reflection of $I$ with respect to $BC$, then $I', P, M$ are collinear (lie on the same line).

Please help, I am lost, I don't have any idea! Thanks a lot in advance! Anyway, this problem is based on a problem connected to the reflection of the incenter, which appeared in Mathematical Excalibur.

For help, here is a figure:

答:
@Alex Zhao's proof is incorrect, but I think I've managed to make the idea work.  As in the original, this is just a big angle chase.

Let $K$ be the intersection of $(I)$ and $BC$ closest to $C$.

The triangle $ABK$ is symmetric about $BI$.$\qquad(*)$

Therefore $$
\begin{aligned}
\measuredangle IKB =\measuredangle BAI=\measuredangle IAC&\implies\measuredangle IAC+\measuredangle CKI=\pi\\
&\implies CKIA \text{ is cyclic}\\
&\implies \measuredangle KIA=\pi-\measuredangle ACB.\qquad(**)
\end{aligned}
$$

We claim $\measuredangle KPB+\measuredangle KI'B=\pi$, because
$$\begin{aligned}
\measuredangle KPB
&\overset{}=\measuredangle APB-\measuredangle APK\\
&= \pi-\measuredangle ACB-\measuredangle KIA/2\\
&=(\pi-\measuredangle ACB)/2 \qquad\text{ (by $(**)$)}\\
&=(\measuredangle CAB+\measuredangle CBA)/2\\
&=\measuredangle IAB+\measuredangle IBA\\
&=\pi-\measuredangle AIB\\
&=\pi-\measuredangle BIK\qquad\text{ (by $(*)$)}\\
&=\pi-\measuredangle KI'B.
\end{aligned}
$$
Thus $PBI'K$ is cyclic.

Therefore
$$
\begin{aligned}
\measuredangle BPI'
&=\measuredangle BKI'\\
&=\measuredangle IKB\\
&=\measuredangle BAI\\
&=\measuredangle BAM\\
&=\measuredangle BPM.
\end{aligned}
$$
Thus $P,I',M$ are collinear and we're done.


问:
Output of a absolute value function?
In the book Theory of computation by Michael Sipser.
"For example, the absolute value function ${\text{abs}}$ takes a number $\boldsymbol x$ as input and
returns $\boldsymbol x$ if $\boldsymbol x$ is positive and $\boldsymbol {-x}$ if $\boldsymbol x$ is negative. Thus $\text{abs}\boldsymbol{(2)}\boldsymbol = \text{abs}\boldsymbol{(-2)} \boldsymbol{=
2.}$". But absolute value function always give positive output then how does the book tell that function $\text{abs}$ returns $x$ if $x$ is positive and $-x$ if $x$ is negative. How is that even possible.
Thanks in advance.

答:
But absolute value function always give positive output then how does the book tell that function abs returns x if x is positive and -x if x is negative.

If I am interpreting your query correctly, it actually represents two questions:

If (for example) abs($x$) = 2, what are the possible values of $x$?

How do you know that if abs($x$) = $s \times x$, where $s \in \{1, -1\},$ then when $x > 0, s = 1$ is used and when $x < 0, s = -1$ is used?

Answers

It is clear from the definition that abs($x$) = $s \times x$, where $s$ is either -1, or +1.  Therefore, if abs($x$) = 2, then $x \in \{2, -2\}.$

Since the absolute value function always returns a non-negative number, when $x > 0$, the scalar that is applied must be $s = 1$, rather than $s = -1.$

Similarly, when $x < 0$, the scalar that is applied must be $s = -1$.  Otherwise, the result would not be non-negative.

Title: Output of a absolute value function?


问:
finding a probability density function of a random variable given a relation between another random variable
I am trying to find a pdf for a random variable $X$ where $X=-2Y+1$ and $Y$ is given by $N(4,9)$

Here is my attempt:

we know $\mu=4$ and $\sigma=3$. so that the normal distribution of $Y$ is given by $\frac{1}{3\sqrt{2\pi}}e^\frac{-(y-4)^2}{18}$

We can differentiate the cumulative function of $X$ to get the pdf for $X$.

cdf of $X = P(X<x)$ = $P(-2Y+1<x)=P(Y<\frac{-(x-1)}{2})=\int_{-\infty}^{\frac{-(x-1)}{2}}\frac{1}{3\sqrt{2\pi}}e^\frac{-(y-4)^2}{18}dy$

so $\frac{d}{dx}(\int_{-\infty}^{\frac{-(x-1)}{2}}\frac{1}{3\sqrt{2\pi}}e^\frac{-(y-4)^2}{18}dy)=f(x)$, which is the density function for $X$

$f(x)=-\frac{1}{6\sqrt{2\pi}}e^\frac{-(\frac{-(x-1)}{2}-4)^2}{18}$

Is this a correct way to approach the problem? I feel like my answer is very funky.

