问:
Smooth lower bound on a continuous function
Let $M$ be smooth manifold, and $f$ be a real valued non negative  continuous function on $M$. Must there exist  $g$ that is  real valued non negative  smooth function on $M$ such that $0\leq g \leq f$ and such that $g$ vanishes iff $f$ vanishes ?

Note : $M$ is Hausdorff and second countable.

答:
Let $\{(\psi_\alpha, U_\alpha)\}_\alpha$ be a partition of unity on $M$.

Clearly, we have that $f\psi_\alpha$ is continuous on $M$ with support in $U_\alpha\subseteq M$. This effectively reduces the problem to finding a smooth approximation of a compactly supported function in $\mathbb{R}^n$, i.e. if $\varphi_\alpha: V_\alpha\subseteq \mathbb{R}^n\rightarrow U_\alpha \subseteq M$ is a coordinate patch, then we are just finding a smooth approximation to the function $(f\psi_\alpha)\circ \varphi_\alpha: V_\alpha \rightarrow \mathbb{R}$. This is a standard result in real analysis, so I will skip the detail. Hence, we can fine a $\tilde g_\alpha: V_\alpha \rightarrow \mathbb{R}$ such that
\begin{align}
0\le \tilde g_\alpha \le (f\psi_\alpha)\circ \varphi_\alpha.
\end{align}

In effect, we can define $g_\alpha:U_\alpha\rightarrow \mathbb{R}$ by $g_\alpha = \tilde g_\alpha\circ \varphi_\alpha^{-1}$. Notice that $\operatorname{supp} g_\alpha \subset \operatorname{supp}(f\psi_\alpha)$ by construction. Finally, define
\begin{align}
0\le g = \sum_\alpha g_\alpha \le \sum_\alpha f\psi_\alpha = f.
\end{align}


问:
$01^*01^*(0 \lor 1)^* \lor 10^*10^*1(0 \lor 1)^*$ is wrong
I have the following finite automata. I would like to know its regular expression. I gave $01^*01^*(0 \lor 1)^* \lor 10^*10^*1(0 \lor 1)^*$, but it seems my answer is wrong. Can you tell me why?

答:
It seems (almost) correct to me. I guess what you meant was
$$01^*01^∗\color{red}{0}(0∨1)^∗∨10^∗10^∗1(0∨1)^∗$$

ContentLicense: CC BY-SA 4.0


问:
Given that $p>n+1$, how to prove that: $\sum_{k=1}^{p-n}\frac1{(n+1)^k}$ is less than $\frac 1n$.
Given that $p>n+1$, how to prove that:$$
\sum_{k=1}^{p-n}\frac1{(n+1)^k}$$ is less than $\frac 1n$.

I tried to show to for every value of $k$ the result is smaller than $\frac 1 {n^2}$ but it didn't work

答:
This is a geometric series with $$a_1 = r = \frac{1}{n + 1}$$
Applying the formula $$S_n = \frac{a_1 (1 - r^n)}{1 - r}$$
we get
$$S_{p - n} = \frac{1}{n + 1} \cdot \frac{1 - (n + 1)^{n - p}}{1 - (n + 1)^{-1}} = \frac{1 - (n + 1)^{n - p}}{n} < \frac{1}{n}$$
This assumes that $(n + 1)^{n - p} > 0.$


问:
Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$
I'm trying to prove rigorously that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$.  Where $f$ is the pdf of the random variable $X$.

I can't find a proof on the wikipedia article, or if it's there then it's disguised enough that I can't recognize it.  Basically what I've got is a sort of semi-rigorous (and probably incorrect) proof of the equality, but maybe somebody could help me flesh out the details.

Proof:

Beginning from the definition that $E[X]:=\int_{\Omega}X\;dP$.

Given a random variable $X:\Omega\rightarrow\mathbb{R}$, we have the induced measure on $(\mathbb{R}, \mathscr{B}(\mathbb{R}))$ given by $P(\{X\in A\})$.  Then by the Radon-Nikodym theorem there exists a measurable function $f:\mathbb{R}\rightarrow [0,\infty)$ such that $$P(\{X\in A\}) = \int_Afd\mu,$$

where $d\mu$ is the Lebesgue measure.  From here it gets a bit hand-wavy.  Basically since this measure was induced by the random variable $X$, then on $\mathbb{R}$ this random variable is simply given by the identity function $g(x)=x$.  And thus by definition we write the expected value of $g$ with respect to our Radon-Nikodym produced measure in the form $$\int_{-\infty}^{\infty}x\;d\Big(\int_Afd\mu\Big).$$

Now by the Fundamental Theorem of Calculus, this becomes $$\int_{-\infty}^{\infty}xf(x)\;dx.$$

What does everyone think about this?

