问:
Is there a text book containing a self-contained and complete proof of the Jordan Curve theorem?
I seem to remember (in my undergraduate years) encountering a book on complex analysis which contained a proof of the Jordan Curve Theorem, building up from first principles - so self-contained and complete. I am now looking for such a proof, and haven't been able to find one. Does anyone know of such a book, or alternatively an online resource which starts at the beginning and finishes at the end? The level I'm looking for is roughly a demanding first course in complex analysis.

Edit: Does Beardon's book "Complex Analysis: The Argument Principle in Analysis and Topology" do this? I can't find a review or synopsis.

答:
I don't know if this is what you are looking for, but I found the following resources:

https://luisto.fi/misc.shtml Click on "An essay about the jordan curve theorem". It is in the spirit of complex analysis.

"Introduction to the theory of functions of a complex variable" by Wolfgang J. Thron.

Hope this helps!


问:
A cardinality of sets question
Sometimes I don't know if I get questions about the cardinality of sets right. For example, what are the cardinalities of the next sets: $A=P(\mathbb{R}$\ $\mathbb{Q})$,$B=P(\mathbb{N}\cup\{\mathbb{Z}\})$, $C=P(\mathbb{N})\cup{\mathbb{Z}}$, $D=(0,1)$\ $\mathbb{Q}$, $E=\{x\in \mathbb{R}:x^{3}\in \mathbb{Z}\}$.

In addition, is there a way to assure my answer? because sometimes I feel that I am doing right, but apparently I find otherwise.

答:
Here are some pointers:

$A$: $\Bbb R$ is uncountable and $\Bbb Q$ is countable, so $\Bbb R\setminus\Bbb Q$ has the same cardinality as $\Bbb R$, and its power set therefore has the same cardinality as $\wp(\Bbb R)$.

$B$: $\Bbb N\cup\{\Bbb Z\}$ contains just one more element than $\Bbb N$, so it has the same cardinality as $\Bbb N$, and $\wp(\Bbb N\cup\{\Bbb Z\})$ therefore has the same cardinality as $\wp(\Bbb N)$.

$C$: $\wp( \mathbb N)$ is uncountable, and $\Bbb Z$ is only countable, so their union has the same cardinality as the bigger one.

$D$: See my comments on $\Bbb R\setminus\Bbb Q$ in the comment above on $A$.

$E$: The function $f(x)=\sqrt[3]x$ injective (one-to-one), and $E=f[\Bbb Z]$.

ContentLicense: CC BY-SA 4.0

Title: A cardinality of sets question


问:
Roots and discrete logarithms
I need to prove the following:

There are given numbers $p,a,b$, where $p$ is prime, $a, b \in \mathbb{N}$ and $a$, $p$ - coprime.
There also exist $r$ and $l$ such that $r^2 \equiv a\pmod p$ and $b^l \equiv a\pmod p$.
How can I prove, that there exists an $z \in \mathbb{Z}$ with $z^2 \equiv a\pmod p$ AND $b^z \equiv a\pmod p$?

I thought: if $r^2 \equiv a\pmod p$ and $b^l \equiv a\pmod p$ and $z^2 \equiv a\pmod p$ and $b^z \equiv a\pmod p$, then $r^2 = z^2 = b^l = b^z$, so $r = z = l = 2 \in \mathbb{Z}$. Is this correct? (I think this only holds true if I calculate in a certain Ring or Field, but I'm not sure.) The exercise points out at roots and discrete logarithms, but I don't know how to apply them here.

Thanks in advance :D

答:
The congruence $z^2\equiv a\pmod p$ holds whenever $z\equiv r\pmod p$.

Because $b^{p-1}\equiv1\pmod p$ the congruence $b^z\equiv a\pmod p$ holds whenever $z\equiv \ell\pmod {p-1}$.

The Chinese Remainder Theorem is your friend, and gives you the claim.

Title: Roots and discrete logarithms


问:
Finding the left hand side limit with absolute value
I have the following limit problem, and I need to find the limit as x approaches 1 from the left hand side, however, I can't get to the correct answer. I always land up with the answer of 2 but the answer should be -2. Can anyone help?

Here is the Question:

```<code>lim(as x approaches 1 from the left-hand side) (x^2 - 1)/abs(x-1)
</code>```

This is what I have done:

```<code>lim (x^2 - 1)/abs(x-1) = lim (x-1)(x+1)/(x-1) = lim (x+1) = 1+1 = 2
</code>```

答: You've missed out the "abs" from the denominator in your first step. That's all.</p>


问:
What kind of transformation is that?
We have the glide reflection
\begin{equation*}\kappa \begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}x+2\\ -y+2\end{pmatrix}\end{equation*}
and the rotation
\begin{equation*}\delta \begin{pmatrix} x\\ y\end{pmatrix}=\begin{pmatrix} -x +2\\ -y+2\end{pmatrix}\end{equation*}
The composition of these maps is
$$\kappa\circ\delta\begin{pmatrix}x \\ y\end{pmatrix}= \kappa \left (\delta \begin{pmatrix}x \\ y\end{pmatrix}\right )=\kappa \begin{pmatrix} -x +2\\ -y+2\end{pmatrix}=\begin{pmatrix} -x +4\\ -y+4\end{pmatrix}$$
What kind of transformation is that?

答:
As I pointed out in my comment, the right map is
$$(\kappa \circ \delta)(x,y) = (-x+4,y).$$
Note that you can write $(\kappa \circ \delta)(x,y) = (-x,y)+(4,0)$, or equivalently using matrices
$$(\kappa \circ \delta)
\left(
\begin{matrix}
x \\
y
\end{matrix}
\right)
= \left(
\begin{matrix}
-1 & 0 \\
0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
x \\
y
\end{matrix}
\right)+
\left(
\begin{matrix}
4 \\
0
\end{matrix}
\right)
= A
\left(
\begin{matrix}
x \\
y
\end{matrix}
\right)
+ b.
$$
So $\kappa \circ \delta$ îs an affine transformation of $\mathbb R^2$ and is a composition of an improper ($\det A = -1$) rotation $A \in \mathrm{O}(2)$ and a translation by a vector $b = (4,0)$ (horizontal translation in positive direction). The transformation fixes the line given by points $(2,y)$, because $(\kappa \circ \delta)(2,y) = (2,y)$ for all $y \in \mathbb R$.

Title: What kind of transformation is that?


问:
How do get combination sequence formula?
What would be a closed-form formula that would determine the ith value of the sequence

1, 3, 11, 43, 171...

where each value is one minus the product of the previous value and 4?

Thanks!

答:
The general formula is:

$$a_n=\frac{2}{3}4^n+\frac{1}{3}$$

This can be derived using the trick of converting a non-homogeneous difference equation, which in this case is $a_n=4a_{n-1}-1$, into a homogeneous one of a higher degree, which in this case is $a_{n+1}=5a_n-4a_{n-1}$. This has characteristic polynomial $x^2-5x+4=(x-4)(x-1)$, and therefore
$$a_n=\alpha 4^n+\beta 1^n$$
for some $\alpha, \beta$. Solving the system $$1=a_0=\alpha+\beta$$
$$3=a_1=4\alpha+\beta$$ gives $\alpha=\frac{2}{3}$ and $\beta=\frac{1}{3}$.

ContentLicense: CC BY-SA 3.0


问:
Examples of a non-finitely generated $\mathbb Z$-module
I am looking for a couple of examples of a $\mathbb Z$-module $M \neq 0$ with
$$
\mathrm{Hom}_{\mathbb Z}(M, \mathbb Z) = 0 \\
\mathrm{Ext}^1_{\mathbb Z}(M, \mathbb Z) = 0
$$

If $M$ was finitely generated this would not be possible since $M$ would be the direct sum of a free part and a torsion part.

To put this into context: I would like to realize $M$ as the first homology group of a CW complex $X$. My ultimate goal would be to construct a complex $X$ such that its first homology group is nonzero but all cohomology groups (except in degree 0) vanish. The above constraints arise by looking at the exact sequence of the universal coefficient theorem in cohomology.

答:
I dare say there does not exist such $M$. Indeed, $\mbox{Ext}^1_\mathbb{Z}(M,\mathbb{Z})$ arises from the short exact sequence
$$
0 \to F_1 \stackrel{r}{\to} F_0 \to M \to 0
$$
(where $F_0$ and $F_1$ are free) as the defect of exactness of the $\mbox{Hom}(-,\mathbb{Z})$ functor :
$$
0 \to \mbox{Hom}(M,\mathbb{Z}) \to \mbox{Hom}(F_0,\mathbb{Z}) \to \mbox{Hom}(F_1,\mathbb{Z}) \to \mbox{Ext}^1_\mathbb{Z}(M,\mathbb{Z}) \to 0
$$
Now if $\mbox{Ext}^1_\mathbb{Z}(M,\mathbb{Z})=0$ and $\mbox{Hom}(M,\mathbb{Z})=0$, then this reduces to
$$
0\to \mbox{Hom}(F_0,\mathbb{Z}) \to \mbox{Hom}(F_1,\mathbb{Z}) \to 0
$$
which means that $\mbox{Hom}(r,\mathbb{Z}):\mbox{Hom}(F_0,\mathbb{Z}) \to \mbox{Hom}(F_1,\mathbb{Z})$ is an isomorphism, in particular it's injective, which means that $r$ is surjective, and hence $M=0$.</p>


问:
Understanding Corollary 2.3.2 from Bickel's Mathematical Statistics.
I am trying to understand the Corollary 2.3.2 from Bickel's book mathematical statistics.

The corollary says

Consider the exponential
family$$p(x,\theta)=h(x)\exp\left\{\sum_{j=1}^kc_j(\theta)T_j(x)-B(\theta)\right\},x\in\mathcal{X},\theta\in\Theta.$$
Let $C^0$ denote the interior of the range of
$(c_1(\theta),c_2(\theta),\dots,c_k(\theta))^T$ and let $x$ be the
observed data. If the equations $$E_\theta T_j(X)=T_j(x),j=1,\dots,k$$
have a solution $C(\hat{\theta})\in C^0,$ then it is the unique MLE of
$\theta$.

I think the conclusion of this corollary is useful, as it is not easy to show the existence and uniqueness of a curved exponential family in general. However, the book does not given any example of using this corollary.

I am trying to understand this corollary. Can someone give an example showing the usefulness of it?

答:
After some exploration, I think one example is that when we proving the uniqueness of the MLE of the bivariate normal distribution. In the bivariate normal distribution, we have five parameters $\theta = (\mu_1,\mu_2,\sigma_1, \sigma_2,\rho)\in({\mathbb{R},\mathbb{R},\mathbb{R}^+,\mathbb{R}^+,(0,1)})=\Theta$. We could use the Hessian of the p.d.f being negative definite to show that $\hat{\theta}_{MLE}$ is unique. But calculating the second derivative with respect to five parameters is a pain. However, using the corollary, if we could show that $c(\hat{\theta}_{MLE})$ lays in the interior of $c(\Theta)$, we get the uniqueness of the $\hat{\theta}_{MLE}$.

Please note however, this corollary cannot be used for cases like $\mathcal{N}(\theta,\theta^2)$, from which $c(\Theta)$ is a parabola in $\mathbb{R}^2$, that does have an "interior" of a ball in $\mathbb{R}^2$. Thus, this corollary does not apply.


问:
How to translate word problem to mathematical statement
I have this math problem.

...$x$ animals are positioned at random points of a circle with 1 kilometer
circumference. Every animal has a certain amount of calories stored in fat. Once an animal $m$
reaches another $n$, it can eat $n$ and gain 100% of $n$’s fat. There is enough fat distributed between all $x$ animal to fuel any single one of them for 1 kilometer. Prove that there is an animal
that can run around the circle clockwise, eating everyone as he goes, and run out of fat
just as he reaches his starting point.

I can intuitively decipher that it requires some form of induction to prove. That is for if we assume n animals can do it, then n+1 animals can do it too. But I'm unable to come up with a mathematical statement $P(n)$ to induct on, that describes the question.

答: HINT: First show that there must be at least one animal, $A_1$, among the $n+1$ who has enough calories to reach the next animal, $A_2$. Remove $A_2$, and give $A_1$ all of $A_2$’s calories in addition to its own. Apply the induction hypothesis to the remaining $n$ animals to conclude that one of them can complete the circle, and show that the same animal can complete the circle of $n+1$ animals.


问:
Supremum of Infimum and Infimum of Supremum
Let $X$ and $Y$ be nonempty sets, and let $f:X\times Y \rightarrow \mathbb{R}$ be a bounded function.

a) Prove that $$\sup_{y\in Y}\left(\inf_{x\in X}f(x,y)\right)\leq \inf_{x\in X}\left(\sup_{y\in Y}f(x,y)\right)$$

b) Give an example (with proof) where the inequality is strict.

I am stumped. First of all, I am having a hard time understanding the notation. The way I understand it, in the most basic terms, we have that the least upperbound of the greatest lower bound of the function is less than or equal to the greatest lower bond of the least upper bound of the function. But does this even make sense? sup and inf are unique, so wouldn't $$\sup_{y\in Y}\left(\inf_{x\in X}f(x,y)\right) = \inf_{x\in X}f(x,y)?? $$ and likewise for the inf of the sup...am I missing something? Plus I'm not sure what implications the subscripts $x\in X$ and $y\in Y$ carry.

Please help me get an understanding of what the notation means and let me know if my reasoning is off.

