问:
Proof or reject condition number if square matrix
I want to proof(if true) or reject:
For all A Square matrix nxn K(10A) = 10K(A), i think it's wrong since:
K(cA)=K(A) (refe Show property of condition number)

if K(cA) == K(A) which is != from 10K(A), is my proof is correct?

Body:
Your proof is correct. To go into a bit more detail,
$$
\kappa(10A)=||10A|| \cdot ||(10A)^{-1}||=10||A|| \cdot 10^{-1}||A^{-1}||=||A|| \cdot ||A^{-1}|| = \kappa(A) \neq 10 \kappa(A).
$$


问:
Rationale behind truth values
I originaly asked a question on Programmers.SE to know why $0$ was consider $\text{false}$ and all the other [integral] values were considered $\text{true}$. That was a huge debate and many said it was a legacy from Boolean algebra where $0$ is indeed $\text{false}$ and $1$ is $\text{true}$.

Somebody suggested I go further and ask here why this is actually the case in Boolean algebra. So here is the question: what is the rationale for $0$ to be $\text{false}$ and $1$ to be $\text{true}$ and not the other way around in Boolean algebra?

答:
The numbers you use essentially don't matter. But if you want to represent your $2^4$ truth functions (see Wikipedia) using arithmetic, then $0,1$ come in handy. This is because their properties of being the additive and multiplicative neutral element simplifies some computations.

You can represent the functions using any numbers, really. If $a$ can be a number representing $\text{true}$ or another number representing $\text{false}$, then

$$\text{NOT}(a):=\text{true}+\text{false}-a$$

works out for defining the negation. For example

$$\text{NOT}(\text{true}):=\text{true}+\text{false}-\text{true}=\text{false}.$$

Here follows a nice graphic showing all the general constructions. As examples, the use of $\{0,1\}$ and also $\{-1,1\}$ is demonstated. Notice how using $0$ "for $\text{false}$" eliminates all the terms involving the number $s_0$, making the $\{0,1\}$ column specifically short and simple for computations.

ContentLicense: CC BY-SA 3.0

Title: Rationale behind truth values


问:
Special Case for Bézout Theorem
Over an algebraically closed field one has that by Bézout, two hypersurfaces in projective $n$-space always intersect. Is there an elementary proof for this statement?

答:
Yes, dimension theory does it (assuming $n>1$). One of the most basic results about intersections in projective space is that if $X,Y\subset\Bbb P^n_k$ are irreducible of codimension $r,s$ respectively, then every irreducible component of $X\cap Y$ has codimension at most $r+s$, and if $r+s\leq n$ this intersection is guaranteed to be nonempty. For a proof, see Hartshorne theorem I.7.2.

Alternatively, if you know that the complement of a hypersurface is affine (use the appropriate Veronese embedding to see this) and no affine variety contains a projective variety of positive dimension, you can see that for projective hypersurfaces $X,Y$, $\Bbb P^n_k\setminus X$ cannot contain $Y$ - this means $X\cap Y$ is nonzero, though it doesn't give you the result on dimensions like the previous paragraph.

ContentLicense: CC BY-SA 4.0

Title: Special Case for Bézout Theorem


问:
Prove that $(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})<3$
I have tried to use induction, but after I assume that P(n) is true, I can't go further to prove that P(n+1) is true as well. I also have tried to find an intermediate inequality, but I can't figure out which inequality I should start from.

Something that seemed to be useful was taking P(n) and multiplying it by $(1+\frac{1}{(n+1)^3})$, therefore I have come to this

$(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})<3 | \times(1+\frac{1}{(n+1)^3})$

$(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})(1+\frac{1}{(n+1)^3})<3(1+\frac{1}{(n+1)^3})$

but, as anyone could imagine, I came to contradiction because I've tried to prove that $3(1+\frac{1}{(n+1)^3})<3$ which is false.

Any help it would be useful.

答: Using the fact $1+x\le e^x$ for all real $x,$ we have $$\left(1+ \frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\cdots\left(1+\frac{1}{n^3}\right)\le \dfrac{9}{4}\exp\left(\sum_{k=3}^{n}\dfrac{1}{k^3}\right).$$ Now use the fact that $$\sum_{k=1}^{\infty}\dfrac{1}{k^3}\lt\dfrac{\pi^2}{7}.$$


问:
Integral of $(x+c)^n e^{-x^2}$
As part of some convolution, I am desperately trying to solve the following integral:

$$
\int_a^b (x+c)^ne^{-x^2} dx \,\,\,\, \text{with} \,\,\,\, n \in \mathbb{N}.
$$

I am looking forward to any input! Many thanks!

答:
lets start off with:
$$I=\int_a^b (x+c)^n e^{-x^2}dx$$
let $u=x^2\Rightarrow du=2xdx\therefore dx=u^{-1/2}du$ and so:
$$I=\int_{\sqrt{a}}^{\sqrt{b}}u^{-1/2}(u^{1/2}+c)^ne^{-u}du$$
and now we have to expand this:
$$u^{-1/2}(u^{1/2}+c)^n=u^{-1/2}\sum_{k=0}^n{n\choose k}(u^{1/2})^kc^{n-k}=\sum_{k=0}^n{n\choose k}u^{(k-1)/2}c^{n-k}$$
so if we bring this into the integral and interchange the integral and the summation we get:
$$I=\sum_{k=0}^n{n\choose k}c^{n-k}\int_{\sqrt{a}}^{\sqrt{b}}u^{(k-1)/2}e^{-u}du$$
and we know that the incomplete gamma function is defined as:
$$\gamma(s,x)=\int_0^xt^{s-1}e^{-t}dt$$
so this would make:
$$I=\sum_{k=0}^n{n\choose k}c^{n-k}\left[\gamma\left(\frac{k+1}{2},\sqrt{b}\right)-\gamma\left(\frac{k+1}{2},\sqrt{a}\right)\right]$$

Title: Integral of $(x+c)^n e^{-x^2}$


问:
Theorems you wish you knew in complex analysis
I recently stumbled upon a theorem of Landau, which states

Let $f(z)=\sum a_nz^n$ such that $f'(0)\neq 0$ and such that $f^{-1}(\{0,1\})=\emptyset$. The radius of
convergence of $f$ is bounded by a constant $C(f(0),f'(0))$

The theorem is not hard to prove once one knows the Ahlfors-Schwarz-Pick lemma, but it is an extremely surprising statement, and a quite elementary one. In particular, one can easily state and explain it to a student taking a first course in complex analysis. A similar situation occurs with Picard's theorems and Schottky's theorem.

My question is: What other "relatively unknown" theorems do you think fit this description?

Before this question gets closed as "opinion based", let me state some criteria for what I am looking for:

The statement is elementary (as explained before)

The theorem is not usually taught (or stated) in a first course in complex analysis

The proof requires nontrivial machinery from more advanced math(e.g. A-S-P for Landau's theorem)

答:
A comment that got long:

In terms of usefulness I would name two such results:

1.(edited as per comments as is known also as Bloch's Heuristic principle) the Bloch/Robinson principle, namely that if a "property" of holomorphic or meromorphic functions would imply such functions to be constant when defined in the full plane, then a family of holomorphic/meromorphic functions satisfying that property on some domain is normal (eg bounded, missing two values for analytic, three values for meromorphic etc)

(can be made rigorous as a Theorem of Zalcman by defining the notion of property as usual on "elements" $(f,U)$ with some obvious properties of extension, affine invariance etc)

one freely available reference paper is on arxiv - W Bergweiler Bloch's principle, (pdf link) later published in Computational Methods and Function Theory,6 2006)-

2.An analytic function defined on the interior of a Jordan curve, continuous on the closed domain and injective on the curve is then injective on the full closed domain (note that the curve may be weird like an Osgood curve so $f$ can be far from differentiable on $J$, $f'$ can be quite unbounded inside $J$ etc; also another application is if $f$ is analytic beyond the boundary but has a critical point there; then it is of course not injective around the critical point, but if it is injective on the boundary, it is still injective inside which is a bit counterintuitive; $z^2-2z$ on the unit disc is a good example here)
(as noted in the comments this is known as Darboux-Picard Theorem)

In terms of "coolness" results like:

if two meromorphic functions in the plane share five values (or two entire functions share four finite values as they then share infinity as an empty set of course) they are equal - here share a value $a$ means that the sets $f(z)=a, g(z)=a$ are equal including multiplicity
($e^z, e^{-z}$ share $0,1,-1, \infty$)

an entire function can have at most two values for which $f(z)=a$ implies $f'(z)=0$ (in other words there are at most two values $a$ for which all the roots of $f(z)-a$ are multiple), while for plane meromorphic functions the number is $4$ where of course we include $\infty$ now so poles too)

(elliptic Weierstrass and $\sin, \cos$ show that the result is optimal)


问:
Graph on argand plane
{$i+\bar z:1<|z|<4, z \in \mathbb{C}$}. If I want to shade this region in Argand plane, is it from z=i+4 to z=i-4?

How can I justify this answer?

答: The region defined by $1< |z| < 4$ is the open annulus. Conjugating the annulus (i.e., $\bar{z}$ flips the annulus across the horizontal axis, which in this case has no effect. Then, adding $i$ translates the annulus.

Title: Graph on argand plane


问:
Exponential function 1/x = e
Why is the following identity true?

$$\lim_{x\to0}(1+\sin2x)^{1/x}=\lim_{x\to0}\exp\left(\frac{1}{x}\ln(1+\sin2x)\right)$$

答:
Let $f(x)=1+\sin2x$ and $g(x)=\dfrac1x$. We want to show that $f(x)^{g(x)}=\mathrm{exp}\big(g(x)\cdot\ln f(x)\big)$.

We start by observing that $\exp(ab)=(\exp a)^b$. This follows from the fact that $y^{ab}=\left(y^a\right)^b$.

Therefore:

\begin{align*}
\mathrm{exp}\big(g(x)\cdot\ln f(x)\big)
&=\mathrm{exp}\big(\mathrm{ln}\,f(x)\big)^{g(x)} \\
&=f(x)^{g(x)}
\end{align*}

And we are done.

Title: Exponential function 1/x = e


问:
Vector Field, Curl and Divergence. Simple computing and proving:

So I went through (a) and showed that curl( v x F) = 2v. (doing cross products twice, which comes out pretty simple.)

Now i am stuck at (b), it says use the results from part (a) to compute:
line integral <2,3,-4> x F * dr..

how should i do (b) and (c)?

答: (c) the curl of a vector field has no divergence, or $\nabla \cdot (\nabla \times \textbf{F})=\textbf{0}$</p>


问:
Profit and loss question
A shopkeeper buys $240$ pears for $60$ dollars . Some of these pears are bad and they are thrown away. The remaining pears are sold at $35$ cents each , thus making a profit of $19.80$ dollars . Calculate the percentage of pears that are thrown away.

My workings -

Let $x$ be the number of pears thrown away

Selling price of remaining pears = $(240-X)(0.35) = (84 - 0.35x) $

Profit = selling price of the remaining pears - cost price

$19.80 = ( 84 - 0.35x)$ - cost price

I'm not sure what to put into the 'cost price of remaining pears' section for the remaining to find the value of 'X'

答:
The remaining pears are worth $35$ cents each. The profit from them was $79.80$ dollars (because we can add the $60$ dollars he paid. We can divide:

$$\frac{7980}{35} = 228$$

So they kept $228$ pears, and $240 - 228 = 12$ pears that were thrown away.

$12$ is $\frac{1}{20}$ of $240$, or $5$%.

So $\boxed{5\text{%}}$ of the pears were thrown away.

Title: Profit and loss question


问:
What is a simple formula that can create the patterns $+1 +2 +3 +4$, or $+1 +3 +5 +7$, or $+1 +4 + 7 +10$, etc.
Originally I was looking for some simpler more intuitive way to appreciate the value of squaring something. I looked for how much it was increasing by every time I add $+1$ to the size of what was being squared. so in other words $3$ squared is $+5$ more than $2$ squared, and $4$ squared is $+7$ more than $3$ squared, and $5$ squared is $9$ more than $4$ squared. You can see that whenever you increase the number being squared by $1$, the amount increased itself increases by $+2$ more than the previous increase. ie, the increase itself increases by a rate of $+ 2$.

So

$$- 1^2 - 0^2 = +1$$
$$- 2^2 - 1^2 = +3$$
$$- 3^2 - 2^2 = +5$$
$$- 4^2 - 3^2 = +7$$
$$- 5^2 - 4^2 = +9$$

and

the difference between the $+3$ and $+1$ is $2$

the difference between the $+5$ and $+3$ is $2$

the difference between the $+7$ and $+5$ is $2$

the difference between the $+9$ and $+7$ is $2$  etc...

So I say, it's "increase bonus" is at a rate of +2. That's definitely not a correct mathematical term, but in my ignorance that's what i'm calling it.

This struck me as very interesting, because I used to make little wc3 mods as a kid and they gave tools that made same pattern. What I became interested in, was what if I wanted it to increase by a pattern of say +1, +2, +3, +4, +5. (for an increase bonus of +1). Or what if I wanted it to be +1, +4, +7, +10, +13 (an increase bonus of +3)?

My main question is, is there a formula where I can easily change the "increase bonus" to whatever rate I want? While I can do this manually, I don't know how to represent it as a simple, calculable formula.

My side question is... what is this formula / branch of mathematics actually officially called?