答:
I'll try to provide a little more general result. Let $(\Omega,\ \mathcal{E},\ P)$ be a probability space and let $X\colon \Omega\longrightarrow \mathbb{R}$ be a random variable, i.e for each $I\in \mathcal{B}$, $X^{-1}(I)\in\mathcal{E}$, where $\mathcal{B}$ is the usual Borel $\sigma-$algebra on $\mathbb{R}$. Let us write $\mu:=\mu_{X}$ for the probability distribution of $X$, i.e for the measure defined on $\mathcal{B}$ by $\mu(I):=P(X^{-1}(I))$ for each $I\in\mathcal{B}$. Then the following holds.

Theorem (Abstract-Concrete Formula): Let $\phi\colon\mathbb{R}\longrightarrow \mathbb{R}$ be a borelian function, i.e $\phi^{-1}(I)\in\mathcal{B}$ for every $I\in\mathcal{B}$, and write $\phi(X)$ for the composition $\phi\circ X$. Suppose at least one between the integrals $$\int_{\Omega} \phi(X)\ dP\quad\text{and}\quad \int_{\mathbb{R}}\phi (x) \ d\mu $$
  exists (resp. exists and it is finite). Then also the other one exists (resp. exists and it is finite) and it holds that $$\int_{\Omega} \phi(X)\ dP=\int_{\mathbb{R}}\phi (x) \ d\mu\ .$$
  In particular, $\phi(X)$ is summable with respect to $P$ if and only if $\phi$ is summable with respect to $\mu$.

(When I say that a Lebesgue integral for a measurable function exists, I allow that it is not finite). The proof of this fact is quite straightforward but it requires some measure theory results such as the approximation theorem with simple functions and the Lebesgue's monotone convergence theorem. Indeed, suppose first that $\phi$ is a (finitely) simple and positive function. Then also $\phi(X)$ is simple (and positive) and (therefore) both mentioned integrals always exist. Writing $\phi=\sum_{i=1}^{n} c_{i}1_{E_{i}}$, where $n=\vert \phi(\mathbb{R})\vert$, $\phi (\mathbb{R})=\{c_{1},\cdots,c_{n}\}$ and $E_{i}:=\phi^{-1}(\{c_{i}\})$, we get $$\int_{\mathbb{R}}\phi (x) \ d\mu=\sum_{i=1}^{n}c_{i}\mu (E_{i})=\sum_{i} c_{i}P(X^{-1}(E_{i}))=\sum_{i} c_{i}P(X^{-1}(\phi^{-1}(\{c_{i}\})))=\int_{\Omega} \phi(X)\ dP.$$ Assume now $\phi$ is a non-negative borelian function. Then there exists a non-decreasing sequence $(\phi_{n})_{n\in \mathbb{N}}$ of simple, positive functions such that $\lim\limits_{n\to \infty}\phi_{n}(x)=\phi (x)$ for every $x\in\mathbb{R}$. By monotone convergence theorem we get immediately that $$\int_{\mathbb{R}}\phi (x) \ d\mu=\int_{\mathbb{R}}(\lim_{n\to\infty}\phi_{n} (x)) \ d\mu =\lim_{n\to\infty} \int_{\mathbb{R}}\phi_{n} (x) \ d\mu=\lim_{n\to\infty} \int_{\Omega}\phi_{n} (X) \ dP=\int_{\Omega} \phi(X)\ dP.$$ Finally, suppose only $\phi\colon \mathbb{R}\longrightarrow\mathbb{R}$ is a borelian function, with no further restrictions. Then one can write $\phi=\phi^{+}-\phi^{-}$ (here for notation). Suppose $\int_{\mathbb{R}}\phi (x) \ d\mu$ exists: then at least one between $\int_{\mathbb{R}}\phi^{+} (x) \ d\mu$ and $\int_{\mathbb{R}}\phi^{-}(x) \ d\mu$ (say, the first one) must be finite and hence also $\int_{\Omega}(\phi(X))^{+} \ dP$ is finite, i.e $\int_{\Omega}\phi(X)\ dP$ exists. It is now clear that we can conclude with our thesis.