答:
The proof for part a):

By the definition of $\sup$, $$\sup_{y\in Y} \left(\inf_{x \in X} f(x,y)\right) - \epsilon <  \inf_{x \in X} f(x,y^\ast)$$ for some $y^\ast \in Y$. Also, $$ \inf_{x \in X} f(x,y^\ast) \leq \inf_{x \in X} \left( \sup_{y \in Y} f(x,y) \right), $$ because $f(x,y^\ast) \leq \sup_{y \in Y} f(x,y)$ for every $x$. Combine the two above we have $$\sup_{y\in Y} \left(\inf_{x \in X} f(x,y)\right) - \epsilon < \inf_{x \in X} \left( \sup_{y \in Y} f(x,y) \right).$$ This holds for any $\epsilon>0$. Therefore, $$\sup_{y\in Y} \left(\inf_{x \in X} f(x,y)\right) \leq \inf_{x \in X} \left( \sup_{y \in Y} f(x,y) \right). $$ $$\tag*{$\blacksquare$}$$


问:
Intersection lattice of a real braid arrangement.
In $\mathbb{R}^d$, the collection of hyperplanes $\mathcal{H}=\left\{x_{j}=x_{k}: 1 \leq j<k \leq d\right\}$ is called the $d$-dimensional real braid arrangement. I’m interested in the intersections of those hyperplanes, specifically with the added order by reverse inclusion ($x \preceq y \Leftrightarrow x \supseteq y$).

As an example, for $d = 3$, in the arrangement we have the three hyperplanes given by $x_1 = x_2, x_1 = x_3$, and $x_2 = x_3$. Rewriting them as tuples (i.e. $(x,x,x_3),(x,x_2,x),(x_1,x,x)$, respectively), it is easy to see that any two hyperplanes intersect at a line, which all coincide. The intersection lattice would looks like this:

(If we intersect just $1$ hyperplane, we get that hyperplane back. Intersecting $2$ hyperplanes gives us a line. By the same idea, the result of intersecting $0$ hyperplanes is defined to be $R^d$.)

This is a lot harder in dimension $4$ and up. I’ve computed that there are $d \choose 2$ hyperplanes for each dimension $d$, so the first line of thinking to consider all $2^{d \choose 2}$ sets of hyperplanes. But going this way there are some questions I’m still not able to answer:

Do we have intersections of every dimension from $1$ to $d - 2$?

If so, do some of them coincide, like how the three lines in $\mathbb{R}^3$ do?

In general I’m just looking for a systematic way to think about this problem. Any suggestion would be appreciated!

答:
These intersections are just the equivalence relations on a set with $d$ elements.  Namely, for any equivalence relation $\sim$ on $\{1,\dots,d\}$, there is an intersection consisting of all elements $x\in\mathbb{R}^d$ such that $i\sim j$ implies $x_i=x_j$, and every intersection has this form for a unique $\sim$.  So you are just asking about the lattice of equivalence relations on $\{1,\dots,d\}$, ordered by inclusion.

So for your first question, an intersection of dimension $e$ is just an equivalence relation with $e$ equivalence classes.  This can easily be achieved for any $e$ from $1$ to $d$: just partition the set into $e$ subsets.  Two collections of hyperplanes have the same intersection iff the corresponding collections of ordered pairs of elements of $\{1,\dots,d\}$ generate the same equivalence relation.  So, for any equivalence relation that has an equivalence class of size greater than 2, there is more than one way to write it as an intersection of hyperplanes: for instance, if $\{i,j,k\}$ is one of the equivalence classes, you could write use any two of the hyperplanes corresponding to two of $i,j,$ and $k$ to make them all equivalent.


问:
Modified weight 2 Eisenstein series is a modular form for $\Gamma_0(N)$
I'm doing exercise 1.2.8(e) in Diamond & Shurman's A First Course in Modular Forms. The problem is to show that $G_{2,N}(\tau) := G_2(\tau)-NG_2(N\tau)$ is in $M_2(\Gamma_0(N))$. To show this, I need to argue that $G_{2,N}$ satisfies $G_{2,N}[\gamma]_2=G_{2,N}$ where $f[\gamma]_2(\tau):=j(\gamma,\tau)^{-2}f(\gamma(\tau))$ is the weight-2 operator for every $\gamma\in\Gamma_0(N)$, and that $G_{2,N}$ is holomorphic on $\mathcal{H}$ and holomorphic at the cusps.

So far, I proven that $G_2(\tau)$ satisfies the transformation formula $$G_2[\gamma]_2(\tau) = G_2(\tau)-\frac{2\pi ic}{c\tau+d},\;\;\;\;\gamma = \begin{bmatrix}a&b\\c&d\end{bmatrix}$$
and that
$$\frac{\pi}{j(\gamma,\tau)^2\Im(\gamma(\tau))}=\frac{\pi}{\Im(\tau)}-\frac{2\pi ic}{c\tau+d},$$
and I've concluded that $G_2(\tau) - \frac{\pi}{\Im(\tau)}$ is weight-2 invariant under $SL_2(\mathbb Z)$.

I have no idea how to conclude that $G_{2,N}(\tau)$ is a modular form for $\Gamma_0(N)$. Specifically, how does the weight-2 operator act on $G_{2,N}$? Is it
$$G_{2,N}[\gamma]_2(\tau) = (c\tau+d)^{-2}(G_2(\gamma(\tau))-NG_2(\gamma(N\tau)))$$
or
$$G_{2,N}[\gamma]_2(\tau) = (c\tau+d)^{-2}(G_2(\gamma(\tau)) - NG_2(N\gamma(\tau))?$$
In either case, $N\gamma(\tau)$ and $\gamma(N\tau)$ are not of the form $\alpha(\tau)$ for some $\alpha\in SL_2(\mathbb Z)$, so how should $G_2(\tau)$ transform?

Moreover, I don't know how to verify that $G_{2,N}$ is holomorphic on $\mathcal{H}$ or at the cusps. I would greatly appreciate detailed answers to both of these.

P.S. This is not homework, I'm simply trying to better understand computational aspects of modular forms.

答:
Note that $\tau \longmapsto NG_2(N\tau)$ is the function $G_2[\gamma]_2$ with $\gamma=\begin{bmatrix}N & 0\\0&1 \end{bmatrix}$. Moreover, if $\gamma’=\begin{bmatrix}a &b\\c&d\end{bmatrix} \in \Gamma_0(N)$, then $$\gamma\gamma’=\begin{bmatrix} Na & Nb \\ c & d\end{bmatrix}=\begin{bmatrix} a & Nb \\ c/N & d\end{bmatrix}\gamma:=\gamma’’\gamma,$$ and $\gamma’’ \in SL_2(\mathbb{Z})$, so that $G_{2,N}[\gamma’]_2(\tau)=G_2(\tau)-\frac{2i\pi c}{c\tau+d}-(G_2[\gamma’’]_2)[\gamma]_2(\tau)=G_2(\tau)-\frac{2i\pi c}{c\tau+d}-NG_2[\gamma’’]_2(N\tau)=G_2(\tau)-\frac{2i\pi c}{c\tau+d}-NG_2(N\tau)+N\frac{2i\pi (c/N)}{(c/N)(N\tau)+d}=G_{2,N}(\tau)$.

Edit: Now let’s address the holomorphicity issues. First, we show that $G_2[\gamma]_2$ is holomorphic everywhere (including at infinity) for any $\gamma \in SL_2(\mathbb{Z})$. Given the transformation formula, it’s enough to show that $G_2$ is holomorphic everywhere and at infinity.

It’s enough to show that $G_2(\tau)$ is holomorphic  on the half plane and bounded at infinity. But $G_2(\tau)=2\zeta(2)+2\sum_{c \geq 1}{f(c\tau)}$ where $f(z)=\sum_{d \in \mathbb{Z}}{\frac{1}{(z+d)^2}}$. As $f$ is the sum of a (locally) normally convergent series of holomorphic functions, $f$ is holomorphic.

Now, if $z$ has imaginary part at least $A > 0$ and real part between $-1$ and $1$, then $$f(z)=\sum_{d \in \mathbb{Z}}{\int_{-1/2}^{1/2}{\int_0^t{\frac{-2du}{(z+d+u)^3}}\,dt}},$$ which shows that $|f(z)| \leq 2\sum_{d \in \mathbb{Z}}{\sup_{t \in (d-1/2,d+1/2)}\, |\frac{1}{(z+t)^3}|}$.

The term for $d=0$ is at most $\frac{1}{A^3}$, and the term for $d \geq 1$ is at most $\frac{1}{(A^2+(|d|-1/2)^2)^{3/2} } \leq \frac{8}{(A^2+d^2)^{3/2}} \leq \frac{16}{A^3+|d|^3}$ by convexity.

It follows that $|f(z)| \leq 2A^{-3}+32\sum_{d \geq 1}{(A^3+d^3)^{-1}}$. By separating the cases $d \geq A$ and $d \leq A$, we get finally that $|f(z)| \leq C(A^{-3}+A^{-2})$ where $C$ is a numerical constant.

Since $f$ is $1$-periodic, it follows that $|f(z)| \leq \frac{C}{|Im(z)|^3}+\frac{C}{|Im(z)|^2}$ and therefore the series $\sum_{c \geq 1}{f(cz)}$ converges normally on all $\{Im z > A\}$, $A>0$.

It follows from this and the bound on $f$ that that series is a holomorphic function of $z$ and goes to zero as $Im(z)$ goes to infinity, so that $G_2(z)$ is holomorphic on the half-plane and at the cusps.

Thus, $G_{2,N}$ is also holomorphic on the half-plane, $\Gamma_0(N)$-invariant and bounded (hence holomorphic) at infinity. Moreover, it has a well know Fourier expansion with coefficients in $O(n^2)$, so by Proposition 1.2.4 it is modular.


问:
Is product of a monoidal category $C$ with itself a part of $C$?
If $C$ is a monoidal category, where $⊗ : C \times C \to C$, is $C \times C$ considered part of the category $C$ or is it a separate category (a product category)?

答:
Given categories $\mathcal{C}$ and $\mathcal{}$, we can form a new category $\mathcal{C}\times \mathcal{D}$, as follows:

The objects are pairs $(C,D)$ where $C$ is an object in $\mathcal{C}$ and $D$ is an object in $\mathcal{D}$.

The morphisms are defined by
$$\operatorname{Hom}((C_1,D_1),(C_2, D_2))=\operatorname{Hom}(C_1, D_1)\times \operatorname{Hom}(C_2, D_2)$$
and are composed in each component individually.

In your definition, $\mathcal{D}=\mathcal{C}$.


问:
Is $F(2)-F(-2)$ the same thing as $f(2)-f(-2)$?
 I thought that they were the same thing and that I would just use the graph numbers but the answer is not $4$ so apparently I'm wrong.

Based on the picture, we have $f$ to be the linear function on $[-2,-1]$ given by $f(-2)=-2$ and $f(-1)=0$. And $f$ to be the linear function on $[-1,2]$ given by $f(-1)=0$ and $f(2)=2$.

答:
There is no unique antiderivative of $f$, but for any number $a \in (-2,2)$ the quantity
$$F(x) = \int_a^x f(t)\,dt$$
is the area (with sign) of the portion of plane between the graph of $f$ and the horizontal axis. Thus
\begin{align*}
F(2)-F(-2) & = \int_a^2 f(t)\,dt - \int_a^{-2} f(t)\, dt \\
& = \int_a^2 f(t)\,dt + \int_{-2}^a f(t)\, dt \\
& = \int_{-2}^2 f(t)\, dt.
\end{align*}
This area is the sum of the areas (with sign) of two triangles and corresponds to $-1+3=2$. On the other hand, $f(2)-f(-2) = 4$.


问:
Solving a quadratic matrix equation with transpose matrix
Let $X$ be an unknown matrix and $A$ a known matrix, and suppose the problem is well posed in terms of dimensions.

Is there any way to solve the matrix equation $X^TX=A$, analytically or using numerical methods?

答:
The matrix $A$ is symmetric. Thus there exists a unitary matrix $U$ such that $UAU^T=D=diag(d_{i})$ is diagonal with real non-negative entries (source: https://en.wikipedia.org/wiki/Symmetric_matrix#Complex_symmetric_matrices).

Thus,
$$ A = X^TX = (U^\dagger \sqrt{D})(\sqrt{D}U^*). $$

Inspection tells you that
$$ X = \sqrt{D}U^*. $$

Note that this is not unique. For any orthogonal matrix $R$, $\tilde X=RX$ is also a solution.


问:
$K_9$ edge colouring
Assume we have $K_9$ graph and $4$ colours. We want to colour its edges and obtain following property:
If we change colours in our colouring, such graph will be isomorphic to the previous one (before colour change). Moreover, such graph is unique respect to automorphism group.
How to do it?

So far, I guessed that every vertice should be connected with 2 edges in all 4 colours (for example 2 blue, 2 green, 2 red, 2 yellow).

答:
Here is an example of how to achieve most of what you are describing - except I don't really understand your requirement "such graph is unique respect to automorphism group." Each colour is a set of $3$ $C_3$ graphs. ($C_3$ is a triangle so it is also $K_3$).

My attempt to meet the requirement by decomposing into $C_9$ graphs ran into trouble. If this is possible I would like to see one - update:@ThomasLesgourgues was kind enough to point me at the Wikipedia for Hamiltonian Decomposition which shows how $K_9$ has such a set of spanning $C_9$ factor graphs. and if not, it would be great to establish why putting in three such $\require{enclose}\enclose{horizontalstrike}{C_9}$ cycles results in the last being $\enclose{horizontalstrike}{3\times C_3}$.

Title: $K_9$ edge colouring


问:
Integrability of Fourier Transform of Lipshitz functions
If a function $f$ is Lebesgue integrable and satisfies Lipschitz condition, must its Fourier transform $\hat{f} $ be Lebesgue integrable or not?

There's a theorem about Fourier series, saying the Fourier series of periodic Lipschitz function must be absolutely convergent, and I'm just thinking about if it has a similar analog in Fourier transform.

I have proven that the improper integral $$\displaystyle{\int _{- \infty }^{\infty }\hat {f} (\xi )e^{2 \pi i \xi x} d \xi }$$ is uniformly convergent for $x\in \mathbb{R}$.

My thought is to find a sequence of Lipschitz integrable functions, with $L^1$-norm and Lipschitz norm both $\le 1$, whose Fourier transform have an increasing $L^1$-norm, then by the closed graph theorem, there must exist an integrable Lipschitz function whose Fourier transform is not integrable.

But I find that kind of construction extremely difficult to be accomplished. Anyone has ideas?

