答: The row player has an expected payout of $0.5$ and the column player has an expected payout of $-0.5$, so if you judge the winner on their expected payout from playing the mixed strategies the row player is the winner.

Title: Who is the winner in the zero-sum game?

Body:
I have a zero-sum game as described in the table below:

Using minimax = maximin approach I get that the first player should play A with $p^*=\frac{5}{12}$ and B with $p^*=\frac{7}{12}$.

The second player should then mix A and B with $q^*=\frac{1}{2}$.

I found the value of the game to be: $v=0.5$.

The question is asking to determine the winner. And since, the game is zero-sum, I know that
one player wins and the other loses (what is one player's gain is the
other's loss etc.). But how does this apply to this situation?
Basically, every time they play the same action, player 2 wins and
when they play different actions - player 1 wins. But how to
determine the overall winner? I think it is based on the value of the
game. My guess is that when the value of the game, v is positive,
then the first player is the winner? Is that correct?

$$\begin{array}{c|c|c|c|}
 & \text{A} & \text{B} & \text{min} \\ \hline
\text{A} & -3 & 4 & -3\\ \hline
\text{B} & 3 & -2 & -2 \\ \hline
\text{max} & 3 & 4 \\ \hline
\end{array}$$


问:
Nonlinear differential equation. $p^3-4px^4+8yx^3=0$
Nonlinear differential equation.
Let $p=\dfrac{dy}{dx}$
$$p^3-4px^4+8yx^3=0$$

I tried to make $y$ alone and take derivative wrt. $x$ as

$$y=\dfrac{p^3}{-8x^3}+\dfrac{xp}{2}$$

then

$$y'=p=\dfrac{-3}{8x^3}p^2p'+\dfrac{3}{8x^4}p^3$$

but cannot solve for $p$

答:
$$p^3-4px^4+8yx^3=0$$
$$(y')^3-4y'x^4+8yx^3=0$$
Change the variable $x$ ( $ \dfrac {dy}{dx} \to 2x\dfrac {dy}{dx^2}$) :
$$(\dfrac {dy}{dx^2}2x)^3-8x^5\dfrac {dy}{dx^2}+8yx^3=0$$
$$y'^3-uy'+y=0$$
$$y=uy'-y'^3$$
Where $u=x^2$.

This is Clairaut's differential equation.

ContentLicense: CC BY-SA 4.0


问:
Guides/tutorials to learn abstract algebra?
I recently read up a bit on symmetry groups and was interested by how they apply to even the Rubik's cube. I'm also intrigued by how group theory helps prove that "polynomials of degree $\gt4$ are not generally solvable".
I love set theory and stuff, but I'd like to learn something else of a similar type. Learning about groups, rings, fields and what-have-you seems like an obvious choice.

Could anyone recommend any informal guides to abstract algebra that are written in (at least moderately) comprehensible language? (PDFs etc. would also be nice)

答:
I can highly recommend "A Book of Abstract Algebra", by Charles C. Pinter. You'll learn about groups, rings and fields. You will also learn enough Galois Theory to understand why polynomials of degree higher than $4$ are, in general, not solvable by radicals.

It is 'formal' in the sense that it is rigorous, but the author is also very good at explaining the intuition behind all ideas. It is much less dense than most Abstract Algebra books, and, in my opinion, and excellent introduction to the subject.

Furthermore, it isn't expensive and it contains solutions to numerous exercises. See the amazon page of the this book for more positive reviews. Again, highly recommended!

Added: once you finished this book, you're ready for more advanced treatments of abstract algebra. After Pinter's book, you could try "A First Course in Abstract Algebra" by John B. Fraleigh. After that one, a great option is "Abstract Algebra" by Dummit and Foote. This is quite an advanced textbook, but a good one nevertheless. Once you've worked your way through these books (I advise you not to just read through them, but actually soak up the information by doing the exercises and reading actively) you will have a strong basis of knowledge in abstract algebra. By then you can tackle more advanced topics.

ContentLicense: CC BY-SA 3.0


问:
Is $y$ to the power of $\frac{1}{\infty}$ equals 1?
Is

$$y^{\frac{1}{\infty}} = 1 \quad ? $$

If that's correct. $\log_1 (1)$ will be solved.

Body: For $y\in(0,\infty)$, $\lim_{x\to\infty} y^{1/x}=1$. For $y=0$, $\lim_{x\to\infty}y^{1/x}=0$.


问:
Show that if $x \geq y$, then $(e^{\frac{x}{2}}-e^{\frac{y}{2}})^2 \leq \frac{(x-y)^2}{8}(e^x+e^y)$
I tried to naively fix $y$ and then study the sign of

$$f:x \mapsto \frac{(x-y)^2}{8}(e^x+e^y)-(e^{\frac{x}{2}}-e^{\frac{y}{2}})^2 $$

but it gets ugly pretty fast.

答:
Let
\begin{align}
 f(t)&=\frac{t^2}8-1 + \text{sech}\frac t2\\
f’(t)&= \frac t4-\frac12 \tanh\frac t2\text{sech}\frac t2\\
f’’(t)&= \frac14(1+\text{sech} \frac t2 -2 \text{sech}^3 \frac t4)\\
&= \frac14[(1-\text{sech}^3 \frac t2)+ \text{sech} \frac t2
(1-\text{sech}^2 \frac t2)]>0
\end{align}
Note $f(0)=f’(0) =0$. Thus, $f(t)$ is convex with the minimum value $0$, meaning $f(t)\ge0$. Then, substitute $t=x-y$ into $f(t)$ above to get
$$f(x-y) = \frac{(x-y)^2}8-1+\frac{2e^{\frac{x-y}2}}{e^{x-y}+1} \ge0
$$
and rearrange the inequality to obtain
$$(e^{\frac{x}{2}}-e^{\frac{y}{2}})^2 \leq \frac{(x-y)^2}{8}(e^x+e^y)$$


问:
Are there packages/databases for incidence matrices of abstract polytopes
I'd like to do computations with incidence matrices of abstract polytopes.
There's a database mentioned in this older post :

what abstract regular polyhedra exist?

but I don't see extract the incidence matrices from what's provided. I can find vertix-edge graphs
for a number of cases, but I need the full relations (vertix-edge, edge-face,...).
Are there any software packages (for GAP, Sage, ...) that have such a database and some utilities
to work with it? Even a small set of familiar examples (paltonic solids,...) would be good to start with.

答:
The incidence matrices in its generality, i.e. way beyond the regular solids only, is essentially my own elaboration. You might be welcome to look into my website at https://bendwavy.org/klitzing/home.htm which provides to my knowledge the most encompassing list of elaborated incidence matrices, even if all those have been worked out manually in the sequel of half a lifetime.

--- rk


问:
Analytic method for number theory-do we have to assert second-order logic?
I am an undergraduate. I am just starting to study logic and analytic number theory at the same time, so please forgive me if I made an elementary misunderstanding.

A lot of theorem in number theory can be stated completely in term of first order logic added with Peano axiom. But they were proven using analysis, and for many of them we do not know of any elementary method to do so, which means that second order logic needed to be used.

So I have 3 questions:

-Theorem in number theory need first order logic to be stated, so axiom of first order logic are assumed. But since their proof use second order logic, does that means that these theorems are, essentially, proved by asserting new axioms (ie. axioms of second order logic)? Would this pose any problems?

-Is there any theorem in number theory that is proved, and is also proved that all possible proof of that theorem must use more than first order logic?

-If we do not accept the axioms of second order logic, is it possible to construct a system of number satisfying Peano arithmetic such that: there is no possible way of using analysis? some of those theorem proved using analysis are false?

答:
mathematical proofs are generally/mostly thought of "living" in zfc (zermlo fraenkel & choice) set theory. If principles are used that go beyond that theory, then typically this needs to be made explicit. So, proving an "arithmetical" formula with means of zfc is a valid mathematical argument.

the field in which people try to find out which portion of analysis(i.e. second order number theory) are necessary to prove some statements in arithmetic is called reverse mathematics. See also wikipedia

to prove that a given part of second order arithmetic is necessary to prove a theorem at hand, one uses the theorem to prove the axioms of the given system - thus the name "reverse" mathematics.


问:
Find the line integral for the vector field: $F=x\hat{e}_x+z\hat{e}_y+x\hat{e}_z$
We have a vector field: $F=x\hat e_x+z\hat e_y+x\hat e_z$

I have to find the line integral:

$\displaystyle\oint_C \boldsymbol F \cdot d \boldsymbol l$

where the contour $C$ consists of: a straight line from the origin to $(0, 0, b)$; a
straight line from $(0, 0, b)$ to $(0, b, 0)$; and a straight line from $(0, b, 0)$ to the
origin,

答:
There seems to be a few mistakes.

In a vector field (not scalar), the line integral is given by

$I = \displaystyle \int_C \vec{F} \cdot d\vec{r} = \int_C \vec{F}(\vec{r}(t)) \cdot r'(t) \, dt$

a) Straight line from $(0,0,0)$ to $(0,0,b)$ can be parameterized as $\vec{r}(t) = (0,0,b)t \,$ where $0 \leq t \leq 1$.

$x = 0, y = 0, z = bt$

Vector field is $(x,z,x) = (0,bt,0)$

Line integral is $\displaystyle \int_0^1 (0,bt,0) \cdot (0,0,b) \,dt = 0 \,$ as the dot product is zero.

b) Straight line from $(0,0,b)$ to $(0,b,0)$ can be parameterized as $\vec{r}(t) = (0,0,b) + (0,b,-b)t \,$ where $0 \leq t \leq 1$.

$x = 0, y = bt, z = b-bt$

Vector field is $(x,z,x) = (0,b-bt,0)$

Line integral is $\displaystyle \int_0^1 (0,b-bt,0) \cdot (0,b,-b) \,dt = \int_0^1 (b^2-b^2t) \,dt = \frac{b^2}{2}$

Can you do similarly for straight line from $(0,b,0)$ to $(0,0,0)$? (it should be zero)

So your line integral over the given path is $\frac{b^2}{2}$.


问:
Computation of antiderivative at given points

Given $f(x)=
\left\{
\begin{array}{ll}
      -sin(x) & x \geq 0 \\
      1-x^{2} & x<0
\end{array}
\right.$

such that $F(\frac{\pi}{2})=0$. Calculate $F(\pi)+F(-1)$

The given function is discontinuous at x=0 so the integral can be split into two and can be calculated separately. With this approach i am able to get the integrals as $cos(x)+c_{1}$ for x>=0 and $x-x^{3}/3+c_{2}$ for x<0. At x=$\pi/2$, i am finding the value of $c_{1}$ which comes out to be 0. So at x=0 i am finding $c_{2}$ by using $cos(x)+c_{1}=x-x^{3}/3+c_{2}$.

With $c_{1}$=0 and $c_{2}$=1, i am plugging in the values $\pi$ and -1 respectively in $F(x)$ and am getting as $\frac{-2}{3}$.

Can you please let me know if the approach to the above formulation is correct?

答:
$ f $ is continuous at $ I= [0,+\infty) $, so it has an antiderivative at $ I $ given by
$$(\forall x\ge 0)\;\;F_1(x)=\cos(x)+C_1$$

By the same, it has an antiderivative at $(-\infty,0) $ given by
$$(\forall x<0)\;\; F_2(x)=x-\frac{x^3}{3}+C_2$$

$ F $ is continuous at $ 0 $, if $C_2=C_1+1$.

$ F(\frac{\pi}{2})=0 $ if $ C_1=0 $.

So, $$F(-1)+F(\pi)=-1+\frac 13+1-1=\frac{-2}{3}$$


问:
Sum of reciprocals of odd squares and integral of $\log(x)/(1-x^2))$
I would like to prove that $\int_{0}^{1}\frac{\log(x))}{1-x^2}\,dx=-\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}. $

I know that the RHS is equal to $-\frac{\pi^2}{8} $.

Additionally, I have tried to make the substitution $x=\sin(u)$ to the integral, which leaves with $\int_{u=0}^{\pi/2} \frac{\log(\sin(x))}{\cos(x)}\,dx $ but I can't seem to proceed from here.

答:
Using partial fraction expansion reveals

$$\frac1{1-x^2}=\frac12\left(\frac{1}{1-x}+\frac1{1+x}\right)$$

Next using the Taylor series for $\log(x)$ and owing to uniform convergence, we can integrate term by term to arrive at

$$\begin{align}
\int_0^1 \frac{\log(x)}{1-x}\,dx&=-\sum_{n=1}^\infty (-1)^{n-1}\int_0^1 \frac{(x-1)^{n-1}}{n}\,dx\\\\
&=\sum_{n=1}^\infty  (-1)^{n-1}\frac{(-1)^{n}}{n^2}\tag1
\end{align}$$

Next, using integration by parts on the integral $\int_0^1\frac{\log(x)}{1+x}\,dx$ we find that

$$\begin{align}
\int_0^1 \frac{\log(x)}{1+x}\,dx&=-\int_0^1 \frac{\log(1+x)}{x}\,dx\\\\
&=-\sum_{n=1}^\infty (-1)^{n-1} \int_0^1 \frac{x^{n-1}}{n}\,dx\\\\
&=-\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\tag2
\end{align}$$

Finally, adding $(1)$ and $(2)$ and dividing by $2$ yields

$$\begin{align}
\int_0^1 \frac{\log(x)}{1-x^2}\,dx&=-\frac12\sum_{n=1}^\infty \left(\frac{1+(-1)^{n-1}}{n^2}\right)\\\\
&=-\sum_{n=1}^\infty\frac1{(2n-1)^2}
\end{align}$$

as was to be shown!


问:
Sum of two random variables with $Geo(p)$ distribution
I need to prove that the following expression is a probability function

$$P(X+Y=k)={k+1\choose k}(1-p)^k p^2$$ with $$k=0,1,...$$

First of all is easy to verify that $f(x) \ge 0$, the problem comes when verifying that the sum adds up to 1, so far I have:

\begin{align}
P(X+Y=k) &= \sum_{x=0}^{k} P(X=x, Y=k-x)
\\& = \sum_{x=0}^{k} P(X=x) P(Y=k-x)
\\& = \sum_{x=0}^{k} (1-p)^xp(1-p)^{k-x}p
\\& = (k+1) (1-p)^k p^2
\end{align}

But now I'm stuck proving that that function sums 1, I imagine I must use the definition of $geo(p)$ but I don't really know how, can someone give me a hint on what to do next?

