问:
Which of these pairs of complex numbers are in the same orbit?
Consider the action of $\mathbb{Z}_4$ on the complex plane by rotations. Which of these pairs of complex numbers are in the same orbit?

A: $1$ and $e^{i\pi/3}$

B: $1$ and $e^{-i\pi/4}$

C: $e^{-i\pi/2}$ and $e^{i\pi/2}$

D: $e^{i\pi/2}$ and $e^{-3i\pi/4}$

So I know for them to be in the same orbit they must satisfy the equation:
$$
e^{2ik\pi/n} \times z
$$
where $n$ is $4$ in my case and z is each complex number in such a way so that the results are connected. But I'm not sure what to do exactly...

答:
I'll say $z \sim w$ if they lie in the same orbit under this action. Per your definition, $z \sim w$ iff there is some $k$ for which $z e^{2 \pi i k/4} = w$. Let $S = \{e^{2 \pi i k/4} : k \in \mathbb Z\}$. $S$ only has 4 elements, as it only depends on $k$ modulo 4. Indeed, $S = \{1, e^{\pi i /2}, e^{\pi i}, e^{\pi i 3/2}\}=\{1, i, -1, -i\}$.

Now, we have equivalently that $z \sim w$ iff $z/w \in S$. This gives us a way to approach the problem: divide the two values and see if they are any of the 4 elements in $S$.

a) $e^{i \pi / 3}$ is not in $S$ so these are not in the same orbit.

b) $e^{-i \pi / 4}$ is not in $S$ so these are not in the same orbit.

c) $e^{i \pi / 2}/e^{-i \pi / 2} = e^{i \pi} \in S$ so these are in the same orbit.

d) $e^{i \pi / 2}/e^{-3 i \pi / 4} = e^{i \pi/4} \notin S$ so these are not in the same orbit.


问:
Line Integral $\int_{C} \frac{x dy - y dx}{x^{2}+y^{2}}$ over circle with centre not at origin
How can I calculate the Line Integral $\int_{C} \overrightarrow{B}\cdot d\overrightarrow{r}$, where $\overrightarrow{B}$ is given by:

$\overrightarrow{B}=\frac{(-y,x)}{x^2+y^2}$

and $C$ is a circle in the xy-plane, with radius $a$ and centre at $(x_0,y_o,0)$?

My friend suggested that the answer would depend on either the radius, the centre of the circle or both, but I am having a hard time solving this problem. I really need help with clearing things up here...

答:
The vector field $B$ is defined everywhere except at the origin. We can calculate its curl:

$$ \begin{aligned}
\operatorname{curl} B
  &= \frac{\partial}{\partial x} \left(\frac{x}{x^2 + y^2}\right) - \frac{\partial}{\partial y} \left(\frac{-y}{x^2 + y^2} \right) \\
  &= \frac{2x^2 - (x^2 + y^2)}{(x^2 + y^2)^4} + \frac{2y^2 - (x^2 + y^2)}{(x^2 + y^2)^4} \\
  &= 0.
\end{aligned} $$

Therefore by Green's theorem, whenever $C$ is a circle oriented counterclockwise not enclosing the origin and $D$ is the closed disc bounded by $C$, we have
$$ \int_C B \cdot dr = \iint_D (\operatorname{curl} B)  dA = 0.$$
This equation makes sense because $D$ does not contain the origin. If we were to try to apply it in the case where $C$ encloses the origin, then $D$ contains the origin and so in the second integral we are trying to integrate over a place where the function is undefined, which doesn't make sense.

In the case that $C$ does enclose the origin, we can take a different approach. First let's try integrating $B \cdot dr$ over a circle $C_R$ of radius $R$ centred on the origin. Let $f(\theta) = (R \cos \theta, R \sin \theta)$.
$$ \begin{aligned}
\int_{C_R} B \cdot dr
  &= \int_0^{2 \pi} B(f(\theta)) \cdot f'(\theta) d \theta \\
  &= \int_0^{2 \pi} \frac{(-R \sin \theta, R \cos \theta)}{R^2} \cdot (- R \sin \theta, R \cos \theta) d \theta \\
  &= \int_0^{2 \pi} 1 \,\,d \theta\\
  &= 2 \pi.
\end{aligned}$$
Notice that the result does not depend on the radius $R$. More is true in fact: it does not even depend on the position (or shape) of the circle, only on the fact that the circle encloses the origin. This can be seen as follows: if $C$ is a closed path enclosing the origin, then we can take $R$ small enough so that $C_R$ is contained entirely inside $C$. We know the path integral over $C_R$, then applying Green's theorem as before to the region enclosed by $C$ minus the region enclosed by $C_R$ (thereby removing the problematic origin point) will give that $\int_C B \cdot dr = \int_{C_R} B \cdot dr = 2\pi$.

Therefore the answer is either $0$ or $2 \pi$, depending on whether or not the curve $C$ encloses the origin.

ContentLicense: CC BY-SA 4.0


问:
A group as the union of finitely generated subgroups
Suppose a group $G$ satisfies

(1) $G=\bigcup_{i\ge 1} G_i$, where $G_i \le G_{i+1}$ and all are subgroups of $G$;

(2) there exists a number $m$ such that each $G_i$ is generated by at most $m$ elements.

Can we prove that $G$ is finitely generated?

答: This is false. Consider, for instance, $G = \mathbb Q$ and $G_i = \frac{1}{i!} \mathbb Z$. Each $G_{i!}$ is generated by $1/{i!}$, but $\mathbb Q = \bigcup \frac{1}{i!} \mathbb Z$. Furthermore, I claim that $\mathbb Q$ is not finitely generated. Indeed, take a finite subset $\emptyset \neq S \subseteq \mathbb Q$. Let $n$ be the least common multiple of the denominators of the elements of $S$. Then $\langle S\rangle \subseteq \frac{1}{n} \mathbb Z \neq \mathbb Q$, so $\mathbb Q$ is not finitely generated.


问:
Metric tensor derivative identity
In this paper, the following identity is used without proof nor reference:

$$g^{\alpha \mu} g^{\beta \lambda} \dot{g}_{\alpha \beta} \equiv -\dot{g}^{\mu \lambda}$$

where $g$ is metric tensor and $\dot{g}$ is derivative of $g$.

How can this identity derived?

答:
I found out that this is from the matrix identity,
$$\partial X^{-1} = - X^{-1} (\partial X) X^{-1}$$

Proof for this identity is in here.

With this, we have
$$X (\partial X^{-1}) X = - \partial X$$

By substituting $X$ with $g^{\mu \lambda}$, we get the identity!

Title: Metric tensor derivative identity


问:
Partial isometry and perturbation of a $C^*$ algebra.
Let $A$ be a $C^*$algebra and $I$ be an ideal. Suppose that $I$ admits an approximate identity consisting of projections in $I$. Let $u \in U(A/I)$. Then is there a partial isometry $v \in A$ such that $\pi(v)= u$, where $\pi$ is the quotient map?

I think if we use perturbation we can find such partial isometry but I couldn't. Could anybody help me out?

答:
Given $u$ in $U(A/I)$, choose any $w$ in $A$ such that $\pi (w)=u$.  One then has that
$$
  \pi (w^*w-1)=0,
  $$
so $w^*w-1\in I$.   Given any $\varepsilon >0$, we then have by hypothesis that there exists a projection $p$ in $I$ (chosen e.g. from
an approximate identity) such that
$$
  \varepsilon >\|(w^*w-1) - p(w^*w-1)\| = $$ $$ =
  \|(1-p)(w^*w-1)\| = \|p^\perp (w^*w-1)\|,
  $$
where  $p^\perp  = 1-p$.
Letting $y=wp^\perp $, observe that
$$
  \pi (y) = \pi (wp^\perp ) = \pi (w-wp)=\pi (w) = u,
  $$
and moreover that
$$
  \|y^*y-p^\perp \| =
  \|p^\perp w^*wp^\perp -p^\perp \| = $$ $$ =
  \|p^\perp (w^*w-1)p^\perp \| \leq
  \|p^\perp (w^*w-1)\|\|p^\perp \| <\varepsilon .
  $$

Since $p^\perp$ is idempotent, anything nearby is approximately idempotent so,  by choosing  $\varepsilon $
sufficiently small, we may assume that
$$
  \|(y^*y)^2 - y^*y\| < \frac14,
  $$
which in turn implies that $\frac12$ cannot be in the spectrum of $y^*y$.

Let
$$
  f:\mathbb R \setminus \left\{\textstyle \frac12\right\} \to \mathbb R
  $$
be the continuous function vanishing identically on $(-\infty ,1/2)$, and such that
$f(t)= t^{-1/2}$, for  $t\in (1/2, \infty )$.  Clearly
$$
  tf(t)^2 = 1_{(\frac12,\infty )},
  $$
which is a real  idempotent function, hence
$$
  f(y^*y)\ y^*y \ f(y^*y)
  $$
is a self-adjoint idempotent element, i.e. a  projection.  Therefore
$$
  v:= y  f(y^*y)
  $$
is a partial isometry thanks to the fact that $v^*v$ is a projection.  To conclude we claim that $\pi (v)=u$, and this is
easy to see since
$$
  \pi (v) = \pi (y)  \pi (f(y^*y)) = \pi (y)  f(\pi (y^*y)) = $$ $$ =
  u  f(u^*u) =   u f(1) = u.
  $$


问:
Proving that $\frac{\sigma_{n-1}}{\omega_n} = n$ in $\mathbb{R}^n$
If $\sigma_{n-1}$ was the surface area of the unit sphere in $\mathbb{R}^n$ and $w_{n}$ was the area of the unit ball in $\mathbb{R}^n$, my lecture notes prove that $$\frac{\sigma_{n-1}}{\omega_n} = n$$ by the proof $$\omega_n = \int_{B_1(0)} dx$$$$=\int_0^1\sigma_{n-1} z^{n-1} dz$$$$=\sigma_{n-1}\int_0^1z^{n-1}dz$$$$=\frac{\sigma_{n-1}}{n}$$ but I don't understand how we got from the 1st line to the second line, namely that $$ \int_{B_1(0)} dx =\int_0^1\sigma_{n-1} z^{n-1} dz$$ could someone please explain this to me? Thanks!</p>

答:
As @robjohn said, fill the unit ball in $\mathbb R^n$ up with concentric shells of radius $r$, $0<r\le 1$. The "surface area" of the sphere of radius $r$ is $\text{area}(S_r)=r^{n-1}\omega_n$, and so $$\text{volume}(B_1(0)) = \int_0^1 \text{area}(S_r)\,dr = \int_0^1 r^{n-1}\omega_n\,dr\,.$$
(Note that $k$-dimensional volume scales by a factor of $r^k$ when you stretch the object by a factor of $r$.)</p>

ContentLicense: CC BY-SA 3.0


问:
Limit of $(1-\frac{1}{2^n})^n$.
What is the limit of the sequence $(1-\frac{1}{2^n})^n$?

Let $a_n:=(1-\frac{1}{2^n})^n$ then $\ \ln \ a_n=n\ \ln(1-\frac{1}{2^n})$

$\implies \lim \limits_{n\rightarrow \infty} \ln\ a_n=\lim\limits_{n \rightarrow \infty} n\ \ln(1-\frac{1}{2^n})$

$\implies \ln \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \frac{\ln(1-\frac{1}{2^n})}{\frac{1}{n}}$

$\implies \ln \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \frac{\frac{ln2}{2^n(1-2^n)}}{\frac{-1}{n^2}}  [\frac{0}{0} \ \ \text{form}]$

I am stuck here, what should I do after this?

答:
Continuing where you left:
$$  \frac{ \frac{\ln2}{2^n (1 -2^n)}  }{\frac{-1}{n^2}} = - \frac{n^2 \ln2}{2^n(1-2^n)} = \frac{n^2 \ln2}{2^n(2^n - 1)}$$

It's pretty obvious / not hard to prove that this tends to $0$ as $n$ tends to infinity ("exponent beats polynomial") and so your limit should go to $e^0 = 1$

Another approach is to "fix" the exponent so it would suit the limit of $e$:

$$ \lim_{n \rightarrow \infty}( 1 - \frac{1}{2^n})^n = \lim_{n \rightarrow \infty}( \overbrace{(1 + \frac{-1}{2^n})^{2^n}}^{\text{1/e}})^{ \overbrace{\frac{n}{2^n}}^{\text{Tends to 0}}}$$

Title: Limit of $(1-\frac{1}{2^n})^n$.


问:
Eigenvalues of a linear transformation in $\mathbb{R^3}$

Let $T: \mathbb{R^3} \rightarrow \mathbb{R^3}$ be the rotation of $\mathbb{R^3}$ around the $x_{2}$-axis by 90 degrees. Find all real eigenvalues of T without computing its standard matrix. Explain your reasoning.

So, for this problem, I initially thought this might be some interesting application of one of the properties of determinants, specifically the property $\det(AB) = \det(A)\det(B)$, with the $x_{2}$ rotational matrix in $\mathbb{R^3}$, with the following as $A$:\begin{vmatrix}
\cos(\theta) & 0 & -\sin(\theta)\\
0 & 1 & 0\\
\sin(\theta) & 0 & \cos(\theta)
\end{vmatrix}
and some matrix with variables being $B$.

When that didn't work for me, I tried to simply solve for the eigenvalues after applying the transformation, by doing:

$\det\left(\begin{vmatrix}
\cos(\theta) & 0 & -\sin(\theta)\\
0 & 1 & 0\\
\sin(\theta) & 0 & \cos(\theta)
\end{vmatrix}*\begin{vmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{vmatrix}-\begin{vmatrix}
\lambda & 0 & 0\\
0 & \lambda & 0\\
0 & 0 & \lambda
\end{vmatrix}\right)$

But I think I'm simply overcomplicating it now. Can someone help me figure out what I'm missing?

答:
Maybe I'm confused. If you are searching the eigenvalues, you have to solve $$Tx=\lambda x$$ where $x\in\mathbb{R}^3$ and $x\neq 0$. This is equivalent to find $\lambda$ such that$$\begin{vmatrix}
\cos(\theta)-\lambda & 0 & -\sin(\theta)\\
0 & 1-\lambda & 0\\
\sin(\theta) & 0 & \cos(\theta)-\lambda
\end{vmatrix}=0$$ then $$(\cos(\theta)-\lambda)[(1-\lambda)(\cos(\theta)-\lambda)]+\sin(\theta)[\sin(\theta)(1-\lambda)]=0$$ $$(1-\lambda)(\cos(\theta)-\lambda)^2+(1-\lambda)\sin^2(\theta)=0$$ $$(1-\lambda)[(\cos(\theta)-\lambda)^2+\sin^2(\theta)]=0$$ $$(1-\lambda)(\lambda^2-2\lambda\cos(\theta)+1)=0$$
and you know that $\theta=\frac{\pi}{2}$. Maybe this could help you.