Corollary: In the previous situation, if $E[X]<+\infty$, then $$E[X]:=\int_{\Omega} X\ dP=\int_{\mathbb{R}} x\ d\mu.$$ In particular, if $\mu$ is absolutely continuous with respect to Lebesgue's Measure (on $\mathbb{R}$) and has density $f$, then we get $E[X]=\int_{\mathbb{R}} xf(x)\ dx$ (by (a straightforward consequence of) Radon-Nikodym theorem).

ContentLicense: CC BY-SA 3.0


问:
Set theory. Let A be a set. Then A⊆P(∪A).
Let A be a set. Then A⊆P(∪A).

I already tried to do this question and lost points. can someone please show me how to correctly solve this problem so I can be prepared for the midterm.

答:
Just follow the definitions carefully. You need to show that if $a\in A$, then $a\in\wp\left(\bigcup A\right)$, i.e., $a\subseteq\bigcup A$. To do this, you need to show that if $a\in A$, and $x\in a$, then $x\in\bigcup A$. What is the definition of $\bigcup A$? Something is an element of $\bigcup A$ if and only if … what?

Your attempt contains several errors. First, it’s not generally true that $A=\bigcup A$. If $A=\{\varnothing\}$, for instance, then $A\ne\varnothing$, but $\bigcup A=\varnothing$ (why?). Thus, you cannot conclude that $\wp(A)=\wp\left(\bigcup A\right)$. Finally, $A$ is always an element of $\wp(A)$, but it is not in general a subset of $\wp(A)$: that happens only when every element of $A$ is also a subset of $A$, which is by no means always the case.


问:
Categorical products are associative
Let $\mathbf{C}$ be a category with products, and let $A, B, C \in \mathbf{C}$. I wish to show that there exists a morphism $h: (A \times B) \times C \to A \times (B \times C)$ which is an isomorphism.

I believe I can get that we have morphisms $g_1:(A \times B) \times C \to A \times (B \times C)$ and $g_2:A \times (B \times C) \to (A \times B) \times C$ and that they are the unique morphisms satisfying the universal property for products.

However, I don't see how to show that one/either of them is an isomorphism. We need to prove that their compositions are identity morphisms on those product objects. This amounts to showing that for any objects $U, V, W, X$ and arrows
$$
\begin{aligned}
f_U&: U \to (A \times B) \times C \\
f_V&: (A \times B) \times C \to V \\
f_W&: W \to A \times (B \times C) \\
f_X&: A \times (B \times C) \to X \\
\end{aligned}
$$
we have
$$
\begin{aligned}
(g_2 \circ g_1) \circ f_U &= f_U \\
f_V \circ (g_2 \circ g_1) &= f_V \\
(g_1 \circ g_2) \circ f_W &= f_W \\
f_X \circ (g_1 \circ g_2) &= f_X.
\end{aligned}
$$
The first two equalities above assert that $g_2 \circ g_1 = \mathrm{id}_{(A \times B) \times C}$, and the second pair asserts that $g_1 \circ g_2 = \mathrm{id}_{A \times (B \times C)}$. From here, though, I am stuck.

答: See 1.5 "Diagram Chasing" here https://compose.ioc.ee/categoryTheory2020/week3/week3.pdf</p>

Title: Categorical products are associative


问:
If $E$ is a normed vector space and $\tau$ is a vector topology on $E$ s.t. $f$ is $\tau$-continuous on every closed ball, is $f$ $\tau$-continuous?
Let $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$, $E$ be a normed $\mathbb K$-vector space, $\tau$ be a vector topology on $E$ such that $\overline B_1(0):=\{x\in E:\left\|x\right\|_E\le1\}$ is $\tau$-compact and $f:E\to\mathbb C\setminus\{0\}$ be $\tau$-continuous and uniformly $\left\|\;\cdot\;\right\|_E$-continuous.

