问:
Jordan normal form of $\left(\begin{smallmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{smallmatrix}\right)$
I want to find the Jordan normal form of $A=\begin{pmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{pmatrix}$, but somewhere I think that I make a mistake (I am quite new to the computation of the Jordan normal form).

I could find that $\lambda=3$ is $A$'s only eigenvalue and its algebraic multiplicity is $3$.

Afterwards, I solved the system $(A-3I_3)X=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ and I got that $x=\begin{pmatrix} -a-b \\ a \\ b \end{pmatrix}$ for some parametres $a, b\in \mathbb{C}$.

I picked the eigenvectors $X_1=\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ and $X_2=\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$ and then I got stuck. I tried solving both the equations $(A-3I_3)X'=X_1$ and $(A-3I_3)X''=X_2$, but these have no solutions, so I can't find any generalised eigenvectors. What should I do?

答:
After you get the Jordan form (as psidaga's answer did)
\begin{align*}
J = \begin{pmatrix}
3 & 1 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix},
\end{align*}
in order to determine the similarity matrix $P$ such that $P^{-1}AP = J$, it is is equivalent to solve the system
\begin{align*}
& A\alpha_1 = 3\alpha_1, \tag{1} \\
& A\alpha_2 = \alpha_1 + 3\alpha_2, \tag{2} \\
& A\alpha_3 = 3\alpha_3,
\end{align*}
where $\alpha_1, \alpha_2, \alpha_3$ are three columns of $P$. As you did, you can fix $\alpha_3$ to be $(-1, 0, 1)^T$. To determine $\alpha_1$ and $\alpha_2$, note that $A - 3I_{(3)}$ is nilpotent with nilpotent index $2$, i.e., $(A - 3I_{(3)})^2 = 0$, hence any non-zero vector $\alpha_2$ that is not already an eigenvector and $\alpha_1 := (A - 3I_{(3)})\alpha_2$ would satisfy equations $(1)$ and $(2)$, since $(A - 3I_{(3)})\alpha_1 = (A - 3I_{(3)})^2\alpha_2 = 0$. For example, let $\alpha_2 = (1, 0, 0)^T$, then $\alpha_1 = (1, -2, 1)^T$. It can be easily verified that
\begin{align*}
P = \begin{pmatrix}
1 & 1 & -1 \\
-2 & 0 & 0 \\
1 & 0 & 1
\end{pmatrix}
\end{align*}
is invertible and satisfies $A = PJP^{-1}$.

Of course, $P$ is not unique, depending on what $\alpha_2$ you are choosing.


问:
Last Two digits of ${14}^{{14}^{14}}$
How to calculate the last two digits of ${14}^{{14}^{14}}$? With the help of any method. I have tried and have got the last digit to be $6$. But not sure.</p>

答:
Review the theory and techniques found here.

We write as true (no calculator is necessary),

$\; 14^{14} \equiv (14^2)^7 \equiv 16 \times 16^2 \times 16^2  \times 16^2 \equiv 16 \times 16 \times 16  \times 16 \equiv 16 \pmod{20}$

So (again using mental calculations),

$\; \displaystyle{ {14}^{{14}^{14}} \equiv 76 \times {14}^{16} \equiv 76 \times ({14}^{2})^8 \equiv 76 \times ({96}^{2})^4 \equiv ({16}^{2})^2 \equiv 36 \pmod{100}}$

ContentLicense: CC BY-SA 3.0

Title: Last Two digits of ${14}^{{14}^{14}}$

ContentLicense: CC BY-SA 4.0


问:
Joint density of independent gamma distributions
Let $X,Y$ be independent r.v such that $X \sim \Gamma(\alpha, \lambda)$ and $Y \sim \Gamma (\beta, \lambda)$. Which is the joint density of $(X,X+Y)$? And the conditional density $g_{X\mid X+Y}(x\mid z)$ of $X$ given $X+Y = z$?

I know that if $X,Y$ are independent Gamma distribution then $X+Y \sim \Gamma (\alpha + \beta, \lambda)$ but how should I compute the joint density? Of course $X$ and $X+Y$ are not independent at this point.. thanks

EDIT:

Based on your answer I managed to get back at some theory and this is what I've come to so far:

basically we have $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ which maps $(X,Y)^T \mapsto (X,X+Y)^T$. It is easy to see that this is the linear transformation associated to the matrix
$$A =\begin{bmatrix}1 & 0\\1 & 1\end{bmatrix}.$$
Now we find the inverse of transformation, as you stated
$$
\begin{cases}
x = u\\
y= z-u\\
\end{cases}
$$
finding that $\phi^{-1}(x,y) = (u,z-u)$ and so the joint density should be $f(\phi^{-1}(x,y))|\det A^{-1}|$ (with last term equals to $1$ and this is OK).

The only doubt that now I have is: density $f$ refers to which random variable? How this get substituted at this point? thanks, my main interest is always about understanding the whole process, and not so much the ending result.

Ok I got it.

If we rewrite the joint density of initial vector $(X,Y)$ we know $X,Y$ are independent and therefore the joint density is the product of marginals:

$$f(x,y)= \frac{\lambda^{\alpha + \beta}}{\Gamma(\alpha + \beta)}x^{\alpha-1} e^{-\lambda x} y^{\beta-1}e^{-\lambda y}$$

and then we get

$$f(u,z-u)= g(u,z) = \frac{\lambda^{\alpha+ \beta}}{\Gamma(\alpha + \beta)}u^{\alpha-1}(z-u)^{\beta-1}e^{-\lambda z}$$

This of course iff $z \ge u$ since it is a density. Correct?

答:
setting

$$
\begin{cases}
z=x+y \\
u=x
\end{cases}$$

that is

$$
\begin{cases}
x=u \\
y=z-u
\end{cases}$$

The Jacobian is evidently 1 thus the joint distribution is

$$f_{UZ}(u,z)=\frac{\lambda^{\alpha+\beta}}{\Gamma(\alpha)\Gamma(\beta)}u^{\alpha-1}(z-u)^{\beta-1}e^{-\lambda z}\cdot\mathbb{1}_{(0;\infty)}(u)\cdot\mathbb{1}_{(u;\infty)}(z)$$

$(U,Z)$ bivariate support is an infinite triangle (over or under the line $U=Z$ as you set the axes) because

$0<Y<\infty$ or

$0<Z-U<\infty$

that means $Z>U$

Now I think you can proceed by yourself


问:
How To Prove This Is Continuous Or Discontinuous
How do I prove this is continuous/discontinuous?

A function $f(x)$ (can be any function, in general) is defined on the interval $[3,7]$, where $f(3)=2$ and $f(7)=4$. The range of $f(x)$ on this interval is $$2\leq f(x)\leq 4$$ for all $x \in [3,7]$.

$f$ has unique values on $[2,4]$ (so if $b$ is in $[2,4]$ there is only one $c$ in $[3,7]$ such that $f(c)=b$)

How should I prove that this is continuous/not continuous? I think using the intermediate value theory, we can assume that f(x) is continuous first and then use say that there is a point $c$ and  $$f(c) = \alpha$$ and $f(a) < \alpha < f(b)$. What do I do with the point next? The function $f(x)$ can be anything so I cannot use a single value for it. Please can someone help.

答: If you can prove that for every alpha between f(a) and f(b) the function takes on a value, then you have proved the intermediate value theorem. In this cases, what I look for is a counter example. What you want is to look for all the f(c). That would be tedious. Just look for a counter example, such that f(d) = $\beta$ and $\beta$ does not lie in the interval between f(a) and f(b). If you cannot find such d, then from the intermediate value theorem, it is continuous.


问:
Finding the limit of the integral of some uniformly convergant sequence of functions
Let $\{f_n\}_{n\in \mathbb N}$ be a sequence of uniformly convergent functions on the interval [0,1]. Find the limit of the following integral: $$\lim_{n\to \infty} \int_0^1nf_n(t)e^{-nt}\,dt$$

Here is my thought process so far. In isolation l'hospital's rule would indicate that  $\lim_{n\to \infty} \frac{n}{e^{nt}}$ would tend to zero, and so I expect that this exponential term will be more dominating.

Normally Lebesgue's Dominated convergence theorem (or at the very least his monotone convergence theorem) would allow us to say that this uniformly convergent sequence of functions would converge to f...however I am guessing the surrounding terms sufficiently mess this up to the point where we no longer have convergence. My guess is the point of this question is to prove that this is the case.

We know that  $\lim_{n\to \infty}\{f_n\}_{n\in \mathbb N}$ is measurable, but If I can show that
$g_n= \frac{n}{e^{nt}}$ is not measurable for $t\in[0,1]$ (because it tends to zero), then can I say this limit does not exist? I'm pretty confused, as is abundantly clear hahahaha.

答:
Let $(f_n)$ be a sequence of continuous functions that converges uniformly to $f$ on $[0,1]$.  By the triangle inequality, we have

$$\left|\int_0^1 n e^{-nt} f_n(t) \, dt - f(0) \right|\leqslant \left|\int_0^1 n e^{-nt} (f_n(t) - f(t)) \, dt \right|+ \left|\int_0^1 n e^{-nt} f(t) \, dt - f(0) \right|$$

For the first term on the RHS, we have

$$\left|\int_0^1 n e^{-nt} (f_n(t) - f(t)) \, dt \right| \leqslant \int_0^1 n e^{-nt}|f_n(t) - f(t)|\, dt$$

As $f_n \to f$ uniformly, for any $\epsilon > 0$ there exists $N_1 \in \mathbb{N}$ such that if $n > N_1$, then $|f_n(t) - f(t)| < \frac{\epsilon}{2}$ for all $t \in [0,1]$, and

$$\left|\int_0^1 n e^{-nt} (f_n(t) - f(t)) \, dt \right| \leqslant \frac{\epsilon}{2} \int_0^1ne^{-nt} \, dt = \frac{\epsilon}{2} (1 - e^{-n}) < \frac{\epsilon}{2} $$

Thus, for all $n > N_1$,

$$\left|\int_0^1 n e^{-nt} f_n(t) \, dt - f(0) \right|\leqslant \frac{\epsilon}{2}+ \left|\int_0^1 n e^{-nt} f(t) \, dt - f(0) \right|$$

Since $f$ is the uniform limit of a sequence of continuous functions it is continuous, and the proof given here shows there exists $N_2$ such that if $n > N_2$

$$\left|\int_0^1 n e^{-nt} f(t) \, dt - f(0) \right| < \frac{\epsilon}{2}$$

Altogether this proves that

$$\lim_{n \to \infty}\int_0^1 n e^{-nt} f_n(t) \, dt = f(0)$$


问:
Implicit function graphing
I came across the function,
$$z=\frac{4xy(x^2-y^2)}{x^2+y^2}$$
and I had to draw the level curve for this equation. Clearly the equation is already explicitly expressed in the form $z=f(x,y)$ and therefore for every value of $z$, the "shape" of the curve would be the same. We assume $z$ to be some constant $c$ and proceed. The main problem arises when we try to graph the following function which is now implicit in $x$ and $y$:
$$c=\frac{4xy(x^2-y^2)}{x^2+y^2}$$

MY ATTEMPT:
Firstly I tried proceeding normally. The function does not seem to be defined at the origin $(0,0)$. Now I tried checking behaviour at infinity.
$$c=\lim_{x\to\infty}\frac{4xy(x^2-y^2)}{x^2+y^2} $$
Now the very first question that arises is how do we deal with $y$? We do not know if the limit of $y$ as $x$ approaches $\infty$ is $0$, $\infty$ or finite. Therefore we cannot comment on how the function behaves at either infinity.
The next thing that we do is look at the derivative. The derivative here is:
$$y'= \frac{12xy^2+4y^3-12x^2y+2cx}{4x^3-2cy}$$
Clearly this derivative is also difficult to handle because of the implicit nature of the function. The only  takeaway from it is the fact that there exists a vertical asymptote at those points where
$$y=\frac{2x^3}{c}\;\;\;\;\;\;...(1)$$
which again being tedious gives this:
$$64x^8+(8c-8c^2)x^4+c^3=0$$
which anyone would discard at first look. If we just had to solve equation $(1)$ and our curve we could have tried approximating and playing around with graphs but that does not work here because the graph is what we actually want. Clearly I have ventured the wrong way, correct it is, but too rigorous to be solved.

MY 2nd ATTEMPT:
Another train of thought that arises is to try polar coordinates, which, by the way, was also suggested in the book. Doing so, we get the curve to be:
$$c= r^2sin(4\theta)$$
Now the problem here is that the polar system is a lot less intuitive compared to the Cartesian coordinate system as the variables deal with distance from the origin and angle with the x-axis. I still do not  see a clear way out.

So how do we deal with this? I know that a graphing calculator is the simplest solution, but then I am not a fan, especially in a question that deals specifically with "sketching" a graph. The book is old enough that there weren't any easily available graphing calculators around like we have today. So, could someone describe the technique to sketch this curve or is it that there isn't any?

P.S: I have seen some Stack Exchange questions and browsed the internet but those were simpler ones like $x=f(y)$ or $y^2=f(x)$ type, so I won't be linking them here.

答:
As mentioned in the hint, you can write this as

$$ c = r^2 \sin(4\theta)$$

For $c=0$, $r=0$ or $\sin(4\theta) = 0$ when $\theta$ is a multiple of $\pi/4$.  That is, the level curve for $z=0$ consists of the $x$ and $y$ axes and the lines $y=x$ and $y=-x$.

For $c \ne 0$, $r = \sqrt{c/\sin(4\theta)}$.  Of the eight intervals $(0, \pi/4), (\pi/4, \pi/2), \ldots, (7\pi/4, 2\pi)$, there are four where $\sin(4\theta)$ has the same sign as $c$.  This gives you curves that are closest to the origin at the middle of the interval, and go off to $\infty$ as you approach the endpoints.  The curves are symmetric under rotation by $\pi/2$.