答:
A polynomial of degree $n$ will have the $n$th finite difference equal to $n!$ times the leading coefficient of the polynomial.  In your example, you have
\begin{array}{|c|c|c|c|}
\hline
x & x^2 & \text{1st difference} & \text{2nd difference} \\
\hline
1 & 1 & & \\
2 & 4 & 3 & \\
3 & 9 & 5 & 2 \\
4 & 16 & 7 & 2 \\
5 & 25 & 9 & 2 \\
\hline
\end{array}

If you want the bonus to be $+3$ instead, you can use $\frac32x^2$ instead of $x^2$:
\begin{array}{|c|c|c|c|}
\hline
x & \frac32x^2 & \text{1st difference} & \text{2nd difference} \\
\hline
1 & \frac32 & & \\
2 & 6 & \frac92 & \\
3 & \frac{27}2 & \frac{15}2 & 3 \\
4 & 24 & \frac{21}2 & 3 \\
5 & \frac{75}2 & \frac{27}2 & 3 \\
\hline
\end{array}

Or, if you want instead the third finite difference to be $+3$, you can use $\frac12x^3:$

\begin{array}{|c|c|c|c|c|}
\hline
x & \frac12x^3 & \text{1st difference} & \text{2nd difference} & \text{3rd difference}\\
\hline
1 & \frac12 & & & \\
2 & 4 & \frac72 & & \\
3 & \frac{27}2 & \frac{19}2 & 6 & \\
4 & 32 & \frac{37}2 & 9 & 3\\
5 & \frac{125}2 & \frac{61}2 & 12 & 3\\
\hline
\end{array}


问:
Show that the only solution to $\phi(n) =n-2$ is $n=4$
Came across this question in Number Theory.

Let $\phi$ denote Euler's totient function;  Show that the only
  solution to $\phi(n) =n-2$ is $n=4$

My workings so far have included, firstly proving that $4$ is indeed a solution, I have also noted that the solution cannot be prime because $\phi(p)=p-1$.

From here, could I note that my solution must be composite, and the only way in which it could have only 2 numbers which are not coprime would be if n is a square number?

Not sure how to prove this completely though.

Also, could I incorporate the fact that  $\phi(p^k)=p^{k-1}(p-1)$ in any way?

All help greatly appreciated. Thanks.

答:
The equation $\phi(n) = n - 2$ means that there are $n - 2$ numbers in $\{1, 2, \dots, n\}$ that are relatively prime to $n$. In other words, there are exactly two numbers in $\{1, 2, \dots, n\}$ that are not relatively prime to $n$. One of them of course is $n$ itself. So there is exactly one other number, in $\{2, \dots, n-1\}$, which shares any factors in common with $n$.

Now as you observed, $n$ cannot be prime (because then there would be none). So $n$ is composite, and therefore has some prime factor $p$. This is one number with which $n$ shares a common factor, and if $n > 2p$, then $2p$ is another such number. So $n = 2p$. We now know that $n$ is even, i.e., $2$ is a prime factor $p$ of $n$, and from $n = 2p$, we get $n = 2\times2 = 4$.


问:
Let $A$ a Lebesgue measurable and its measure is 0, prove the Lebesgue measure of $A^2 = 0$
Let $A \subset \mathbb{R}$ and $A$ is bounded with $m(A)=0$. ($m$ is lebesgue measear). Set $E =\{x^2:x \in A \}$.

Prove $E$ is measurable and its measure is $0$.

答:
Let $\varepsilon >0$ and $((a_n,b_n))_{n\geqslant 0}$ such that $A\subset\bigcup_{n\geqslant 0}(a_n-b_n)$, the $(a_n,b_n)$ are pairwise disjoint and $\sum_{n\geqslant 0}(b_n-a_n)\leqslant\varepsilon$. Let $n\geqslant 0$, if $a_n\geqslant 0$ then $(a_n,b_n)^2=(a_n^2,b_n^2)$, if $b_n\leqslant 0$ then $(a_n,b_n)^2=(b_n^2,a_n^2)$ and if $a_n b_n\leqslant 0$, then $(a_n,b_n)^2=(0,\max(a_n^2,b_n^2))$. But $A^2\subset\bigcup_{n\geqslant 0}(a_n,b_n)^2$ and there is at most one $n$ such that $a_nb_n\leqslant 0$, if such a $n$ exists then $m((a_n,b_n)^2)=\max(a_n^2,b_n^2)=\max(|a_n|,|b_n|)^2\leqslant (b_n-a_n)^2\leqslant\varepsilon^2$. Thus if $M>0$ is such that $A\subset[-M,M]$, then $|a_n+b_n|\leqslant 2M$ for all $n$ and
$$ \sum_{n\geqslant 0}m((a_n,b_n)^2)\leqslant \varepsilon^2+\sum_{\substack{n\geqslant 0}}|b_n^2-a_n^2|\leqslant \varepsilon^2+2M\sum_{n\geqslant 0}(b_n-a_n)\leqslant(\varepsilon+2M)\varepsilon $$
Thus $m(A^2)=0$.


问:
Matrix Difference Equations Using Undetermined Coefficients?

$\begin{cases}x[n + 1] = -x[n] + 3 \\ y[n + 1] = -y[n] - e^{-n}\end{cases}$

Although I realize this system happens to be decoupled, I want to solve it using a general technique which doesn't depend on this contingency.  I first solve the homogeneous system, which has coefficient matrix $\begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$, which in turn has eigenvalue $-1$ with all of $\mathbb{R}^2$ as its eigenspace, so the homogeneous solution should be $\vec{x_h} = A(-1)^n\begin{bmatrix}1 \\ 0\end{bmatrix} + B(-1)^n\begin{bmatrix}0 \\ 1\end{bmatrix}$.

Based on how undetermined coefficients works for systems of ODEs, I guess that $\vec{x_p} = \vec a + \vec b e^{-n}$ will be the form of the particular solution.  Then the system becomes $\vec a + \vec b e^{-(n + 1)} = \begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}(\vec a + \vec b e^{-n}) + \begin{bmatrix}3 \\ -e^{-n}\end{bmatrix}$.  Expanding, $\begin{bmatrix}a_1 + \frac{b_1}{e} e^{-n} \\ a_2 + \frac{b_2}{e} e^{-n}\end{bmatrix}$ = $\begin{bmatrix}3 - a_1 - b_1 e^{-n} \\ -a_2 - b_2 e^{-n} - e^{-n}\end{bmatrix}$, leading to $\begin{cases} a_1 = 3 - a_1 \\ \frac{b_1}{e} = -b_1 \\ a_2 = -a_2 \\\frac{b_2}{e} = -b_2 - 1\end{cases}$ with solution $a_1 = \frac{3}{2}, b_1 = 0, a_2 = 0, b_2 = \frac{e}{e + 1}$, and finally the general solution $\vec x = A(-1)^n\begin{bmatrix}1 \\ 0\end{bmatrix} + B(-1)^n\begin{bmatrix}0 \\ 1\end{bmatrix} + \begin{bmatrix}\frac{3}{2} \\ \frac{1}{e^n + e^{n - 1}}\end{bmatrix}$.

I couldn't make this match Wolfram's answer exactly, and the graphs aren't identical.  There isn't much information on this topic on Youtube, etc., so I just did what felt natural based on solving matrix ODEs.
Does this undetermined coefficients technique from matrix ODEs transfer over to matrix difference equations in the above manner, as one would expect?  If so, did I make the correct particular solution guess (a sum of the forms one would guess for each individual equation in the system)?  I wasn't sure if the $3$ forcing term would "resonate" with the $(-1)^n$ characteristic mode, requiring an extra term in the guess, but $(-1)^n$ technically isn't constant, and the coefficient system turned out consistent, so I suppose this should be unnecessary.  Where did I go wrong?

答:
$\def\A{{\bf A}}
\def\B{{\bf B}}
\def\C{{\bf C}}
\def\M{{\bf M}}
\def\I{{\bf I}}
\def\O{{\bf 0}}
\def\x{{\bf x}}
\def\y{{\bf y}}$There is a minor sign error in your work. Note that $b_2 = -e/(1+e)$. Taking this into account, you will arrive at the correct solution.
Below we consider the technique in a fairly general way.

Consider a vector of discrete functions $\y_n$ such that
\begin{align*}
\y_{n+1}-\y_n = \M\y_n,\tag{1}
\end{align*}
where $\M$ does not depend on $n$.
For example, if
$$\y_n = \left[\begin{array}{cccccc}
1&n&n^2&n^3&n^4&\cdots
\end{array}\right]^T$$
then such an $\M$ exists,
$$\M = \left[\begin{array}{cccccc}
0&0&0&0&0&\cdots \\
1&0&0&0&0&\cdots \\
1&2&0&0&0&\cdots \\
1&3&3&0&0&\cdots \\
1&4&6&4&0&\cdots \\
\vdots& & & & &\ddots
\end{array}\right]$$
If
$$\y'_n = \left[\begin{array}{cccccc}
r^n&nr^n&n^2r^n&n^3r^n&n^4r^n&\cdots
\end{array}\right]^T$$
then
$$\M' = \left[\begin{array}{cccccc}
r-1&0&0&0&0&\cdots \\
r&r-1&0&0&0&\cdots \\
r&2r&r-1&0&0&\cdots \\
r&3r&3r&r-1&0&\cdots \\
r&4r&6r&4r&r-1&\cdots \\
\vdots& & & & &\ddots
\end{array}\right]$$
We can consider combinations of such discrete functions such as
$$\y_n = \left[\begin{array}{cccc}
1&n&r^n&nr^n \\
\end{array}\right]^T$$
which results in a block diagonal form for $\M$,
$$\M = \left[\begin{array}{cccccc}
0&0&0&0 \\
1&0&0&0 \\
0&0&r-1&0 \\
0&0&r&r-1
\end{array}\right]$$

Suppose that we wish to solve
\begin{align*}
\x_{n+1} &= \A\x_{n} + \B\y_n\tag{2}
\end{align*}
where $\A,\B$ are constant matrices and where $\y_n$ is a discrete function satisfying (1).
We search for homogeneous and particular solutions, $\x_n^h,\x_n^p$, such that
\begin{align*}
\x_n &= \x_n^h + \x_n^p.
\end{align*}
Clearly $\x_n^h = \A^n \x_0^h$.
Assume that $\x_n^p = \C\y_n$ where $\C$ is independent of $n$.
This implies that
$$(\C\M+(\I-\A)\C-\B)\y_n = \O.$$
We assume that the elements of $\y_n$ are independent discrete functions of $n$.
Thus, we must solve
\begin{align*}
\C\M+(\I-\A)\C&=\B.\tag{3}
\end{align*}
This can be solved using the vec trick, for example.
A unique solution will exist if $\M$ and $\A-\I$ share no eigenvalues.

For the original question we have
\begin{align*}
\y_n &= \left[\begin{array}{c}
1 \\
e^{-n}
\end{array}\right] \\
\M &= \left[\begin{array}{cc}
0&0\\
0&\frac 1 e -1
\end{array}\right] \\
\A &= -\I \\
\B &= \left[\begin{array}{cc}
3&0\\
0&-1
\end{array}\right]
\end{align*}
Equation (3) becomes
\begin{align*}
\left[\begin{array}{cc}
2c_{11} & \frac{1+e}{e}c_{12} \\
2c_{21} & \frac{1+e}{e}c_{22}
\end{array}\right]
&= \left[\begin{array}{cc}
3&0\\
0&-1
\end{array}\right]
\end{align*}
which is easily solved, with the result
\begin{align*}
\C &=
\left[\begin{array}{cc}
\frac32 & 0 \\
0 & -\frac{e}{1+e}
\end{array}\right]
\end{align*}
Thus,
\begin{align*}
\x_n &= \left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right]^n
\left[\begin{array}{c}
x_0^h \\
y_0^h
\end{array}\right]
+ \left[\begin{array}{cc}
\frac32 & 0 \\
0 & -\frac{e}{1+e}
\end{array}\right]
\left[\begin{array}{c}
1 \\
e^{-n}
\end{array}\right] \\
&= \left[\begin{array}{c}
(-1)^n x_0^h + \frac32 \\
(-1)^n y_0^h - \frac{e}{1+e}e^{-n}
\end{array}\right]
\end{align*}


问:
Non-degenerate sub $C^*$-algebra of the compact operators.
Let $B_0(H)$ be the compact operators on the Hilbert space $H$ and let $B \subseteq B_0(H)$ a $C^*$-subalgebra that acts non-degenerately on $H$. Let $\{p_i: i \in I\}$ be a maximal family of pairwise orthogonal minimal projections. The existence of this family is ensured by Zorn's lemma. Is it true that
$$H = \bigoplus_i p_i (H)$$

Attempt:

Write $K$ for the direct sum. If $K$ is a proper subspace, we may fix a non-zero $\xi \in K^\perp$. The idea is now to use this vector to construct a minimal projection that will contradict maximality, but I'm not sure how to use the non-degeneracy to do this.

答:
Let
$
  p = \sum_{i\in I}p_i,
  $
where the sum is known to converge strongly.  Then, for every  $a$ in $B$, notice that
$$
  pa = \sum_{i\in I}p_ia,
  $$
where the convergence is now in norm since right-multiplication by compact operators turns strong convergence (of
bounded nets) into norm
convergence.  Therefore $pa\in B$.

Now, assuming by contraction that $p\neq 1$, and
since $B$ is non-degenenerate, there must be some element $a$ in $B$ such that $(1-p)a\neq 0$, and hence also that
$$
  c:= (1-p)aa^*(1-p) \neq 0.
  $$
Since $c$ is a self-adjoint compact operator in $B$, any one of its spectral projections,  say $q$,  also lie in $B$.
It is also clear that $q\perp p_i$ for every $i$, and since $q$ is necessarily finite dimensional, we may choose a
minimal projection among the projections of $B$ dominated by $q$.  The existence of  such a projection then contradicts
the maximality of the originaly chosen family.


