问:
How to prove that $fg$ and $gf$ are order embeddings?
From Wikipedia, we know that

Given two posets $(S,\preceq_S)$ and $(T,\preceq_T)$, a function $f:S\to T$ is an order embedding if $f$ satisfies:
$$
\forall x,y\in S, \quad x \preceq_S y \Leftrightarrow f(x) \preceq_T f(y)
$$

Similarly, we can define a reverse embedding:

Given two posets $(S,\preceq_S)$ and $(T,\preceq_T)$, a function $f:S\to T$ is an reverse embedding if $f$ satisfies:
$$
\forall x,y\in S, \quad x \preceq_S y \Leftrightarrow f(y) \preceq_T f(x)
$$

Now, we return to the question, given two posets $(S,\preceq_S)$ and $(T,\preceq_T)$, $f:S\to T$, $g:T\to S$ are both reverse embeddings, how to prove that $fg$ and $gf$ are non-decreasing order embeddings if and only if $x\preceq_S g(y)\Leftrightarrow y\preceq_T f(x)$?

Note: $fg$ and $gf$ are non-decreasing order embeddings means:
$$
x \preceq_S gf(x)\quad \mathrm{and}\quad y \preceq_T fg(y)
$$

I thought for a long time, but only figured out how to prove the necessity:

From the question we know $fg$ and $gf$ are non-decreasing order embeddings, and because $f$ and $g$ are both reverse embeddings, we could obtain that
$$
x\preceq_S g(y) \Rightarrow fg(y) \preceq_T f(x) \Rightarrow y \preceq_T fg(y) \preceq_T f(x) \Rightarrow y \preceq_T f(x)
$$
Similarly,
$$
y \preceq_T f(x) \Rightarrow gf(x) \preceq_S g(y) \Rightarrow x \preceq_S gf(x) \preceq_S g(y) \Rightarrow x \preceq_S g(y)
$$

I still haven't come up with proof of sufficiency, could you help me solve it?

Thank you!

答:
Assume $x\leq_S g(y)\Leftrightarrow y\leq_T f(x)$ for all $x\in S$ and $y\in T$.

For all  $x\in S$, since $f(x)\leq_T f(x)$, we get $x\leq g(f(x))$ [applying $y\leq_T f(x)\Rightarrow x\leq_S g(y)$ with $y = f(x)$].

For all  $y\in T$, ssince $g(y)\leq_S g(y)$, we get $y\leq f(g(x))$ [applying $x\leq_S g(y)\Rightarrow y\leq_T f(x)$ with $x = g(y)$].

There's something very strange happening with the terminology in your question. You define "$f\colon S\to T$ is an order embedding" to mean "$x\leq_S y\Rightarrow f(x)\leq_T f(y)$". But then you seem to redefine "$f\colon S\to S$ is an order embedding" to mean "$x\leq_S f(x)$". Not every order embedding satisfies the latter property. I would call an order embedding satisfying the property "non-decreasing".

So the correct statement would be: If $f\colon S\to T$ and $g\colon T\to S$ are reverse embeddings, then $gf\colon S\to S$ and $fg\colon T\to T$ are non-decreasing order embeddings if and only if for all $x\in S$ and $y\in T$, $x\leq_S g(y)\Leftrightarrow y\leq_T f(x)$.


问:
Test whether a given function is polynomial
You have a black box function to which you can give real number inputs and from which you can receive real number outputs. How would you test whether it is likely to be a polynomial?

One expensive idea is to use finite differences:

Choose a maximum degree n of the "polynomial" you are testing.

Choose a consecutive sequence with random step size, and evaluate the function there to get an output sequence. E.g., $$[ 2, 2.3, 2.6, 2.9,\dots] \to [ 4.81, 5.02, 5.05, 4.90,\dots]$$

Using the output sequence as S[0], define S[n] so that its k^th entry S[n][k] = S[n-1][k+1]-S[n-1][k]. E.g. S[1] = [5.02-4.81,5.05-5.02,4.90-5.05,...] = [0.21,0.03,-0.15,...]

If the function is a polynomial (of degree at most n), then the sequence S[n+1] should be all zeros.

Some issues about programming this method:

Would be expensive for large n

If S[0] has large values, computer arithmetic will produce bad results for S[1] and beyond.

答:
I don't know much about "likely". But $N+1$ distinct points uniquely determine a degree $\le N$ polynomial by Lagrange interpolation. If the obtained points are $(x_i, y_i)$, for $i=0, \ldots N$, then first define:

$$L_{N, j}(x) = \prod_{k \ne j}^{N}\frac{x-x_k}{x_j-x_k}$$
(which is clearly a polynomial of degree $N$). The unique polynomial passing through those points is $$p(x) = \sum_{j=0}^{n}y_j L_{N, j}(x),$$
which is a polynomial of degree at most $N$. If you compute one more point from your black box, then you can compare it with this polynomial to see whether it fits.

By the same argument, there is no way to deterministically rule out polynomials given a finite set of input values, since the polynomial described above can always be constructed.

If you choose a constant step size (like $1$), your $p(x_{N+1})$ may be easy to calculate. In that case, you would need to only compute

$$\sum_{j=0}^{n}y_{j}\prod_{j \ne k}(x_{N+1} - x_j) = \sum_{j=0}^{n}y_{j}\prod_{j \ne k}(N+1-j)$$

which doesn't sound over-complex.

ContentLicense: CC BY-SA 4.0


问:
Questions about Moore–Penrose inverse
I have some questions about Moore–Penrose inverse.

Let $A, P\in \mathbb{R}^{d\times d}$. Suppose $A$ is positive definite and $P$ is a projection matrix with $P^2=P, P^\top=P$. I try to prove that $A^{-1}-(PAP)^{-}$ is semi-positive definite. Here $B^-$ denotes the Moore–Penrose inverse of $B$. (I can prove it using some statistical methods. But how to verify it directly.)

Suppose $A_n\to A$ with $A_n$'s and $A$ being semi-positive definite. Do we have $A_n^- \to A^-$? (It holds when $A$ is positive definite.)

As pointing out by user1551, 2 is not true in general. Actually, I try to answer the following question. Suppose $A_n\to A$ with $A_n$'s and $A$ being positive definite, do we have $(PA_nP)^- \to (PAP)^-$? Here $P$ is a projection matrix with $P^2=P$ and $P^\top=P$.

Thanks.

答:
Yes. By a change of orthonormal basis, we may assume that $P$ and $A$ are partitioned as
$$
P=\pmatrix{I&0\\ 0&0}\text{ and }A=\pmatrix{X&Y\\ Y^T&Z}.
$$
Let $S$ be the Schur complement of $X$ in $A$ (i.e. $S=Z-Y^TX^{-1}Y$). Then
\begin{aligned}
A^{-1}
&=\pmatrix{X^{-1}+X^{-1}YS^{-1}Y^TX^{-1}&-X^{-1}YS^{-1}\\ -S^{-1}Y^TX^{-1}&S^{-1}}\\
&=\pmatrix{X^{-1}&0\\ 0&0}+\pmatrix{X^{-1}YS^{-1}Y^TX^{-1}&-X^{-1}YS^{-1}\\ -S^{-1}Y^TX^{-1}&S^{-1}}\\
&=(PAP)^-+\pmatrix{X^{-1}YS^{-1/2}\\ -S^{-1/2}}\pmatrix{S^{-1/2}Y^TX^{-1}&-S^{-1/2}}\\
&\succeq(PAP)^-.
\end{aligned}

No. E.g. when $\{A_n\}_{n\in\mathbb N}$ is a sequence of positive real numbers that converges to $A=0$, we have $A_n^-\to+\infty\ne0=A^-$ as $n\to\infty$.

Yes. Since $A_n\to A$, we have $PA_nP\to PAP$. As $P$ is an orthogonal projection and both $A_n$ and $A$ are positive definite, the range of $P$ is an invariant subspace as well as the column spaces of both $PA_nP$ and $PAP$, and the latter two matrices are nonsingular on the range of $P$. It follows that the inverse of $PA_nP$ on the range of $P$ converges to the inverse of $PAP$ on the range of $P$. Hence $(PA_nP)^-\to(PAP)^-$. In terms of block matrices, we mean the assumption
$$
A_n=\pmatrix{X_n&Y_n\\ Y_n^T&Z_n}\to A=\pmatrix{X&Y\\ Y^T&Z}
$$
implies that $X_n\to X$ and in turn
$$
(PA_nP)^-=\pmatrix{X_n^{-1}&0\\ 0&0}\to (PAP)^-=\pmatrix{X^{-1}&0\\ 0&0}
$$
when $n\to\infty$.

Title: Questions about Moore–Penrose inverse


问:
What's the probability that the boy loves pizza?
Problem

A survey of boy-girl couples suggests that in 20% of couples both the boy and the girl love pizza, in 5% of couples the girl loves pizza, but the boy doesn't, in 10% couples the boy loves pizza, but the girl doesn't.

If a boy is chosen randomly and he loves pizza, what's the probability that the girl loves pizza too?

Solution attempt

$A$ - the boy loves pizza,
$B$ - the girl loves pizza

Then we know that $P(AB)=0.2$, we need $P(B|A)$, but I guess we don't know $P(A)$, we only know $P(\overline{A}B)$ and $P(A\overline{B})$, but I don't know what to do with that information.

How to solve that problem?

答: Note $P(A)=P(AB)+P(A\bar{B})=0.2+0.1=0.3$. You can do the rest.


问:
Absolute value of sum of absolute values
Wondering if it's generally true that:

$$||a|+| b||=|a|+|b|$$

If I assumme the contrary, that is $||a|+| b|| = m$, $|a|+|b|=k$, and $k \neq m$,

then $|a|+| b| = m$ or $|a|+| b| = -m$. But since $k \neq m$, this is a contradiction. So it has to be the case that they're equal.

Does this reasoning make sense ?

答:
For any real number $x,$ the absolute value or modulus of $x$ is defined as
$$|x| = \left\{
   \begin{array}{rl}
     x, & \text{if }  x \geq 0 \\
     -x, & \text{if } x < 0
   \end{array}\right.$$
Letting $x=|a|+|b|$ we see that since $|a|,|b|\ge 0$
$$|x| = ||a|+|b||=x=|a|+|b|$$
because both $|a|\ge 0$ and $|b|\ge 0$.


答:
Hint:

In $100$ couples:

How many couples do you expect to see where both the boy and the girl love pizza?

How many couples do you expect to see where the boy loves pizza but the girl does not?

How many couples do you expect to see where the boy loves pizza?

In what proportion of the couples where  the boy loves pizza do you expect to see the girl loves pizza too?

More formally, the answers to these questions are $100 \,\mathbb P(AB)$, $100\, \mathbb P(A\overline B)$, $100 \,\mathbb P(A)$ and $\mathbb P(B \mid A)= \frac{100 \mathbb P(AB)}{100 \mathbb P(A)} = \frac{\mathbb P(AB)}{ \mathbb P(AB)+ \mathbb P(A\overline B)}$, all of which you can calculate

Title: What's the probability that the boy loves pizza?

Body:
Problem

A survey of boy-girl couples suggests that in 20% of couples both the boy and the girl love pizza, in 5% of couples the girl loves pizza, but the boy doesn't, in 10% couples the boy loves pizza, but the girl doesn't.

If a boy is chosen randomly and he loves pizza, what's the probability that the girl loves pizza too?

Solution attempt

$A$ - the boy loves pizza,
$B$ - the girl loves pizza

Then we know that $P(AB)=0.2$, we need $P(B|A)$, but I guess we don't know $P(A)$, we only know $P(\overline{A}B)$ and $P(A\overline{B})$, but I don't know what to do with that information.

How to solve that problem?


问:
Is $\sum_{n=1}^{\infty} {\frac{(-1)^n}{x+2^n}} $ in $(-2,\infty)$ uniformly convergent?

Is $\sum_{n=1}^{\infty} {\frac{(-1)^n}{x+2^n}} $ in $(-2,\infty)$ uniformly convergent?

I started by checking if it is pointwise convergent, because if it wasn't then especially it is not uniformly conveergent. But by Leibnitz, it is convergent.

But I got stuck on proving that it is uniformly convergent. Any help would be appreciated!

答: Hint: Use Weierstrass M-test with $$ M_n = \frac{1}{2^n-2}.$$</p>

ContentLicense: CC BY-SA 3.0


问:
How to get $2\langle \nabla u, \partial_t \nabla u \rangle=\frac{d}{dt}\|\nabla u\|^2_{L^2(U)}$
In the book "Partial Differential Equations"-Lawrence C. Evan, Section 5.2, we get $2\langle u'(t), u(t)\rangle=\frac{d}{dt}\|u(t)\|^2_{L^2(U)}$ when $u\in L^2(0,T; H^1_0(U))$ and $u'\in L^2(0,T; H^{-1}(U)).$

When will we get the formula

$$2\langle \nabla u(.,t), \partial_t \nabla u(.,t) \rangle=\frac{d}{dt}\|\nabla u(.,t)\|^2_{L^2(U)}~~?$$

答:
Denote $\frac{du}{dt}$ by $u'$, if $u,u' \in L^2(0,T;H_0^1(U))$ then
$$\left(\|\nabla u\|_{L^2(U)}^{2}\right)'=\langle \nabla u, \nabla u \rangle'=\langle \nabla u', \nabla u \rangle+\langle \nabla u, \nabla u' \rangle=2\langle \nabla u, \nabla u' \rangle,$$
since $\langle f, g \rangle'=\langle f', g \rangle +\langle f, g' \rangle$, where $\langle  \cdot, \cdot \rangle$ denotes the inner product in $L^2(U)$.


问:
Is the quotient field of $\mathbb{Z}[\sqrt{-3}]$ galois over $\mathbb{Q}$ and what is the Galois group?
I proved that $\mathbb{Z}[\sqrt{-3}]$ is isomorphic to $\mathbb{Z}[x]/(x^{2}+3)$.

Because $f = (x^{2}+3)$ is irreducible by Eisenstein. Is that enough to conclude that it splits over $\mathbb{Q}$? If f spits it would mean that $\mathbb{Z}[x]/(x^{2}+3)$ is galois, doesn't it?

Now I want to find the Galois group, but for that I need the automorphisms which act on the roots. Now I'm not sure if I should look at $[\mathbb{Z}(\sqrt{-3}):\mathbb{Z}]=2$, $[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}]=2$, $[\mathbb{Z}(\sqrt{3}, i):\mathbb{Z}]=4$ or $[\mathbb{Q}(\sqrt{3}, i):\mathbb{Q}]=4$.