Title: Implicit function graphing


问:
Do we get predicative ordinals above $\Gamma_0$ if we use hyperexponentiation?
I am trying to understand the Veblen hierarchy but I still find it confusing.
The Feferman–Schütte ordinal, $\Gamma_0$, can be described as the set of all ordinals that can be written as finite expressions, starting from zero, using only the Veblen hierarchy and addition. The Veblen hierarchy starts with the function $\omega^\alpha$ and uses recursion to generate larger and larger functions.
But I am not sure why we start with $f(\alpha)=\omega^\alpha$, or why we restrict the definition of $\Gamma_0$ to Veblen hierarchy and addition. Would we  obtain recursive ordinals larger than $\Gamma_0$ if we start the hierarchy with some faster growing function? For instance we can use hyperoperators to define  $f(\alpha)=(\alpha\uparrow ^\alpha \alpha)^{(\alpha)}$ (where the upperscript $(\alpha)$ means $\alpha$-times composition) which is immensely huge. We could also include this new hierarchy plus addition and any hyperoperator to define the limit $\Gamma_0$. But it is not clear to me if doing it this way results in the same $\Gamma_0$, or in a larger (but still predicative?) one.</p>

答:
This is mostly a more extended version of my comment. As for a book or published paper where $\Gamma_0$ is carefully and rigorously obtained, probably the best I know of is:

Wolfram Pohlers, Proof Theory. The First Step into Impredicativity, 2008.

However, the audience and goals of Pohlers' book are quite distant (in my opinion) from the basic ideas behind normal functions on ordinals and fixed-points of normal functions, which I think have not had enough exposition about in the mathematical literature. For example, pick up any of the dozens of introductory set theory books and many more related books (in real analysis, topology, etc.) in which ordinals are discussed, and you'll see nearly identical discussions of how the countable ordinals include $\epsilon_0$ and beyond (but no author seems to vary on this and indicate, even if briefly, some of the interesting things that lie in the "and beyond" realm), and if you pick up any of the more advanced books and papers in set theory and in proof theory where large countable ordinals are discussed, you're in for a whirlwind ride in which $\epsilon_0$ and $\Gamma_0$ are barely mentioned as bacteria on a microscope slide in the quest to reach the distant quasars of ever more exotic recursive ordinal notations, which are cryptically pitched to an audience of a few specialists. There seem to be no middle-ground expositions of this topic.

Maybe A short introduction to Ordinal Notations by Harold Simmons comes close to being a middle-ground exposition.

However, I think much better is the following, which is not written with applications of large ordinals to proof theory as its end goal. Instead, the focus is on the ride up through the ordinals.

John Baez, This Week's Finds in Mathematical Physics (Week 236; 26 July 2006)

Also, look up diagonal intersection. I believe this is the operation that takes you from ordinals defined via the $\alpha$'th order hyper-operation (applied to $\beta$) to $\Gamma_{0}.$

Finally, below are some things I've posted elsewhere that may be of use. Incidentally, I never continued the treatment of ordinal numbers I began in the first post below, as I wound up getting busy at work. I also started to think that it wasn't such a good idea to spend so much time preparing ASCII expositions of things that I had planned to write up more formally in LaTeX and distribute in some manner (publication, math arXiv site, or perhaps just toss out like I did with this).

Renfro, ORDINAL NUMBERS #1, sci.math, 29 October 2006. [Last version of this essay. See the discussion at the end concerning ${\omega}^{\omega}$ being simultaneously the ${\omega}^{\omega}$'th ordinal and the ${\omega}^{\omega}$'th limit ordinal.]

Renfro, Transfinite Aleph and Beth ordinal sequences, sci.math, 23 November 2003.

Renfro, fixed points with ordinal maps, sci.math, 30 November 2008. [See also the follow-up on 2 December 2008.]


问:
I have to check the convergence of the following series
I have to check the convergence of the following

$$\sum_{n=1}^\infty \frac{\ln^5 n}{n^2}, \quad \sum_{n=1}^\infty \frac{\ln n + \cos n}{n^2}$$

and I need some clues regarding how to approach them.

For the first one, I tried using the integral test and eventually I got an answer, but I wanted to know if there was a simpler way that didn't involve integrating by parts 6 times.
For the second one, I tried looking for another series to try and apply the limit test, but I was unsuccessful.

答:
For any $\alpha>0$ eventually

$$n^\alpha\ge \log n$$

then for the first one use that eventually

$$ \log^5n\le n^{\frac12}$$

and for the second one

$$ \log n+\cos n\le n^{\frac12}$$

then refer to direct comparison test.


问:
How to show a curve lies in a plane given 3 constant vectors
How does one show that given 3 non-parallel vectors in $\mathbb{R}^3$, a curve with parameterisation:
$$\textbf{s}(t)=(x_1t^2+y_1t+c_1,x_2t^2+y_2t+c_2,x_3t^2+y_3t+c_3)$$ lies in a plane?

Additionally, how do we find the equation of the plane?

I know that any two non-parallel vectors lie on a plane but have no idea when it comes to 3...

答:
Note that $ s(t) = \begin{pmatrix}
x_1 & y_1 & c_1\\
x_2 & y_2 & c_2\\
x_3 & y_3 & c_3 \end{pmatrix} \begin{pmatrix} t^{2} \\
t\\
1 \end{pmatrix}$, and the curve $ \begin{pmatrix} t^{2} \\
t\\
1 \ \end{pmatrix} $ lies in a plane, and the image of a plane under a linear transformation is a plane.


问:
Proof of $\log(n!) < (n-1)\log(n)$

Prove that $\log(n!) \leq (n-1)\log(n)$ directly and by induction.

I'm having a hard time with this one. I tried:

$$\log(n!) + \log(n+1) \leq (n-1)\log(n+1) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n) - \log(n) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n) - \log(n) + \log(n+1) \leq n\log(n+1) - \log(n) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n+1) - \log(n) + \log(n+1) \leq n\log(n+1) + \log(n+1)$$
$$\log((n+1)!) \leq (n+1)\log(n+1)$$

But that's not what is asked, and I can't find a way of getting there.

答:
By induction you have,

$$\log(n!)+\log(n+1)\le (n-1)\log(n)+\log(n+1)\to$$
$$\log[(n+1)!]\le (n-1)\log(n)+\log(n+1)$$

Remmember that $\log()$ is a strictly increasing function, so $\log(n)< \log(n+1)$. It means,

$$(n-1)\log(n)\le (n-1)\log(n+1)\to$$
$$(n-1)\log(n)+\log(n+1)\le (n-1)\log(n+1)+\log(n+1)=n\log(n+1).$$

Title: Proof of $\log(n!) < (n-1)\log(n)$


答: $s(t)$ is a plane curve iff $\det(s'(t),s''(t),s'''(t))=0$.

Title: How to show a curve lies in a plane given 3 constant vectors

Body:
How does one show that given 3 non-parallel vectors in $\mathbb{R}^3$, a curve with parameterisation:
$$\textbf{s}(t)=(x_1t^2+y_1t+c_1,x_2t^2+y_2t+c_2,x_3t^2+y_3t+c_3)$$ lies in a plane?

Additionally, how do we find the equation of the plane?

I know that any two non-parallel vectors lie on a plane but have no idea when it comes to 3...


问:
non-recursive function

Give a direct proof that the set $\{x|\Phi_x(1) \downarrow\}$ (which is a set of program numbers that halt on input $1$) is not recursive.

I've got an idea that indirect proof must work. Assuming that the set is recursive and ... But I have no idea how to go on from here.

The course source is Computability, Complexity, and Languages, Second Edition: Fundamentals of Theoretical Computer Science by Davis. So I'm familiar with notation used there and understand it better, if possible.

答:
The following is not a (completely) formal proof, the reason is that there are many different model for computation and I don't not which one to choose. Anyway I'm pretty much confident that the argument can be easily applied in every model you choose.

For start let's remind that your set is recursive if and only if the predicate $\Phi_x(1)\downarrow$ is decidable.

The key for the proof is that the function which send every $n \in \mathbb N$ in the number representing the function

$$f(x) = \Phi_n(n)$$

is total recursive.

So for every $x \in \mathbb N$ we have that
$$\Phi_{f(x)}(1) \downarrow \iff f(x)(1) = \Phi_x(x) \downarrow$$
so the predicate $\Phi_{f(x)}(1)\downarrow$ is equivalent to the predicate $\Phi_x(x)\downarrow$. Because $f$ is total recursive the predicate $\Phi_{f(x)}(1)\downarrow$ is decidable too. But we proved that this predicate is equivalent to the predicate $\Phi_x(x)\downarrow$ and so it should be decidable too but we know that this cannot be because of the halting problem.

Hope this help, feel free to ask if something is not clear.

Title: non-recursive function


问:
How does one perform the transformation on the Leibniz-Madhava series to arrive at this Nilakantha accelerated series for Pi?
$$\pi=\displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n}{2n+1}$$
to
$$\pi=\sqrt12(\dfrac{1}{1*3^0}-\dfrac{1}{3*3^1}+\dfrac{1}{5*3^2}-\dfrac{1}{7*3^3}+...)$$

After reading through several books and PDFs regarding series acceleration, I cannot for the life of me actually come to the logical flow necessary to transform the original series into the latter series. I set off several years ago to understand this transformation. Over the years I've learned so much about transformations and can accelerate many series for constants through many glorious methods, however I've never learned how the series that got me started can be derived. I cannot find a direct derivation anywhere.

Can someone please illuminate me?

Body:
let $\pi_n=\sum_{i=0}^{n}\frac{(-1)^n}{2n+1} $

$$\pi_n=\sum_{i=0}^{n}\int_0^1(-1)^nx^{2n}dx $$

$$\pi_n=\int_0^1\sum_{i=0}^{n}(-x^2)^ndx $$
$$\pi_n=\int_0^1\frac{1-(-x^2)^{n+1}}{1+x^2} $$
$$\pi_n=\int_0^1\frac{1}{1+x^2} +R_n$$
with $R_n=\int_0^1\frac{(-x^2)^{n+1}}{1+x^2}$.
You can see that $|R_n|<\int_0^1x^{2n+2}\leq \frac{1}{2n+3} \rightarrow 0 $ when n goes to infinity.
And you have :

$$\pi_n=[arctan(x)]_0^1+R_n=\frac{\pi}{4}+R_n $$

so :
$$\pi=4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} $$

you can use the same reasoning for the second sum, (I'll skip some details about the convergence, you can use partial sums as I did above) :

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}=\sum_{n=0}^{\infty}\int_0^1\left(\frac{-x^2}{3}\right)^n dx$$

$$ \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}= \int_0^1 \frac{1}{1+\frac{x^2}{3}}dx$$
$$ \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}= \sqrt{3}\int_0^{\frac{1}{\sqrt{3}}} \frac{1}{1+y^2}dy$$
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}=\sqrt{3}\times arctan(\frac{1}{\sqrt{3}}) $$
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}=\frac{\pi}{\sqrt{12}} $$

So you get the final result :
$$\boxed{ \pi=\sqrt{12}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}}$$


问:
Rearranging linear system with complex conjugate matrices
I have three complex matrices $X$, $A$ and $B$. $A$ and $B$ are diagonal matrices, while $X$ is a full non-symmetric matrix.

I have an equation

$$ Y = XAX^T + \bar{X}B\bar{X}^T $$

where the overbar indicates complex conjugate.

I know the value of $C=(A+B)$. How can I express $Y$ in terms of $C$?

In other words, I'm trying to obtain something in the form

$$ Y = ? (A+B) ? $$

Is this possible? If yes, how?

答: This is hopeless. E.g. when $B=-A$, a product of the form $?(A+B)?$ is always zero, but $XAX^T+\bar{X}B\bar{X}^T=XAX^T-\bar{X}A\bar{X}^T$ can be easily nonzero.


问:
Evaluate $\prod_{k=1}^{\infty}\left(2\cos\left(\frac{\pi}{3^{k+1}}\right)-1\right)$ from polynomial
I found the problem online (without solution) that to evaluate $$\prod_{k=1}^{\infty}\left(2\cos\left(\frac{\pi}{3^{k+1}}\right)-1\right)$$
given that we know how to solve polynomial
$$4x^3-3x+1 = 0$$

Right now I know that $$0<\prod_{k=1}^{\infty}\left(2\cos\left(\frac{\pi}{3^{k+1}}\right)-1\right)\leq 1$$ due to the fact that $\lim_{k\rightarrow\infty}\cos\left(\frac{\pi}{3^{k+1}}\right) = 1$ so we should have something like $0<(0.8...)...(\text{almost 1})...<1$ (my guess is that our answer is around $0.8$ to $0.9$) and I also know that the solution for the polynomial is $x = -1, \frac{1}{2}$.

One thing I found out interesting is that the polynomial can be express as $(x+1)(2x-1)^2$ in which is quite the same with our product if we let $x = \cos\left(\frac{\pi}{3^{k+1}}\right)$.

So how can I solve this problem using the polynomial? Could anyone provides me a hint or some guideline to solve this problem?