答:
This is a really nice question, I was surprised after thinking for a couple of hours and none of the standard examples in a harmonic analysis course fitting the description. In any case, I will try to present what can and what cannot be done to the integrability of the Fourier transform of a Lipschitz function.

Notice also that I will be employing standard analysis notation (such as $\lesssim$ and $O(g)$) as well as some standard theorems in Fourier analysis, such as van der Corput's lemma for bounding oscillatory integrals (also known as 'stationary phase method'). If you are not familiar with those, I would gladly refer you to Stein's "Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals", whose chapter VIII contains all that we shall use here.

First of all, let us notice that there are a couple of things that can be proved with slightly different assumptions, or slightly more restrictive results that can be proved with the same assumptions:

1.1. If instead of $f$ being Lipschitz, one requires the different condition that $\|f(x+h)-f(x)\|_{L^2(dx)} = O(h^{\alpha}),$ for some $\alpha > 1/2,$ then the same conclusion of Bernstein's theorem holds. That is, if the above condition is satisfied and $f \in L^1(\mathbb{R}),$ then $\widehat{f} \in L^1(\mathbb{R}).$ A proof of this fact can be found, for instance, in the book "Introduction to the Theory of Fourier Integrals", by Titchmarsh.

As a by-product, notice that if $f \in L^1(\mathbb{R})$ is Lipschitz and has compact support, then the condition in Titchmarsh's theorem above is trivially satisfied from the Lipschitz condition, and thus $\widehat{f} \in L^1(\mathbb{R}).$ This shows that, if a counterexample exists, it cannot have compact support.

1.2. On the other hand, notice that $f \in L^1, f' \in L^{\infty}$ implies directly that $f \in L^{\infty},$ and thus, by interpolation, $f \in L^p, \forall p \ge 1.$ By the Hausdorff--Young inequality, we have that $\widehat{f} \in L^q, \forall q \ge 2.$ In fact, this can be significantly improved:

Claim: Let $f \in L^1(\mathbb{R})$ be Lipschitz, i.e., $f' \in L^{\infty}.$ Then $\widehat{f} \in L^p(\mathbb{R}), \forall p > 1.$

Proof: In Grafakos's "Modern Fourier Analysis", the following result can be found as a by-product of a characterization of Lipschitz spaces: 'Let $\Delta_j(f) = (\psi(\xi/2^j) \widehat{f})^{\vee},$ where $\psi$ is a smooth bump supported on $[1/2,4]$ and equal to 1 on $[1,2].$ If $f$ is Lipschitz, then

$$\sup_{j \ge 0} 2^j \|\Delta_j(f)\|_{\infty} < + \infty.'$$

We will not prove this here, but by assuming such a result, we have the following: for $\psi$ as in the statement of the Littlewood-Paley characterization of Lipschitz functions, we have

$$\int_{2^j}^{2^{j+1}} |\widehat{f}(\xi)|^p \, d\xi \le \int_{\mathbb{R}} |\psi(\xi/2^j) \widehat{f}(\xi)|^p \psi(\xi/2^j) \, d \xi \lesssim 2^{(1-\frac{p}{2})j} \|\Delta_j(f)\|_2^{p},$$

where in the last inequality we used Hölder and Plancherel. Now, by interpolation we have

$$\|\Delta_j(f)\|_2 \le \|\Delta_j(f)\|_{\infty}^{1/2} \|\Delta_j(f)\|_1^{1/2}.$$

From the definition of $\Delta_j,$ we have $\|\Delta_j(f)\|_1 \lesssim \|f\|_1,$ and from the characterization of Lipschitz functions mentioned before,
$\|\Delta_j(f)\|_{\infty} \lesssim 2^{-j}.$ Putting all those considerations together, we obtain

$$ \int_{2^j}^{2^{j+1}} |\widehat{f}(\xi)|^p \, d\xi \lesssim 2^{(1-p)j},$$

and the right-hand side above is summable in $j \ge 1$ if $p>1,$ as desired. $\square.$

Thus, the Fourier transform of any such counterexample has to integrate to any order $p>1.$

With these intuitions at hand, we can finally provide our counterexample: let $\psi_0(x)$ be a smooth function satisfying $\psi_0 = 1$ if $x > 20,$ and $\psi_0 = 0$ if $x < 10.$ Define then the Fourier transform of our conterexample to be

$$ \widehat{f}(x) = \frac{e^{i\frac{x^2}{2\log x}}}{x \log x} \psi_0(x).$$

Then we need to prove that

$$f(\xi) = \int_{\mathbb{R}} \frac{e^{i\frac{x^2}{2\log x} - i\xi x}}{x \log x} \psi_0(x) \, dx$$

is Lipschitz and integrable.

2.1. $f$ is Lipschitz: it is not too hard (even though it is a bit technical) to show that $f$ is Lipschitz if and only if the integrals

$$ J_N(\xi) = \int_{10}^N \frac{e^{i\frac{x^2}{2 \log x}}}{\log x} \psi_0(x) e^{-i \xi x} \, dx $$

are uniformly bounded on $\xi, N.$ Define then the phase function $\phi(x,\xi) = \frac{x^2}{2\log x} - \xi x.$ Suppose that, for some $x_{\xi} \in \mathbb{R},$ we have $\partial_x \phi(x_{\xi},\xi) = 0.$ As $\partial_x \phi(x,\xi) = \frac{x}{\log x} + \frac{x}{2 \log^2 x} - \xi,$ then it is also not hard to show that $x_{\xi} \gtrsim |\xi| \log |\xi|.$ We split then the integral defining $J_N(\xi)$ into the one over the interval $[5,N] \setminus [\xi^{1/2},\xi^{3/2}],$ denoted by $J_N^1(\xi),$ and the one over $[\xi^{1/2},\xi^{3/2}],$ denoted by $J_N^2(\xi).$ Thus, $J_N(\xi) = J_N^1(\xi) + J_N^2(\xi).$

For $J_N^1,$ we use that the derivative of the phase satisfies $|\partial_x\phi(x,\xi)| \gtrsim \max\{|\xi|,\frac{x}{\log x}\},$ while for $x$ sufficiently large (10 should do for this) $\partial_{xx}^2 \phi(x,\xi) > 0,$ so we are in position to apply van der Corput's lemma, which then implies that

$$|J_N^1(\xi)| \lesssim \frac{1}{|\xi|}.$$

For the analysis of $J_N^2,$ we notice that the interval $[\xi^{1/2},\xi^{3/2}]$ contains the zero $x_{\xi}$ of $\partial_x \phi(x,\xi),$ and in that interval it also holds that $\partial_{xx}^2 \phi(x,\xi) \gtrsim \frac{1}{\log \xi}.$ Using van der Corput's lemma again, we have that

$$|J_N^2(\xi)| \lesssim \frac{(\log \xi)^{1/2}}{\log \xi} \lesssim \frac{1}{(\log \xi)^{1/2}}.$$

This finishes the case $\xi \gg 1.$ For the small $\xi$ case, the analysis here carries out even easier than what is written here. Thus, $f$ is Lipschitz.

2.2. $f$ is integrable: The idea is basically the same as here, but integrating by parts first. Indeed, we need to prove some uniform estimates on the functions

$$f_N(\xi) := \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{x \log x} \psi_0(x) \, dx.$$

Integrating by parts with factors $u = \frac{e^{i\frac{x^2}{2\log x}}}{x \log x} \psi_0(x)$ and $dv = e^{-i \xi x} dx.$ This gives that $f_N(\xi)$ is equivalently equal to

$$\frac{1}{i\xi}\frac{ e^{i( \frac{N^2}{2\log N} - \xi N)}}{N \log N} + $$
$$ \frac{1}{i \xi} \int_{10}^N \partial_x \left(\frac{x^2}{2\log x}\right) x \log x \frac{e^{i(\frac{x^2}{2\log x} - \xi x)} }{x^2 \log^2 x} \, \psi_0(x) dx + $$
$$ \frac{1}{ i\xi} \int_{10}^N \frac{\log x + 1}{x^2 \log^2 x} e^{i(\frac{x^2}{2 \log x} - \xi x)} \psi_0(x) \, dx + $$
$$ \frac{1}{i\xi} \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{x \log x} \psi_0'(x) \, dx =: \frac{1}{i\xi} (I_1(N,\xi) + I_2(N,\xi) + I_3(N,\xi) + I_4(N,\xi)).$$

One notices quite immediately that $I_1(N,\xi) \to 0$ as $N \to \infty,$ so we do not need to worry about that term when taking limits. Also, $\frac{\psi_0'(x)}{x \log x}$ has support on $[10,20]$ and is smooth there, while the derivative of the phase $\partial_x\phi(x,\xi)$ does not vanish for $\xi$ small, and neither does it on the support for $\xi$ sufficiently large. Thus, this term is bounded and decays as fast as one wishes - i.e., $|I_4(N,\xi)| = O_R(\xi^{-R})$ when $\xi \to \infty,$ for any $R > 0.$ Thus $I_4(N,\cdot) \in L^1(\mathbb{R})$ uniformly on $N \to \infty.$

We then analyze the remaining terms $I_2,I_3.$ In fact, the analysis of $I_2$ is much more delicate than that of $I_2,$ so we perform only the former and defer the latter to whoever wishes to fill in the gaps.

Rewrite $I_2(N,\xi)$ as

$$\int_{10}^N \left( \frac{x}{\log x} + \frac{x}{2 \log^2x} \right) \frac{x \log x}{x^2 \log^2 x} e^{i(\frac{x^2}{2\log x} - \xi x)} \psi_0(x) \, dx = $$
$$ = 2 \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{ (\log x)^2} \psi_0(x) dx + \int_{10}^N \frac{e^{i(\frac{x^2}{2\log x} - \xi x)}}{(\log x)^3} \psi_0(x) \, dx.$$

These resemble a lot the integrals we had to estimate for proving $f$ is Lipschitz. In fact, the very same argument is applicable in these cases as well, where one obtains (after redoing the same computations with van der Corput and what not)

$$|I_2(N,\xi)| \lesssim \frac{1}{(\log |\xi|)^{3/2}}.$$

As mentioned before, the analysis of $I_3$ is in fact easier, and doing the computations yields

$$|I_3(N,\xi)| \lesssim \frac{1}{|\xi|}.$$

Notice that all such estimates were made uniformly on $N \to \infty.$ Putting all together and taking limits, one obtains that, for $|\xi| \gg 1,$

$$|f(\xi)| \lesssim \frac{1}{|\xi| (\log|\xi|)^{3/2}}.$$

As the right-hand side is integrable and $f$ was already seen to be Lipschitz, this implies $f \in L^1(\mathbb{R}),$ as desired.


答:
Since the $\text{LHS}$ is symmetric in $a,b,c$, and the $\text{RHS}$ can be written as
$$
\left(\frac{(a+b+c)-3a}{2}\right)^3
$$
it suffices to prove the inequality with the additional assumption $a\le b\le c$.

Replacing $b,c$ by $a+x,a+x+y\;$respectively (where $x,y\ge 0$), we get
\begin{align*}
\text{LHS}-\text{RHS}
&=
\Bigl({\small{\frac{1}{8}}}\Bigr)
(
24ax^2+24axy+24ay^2+8x^3+12x^2y+18xy^2+7y^3
)
\\[4pt]
\end{align*}
all terms of which are nonnegative, thus proving the required inequality.

Title: Proof then $a^3+b^3+c^3-3abc\geq \left(\displaystyle\frac{b+c}{2}-a\right)^3$

Body:
If $a,  b,  c$ are non-negative numbers, prove  then

$a^3+b^3+c^3-3abc\geq \left(\displaystyle\frac{b+c}{2}-a\right)^3$

$8a^3+8b^3+8c^3-24abc\geq\displaystyle(b^3+3b^2c+3bc^2+c^3-6a(b^2+2bc+c^2)+12a^2(b+c)-8a^3)$

$16a^3+7b^3+7c^3\geq 3b^2c+3bc^2-6ab^2+12abc-6ac^2+12a^2b+12a^2c$

$16a^3+7b^3+7c^3+6ab^2+6ac^2\geq 3b^2c+3bc^2+12abc+12a^2b+12a^2c$

$AM-GM$

$\frac{6a^3+6ab^2}{2}\geq 6\sqrt{a^4b^2}$

$6a^3+6ab^2\geq 12a^2b...(1)$

$\frac{6a^3+6ac^2}{2}\geq 6\sqrt{a^4c^2}$

$6a^3+6ac^2\geq 12a^2c...(2)$

$\frac{4a^3+4b^3+4c^3}{3}\geq 4^3\sqrt{a^3b^3c^3}$

$4a^3+4b^3+4c^3\geq 12abc...(3)$

$\frac{b^3+b^3+c^3}{3}\geq ^3\sqrt{b^6c^3}$

$2b^3+c^3\geq 3b^2c...(4)$

$\frac{b^3+c^3+c^3}{3}\geq ^3\sqrt{b^3c^6}$

$b^3+2c^3\geq 3bc^2...(5)$

$(1)+(2)+(3)+(4)+(5)$, we get

$16a^3+7b^3+7c^3+6ab^2+6ac^2\geq 3b^2c+3bc^2+12abc+12a^2b+12a^2c$

Proved.


问:
Show that $f(x) = 2x^2+4$ is reducible over $\mathbb{C}$
I know that this question has already been asked here before, but there's just one part that I didn't get. By definition a polynomial is irreducible whenever we can write $f(x) = g(x)h(x)$ and $g(x)$ or $h(x)$ is a unit. In our case we have

$f(x) = 2(x^2+2)$, but $2$ is a unit over $\mathbb{C}$, right? so why is $f(x)$ reducible? Am I using the right definition?

答:
The definition is correct: irreducible means $\it{whenever}$ we have a decomposition $f = g h$, then $g$, or $h$ is a unit.  You only showed one decomposition. But there is another one
$$2 x^2 + 4 = (2 x + i \,2\sqrt{2})(x-i \sqrt{2})$$
where neither $g$, nor $h$ is a unit.