If anyone could clear those things up for me, it would be a lot of help.

答:
$X^2+3$ is irreducible over $\Bbb{Q}$ so $\Bbb{Q}[x]/(x^2+3)$ is a field, isomorphic to $\Bbb{Q}(\sqrt{-3})$.

$X^2+3=(X+\sqrt{-3})(X-\sqrt{-3})$ splits completely over $\Bbb{Q}(\sqrt{-3})$ so $\Bbb{Q}(\sqrt{-3})/\Bbb{Q}$ is Galois.

For each root there is an automorphism sending $\sqrt{-3}$ to that root.

Therefore, the Galois group has two elements: the identity, and $\sigma(a+b\sqrt{-3})=a-b\sqrt{-3}$.

Both restrict to automorphisms of the subring $\Bbb{Z}[\sqrt{-3}]$.

In some contexts we can say that $\Bbb{Z}[\sqrt{-3}]/\Bbb{Z}$ is Galois.

Interestingly $\Bbb{Z}[\frac{1+\sqrt{-3}}{2}]$ is a PID while $\Bbb{Z}[\sqrt{-3}]$ is not even a Dedekind domain (the ideal $(2,\sqrt{-3}-1)$ is not principal and not inversible).


问:
In a velocity-time graph, if the velocity is negative and the slope is negative, is it still accelerating?
So in a velocity-time graph, a negative velocity means the speed in the opposite direction. So from -10 m/s to -30 m/s, it is accelerating in the opposite direction. However the slope is negative, so is it still accelerating?</p>

答:
No.

To illustrate take a simple example with velocity $v<0$ during $0<t<1.5 $

$$ v=t^2- 3t$$

the acceleration

$$ a=\dfrac{dv}{dt}=2 t-3\;<0 \text{ which is decelerating in the above interval.}$$


问:
What is the value of $a+b$ Satisfies these two equations.
Let $a,b\in \mathbb R$ Such that :
$$\cases{a^3-3a^2+5a-17=0 \\ b^3-3b^2+5b+11=0}$$

Find the value of $a+b$

I’ve tried to add the two equation and factoring $a+b$ but it didn’t help me.
I think that letting $f(x)=x^3-3x^2+5x-17$ And $g(x)=b^3-3x^2+5x+11$ Could help . claim :$a+b=2$

Body:
Let $a=c+1, b=d+1$

Then the equations becomes

$$c^3+2c-14=0$$

$$d^3+2d+14=0$$

Add them up,

$$c^3+d^3+2c+2d=0\iff (c+d)(c^2-cd+d^2)+2(c+d)=0\\(c+d)(c^2-cd+d^2+2)=0$$

Since $c^2-cd+d^2+2$ clearly $>0$ we know $c+d=0$ so $a+b=2$


问:
How to anti-symmetrize product of two tensors?
Say I have an anti-symmetric tensor $f_{ab}$ and two other tensors $M^{ij}$, $N^{kl}$ . Now consider the contraction $f_{ab}M^{ai}N^{bj}$, since we know $f_{ab}$ is anti-symmetric we know that for any tensor that is contracted with it, only it's anti-symmetric part will contribute to the contraction. Now my question is, wich of the following equations is true?

$$f_{ab}M^{ai}N^{bj} = f_{ab} \times  \frac{1}{2}(M^{ai}N^{bj} - M^{bi}N^{aj}) $$ or
$$f_{ab}M^{ai}N^{bj} = f_{ab} \times  \frac{1}{2}(M^{ai}N^{bj} - M^{bj}N^{ai})$$

A short explanation of why whichever is true holds would be appreciated.

答: Note$$f_{ab}M^{ai}N^{bj}=-f_{ba}M^{ai}N^{bj}=-f_{ab}M^{bi}N^{aj}.$$And averaging the first and last expressions,$$f_{ab}M^{ai}N^{bj}=f_{ab}\times\tfrac12(M^{ai}N^{bj}-M^{bi}N^{aj}),$$in agreement with your first attempt.


答:
Let's denote $a+b = x$, and substitute $b = x - a$ in the second equation

\begin{cases}
a^3-3a^2 + 5a - 17=0 \\
(x-a)^3 - 3(x-a)^2 + 5(x-a) + 11 = 0
\end{cases}

\begin{cases}
a^3-3a^2 + 5a - 17=0 \\
-a^3 + 3 a^2 x - 3 a^2 - 3 a x^2 + 6 a x - 5 a + x^3 - 3 x^2 + 5 x + 11 = 0
\end{cases}

Add the first equation to the second equation:
\begin{cases}
a^3-3a^2 + 5a - 17=0 \\
3 a^2 x - 6 a^2 - 3 a x^2 + 6 a x + x^3 - 3 x^2 + 5 x - 6 = 0
\end{cases}

Notice that the second equation may be rewritten as:
$$x(3a^2-3ax+x^2-x+3) - 2(3a^2-3ax+x^2-x+3) = 0$$
$$(x-2)(3a^2-3ax+x^2-x+3) = 0$$

We therefore get that $a+b = x = 2$.

Title: What is the value of $a+b$ Satisfies these two equations.

Body:
Let $a,b\in \mathbb R$ Such that :
$$\cases{a^3-3a^2+5a-17=0 \\ b^3-3b^2+5b+11=0}$$

Find the value of $a+b$

I’ve tried to add the two equation and factoring $a+b$ but it didn’t help me.
I think that letting $f(x)=x^3-3x^2+5x-17$ And $g(x)=b^3-3x^2+5x+11$ Could help . claim :$a+b=2$


问:
$f(z)= (s-1) \zeta(s)$. Prove that $\log |f(0)|=0$
I am reading Balazard, Saias and Yor Paper on the RH where they consider a function$$  f(z)= (s-1) \zeta(s)$$  where $ \ s=\frac{1}{1-z}$ and $\zeta $ denotes the Zeta function

$$\text{Claim:} \qquad \log|f(0)|=0$$
My Try-

$$\lim_{s \to 1} (s-1)\zeta(s)=1$$
$$\lim_{z \to 0} f(z)=1$$
$f$ is continuous at $0$ as it is analytic near $0$
$$f(0)=\lim_{z \to 0} f(z)=1$$
$$\log|f(0)|=0$$
Thank you for your help.

答:
No mistake, the reasoning is fine.

$(s-1)\zeta(s)$ is analytic at $s=1$ and $1/(1-z)$ is analytic at $0$ whence the composition is analytic at $z=0$.


问:
Integrating Chebyshev polynomial of the first kind
I'm trying to evaluate the integral of the Chebyshev polynomials of the first kind on the interval $-1 \leq x \leq 1 $ .
My idea is to use the closed form
$$T_n(x) = \frac{z_1^n + z_2^{-n} }{2}$$
where $z_1 = (x + \sqrt{x^2 - 1})$ and $z_2 = (x - \sqrt{x^2 - 1})$,
giving the following integral:
$$
\int _{-1}^{1}\!1/2\, \left( x+\sqrt {-1+{x}^{2}} \right) ^{n}+1/2\,
 \left( x-\sqrt {-1+{x}^{2}} \right) ^{n}{}{dx}
$$
I'm stuck at integrating $z_1$ and $z_2$. I tried integrating by parts $n$ times, but i'm looking for a general formula. My calculus is pretty rusty so i'm not sure if this is the way to go. Any tips?
Thanks a lot.

答:
It is actually not that hard. You can derive a lot of relations on the wiki page yourself by substituting $\cos\theta$ for $x$ and use the defining relation of Chebyshev polynomials:

$$T_n(\cos\theta) = \cos( n\theta)$$

For example, one have:
$$\begin{align}\int T_n(x) dx = & \int T_n(\cos\theta) d\cos \theta\\
= & -\int \cos(n\theta)\sin\theta d\theta\\
= & -\frac12 \int \left(\sin((n+1)\theta) - \sin((n-1)\theta)\right)d\theta\\
= & \frac12 \left(\frac{\cos((n+1)\theta)}{n+1} - \frac{\cos((n-1)\theta)}{n-1}\right) + \text{const.}\\
= & \frac12 \left(\frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)}{n-1}\right) + \text{const.}
\end{align}
$$


问:
Stereographic projection: line element
We assign coordinates ($\rho, \phi$) to each point in the surface of a sphere, where $\rho$ is the distance from the south pole of the sphere to the point where a straight line passing through both the south pole and the point in question intersects the tangent plane to the north pole, and $\phi$ is the usual azimuthal angle. Show that the line element for the surface of the sphere in these coordinates is $ds^2=\frac{d\rho^2}{(1+\frac{\rho^2}{a^2})^2}+\frac{\rho^2d\phi^2}{1+\frac{\rho^2}{a^2}}$.

My attempt was as follows:

In cartesian coordinates, $ds^2=dx^2+dy^2+dz^2$. I placed the origin of coordinates in the center of the sphere and differenciated the equation defining the sphere (we assume radius a): $x^2+y^2+z^2=a$ to get $2xdx+2ydy+2zdz=0$. From this last one I got a constraint over $dz$ and I plugged it into the equation for $ds$, obtaining the following expression: $ds^2=dx^2+dy^2+\frac{(xdx+ydy)^2}{a^2-(x^2+y^2)}$.

Once there, I tried looming for a relation between x, y and $\rho$, which I found to be by using a bit of trigonometry and finding the intersection point of the sphere surface and the line coming from the pole: $x=(\frac{2a}{\rho})^2\sqrt{\rho^2-(2a)^2}cos(\phi)$ and $y=(\frac{2a}{\rho})^2\sqrt{\rho^2-(2a)^2}sin(\phi)$, but differenciating these and plugging them into the expression I had found for $ds^2$ leads me to some terrible algebra which I think is not taking me to the correct solution...

答:
OK, this problem is a mess. Now that we have one of the standard stereographic projection mappings, we know that the mapping is conformal, and so the formula as given must be incorrect.

Let $a$ be the diameter of the sphere, and let $(\psi,\phi)$ be the usual spherical coordinates (with $\psi$ the angle from the vertical and $\phi$ the polar angle, as given in the problem). Then we all know that
$$ds^2 = \big(\frac a2\big)^2(d\psi^2 + \sin^2\psi\,d\phi^2).$$
We observe that we have a right triangle with legs $a$ and $\rho$, and angle $\psi/2$ at the south pole. Thus,
$$\rho = a\tan\big(\frac\psi 2\big) \quad\text{and}\quad \sin\psi = 2\sin\big(\frac\psi 2\big)\cos\big(\frac\psi 2\big)=\frac{2a\rho}{a^2+\rho^2}.$$
Now easy computation gives
$$\frac{d\rho}{1+\big(\frac\rho a\big)^2}=\frac a2\,d\psi.$$
Thus,
\begin{align*}
ds^2 &= \big(\frac a2\big)^2(d\psi^2 + \sin^2\psi\,d\phi^2) = \frac{d\rho^2}{\big(1+(\frac\rho a)^2\big)^2} + \frac{\rho^2\,d\phi^2} {\big(1+(\frac\rho a)^2\big)^2} \\
&= \frac1{\big(1+(\frac\rho a)^2\big)^2}(d\rho^2 + \rho^2\,d\phi^2),
\end{align*}
which shows that the metric is conformally equivalent to the usual metric on the tangent plane at the north pole, as expected.

Title: Stereographic projection: line element


问:
Compute $E[\alpha]$ where $\alpha = \inf\{i > 0 : X_i > X_0\}$ and $\{X_i\}_{i=0}^{\infty}$ are IID.
If $\{X_i\}_0^{\infty}$ are IID discrete, then I'm quite sure $E[\alpha] = \infty$ because
$$
E[\alpha] = \sum_{\{x_j \}_{j=0}^{N}} E[\alpha | X_0 = x_j] P(X_0 = x_j) = \infty = E[\alpha | X_0 = \sup\{x_j \}_{j=0}^{N}]\\
$$

In this set of values, you will eventually condition on $X_0$ being the maximum value, and as long as the probability of obtaining that maximum value is non-zero, then $E[\alpha | X_0 = \sup\{x_j \}_{j=0}^{N}]$ is necessarily $\infty$ because you will never find another $X_i$ such that $i > 0$ and $X_i > X_0$.

Now what if $\{X_i\}_0^{\infty}$ are IID continuous. I feel the same logic should apply, but for continuous distributions, there should be a probability of zero that $X_0$ realizes the maximum value (or any value), which would seem to mean $E[\alpha]$ is finite.

Is this where the "almost surely" comes in? Which would mean we could assume that there is a non-zero probability that $X_0$ realizes the maximum value of the continuous distribution, in which case the logic applied in the discrete case will still apply here.

答:
Note that, conditional on $X_0$, the events $\{X_i > X_0\}_i$ are independent and all have conditional probability $1-F(X_0)$. Thus, conditional on $X_0$, one has that $\alpha$ is a geometric random variable with chance of success $p(X_0) = 1-F(X_0)$.
$$
E[\alpha] = E[E[\alpha | X_0]] = E\left[ \frac{1}{p(X_0)} \right] = E\left[\frac{1}{1-F(X_0)}\right].
$$

Indeed, if $X_0$ is discrete with finite support or otherwise has positive probability of achieving its supremum, then there is a positive chance that $F(X_0) = 1$, so that the expectation is infinite.

In the continuous case we can easily see the expectation is infinite. We can simplify the expectation further if the distribution is continuous, by the Stieltjes change of variables formula for continuous finite variation processes,
$$
\begin{align}
&= \int_\mathbb{R} \frac{1}{1-F(x)}dF(x) \\
 &= (-\log(1-F(\infty))) - (-\log(1-F(-\infty)))\\
&= \infty-0.
\end{align}
$$
Thus in the continuous case the result still holds.

In the general case, we can consider arbitrary $y_n$ increasing and we find using that $1/(1-x)$ is increasing on $[0,1]$ that

$$
\begin{align}
E[\alpha] &\geq \sum_n E\left[\frac{1_{X_0 \in (y_n, y_{n+1}]}}{1-F(X_0)}\right]\\
& \geq \sum_n \frac{1}{1-F(y_n)}P(y_n< X_0 \leq y_{n+1})\\
& = \sum_n \frac{F(y_{n+1})-F(y_n)}{1-F(y_n)}
\end{align}
$$

Assuming that $X_0$ does not achieve its essential supremum with positive probability (that case was already covered), we can choose the $y_n$ cleverly to make the sum infinite.
Choose any $y_0$ such that $0 < F(y_0) < 1$ and recursively define $y_{n+1}$ to be such that $$1> F(y_{n+1}) - F(y_{n}) > \frac{1}{2}(1-F(y_n)).$$ We can always do this since the sup is never attained and $F(y)-F(y_n) \to 1-F(y_n)$ as $y \to \infty$.