答:
Well, here's how to use the polynomial and the factorization you found (thanks for doing the dirty work! ;).
You can write the triplication formula also in this way:
$$\cos3\theta=4\cos^3\theta-3\cos\theta.$$ With $$c_k=\cos\frac\pi{3^{k+1}},$$ you'll have
$$4\,c^3_k-3\,c_k=c_{k-1},$$ and adding $1$ on both sides,
$$(c_k+1)(2\,c_k-1)^2=c_{k-1}+1,$$ i.e.
$$2\,c_k-1=\sqrt{\frac{c_{k-1}+1}{c_k+1}}.$$ So we get another telescoping product, and its value is $$\sqrt{\frac{c_0+1}{c_\infty+1}}=\sqrt{\frac{\cos\frac\pi3+1}{2}}=\cos\frac\pi6=\frac{\sqrt{3}}2.$$


问:
How to calculate the determinant of a matrix using Laplace?
How to calculate the determinant using Laplace?
$$
\det \begin{bmatrix}
       0 & \dots & 0 & 0 &   a_{1n}           \\[0.3em]
       0 & \dots & 0 & a_{2,n-1} &   a_{2n}   \\[0.3em]
       \dots & \dots & \dots & \dots & \dots             \\[0.3em]
       a_{n1} & \dots & a_{n,n-2} & a_{n,n-1} & a_{nn}  \\[0.3em]
     \end{bmatrix}
$$
I think it's something like:
$$
(a_{n1} * ... *a_{1n}) * (-1)^{n(n+1)}
$$
But I'm not sure about it.</p>

答:
Interchanging two columns of a matrix changes the sign of determinant. Therefore
$$
\det \begin{bmatrix}
       0 & \dots & 0 & 0 &   a_{1n}           \\[0.3em]
       0 & \dots & 0 & a_{2,n-1} &   a_{2n}   \\[0.3em]
       \dots & \dots & \dots & \dots & \dots             \\[0.3em]
       a_{n1} & \dots & a_{n,n-2} & a_{n,n-1} & a_{nn}  \\[0.3em]
     \end{bmatrix}
=\\
=(-1)^{\frac{n(n-1)}2}
\det \begin{bmatrix}
       a_{1n} & 0 & 0 & \dots & 0 &              \\[0.3em]
       a_{2n} & a_{2,n-1} & 0 & \dots & 0        \\[0.3em]
       \dots & \dots & \dots & \dots & \dots             \\[0.3em]
        a_{nn} & a_{n,n-1} & a_{n,n-2} & \dots & a_{n1}     \\[0.3em]
     \end{bmatrix}
$$
(We have made switched the neighboring row $(n-1)$-times to get the last column to the first place. Then $(n-2)$-times for the column before. Altogether we need $1+2+\dots+(n-1)=\frac{n(n-1)}2$ transpositions.)

Now we have a lower triangular matrix and the determinant is precisely the product of the elements on the diagonal. So the determinant is
$$=(-1)^{\frac{n(n-1)}2} a_{1,n}a_{2,{n-1}}\cdots a_{n,1}.$$

This is the same result as the one given by egreg, since
the number $\frac{n(n+3)}2-\frac{n(n-1)}2=\frac{4n}2=\frac2n$ is even.


答:
\begin{align}
\log(n!) &\leq (n-1)\log(n) \\
\iff \log(n) + \log(n!) &\leq \log(n^n) \\
\iff \log(nn!) &\leq \log(n^n)
\end{align}

Since $\log$ is an increasing function, and $nn!\leq n^n$, the third inequality holds true for all $n \in \mathbb{N}$. I would try proving that $nn! \leq n^n$ myself, and then you can compare with my solution:

\begin{align}
nn! &\leq n^n \iff n! \leq n^{n-1} \\[6pt]
n! &= \underbrace{n \cdot (n-1) \cdot (n-2) \cdot (n-3) \cdot \ldots \cdot 2}_{n-1 \text{ terms}} \\[6pt]
n^{n-1}&=\underbrace{n \cdot n \cdot n \cdot n \cdot \ldots \cdot n}_{n-1 \text{ terms}}
\end{align}

The case $n=1$ is a mild annoyance, but can be dealt with easily.

Title: Proof of $\log(n!) < (n-1)\log(n)$

Body:
Prove that $\log(n!) \leq (n-1)\log(n)$ directly and by induction.

I'm having a hard time with this one. I tried:

$$\log(n!) + \log(n+1) \leq (n-1)\log(n+1) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n) - \log(n) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n) - \log(n) + \log(n+1) \leq n\log(n+1) - \log(n) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n+1) - \log(n) + \log(n+1) \leq n\log(n+1) + \log(n+1)$$
$$\log((n+1)!) \leq (n+1)\log(n+1)$$

But that's not what is asked, and I can't find a way of getting there.


问:
Question regarding a series that contains logarithms
$$\sum_{n=1}^\infty\bigl(3\log(n^2+1)-2\log(n^3+1)\bigr)$$

I tried the limit comparison test with $\sum_{n=1}^\infty\frac1{n^2}$ and then applied L'Hopital's Rule. What I ended up was a limit which was a real number and as the series of $\frac1{n^2}$ converges, my original series converges as well. However, I was wondering if there is another way of solving the problem, one that involves fewer computations.

答:
Similar to TheSilverDoe's answer, we have that eventually

$$3\log(n^2+1)-2\log(n^3+1)=\log\left(\frac{(n^2+1)^3}{(n^3+1)^2}\right)=\ln\left( 1+\frac{3n^4-2n^3+3n^2}{n^6+2n^3+1}\right) \le$$

$$\le \frac{3n^4-2n^3+3n^2}{n^6+2n^3+1} \le \frac{4n^4}{\frac12n^6}=8\frac1{n^2}$$

and we can conclude by direct comparson test.


问:
Determine a basis of the $\mathbb{Z} / 3\mathbb{Z}$-vector space $V$
I guys, I have this vector space:
$$V = \{(x_1, x_2, x_3, x_4)^T ∈ (\mathbb{Z}/3\mathbb{Z})^4 | x_1 + x_2 + x_3 + x_4 = [0]\}$$
of this I have to determine a basis of the $\mathbb{Z}/3\mathbb{Z}$-vector space $V$.

So my problem now is how to proceed to do this, because I don't know how to treat $\mathbb{Z}/3\mathbb{Z}$.

答:
Note that the equation dictates that you can choose $x_1, x_2, x_3 $ to your liking, and $x_4 = -x_1 - x_2 - x_3$, Therefore every vector in V is of the form $\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
-x_1 -x_2 -x_3 \end{pmatrix} $. Can you find a basis now? not that there is nothing special about the field we are working over.


问:
What do Greek Mathematicians use when they use our equivalent Greek letters in formulas and equations?
Like for example, it's common to use the Greek letter $\theta$ to represent an angle right? So what would a Greek person doing math use to represent an angle? Would they also use $\theta$? Or is there another notation that they would use in order for them to use their letters like we do? Such as if we say $A\geq B$, would a Greek student, mathematician, or whoever say: $\alpha \geq \beta$ or is there something else they say? It just seems like the Greek letters from a non-Greek point of view have so much meaning to us, but then how do they percieve their letters used in mathematics?</p>

答:
I am Greek.

We use all the letters, Greek or Latin. But we pronounce some of them in a different way.

For example the letter $\mu$ is pronounced "me". The letter $\beta$ as "vita". The angles are almost always named as $\theta, \phi, \omega$.

If somebody writes $A\ge B$ we read the same as $\alpha\ge \beta$.

The only problem is when you want a student to understand that $\chi \psi$ denotes $\chi\cdot \psi$ but $\ln a$ has nothing to do with $l\cdot n \cdot a$.

Finally, I wanted to add that actually $\pi$ is not pronounced like the food - for example (apple-pie) - but like "me" with the letter p at the beginning (that is, "pe").


问:
For every natural number $n$: $1 + 3 + 5 + \cdots + (2n− 1) = n^2$, prove by smallest counterexample?
I am confused with the concept of "smallest counterexample". So first we have to prove base $n = 1$ so $(2-1) = 1^2$ which is true. But now what do I do?

答:
There not exist counterexamples indeed

$$n^2-(n-1)^2= 2n-1 \implies n^2 =(n-1)^2+(2n-1)$$

To proceed by induction, for the inductive step, just use that assuming true as hypotesis

$$1+3+5+\ldots+(2n-1)=n^2 $$

we obtain

$$ 1+3+5+\ldots+(2n-1)+(2n+1)=n^2+2n+1=(n+1)^2$$

Assume that the identity holds for $n=1$ up to a minimum value $n=k$ such that

$$1 + 3 + 5 + \cdots +(2k-3)+ (2k− 1) \neq k^2$$

but by hypotesis we have that

$$1 + 3 + 5 + \cdots +(2k-3) =(k-1)^2$$

and

$$(k-1)^2+ (2k− 1) =k^2-2k+1+2k-1= k^2$$

which is a contradiction therefore such minimum value doesn't exist and the identity is always true.


问:
Is $\sum_{n=1}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$ convergent? Is it absolute or conditional convergence?
Hello I've come across this problem my teacher gave me. Consider the infinite series $\sum_{n=1}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$. It's obvious that when $n = 1$, the value doesn't exist. But I asked my teacher and she said we can separate the $\frac{1}{\ln(1)}+\sum_{n=2}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$, and since $\sum_{n=2}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$ is conditional convergent, then the series converges as well.

I've proven that the sequence $\frac{n^{1/n}}{\ln(n)}$ continuous, positive, and decreasing over $(1, \infty)$. (I checked the derivative of it and the value is $\leq 0$).

Next, I used the limit comparison test, to compare $\frac{n^{1/n}}{\ln(n)}$ and $\frac{1}{\ln(n)}$, which diverges, meaning $\frac{n^{1/n}}{\ln(n)}$ also does.

Someone asked this question a while ago. But I remember somewhere in the lecture, there was once this theorem about if an element is removed/added from a convergent series (or something similar) the corresponding series would be convergent as well.

Does a series converge if its initial value is undefined?

Update: Can I actually use the Integral Test for this one?

Update 2: Sorry had a typo for $\frac{1}{\ln(0)}$

Thank You.

答:
As noticed in the comments the series should be written as

$$\sum_{n=2}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$$

since for $n=1$ the expression is meaningless.

For the other issue, we have that for any convergent series since

$$\sum_{n_0}^\infty a_n=\sum_{n_0}^{n_1} a_n+\sum_{n_1+1}^\infty a_n$$

and the summation $\sum_{n_0}^{n_1} a_n$ is finite we have that

$$\sum_{n_0}^\infty a_n\quad \text{converges (or diverges)}\iff \sum_{n_1+1}^\infty a_n \quad \text{converges (or diverges)}$$

Your discussion for convergence is fine.


答:
Given any statement of the form "For all natural numbers $n$, [...]", if there are any counterexamples, then there is necessarily a smallest counterexample. We assume that there is a counterexample, and therefore a smallest counterexample, and call it $k$.

The goal is to reach a contradiction and this conclude that there cannot be any counterexamples, and therefore that the statement is true for all natural numbers.

There are now two possibilities. Either $k$ is the smallest possible natural number (whether that's $0$ or $1$ is a matter of definition), or it isn't. The former is easily disproven, that's the base case.

So the statement makes sense and is true for $n=k-1$. Which is to say, we have
$$
1+3+\cdots+(2k-3)=(k-1)^2
$$
What does that say about our smallest counterexample? Well, we have
$$
1+3+\cdots+(2k-3)+(2k-1)=(k-1)^2+(2k-1)\\
=k^2-2k+1+2j-1=k^2
$$
But this goes against $k$ being the smallest counterexample. So we have reached a contradiction, and finished the proof.

PS. This way of writing a proof by contradiction isn't actually necessary. We don't need the contradiction here to conclude. If we remove the initial assumption-for-contradiction, this proof simply shows directly (by way of induction) that the statement is true for all $n$. But this "smallest counterexample" way of thinking may be an approachable way to learn induction in the first place.

Title: For every natural number $n$: $1 + 3 + 5 + \cdots + (2n− 1) = n^2$, prove by smallest counterexample?

Body: I am confused with the concept of "smallest counterexample". So first we have to prove base $n = 1$ so $(2-1) = 1^2$ which is true. But now what do I do?


问:
I need a proof for a scalar product - no numbers allowed
Hello I simply cant explain to myself why this equation holds. Lets say we have an orthonormal basis $\vec{i}, \vec{j}$ and 2 2-D vectors in this basis which are:

\begin{align}
\vec{a} &= a_1\vec{i} + a_2 \vec{j}\\
\vec{b} &= b_1\vec{i} + b_2 \vec{j}
\end{align}

I know that I can calculate the norms of vectors $\left|\vec{a}\right|$ and $\left|\vec{b}\right|$ like this:

\begin{align}
\left|\vec{a}\right| = \sqrt{{a_1}^2 + {a_2}^2}\\
\left|\vec{b}\right| = \sqrt{{b_1}^2 + {b_2}^2}
\end{align}

When I try to justify the equation $\vec{a} \cdot \vec{b} = \left|\vec{a}\right|\left|\vec{b}\right| \cos \alpha = a_1b_1 + a_2b_2$ for a scalar product I just can't seem to do it. I can only do it in for the easiest case $\vec{a} \cdot \vec{a}$:

\begin{align}
\vec{a} \cdot \vec{a} &= \left|\vec{a}\right| \left|\vec{a}\right| \overbrace{\cos \alpha}^{=1}\\
\vec{a} \cdot \vec{a} &= \left|\vec{a}\right| \left|\vec{a}\right|\\
\vec{a} \cdot \vec{a} &= \sqrt{{a_1}^2+{{a_2}^2}}\sqrt{{a_1}^2+{{a_2}^2}}\\
\vec{a} \cdot \vec{a} &= {a_1}^2+{{a_2}^2}
\end{align}

Question: How do we do it for a $\vec{a} \cdot \vec{b}$? Lets say for a start that $\alpha = 0$. What about if $\alpha \neq 0$? Please provide a similar derivation as I did for $\vec{a} \cdot \vec{a}$ using norms.

答:
Let
$$\mathbf{a}=|a|\cos\theta \mathbf{i}+|a|\sin\theta \mathbf{j}$$
and
$$\mathbf{b}=|b|\cos\phi \mathbf{i}+|b|\sin\phi \mathbf{j}$$

Now, we have
$$
\mathbf{a\cdot b} = |a||b|\cos\theta\cos\phi + |a||b|\sin\theta\sin\phi
$$
Now use a trigonometric identity to simplify, with $\alpha=\theta-\phi$.

At 71GA's request, I'm elaborating on the relationship between my expression of $\mathbf{a}$ and the cartesian expression.