答:
Alternatively, one may use the very well known (and easy) fact that

If $S \sim N(0,1)$ , then $T = aS + b \sim N(b , a^2)$ .

Here, as $Y \sim N(4,9)$ , so $Y = 3S + 4$ . Therefore, $$X = -2Y +1 = -2(3S + 4) + 1 = -6S - 7$$ Thus, $X \sim N(-7 , 36)$ .


问:
Find the GS to the following InHomo System
$x_1' = x_1+3x_2 + e^{-2t}$

$x_2'= x_1-x_2$

is the InHomo System.

This is using eigenvalues and eigenvectors to solve this Inhomo DE.

The book gives me the answers, I just need help understanding it.

I need help solving for the unknown vectors. I don't understand how the book got the answers below. If anyone can help me with this I would really appreciate it. I don't understand how they got $E=-\cfrac{1}{4}\begin{pmatrix}
1 \\
0
\end{pmatrix}$

$G=\cfrac{1}{4}\begin{pmatrix}
1 \\
-1
\end{pmatrix}$

The book completely skips this part and I would like help understanding how they got those vectors.

Thank You For Any Help.

答:
Did you try to solve the given equations ?
$$-2G=AG$$
$$(A+2I)G=0$$
$$\left(\pmatrix {1 & 3 \\ 1&-1}+ \pmatrix {2&0 \\0&2} \right)G=0$$
$$\pmatrix {3 & 3 \\ 1&1} \pmatrix {g_1 \\g_2}=0$$
$$\implies g_1=-g_2$$
$$\implies G=k\pmatrix {1 \\-1}$$
Plug this in the first equation:
$$-2E+G=AE+ \pmatrix {1 \\0}$$
$$(A+2I)E=G- \pmatrix {1 \\0}$$
$$(A+2I)E= \pmatrix {k-1 \\ -k}$$
$$\pmatrix {3 & 3 \\ 1&1} \pmatrix {e_1 \\e_2}= \pmatrix {k-1 \\ -k}$$
$$-3k=k-1 \implies k =\dfrac 14$$
$$e_1+e_2 =-\dfrac 14$$
If you choose  $e_2=0$ then $e_1=-\dfrac 14$. and $E=-\dfrac 14 \pmatrix {1 \\0}$.


问:
Proving $\lim _{n\to\infty} \sqrt{n} \int_{1}^{\infty} e^{-n x^{2}} d x=0$
How to prove the following?
\begin{equation}
\lim _{n \rightarrow \infty} \sqrt{n} \int_{1}^{\infty} e^{-n x^{2}} d x=0
\end{equation}

答:
$$I_n= \sqrt{n} \int_{1}^{\infty} e^{-n x^{2}} dx$$ Let $x=\frac{t}{\sqrt{n}}$ to make
$$I_n=\int_{\sqrt n}^{\infty}e^{-t^{2}} dt=\frac{\sqrt{\pi }}{2}  \left(1-\text{erf}\left(\sqrt{n}\right)\right)$$

Using the asymptotics of the error function
$$\text{erf}\left(\sqrt{n}\right)=1-\frac{e^{-n}}{\sqrt\pi }\left(\frac{1}{\sqrt{ n} } -\frac{1}{2n\sqrt{ n} }+\cdots\right)$$
$$I_n\sim \frac {e^{-n}} {2\sqrt{ n} }$$ which is a very good approximation of $I_n$ even if $n$ is small (for $n=3$, the exact value is $0.00001958$ while the asymptotics gives $0.00002057$).


问:
Proving that a subset $\mathscr{H}$ is dense in $\mathbb{R}$

Consider the subset $\mathscr{H} =\displaystyle\left\{\frac{a}{10^{b}} \ |\: a,b\in\mathbb{Z}\: \right\}$. Then is $\mathscr{H}$ dense in $\mathbb{R}$.

To prove that $\mathscr{H}$ is dense I have to show that given $x,y \in\mathbb{R}$  there is a $m,n\in\mathbb{Z}$ such that $x < \frac{m}{10^{n}} <y$.

Thats it I am hopelessly stuck.