问:
Calculate max/min of $x_1 x_2+y_1 y_2+z_1 z_2+w_1 w_2$
What is a good way to calculate max/min of
$$x_1 x_2+y_1 y_2+z_1 z_2+w_1 w_2$$
where $x_1+y_1+z_1+w_1=a$ and $x_2+y_2+z_2+w_2=b$ and $x, y, z, w, a, b \in \mathbb{N} \cup \{0 \}$, and please explain your answer (how your result comes out).</p>

答: I take it your natural numbers start at $1$. To maximize, let $x_1=a-3$, $x_2=b-3$, all other variables $1$. To minimize, $x_1=a-3$, $y_2=b-3$, all other variables $1$. </p>


问:
Particular solution of $r^2y'' + 2ry' - 6y = -ar^2$ when finite at 0
I am trying to find a specific solution for  the ODE: $$r^2y'' + 2ry' - 6y = -ar^2$$
with a being a positive constant. The complementary solution I have found is $$y^h=c_1r^2+c_2/r^3$$
but in this case, I know that $c_2 = 0$ because the solution is known to be finite at $r=0$, so that $$y^h=c_1r^2$$
However, I cannot seem to be able to find a complementary solution. Using an online ODE solver that does not offer step by step instructions, I found that the particular solution is
$$-\frac{ar^2\ln \left(r\right)}{5}+\frac{ar^2}{25}$$
but this solution is problematic since it results in $y$ not being finite at $r=0$.

I would really appreciate some further insight on obtaining the particular solution of this ODE

答:
With $y_h(r) = \frac{c_1}{r^3}$, assuming as a particular solution

$$
y_p = \frac{c_1(r)}{r^3}
$$

(Variation of constants method) after substitution we have

$$
r c''_1(r)-4c'_1(r)+ar^4=0
$$

now calling $u(r) = c_1'(r)$ we follow with

$$
r u'(r) - 4u(r) + a r^4 = 0
$$

here $u_h(r)$ is easily obtained giving

$$
u_h(r) = c_2 r^4
$$

now applying the same procedure making $u_p(r) = c_2(r)r^4$ and substituting we obtain easily

$$
u(r) = c_2r^4-a r^4\ln r = c'_1(r)
$$

and after integration we have

$$
c_1(r) = \frac{a r^5}{25}-\frac{1}{5} a r^5 \ln (r)+\frac{c_2 r^5}{5}
$$

and finally choosing $c_2=0$ (it is a particular solution)

$$
y_p(r) = \frac{c_1(r)}{r^3} = \left(\frac{a r^5}{25}-\frac{1}{5} a r^5 \ln (r)\right)\frac{c_1}{r^3}
$$


问:
Proving $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\ldots\left(1+\frac{1}{n^3}\right)<3$ for all positive integers $n$

Prove that $\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\ldots\left(1+\dfrac{1}{n^3}\right)<3$   for all positive integers $n$

This problem is copied from Math Olympiad Treasures by Titu Andreescu and Bogdan Enescu.They start by stating that induction wouldn't directly work here since the right hand side stays constant while the left increases.They get rid of this problem by strengthening the hypothesis.$$\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\ldots\left(1+\dfrac{1}{n^3}\right)\le3-\dfrac{1}{n}$$
and then proceed by induction.

The problem is that I can't find a motivation for the above change.I mean,we could have subtracted a lot of things from the RHS but what should nudge us to try $\dfrac{1}{n}$?The rest of the proof is quite standard,but I can't see how I am supposed to have thought of it.Is it just experience?Or is it a standard technique?A little guidance and motivation will be appreciated.

答:
Just to give a different approach (similar to Jack D'Aurizio's), note that, for $k\gt1$, we have

$$\ln\left(1+{1\over k^3}\right)\lt{1\over k^3}\lt{1\over k^3-k}={1\over2}\left({1\over k-1}-{2\over k}+{1\over k+1}\right)$$

It follows that

$$\begin{align}
\ln\left(\left(1+{1\over1^3}\right)\left(1+{1\over2^3}\right)\left(1+{1\over3^3}\right)\cdots\right)
&=\ln2+\sum_{k=2}^\infty\ln\left(1+{1\over k^3}\right)\\
&\lt\ln2+{1\over2}\sum_{k=2}^\infty\left({1\over k-1}-{2\over k}+{1\over k+1}\right)\\
&=\ln2+{1\over2}\left({1\over2-1}-{2\over2}+{1\over3-1} \right)\quad\text{(after telescoping)}\\
&=\ln2+{1\over4}\\
&={1\over4}\ln(16e)\\
&\lt{1\over4}\ln81\\
&=\ln3
\end{align}$$


问:
Are these 2 graphs isomorphic?

They meet the requirements of both having an $=$ number of vertices ($7$).

They both have the same number of edges ($9$).

They both have $3$ vertices of degree $2$ and $4$ of degree $3$.

However, graph two has $2$ simple circuits of length $3$ whereas graph one has only $1$ of length $3$.

Is this not a valid method for checking isomorphism?

My guide says that these two figures are isomorphic.

答: Yes. In the second graph, slide node 4 out of the square and observe that nodes 1,2,3,7,4,5 now form a hexagon, with node 6 inside it. Node 6 now corresponds to node D. For rigorous details, see Daniel Rust's answer. You can check that his construction is indeed an isomorphism by applying the definition: that for any $a$, $b$ in the first graph, $f(a)$ is connected to $f(b)$ if and only if $a$ is connected to $b$.

Title: Are these 2 graphs isomorphic?


问:
Limit proof for a function
Regarding: If there exists an $L∈R$ such that $lim_{x→a}f(x)=L$ for every $a∈R$, then $f(x)=L$ for every $x∈R$

I would like to make a counter-example for the question.

I chose a function, and want to show that it has a Limit:

$f(x) = \begin{cases}L&x\ne a\\L/2 &x=a\end{cases}$

I started with:

Let $ϵ>0$. We choose $δ= $

Let $x$ such that $0<|x−a|<δ $

So $|f(x)-L|< $

I don't know how to chose δ.

I am looking for something like $|a-x|/2$ , but I am a bit confused...

Any help will be awesome!

答:
Since $|x-a|>0$ you know $x\neq a$, hence $f(x)=L$. Thus $|f(x)-L|=0<\epsilon$.

In summary, any choice of $\delta$ is sufficient.

Edit: As Bungo pointed out, this is just the first part of proving the counterexample. You need also to prove $\displaystyle\lim_{x\to b}f(x)=L$ for every $b\in\mathbb R$, not just the special case $b=a$. I believe $\delta=|b-a|$ should work.

Title: Limit proof for a function


问:
Is this improper integral answer correct?
So I'm working on improper integrals and con/divergence and want some assurance that I've  done the following correctly.

$\int^∞_{-∞}cos(\pi t)$

As far as I'm aware this is convergent if and only if
$\int^∞_{0}cos(\pi t)$ and
$\int^0_{-∞}cos(\pi t)$ are both convergent.

So starting by working out the indefinite integral...

$\int cos(\pi t)$ = $\frac{1}{\pi}sin(\pi t)+C $

So if we work on the indefinite integral $\int^0_{-∞}cos(\pi t)$, we have...

$\int^0_{-∞}cos(\pi t)$ = $lim_{t->-∞}\int^0_t cos(\pi t)$

= $lim_{t->-∞}\frac{1}{\pi}[sin(\pi t)]^0_t$

= $lim_{t->-∞}\frac{1}{\pi}[0-sin(\pi t)]$

= $lim_{t->-∞}\frac{-sin(\pi t)}{\pi}$

Since the sin function is periodic and doesn't tend towards a limit as t approaches negative infinity, then the limit does not exist. Hence the integral is divergent.

By this logic, the full integral
$\int^∞_{-∞}cos(\pi t)$
is also divergent and we don't need to bother calculating or testing the integral for positive infinity.

I'm still a little unsure about this topic so I'm looking for reassurance as to if I am correct, or if I have done something wrong?

答:
let us take an odd function, that is $f(-x)=-f(x)$. Now define the integral:
$$I(n)=\int_{-n}^nf(x)dx=\int_{-n}^0f(x)dx+\int_0^nf(x)dx$$
$$I(n)=\int_n^0f(-x)(-dx)+\int_0^nf(x)dx$$
$$I(n)=\int_n^0f(x)dx+\int_0^nf(x)dx$$
$$I(n)=-\int_0^nf(x)dx+\int_0^nf(x)dx=0$$
so as $n\to\infty$ ,$I(n)=0$ even if over the positive or negative regions the integral is not convergent, however for the case of an even function these positive and negative regions do not "cancel" out and so it is required that both the positive and negative regions converge independently


问:
The shortest distance between a vertex in a simplex and a non-adjacent face
Provided a simplex defined by four vertices $(v_1,v_2,v_3,v_4)$, and known edge lengths, for a given vertex $v_i$, how can I calculate the shortest distance between $v_i$ and the non-adjacent face of the simplex?  To provide an example, if $v_i=v_1$, I would want to calculate to shortest distance to the triangle defined by the vertices $(v_2,v_3,v_4)$.</p>

答:
If you drop a perpendicular from $v_1$ to the plane spanned by $v_2, v_3, v_4$, it will land either within the triangle $T=v_2v_3v_4$ or outside it. If it lands within, all three of the other faces of the simplex have acute angles adjacent to $T$, while if it lands outside, two of them will have an obtuse angle adjacent to $T$. You can therefore use the law of cosines to distinguish between these two possibilities while knowing only the side lengths of your simplex.

If it lands within the triangle:

The shortest distance from $v_1$ to the opposite face is the length of the perpendicular: i.e., an altitude of the simplex. So it is given by $3V/A$, where $A$ is the area of the opposite face and $V$ the volume of the simplex. If you know all the side lengths of the simplex, you can compute $A$ via Heron's formula, and $V$ via its 3-dimensional generalization.

If it lands outside the triangle:

The shortest distance from $v_1$ to the opposite face will either be along an edge or across the acute face. The shortest distance across a face is given by $2A/b$, where $A$ is the area of that face (again attainable via Heron's formula) and $b$ the length of the opposite edge. So compare this to each of the three edge lengths; the answer you want will be the shortest of those four quantities.


问:
I'm confused in what to do to demonstrate downard concavity on an interval.
Let $f(x) = -x^4 + 40x^3 - 400x^2 + 1000x$ which we know $f''(x) = -12x^2 + 240x - 800$. We have an interval $[0,4]$.

I know that if $f''(x) \leq 0 \ \forall x$ then $f(x)$ is downward concave. However it seems that I can't just compute $f''(0)$ and $f''(4)$ and call it a day. It seems tedious or down right impossible to compute the second derivative for every point in the interval. What should I do instead?

答:
$$(x-10)^2 = x^2 - 20 x + 100$$
$$-12(x-10)^2 = -12x^2 + 240 x - 1200$$

$$ 400-12(x-10)^2 = 400-12x^2 + 240 x - 1200 =-12x^2 + 240 x - 800 $$

You want to know when $ 400-12(x-10)^2 = 0,$ also when it is positive and when negative

If
$$ 400 = 12(x-10)^2 $$
then
$$\frac{400}{12} = \frac{100}{3} = (x-10)^2 $$
so that
$$ x-10 = \pm \frac{10}{\sqrt 3}  =  \pm \frac{10 \sqrt 3}{ 3}  $$
$$ x = 10  \pm \frac{10 \sqrt 3}{ 3}  $$

To find out about positive or negative, just check what you get for $x=10$ ,
that gives the sign in the middle interval. For outside, we know the larger root is below $20,$ the smaller root is above $0,$  so check at $x=20$ and at $x=0$


问:
differential of homogeneous function
I am trying to calculate the dy of:

$y+xy=x$

I know the formula $\frac{dy}{dx}=f(x)'$

My logic says to do so: $y(1+x)=x$ ->$y=\frac{x}{\left(1+x\right)}$

$dy=(\frac{x}{\left(1+x\right)})'dx$ -> $dy=\frac{1}{\left(1+x\right)^2}dx$

But my textbook has the answer:

$dy=\frac{1-y}{1+x}dx$

Where am I wrong?

Thank you

答:
Same as your answer...

$\frac{1}{(1+x)^2}=\frac{1}{1+x} \cdot \frac{1}{1+x}$

and in the equation $y+xy=x$ we have:

Subtract $xy$ on both sides: $y=x-xy$

Factor the right hand side: $y=x(1-y)$

Divide both sides by $1-y$: $\frac{y}{1-y}=x$

So we can write your answer as:

$\frac{1}{(1+x)^2}=\frac{1}{1+x} \cdot \frac{1}{1+x}$

$\frac{1}{(1+x)^2}=\frac{1}{1+\frac{y}{1-y}} \cdot \frac{1}{1+x}$

$\frac{1}{(1+x)^2}=\frac{1-y}{1-y+y} \cdot \frac{1}{1+x}$ (I multiplied first fraction in product by $1-y$)

$\frac{1}{(1+x)^2}=\frac{1-y}{1} \cdot \frac{1}{1+x}$

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

--------------Also if you wanted the book's answer------------

Just differentiate as is instead of putting it in explicit form and then differentiating.
$y+xy=x$

$y'+xy'+1y=1$ I did the $xy$ term using product rule...

Now let's solve for $y'$

$y'(1+x)+y=1$

$y'(1+x)=1-y$

$y'=\frac{1-y}{1+x}$

Title: differential of homogeneous function


问:
How do I prove $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$?
How do I prove this inequality with sequential Fibonacci numbers: $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$ (i.e. $\frac{5}{3}>\frac{3}{2}$)? I know these formulas alternately overestimate and underestimate the golden ratio as $n$ gets large.