问:
A Level Differential Equation
I have been trying to solve this problem for 15 minutes and I’m not any closer to a solution (in an exam, we are expected to do questions like this in 6 minutes or less which is why I’m asking now). I can’t reach the expression shown in the question. Any help would be greatly appreciated. Just to clarify, you also need to find the value of k

So far I have tried to integrate (1-x^2)^-0.5 dx and to perform the same action for the other term, but I get an answer containing arc sins and sins, and I’m not sure if this is the right course of action or not, it’s just that it doesn’t help me get to the expression required in the question. I keep getting y = sin(C+arcsin(x)).

答:
You are almost there. Let $x=\sin\theta$ so that $\arcsin x=\theta$. Then
$$y=\sin(C+\theta)=\sin C\cos\theta+\cos C\sin\theta=\sin C \sqrt{1-x^2}+x\cos C$$

Notice that if we make $\sin C=\frac12$ then $\cos C=\frac12\sqrt3$ so
$$2y=\sqrt{1-x^2}+\sqrt3 x$$

Comparing with the solution given we see that $k=3$.

Check: When $y=\frac12\sqrt3, x=\frac12$ then LHS$=\sqrt3$ and RHS$=\frac12\sqrt3+\frac12\sqrt3=\sqrt3$ so LHS=RHS.

Title: A Level Differential Equation


问:
Prove or disprove the membership of elements of submodules
Let $M$ be $R$-module and $N$ be submodule of $M$. Let $x,y\in M$. Prove or disprove:

(1) If $x,y\in N$ then $x+y\in N$.

(2) If $x\in N$ and $y\not\in N$ then $x+y\not\in N$.

(3) If $x\not\in N$ and $y\not\in N$ then $x+y\in N$.

I try as below

(1) Obviously, since $N$ be submodule of $M$ then $x+y\in N$.

(2) I doubt it. I take example $M=\mathbb{Z}$:$\mathbb{Z}$-module and the submodule is $N=\mathbb{2Z}$:$\mathbb{Z}$-module. Now, take $2n+1\not\in \mathbb{2Z}$ and $2m\in \mathbb{2Z}$, $m,n\in\mathbb{Z}$.
\begin{align}
(2n+1)+2m=4n+1=2(n+m)+1\not\in\mathbb{2Z}.
\end{align}
Generally, does $x+y\not\in N$ holds?

(3) I doubt it. I take example $M=\mathbb{Z}$:$\mathbb{Z}$-module and the submodule is $N=\mathbb{2Z}$:$\mathbb{Z}$-module. Now, take $2n+1\not\in \mathbb{2Z}$ and $2m+1\not\in \mathbb{2Z}$, $m,n\in\mathbb{Z}$.
\begin{align}
(2n+1)+(2m+1)=2(n+m)+2=2((n+m)+1)\in\mathbb{2Z}.
\end{align}
Generally, does $x+y\in N$ holds?

答:
(1) is certainly correct.

For (2) you gave an example, but you can prove it in general. Indeed, if $x + y \in N$ then $y = (x + y) - x \in N$, a contradiction.

For (3) your example doesn't make sense. You found $x, y \notin N$ and showed $x + y \in N$, but since you said you doubted it you want to show exactly the opposite. Here's an example: take the $\mathbb Z$ module $M = \mathbb Z \times \mathbb Z$ and let $N = \{(n, 0) : n \in \mathbb Z\}$. Then $(0, 1), (0, 2) \notin N$ and $(0, 1) + (0, 2) = (0, 3) \notin N$.


答:
Looks good to me.

I would say, though, that if you were writing this on a homework, the first part of this answer which contains the majority of the work is simply the scratch work, and you should try to flesh out the proof on the bottom a little bit more, in the standard way for these types of proofs.

Title: Proof by $\epsilon - \delta$ definition that $\lim_{x \to 1} \frac{x^2-1}{x^3-1} = \frac{2}{3}$

Body:
$$\lim_{x \to 1} \frac{x^2-1}{x^3-1} = \frac{2}{3}$$

I want to know if my proof is correct. If not, please show me how to fix it, if possible.

Let $\epsilon > 0$.

$\left| \dfrac{x^2-1}{x^3-1}-\dfrac23 \right|=\left| \dfrac{3x^2-3-2x^3+2}{3x^3-3} \right|=\left| \dfrac{-2x^3+3x^2-1}{3x^3-3}\right|=\left|\dfrac{(x-1)^2(2x+1)}{3(x-1)(x^2+x+1)} \right|=\dfrac{|x-1||2x+1|}{3|x^2+x+1|}$

Assume $\delta < 1$. Then $|x-1|<1$ implies:

(i) $-2 <  0 < x < 2 \implies |x| < 2$

(ii) $-5<-3<2x+1<5 \implies |2x+1|<5$

(iii) $-3<-1<x+1<3 \implies |x+1|<3$

(iv) $|x^2+x+1| = |x(x+1)+1| \le |x||x+1|+1 < 2.3+1 = 7 $

So, make $\delta = \min \{1, \dfrac{21\epsilon}{5} \}$. Then $|x-1|<\delta \implies \dfrac{|x-1||2x+1|}{3|x^2+x+1|}<\epsilon$.


问:
Extending a Group Character to a Dirichlet Character
In https://math.mit.edu/classes/18.785/2015fa/LectureNotes17.pdf, just after Definition 17.4 on page 7, the author states that "Every $m$-periodic Dirichlet character $\chi$ restricts to a group character on $(\mathbb{Z}/m\mathbb{Z})^\times$; conversely, every group character $\chi$ of $(\mathbb{Z}/m\mathbb{Z})^\times$ can be extended to a Dirichlet character $\chi$ by defining $\chi(n)=0$ for $n\not\in(\mathbb{Z}/m\mathbb{Z})^\times$." In Remark 17.5 on the same page the author clarifies what is meant by $n\not\in(\mathbb{Z}/m\mathbb{Z})^\times$ - it means the image of $n$ under the quotient map $\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}$ is not in $(\mathbb{Z}/m\mathbb{Z})^\times$.

But what confused me is that, in Definition 17.4, the author defines a Dirichlet character to be simply a period, completely multiplicative arithmetic function. So why is it that a Dirichlet character can only be viewed as an extension of a character of $(\mathbb{Z}/m\mathbb{Z})^\times$, as opposed to an extension of a character simply of $\mathbb{Z}/m\mathbb{Z}$?

I have thought about this all day, and attempted to find a proof as to why $(\mathbb{Z}/m\mathbb{Z})^\times$ rather than $\mathbb{Z}/m\mathbb{Z}$ is needed, but have been unsuccessful. What I'd really like to see is a concrete example that shows how Definition 17.4 would NOT be satisfied if $\chi$ were an extension of $\mathbb{Z}/m\mathbb{Z}$ rather than $(\mathbb{Z}/m\mathbb{Z})^\times$.

Any help is greatly appreciated!

答:
$\mathbf Z/ m\mathbf Z$ is a group under addition with $m$ elements, so a group homomorphism
$$\mathbf Z/ m\mathbf Z \to \mathbf C^\times$$
would not give a multiplicative function when viewed as a function from $\mathbf Z\to \mathbf C$, it would be an additive function instead.

To give an example we have a character
$\mathbf Z/2\mathbf Z \to \mathbf C^\times$
given by $1\mapsto -1$ and $0\mapsto 1$, here this is no longer a multiplicative function when viewed as a function
$$\chi:\mathbf Z \to \mathbf C^\times$$
as $-1= \chi(1) \ne \chi (1) \cdot \chi(1) = 1$.


问:
Proving that an infinite Euclidean space is linear vector space
Assuming real quantities, with the scalar product defined as

$$\langle x,y \rangle = x_{1}y_{1}+x_{2}y_{2}+\cdots, $$

where $x,y$ are vectors in the real infinite Euclidean space $\mathbb{E}_{\infty}$, and assuming that all vectors in such space are by definition of finite length (e.g. $|x|=\langle x,x \rangle=x^{2}_{1}+x^{2}_{2}+\cdots$ converges to a finite value), how can we prove that the product  $\langle x,y \rangle$ will also exist and be finite?  (In other words, we prove that $\mathbb{E}_{\infty}$ will be a linear vector space.)

I am especially interested if there are two approaches of proof: (1) one that is simple and does not use Cauchy-Schwarz inequality (if possible), and (2) another one that uses the Cauchy-Schwarz inequality. This will help me understand the differences and compare with literature (e.g. Friedman's book, ch-1).

Finally, my attempt for the former was to use the D'Alambert (ratio) test to say that, if $|x|,|y|$ converge, then we know that $\lim_{n \rightarrow \infty} |\frac{x_{n+1}}{x_{n}}|^{2}<1$ and $\lim_{n \rightarrow \infty} |\frac{y_{n+1}}{y_{n}}|^{2}<1$, thus $\lim_{n \rightarrow \infty} |\frac{x_{n+1}}{x_{n}}|<1$ and $\lim_{n \rightarrow \infty} |\frac{y_{n+1}}{y_{n}}|<1$.  Then $\lim_{n \rightarrow \infty} |\frac{x_{n+1}}{x_{n}}\frac{y_{n+1}}{y_{n}}|<1$ gives the absolute convergence of the series for $\langle x,y \rangle$. But I am told this is not a rigorous answer because the convergence of a given series does not tell us anything about the ratio test.  Does this also mean that the ratio test for absolute convergence is sufficient but not necessary?  Any elaboration on this would be appreciated.

答:
The ratio test only says that if $\lim_{n\rightarrow\infty}\left|\frac{a_n}{a_{n+1}}\right| < 1$ then $\sum_{n=1}^{\infty}a_n$ converges absolutely; the converse is false - for instance, $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges despite the fact that the ratio of consecutive terms tends to $1$. Thus, it's not much good here - all you are given is that some sequences converge, but nothing about how quickly they do so.

A better thing to note is that, if you have some sequence of $a_n$, then $\sum_{n=1}^{\infty}a_n$ converges absolutely if and only if there is some upper bound $B$ such that $\sum_{n=1}^N |a_n| \leq B$ for all $N$ - this is, more or less, the monotone convergence theorem for real numbers. This is a more helpful characterization because it is bidirectional - you can use it to transform your givens into some statement about finite sums and then to transform a statement about finite sums back into a statement about limits.

More clearly, your goal could be:

Suppose that there are $B_1$ and $B_2$ such that $\sum_{n=1}^Nx_n^2\leq B_1$ and $\sum_{n=1}^Ny_n^2\leq B_2$ for all $N$. Show that there is some $B$ such that $\sum_{n=1}^N|x_ny_n| \leq B$ for all $N$.

The motivation for transforming the goal thusly is that we have gotten rid of most the analytical difficulties and turned this into an algebra problem.

The most straightforward thing to do would be to say:

Let $B=B_1+B_2$. Note that for any $n$, we have either that $|x_n|\leq |y_n|$ or $|y_n|\leq |x_n|$. In the former case, note $|x_ny_n| \leq y_n^2$ and in the latter $|x_ny_n| \leq x_n^2$. In any case, $|x_ny_n| \leq x_n^2+y_n^2$. Therefore, $\sum_{n=1}^N|x_ny_n| \leq \sum_{n=1}^Nx_n^2 + \sum_{n=1}^N y_n^2 \leq B_1+B_2$ as desired, so $\sum_{n=1}^Nx_ny_n$ converges absolutely.

If you wanted to use Cauchy-Schwarz, you could use it here too:

Let $B=\sqrt{B_1B_2}$. For any $n$, consider the vectors $(|x_1|,\ldots,|x_n|)$ and $(|y_1|,\ldots,|y_n|)$. By the Cauchy-Schwarz inequality applied to these vectors, we have
$$\sum_{n=1}^N |x_ny_n| \leq \sqrt{\sum_{n=1}^N x_n^2 \cdot \sum_{n=1}^N y_n^2} \leq \sqrt{B_1B_2}.$$

This gives the algebraic result you need just as well - and gives a somewhat tighter bound on the ultimate $\sum_{n=1}^{\infty}|x_ny_n|$, although this is a bit redundant since you could prove the Cauchy-Schwarz inequality as soon as you know that this space is in fact an inner product space. Note that the proofs are essentially the same - after transforming the analytical statement into an algebraic one, we just need some algebra to fill in a gap and can do this in a number of ways.


问:
Residue Calculus with different enclosing methods of contour
Consider the following integral: $$\frac{1}{2\pi i }\oint_C f(z) dz$$ with $$f(z)= \frac{z^2+1}{z^3}$$ and $C$ is the unit circle centered at the origin.

Consider two ways to do it:

Enclosing the singular points inside $C$:
This is actually its residue at $z=0$ (order = $3$). So $$\text{Res}(f(z),z=0)=\frac{1}{(3-1)!}\frac{d^2}{dz^2}(z^2+1)\bigg|_{z=0}=1$$

Enclosing the singular points outside $C$ (by including the point at infinity):

From the textbook I read, 1. and 2. should give the same answer. However, I have no idea how to do 2.?

Can anyone please let me know how to deal with 2.?

答:
The function $f(z) = \frac{z^2+1}{z^3}$ is holomorphic at $\infty$, yet still has a residue there.  There are no other singular points outsize $C$ to consider.

$$ -\frac{1}{2\pi i} \oint_{C} f(z)\,dz=\text{Res}_{z=\infty} f(z) = -\lim_{z\to \infty} zf(z)= -1.$$

Here is a source defining the residue at infinity.

http://www.stat.phys.titech.ac.jp/~yokoyama/note5.pdf


答:
Here it is another way to solve it for the sake of curiosity.