Then we have

$$
E[\alpha] \geq \sum_n \frac{1}{2} = \infty.
$$

In summary, $E[\alpha] = \infty$ for all possible distributions of $X_0$ on $\mathbb{R}$.


问:
Intersection points of four lines
Let a be perpendicular to b, and the intersection point on a and b be called "S", further, let c be perpendicular to d and the intersection point be "T", further a $ \nparallel$ c, a $ \nparallel$ d. Show that these 4 lines have 6 intersection points.

So I know that n lines can have at most $\frac{n^2-n}{2}$ , so in that case (16-4)/2=6, but how can I argue that this is indeed the case here?

答:
The number of points would reduce from $6$ when one or more pair of lines are coincident or parallel or atleast three lines are concurrent.

Probably one can assume $a,b,c,d$ are distinct to rule out coincidence.

Then $a \nparallel c \,$, $\;a \nparallel d$ can be used to show no two of four lines are parallel.

But this is still not enough since it can happen that one of $c,d$ passes through $S$ i.e.,  for instance if $a,b,c$ concur at $S$ which is distinct from $T$ then there will be less than $6$ intersection points (four in this case, as can be seen by drawing a quick diagram).

Title: Intersection points of four lines


问:
Questions about expressing each of the following statements in formal language and negating each of them respectively
I am trying to express the following statements in formal language and negating each of them.

The following is my attempted solution.

The universe of each statements is the set of real numbers and both $\varepsilon$ and $\delta$ denote positive real numbers.

Let $a$ be a fixed element of $\mathbb{R}$. For every $\varepsilon$ there exist $\delta$ such that for every $x \in \mathbb{R}$, if $|x-a| < \delta$, then $|x^{2} - a^{2}| < \varepsilon$.

For every $\varepsilon$, there exists an integer $N$ such that $1/n$ < $\varepsilon$ for all $n \geq N$.

Each of the followings is the expression in formal language for each of the statements respectively:

$\forall \varepsilon,(\varepsilon \in \mathbb{R}^{+} \rightarrow \exists \delta ,(\delta \in \mathbb{R}^{+} \wedge \forall x,(|x-a|<\delta \rightarrow |x^{2} - a^{2}|<\epsilon))).$

$\forall \varepsilon,(\varepsilon \in \mathbb{R}^{+} \rightarrow \exists N,(N\in \mathbb{Z} \wedge \forall n,(n \geq N \rightarrow 1/n < \varepsilon))).$

The respective negation of the each of the statements are the following:

$\exists \varepsilon ,(\varepsilon \in \mathbb{R}^{+} \wedge \forall \delta ,(\delta \in \mathbb{R}^{+} \rightarrow \exists x,(|x-a| < \delta \wedge |x^2 - a^2| \geq \varepsilon))).$

$\exists \varepsilon ,(\varepsilon \in \mathbb{R}^{+} \wedge \forall N,(N \in \mathbb{Z} \rightarrow \exists n,(n \geq N \wedge 1/n \geq \varepsilon))).$

Each of the above negations of the original statements can be expressed as plain English respectively as follows:

Let a be a fixed element of $\mathbb{R}$. There is a positive real number $\varepsilon$ and for every $\delta$, if $\delta$ is a positive real number, then there exists $x$ such that $|x-a| < \delta$ and $|x^2 - a^2| \geq \varepsilon$.

There exists a positive real number $\varepsilon$ and for every $N$, if $N$ is an integer, then there exists an $n$ such that $n \geq N$ and $1/n \geq \varepsilon$.

My questions are: Are all my expressions correct , and why in all of the cases, when one negates each of the statements, one does not negate: the definition "Let $a$ be a fixed element of $\mathbb{R}."$, $\varepsilon \in \mathbb{R}^{+}$, $\delta \in \mathbb{R}^{+}$, and $N \in \mathbb{Z}$?

All statements are form:

Daepp, U. and Gorkin, P., 2011. Reading, writing, and proving. 2nd ed. New York: Springer, p.46.

答:
I would call these formulations of the statements and their negations correct. The extra subdomain quantifiers you mention could be avoided but you’d get a longer statement and it wouldn’t really affect the negation, in that $\forall \varepsilon \in \Bbb R^+ \phi$would be negated as $\exists \varepsilon \in \Bbb R^+ \lnot \phi$ etc. So leave them in I’d say.

The $a$ is what’s formally called a parameter of the formula; it need not be negated.


问:
strong math induction, base case fails?
For the sequence, a0,a1,a2,...,ak  , a0=5, a1=11, ak=7ak-1 - 10ak-2 for k $\ge$2. Prove an = $3*2^n +2*5^n$, n $\ge$ 0.

The first base case of n=0 works but for n=1,a0 is 11, but the base case is 16 when substituted in. Does this mean that the proof has failed?

答:
As The statement is not true for $n=1$ then the statement is just false.  Also it is not true for $a_2= 77-50 = 22$.

But if $a_1 = 16$ and the $11$ was a typo we get $a_2 = 7*16-50 = 62 = 3*2^2 + 2*5^2$ and the statement is verified for $n = 2$.

If so maybe the statement is true if we state $a_1 = 16$ instead.

Can we do an induction step?:

If $a_k = 3*2^k + 2*5^k$ and $a_{k-1} = 3*2^{k-1} + 2*5^{k-1}$ then does

$a_{k+1} = 7(3*2^k + 2*5^k) - 10(3*2^{k-1} + 2*5^{k-1}) \overbrace{=}^? 3*2^{k+1} + 2*5^{k+1}$?

That seems like a simple thing to verify (or refute) by simple algebra.

Title: strong math induction, base case fails?


问:
order of parameters in differential forms
I am completely mystified about parameters in differential forms and wherever I looked it's just glossed over.

Take for example the $2-$form $\omega = dx\land dy$ in $\mathbb{R}^n$.

Usually it's just defined like that, but given two vectors $(x,v_1), (x,v_2)$ in the tangent space at $x$, it shold be that $dx \land dy = dy \land dx$, since
$$(dx \land dy) (x)((x,v_1), (x,v_2)) = $$
$$- (dy \land dx)(x)((x,v_2)(x,v_1))=$$
$$ (dy \land dx)(x)((x,v_1)(x,v_2))=$$

According to my book, tough, $dx \land dy = - dy \land dx$.
Where am i wrong?

答:
First of all, you're writing incorrect equations. Once you evaluate individual terms, there are no more wedges. The correct equation, using your (cumbersome!) notation, is
\begin{multline*}
(dx\wedge dy)((x,v_1),(x,v_2)) = dx((x,v_1))dy((x,v_2)) - dx((x,v_2))dy((x,v_1)).
\end{multline*}
And of course from this you see that, in fact, $dx\wedge dy = -dy\wedge dx$.


问:
Find the eventually periodic points of the function $f(x)=\lvert x-1 \rvert$
Let $\emptyset\neq X\supset A$,  $f:A\rightarrow X$,

a point $x\in A$ is eventually periodic point  of $f$, if :
$$\exists n_0\in\mathbb N \exists N\in\mathbb N: f^{n+N}(x)=f^n(x) \quad \forall n\ge n_0.$$

Note: $f^n(x) := \underbrace{f\circ f\circ f\circ \dots \circ f}_{ \text{$n$ times }}(x).$

I don't  quite understand the following proof.

$$\text{Let } f:\mathbb R \rightarrow \mathbb R : f(x)= \lvert x-1\rvert,$$
$$f(0)=1,f(1)=0\implies 0 \text{ and } 1 \text{ are periodic points of $f$.}$$
$$f(-1)=2, f(-k)=k+1, f(k)=k-1 ,\forall k\ge 2,k\text{ integer.} \implies $$
$$\implies \text{every integer point is eventually periodic point of $f$ .}$$

I understand the part that $1$ and $-1$ are periodic points, but im a bit confused about the rest of the proof.

答:
Consider the following diagram:
$$
\begin{matrix}
&&&& -1 && -2 && -3 && -4 & \cdots \\
&&&& \downarrow && \downarrow && \downarrow && \downarrow \\
0 & \longleftrightarrow & 1 & \longleftarrow & 2 & \longleftarrow & 3 & \longleftarrow & 4 & \longleftarrow & 5 & \longleftarrow & \cdots
\end{matrix}
$$

If $f$ is repeatedly applied to an integer $k$, we eventually end reach $1$. From that point, the sequence $f^n(k)$ becomes periodic: 1, 0, 1, 0, 1, 0, ...

For a formal proof, because $1$ is eventually periodic and $f(k)=k-1$ for positive $k$, it follows by induction that all positive integers are eventually periodic.

From $f(-k) = k+1$ for positive $k$ it then also follows immediately that all negative integers are eventually periodic.


问:
Absolute value of vector components
When is it possible to assume that for any vector $\vec{x}=(x_1,\dots, x_n)\in\mathbb{R}^n$;

$$|x_{i}| \leq \|\vec{x}\|$$

For the case in  $\mathbb{R}^2$:

$$|x_2| \leq \|\vec{x}\|$$

$$\left|x_2\right| \leq x_2 \cdot \sqrt{\left(\frac{x_1}{x_2}\right)^2+1}$$

Define $\alpha = \sqrt{\left(\frac{x_{1}}{x_2}\right)^2+1}$

Then $|x_i| \leq \|\vec{x}\| \iff \alpha >1$.

Is there a more general result ? Does this reasoning hold ?

答:
Let $X=\mathbb{R}^{N}$. Then, for $x\in X$, the (square of the) $2$-norm is defined as:
\begin{equation*} \|x\|_{2}^{2}=\sum_{i=1}^{N}|x_{i}|^{2}\geq |x_{i}|^{2} \text{ for a given index } i \qquad x=(x_{1},\ldots,x_{N}) \end{equation*}
so taking the square root gives $\|x\|_{2}\geq |x_{i}|$ for all $i\in\{1,\ldots,N\}$. Next, it is a standard result that every norm on a finite-dimensional space (like $X$ in this case) is equivalent, so given another norm, say $\|\cdot\|'$, there are constants $c_{1},c_{2}>0$ such that
\begin{equation*} c_{1}\|x\|'\leq\|x\|_{2}\leq c_{2}\|x\|' \end{equation*}
for all $x\in X$. In particular, it would follow that
\begin{equation*} |x_{i}|\leq\|x\|_{2}\leq c_{2}\|x\|' \end{equation*}
so you can bound individual components by the overall norm for any norm on $X$, up to a constant dependent on the norm (with this constant being $1$ for the $2$-norm).

Title: Absolute value of vector components


问:
Find the maximal interval of existence without actually solving the ODE.
$$y^{\prime}=-\frac{4}{x^{2}}-\frac{1}{x} y+y^{2}, y(1)=1$$

This is a Ricatti equation, $f(x,y)$ is a polynomial in $y$ so it's locally Lipschitz and thus there exists a unique solution to the IVP above, $y_p = \pm \frac{2}{x}$ is a particular solution, we can go ahead and solve it using the change of variable $y = y_p +u$ but is there a way to find the maximal interval of existence without actually doing that?

答:
You can transform and combine
$$
x(xy)'=(xy)^2-4.
$$
As the initial point is between the stationary solutions $xy(x)=\pm 2$, the solution exists for the full interval $(0,\infty)$.


答:
Since the map (defined for $z\ne0$)
$$
z\mapsto \frac{1}{z}=\frac{1}{|z|^2}\bar{z}
$$
sends positive imaginary part to negative and conversely, while $z\mapsto z+3$ doesn't change the imaginary part, if you start from $z\in H^{+}$, you have, in sequence
\begin{align}
&z\in H^{+}\\
&\frac{1}{z}\in H^{-}\\
&\frac{1}{z}+3=\frac{3z+1}{z}\in H^{-}\\
&\biggl(\frac{3z+1}{z}\biggr)^{-1}=\frac{z}{3z+1}\in H^{+}
\end{align}
Similarly if you start from $z\in H^{-}$.

Therefore (1) is true. This of course implies that (2) is false, because $H^{+}\cap H^{-}=\emptyset$.

In order to exclude (3) and (4) you need to find a counterexample.

But, if $z=-1/3+i\in H^{+}$, $z/(3z+1)=1-i/3\notin L^{-}$, which excludes (3). Similarly to exclude (4).

Title: $f(z)={z\over 3z+1}$ maps

Body:
Define

$H^{+}=\{z:y>0\}$

$H^{-}=\{z:y<0\}$

$L^{+}=\{z:x>0\}$

$L^{-}=\{z:x<0\}$

$f(z)={z\over 3z+1}$  maps

$1.$ $H^+\to H^+$ and $H^-\to H^-$

$2$. $H^+\to H^-$ and $H^-\to H^+$

$3.$ $H^+\to L^-$ and $H^-\to L^+$

$4.$ $H^+\to L^+$ and $H^-\to L^-$

If I take $z\in H^{+}$ then ${1\over z}\in H^{-}\Rightarrow {1\over z}+3={3z+1\over z}\in H^{-}$,also real line goes to real line, Now what can I conclude?please help.


问:
Every noncollinear $3$ points of $S\subset\mathbb{R}^2$ contains inside of it another point of $S$. Is $S$ is a dense subset of some convex hull?

Proposition: Suppose $S \subset \mathbb{R}^2$ contains at least $3$
noncollinear points. Suppose further that for every noncollinear $3$
points of $S$, the triangle formed by those $3$ points contains
inside of it - and not on it's boundary - another point of $S$.

Is $S$ a dense subset of some convex hull $H \subset \mathbb{R}^2$?

The converse statment:

$S$ is a dense subset of some convex hull $\implies$ every noncollinear $3$ points of $S$ contains inside it's triangle - and not on the boundary - another point of $S$,

is obviously true due to the definition of "dense subset of $\mathbb{R}^2:$ inside every triangle there will be a point of $S.$

$$$$

I think the proposition is true: if not, there exists a neighbourhood $N_x(r)$ (circle) around some point $x \in (\mathbb{R}^2 - S)$ inside the convex hull of $S$ such that (at least) $3$ points of $S$ lie on the circumference of this circle, but no member of $S$ lies inside $N_x(r)$, and this would contradict the property in the question. However, I'm not 100% certain this is correct and I'm not sure how to formalise this.