Since $f$ is uniformly $\left\|\;\cdot\right\|_E$-continuous, there is a $\delta>0$ with $$|f(x)-f(y)|<\frac\varepsilon2\tag1$$ for all $x,y\in E$ with $\left\|x-y\right\|_E<\delta$. Let $r>0$, $k\in\mathbb N$, $0=t_0<\cdots<t_k=1$ with $\max_{1\le i\le k}|t_{i-1}-t_i|<\frac\delta r$. Then $$|f(t_{i-1}x)-f(t_ix)|<\frac\varepsilon2\tag2$$ for all $1\le i\le k$ and $x\in\overline B_r(0)$ and $$\frac{f(t_ix)}{f(t_{i-1}x)}\in B_{\frac12}(1)\subseteq\mathbb C\setminus(-\infty,0]\tag3$$ for all $x\in\overline B_r(0)$.

By $(3)$, $$g_r(x):=\sum_{i=1}^k\ln\frac{f(t_ix)}{f(t_{i-1}x)}\;\;\;\text{for }x\in\overline B_r(0)$$ is well-defined and $\ln$ is continuous on $\mathbb C\setminus(-\infty,0]$. Note that $$g(x):=g_r(x)\;\;\;\text{for }x\in E\text{ and }r>0\text{ with }x\in\overline B_r(0)$$ is a well-defined function on $E$.

Question: Are we able to conclude that $g$ is $\tau$-continuous and $\left\|\;\cdot\;\right\|_E$-continuous?

It should be obvious that for any $r>0$, the function $g_r$ is continuous with respect to the subspace topologies induced by $\tau$ and $\left\|\;\cdot\;\right\|_E$ on $\overline B_r(0)$. But is this enough to conclude?

答:
Let $H$ be an infinite dimensional Hilbert space and let $D\subseteq H$ be a proper dense subspace.   Consider the locally convex
topology $\sigma (H, D)$ on $H$ given by the family of semi-norms
$\{p_\eta \}_{\eta \in D}$, where
$$
  p_\eta (\xi ) = |\langle \xi , \eta \rangle |, \quad (\xi \in H).
  $$

It is easy to see that $\sigma (H, D)$ is a Hausdorff topology, coarser than the weak topology  (notice that the
weak topology is what you get by choosing $D=H$).

Consequently the identity map
$$
  \text {id} : \big (\bar B_1(0), \sigma (H, H)\big ) \to \big (\bar B_1(0), \sigma (H, D)\big )
  $$
is continuous and, since its  domain  is known to be  compact, while its
co-domain is clearly Hausdorff, we conclude that it is a homeomorphism, which means that $\sigma (H,D)$ coincides with
$\sigma (H,H)$ on $\bar B_1(0)$.

In particular any norm-continuous linear functional is continuous on $\big (\bar B_1(0), \sigma (H, D)\big )$, but it is well known
that the continuous functionals on $H$ relative to $\sigma (H, D)$ are only those of the form
$$
  \xi \mapsto \langle \xi ,\eta \rangle ,
  $$
with $\eta \in D$.

In conclusion, if $\zeta $ is any vector in $H\setminus D$, then the linear functional
$$
  \xi \mapsto \langle \xi ,\zeta \rangle
  $$
is continuous relative to $\sigma (H,D)$ on $\bar B_1(0)$, and hence also on any bounded set, but it is not continuous on the whole
of $H$ with this topology.


问:
Let $q(x)=1+x+x^2+...+x^n$. Prove that if $p_1(x),..,p_n(x)$ is any basis for $W$, then $p_1(x),...,p_n(x),q(x)$ is a basis for $P_n(\mathbb{R})$
Let $W$ be a subspace of $P_n(\mathbb{R})$ be defined by $W = \{p(x) \in P_n (\mathbb{R}) |p(1)=0 \}$. Let $q(x)=1+x+x^2+...+x^n$. Prove that if $p_1(x),..,p_n(x)$ is any basis for $W$, then $p_1(x),...,p_n(x),q(x)$ is a basis for $P_n(\mathbb{R})$

My Attempt

First, construct a basis for $W$: $\{(x-1),(x^2-1),...,(x^n-1)\}$. Since both are basis, for $a_1...,a_n$ there is a  $b_1,...,b_n$ in $\mathbb{R}$:

$$a_1(x-1) + ... + a_n(x^n-1)=b_1p_1(x)+...+b_np_n(x)$$

Now I need to show that $p_1(x),...,p_n(x),q(x)$ is both spanning and linearly independent. Let $k \in \mathbb{R}$:

$$b_1p_1(x) + ... + b_np_n(x) + kq(x) = a_1(x-1)+ ... + a_n(x^n-1) + k(1+x+...+x^n) \\
= -a_1 -....-a_n + k + x(a_1+k) + ... + x^n(a_n + k)$$

Hence it must be spanning. Also, the only solution to $= \vec{0}$ is $a_1= ... =a_n=k=0$. Hence is a basis.