To begin with, notice that
\begin{align*}
\frac{|x-1||2x+1|}{3|x^{2} + x + 1|} = \frac{|x-1||(2x - 2) + 3|}{3|x^{2} + x + 1|} \leq \frac{2|x-1|^{2} + 3|x-1|}{3|x^{2}+x+1|}
\end{align*}

Since we have that
\begin{align*}
x^{2} + x + 1 = \left(x^{2} + x + \frac{1}{4}\right) - \frac{1}{4} + 1 = \left(x + \frac{1}{2}\right)^{2} + \frac{3}{4} \geq \frac{3}{4} > 0 \Rightarrow 0 < \frac{1}{x^{2}+x+1} \leq\frac{4}{3}
\end{align*}

we can conclude that
\begin{align*}
\frac{|x-1||2x+1|}{3|x^{2} + x + 1|} \leq \frac{8|x-1|^{2} + 12|x-1|}{9} \leq \frac{8\delta^{2} + 12\delta}{9} = \varepsilon
\end{align*}

Thus for every $\varepsilon > 0$ there corresponds a $\displaystyle\delta = \frac{-3 + 3\sqrt{1+2\varepsilon}}{4} > 0$ such that
\begin{align*}
0 < |x - 1| \leq \delta \Rightarrow \left|\frac{x^{2}-1}{x^{3}-1} - \frac{2}{3}\right| \leq \varepsilon
\end{align*}
and we are done.

Hopefully this helps!


问:
Generalization of the Cartesian basis in Hilbert space
Given the Hilbert space of continuous functions $H = \{ u : [0,1] \to \mathbb{R}\}$ with
inner product : $ = \int_0^1 u(x)w(x)dx$,

is there a generalization of a Cartesian basis ?

Intuitively, what I want to have is a basis that "picks up" only the value of $u(x)$ at $x$ each time, like (1,0) and (0,1) in $\mathbb{R}^2$

I was thinking something like $B = \{ \delta(x-l), l \in [0,1] \}$, but this is not continuous.

Thanks a lot,

Kostas

答:
I suppose a precise statement of what the OP has in mind would be:

Is there a basis $\{e_x\}_{x\in[0,1]}$ of $H$ such that, once any $f\in H$ is expanded in that basis, the coefficient of each $e_x$ is precisely $f(x)$?

In other words
$$
  f= \sum_{x\in[0,1]} f(x)e_x.
  $$

The difficulty here is the fact that this will often have uncountably many nonzero summands.  While infinite sums make perfectly good sense in normed spaces, no generalization of the theory of series allows for summing uncountably many nonzero terms.

The short answer is thus that no such basis exist.  Nevertheless the intuition behind the desire for the existence of a basis like that should not be totally ignored since it underlies many important developments in Analysis.  A similar situation is expressed by the highly intuitive although totally meaningless expression
$$
\int_a^bf(x)\, dx = \sum_{x=a}^b f(x)\Delta_x.
$$

This idea pops up again in the subtle difference between spectral values, on the one hand, and eigenvalues of a self-adjoint operator on an infinite dimensional space.

In the theory of unitary group representations this also shows up in the delicate problem of decomposing a representation as the direct sumof irreducible ones.


问:
Allowed values of $x$ in $x \sin 6 y-\left(3 x^2-5\right) \sin ^23 y=0$ when $y=0$? Can someone please check the given explanation?
I have this two-variable equation
$$x \sin 6 y-\left(3 x^2-5\right) \sin ^23 y=0$$
for $0\leqslant x,y<\infty$, and I want to check which values of $x$ satisfy this equation when $y=0$.
From the above equation, I see that when $y=0$, it is fulfilled for all  $0\leqslant x<\infty$.

Then, I rewrite the equation as
$$\frac{x}{3 x^2-5}=\frac{1}{2} \tan (3 y),$$
from here, I see that when $y=0$, there are only two solutions for $x$, i.e. $x=0$ and $x\to \infty$.

Then, what can I say in general about the solutions of $x$ when $y$ is zero?

And what can we say about the point $x=\sqrt{\frac{3}{5}}$? Is it an acceptable solution for $3y=\frac{\pi }{2}$? Since from the first equation, it can be a solution, but I have doubts about it from the second form of the equation.

答:
When you "rewrite" an equation by multiplying or dividing by some factor, the new equation is equivalent to the old when that factor is not $0$.  When it is $0$, the two are generally not equivalent.

So, your first equation
$$ x \sin 6 y - (3 x^2 - 5) \sin^2 3 y $$
is indeed always true when $y = 0$, or more generally when $\sin 3 y = 0$.  When you divide by $(3 x^2 - 5) \sin 6 y$ to get your second equation

$$ \frac{x}{3x^2-5} = \frac{1}{2} \tan 3y $$

this is equivalent to the first as long as $3 x^2 - 5 \ne 0$ and $\sin 6 y \ne 0$.  When
$\sin 3y = 0$, the right side is $0$ but the left is not (unless $x=0$).  When $3x^2 - 5 = 0$ the left side is undefined but the right side is defined (unless $\cos 3 y = 0$).
When $\cos 3y = 0$ the right side is undefined but the left side is defined (unless $3 x^2-5 = 0$).


问:
How do you show Tonelli’s theorem can fail in the σ-nonfinite case, and also that product measure does not need to be unique.
For this, I showed that
If we have the set $I=[0,1]$ equipped on one side with the Borel set and Lebesgue measure λ for X and on the other side the same set $I=[0,1]$ equipped with the power set and the counting measure m for Y.

The diagonal set Δ={$(x,x) : x∈[0,1]$} is a closed subset of $I_2$. Hence
Δ$\in(I_2)=\mathcal{B}(I)×\mathcal{B}(I)⊂\mathcal{B}(I)×\mathcal{B}(I)$.
Therefore, the characteristic map $1_Δ$ is measurable for the measure induced on $X×Y$. However, we have on one side
$\int_{I}(\int_{I}1_Δ(x,y)dm(y))dλ(x)=\int_{I}(\int_{I}1_{x}(y)dm(y))dλ(x)=\int_{I}m({x})dλ(x)=λ(I)=1$

and on the other side

$\int_{I}(\int_{I}1_Δ(x,y)dλ(x))dm(y)=\int_{I}(
\int_{I}1_{y}(x)dλ(x))dm(y)=\int_{I}λ({y})dm(y)=\int_{I}0 dm(y)=0$

These are not equal and Tonelli's Theorem failed. Is this right?

Also, in the case for uniqueness. I need to show that there is more than one measure $\mu$ on $\mathcal{L}[0,1]\times 2^{[0,1]}$ with the property that $\mu(E\times F)=\mu(E)$#$(F)$ (where # is the counting measure of F) for all $E\in \mathcal{L}([0,1])$ and $F\in 2^{[0,1]}$. How should I start with this?

答:
$1$. Yes, your proof that Tonelli's theorem conclusion fails in this case is correct. The only minor correction is:

The diagonal set $\Delta=\{(x,x) : x∈[0,1]\}$ is a closed subset of $I_2$. Hence
$\Delta\in\mathcal{B}(I_2)=\mathcal{B}(I)×\mathcal{B}(I)\subset \mathcal{B}(I)×2^I$.

$2$. Now, let us show that there is more than one measure $\mu$ on $\mathcal{B}([0,1])\times 2^{[0,1]}$ with the property that $\mu(E\times F)=\lambda(E)\#(F)$ (where $\lambda$ is the Lebesgue measure and  $\#$ is the counting measure) for all $E\in \mathcal{B}([0,1])$ and $F\in 2^{[0,1]}$.

Let $\mu_1$ be the the product measure $\lambda \times \#$. It is immediate that for all $E\in \mathcal{B}([0,1])$ and $F\in 2^{[0,1]}$,
$$\mu_1(E\times F) = (\lambda \times \#) (E\times F)= \lambda(E)\#(F)$$

Let $\mu_2$ be a measure defined on $\mathcal{B}([0,1])\times 2^{[0,1]}$ by, for all $A \in \mathcal{B}([0,1])\times 2^{[0,1]}$,

$$\mu_2(A) = \sum_{y\in[0,1]} \lambda(([0,1]\times\{y\}) \cap A)$$

It is easy to see that $\mu_2$ is in fact a measure. Moreover, for all $E\in \mathcal{B}([0,1])$ and $F\in 2^{[0,1]}$,
$$\mu_2(E\times F) = \sum_{y\in[0,1]} \lambda(([0,1]\times\{y\}) \cap (E\times F))=\sum_{y\in F}\lambda(E)=\lambda(E)\#(F)$$

Lets us prove that $\mu_1\neq \mu2$.

Consider the diagonal set $\Delta=\{(x,x) : x∈[0,1]\}$ again.

Note that, if $E\times F$ is a rectangle such that $\mu_1(E\times F)$ is finite, then $\lambda(E)=0$ or $F$ is a finite set.  So $\Delta$ can not be covered by any countable collection of rectangles $E\times F$ with $\mu_1(E\times F)$ finite.

So $\mu_1(\Delta)=+\infty$ (in fact, we proved more: $\Delta$ is not even $\mu_1$-$\sigma$-finite).

On the other hand,

$$\mu_2(\Delta) = \sum_{y\in[0,1]} \lambda(([0,1]\times\{y\}) \cap \Delta)= \sum_{y\in[0,1]} \lambda(\{y\})=0$$


问:
An undefined notation $P(\{x\in \mathcal{X}:g(x)\in A\})\stackrel{?}{=} P(\{x\in S:g(X(x))\in A\})$ in Casella and Berger's Statistical Inference
I'm reading Casella and Berger's Statistical Inference. In the beginning of Chapter 2, they said that:

Formally, if we write $y=g(x)$, the function $g(x)$ defines a mapping from the original $\color{red}{\text{sample space}}$ of $X$, $\mathcal{X}$, to a new sample space, $\mathcal{Y}$, the sample space of the random variable $Y$. That is,
$$g(x):\mathcal{X}\to \mathcal{Y}.$$

On the next page, they said that:

$$
\begin{array}{lll}
P(Y\in A)
&=& P(g(X)\in A) \\
&=& P(\{x\in \mathcal{X}: g(x)\in A\}) \\
&=& P(X\in g^{-1}(A))\tag{*}
\end{array}$$

My question is about the notation $P(\{x\in \mathcal{X}: g(x)\in A\})$.
I want to check my following guesses:

The notation $\mathcal{X}$ should be the $\color{red}{\text{image}}$ of the random variable $X$, not the $\color{red}{\text{sample space}}$ of the random variable $X$. Am I right?

I don't think it was a typo since the authors mentioned that again on page 50:

When transformations are made, it is important to keep track of the sample spaces of the random variables; otherwise, much confusion can arise. When the transformation is from $X$ to $Y=g(X)$, it is most convenient to use
$$\mathcal{X}=\{x:f_X(x)>0\}\text{ and }\mathcal{Y}=\{y:y=g(x)\text{ for some }x\in \mathcal{X}\}.$$

$P(\{x\in \mathcal{X}: g(x)\in A\})$ should be defined as $\sum\limits_{g(x)\in A}P(X=x)$ or $\int_{g(x)\in A}P(X=x)$.

Suppose that the random variable $X$ is from $S$ to $\mathcal{X}$. In rigorous, (*) should be written as
$$
\begin{array}{lll}
P(Y\in A)
&=& P(g(X)\in A) \\
&=& P(\{x\in \color{red}{S}: g(\color{red}{X}(x))\in A\}) \\
&=& P(X\in g^{-1}(A))
\end{array}$$

答:
I've confirmed this question with the author.
It is indeed
$$P_X(\{x\in \mathcal{X}:g(x)\in A\})=P(\{s\in S:g(X(s))\in A\}),$$ where $X$ is a random variable from $S$ to $\mathcal{X}\subseteq \mathbb{R}$.
Thanks for all comments.


问:
Measurability and convergence of sequence of functions
Let $(E,d),(X,d')$ be metric space, $(f_n)_n$ be a sequence of measurable function from $(X,\mathcal{B}(X))$ to $(E,\mathcal{B}(E)),$ $f:X \rightarrow E,$ $(\mathcal{B}(X),\mathcal{B}(E))$- measurable. Let $Y$ be the set of $x \in X$ such that for every sequence $(x_n)_n$ of elements of $X$ converging to $x,\lim_nf_n(x_n)=f(x):$ $$Y= \left\{x \in X; \forall (x_n)_n \in X^\mathbb{N},\lim_nx_n=x \implies \lim_n f_n(x_n)=f(x)\right\}$$

Is it true that $Y \in \mathcal{B}(X)?$

The problem is that for every $(x_n)_n \in X^{\mathbb{N}}$ is not countable, notice if $f_n=f,$ then $Y$ is the set of continuity of $f.$ Is there any property which describe $Y$ (continuity, uniform convergence...) so that the problem becomes easier? If not, how can we prove-disprove it?

答:
Main Idea

Because my construction is somehow not friendly to follow, I'll give a broad summarization of my idea.

So, in our main question, we have to deal with a sequence of functions. Not only that, the sequence convergence is also involved. Given the complexity of directly dealing with that object. I manoeuvre a new sequence permitting us to translate all of our necessary information into it, hence also our question. Then after translation, our analysis of the sequence of functions is translated into an analysis of a specific function.

So you can see in next section, under my transformation, the continuous point of the new function $F$ is "roughly" the point of interest in our main question. And as we know, the set of continuous point of a function is measurable, hence we have the conclusion.

The presence of "Roughly" in previous paragraph is due to the fact that those continuity point of $f$ is also the continuity point of $f$ ( I have proved the vice versa in main answer).
So, considering the set of countinuity of $F$ is entirely true. However, some modifications can be made in the proof of the measurability ( for the set of continuity) in order that we can have the desired property.

The organization of my answer is follows:

The construction of new space

The construction of new function $F$ that contains all information of $(f_n)$

The equivalence of the set of continuity of $F$ and our set of interest ( intersects with the set of continuity of $f$)

The modication needed for the measurability of new set
$$\tilde{Z}:= \{ p=(\infty, x) : \forall p_n \in P \text{ such that}: p_n \rightarrow p \text{ and } p_W(p_n)<+\infty \text{ we have } F(p_n) \rightarrow F( p) \}$$

Main Answer

Here is my assumption on $X$:

$X$ is separable (1)

For the convenience of showing the main idea of my proof, let's change $Y$ for now to $Y'=Y\cap V$, where $V$ is the set of continuity of $f$

Measurability of $Y'$

Let :

$W:=\overline{\mathbb{N}^*} =\mathbb{N^*}\cup \{ \infty\} $, the extended set of positive natural numbers.