Body:
With $q:=1-p$, note that
\begin{align}
S &= \sum_{k = 0}^\infty (k+1) q^k
\\
S &= 1 + 2 q + 3 q^2 + \cdots
\\
S &= (1 + q + q^2 + \cdots) + q(1 + 2q + 3q^2 + \cdots)
\\
S &= \frac{1}{1-q} + qS
\\
S &= \frac{1}{(1-q)^2}
\end{align}


答:
You want to show that
$$
\sum_{k=0}^{\infty} P(X+Y=k) = 1
$$
Note that
\begin{align*}
\sum_{k=0}^{\infty} P(X+Y=k)
&=p^2\sum_{k=0}^{\infty} (k+1)(1-p)^k\\
&=p^2 + p^2\sum_{k=1}^{\infty} (k+1)(1-p)^k\\
&=p^2 + p^2 \left [\sum_{k=1}^{\infty} k(1-p)^k + \sum_{k=1}^{\infty} (1-p)^k \right ]\\
&=p^2 + p^2 \left [\frac{1-p}{p^2} + \frac{1-p}{p} \right ]\\
&= 1
\end{align*}

The first summation is evaluated by using this result, and the second is a standard geometric series

Title: Sum of two random variables with $Geo(p)$ distribution

Body:
I need to prove that the following expression is a probability function

$$P(X+Y=k)={k+1\choose k}(1-p)^k p^2$$ with $$k=0,1,...$$

First of all is easy to verify that $f(x) \ge 0$, the problem comes when verifying that the sum adds up to 1, so far I have:

\begin{align}
P(X+Y=k) &= \sum_{x=0}^{k} P(X=x, Y=k-x)
\\& = \sum_{x=0}^{k} P(X=x) P(Y=k-x)
\\& = \sum_{x=0}^{k} (1-p)^xp(1-p)^{k-x}p
\\& = (k+1) (1-p)^k p^2
\end{align}

But now I'm stuck proving that that function sums 1, I imagine I must use the definition of $geo(p)$ but I don't really know how, can someone give me a hint on what to do next?


问:
What does Hinich mean by "homotopy" and "contractible"?
In "Homological Algebra of Homotopy Algebras", Hinich talks about homotopies and contractible (cochain) complexes. More specifically, I want to look at the following two points:

§4.3, see also §6.5. Let $(V,d)\in C(k)$ be an unbounded (cochain) complex over a ring $k$ and $\alpha:V\rightarrow V$ be a map of chain complexes of degree $0$. Hinich implies that the definition of a homotopy between $\text{id}_V$ and $\alpha$ is a map $h:V\rightarrow V$ of degree $-1$ such that
$$
dh = \text{id}_V-\alpha.
$$
This doesn't seem to match the usual definition of a chain homotopy: a map $h$ of degree $-1$ such that $dh+hd=\text{id}_V-\alpha$. Am I missing something or badly interpreting Hinich's statement?

§4.4, see also §6.6. Hinich introduces $(X,d_X)\in C(k)$, a contractible complex. While he does not clearly say what this means, I think* he means that there exists a "homotopy" $h$ between $\text{id}_X$ and $0$, that is, $d_Xh=\text{id}_X$. This is strange, as it would imply that the differential $d_X$ is surjective.

*The reason for this is that Hinich introduces another complex $A^\#$, defines the complex $(V,d)$ by $V=A^\#\oplus X$ and the map $\alpha:V\rightarrow V$ as $\alpha:A^\#\oplus X\rightarrow A^\#\rightarrow A^\#\oplus X$, and then says that there exists a homotopy $h$ that vanishes on $A^\#$ such that $dh=\text{id}_V-\alpha$ (it is actually written $dh=\alpha$, but I believe this is a mistake; compare with the definition of homotopy and with the similar statement in §6.6). Restricting the equation to $X$, we get $d_Xh=\text{id}_X$.

I'm not too confident in my interpretation of Hinich's statements, so I would be grateful if anyone could clarify these issues that I have :)

答:
Concerning your first question: d(h) = d∘h + h∘d (by definition), so Hinich's definition is the standard one.

For the second question, having d(h) = id defines a contractible chain complex,
since d(h) = d∘h + h∘d = id.


问:
The projection of $v_1-v_2$

Let $v_1$ and $v_2$ be two orthogonal vectors in $\mathbb{R}^4$ in respect with standard dot product and $U=span\{v_1,v_2\}$. Prove that the projection of $v_1 - v_2$ onto $U$ is $v_1 - v_2$.I know that $\text{Proj}_U(v_1-v_2) = \frac{<v_1,  v_1-v_2>}{ \|v_1^2\|} v_1 + \frac{<v_2 , v_1-v_2>}{\|v_2^2\|} v_2 $.In this point I am stuck because I can't prove what I have asked .

答:
If $v_1$ and $v_2$ are orthogonal, then the projection of a vector $w$ onto $U=\text{span}\{v_1, v_2\}$ is

$$\text{Proj}_U(w) = \frac{v_1 \cdot w}{v_1 \cdot v_1} v_1 + \frac{v_2 \cdot w}{v_2 \cdot v_2} v_2.$$
(Check your notes for something to this effect.)

You can then check that if $w$ is a linear combination of $v_1$ and $v_2$, then $\text{Proj}_U(w)=w$ by plugging it into the above formlua.

Title: The projection of $v_1-v_2$


问:
Combinatorial proof that $\sum_{i=0}^n 2^i\binom{n}{i}i!(2n-i)! = 4^n(n!)^2$
I'm looking for a combinatorial proof that $$\sum_{i=0}^n 2^i\binom{n}{i}i!(2n-i)! = 4^n(n!)^2.$$

My thoughts so far: the RHS counts the number of pairs of permutations on $n$ elements along with an $n$-tuple whose entries come from 4 choices.
The LHS might count the same thing but partitioned into cases somehow.

答:
Alternatively, a more combinatorial flavor approach:

Multiply both sides by $\binom{2n}{n}$ so you get
$$\sum_{i=0}^n 2^i\binom{2n}{\color{red}{n}}\binom{\color{red}{n}}{i}i!(2n-i)! = 4^n(n!)^2\frac{(2n)!}{n!^2},$$
then, using that $\binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c},$ we get
$$\sum_{i=0}^n 2^i\binom{2n}{i}\binom{2n-i}{n-i}i!(2n-i)! = 2^{2n}(2n)!,$$ cancelling the factorials, we get
$$\sum_{i=0}^n 2^i\binom{2n-i}{n} = \underbrace{2^0\binom{2n-0}{n}}_{*_1}+2\sum_{i=1}^n 2^{i-1}\binom{2n-i}{n}=2^{2n}.$$
This last expression can be proved as follows: Consider a binary string $x\in \{0,1\}^{2n},$ is clear that either $\underbrace{|x|_0= |x|_1}_{*_1}$ or $|x|_0\neq |x|_1,$  where $|x|_a$ is the number of symbols equal to $a$ in the string.
If it is the first case(i.e., $*_1$), then there are $\binom{2n}{n}=2^0\binom{2n-0}{n}$ ways to do this. If not, then is clear that at some point(going from left to right) in the string, there is a symbol(we have $2$ ways to choose which symbol) that will occur $n+1$ times, call this point $2n-i+1$ with $1\leq i\leq n.$ You need to pick where are the remaining $n$ symbols to the left in $\binom{2n-i}{n}$ and it does not matter what happens to the right so there are $2^{2n-(2n-i+1)}=2^{i-1}$ ways to do this. By double counting, the LHS is equal to the RHS.

Going backwards, and considering the combinatorial interpretation of each step, one can construct a story proof around this with the subject being signed permutations of $[2n].$ For example, when you divide by the $\binom{2n}{n}$ you are saying that instead of considering any permutation, you are going to consider just permutations that the first $n$ are in the first $n$ and respectively the last $n.$ So, essentially, you have in the RHS colored permutations of tuples of $n!,$ in the LHS the same, but taking into account where do you see the $n+1-$th occurrence of the most frequent symbol in the coloring.


问:
Calculating the sup and inf of the expression $\frac{\sqrt{n^2+k}-n}{\sqrt{n^2+2k}-n}$, where $n,k \in \mathbb{N}$
I need to calculate supremum and infimum of a set given by:
$$\frac{\sqrt{n^2+k}-n}{\sqrt{n^2+2k}-n}$$

for $n, k \in \mathbb{N} \setminus \{0\}$

So, I see that: $\frac{\sqrt{n^2+k}-n}{\sqrt{n^2+2k}-n}= \frac{\sqrt{1+\frac{k}{n^2}}-1}{\sqrt{1+\frac{2k}{n^2}}-1}$

I know for sure that the expression is bigger when $\frac{k}{n^2}$ is smaller because there are 2 of these in denominator and just one in nominator. So I know that I will get:

superemum for $k = 1$, $n \implies \infty$

infimum for $n = 1$, $k \implies \infty$

I get:

INF = $\lim_{k \to +\infty} \frac{\sqrt{1+k}-1}{\sqrt{1+2k}-1} = \frac{1}{2}$

SUP = $\lim_{n \to +\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n^2+2}-n} = ?$

And that's it. I cant find that limit. If I just take $n^2$ out, I get $\frac{0}{0}$. If I make it a difference of squares I get $\infty - \infty$ in nominator.

Body:
First we rewrite the ratio as
$$R(n,k):=\frac{\sqrt{n^2+k}-n}{\sqrt{n^2+2k}-n} = \frac{1 \cdot (\sqrt{n^2+2k}+n)}{2 \cdot (\sqrt{n^2+k}+n)} = \frac{\sqrt{1+\frac{2k}{n^2}}+1}{2(\sqrt{1+\frac{k}{n^2}}+1)}$$

Note that
$$
\frac{\sqrt{1+\frac{2k}{n^2}}+1}{2(\sqrt{1+\frac{k}{n^2}}+1)} > \frac{\sqrt{1+\frac{k}{n^2}}+1}{2(\sqrt{1+\frac{k}{n^2}}+1)} = \frac 12 \tag1
$$
and
$$
\lim_{n\to\infty, k=1}\text{LHS of }(1)=\frac{1}{2}
$$

On the other hand
$$
\frac{\sqrt{1+\frac{2k}{n^2}}+1}{2(\sqrt{1+\frac{k}{n^2}}+1)} < \frac{\sqrt{2+\frac{2k}{n^2}}+\sqrt 2}{2(\sqrt{1+\frac{k}{n^2}}+1)} = \frac{\sqrt 2}{2} \tag2
$$
$$
\lim_{n \to \infty, k=n^2} \text{ LHS of (2) } = \frac{\sqrt{1+2n^2}+1}{2(\sqrt{1+n^2}+1)} = \lim_{n \to \infty} \frac{\sqrt{\frac{1}{n^2}+2}+\frac 1n}{2(\sqrt{\frac{1}{n^2}+1}+\frac 1n)} = \frac{\sqrt 2}{2}.
$$

Therefore $\inf R(n,k)=\frac 12, \sup R(n,k)=\frac{\sqrt 2}{2}$


问:
Determine the points where $f$ is has a local minimum/maximum. Multivariable calculus question.
This is not homework, but it is in my book and I find it hard to solve:

Determine the points where $f$ is has a local minimum/maximum.
  Determine if it strong/weak and absolute/relative and interior
  extremum or boundary extremum.

$$f(x,y)= (x^2-y^2)e^{-x^2-y^2}$$
  The domain of $f$ is $E=\{(x,y): x^2+y^2 \leq 4 \}$

A worked solution would be really appreciated. Or if someone knows some worked solutions of exercises like these (somewhere on the internet), that would be great as well. Or a step-by-step plan how to solve exercises like these. In other words, any help is really appreciated.

答:
Well, on the boundary of $E$, we have $f(x,y)=(2x^2-4)e^{-4}$ (why?). This shouldn't be difficult to maximize/minimize, bearing in mind that $-2\le x\le 2$.

To find critical points on the interior of $E$, start by finding the first partial derivatives of $f$, and setting both to $0$. Solve this system, and see what solutions $(x,y)$ have the property that $x^2+y^2<4$. Once any such critical points have been found, find all the second partials of $f$, and use the second partial derivative test to determine which of them (if any) are extrema.

Classify all the extrema as directed.


问:
Tensor products and sum
We know that tensor product commutes with direct sum. I am wondering if tensor product commutes with sum (or finite sum.)

I.e.) $N\bigotimes (\sum M_i)=\sum(N\otimes M_i)$

$N,M_i$ are $A$ module. ($A$ is commutative ring with unit)

I couldn't find a counter example. Please lead me to find one (if the statement is wrong).

答:
Yes. Tensor products are left adjoint to taking homs, and so they commute with arbitrary colimits. In particular

$$N \otimes \left ( \sum M_\alpha \right ) = \sum (N \otimes M_\alpha)$$

I'm sure there is a more pedestrian way to see this as well, but as some pople say: "category theory exists to make obvious things obviously obvious". I think this is a good example of that principle in action!

After searching, I found a previous question that avoids abstract nonsense.

I hope this helps ^_^

Title: Tensor products and sum


答:
From
$$ p=\dfrac{-3}{8x^3}p^2p'+\dfrac{3}{8x^4}p^3$$
one has
$$ p'=\frac1xp-\frac{8x^3}{3}p^{-2}. $$
This is a Bernoulli equation. Letting $u=p^{3}$ and then
$$ u'=3p^{2}p'=3p^{2}\bigg(\frac1xp-\frac{8x^3}{3}p^{-2}\bigg)=\frac{3}{x}u-8x^3 $$
which is a linear DE and you can do the rest.