Body:
$$\begin{align*}
f_{2n+1}f_{2n-1} - f_{2n}^2
&= \det\pmatrix{f_{2n+1}&f_{2n}\\f_{2n}&f_{2n-1}}\\
&= \det\pmatrix{f_{2n+1}-f_{2n}&f_{2n}\\f_{2n}-f_{2n-1}&f_{2n-1}}&&C_1 \to C_1-C_2\\
&= \det\pmatrix{f_{2n-1}&f_{2n}\\f_{2n-2}&f_{2n-1}}\\
&= \det\pmatrix{f_{2n-1}&f_{2n}-f_{2n-1}\\f_{2n-2}&f_{2n-1}-f_{2n-2}} && C_2 \to C_2-C_1\\
&= \det\pmatrix{f_{2n-1}&f_{2n-2}\\f_{2n-2}&f_{2n-3}}\\
&= \vdots\\
&= \det\pmatrix{f_{3}&f_{2}\\f_{2}&f_{1}}\\
&= \det\pmatrix{2&1\\1&1}\\
&= 1\\
&> 0\\
\frac{f_{2n+1}}{f_{2n}}&> \frac{f_{2n}}{f_{2n-1}}
\end{align*}$$

(for $n \ge 1$, so that $f_{2n-1}$ and $f_{2n}$ are both positive)


问:
Is $w$ a vector space?
Let $W$ be the set of all solutions$(X,y,i,r)$ such that $a+b=m^2$. is $w$ a vector space?
Can anybody do the whole thing for me and explain shortly every step? I have to do this kind of lots of question today.</p>

答: Two answers have already shown that your $W$ is not a vector space because it's not closed under addition.  It seems easier to reach the same conclusion by showing it's not closed under multiplication by scalars, say by $2$, or by $-1$.</p>

Title: Is $w$ a vector space?


问:
$|2-|1-x||=|x^2-1|$ equation, incorrect solution

When I solved this equation I got $x_1 = -1$, $x_2 = 0$ and $x_3 = \frac{-1+\sqrt{17}}{2}$

I solved it like this: I first sketched a graph and I could see that the first two solutions were trivial, I got the third one by solving $x^2-1 = -(x-3)$ for $x > 1$, since I could see from my graph that there was an intersection.

Wolfram alpha only displays the first two solutions and doesn't mention the third one, is my solution to the whole problem incorrect?
I also have a question, how do you view if someone solves a problem by sketching a graph and solves it from there and doesn't do all the required calculation work to show that the expression inside every absolute value can be negative and what happens if it is positive and does everything from there.

Thank you for your answers.

答:
Alpha does all computations in the complex and it turns out that the solution includes a whole locus with a terribly complicated expression (because the equation is equivalent to a quartic one). But that locus certainly includes the third real root.

Now the following Alpha request will give you the three real roots: https://www.wolframalpha.com/input/?i=real+roots+of+abs%282-abs%28x-1%29%29-abs%28x%5E2-1%29


答: Your method is very good but it is not considered a rigorous proof. You can use the graph to turn this into a rigorous proof. We see on the graph that there are only three cases to study: $x\lt -1$ and  $-1\le x \le 1$ and $1 \le x$. Each of these three cases is very easy to study.

Title: $|2-|1-x||=|x^2-1|$ equation, incorrect solution

Body:
When I solved this equation I got $x_1 = -1$, $x_2 = 0$ and $x_3 = \frac{-1+\sqrt{17}}{2}$

I solved it like this: I first sketched a graph and I could see that the first two solutions were trivial, I got the third one by solving $x^2-1 = -(x-3)$ for $x > 1$, since I could see from my graph that there was an intersection.

Wolfram alpha only displays the first two solutions and doesn't mention the third one, is my solution to the whole problem incorrect?
I also have a question, how do you view if someone solves a problem by sketching a graph and solves it from there and doesn't do all the required calculation work to show that the expression inside every absolute value can be negative and what happens if it is positive and does everything from there.

Thank you for your answers.


问:
Derive the Newton-cotes formula for $\int f(x) dx$ using nodes $o, \frac13, \frac23, \1$.
Can someone please let me know if I am on the right track here. Obviously, I still need to perform the integrations as well as evaluate the integrations(from 0 to 1) to determine the coefficients. What I gather from my notes, and the text book, this is the procedure for equally spaced nodes but I'm not sure as there are several methods (trap. rule, Simpson's rule, composite trap. rule, etc. Any advice would be greatly appreciated!!

$\int^1_0 p_3(x)dx = f(a) \int^1_0\frac{(x-\frac13)(x-\frac23)(x-1)}{(0-\frac13)(0-\frac23)(0-1)}dx + f(b)\int^1_0\frac{(x)(x-\frac23)(x-1)}{(\frac13-0)(\frac13-\frac23)(\frac13-1)} + f(c)\int^1_0\frac{(x)(x-\frac13)(x-1)}{(\frac23-0)(\frac23-\frac13)(\frac23-1)} + f(d)\int^1_0\frac{(x)(x-\frac13)(x-\frac23)}{(1-0)(1-\frac13)(1-\frac23)}$

答:
You can crunch those integrals, or you can reason that the approximation must be exact for $f = 1, x, x^2, x^3$ and get a linear system for the weights:
Write the approximation as
$$\int_{0}^{1}p_3(x)\,dx = \sum_{i = 1}^{4}w_if(x_i)$$
where $x_1 = 0$, $x_2 = 1/3$, $x_3 = 2/3$, $x_4 = 1$ and the $w_i$ are to be determined. The approximation must be exact for $f = 1, x, x^2, x^3$, so that gives the system of equations
$$
\begin{bmatrix}
1 & 1 & 1 & 1 \\
x_1 & x_2 & x_3 & x_4 \\
x_1^2 & x_2^2 & x_3^2 & x_4^2 \\
x_1^3 & x_2^3 & x_3^3 & x_4^3 \\
\end{bmatrix}
\begin{bmatrix}
w_1 \\
w_2 \\
w_3 \\
w_4 \\
\end{bmatrix}
=
\begin{bmatrix}
1 \\
1/2 \\
1/3 \\
1/4 \\
\end{bmatrix}
$$
which you can solve to find the $w_i$. Both your approach and this approach can generate the weights for any choice of distinct nodes; there is nothing special about equally spaced nodes.

Edit on why it should be exact on $1, x, x^2, x^3$:

The integration method is based on calculating the interpolation polynomial $p_3(x)$ and integrating it. When you interpolate $4$ points from a function $f$ that is a polynomial of degree $3$ or less, the interpolation polynomial $p_3$ matches $f$ identically (by uniqueness of the interpolating polynomial).

One way to see why exactness makes the approximation good is to consider the approximation on a small interval $[-h, h]$:
$$\int_{-h}^{h}f(x)\,dx \approx 2h\sum_{i = 1}^{N}w_if(x_ih).$$
Define the error in the approximation for a function $f$ as
$$E(f) := \int_{-h}^{h}f(x)\,dx - 2h\sum_{i = 1}^{N}w_if(x_ih)$$
Note that $E$ is a linear map. Writing $f$ as a Taylor series around $0$ gives
\begin{align}
E(f) &= E\left(\sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n\right) \\
&= \sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}E(x^n)
\end{align}
You can verify that $E(x^n) = O(h^{n + 1})$. Hence if $k$ is the smallest integer for which $E(x^k) \neq 0$, then the error is
$$E(f) = \frac{f^{(k)}(0)}{k!}E(x^k) + O(h^{k + 2}) = O(h^{k + 1}).$$
So exactness on polynomials increases the accuracy of the approximation on all (smooth enough) functions.


问:
closedness of a subset of a bipolar set
I do not have a very strong knowledge of bipolar sets and all this stuff. Thus it could be that the question is rather easy. However I was not able to prove by myself the closedness of the following set:

We are looking at the space $L^\infty(\Omega,\mathcal{A},P)$ with a probability measure $P$. Then it is well known that the space $ba(P)$ of bounded finitely additive signed measures on $(\Omega,\mathcal{A})$, which are absolutely continuous with respect to $P$ is the dual space of $L^\infty$. Suppose $C$ is a set in $L^\infty$, which is a convex cone containing $0$ and norm closed. Therefore it is also $\sigma(L^\infty,ba)$-closed. With $C^\circ$ we denote the polar cone of $C$, which is convex and $\sigma(ba,L^\infty)$-closed. Now I'm interested in the following set:

$$K:=\{\mu\in C^\circ:\mu(\Omega)=1\}$$

This set is clearly convex, but why is it again closed in $\sigma(ba,L^\infty)$? Thanks in advance for your help.

答:
Lemma. Let $x\in X$ then the evaluation map
$$
\operatorname{ev}_x:X^*\to\mathbb{C}:f\mapsto f(x)
$$
is weak-$^*$ continuous.

Proof. Fix $\varepsilon>0$ and $f\in X^*$. Consider weak-$^*$ open set $U_{f,\varepsilon}=\{g\in X^*:|g(x)-f(x)|<\varepsilon\}$. Obviously for all $g\in U_{f,\varepsilon}$ we have
$|\operatorname{ev}_x(g)-\operatorname{ev}_x(f)|=|g(x)-f(x)|<\varepsilon$
Hence $\operatorname{ev}_x$ is weak-$^*$ continuous.

For your particuar case evaluation map have the form
$$
\operatorname{ev}_x:ba\to\mathbb{C}:\mu\mapsto\int_\Omega x(\omega)d\mu(\omega)
$$
where $x\in L_\infty$.
Consider set $S=\{\mu\in ba:\mu(\Omega)=1\}$. Since $\operatorname{ev}_x$ is weak-$^*$ continuous, then the set $S=(\operatorname{ev}_{\chi_\Omega})^{-1}(\{1\})$ is weak-$^*$ closed as preimage of closed set $\{1\}\subset\mathbb{C}$. Thus the set $K=C^o\cap S$ is weak-$^*$ closed as intersection of two closed sets.

Title: closedness of a subset of a bipolar set


答:
The characteristic equation $^2−−1=0$ has two roots $\varphi=\frac{1+\sqrt 5}{2}, \psi=\frac{1-\sqrt 5}{2}$. Vieta’s formulas give $\varphi+\psi=1, \varphi \psi = -1$. Then we can rewrite $f_{n+1} = f_n + f_{n-1}$ as
$$f_{n+1}−\varphi f_=\psi (f_−\varphi f_{-1}).$$

Note that $\psi<0$. Then
$$f_{+1}−\varphi f_=\psi^ (f_1− \varphi f_0 )=\psi^ \begin{cases} >0 & \text{ if } n \text{ is even;} \\ <0 & \text{ if } n \text{ is odd.} \end{cases}$$
Hence
$$
\frac{f_{2n+1}}{f_{2n}}>\varphi >\frac{f_{2n}}{f_{2n-1}}
$$

Title: How do I prove $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$?

Body: How do I prove this inequality with sequential Fibonacci numbers: $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$ (i.e. $\frac{5}{3}>\frac{3}{2}$)? I know these formulas alternately overestimate and underestimate the golden ratio as $n$ gets large.


问:
N distinct elements, is a permutation or all possible subsets larger?
Given N distinct elements (where N is large), how do I determine if a permutation of it is larger, or if all possible subsets is larger?

I eyeballed it and it seems like the permutations would be bigger. It seems like this subset is largest: N choose N/2, but this subset is kind of close to $2^{N/2}$, which is clearly a lost smaller than $N!$. Not sure how to prove this though

答:
If you mean whether the size of the set of all possible permutations of an N-element set is bigger than the size of the set of all possible subsets, i.e. whether $N!>2^N$ then the answer is yes, if $N>3$. That it is wrong for $N=1,2,3$ is a simple computation. The actual statement follows by induction:

The induction start: we take $N=4$ and obtain $N!=24>16=2^4$

Now suppose the statement holds for $N-1$, then since clearly $N>2$ we have $$N!=N(N-1)!>2(N-1)!>2\cdot 2^{N-1}=2^N$$ where the last inequality follows from the induction hypothesis.


答: The orders are $2^n$ and $n!$.  But $n!\gt 2^n$ for $n\gt3$, by inspection.  So there are more permutations, in this case.

Title: N distinct elements, is a permutation or all possible subsets larger?

Body:
Given N distinct elements (where N is large), how do I determine if a permutation of it is larger, or if all possible subsets is larger?

I eyeballed it and it seems like the permutations would be bigger. It seems like this subset is largest: N choose N/2, but this subset is kind of close to $2^{N/2}$, which is clearly a lost smaller than $N!$. Not sure how to prove this though


问:
Inequality using Cauchy Schwarz
I’ve recently been looking into applications of the Cauchy Schwarz inequality.

I’ve seen it stated in the form:

$(a_1^2 + a_2^2 + … + a_n^2)(b_1^2 + b_2^2  + … + b_n^2) \ge (a_1b_1 + a_2b_2 + … + a_nb_n)^2$

However I’m struggling to see how it could prove this cyclic sum inequality:

$$\sum_\text{cyc} \frac{x^2}{y + z} \ge \frac{(x+y+z)^2}{2(x+y+z)}$$

Where $xyz = 1$ and $x,y,z$ $\epsilon$ $\mathbb{R}$

Any help?

答:
Assuming $x, y, z \geq 0$, the Cauchy-Schwarz inequality gives

$$
\begin{align*}
&\left((y + z) + (z + x) + (x + y)\right)\left(\frac{x^2}{y + z} + \frac{y^2}{z + x} + \frac{z^2}{x + y}\right) \\
&\geq \left(\sqrt{y + z}\sqrt{\frac{x^2}{y + z}} + \sqrt{z + x}\sqrt{\frac{y^2}{z + x}} + \sqrt{x + y}\sqrt{\frac{z^2}{x + y}}\right)^2 \\
&= (x + y + z)^2
\end{align*}
$$

and you can divide out $(y + z) + (z + x) + (x + y) = 2(x + y + z)$ on both sides to get your inequality.

Title: Inequality using Cauchy Schwarz


问:
get the distribution function
I have the following density function:
$ f(x) = \left\{
  \begin{array}{l l}
    cx^2+|x| & \quad \text{if -1/2<x<1/2}\\
    0 & \quad \text{otherwise}
  \end{array} \right.$

we know that $\int_{-\infty}^{\infty}{f(x)dx = 1}=\int_{-1/2}^{1/2}{cx^2+|x|dx = 1}$ and I get $c=9$.