$d_W$ denotes a metric on $W$ which is defined as $d_W(m,n)=\left| \frac{1}{m}-\frac{1}{n}\right|$

For consitency, $d_X$ denotes the metric of $X$ instead of $d'$

$f_{\infty} := f$

$P:=W\times X$ denote the product metric space with the product metric $d_P:= d_W+d_X$

Remark 1: The metric $d_P$ is equivalent to the usual product metric. I use that one instead of the usual one just because of its convenience for the thought process.

Remark 2 $W$ is a separable space  ( though it is not separated).

Remark 3 Because $W$ and $X$ are seperable, we have $\mathcal{B}(P)= \mathcal{B}(W)\times  \mathcal{B}(X)$

The following is the main argument of our proof.

Let $F$ be a function on $P$ taking value in $E$, defined as follows:

$$ \begin{align} F:& W\times X & \longrightarrow & E \\ &(n,x) & \longmapsto & f_n(x)\end{align} $$

For all measurable set $B \in \mathcal{B}(\mathbb{R})$, we have:

$$ F^{-1}(B)= \bigcup_{ n \in \overline{N^*}} \left( \{n\}\times f^{-1}_n(B)  \right)$$

So $F$ is measurable, thus, its set of continuity, denoted by $Z$ for now, is measurable.

Let $U \subset X$ be the set that satisfies : $ \{\infty\} \times U = Z\cap \left( \{\infty\}\times X \right)$

Clearly, the $RHS$ of equality above is measurable. And as the Borel which is defined on $P$ is a product Borel, thus its projection $p_X: P \rightarrow X$ is measurable, which implies the measurability of $U$.

Besides, the definition of $U$ can be requoted as follows:
$$U= \left\{ x \vert \forall  ( n_k, x_{k}) \xrightarrow{ d_P} (\infty, x) \text{ we have } f_{n_k}(x_k) \rightarrow f(x) \right\} $$
I'll prove that $U=Y'$

As the condition for a point to be in $U$ is stronger than the one for $Y'$, we have the inclusion $ U \subset Y'$.

On the other hand, if $x$ is a point in $Y'$ and $(p_k)=( n_k, x_{k})$ is an abitrary sequence of points in $P$ such that $ p_k \longrightarrow (\infty, x)$ .

We can either choose a subsequence $(p_{k_m})$ that converges to the same limit and that:

either, $n_{k_1}<n_{k_2}<n_{k_3}<...$.

or $+\infty= n_{k_1}=n_{k_2}=n_{k_3}=...$

Remark Such a choice is possible due to the way we define $d_W$

In the first case, we fill that subsequence so have a new sequence $(q_i)=(i,y_i)$ defined as follows:

$y_i= x_{k_m} $ if $n_{k_m} \le i < n_{k_{m+1}}$

$y_i$ be any point if $i<n_{k_1}$

$\Rightarrow  y_i \rightarrow x$ (due to the convergence of $(p_{k_m})$)

Due to the property of $Y'$, we imply that :
$$\begin{align} & F(q_i) &\longrightarrow & F(( \infty, x))\\ \Rightarrow & F(p_{k_m}) &\longrightarrow & f(x) \end{align} $$

The second case, the limit is achieved by definition of $V$.

So, for any sequence $( F(p_k) )$ , we can find a subsequence that converges $f(x)$.
Thus $F(p_k) \longrightarrow f(x)$ (In fact, we have to take a subsequence of an arbitrary subsequence but it doesn't matter because of our freedom in choosing $(p_k)$ ) .

$\rightarrow x \in U$

$\rightarrow Y' \subset U$
Thus $Y'=U$.

Conclusion : $Y'$ is measurable

Discussion :

We see that we can even prove that $Y'$ is $G_\delta$

The assumption is rather weak because most of the spaces with which we usually deal are separable. (The separability permit us to construct a lot of structures on topological space)

The assumption is essential for our proof because we need a link between the product Borel and the Borel of product metric.

Measurability of Y

As shown above, the continuity of $F$ at a point $p=(\infty,x) \in P$ also contains the continuity of $f$ at point $x$. This hinders us from having our desired conclusion. However, with some modifications in the proof for the measurability of $Z$, we can relax that constraint.

Let :
$$ U_n = \{ p \in P : \exists \delta_p >0 s.t \forall y ,z \in B(p,\delta_p) \text{ and } p_W(y)< +\infty, p_W(z)<+\infty \text{ we have: } \\ d_X(F(y),F(z))<1/n \}$$

where $p_W: P \rightarrow W$ is the projection on $W$.

We can see that this set is open because for any $p \in U_n$ and for any $w \in B(p, \delta_p)$, $w$ is in $U_n$ with $\delta_{w}:= \delta_p- d_P(p,w)>0$.

Then, let:

$\tilde{Z}:= \bigcap_{n \ge 1} U_n$

$\tilde{U} := p_X\left( \tilde{Z}\cap (\{\infty\}\times X ) \right)$

By the same procedure as above, we can prove that $\tilde{U}=Y$.


问:
Banach spaces $X^*=Y^*$ implies $X=Y$?
Let $X$ and $Y$ be two Banach space. If $X^*$ is isometry to $Y^*$, then $X$ is isometry to $Y$? Here $^*$ means the dual space.

If one of $X$ and $Y$ is reflective,the answer is affirmative. Here is my proof. (“=” means isometry)Suppose $X$ is reflective, then $Y^*=X^*$ which is reflective, so $Y^{**}$ is reflective. The classical embedding $Y \mapsto Y^{**}$ shows $Y$ is (isometry to) a closed subspace of $Y^{**}$. Hence $Y$ is reflective by the result that any closed subspace of reflexive space is reflective. So $Y=X^{**}=Y^{**}=Y$.

For non-reflective case, I try to prove but failed. And I guess this claim is false. But It seems to be difficult to give counter examples.
Can someone give me some hint or references? Thanks!

答:
There exists a Banach space $X$ such that $X^*$ is isometrically isomorphic to $\ell¹$, but $X$ is not even isomorphic to $c_0$. In fact these spaces are actively studied in a subfield of Functional Analysis called "pre-duals of $\ell^1$".  See, for example:

[1] Bessaga, Czeslaw; Pełczyński, Aleksander, Spaces of continuous functions. IV: On isomorphical classification of spaces of continuous functions, Stud. Math. 19, 53-62 (1960). ZBL0094.30303.

[2] Benyamini, Y.; Lindenstrauss, J., A predual of (l_1) which is not isomorphic to a C(K) space, Isr. J. Math. 13, 246-254 (1972). ZBL0253.46044.

Title: Banach spaces $X^*=Y^*$ implies $X=Y$?


问:
"Conical Closure" of an Ideal?
Let $\mathbb{F}$ be an algebraically closed field and let $I\subsetneq\mathbb{F}[x_1,\ldots,x_n]$ be a radical ideal. We can send $I$ to a homogeneous radical ideal in three steps:
$$I \mapsto V(I) \mapsto \mathrm{Cone}(V(I)) \mapsto I(\mathrm{Cone}(V(I))).$$
Here $V(I)\subseteq\mathbf{F}^n$ is the zero set of the ideal $I$, $\mathrm{Cone}(V(I))\subseteq\mathbf{F}^n$ is the cone over this set, i.e., $$\mathrm{Cone}(V(I))=\{\lambda\mathbf{p}: \mathbf{p}\in V(I), \lambda\in\mathbb{F}\},$$ and $I(\mathrm{Cone}(V(I)))$ is the ideal of polynomials that vanish on this cone. (This final ideal is homogeneous since it is the ideal of a conical set and it is radical by the Nullstellensatz.) Let us define $\mathrm{Cone}(I):=I(\mathrm{Cone}(V(I)))$.

Question: How is $\mathrm{Cone}(I)$ described in terms of $I$? If $I=(f_1,\ldots,f_m)$ then is it true that $\mathrm{Cone}(I)$ is generated by the homogeneous parts of the polynomials $f_i$?

答:
The correspondence between ideals and closed subsets reverses inclusion. Therefore, the expected ideal should not be the ideal $I^+$ generated by homogeneous parts of generators of $I$, as such ideal would contain $I$, but rather
$$\mathrm{Cone}(I)=I^{-}:=(\{f \in I\; |\;f \text{ is homogeneous}\}) \subseteq I.$$
Indeed, as $I^-$ is the biggest homogeneous ideal contained in $I$, $V(I^-)$ is the smallest conical subset containing $V(I)$, which is $\mathrm{Cone}(V(I))$.

(If you prefer to talk strictly about varieties, i.e. $I$ is assumed to be a radical ideal, note that then $I^-$ is radical too, so it works out.)

Title: "Conical Closure" of an Ideal?


问:
Uniqueness of Seifert graphs
If we make the bands and disks of a Seifert surface really small and really thin the surface collapses to a graph. It is called a Seifert graph.

If it is not a directed and weighted graph, can we say that for every unique Seifert graph there is a Seifert surface and vice versa?

If it is not a directed but weighted graph, can we say that for every unique Seifert graph there is a Seifert surface and vice versa?

If it is a both directed and weighted graph, can we say that for every unique Seifert graph there is a Seifert surface and vice versa?

答: With regards to your first question, I will assume that you are taking your Seifert graph to be embedded in $\mathbb R^3$. This can correspond to multiple Seifert surfaces by taking any Seifert surface with this graph and then adding lots of full twists into one of the bands. I am not sure what you expect the weights and directions to mean for the Seifert surface, so I am not sure how to respond to questions 2 and 3.</p>

Title: Uniqueness of Seifert graphs


问:
The product of which 6 primes is 201-times larger than their sum?
I have been able to find the answer via exhaustion. I defined $a\le b\le c\le d\le e\le f$ as the primes and solved an inequality for each variable. The answer is $2,2,3,3,7,67$. However, the number of cases to solve got quite large and I was wondering if there is any neat solution to this problem.

答:
Start with the observation of J. W. Tanner that two of the primes must be $3,67$. Note that if all of the primes are odd, their product is odd, but their sum is even, so we can conclude that one of the primes must be $2$.

Restate the problem as $2\cdot 3\cdot 67\cdot p_1p_2p_3=201(p_1+p_2+p_3+72)$ from which $2\cdot p_1p_2p_3=(p_1+p_2+p_3+72)$. Once again, if all unidentified primes are odd then LHS is even but RHS is odd, so there is another factor of $2$ present. This yields $4\cdot p_1p_2=(p_1+p_2+74)$. We can quickly convince ourselves that $p_1\ne p_2$ and $p_1\equiv p_2 \bmod 4$.

The smallest pair of primes that are $\equiv 1 \bmod 4$ are $5,13$, but $4\cdot 5\cdot 13>5+13+74$ and looking at larger primes only makes the inequality greater. The smallest pair of primes that are $\equiv 3 \bmod 4$ are $3,7$, and this provides a solution. The next larger pair of primes that are $\equiv 3 \bmod 4$ are $7,11$, and also in this case, $4\cdot 7\cdot 11>7+11+74$ and looking at larger primes only makes the inequality greater. So the remaining primes are $3,7$.

The complete list is as OP found: $2,2,3,3,7,67$.


问:
If σ does not have any relation symbols, then h is a σ-embedding if and only if it is one-to-one.
Let $A, B$ be $σ$-structures and let $h:A→B$ be a $σ$-homomorphism. Prove that if $σ$ does not have any relation symbols, then $h$ is a $σ$-embedding if and only if it is one-to-one.

Definition 1: A $σ$-embedding is a $σ$-homomorphism  for which $h$ is a $σ$-isomorphism from $A$ to $B’$ with $B’$ being the substructure of $B$ formed by $h[A]$.

Definition 2: A $σ$-isomorphism is a $σ$-homomorphism that has a 2-sided inverse $h^{-1}$, which is also a $σ$-homomorphism.

From those definitions, apparently the rest of the proof is trivial. However I'm not sure... First of all I'm confused about why it only needs to be one-to-one, why not onto? Is it because we excluded the relation symbols for which the property is only one sided?

答:
The map $h$ need not be onto for this problem as the definition of an embedding is that it is an isomorphism onto its image $h[A]$. The size of the "ambient" structure $B$ doesn't matter - only how $A$ relates to $h[A]$.

One side of this proof is easier. Suppose that $h$ is an embedding. Then it is an isomorphism onto $h[A]$, i.e. there is an inverse map $h^{-1}: h[A] \longrightarrow A$. Hence, if $h(a_1) = h(a_2)$ then $a_1=h^{-1}(h(a_1))=h^{-1}(h(a_2))=a_2$ and $h$ is one to one. Note here that we didn't need $h^{-1}$ to be a $\sigma$-homomorphism; this works equally well if we forget the $\sigma$-structure and consider these only as sets.

For the converse, suppose that $h$ is one to one. Then I claim that there is an inverse $h^{-1}: h[A] \longrightarrow A$. Indeed, it is not hard to see that there is a set map $h^{-1}$. For any $b \in h[A]$, there is a unique element $a \in A$ such that $h(a) = b$. This is precisely the definition of $h$ being one to one. Hence, we can simply define $h^{-1}(b) = a$ for $b \in h[A]$. We are not done yet, however, as we need to show that $h^{-1}$ is in fact a $\sigma$-homomorphism. And here is where we need to use the assumption that $\sigma$ has no relation symbols.

This assumption tells us that $\sigma$ consists solely of function symbols, so indeed let's let $f$ be an $n$-ary function symbol. Consider $h^{-1}(f(b_1, \dots, b_n))$ for $b_i \in h[A]$. Write $h(a_i) = b_i$ for all $i$. Then observe that
$$
\begin{align*}
h(h^{-1}(f(b_1, \dots, b_n))) &= f(b_1, \dots, b_n)\\
&= f(h(a_1), \dots, f(h(a_n)))\\
&=h(f(a_1, \dots, a_n)),
\end{align*}
$$
as we assumed that $h$ was a $\sigma$-homomorphism. Furthermore, as $h$ is one to one, this equation tells us that $h^{-1}(f(b_1, \dots b_n)) = f(a_1, \dots, a_n) = f(h^{-1}(b_1), \dots, h^{-1}(b_n))$. Thus, $h^{-1}$ is a $\sigma$-homomorphism and $h$ is a $\sigma$-embedding. This notation is a bit off in the case where $f$ is a $0$-ary function, i.e. a constant, but the same proof holds.