Is this a valid proof?

$$$$

答:
It is correct but you don't have to find a basis for $W$ first.

Linear dependence: assume $$0 = \alpha_1p_1(x) + \cdots + \alpha_np_n(x) + \beta q(x)$$
and plug in $x = 1$. We get $0 = \beta q(1) = (n+1)\beta$ so $\beta = 0$. Therefore
$$0 = \alpha_1p_1(x) + \cdots + \alpha_np_n(x)$$
so since $\{p_1, \ldots, p_n\}$ is a basis for $W$ it follows $\alpha_j = 0$ for $1 \le j \le n$.

Span: let $p \in P_n(\Bbb{R})$. Then notice that $p(x) - \frac{p(1)}{n+1}q(x) \in W$ so there exist scalars $\alpha_1, \ldots, \alpha_n \in \Bbb{R}$ such that
$$p(x) - \frac{p(1)}{n+1}q(x) = \alpha_1p_1(x) + \cdots + \alpha_np_n(x)$$
so rearranging gives
$$p(x) = \alpha_1p_1(x) + \cdots + \alpha_np_n(x) + \frac{p(1)}{n+1}q(x).$$


问:
Piecewise integral $f(x)=\begin{cases}x, 0\leq x \lt 1\\x-2, 1 \leq x \leq 2\end{cases}$
I've got stuck on a simple exercise after reading the first chapter on integrals in Spivak's Calculus. Decide if $f(x)$ is integrable and calculate the integral:

$$f(x)=\begin{cases}x, 0\leq x \lt 1\\x-2, 1 \leq x \leq 2\end{cases}$$

From the graph of the function I see that it's clearly $0$.

Also one of the theorems seems to come useful: if $f$ is integrable on $[a,c]$ and on $[c,b]$, then it's integrable in $[a,b]$ and $\int_a^b{f}=\int_a^c{f}+\int_c^b{f}$ for $a<c<b$.

Let $c=1$, it would seem that $$\int_0^2{f(x)dx}=\int_0^1{xdx}+\int_1^2{(x-2)dx}=\int_0^2{xdx}-\int_1^2{2dx}=2-2=0$$

The definition of the integral given in the chapter defines it in terms of a function $f$ on a closed interval $[a,b]$ as the the common value that equals the supremum of the set of all lower sums and the infimum of the set of all upper sums. But here I have $f(x)=x, x \in [0,1)$, it's not closed and $\int_0^1{xdx}$ doesn't really make sense. At this point I don't know anything about integrals defined on half-open intervals. And $f(x)$ on $[0,1]$ is not $x$, but also $x-2$ at $x=1$.

How to approach this problem knowing only the definition of the integral and the basic theorems like that above?

答: Indeed,$$\int_0^2f(x)\,\mathrm dx=\int_0^1f(x)\,\mathrm dx+\int_1^2f(x)\,\mathrm dx.$$On the other hand, if $x\in[0,1]$, then$$f(x)=\begin{cases}x&\text{ if }x<1\\-1&\text{ if }x=1.\end{cases}$$So, on $[0,1]$, $f(x)=g(x)-2h(x)$, with $g(x)=x$ and$$h(x)=\begin{cases}0&\text{ if }x<1\\1&\text{ if }x=1.\end{cases}$$But Spivak proves, right after theorem 2 of that same chapter, that a function similar to $h$ is integrable and that its integral is $0$. So, since $f|_{[0,1]}$ is the sum of integrable functions, $f|_{[0,1]}$ is integrable too.


问:
Cardinality of the set of divergent sequences
How to find the cardinality of the set of divergent sequences? Let's name this set $A$.

I know that cardinality of the set of sequences equals $2^{\lvert \mathbb{N}\rvert}$, so $\lvert A\rvert\le 2^{\lvert \mathbb{N}\rvert}$.
How to strictly prove to the other side?

Thank you for your time.