问:
Mean and kinda almost sure convergence relation
Considering:

Almost sure convergence as $X_n \xrightarrow[]{\text{a.s}} X \Leftrightarrow P(\lim_{n \rightarrow \infty}X_n=X)=1$ and

Mean $L_p$-convergence as $X_n \xrightarrow[]{\text{mean}}  X \Leftrightarrow \mathbb{E}|X_n-X|^p \rightarrow 0$

Are there any examples of such Random variables that they converge in terms of mean and do not converge in terms of 'almost sure' convergence and vice-versa - converging a.s (in terms of definition listed above) and do not converging in terms of mean.

May you help me with this one? (Doing this as a part of my types of RV convergence implication tree proof building - I've used infamous Riss model and similar during proofing, however don't have any ideas for this two - I want to prove that mean (so-called $L_p$) convergence do not necessarily imply a.s convergence and vice versa).

答:
Consider
$$X_n=\begin{cases}n^3 & \text{with probability }n^{-2} \\ 0 & \text{with probability }1-n^{-2}.\end{cases}$$
Then $X_n\to0$ almost surely: indeed, $\sum\mathbb P(\lvert X_n\rvert>\varepsilon)\leq\sum n^{-2}$ which converges, so we can apply Borel-Cantelli. But $\mathbb E[X_n]=n\to\infty$.

Now consider independent rvs
$$Y_n=\begin{cases}1 & \text{with probability }n^{-1} \\ 0 & \text{with probability }1-n^{-1}.\end{cases}$$
Then $\mathbb E[Y_n]=1/n\to0$, but $Y_n$ doesn't converge a.s. Indeed, as $\sum\mathbb P(Y_n=1)=+\infty$ and the events $\{Y_n=1\}$ are independent, the second Borel-Cantelli lemma gives that $$\mathbb P\left(\limsup_{n\to\infty}\{Y_n=1\}\right)=1,$$
so we can't have $Y_n\to0$ almost surely.


问:
Probability of two disjoint events but the probability of one event is unknown
I'm given that A and B are two disjoint events where $P(A) = 0.4$.

The first part of the problem asked if $P(B) = 0.8$ is possible which I answered as no since that would make $P(A \cup B)$ bigger than the sample space $P(\Omega)$.

However, the second part then asks me to find $P(A \cap B^c)$. I'm really confused on where to start since I feel like it's unsolvable since I don't actually know what $P(B)$ is. I thought it might of just been 0, but since they're disjoint $A$ is guaranteed to be a subset of $B^c$ which means the answer is a non-zero number.

I'm not sure if I'm thinking about this right and feel like I've hit a dead end from a mistake I've made.

答: As $A$ and $B$ are disjoint events, we have that $A \subset B^{c}$, so $A \cap B^{c} = A$.


答:
It might help to draw a diagram:

$A$ and $B$ are disjoint, so they have no overlap - meaning that $A$ sits entirely inside $B^c$ - meaning that $A \cap B^c = A$.

Alternatively, knowing that $P(A) = P(A \cap B) + P(A \cap B^c)$ for any events $A, B$, you can just plug in the values you already know to get the right answer.

Title: Probability of two disjoint events but the probability of one event is unknown

Body:
I'm given that A and B are two disjoint events where $P(A) = 0.4$.

The first part of the problem asked if $P(B) = 0.8$ is possible which I answered as no since that would make $P(A \cup B)$ bigger than the sample space $P(\Omega)$.

However, the second part then asks me to find $P(A \cap B^c)$. I'm really confused on where to start since I feel like it's unsolvable since I don't actually know what $P(B)$ is. I thought it might of just been 0, but since they're disjoint $A$ is guaranteed to be a subset of $B^c$ which means the answer is a non-zero number.

I'm not sure if I'm thinking about this right and feel like I've hit a dead end from a mistake I've made.


问:
Why is it necesarry to normalize certain vectors in phong reflection model?
To be specific when we do dot product between light vector and normal vector in difusse part - we normalize normal vector.

And dot product between reflection vector and eye position vector in specular part - we again normalize normal vector when we are calculating reflection vector and also view direction vector.

All i was able to find is that we do not care about their magnitude or position, only direction, but why is it so? Why is it mistake to not normalize them?

"When calculating lighting we usually do not care about the magnitude of a vector or their position; we only care about their direction. Because we only care about their direction almost all the calculations are done with unit vectors since it simplifies most calculations (like the dot product). So when doing lighting calculations, make sure you always normalize the relevant vectors to ensure they're actual unit vectors. Forgetting to normalize a vector is a popular mistake." - https://learnopengl.com/Lighting/Basic-Lighting

https://en.wikipedia.org/wiki/Phong_reflection_model

答: They need to be normalized in order so dot product returns cos value <-1, 1> which will be multiplied with color intensity. If it would not be in that interval, color will not be right.


问:
Cartesian product of two collections of sets?
If we have
$\mathcal{G} = \{\phi, \{0\}, \{1\}, \{1,2\} \},$
would $\mathcal{G} \times \mathcal{G}$ have the form $\{\phi, (\{0\}, \{0\}), (\{0\}, \{1\}), (\{0\}, \{0, 1\}), \dots \}$?

In that case, what would an example of $(w_1, w_2) \in B \in \mathcal{G} \times \mathcal{G}$ be?

My working:
Take for example $B = (\{0\}, \{0, 1\})$. This is not a set; so, I don't understand how one could take an element $(w_1, w_2) \in B$.

P.S.: I am taking a Probability Theory class now and even though I did a Measure Theory class last term, I am finding things very difficult. But I really want to get good at this stuff. Would there be any reading material you'd recommend? (I find the lectures hard to follow.)

EDIT:
I encountered this statement in the book Theory of Probability and Random Processes by Koralov. In Definition 2.1, he states:

A finite-dimenstional cylinder is a set of the form $$A = \{\omega: (\omega_{t_1}, \dots, \omega_{t_k}) \in B \},$$ where $t_1, \dots, t_k
\geq 1$, and $B \in \mathcal{G} \times \dots \times \mathcal{G}$ (k
times).

答:
The context of the definition that you give in your edit is quite important. The 'Cartesian Product' can then be interpreted in a different way.

First let's denote $G$ as the set of the $\sigma$-algebra $\mathcal{G}$, i.e. $\mathcal{G} = (G,\mathcal{G})$. Further set  $\mathcal{F} := \mathcal{G} \times \dots \times \mathcal{G}$. If you think of $\mathcal{F}$ as being a $\sigma$-Algebra on the power set of the Cartesian Product of $G$ instead of the Cartesian Product itself, then the expression

$$(\omega_{t_1}, \dots, \omega_{t_k}) \in B \in \mathcal{F} \ \big(=\mathcal{G} \times \dots \times \mathcal{G}\big)$$
makes much more sense.

Because then $\mathcal{F}$ is a subset of $\mathcal{P}(G^{k})$ (where $G$ is the set of the $\sigma$-algebra $\mathcal{G}$) and hence every member $B$ of it is a subset of the Cartesian Product $G \times \dots \times G$.

In fact, in the context of your definition in the book this "$\mathcal{G} \times \dots \times \mathcal{G}$" is - by abuse of notation - a shorthand for the Product $\sigma$-algebra of $k$-times the $\sigma$-algebra $\mathcal{G}$.

Sometimes this is denoted $\mathcal{G} \otimes \dots \otimes \mathcal{G}$ to prevent confusion with the Cartesian Product.

(And the book then goes on to make a connection between the $k$-times Product $\sigma$-algebra and the $n$-times Product $\sigma$-algebra for $k \leq n$. As seen here on page 25.)

About your P.S.: I usually recommend Klenke's Probability Theory which starts with some detailed measure theory then covers the basic theory, stochastic processes, martingales up to stochastic differential calculus. (Plus I got the notation of the Product $\sigma$-algebra from there.)

But you should have a look at many books to find one that might suit you and your class best.


问:
Probability question SAT Math Level 2
I stumbled upon this question on the SAT test(Math Level 2) and I do not know how to solve.

I'd appreciate if someone will explain to me how to think about it as probability is not my strong suit.

"A meteorologist reports that there is 30% probability of rain and no sun. If there is a 40%
probability of no rain, then the probability of both rain and sun is

(A) 0.16
(B) 0.24
(C) 0.30
(D) 0.50
(E) 0.60

答:
There are a couple of steps involved, but the main things you need to consider are:

The probability of event $A$ and the probability of event $A^c$ (the complement of $A$, or "the event that $A$ doesn't happen") sum to 1, i.e. $P(A) + P(A^c) = 1$. For example, the probability of "sun" and "no sun" add up to 1.

For any two events $A$ and $B$, the probability of $A$ and $B$ and the probability of $A$ and $B^c$, sum to the probability of $A$, i.e. $P(A \cap B) + P(A \cap B^c) = P(A)$. For example, the probability of "rain and no sun" plus the probability of "rain and sun" add up to the probability of "rain".

Title: Probability question SAT Math Level 2


问:
Bayes rule with multiple conditions
I am wondering how I would apply Bayes rule to expand an expression with multiple variables on either side of the conditioning bar.

In another forum post, for example, I read that you could expand $P(a,z \mid b)$ using Bayes rule like this
(see Summing over conditional probabilities):
$$P( a,z \mid b) = P(a \mid z,b) P(z \mid b)$$

However, directly using Bayes rule to expand $P(a,z \mid b)$ doesn't seem to be the right way to start out:

$$P(a,z\mid b) = { P(b\mid a,z)P(a,z) \over P(a)}$$

答:
Since the op mentioned expand an expression with multiple variables on EITHER side of the conditioning bar., here is my attempt to derive P(a|z,b) using Bayesian rule, assuming that A is a discrete variable:

$$\begin{align}
P(A=a1|z,b) &= \frac{P(a_1, z, b)}{P(z,b)} \\
         &= \frac{P(a_1,b|z) P(z)}{P(b|z)P(z)} \\
         &= \frac{P(a_1,b|z)}{P(b|z)} \\
         &= \frac{P(b|a_1,z)P(a_1|z)}{P(b|z)} \\
         &=  \frac{P(b|a_1,z)P(a_1|z)}{P(b|a_1,z)P(a_1|z) + P(b|a_2,z)P(a_2|z) + ...}
\end{align}$$

Compare this with the "single variable" version where

$$\begin{align}
P(A=a_1|b) &= \frac{P(a_1,b)}{P(b)} \\
         &= \frac{P(b|a_1)P(a_1)}{P(b)} \\
         &=  \frac{P(b|a_1)P(a_1)}
                  {P(b|a_1)P(a_1) + P(b|a_2)P(a_2) + ..}
\end{align}$$

Can we think of Z as a extra "condition" that's added to both the numerator and denominator?

Title: Bayes rule with multiple conditions


问:
Prove if $f : \mathbb{R}\to\mathbb{R}$ is continuous and $S \subset \mathbb{R}$ is compact, then the image is also compact.
I am not 100% sure about my proof, ideally I would like some hints/confirmation on my proof.

Proof:

Since I want to show $f(S)$ is compact I need to show it is closed and bounded.

Closure

Consider a sequence $(x_n) \subset S$ where  $x_n \to L \in S$ (by compactness). Since $L \in S$ we have that since the function is continuous $f(L) \in f(S)$.

Then using sequential continuity on this sequence  $f(x_n) \to f(L) \in f(S)$. Since we can represent any sequence in $f(S)$ as the image of any given sequence in $S$ the set $f(S)$ is closed.

Bounded

Since $S$ is bounded and $f$ continuous, suppose $s^*$ maximises the function $f$ on $S$ - if $f(s^*) \to \infty$ $f$ is not continuous. Since $max(S)$ and $min(S)$ are finite numbers, we cannot have divergence to infinity of the function over the elements of $s$, so $f(s^*)$ is the upper bound of $f(S)$. Repeating for the lower bound I have that S is bounded.

Is my proof correct. I have concerns that my proof for bounded is not very well worded, also I feel like my proof for closure may miss certain cases.

答:
Nothing to do with $\Bbb R$. Let $f: X \to Y$ be a continuosu function between spaces and $S \subseteq X$ compact.

Let $\mathcal{U}$ be an open cover of $f[S]$. Then $\{f^{-1}[U]\mid U \in \mathcal{U}\}$ is an open cover of $S$ by continuity of $f$ and so has a finite subcover $f^{-1}[U_1],\ldots f[U_n]$. Then $\{U_1, \ldots, U_n\}$ is a finite subcover of $\mathcal{U}$. Hence $f[S]$ is compact.

This is the essence.


问:
Limit by definition when the sign of the variable is unknown
I know that for example if we need to prove by definition ( $\varepsilon - \delta $  ) that:

$$ \lim_{x \to 0^+} \frac{1}{x}  = \infty$$

We need to find $\delta$ such that for all $N$ we have: if $0 < x - 0 < \delta \implies f(x) > N$

So we can do:

$$  \frac{1}{x} >N \implies  x < \frac1N $$

Thus choosing $$ \delta = \frac{1}{N}$$

Solves the problem.

But, what I am stuck with, is what do we do when $x$ is unknown, thus flipping (or not) twice the whole inequality making it impossible to make the form of $x< \text{something}$

For example, prove that:

$$ \lim_{x \to (- \frac{\pi}{2})^+} \frac{1}{\cos x} = \infty$$

We do:

$$ \frac{1}{ \cos x} > N \\ \cos(x) < \frac1N \\ \text{Illegal move?} \\ x > \arccos( \frac1N) \\ \implies \delta = \arccos( \frac1N)$$

I know that $\arccos$ is monotonically decreasing thus we flip the inequality, but then we can't get to a point when we have $x< g(N)$ we only have $x > g(N)$

Another related question: what do we do when we don't have a monotonic function? we can't just take the inverse to isolate $x$ and keep the inequality the same can't we?

答:
"We need to find $\delta$ such that for all $N$ ...". No, you need to switch the quantifiers.

For every $N>0$ you need to find $\delta>0$ such that $0<x<\delta$ implies $f(x)>N$.

Giving an $\epsilon$-$\delta$ proof is, although similar to, not quite the same as solving inequalities.