Title: Nonlinear differential equation. $p^3-4px^4+8yx^3=0$

Body:
Nonlinear differential equation.
Let $p=\dfrac{dy}{dx}$
$$p^3-4px^4+8yx^3=0$$

I tried to make $y$ alone and take derivative wrt. $x$ as

$$y=\dfrac{p^3}{-8x^3}+\dfrac{xp}{2}$$

then

$$y'=p=\dfrac{-3}{8x^3}p^2p'+\dfrac{3}{8x^4}p^3$$

but cannot solve for $p$


答:
You can observe that your numbers can also be written as
$$
\dfrac{\dfrac{k}{n^2}\Bigl(\sqrt{1+\dfrac{2k}{n^2}}+1\Bigr)}
{\dfrac{2k}{n^2}\Bigl(\sqrt{1+\dfrac{k}{n^2}}+1\Bigr)}
=
\frac{1}{2}\dfrac{\sqrt{1+\dfrac{2k}{n^2}}+1}
{\sqrt{1+\dfrac{k}{n^2}}+1}
$$
If you take $k=rn^2$, you get
$$
\frac{1}{2}\frac{\sqrt{1+2r}+1}{\sqrt{1+r}+1}
$$
whose limit for $r\to\infty$ is $\sqrt{2}/2$.

Can you prove this is the supremum?

If you take $k=0$ (I know it's invalid, but it's certainly a starting point with the new expression, isn't it?), you get $1/2$. Can you prove this is the infimum?

Title: Calculating the sup and inf of the expression $\frac{\sqrt{n^2+k}-n}{\sqrt{n^2+2k}-n}$, where $n,k \in \mathbb{N}$

Body:
I need to calculate supremum and infimum of a set given by:
$$\frac{\sqrt{n^2+k}-n}{\sqrt{n^2+2k}-n}$$

for $n, k \in \mathbb{N} \setminus \{0\}$

So, I see that: $\frac{\sqrt{n^2+k}-n}{\sqrt{n^2+2k}-n}= \frac{\sqrt{1+\frac{k}{n^2}}-1}{\sqrt{1+\frac{2k}{n^2}}-1}$

I know for sure that the expression is bigger when $\frac{k}{n^2}$ is smaller because there are 2 of these in denominator and just one in nominator. So I know that I will get:

superemum for $k = 1$, $n \implies \infty$

infimum for $n = 1$, $k \implies \infty$

I get:

INF = $\lim_{k \to +\infty} \frac{\sqrt{1+k}-1}{\sqrt{1+2k}-1} = \frac{1}{2}$

SUP = $\lim_{n \to +\infty} \frac{\sqrt{n^2+1}-n}{\sqrt{n^2+2}-n} = ?$

And that's it. I cant find that limit. If I just take $n^2$ out, I get $\frac{0}{0}$. If I make it a difference of squares I get $\infty - \infty$ in nominator.


问:
How to show that $A=B-C$
How to show that for a real symmetric matrix $A,~A$ can be written as $A=B-C$ where $B,C$ are positive definite real symmetric matrices?

Please help me ! I'm clueless.

答: Let $C=cI$ where $I$ is the identity matrix and $c\gt0$ is chosen so that for each eigenvalue $\lambda$ of $A$, $c+\lambda\gt0$. </p>

Title: How to show that $A=B-C$


问:
Correlation of linearly related variables
I was reading an article in which they show a correlation table that looks very suspicious to me. I am not a mathematician, so I would like extra advice on this issue. I can summarize the problem as follows.

Consider these 2 sets of random variables: $\{X_i, Y_i, Z_i\}$ for $i \in \{1,2\}$.

We know that for all $i$, $Z_i=1-X_i-Y_i$.

The table presented in the paper, shows 3 values of $\rho_K$ for $K \in \{X,Y,Z\}$, where $\rho_k$ is the Pearson correlation  coefficient between $K_1$ and $K_2$.

In their paper, both $\rho_X$ and $\rho_Y$ are high (above 0.7) and statistically significant, but $\rho_Z$ is low and not significant.

Is this even possible?? I ran a few simulations in R and the simplest scenario with independent Xs and Ys and definitely I end up with a high $\rho_Z$. I also tried to linearly relate (in my simulations) $X_i$ to $Y_i$ and still end up with high and significant $\rho_Z$. What am I missing?

Thanks!

答:
Yes, it's possible. Even more extreme scenarios are possible.
Start with $Z_1$ and $Z_2$ that are uncorrelated, so $\rho_Z = 0$.
Let $W$ be some random variable with large variance.  Let $X_i = 1-W$ and
$Y_i = 1 - X_i - Z_i = W - Z_i$.  Then $\rho_X = 1$ and $\rho_Y$ is close to $1$.


问:
How do I prove that $\sum _{n=1}^{\infty }\:\ln\left(1+\frac{1}{n^2}\right)$ converges without applying the series integral test?
I am at a loss as to how I can prove that the sum $\sum_{n=1}^{\infty }\:\ln\left(1+\frac{1}{n^2}\right)$ converges. At first I tried the ratio test, but that came out inconclusive. Then I tried it by showing that the sequence of partial sums is bounded, to no avail. If anyone could give me a hint on this, it would be much appreciated!

答: HINT: $0 \le \ln(1+n^{-2}) \le n^{-2}$ for all integers $n$.


问:
Simple probability question that I can't figure out
Given 10 cards with numbers 1,2,3...10, 2 distinct cards are randomly chosen and duplicated, and then the 12 cards are shuffled.

What is the probability that the numbers 1,2,3,4 are the numbers on the first 4 cards (in that order)?

Given that the above is true, what is the probability that 1 was duplicated?

答:
$\color{Blue}{\textbf{Part 1 }}$

Heya don't beat yourself up this question is definitely very fiddly. We are going to solve it together. We are going to solve the problem using conditioning. Here are the four steps we will take:

Define some new events based on what cards are duplicated.

Work out the probability of these events.

Work out the probability of success conditioned on these events.

Put this all together using the law of total probability

$\textbf{Step 1:}$

Define $A_0,A_1$ and $A_2$ the events that respectively when duplicating we get $0,1$ or $2$ of the "helpful cards", the $1,2,3 $ or $4$. I.e if we duplicate a 3 and a 7 then $A_1$ has occured. Instead if we duplicated the $1$ and $4$ then $A_2$ occurred. There is a term for these three events and mathematicians call them a "disjoint partition". This is because 1 of them and exactly 1 of them must happen.

$\textbf{Step 2:}$

We know wish to work out the probability of the $A_i$ events.

$\mathbb{P}(A_2)=\frac{4}{10}\cdot\frac{3}{9} = \frac{12}{90}$

$\mathbb{P}(A_0)=\frac{6}{10}\cdot\frac{5}{9} = \frac{30}{90}$

$\mathbb{P}(A_1)=1-\mathbb{P}(A_2)-\mathbb{P}(A_0)= \frac{48}{90} $

$\textbf{Step 3:}$

We now wish to work out the probability of success in 3 separate circumstances. Let us call $S$ success. We will first calculate $\mathbb{P}(S|A_0)$, this is the probability of success given we didn't manage to duplicate any helpful cards. And as such there is exactly one and only one combination of the initial first four cards for success. We must draw the $1$ followed by the $2$ followed by the $3$ followed by the $4$. And as such we have only $1$ choice for the first 4 cards and after this we then have $8!$ ways to order the remaining cards and hence the number of successful combinations is $\frac{8!}{12!} = \mathbb{P}(S|A_0)$

Now onto $\mathbb{P}(S|A_1)$.

Instead of having only $1$ successful combination for our initial four cards  we now have $2$. Let us say the duplicated card was the $3$. And call the original three $ 3_O$ and the duplicated three $3_D$. Our two successful first four card combinations are $1$ followed by the $2$ followed by the $3_O$ followed by the $4$ and $1$ followed by the $2$ followed by the $3_D$ followed by the $4$. Hence we have twice as many successful combinations and as such $\mathbb{P}(S|A_1) = 2\cdot \mathbb{P}(S|A_0) = \frac{2\cdot8!}{12!}$

And finally $\mathbb{P}(S|A_2)$.

You might have guessed it but we know have four successful combinations for our initial four cards. There are now two cards that we have 2 different choices for. For sake of a nice example let us say the 2 and 4 have been duplicated. The four successful initial combinations are: $(1,2_O,3,4_O),(1,2_O,3,4_D),(1,2_D,3,4_O),(1,2_D,3,4_D)$

And hence $\mathbb{P}(S|A_2)= 4\cdot \mathbb{P}(S|A_0) = \frac{4\cdot8!}{12!}$

$\textbf{Step 4:}$

The law of total probability states that if $A_1, A_2 , A_3, ..., A_n$ are a disjoint partition then for any event $X$ we have $\mathbb{P}(X) = \sum\limits_{i=1}^n\mathbb{P}(X|A_i)\cdot\mathbb{P}(A_i)$

We have our disjoint partition and their probabilities and we have our probability of success conditioned on this partition so let us put this all together:

$\mathbb{P}(S) = \sum\limits_{i=1}^n\mathbb{P}(S|A_i)\cdot\mathbb{P}(A_i) = \mathbb{P}(S|A_0)\cdot\mathbb{P}(A_0) + \mathbb{P}(S|A_1)\cdot\mathbb{P}(A_1) + \mathbb{P}(S|A_2)\cdot\mathbb{P}(A_2) =$

$\frac{8!}{12!} \cdot \frac{30}{90} + \frac{2\cdot8!}{12!} \cdot \frac{48}{90} + \frac{4\cdot8!}{12!} \cdot \frac{12}{90} =\frac{8!}{90\cdot 8!}(30+96+48) = \frac{174\cdot8!}{90\cdot 12!} \approx 0.0162 \% $ This is the answer to your first question.

$\color{Blue}{\textbf{Part 2 }}$

Now we will move onto the next part. The probability that a 1 was duplicated given we succeeded. We will do this in three steps:

Use Bayes Formula to work out the probability of which $A_i$ occurred given success

Work out the probability of a $1$ being duplicated given what $A_i$ occurred

Use the law of total probability to finish the problem

$\textbf{Step 1:}$

Bayes formula gives us $\mathbb{P}(A_i|S) = \frac{\mathbb{P}(S|A_i)\cdot\mathbb{P}(A_i)}{\mathbb{P}(S)}$ We have already worked out each of these values so I will just list them here for you, all the factorials cancel out so the answers are nice and simple:

$\mathbb{P}(A_0|S) = \frac{5}{29}$

$\mathbb{P}(A_1|S) = \frac{16}{29}$

$\mathbb{P}(A_2|S) = \frac{8}{29}$

$\textbf{Step 2:}$

Call $D$ the event that a $1$ was duplicated. We know would like to find out the probability a $1$ was duplicated under each of the $A_i$

$\mathbb{P}(D|A_0)$ is clearly $0$ as $A_0$ means none of $1,2,3$ or $4$ were duplicated.

$\mathbb{P}(D|A_1)$ is clearly $\frac{1}{4}$ as under $A_1$ only one of $1,2,3,4$ were duplicated each with equal likelihood

$\mathbb{P}(D|A_2)$ is clearly $\frac{1}{2}$ as under $A_1$ two of $1,2,3,4$ were duplicated each with equal likelihood

$\textbf{Step 3:}$

Again using the law of total probability we have: $\mathbb{P}(D|S) = \mathbb{P}(D|A_0)\cdot\mathbb{P}(S|A_0) + \mathbb{P}(D|A_1)\cdot\mathbb{P}(S|A_1) + \mathbb{P}(D|A_2)\cdot\mathbb{P}(S|A_2) = $

$0 \cdot \frac{5}{29} + \frac{1}{4} \cdot \frac{16}{29} + \frac{1}{2} \cdot \frac{8}{29} = \frac{24}{116}  =\frac{6}{29} \approx 21\% $

I hope this helped. This was a fiddly question so no worries for having found it hard. If you would like anything made more clear let me know.


问:
Monte Carlo simulation: confidence interval of the ratio of two beta distributions
I am trying to estimate the accuracy of a set of Monte Carlo simulation where my result is

\begin{equation}
C=1-\frac{P_X(1)}{(P_Y(1))^2}
\end{equation}

and $X$ and $Y$ are the results of two separate experiments, so they are independent. In particular, both simulations only outputs zeros or ones so what I am measuring is the probability of getting a 1 in each of the two indipendently. To give an example, it's as if as you have an unfair coin and you are trying to determine its balancing by throwing it many times.

I already know that the conjugate prior of $X$ and $Y$ is a beta distribution. I am trying to compute how accurate my results can be given a sample of size $N$, and I would be satisfied if I could find analytically the variance $\text{Var}(C)$.

I started by simplifying a bit and assume beta symmetric, which yields $\text{Var}(X) = \frac{\mu (1-\mu)}{1 + N_r}$ where $\mu$ is mean and $N_r$ is the size of the sample. Similarly, $\text{Var}(Y^2)$ itself shouldn't be a problem. If it was by itself, considering that the sample size is large, I could approximate it with a Gaussian and compute the variance of its square as in Mean and variance of Squared Gaussian: $Y=X^2$ where: $X\sim\mathcal{N}(0,\sigma^2)$? .

However, I already know the variance and average of the inverse of a normal do not exist, so how do you compute the variance on this ratio? Also, if you think that trying to compute the variance is not the right approach, feel free to suggest something better.

答:
If I correctly understand the things posted in comments, you have $X,Y\sim\text{i.i.d.} \operatorname{Uniform}(0,1)$ and two sequences of Bernoulli random variables, both conditionally i.i.d. given $X,Y,$ with respective probabilities $X$ and $Y$ of being equal to $1.$

The conditional distributions of $X,Y$ given the numbers of $1\text{s}$ and $0\text{s}$ for both of those sequences are given by
\begin{align}
& \Pr(X\in A\mid\alpha,\beta,\gamma,\delta) = \int_A \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1} (1-x)^{\beta-1}\, dx \text{ for } A \subseteq[0,1] \\[8pt]
& \Pr(Y\in A\mid\alpha,\beta,\gamma,\delta) = \int_A \frac{\Gamma(\gamma+\delta)}{\Gamma(\gamma)\Gamma(\delta)} x^{\gamma-1} (1-x)^{\delta-1}\, dx \text{ for } A \subseteq[0,1] \\[6pt]
& \text{where } \alpha,\beta \text{ are the respective numbers of 1s and 0s from the first} \\
& \text{sequence and } \delta,\gamma \text{ from the second.}
\end{align}
(This result is central to the famous posthumous paper of Thomas Bayes published in 1763.)