答: Hint: What do elements of $\mathscr{H}$ look like in decimal notation?</p>


问:
How to compute this double integral involving the gamma function?
Assuming that $a > 0,\ {\tt Mathematica}$ says that
$$\!\!\!
2\int_{0}^{\infty}\!\!\!\int_{0}^{\infty}
\frac{v^{a}\,\left(u + v\right)^{a - 1}\,\,
\mathrm{e}^{-u -2 v}}{\Gamma^{2}\left(a\right)} \,\mathrm{d}u\,\mathrm{d}v =
a - \frac{\Gamma\left(a + 1/2\right)}{\Gamma\left(1/2\right)\Gamma\left(a\right)}.$$

How can we arrive at the solution $?$.
I don't see the proper change(s) of variables.

Note: $\displaystyle\Gamma\left(a\right) =
\int_{0}^{\infty}t^{a - 1}\,\,\mathrm{e}^{-t}
\,\mathrm{d}t$.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{2\int_{0}^{\infty}\int_{0}^{\infty}
{v^{a}\pars{u + v}^{a - 1}\,
\expo{-u - 2v} \over \Gamma^{2}\pars{a}}\dd u\,\dd v}
\\[5mm] = &\
{2 \over \Gamma^{2}\pars{a}}
\int_{0}^{\infty}\int_{v}^{\infty}
v^{a}\expo{-v}u^{a - 1}\,\expo{-u}\,\,\dd u\,\dd v
\\[5mm] = &\
{2 \over \Gamma^{2}\pars{a}}
\int_{0}^{\infty}\int_{0}^{\infty}\
\underbrace{\bracks{\int_{-\infty}^{\infty}
{\expo{\ic k\pars{u - v}} \over k - \ic 0^{+}}{\dd k \over 2\pi\ic}}}
_{\ds{\Theta\pars{u - v}}}\ \times
\\[2mm] &\
v^{a}\expo{-v}u^{a - 1}\,\expo{-u}\,\,\dd u\,\dd v
\end{align}
$\ds{\Theta}$ is the Heaviside Theta
( or Heaviside Step ) Function.

Then,
\begin{align}
&\bbox[5px,#ffd]{2\int_{0}^{\infty}\int_{0}^{\infty}
{v^{a}\pars{u + v}^{a - 1}\,
\expo{-u - 2v} \over \Gamma^{2}\pars{a}}\dd u\,\dd v}
\\[5mm] = &\
{2 \over \Gamma^{2}\pars{a}}
\int_{-\infty}^{\infty}{1 \over k - \ic 0^{+}}
\underbrace{\bracks{\int_{0}^{\infty}v^{a}
\expo{-\pars{1 + \ic k}v}\,\,\dd v}}
_{\ds{\pars{1 + \ic k}^{-a - 1}\,\,\Gamma\pars{a + 1}}}
\ \times
\\[2mm] &\
\underbrace{\bracks{\int_{0}^{\infty}u^{a - 1}
\expo{-\pars{1 - \ic k}u}\,\,\dd u}}
_{\ds{\pars{1 - \ic k}^{-a}\,\,\Gamma\pars{a}}}
{\dd k \over 2\pi\ic}
\\[5mm] = &\
{2 \over \Gamma^{2}\pars{a}}\,\Gamma\pars{a + 1}
\Gamma\pars{a}\ \times
\\[2mm] &\
\int_{-\infty}^{\infty}
{\pars{1 + \ic k}^{-a - 1}\,\,\pars{1 - \ic k}^{-a} \over
k - \ic 0^{+}}\,{\dd k \over 2\pi\ic}
\\[5mm] = &\
2a\left[%
\mrm{P.V.}\int_{-\infty}^{\infty}
{\pars{1 + k^{2}}^{-a}\,\pars{1 + \ic k}^{-1} \over k}\,{\dd k \over 2\pi\ic}\right.
\\[2mm] &\
\left.+
\int_{-\infty}^{\infty}
\pars{1 + k^{2}}^{-a}\,\pars{1 + \ic k}^{-1}
\bracks{\ic\pi\,\delta\pars{k}}{\dd k \over 2\pi\ic}\right]
\\[5mm] = &\
2a\int_{0}^{\infty}
{1 \over \pars{1 + k^{2}}^{a}\,k}
\pars{{1 \over 1 + \ic k} - {1 \over 1 - \ic k}}
\,{\dd k \over 2\pi\ic}
\\[2mm] &\
+ a
\\[5mm] = &\
a - {2a \over \pi}\
\underbrace{\int_{0}^{\infty}{\dd k \over \pars{1 + k^{2}}^{a + 1}}}
_{\ds{\root{\pi}\Gamma\pars{a + 1/2} \over 2\Gamma\pars{a + 1}}}
\\[5mm] = &\
\bbx{a - {\Gamma\pars{a + 1/2} \over \Gamma\pars{1/2}\Gamma\pars{a}}}\quad
\pars{\substack{\mbox{Note that}\\[1mm]
\ds{\root{\pi} = \Gamma\pars{1 \over 2}}}}\\ &
\end{align}