In your second example, given $N>0$, you need $\delta>0$ such that
$$
-\frac{\pi}{2}<x<-\frac{\pi}{2}+\delta\Rightarrow \frac{1}{\cos(x)}>N
$$

This comes immediately from

the function $\cos(x)$ is continuous and particularly at $x=-\frac{\pi}{2}$

$\cos(-\frac{\pi}{2})=0$

$\cos(x)>0$ on $(-\frac\pi2,0)$


问:
Why is the variance of this random variable one?
I stumbled across the following problem in the paper "STOCHASTIC SPANNING TREE PROBLEM" of Ishii.
There is a random variable defined by
$$
\frac{\sum_j (c_j x_j - \mu_jx_j)}{(\sum_j\sigma_j^2x_j^2)^{1/2}},
$$
where each $c_j$ is distributed according to a normal distribution with mean $\mu_i$ and variance $\sigma_j^2$. It is said that this variable is distributed according to the standard normal distribution and I try to proof this for myself.

For the expected value I get zero, just by using its linearity and then the sums in the numerator cancel out.
Now to the variance: I try to use the property that $\text{Var}(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2$ where I already know that the second term is zero.
Then I square the fraction and aim to get a nominator that cancels out to one with the denominator. Since the denominator is constant I drag it out of the expected value and get
$$
\frac{\mathbb{E} \left[ \left(  \sum_j (c_j x_j) - \sum_j(\mu_jx_j)  \right)^2 \right]}{\sum_j\sigma_j^2x_j^2}
$$
which I expand to
$$
\frac{\mathbb{E} \left[ \left(  \sum_j c_j x_j \right)^2  -2 \left(  \sum_j c_j x_j \right)\left(  \sum_j \mu_j x_j \right) + \left(\sum_j\mu_jx_j  \right)^2 \right]}{\sum_j\sigma_j^2x_j^2}
$$
where I again use the linearity of the expected value to get
$$
\frac{\mathbb{E} \left[ \left(  \sum_j c_j x_j \right)^2 \right] - \left(\sum_j\mu_jx_j  \right)^2}{\sum_j\sigma_j^2x_j^2} = \frac{\mathbb{E} \left[ \left(  \sum_j c_j x_j \right)^2  - \left(\sum_j\mu_jx_j  \right)^2\right]}{\sum_j\sigma_j^2x_j^2}
$$
and now I am stuck and don't know what I miss.

I hope somebody can bring ligth into the dark.

Thank you in advance,

T3 K14

答:
Independence of $c_j$ is needed.

$(\sum_jc_jx_j)^2=\sum_jx_j^2c_j^2+\sum_j\sum_{i\ne j}x_ix_jc_ic_j$

$(\sum_j\mu_jx_j)^2=\sum_jx_j^2\mu_j^2+\sum_j\sum_{i\ne j}x_ix_j\mu_i\mu_j$

For $i\ne j$, $E(c_ic_j)=\mu_i\mu_j$ (independence) so double sum over $i,j$ is zero.

$E(x_j^2(c_j^2-\mu_j^2))=x_j^2(E(c_j^2)-\mu^2)=x_j^2\sigma_j^2$, so sum over $j$ in the numerator is identical to the denominator.


问:
Show that $g(z)=(z-2i)^2-4e^{iz}$ has two zeros in the upper half complex plane
Just finished my exam in Complex Analysis and while I know that they wanted me to do, I just can't find the right function to compare with.

I have shown previous to this that $g$ only has simple zeros and now I have to prove that $g(z)$ has exactly two zeros in $\{w\in\mathbb{C}\vert \text{Im}(w)>0\}$. One of the hints tells us to look at the simple curve that runs from $[-R,R]$ and then in a semicircle with $Re^{it}$ with $t\in[0,\pi]$ and use Rouche's theorem.

I know that the function should have two simple zeros, since $g(z)$ only has simple zeros. I thought of a function like $f(z)=(z-i)(z-2i)$ because it has two zeros in the upper half plane, but I just never moved on from here.

What to do?

答:
This one is not that hard once you notice that $|e^{iz}|<1$ in the upper half-plane so this means that all zeroes of $g$ in the upper plane must be inside the disc $|z-2i| < 2$, while on the circle $C:|z-2i|=2$, the only possible solution would be at $z=0$ which doesn't work since $g(0)=-8$

But then considering $f(z)=(z-2i)^2$ we have that $|f|=2$ on the circle $C$, while for $t \ge 0$ we have that $tf+g=(t+1)(z-2i)^2-4e^{iz}$ doesn't vanish on the circle either as for $t>0, |(t+1)(z-2i)^2|=4(t+1)>4 \ge |4e^{iz}|$ and $t=0$ is the case above ($g$ doesn't vanish on the circle).

Then by Rouche (or just noticing that $\lambda g +(1-\lambda)f, 0 \le \lambda \le 1$ is a homotopy from $f$ to $g$ that doesn't vanish on the circle $C$ by the above), it follows that $f,g$ have the same number of zeroes inside the disc and that is clearly $2$, so we are done!


答:
$P$(no rain$)\space=0.4 \implies P($rain$)=0.6$

$P$(rain and no sun)$\space =0.3$

$\implies P$(rain and sun)$\space =P$(rain)$\space- P($rain and no sun$) \space=0.6-0.3=0.3$

Title: Probability question SAT Math Level 2

Body:
I stumbled upon this question on the SAT test(Math Level 2) and I do not know how to solve.

I'd appreciate if someone will explain to me how to think about it as probability is not my strong suit.

"A meteorologist reports that there is 30% probability of rain and no sun. If there is a 40%
probability of no rain, then the probability of both rain and sun is

(A) 0.16
(B) 0.24
(C) 0.30
(D) 0.50
(E) 0.60


问:
Proving that $\bigcup_{M \in A} \{M_{ij}\} \cap \bigcup_{M \in B}\{M_{ij}\} = \bigcup_{M \in A \cap B} \{M_{ij}\}$
Let $A, B$ be two sets of $n \times n$ matrices, is it true that:

$$\forall i,j \in \{1,...,n\}: \bigcup_{M \in A} \{M_{ij}\} \cap \bigcup_{M \in B}\{M_{ij}\} = \bigcup_{M \in A \cap B} \{M_{ij}\}$$

I was able to prove that:

$$\bigcup_{M \in A \cap B} \{M_{ij}\} \subseteq \bigcup_{M \in A} \{M_{ij}\} \cap \bigcup_{M \in B}\{M_{ij}\}$$

But I'm having some trouble proving the opposite.

I started with: Let $i,j \in \{1,...,n\}$ and  $x \in \bigcup_{M \in A} \{M_{ij}\} \cap \bigcup_{M \in B}\{M_{ij}\}$, then:

$x \in \bigcup_{M \in A} \{M_{ij}\} \wedge x \in \bigcup_{M \in B}\{M_{ij}\}$ meaning that $\exists M \in A:M_{ij} = x$ and $\exists M' \in B: M'_{ij}=x$, but Now to colcude that $\exists K \in A \cap B: K_{ij}=x$ I would need that $M=M'$.

Is there a way to prove this or is this simply not true?

答:
As you suspected, the converse containment doesn't hold. To construct a simple counterexample, just take the sets $A$ and $B$ to be singleton sets with a matrix that has the same entry in $(i,j)$ spot and different entry in some other.

As an explicit counterexample, consider $A=\left\{\begin{bmatrix}0&0\\0&0\end{bmatrix}\right\}$ and $B=\left\{\begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}$

Then $A\cap B=\emptyset$ and thus your LHS of the claim is the set $\{0\}$ for all $(i,j)$ except $(2,2)$ and the RHS is always $\emptyset$

A better counterexample:

To construct a counterexample for some $(i,j)$, make the sets $A,B$ such that $\bigcup_{x\in A} x_{ij}=\bigcup_{x\in B} x_{ij}$ but other entries are changed in the two matrices to make them not be in $A\cap B$

$$A=\left\{\begin{bmatrix}\color{red}0&0\\0&0\end{bmatrix}, \begin{bmatrix}\color{red}1&\underline{0}\\0&0\end{bmatrix}\right\}\\ B=\left\{\begin{bmatrix}\color{red}0&0\\0&0\end{bmatrix},\begin{bmatrix}\color{red}1&\underline{1}\\0&0\end{bmatrix}\right\}$$

Then, for $(i,j)=(1,1)$, the LHS of your claim is $\{\color{red}0,\color{red}1\}$ but the RHS of
your claim is $\{0\}$ because $A\cap B$ is only the singleton set with zero matrix.

To keep both sets of the claim non-empty, I let the zero matrix be in both $A$ and $B$ but changed the $(1,2)$-entry of the second element of $A,B$ keeping the $(1,1)$-entry same.


问:
Find the value of $\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}$
Can this limit be solved without using L'Hopital's rule :

$$\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}$$

Answer of this limit is : $1$

答:
$\displaystyle\lim_{x\to\infty }\left(\tan^{-1}\left(\frac{1}{x}\right) \right )^{\frac{1}{x}}=\lim_{x\to\infty }e^{\frac{\ln\left(\tan^{-1}\left(\frac{1}{x}\right)\right)}{x}}=\Bigg[\tan^{-1}\left(\frac{1}{x}\right)=u\Bigg/\tan\implies \frac{1}{x}=\tan u\Bigg]$
$$x\to\infty\implies u\to 0^+$$

then $$\lim_{x\to\infty }\left(\tan^{-1}\left(\frac{1}{x}\right)\right)^{\frac{1}{x}}=\lim_{u\to 0^+}u^{\tan (u)}=\lim_{u\to 0^+}e^{\tan(u)\ln(u)}$$

$$\lim_{u\to 0^+}e^{\tan(u) \ln(u)}=\lim_{u\to 0^+}e^{\frac{\tan(u)}{u} u\ln(u)}$$

we know that $$\lim_{u\to 0^+}\frac{\tan(u)}{u}=1\;\&\;\lim_{u\to 0^+}u\ln(u)=0$$

so $$\lim_{u\to 0^+}e^{\frac{\tan(u)}{u} u\ln(u)}=e^{0}=1$$

so $$\lim_{x\to\infty }\left(\tan^{-1}\left(\frac{1}{x}\right)\right)^{\frac{1}{x}}=1$$


问:
Prove that $\sigma(am) < \sigma(am + 1)$ for infinitely many positive integers $m$
Given a positive integer $a$, prove that $\sigma(am) < \sigma(am + 1)$ for infinitely many positive integers $m$.

($\sigma(n)$ is the sum of all positive divisors of the positive integer number $n$.)

I have two problem when trying to use the hint.

Let $p_1,p_2,...,$ be the primes which do not divide $a$. Let $k$ be the minimum integer such that $\dfrac{2\sigma(a)}{a} < \prod_{i=1}^k \left (1 + \dfrac{1}{p_i} \right )$. How can I prove $\prod_{i=1}^\infty \left (1 + \dfrac{1}{p_i}\right )$ diverges to show $k$ exist.

By Dirichlet's Theorem there exists infinitely many primes $q$ such that $q \equiv -a^{-1} \pmod{p_i}$ for $1 \le i \le k$. Choose $m=q$, then
$\sigma(aq+1) \ge (aq+1) \cdot \prod_{i=1}^k \left (1 + \dfrac{1}{p_i}\right ) > \dfrac{2(aq+1)\sigma(a)}{a} > 2q\sigma(a) > \sigma(aq)$.

The problem is done after that, but how can I prove $q$ exists without using Dirichlet's Theorem (which are too much for a High School Olympiad problem).

答:
This is only an idea for how one might proceed without using Dirichlet

Let $p_1,p_2,...$ be the primes, in numerical order, which do not divide $a$. For some integer $N$, let $s$ be the multiplicative inverse of $p_1p_2... p_N$ modulo $a$, where $ 1\le s<a$. Then there is an integer $m$ such that
$$am+1\equiv p_1p_2... p_Ns.$$
Then $m< p_1p_2... p_N$ and is only divisible by primes which either divide $a$ or are $p_i$ for $i>N$.

We have $\sigma(am)<am \prod_{q|a} \frac{q}{q-1}\prod_{p_i|m} \frac{p_i}{p_i-1} $ and $\sigma(am+1)>(am+1)\prod_{i=1}^{N}\frac{p_i+1}{p_i}, $ where $\prod_{p_i|m} {p_i}< p_1p_2... p_N$.

Now $\prod_{q|a} \frac{q}{q-1}$ is fixed. The $p_i$ which divide $m$ are larger and fewer than the $p_i,1\le i \le N$. Let $N$ tend to infinity ....?


问:
〈b, a, b〉 and 〈a, b, a〉: For what non zero values of a and b are these two vectors parallel?
Looking for a hint on how to solve this problem.
I know for the vectors to be parallel, the cross product must equal the zero vector, but I'm unsure on how to use that information to solve for values.

$\langle b, a, b\rangle \times\langle a, b, a\rangle$.

For what non zero values of $a$ and $b$ are these two vectors parallel?

答: You have the right idea. Take the cross product and simplify the resulting expression. See if you can solve for $a$ in terms of $b$.


问:
Show that $Y_t \int_0^t \frac{1}{Y_s} ds$ and $\int_0^t Y_s ds$ are equal in distribution
Let $B=(B_t)_{t \geq0}$ be a standard Brownian motion and let the two stochastic processes $X=(X_t)_{t \geq0}$ and $Y=(Y_t)_{t \geq0}$ solve the following two stochastic differential equations

$$dX_t = (1+3X_t)dt+ (1+5X_t)dB_t$$

and

$$dY_t = 3Y_tdt+ 5Y_tdB_t$$

respectively. Also we have $X_0 = Y_0 =1$. The task is to show that

$$X_t- Y_t(1+\int_0^t\frac{1}{Y_s}dB_s) \stackrel{d}{=} -4\int_0^tY_s ds.$$

I have already shown using Ito's lemma that

$$X_t = Y_t(1+\int_0^t\frac{1}{Y_s}dB_s -4\int_0^t \frac{1}{Y_s}ds)$$

thus to finish the problem I need to show that

$$Y_t \int_0^t \frac{1}{Y_s} ds \stackrel{d}{=} \int_0^t Y_s ds.$$

I'm not really sure how to go about doing this? The wording is actually equal in law but I assume this means equal in distribution?