Consider the vector in the form
$$
\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}
$$
We now define
$$|\mathbf{a}|^2 = a_1^2+a_2^2$$
Note that this forms a right triangle relationship. For vectors with positive $a_1$, we may use the formed right triangle to determine the angle of the vector from the positive $x$-axis, $\theta$. Namely,
$$
\tan(\theta) = \frac{a_2}{a_1}
$$
This relationship also applies if $a_1$ is negative or zero, but the angle itself must be adapted to suit. We may then solve for $a_1$ and $a_2$, by noting that
\begin{align}
\frac{|\mathbf{a}|^2}{a_1^2}&=1+\frac{a_2^2}{a_1^2}\\
&=1+\tan^2(\theta)\\
&=\sec^2(\theta)
\end{align}
and so, we have
$$
a_1^2 = |\mathbf{a}|^2\cos^2(\theta)
$$
and, by appropriate choice of angle, you have
$$
a_1 = |\mathbf{a}|\cos(\theta)
$$
Substituting back into our equation for $\tan(\theta)$, we have
$$
a_2 = |\mathbf{a}|\sin(\theta)
$$
As such, we have
$$
\mathbf{a} = |\mathbf{a}|\cos\theta\mathbf{i}+|\mathbf{a}|\sin\theta\mathbf{j}
$$
where $\theta$ is the angle of the vector from the positive $x$-axis. And we can repeat this for vector $\mathbf{b}$, with $\phi$ as the angle. Thus, the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ is $\alpha=\theta-\phi$, or is different from this by $2\pi$ (in which case the relation $\mathbf{a\cdot b}=|\mathbf{a}||\mathbf{b}|\cos(\alpha)$ is unchanged by the addition or subtraction of $2\pi$).


问:
If $\Bbb E[Z_n]\xrightarrow[n\rightarrow\infty]{} 0$ then can we say $Z_n=0$ with high probability?
This is in the context of subgraph containment problem. Say I managed to prove that a certain subgraph $H$ (like a triangle or a small cycle) of an Erdos-Renyi graph $G$ on $n$ verticies with edge probability $p$ ($G\sim\text{ER}(n,p)$) satisfies $$\Bbb E[\text{# times }H\text{ appears in }G ]=1/n$$ which goes to zero ($p$ being specified in the question as some function of $n$).

Is this enough to say that "$G$ does not contain $H$ with high probability"?

答:
Yes it is, almost. I have seen it written that way. The best way to put it is that the probability that $H$ appears goes to 0 as $n$ increases. Or the $Z_n$s go to 0 with high probability, with $Z_n$ as you defined. Note the following claim:

Claim 1: Let $Y$ be a random variable that takes on nonegative integral values. Then
$\Bbb{E}[Y] \ge \Bbb{P}[Y > 0]$.

This claim is easy to show. So if you are given a series $\{Y_n\}$ of rvs where $Y_n$s are a sequence of r.v.s as in the hypothesis of Claim 1 where $\Bbb{E}[Y_n] \le n^{-1}$ it follows that
the $Y_n$s go to 0 with high probability as well.

As $\#\{$times $H$ appears in $G\}$ is a random variable that takes nonegative integers, it follows from Claim 1 that

$$\frac{1}{n} \ge \Bbb E[\text{# times }H\text{ appears in }G ] \ge \Bbb P[\text{# times }H\text{ appears in }G ]$$


答:
Close to being an integral test

$$f(t)=t^{1/t}/\log t,\qquad f'(t)=O(1/t^2)+O(1/(t\log^2 t))$$
$$f(2n)-f(2n+1)=\int_{2n}^{2n+1}f'(t)dt=O(1/(n\log^2 n))$$

Thus $\sum_{n\ge 1} f(2n)-f(2n+1)$ converges absolutely.

Title: Is $\sum_{n=1}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$ convergent? Is it absolute or conditional convergence?

Body:
Hello I've come across this problem my teacher gave me. Consider the infinite series $\sum_{n=1}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$. It's obvious that when $n = 1$, the value doesn't exist. But I asked my teacher and she said we can separate the $\frac{1}{\ln(1)}+\sum_{n=2}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$, and since $\sum_{n=2}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$ is conditional convergent, then the series converges as well.

I've proven that the sequence $\frac{n^{1/n}}{\ln(n)}$ continuous, positive, and decreasing over $(1, \infty)$. (I checked the derivative of it and the value is $\leq 0$).

Next, I used the limit comparison test, to compare $\frac{n^{1/n}}{\ln(n)}$ and $\frac{1}{\ln(n)}$, which diverges, meaning $\frac{n^{1/n}}{\ln(n)}$ also does.

Someone asked this question a while ago. But I remember somewhere in the lecture, there was once this theorem about if an element is removed/added from a convergent series (or something similar) the corresponding series would be convergent as well.

Does a series converge if its initial value is undefined?

Update: Can I actually use the Integral Test for this one?

Update 2: Sorry had a typo for $\frac{1}{\ln(0)}$

Thank You.


答:
To even be allowed to consider "$\sum_{n=0}^{+\infty} a_n$", the first thing is that all the $a_n$'s must be defined for $n \geq 0$. In your case, this is not satisfied, therefore the series has no sense.

If all is well-defined, we can say that  ($\sum_{n=0}^{+\infty}$ converges) iff (for all $k \in \mathbb{N}$, $\sum_{n=k}^{+\infty}$ converges) : that means that the first terms are not important with regard to the convergence or the divergence of the series.

Finally, to your specific series (considering that the sum begins at $n=2$), you can see that $$\left| \frac{(-1)^n n^{1/n}}{\ln(n)}\right| \sim \frac{1}{\ln(n)}$$

so by comparison, the series is not absolutely convergent. But by the special criteria for alternating series, it is convergent. So the series is conditionally convergent.


问:
Time average of product of two functions
Suppose the functions $e(x)$ and $h(x)$ with Fourier transforms $E(k)$ and $H(k)$. What is the time average A of $e(x)h(x)$ in function of their Fourier transforms $E(k)$ and $H(k)$?

My attempt:
\begin{align}
A &= \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X} e(x)h(x) dx \\
&= \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X} \int\limits_{-\infty}^{\infty} E(k)e^{ikx} dk  \int\limits_{-\infty}^{\infty} H(k')e^{ik'x} dk' dx \\
&=  \int\limits_{-\infty}^{\infty}  \int\limits_{-\infty}^{\infty}E(k) H(k')\left[ \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X}e^{i(k+k')x} dx \right] dk dk'  \,,
\end{align}

using Wolfram Alpha
$$\lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X}e^{i(k+k')x} dx = 0\,,$$

and thus the average $A=0$ for all functions $e(x)$ and $h(x)$... Obviously, this cannot be the case, so I have made a mistake somewhere but I cannot find it...

Body: $$  \int_{-X}^{X} \; \mathrm{e}^{\mathrm{i} (k + (-k))x} \,\mathrm{d}x = 2X  \text{.}  $$


问:
The parking problem riddle
Assume a street of 300 meters, that you can park your car alongside the pavement. Assume that there is a big parking problem in the area. Assume that the pavement is continuous, without interruptions, and that you can park alongside all of it. Assume that the length of a car is 3 meters long. Assume, for simplicity, that cars can park without space between them (bumper to bumper). Assume, that when a car comes to the street if chooses an equally random parking space (please try to express this randomness) from the free spaces left. Therefore, it may "ruin" parking places for other cars. Please try to determine what is the expectancy of cars the can park alongside the street. </p>

答:
According to http://mathworld.wolfram.com/RenyisParkingConstants.html, if $M(n)$ is the expected number of cars of length 1 to fit onto a stripe of length $n$, you have $$
  m := \lim_{n\to\infty} \frac{M(n)}{n} \approx 0.7476 \text{.}
$$

In other words, the expected density (i.e., the total length of the parked cars divided by the length of the pavement) goes to $0.7476\ldots$ as the pavement length goes to infinity. In your case that leads to an estimate for the expected number $N$ of 3m long cars on a 300m long pavement of $$
  \frac{3N}{300} \approx 0.7476 \implies N \approx 74.76 \text{.}
$$

A more precise estimate from that page is $$
  M(n) \approx mn + m - 1 \text{,}
$$
again for the expected number of 1m long cars on a stripe of length $n$. To apply that to your question, one has to use that the expected number $N$ of  3m long cars on a 300m long stripe is the same as the expected number of 1m long cars on a 100m long stripe, which yields $$
  N \approx M(100) \approx 74.51 \text{.}
$$

Title: The parking problem riddle


答:
$\log(n!)$

$=\sum_{m=1}^{n}\log(m)$

$=\log(1)+\sum_{m=2}^{n}\log(m)$

$=\sum_{m=2}^{n}\log(m)$

$\log(m)$ is an increasing function

So

$\sum_{m=2}^{n}\log(m)$

$\le \sum_{m=2}^{n}\log(n)$

$ = (n-2+1)\log(n)$

$ = (n-1)\log(n)$

For induction proof:

Base case: $n=1$

$\log(1!) = 0$.

$(1-1)\log(1) = 0$.

So $\log(n!) \le (n-1)\log(n)$ is true for $n=1$.

Assume true for $n=k$

$\log(k!) \le (k-1)\log(k)$

Now prove true for $n=k+1$

$\log((k+1)!)$

$=\log(k+1) + \log(k!)$

$\le \log(k+1) + (k-1)\log(k)$ (by inductive assumption)

$\le \log(k+1) + (k-1)\log(k+1)$ (again since log is an increasing function)

$= ((k+1)-1)\log(k+1)$

So it's true for $n=k+1$

Title: Proof of $\log(n!) < (n-1)\log(n)$


问:
Limit of $(3n+1)^{1/n}$
I am working through a question that requires me to prove that a certain limit tends to 1. I have reached an endpoint where I have:

$$\lim\limits_{n\to \infty} (3n+1)^{1/n}$$

I know that and can use the fact that $\lim\limits_{n\to \infty}( n^{1/n}) = 1$

However, I am not sure if I can just use this fact to say that it applies for $(3n+1)^{1/n} $ and I also am stuck on proving it.

答:
As a hint you apply $\ln$ to both sides of limit.

As a second hint rewrite like below
$$ \lim_{n\to \infty} (3n+1)^{1/n}=\\\lim_{n\to \infty}\sqrt[n]{3n+1}=\lim_{n\to \infty}\sqrt[n]{3}\sqrt[n]{n+\frac13}=\\
\lim_{n\to \infty} \sqrt[n]{3}\lim_{n\to \infty}\sqrt[n]{n+\frac13}=1\times1$$

Title: Limit of $(3n+1)^{1/n}$


问:
Min cost max flow optimization problem
Let $I$ be the set of customers and let $J_i \subseteq J$ be the set of items from which customer $i \in I$ wants to buy only one, where $J$ are the items. Denote $w(i,j) \geq 0$ as the price customer $i$ is willing to pay for item $j$. Assume $|I| = |J| = n$.

I want to find an algorithm which computes the maximum number of customers that get an item at the minimum total cost. I'd like to do this using a min-cost max flow theorem. However, I cannot seem to succeed in constructing a graph for this problem where we can both minimize and maximize functions. Can anyone give a suggestion how I should make this graph?

答: OP was under the impression that the cost in a min-cost-flow problem should be at most the capacity. This is not the case. As such, the solution is a straightforward application of the min-cost-max-flow algorithm.</p>

Title: Min cost max flow optimization problem


问:
Linear dependence of a set vs a family of vectors
In one of my lectures, after defining linear dependence as such:

Let R be a ring and M an R-module. A finite family $(x_i)_{i=1, ..., n}$ of elements in M are called linearly dependent if there exists $(a_i)_{i=1, ..., n}$ in R, not all zero, such that $a_1x_1 + ... +a_nx_n = 0$.

They followed it with a remark:

If a family contains repetition, it is linearly dependent: Indeed, if $x_{i_0}=x_{i_1}$ with $i_0<i_1$ then take $a_i$ such that

$a_i=0 \text{ for } I \notin \{i_0, i_1\}$

$a_{i_0} = 1 \text{ and } a_{i_1} = -1$

Together with the above definition, this makes sense mathematically. However, intuitively it does not, since I imagine two linearly dependent vectors in, say for example, 2 dimensions as on the same line. With this, if you have two linearly independent vectors, and just add the first one again to the family, suddenly they're all linearly dependent? How does this make sense intuitively?

I think my confusion here lies with the concept of a "family". Since in a set there are no repeats, this problem wouldn't occur. So why is linear dependence different for a set and a family, if they have the same elements and the family only has a few repeats?

答:
I think you're thinking of linear dependence / independence as a property of the individual vectors, but it's a property of the family of vectors.

If you've got a bowl of M&Ms that are all different colors-- say, red, blue, and yellow-- you might call it an "all different colors" bowl of M&Ms. But if you put another blue M&M in the bowl, now you don't have an "all different colors" bowl of M&Ms anymore. It's not that the individual M&Ms now lack the "all different colors" property, it's that the bowl does.

Similarly, if you've got a family of linearly independent vectors, and you add a duplicate, it's not the vectors that become linearly dependent, it's the family of vectors that becomes linearly dependent.


答: Per the hint in comments, we squeeze the limit expression between $(3n)^{1/n}$ and $(4n)^{1/n}$, both of which evaluate to $1$, so the limit is $1$. (It's easier to show that $\lim_{n\to\infty}a^{1/n}=1$ for $a\ge1$.)

Title: Limit of $(3n+1)^{1/n}$

Body:
I am working through a question that requires me to prove that a certain limit tends to 1. I have reached an endpoint where I have:

$$\lim\limits_{n\to \infty} (3n+1)^{1/n}$$

I know that and can use the fact that $\lim\limits_{n\to \infty}( n^{1/n}) = 1$

However, I am not sure if I can just use this fact to say that it applies for $(3n+1)^{1/n} $ and I also am stuck on proving it.


答:
I'm more familiar with the vector space setting, but intuitively speaking, linear independence means that the family of vectors is minimal out of all families of vectors with the same span. That is, if you remove any single vector from the family, you'll reduce the span.

For example, two linearly independent vectors will span a plane, but removing either of the vectors will reduce the span to a single line. This makes them linearly independent. We can easily find three vectors to span a plane, but inevitably we only really need two of the vectors; one of the three can be safely removed, and the remaining two vectors will span the same plane.

If you have a family of vectors with repetitions, then removing repetitions can never damage the span, so instantly this precludes the family from remaining linearly independent.