The distribution function should be:
$ F(x) = \left\{
  \begin{array}{l l}
    3x^3-x^2/2 & \quad \text{if -1/2<x<0}\\
    3x^3+x^2/2 & \quad \text{if 0 <= x<1/2}\\
    0 & \quad \text{otherwise}
  \end{array} \right.$

If i want to test if $F(x)->1, \text{as} \space x->\infty$ it is not $1$, is the distribution function right?

答:
The cdf is not right. We will use the fact that
$$F(x)=\int_{-\infty}^x f(t)\,dt,$$
where $f(t)$ is the density function.

Thus $F(x)=0$ for $x\lt -\frac{1}{2}$.

For $-\frac{1}{2}\le x\le 0$, we have
$$F(x)=\int_{-1/2}^x (9t^2-t)\,dt.$$
The antiderivative was computed correctly, but the substitution of endpoints was not. Once you compute an antiderivative $G(t)$, you need to find $G(x)-G(-1/2)$. You should end up with $\frac{1}{2}$ more than what you actually got.

For $0\lt x\le \frac{1}{2}$,
$$F(x)=\frac{1}{2}+\int_0^x (9t^2+t)\,dt.$$
Finally, $F(x)=1$ for $x\gt \frac{1}{2}$.

Title: get the distribution function


问:
Endomorphism ring of elliptic curve over $\overline{\mathbb{F}_p}$
I want to prove that the endomorphism ring of elliptic curve $E$ over $\overline{\mathbb{F}_p}$ is not isomorphism to $\mathbb{Z}$.

We can find a $q=p^n$ such that $E$ is defined over $\mathbb{F}_q$. I have proved that the Frobenius endmorphism $(x,y)\rightarrow (x^q,y^q)$ is not equal to $[m]$ for any $m\in\mathbb{Z}$, thus the injective map $\mathbb{Z}\rightarrow\operatorname{End}E$ given by $m\rightarrow[m]$ is not surjective, thus this map is not an isomorphism, but can there exist other map such that $\mathbb{Z}\rightarrow\operatorname{End}E$ is an isomorphism?

For example, the map $\mathbb{Z}\rightarrow\mathbb{Z}$ given by $x\rightarrow 5x$ is injective and not surjective, but $\mathbb{Z}\cong\mathbb{Z}$.

Thanks.

答:
Depends on whether you want to use deep results. The way I see it depends on work of Deuring from 1941, and other general facts not elementary but still not as deep as Deuring:

In case the Frobenius endomorphism $(x,y)\mapsto(x^q,y^q)$ is not an integer, you’re done. If it is an integer, it will be $[q^{1/2}]$, and since a power of $p$ is purely inseparable, your elliptic curve is supersingular. But Deuring showed that a supersingular elliptic curve in characteristic $p$ has an endomorphism ring that is an order in (i.e. free over $\Bbb Z$ of same dimension as) the central division algebra over $\Bbb Q$ of dimension four, ramified only at $p$ and infinity.

So in this interesting case, the endomorphism ring is much larger than $\Bbb Z$.

I hope that someone else can give a more elementary way of seeing what you want — I don’t consider this answer to be at all satisfactory.


问:
Why are the roots of $y^2 - iy + 2 = 0$ not complex conjugates?
So I have the equation $y^2 - iy + 2 = 0$. I use the quadratic formula to solve it, and the solutions I got are $2i$ and $-i$. Why aren't the solutions here complex conjugates? Why didn't the quadratic formula produce complex conjugates as solutions in this case?

答:
The 'conjugates rule' only applies when the coefficients of the quadratic are real. Say the roots of a quadratic $f(z)$ are $\alpha=a+bi$ and $\beta=c+di$. Then, by the factor theorem, the factorisation of $f(z)$ must be
$$
(z-\alpha)(z-\beta)=(z^2-(\alpha+\beta)z+\alpha\beta) \, .
$$
This means that in order for $f(z)$ to have real coefficients,
$$
\alpha + \beta \in \mathbb{R} \text{ and } \alpha\beta \in \mathbb{R} \, .
$$
The only time this is true when $\alpha$ and $\beta$ are both real, or when $\alpha$ and $\beta$ form a complex conjugate pair; it is a good exercise to verify this. Otherwise, $f(z)$ would have non-real coefficients, and there would be no requirement that its roots form a complex conjugate pair. I can easily think up a quadratic where the roots don't form a complex conjugate pair. For example, $\alpha = 5+7i$ and $\beta=2+3i$ gives the equation
$$
z^2-(7+10i)z+(-11+29i) = 0 \, .
$$

Throughout this answer, I assume that the leading coefficient of the quadratic is $1$. However, this is okay, because any quadratic equation of the form
$$
az^2+bz+c=0
$$
can be turned into
$$
z^2+Bz+C = 0
$$
if we divide through by $a$.


答:
Write $a:=y+z,\,b:=z+x,\,c:=x+y$ (a common strategy with $3$-variable cyclic-symmetry problems, especially if $y+z$ etc. appears), so the conjectured inequality is $(a+b+c)\sum_\text{cyc}\frac{x^2}{a}\ge(x+y+z)^2$. Now take $a_1=\sqrt{a},\,b_1=\frac{x}{\sqrt{a}}$ etc.

Note this proof never ever used $xyz=1$. Multiplying each of $x,\,y,\,z$ by $\lambda>0$ also multiplies both sides of your inequality by $\lambda$ and preserves its truth, and so whatever proof we obtain should only need to use $xyz>0$.

Title: Inequality using Cauchy Schwarz

Body:
I’ve recently been looking into applications of the Cauchy Schwarz inequality.

I’ve seen it stated in the form:

$(a_1^2 + a_2^2 + … + a_n^2)(b_1^2 + b_2^2  + … + b_n^2) \ge (a_1b_1 + a_2b_2 + … + a_nb_n)^2$

However I’m struggling to see how it could prove this cyclic sum inequality:

$$\sum_\text{cyc} \frac{x^2}{y + z} \ge \frac{(x+y+z)^2}{2(x+y+z)}$$

Where $xyz = 1$ and $x,y,z$ $\epsilon$ $\mathbb{R}$

Any help?


问:
Transforming a Weibull Distribution
X follows a Weibull distribution (k, $\lambda$), but I want to find the expected value of Y where
$$y = A e^{(-Bx)}$$
So I'm trying to solve:
$$E[Y] = {\int_0^\infty}A e^{(-Bx)}{f(x)}{dx}$$
where $f(x)$ is the pdf of the Weibull distribution.
$$f(x)=\frac {k}{\lambda} (\frac {x}{\lambda})^{k-1}e^{-(x/\lambda)}$$

This is as far as I've gotten:
$$E[Y] = A \frac {k}{\lambda^k} {\int_0^\infty} x^{k-1}(e^{-x})^{(\frac {1}{\lambda}+B)}{dx}$$

I can see the form of the Gamma function in there, but I'm not sure how to isolate it.

Note: I actually know the expected value of Y that I want (as well as the values of A, k, and $\lambda$). I'm hoping the solution here will be easy enough to invert and then solve for B.

答: The moment generating function of a Weibull distribution seems to suggest something like $$\mathbb E[Y] = A\sum_{n=0}^\infty \frac{(-B)^n\lambda^n}{n!}\Gamma(1+n/k)$$ at least for $k\geq1$, which does not seem likely to give a simple expression for $B$

Title: Transforming a Weibull Distribution


问:
divergent vs undefined integrals/summations
I have tagged this as a soft questions because it is more a question of language/notation but for certain summations/integrals these terms seem to be used interchangably and I would like some clarity on their actual meanings.

As a first example I have the sum:
$$S=\lim_{n\to\infty}\sum_{n=0}^n(-1)^n=1-1+1-1...$$
If we stop this at a given point (as long as the first value is taken as 1) this can either take the value of $1$ or $0$, both of which are finite values, but I have all kinds of manipulations of this series and there are ways of making this series take really any value you choose. This suggests that its value is undefined, but is it finite or would it be considered convergent?

Another example is the integral:
$$I=\lim_{n\to\infty}\int_0^n\sin(x)dx$$
obviously it doesn't matter particularly if this is cosine or sine or any other periodic wave satisfying the following conditions:
$$f(x+T)=f(x)$$
$$\int_a^{a+T}f(x)dx=0$$
but in a similar way to the first function we can break these integrals off at any point and it will have a finite value:
$$I(n)=\int_0^n \sin(x)dx=\int_0^{2\pi k}\sin(x)dx+\int_{2\pi k}^n\sin(x)dx$$
where $2\pi k<n<2\pi(k+1)$ and $k\in\mathbb{N}$. Clearly this first integral is equal to $0$ and at most this second integral can equal $1$ so it seems fair to say that:
$$|I(n)|\le1$$
for any finite $n$. However what happens when $n\to\infty$?

答:
A sequence that does not converge is said to diverge. This is the definition of divergent sequence. So asking, “does this sequence diverge?” is automatically answered when you answer the question, “does this sequence converge?”. Your first series does not converge, i.e. it diverges, so your proposed number $S$ does not exist.

Recall that a series is just a sequence of partial sums, and the order of terms in a series does matter unless there are either finitely many positive or finitely many negative terms.

Your first series is known as Grandi’s Series, by the way.

I always find it useful to think of a series as a limit of a sequence of partial sums, which is discussed in the introduction to the Wikipedia entry on series. So what the series $$1-1+1-1...$$ really is, is the limit if it exists, as $n \to \infty$ of the sequence (of partial sums): $(1,0,1,0,1,0,...)$. Clearly this sequence, which "is basically" your series, does not converge, i.e. it diverges.

Note that the above series is not the same as $$-1+1-1+1-...,$$ whose corresponding sequence of partial sums is $(-1,0,-1,0,-1,0,...)$, which is different to the previous sequence. [It is true that this series/sequence of partial sums diverges because again, the sequence of partial sums does not converge to anything.]

In other words, in a series, the order of terms matters.

$$$$

Furthermore, there is in fact a theorem called the Riemann Series Theorem, which states that for any number $L$, the terms of a conditionally convergent series can be rearranged so that the series is equal to $L$. For example, this says that you can rearrange the terms in the sum of the alternating harmonic series to make the sum equal to any number, including very large negative or very large positive numbers. Personally I find this fact remarkable. This is proven in Rudin's PMA Theorem $3.54$.

$$$$

And whilst Grandi’s Series does not converge in the usual sense with partial sums, there are non-standard summation methods under which it converges: See: https://en.m.wikipedia.org/wiki/Summation_of_Grandi%27s_series

It is clear that the Cesàro Sum of Grandi’s series converges to $\frac12$ and you could also look into the Abel Sum and other methods of summing the series if you want...

Now to answer some of your questions:

"...but I have all kinds of manipulations of this series and there are ways of making this series take really any value you choose." This is false: no Grandi's series does not converge, and in fact it is easy to see that no rearrangement of Grandi's series converges.

"This suggests that its value is undefined, but is it finite or would it be considered convergent?" I think the term you're looking for is "bounded", and yes, Grandi's series is bounded. Obviously, some rearrangements of Grandi's series will be unbounded, e.g. $-1+1+1-1+1+1-1+1+1-1...$ will tend to $+ \infty$, thus being an unbounded sequence of partial sums.

I haven't spent much time thinking about the second part of your question: the sine/cosine stuff, but I get the feeling that my answer can be applied to that as well... in fact, it is answered by: https://en.wikipedia.org/wiki/Integral_test_for_convergence , which says that a series converges $ \iff$ it's corresponding integral converges, so long as the series and the integral have the same limits. So really, it's the contrapositive to this theorem that answers the second half of your question.


问:
Proving that if $\gcd(m, n) = 1$, and if $d | mn$, then there exist unique numbers $a$ and $b$ such that $a|m$, $b|n$, and $d = ab$.
What do I know?

If d | mn, there exist an integer k such that dk = mn.

I also know that because gcd(m, n) = 1 there exist some integers x and y such that mx + ny = 1.

I am having trouble to prove the statement because I don't even know how to start. Am I missing a key insight?

答:
Since $(m,n)=1$, $\exists x,y:$
$$
1=mx+ny\tag1
$$
Multiply $(1)$ by $d\frac{(m,d)(n,d)}{(m,d)(n,d)}$:
$$
d=(m,d)(n,d)\left(\frac{d}{(n,d)}\frac{m}{(m,d)}x+\frac{d}{(m,d)}\frac{n}{(n,d)}y\right)\tag2
$$
$(2)$ says that $(m,n)=1\implies(m,d)(n,d)\mid d$.

Since $d\mid mn$, we have
$$
d=(mn,d)|(m,d)(n,d)\tag3
$$
$(3)$ is true because
$$
\overbrace{(mx_1+dy_1)}^{(m,d)}\overbrace{(nx_2+dy_2)}^{(n,d)}=\overbrace{mn(x_1x_2)+d(nx_2y_1+dy_1y_2+mx_1y_2)}^\text{divisible by $(mn,d)$}\tag4
$$
$(3)$ says that $d\mid mn\implies d\mid(m,d)(n,d)$.

Therefore, $(2)$ and $(3)$ say that $(m,n)=1$ and $d\mid mn$ imply that
$$
d=\overset{a\\\raise{3pt}{\downarrow}}{(m,d)}\overset{b\\\raise{3pt}{\downarrow}}{(n,d)}\tag5
$$
Since $a$ is the greatest factor of $d$ that divides $m$ and $b$ is the greatest factor of $d$ that divides $n$, neither can be greater. Their product is $d$, so neither can be less. Thus, they are unique.