As an estimate of $C = 1 - \dfrac X {Y^2}$ my first thought is
\begin{align}
& \operatorname E(C\mid\alpha,\beta,\gamma,\delta) \\[8pt]
= {} & \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)} \cdot \frac{\Gamma(\delta+\gamma)}{\Gamma(\delta)\Gamma(\gamma)} \iint\limits_{[0,1]^2} \left( 1 - \frac x{y^2} \right) x^{\alpha-1} (1-x)^{\beta-1} y^{\gamma-1} (1 - y)^{\delta -1} \, d(x,y) \\[8pt]
= {} & 1 - \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)} \cdot \frac{\Gamma(\delta+\gamma)}{\Gamma(\delta)\Gamma(\gamma)} \cdot \iint\limits_{[0,1]^2} x^\alpha (1-x)^{\beta-1} y^{\gamma-3} (1 - y)^{\delta -1} \, d(x,y) \\[8pt]
= {} & 1 - \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)} \cdot \frac{\Gamma(\delta+\gamma)}{\Gamma(\delta)\Gamma(\gamma)} \left( \frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma(\alpha+1+\beta)} \cdot\frac{\Gamma(\gamma-2) \Gamma(\delta)}{\Gamma(\gamma-2+\delta)} \right) \\[8pt]
= {} & 1 - \frac{(\gamma+\delta-2)(\gamma+\delta-1) \alpha }{(\alpha+\beta)(\gamma-2)(\gamma-1)}.
\end{align}


答:
You know that $$\lim_{n\to +\infty}n^2\ln(1+\frac{1}{n^2})=1$$

So, for $ n $ great enough
$$n^2\ln(1+\frac{1}{n^2})\le 2$$

or

$$\ln(1+\frac{1}{n^2})\le \frac{2}{n^2}$$

Conclude by comparison test.

Title: How do I prove that $\sum _{n=1}^{\infty }\:\ln\left(1+\frac{1}{n^2}\right)$ converges without applying the series integral test?

Body: I am at a loss as to how I can prove that the sum $\sum_{n=1}^{\infty }\:\ln\left(1+\frac{1}{n^2}\right)$ converges. At first I tried the ratio test, but that came out inconclusive. Then I tried it by showing that the sequence of partial sums is bounded, to no avail. If anyone could give me a hint on this, it would be much appreciated!


问:
Probability of sequence
The question is as follow: Let $(X_{n})_{n}$ be a sequence of Random variable that is independent with the probability

$P(X_{n}=1)=1-P(X_{n}=0)=\frac{1}{n}$

Show that $P\left( \underset { n\rightarrow \infty  }\liminf{ X }_{ n }=0 \right) =1$

I tried to use the Borel-Cantelli Theorem to prove this, however I obtain that

$P\left( \underset { n\rightarrow \infty  }\liminf{ X }_{ n }=0 \right) =0$

${ F }_{ n }:=\left\{ { X }_{ n }=0 \right\} $

$\sum _{ n=1 }^{ \infty  }{ P\left( { F }_{ n }^{ C } \right)  } =\sum _{ n=1 }^{ \infty  }{ 1-P\left( X_{ n }=0 \right)  } =\sum _{ n=1 }^{ \infty  }{ \frac { 1 }{ n }  } =\infty$

By Borel-Cantelli, we know that $P\left( { \underset { n\rightarrow \infty  }\limsup  }{ F }_{ n }^{ C } \right) =1$

Since ${ \left( { \underset { n\rightarrow \infty  }{ \liminf }  }{ F }_{ n } \right)  }^{ C }=\left( { \underset { n\rightarrow \infty  }{\limsup}  }\quad { F }_{ n }^{ C } \right) $,

(ie) $P\left( { \underset { n\rightarrow \infty  }{ \liminf }  }\quad { F }_{ n } \right) =1-P\left( { \underset { n\rightarrow \infty  }{ \limsup }  }\quad { F }_{ n }^{ C } \right)$,

I obtain that $P\left( { \underset { n\rightarrow \infty  }{ \liminf }  }\quad { X }_{ n }=0 \right) =0$

Since this is the opposite of what I want to prove and the statement is said to be true, I can't see where I went wrong in my argument.

答: Note $1-P(X_n=0)=1/n$ implies $P(X_n=0)=1-1/n$ or $P(X_n=0)=(n-1)/n$ and apply Borel-Cantelli.  Hope this helps.</p>

Title: Probability of sequence


问:
Measure on Borel sets
If $\mu_1,\mu_2$ are two Borel measures with $\mu_1([a,b])\le \mu_2([a,b])$ for all $a<b$, how we can prove that $\mu_1(A)\le \mu_2(A)$ for all Borel sets $A$? (All sets are from $\mathbb{R}$)
Unfortunately, the set of intervals $[a,b]$ is not even a semi-ring, so I don't even know how to start here. Would be nice if someone has some ideas about it.
Thanks in advance.

答:
Disclaimer: This answer is much longer than a solution to this problem needs to be because I'm explaining everything in detail -- you can easily write this down in 2-3 lines.

I don't know if you're familiar with the Monotone Class Lemma, but this seems like a standard application. Anything that relates the properties of measures that hold under limits (in this case, weak inequality) is usually solvable using this lemma. So, for future reference, if you're up against a problem that asks gives you a property of measures that holds for a certain class of "nice" sets (in this case, intervals), and th eproperty holds under limits (in this case, the inequality), then MCL is probably a useful tool.

Step 1. Notice that this holds for half open half closed intervals of the form $(a,b]$. Now that $(a,b] = \bigcup_{n \in \mathbb{N}} [a+\frac{1}{n}, b]$ so,
$$
\mu_{1}((a,b]) = \lim_{n \to \infty} \mu_{1}([a + \frac{1}{n}, b]) \leq lim_{n \to \infty} \mu_{2}([a+\frac{1}{n}, b]) = \mu_{2}((a,b])
$$

Step 2. Recall the Monotone Class Lemma (or look it up, if you aren't familiar with it): If $\mathcal{A}$ is an algebra then the $\sigma$-algebra generated by $\mathcal{A}$, denoted $\sigma(\mathcal{A})$, is equal to the smallest monotone class containing $\mathcal{A}$, denoted $\mathcal{M}(\mathcal{A})$. A monotone class is a collection of sets which is closed under increasing unions and decreasing intersections.

Step 3. We will now apply this result. Our algebra $\mathcal{A}$ will be the algebra generated by half open half closed intervals $(a,b]$ (which, you may recall, consists of disjoint unions of half-open half-closed intervals). If we can show that the set
$$
B = \{A \in \mathcal{B}: \mu_{1}(A) \leq \mu_{2}(B)\}
$$
is a monotone class which contains $\mathcal{A}$, then it must contain $\mathcal{M}(A)$, so by the Monotone Class Lemma is must contain $\sigma(\mathcal{A})$, i.e. all Borel sets. But this is fairly straightforward, since if $\{A_{n}\}_{n \in \mathbb{N}}$ is a collection in $B$ such that $A_{n} \uparrow A$, then $\mu_{1}(A) = \lim_{n \to \infty} \mu_{1}(A_{n}) \leq \lim_{n \to \infty} \mu_{2}(A_{n}) = \mu_{2}(A)$ (and same for decreasing sequences), which shows that $A \in B$. This shows that $B$ is a monotone class and we're done.

Title: Measure on Borel sets


问:
Find the transfer function and find the stationary solution associated with $u(t) = \cos(t)$
Problem

Given the system
$$\mathbf{x}'(t)= \begin{bmatrix}2 & -1 \\ 3 & 1 \end{bmatrix}\mathbf{x}(t)+\begin{bmatrix}0 \\-1 \end{bmatrix}u(t) $$
$$y(t) = 2x_1(t)-x_2(t) $$

Find the transfer function for the linear system.

Find the stationary solution associated with $u(t)=\cos(t)$

My attempt

1)
The transfer function is easily found with the formula:
$$H(s)=-\mathbf{d}^T(\mathbf{A}-s\mathbf{I})\mathbf{b}$$
We see that $-\mathbf{d}^T= \begin{bmatrix} -2 \\1 \end{bmatrix}$, $\mathbf{A}=\begin{bmatrix}2 & -1 \\ 3 & 1 \end{bmatrix}$, $\mathbf{b}=\begin{bmatrix}0 \\-1 \end{bmatrix}$, and $\mathbf{I}$ is the identity matrix. Plugging in these values in the formula and using Maple gives
$$H(s)=\frac{s}{s^2-3s+5},s^2-3s+5 \neq0 $$

2)

In my book, it says that the stationary solution associated with $u(t)=e^{st}$ is $y(t)=H(s)e^{st}$.

But what do I do when $u(t)=\cos(t)$?

I have the solution to this problem, and it says that the stationary solution is $y(t)=-\frac{3}{25}\cos(t)-\frac{4}{25}\sin(t)$.

Can someone show me how they arrive at that answer? It seems like you have to plug in a value for $s$, but I'm not sure what.

答:
if you apply a cosine or a sine to a transfer function, the resulting stationary output will be also a cosine or a sine with an altered phase and magnitude. This is because this $s$ can be seen as a frequency variable $s = jw$. This means that if the input of the following form:
$$u(t) = R\text{cos}(wt)$$
the stationary output is something like this:
$$R|H(jw)|cos(wt+\angle H(jw))$$
Hope this helps


问:
Periodicity of solution to initial value problem
Consider the ODE $$\begin{cases}
x''(t)+x(t)+x(t)^3=0\\
(x(0),x'(0))=(x_0,0)\\
\end{cases}$$ where $x(t)$ takes values in $\mathbb{R}$ ,Prove that for all $x_0\in\mathbb{R}$ ,the solution of the above system is periodic.
It's a question of S.-T.Yau College Student Mathematics Contest in 2020.
I assume $p=x'(t)$ , and  get the equation $$\frac{dp}{dx}p+x+x^3=0$$ Then know $$2p^2+2x^2+x^4=2x_0^2+x_0^4$$ It's too complicated to solve this equation, and I don't know what to do afterward. How to get periodicity without solving the equation?

答:
You don't have to solve. Note that
$$2p^2+2x^2+x^4=2x_0^2+x_0^4$$
is a closed curve and hence $(x,p)$ is periodic.


问:
Formula for the angle bisector length with area and two angles
How can I proof $w_\gamma
=\dfrac{2F}{c\cos \left(\frac{\alpha -\beta }{2}\right)}$ ?

Where $w_\gamma$   is the length of the angle bisector of the side $c$

and  $\alpha$, $\beta$ are the angles of the vertices $A$, $B$

and $F$ is the area of the triangle.

I have no idea how to start.

答:
Without loss of generality, let's assume that $|AC|>|BC|$.
Then

\begin{align}
\triangle CDH_c:\quad
w_{\gamma}=|CD|
&=
\frac{|CH_c|}{\cos\angle H_cCD}
\tag{1}\label{1}
\\
&=\frac{|CH_c|}{\cos(\tfrac12\gamma-(90^\circ-\beta))}
\tag{2}\label{2}
\\
&=\frac{|CH_c|}{\cos(\tfrac12(180^\circ-\alpha-\beta)-(90^\circ-\beta))}
\tag{3}\label{3}
\\
&=\frac{|CH_c|}{\cos\tfrac{\beta-\alpha}2}
=\frac{|CH_c|}{\cos\tfrac{\alpha-\beta}2}
\tag{4}\label{4}
.
\end{align}

As @player3236 pointed out in a comment,

\begin{align}
|CH_c|&=\frac{2F}c
\tag{5}\label{5}
.
\end{align}

Substitution of \eqref{5} into \eqref{4}
provides the desired result

\begin{align}
w_{\gamma}&=|CD|
=\frac{2F}{c\cdot\cos\tfrac{\alpha-\beta}2}
\tag{6}\label{6}
.
\end{align}

Note that since $\cos\tfrac{\alpha-\beta}2=\cos\tfrac{\beta-\alpha}2$,
the expression \eqref{6} holds also if $|AC|<|BC|$,
and it also holds if $|AC|=|BC|$ too.


问:
solving $1+\frac{1}{x} \gt 0$
In solving a larger problem, I ran into the following inequality which I must solve:

$$ 1+\frac{1}{x} \gt 0.$$

Looking at it for a while, I found that $x\gt 0$ and $x\lt -1$ are solutions. Please how do I formally show that these are indeed the solutions.

答: You asked how to show this formally, so a picture is possibly not very useful, but I include one because I need practice with tikz. Vadim's answer above is excellent.

Title: solving $1+\frac{1}{x} \gt 0$


Body:
Mean value theorem:

$\log (1+1/n^2)=$

$\dfrac{\log (1+n^2)-\log (n^2)}{1}= \dfrac{1}{c}\le \dfrac{1}{n^2},$ where $c \in [n^2,n^2+1].$


问:
Examples of (families of) functions that are badly approximatable by polynomials but well approximatable by splines
As the title says, I am looking for concrete examples of functions which are badly approximatable by polynomials (i.e. a slow convergence rate), but well approximatable by B-splines (hopefully, irrespective of the choice of knots, but I guess that is not generally possible?). I am looking at the problem from a least-squares setting. That is, I have a random collection of samples of the function, and I am solving an overdetermined problem using either a B-spline basis, or a polynomial basis (e.g. Chebyshev).

thanks!

答: For smooth functions, e.g., differentiable functions, both spline and polynomial approximations give the same degree of approximation. For very smooth functions, e.g., analytic functions, polynomials are better. Polynomial approximation does not exhibit saturation like spline approximation. As far as finding these approximations using random collection of samples, the question needs to be made more precise. Assuming that the approximation is taking place on $[-1,1]$ and the samples are distributed properly, there are ways to construct both polynomial and spline approximations to yield asymptotically the same degree of approximation. For polynomial approximation, see my paper Ja'en J. of Approx., {\bf 1} (1) (2009), 1--25, for spline approximation, see my paper J. Math. Anal. and Appl. {\bf 251} (2000), 356-363. In both cases, it is not necessary to solve a system of equations. Spline approximation will usually work better if the function is equal to 0 on parts of the interval because we can manage the approximation to be 0 there as well for the most part. The same can be achieved with polynomials using my localized kernels, but of course, the approximation will not be identically zero on any interval. When the samples are not dense on the interval, we don't expect a good approximation on the whole interval, but see my paper Journal of Computational Physics, {\bf 249}  (2013) 96--112 for a procedure we have called Minimum Sobolev Norm interpolation.