By the way, I think it's always good practice to try to break the conclusion by breaking the assumptions. That is, if we had relation symbols in $\sigma$ could we find counterexamples to this claim? Here's an example: take $\sigma$ to contain a single binary predicate $P$. Let $A$ and $B$ both equal $\mathbb N$ as sets. For the $\sigma$-structure $A$, we say that $P_A(a_1, a_2)$ iff $a_1 < a_2$. For the $\sigma$-structure $B$ we say that $P_B(b_1, b_2)$ is always true. Then the identity map $id: A \longrightarrow B$ is a $\sigma$-homomorphism and it is one to one, but it is not an embedding. Indeed, the inverse (which is still the identity) does not preserve $P$, as $P_B(1, 0)$ is true but $P_A(1, 0)$ is not.


问:
function in $ C^1$ has same sign in two adjacent intervals then $x_0$ must be at least a double zero
I find the following claim in a book: Suppose $f\in C^1[a,b]$ and $x_0\in (a,b)$ is a zero of $f$, i.e., $f(x_0)=0$. Assume that $f(x)>0$ for all $x\in[a,b]\setminus \{x_0\}$. Then $x_0$ must be at least a double zero of $f$.

My try: If $x_0$ is a single zero, then $f(x)=(x-x_0)g(x)$ with $g(x_0)\neq 0$. Since $g$ is also continuous then either $g(x)>0$ or $g(x)<0$ in a small neighbor of $x_0$.  If $g(x)>0$ in a neighbor of $x_0$. Then $f(x)=(x-x_0)g(x)<0$ when $x<x_0$ which is a contradiction.

My question is why do we need the assumption $f\in C^1[a,b]$ ? It seems that $f\in C[a,b]$ is enough to let the claim be true.

答:
I'm not sure about the context, but what is exactly meant here by double zero? The function

$$f(x) = \begin{cases}e^{-\frac{1}{x^2}} & x \neq 0 \\ 0 & x = 0 \end{cases}, x \in [-1, 1]$$

has only $1$ zero at the origin and is positive everywhere else. What is meant here exactly by the order of the $0$ at the origin here? The first non-zero derivative at $x_0$?

The reason we're dealing with $C^{1}$ is to make sure that the notion derivative is even defined for the function. Otherwise, the function $g(x)=|x|, x \in [-1, 1]$ would also only have one zero at $x=0$ but it is not even differentiable at that point.

The way I would prove this question is to assume the opposite(as you did) and conclude that $\exists \epsilon > 0$ where $f(x_0 - \epsilon) < 0$ if the first derivative were non-zero, which would contradict the initial assumption that the function is always non-negative.

edit: By the opposite I mean "suppose the function has a non-zero derivative at $x_0$".


答:
Say we take two $\sigma$-structures $A$ and $B$ and $h\colon A\to B$ an injective $\sigma$-homomorphism.

Then $h(A)$ contains all constants in your signature and is closed under functions in your signature: if $x\in h(A)$, then $x=h(a)$ for some $a\in A$. If $f$ is a function symbol in your signature, then
$$f^B(x)=f^B(h(a))=h(f^A(a))$$
also belongs to $h(A)$.

So how would we go on to prove that the inverse $h^{-1}\colon h(A)\to A$ is a morphism? Well, of course it preserves the interpretation of constants. For a $1$-ary function symbol $f$, we have
$$h\circ f^A=f^B\circ h$$
so composing with $h^{-1}$ both on the left and on the right implies that
$$f^A\circ h^{-1}=h^{-1}\circ f^B$$
which means that $h^{-1}$ is also a $\sigma$-homomorphism. A similar argument holds for general $n$-ary function symbols.

But the problem is that the $\sigma$-homomorphism condition on a relation symbol $R$ (which I assume to be a binary relation just for simplicity) is that
$$(h\times h)(R^A)\subseteq R^B$$,
which only translates to
$$R^A\subseteq (h^{-1}\times h^{-1})(R^B).$$
The inclusion is on the opposite direction than the one that would be necessary for $h^{-1}$ to preserve $R$!!! That's where things could go awry.

For an specific example, take the signature $\sigma=(R)$ with only one binary relation symbols. Take your favourite set $X$ with at least two elements, and consider the two interpretations of $\sigma$ as follows:
$$A=(X,\mathrm{eq}),\qquad B=(X,\mathcal{P}(X)),$$
where "$\mathrm{eq}$" is the equality relation on $X$ and $\mathcal{P}(X)$ is the  finest relation on $X$ (everyone is related to everyone). Then the identity $A\to B$ is obviously a $\sigma$-homomorphism, but it is not a $\sigma$-isomorphism.

Title: If σ does not have any relation symbols, then h is a σ-embedding if and only if it is one-to-one.

Body:
Let $A, B$ be $σ$-structures and let $h:A→B$ be a $σ$-homomorphism. Prove that if $σ$ does not have any relation symbols, then $h$ is a $σ$-embedding if and only if it is one-to-one.

Definition 1: A $σ$-embedding is a $σ$-homomorphism  for which $h$ is a $σ$-isomorphism from $A$ to $B’$ with $B’$ being the substructure of $B$ formed by $h[A]$.

Definition 2: A $σ$-isomorphism is a $σ$-homomorphism that has a 2-sided inverse $h^{-1}$, which is also a $σ$-homomorphism.

From those definitions, apparently the rest of the proof is trivial. However I'm not sure... First of all I'm confused about why it only needs to be one-to-one, why not onto? Is it because we excluded the relation symbols for which the property is only one sided?


问:
Proving the complex function $f(n) = n^2 + 2$ is onto
Here's the question,

For the function:

$$ f: \mathbf{C} \to \mathbf{C} $$
$$ f(n) = n^2 +2 $$
Is $ f $ onto? Justify your answer.

So this was my approach
$$ a+bi = (c+di)^2 +2 $$
$$ a+bi = c^2 -d^2 +2cdi + 2 $$
$$ a = c^2 -d ^2 + 2 $$
$$ b = 2cd $$

At this point, I know I must prove that for any $a$ and $b$ there exists a $c$ and $d$ but I do not know how.

答:
As noted in the comments, you probably meant to write $\mathbf C$ rather than $\mathbf Z$. It is decidedly not onto for $\mathbf Z$, for instance, $-10$ is not in the image. However, it is onto for $\mathbf C$. I think the easiest way to understand this is with the polar form of a complex number.

For $w \in \mathbf C$ we can write $w = r e^{i \theta}$ where $r, \theta \in \mathbb R$ and $r > 0$. $r$ is the magnitude of $w$ and $\theta$ is the angle it makes with the positive real axis. If you're not comfortable with complex exponentials, just define $e^{i \theta} = \cos(\theta) + i \sin(\theta)$. Now, for $z \in \mathbf C$, write $z - 2 = r e^{i \theta}$. Then $(\sqrt{r} e^{i \theta/2})^2 = re^{i \theta} = z - 2$, so $f(\sqrt{r} e^{i \theta/2}) = z$ so $f$ is onto. Geometrically, $\sqrt{r} e^{i \theta/2}$ is the resulting of square rooting the magnitude and cutting the angle with the positive real axis in half.


问:
Does anyone know what this theorem is?
During a computational complexities class, the professor stated the following theorem, without naming it.

Let $G$ be a graph, with adjacency matrix $A$, and clique-number $k$. i.e. the largest clique in $G$ has size $k$, then

$$ 1-\frac{1}{k} = \max\left\{\sum_{1\leq i,j\leq n}A_{i,j}p_ip_j \ \Big|\ \forall i\ p_i\geq 0,\sum_i p_i =1\right\}$$

Does anyone know what theorem this is?

答:
This is related to Turán's theorem, which says that an $n$-vertex graph with clique number $k$ has at most $(1 - \frac1k)\frac{n^2}{2}$ edges.

Specifically, Turán's theorem follows from setting $p = (\frac1n, \frac1n, \dots, \frac1n)$ in the result you're quoting. In that case, the sum $\sum_{i=1}^n \sum_{j=1}^n A_{ij} p_i p_j$ contributes $\frac{2}{n^2}$ for every edge, so it is $\frac{2m}{n^2}$ for an $m$-edge graph. If this sum can be at most $1 - \frac1k$, then we conclude $m \le (1 - \frac1k) \frac{n^2}{2}$.

The general result that $1 - \frac1k$ is the maximum over all nonnegative vectors $p$ with sum $1$ doesn't have a name. We usually prove it in the course of proving Turán's theorem;
see this question on MSE for some discussion. This can be done in an ad-hoc way or using Lagrange multipliers.

Title: Does anyone know what this theorem is?


问:
Why is the gcd of $61+35\sqrt{3} $ and $170+32\sqrt{3}$ is $19 + 11\sqrt{3}$?
As a completion or correction to the solution of my question here:

Find gcd of $a = 170 + 32\sqrt{3}$ and $b = 61 + 35\sqrt{3}.$ Then find $f,g \in \mathbb{Z}[\sqrt{3}]$ such that $af + bg = d$ using norm function.

I was told by my professor that:  the gcd of $61+35\sqrt{3} $ and $170+32\sqrt{3}$ is $19 + 11\sqrt{3}$ and that we should not use the norm function and that we should use the ordinary procedure for calculating the gcd to get it. And my professor added that you will even get that  $$19 + 11\sqrt{3} = (61 + 35\sqrt3) (718 - 413\sqrt{413}) +(170+32\sqrt{3})(-3 + \sqrt{3}),$$ As I mentioned in the above post I do not know how to complete after this step:
$$\frac{170 + 32 \sqrt{3}}{61 + 35 \sqrt{3}} = \frac{3505}{23} - \frac{1999 \sqrt{3}}{23},$$ specifically could anyone show me what should be the remainder in my previous line please?

答:
Division with remainder for number of these forms should be done as follows:

$$\frac{170 + 32 \sqrt{3}}{61 + 35 \sqrt{3}} = \frac{3505}{23} - \frac{1999 \sqrt{3}}{23}$$

This would be the fraction. Now to get the integral quotient, take the closest integer for the coefficients of the rational and irrational par

$$\frac{3505}{23} =152 + \frac{9}{23} \\
\frac{1999}{23} = 86 + \frac{21}{23} = 87 -\frac{2}{23}$$

Therefore, the integral quotient is
$$152 - 87 \sqrt{3}$$
and so
$$170 + 32 \sqrt{3} = (61 + 35 \sqrt{3})(152- 87 \sqrt{3})+ 33 + 19 \sqrt{3}$$

So $33 + 19 \sqrt{3}$ is the remainder.

Now divide $61 + 35 \sqrt{3}$ by $33 + 19 \sqrt{3}$. We have
$$\frac{61 + 35 \sqrt{3}}{33 + 19 \sqrt{3}} = 3 - \frac{2}{3}\sqrt{3}$$.  Therefore the integral quotient is
$3 - \sqrt{3}$.  The remainder is
$$(33 + 19 \sqrt{3}) \cdot\frac{\sqrt{3}}{3}=19 + 11 \sqrt{3}$$

Notice that the last remainder $19 + 11 \sqrt{3}$ divides the last divider $33 + 19 \sqrt{3}$. Therefore, we stop here, the $\gcd$ is $19 + 11 \sqrt{3}$.

Note:  the division process is very similar to the usual division for integers.


问:
Algorithm for finding many gcd(a,b) over a range of a?
I have a procedure where I calculate gcd(a,b) many times, and it can be computationally expensive. I am trying to instead create an array containing gcd(a,b) values by using a factor sieve of some kind, but I don't know a good way to do it.

For instance if I am looking at $a=24$ then I'd be looking for a way to calculate many gcd(24,b) values in one loop or something similar.

I hope I am making sense!

Body:
In APL (NARS2000, free) :

      size ← 15

      2 2⍴(0)((1,size)⍴⍳size)((size,1)⍴⍳size)(gcd_sieve size)
  0   1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

  1   1 1 1 1 1 1 1 1 1  1  1  1  1  1  1
  2   1 2 1 2 1 2 1 2 1  2  1  2  1  2  1
  3   1 1 3 1 1 3 1 1 3  1  1  3  1  1  3
  4   1 2 1 4 1 2 1 4 1  2  1  4  1  2  1
  5   1 1 1 1 5 1 1 1 1  5  1  1  1  1  5
  6   1 2 3 2 1 6 1 2 3  2  1  6  1  2  3
  7   1 1 1 1 1 1 7 1 1  1  1  1  1  7  1
  8   1 2 1 4 1 2 1 8 1  2  1  4  1  2  1
  9   1 1 3 1 1 3 1 1 9  1  1  3  1  1  3
 10   1 2 1 2 5 2 1 2 1 10  1  2  1  2  5
 11   1 1 1 1 1 1 1 1 1  1 11  1  1  1  1
 12   1 2 3 4 1 6 1 4 3  2  1 12  1  2  3
 13   1 1 1 1 1 1 1 1 1  1  1  1 13  1  1
 14   1 2 1 2 1 2 7 2 1  2  1  2  1 14  1
 15   1 1 3 1 5 3 1 1 3  5  1  3  1  1 15

And copy + paste this code :

      ⎕cr'gcd_sieve'
gcdmatrix ← gcd_sieve size;i;temp
gcdmatrix ← (size,size)⍴1
:for i :in 1+⍳¯1+⌊size÷2
  gcdmatrix[i×temp∘.,temp←⍳⌊size÷i] ← i
:endfor
gcdmatrix[⊂[2]((⌈size÷2),2)⍴2/temp] ← temp ← (⌊size÷2)+⍳⌈size÷2

Example

      gcd_array ← gcd_sieve 15

      gcd_array[8;12]
4
      gcd_array[12;6]
6
      gcd_array[7;5]
1

Have a nice day.


问:
The rank of torsion free nilpotent group
Let $N$ be a torsion free finitely generated nilpotent group. We define the rank of $N$ to be the natural number $m$ such that there exists a finite subnormal series
$$
N=N_m \vartriangleright N_{m-1} \vartriangleright \cdots \vartriangleright N_1 \vartriangleright N_0=\{e\}
$$
such that $N_j/N_{j-1} \cong \mathbb Z$ for each $1 \le j \le m$.

Can we show the rank of $N$ is exactly the minimal number of generating elements of $N$?

答:
I don’t think so. Take the free nilpotent group of class $2$ and rank $2$, $G$, which consists of elements of the form $x^ay^b[y,x]^c$, with multiplication given by
$(x^ay^b[y,x]^c)(x^ry^s[y,x]^t)= x^{a+r}y^{b+s}[y,x]^{c+t+rb}.$
I claim the rank as you define it is $3$, even though it is $2$-generated.