You can also think about it in terms of null linear combinations. If $v_1, \ldots, v_n$ is a family of vectors with $v_1 = v_2$, then
$$1v_1 + (-1)v_2 + 0 v_3 + \ldots + 0 v_n = 0$$
is a non-trivial linear combination. If you prefer to think of one vector as a linear combination of the others, then $v_1 = 1v_2 + 0v_3 + \ldots + 0v_n$ is such a relationship.

From every which way, it makes sense for repeated vectors to be linearly dependent.

Title: Linear dependence of a set vs a family of vectors

Body:
In one of my lectures, after defining linear dependence as such:

Let R be a ring and M an R-module. A finite family $(x_i)_{i=1, ..., n}$ of elements in M are called linearly dependent if there exists $(a_i)_{i=1, ..., n}$ in R, not all zero, such that $a_1x_1 + ... +a_nx_n = 0$.

They followed it with a remark:

If a family contains repetition, it is linearly dependent: Indeed, if $x_{i_0}=x_{i_1}$ with $i_0<i_1$ then take $a_i$ such that

$a_i=0 \text{ for } I \notin \{i_0, i_1\}$

$a_{i_0} = 1 \text{ and } a_{i_1} = -1$

Together with the above definition, this makes sense mathematically. However, intuitively it does not, since I imagine two linearly dependent vectors in, say for example, 2 dimensions as on the same line. With this, if you have two linearly independent vectors, and just add the first one again to the family, suddenly they're all linearly dependent? How does this make sense intuitively?

I think my confusion here lies with the concept of a "family". Since in a set there are no repeats, this problem wouldn't occur. So why is linear dependence different for a set and a family, if they have the same elements and the family only has a few repeats?


问:
Length of a circuit
I am reading an article about "Longest Paths in Digraphs". In the proof there is a step that they considered as a trivial step, it seems easy but I am not being able to write or concieve an exact proof for it.

We have a strong digraph $D$, that has minimum in degree $h$ and minimum out degree $k$. Let $C$ be the longest circuit in $D$ of length $c$; then $c\ge \max(h,k)+1$.

Why is that true?

I think it is solved by contradiction and we can find a larger circuit.

Suppose to contradiction that  $c< \max(h,k)+1$ and suppose without loss of generality that  $\max(h,k)=k$, then $c \le k$ which means that every vertex on the circuit has at least $1$ out neighbor outside $C$.

But how can we continue to find a larger circuit?

答:
Let us assume here that $k \ge h$ as you did and let $P$ be a path of length $c \ge \max(h,k)+1 = k+1$. Your line of reasoning shows that there is indeed such a path $P$. We use this to show how to get a circuit $C$ with at least $k+1$ vertices as follows:

Now let us write (*) $P=y_1y_2\ldots y_c$ where $y_1$ is the first vertex in $P$ and where $y_c$ is the last vertex in $P$.

If $y$ has a neighbor $y'=y_{c+1}$ outside of $P$ then add the arc $y_cy' =y_cy_{c+1}$ to $P$, now $P$'s length is $c+1$ instead of $c$ as it was before.

If all of $y$'s neighbours are in $P$, then let $j$ be the smallest integer
such that $y_cy_j$ is an arc in $G$, where $y_j$ is as in (*). Then $j \le c-k$ [why is this] so $C= y_jy_{j+1} \ldots y_cy_j$
is a circuit of length at least $k+1$.

Now note that 1. can be true for at most $n-k$ times as the number of vertices in any path $P'$ in $G$ is of course upper-bounded by the number of vertices in the digraph $G$. So you will indeed obtain a circuit of length at least $k+1$ after adding at most $n-k-1$ vertices to $P$ as in 1. above.

I trust that you can handle the case where $h > k$.

Title: Length of a circuit


问:
Example of a c.u.b. set
Let $\kappa$ a cardinal of cofinality $\omega$; let $C \subseteq \kappa$ be a unbounded countable subset. Why is then $C$ closed (and thus a c.u.b.)? This means that if $\delta < \kappa$ is a limit ordinal, and $C \cap \delta$ is unbounded in $\delta$, then $\delta \in C$.

I doubt that this is true; however, it's a short statement in Kunen's set theory book.

答:
What Kunen meant was that if $C$ is the range of a countable increasing unbounded sequence in $\kappa$, that is, $C$ is unbounded and has order type $\omega$, then it will be vacuously closed, since it has no limit points below $\kappa$.

This situation shows that the concept of club on cardinals of countable cofinality doesn't work as we want.

ContentLicense: CC BY-SA 2.5

Title: Example of a c.u.b. set


问:
Proof that the sample mean is the "best estimator" for the population mean.
I've always heard that the sample mean $\overline{X}$ is "the best estimator" for the population mean $\mu$. But is that always true regardless of the population distribution? is there any proof for that?
For example let's suppose for an unknown population, we have three samples, say $X_1$, $X_2$, $X_3$. Based on what I've heard (Not necessarily true) the estimator defined as the following:
$$\frac{1}{3}(X_1+X_2+X_3)$$
is always preferable to, for instance:
$$\frac{1}{6}(X_1+X_3)+\frac{2}{3}X_2$$
or
$$\max(X_1, X_2, X_3)$$
But in what sense is it better? and why?

答:
No

No, the sample mean is not always the best estimator. See for instance (besides the example of the mean of a uniform distribution by StubbornAtom) the case of a sample from a Laplace distributed population in the image below (code to compute that image is in this q&a)

In this image you see that the sample median can be a better estimator than the sample mean. Note that the distribution is more concentrated around the true location parameter (in this example this is 0).

Yes

Yes, the sample mean is the best unbiased linear estimator. This is a corrolary of the Gauss Markov theorem, which states that the least squares estimator is the unbiased linear estimator with the lowest variance (best linear unbiased estimator BLUE). The corrolary follows from this theorem because the sample mean is the least squares estimator of a population mean.

That the sample mean is BLUE does not contradict that the best unbiased estimators for the Laplace distribution are the median and for the uniform distribution the maximum. This is because those estimators are not linear estimators.

Linear estimator

A linear estimator is a linear function of the observed $Y$ e.g. in OLS the estimator is a matrix multiplied with the observed values  $\hat\theta = M \cdot Y$ where the matrix is $M = (X^TX)^{-1}X^T$)$.

The sample mean $\frac{1}{3} Y_1 + \frac{1}{3} Y_2 + \frac{1}{3} Y_3$ is a linear estimator but the maximum $\text{max}(Y_1,Y_2,Y_3)$ is not.

Note that an estimator like  $\frac{1}{6} Y_1 + \frac{1}{6} Y_2 + \frac{2}{3} Y_3$ may actually be possibly best as well. The Gauss Markov theorem is only true for independent errors with equal variance.

For the more general case, if you are considering observations with varying errors and potentially correlated, then you are considering generalized least squares which is also BLUE. For this case you do not consider the simple mean but instead a weighted mean.


Title: Limit of $(3n+1)^{1/n}$

Body:
We can use that

$$1=(1)^{1/n}\le (3n+1)^{1/n}\le (4n)^{1/n}=4^\frac1n \cdot n^\frac1n$$


问:
Angle bisectors of triangle and related circumcircles

Let the interior and exterior bisectors of the angles of $\triangle ABC$ meet their respective opposite sides in $X$, $Y$, $Z$ and $X'$, $Y'$, $Z'$. Show that the circles $(AXX')$, $(BYY')$, and $(CZZ')$ have 2 common points.

I have shown that the points $X'$, $Y'$, $Z'$ are collinear using Menelaus' theorem, but I can't proceed further. Is inversion a possible method?

答:
Hint: These are Apollonian circles. There is a statement and proof in following reference:

College Geometry

An introduction to the modern geometry of the triangle and the circle

Author: Vattan Altshiller Court

ISBN 964-318-222-3


Title: Limit of $(3n+1)^{1/n}$

Body:
Answer:

$(3n+1)^{\frac{1}{n}} =e^{\frac{1}{n}ln(3n+1)}=e^{\frac{1}{n}ln(n(3+\frac{1}{n} )) }=e^{\frac{1}{n}ln(n) +\frac{1} {n} ln(3+\frac{1}{n} )}$

So :

$\lim_{n\to+\infty} (3n+1)^{\frac{1}{n}}=\lim_{n\to+\infty} e^{\frac{1}{n}ln(n) +\frac{1} {n} ln(3+\frac{1}{n} )}=1$

Because

$\lim_{n\to+\infty} \frac{ln(n)}{n} =0$

And

$\lim_{n\to+\infty}\frac{1} {n}ln(3+\frac{1}{n} ) =0$


问:
Finding the length of a curve of intersection between a parabolic cylinder and a surface.
If $A$ is the curve of intersection of the parabolic cylinder $x^2 = 2y$ and the
surface $3z = xy$, then what is the length of $A$ from the origin to the point (6, 18, 36)?

What I have done so far:

I know to parametrise the cylinder to get $x$ and $y$ in terms of $sin$ and $cos$ however I am unsure of how to do this.

Body:
$$x^2=2y\implies y=x^2/2$$
$$3z=xy=x^3/2$$
Let $x=t$. Then $A$ from the origin to $(6,18,36)$ is parametrised as
$$(t,t^2/2,t^3/6)\qquad t\in[0,6]$$
The arc length is
$$\int_0^6\sqrt{1^2+t^2+(t^2/2)^2}\,dt=\frac12\int_0^6\sqrt{t^4+4t^2+4}\,dt$$
$$=\frac12\int_0^6(t^2+2)\,dt=\frac12\left[\frac13t^3+2t\right]_0^6=42$$


问:
Looking for a function $f$ that solves $f'(x)= \alpha(x) f(x)$ for a continuous $\alpha$ and starts off positive but then turns negative
I am looking for a continuously differentiable function $f:[-1,1]\to\mathbb R$ (note that I don't care about the interval $[-1,1]$ in particular; it can be replaced by any interval $[a,b]$ with $a<b$ of your choice) that satisfies

$f(t)>0$ for all $-1\le t<0$,

$f(0)=0$,

$f(t)<0$ for all $0<t\le 1$

as well as

$$f'(t)=\alpha(t) f(t)$$

for some continuous $\alpha:[-1,1]\to\mathbb R$.

Essentially I am looking for something like this:

My attempts: The most obvious attempt is  $f(t)=-t^3$. However, one notices that we need to have $\alpha(t)=\frac{3}{t^2}$ for $t\not=0$, which cannot be extended to a continuous function. So $f(t)=t^3$ doesn't work.

In fact, if $f$ is any polynomial with $f(0)=0$, we know that $\frac{f'}{f}$ diverges at $0$ so that $\alpha=\frac{f'}f$ can't be continuous....

Next try, $f(t)=-\exp\left(-\frac1t\right)$ for $t>0$ and $f(-t)=-f(t)$: This fails immediately as for $t\not=0$, $f'(t)=\frac{f(t)}{t^2}$, so that, once again, $\alpha(t)=\frac{f'(t)}{f(t)}=\frac{1}{t^2}$ can't be made continuous at $t=0$.

Final attempt: I define $g(t)=2 \operatorname W(\exp(t))-t-1$, where $\operatorname W$ is the Lambert W function. This is a rotation of the function $t\mapsto\exp(t)$ by 45 degrees and it looks like this:

Then I define $$f(t)=\begin{cases}
g(t+1) & t\le 0,\\ -g(t+1) & t>0.
\end{cases}$$

It looks like this:

However, once again we have that $\lim_{t\to0}\frac{f'(t)}{f(t)}$ is non-existant so that $\alpha$ can't be continuous at $0$.

So my question: Does a function $f$ as above even exist? If yes, can you give me an example? If not, why doesn't such a function exist??

答:
Any function that is solution will have the form $f(x) = f(0) \exp\left(\int_0^xa(t)\mathrm{d}t\right)$ and thus will have a constant sign (the sign of $f(0)$).
Moreover, because of Cauchy-Lipschitz theorem, any solution that is vanishing at zero will be vanishing everywhere. See "first order linear differential equation" or any related topic.


问:
Explicit value for series
We consider the sum
$$
\sum_{n_{1},\, n_{2}\ \in\ \mathbb{Z}}\,\,\,
\frac{16}{(2 n_1 - 2 \nu n_2 + 1)^2 (2 n_1 + 2 \nu - 2 \nu n_2 - 1)^2}
$$
with $\nu = (-1)^{1/3}$ and $\mathrm{i}$ is the imaginary unit.

I am wondering if there is a closed form expression for that double series.

Notice, that it definitely converges, as the denominator essentially behaves like
$\left\Vert\,{n}\,\right\Vert^{4}$.

答:
Let $L=\Bbb{Z}+e^{2i\pi /3}\Bbb{Z}$, your sum is $f(1/2)$ with
$$f(z)=\sum_{l\in L} \frac1{(z+l)^2(z+l+e^{2i\pi/3}-1)^2}$$
Which is meromorphic $L$-periodic with a pole of order $2$, with $a=e^{2i\pi/3}-1$
we get  $$f(z)=\frac{2}{a^2}\wp_L(z)+C$$

Next, substracting from $\wp_L'(z)^2-4\wp_L(z)^3$ a sum over its poles to get a bounded entire function, we have $$\wp_L'(z)^2=4\wp_L(z)^3-\frac43 \pi^4 E_4(e^{2i\pi /3}) \wp_L(z)-\frac8{27} \pi^6 E_6(e^{2i\pi /3})$$ where $\frac43 \pi^4 E_4(e^{2i\pi /3}),\frac8{27} \pi^6 E_6(e^{2i\pi/3})$ are given in term of the Laurent coefficients of $\wp_L(z)-z^{-2}$. Because $L=e^{2i\pi /3}L$ we have $E_4(e^{2i\pi /3})=0$. And $\wp_L'(z)$ is odd $L$-periodic thus $\wp_L'(1/2)=\wp_L'(-1/2)=0$.