问:
Example of application of stochastic calculus to discrete probability
Integrals extend the concept of sum in the continuous domain, and the usefulness of such extension with respect to the generalized concepts (i.e. discrete mathematics) can be easily illustrated with some examples, such as providing bounds on the value of a series by upper and lower bounding it with appropriate integrals.

Stochastic calculus extends the concept of sum of random variables in a similar fashion. Can we also provide a corresponding motivating example of an application of stochastic calculus in a discrete-probability setting? In other words, can we prove something about a discrete process by the means of stochastic calculus, that we could not easily prove otherwise?

答:
Let me have a go, although this might not be exactly what you were after.

Stochastic integral can be intuitively interpreted as an outcome of a gambling strategy, where the integrand is the strategy and the integrator is the (random) outcome of the game being played.

Example: A gambling game involves a coin flip. If the outcome is heads the bettor wins, if it's tails, the bettor loses. In this case the "integrator" could be modeled via $X_t=\left\{-1,1\right\}$ with probability $\frac{1}{2}$. My gambling strategy as a function of the game outcomes is to bet 2 dollars every time (so that the "integrand" is a constant $2$). The outcome of this game, that I denote $I(t)$ after 10 rounds will be:

$$I(t=10):=\sum_{i=1}^{i=t}2X_i$$

Now what if I am interested in computing the outcome of investing 100 dollars into a specific stock (say the stock costs 100 dollars, so that I purchase one unit of it). In discrete space, suppose the stock can go up or down every unit of time by a certain percentage amount (where this percentage amount is represented by $\sigma$). After $t$ units of time, the stock value would be:

$$S_t=S_0+\sum_{i=1}^{i=t}\sigma S_{i-1} X_i$$

What if the percentage amount by which the stock can go up or down isn't fixed, but rather dependent on all the incoming news and events that affect the wider economy? The sum of all these incoming random events could be modeled as a normally-distributed random shock with a Brownian motion increment, improving the model:

$$S_t=S_0+\sum_{i=1}^{i=t}\sigma S_{i-1}(W_{i}-W_{i-1})$$

However, time is (arguably) continuous, and when the stock market is open, people can buy & sell stocks at any time (prices can move at tiny increments called ticks that make the prices (almost) continuous). How to model this "continuity" to get a more realistic stock price model?

How about via Ito integral:

$$S_t=S_0+\int_{h=0}^{h=t}\sigma S_h dW_h=S_0+\lim_{n\to\infty}\sum_{i=t_1}^{i=t_n}\sigma S_{i-1}( W_{i}-W_{i-1})$$

Above, the stock is now exposed to "continuous shocks" driven by the Brownian motion integrator (reflecting the continuously incoming random news and events affecting the economy). The above equation has the well-known solution (using Ito's lemma):

$$S_t=S_0e^{-0.5 \sigma^2 t + \sigma W_t}$$

The continuous model is "more realistic" and it allowed us reach the conclusion that the stock price at a specific future point in time converges to a log-normally distributed random variable, with the expected value centered on the "starting value" $S_0$. Furthermore, log-normal distribution is nice also because it has a positive domain, and stock prices cannot go negative, since zero price means bankruptcy. Also, the log-normal solution above is a Markov process, meaning that future stock prices only depend on last observed value, rather than the entire path taken to reach that value.

It is nice to get all these properties as a continuous limit of the discrete model.


问:
Entire sequence $g_{n} \rightarrow g$ with $g_{n}$ having only real zeros
Say we have a sequence of entire functions $g_{n}$, and $g_{n} \rightarrow g$ uniformly on compact sets in $\mathbb{C}$. I want to show that if each $g_{n}$ has only real zeros, then all the zeros of $g$ must also be real.

My best guess was to assume $g$ has a non real zero $w$ and take a compact set $K$ containing no real numbers and containing $w$. Then $g_{n}$ never vanishes on $K$ yet $g$ does. What if we look at the the sequence of functions $\frac{1}{g_{n}}$ analytic on $K$. Is it true $\frac{1}{g_{n}} \rightarrow \frac{1}{g}$ uniformly on compact sets? Because then we would get $\frac{1}{g}$ is analytic in $K$, a contradiction. Thanks for the help!

答:
As pointed out by Tau, you must assume that the limit $f$ is non zero.

Then one can use Rouché's theorem:

Theorem. Let $f,g$ be two holomorphic functions in a neighborhood of $\overline {D(a,r)}$. We suppose that for $|z-a|=r$, we have $|f(z)-g(z)| < |f(z)|$. Then $f$ and $g$ have the same number of zeroes in $D(a,r)$.

Now, in your situation, you choose $a$ a non-real zero of $f$, and $r$ such that $f(z)$ is never zero on the circle $|z-a|=r$, therefore $|f(z)| > \delta >0$ there. Then for $n$ sufficiently big, you will have $|f(z)-f_n(z)| < \delta < |f(z)|$ on $|z-a|=r$.
Up to choosing a smaller $r$ so that $\overline D(a,r)$ does not intersect the real axis, you're done.


问:
Radius of convergence - ratio test for power series/real numbers
Having trouble with when to apply the ratio test for power series and when to apply the ratio test for real numbers.

For example, find radius of convergence of these....

$\sum_{n=0}^{\infty}(-1)^n (x+3)^n$

For this I would use ratio test for power series and get radius of convergence = 1, fine.

but for

$\sum_{n=0}^{\infty}n!(n+1)!(x+1)^{2n}$

Why do we need to use the ratio test for real numbers, and define for a fixed $x$? and why can't we use the ratio test for power series?

Is it because the $(x+1)^{2n}$ part is $2n$ and not $n$?

答:
Let $$\sum _{n=0} ^{\infty} n!(n+1)!(x+1)^{2n}$$ wants be convegent, so according to tests, we should have:

$$\lim_{n\to \infty}{\Big|\frac{(n + 1)! (n+2)! (x+1)^{(2n + 2)}}{n!(n+1)!(x+1)^{2n}}\Big|}<1$$ So,

$$\lim_{n\to \infty}{|(n + 1)(n + 2)x^2|}<1$$

Since for each $x\ne0$ we see $\lim_{n\to \infty}{|(n+1)(n+2)x^2|}=\infty$ so only for $x=0$ our series will be convergent and hence radius of convergence is equal to zero.


问:
Integral calculations with exponentials ($\int_{-\infty}^{+\infty}e^{ax^2+bx+c-1}$)
I have a probability distribution function (pdf) $f(x)$ wich has the following form:

$$f(x)=e^{ax^2+bx+c-1}$$

Let's assume that the pdf has a mean value $\mu$ and a variance value $\sigma^2$.

I want to prove that this pdf is the Gaussian function. So, I want to show that:

$$f(x)=\frac{1}{\sqrt{2\pi}\sigma}exp\left \{\frac{-(x-\mu)^2}{2\sigma^2}\right \}$$

I consider finding the parameters $a$, $b$ and $c$ by making use of the following equations:
$$\int_{-\infty}^{+\infty}f(x)=1$$
$$\int_{-\infty}^{+\infty}xf(x)=\mu$$
$$\int_{-\infty}^{+\infty}x^2f(x)=\sigma^2 + \mu^2$$

However, I find diffuculty on calculating these integrals since I am not very familiar with such exponential integrations that have a polynomial exponent with a 2nd or higher degree.

For instance, how do I have to approach the first one ($\int_{-\infty}^{+\infty}f(x)=1$)? Do I have to make use of the Gaussian integral somehow? ($\int_{-\infty}^{+\infty}e^{-x^2}=\sqrt{\pi}$). It seems to me that the integral is divergent.

For the other two equations do I have to make use of integration by parts and if so how do I have to approach it?

Any help is appreciated!Thanks in advance!

答:
First, note that $a<0$ or else the integral would diverge. Instead of treating the three equations separately, we can combine them to get some easier integrals, Let

$$A=\int_{-\infty}^\infty f(x)\ dx,\quad  B= \int_{-\infty}^\infty xf(x)\ dx,\quad C=\int_{-\infty}^\infty x^2 f(x)\ dx$$

Notice that $2a B+b A=0$ because

$$\int_{-\infty}^\infty (2ax+b)e^{a x^2+bx +c-1}\ dx = \bigg [e^{a x^2+bx +c-1}\bigg]_{-\infty}^{\infty}=0$$

Which gives you the necessary condition $2a\mu+b=0$ or $a=\frac{-b}{2\mu}$ (This means that $b>0$). Now we can write $f$ as

$$f(x)= \exp\left(-\frac{b}{2\mu}\left(x^2-2\mu x+\mu^2\right)+c-1+\frac{b\mu}{2}\right)= d \exp\left(-\frac{b}{2\mu}\left(x-\mu\right)^2\right)$$

(I have replaced $e^{c-1+\frac{b\mu}{2}}$ by d). You can use $A$ to find that $d= \frac{1}{\sqrt{2\pi}} \sqrt{\frac{b}{\mu}}$. Now the last piece of the puzzle is $b$. Notice that

$$\sigma^2 =\int_{-\infty}^{\infty}  d(x^2-\mu x) \exp\left(-\frac{b}{2\mu}\left(x-\mu\right)^2\right)\ dx = d \left(\frac{\mu}{b}\right)^{3/2}
\sqrt{2\pi}=\frac{\mu}{b}$$

Now, $b=\frac{\mu}{\sigma^2}$, and your desired resuult follows immediately from this.


问:
Limit of an integral with parameter
I am studying integrals with parameters and I have come across a problem. In a book there is a solved example:
\begin{equation}
\lim_{R \to \infty} = \int_0^{\pi/2}e^{-R\sin x} \, dx
\end{equation}
And it is solved in a way where we simplify the integral with taking upper approximations:
\begin{equation}
 \int_0^{\pi/2}e^{-R\sin x} \, dx \leq\int_0^{\pi/2}e^{-(R/2) x} \, dx
\end{equation}
And then the integral is solved and limit calculated. My question is, when can I make these upper approximations to simplify the integral?

答:
Hint

Use the useful inequality

$$(\forall x\in[0,\frac{\pi}{2}])\;\;\;  \frac{2}{\pi}x\le \sin(x)$$

and

$$e^{-R\sin(x)}\le e^{-\frac{2R}{\pi}x}$$

Thus

$$0\le\int_0^{\frac{\pi}{2}}e^{-R\sin(x)}dx\le \int_0^{\frac{\pi}{2}}e^{-Ax}dx$$

$$=\frac{-1}{A}\Bigl[e^{-Ax}\Bigr]_0^{\frac{\pi}{2}}$$
$$=\frac{\pi}{2}\frac{e^{-R}-1}{R}$$

where $ A=\frac{2R}{\pi}$.

Your limit is zero.

Title: Limit of an integral with parameter


问:
Looking for a function that approximates a parabola
I have a shape that is defined by a parabola in a certain range, and a horizontal line outside of that range (see red in figure).

I am looking for a single differentiable, without absolute values, non-piecewise, and continuous function that can approximate that shape. I tried a Gaussian-like function (blue), which works well around the maximum, but is too large at the edges. Is there a way to make the blue function more like the red function? Or, is there another function that can do this?

答:
A "meta-answer", if I may.

So, you have a smoothness constraint. But - don't take it farther than you need to. That is, there's a natural (?) tendency to assume that if you're asked for a smooth approximation than it has to be infinitely-differentiable. So, you went for a Gaussian. I'm not saying that's a bad choice, but - you should certainly consider a solution which "just" meets the constraint and not much beyond that.

What immediately comes to my mind is "nearly-cheating": taking the constraint-breaking piecewise function (say its pieces are $p_1$ and $p_2$) and smoothing its pieces out, like so:
$$
f(x) = (1-t(x)) \cdot p_1(x) + t(x) \cdot p_2(x)
$$
with some function $t : \mathbb{R} \to [0,1]$. that is, $p_1$ will gradually transition into $p_2$ instead of just switching to it.

In your case, a super-simple $t(x)$ would be the following piecewise linear affair:
$$
t(x) =
\begin{cases}
0 & x < -100 \\
\frac{x+100}{100} &- 100 \leq x < 0 \\
\frac{100-x}{100} &0 \leq x < 100 \\
0 & x \geq 100
\end{cases}
$$
(seeing how your parabola segment crosses at -50 and 50.)

This is of course not the greatest choice of $t$. You can either narrow the linear pieces going from 0 to 1 and back; or you could add more pieces; or you could make it a spline of higher degree. Narrowing the transition segment is fine - since the smoothed function will still meet your constraints (you were not limited on the absolute value of the derivative). You will then have a fifth, central linear piece with value 1.

If you wanted to play nicer, you might want to increase the degree of the piecewise linear functions, to get a higher-degree spline. When you think in that direction, you might notice that what you have already is a degree-1 Bernstein polynomial - a basis for the space of polynomials from [0,1] to $\mathbb{R}$. So, IIANM your smoothing coefficients which generalizes into making Bezier splines of your chosen degree of smoothness.


答:
This is not a simplification of the integral, but rather an inequality that relates the integral to a simpler integral.

The reason it can be used to find the limit is this:
$$
0 \le \int_0^{\pi/2}e^{-R\sin x} \, dx \leq\int_0^{\pi/2}e^{-(R/2) x} \, dx \to0 \text{ as } R\to+\infty.
$$
The point is that if a function of $R$ is between $0$ and another function of $R$ that approaches $0$, then that function that is squeezed between those two also approaches $0.$

Title: Limit of an integral with parameter

Body:
I am studying integrals with parameters and I have come across a problem. In a book there is a solved example:
\begin{equation}
\lim_{R \to \infty} = \int_0^{\pi/2}e^{-R\sin x} \, dx
\end{equation}
And it is solved in a way where we simplify the integral with taking upper approximations:
\begin{equation}
 \int_0^{\pi/2}e^{-R\sin x} \, dx \leq\int_0^{\pi/2}e^{-(R/2) x} \, dx
\end{equation}
And then the integral is solved and limit calculated. My question is, when can I make these upper approximations to simplify the integral?