答:
Your distribution function is not correct. It should be
$$
F(x) = \left\{
  \begin{array}{l l}
0,\quad &x<-\frac{1}{2}\\
3x^3-\frac{x^2}{2}+\frac{1}{2} \quad &-\frac{1}{2}<x<0\\
    3x^3+\frac{x^2}{2}+\frac{1}{2} & \quad 0 \le x < \frac{1}{2}\\
    1 & \quad \frac{1}{2}\le x
  \end{array} \right.
$$</p>

Title: get the distribution function

Body:
I have the following density function:
$ f(x) = \left\{
  \begin{array}{l l}
    cx^2+|x| & \quad \text{if -1/2<x<1/2}\\
    0 & \quad \text{otherwise}
  \end{array} \right.$

we know that $\int_{-\infty}^{\infty}{f(x)dx = 1}=\int_{-1/2}^{1/2}{cx^2+|x|dx = 1}$ and I get $c=9$.

The distribution function should be:
$ F(x) = \left\{
  \begin{array}{l l}
    3x^3-x^2/2 & \quad \text{if -1/2<x<0}\\
    3x^3+x^2/2 & \quad \text{if 0 <= x<1/2}\\
    0 & \quad \text{otherwise}
  \end{array} \right.$

If i want to test if $F(x)->1, \text{as} \space x->\infty$ it is not $1$, is the distribution function right?


问:
$X-P$ is countable, if $P$ if the set of condensation points and $X$ is a separable metric space.
I am studying basic Topology on my own, and I would like to have a check on my proof of the statement below.

(Note: there is a similar exercise on Baby Rudin, Ex. 27, Ch. 2, applied to $\mathbb R^k$, but my line of reasoning is different.)

Let $(X,d)$ be a separable metric space, and $P$ be the set of condensation points of $X$. Show that $X-P$ is countable.

Proof. I will make use of Lindelöf covering theorem, and show that if $X-P$ were uncountable then there would be an open covering $\mathcal N$ of $X-P$ with no countable subcovering, thus contradicting the separability of $X$.

Suppose then $X-P$ is uncountable. Take an element $s \in X-P$ and a neighborhood $N(x)$ of $x$ that contains only countably many points of $X$ (this is possible since $s \not \in P$). Add $N(x)$ to the collection of open sets $\mathcal N$. This "leaves out" uncountably many points of $X-P$. So we can pick another one and add to $\mathcal N$ a neighborood of such point, containing only a finite subset of $X$.

By repeating this procedure we create the open cover $\mathcal N$ which is uncountable (the countable union of countable sets is in fact countable), and such that no countable subcollection of $\mathcal N$ covers $X-P$. This (by Lindelöf theorem) contradicts separability of $X$.

答:
The construction of $\mathcal{N}$ can be tightened up considerably, but the argument is basically correct.

Your argument is a very informal and slightly hand-wavy version of a proof using transfinite recursion to construct, one set at a time, an open cover of $X\setminus P$ with no countable subcover. This is intuitively satisfying and can be made rigorous, but to do so requires some basic knowledge of infinite ordinal numbers or at least some experience working with uncountable well-ordered sets.

It’s actually easier, if perhaps less intuitive, to construct an open cover all at once and use it to show directly that $X\setminus P$ must be countable.

Proof: By definition each $x\in X\setminus P$ has a countable open nbhd $N(x)$. Let $\mathcal{N}=\{N(x):x\in X\setminus P\}$; $\mathcal{N}$ is an open cover of $X\setminus P$, so it has a countable subcover $\mathscr{C}$. Then $X\setminus P\subseteq\bigcup\mathscr{C}$, so $|X\setminus P|\le\left|\bigcup\mathscr{C}\right|$. And $\bigcup\mathscr{C}$ is the union of countably many countable sets, so it is countable, and $X\setminus P$ is therefore countable as well.


问:
Question about uniform convergence interval
I want to determine in what interval this series has uniform convergence

$$\sum_{n=0}^{\infty}(1-x)^n $$

Finding the radius of convergence is easy, since $|1-x|<1$

$$-1<1-x<1 $$
$$0<x<2 $$

The convergence radius is therefore $\rho=1$. To determine the range of uniform convergence, I use this proposition from my text book.

Proposition 5.30 Let $\sum_{n=0}^\infty c_nx^n$ be a power series with radius of convergence $\rho>0$.  Then $\sum_{n=0}^\infty c_nx^n$ is uniformly convergent on any interval of the form $[-r,r]$, where $r\in]0,\rho[$.

It tells me to choose an $r$ from the interval, $]0,1[$. Then the series is convergent in $[-r,r]$, but I don't see how this can be true.

Say I choose $r=0.5$, then the series is unformly convergent in $[-0.5,0.5]$.

But now the series doesn't even converge, since $|1+0.5|>1$, and it becomes divergent.

Origin of this question

This question was originally a multiple choice question, asking in what interval $\sum_{n=0}^{\infty}(1-x)^n $ is uniformly convergent. There were, among others, these two choices

$\frac{1}{4} \leq  x \leq \frac{3}{4}$

$0<x<1$

It turns out that the correct answer is $\frac{1}{4} \leq  x \leq \frac{3}{4}$, but I don't see why the second choice isn't correct. We know that $x$ can come arbitrarily close to $0$, and still ensure uniform convergence. We have also seen (from comments on John's answer) that $0.1 \leq \leq 1.9 x$ ensures uniform convergence. So why can't we say that $0<x<1$ ensures uniform convergence?

答:
Your series does not have the form indicated in the theorem -- the theorem involves powers of a single variable; your series involves powers of $(1-x)$.

If you let $y = 1-x$, then you can apply the theorem to the series
$$
\sum y^n
$$
to conclude an interval of uniform convergence in $y$; you can then use this to say for which $x$ the original series converges uniformly.


问:
Exponential moments of expected return time in an irreducible Markov chain.
Let $\left( X_n\right)_{n\geqslant0}$ be a Markov chain on a finite set $E$. For all $x\in E$, we define the return time to $x$ by
$$\tau_x=\inf\left\{n\geqslant1\colon X_n=x\mid X_0=x\right\}.$$

If the chain is irreducible, we know that for any $x\in E$, we have $\mathbb{E}_x\left[\tau_x\right]<\infty$.

Show that there exists $\varepsilon>0$ such that $$\mathbb{E}_x\left[\exp\left(\varepsilon\tau_x\right)\right]<+\infty$$

Deduce that for all $p\geqslant1$, $$\mathbb{E}_x\left[\left(\tau_x\right)^p\right]<+\infty$$

The second part is easy if we have the first part, but I do not know how to get this one...

Any hints ?

答:
Hints: (i) Let $p_{n} = Prob\{\tau_{x} = n\}$. Then $E \exp(\epsilon \tau_{x}) = \sum_{n=1}^{\infty} p_{n} \exp(\epsilon n)$.

(ii) $\exp(\epsilon n) = [\exp(\epsilon)]^{n}$.

(iii) $\exp(\epsilon) > 1$ for any $\epsilon > 0$.

The rest depends on what you have learned about $p_{n}$.


问:
Proof of divisibility of 3 sequential cubic numbers by induction
The task is to prove that the sum of three subsequent cubic numbers is divisible by 9.
My approach looks like this:

$n^3+(n+1)^3+(n+2)^3 = 9M$

Proof that the equation is true for $n=1$

$1^3+2^3+3^3 = 1+8+27 = 36;$

$ 36/9 = 4$

Assume that the equation is true for $n<=k$

$k^3+(k+1)^3+(k+2)^3 = 3k^3+9k^2+15k+9$

Show that the equation is true for $n = k+1$

$(k+1)^3+(k+2)^3+(k+3)^3=3k^3+18k^3+42k+36$

And this is the point where I am stuck. With a simpler formula I would simply replace the $k$ with $9M $ but how can you do it in this case?

答:
Here is a solution without induction.

Let $n-1$, $n$, and $n+1$ be three consecutive numbers.  As such, we have $$(n-1)^3+n^3+(n+1)^3=3n(n^2+2).$$

Divisibility by $9$ requires divisibility by $3$, twice.  Hence, it suffices to show that either $n$ or $n^2+2$ is a multiple of $3$.

For $n\equiv0\pmod3$, we have $$n=3k.$$
For $n\equiv\pm1\pmod3$, we have $$n^2+2=(3k\pm1)^2+2=9k^2\pm6k+3=3(2k^2\pm2k+1).$$


答:
Make use of the built-in $RealAbs$ to get correct results from WolframAlpha.

Same results with

$Reduce[RealAbs[2 - RealAbs[1 - x]] == RealAbs[x^2 - 1]]$

$x == -1 || x == 0 || x == \frac{1}{2} (-1 + \sqrt{17})$

There is no general rule. I won't give one. Follow the ideas from Your instructours as close as possible. Read problems carefully. Mathematica can be done sorrow and comprehensive. Sometimes it is really hard to get best practices solutions on the internet. RealAbs or Abs are built-in of different signature. The signature is something that makes object oriented functions different even if the name of the functions are the same. A concept from inheritence. For mathematicians at the starter level this is very strange. RealAbs take only reals, while Abs expects complexes as input and treats the input that methodology process along.

That is nothing really complicated as long as one knows that differences and built-ins exist. But no mathematician is commonly informed that way.  This problem is not included even in complexity theory. It is a jump in thoughts and knowledge that is not connected. There is no work around Abs else than using Assumption or Assuming even this tread the problem different and new problems arise.

Gather as much information as possible and search for problems similar to yours untill you are satisfied or your instructors and judges are satisfied. You can not cover everything at first attempts and the path is the target or goal.

A graphics has only illustrative purposes unless it is required by the problem. It guides a solution or represents the solution. Mathematica is often quite concise and very little proving in the output. This is all left to the interpretation of the operator of the packages.