Edit:

I have shown that $$Y_s \sim \text{Lognormal}(-\frac{19s}{2}, 25s)$$ and $$\frac{Y_t}{Y_s} \sim \text{Lognormal}(-\frac{19(t-s)}{2}, 25(t-s)).$$ Is this sufficient?

答:
The equality in law
$$
\int_0^t \frac{Y_t}{Y_s} ds \stackrel{d}{=} \int_0^t Y_s ds
$$
is indeed true.

Just note that (using simple calculus) the rhs side is equal to $\int_0^t Y_{t-s} ds$

Now, the processes $s \mapsto \frac{Y_t}{Y_s} $ and
$s \mapsto Y_{t-s} $ are equal in law (do you see why?), so their time integrals will also be equal in law.


问:
About a detail of the proof of $\mathsf{GCH}$ via iterated ultrapower
I am reading the proof of $L[U] \models \mathsf{GCH}$ given on Mitchell's Beginning Inner Model Theory at the handbook, but I confronted a detail how to fill the details.

Mitchell's proof goes as follows: let $\kappa$ be a measurable cardinal with a normal measure $U$. We only consider the case $\lambda<\kappa$, and we will show that $2^\lambda=\lambda^+$.

Let $x\subseteq \lambda$. Take $\tau$ such that $x\in L_\tau[U]\models \mathsf{ZFC^-}$. Take a Skolem hull $X\prec L_\tau[U]$ such that $\lambda\cup\{x\}\subseteq X$ and $|X|=\lambda$. If $M_x\cong X$ be the transitive collapse, then $M_x=L_{\alpha_x}[U_x]$ for some ordinal $\alpha_x$ and a normal filter $U_x$.

We claim that $\{z\subseteq \lambda\mid z<_{L[U]}x\}\subseteq M_x$. Consider the iterated ultrapower $\operatorname{Ult}_\kappa (M_x,U_x)$ and $i^{U_x}_\kappa: M_x\to \operatorname{Ult}_\kappa (M_x,U_x)$. Then

$\operatorname{Ult}_\kappa (M_x,U_x)=L_{\alpha_x'}[i^{U_x}_\kappa(U_x)]$ for some $\alpha_x'$, and

$i^{U_x}_\kappa(U_x)\subseteq U$, since $i^{U_x}_\kappa(U_x)$ is 'generated' by a club set $\{i_\nu^{U_x}(\kappa):\nu<\kappa\}$.

Since $i_\lambda^{U_x}$ is the identity over $\mathcal{P}^{M_x}(\lambda)$, we have $\{z\subseteq\lambda\mid z<_{L[U]}x\}\subseteq \mathcal{P}^{L_{\alpha_x'}[U_x]}(\lambda)\subseteq M_x$.

I see that why $i_\lambda^{U_x}$ is the identity over $\mathcal{P}^{M_x}(\lambda)$, but I have no idea that why the inclusions hold.
The second inclusion has some oddity since $\alpha_x'$ apparently bears no relation with $U_x$. I guess $U_x$ is a typo of $i_\kappa^{U_x}(U_x)$, but is it really a typo?
I also have no idea about how to show the first inclusion. The proof has not used $i^{U_x}_\kappa(U_x)\subseteq U$, so I guess it might be related. I would appreciate your help.

答:
First, I believe you are right that $U_x$ is a typo and should be $i^{U_x}_\kappa(U_x)$ instead.

I don't know how the proof should go without some kind of comparison lemma. In particular, I am not sure if this is the sketch that the article has in mind. But here it goes: with the comparison lemma, we iterate $L[U]$ and $M_x$ far enough so that they become comparable as some $L[D]$ and $L_\xi[D]$ respectively.

So now let $z<_{L[U]}x$ where $x\in M_x$. The iterate $L_\xi[D]$ of $M_x$ has the same rank $\lambda+1$ stuff as $M_x$, so in particular $x\in V_{\lambda+1}\cap M_x = V_{\lambda+1}\cap L_\xi[D]$. Also, the embedding from $L[U]$ to $L[D]$ fixes all elements from $V_\kappa$ and since $\kappa>\lambda$, we have that $z<_{L[D]}x$ in $L[D]$ as well. But now it's straightforward that if $x\in L_\xi[D]$ and $z<_{L[D]}x$, then $z\in L_\xi[D]$, by general properties of the definable well-ordering of structures like $L[D]$.


问:
$\sum a_n$ be convergent but not absolutely convergent, $\sum_{n=1}^{\infty} a_n=0$
Let $\sum a_n$ be convergent but not absolutely convergent, $\sum_{n=1}^{\infty} a_n=0$,$s_k$ denotes the partial sum then could anyone tell me which of the following is/are correct?

$1.$ $s_k=0$ for infinitely many $k$

$2$. $s_k>0$ and $<0$ for infnitely many $k$

3.$s_k>0$ for all $k$

4.$s_k>0$ for all but finitely many $k$

if we take $a_n=(-1)^n{1\over n}$ then $\sum a_n$ is convergent but not absolutely convergent,but I don't know $\sum_{n=1}^{\infty} a_n=0$? so I am puzzled could any one tell me how to proceed?

答:
None of them are necessarily true.

We can easily compute a series from its partial sums, so let's specify the $s_k$.

Define
$$
s_k=\left\{\begin{array}{}
-\frac1k&\text{if $k$ is odd}\\[4pt]
-\frac1{k^2}&\text{if $k$ is even}
\end{array}\right.
$$
Then $a_1=-1$ and for $k\gt1$,
$$
a_k=\left\{\begin{array}{}
\frac1{(k-1)^2}-\frac1k&\text{if $k$ is odd}\\[4pt]
\frac1{k-1}-\frac1{k^2}&\text{if $k$ is even}
\end{array}\right.
$$
Show that this series is not absolutely convergent, its sum is $0$, and it fails to satisfy any of the conditions.


问:
Exhibit an example of a vector space $V$ with non-equal subspaces $U_1, U_2, U_3$ such that $U_1 \oplus U_3 = U_2 \oplus U_3$.
Can someone give me a hint? Not an answer.

Exhibit an example of a vector space $V$ with non-equal subspaces $U_1, U_2, U_3$ such that $U_1 \oplus U_3 = U_2 \oplus U_3$.

$\oplus$ means direct sum.

How does $U_1 \neq U_2$ if $U_1 \oplus U_3 = U_2 \oplus U_3$?

Body:
Consider the usual vector space $V=\mathbb{R}^{2}.$ Let $e_{1}=(1,0)$
and $e_{2}=(0,1)$. Let $U_{1}=\mbox{span}\{e_{1}\},$ $U_{2}=\mbox{span}\{e_{1}+e_{2}\}$,
and $U_{3}=\mbox{span}\{e_{2}\}$, then $U_{1}\oplus U_{3}=U_{2}\oplus U_{3}=V$.


问:
Carmichael lambda function proof

Let $\lambda$ be the Carmichael Lambda function, as defined on wikipedia. Let $a,b\in\mathbb{Z}.$ Prove the following

If $a \mid  b$ then $\lambda(a) \mid\lambda(b).$

$\lambda(\operatorname{lcm}(a,b)) = \operatorname{lcm}(\lambda(a), \lambda(b))$.

I think the first statement can be shown using the unique factorization theorem and a case-by-case proof for the values of $\lambda$ on prime powers.

Wikipedia says that the second statement follows immediately from the recursive definition, but I don't think this is so; there's more to it than that. However, I'm not really sure how to prove the second statement by using the first definition more closely; if $a$ and $b$ are coprime then maybe using the unique factorization theorem will help but I'm not sure how to extend this result to when $a$ and $b$ are not coprime. I'd probably start by using this result for $\frac{a}{\gcd(a,b)}$ and $\frac{b}{\gcd(a,b)}.$

答:
Hint $ $ Like analogous proofs for  Euler $\phi$ (totient) function, by multiplicativity we can reduce to the case of prime powers, where the proofs are straightforward. Alternatively, and more conceptually, we can use the universal properties of lcm and $\lambda$ (= universal order or expt). Let's prove $(2)$ in this manner. First, by $(1)$ and the lcm universal property we have, with $[x,y] := {\rm lcm}(x,y)$

$$a,b\mid [a,b]\,\Rightarrow\, \lambda(a),\lambda(b)\mid \lambda([a,b])\,\Rightarrow\, [\lambda(a),\lambda(b)]\mid \lambda([a,b])\qquad$$

For the reverse divisibility, if $(x,[a,b])=1$ then $(x,a)=1=(x,b),\,$ so $\,x^{\lambda(a)}\equiv 1\pmod{\!a},\,$ $\,x^{\lambda(b)}\equiv  1\pmod{\!b},\,$ so $\,x^{[\lambda(a),\lambda(b)]}\equiv 1\pmod{[a,b]},\,$ thus $\,\lambda([a,b])\mid [\lambda(a),\lambda(b)],\,$ by the universal analog of the Order Theorem, which shows that the minimal universal expt $\,\lambda\,$ divides every universal expt.

Title: Carmichael lambda function proof


问:
$\sum_{k = 1}^{\infty}\frac{1}{k^c} < 1 + \sum_{k = 1}^{\infty}\big(\frac{1}{2^\omega}\big)^k$. Why is this inequality true?
The proof to the generalized harmonic series has the following inequality:

$$\sum_{k = 1}^{\infty}\frac{1}{k^c} < 1 + \sum_{k = 1}^{\infty}\big(\frac{1}{2^\omega}\big)^k$$

with $c > 1$ and $c = 1 + \omega$, $\omega > 0$.

I am aware that the generalized harmonic series can be proven alternatively. But I don't see the above inequality. Why is that true? I tried to list out a few of the first terms and compare them term-by-term.

LHS: $ 1 + \frac{1}{2\cdot2^{\omega}} + \frac{1}{3\cdot3^{\omega}} + \frac{1}{4\cdot4^{\omega}} + \frac{1}{5\cdot5^{\omega}} + \dots$

RHS: $ 1 + \{\frac{1}{2^{\omega}} + \frac{1}{2^{2\omega}} + \frac{1}{2^{3\omega}} + \frac{1}{2^{4\omega}} + \dots\} = 1 + \{\frac{1}{2^{\omega}} + \frac{1}{4^{\omega}} + \frac{1}{8^{\omega}} + \frac{1}{16^{\omega}} + \dots\}$

Starting from $k = 5$, however, the term of LHS is not necessarily smaller than that of RHS.

Body:
Group by packets of size $2^n$:
$$\sum_{k=1}^{+\infty}\frac 1 {k^c}=\sum_{n=0}^{+\infty}\left(\sum_{m=2^n}^{2^{n+1}-1}\frac 1 {m^c}\right)< \sum_{n=0}^{+\infty}\frac{2^n}{(2^n)^c}=\sum_{n=0}^{+\infty}\frac 1 {2^{\omega n}}$$


答:
You don't really need to use the cross-product.

 If two vectors are parallel, then one is a scalar multiple of the other.

In this case, that means $b/a=a/b$, which implies $b^2=a^2$, which implies $b=\pm a$.

Title: 〈b, a, b〉 and 〈a, b, a〉: For what non zero values of a and b are these two vectors parallel?

Body:
Looking for a hint on how to solve this problem.
I know for the vectors to be parallel, the cross product must equal the zero vector, but I'm unsure on how to use that information to solve for values.

$\langle b, a, b\rangle \times\langle a, b, a\rangle$.

For what non zero values of $a$ and $b$ are these two vectors parallel?


问:
Function is either nowhere zero or a constant function.

Hurwitz's theorem:

Let $\{ f_n\}$ be a sequence of analytic functions on an open set $D$, which converges uniformly to a function $f$ on compact subsets of $D$. If the functions $f_n$ are nowhere zero, then $f$ is either identically zero or nowhere zero.

Proof by contradiction:

Suppose $f (z) = 0$ for some $z ∈ D$ and that $f(z) \not \equiv 0$. Take a small curve $\gamma \subset D $, centered at $z$, then $f(z) \neq 0$ on $\gamma$. Thus,
$$
\frac{f'_n}{f_n} \to \frac{f'}{f} \text{ on } \gamma
$$
and therefore,
$$
\frac{1}{2\pi i}\oint_{\gamma}\frac{f'(z)}{f(z)}dz=\lim_{n\to\infty}\frac{1}{2\pi i}\oint_{\gamma}\frac{f_n'(z)}{f_n(z)}dz=0 \tag{*}
$$
However,
$$
\frac{1}{2\pi i}\oint_{\gamma}\frac{f'(z)}{f(z)}dz \geq 1
$$
according to the argument principle. Thus, we have a contradiction and therefore $f(z) \equiv 0$ or nowhere zero.

My question is whether the above proof for Hurwitz's theorem shows that it satisfies constant functions as well.

答: Well the theorem applies to functions that are either identically zero or nowhere zero. Any constant function is either the zero function (identically zero) or not the zero function (nowhere zero).</p>


问:
Can finite induction be independent of mathematical induction?
Suppose we have a property $P(n)$ pertaining to natural numbers and $P(n) \to P(n+1)$.
We know by mathematical induction that $P(N)$ implies $P$ being true for all natural numbers greater than or equal to $N$.

It is an obvious consequence of this that for a finite number of natural numbers greater than or equal to $N$, $P$ is also true. I am wondering, however, that whether this "finite induction" can be applied even if we didn't know mathematical induction principle. This might seem trivial, just like $P(1) \to P(2)$, $P(2) \to P(3)$ and so on, but I don't know any logical rules approving this...

答:
Induction will hold an any set $X$ (possibly finite) with $x_0\in X$ (the "first" element) and $f: X \to X$ (the "successor" function) such that:

$~~~~~~X=\{ x_0, f(x_0), f(f(x_0)), \cdots \} $

We have:

$~~~~~~\forall P\subset X: [x_0\in P \land \forall a\in P:[ f(a)\in P] \implies P=X]$


问:
Area bounded with curves
I need to calculate area bounded with curves $(x^2+y^2)^2\le 3(x^2-y^2), x^2+y^2\le \sqrt{2}x$. Here we have inside of leminscate and circle. I think i should use that $P=\iint dxdx$ but not sure how.