Thanks in advance.

答:
It's false. Take, for example, take two points $u, v \in \Bbb{R}$ with distance $3$ between them, and let
$$S = B(u; 1) \cup B(v; 1),$$
i.e. the union of the two open unit balls with centres $u$ and $v$. The closure is just the union of the closed unit balls:
$$B[u; 1] \cup B[v; 1],$$
which is clearly not convex.

But, $S$ has the desired property. If we take non-colinear $x, y, z \in S$, then two of the points must lie in the same open ball, which means the line segment between them lies in the convex open subset of $S$, which we can perturb slightly to get something in the interior of the triangle $\operatorname{conv}\{x, y, z\}$.


问:
Is there a name for a graph in which every vertex is both a central vertex and a peripheral vertex
Is there a name for a graph for which every vertex
is simultaneously in the center and in the periphery of the graph?

If I'm not mistaken the graph representing the states of the Rubik's cube
whose edges are the twists of the cube is such a graph.  I.e., for such a graph there is no special state.    I suppose the complete graph $K_n$ is such a graph as well.

答:
Such a graph is called a self-centered graph.

This is defined in the linked post as a graph whose diameter equals its radius; therefore the maximum eccentricity of any vertex equals the minimum eccentricity.

Note that the Rubik's cube graph has a much stronger property. It is vertex-transitive: there is an automorphism that maps any vertex to any other vertex. In a sense, no graph property can distinguish a vertex from any other. A self-centered graph is one where vertices cannot be distinguished by eccentricity alone, but there might still be other distinguishing characteristics.


问:
$\lim_{x\to +\infty} \ln (1+x) \cdot \sin (\frac{1}{\ln(2x-1)}) $
I have to solve the limit
$$\lim_{x\to +\infty} \ln (1+x) \cdot \sin \left(\frac{1}{\ln(2x-1)}\right)  $$

applying Taylor's series.
I can change the variable:
$$\lim_{y\to o^{+}} \ln \left(1+\frac{1}{y}\right) \cdot \sin \left(\frac{1}{\ln(2\frac{1}{y}-1)}\right) =  \lim_{y\to o^{+}} \ln \left(\frac{y+1}{y}\right) \cdot \sin \left(\frac{1}{\ln(\frac{2-y}{y})}\right) =  \lim_{y\to 0^{+}} (\ln (y+1)-\ln y) \cdot \sin \left(\frac{1}{\ln (2-y) - \ln y}\right) $$

But I have problems with the taylor's series of $\ln x $ in $0$

Then I tried with:
$$\ln (1+x) \cdot \sin \left(\frac{1}{\ln(2x-1)}\right) \sim
  \ln (1+x) \cdot \frac{1}{\ln(2x-1)}= \ln (1+\tfrac{1}{y}) \cdot \frac{1}{\ln(2\frac{1}{y}-1)}= \frac{\ln (1+\frac{1}{y})}{\ln (\frac{2}{y}-1)} =  \frac{\ln(y+1)- \ln y}{\ln(2-y)- \ln y} \sim   \frac{- \ln y}{\ln(2)- \ln y}  \sim 1 $$

and it is indeed  the suggested solution in my book. But I haven't applied taylor's series.
Can someone give some hint how to procede with the other method?

答:
I think what your exercise meant is : do a Taylor series expansion, up to the first order, of $t\to \sin(t)$. You directly wrote an equivalent, which of course amounts to the same, since $\sin 0=0$.
But if you want to be very Taylor series compliant, you can choose as a variable $y=\ln(2x-1)^{-1}$. Then $y\to 0$ as $x\to \infty$.

$$
\ln (x+1) = \ln ( \frac12 \left( 2x -1 +1\right) +1 ) = \ln\left( \frac{\left( 2x -1\right) +3}{2}\right)
$$
which becomes
$$
\ln (x+1) = \frac1y -\ln(2) + \ln \left( 1+ 3 \exp\left(-\frac1y\right) \right).
$$
And your problem is
$$
\left(\frac1y -\ln(2) + \ln \left( 1+ 3 \exp\left(-\frac1y\right) \right)\right)\sin y
$$
as $y\to0$. Then an expansion of $\sin y=0+y+o(y)$...and surprise, surprise, the result you already found appears.


问:
Issues with $\lim_{n\rightarrow\infty}(a_n*b_n) = \lim_{n\rightarrow\infty}(a_n) * \lim_{n\rightarrow\infty}(b_n)$
A few months ago I was answering a question of someone about limits.
And I use what I have been taught at my high school, the book that we learn from it says the following:

Translation:

limit = نهاية

the product = الجداء

Specifically, I talk about the laws when one sequence at least diverges to infinity.
But one argues with me that this is actually wrong.

Now I am reading Rudin's "Principles of Mathematical Analysis" and he approves that law is correct when the two sequences involved in the product converges

I see people say that a sequence converges to infinity but from what I read a sequence that goes to infinity diverges not converges, and Rudin didn't talk about this case (at least in the first three chapters).

So can we say that a sequence that goes to infinity converges and therefore the law applied, or the law apply for another reason, or there some wrong here?

答:
Usually, one says that a sequence $(a_n)_{n\in\Bbb N}$ of real numbers converges if it converges to a real number $l$. Then we say that the limit of the sequence is $l$, aand this is expressed by $\lim_{n\to\infty}a_n=l$.

However, this does not mean that this is the only case in which we can talk about the limit of the sequence. There are also the notions of the limit of a sequence being $\infty$ or $-\infty$, which is expressed by $\lim_{n\to\infty}a_n=\infty$ and by $\lim_{n\to\infty}a_n=-\infty$ respectively. However, such a sequence does not converge.

And, yes, it is true that, for instance, if $\lim_{n\to\infty}a_n=\infty$ and if $\lim_{n\to\infty}b_n=2$, then $\lim_{n\to\infty}(a_nb_n)=\infty$.


问:
What is the value of $a+b$ Satisfies these two equations.
Let $a,b\in \mathbb R$ Such that :
$$\cases{a^3-3a^2+5a-17=0 \\ b^3-3b^2+5b+11=0}$$

Find the value of $a+b$

I’ve tried to add the two equation and factoring $a+b$ but it didn’t help me.
I think that letting $f(x)=x^3-3x^2+5x-17$ And $g(x)=b^3-3x^2+5x+11$ Could help . claim :$a+b=2$

答:
Consider the arithmetic mean of two expressions
$$f(x)=x^3-3x^2+5x-3$$

Then $f(x)=(x-1)((x-1)^2+2)$ has only root at $x=1$. This means graph of $f(x)$ is symmetric about the point $(1,0)$. It increases monotonically like graph of $y=x^3$ from $-\infty$ to $+\infty$.

Now $a,b$ are such that $f(a)=14$, $f(b)=-14$. So $(a,f(a))$ and $(b,f(b))$ are symmetrically located about $(1,0)$. Hence $$\frac{a+b}{2}=1 \quad ; \quad \frac{f(a)+f(b)}{2}=0$$


问:
Question about new proof in number theory?
Very recently, Yitang "Tom" Zhang from the University of New Hampshire made a huge step toward proving the twin prime conjecture. He proved that there are infinitely many pairs of prime separated by a constant distance, and that distance is less that $7 \cdot 10^7$. For example, in the special case of the twin prime conjecture, pairs of primes are separated by 2, such as 3 and 5, 5 and 7, 11 and 13, etc. Now people are saying that they might be able to use his methods to prove that the difference is lower than 16. My question is, does his proof limit the possibility of these pairs of primes to only one pair, or does it potentially allow for all the 7 million possibilities? Thanks! By the way this is where I found out about this: https://www.youtube.com/watch?v=vkMXdShDdtY</p>

答: The proof does not give any specific $k$ for which $(n,n+k)$ is prime infinitely often. But such a $k\le7\cdot10^7$ does exist by his theorem. In fact the theorem even implies that there is a $k$ such that there are infinitely many consecutive primes $(n,n+k).$


问:
Let us determine the Doob's decomposition of $(X_n^3)$, where $X_n$ is the one-dimensional symmetric random walk.
I was completely stuck. I know, that $X_n^3=M_n+A_n$, where $M_n$ is a martingale. Please help me :(

答:
Apply the binomial theorem to one step of the random walk,
$$
X_{n+1}^3=(X_n+\Delta X_n)^3=(M_n+A_n)+3X_n^2ΔX_n+3X_n +ΔX_n,
$$
using $ΔX_n^2=1$. So with $M_{n+1}=M_n+(3X_n^2+1)ΔX_n$ and $A_{n+1}=A_n+3X_n$ you get a martingale and a predictable process.


答:
Assuming the notation $ \Vert \vec x \Vert $ denotes the standard norm in $ \mathbb R^n $, the answer to your first question is always. The argument is as follows. Let $ x $ be in $ \mathbb R^n $ and $ x_i $ be its $i^{th} $ component. Then,
\begin{align*}
     | x_i |
          & = \sqrt{ |x_i|^2 }
          \\
          & \le  \sqrt{ |x_i|^2 + \sum_{j \neq i} |x_j|^2}
          \\
          & = \Vert \vec x \Vert
          .
\end{align*}

The key step here is the inequality above, which relies on the fact that the function $ f(t) = \sqrt{t} $ is increasing.

The argument you provide is a good place to start for the $ \mathbb R^2 $ case, but it requires some fixes: (1) exclude the $ x_2 = 0 $ case in order to divide by $ x_2 $, (2) replace the factor $ x_2 $ with $|x_2|$ on the right (remember: $ \sqrt{z^2} = |z| $ for real $ z $), (3) replace $ \alpha > 1 $ with $ \alpha \ge 1 $, and (4) state the connection between the statements you provide. After implementing these fixes, you can then establish that the inequality is always true in $ \mathbb R^2 $ by observing that $ \alpha \ge 1 $ always holds. This final bit follows also from the fact that $ f(t) = \sqrt{t} $ is increasing:
\begin{align*}
     \alpha
          & = f\left( 1 + \left(\frac{x_1}{x_2}\right)^2 \right)
          \\
          & \ge  f(1)
          \\
          & = 1
          .
\end{align*}

Title: Absolute value of vector components

Body:
When is it possible to assume that for any vector $\vec{x}=(x_1,\dots, x_n)\in\mathbb{R}^n$;

$$|x_{i}| \leq \|\vec{x}\|$$

For the case in  $\mathbb{R}^2$:

$$|x_2| \leq \|\vec{x}\|$$

$$\left|x_2\right| \leq x_2 \cdot \sqrt{\left(\frac{x_1}{x_2}\right)^2+1}$$

Define $\alpha = \sqrt{\left(\frac{x_{1}}{x_2}\right)^2+1}$

Then $|x_i| \leq \|\vec{x}\| \iff \alpha >1$.

Is there a more general result ? Does this reasoning hold ?


问:
coefficients of $Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XZ²+a_6Z³$ over $K$
Elliptic curve over $K$ is defined as 'genus one smooth curve'. Using Rieman-roch theorem, we can deduce Weierstrass equation form
$Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XZ²+a_6Z³$ over $K$ .

I wonder why coefficients's number are arranged $1,3,2,4,6$. Why not $1,2,3,4,5,6$?

Please tell me the background.

Thank you in advance.

答:
It is due to homogeneous weights of the variables $\,X,Y,Z.\,$
Define $$ X:=x\,t^2,\quad Y:=y\,t,\quad Z:=z\,t^4 \tag{1} $$
where $\,t\,$ is a formal variable counting weight.
The equation of the curve is
$$ Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XZ^2+a_6Z^3. \tag{2}$$
Divide both sides by $\,t^6\,$ and use equation $(1)$ to get
$$ y^2z+a_1xyz\,t+a_3yz^2t^3=x^3+a_2x^2z\,t^2+a_4xz^2t^4+a_6z^3t^6. \tag{3}$$
The $\,a_1,a_2,a_3,a_4,a_6\,$ subscripts come from the exponent
of $\,t.\,$

An alternative, perhaps simpler approach is to define
(using Jacobian coordinates)
$$ X := x/u^2,\quad Y := y/u^3,\quad Z :=1 \tag {4} $$
where $\,u\,$ denotes weight with $\,x\,$ of weight $2$ and
$\,y\,$ of weight $3$.

Multiply both sides of equation $(2)$
by $\,u^6\,$ and use equation $(4)$ to get
$$ y^2+a_1xy\,u+a_3y\,u^3=x^3+a_2x^2\,u^2+a_4x\,u^4+a_6\,u^6. \tag{5}$$
This is close to equation $(3)$ with the same
interpretation but now with exponents of $\,u.$


问:
find a group with the property :
a)find a nontrivial  group $G$ such that $G$ is isomorphic to $G \times G $

what i'm sure is that $G$ must be infinite ! but have now idea how to get or construct such group
i chose many $G$'s but all of the homomorphism was not injective

b) an infinite group in which every element has finite order but for each positive integer n there is an element of order n

the group  $G = (Z_1 \times Z_2 \times Z_3 \times Z_4 \times ...) $ satisfies the conditions except the one which says that every element have finite order .

how can we use this group to reach the asked group ?

答: For the second problem, you can use the subgroup of  $\mathbb{Z}_1\times \mathbb{Z}_2\times \cdots\times \mathbb{Z}_n \times \cdot$ consisting of all sequences $(a_1,a_2,a_3,\dots)$ such that  all but finitely many of the $a_i$ are $0$.</p>

Title: find a group with the property :


问:
Conditional expectation of second order statistic
Let $X_1, X_2$ be two independent, non-negative random variables with finite means from the same distribution function $F(x)$ and density function $f(x)$.
Denote their order statistics with $0 < Y_1 < Y_2$.

I want to compute $E(Y_2 | Y_1 = y_1)$ or to be more specific I want to show for $y_1 > 0$:
\begin{align*}
E(Y_2 | Y_1 = y_1) = y_1 + \frac{\int_{y_1}^{\infty} 1 - F(u) du}{1 - F(y_1)}.
\end{align*}
The joint density function of $Y_1,Y_2$ is given by
\begin{align*}
f_{Y_1, Y_2}(y_1,y_2) = 2 f(y_1) f(y_2) \chi_{y_1 < y_2},
\end{align*}
where $\chi$ is the indicator function.
The conditional expectation above can be computed by
\begin{align*}
\frac{\int_0^{\infty} y_2 f_{Y_1,Y_2}(y_1,y_2) dy_2} {\int_0^{\infty} f_{Y_1,Y_2}(y_1,y_2)  dy_2}
= \frac{\int_{y_1}^{\infty} y_2 f(y_1) f(y_2)  dy_2} {\int_{y_1}^{\infty} f(y_1) f(y_2) dy_2}
= \frac{\int_{y_1}^{\infty} y_2 f(y_2)  dy_2} {\int_{y_1}^{\infty} f(y_2) dy_2}
= \frac{\int_{y_1}^{\infty} y_2 f(y_2)  dy_2} {1 - F(y_1)}
\end{align*}
At this point I am stuck.