Indeed, if $G/N$ is isomorphic to $\mathbb{Z}$, then $N$ contains the commutator subgroup. Since $G^{\rm ab}$ is free abelian of rank $2$, then $N$ properly contains $[G,G]$, and is itself free abelian of rank $2$. Thus, it has rank $2$, so we need at least two more steps to drop down to the trivial group. This gives a “rank” as per your definition of $3$.

Replacing class $2$ with class $c$, you should be able to get arbitrarily large “rank” while still having the group be $2$-generated.


问:
Use definition of differentiability to prove $g:\mathbb{R}^2 \to \mathbb{R}:(x,y) \to x^2 y^2$ is differentiable at any point
By using the definition of a differentiable function prove
$g:\mathbb{R}^2 \to \mathbb{R}:(x,y) \to x^2 y^2$ is differentiable at any point $(x_0, y_0) \in \mathbb{R}^2$ and check this answer by computing Jacobi Matrix = $g'$ at $(x_0, y_0)$.

Here, I know the definition of a differentiable function and to set up the question using the formula that needs to be satisfied
$f(x,y)-f(x_0,y_0)= L((x,y)-(x_0,y_0)) + r((x,y),(x_0,y_0))$.
I believe this is the right setup but I haven't done a question before in $\mathbb{R}^2$ and when I do my calculations I get stuck at working out what is L and r despite in the second part I can work out the jacobi matrix $g'=(2xy^2\ \ 2yx^2)$ and so $g'(x_0,y_0) =(2x_0y_0^2\ \ 2y_0x_0^2)$.

Thanks in advance as I have really bit a brick wall with this one :)

答: First, note that every linear map $L:\mathbb{R}^2\rightarrow\mathbb{R}$ has the form $L(x,y)=ax+by$, where $a=L(1,0),\,b=L(0,1)$. Now, if $f$ is differentiable at $(x_0,y_0)$, show that $a=\frac{\partial f}{\partial x}(x_0,y_0),\,b=\frac{\partial f}{\partial y}(x_0,y_0).$ Hence, in your particular case, you get $L$ and $r$ will be $f(x,y)-f(x_0,y_0)-L(x-x_0,y-y_0)$.


问:
Algebra proof within relation problem
I need to show an equivalence relation on $ℕ$ defined as follows:

$x$ ~ $y$ ⇔ $x \times y$ is a square

Reflexivity: $x \times x$ is a square, so $x$ ~ $x$, so ~ is reflexive

Symmetry: $x$ ~ $y$ implies $y$ ~ $x$, $x\times y$ is a square, so is $y \times x$

But for transitivity I need to show that if $xy$ and $yz$ are both squares, then $xz$ is also a square.

So I did this:

$xy = a^2$ and  $yz = b^2$, then $xz = \frac{a^2\times b^2}{y^2}$ and $xz = (\frac{ab}{y})^2$

But I don't think this works because there is a fraction. How else can I show that $xz$ is a square?

答:
$$ xz y^2 = a^2 b^2 $$

So we know $a^2 b^2$ is divisible by $y^2$. In other words, $\dfrac{ab}{y}$ is an integer. So your solution is correct.

Title: Algebra proof within relation problem


答:
Let $x = \prod_{i=1}^n p_i^{c_i}$, $y = \prod_{j=1}^n p_j^{d_j}$ and $z = \prod_{k=1}^n p_k^{f_k}$, where $P = \{p_i \mid 1\leq i \leq n\}$ is a set of primes containing the primes that divide any of $x$, $y$, or $z$.  Then, since $xy$ and $yz$ are squares, writing the factorizations of $xy$ and $yz$ into primes in $P$, we obtain the relations \begin{align*}
c_i + d_i &\cong 0 \pmod{2}  \\
d_i + f_i &\cong 0 \pmod{2}
\end{align*}
for each $i \in [1,n]$.  Then \begin{align*}
(c_i + d_i) + (d_i + f_i) &\cong 0 \pmod{2}  \\
c_i + 2d_i + f_i &\cong 0 \pmod{2}  \\
c_i + f_i &\cong 0 \pmod{2}  \text{,}
\end{align*}
which says $xz$ is a square.

Title: Algebra proof within relation problem

Body:
I need to show an equivalence relation on $ℕ$ defined as follows:

$x$ ~ $y$ ⇔ $x \times y$ is a square

Reflexivity: $x \times x$ is a square, so $x$ ~ $x$, so ~ is reflexive

Symmetry: $x$ ~ $y$ implies $y$ ~ $x$, $x\times y$ is a square, so is $y \times x$

But for transitivity I need to show that if $xy$ and $yz$ are both squares, then $xz$ is also a square.

So I did this:

$xy = a^2$ and  $yz = b^2$, then $xz = \frac{a^2\times b^2}{y^2}$ and $xz = (\frac{ab}{y})^2$

But I don't think this works because there is a fraction. How else can I show that $xz$ is a square?


问:
Stability under supremum of sets of social choice function with single peaked preferences
Here is a question emerging from reading Moulin, H. (1980). On strategy-proofness and single peakedness. Public Choice, 35(4), 437–455.

The setting is as follows:

A non-empty finite set of individuals $N$ indexed by $i=1,\dots,n$.

The set of alternatives is the real line $\mathbb{R}$.

Agents are endowed with single-peaked preferences represented by some utility function $u\in S$ over $\mathbb{R}$, where $S$ the set of utility representation of single-peakened preferences is defined as follows:

$u\in S$ if and only if there exists an alternative $x(u)$, the "peak" of $u$ such that for all $x,y\in \mathbb{R}$

$x\leq y < a \Rightarrow u(x)\leq u(y) < u(a)$

$a <x \leq y \Rightarrow u(a) > u(x) \geq u(y)$.

A social choice function $C:\mathbb{R}^n\rightarrow \mathbb{R}$ associates every profile of peakes $(x(u_1),\dots,x(u_n))$ with a chosen alternative.

A social choice function is said to be strategy-proof if for all $i\in N$, $u_i\Big(c\big(x(u_i),x(u_{-i})\big)\Big) \geq u_i\Big(c\big(x(u_i'),x(u_{-i}\big)\Big)$, for every $u_i' \in S$, where $x(u_{-i})$ is the $(n-1)$-tuple of peaks announced by agents other than $i$.

Here is my question. At the end of the article, Moulin argues that

"$\Sigma_n$, the set of strategy-proof voting schemes [i.e. social choice functions], is stable by the operation of supremum and infimum (not necessarily finite)."

Could anyone help me understanding what stability under supremum and infimum operation actually means (in this context and in general)? I've googled the concept but could not find anything helpful. Two questions in particular:

Does "stability of a set under some operation" simply means that the set of is unchanged when the operation is applied to it?

How do you apply the supremum and infimum operation to functions? I understand that functions are sets but how does the supremum/infimum operator apply to set which are cartesian products of other sets?

答:
To complement Michael Greinecker's helpful comment (and to be sure that I understand correctly):

A property shared by a family of functions is stable under some operation if applying this operation yields a family of functions which also have this property

A more detailed definition of pointwise suprema can be found at http://www.proofwiki.org/wiki/Definition:Pointwise_Supremum


问:
The evaluation map $\phi_a : \mathbb{R}[x] \to \mathbb{R}$, such that $\phi_a(f) = f(a)$ for some fixed $a \in \mathbb{R}$, is a ring homomorphism.

The evaluation map $\phi_a : \mathbb{R}[x] \to \mathbb{R}$, such that
$\phi_a(f) = f(a)$ for some fixed $a \in \mathbb{R}$, is a ring
homomorphism. Therfore it's kernel is an ideal in $\mathbb{R}[x]$. All
ideals in $\mathbb{R}[x]$ are principal ideals. Find the polynomial
which generates this ideal.

SOL: $f(x) = (x-a)$ is the lowest degree non-zero polynomial which maps to $0$ hence it generates the entire ideal. Any polynomial in $\mathbb{R}[x]$ which vanishes at $a$ has a linear term in factor equal to $f(x)$.

I don't understand the question.

Q1. Generate what ideal? Find the polynomial that generates which ideal is it asking?
All

Q2. Why does it say that ideals in \mathbb{R}[x] are principal ideals?

Q3. In solution, why does $(x-a)$ generate the entire ideal? What is entire ideal? All ideals in $\mathbb{R}[x]?$

答:
Q1: kernels of ring homomorphisms are ideals. The question discusses $\ker(\phi_a)$. That is this ideal. I would rather state "find a polynomial which generates this ideal", since the polynomial is not unique, but they are all the same up to multiplication by some non-zero real.

Q2: I don't like the way the question is stated. I assume that the question is just giving out a fact, which can be assumed without proof (and which is actually true): Every ideal of $\mathbb{R}[x]$ is generated by some single polynomial (i.e., it is of the form $\langle p(x)\rangle=\left\{p(x)q(x):q(x)\in\mathbb{R}[x]\right\}$ for some polynomial $p(x)$).

Q3: I assume you have the solution at hand (which we don't). The polynomial $(x-a)$ indeed generated the ideal $\ker(\phi_a)$ (this is related to question Q1), that is, every polynomial in $\ker(\phi_a)$ is of the form $(x-a)q(x)$ for some polynomial $q(x)$, which is to say that $a$ is a root of any polynomial in $\ker(\phi_a)$ (is this obvious?).


答:
Recall that the kernel of a ring homomorphism $f\colon R \to S$ is an ideal in $R$. You are asked for the polynomial generating $\ker \phi_a$.

Univariate polynomial rings $k[x]$ are PIDs. They are Euclidean domains (you can carry out division with remainder), hence PIDs.

The claim is the ideal generated by $x-a$, which we denote $(x-a)$, equals $\ker \phi_a$. So there are two things to show: first that elements in $(x-a)$ map to zero under $\phi_a$, and second that all elements of $\ker \phi_a$ are of this form. That is what they mean when they say $(x-a)$ generates the entire ideal ($\ker \phi_a$). As for why this is the case: again, carry out division with remainder. Do you recall the factor theorem?

Body:
The evaluation map $\phi_a : \mathbb{R}[x] \to \mathbb{R}$, such that
$\phi_a(f) = f(a)$ for some fixed $a \in \mathbb{R}$, is a ring
homomorphism. Therfore it's kernel is an ideal in $\mathbb{R}[x]$. All
ideals in $\mathbb{R}[x]$ are principal ideals. Find the polynomial
which generates this ideal.

SOL: $f(x) = (x-a)$ is the lowest degree non-zero polynomial which maps to $0$ hence it generates the entire ideal. Any polynomial in $\mathbb{R}[x]$ which vanishes at $a$ has a linear term in factor equal to $f(x)$.

I don't understand the question.

Q1. Generate what ideal? Find the polynomial that generates which ideal is it asking?
All

Q2. Why does it say that ideals in \mathbb{R}[x] are principal ideals?

Q3. In solution, why does $(x-a)$ generate the entire ideal? What is entire ideal? All ideals in $\mathbb{R}[x]?$


问:
Proof for Perfect $n^{th}$ power.
A positive integer, $k$, is a perfect nth power if, there exist  a positive integer a such that $k=a^n$, where $n$ is also a positive integer.

Question is:

How can I prove, for all positive integers $k, m, n$ (where $m$ and $n$ are coprime), if $k$ is a perfect $m^{th}$ power and a perfect $n^{th}$ power, then it is a perfect $mn^{th}$ power. This looks a lot like perfect squares, but it is more than perfect squares. So for example $16 = 2^4 = 4^2$

I'm not sure how to do this. I tried some patterns, is it possible to use gcd? I know that if $gcd(a, b)=1$, then $gcd(a^n, b^n)=1$. But I don't know how this can help either.

答:
Hint: $(m,n)=1 \Rightarrow \exists x, y \in \mathbb{N}, mx-ny=1.$

Can you start from here?

 If $k=a^m=b^n$, then $a=a^{mx-ny}=\left(\frac{b^x}{a^y}\right)^n.$

Title: Proof for Perfect $n^{th}$ power.


问:
Divisibility of the function $\,f(n)=2n^2-1$
Let's consider the following arithmetic function:
$$f(n)=2n^2-1,\;\;\;\;\;\;n\in \mathbb N.$$
Prove (or eventually disprove) that either $\,f(n)\,$ is prime or, otherwise, a natural $\,1\lt m\lt n\,$ exists s.t. $\,f(m)\,|\,f(n)$.

Many thanks.

答:
Disprove: $n = 29, 32, 53, 55, 57, 60, 67,\dots$ are all counterexamples to the assertion.

E.g. for $n = 29$: $f(n) = 1681 = 41^2$.

If there exists $1 < m < n$ such that $f(m) \mid f(n)$, then necessarily $f(m) = 41$, and hence $m^2 = 21$, which is impossible.


答:
If you know the fundamental theorem of arithmetic, then you should be able to show the following lemma:

A positive integer $k$ is a perfect $n$-th power if and only if for every prime number $p$, the exponent of $p$ appearing in the factorization of $k$ is a multiple of $n$.

Now if $k$ is a perfect $m$-th power and a perfect $n$-th power, then by the lemma, for every prime $p$, the exponent of $p$ appearing in the factorization of $k$ is simultanously a multiple of $m$ and a multiple of $n$.

Since $m$ and $n$ are coprime, that exponent is a multiple of $mn$. This being true for every $p$, using the lemma again tells us that $k$ is a perfect $mn$-th power.

Title: Proof for Perfect $n^{th}$ power.

Body:
A positive integer, $k$, is a perfect nth power if, there exist  a positive integer a such that $k=a^n$, where $n$ is also a positive integer.

Question is:

How can I prove, for all positive integers $k, m, n$ (where $m$ and $n$ are coprime), if $k$ is a perfect $m^{th}$ power and a perfect $n^{th}$ power, then it is a perfect $mn^{th}$ power. This looks a lot like perfect squares, but it is more than perfect squares. So for example $16 = 2^4 = 4^2$

I'm not sure how to do this. I tried some patterns, is it possible to use gcd? I know that if $gcd(a, b)=1$, then $gcd(a^n, b^n)=1$. But I don't know how this can help either.


问:
Can extending a finite ground field make modules isomorphic?
$\def\Hom{\mathrm{Hom}}$Let $k$ be a field, $A$ a $k$-algebra and let $M$ and $N$ be $A$-modules, finite dimensional over $k$. Let $K$ be an extension of $k$, so $A \otimes K$ is a $K$-algebra and $M \otimes K$ and $N \otimes K$ are $A \otimes K$ modules.