For such a problem. Ask Wolfram Alpha first about each of the functions on both sides of the equal. Then use the intevals given there for a case study. Then solve the equation on this intervals.

```<code>PiecewiseExpand[
 RealAbs[2 - RealAbs[1 - x]] == RealAbs[x^2 - 1], , -2 < x < 2]

(-x + x^2 == -2 && x > 3) || (-x + x^2 == 2 &&
   x == 1) || (-x + x^2 == 2 && x == -1) || (-x + x^2 == 2 && x <= 1 &&
    x + x^2 == 0) || (x > -1 && x < 1 && x + x^2 == 0) || (x > 1 &&
   x <= 3 && x + x^2 == 4) || (x < -1 && x + x^2 == 0)
</code>```

Solve on each interval and the put it all together.

Body:
Make use of the built-in $RealAbs$ to get correct results from WolframAlpha.

Same results with

$Reduce[RealAbs[2 - RealAbs[1 - x]] == RealAbs[x^2 - 1]]$

$x == -1 || x == 0 || x == \frac{1}{2} (-1 + \sqrt{17})$

There is no general rule. I won't give one. Follow the ideas from Your instructours as close as possible. Read problems carefully. Mathematica can be done sorrow and comprehensive. Sometimes it is really hard to get best practices solutions on the internet. RealAbs or Abs are built-in of different signature. The signature is something that makes object oriented functions different even if the name of the functions are the same. A concept from inheritence. For mathematicians at the starter level this is very strange. RealAbs take only reals, while Abs expects complexes as input and treats the input that methodology process along.

That is nothing really complicated as long as one knows that differences and built-ins exist. But no mathematician is commonly informed that way.  This problem is not included even in complexity theory. It is a jump in thoughts and knowledge that is not connected. There is no work around Abs else than using Assumption or Assuming even this tread the problem different and new problems arise.

Gather as much information as possible and search for problems similar to yours untill you are satisfied or your instructors and judges are satisfied. You can not cover everything at first attempts and the path is the target or goal.

A graphics has only illustrative purposes unless it is required by the problem. It guides a solution or represents the solution. Mathematica is often quite concise and very little proving in the output. This is all left to the interpretation of the operator of the packages.

For such a problem. Ask Wolfram Alpha first about each of the functions on both sides of the equal. Then use the intevals given there for a case study. Then solve the equation on this intervals.

PiecewiseExpand[
 RealAbs[2 - RealAbs[1 - x]] == RealAbs[x^2 - 1], , -2 < x < 2]

(-x + x^2 == -2 && x > 3) || (-x + x^2 == 2 &&
   x == 1) || (-x + x^2 == 2 && x == -1) || (-x + x^2 == 2 && x <= 1 &&
    x + x^2 == 0) || (x > -1 && x < 1 && x + x^2 == 0) || (x > 1 &&
   x <= 3 && x + x^2 == 4) || (x < -1 && x + x^2 == 0)

Solve on each interval and the put it all together.


问:
How do I prove $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$?
How do I prove this inequality with sequential Fibonacci numbers: $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$ (i.e. $\frac{5}{3}>\frac{3}{2}$)? I know these formulas alternately overestimate and underestimate the golden ratio as $n$ gets large.

Body:
The base case of OP's equality is, for $n=1$,

$$\frac21>\frac11.$$

Assume that OP's inequality holds for some $n = k$, $k \ge 1$.

$$\frac{f_{2k+1}}{f_{2k}} > \frac{f_{2k}}{f_{2k-1}}$$

Then for $n=k+1$, using the mediant inequality twice,

$$\begin{array}{rcrclcl}
\dfrac{f_{2k+1}}{f_{2k}} &>&&&&& \dfrac{f_{2k}}{f_{2k-1}}\\
\dfrac{f_{2k+1}}{f_{2k}} &>&&& \dfrac{f_{2k+1}+ f_{2k}}{f_{2k}+f_{2k-1}} &> &\dfrac{f_{2k}}{f_{2k-1}}\\
\dfrac{f_{2k+1}}{f_{2k}} &>&&& \dfrac{f_{2k+2}}{f_{2k+1}}\\
\dfrac{f_{2k+1}}{f_{2k}} &>& \dfrac{f_{2k+1}+f_{2k+2}}{f_{2k}+f_{2k+1}} &>& \dfrac{f_{2k+2}}{f_{2k+1}}\\
&&\dfrac{f_{2k+3}}{f_{2k+2}} &>& \dfrac{f_{2k+2}}{f_{2k+1}}
\end{array}$$

(The intermediate step also proves the other half of the inequality pair:

$$\frac{f_{2n+2}}{f_{2n+1}} < \frac{f_{2n+1}}{f_{2n}})$$


答:
If you want a proof with induction:

Your proof is almost correct, you're just missing the final step:

$3k^3+18k^2+42k+36 = (3k^2 + 9k^2 +15k + 9) + (9k^2 +27k + 27)$.

Note that the former is divisible by 9 by our induction hypothesis, and the latter can be written as $9(k^2 + 3k + 3)$, which is also divisible by 9.

Since the sum of two numbers divisible by 9, is also divisible by 9, we have that

$3k^3 + 18k^2 + 42k+36$ is also divisible by 9.

Title: Proof of divisibility of 3 sequential cubic numbers by induction

Body:
The task is to prove that the sum of three subsequent cubic numbers is divisible by 9.
My approach looks like this:

$n^3+(n+1)^3+(n+2)^3 = 9M$

Proof that the equation is true for $n=1$

$1^3+2^3+3^3 = 1+8+27 = 36;$

$ 36/9 = 4$

Assume that the equation is true for $n<=k$

$k^3+(k+1)^3+(k+2)^3 = 3k^3+9k^2+15k+9$

Show that the equation is true for $n = k+1$

$(k+1)^3+(k+2)^3+(k+3)^3=3k^3+18k^3+42k+36$

And this is the point where I am stuck. With a simpler formula I would simply replace the $k$ with $9M $ but how can you do it in this case?


问:
Structure Tensor definition, what is the point?
The structure tensor is defined like this
 where $(G^i_x, G^i_y)$ the i’th gradient in a square area around a point.

I know that structure tensors are used in image analysis - here one could use the structure tensor when defining corners in an image. Corners are characterized by having sharply changing gradient orientations at some point(s). Thus, to me, it makes sense to look at the diagonal of the structure tensor(the sums over gradients wrt either x or y).

However, what is the point of having $\sum_{i=1}^{n}{G^i_x, G^i_y}$? Isn't it enough to consider the elements in the diagonal?

So why don't we just use a matrix defined like this:

$$
        \mathbf{J} =
        \left[\begin{array}{r}
             \sum_{i=1}^{n}{G^i_x}^2\\
             \sum_{i=1}^{n}{G^i_y}^2
        \end{array}\right]
$$
instead ?

答:
Consider the central pixel in the following two images:

At that central pixel, the structure tensors are:
$$
\begin{pmatrix} 102.82 & 102.83 \\ 102.83 & 102.82 \end{pmatrix}
\qquad
\begin{pmatrix} 102.82 & 9.9095 \\ 9.9095 & 102.82 \\ \end{pmatrix}
$$

The two diagonal components of the structure tensor are equal in both cases (as you can tell I adjusted the contrast in one of the images to make it so). The only way of distinguishing these two geometries is through the off-diagonal element.


问:
Divergent sequence with only one accumulation point
I'm trying to  find some examples of divergent sequences that have a unique accumulation point (in the context of $R^n$, with the usual topology). I know that if a sequence converges, then it has a unique accumulation point, but the inverse sentence is not true. I found one using $n$ and $1/n$, but I can't find another example.

答:
Let $\langle a_n:n\in\Bbb N\rangle$ be any convergent sequence, and let $\langle b_n:n\in\Bbb N\rangle$ be any divergent sequence with no accumulation point. (For instance, $\langle b_n:n\in\Bbb N\rangle$ could be any monotonic sequence diverging to $\infty$ or $-\infty$.) For each $n\in\Bbb N$ let $c_{2n}=a_n$ and $c_{2n+1}=b_n$, so that

$$\langle c_n:n\in\Bbb N\rangle=\langle a_0,b_0,a_1,b_1,a_2,b_2,\ldots\rangle\,;$$

then $\langle c_n:n\in\Bbb N\rangle$ diverges and has exactly one accumulation point, namely, the limit of the convergent sequence $\langle a_n:n\in\Bbb N\rangle$.


答: For a degree-$n$ polynomial function $p(z)=\sum_{i=0}^np_iz^i$ with $p_n\ne0$, define $\bar{p}(z):=\sum_ip_i^\ast z^i$, which is the same polynomial iff each $p_i$ is real. Then $\bar{p}(z^\ast)=\sum_ip_i^\ast z^{\ast i}=(p(z))^\ast$, and $p(z)=0\iff\bar{p}(z^\ast)=0$. The complex-conjugates rule for polynomials with real coefficients is a special case of this.

Title: Why are the roots of $y^2 - iy + 2 = 0$ not complex conjugates?

Body: So I have the equation $y^2 - iy + 2 = 0$. I use the quadratic formula to solve it, and the solutions I got are $2i$ and $-i$. Why aren't the solutions here complex conjugates? Why didn't the quadratic formula produce complex conjugates as solutions in this case?


问:
K-theory proof of index theorem - some minor confusion
I am trying to understand the general approach to the $K$-theory proof of the Atiyah-Singer index theorem, using this https://arxiv.org/pdf/math/0504555.pdf paper. I ran into some confusion on page 29, where the following is stated:

"It only remains to show that the analytic index commutes with the Thom isomorphism $\phi:K(X)\to K(V)$ where $V$ is a complex vector bundle over $X$. [...] This problem is considerably simplified if we consider trivial bundles which can be expressed as the product $V = X \times\mathbb{R}^n$."

On the same page, it goes on to consider a vector bundle $Y$ which seems to be the associated bundle of some principal $G$-bundle, but the author again considers $P\times_{O(n)} \mathbb{R}^n$, that is, a real vector bundle. I don't quite understand how this makes sense, if we want to prove something for complex vector bundles. I get that we can view a complex vector bundle as a real vector bundle by just "forgetting" about the complex structure, but since the Thom isomorphism (at least in the paper) is only defined for complex vector bundles, I think I am missing something more important. I cannot quite put my finger on it, so if someone could explain the construction on page 29, that would be greatly appreciated.

答: It seems that this construction was made for real vector bundles because every complex vector bundle can be regarded as a real vector bundle when discarding the complex structure. I am having some trouble justifying this, since we need to add the complex structure again for the Thom isomorphism, and I would like to hear why we don't use $U(n)$-vector bundles instead, since $U(n)$ is also a compact Lie group. Can we not form any complex vector bundle in this way, like we can form any real vector bundle as the associated bundle of some principal bundle?


问:
Lebesgue Integration, Uniform Convergence
Good evening/afternoon/morning!

I am in a situation here where I think I have found a solution but it all feels too simple and good to be true. Here is our problem:

Let  $f_n(x):[0,1] \to [0,1]$ for $n\in\mathbb{N}$ by :

$f_n(x)= \begin{cases}
      nx & x\in[0,\frac{1}{n}] \\
      1 & x \in(\frac{1}{n},1]
   \end{cases}$

Let $f$ be the Pointwise Limit of $f_n$ Show that there is no set
$B ⊂ [0, 1]$ of Lebesgue measure $0$ so that $f_n$ to $f$ converges uniformly
on the set $[0, 1] \backslash B$ (= $[0,1]\cap B^c$)

Here is a sketch of the "fantastic" solution I think I have found:

Firstly $f = 1$ everywhere but $f(0) = 0 $

Let $I:= [0,1] \backslash B$ be the domain in which $f_n$ converges uniformly. Uniform convergence (which is the same limit of pointwise) implies Riemann Integrability which inturn implies a Lebesgue Integral with the same value. Also due to the uniform convergence the limit of the integral is the integral of the limit and hence:

$ \lim_{n \to \infty}\int_If_ndx=\int_Ifdx = \int_Ifd\lambda=\int_I1\cdot d\lambda = \lambda(I)$

And hence all I would need to do in theory would be to show that $ \lim_{n \to \infty}\int_If_ndx < 1 $ And im done as $I$ and $B$ are disjoint and $\lambda(B) = 1-\lambda(I) > 0$

Any pointers or help would be greatly appreciated. Also any way to show $ \lim_{n \to \infty}\int_If_ndx < 1 $ would be amazing!

答:
You can argue in this way.

Assume, by the way of contradiction, that there is $B\subset[0,1]$ with $m(B)=0$ and such that $f_n\to f$ uniformly on $[0,1]\setminus B$.

For any $n\in\Bbb N$, let $I_n=\left(0,\frac{1}{n^2}\right)$. Since $m(I_n)=\frac{1}{n^2}>0$ and $m(B)=0$, then $m(I_n\setminus B)=m(I_n)-m(I_n\cap B)=\frac{1}{n^2}-0>0$; and, in particular, $I_n\setminus B\ne\varnothing$ and there is $x_n\in I_n\setminus B$. So we have constructed a sequences $(x_n)_n\subset[0,1]\setminus B$ with $0<x_n<\frac{1}{n^2}$, in particular, $0<x_n<\frac{1}{n}$. Hence
$$\sup_{x\in [0,1]\setminus B} |f_n(x)-f(x)| \ge |f_n(x_n)-f(x_n)|=\left|n x_n-1\right|\to1\ (n\to\infty)$$ (observe that $0<nx_n<n\frac{1}{n^2}=\frac{1}{n}$)wich is a contradiction with $f_n\to f$ uniformly on $[0,1]\setminus B$.


问:
Sobolev inequalities for $W^{1,s}(Q)$ and $W^{2,s}(Q)$
Let $Q$ a bounded regular domain of $\mathbb{R}^3$, and $s>3$.

Let $u \in W^{1,s}(Q)$ (i.e $u, \nabla u$ are in $L^s(Q)$).

I know I have the following sobolev continuous embedding : $W^{1,s}  \hookrightarrow C^{0,1-3/s}(Q)$ and therefore I can write :
$$
\sup_{x \in Q} |u| \leq C \|u\|_{W^{1,s}(Q) }
$$
However, it is unclear for me, denoting $m=\frac{1}{|Q|}\int_{Q} u \ \mathrm{d}x$, that i have the following inequality :
$$
\sup_{x \in Q} |u-m| \leq C_1 \|\nabla u\|_{L^s(Q) } \label{1}\tag{1}
$$
It seems like some Poincaré inequality was used but I don't really see how. Do you have any ideas on how to proof this inequality ?

Additional questions

If I take $u \in W^{2,s}(Q)$, can i proof the following inequality ?

$$\sup_{x \in Q} |u-m| \leq C_1 \|\nabla^2 u\|_{L^s(Q) } \label{2}\tag{2}$$

Do I need to add any hypothesis on $u$ to prove \eqref{1} and \eqref{2}? I was thinking something like $\gamma_0 (u)=0$ on $\partial Q$...

Thanks for your help.

答:
Let me prove \eqref{1} for functions $u$ satisfying $\int_\Omega u=0$. Assume the claim does not hold. Then for each $n$ there is $(u_n)$ in $W^{1,s}$ such that
$$
\|u_n\|_{L^\infty} > n \|\nabla u_n\|_{L^s}.
$$
Hence, $u_n\ne0$, and we can assume
$\|u_n\|_{L^\infty}=1$. This implies $\nabla u_n\to 0$. In addition, $(u_n)$ is bounded in $W^{1,s}$. Thus $(u_n)$ is bounded in that Hoelder norm. By Arzela-Ascoli, we have after - possibly extracting a subsequence - $u_n \to u$ in $C(\bar Q)$. By the convergence of $\nabla u_n$, we have $\nabla u=0$, so $u$ is constant. However, along the sequence $\int_\Omega u_n=0$, so $u=0$. This implies $\nabla u_n \to 0$ in $L^s$ and $u_n\to 0$ in $W^{1,s}$, which is a contradiction to $\|u_n\|_{L^\infty}=1$.

The second claim is not true: take $Q=B_1(0)$, $u(x) = c^Tx$ with $c\ne 0$. Then $m=0$, $\nabla ^2u=u$, but $\|u\|_{L^\infty}= |c|$. If one would try the above proof, then one finds that the limit is not a constant function but piecewise linear. The condition $m=0$ only fixes one degree of freedom, but there are $3$ more. One needs additional conditions to conclude $u=0$.


问:
Homotopy between two closed curves in the punctured plane.
I’m looking at the following lemma.

Suppose $\gamma_{0}, \gamma_{1}: \mathbb{S}^{1} \rightarrow \mathbb{R}^{2} \backslash\{0\}$ are two
closed curves in the punctured plane. If