Whence
$$\wp_L(1/2) =  e^{2i\pi n/3} (\frac14 \frac8{27} \pi^6 E_6(e^{2i\pi /3}))^{1/3}$$ Finally, this is where we need the whole theory of complex tori/elliptic curves/modular forms, $$E_6(e^{2i\pi /3})=(-1728 \Delta(e^{2i\pi/3}))^{1/2}=\pm i 3\sqrt{3} 2^3 (2\pi)^6 \eta(e^{2i\pi/3})^{12}$$ $$= \pm i 3\sqrt{3} 2^3 (2\pi)^6  (e^{-\frac{\pi i}{24}}\frac{\sqrt[8]{3}  \Gamma \left(\frac{1}{3}\right)^{3/2}}{2 \pi })^{12}$$

Computation of $C$:

$\sum_m \sum_n$ means $\lim_{M\to \infty} \sum_{m=-M}^M \lim_{N\to \infty} \sum_{n=-N}^N$.

From $\sum_n \frac1{(z+\nu^t n)^2}=\nu^{-2t}\frac{\pi^2}{\sin^2(\pi \nu^{-t} z)}$ and the absolute convergence of $\sum_m \frac{\pi^2}{\sin^2(\pi (\nu^{-t} z+m \nu))}$

then $$g(z)=\sum_{t=0}^2 \sum_m \sum_n \frac1{(z+\nu^t (n+m \nu))^2}+\frac1{(z+\nu^t(n+m \nu)+a)^2}$$ $$=2\sum_{t=0}^2  \sum_m \sum_n \frac1{(z+\nu^t (n+m \nu))^2}$$ $$
=2\sum_m \sum_n \sum_{t=0}^2 \frac1{(z+\nu^t (n+m \nu))^2}-\frac{1_{(n,m)\ne (0,0)}}{(\nu^t (n+m \nu))^2}=6 \wp_L(z)$$

From $\sum_n \frac1{z+n}-\frac1{z+n+b}=\frac{2i\pi}{e^{2i\pi z}-1}-\frac{2i\pi}{e^{2i\pi (z+b)}-1}$ then
$$h(z)=\sum_{t=0}^2 \sum_m \sum_n \frac1{z+\nu^t (n+m\nu)}-\frac1{z+\nu^t (n+m\nu)+a}$$ $$=\sum_{t=0}^2 \nu^{-t}\sum_m \frac{2i\pi}{e^{2i\pi (\nu^{-t} z+m\nu )}-1}-\frac{2i\pi}{e^{2i\pi (\nu^{-t} (z+a)+m\nu )}-1}=2i\pi (\nu^0+\nu^{-1}-2\nu^{-2})=-6i\pi \nu$$
and hence $$3 f(z)=
3\sum_{l\in L} \frac1{a^2 (z+l)^2} - \frac2{a^3 (z+l)} + \frac1{a^2 (a + z+l)^2} + \frac2{a^3 (a + z+l)}$$ $$=\sum_{t=0}^2\sum_{l\in L} \frac1{a^2 (z+\nu^t l)^2} - \frac2{a^3 (z+\nu^t  l)} + \frac1{a^2 (a + z+\nu^t  l)^2} + \frac2{a^3 (a + z+\nu^t  l)}$$ $$=
\frac1{a^2} g(z)-\frac2{a^3} h(z)=\frac6{a^2} \wp_L(z)+\frac2{a^3} 6i\pi \nu$$

Title: Explicit value for series


问:
If $A$ is an open set in $\mathbb{R^n}$ is $v + A = \{v + x: x\in A \}$ an open set $\forall v \in \mathbb{R^n}$?
I'm trying to prove that if $A$ is an open set in $\mathbb{R^n} \Rightarrow$ $\forall v \in \mathbb{R^n}$ : $v + A = \{v + x: x \in A \}$ is an open set in $\mathbb{R^n}$

I know that a set $A$ is an open set if
$$
\forall x \in A, \exists \epsilon \gt 0 : B(x, \epsilon) \subset A
$$

The proof seems trivial but I dont know how to prove it.Any hints or suggestions?

答:
Assume that $A \subset \mathbb{R}^n$ is an open set and define $v+A = \{v+x \, |  \, x \in A \}$ for a $v \in \mathbb{R}^n$. We will show that $v+A$ is open. Let $x \in v + A$. Observe that $x-v \in A$, and that consequently there exits an $\epsilon > 0$ such that $B(x-v, \epsilon) \subset A$. Now, its easy to see (and check if you want to) that $B(x, \epsilon) \subset v + A$.

The geometric idea here is that $v+A$ is simply a translation of the set $A$, so that if $B(x, \epsilon) \subset A$, the corresponding translated ball $B(x+v, \epsilon) \subset v+A$.


问:
Let $f \colon A \rightarrow B$ be a function and $S$ a subset of $A$.
Let $f \colon A \rightarrow B$ be a function and $S$ a subset of $A$.

Are the following containments always valid?.

$(1)    f^{-1} (f(S))⊆ S$

$(2)     S⊆f^{-1} (f(S))$

Attempt

$S ⊆ A$

$f^{-1}(S) = \{a ∈ B : f(a) ∈ S\}$

$ a ∈ f^{-1}(S) ⟺ f(a) ∈ S$

$ f^{-1}(S) ⊆ B$

I just need a little help finishing the proof. Am i wrong in the attempt? Thanks.

答:
Recall that for a function $ f: X \rightarrow Y $, and two subsets $A\subseteq X, B\subseteq Y$ that

$$ (f \circ f^{-1})(B) = I_Y(B)=B \\ (f^{-1} \circ f)(A) = I_X(A)=A $$

Where $I_X, I_Y$ are the identity functions. Do you think you can continue from here?


问:
Integrals of quantile function
Suppose $X = [0,1]$ and we have two cumulative distribution functions $F$ and $G$ on $X$ of two random variables.  I've just read the claim $$\int_0^1 |F^{-1}(x) - G^{-1}(x)|dx = \int_0^1|F(y) - G(y)|dy$$

where the only reason described is "by Fubini's theorem".  This is insanity to me.  I know that under the uniform Lebesgue measure on $[0,1]$, $F^{-1}(x)$ has law $F$ and similarly for $G$ so the left hand integral is the expected value of the absolute value of the difference between two random variables with cdfs $F$, $G$, but I have no clue how to relate this to the integral on the right hand side.  Please help me if you can.

答: Intuitively it should be simple: it amounts to saying that the green areas below are equal in the two charts, as there is simply a reflection in the bottom-left to top-right diagonal to achieve the inverse of the cumulative distribution functions

Title: Integrals of quantile function


问:
Is $\sum_{n=1}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$ convergent? Is it absolute or conditional convergence?
Hello I've come across this problem my teacher gave me. Consider the infinite series $\sum_{n=1}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$. It's obvious that when $n = 1$, the value doesn't exist. But I asked my teacher and she said we can separate the $\frac{1}{\ln(1)}+\sum_{n=2}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$, and since $\sum_{n=2}^\infty (-1)^n \frac{n^{1/n}}{\ln(n)}$ is conditional convergent, then the series converges as well.

I've proven that the sequence $\frac{n^{1/n}}{\ln(n)}$ continuous, positive, and decreasing over $(1, \infty)$. (I checked the derivative of it and the value is $\leq 0$).

Next, I used the limit comparison test, to compare $\frac{n^{1/n}}{\ln(n)}$ and $\frac{1}{\ln(n)}$, which diverges, meaning $\frac{n^{1/n}}{\ln(n)}$ also does.

Someone asked this question a while ago. But I remember somewhere in the lecture, there was once this theorem about if an element is removed/added from a convergent series (or something similar) the corresponding series would be convergent as well.

Does a series converge if its initial value is undefined?

Update: Can I actually use the Integral Test for this one?

Update 2: Sorry had a typo for $\frac{1}{\ln(0)}$

Thank You.

答:
The series is an alternating series for $a_n=\frac{n^{1/n}}{\ln n}$. What you nee is to check if $a_n$ is decreasing and $\lim_{n\to\infty}a_n=0$.

(1). Since $n^{1/n}$ is decreasing when $n\ge 3$ and $\ln n$ is increasing, one has that $a_n$ is decreasing for $n\ge 3$.

(2). It is easy to check $\lim_{n\to\infty}a_n=0$.

Then by the Alternating Series Test, $\sum_{n=2}^\infty(-1)^na_n$ converges.


问:
Positive definite 2-sequence from positive definite 1-subsequences?
If $p = (p_{n,m})_{n,m \geq 0}$ is a 2-sequence for which it is known that all the 1-subsequences $s^{(ab)} = (p_{ka,kb})_{k \geq 0}$ are positive semidefinite, for arbitrary $a,b \in \mathbb{N}_0$, can we conclude that $p$ is a positive semidefinite 2-sequence?

答: No. A counter-example is the 2-sequence $p$ with $p_{2k,k} = 1$ for all $k \geq 0$ and $p_{n,m} = 0$ elsewhere. The 1-subsequences $s^{(ab)}$ are all positive semidefinite (since they are either $(1,0,0,\cdots)$ or $(1,1,\cdots)$) but $p$ is not, because for the three pairs of indices $(n_1,m_1) = (0,0), (n_2,m_2) = (1,0)$ and $(n_3,m_3) = (1,1)$ and the three-vector $c = (1,1,-1)$ we have $$\sum_{i,j=1}^3 c_i c_j p_{n_i+n_j,m_i+m_j} = -1 < 0.$$ This violates the requirement for the positive semidefiniteness of a sequence (e.g. Nussbaum, Ark. Mat. 6, 10 (1965)).


答:
An alternative approach. Let $s: \mathbb{R}\times \mathbb{R} \to \mathbb{R}$ be the sum function:
$$s(v,u) := v+u$$
It is easy to see that $s$ is continuous. Thus, for a fixed $v \in \mathbb{R}$, the associate function $s_{v}: \mathbb{R}\to \mathbb{R}$ defined by:
$$s_{v}(u) := v+u$$
is not only continuous, but a homeomorphism (that is, it is a continuous bijection and its inverse is again continuous). Thus, if $A$ is an open set:
$$s_{v}(A) = v+A$$
must be open.

Title: If $A$ is an open set in $\mathbb{R^n}$ is $v + A = \{v + x: x\in A \}$ an open set $\forall v \in \mathbb{R^n}$?

Body:
I'm trying to prove that if $A$ is an open set in $\mathbb{R^n} \Rightarrow$ $\forall v \in \mathbb{R^n}$ : $v + A = \{v + x: x \in A \}$ is an open set in $\mathbb{R^n}$

I know that a set $A$ is an open set if
$$
\forall x \in A, \exists \epsilon \gt 0 : B(x, \epsilon) \subset A
$$

The proof seems trivial but I dont know how to prove it.Any hints or suggestions?


问:
Gradient iff conditions on vector-valued Lipschitz function on $R^n$
Suppose that $f\in C^1(\mathbb{R})$.

$f\colon \mathbb{R}\to \mathbb{R}$ is Lipschitz if and only if $|f'|$ is bounded on $\mathbb{R}$.

Is it still true in multidimensional and vector-valued situation?

Say, $f\colon \mathbb{R}^n \to \mathbb{R}$. Is it true that $||\nabla f||_2$ bounded $\Longleftrightarrow f$ is Lipschitz?

Say, $f\colon \mathbb{R}^n \to \mathbb{R}^m$. Is it true that $||Jf||_2$ bounded $\Longleftrightarrow f$ is Lipschitz, where $Jf$ is the Jacobian of $f$ and $\|\,\|_2$ is matrix induced norm in L2 sense?

If not, what are some good iff/if conditions by using gradients?

Example, let $f(\mathbf{x}) = x_1x_2$, where $\mathbf{x} = [x_1\,x_2]^T\in \mathbb{R}^2$. Is this $f(\mathbf{x})$ Lipschitz? It intuitively looks like Lipschitz, but its gradient is unbounded.

答:
The idea that a function is Lipschitz iff its grandient is bounded is correct, but to state rigorously there are a couple of technical details that you have to take into account, which naturally lead to the notion of Sobolev space $W^{k,p}$. Indeed, one can show that if $U$ is a bounded domain with Lipschitz boundary then a function is Lipschitz if and only if it belongs to $$W^{1,\infty}(U):=\left\{u\in L^1_{\text{loc}}(U):u\in L^\infty(U), \nabla u\in L^\infty(U)\right\}$$where $\nabla u$ is understood in the distributional sense. In particular, the norm in your first question should be the $L^\infty$ norm instead of the euclidean one.

I recomend Evans' "Partial Differential Equations'', chapter 5 in general, section 5.8.2, b) for this questions in particular and the references given there for further investigations.

Hope it helps.

OwnerDisplayName: anon


问:
$H \odot K$ complete implies $\dim H$ or $\dim K <\infty$
Let $H,K$ be non-zero Hilbert spaces. Then the algebraic tensor product $H \odot K$ is an inner product space for the unique inner product determined by
$$\langle h \otimes k, h' \otimes k'\rangle = \langle h, h'\rangle_H \langle k, k' \rangle_K$$

I am trying to prove that $H \odot K$ is a Hilbert space (i.e. $H \odot K$ is complete for the norm induced by this inner product) implies that $\dim H <\infty$ or $\dim K < \infty$.

Let us prove the contrapositive, i.e. so suppose that $H$ and $K$ are both infinite dimensional. How can I show that there is a Cauchy sequence in $H \odot K$ that does not converge?

答:
It is very easy to exhibit a non-converging Cauchy sequence in $H\odot K$, but actually proving that it does not
converge is a bit more difficult.  The approach suggested by @Max, although somewhat  sophisticated, is perhaps the easiest
way to pin down all of the details.  Here is a pedestrian way to describe this method.

For each pair of vectors $(x, y)\in H\times K$, consider the bounded linear operator
$$
  T_{x, y}:H\to K
  $$
given by
$$
  T_{x, y}(z) =  \langle x,z\rangle y, \quad\forall z\in H.
  $$
(For simplicity I am assuming that we are working with real vector spaces but this can be fixed in the complex case by
seeing   $T_{x, y}$ as an operator defined on $\bar H$, namely the Hilbert space made out of $H$ by adopting the new scalar
multiplication operation $\lambda \cdot x:= \bar \lambda x$).