I would like to say that, if $M \otimes K \cong N \otimes K$, then $M \cong N$.

When $k$ is infinite, there is a very easy proof. The $K$-vector space $\Hom_{A\otimes K}(M\otimes K, N \otimes K)$ is simply $\Hom_{A}(M,N) \otimes K$. Saying $M \cong N$ means that there is a matrix in $\Hom_{A}(M, N)$ with nonzero determinant. If the polynomial $\det( \ )$ is nonzero somewhere after extending scalars, then it is already nonzero over $k$.

But, over a finite field, this argument is broken. If $M$ and $N$ were dimension $3$ over $k = \mathbb{F}_2$, and $\Hom(M,N)$ consisted of those matrices of the form $\left( \begin{smallmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & x-y \end{smallmatrix} \right)$, then this matrix is singular for any $(x,y) \in k^2$, but in extension fields it can be invertible.

Nonetheless, I think I have an argument that the statement is true for finite fields as well. Has anyone seen this statement before? Is there an easy proof I missed?

答: This question was answered in comments: This is called the Noether-Deuring theorem and a proof can be found at the MO link.</p>


问:
Distance between the curves from a particular point
The distance between the points $P(u,v)$ and the curve $x^2+4x+y^2=0$ is the same as the distance between the points  $P(u,v)$ and $M(2,0)$. If $u$ and $v$ satisfy the relation $u^2-\frac{v^2}{q}=1$, then $q$ is greater than or equal to

(A) 1

(B) 2

(C) 3

(4) 4

My approach is as follow Curve $x^2+4x+y^2$ represent a circle with centre $(-2,0)$ and radius $2$.

The parametric equation of the circle is $-2cos\theta+2$ and $2sin\theta$

$D^2=(u-2)^2+v^2$

Also $D^2=(u+2cos\theta-2)^2+(v-2sin\theta)^2$

How we will proceed from here

答: The distance between the circle and $P(u,v)$ is the distance between $P$ and the centre minus the radius. So we have$$\left|\sqrt{(u+2)^2+v^2}-2\right|=\sqrt{(u-2)^2+v^2}$$Squaring,$$\begin{align*}&(u+2)^2+v^2+4-4\sqrt{(u+2)^2+v^2}=(u-2)^2+v^2\\\implies&2u+1=\sqrt{(u+2)^2+v^2}\\\implies&4u^2+4u+1=(u+2)^2+v^2\\\implies&u^2-v^2/3=1.\end{align*}$$


问:
$z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . Show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements.
As in the title: $z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . I need to show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements. I know how to do it "non creative" way. So, I transform both complex numbers into trigonometric form and divide:
$$\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}} = \frac{2[\cos(\frac{7}{4}\pi)+\sin(\frac{7}{4}\pi)i]}{2[\cos(\frac{5}{6}\pi)+\sin(\frac{5}{6}\pi)i]}= \cos \left(\frac{11}{12}\pi \right)+\sin \left( \frac{11}{12}\pi \right)i$$

Now, I see that this is $165^{\circ}$, and that module $= 1$. The problem is, that the only way I see to find the number of elements of the X set is to calculate degrees:
$$165^{\circ}, 330^{\circ}, 135^{\circ}, 300^{\circ}, 105^{\circ}, 270^{\circ}, 75^{\circ}, 240^{\circ}, 45^{\circ}, 210^{\circ}, 15^{\circ}, 180^{\circ}, 345^{\circ}, 150^{\circ}, 315,^{\circ} 120^{\circ}, 285^{\circ}, 90^{\circ}...$$

So I feel like there has to be some more effective way to solve that problem and here comes my question. Do you know any?

答:
Hint

Calculate separatelly
$$\frac{\sqrt{2}(1-i)}{2}$$
and
$$\frac{1−i\sqrt{3}}2$$


问:
Proving symmetry and transitivity of $=$ from axioms
I begin with the following axioms for $=$ (equality):

$\forall x (x=x)$

$\forall x_1...\forall x_n\forall y_1...\forall y_n ((x_1=y_1 \land ...\land x_n=y_n)\to (p^nx_1...x_n \leftrightarrow p^ny_1...y_n))$ for each n-ary predicate symbol $p^n$ in the first order language

$\forall x_1...\forall x_n\forall y_1...\forall y_n ((x_1=y_1 \land ...\land x_n=y_n)\to (f^nx_1...x_n = f^ny_1...y_n))$ for each n-ary function symbol $f^n$ in the first order language

To prove $\forall x\forall y (x=y \to y=x)$ here's what I'm thinking:

First, I shall use Axiom 2 with $x_1 = x, x_2 = x, y_1 = x, y_2 = y$ to get $\forall x\forall y ((x=x \land x=y) \to (p^2xx \leftrightarrow p^2xy))$. Now, am I free to define $p^2ab$ as $b=a$ for all $a,b$? If yes, then I think I'm done (using Axiom 1, additionally to eliminate occurrences of $x=x$).

To prove $\forall x\forall y\forall z ((x=y\land y=z) \to (x=z))$:

The obvious choice seems to be axiom 3 this time? $\forall x\forall y\forall z ((x=y\land y=z) \to (f^2xy=f^2yz))$. How do I define $f^2$ to get the result? Are there better ways of proving this?

答:
For transitivity, use the instance of axioms schema 2 where $n=2$ and $p^2t_1t_2\equiv t_1=t_2$, and specialize to $x_1\leftarrow x$, $x_2\leftarrow z$, $y_1\leftarrow y$, $y_2\leftarrow y$ to arrive at
$$ (x=y\land z=y)\to (x=z\leftrightarrow y=y)$$
and use axiom 1 and symmetry, and generalize.


答:
Note that $$z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}} =\frac{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i}{\frac{1}{2}−\frac{\sqrt{3}}{2}i}=\frac{e^{-\frac{\pi}{4}i}}{e^{-\frac{\pi}{3}i}}=e^{\frac{\pi}{12}i}$$

Thus $$z^n=e^{\frac{n\pi }{12}i}$$

The proof follows from the fact that $z^{24}=1$

Title: $z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . Show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements.

Body:
As in the title: $z =\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}}$ . I need to show that the X set =$ \{z^n : n ∈ Z\} $ is finite and find the number of its elements. I know how to do it "non creative" way. So, I transform both complex numbers into trigonometric form and divide:
$$\frac{\sqrt{2}(1-i)}{1−i\sqrt{3}} = \frac{2[\cos(\frac{7}{4}\pi)+\sin(\frac{7}{4}\pi)i]}{2[\cos(\frac{5}{6}\pi)+\sin(\frac{5}{6}\pi)i]}= \cos \left(\frac{11}{12}\pi \right)+\sin \left( \frac{11}{12}\pi \right)i$$

Now, I see that this is $165^{\circ}$, and that module $= 1$. The problem is, that the only way I see to find the number of elements of the X set is to calculate degrees:
$$165^{\circ}, 330^{\circ}, 135^{\circ}, 300^{\circ}, 105^{\circ}, 270^{\circ}, 75^{\circ}, 240^{\circ}, 45^{\circ}, 210^{\circ}, 15^{\circ}, 180^{\circ}, 345^{\circ}, 150^{\circ}, 315,^{\circ} 120^{\circ}, 285^{\circ}, 90^{\circ}...$$

So I feel like there has to be some more effective way to solve that problem and here comes my question. Do you know any?


问:
The probability a survey participant is watching a certain channel(s)
There are 3 TV channels: A, B, C. From a survey conducted among viewers the following probabilities were found:

$$P(watch\space no\space channel) = P(A^c\cap B^c \cap C^c) = 0.3$$

$$P(watch\space exactly\space 2\space channels) = P(A\cap B \cap C^c)+P(A\cap B^c \cap C) + P(A^c\cap B\cap C) = 0.25$$

$$P(watch\space exactly\space 1\space channel) = P(A\cap B^c \cap C^c)+P(A^c\cap B \cap C^c) + P(A^c\cap B^c\cap C) = 0.37$$

$$P(A\cup B) = 0.56$$

$$P(A\cap B|A\cup B) = 0.25$$

$$P(A^c \cap B^c \cap C|B\cup C) = \frac{2}{9}$$

I need to fill up the corresponding Venn diagram.
so from item (1) I got that

$$P(A\cup B \cup C) = 1 - P(A^c\cap B^c \cap C^c) = 0.7$$

then from (4) I got

$$P(A^c \cap B^c \cap C) = P(A\cup B \cup C) - P(A\cup B) = 0.7 -
    0.56 = 0.14$$

and from (6) I get

$$P(A^c \cap B^c \cap C|B\cup C) = \frac{P((A^c \cap B^c \cap C)
    \cap (B\cup C))}{P(B\cup C)} = \frac{P(A^c \cap B^c \cap C)}{P(B\cup
    C)}=\frac{2}{9}\Longrightarrow P(B\cup C) = (9 \cdot 0.14) / 2 =
    0.63$$

$$\Longrightarrow P(A\cap B^c \cap C^c) = P(A\cup B \cup C) - P(B\cup C) = 0.7 -
    0.63 = 0.07$$

and from (3), (8) and (10) I get

$$P(A^c\cap B \cap C^c) = 0.16$$

from (2), (3) and (7) I get

$$P(A\cap B \cap C) = 0.7 - 0.37 - 0.25 = 0.08$$

from (4) and (5) I get

$$P(A\cap B|A\cup B) = \frac{P((A\cap B)\cap (A\cup B)}{P(A\cup B)} = \frac{P(A\cap B)}{P(A\cup B)} = \frac{P(A\cap B)}{0.56} = 0.25 \Longrightarrow P(A\cap B) = 0.14$$

and from (12) and (13) I get

$$P(A\cap B \cap C^c) = P(A\cap B) - P(A\cap B \cap C) = 0.14 - 0.08 = 0.06$$

I still need to calculate $P(A\cap B^c \cap C)$ and $P(A^c\cap B \cap C)$ but I couldn't find a way to do it.

答:
The corresponding Venn diagram has eight elementary domains, but in the problem are provided only six equalities. Together with condition that the sum of all probabilities is one, we have seven equalities, that is an underdetermined system.

We show that the problem has no unique solution by considering the set $V$ of elementary domains (named like $A^*\cap B^*\cap C^*$) as a set of elementary events of a probability space with the probabilities taken from your solution and satisfying all problem conditions. Namely, we put

$P(A\cap B \cap C)=0.08,$

$P(A\cap B \cap C^c)=0.06,$

$P(A\cap B^c \cap C)=x,$

$P(A\cap B^c \cap C^c)=0.07,$

$P(A^c\cap B \cap C)=0.19-x,$

$P(A^c\cap B \cap C^c)=0.16,$

$P(A^c\cap B^c \cap C)=0.14,$

$P(A^c\cap B^c \cap C^c)=0.3.$

It is easy to check that for any $0\le x\le 0.19$ all problem conditions are satisfied.


问:
If $\limsup_{x\to\infty} (1-F(x)+F(-x))\le \varepsilon$, how to prove that $F$ is a distribution function?
I read Rick Durrett's probability and was confused by the proof of theroem that states the limit of the tight distribution sequence is a distribution function of a probability measure. (just after Helly's selection theorem)

It proves that $\limsup_{x\to\infty} (1-F(x)+F(-x))\le \varepsilon$, and says since $\varepsilon$ is arbitrary, it follows that $F$ is a distribution function.

But why? It only means that $\limsup_{x\to\infty} (F(x)-F(-x))\to 1$, does not mean $\lim_{x\to+\infty}F(x)=1,\lim_{x\to-\infty}F(x)=0$, and $F$ is nondecreasing, right continuous.（Which is the definition of a distribution function.）

I could not prove it, can anyone tells me why?

答: By assumption, $F$ is the pointwise limit of a sequence of CDFs $F_n$ (namely, the subsequence whose existence is implied in the Helly selection theorem). This automatically implies that $F$ is nondecreasing and right continuous and that $0 \le F \le 1$. However, it is possible that $\lim_{x \to\infty} F(x) < 1$ and that $\lim_{x \to -\infty} F(x) > 0$ (this is demonstrated by the counterexample stated after Helly's selection theorem). The condition in your post would imply this is not the case for a tight sequence of CDFs, and that is the remaining condition for establishing that $F$ is a CDF.


问:
Is this logic proof correct?
I wanted to prove that $\forall x \in \mathbb Z : x+x^2$  is even (1).

This is the same as $\forall x \in \mathbb Z \Rightarrow x+x^2$ is even.

So applying the contrapositive : $x+x^2$ is odd $\Rightarrow \exists x \not\in \mathbb Z$.

So if I prove that then I prove (1).

But since the quantifier is the consequent the $x$ of the  $\exists$ is not the same $x$ of the antecedent, basically I can prove both apart and I don't have to prove each $x$, then

(a) $1,302775638+1,302775638^2 = 3$  (odd)

(b) $5,7 \not\in \mathbb Z$ (not the same $x$)

I also taught that even if its the same $x$ because you have to particularize the quantificators before applying properties, I still just have to find a valid case to prove the contrapositive.

For the case when the antecedent ($x+x^2$ is odd) is false then by definition of the conditional, the truth value is true.

So I proved the contrapositive proving the original (1).

What do you think about this proof?

答:
Your attempt does not make sense. The biggest problem is your misunderstanding of how the quantifiers work.

Expressions like '$\forall x \in \mathbb Z$' and $'\exists x \in \mathbb N$' are not propositions (statements) by themselves, so you cannot use them as either part of a conditional like you have.

Your first statement (1) is '$\forall x \in \mathbb Z : x + x^2 \text{ is even}$'. This can be rewritten correctly as

$$\forall x : (x \in \mathbb Z \rightarrow x + x^2 \text{ is even})$$

The contrapositive of the part in brackets would be '$(x + x^2 \text{ is not even} \rightarrow x \not\in \mathbb Z)$'. So the following statement is equivalent to (1):

$$\forall x : (x + x^2 \text{ is not even} \rightarrow x \not\in \mathbb Z)$$

If you found a counterexample, it would be a number $x$ such that $x + x^2$ was even, but $x$ was an integer: this would then be sufficient to show that (1) is false. But a single example is not enough to show that (1) is true, even using a contrapositive.