Body:
\begin{align}
\int_0^{y_1}(1-F(u))du&=\left(u(1-F(u)\right)\left|_{0}^{y_1}\right.+\int_0^{y_1}uf(u)du\\
&=y_1(1-F(y_1))+\int_0^{y_1}uf(u)du
\end{align}
further $\int_{0}^{\infty} (1-F(u))du=\int_0^{\infty}uf(u)du$ hence
\begin{align}
\int_{y_1}^{\infty} (1-F(u))du&=\int_0^{\infty}uf(u)du-\int_0^{y_1}uf(u)du\\
&=\int_0^{\infty}uf(u)du-y_1(1-F(y_1))-\int_0^{y_1}uf(u)du\\
&=\int_{y_1}^{\infty}uf(u)du-y_1(1-F(y_1))
\end{align}

implying that $$\frac{\int_{y_1}^{\infty} u f(u)  du} {1 - F(y_1)} =\frac{\int_{y_1}^{\infty} (1-F(u))du+y_1(1-F(y_1))} {1 - F(y_1)}$$

as you are after.


问:
Patterns of zeros in nonsingular $0,1$ matrices
I am reviving a question that the OP, Svyashennik Sanya, deleted.  Since it will not be visible to all users, I restate it here.

It is true that a non-singular $n\times n$ real matrix, all of whose elements are either $0$ or $1$, always has an $(n-1)\times(n-1)$ submatrix with a transversal all whose elements are $0$?  By a "transversal" I mean a set of $n-1$ elements comprising one from each row and one from each column of the minor.  (The OP used the term "diagonal", but I like "transversal" better.)

I verified this for $n\leq5$ with an exhaustive, brute-force python script.  The $n=6$ case would take too long to run.  (The OP claimed, in a comment, to have proved it when the the original matrix has a row or column of all $1$'s, but I didn't understand the proof.  That comment was deleted before the post was deleted, so I presume the proof was incorrect.)

I don't know if Svyashennik Sanya deleted the question because he or she found a counterexample or a simple proof, or just because no one had answered.  (Well, I answered, but I quickly realized that I'd misunderstood the question and deleted my answer.)  If the former, and you're reading this, please undelete your question and answer it.

I can't think of any reason why this should be true, but the fact that it's true for $n\leq 5$ gives me pause.

I've been playing with this problem for the last four days, and I've made no progress.  Of course, one immediately thinks of the Laplace expansion by minors, but that doesn't seem to lead anywhere.

Any ideas?

EDIT

Perhaps I should mention that one idea I had was to show that a square, $0,1$ matrix with no $0$-transversal is singular. Then all the $(n-1)\times(n-1)$ minors are $0$ and the original matrix has determinant $0$.  Sadly, this isn't true.  Any $2\times2$ matrix with three $1$'s and one $0$ is a counterexample.

答: If $M$ is a nonsingular $n \times n$ matrix and $J$ denotes the $n \times n$ all-ones matrix, then $J - M$ has rank at least $n-1$, so it must have an invertible $(n-1) \times (n-1)$ minor.  This minor must have an all-ones transversal, which translates to an all-zeroes transversal in a minor of $M$.


问:
Pseudovectors in Geometric / Clifford Algebra
I am studying geometric algebra and I am confused about why pseudovectors are written as single vectors with an $i$ in front.  In other words, for basis vectors $\gamma_{\mu}$ where $\mu = 1,2,3,4$, there are 4 pseudovectors, written as $i \gamma_{\mu}$.

I know that in general, for a space of dimension $N$, pseudovectors are elements of the set of $(N-1)$ fold wedge products, namely $\wedge^{N-1}\mathcal{C}_N$ where $\mathcal{C}$ denotes a Clifford algebra from dimension $N$.

So from this, for a 4 dimensional space, the pseudovectors are trivectors, for example $(\gamma_1\wedge \gamma_2 \wedge \gamma_3)$.  Where does the $i$ come in?  I know that the pseudoscalar is written as an $N$ fold wedge product, so in this case would be a quad-vector, but I don't see how that can be worked in.

I can't seem to find any literature that explains this directly.  I did find something that says there is a one-to-one mapping between $(N-1)$ fold wedge products, and single vectors, but it didn't go much into exactly how this mapping works.

答:
"I know that in general, for a space of dimension N, pseudovectors are elements of the set of (N−1) fold wedge products."

I think that this is wrong: it should be N fold wedge products. Thus in a plane pseudovectors are bivectors, wedge products of 2 vectors.

"Aren't bivectors, like torque, pseudovectors in 3D though?" No.


问:
Counting number of solutions with restrictions
I want to count the number of non-negative integer solutions to an equation such as

$$x+5y+8z=n$$

I can do this using generating functions; for example, the answer here is

$$[x^n]\frac{1}{(1-x)(1-x^5)(1-x^8)}$$
where $[x^n]$ is the coefficient of $x^n$.

But what if I add a restriction between variables such as $y \le z$? I have no idea how to count the number of solutions with this additional constraint. Any suggestions?

答:
If you want $y\leq z$, use the substitution $z = y + z'$, and you're solving $x + 5y + 8(y+z') = n$, which is equivalent to non-negative integer solutions for $x + 13y + 8 z' = n$.

If you want $x\leq y \leq z$, a similar method will work. You could also do this for $2y \leq z$.

Of course, with more restrictions, this can get harder to clearly do. It could be difficult to solve subject to $x \leq y, z \leq 2y, x \geq \pi z$ (assuming any solutions exist).


问:
writing a differential equation but i dont understand the answer someone gave if someone could help explain?
they used compound interest formula for doing constant rate of population growth. the gave this answer for the differential equation. i want to understand what they did/ how they did it.

A differential equation including initial conditions.

$$P = P(0) (1+r)^t$$

$$\frac{dP}{dt} = P(0)  \cdot \ln{(1+r)}  \cdot (1+r)^t$$

$$\frac{dP}{dt} = k (1+r)^t$$

can someone explain what they did each step I feel like I'm missing something

答:
\begin{align*}
\frac d{dt} a^t = {} & \lim_{h\to0} \frac{a^{t+h} - a^t} h \\[2pt]
& \text{(definition of “derivative”)} \\[10pt]
= {} & \lim_{h\to0}\left( a^t\cdot \frac {a^h-1} h \right) \\[10pt]
= {} & a^t\cdot\lim_{h\to0} \frac{a^h -1 } h \\[2pt]
& \text{since $a^t$ does not depend on $h$} \\[10pt]
= {} & \big(a^t \times\text{constant}\big) \\
& \text{since the limit does not depend on $t$}
\end{align*}

Just what the “constant” is can be seen via the chain rule if one knows that the constant is $1$ when $a=e\approx 2.71828\ldots\,$:

\begin{align}
\frac d {dt} a^t = {} & \frac d {dt} e^{t\ln a} = e^{t\ln a}\cdot \frac d {dt} (t\ln a) \\[10pt]
= {} & a^t \ln a.
\end{align}

And here you see what is “natural” about $e$: It is the base for which this “constant” is $1.$


问:
Proof by counterexample

If $n$ is prime, then $\sqrt{n}$ is irrational. Prove this statement.

If I were to prove this using proof by contradiction, I would do:

Suppose $n$ is prime and $\sqrt{n}$ is rational. Let $\sqrt{n}=\frac{a}{b}$ where $a,b\in \mathbb{R}$ and have no common factor other than 1.

And then I would go on and get a contradiction like $a$ is even and $b$ is even.

However, I was wondering if I could also use proof by counterexample:

($n$ is prime) $\Rightarrow$ ($\sqrt{n}$ is irrational)

The negation of this statement is:

($n$ is prime) $\land$ ($\sqrt{n}$ is rational)

A counter example to this is $n=2$.

Therefore, this statement is disproved and hence, its negation is proven to be true.

Is this also a valid proof?

答:
The claim is in plain text

"For every prime $n$, $\sqrt{n}$ is irrational."

The negation is in plain text

"There is some prime $n$, such that $\sqrt{n}$ is rational."

This is not a statement about all primes, but only about some prime. Therefore we cannot disprove it by a single counterexample.

To see the flaw : Assume $\sqrt{5}$ would be rational. Then the statement would be false since $5$ is prime, but $\sqrt{5}$ would not be irrational. You see, that the case $n=2$ is not enough.

Title: Proof by counterexample


问:
Proving two limits equal to each other?
I have come across this proof that is a bit tricky to me. The value of An is making it pretty hard.

Prove if $\lim\limits_{n\to \infty} S_n = S$, then $\lim\limits_{n\to \infty} A_n = S$ where $A_n = \frac{S_1 + S_2 + \cdots + S_n}{n}$ ?

I was thinking to use $\varepsilon-N$ definition but am pretty stuck.

答:
We can see this problem as an application of $Cesaro$ criterion.

Let  $(S_{n})$ be a sequence of real numbers which converges to $S$, define

$${\displaystyle A_{n}:=\sum _{m=1}^{n}S_{m}=S_{1}+\dots +S_{n},\quad b_{n}:=n}$$
then $\displaystyle (b_{n})$ is strictly increasing and diverges to $\displaystyle +\infty $ . Now we compute

$$\displaystyle \lim _{n\to \infty }{\frac {A_{n+1}-A_{n}}{b_{n+1}-b_{n}}}=\lim _{n\to \infty }S_{n+1}=\lim _{n\to \infty }S_{n}=S$$

and so

$$\displaystyle \lim _{n\to \infty }{\frac {S_{1}+\dots +S_{n}}{n}}=\lim _{n\to \infty }S_{n}$$

So in general given any sequence $\displaystyle (S_{n})_{n\geq 1}$ of real numbers, suppose that

$$\displaystyle \lim _{n\to \infty }S_{n}$$ exists (finite or infinite), then
$$\displaystyle \lim _{n\to \infty }{\frac {S_{1}+\dots +S_{n}}{n}}=\lim _{n\to \infty }S_{n}=S$$

Title: Proving two limits equal to each other?


问:
What are a , b and c?
$$y = ax^2 + bx + c$$

which is tangent at the origin with the line $y=x$, It is also tangential with the line $y=2x + 3$. Determine the function! Draw a figure!

My main question is this solvable? I am doubtful?

答:
Your problem is now that you have $y = ax^2 + x$ tangential to $y = 2x + 3$.  This means you have some number $n$ where $an^2 + n = 2n + 3$ (they meet at a point) and $2an + 1 = 2$ (they meet tangentially).

You have two equations and two unknowns; I'm sure you can solve from here.

Title: What are a , b and c?


答: Hint for b: Consider a quotient of $\mathbb{Q}$.</p>

Title: find a group with the property :

Body:
a)find a nontrivial  group $G$ such that $G$ is isomorphic to $G \times G $

what i'm sure is that $G$ must be infinite ! but have now idea how to get or construct such group
i chose many $G$'s but all of the homomorphism was not injective

b) an infinite group in which every element has finite order but for each positive integer n there is an element of order n

the group  $G = (Z_1 \times Z_2 \times Z_3 \times Z_4 \times ...) $ satisfies the conditions except the one which says that every element have finite order .

how can we use this group to reach the asked group ?


问:
What is a mathematical structure?.
So I'm a novice to abstract algebra and topology and I've just started trying to learn what an isomorphism is however I only think I  understand the defintion of order isomorphism.

What i don't grasp is how an isomorphism is not defined exactly the definitions online  seem to  change between mathematical structures.so for example an order isomorphism between 2 ordered set $(X,\leq)$ and $(Y,\leq)$ is  a bijective order preserving function that preserves all the properties of the order relation.

However what I don't understand is the term mathematical structure for example I read that $(X,R)$ where $R$ is a endorelation is a relational structure. But is it then not a mathematical structure can it not have an isomorphism?.

答:
Isomorphisms do in fact make sense for relational structures, and an extremely broad notion of isomorphism is provided by model theory (extending that provided by universal algebra, which handles operational structures). Specifically:

A structure (more precisely: first-order structure) $\mathfrak{X}$ is a set $X$ equipped with some finite-arity relations and functions. Generally $X$ is required to be nonempty, but this isn't a huge issue here. The relations and functions involved are also labelled in an appropriate way, and the set of labels of the structure is called the structure's signature. For a symbol $S$ in the signature of $\mathfrak{X}$, we write "$S^\mathfrak{X}$" for the function/relation on $X$ corresponding to $S$.

Given two structures $\mathfrak{A},\mathfrak{B}$ in the same signature $\Sigma$ with underlying sets $A,B$, an isomorphism $f:\mathfrak{A}\cong\mathfrak{B}$ is a bijection from $A$ to $B$ such that

for each $n$-ary relation symbol $R$ in the signature $\Sigma$ and each tuple $(a_1,...,a_n)\in A^n$ we have $$R^\mathfrak{A}(a_1,...,a_n)\iff R^\mathfrak{B}(f(a_1),...,f(a_n)),$$ and

for each $n$-ary function symbol $F$ in the signature $\Sigma$ and each tuple $(a_1,...,a_n)\in A^n$ we have $$f(F^\mathfrak{A}(a_1,...,a_n))=F^\mathfrak{B}(f(a_1), ..., f(a_n)).$$

For example, linear orders can be thought of as structures in a signature consisting of a single binary relation symbol $\triangleleft$. The relevant notion of an isomorphism between linear orders $(A,\triangleleft^\mathfrak{A})$ and $(B,\triangleleft^\mathfrak{B})$ is - as expected - a bijection $f:A\rightarrow B$ such that $x\triangleleft^\mathfrak{A}y\iff f(x)\triangleleft^\mathfrak{B}f(y)$.

There are various ways to generalize the notion of first-order structure. Most obviously, we can introduce sorts (this actually isn't really that different foundationally, since there isn't much distinguishing sorts from unary predicates until we look at more technical questions) or allow functions/relations to have infinite or even unbounded arities (this winds up resulting in a very nasty general theory, but still makes perfect sense at a foundational level). The notion of isomorphism similarly generalizes directly. For example, if we construe topological spaces as two-sorted structures (sorts being points and open sets) with an unbounded-arity operation on the set sort (arbitrary union) and a binary relation (elementhood) connecting the two sorts themselves, the resulting notion of isomorphism coincides exactly with homeomorphism.