答:
Hint:

For any non-rational $\,r\in\Bbb R\,$ , the sequence $\;\{a_{r,n}\}_{n\in\Bbb N}:= \{rn\}_{n\in\Bbb N}\;$ is divergent


问:
Solving multivariate quadratic equations over the integers
I am looking for a method (if it exists) to solve over the integers the following sum of squares equation:

$$ x_1^2 + x_2^2+x_3^2 + \cdots + x_n^2 = m,$$ with $m \in \mathbb{N}.$

Someone has any idea about books, articles dealing with this kind of problem?

Thanks in advance!

答:
If you just want sequential integers:
$\qquad m=(1^2+2^2+3^2+\dots,n^2)=\sum_{i=1}^n i^2=\dfrac{n(n+1)(2n+1)}{6}. \tag{1}$

For other sums of squares, you must know what the individual values or their limits, e.g.

$5^2+6^2+7^2+8^2\quad
=\sum_{i=1}^8 i^2-\sum_{i=1}^4 i^2\quad
\text{using the formula above} \tag{2}$

If you know how many square are identical, just multiply by that number those squares.

If you know sums that are discontinuous in spots, calculate the sequential parts separately as in equation
$\space (2)\space$ above.

For "sum" formulas up to $x^{10}$ look here.


问:
matrix for this system of non-linear equation?
Let's consider the product of a square matrix and a vector: $
M{\times}v =
\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}
{\times}
\begin{bmatrix} a \\ b \\ c \end{bmatrix}=
\begin{bmatrix} 1 {\times} a + 2 {\times} b + 3 {\times} c \\
                4 {\times} a + 5 {\times} b + 6 {\times} c \\
                7 {\times} a + 8 {\times} b + 9 {\times} c \end{bmatrix}$

How can I write $M$ such that $M{\times}v=
\begin{bmatrix} 1 {\times} a + 2 {\times} b + a {\times} b \\
                4 {\times} a + 5 {\times} b + a {\times} c \\
                7 {\times} a + 8 {\times} b + b {\times} c \end{bmatrix}$

The difference is in the third column where it's not just some constant times $c$ but the product of two other inputs.

Update

Ben suggested using the input variables $a$ and $b$ directly in the matrix $M$, but then $M$ wouldn't be made of constants only (like in the first example)

答:
You can't. Define $$v_1 = \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right]$$
and $$v_2 = \left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right]$$
By your definition: $$M v_1 = \left[ \begin{array}{c} 1 \\ 4 \\ 7 \end{array} \right]$$
and $$M v_2 = \left[ \begin{array}{c} 2 \\ 5 \\ 8 \end{array} \right]$$
However, $$M (v_1 + v_2) = \left[ \begin{array}{c} 4 \\ 9 \\ 15 \end{array} \right] \neq M v_1 + M v_2$$
Remember that matrix multiplication by a vector is a linear operator.


问:
What is the connection between different definitions for theta functions?
In his 2002 thesis, Zwegers defines theta functions associated with definite and indefinite quadratic forms. For example, if $Q:\mathbb{R}^r\to \mathbb{R}$ is a positive definite quadratic form with associated bilinear form $B(x,y) = Q(x+y)-Q(x)-Q(y)$  then the theta series associated with $Q$ is

$$\Theta(z;\tau):=\sum_{n\in \mathbb{Z}^r}e^{2\pi i(Q(n)\tau+B(n,z))}=\sum_{n\in\mathbb{Z}^r}q^{Q(n)+B(n,z)}.$$

$\Theta(0,\tau)$ is a modular form of weight $r/2$.

On the other hand, Ono defines theta functions in terms of Dirichlet characters, and cites a theorem of Serre and Stark that such theta functions form a basis for a space of weight 1/2 cusp forms. For example, if $\chi_0$ is the trivial character then

$$\theta_0(z)=\theta(\chi_0,0,z):=\sum_{n\in \mathbb{Z}}q^{n^2} = \sum_{n\in\mathbb{Z}}e^{2\pi in^2z}.$$

Is there a connection between the quadratic form definition and the Dirichlet character definition for theta functions? Is there, for example, a sort of dictionary which allows you to pass from the Dirichlet definition for a theta function to the quadratic form definition?