The last integral can be evaluated with
Ramanujan's Master Theorem. Namely,
\begin{align}
&\int_{0}^{\infty}
{\dd k \over \pars{1 + k^{2}}^{a + 1}} =
{1 \over 2}\int_{0}^{\infty}
{k^{\color{red}{1/2} - 1} \over
\pars{1 + k}^{a + 1}}\,\dd k
\end{align}
and
\begin{align}
&{1 \over \pars{1 + k}^{a + 1}} =
\sum_{n = 0}^{\infty}{-a - 1 \choose n}k^{n}
\\[5mm] = &\
\sum_{n = 0}^{\infty}{a + n \choose n}
\pars{-1}^{n}k^{n} =
\sum_{n = 0}^{\infty}
\color{red}{\Gamma\pars{a + 1 + n} \over \Gamma\pars{a + 1}}{\pars{-k}^{n} \over n!}
\end{align}
Then,
\begin{align}
&\int_{0}^{\infty}
{\dd k \over \pars{1 + k^{2}}^{a + 1}} =
{1 \over 2}\Gamma\pars{1 \over 2}\,
{\Gamma\pars{a + 1 - 1/2} \over
\Gamma\pars{a + 1}}
\\[5mm] = &
\root{\pi}\Gamma\pars{a + 1/2} \over 2\Gamma\pars{a + 1}
\end{align}


问:
Let $f(x,y,z)=e^{-x-y-z},x>0,y>0,z>0$, and $=0$ otherwise, be the joint PDF of $(X,Y,Z)$. Compute $P(X=Y<Z)$.
Let $$f(x,y,z)=\begin{cases}e^{-x-y-z}, \ \ \text{$x>0,y>0,z>0$} \\0,\ \ \text{otherwise}\end{cases}$$
be the joint PDF of $(X,Y,Z)$. Compute $P(X=Y<Z)$.

According to the textbook, the answer is $0$, but I am unsure how to proceed.

答: It is zero, because for continuous random variables events like “$X = \text{something}$” have zero probability. More precisely, to calculate given probability, you have to integrate joint PDF over region $\{(x, y, z) : x=y, y<z\}$ which has zero measure (volume) in $\mathbb{R}^3$.


问:
Vectorization of a symmetric hollow matrix
I would like to cast some problems involving a symmetric hollow (diagonal entries equal to $0$) matrix in an equivalent form represented by the vectorization of the matrix. I know that if the matrix is only symmetric, then an effective way is to use the half-vectorization of the matrix and the use of duplication and elimination matrices. I would like to find something similar to the case of hollow symmetric matrices.

I have a real-valued function $f$ with argument a symmetric hollow matrix $\mathbf{A}$. Because all the information of the matrix is stored in the strictly lower (or upper) triangular part of the matrix, call it $\mathbf{A}_1$ , I guess there should be a way to rewrite the function by using the vectorization of the matrix $\mathbf{A}_1$, i.e. $\mathbf{a}=\mathbf{A}_1$.

For instance, considering a symmetric matrix $\mathbf{B}$:

$$ f(\mathbf{A})= \operatorname{trace}(\mathbf{A}\mathbf{B}) = 2 \mathbf{a}^{\top}\mathbf{b}$$

This is immediate with this particular function, but how can I generalize to other functions? For instance, what about $ f(\mathbf{A})= \operatorname{trace}(\mathbf{A}^\top\mathbf{A}\mathbf{B})$.
Can in general also the product $\mathbf{A}\mathbf{A}$ be rewritten with a particular reshape of the vectors product?