It is easy to see that the map
$$
  (x, y)\in H\times K\mapsto   T_{x, y}\in  \mathscr {B}(H,K)
  $$
is bilinear so, by the universal property of (algebraic) tensor products, there exists a unique linear map
$$
  T :H\odot K\to  \mathscr {B}(H,K),
  $$
such that $T (x\otimes y) = T_{x,y}$.
We next claim  that
$$
  \langle T (\xi )x,y\rangle  =  \langle \xi ,x\otimes y\rangle ,
  \tag {1}
  $$
for every $\xi $ in $H\odot K$, $x\in  H$, and $y\in  K$. To prove this write  $\xi =\sum_{i=1}^nx_i\otimes y_i$, and notice that
$$
  \langle T (\xi )x,y\rangle  =
  \sum_{i=1}^n   \langle T_{x_i,y_i}(x),y\rangle  = $$ $$ =
  \sum_{i=1}^n   \langle \langle x_i,x\rangle y_i,y\rangle  =
  \sum_{i=1}^n   \langle x_i,x\rangle \langle y_i,y\rangle  = $$ $$ =
  \sum_{i=1}^n   \langle x_i\otimes y_i,x\otimes y\rangle  =
  \langle \xi ,x\otimes y\rangle,
  $$
proving the claim.

Note that the range of each $T_{x,y}$  is the one-dimensional space spanned by $y$, so $T_{x,y}$ has rank 1.
Furthermore, every $\xi $ in $H\odot K$  may be writen as  $\xi =\sum_{i=1}^nx_i\otimes y_i$, so
$$
  T (\xi ) = \sum_{i=1}^n T_{x_i,y_i},
  $$
so we see that   $T (\xi )$ has rank at most $n$, hence finite, for every $\xi $ in $H\odot K$.

The final contradiction will be achieved by proving that, if $H$ and $K$ are infinite dimensional, and $H\odot K$ is  complete, there exists a vector $\xi $ in $H\odot K$  such that $T (\xi )$ has infinite rank.

Let us therefore assume from now on that $H$ and $K$ are both infinite dimensional, so we may find (necessarily infinite)
orthonormal bases
$\{e_i\}_{i\in  I}$, and $\{f_j\}_{j\in  J}$, for $H$ and $K$ respectively, and it is well known that $\{e_i\otimes f_j\}_{(i, j)\in  I\times J}$ is an
orthonormal basis for $H\otimes K$.

Assuming by contradiction that $H\odot K$ is complete,  observe that  $H\odot K$ coincides with its completion,  so   $H\odot K = H\otimes K$.

Choose infinite subsets $\{i_n:n\in {\mathbb N}\}\subseteq I$, and $\{j_n:n\in {\mathbb N}\}\subseteq J$, and let
$$
  \xi  = \sum_{n=1}^\infty  {1\over n^2} (e_{i_n}\otimes f_{j_n}).
  $$
This is well defined because the series is clearly absolutely convergent and we are working in a complete space!

Using (1) we have that
$$
  T (\xi )e_{i_n} =
  \sum_{j\in J} \langle T (\xi )e_{i_n}, f_j\rangle f_j =
  \sum_{j\in J} \langle \xi , e_{i_n}\otimes f_j\rangle f_j =
  \langle \xi , e_{i_n}\otimes f_{j_n}\rangle f_{j_n} =
  {1\over n^2} f_{j_n}.
  $$
This shows that ${1\over n^2} f_{j_n}$ lies in the range of $T (\xi )$, and hence that $T (\xi )$ has infinite rank,  a contradiction.


问:
A construction using straightedge and compass
Given a circle, it's easy to contruct its center.

The question is: given an ellipse, draw the foci.

I don't know whether it's possible to do this using only straightedge and compass.

答:
Let $\lambda$ a given ellipse, let's draw its foci.

See the figure below:

Draw two parallel chords $AB$ and $DC$.

Find out $M$ and $N$ (midpoints of $AB$ and $DC$ respectively).

Draw the straightline $r$ through $M$ and $N$.

Let $\{E, E'\} = r \cap \lambda$, find out $O$ the midpont of $EE'$.

Draw an arc $\mu$ centered at $O$ and radius $AO$, such that $\{A, A'\}= \mu \cap \lambda$.

Draw the straightline $s$ through $A$ and $A'$.

Draw a straightline $t$ such that $O \in t$ and $t\parallel s$.

Draw $u$ (the perpendicular bisector of $AA'$).

Let $\{G, G'\} = u \cap \lambda$, and $\{J, J'\} = t \cap \lambda$, $GG'$ and $JJ'$ are the axes of symmetry of $\lambda$. Now draw an arc $\nu$ centered at $G$ and radius $JO$, such that $\{F1, F2\}= \nu \cap t$. $F1$ and $F2$ are the desired foci of $\lambda$.


问:
recursive sequence proofs
I'm kinda new to recursive sequences and I'm struggling with an excercise. I apologize in advance for the long question and my lack of knowledge on how to approach such problems.

Let $x_n$ be defined such that $x_1 = 1$, $x_{n+1} = \frac{x_n^2+1}5$

And there there are 4 things I have to prove/show,

(1) Show that $x_n$ $\leq$ $1$ for all $n$.

I'm not sure if I can prove this directly or by induction? What would be best for this? I don't know how to approach this when it's defined recursively.

(2) Show that $\lvert x_{n+1} - x_n \rvert$ $\leq (\frac{2}{5})^{n-1}$

I'm not sure but could I use $\lvert\frac{x_{n+1}^2+1}{5}$ $-$ $\frac{x_n^2+1}{5}\rvert$$\leq (\frac{2}{5})^{n-1}$ and then try to solve this. How do i deal with the $x_n$?

(3) Show that $x_n$ is a Cauchy-sequence.

I know that a Cauchy sequence is defined as : For every positive real number $\varepsilon$, there is a positive integer $N$ such that for all natural numbers $m, n \gt N$: $\lvert x_m - x_n \rvert \lt \varepsilon$. Can I apply this definition directly in this sequence and the go on from there?

(4) Calculate the limit of $x_n$.

This should be straightforward if it weren't a recursively defined sequence.

If I however, calculate the first few elements of the sequence I get:

$x_1 = 1$, $x_2 = \frac{2}{5}$, $x_3 = \frac{29}{125}$, $x_4 = \frac{841}{15625}$ $\ldots$

This approaches $0$ (I'm guessing) since the second term is smaller than $1$. And since a square of a number less than $1$ is always smaller than the original number, $\lim \limits_{n \to \infty}$ $= 0$.

How can I prove this correctly without just trying different numbers?

Thank you in advance!

答:
Since $x_1 \leq 1$, we just need to prove that, if $x_n \leq 1$, then $x_{n+1}\leq 1$. This is straightforward because we have that  $x_{n+1} = \dfrac{x_n^2+1}{5}\leq \frac 25 \leq 1$. Also, considering that $x_n >0$, we have established that $0 < x_n \leq 1, \forall_n$.

$$|x_{n+1} - x_n| = \left| \frac{x_n^2+1}{5} - \frac{x_{n-1}^2+1}{5} \right|=\left| \frac{(x_n-x_{n-1})(x_n+x_{n-1})}{5}\right| \leq \frac 25 |x_n -x_{n-1}|$$
Using this inequality recursively, one can realize that
$$|x_{n+1}-x_n|\leq (2/5)^{n-1}|x_2-x_1| \leq (2/5)^{n-1}.$$

The sequence is Cauchy if, for every $\varepsilon >0$, there is some order $p$ such that $|x_n-x_m| < \varepsilon$ for all $n \ge m \ge p$.
Now,
\begin{align*}
|x_n-x_m|=&|x_n - x_{n-1}+x_{n-1} -x_{n-2}+x_{n-2} + \cdots -x_m|\\
\leq & |x_n-x_{n-1}| + |x_{n-1}-x_{n-2}| + \cdots +|x_{m+1}-x_m|\\
\leq & (2/5)^{n-1} + (2/5)^{n-2} + \cdots +(2/5)^{m-1}\\
=& \dfrac{1-(2/5)^n}{1-\frac 25}\times (2/5)^m = \frac 53 (1-(2/5)^n) (2/5)^m \leq \frac 53 (2/5)^m
\end{align*}
so you just choose $p$ in such a way that $\frac 53 (2/5)^p < \varepsilon$.

If the limit exists, and we know it exists because Cauchy sequences in $\mathbb{R}$ are convergent, you have that
$$
x_{n+1} = \frac{x_n^2+1}{5} \Rightarrow L=\frac{L^2+1}{5}
$$
and so $L = \frac{1}{2}
   \left(5\pm \sqrt{21}\right)$. Knowing that $L \in [0,1]$, we must choose the "-" sign and conclude that $L=\frac{1}{2}
   \left(5-\sqrt{21}\right)$.

Title: recursive sequence proofs


问:
Good way to extrapolate time series data based on min/max/mean and sample size
I have a small set of annual data on the operations cost of solar power plants that I'd like to extrapolate into future years for some basic modelling. For each year 2011 through 2018, I have:

The mean operating cost of each plant in dollars per kilowatt of capacity

The min and max operating costs in the sample

Number of plants included in the sample (with additional plants added to the sample each year)

Total combined capacity (in kilowatts) of all plants in the sample

The data looks something like this:
$$\begin{array}{c|c|c|}
 & \text{Mean} & \text{Min} & \text{Max}& \text{Count}& \text{Capacity} \\ \hline
\text{2011} & $65 & $53 & $71 & 9 & 199 \\ \hline
\text{2012} & $34 & $22 & $44 & 19 & 326 \\ \hline
\text{2013} & $47 & $34 & $54 & 29 & 463 \\ \hline
\text{2014} & $26 & $18 & $44 & 31 & 476 \\ \hline
\text{2015} & $23 & $16 & $54 & 50 & 673 \\ \hline
\text{2016} & $24 & $19 & $42 & 56 & 848 \\ \hline
\text{2017} & $27 & $20 & $39 & 60 & 995 \\ \hline
\text{2018} & $30 & $19 & $68 & 76 & 1,227 \\ \hline
\end{array}$$

What would be a reasonable way to estimate the mean operating costs for years in 2019 and forward?

答:
A couple of caveats.

We are assuming that the plant technologies are similar and older years and newer years are comparable

There may be specific operating costs that are higher for older plants than newer plants (eg: repairs and maintenance). Since the sample data does not provide any way for identifying this reliably, we are ignoring it.

A simple method might be assigning weightage for the time periods.

Let us say the weight $W_t$ we want to assign for time period $t$ is

inversely proportional to how recent the data point is (ie., older years get lesser weightage). The rationale for this is we are biasing the costs with the recency effect considered. Any operational optimizations or cost increases carried out through the most recent years are given higher weightage.

inversely proportional to the $\sqrt n_t$ where $n_t$ is the count of plants in the sample for that time period. The rationale is that margin of error in a sample is inversely proportional to the square root of the size of the sample (in a normal distribution).

inversely proportional to the $\sqrt C_t$ where $C_t$ is the total capacity for that period. Rationale is same as above.

So, we have

$$W_t \propto {t \over \sqrt {n_t \times C_t}}$$

$$W_t = {k \times t \over {\sqrt {n_t \times C_t}}}$$ where $k$ is a constant.

The relative weightage (a real value between 0 and 1) is then given by

$$w_t = {W_t \over \sum_j W_j}, j > 0$$

Note that the proportionality constant $k$ is cancelled when we take the relative weightage.

$j$ in the denominator is the index of the time period and we take $j > 0$ to avoid multiplying by 0 causing the term to vanish.

Then we can then predict the average operating cost per KW for future years using a weighted average of costs of prior years with weights calculated as given above.

This will keep your model simple and explainable and not reliant on any hypothesis that is difficult to prove.

I would also suggest using quarterly data, if available. You will get higher accuracy of prediction with 4x data points and also factor any seasonal impacts of costs without complicating the model too much.


问:
Demonstrate Risk Formula in Markowitz Asset Pricing Model
there!

I'm trying to demonstrate Markowitz Model's formula for the risk of a portfolio with $N$ assets. Considering that the risk can be measured in terms of the variance of the portfolio ($\sigma^2_P$), the percentage of the $i$-th asset ($i=1,2,\ldots,N$) on the portfolio value ($X_i$), the variance of the $i$-th asset ($\sigma^2_i$) and the covariance between the $j$-th and the $i$-th asset ($\sigma_{ij}$), $j \ne i$.

I tried to start the demonstration by assuming that the return of the portfolio is given by $R_P = \sum_{i = 1}^N X_i \cdot R_i$ and that the expected return of the portfolio if given by $E(R_P) = \sum_{i=1}^N X_i \cdot E(R_i)$.

After that, I remembered that $Var(X) = E\left[(X-E(X))^2\right]$, substituted $X = R_P$, which gave me the expression:

$\displaystyle \sigma^2_P = E\left[\left(R_P - E(R_P)\right)^2\right] = E\left[\left(\sum_{i=1}^{N} X_i \cdot R_i - \sum_{i=1}^{N} X_i \cdot E(R_i)\right)^2\right]$

I need to get to the expression:

$\displaystyle \sigma^2_P = \sum_{i=1}^N X_i^2\cdot\sigma^2_i + \sum_{i=1}^N\sum_{j=1}^N X_i \cdot X_j \cdot \sigma_{ij}$

Any ideas on how to get there? Thanks.