$$
\left\|\gamma_{0}(z)-\gamma_{1}(z)\right\| \leq\left\|\gamma_{0}(z)\right\|
$$
for all $z \in \mathbb{S}^{1}$, then $\gamma_{0} \simeq \gamma_{1}$ in $\mathbb{R}^{2} \backslash\{0\} .$

The lecturer provides the following analogy to illustrate the idea behind the lemma.

Suppose you are walking your dog and you pass near a tree. To ensure that you don't get the leash tangled around the tree, it suffices to adjust its length so that, at any instant, it is shorter than your distance to the tree. Mathematically, we prove this "dog-on-a-leash" lemma simply by considering a linear homotopy.

I can see how the two curves $\gamma_0$ and $\gamma_1$ satisfying the condition in the statement of the lemma would forbid $\gamma_1$ to “wrap around” the missing point. But beyond that I don’t see how the analogy translates to a linear homotopy. Some more elaboration on the idea and the proof would be appreciated.

答:
When you hear or read "linear homotopy" your mind should automatically and immediately go to $H:S^1\times[0,1]\to \Bbb R^2$ given by
$$
H(z,t)=(1-t)\gamma_1(z)+t\gamma_0(z)
$$
because that's what "linear homotopy" means. It is often the simplest homotopy between two maps, and therefore also often the first thing to try when you want a homotopy between two maps. At least as long as the codomain of the two maps is a vector space or a subset of a vector space.

And indeed, for any fixed $z$, the given requirement on $|\gamma_1(z)-\gamma_0(z)|$ means that $H(z,t)\neq0$ for all $t\in[0,1]$, and we do have a valid homotopy.

PS. This exact argument lies at the heart of my personal favourite proof of the fundamental theorem of algebra: for any degree-$n$ (monic) polynomial $p$, if $S^1$ is a large enough circle around the origin, and $\gamma_1(z)=p(z)$ and $\gamma_0(z)=z^n$, the two therefore go around the origin the same number of times. And unless $0$ is a root of $p$, a small enough circle around the origin is mapped by $p$ to a circle that doesn't go around the origin. So some circle in-between needs to cross the origin, and $p$ has a root.


问:
Show that $\frac {a_1^2}{a_2}+\frac {a_2^2}{a_3}+...+\frac {a_n^2}{a_1}\geq a_1+a_2+...+a_n$ using AM-GM.

Given $a_1,a_2,...,a_n$ be positive reals. Show that $\displaystyle\frac {a_1^2}{a_2}+\frac {a_2^2}{a_3}+...+\frac {a_n^2}{a_1}\geq a_1+a_2+...+a_n$ using AM-GM.

I know how to slve it using rearrangement inequality, but I can't. How should I apply AM-GM? Thanks.

Body:
$$\dfrac{a^2_{1}}{a_{2}}+a_{2}\ge 2a_{1}$$
$$\dfrac{a^2_{2}}{a_{3}}+a_{3}\ge 2a_{2}$$
$$\cdots\cdots$$
$$\dfrac{a^2_{n}}{a_{1}}+a_{1}\ge 2a_{n}$$

add all inequalities and you're done!


答:
If using hyperreal numbers, there is a way to do the summation so that you get an actual value for the divergent series (though often it is a specific hyperreal infinite value).  The paper (coauthored by me) is "Hyperreal Numbers for Infinite Divergent Series".

The reason why you can regroup it and get different answers is that, in a hyperreal sense, you are actually changing the number of items being summed.  That is, if $\omega$ is the unit infinity, having $\omega - 1$ terms or $\frac{\omega}{2}$ terms is a different number of terms than the original.

The easiest way to find a value, that we found (see Section 9), is to take a discrete integral (i.e., a symbolic summation), and then replace all occurrences of $-1^\omega$ with zero.  This is not proven in the paper, but we have an unpublished proof of it, and the paper cites a proof that $\sin(\omega) = 0$ in the surreal numbers.

In this case, the sum is $\frac{1}{2}$, which matches the Cesaro summation.

In short, using hyperreal numbers allows you to do things that you couldn't do when you restrict yourself to the reals.

Title: divergent vs undefined integrals/summations

Body:
I have tagged this as a soft questions because it is more a question of language/notation but for certain summations/integrals these terms seem to be used interchangably and I would like some clarity on their actual meanings.

As a first example I have the sum:
$$S=\lim_{n\to\infty}\sum_{n=0}^n(-1)^n=1-1+1-1...$$
If we stop this at a given point (as long as the first value is taken as 1) this can either take the value of $1$ or $0$, both of which are finite values, but I have all kinds of manipulations of this series and there are ways of making this series take really any value you choose. This suggests that its value is undefined, but is it finite or would it be considered convergent?

Another example is the integral:
$$I=\lim_{n\to\infty}\int_0^n\sin(x)dx$$
obviously it doesn't matter particularly if this is cosine or sine or any other periodic wave satisfying the following conditions:
$$f(x+T)=f(x)$$
$$\int_a^{a+T}f(x)dx=0$$
but in a similar way to the first function we can break these integrals off at any point and it will have a finite value:
$$I(n)=\int_0^n \sin(x)dx=\int_0^{2\pi k}\sin(x)dx+\int_{2\pi k}^n\sin(x)dx$$
where $2\pi k<n<2\pi(k+1)$ and $k\in\mathbb{N}$. Clearly this first integral is equal to $0$ and at most this second integral can equal $1$ so it seems fair to say that:
$$|I(n)|\le1$$
for any finite $n$. However what happens when $n\to\infty$?


问:
Defining a right $A$-module structure on $\text{Hom}_\mathbb{Z}(E,G)$
Let $A$ be a ring. Suppose $E$ is a right $A$-module, $F$ a left $A$-module and $G$ a $\mathbb{Z}$-module. Is there a way to define a right $A$-module structure on $\text{Hom}_\mathbb{Z}(E,G)$?

This would equivalent to defining a ring homomorphism $\phi:A^{\text{op}}\rightarrow\mathcal{E}$, where $\mathcal{E}$ is the endomorphism ring of $\text{Hom}_\mathbb{Z}(E,G)$. For $a\in A$, what should $\phi_a:\text{Hom}_\mathbb{Z}(E,G)\rightarrow \text{Hom}_\mathbb{Z}(E,G)$ be?

答: You define $F$ but it doesn't appear in your question so I'm not sure you've asked the question you meant to ask. The general pattern is that $\text{Hom}$ is a contravariant functor in its first argument so it switches left and right module structures: $\text{Hom}_{\mathbb{Z}}(E, A)$ (for $A$ an abelian group; IMO it's good practice to reserve $G$ for a not-necessarily-abelian group) naturally acquires a left $A$-module structure by precomposition, and similarly $\text{Hom}_{\mathbb{Z}}(F, A)$ naturally acquires a right $A$-module structure by precomposition.


问:
Show that the Möbius transformations forms a group!
I'm wondering how to prove the associativity and identity to prove that the Möbius transformations forms a group.

A Möbius transformation is a complex function of the form $M(z)=\dfrac{az+b}{cz+d}$.

Thank you in advance :)

答:
Since you are explicitly asking for the direct proof:

concerning the associativity, this is just the same proof as for the associativity of any group of functions. So consider three moebius transforms $M_1,M_2,M_3$ then for any complex number $z$ we have $$M_1\circ( M_2\circ M_3)(z)=M_1((M_2\circ M_3)(z))=M_1(M_2(M_3(z)))=(M_1\circ M_2)(M_3(z))=(M_1\circ M_2)\circ M_3(z)$$ Note that we did not use any of the properties of the moebius transforms since the associativity just follows from the definition of the concatenation of functions.

For the identity consider the element $M(z)=id(z)=z$, i.e. choose $a=d=1,b=c=0$ then clearly $ad-bc\neq 0$ so this is a moebius transform and you can easily verify that this is the identity for your group.

I guess you already figured out the formula for the inverse transformation since you were only asking for the other two properties, but still for completeness: it is easy to check that for any moebius transform $M(z)=\frac{az+b}{cz+d}$ we have an inverse given by $M^{-1}(z)=\frac{dz-b}{-cz+a}$ and this also is a moebius transform

We actually also have to show that the concatenation of two moebius transforms is again a moebius transform: but $$M_1\circ M_2(z)=M_1(\frac{a_2z+b_2}{c_2z+d_2})=\frac{a_1(\frac{a_2z+b_2}{c_2z+d_2})+b_2}{c_1(\frac{a_2z+b_2}{c_2z+d_2})+d_2}=\frac{(a_1a_2+b_1c_1)z+(a_1b_2+b_1d_2)}{(a_2c_1+c_2d_1)z+(b_2c_1+d_1d_2)}$$ Now denoting the coefficients of the concatenation by $a',b',c',d'$ we have $a'd'-c'b'\neq 0$ which can either be verified by direct calculation or the fact that $0\neq det(\left(\begin{array}{l}a_1&b_1\\c_1&d_1\end{array}\right)\left(\begin{array}{l}a_2&b_2\\c_2&d_2\end{array}\right))=\det(\left(\begin{array}{l}a'&b'\\c'&d'\end{array}\right))$so this claim also holds.

So the set of moebius transforms forms a group under concatenation.

Note that it actually suffices to prove 3 and 4 as this shows that the moebius transforms form a subgroup of the group of all bijections of the extended complex numbers to themselves. But since you were explicitly asking for the first two steps i provided them too :)

Lg Mo


问:
Given n distinguishable d-dimensional points what is the largest number of different ways they can be linearly separated?
Suppose we have $n$ distinguishable points in $\mathbb{R}^d$. What is $f(n, d)$, the largest number of different ways we can separate them using a single hyperplane? I don't consider swapping the 'left' and 'right' side of the plane to be different.

I found the following question for the $d = 2$ case, so $f(n, 2) = \binom{n}{2} + 1$.

You may assume that the points are in a position allowing the most number of separations. For $d = 2$ it is shown this doesn't matter (beyond no collinearity), but I don't know if this also holds for higher dimensions (with points in general position).

答:
Assuming the $n$ points in general position, as shown in the paper "The number of partitions of a set of N points in k dimensions induced by hyperplanes" by E. F. Harding, the function you are looking for is:

$$f(n,d) = \sum_{k=0}^{d}{n-1 \choose k}$$


答:
I don't know how much you've done on your own, but to help subsequent users, I'll put the full answer. Stop after each hint and try and do it yourself.

Hint 1:
Prove that the Möbius transformations are bijective from the extended complex numbers to themselves.

Hint 2:
The Möbius transformations are therefore a subset of the set of bijections on the extended complex numbers.

Hint 3:
The set of all bijections of the complex numbers is a group.

Hint 4:
Composition of two Möbius transformations and inversion of one Möbius transformation is another Möbius transformation.

Hint 5:
A closed subset of a group is a group.

Hint 6:
The set of Möbius transformations is closed.

Title: Show that the Möbius transformations forms a group!

Body:
I'm wondering how to prove the associativity and identity to prove that the Möbius transformations forms a group.

A Möbius transformation is a complex function of the form $M(z)=\dfrac{az+b}{cz+d}$.

Thank you in advance :)


问:
What is the probability that an event will happen when the probability decreases exponentially?
So I am just a middle student who started studying probability in my stats class and I see the questions on this site are little advanced so I don't hope this one is too basic.

Anyways lets say that the probability of an event happening was some number(like 1%), and each time a trial is run the of the event happening decreases by half(so 1% probability for the first trial, .5% probability for the second, .25% probability, and so on).
What is the probability that the event will happen after 1 trial, 10 trials, 100 trials, and an infinite amount of trials?

答:
Well, I don't think you'll understand most of the answers same as me but well, genereally this is what will happen first one is 1% then 0.5% then 0.25% so the probability here is

Pn= 1% * 0.5^n

can't use the maths symbols as I'm new here but there you go, n refers to the number of trial, 1% is your starting point, 0.5 is the chance decrease( n times)


问:
How Can I apply Arzela-Ascoli to show that $f_n \to f$ uniformly?
I have that $M >0$ and $f_n:[0,1] \to R$ is a sequence of Lipschitz functions with common Lipschitz constant $M$, i.e. $|f_n(x)-f_n(y)| \le M|x-y|$ and I have that $f_n \to f$ pointwise.

How can I apply Arzela Ascoli theorem to show that $f_n \to f$
Since $f_n \to f$ pointwise, I can show that $f$ is Lipschitz continuous by doing the following

$|f_n(x)-f_n(y)| \le M|x-y| \quad \forall n $ and $x,y \in [0,1]$

Since $f_n(x) \to f(x)$ and $f_n(y) \to f(y)$ when $n \to \infty$, it follows that

$|f_n(x) - f_n(y)| \to |f(x)-f(y)|$ and from $|f_n(x)-f_n(y)| \le M|x-y|$ we get that

$|f_n(x)-f_n(y)|\le M|x-y|$.

Now I dont know how to show that it follows that $f_n \to f$ uniformly and the hint is to use Arzela-Ascoli theorem. Can somebody help me?

答:
Theorem 1. (Arzela Ascoli) Consider a sequence of real-valued continuous functions $\{ f_n \}_{n \in \mathbb N}$ defined on a closed and bounded interval $[a, b]$ of the real line. If this sequence is uniformly equicontinuous and uniformly bounded. Then there exists a subsequence $\{ f_{n_k} \}_{k \in \mathbb N}$ that converges uniformly.

Theorem 2. (Subsequence Principal)A sequence $\{ x_n\}$ in topology space $X$ converges to $x^*$ iff every its subsequence has a further subsequence (sub-sub-sequence) converging to $x^*$.

Step 1. We prove that: $\{ f_n \}_{n \in \mathbb N}$ is uniformly equicontinuous and uniformly bounded. Indeed, the uniformly equicontinuous holds  directly thanks to the "common Lipchitz" assumption. Regarding the uniform boundedness, observe that $|f_n(x)|\leq |f_n(0)|+M$ and $|f_n(0)|$ is bounded since $f_n(0)$ converges to $f(0)$. Applying Theorem 1, we conclude that there exists a subsequence $\{ f_{n_k} \}_{k \in \mathbb N}$ that converges uniformly to a function, say $h$.

Step 2. As $f_n$ converges to $f$ pointwise and $f_{n_k}$ also converges to $h$ poitwise, then $h=f$. Thus, $\{ f_{n_k} \}_{k \in \mathbb N}$ uniformly converges to $f$.

Step 3. Repeat the arguments in Step 1 and Step 2 for arbitrary subsequence of $\{ f_{n} \}_{n \in \mathbb N}$, we will see that this subsequence has a further subsequence uniformly converges to $f$. Applying the Theorem 2, we conclude that $f_n$ converges to $f$ uniformly.


问:
Conditional Probability (Exam Pass/Fail)
A student needs to take and pass three exams to pass a course. If she fails an exam she cannot
take the subsequent one(s) and fails the course. The probabilities of passing the exams are 0.9,
0.75 and 0.6 consecutively. Given that the student did not pass the course, what is the probability
that she failed the th exam,  = 1,2,3?

I tried to solve it with conditioning on failing for all i values. But seems like I am missing something.

Thanks for your help.

答: Fail first exam $p_1=0.1$, fail second exam $p_2=0.9\times 0.25=0.225$, fail third exam $p_3=0.9\times 0.75 \times 0.4=0.27$. Total $=0.595$  As you must see, failure on a later exam results only if previous exams passed.