Thank you

EDIT

If I have  function $f(\mathbf{A})$, can I compute in this case gradient and Hessian in the hollow-half-vec space, and being able to compute back the associated matrices? As a related subquestion, is the $\operatorname{vechh(\cdot)}$ an invertible operation?
Of course, once that I have $\mathbf{a}=\operatorname{vechh(\mathbf{A})}$ I can easily reshape the vector in a way to get back $\mathbf{A}$. However, how is this mathematically represented? I should be able to do something like:

$$ \mathbf{A}= \operatorname{vec}^{-1} (\mathbf{D}\mathbf{H}^{-1}\mathbf{a}) \mathbf{P}^\top= \operatorname{vec}^{-1} (\mathbf{D}\mathbf{H}^{-1}\mathbf{H}\mathbf{E}\operatorname{vec}(\mathbf{AP})) \mathbf{P}^\top $$

where $\mathbf{D}$,$ \mathbf{E}$ are the duplication and elimination matrices, respectively, and $\mathbf{H}^{-1}$ is the inverse of the "hollow" matrix $\mathbf{H}=[\mathbf{0}_{l\times n}  \mathbf{I}_l]$.

As a case study, we could take $f(\mathbf{A})= \operatorname{trace}(\mathbf{A}\mathbf{B}) = 2 \mathbf{a}^{\top}\mathbf{b}$, with $\mathbf{A}$, $\mathbf{B}$ symmetric matrices, $\mathbf{A}$ hollow. It would be nice to have a methodology to express each function $f(\mathbf{A})$ in an equivalent form $f(\mathbf{a})$.

答:
$\def\m#1{ \left[\begin{array}{r}#1\end{array}\right] }$
Given a hollow matrix
$$\eqalign{
A &\in {\mathbb R}^{n\times n} \\
}$$
the hollow-half-vec operation is analogous to the more familiar half-vec operation. Both can be described in terms of the standard vec operator
$$\eqalign{
{\rm vechh}(A) &= E_h\cdot{\rm vec}(A) \quad&\sim\quad
{\rm vech}(A) &= E\cdot{\rm vec}(A) \\
{\rm vec}(A) &= D_h\cdot{\rm vechh}(A) \quad&\sim\quad
\;\;{\rm vec}(A) &= D\cdot{\rm vech}(A) \\
}$$
where $E_h\in{\mathbb R}^{\ell\times n^2}$ is the hollow-elimination matrix. It is a sparse binary matrix $\,\ell = \tfrac 12(n^2-n)\,$. There is one non-zero element in each row, whose column index corresponds to the index of the conserved element in ${\rm vec}(A)$.

The hollow-duplication matrix is $D_h\in{\mathbb R}^{n^2\times\ell}$ is also a sparse binary matrix, whose elements are such that $E_hD_h=I$ and whose columns sum to two, i.e. $\;\tfrac 12D_h^T{\tt1} = {\tt1}.\;$
Again, this is analogous to the half-vec case where $ED=I\,$ (however the column sums of $D$ vary between one and two).

Interestingly, the pseudoinverse $D_h^+$ can serve as an elimination matrix, although $E_h^+$ fails as a duplication matrix, i.e.
$$\eqalign{
{\rm vechh}(A) &= D_h^+\cdot{\rm vec}(A) \quad&\sim\quad
{\rm vech}(A) &= D^+\cdot{\rm vec}(A) \\
{\rm vec}(A) &\ne E_h^+\cdot{\rm vechh}(A) \quad&\sim\quad
\;\;{\rm vec}(A) &\ne E^+\cdot{\rm vech}(A) \\
D_h^+D_h &= I \quad&\sim\quad \quad D^+D &= I \\
}$$
$\Big[\,$In the hollow case, the calculation is particularly easy since
$D_h^+ = \tfrac 12D_h^T\;\Big]$

For example, for $\;n=4,\;\ell=6$
$$\eqalign{
&E_h[1,2] &= 1,\qquad &D_h(2,1) &= 1,\quad &D_h(5,1) &= 1 \\
&E_h[2,3] &= 1,\qquad &D_h(3,2) &= 1,\quad &D_h(9,2) &= 1 \\
&E_h[3,4] &= 1,\qquad &D_h(4,3) &= 1,\quad &D_h(13,3)&= 1 \\
&E_h[4,7] &= 1,\qquad &D_h(7,4) &= 1,\quad &D_h(10,4)&= 1 \\
&E_h[5,8] &= 1,\qquad &D_h(8,5) &= 1,\quad &D_h(14,5)&= 1 \\
&E_h[6,12]&= 1,\qquad &D_h(12,6)&= 1,\quad &D_h(15,6)&= 1 \\
}$$
The indices of the vec and vechh vectors mapped onto
$4\times 4$ matrices helps elucidate the components
given above.
$$\eqalign{
\m{1&5&9&13\\2&6&10&14\\3&7&11&15\\4&8&12&16} \qquad\qquad
\m{0&1&2&3\\1&0&4&5\\2&4&0&6\\3&5&6&0}
\\
}$$