问:
Cayley graphs of product of groups
From this question, we have that the Cayley graph of direct product of two groups is a cartesian product of some cayley graphs on the factor groups. But, I do not see this translation easily.

Specifically, let us say I take the group $\mathbb{Z}_2\times\mathbb{Z}_8$ and the cycle graph on it. The cycle graph can be written as a cayley graph by taking any non-involution element and its inverse as the generating set in the group $\mathbb{Z}_2\times\mathbb{Z}_8$. But, how is it a cartesian product of some graphs on the indivdual groups $\mathbb{Z}_2$ and $\mathbb{Z}_8$? Is the ususal definition of cayley graphs not used while making the product graph? That is, can the cycle on the product group elements be written as a cartesian product of two $1$-factors on the individual factor group elements?  Any hints? Thanks beforehand.

答:
$\mathbb{Z}/2 \times \mathbb{Z}/8$ is not cyclic, so it needs at least $2$ generators to get the whole group. So if you take the "cycle graph" as you've defined it, you will not get a cayley graph at all.

If instead you take the cayley graph with two generators (one of $\mathbb{Z}/8$ and one of $\mathbb{Z}/2$) you'll find you do get the product of the respective cayley graphs.

I hope this helps ^_^

Title: Cayley graphs of product of groups


问:
When are endomorphisms of a module only isomorphisms or zero?
Let $k$ be a field, $A$ a $k$-algebra of finite length and $M$ an $A$-module of finite length. When does it happen, that all endomorphisms of $M$ are either an isomorphism or zero? It is easy to see that if $M$ is simple, then it happens and if it happens, then $M$ must be indecomposable.

答: Modules whose endomorphism algebra is a division ring are called bricks.  As you mentioned, simple modules are bricks, but the converse is far from true.  They are related to torsion pairs and $\tau$-tilting theory.  While I will not attempt to outline the theory here, I can perhaps point to this paper by Demonet, Iyama and Jasso and this other paper by Asai on semibricks.


问:
Dimension and Basis of a Vector Subspace
Given $F=\{(x,y,z,t)\in \mathbb{R}^4: x+t=0\}$ and $G=\{(x,y,z,t)\in \mathbb{R}^4:y=-z, x=t\}$, how can you prove they are subspaces of $\mathbb{R}^4$? Besides, if you want to find the basis and the dimension of $F$, $G$, $F\cap G$ and $F\cup G$, how would you proceed?

答:
It is helpful to rewrite the elements of $F$ and $G$ so that their elements are clearer. For example, $$F = \{(x,y,z,-x)\;|\;x,y,z \in \mathbb{R}\}\;\text{and}\;G = \{(x,y,-y,x)\;|\;x,y \in \mathbb{R}\}.$$ I'll leave it to you to show that $F$ and $G$ are subspaces of $\mathbb{R}^4$ and that $\{(1,0,0,-1), (0,1,0,0), (0,0,1,0)\}$ is a basis of $F$ and that $\{(1,0,0,1), (0,1,-1,0)\}$ is a basis of $G$. The thought process behind finding these is the idea that $F$ depends only on the first three variables whereas $G$ depends only on the first two by definition.

We need to be careful about $F \cup G$ because it is not actually a subspace. For example, $(1,0,0,1), (1,0,0,-1) \in F \cup G$ whereas their sum $(2,0,0,0)$ is not. $F \cap G$ is a subspace and it's given by $$F \cap G = \{(0,y,-y,0)\;|\;y \in \mathbb{R}\}$$ with a one element basis $\{(0,1,-1,0)\}$.


问:
properties of recursively enumerable sets
$A \times B$ is an r.e.(recursively enumerable) set,  I want to show that $A$ (or $B$) is  r.e. ($A$ and $B$ are nonempty)

I need to find a formula. I've got an idea that I should use the symbolic definition of an r.e. set. That is, writing a formula for the function that specifies $A$ or $B$, assuming a formula exists for $A \times B$. I guess I must use Gödel number somewhere.

I should've mentioned that I asked this question over there (Computer Science) but since I am more interested in mathematical arguments and looking for formulas, I'd ask it here as well.

答:
I don't know why you request some formula.

Just take a Turing machine $M$ that enumerates $A\times B$, but instead of outputting some tuple $(a,b)$, it only outputs $a$, resp. $b$. This modified machine $M'$ enumerates $A$, resp. $B$.