答:
See the picture below:

Expressing the upper half of the region that you mentioned in polar coordinates, we get:

$r^4\geqslant3r^2\cos(2\theta)\bigl(\iff r^2\geqslant3\cos(2\theta)\bigr)$;

$r^2\leqslant\sqrt2\,r\cos(\theta)\bigl(\iff r\leqslant\sqrt2\cos(\theta)\iff r^2\leqslant2\cos^2(\theta)\bigr)$

We have\begin{align}3\cos(2\theta)\geqslant2\cos^2(\theta)&\iff6\cos^2(\theta)-3\geqslant2\cos^2(\theta)\\&\iff\cos(\theta)\geqslant\frac{\sqrt3}2\\&\iff\theta\geqslant\frac\pi6.\end{align}On the other hand, if $\theta\in\left[\frac\pi4,\frac\pi2\right]$, then $3\cos(2\theta)\leqslant0$. So, compute$$\int_{\pi/6}^{\pi/4}\int_{\sqrt{3\cos(2\theta)}}^{\sqrt2\cos(\theta)}\rho\,\mathrm d\rho\,\mathrm d\theta+\int_{\pi/4}^{\pi/2}\int_0^{\sqrt2\cos(\theta)}\rho\,\mathrm d\rho\,\mathrm d\theta.$$You should get$$\frac\pi6-\frac{3-\sqrt3}4$$and therefore the area that you're after is$$\frac\pi3-\frac{3-\sqrt3}2\approx0.413223.$$

Title: Area bounded with curves


答:
Cross product will work.  What is the cross product? $(a^2-b^2,0, b^2-a^2) = \bf 0$

For what values of $a,b$ does $a^2 - b^2 = 0$?

$a = \pm b$

Alternatively (and I think more simply)

The vectors are parallel if:

$(a,b,a) = \lambda (b,a,b)$

giving

$a  = \lambda b\\
b = \lambda a$

and

$\lambda = \pm 1$


问:
Uniqueness of an Isomorphism
I need to prove the following:

The numbers $5$ and $7$ are coprime, so there exists an Isomorphism $f\ : \mathbb{Z}/3\mathbb{Z}\ \times\ \mathbb{Z}/4\mathbb{Z}\ \rightarrow\  \mathbb{Z}/35\mathbb{Z}$. (I have found that isomorphism and wrote it as a table: $([a]_3, [b]_4)\ \rightarrow\ [4a + b]_{12}$)). How can I prove that this Isomorphism is unique, that there is no other isomorphism  $f'\ : \mathbb{Z}/3\mathbb{Z}\ \times\ \mathbb{Z}/4\mathbb{Z}\ \rightarrow\  \mathbb{Z}/12\mathbb{Z}$, with $f'\ =\ f$? The Chinese remainder theorem doesn't mention anything about uniqueness, nor does Bezout's identity.

Thanks in advance :D

答:
Note: The map you propose is not an isomorphism! It’s not even a ring homomorphism.

You send $([1]_3,[1]_4)$ to $[5]_{12}$, but the square of $[5]_{12}$ is $[1]_{12}$, and the square of $([1]_3,[1]_4)$ is itself. So your map does not send products to products. Altenrnatively, you send $([1]_3,[0]_4)$ to $[4]_{12}$. But the product of $([1]_3,[1]_4)$ and $([1]_3,[0]_4)$ is $([1]_3,[0]_4)$, so the product of their images should be $[4]_{12}$, and yet $[5]_{12}\times[4]_{12}=[20]_{12}=[8]_{12}\neq[4]_{12}$.

To prove uniqueness (once you find the right isomorphism), the key is whether $\mathbb{Z}/12\mathbb{Z}$ has nontrivial automorphisms or not. If the only automorphism is the identity, and $\phi,\psi$ are two isomorphisms, then since $\phi\circ\psi^{-1}$ is an automorphism of $\mathbb{Z}/12\mathbb{Z}$ it must equal the identity, hence $\phi=\psi$. Conversely, if $\mathbb{Z}12\mathbb{Z}$ has a non-identity automorphism $\theta$, and $\phi$ is an isomorphism, then $\theta\circ\phi$ is also an isomorphism, but $\theta\phi\neq\phi$ since $\phi$ is invertible, but $\theta\neq\mathrm{id}$.

If you require your ring morphisms to be unital, then any automorphism must send $[1]$ to $[1]$, and hence by induction if $\theta$ is an automorphism then $\theta([n])=[n]$ for all $n$ and the result follows.

If you do not require your ring morphisms to be unital, then since $[1][1]=[1]$, then $\theta([1])$ must be an idempotent. Moreover, because $\theta$ is also a group isomorphism, it must send $[1]$ to an additive generator of $\mathbb{Z}/12\mathbb{Z}$. The additive generators are $[1]$, $[5]$, $[7]$, and $[11]$, and the only idempotent among them is $[1]$ (the rest have $[5][5]=[7][7]=[11][11]=[1]$). So the automorphism must still send $[1]$ to $[1]$ and the result follows as before.

Title: Uniqueness of an Isomorphism


问:
Proof that a Cartesian category is monoidal
If $\mathcal C$ is a category with products and a terminal object, then $\mathcal C$ is monoidal.

This seems obvious, but wherever I look for a proof or a reference it simply states that the proof is "pretty involved".

I found this 8 years old question on MathOverflow:

https://mathoverflow.net/questions/111193/monoidal-structure-on-a-category-with-products-and-with-terminal-object

The answer given there uses the fact that Set is monoidal, along with the Yoneda lemma.
Is there any other reference that proves it more "generally" (i.e without resorting to Set), or a standard reference?

答:
To define the product functor we need to fix a binary product, together with the projections, for each pair of objects.

The first thing to verify here is that it indeed defines a functor, i.e. uniquely determines the image of any pair of morphisms (in a functorial way).

I guess, the most problematic parts of the statement are the coherence conditions.

To get the proof smoother, let's also fix ternary and quarternary products (and nullary product as fixing the terminal object $1$, and unary products with the identity as projection).

For example, if we write $(a\times b\times c)\times d$, that is a binary product whose first term is a ternary product, and by the universal property we can easily see that it has a unique isomorphism from the quarternary product $a\times b\times c\times d$ that commutes with our fixed projections.

Similarly, for any act of 'putting in a pair of parenthesis' into a product of objects there belongs a unique isomorphism.

Now we define the associators $\alpha_{a,b,c}$ as the composite $(a\times b)\times c\longleftarrow a\times b\times c\longrightarrow a\times (b\times c)$ of the above isomorphisms.

Once again using the universal property of products, one can prove that if we put two pairs of parentheses in a product through the above isomorphisms, then the order doesn't matter: we'll receive commutative squares (like in the picture below).

To conclude the coherence pentagon condition, I found this picture among my old notes (on Leinster's unbiased bicategories), where the product sign $\times$ is omitted (so for example the top arrow is read as $((a\times b)\times c)\times d\overset{\alpha_{a,b,c}\times d}\longrightarrow (a\times(b\times c))\times d$).

The coherence conditions for unit are similar but easier, they might be thought as the case when we put an empty pair of parentheses into a product, like in "$ab\to ab()$" which would mean $a\times b\longrightarrow a\times b\times 1$.


问:
Prove/disprove $(\int_0^{2 \pi} \!\!\cos f(x) \, d x)^2+(\int_0^{2 \pi}\!\!\! \sqrt{(f'(x))^2+\sin ^2 f(x)} \, dx)^2\ge 4\pi^2$

Let $f(x)$ be a differentiable function on $[0,2\pi]$ s.t. $0\leq f(x)\leq 2\pi$ and $f(0)=f(2\pi)$. Prove or disprove that
$$
\left(\int_0^{2 \pi} \cos f(x) \,d x\right)^2+\left(\int_0^{2 \pi} \sqrt{(f'(x))^2+\sin^2 f(x)} \, d x\right)^2 \geq(2 \pi)^2
$$

It seems that when $f$ is an arbitrary constant, the left side equals $(2\pi)^2$ and seems to be the minimum. But how can I show that there's no other $f$ that makes the left side equal (or be less than) $(2\pi)^2$?

A geometric interpretation of the inequality has been found: Consider a closed curve on a sphere: $C=\{(\cos x\cdot\sin f(x),\,\sin x\cdot\sin f(x),\,\cos f(x))\mid x\in[0,2\pi)\}$, we have its perimeter $\displaystyle L=\int_0^{2\pi}\sqrt{(f'(x))^2+\sin^2 f(x)}\,dx$ and its area $\displaystyle S=2\pi-\int_0^{2\pi}\cos f(x)\,dx$. From spherical isoperimetric inequality $L^2\ge S(4\pi-S)$, we have $(2\pi-S)^2+L^2\ge(2\pi)^2$, and the equality holds iff $C$ is any circle on the sphere. In this way we get the original inequality in the sense of geometry.

Now the question is, how to prove the inequality with only pure analysis methods?

答:
Maybe I am simplifying this too much, but if f(x) is not constant, then $$\sqrt{(f'(x))^2 + (\sin f(x))^2} > |\sin f(x)|$$ because $$f'(x) \neq 0$$ for some value(s) in the interval of integration. This leads to the following:
$$\left(\int_0^{2\pi} \cos f(x) \: dx \right)^2 + \left( \int_0^{2\pi} \sqrt{(f'(x))^2 + \sin^2 f(x)} \: dx \right)^2 $$
$$ \geq \left(\int_0^{2\pi} \cos f(x) \: dx \right)^2 + \left( \int_0^{2\pi} \sqrt{\sin^2 f(x))^2} \: dx \right)^2 = (2 \pi)^2$$


问:
Proving $\mathcal{L}$-structures can be elementarily equivalent, but not isomorphic
I'm reading through Marker's Model Theory: An Introduction, and he defines two $\mathcal{L}$-structures $\mathcal{M}, \mathcal{N}$ to be isomorphic if there exists a bijection $\eta : M \to N$ which respects the interpretations of the various symbols of the language $\mathcal{L}$. He also defines two structures to be elementarily equivalent if they have the same full theory (i.e. set of true sentences). He shows that isomorphic structures are elementarily equivalent, but not vice versa. My intuition is that isomorphism is strictly stronger than elementary equivalence, and I have an idea of how to construct an example. If $\mathcal{L} = \emptyset$ is the empty language of pure sets, then the $\mathcal{L}$-structures are just sets and any bijection is an isomorphism between structures.

However, I suspect that any two infinite sets are elementarily equivalent. The atomic formulas in the language are exactly the formulas $v_i = v_j$ for variables $v_i, v_j$. My intuition is that any sentence is either (a) contradictory, i.e. is true for no $\mathcal{L}$-structures, (b) describes the existence of a subset of a cardinality of some finite cardinality, or (c) describes the nonexistence of a subset of some cardinality $n$. Statements of type (a) would never be true of an infinite set, statements of type (b) would be true of all infinite sets, and statements of type (c) would never be true of an infinite set. My reason for thinking these are the only statements is that a formula can only every describe some combination of equalities or inequalities between finitely many elements, since it only has finitely many variable symbols in it. But I have no idea how to turn this intuition into a fleshed out proof.

I've only finished the first chapter of the book, so I suppose it's possible that I'm asking a question that requires way more technology than I have so far. How can I turn my proof idea into an actual proof? Or am I off base altogether?

Thanks!

答:
You have the exact right idea!

There are a few ways to make this precise. One way is to show that the theory of infinite sets is complete. This means that, up to elementary equivalence, there is only one model.

But one way to show completeness is to use the Łoś–Vaught Test, which says that a theory is complete whenever there is exactly one model of some infinite cardinality (and there's no finite models).

So for your example, it's clear that any two models of the same cardinality are isomorphic. And if you add axioms saying "I have infinitely many elements" then your theory satisfies the conditions for the Łoś–Vaught test.

Then the theory of infinite sets is complete, and any two models are elementary equivalent.

Another way to make your exact idea precise is with the notion of an Ehrenfeucht–Fraïssé Game. These say, basically, that if you can "finitely simulate" any formula from one model in another, then your two models are elementary equivalent.

The breakdown of possible formulas that you've outlined can quickly be converted into a formal Ehrenfeucht–Fraïssé proof. The reason I led with the completeness proof is because I encourage you to do the conversion yourself! I'm sure that now you know the name of this technique you'll be able to flesh out your idea!

Of course, you can always ask here again if you get stuck.

I hope this helps ^_^


问:
proving the symmetric and transitive properties of an equivalence relation
Let $S$ be a non-empty set, $A$ be the set of all functions from $S$ to $S$, and $B$ be the set of all bijective functions from $S$ to $S$. Define the following relation on A:

$$f \sim g \iff \exists h \in B \mid fh=g$$

where $fh$ is the composition of $f$ and $h$. Is this an equivalence relation on $A$?

This is what I have so far:

$ah=a ⇒  a^{-1} ah=a^{-1} a ⇒ eh=e⇒ h=e$ and $e \in B$ therfore $a \sim a$
Because $e$ is the identity element and all sets have an identity element, therefore it is reflexive.

$$ah=b ⇒ a^{-1} ah=a^{-1} b⇒ eh=a^{-1} b⇒ h=a^{-1} b$$
$$bh=a⇒ b^{-1} bh=b^{-1} a⇒ eh=b^{-1} a⇒ h=b^{-1} a$$
But I do not know how to show that since $a \sim b$ then $b \sim a$. I think I have a start but I’m not sure how to show this.

I am also not sure how to show that it is transitive because I keep solving for h but don’t know how to show this.

答:
For symmetry show that $ah=b$ implies $bh^{-1}=a$ .

For transitivity show that $ah_1=b$ and $bh_2=c$ implies $ah=c$ where $h=h_1 h_2$.


答:
There is nothing that says that $a$ is bijective or that $a^{-1}$ exists.

But $h$ is bijective and $h^{-1}$ exists

Reflexive $a\sim a$

Let $h$ be the identity function.