Title: What is a mathematical structure?.


问:
How would I find the integral $\int_0^1 \frac{x^2+x+2}{(x+1)(1+x^2)}\,\mathrm{d}x$?
The goal is to find this integral: $$\int_0^1\frac{x^2+x+2}{(x+1)(1+x^2)}\,\mathrm{d}x$$
I have been on this question for a long time and I am halfway through solving it:

As shown above, using partial fraction decomposition with two linear factors, I have found that $$\int_0^1 \frac{x^2+x+2}{(x+1)(1+x^2)}\,\mathrm{d}x=\int_0^1 \frac{1}{x+1}\,\mathrm{d}x+\int_0^1\frac{1}{1+x^2}\,\mathrm{d}x$$
Where do I go from here? Any help is appreciated.

答:
For $\int \frac{1}{x+1} \, dx$ you should spot that since the derivative of $\log|x|$ is $1/x$, the derivative of $\log|x+1|$ is
$$
\frac{1}{x+1} \cdot \frac{d}{dx}(x+1)=\frac{1}{x+1}\cdot 1 = \frac{1}{x+1}
$$
using the chain rule. Hence,
$$
\int \frac{1}{x+1} \, dx = \log\left|x+1\right|+C \, ,
$$
and you can apply the fundamental theorem of calculus to find the definite integral. For $\int_{0}^{1} \frac{1}{1+x^2} \, dx$, it helps to simply know that the derivative of $\arctan x$ is $\frac{1}{1+x^2}$. Failing that, you can make the substitution $x=\tan\theta$. If you do make this substitution, remember to change the integral bounds.


答:
There is a whole theory of structures in logic, but I'll try to give a quick working mathematician's version.  A structure consists of one or more sets, zero or more functions between those sets, zero or more relations and predicates on those sets, zero or more constants chosen from those sets, and zero or more properties that these data all need to satisfy.

For example, a poset is consists of

a set $X$.

a binary relation on $X$ (which I guess is also known as an endorelation on $X$).  Call it $\leq$ for this definition.

the fact that $\leq$ is reflexive, antisymmetric, and transitive.

We usually write the data of a structure as a tuple.  In the case of posets, we might write that $(X,\leq)$ is a poset to say that $X$ is the set and $\leq$ is the relation.

The idea of an isomorphism of structures is a collection of bijections of all the sets under consideration with the property that they "preserve" the functions, relations, predicates, and constants.

In the case of posets, if $(X,\leq)$ and $(Y,\preceq)$ are posets, then an isomorphism between them consists of a bijection $f:X\to Y$ and the property that for all $x_1,x_2\in X$, then
$$x_1\leq x_2 \text{ if and only if } f(x_1)\preceq f(x_2).$$

The theory of mathematical structures is what lets you know how to come up with the definition of an isomorphism for any given structure.  Here's another structure: a semigroup consists of

a set $S$

a function $\mu : S\times S\to S$ called multiplication

the fact that for all $x,y,z\in S$, then $\mu(x,\mu(y,z))=\mu(\mu(x,y),z)$ ("associativity").

If $(S,\mu)$ and $(S',\mu')$ are both semigroups, then an isomorphism would be a bijective function $f:S\to S'$ such that for all $x,y\in S$,
$$f(\mu(x,y)) = \mu'(f(x), f(y)).$$

Title: What is a mathematical structure?.

Body:
So I'm a novice to abstract algebra and topology and I've just started trying to learn what an isomorphism is however I only think I  understand the defintion of order isomorphism.

What i don't grasp is how an isomorphism is not defined exactly the definitions online  seem to  change between mathematical structures.so for example an order isomorphism between 2 ordered set $(X,\leq)$ and $(Y,\leq)$ is  a bijective order preserving function that preserves all the properties of the order relation.

However what I don't understand is the term mathematical structure for example I read that $(X,R)$ where $R$ is a endorelation is a relational structure. But is it then not a mathematical structure can it not have an isomorphism?.


问:
when does $\det(AB^T+BA^T)\le \det(AA^T+BB^T)$ hold?
When does the following matrix inequality hold?

$$\det(AB+B^TA^T)\le \det(AA^T+BB^T)$$

$A$ and $B$ are any real matrices.

My reply gives a counter example. The question is under what condition does that hold?

[Update]

The inequality actually I would like to prove is not the above one. It should be

$$\det(AB^T+BA^T)\le \det(AA^T+BB^T)$$

I also changed the title. Thanks julien.

Body:
I found a counter example by Matlab. I know that is cheap..... :-|

$A = 0.4582 \quad  0.1567  \quad  0.0519\\
     0.8204  \quad  0.9111  \quad  0.0512 \\
     0.9831  \quad  0.1020  \quad  0.6691$

$B = 0.8492  \quad  0.0900 \quad  0.0787\\
    0.8890  \quad  0.0126  \quad  0.7477\\
    0.7770  \quad  0.2320 \quad   0.0607$

f = @(A,B)(det(A*A'+B*B')-det(A*B+B'*A'))

f(A, B) gives -1.9356


答:
Given
$P = P_o (1+r)^t$

I'll apply natural logarithm on both sides of the equation

$\ln P = \ln (P_o (1+r)^t)$

$\ln P = \ln P_o + t\ln (1+r)$

Then I'll differentiate the equation both sides

$\frac{dP}{P} = 0 + \ln (1+r) dt$

$\frac{dP}{P} = \ln (1+r) dt$

$\frac{dP}{dt} = P\ln (1+r)$

but $P = P_o (1+r)^t$

thus
$\frac{dP}{dt} = P_o (1+r)^t\ln (1+r)$

$\frac{dP}{dt} = P_o\ln (1+r) (1+r)^t$

but $P_o\ln (1+r) = k$ where by $k$ is constant

therefore
$\frac{dP}{dt} = k (1+r)^t$

Title: writing a differential equation but i dont understand the answer someone gave if someone could help explain?

Body:
they used compound interest formula for doing constant rate of population growth. the gave this answer for the differential equation. i want to understand what they did/ how they did it.

A differential equation including initial conditions.

$$P = P(0) (1+r)^t$$

$$\frac{dP}{dt} = P(0)  \cdot \ln{(1+r)}  \cdot (1+r)^t$$

$$\frac{dP}{dt} = k (1+r)^t$$

can someone explain what they did each step I feel like I'm missing something


问:
In a velocity-time graph, if the velocity is negative and the slope is negative, is it still accelerating?
So in a velocity-time graph, a negative velocity means the speed in the opposite direction. So from -10 m/s to -30 m/s, it is accelerating in the opposite direction. However the slope is negative, so is it still accelerating?</p>

答: A physicist would use the word "accelerating" to mean the change in velocity, whether positive or negative, i.e. whether speeding up or slowing down.</p>


问:
Prove or disprove these two statements

Let $V$ be a vector space over a field $F$ and $v_1,v_2,v_3,v_4$ four vectors that are different from zero then:

a) if $\{v_1,v_2,v_3,v_4\} $ are linearly dependent and $\{v_1 + v_2 + v_3 + v_4, v_2 + 2v_3, v_3 - v_4, v_2 + v_3+kv_4 \}$ are linearly dependent then $k=1$.

My thoughts: I used the theorem that if we get zero rows while reducing a matrix to it's row echelon form then the rows are linearly independent. Here's my matrix after subtracting row $2$ from row $4$:

$$\begin{bmatrix}1&1&1&1\\0&1&2&0\\0&0&1&-1\\0&0&-1&k\end{bmatrix}$$

Now I answered that in order to get a zero row we need $k=1$.

But well that was wrong, I guess I'm missing some fundamental knowledge about linear dependence but I can't get what part is my mistake, is it using the matrix? or I shouldn't even bother since the vectors are linearly dependent?

b) if $\{ v_1, v_2, v_3\}$ , $\{v_3,v_4\}$ are linearly independent sets and $v_4 \in sp \{v_1, v_3\}$ then $\{ v_2, v_3, v_4 \}$ are linearly independent.

This statement is true, I sat for too long trying to come out with a counter example and I failed, and while doing it I tried to prove, but I had no idea where to start the proof and how to construct it.

I would really appreciate any help and feedback, thanks in advance!

答:
For part (b): I suspect the condition is supposed to be $\{v_3,v_4\}$ being linearly independent (you have $\{v_3,v_3\}$); it's the only pair that makes sense, since any other pair's independence is subsumed in the assumption that $\{v_1,v_2,v_3\}$ is linearly independent. And if we don't have that second condition, we get a counterexample by letting $v_4=v_3$. So, assuming that:

Because $v_4\in\mathrm{span}(v_1,v_3)$, then we can write $v_4=\alpha v_1+\beta v_3$ for some scalars $\alpha$ and $\beta$. We also know $v_4\neq0$, so $\alpha$ and $\beta$ are not both equal to zero. And since $\{v_3,v_4\}$ is linearly independent, $\alpha$ must be nonzero (otherwise, $v_4=\beta v_3$ and $\{v_3,v_4\}$ would be linearly dependent).

Now let's take a linear combination of $v_2$, $v_3$, and $v_4$ that is equal to $0$:
$$r_2v_2 + r_3v_3 + r_4v_4=0.$$
We want to prove that $r_2=r_3=r_4=0$.

Plugging in the value of $v_4$, we get
$$\alpha r_4v_1 + r_2v_2 + (r_3+\beta r_4)v_3 = 0.$$
Since we know that $\{v_1,v_2,v_3\}$ is linearly independent, this means that $\alpha r_4=0$, $r_2=0$, and $r_3+\beta r_4=0$.

Since $\alpha r_4=0$, either $\alpha=0$ or $r_4=0$. But we know from the fact that $\{v_3,v_4\}$ is linearly independent that $\alpha\neq 0$, hence $r_4=0$. And then the last equation becomes $r_3=0$, proving that $r_2=r_3=r_4=0$, which is what we wanted to prove.

Title: Prove or disprove these two statements


问:
Solve the following Bernoulli ODE
Solve the following Bernoulli ODE: $\frac{dy}{dx}+x·y=xy^{-1}$

I have managed to find the integrating factor which is $2xdx$. Integrating this results in $e^{x^2}$.
I am unsure on how to proceed in obtaining the final answer, which is $y(x)=\sqrt{1+Ce^{-x^2}}$

答:
This is the differential you have in the pic:
$$\frac{dy}{dx}+x·y=x·y^{-1}$$
$$\frac{dy}{dx^2}\dfrac {dx^2}{dx}+x·y=xy^{-1}$$
$$2\frac{dy}{dx^2}+y=y^{-1}$$
$$2y\frac{dy}{dx^2}+y^2=1$$
$$\frac{dy^2}{dx^2}+y^2=1$$
$$u'+u=1$$
Where $u=y^2$. This is separable. Integrate.
$$\int \dfrac {du}{u-1}=-\int dx^2$$

Title: Solve the following Bernoulli ODE


问:
Inconclusive convergence tests and Divergence Test
What happens when the series is alternating and the tests that I know are inconclusive?

$\sum_{n=1}^\infty (-1)^n (n/(n+5))$.

The $b_n$ series is not decreasing, so I can't use the alternating series test (I took the derivative of $f(x)=x/(x+5)$ - slope is positive everywhere).

When I do the ratio test I get $=1$, so inconclusive.

So - am I allowed to simply ignore the alternating portion of the series and apply the divergence test which allows me to say lim as $n$ goes to $\infty$ of $a_n=1$ and since the limit does NOT $=0$, then the series diverges?

I thought I could only use the divergence test on non-alternating series!

答: Since the limit $\lim_{n\to\infty}(-1^n)\frac n{n+5}$ does not exist, then, in particular, you don't have $\lim_{n\to\infty}(-1^n)\frac n{n+5}=0$, and therefore the series diverges.


问:
Does the following integral converge or diverge and can it be solved?
I have been stuck solving the following integral for some time now: $$\int_0^{2R} \frac{\sin^2\left(\frac{x}{4R}\pi\right)R^2}{\left(\sqrt{\left(d+Z\right)^2+\sin^2\left(\frac{x}{4R}\pi\right)}\right)^3}\,\mathrm{d}x$$
If it isn't possible to solve, is there a way to show whether this integral converges or diverges for $\lim_\limits{R\to\infty}$?

Any help or suggestions on how to solve this are appreciated.

答:
First note that your integral diverges for $d + Z = 0$. For $d + Z \neq 0$ we can write
\begin{align}
I (R,d,Z) &\equiv \int \limits_0^{2R} \frac{R^2 \sin^2 \left(\frac{\pi x}{4R}\right)}{\left[(d+Z)^2 + \sin^2 \left(\frac{\pi x}{4R}\right)\right]^{3/2}} \, \mathrm{d} x \stackrel{x = \frac{4 R t}{\pi}}{=} \frac{4 R^3}{\pi} \int \limits_0^{\pi/2} \frac{\sin^2(t)}{[(d+Z)^2 + \sin^2(t)]^{3/2}} \, \mathrm{d} t \\
&\equiv \frac{4 R^3}{\pi} f(\lvert d + Z \rvert) \, .
\end{align}
Here we have defined the function
$$ f \colon (0,\infty) \to (0,\infty) \, , \, f(a) = \int \limits_0^{\pi/2} \frac{\sin^2(t)}{[a^2 + \sin^2(t)]^{3/2}} \, \mathrm{d} t \, .$$
Since $f$ is always positive, we immediately see that
$$ \lim_{R \to \infty} I (R,d,Z) = \lim_{R \to \infty} \frac{4 R^3}{\pi} f(\lvert d + Z \rvert) = \infty \, .$$

In order to compute your integral we need to find $f$. This is somewhat tricky and involves elliptic integrals, but it can be done: For $a > 0$ we have
\begin{align}
f(a) &= \int \limits_0^{\pi/2} \frac{1 + a^2 - \cos^2(t) - a^2}{[1 + a^2 - \cos^2(t)]^{3/2}} \, \mathrm{d} t = \int \limits_0^{\pi/2} \left[\frac{1}{\sqrt{1 + a^2 - \cos^2(t)}} - \frac{a^2}{[1 + a^2 - \cos^2(t)]^{3/2}}\right] \mathrm{d} t \\
&= \int \limits_0^{\pi/2} \left[\frac{1}{\sqrt{1 + a^2} \sqrt{1 - \frac{\cos^2(t)}{1+a^2}}} - \frac{1}{(1+a^2)^{3/2}} \left(\frac{a^2}{\left(1 - \frac{\cos^2(t)}{1+a^2}\right)^{3/2}} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\sin(t) \cos(t)}{\sqrt{1 - \frac{\cos^2(t)}{1+a^2}}}\right)\right] \mathrm{d} t \\
&= \frac{1}{\sqrt{1+a^2}} \int \limits_0^{\pi/2} \left[\frac{1}{\sqrt{1 - \frac{\cos^2(t)}{1+a^2}}} - \sqrt{1 - \frac{\cos^2(t)}{1+a^2}}\right] \mathrm{d} t \\
&= \frac{\operatorname{K} \left(\frac{1}{\sqrt{1+a^2}}\right) - \operatorname{E} \left(\frac{1}{\sqrt{1+a^2}}\right)}{\sqrt{1+a^2}} \, ,
\end{align}
where $\operatorname{K}$ and $\operatorname{E}$ are the complete elliptic integrals of the first and second kind, respectively. Note that the integral of the derivative term in the second line vanishes, so we are allowed to add it without changing the result.