I'm not sure exactly what you meant in your sentence about particularizing the quantifiers. You can use the facts that $(P \rightarrow Q)$ is equivalent to $\neg(P \land \neg Q)$ and that $\forall x \neg$ is equivalent to $\neg \exists x$ to rewrite (1) in other ways:

$$\forall x : \neg (x + x^2 \text{ is not even} \land x \in \mathbb Z)$$

$$\neg \exists x : (x+x^2 \text{ is not even} \land x \in \mathbb Z)$$

But none of these allow you to prove what you want using a single example.

(It's not part of your question, but you can prove what you want very easily by considering the two cases where $x$ is odd or $x$ is even.)

Title: Is this logic proof correct?


答:
You are trying to prove that:

$$\forall x \in \mathbb{Z}: x^2+x  \ \text{is even}$$

Let $x \in \mathbb{Z}$. Suppose that $x$ is even. Then, we are done because $x^2+x = x(x+1)$ and that has to be divisible by $2$ if $x$ is even (you can write out explicitly why this is for your own understanding).

Let $x$ be odd. Then, $x+1$ is even. So, again, $x(x+1)$ is divisible by $2$. You can write this out explicitly if you wish.

I'm not sure why you found the need to use the contrapositive. I think I would advise you to try looking for direct proofs to a given statement first. Like, when you get a statement, just try to prove it directly. If that doesn't work out, then try to find a proof by contradiction/contraposition.

One other thing that I've noticed is that you seem to be using terminology from logic quite a bit. While I find this rather laudable and I do wish more people made use of notation and concepts from logic when writing out proofs, it is rare that someone will talk about "truth values" or whatnot when writing out proofs (at least, I don't see that being the case very often with topics that aren't within logic).

So, I'd say that you should have a conceptual understanding of the logic that you're making use of but you don't actually have to speak in terms of truth values and so on. Of course, you can still do it if you want to but it's not necessary and, in some cases, it may obfuscate the idea you're trying to get across.

Detailed Analysis of your argument:

So, I'm just going to comment on a few things that I personally found a little strange. I'm not going to comment on typesetting issues or whatever because that is something that is easily fixed. I will quote you rather extensively here.

But since the quantifier is the consequent the $x$ of the $\exists$ is not the same $x$ of the antecedent, basically I can prove both apart and I don't have to prove each $x$.

You're kind of misusing terminology here from logic. When we have two propositions $p$ and $q$, then the implication $p \implies q$ is where we typically use the words "antecedent" and "consequent". $p$ is usually called the antecedent with respect to the given implication and $q$ is called the consequent with respect to the given implication.

In this instance, there is no implication. Now, you can turn this into an implication if you want by the following:

$$x \in \mathbb{Z} \implies x^2+x \ \text{is even}$$

If you want to show the contrapositive of the implication, then you need to show that:

$$x^2+x \ \text{is odd} \implies x \notin \mathbb{Z}$$

In other words, you assume that $x^2+x$ is odd and then you actively show that $x \notin \mathbb{Z}$.

I also tought that even if its the same x because you have to particularize the quantificators before applying properties, i still just have to find a valid case to prove the contrapositive

I'm not sure what you mean by "particularizing the quantifiers". There is no such notion that I have heard of. Did you mean "choosing an arbitrary but fixed $x$" or something of the like?

If you observe the argument that I presented above, you'll notice that I did choose an $x \in \mathbb{Z}$ for each case. However, I kept it entirely ambiguous what this $x$ was, other than assuming odd and even-ness for each case. In other words, I'm not saying that $x = 3$ or something. I'm saying that we should allow $x$ to be, generically, any odd number.

a) 1,302775638+1,302775638^2 = 3 (odd) b) 5,7 ∉ ℤ (not the same x)

I'm not exactly sure what you're doing here. I think it's common in Europe to use $,$ as the decimal point. But I'm not exactly sure how that sum on the left is equal to 3? Besides, I'm not even really sure what that's supposed to accomplish.

No one is saying, really, that if $x \notin \mathbb{Z}$, then $x^2+x$ is even. In fact, if you just looked at, say, $x = 1.5$, then this is clearly not true. However, the statement that's being made is that if $x \in \mathbb{Z}$, then $x^2+x$ is even.

So, you can't object to it and say "Ah but it doesn't work when $x = 1.5$". That isn't a valid objection because no one made the claim that it did work for numbers of that form.

So, in short, your argument doesn't work. I have, of course, given an indication on how to proceed.

Title: Is this logic proof correct?

Body:
I wanted to prove that $\forall x \in \mathbb Z : x+x^2$  is even (1).

This is the same as $\forall x \in \mathbb Z \Rightarrow x+x^2$ is even.

So applying the contrapositive : $x+x^2$ is odd $\Rightarrow \exists x \not\in \mathbb Z$.

So if I prove that then I prove (1).

But since the quantifier is the consequent the $x$ of the  $\exists$ is not the same $x$ of the antecedent, basically I can prove both apart and I don't have to prove each $x$, then

(a) $1,302775638+1,302775638^2 = 3$  (odd)

(b) $5,7 \not\in \mathbb Z$ (not the same $x$)

I also taught that even if its the same $x$ because you have to particularize the quantificators before applying properties, I still just have to find a valid case to prove the contrapositive.

For the case when the antecedent ($x+x^2$ is odd) is false then by definition of the conditional, the truth value is true.

So I proved the contrapositive proving the original (1).

What do you think about this proof?


问:
Maximizing the volume of a rectangular prism
A rectangular prism has a surface area of $300$ square inches.  What whole number dimensions give the prism the greatest volume?

This is a math olympiad problem. It involves the volume and surface area of a prism.

Please, find a good solution.

Thank You!

答:
If we are assuming a fully rectangular prism---right angles all around---then there actually are only two candidates: $(h,w,l)=(20,5,2)$ and $(h,w,l)=(12,9,2)$. The former has a volume of 200, the latter has a volume of 216. So $(12,9,2)$ it is.

I solved this by exhaustive search. Here's my very dumb MATLAB code. Certainly it's not the most efficient but sometimes life calls for quick and dirty solutions. At least I limit the loops to 75, since the largest single dimension to achieve a surface area of 300 has to be less than that: $(75,1,1)$ gives us a surface area of 302:

```<code>for h = 1 : 75,
    for w = 1 : h,
        for l = 1 : w,
            if 2*(h*w+h*l+w*l) == 300,
                [h,w,l,h*w*l]
            end
        end
    end
end
</code>```

So how would you do this without code? Well, I'd probably begin an exhaustive search, frankly, but with some efficiency. Let's rewrite the surface area formula as follows:
$$2(hw+hl+wl)=300 \quad\Longrightarrow\quad hw+l(w+h)=150
\quad\Longrightarrow\quad (h+l)(w+l)=150+l^2$$
So for a fixed value of $l$, $150+l^2$ must admit an integral factorization $pq$ with $p,q>l$.

Consider $l=1$: we have $pq=151$. Since 151 is prime, so there are no
solutions. Continuing:

$l=2$, $pq=154$, prime factors 2, 7, and 11.
Since $p,q>2$, our choices are $(p,q)\in\{(22,7),(14,11)\}$ or
$(h,w,l)\in\{(20,5,2),(12,9,2)\}$. These are the two answers we found
above.

$l=3$: $pq=159$, prime factors 3 and 53; since $p,q>3$, there
are no solutions.

$l=4$: $pq=166$, prime factors 2 and 83; no
solutions with $p,q>4$.

$l=5$: $pq=175$, factors 5, 5, and 7. The only solution
with $p,q>5$ is $(p,q)=(25,7)$, which gives us $(h,w,l)=(20,2,5)$, a
permutation of our previous solution (and not even the optimal one, at that!).

$l=6$: $pq=186$, factors 2, 3,
31; no solutions with $p,q>6$.

$l=7$: $pq=199$, prime, no solutions.

I could continue, but since this is a contest problem I probably would have taken a chance on $(12,9,2)$ pretty much as soon as I acquired it. Even if I had time I doubt I would have bothered to proceed past $l=4$ or $l=5$.

Body:
If we are assuming a fully rectangular prism---right angles all around---then there actually are only two candidates: $(h,w,l)=(20,5,2)$ and $(h,w,l)=(12,9,2)$. The former has a volume of 200, the latter has a volume of 216. So $(12,9,2)$ it is.

I solved this by exhaustive search. Here's my very dumb MATLAB code. Certainly it's not the most efficient but sometimes life calls for quick and dirty solutions. At least I limit the loops to 75, since the largest single dimension to achieve a surface area of 300 has to be less than that: $(75,1,1)$ gives us a surface area of 302:

for h = 1 : 75,
    for w = 1 : h,
        for l = 1 : w,
            if 2*(h*w+h*l+w*l) == 300,
                [h,w,l,h*w*l]
            end
        end
    end
end

So how would you do this without code? Well, I'd probably begin an exhaustive search, frankly, but with some efficiency. Let's rewrite the surface area formula as follows:
$$2(hw+hl+wl)=300 \quad\Longrightarrow\quad hw+l(w+h)=150
\quad\Longrightarrow\quad (h+l)(w+l)=150+l^2$$
So for a fixed value of $l$, $150+l^2$ must admit an integral factorization $pq$ with $p,q>l$.

Consider $l=1$: we have $pq=151$. Since 151 is prime, so there are no
solutions. Continuing:

$l=2$, $pq=154$, prime factors 2, 7, and 11.
Since $p,q>2$, our choices are $(p,q)\in\{(22,7),(14,11)\}$ or
$(h,w,l)\in\{(20,5,2),(12,9,2)\}$. These are the two answers we found
above.

$l=3$: $pq=159$, prime factors 3 and 53; since $p,q>3$, there
are no solutions.

$l=4$: $pq=166$, prime factors 2 and 83; no
solutions with $p,q>4$.

$l=5$: $pq=175$, factors 5, 5, and 7. The only solution
with $p,q>5$ is $(p,q)=(25,7)$, which gives us $(h,w,l)=(20,2,5)$, a
permutation of our previous solution (and not even the optimal one, at that!).

$l=6$: $pq=186$, factors 2, 3,
31; no solutions with $p,q>6$.

$l=7$: $pq=199$, prime, no solutions.

I could continue, but since this is a contest problem I probably would have taken a chance on $(12,9,2)$ pretty much as soon as I acquired it. Even if I had time I doubt I would have bothered to proceed past $l=4$ or $l=5$.


问:
Show function is continuous with respect to metric space
Show that the function $Ψ: (X,d_\infty) → (R,d),$ defined by $Ψ(f ) = \arctan(f (3/4))$ is continuous.

When $d_∞(f , g ) = \max_{x∈[0,1]}|f (x)− g (x)|$ and $d$ is the standard metric on $R$, i.e., $d(x,y)=|x-y|$

I know that arctan is freely continuous and that I'm meant to use that fact? but I'm unsure how to start it

答:
Define $\phi : (C[0,1], d_\infty) \rightarrow (\mathbb{R} ,d)$ as $\phi (f)= f(\frac{3}{4})$.
$\phi$ is continuous because $$d(\phi (f), \phi (g))=|f(\frac{3}{4})-g(\frac{3}{4})| \leq \max_{x\in[0,1]}|f(x)-g(x)|=d_\infty(f,g).$$

Now $\Psi =\arctan \circ \phi$, so $\Psi$ is continuous as composition of two continuous functions.


Title: How to simplify $\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$?

Body:
How to simplify this?
$$\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$$</p>

Body:
Let
$$x=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$$
Take the cubic power of both sides
$$x^3=2a - (2a-1)x$$
or
$$(x-1)(x^2+x+2a)=0$$
Thus
$$x=1=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$$


问:
Exercise about vector fields on smooth manifolds
Let $V$ be a vector field on $\mathbb{R}^2$. If $[\frac{\partial}{\partial x},V]=V=[V,\frac{\partial}{\partial y}]$, determine $V$.

The first way to look at this is by observing vector fields primarily as functions from $\mathfrak{F}(\mathbb{R}^2)$ to $\mathfrak{F}(\mathbb{R}^2)$. Obviously, if $D=\frac{\partial}{\partial x}+\frac{\partial}{\partial y}$, then $[D,V]=0$.
Writing it out this way: $$\frac{\partial}{\partial x}(V(f))-V(\frac{\partial}{\partial x}f)=V(f)=V(\frac{\partial}{\partial y}f)-\frac{\partial}{\partial y}(V(f))$$, we can also see that $$\frac{\partial}{\partial x}(V(f))=V(f+\frac{\partial}{\partial x}f)$$
$$\frac{\partial}{\partial y}(V(f))=V(\frac{\partial}{\partial y}f-f)$$.

Frankly, I don't know what to do with this. Jacobi identity doesn't give me anything new.

The second way to look at this is to observe these equations locally, at some point $p\in M$. Then we'd be able to write $V$ as $V=\alpha \frac{\partial}{\partial x}+\beta \frac{\partial}{\partial y}$ (or at least I hope we can do this?). However, observed locally, given equations become $0=V=0$, which looks like it's incorrect, because it would mean I have more data than I need.

Other idea was to use the fact that we're working in $\mathbb{R}^2$ and that this is a manifold with one chart basically, but this comes down to the previous idea and the result is the same.

I feel like $[D,V]=0$ is significant here, but I don't know what to do with it. Also, it seems that my attempt at observing things locally is completely flawed. Can we even use the fact that partial derivatives constitute a basis of tangent space the way I've used it?

答:
$\def\del{\partial}$
$\def\ddx{\frac{\del}{\del x}}$
$\def\ddy{\frac{\del}{\del y}}$
Write $V = f\frac{\partial}{\partial x} + g\frac{\partial}{\del y}$. Then for an arbitrary smooth $\phi$ we have
\begin{align*}
[V, \ddx]\phi &= f\phi_{xx} + g\phi_{xy} - \ddx(f\phi_x + g\phi_y) \\
&= - f_x\phi_x - g_x\phi_y,
\end{align*}
so
$$
[V, \ddx] = -f_x\ddx - g_x\ddy.
$$
Similarly,
$$
[\ddy, V] = f_y\ddx + g_y\ddy.
$$
Both are equal to $V$, so we have
$$
-f_x = f_y = f
$$
and
$$
-g_x = g_y = g.
$$
This system of partial differential equations is solved by $f(x, y) = C_1\exp(y-x)$ and $g(x, y) = C_2\exp(y-x)$, so
$$
V = C_1\exp(y-x) \ddx + C_2\exp(y-x)\ddy.
$$