答:
We can write

$$\left(\sum_{i=1}^{N} X_i \cdot R_i - \sum_{i=1}^{N} X_i \cdot E(R_i)\right)^2 = \left(\sum_{i=1}^{N} X_i \,[R_i -  E(R_i)]\right)^2 = \\ \sum_{i=1}^{N}\sum_{j=1}^N X_iX_j \, [R_i -  E(R_i)][R_i -  E(R_i)] \\ = \sum_{i=1}^NX_i^2[R_i -  E(R_i)]^2 + \underset{i \neq j}{\sum_{i=1}^{N}\sum_{j=1}^N} X_iX_j \, [R_i -  E(R_i)][R_i -  E(R_i)] $$

The variance $\sigma_i^2$ and covariance $\sigma_{ij}$ ($i\neq j$) are defined as

$$\sigma_i^2 = E\left([R_i -  E(R_i)]^2\right), \quad \sigma_{ij} = E\left([R_i -  E(R_i)][R_j -  E(R_j)]\right)$$

Thus,

$$\sigma_p^2 = E\left(\sum_{i=1}^NX_i^2[R_i -  E(R_i)]^2 + \underset{i \neq j}{\sum_{i=1}^{N}\sum_{j=1}^N} X_iX_j \, [R_i -  E(R_i)][R_i -  E(R_i)]  \right) \\ = \sum_{i=1}^NX_i^2\, E\left([R_i -  E(R_i)]^2\right) + \underset{i \neq j}{\sum_{i=1}^{N}\sum_{j=1}^N} X_iX_j \, E \left([R_i -  E(R_i)][R_i -  E(R_i)]\right) \\ = \sum_{i=1}^NX_i^2\, \sigma_i^2 + \underset{i \neq j}{\sum_{i=1}^{N}\sum_{j=1}^N} X_iX_j \, \sigma_{ij}$$

Note that the double sum on the RHS is taken over terms where $i \neq j$.


问:
Limit of $\sin x$ as $x$ tends to infinity
I understand that $-1 \leq \sin(x) \leq 1$ for any real $x$. However, the function oscillates and doesn't approach a finite limit as $x$ tends to infinity.

So, what is the mathematically correct statement: the limit is undefined, the limit is indeterminate or the limit approaches infinity?

Further, are the concepts of indeterminate, undefined and infinity different? The terms seem to be used interchangeably though.

Note: I've read the answers on similar posts Limit approaching infinity of sine function and Is limit of $\sin x$ at infinity finite? but couldn't find answers specific to my doubt.

答: There are no "indeterminate limits", only indeterminate forms. An infinite limit has a specific meaning, in that the function is eventually greater or smaller than any finite number. Since the sine does not exhibit this behaviour, its limit at infinity is undefined.</p>


问:
Quadratic equation with multiple parameter
I have the following equation:

$$x^2 +mx+(a−m)^2 =0\text{ ($a$ be there as a fixed parameter)}$$

and I need to calculate for which values of $m$ the equation has exactly one solution for $x.$

I know I have to look the value of $m$ for which the discriminant of this equation is equal to $0.$ But every time when I went this path is got stuck.

Can someone give me some hint I need to search for or look for when trying to calculate this?

Addition:

One thing I forgot to mention is that first of all the goal is to find a value for $m$ and afterwards to calculate the solution(s) fo $x$.

答: $A^2-B^2 =(A+B)(A-B)$, where $A=m$, $B=2(a-m)$. Did you get the hint?


问:
$\lim_{n \to \infty}\sum_1^n \frac{(\log k)^4}{ k^2}$ converges?
Does the  series converge? </p>

<p>$$\lim_{n \to \infty}\sum_1^n \dfrac{(\log k)^4}{  k^2}$$</p>

答: Notice that $\int_1^{\infty}\frac{\ln(x)^4}{x^2}dx = \frac{-24-24 \ln(x) -12 \ln(x)^2-4 \ln(x)^3 -\ln(x)^4}{x}|_1^{\infty}=24$. Since this integral converges your series converges by the integral test.</p>


答:
As you said, the condition is that the discriminant is $0$.
In this case the discriminant is $-4 a^2 + 8 a m - 3 m^2$.
That's a quadratic in $m$: can you solve it?

Title: Quadratic equation with multiple parameter

Body:
I have the following equation:

$$x^2 +mx+(a−m)^2 =0\text{ ($a$ be there as a fixed parameter)}$$

and I need to calculate for which values of $m$ the equation has exactly one solution for $x.$

I know I have to look the value of $m$ for which the discriminant of this equation is equal to $0.$ But every time when I went this path is got stuck.

Can someone give me some hint I need to search for or look for when trying to calculate this?

Addition:

One thing I forgot to mention is that first of all the goal is to find a value for $m$ and afterwards to calculate the solution(s) fo $x$.


问:
The conjugacy class of $H<G$ is equipotent to the set of right/left cosets of $N_G(H)$ in $G$
Here is how I would prove the statement.

I need to find a bijection between $A=\{H^a =a^{-1}Ha|a\in G\}$, the conjugacy class of $H<G$, and $B=\{a*N_G(H)|a\in G\}$, the set of left cosets of $N_G(H)$ in $G$.

Let $f$ be the function $f:A\rightarrow B$ such that $a^{-1}Ha\mapsto a*N_G(H)$.

$f$ is surjective

$\forall a*N_G(H) \in B$, $f(a^{-1}Ha)=a*N_G(H)$

$f$ is injective

Let $a^{-1}Ha$ and $b^{-1}Hb$ be elements of $A$ such that $f(a^{-1}Ha)=f(b^{-1}Hb)$ $\Rightarrow$ $a*N_G(H)=b*N_G(H)$

How do I go on from here? How do I prove that $a=b$ ?

[EDIT]

I thought this was an easier way to prove it.

Let $g:G\rightarrow Cl(H)$ be a map from the group $G$ to the conjugacy classes of $H<G$ such that $x\mapsto H^x=x^{-1}Hx$.

This is clearly a homeomorphism and the kernel of $g$ is $N_G(H)$.

So, for the homeomorphism theorem I have that $\frac{G}{N_G(H)}\cong Cl(H)\Rightarrow card(\frac{G}{N_G(H)})=card(Cl(H))\Rightarrow $ the conjugacy class of $H<G$ is equipotent to the set of right/left cosets of $N_G(H)$ in $G$.

Does this proof work?

答:
Your chosen bijection works more smoothly if you use right cosets $N_G(H)a$.

Define $f:A \to B$ by $f(a^{-1}Ha) = N_G(H)a$. As you said, this is clearly surhective. We have
$$ a^{-1}Ha=b^{-1}Hb \Leftrightarrow ba^{-1}Hab^{-1} = H \Leftrightarrow ba^{-1} \in N_G(H) \Leftrightarrow N_G(H)a=N_G(H)b.$$

Note that the forward implications imply that $f$ is well-defined, and the reverse implications show that it is injective.


答:
For $Ax^2+Bx+C=0$ to have only one solution, $B^2-4AC = 0$ or $B^2 = 4AC$.

So we need $m^2 = 4(a-m)^2 = (2(a-m))^2 \Rightarrow m = \pm(2a-2m) \Rightarrow \color{blue}{m =2a, \frac{2a}{3}}$

To calculate the solution of the equation,

$Ax^2+Bx+C$ has sum of the roots $= \frac{-B}{A}$. Here as the discriminant is zero, both the roots are equal, $x_1 = x_2$.

So, $2x_1 = -\frac{B}{A} = -m \Rightarrow \boxed{x_1 = x_2 =-\frac{m}{2}}$.

The roots are either $\boxed{-a}$ or $\boxed{-\frac{a}{3}}$ corresponding to $m=2a, \frac{2a}{3}$ respectively.

Graph for solutions:

Corresponding to $m=2a$ and $m=\frac{2a}{3}$


问:
Looking for a function $f$ that solves $f'(x)= \alpha(x) f(x)$ for a continuous $\alpha$ and starts off positive but then turns negative
I am looking for a continuously differentiable function $f:[-1,1]\to\mathbb R$ (note that I don't care about the interval $[-1,1]$ in particular; it can be replaced by any interval $[a,b]$ with $a<b$ of your choice) that satisfies

$f(t)>0$ for all $-1\le t<0$,

$f(0)=0$,

$f(t)<0$ for all $0<t\le 1$

as well as

$$f'(t)=\alpha(t) f(t)$$

for some continuous $\alpha:[-1,1]\to\mathbb R$.

Essentially I am looking for something like this:

My attempts: The most obvious attempt is  $f(t)=-t^3$. However, one notices that we need to have $\alpha(t)=\frac{3}{t^2}$ for $t\not=0$, which cannot be extended to a continuous function. So $f(t)=t^3$ doesn't work.

In fact, if $f$ is any polynomial with $f(0)=0$, we know that $\frac{f'}{f}$ diverges at $0$ so that $\alpha=\frac{f'}f$ can't be continuous....

Next try, $f(t)=-\exp\left(-\frac1t\right)$ for $t>0$ and $f(-t)=-f(t)$: This fails immediately as for $t\not=0$, $f'(t)=\frac{f(t)}{t^2}$, so that, once again, $\alpha(t)=\frac{f'(t)}{f(t)}=\frac{1}{t^2}$ can't be made continuous at $t=0$.

Final attempt: I define $g(t)=2 \operatorname W(\exp(t))-t-1$, where $\operatorname W$ is the Lambert W function. This is a rotation of the function $t\mapsto\exp(t)$ by 45 degrees and it looks like this:

Then I define $$f(t)=\begin{cases}
g(t+1) & t\le 0,\\ -g(t+1) & t>0.
\end{cases}$$

It looks like this:

However, once again we have that $\lim_{t\to0}\frac{f'(t)}{f(t)}$ is non-existant so that $\alpha$ can't be continuous at $0$.

So my question: Does a function $f$ as above even exist? If yes, can you give me an example? If not, why doesn't such a function exist??

答: Since $y=0$ is a solution, by the existence and uniqueness theorem, your equation only has zero solution. This is becasue the initial condition is $y(0)=0$.


问:
Induction proof with binomial coefficients and inequalities
I am attempting to prove that for some non-negative integers
p ≥ q ≥ r, ${p \choose r} \geqslant {q \choose r}$.

My base cases were p=q=r=0 and p=q=r=1, and my induction hypothesis is to assume that the statement holds for some arbitrary non-negative values of p, q, and r. I get stuck, of course, at the inductive step. Right now I have:

Consider p + 1, q + 1, e.g.
${p+1 \choose r} ≥ {q+1 \choose r}$
I'm attempting to use Pascal's Identity and the fact that ${(r - 1)! = \frac{r!}{r}}$, but I end up with a very circular argument, where I'm saying that

From ${p+1 \choose r} \geqslant {(q+1) \choose r}$
we get
${p \choose r} + {p \choose r-1} ≥ {q \choose r} + {q \choose r-1}$

then
$$\frac{p!}{r!(p-r)!} + \frac{p!}{(r-1)!(p-r-1)!} ≥ \frac{q!}{r!(q-r)!}+\frac{q!}{(r-1)!(q-r-1)!}$$

And here's where I get stuck, because if this equation reduces to $${p+1 \choose r} \geqslant {q+1 \choose r}$$
then I've just shown something that was a given, and I need to change my inductive step to be more robust? Or would this be sufficient, and I'm overthinking it?

Thank you in advance!

答:
It seems like you're not familiar with double (or even triple) induction. That's fine, you can still induct on 1 variable.

The approach I would take is to show that for a fixed $r$, and for any $ q \geq r$, then we have

$$ { q + 1 \choose r } \geq { q \choose r }. $$

This can be done directly (via comparing the expansion which you wrote), which is my preference. Or it can done via induction (via the binomial identity which you wrote) if you want an induction proof.

Once this is proved, we see that for any $ p \geq q \geq r$, we get the chain of inequalities

$${ p \choose r } \geq { p-1 \choose r } \geq {p-2 \choose r } \geq \ldots \geq {q+1 \choose r } \geq { q \choose r }. $$


答:
OK, calculate the discriminant:
$$m^2-4(a-m)^2=(m+2(a-m))(m-2(a-m))=0$$
This is zero precisely when one of the two factors is zero, so
$$m+2(a-m)=0\implies m=2a$$
$$m-2(a-m)=0\implies m=\frac23a$$


问:
Dynkin pi-system lemma
I have a pretty trivial question which I am really struggling to get to grips with, and I get the feeling I am missing something simple so hopefully I'll be able to clear this issue up here.

The lemma states: If G is a $\pi$-system, then the $d$-system (dynkin-system) generated by G is the $\sigma$-field generated by G.

In the proof I have, we construct the following set:
$D_{1}$ = { B $\in$ $d(G)$ : $\forall$ C $\in$ G, B $\cap$ C $\in$ $d(G)$ }

and we show that $D_{1}$ is infact a $d$-system. I am happy with the remainder of the proof but struggling to see why the whole set $\Omega \in D_{1}$?

By definition of a dynkin-system I know that we must have $\Omega \in d(G)$ so in the set $D_{1}$ I can take element $B = \Omega \in d(G)$ but I can't deduce why the set C must contain $\Omega$ (so that the intersection of $\Omega \cap \Omega = \Omega \in d(G)$) which ultimately will show the first part that $D_{1}$ is infact a d-system.

Any help on the above would be much appreciated!

答:
This definitions are indeed confusing.

Notice that, by definition $G\subseteq d(G)$ because $d(G)$ is the smallest Dynkin system containing $G.$ Notice then that if $x\in G,$ then $x\in d(G).$ So, $\Omega \in d(G)$ by definition and if $C\in G$ then $C\cap \Omega =C\in d(G)$ so $\Omega \in D_1.$

Title: Dynkin pi-system lemma


问:
How to translate "When" in propositional logic?
I'm looking to understand how to represent "When" in a propositional logic.
For example:

1: "I buy food when I get paid." Would the order of this matter, say "When I get paid, I buy food"?

Would this be,  Food -> Paid?

2: "When I go to Walmart to buy food, I don't buy onions"

Would this be   (Walmart & Food) -> -Onions?

3: "I always go to Walmart on Saturday, but I never go any other day"

Would this just be Walmart -> Saturday, or does a statement like this also need to include  -Saturday -> -Walmart?

It might be easier to give answers just using variables such as x,y,z etc, but I made up some examples to try understand it better

答:
"B when A" is "if A, then B".

Thus you are quite right: : "I buy food when I get paid" will be: "If I get paid, then I buy food".

Buy in symbols it is: $\text{GetPaid} \to \text {BuyFood}$.

The second is "When A, then B", and in this case is: "If A, then B".

Thus: $\text {GoWalmart} \to \lnot \text {BuyOnions}$