问:
Efficient ways to calculate Lambert $W_{-1}(z)$ with an arbitrary precision.
I proved and tested that for any number $z\in [-\frac{1}{e}, 0]$ and a number $\epsilon$ the following algorithm returns a number $x$ such that $|W_{-1}(z)-x| < \epsilon$ in time $O\left(\left\lceil\log_{4|W_{-1}(z)|}\left(\frac{1}{\epsilon}\right)\right\rceil\right)$ i.e. it would take $O(1)$ for values of $z$ close to $0$ and it would take $O\left(\log\left(\frac{1}{\epsilon}\right)\right)$ for values of $z$ close to $-\frac{1}{e}$.

```<code>W_{-1}(z, epsilon)
{
    upper = log(z / log(-z));
    lower = upper + log(1 - (1.0 / e));
    x     = log(2.0 * z / (lower + upper));
    while (upper - lower > 2.0 * epsilon)
    {
        if (2.0 * x < lower + upper)
        {
            upper = x;
            lower = log(z / lower);
        }
        else
        {
            upper = log(z / upper);
            lower = x;
        }
        x = log(2.0 * z / (lower + upper));
    }

    return x;
}
</code>```

And I wanted to know whether or not my algorithm was useful or if maybe I should publish it or something. So I tried to search what other algorithms there are to calculate the $-1$ branch of the lambert W function with an $\epsilon$ precision but I didn't find anything (maybe I'm just not very good at searching). What efficient ways there are to calculate the $W_{-1}$ with an arbitrary precision?

答:
This is just fixed-point iteration, which only converges linearly. The bounds are also unnecessary because the iterations from either side will converge in an expected manner, except for $z$ near $-1/e$ and $-0$. It would be better to use something such as Newton's method or Halley's method, which converge much faster.

Newton's method gives the iterations:

$$w_{n+1}=w_n-\frac{w_n-ze^{-w_n}}{w_n+1}\tag0$$

and with the initial estimate

$$w_0=\ln\left(\frac z{\alpha\ln(-z)}\right),\quad\alpha=\frac2{2-\ln(2)}\tag1$$

you can guarantee iterations monotonically converge to the Lambert W function.

To get faster convergence near $-1/e$ you may wish to use instead the initial estimate

$$w_0=-1-\sqrt{2(1+ez)}\tag2$$

which provides an upper bound. Taking one step further with fixed-point iteration, one may instead use

$$w_0=\ln\left(\frac{-z}{1+\sqrt{2(1+ez)}}\right)\tag3$$

which is accurate both as $z\to-1/e$ and $z\to-0$. To obtain another asymptotic term as $z\to-0$, another iteration of fixed-point iteration gives

$$w_0=\ln\left(\frac z{\ln\left(\frac{-z}{1+\sqrt{2(1+ez)}}\right)}\right)\tag4$$

If your concern is having both upper and lower bounds to guarantee an error bound, you may wish to try using the larger step-size

$$w_{n+1}=\min\left\{-1,w_n-\beta\frac{w_n-ze^{-w_n}}{w_n+1}\right\},\quad\beta>1\tag5$$

every so often to see if the solution becomes bounded. Initial bounds may be obtained by noting that $(4)$ is an upper bound while applying $(5)$ to $(4)$ with $\beta=-w_n$ gives a lower bound.

I would suggest, since one can be sure the above converges, to use $(5)$ only once $|w_{n+1}-w_n|$ is sufficiently below the desired precision. The singular initial estimate $(4)$ should also suffice to give rapid convergence.


问:
Axler exercise 8.A.11: diagonalizability

Prove or give a counterexample: if $V$ is a complex vector space and $\dim V = n$ and $T \in \mathcal{L}(V)$, then $T^n$ is diagonalizable.

I have a sketch of a solution, but I'm not convinced that it's sufficiently rigorous.

Consider a counterexample in $V = \mathbb{R}^2$:
\begin{align*}
T & = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \\
T^2 & = \begin{pmatrix} 1 & 2 \\ 0 & 1  \end{pmatrix}.
\end{align*}
The characteristic polynomial of $T^2$ is:
\begin{align*}
p(\lambda) = \det(\lambda I -  T^2) = (\lambda - 1)^2.
\end{align*}
Since $p(\lambda)$ has a repeated root, $T^2$ lacks an eigenbasis, so it has only one linearly independent eigenvector, and is therefore not diagonalizable.

Here is my problem with this solution. If an operator is diagonalizable, then there has to exist a diagonal matrix with respect to some basis, but I just chose the standard basis. How do I establish from this that there is no basis wherein I can find two eigenvectors?

答:
Your reasoning is incorrect. While it is true that if the characteristic polynomial has no repeated roots then the matrix is diagonalizable, the converse is false. A matrix can have a characteristic polynomial with repeated roots and still be diagonalizable. The simplest example is the $n\times n$ identity matrix, which is itself diagonal, but has characteristic polynomial $(-1)^n(x-1)^n$. (Note: I prefer the definition of characteristic polynomial as $\det(A-\lambda I)$ because it requires fewer sign changes when computing the determinant by hand). Of course, you can construct diagonal matrices with whatever characteristic polynomial you want, including one with several multiple roots.

The argument at the point you are in is pretty easy. Note that the matrix for $T^2-I$ is
$$\left(\begin{array}{cc}
0&2\\
0&0
\end{array}\right)$$
which has rank $1$; thus, $N(T^2-I)$ has rank $1$, which means the eigenspace of $1$ has dimension $1$. Thus, there aren’t enough linearly independent eigenvectors for a basis.


问:
Prove that $A \times (B\oplus C) = (A \times B) \oplus (A \times C)$
$A \times (B\oplus C) = (A \times B) \oplus (A \times C)$

$LHS \Rightarrow RHS$

$A \times [(B\land\neg C)\lor (\neg B \land C)]$

{$(a,b): a \in A, b \in (B\land\neg C)\lor (\neg B \land C)$}

and then im stuck here, not sure what else to do with the cartesian product

答:
From where you left off, you have $a \in A,b\in B,b\notin C \vee a\in A, b \notin B, b \in C$.

Hence, $(a,b) \in (A \times B) \vee (A \times C)$ and  $b\in B \Leftrightarrow b\notin C$ so that $(a,b)\in (A \times B) \Leftrightarrow (a,b) \notin (A \times C)$.

Therefore $(a,b) \in (A \times B) \oplus (A \times C)$ and $A \times (B\oplus C) \subseteq (A \times B) \oplus (A \times C).$

The other implication is proven similarly.


问:
System of equation from function containing summation!
A function is given by $Q(p,q)=\sum_{i=1}^{n}(p \cdot t_i+q-y_i)^2$. I found the 2 partial derivatives to be

$\frac{\partial Q}{\partial p}=2 \cdot \sum_{i=1}^{n}(p \cdot t_i+q-y_i)t_i$

$\frac{\partial Q}{\partial q}=2 \cdot \sum_{i=1}^{n}(p \cdot t_i+q-y_i)$

Which I think are correct. Now I have to show that  $\frac{\partial Q}{\partial p}=0$ $\frac{\partial Q}{\partial q}=0$ can be written as a linear system of equations

$\sum_{i=1}^{n}t_i^2p+\sum_{i=1}^{n}t_iq=\sum_{i=1}^{n}t_iy_i$

$\sum_{i=1}^{n}t_ip+nq=\sum_{i=1}^{n}y_i$

I have absolutely no idea on how to show this and I am sure of the fact that it is a simple observation regarding the partiel derivatives.

Any Help Is Much Appreciated.

答:
To make things easier to track you can rename a few things

\begin{align}
A &= \sum_{i=1}^n t_i^2 \\
B &= \sum_{i=1}^n t_i \\
C &= \sum_{i=1}^n t_i y_i \\
D &= \sum_{i=1}^n y_i \\
\end{align}

And you will end up with

\begin{align}
Ap + Bq &=C \\
Bp + nq &= D
\end{align}

And from here is just a few steps, e.g. multiplying the first equation by $B$ and the second by $A$ you get

\begin{align}
ABp + B^2q &=CB \\
ABp + Anq &= AD
\end{align}

now subtract the second from the first

$$
(B^2 - An)q = CB - AD ~~~\Rightarrow~~~ q = \frac{CB - AD}{B^2 - An}
$$

I will leave the other one to you


问:
Analytic method for number theory-do we have to assert second-order logic?
I am an undergraduate. I am just starting to study logic and analytic number theory at the same time, so please forgive me if I made an elementary misunderstanding.

A lot of theorem in number theory can be stated completely in term of first order logic added with Peano axiom. But they were proven using analysis, and for many of them we do not know of any elementary method to do so, which means that second order logic needed to be used.

So I have 3 questions:

-Theorem in number theory need first order logic to be stated, so axiom of first order logic are assumed. But since their proof use second order logic, does that means that these theorems are, essentially, proved by asserting new axioms (ie. axioms of second order logic)? Would this pose any problems?

-Is there any theorem in number theory that is proved, and is also proved that all possible proof of that theorem must use more than first order logic?

-If we do not accept the axioms of second order logic, is it possible to construct a system of number satisfying Peano arithmetic such that: there is no possible way of using analysis? some of those theorem proved using analysis are false?

答: It often happens that number-theoretic statements or proofs that use complex analysis and therefore appear to be second-order over arithmetic can be reformulated in first-order arithmetic.  The techniques used for this include approximating real numbers with rational ones and approximating integrals with Riemann sums.  For example, the Riemann hypothesis can be reformulated in the language of first-order arithmetic.  (I confess that I'm not an expert on such reformulations; in particular, I've never worked through the details of converting the  Riemann hypothesis to first-order.)</p>


问:
Find the transfer function and find the stationary solution associated with $u(t) = \cos(t)$
Problem

Given the system
$$\mathbf{x}'(t)= \begin{bmatrix}2 & -1 \\ 3 & 1 \end{bmatrix}\mathbf{x}(t)+\begin{bmatrix}0 \\-1 \end{bmatrix}u(t) $$
$$y(t) = 2x_1(t)-x_2(t) $$

Find the transfer function for the linear system.

Find the stationary solution associated with $u(t)=\cos(t)$

My attempt

1)
The transfer function is easily found with the formula:
$$H(s)=-\mathbf{d}^T(\mathbf{A}-s\mathbf{I})\mathbf{b}$$
We see that $-\mathbf{d}^T= \begin{bmatrix} -2 \\1 \end{bmatrix}$, $\mathbf{A}=\begin{bmatrix}2 & -1 \\ 3 & 1 \end{bmatrix}$, $\mathbf{b}=\begin{bmatrix}0 \\-1 \end{bmatrix}$, and $\mathbf{I}$ is the identity matrix. Plugging in these values in the formula and using Maple gives
$$H(s)=\frac{s}{s^2-3s+5},s^2-3s+5 \neq0 $$

2)

In my book, it says that the stationary solution associated with $u(t)=e^{st}$ is $y(t)=H(s)e^{st}$.

But what do I do when $u(t)=\cos(t)$?

I have the solution to this problem, and it says that the stationary solution is $y(t)=-\frac{3}{25}\cos(t)-\frac{4}{25}\sin(t)$.

Can someone show me how they arrive at that answer? It seems like you have to plug in a value for $s$, but I'm not sure what.

答:
Write $u(t) = {1 \over 2} (e^{i t} + e^{-i t})$ and get the zero state response
$y(t) = {1 \over 2} (H(i) e^{i t} + H(-i)e^{-i t})$.

Since $H(\overline{s}) = \overline{H(s)}$ we see that
$y(t) = \operatorname{re} H(i) e^{i t} = \operatorname{re} ( (-{3 \over 25} + {4 \over 25}i)(\cos t + i \sin t)$ which simplifies to
$y(t) = -{3 \over 25}\cos t - {4 \over 25} \sin t$.


问:
Covariant functor, and left exact
I'm reading A first course of Homological Algebra by Northcott, and there is something that the author said it was straightforward. But for some reason, I just don't see the straightforwardness of it.

Definition

And let $F:\mathscr{C}_\Lambda \to \mathscr{C}_\Delta$; where $\mathscr{C}_\Lambda$; $\mathscr{C}_\Delta$ be the covariant functor between the category of $\Lambda-$left module, and $\Delta-$left module.

$F$ is said to be left exact iff for every exact sequence of left module: $0 \rightarrow A \xrightarrow{\sigma} B \xrightarrow{\pi} C \rightarrow 0$, the following induced sequence is also exact:
$0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi)} F(C)$

Property

Of course if $F$ is a left exact functor, then it'll preserve monic.

Theorem

Suppose $F:\mathscr{C}_\Lambda \to \mathscr{C}_\Delta$ is a left exact covariant functor, and that $0 \rightarrow A \xrightarrow{\sigma} B \xrightarrow{\pi} C$ is exact in $\mathscr{C}_\Lambda$. Prove that $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi)} F(C)$ is also exact in $\mathscr{C}_\Delta$.

Here's my proof, I'll be very glad if you guys can check my proof, as well as give me some hints to prove it in another easier way (if another way exists).

Proof

Since $F$ preserves monic, the sequence $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi)} F(C)$ is exact at $F(A)$.

$\pi \sigma = 0 \Rightarrow 0 = F(\pi \sigma) = F(\pi)F(\sigma)$

What left is to prove $\ker F(\pi) \subset \mbox{im } F(\sigma)$

Consider the exact sequence: $0 \rightarrow A \xrightarrow{\sigma} B \xrightarrow{\pi} \mbox{im }\pi' \rightarrow 0$, where $\pi'$ is the same as $\pi$ with the codomain contracted to be $\mbox{im } \pi$ since $F$ is left exact, it means that: $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi')} F(\mbox{im }\pi)$ is also exact.

Consider the following chain of homomorphism: $F(B) \xrightarrow{F(\pi')} F(\mbox{im }\pi) \xrightarrow{F(i)} F(C)$

Since $i$ is monic, $F(i)$ is also monic (I think so, but is it right?)

So $\ker F(\pi) = \ker [F(i)F(\pi')] = \ker F(\pi') = \mbox{im }F(\sigma)$.

Is my proof valid? And is there any shorter proof? As the author said that it's straightforward, and I don't really think my proof can be classified to be straightforward.

Thanks guys very much,

Have a good day,

答: Your proof is OK. It will become straight forward with a little bit more experience ;). You have done the only possible thing: Somehow extend the left exact sequence to a short exact sequence, using the image in order to make the last map surjective. By the way, I have given the same proof in SE/292037.</p>

Title: Covariant functor, and left exact


问:
Bounding sum over binomial coefficients
How can I show that $ \sum_{k=0}^{m}\binom{n}{k} \leq (n+1)^m$ for $m \leq n $?

Title: Bounding sum over binomial coefficients

Body:
$(n+1)^m
=\sum_{k=0}^m \binom{m}{k}n^k
$
so if
$n^k\binom{m}{k}
\ge \binom{n}{k}
$
we are done.

$\begin{array}\\
\binom{m}{k}
&=\dfrac{m!}{k!(m-k)!}\\
&=\dfrac{\prod_{j=0}^{k-1}(m-j)}{k!}\\
\dfrac{n^k\binom{m}{k}}{\binom{n}{k}}
&=\dfrac{n^k\prod_{j=0}^{k-1}(m-j)}{\prod_{j=0}^{k-1}(n-j)}\\
&\ge\prod_{j=0}^{k-1}(m-j)
\qquad\text{since } n^k \ge \prod_{j=0}^{k-1}(n-j)\\
&\ge 1\\
\end{array}
$


问:
Help with subsets/subgroups question
Let $S\subseteq G$. Let $K = \bigcap\{H: H \le G\text{ and }S \subseteq H\}$, the intersection of all subgroups of $G$ that contain $S$

Lemma: $\langle S\rangle = K$

Prove that $\langle S\rangle\subseteq K$ and $K\subseteq\langle S\rangle$:

"$S\subseteq\langle S\rangle\le G \implies K \subseteq \langle S\rangle$

Let $x\in\langle S\rangle$. Let $H \le G$ and $S\subseteq H$.

Then $\langle S\rangle\le H \implies x\in H \implies x \in \bigcap H$

$\implies x \in K \implies\langle S\rangle\subseteq K$

Thus $\langle S\rangle = K$"

I don't really understand this proof, largely two steps:

"$S\subseteq\langle S\rangle\le G \implies K \subseteq\langle S\rangle$" (I'm not sure how this implies $K\subseteq\langle S\rangle$)

How is "$x\in H \implies x \in \bigcap H$" implied?

Could someone please elaborate? Any help appreciated

答: For item 1: suppose $s \in S$. Then is $s \in K$? Well, for any $H$ with $ H \le G$ and $S \subset H$ (i.e., any of the $H$s used to define $K$), we have $s \in H$. So $s$ is in the intersection of all such $H$. Hence $s \in K$.

Title: Help with subsets/subgroups question


答:
Just note the notation (even you might know what you are talking about) is not correct

$$A \times [(B\land\neg C)\lor (\neg B \land C)]$$

Is this a set or a proposition? If it's a set
$$A \times (B\oplus C)=A\times[(B\cap C^c)\cup(B^c\cap C)]$$
If it's a proposition, for $(a,b)\in A \times (B\oplus C)$ have
$$a\in A\land b\in[(B\cap C^c)\cup(B^c\cap C)]$$
To show this equation hold, from definition we can write
\begin{align}
(A\times B)\oplus(A\times C)=&\{(a,b):(a,b)\in(A\times B)\land (a,b)\not\in(A\times C)\\
&\lor(a,b)\not\in(A\times B)\land(a,b)\in(A\times C)\}\\
=&\{(a,b):(a\in A\land b\in B)\land(a\in A\to b\notin C)\\
&\lor(a\in A\to b\not\in B)\land(a\in A\land b\in C)\}\\
=&\{(a,b):(a\in A\land b\in B\land b\not\in C)\\
&\lor (b\not\in B\land a\in A\land b\in C)\}\\
=&\{(a,b):a\in A\land [(b\in B\land b\not\in C)\lor (b\not\in B\land b\in C)]\}\\
=&\{(a,b):a\in A\land (b\in B\oplus C)\}\\
=&A \times (B\oplus C)
\end{align}

Title: Prove that $A \times (B\oplus C) = (A \times B) \oplus (A \times C)$

Body:
$A \times (B\oplus C) = (A \times B) \oplus (A \times C)$

$LHS \Rightarrow RHS$

$A \times [(B\land\neg C)\lor (\neg B \land C)]$

{$(a,b): a \in A, b \in (B\land\neg C)\lor (\neg B \land C)$}

and then im stuck here, not sure what else to do with the cartesian product


答:
It will be convenient to let $\mathscr{H}=\{H\le G:S\subseteq H\}$, so that $K=\bigcap\mathscr{H}$.