答:
You know that for a real or complex sequence $ ( u_n)$, we have the equivalence

$$\boxed{\lim_{n\to+\infty}u_n=0\iff \lim_{n\to+\infty}|u_n|=0}$$

So,

$$\lim_{n\to+\infty}|u_n|\ne 0\implies$$
$$\lim_{n\to+\infty}u_n \ne 0 \implies$$
$$\sum u_n \text{ is divergent}$$

Title: Inconclusive convergence tests and Divergence Test

Body:
What happens when the series is alternating and the tests that I know are inconclusive?

$\sum_{n=1}^\infty (-1)^n (n/(n+5))$.

The $b_n$ series is not decreasing, so I can't use the alternating series test (I took the derivative of $f(x)=x/(x+5)$ - slope is positive everywhere).

When I do the ratio test I get $=1$, so inconclusive.

So - am I allowed to simply ignore the alternating portion of the series and apply the divergence test which allows me to say lim as $n$ goes to $\infty$ of $a_n=1$ and since the limit does NOT $=0$, then the series diverges?

I thought I could only use the divergence test on non-alternating series!


问:
help solving a geometric sequence after manipulation
Evaluate the sum. $\sum_{n=2}^\infty \frac{n(-3)^n}{4^{n+1}}$

I have manipulated it to be in a more common form I thought I could solve but I'm still having trouble getting the correct sum. I'm using wolfram to check against. It says the sum evaluates to $\frac{99}{784}=0.12628$

Here is what I have so far.

$\sum_{n=2}^{\infty}\frac{n(-3)^{n}}{4^{n+1}}$

$=\sum_{n=2}^{\infty}\frac{n(-3)^{n}}{4^{n}\cdot4}$

$=\sum_{n=2}^{\infty}\frac{n(-3)^{n}}{4^{n}}\ \cdot\frac{1}{4}$

$=\sum_{n=2}^{\infty}n\left(-\frac{3}{4}\right)^{n}\ \cdot\frac{1}{4}$

$=\frac{1}{4}\sum_{n=2}^{\infty}n\left(-\frac{3}{4}\right)^{n}$

was trying to get the common difference to solve that way but it's not coming out right. How do I properly evaluate the remaining sum?

Body:
$1+x+x^2+x^3+...=\frac{1}{1-x} \iff \sum_{n\geq0}x^n=\frac{1}{1-x}\Rightarrow \sum_{n\geq1}nx^{n-1}=\frac{1}{(1-x)^2}$

$\sum_{n\geq2}n\Big(-\frac{3}{4}\Big)^{n}=-\frac{3}{4}\sum_{n\geq2}n\Big(-\frac{3}{4}\Big)^{n-1}=-\frac{3}{4} \Big( \frac{1}{(1+\frac{3}{4})^2}-1\Big)$


问:
Jordan form of the exponential of a matrix
Given a matrix $A$ which has the form of a Jordan normal matrix:

$$ A = \left(\begin{array}{cccc}
\log(\phi_1) & 1& ...& 0\\
  & \log( \phi_2)& & \\
 &   &  \ddots & 1 \\
0 & & & \log (\phi_n) \\
\end{array}\right)$$

how can we prove that the Jordan normal form of $ \exp(A)= VJV^{-1} $ is $J$:

$$ J = \left(\begin{array}{cccc}
\phi_1 & 1& ...& 0\\
  &  \phi_2& & \\
 &   &  \ddots & 1 \\
0 & & &  \phi_n \\
\end{array}\right)$$

that has the exact same block form as $A$, but only with the exponential of the eigenvalues?

EDIT:

I tried the $2 \times 2$ case to see if it worked and this is what I got:

$$ A = \left(\begin{array}{cc}
\log(\phi_1) & 1\\
 0 & \log( \phi_2) \\
\end{array}\right)$$

We can separate this into to $2$ matricies:

$$ A =  \left(\begin{array}{cc}
\log(\phi_1) & 0\\
 0 & \log( \phi_2) \\
\end{array}\right)
+
\left(\begin{array}{cc}
 0 & 1\\
 0 & 0 \\
\end{array}\right)
$$

Exponentiating:

$$ e^A = \left(\begin{array}{cc}
\phi_1 & 0\\
 0 & \phi_2 \\
\end{array}\right)
 \times
\left[I + \left(\begin{array}{cc}
 0 & 1\\
 0 & 0 \\
\end{array}\right)  \right]
= \left(\begin{array}{cc}
\phi_1  & \phi_1\\
 0 & \phi_2  \\
\end{array}\right)$$

Now, its eigenvalues have all algebraic multiplicity of 1 so the Jordan normal form of $e^A$ is:

$$ J = \left(\begin{array}{cc}
\phi_1  & 0\\
 0 & \phi_2 \\
\end{array}\right)$$

which is not quite what we wanted, I dont know what to do.

答: Hint: If $J$ is the $n\times n$ Jordan block with zero eigenvalue, compute the kernel of $exp(J)$.


问:
Does this proof make sense(The well-ordering principle for N)?
I need to explain the proof given by Des McHale in his "The well-ordering principle for N". I was hoping you could tell me if it makes sense and what else I could add to it. This is what I have so far:

"
In his paper, Des MacHale talks about the Well-Ordering Principle for $\mathbb{N}$. This is a different method to solving proofs. He says that this is a less talked about tool in the mathematicians belt, but is useful to know. He goes on to say that it "appears to be a more natural approach than induction." Using the Well Ordering Principle, also known as WOP, he goes on to solve and prove many theorems. One such theorem that he proves is that $\sqrt2\notin\mathbb{N}$. He uses WOP as a proof for this. In this paper we will attempt to use WOP as well to prove that $\sqrt3\notin\mathbb{Q}$. To put this in English, we want to prove that the square root of three is not a rational number.

Similar to the theorem proving that $\sqrt{2}$ is irrational, to prove that $\sqrt{3}$ is also irrational, it needs to be proven that there is no pair of positive integers (a,b) where $(a/b)^2 = 3$.

Set S can be a set of pairs of integers (a,b) where $(a/b)^2 = 3$

If S is non-empty, according to the Well Ordering Principle, S has to have at least one element c = a + b where $a^2 = 3b^2$.

When $a^2$ is divided by 3, and if $a = 3k + 1$ is odd, $a^2 = 9k^2 + 6k + 1$ which is odd, creating a contradiction.

Following, $a = 3k$ is even where $k\in\mathbb{N}$.

Next, $(3k)^2 = 9k^2 = 3b^2$ so $b^2 = 3k$ is even.

Following our argument before, $b = 2s$ is even too where $s\in\mathbb{N}$.

Because $a^2 = 3b^2$, we have $(3k)^2 = 3(2s)^2$ and then $k^2 = 2s^2$.

However, $a + b = 2(k+s)$, so $k + s < a + b$ which is a contradiction.

This contradiction shows that the set S is empty and therefore there is no pair of positive integers (a,b) where $(a/b)^2 = 3$.

This proves that $\sqrt{3}$ is irrational.

"

Thanks in advance for any help.

答:
I think you are following the argument for $\sqrt{2}$ word for word, just substituting $3$ for $2$ without thinking. This assertion

When $a^2$ is divided by $3$, and if $a=3k+1$ is odd, $a^2=9k^2+6k + 1$ which is
odd, creating a contradiction.

makes no sense. What you know from the preceding line is that $3$ divides $a^2$. You then start thinking about whether $a$ is odd or even, and say it can't be the odd number $3k+1$. But that number need not be odd - what if $k$ were $5$? The real reason $a$ can't be $3k+1$ is that its square $ 9k^2+6k_1$ is not a multiple of $3$. Whether it's odd or even is irrelevant.

You need to think about what form $a$ might have. Odd and even are irrelevant in this argument.


问:
Reversibility of Markov Process and Exponential Distribution of Transition Rates
I am reading the paper Towards Utility-optimal Random Access Without Message
Passing by J. Liu, Y. Yi, A. Proutiere, M. Chiang, H. V. Poor. A sentence in Section 3.2 can be paraphrased as follows:

Because a Markov process is reversible, the stationary distribution
  does not depend on the exponential distribution of the transition
  probabilities (i.e., the income and the service rates) and it only
  depends on the mean of the transition rates.

I have two questions:

Why we need the exponential distribution assumption at all?

Why we can relax the assumption if the process is reversible?

Please describe in detail, since I know almost nothing about Markov processes.

答:
Read the first and second chapters of chatfields book on time series(can be done in 2days). Please proceed to the first three chapters of Hidden Markov Models for Time series.(by Walter Zucchini and Ian MacDonald)-can also be done in 2days.
Go back to your text and there will be no problems.
Anonymous.</p>


问:
Is there a natural intermediate version of PA?
Below, "logic" means "regular logic containing first-order logic" in the sense of Ebbinghaus/Flum/Thomas.

Setup

For a logic $\mathcal{L}$, let $\mathcal{PA}(\mathcal{L})$ be the set of $\mathcal{L}$-sentences in the language of arithmetic consisting of:

the ordered semiring axioms, and

for each $\mathcal{L}$-formula $\varphi(x,y_1,...,y_n)$ the induction instance $$\forall y_1,...,y_n[[\varphi(0,y_1,...,y_n)\wedge\forall x(\varphi(x,y_1,...,y_n)\rightarrow\varphi(x+1,y_1,...,y_n))]\rightarrow\forall x\varphi(x,y_1,...,y_n)].$$

(Note that even if the new logic $\mathcal{L}$ has additional types of variable - e.g. "set" variables - the corresponding induction instances will only allow "number" variables $y_1,...,y_n$ for the parameters.)

For example, $\mathcal{PA}(FOL)$ is just the usual (first-order) PA, and $\mathcal{PA}(SOL)$ characterizes the standard model $\mathbb{N}$ up to isomorphism. Also, we always have $\mathbb{N}\models_\mathcal{L}\mathcal{PA}(\mathcal{L})$.

Note that it's not quite true that $\mathcal{PA}(SOL)$ "is" second-order PA as usually phrased - induction-wise, we still have a scheme as opposed to a single sentence. However, induction applied to "has finitely many predecessors" gets the job done. (If we run the analogous construction with ZFC in place of PA, though, things seem more interesting ....)

Question

Say that a logic $\mathcal{L}$ is PA-intermediate if we have $PA<\mathcal{PA}(\mathcal{L})<Th_{FOL}(\mathbb{N})$ in the following sense:

There is a first-order sentence $\varphi$ such that $PA\not\models\varphi$ but $\mathcal{PA}(\mathcal{L})\models\varphi$.

There is a first-order sentence $\theta$ such that $\mathbb{N}\models\theta$ but $\mathcal{PA}(\mathcal{L})\not\models\theta$.

Is there a "natural" PA-intermediate logic?

(This is obviously fuzzy; for precision, I'll interpret "natural" as "has appeared in at least two different papers whose respective authorsets are incomparable.")

Partial progress

A couple non-examples:

Infinitary logic $\mathcal{L}_{\omega_1,\omega}$ is not PA-intermediate, since in fact $\mathcal{PA}(\mathcal{L}_{\omega_1,\omega})$ pins down $\mathbb{N}$ up to isomorphism (consider $\forall x(\bigvee_{i\in\omega}x=1+...+1\mbox{ ($i$ times)}$). So both second-order and infinitary logic bring in too much extra power.

Adding an equicardinality quantifier ($Ix(\varphi(x);\psi(x))$ = "As many $x$ satisfy $\varphi$ as $\psi$") also results in pinning down $\mathbb{N}$ up to isomorphism (consider "$\neg$ As many $x$ are $<k$ as are $<k+1$"). If we add the weak equicardinality quantifer $Qx\varphi(x) = Ix(\varphi(x);\neg\varphi(x))$, on the other hand, we wind up with a conservative extension of $PA$. That said, the latter is stronger semantically: no countable nonstandard model of $PA$ satisfies $\mathcal{PA}(FOL[Q])$ (to prove conservativity over $PA$ we look at the $\omega_1$-like models).

Fragments of second-order logic don't seem to help either: both $\mathcal{PA}(\Pi^1_1)$ and $\mathcal{PA}(\Sigma^1_1)$ pin down $\mathbb{N}$ (see here).

The proof of Lindstrom's theorem yields a weak negative result: if $\mathcal{L}$ is a logic strictly stronger than first-order logic with the downward Lowenheim-Skolem property, then we can whip up an $\mathcal{L}$-sentence $\varphi$ and an appropriate tuple of formulas $\Theta$ such that $\varphi$ is satisfiable and $\Theta^M\cong\mathbb{N}$ in every $M\models\varphi$. So an example for the above question would have to either allow nasty implicit definability shenanigans or lack the downward Lowenheim-Skolem property. This rules out a whole additional slew of candidates. That said, there is a surprising amount of variety among logics without the downward Lowenheim-Skolem property even in relatively concrete contexts - e.g. there is a compact logic strictly stronger than FOL on countable structures.

答:
This answer recapitulates Ali Enayat's answer to my MO question; I've posted it since he is not on MSE. I've made this answer CW to avoid getting reputation for his work, and if he posts an answer of his own here I'll delete this one.