Symmetric $a\sim b \implies b\sim a$

There exists an $h$ such that $a\circ h = b$

$b\circ h^{-1} = a$

Transitive $a\sim b$ and $b\sim c \implies a\sim c$

$a\circ h = b\\
b\circ g = c\\
a\circ (h\circ g) = c$

The composition of two bijections is a bijection.

Title: proving the symmetric and transitive properties of an equivalence relation

Body:
Let $S$ be a non-empty set, $A$ be the set of all functions from $S$ to $S$, and $B$ be the set of all bijective functions from $S$ to $S$. Define the following relation on A:

$$f \sim g \iff \exists h \in B \mid fh=g$$

where $fh$ is the composition of $f$ and $h$. Is this an equivalence relation on $A$?

This is what I have so far:

$ah=a ⇒  a^{-1} ah=a^{-1} a ⇒ eh=e⇒ h=e$ and $e \in B$ therfore $a \sim a$
Because $e$ is the identity element and all sets have an identity element, therefore it is reflexive.

$$ah=b ⇒ a^{-1} ah=a^{-1} b⇒ eh=a^{-1} b⇒ h=a^{-1} b$$
$$bh=a⇒ b^{-1} bh=b^{-1} a⇒ eh=b^{-1} a⇒ h=b^{-1} a$$
But I do not know how to show that since $a \sim b$ then $b \sim a$. I think I have a start but I’m not sure how to show this.

I am also not sure how to show that it is transitive because I keep solving for h but don’t know how to show this.


问:
Hard problems solving tricks
I'm a math student (last year undergraduate) and I really enjoy solving math problems but sometimes I face problems that I don't see how to solve even after trying every idea I had. When I look up at the solution I understand it perfectly but I can't figure out how the author actually found his solution. Their thinking process is never explained.
Let me take an example:

Let $P \in \Bbb R[X]$. Show that if $\forall x \in [0,1], P(x) \ge 0$ then $\exists A,B,C,D \in \Bbb R[X], P=A^2+XB^2+(1-X)C^2+X(1-X)D^2$.

What I've done:

I tried to look at some special cases ($\deg P = 0,1,2$).

tried to see if there is a structure behind this (by introducing $E= \{A^2+XB^2+(1-X)C^2+X(1-X)D^2 \ | \ A,B,C,D \in \Bbb R[X]  \}$ and looking if $E$ is stable by product) but this was hard.

tried to reinforce the hypothesis by assuming $P \ge 0$, this worked well because I remembered another problem I solved ($P\ge 0 \implies \exists A,B \in \Bbb R[X], P=A^2 + XB^2$). I tried to adapt the proof (what I already tried previously) but it didn't help.

tried to use coefficients (really bad idea here).

tried Lagrange interpolation (not helpful either).

I posted this problem here and @orangeskid advised me to show that, in fact, $P(x)= A^2(x) + x(1-x)D^2(x)$ ie. we could let $B=C=0$, it was a very good hint. Even in a parallel world I'd never have thought of doing so. But why didn't I think of doing this? I don't know, it wasn't obvious for me or maybe I don't see far enough.

Another example:

Let $A, B \in \Bbb Z[X]$ such that $(\gcd(A(n), B(n)))_n$ is periodic. What can we say about $A$ and $B$?

What I tried:

I supposed that $\deg A = \deg B = 1$ even with that $\gcd(A(n),B(n))$ is something that scares me because I don't have a lot of intuition on this sequence. So I applied, Bézout's theorem which says that we can find two sequences $a_n,b_n$ such that $(a_n A(n) + b_n B(n))_n$ is periodic.
And I got stuck here.

I saw a solution here using the resultant of two polynomials, something I never heard of.

There are other examples here, here and here: they all need a little trick, something close to magic (is that cleverness?).

So my questions are how could I find these tricks myself and make them natural? Why the origin of these tricks isn't explained? Should I just learn them "on the job"? Should I be worried (as a student in mathematics) if I don't see them? Thanks for your time!

答:
I've thought a lot about my questions and I think I've found my answer.

The source of my problem is the existence of proofs that aren't natural (to me), and this disturbed me. By natural I mean whose main idea is obvious, or could be found with a little bit of work (on special cases, with a drawing, etc.) or by using a few elementary results of the courses for a student. For example showing that if $u_n \to l$ and $v_n \to l'$ then $u_n + v_n \to l+l'$ is natural. So a proof that is natural (to me) is one that starts from the hypothesis and arrive at the conclusion without using an idea that is far to see from the hypotheses. For example, I don't know natural proofs of the fundamental theorem of algebra or the Schröder-Bernstein theorem. They both use some kind of trick.

I was too accustomed to courses in which one proposal led to another in a quick and smooth way. And I think that I never actually realized that math isn't hard, it's very hard. Tricks are the witness of this. Most courses are designed not to scare students too much (e.g. when something is difficult to prove, we admit it, etc.).

Title: Hard problems solving tricks


问:
Can "integer" be used as an adjective?
Although "integer" in Latin means "whole", as far as I know in English it is used as a noun. Personally, I have not seen an integer "number", but only "integers". My question is that is it wrong to say
k is an integer "number"?

答:
The adjective form is integral, even though this is overloaded with the word integral as it pertains to integration.

As for saying integer number, it's used as a noun here, in the sense that computer science is valid but contains two nouns.

Title: Can "integer" be used as an adjective?


问:
What is the smallest fragment of ZFC that has the same consistency strength as ZFC?
The question in the title is undoubtedly nonsensical, but I am not sure how to state this question properly. Perhaps some examples will help me explain it.

Thanks to Godel and Cohen, we know that ${\sf AC}$ is independent of ${\sf ZF}$. This means that we can feel free to take ${\sf AC}$ as an axiom, because ${\sf ZF}\nvdash\neg{\sf AC}$, so as long as ${\sf ZF}$ is consistent we know ${\sf ZFC}$ is consistent. In other words, we have reduced the problem of proving ${\sf ZFC}$ is consistent to the same question of the weaker theory ${\sf ZF}$.

My question is trying to find out how far we can take this, since I know there are results suggesting the independence of most of the remaining axioms of ${\sf ZF}$. Here's my thought process: Suppose I want to know that my theory $T=A+B+C$ is consistent. If I know $A+B\nvdash\neg C$, then $T$ is consistent if $A+B$ is. If I also know $A\nvdash\neg B$, then $T$ is consistent if $A$ is. And unless $A$ is some trivial logical falsehood like $x\ne x$, the logical axioms alone cannot prove $\neg A$, so if propositional logic is consistent, then $A$ is. But propositional logic is consistent, because it is decidable and so it can be proven consistent with a few truth tables.

Is there some flaw in my argument above, or is there some subset of ${\sf ZF}$ that can not be reduced because we don't know that, say, the axiom of unions is independent from the others? Where does Godel's incompleteness result show its face here? (I may be making some tacit assumption that is equivalent to working a stronger theory, but I don't see it.)

答:
I think a flaw in your argument is that you must consider what the assumptions are when proving that a statement is independent of a theory.  For a non-subtle example, the statement "there is an inaccessible cardinal" is independent of $\mathsf{ZFC}$ because it fails in $V_\kappa$ where $\kappa$ is the least inaccessible cardinal, and holds if there is in fact an inaccessible cardinal.  Nevertheless, the theory "$\mathsf{ZFC} + \text{there is an inaccessible cardinal}$" has higher consistency strength than $\mathsf{ZFC}$ itself.  This happened because we were implicitly assuming (or at least I was implicitly assuming) the consistency of an inaccessible cardinal for the independence proof.

One might think that this issue could be avoided by only considering independence results that can be obtained by inner models and forcing, rather than by large cardinals.  However, one still needs some assumptions to be sure that these inner models and forcing extensions behave as desired.
Although the proofs that, say, $\mathsf{CH}$ and $\mathsf{AC}$ are independent of $\mathsf{ZFC}$ may be fairly robust in this regard, many independence proofs are more sensitive to assumptions.
For example, to show that the Axiom of Replacement is independent from the others, inner model and forcing methods are not going to work because to show that replacement holds in $L$ and in forcing extensions $V[G]$, you need to use replacement in $V$.

We can show that replacement is independent from the others if we assume that it is consistent with the others, just like we can show that the existence of an inaccessible cardinal is independent of $\mathsf{ZFC}$ if we assume that it is consistent with $\mathsf{ZFC}$.  You may encounter similar situations with other axioms, such as Power Set, and this will limit your ability to pare down the collection of axioms in the way you envision.

Of course the question in your title is still valid, once you replace "the smallest fragment" with "a minimal fragment," but perhaps it no longer is the question you want to ask.  If it is, please correct me.


答:
Here are the first couple of steps to illustrate how to proceed.

Step: 1

Your domain of integration should satisfy both inequalities, i.e. you must integrate over the area where both
$$(x^2+y^2)^2 \le 3(x^2-y^2)$$
and
$$x^2+y^2 \le \sqrt{2}x$$
hold. If you plot the curves you'll obtain the following:

Therefore the area to be found is the one that is comprised in both curves, i.e. the area inside the lemniscate in the interval comprised between the origin and the left vertical dashed line plus the area inside the circle delimited by the two vertical dashed lines.

How do we know this is the right integration area?

We test it with points we choose "smartly". First, you plot the equations as in the image below, and then you test certain points to see if they satisfy both your domain conditions (i.e. the inequalities you start with). Since you have 2 inequalities, you have to test (in principle) 2*2 regions. If you look at the picture below, the points sketched are in taken in certain regions delimited by the two curves (the red points are symmetric, so you can test arbitrarily just one of them):

\begin{cases}
p_\text{red} = (1/2,\pm 1/2) \\
p_\text{green} = (1, 0) \\
p_\text{yellow} = (3/2, 0) \\
p_\text{blue} = (2, 0)
\end{cases}

Of course the smart choice is to choose the test points on one of the axes whenever possible, to simplify the calculations as much as possible, but also a random point would do the trick. If you choose the point $p_\text{green}=(1,0)$, you will realise that it is the only one that satisfies both inequalities, and therefore the area mentioned in the previous point is indeed the one we need to compute.

Step: 2

You now need to find the intersection points between the curves, so you know your extrema of integration. For instance, you integrate under the lemniscate between $(0,0)$ and the intersection point, while you integrate under the circle from the latter point until the point where the circle intersects the axis of the abscissae.

How to do it? Simply solve the system
\begin{cases}
(x^2+y^2)^2 = 3(x^2-y^2) \\[2ex]
x^2+y^2 = \sqrt{2}x
\end{cases}

In your case you will get 3 solutions: the trivial $x=0, y=0$ and the two points $p_1 = (x_p, y_p)$ and $p_2 = (x_p, -y_p)$. (I won't provide you with the solution though, so you have to work for it). Another point needed is the intersection between the circle and the axis of the abscissae, let's call it $p_3 = (\sqrt{2},0)$, which is trivial to compute.

You can now observe that the region you want to integrate is symmetric with respect to the axis of the abscissae, so you can limit yourself to find the surface between the lemniscate and the $y=0$ axis in the domain $x\in [0,x_p]$, sum it to the surface between the circle and the $y=0$ axis in the domain $x\in [x_p,\sqrt{2}]$, and finally multiply by 2 (to add the volume below the abscissae axis).

Step: 3

Now that you know your extrema of integration, what do you integrate? Well, you need to find the area between the lemniscate and the $y=0$ axis, therefore you must work on the equation of the lemniscate and transform it into the form $y=f(x)$. Same goes for the circle, that has to be transformed into a function of the form $y=g(x)$. When you perform this step be careful to consider the correct branch, since both functions have multiple values!

Step: 4

When all these steps are in place, your area will be given by:
$$P = 2\,\biggl[ \int_{0}^{x_p} f(x)dx + \int_{x_p}^{\sqrt{2}} g(x)dx\biggr]$$

Hint:
This is not the only way to approach this problem. You could very well transition into polar coordinates instead, given the high level of symmetry of your problem. It is up to you to use the method you are more comfortable with.

I hope this helped!

Title: Area bounded with curves

Body: I need to calculate area bounded with curves $(x^2+y^2)^2\le 3(x^2-y^2), x^2+y^2\le \sqrt{2}x$. Here we have inside of leminscate and circle. I think i should use that $P=\iint dxdx$ but not sure how.


问:
Space which is connected but not path-connected
Consider the following two definitions:

Connected
: A topologiocal space X is connected if it is not the disjoint union of two open subsets, i.e. if X is a disjoint union of two open sets A and B, then A or B is empty set.

Path Connected
: A topological space X is path-connected if any two points in X can be joined by a continuous path.

So far I can picture, I think they should be equivalent. If X is connected iff it is path-connected. But, the Thm I got from lecture is just if X path connected then X is connected (The converse not necessarily true.)

Can anyone give me some picture of space which is connected but not path-connected? Just picture please, cuz I wanna grab the idea. I have already seen some examples like in connected but not path connected?, I cannot grab the idea if it (such space) consist of a single piece, then I couldn't make a continuous path from any two points.

Thank you :)

答:
How about $\mathbb{R}_K$. That is the all open intervals together with intervals of the form

$$(a,b) \setminus K$$

where $K=\{\frac{1}{n} : n \in \mathbb{Z}^+\}$. This space is connected and Hausdorff but not path-connected.


问:
Polygonal decomposition of the klein bottle by pentagons?
How do I prove that there is no regular polygonal decomposition of the klein bottle by pentagons?

Do I use the euler characteristic to prove?

答:
I don´t know what "regular" means when talking about polygonal decompositions, but I´ll write an answer ignoring that word in case it helps. If it´s wrong I´ll delete it.

I don´t see why you couldn´t do what you ask. In the figure you can see a decomposition of a triangle into 4 pentagons. This construction uses as vertices of the pentagons the middle points of the edges of the triangle, but the construction can be easily modified so that you use whatever points you want (1 of each side of the triangle) as vertices of the pentagons. So it seems any triangulation of the klein bottle can be converted into a polygon decomposition in pentagons. You just have to take care to, when you have two triangles adjacent in an edge, use the same point of the edge to divide the triangles into pentagons.