For your first question, $\langle S\rangle$ is by definition a subgroup of $G$ that contains $S$, so $\langle S\rangle\in\mathscr{H}$. $K$ is the intersection of all such subgroups of $G$, so $K$ is a subset of every $H\in\mathscr{H}$. In particular, $K\subseteq\langle S\rangle$.

For your second question, let $x\in\langle S\rangle$. Let $H$ be any member of $\mathscr{H}$; then $x\in\langle S\rangle\le H$, so $x\in H$. This is true for every $H\in\mathscr{H}$, so by definition $x\in\bigcap\mathscr{H}$: that intersection is precisely the set of things that are in every member of $\mathscr{H}$. Thus, $x\in K$, and therefore $\langle S\rangle\subseteq K$.

Title: Help with subsets/subgroups question

Body:
Let $S\subseteq G$. Let $K = \bigcap\{H: H \le G\text{ and }S \subseteq H\}$, the intersection of all subgroups of $G$ that contain $S$

Lemma: $\langle S\rangle = K$

Prove that $\langle S\rangle\subseteq K$ and $K\subseteq\langle S\rangle$:

"$S\subseteq\langle S\rangle\le G \implies K \subseteq \langle S\rangle$

Let $x\in\langle S\rangle$. Let $H \le G$ and $S\subseteq H$.

Then $\langle S\rangle\le H \implies x\in H \implies x \in \bigcap H$

$\implies x \in K \implies\langle S\rangle\subseteq K$

Thus $\langle S\rangle = K$"

I don't really understand this proof, largely two steps:

"$S\subseteq\langle S\rangle\le G \implies K \subseteq\langle S\rangle$" (I'm not sure how this implies $K\subseteq\langle S\rangle$)

How is "$x\in H \implies x \in \bigcap H$" implied?

Could someone please elaborate? Any help appreciated


问:
What is $\frac{(an)!}{n!}$?
How can $\frac{(an)!}{n!}$ be expressed in terms of $a!$, $n!$, $a$, $n$ (and maybe Pochhammers)?</p>

答:
By OP's request, my above comment was that since $m!=(m)_m$, we may write the desired expression as $$\frac{(an)_{an}}{a!}$$

Note: this solution is not the one @lhf commented on, below.

Title: What is $\frac{(an)!}{n!}$?


问:
How would I use the method of differences here to find the sum of this finite series?
I need to use the method of differences to show that $$\sum_ {r=2} ^n \frac {1} {r(r-1)(r+1)} = \frac{1}{4} - \frac{1}{2n(n+1)}$$

I've already found $\frac{1}{r(r-1)(r+1)}$ as partial fractions $\Bigl( \frac{-1}{r}+\frac{1}{2(r-1)}+\frac{1}{2(r+1)} \Bigr)$.

After that, I factorised out $\frac{1}{2}$ to get $$\frac{1}{2}\sum_ {r=1}^n\Bigl(\frac{-2}{r}+\frac{1}{r-1}+\frac{1}{r+1}\Bigr)$$

I then found the sum to be $-2+1+\frac{1}{n}-\frac{2}{n}+\frac{1}{n+1}=\frac{-1}{2}-\frac{2n+1}{2n(n+1)}$.

I then did $$\sum_{r=2}^n\frac{1}{r(r-1)(r+1)}=\sum_{r=1}^n\frac{1}{r(r-1)(r+1)}-\sum_{r=1}^1\frac{1}{r(r-1)(r+1)}$$$$=\frac{-1}{2}-\frac{2n+1}{2n(n+1)}+\frac{1}{2}+\frac{2n+1}{2n(n+1)}$$$$=\frac{3}{4}-\frac{2n+1}{2n(n+1)}$$

I think I've made a mistake somewhere and the $2n+1$ should just be $1$ to get $\frac{1}{4}-\frac{1}{2n(n+1)}$ but I'm not sure where. I could include more details about how I found the sum to be $\frac{-1}{2}-\frac{2n+1}{2n(n+1)}$ if that would help.

答:
Observe we have that
\begin{align}
\frac{1}{r(r-1)(r+1)}=\frac{1}{2r(r-1)}-\frac{1}{2r(r+1)}.
\end{align}


问:
On the existence of solutions to two integer quadratic equations of the same "parity"
Let $a,b,c,d\in\mathbb{Z}$ such that $b\mid a^2+1$, $d\mid c^2+1$, $a\equiv c\,(2), b\equiv d\,(2)$.

I have the equations $$b^2 x^2-2abxy+y^2(a^2+1)=b,\quad d^2 x^2-2cdxy+y^2(c^2+1)=d$$ and I am guaranteed that there exist integer solutions $(x,y)$ to the first one and $(x', y')$ to the second one (Not necessarily the same), but I would like to know if always exist integer solutions $(x_1,y_1)$ of the first and $(x_2,y_2)$ of the second with $x_1\equiv x_2\,(2)$ and $y_1\equiv y_2\,(2)$

Analyzing cases I get to:

$a$ even, $b$ odd then \begin{cases}y \,\text{odd} \Rightarrow x\, \text{even}, \\y\,\text{even} \Rightarrow x \,\text{odd}\end{cases}

So in this case it works.

$a$ odd, $b$ odd then $x$ must be odd but I don't know nothing for $y$.

$a$ even, $b$ even cannot happen

$a$ odd, $b$ even gives me no information.

Thanks in advance!

答:
Following sketch is based on idea that each of the equations must have two solutions that differ in parity, so if you want two equations that also have different parity, this results in four different parity combinations, yet in this case only three parity combinations are possible for any solution...

First let $P=b,Q=-2a,R=\frac{a^2+1}{b}$, after dividing by $b$ we have a diophantine equation
$$
Px^2+Qxy+Ry^2=1\tag{*}.
$$

This is a quadratic diophantine equation, you can search for generic solutions (there are some on this site too). In our case we have discriminant $D=-4$, so we can use for example Lagrange’s Algorithm Revisited:
Solving $at^2 + btu + cu^2 = n$ in the
Case of Negative Discriminant, see especially sections 2 and 4. To summarize, the equation has following solutions by cases:

If $P=1$, $(x,y)=\pm(1,0),\pm(-N,1)$, notice the two solutions have different parity, specifically in second component

If $P=2$, $(x,y)=\pm(\frac{-N+1}{2},1), \pm (\frac{-(N+1)}{2},1)$ for an odd integer $N$, again we can see the parity differs in first component

Otherwise solutions are given by two consecutive convergents in continued fraction of $-\frac{Q}{2P}$ in a form $\pm (x,y)$ and $\pm (Nx+Ry,Px+Ny)$. Now to see that these two will differ in parity is a bit tricky, but can be done by inspecting three cases of parity of first solution $(x,y)$ (see below why only three...), then combine it with equation $(*)$ to show that $(Nx+Ry,Px+Ny)$ will have different parity.

We see that equation $(*)$ has by itself always at least two solutions with different parity.

Notice also that $(*)$ implies we cannot have both $x$ and $y$ even, so there are only three possibilities modulo $2$: $(1,0),(0,1)$ and $(1,1)$.

Finally, for the sake of contradiction assume we have a pair of equations with solutions that do not have same parity. That gives at least two distinct parity combinations from each of them, so at least four different parity possibilities together. That is clearly not possible as we have shown there can be at most three possibilities...


答:
Hint:

The intermediate terms vanish within  a series of $3$ consecutive partial decompositions:
\begin{align}
\frac1{(r-1)(r-2)r}&=\color{blue}{-\frac{1}{r-1}}+\frac{1}{2(r-2)}\color{red}{+\frac{1}{2r} }\\
\frac1{r(r-1)(r+1)}&=\color{red}{-\frac{1}{r}}\color{blue}{+\frac{1}{2(r-1)}}\color{green}{+\frac{1}{2(r+1)}}\\
\frac1{(r+1)r(r+2)}&=\color{green}{-\frac{1}{r+1}}\color{red}{+\frac{1}{2r}}+\frac{1}{2(r+2)}
\end{align}

Title: How would I use the method of differences here to find the sum of this finite series?

Body:
I need to use the method of differences to show that $$\sum_ {r=2} ^n \frac {1} {r(r-1)(r+1)} = \frac{1}{4} - \frac{1}{2n(n+1)}$$

I've already found $\frac{1}{r(r-1)(r+1)}$ as partial fractions $\Bigl( \frac{-1}{r}+\frac{1}{2(r-1)}+\frac{1}{2(r+1)} \Bigr)$.

After that, I factorised out $\frac{1}{2}$ to get $$\frac{1}{2}\sum_ {r=1}^n\Bigl(\frac{-2}{r}+\frac{1}{r-1}+\frac{1}{r+1}\Bigr)$$

I then found the sum to be $-2+1+\frac{1}{n}-\frac{2}{n}+\frac{1}{n+1}=\frac{-1}{2}-\frac{2n+1}{2n(n+1)}$.

I then did $$\sum_{r=2}^n\frac{1}{r(r-1)(r+1)}=\sum_{r=1}^n\frac{1}{r(r-1)(r+1)}-\sum_{r=1}^1\frac{1}{r(r-1)(r+1)}$$$$=\frac{-1}{2}-\frac{2n+1}{2n(n+1)}+\frac{1}{2}+\frac{2n+1}{2n(n+1)}$$$$=\frac{3}{4}-\frac{2n+1}{2n(n+1)}$$

I think I've made a mistake somewhere and the $2n+1$ should just be $1$ to get $\frac{1}{4}-\frac{1}{2n(n+1)}$ but I'm not sure where. I could include more details about how I found the sum to be $\frac{-1}{2}-\frac{2n+1}{2n(n+1)}$ if that would help.


问:
How to prove 'if $A\models B$, then $\models A\Rightarrow B$ ' by contradiction?
I need to find a book or an article with a proof (by contradiction) of the following theorem:

'if $A\models B$, then $\models A\Rightarrow B$'

I need it because I need to check if my proof is ok.

It has to be proof of semantic version, not syntactic.

EDIT: PLEASE CHECK MY PROOF :)

Let's (1) assume that $\alpha \models \beta$. Let's assume (2) that $\not\models \alpha\Rightarrow \beta$. From (2) we have: 'exists interpretation $v$ such that $v(\alpha)=1$ and $v(\beta)=0$'. From (1) we have: 'for any interpretation $v$: if $v(\alpha)=1$, then $v(\beta)=1$'. We remember that from (2) we have 'there exists interpretation $v$ such that $v(\alpha)=1$', so adding this information to (1) we get that for any $v$: $v(\beta)=1$. So we have contradiction: 'there exists $v$ such that $v(\beta)=0$' contradicts 'for any $v$: $v(\beta)=1$'.

答: Let's (1) assume that $\alpha \models \beta$. Let's assume (2) that $\not\models \alpha\Rightarrow \beta$. From (2) we have: 'exists interpretation $v$ such that $v(\alpha)=1$ and $v(\beta)=0$'. From (1) we have: 'for any interpretation $v$: if $v(\alpha)=1$, then $v(\beta)=1$'. We remember that from (2) we have that there exists interpretation $v$ such that $v(\alpha)=1$, so adding this information to (1) we get that for any $v$: $v(\beta)=1$. So we have contradiction: 'there exists $v$ such that $v(\beta)=0$' contradicts 'for any $v$: $v(\beta)=1$'.


问:
Speed of convergence for the secant method
I am studying the following sequence, which comes from the secant method applied to $f : x \mapsto x^3$ between $\left[-\frac{1}{2},1\right]$ :

$$x_{n+1}=x_n - \frac{x_{n}-x_{n-1}}{x_n^3-x_{n-1}^3} x_n^3$$

with $x_0=-\frac{1}{2}$ and $x_1=1$.

I proved that $(x_n)_n$ is increasing and is always strictly negative. I also know that $(x_n)_n$ converges toward $0$.

I want to study the speed of convergence $\underset{n \rightarrow +\infty}{\lim} \frac{|x_{n+1}|}{|x_{n}|}$.

I have the following hint :pose $y_n=-\frac{1}{x_n}$ and study the sequences $(a_n)_n=(y_{n+1}^2-y_n^2)_n$. Then use Césaro's Lemma.

However i don't understant this hint because if I manage to show the convergence of $(a_n)_n$, i will be able to use Césaro's Lemma on $(a_n)_n$ and have a limit for $(y_n^2)_n$ by telescopic summation... which seems weird because $y_n^2=\frac{1}{x_n^2}$ should explose because $x_n \underset{n \rightarrow +\infty}{\longrightarrow} 0$

Any helps or hints are welcomed !

Update I understood how I'll be able to study the speed of convergence using Césaro's lemma. However I still need to find that $(a_n)_n$ is convergent.

答:
I'm not so certain about the hints, but you can observe the ratio of consecutive terms, writing the secant method in symmetric form:

$$x_{n+1}=\frac{f(x_n)x_{n-1}-f(x_{n-1})x_n}{f(x_n)-f(x_{n-1})}=\frac{x_n^3x_{n-1}-x_{n-1}^3x_n}{x_n^3-x_{n-1}^3}$$

Let $t_n=x_{n+1}/x_n,~x_{n+1}=t_nx_n$ to get

$$t_n=\frac{t_{n-1}^2-1}{t_{n-1}^3-1}=\frac{t_{n-1}+1}{t_{n-1}^2+t_{n-1}+1}$$

which converges to $t\approx0.755$, the unique solution of

$$t^3+t^2-1=0$$

and hence

$$\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=t\approx0.755$$


问:
do eigenvectors correspond to direction of maximum scaling?
Does the eigenvector correspond to a direction in which maximum scaling occurs by a given transformation matrix (A) acting upon this vector.
I quote from :
https://math.stackexchange.com/q/243553

No other vector when acted by this matrix will get stretched as much
  as this eigenvector.

Is the above statement always true?... For example let
$$
A = \left( \begin{array}{ccc}
0.578385540014544 & 0.703045745965410 \\
0.477513363789115 & 0.922698950982510 \\
 \end{array} \right)
$$

The largest eigenvalue is 1.35 (approx.)

Now, consider the vector (not eigenvector)
$$
v = \left( \begin{array}{ccc}
-0.538656963091298 \\
-0.842525178326001 \\
 \end{array} \right)
$$

magnitude(v) = 1.0

magnitude(A*v) = 1.373

So this vector(v), which is not the eigenvector of A is scaled by a larger amount (x1.373), compared to the eigenvector which is scaled by x 1.35 (approx.)

Is this just an artifact of numerical precision ? I can easily create more examples of random square transformation matrices (A) where the eigenvector does not correspond to the direction of maximum scaling.

答:
There is something called the matrix norm, $||A||$ which according to Linear Algebra and it's Applications by Strang is defined on $\mathbb R^{n\times n}$ to be $$||A||=\max_{x\neq0}\frac{||Ax||}{||x||},$$
where $||x||=\sqrt{x^Tx}$. From this, $||Ax||\le||A||||x||$. So the norm measures the "amplifying effect" of $A$ on a vector $x$. Equality happens at the optimal $x$. The value $||Ax||/||x||$ is always nonnegative. We square this to get
$$||A||^2=\max_{x\neq0}\frac{x^TA^TAx}{x^Tx}.$$
Matrix $A^TA$ is symmetric. This is known as the Rayleigh quotient, as it is well known that its maximum is attained when $x$ is the eigenvector corresponding to the largest eigenvalue. Then the maximum happens when $x$ is the eigenvector of $A^TA$ corresponding to the largest eigenvalue of that matrix.


问:
Understanding a proof of the Unique Readability Theorem in FOL ( by Shapiro, Classical Logic).
In his proof of the Unique Readability Theorem , Shapiro makes use of a previously proved thesis, namely

Theorem 5. Let α,β be nonempty sequences of characters on our alphabet, such that αβ (i.e α followed by β) is a formula. Then α is not a formula.

At some point in his proof ( 5th bullet bellow)  , Shapiro needs to prove that $(\phi_1\&\phi_2)$ cannot be the same formula as $(\phi_3\lor\phi_4)$.

His argument is that $\phi_1$ cannot be a proper part of $\phi_3$ , nor $\phi_3$ of $\phi_1$.

He justifies this by : Theorem 5.

Could you please tell me in which way Theorem 5 applies here?

Here is the entire proof (https://plato.stanford.edu/entries/logic-classical/#FeatSynt):

Theorem 6. Let θ be any formula of L1K=. If θ is not atomic, then there is one and only one among (2)–(7) that was the last clause applied to construct θ. That is, θ could not be produced by two different clauses. Moreover, no formula produced by clauses (2)–(7) is atomic.

[ Note : the clauses referred to here are the formation rules of the language.]

Proof:

By Clause (8), either θis atomic or it was produced by one of clauses (2)–(7). Thus, the first symbol in θ must be either a predicate letter, a term, a unary marker, or a left parenthesis.

If the first symbol in θ is a predicate letter or term, then θ is atomic. In this case, θ was not produced by any of (2)–(7), since all such formulas begin with something other than a predicate letter or term.

If the first symbol in θ is a negation sign “¬”, then was θ produced by clause (2), and not by any other clause (since the other clauses produce formulas that begin with either a quantifier or a left parenthesis).

Similarly, if θ begins with a universal quantifier, then it was produced by clause (6), and not by any other clause, and if θ begins with an existential quantifier, then it was produced by clause (7), and not by any other clause.

The only case left is where θ begins with a left parenthesis. In this case, it must have been produced by one of (3)–(5), and not by any other clause. We only need to rule out the possibility that θ was produced by more than one of (3)–(5). To take an example, suppose that θ was produced by (3) and (4). Then θ is (ψ1&ψ2) and θ is also (ψ3∨ψ4), where ψ1,ψ2,ψ3, and ψ4 are themselves formulas. That is, (ψ1&ψ2) is the very same formula as (ψ3∨ψ4).

By Theorem 5,ψ1 cannot be a proper part of ψ3, nor can ψ3 be a proper part of ψ1.

So ψ1 must be the same formula as ψ3. But then “&” must be the same symbol as “∨”, and this contradicts the policy that all of the symbols are different. So θ was not produced by both Clause (3) and Clause (4). Similar reasoning takes care of the other combinations.

答:
It's probably clearer to rephrase Theorem 5 as follows:

If $\gamma$ is a proper initial segment of $\sigma$, we can't have both $\gamma$ and $\sigma$ be formulas.

(Here $\sigma$ is $\alpha\beta$ and $\gamma$ is $\alpha$, in the notation of the original phrasing of Theorem 5.)

As an immediate corollary we get:

If $\gamma_1,\gamma_2$ are formulas, then neither is a proper initial segment of the other.

(Otherwise take $\gamma$ to be the shorter and $\sigma$ to be the longer, in the version of Theorem 5 above.)

This is what's being applied to $\psi_1,\psi_3$ in Shapiro's proof.

Strictly speaking, "under the hood" there's one more combinatorial fact being applied (below I'll write concatenation as juxtaposition, following Shapiro):

Suppose $\alpha,\beta_1,\beta_2,\gamma,\delta$ are strings of symbols such that $$\alpha\beta_1\gamma=\alpha\beta_2\delta$$ but $$\beta_1\not\prec\beta_2\quad\mbox{and}\quad\beta_2\not\prec\beta_1.$$ Then $\beta_1=\beta_2$.

Above, $\alpha$ is "$($," $\beta_1=\psi_1,\beta_2=\psi_2$, and $\gamma$ and $\delta$ are "$\wedge\psi_2)$" and "$\wedge\psi_4)$" respectively.