Motivated by the metamathematics of Ramsey's theorem, Macintyre, Simpson, and Schmerl have looked at versions of $\mathsf{PA}$ gotten by adding Ramsey-theory-related quantifiers to first-order logic; see Macintyre and Schmerl/Simpson on the simple Ramsey quantifier and Schmerl on a further extension. The resulting variants turn out to be very nice indeed, having first-order parts corresponding to natural subsystems of the (first-order theory of) second-order arithmetic $\mathsf{Z}_2$.

Additionally, Schmerl also investigated the version of Peano arithmetic corresponding to stationary logic. Here something interesting happens. Schmerl pinned down the set of first-order consequences of this theory together with an additional principle $Det$. It's not clear to me what happens if $Det$ is dropped. Stationary logic is itself a bit subtle, so this is something I'm still thinking about.


问:
Condition for affine independence

Let $E$ be an affine space attached to a $K$-vector space $T$. For a
family $(x_i)_{i\in I}$ of elements of $E$ and any $a\in E$, the set
$$\left\{\sum_{i\in I}\lambda_i(x_i-a)+a\ |\ \lambda\in
 K^{(I)}\land\sum_{i\in I}\lambda_i=1\right\}$$ is the affine linear
variety generated by $(x_i)_{i\in I}$.

Let $(a_i)_{i\in I}$ be a nonempty family of elements of $E$ and $k\in
 I$. This family is said to be affinely independent if and only if the
family $(a_i-a_k)_{i\ne k}$ is linearly independent in $T$.

Now, suppose $(a_i)_{i\in I}$ be a nonempty family of elements of $E$. Suppose that, for every $k\in K$, $a_k$ doesn't not belong to the affine variety generated by $(a_i)_{i\ne k}$. I want to show that this implies that $(a_i)_{i\in I}$ is affinely independent.

Attempt:

Since $I\ne\emptyset$, let $k\in I$. According to the above, it is sufficient to show that the family $(a_i-a_k)_{i\ne k}$ is linearly independent in $T$. So, let $\lambda\in K^{(I-\{k\})}$ such that $\sum_{i\ne k}\lambda_i(a_i-a_k)=0$. Then $\sum_{i\ne k}\lambda_i(a_i-a_k)+a_k=a_k$. This implies that $\sum_{i\ne k}\lambda_i\ne 1$. But how can this help me show that $\lambda=0$?

答:
Suppose to the contrary that $(a_i)_{i\in I}$ is not affinely independent, that is, for some (actually every) $k\in I$, the vectors $a_i-a_k$ are linearly dependent.

This implies that one of them is a linear combination of the other ones:
$$a_j-a_k=\sum_{i\ne j,k}\lambda_i(a_i-a_k)\\
a_j=\sum_{i\ne j,k}\lambda_ia_i+(1-\sum\lambda_i)a_k$$
Or, subtracting any $a\in E$,
$a_j-a=\sum_{i\ne j,k}\lambda_i (a_i-a)+(1-\sum\lambda_i)(a_k-a)$, which shows that $a_j$ is an affine combination of the other ones.

Title: Condition for affine independence


答:
Multiplying by $ y $ , it becomes

$$y\frac{dy}{dx}+xy^2=x$$

or, with $ z=y^2$

$$\frac 12 z' +xz =x$$
Observe that $ z=\color{red}{1}$ is a particular solution.

thus

$$z=C.e^{-x^2}+\color{red}{1}=y^2$$

Title: Solve the following Bernoulli ODE

Body:
Solve the following Bernoulli ODE: $\frac{dy}{dx}+x·y=xy^{-1}$

I have managed to find the integrating factor which is $2xdx$. Integrating this results in $e^{x^2}$.
I am unsure on how to proceed in obtaining the final answer, which is $y(x)=\sqrt{1+Ce^{-x^2}}$


问:
How to find the number of onto and into functions or increasing/decreasing functions, given certain conditions?
I have two related questions that I have a problem in. Here's the 1st one:

What I mean by into functions: Function f from set A to set B is Into function if at least set B has a element which is not connected with any of the element of set A.

1. Number of into/onto functions

Let $A=\{1,2,3,4,5\}$ and $f:A \rightarrow A$ be an into function such that $f(i)≠i,∀$ $i\in A$, then the number of such functions is? (Answer is $980$)

I know the total number of functions is of course $4^5$, but I didn't know how to find the number of into functions, so I looked at the solution for help.

The solution says that the number of required functions is:

Total number of functions − Number of onto functions, and they find the number of onto functions to be $44$. The question then is simple enough but I don't know how to find the number of onto functions here. Seeing the number $44$ I thought it to be the Derangement of $5$, and I'm pretty sure that comes from the fact the its given $f(i) \neq i $, but how do I know that it is only the "onto functions"? Can someone elaborate this for me?

2. Number of increasing/decreasing functions

Number of strictly increasing functions $f: A \rightarrow B$, where $A= \{a_1,a_2,a_3,a_4,a_5,a_6\}$ and $B=\{1,2,3,4,5,6,7,8,9\}$ such that $a_{i+1}>a_i$ and $f(a_i) \neq i$

I do not not how to approach this one. I tried in the following way, I let $a_1$ map to $2$, (as $f(i) \neq i$), so then $a_2$ can map with $3,4,....$ and so on for the other elements upto $a_6$. So the other elements apart from $a_1$ can have $7 \choose5$ ways. I have a doubt here. Does this also consider the case, that, for $a_1=2$ and $a_2=3$, $a_3$ may be $4,5,6$? How does it restrict that the functions obtained here are increasing only? The answer for this one is ${7 \choose5}+{6 \choose 5}+{5 \choose 5}$. Please help me in understanding how and why to go about determining cases like this.

答:
For question $(1)$, you are correct that the number is obtained from derangements.
First, notice that if $S$ is finite and $f:S\longrightarrow S$, then the following are equivalent:

$f$ is bijective

$f$ is injective

$f$ is surjective (onto)

Now, a permutation on some set is a bijection from that set to itself.
Finally, derangement numbers count precisely the permutations $f$ that leave no element fixed, that is, such that $f(i)\neq i$ for all $i$.

For question $(2)$, notice that given any choice of $6$ distinct values from $B$, there is exactly one way to arrange them in increasing/decreasing order.

Now, of course, $1$ cannot be a chosen value from $B$, for then we would violate the condition of $f(a_i)\neq i$ by having to assign $1$ to $a_1$.
Can you show that whenever $1$ is not chosen from $B$, then the arrangement of the values in increasing order will produce a valid assignment for the $f(a_i)$?

In that case, the number of functions will be the number of ways to choose $6$ distinct values from $B\setminus\{1\}$, which is precisely $\binom 86 = 28$.


问:
Approximated number of elastic in a box is 100 with 1 significant figure. How to express this in $x \pm y$ notation?
Question from stackexchange

How many significant figures are there after 96
is correct to 1 significant figure?

From this statement:

It is given that the approximated number of elastic in a box is 100.
If the number is correct to 1

significant figure, what is the smallest possible number of elastic?

Answer: 100

How would one express this in $x \pm y$ notation?:

All x in 100 <= x <= 149 gets mapped to $1 \times 10^2$ = 100.

How to express this in $x \pm y$ notation?

I would have supposed $100 \pm 49$, because 51..149 gets rounded to 100,

but then the statement from the correct answer will be violated:

"smallest possible number is 100".

答: Significant "numbers" is not significant "figures"!  In any case, "96 correct to 1 significant figure" means that it has been rounded to the "ones" place.  And that means the number can be anywhere from 95.5 to 96.5.  [tex]x= 96\pm 0.5[/tex].</p>


问:
Drawing a triangle in a unit circle
This is a question that I derived for a long time ago. It asks if we draw a triangle in a unit circle does all arc lengths $(\alpha ,\beta ,\theta)$ and sides of triangle $(a,b,c)$ can be rational numbers? Intuitively I believe that the rationality does not hold but I can't derive a proof. Do you have any ideas?

答:
$\alpha + \beta + \theta = 2 \pi$, so they can't all be rational.

You might want to write the arc lengths as $\alpha \pi, \beta \pi, \theta \pi$, in which case $\alpha + \beta + \gamma = 2$, and it gets slightly more interesting.

The answer is still no.

(WLOG, we have $0 \leq \alpha, \beta, \gamma \leq 2$.) This arises because $a= 2| \sin \frac{\alpha\pi}{2}|$, which is rational if and only if $\frac{\alpha}{2} \in \{ 0, \frac{1}{6} ,\frac{1}{2}, \frac{5}{6}, 1\} $. This is known as Niven's Theorem.

The only way to get $\frac{\alpha}{2} + \frac {\beta}{2} + \frac{\gamma}{2} = 1$ is then to require that $\alpha\beta\gamma=0$ (slight checking involved), which is the degenerate cases (that I'm ignoring). These cases correspond to either having a point on the circumference, or a chord of the circle.

Title: Drawing a triangle in a unit circle


问:
Existence and uniqueness of maximal solution for first-order non-linear ODE
Let $\alpha \in \mathbb{R}$ and $f$ be a real function defined by $f(u)=-ue^{\alpha u}\ln(\lvert u \rvert)$ if $u \neq 0$ and $f(0)=0$. Let $u_0 \in \mathbb{R}^+$ for the following problem:

\begin{equation}
\begin{cases}
u'(t)=f(u(t)), t\in \mathbb{R}\\
u(0)=0
\end{cases}
\end{equation}

Has the Cauchy problem a unique maximal solution?

I tried:

Let $g$ be a function such that:

\begin{equation}
\begin{cases}
g:\mathbb{R}\times\mathbb{R} \rightarrow \mathbb{R}\\
g:(t,u) \rightarrow f(u(t))
\end{cases}
\end{equation}

I try to verify the Cauchy-Lipschitz theorem. $f$ is continuous so $g$ is also continuous.
Now I need to prove that $g$ is locally Lipschitz with respect to its second variable. I can't manage to prove it. I tried to use the mean value theorem, but without success...
Maybe it does not verify Cauchy-Lipschitz conditions.

答:
The assumptions of the Cauchy-Lipschitz theorem are in fact not fulfilled. Note that $f'(u)\to\infty$ as $u\to 0$ and this implies that $f$ is not locally Lipschitz in the neighborhood of $0$. However, we still can prove that there exists a unique solution of
\begin{equation} \begin{aligned} u'(t) &= f(u), \\
 u(0) &= 0. \end{aligned} \tag{*}\end{equation}

Obviously $u(t)\equiv 0$ is a solution, defined on the whole real line. Therefore we only need to prove that there is no solution starting from $0$ which is not identically $0$. Assume $u$ is a solution and there exists $t_0$ such that $u(t_0)=u_0 \in (0,1)$. Whenever $u(t)\neq 0$ and $1$, we can divide equation (*) by $f(u)$ and obtain
$$ \frac{u'}{f(u)} = 1 $$
and then integrate both sides from $t_0$ to $t$. By the change of variables, we have
$$ \int_{t_0}^t \frac{u'(s)}{f(u(s))}\mathrm{d}s = \int_{u_0}^{u(t)} \frac{\mathrm{d}x}{f(x)}, $$
therefore
$$ \int_{u_0}^{u(t)} \frac{\mathrm{d}x}{f(x)} = t-t_0. $$
Now if $u$ is a solution with $u(0)=0$, in particular $u(t)\to 0$ as $t\to 0$. Thus taking the limit in the integral we need to have
$$ \int_0^{u_0} \frac{\mathrm{d}x}{f(x)} = t_0 <\infty, $$
but $\int_0^{u_0} \frac{\mathrm{d}x}{f(x)} = \infty$, which gives a contradiction.

To show that indeed the integral $\int_0^{u_0} \frac{\mathrm{d}x}{f(x)}$ is divergent, you can use comparison criterion: if $\alpha>0$, then on the interval $(0,u_0)$
$$ \frac{-1}{u\ln u}e^{-\alpha u} \geq \frac{-1}{u\ln u}e^{-\alpha u_0}$$
and if $\alpha\leq 0$ then
$$ \frac{-1}{u\ln u}e^{-\alpha u} \geq \frac{-1}{u\ln u} $$
and the integral of $\frac{-1}{u\ln u}$ is easy to calculate by hand.

The analogous reasoning will work if we assume $u_0\in (-1,0)$.


问:
Condensed SVD decomposition of an outer product

Let $A = uv^{T} \in \mathbb{R}^{m \times n}$. Find the (condensed) SVD decomposition of $A$.

Theorem (Condensed SVD decomposition)

Let $A \in \mathbb{R}^{n \times m}$ be a non-zero matrix of rank $r$. Then there exists $\hat{U} \in \mathbb{R}^{n \times r}, \hat{\Sigma} \in \mathbb{R}^{r \times r} , \hat{V} \in \mathbb{R}^{r \times m}$ such that $\hat{U}$ and $\hat{V}$ are isometries , $\hat{\Sigma}$ is a diagonal matrix with main diagonal entries $\sigma_{1} \ge \dots \ge \sigma_r > 0$ and
$$ A = \hat{U} \hat{\Sigma}\hat{V}^{T}.$$

Attempt at solution

Here the rank of $A$ is $1$ since each column is linearly dependent on $u$. So by the outer product form of the SVD decomposition , we get
$$ A = \sum_{j=1}^{r} \sigma_j u_j v_{j}^{T} = \sigma_1 u_1 v_j^{T} = \hat{U}\hat{\Sigma}\hat{V}^{T}.$$
It follows from the theorem above that
$$ A = \underbrace{\begin{pmatrix} u_1 \dots u_m\end{pmatrix}}_{\widehat{U}} \underbrace{\begin{pmatrix} \sigma_1 \end{pmatrix}}_{\widehat{\Sigma}} \underbrace{\begin{pmatrix} v_1 \\ \vdots \\ v_n\end{pmatrix}}_{\widehat{V}^{T}}.$$
To satisfy the initial equality ,we set $\sigma_1 =1$.

Could someone confirm my reasoning ? I feel like I am missing a step or something is wrong.

答:
When $r=1$, then $\hat{U}$ is a single column and $\hat{V}^\top$ is a single row. You might try $A=uv^\top = u \hat{\Sigma} v^\top$ where $\hat{\Sigma}$ is just the $1\times 1$ matrix $1$, but the problem is that $u$ and $v$ aren't isometries because their columns are not normalized.

However, there is an easy fix.
The SVD of $A=uv^\top$ is $\hat{U} \hat{\Sigma} \hat{V}^\top$ where $\hat{U} = u/\|u\|$, $\hat{V} = v/\|v\|$, and the single entry of $\hat{\Sigma}$ is $\|u\|\|v\|$.