Edit: The definition Ono gives for a theta function $\theta$ in terms of a Dirichlet character $\psi$ is the following:

If $\psi$ is even, then $\theta(\psi,0,z):=\sum_{n\in\mathbb{Z}}\psi(n)q^{n^2}.$

If $\psi$ is odd, then $\theta(\psi,1,z):=\sum_{n\in\mathbb{Z}}\psi(n)nq^{n^2}.$

答:
The theta function attached to the trivial character is the theta function attached to the one dimensional lattice $\Bbb Z$ equipped with the usual quadratic form $q(z) = z^2$.

These are, of course, modular forms of half integral weight, which are automorphic forms on the metaplectic group.

To avoid this complication, a simpler example is to look at an imaginary quadratic field $K$ and an order $\mathcal O \subseteq K$, e.g. the ring of integers of $K$.

We can then attach to any ideal of $\mathcal O$ a theta function, which is given by viewing the ideal as a lattice, equipped with the quadratic form $N_{K/\Bbb Q}$.

If $\chi$ is any ring class character of order $\mathcal O$, then the theta series attached to $\chi$ is in fact a finite linear combination of the various theta series attached to the ideals (one for each ideal class).

Conversely, any theta function attached to a lattice can also be written as a linear combination of theta series attached to characters.

If you are interested in a more general setting of theta series (for $\operatorname{GL}_2$), then you should have a look at the Weil representation. All the local Weil representations pack together to an adelic Weil representation, which then automatically gives rise to theta functions.


问:
Question on basic property of definite integrals
Sorry if this is a trivial question.

We of course have the basic property that if $c\in[a,b]$, $$\int_a^bf(x) \, dx = \int_a^cf(x) \, dx+\int_c^bf(x) \, dx.$$ The question that I have is, is there any way at all to simplify an expression of the form $$\int_a^cf(x) \, dx-\int_c^bf(x) \, dx.$$

Any advice would be greatly appreciated. Thanks.

答:
The most short answer is no, because the $-$ sign can be interpreted as $\int_{b}^{c}f(x) dx$.

It follows of the property $\int_{a}^{b}f(x)dx=-\int_{b}^{a} f(x)dx$

Which is clearly not compartible of the initial proposition for the sum of integrals in some interval.


问:
Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$
Question:

$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $

Answer key:

$\implies  \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $

I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:

$\implies  (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $

$ \implies  (\sin(\theta)-\cos(\theta)-1) \leq0 \because  (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $

Doubt:

I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.

I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too

答:
Imagine radius vector rotation in the four quadrants

At $\theta$ (0,180) sine vanishes

At  $\theta$ (90,270 ) cos vanishes

So the four zones are determined and demarcated as four the quadrants conveniently.

$$F(\theta)=(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) - 2$$

For extremum values only center values of each quad $F(\theta)$ needs to be evaluated/examined that too only for its sign.

We need only to evaluate at odd central values

$$ (2k-1)\dfrac{\pi}{4}$$

First quadrant

$$ 0 (...) -2 \le 0 \text{ okay} $$

Second quadrant

$$(2/\sqrt2)(2-\frac12) >0 \text{ not okay}$$

Third quadrant

$$ (2\sqrt2)(2-1/\sqrt2) \le 0 \text { okay} $$

Fourth quadrant

$$(2 \sqrt2)(....) \le 0 \text { okay} $$

So the correct sign occurs in 1st, 2nd, 4th quadrants only.

$ (\pi/2<\theta<3 \pi/2 ) $ interval is inadmissible.

No need for  a polar graph to be drawn... even if drawn for confirmation it would not be single valued radius in the second quadrant anyways..


问:
General guidance with proofs, fields in linear algebra
I'm taking a linear algebra course and I'm having a lot of issues answering proofs, most of the questions seem "a given" or that they derive from themselves, and I'm just stuck staring at the book.

I'm having a hard time understanding how to approach proofs, what I'm even allowed to "use".

For example,

Prove that

```<code>-(-a) = a
</code>```

Seems a given, it seems to be derived directly from a statement in the book, and I don't know how to further prove something like that.

Another example is

Prove that

```<code>(-a)b = a(-b)= -(ab)
</code>```

In this case, it's hard for me to understand what am I even allowed to do?
how much am I allowed to presume and lean on normal math?
As every property (Commutativity, Associativity, etc..) is defined and tested it's hard to know what you can and cant use.

for example, am I allowed to derive from the question that?

```<code>(-a)b +ab = 0
</code>```

In short, I need guidance in how to mentally approach these kinds of questions, any help would be appreciated.

答:
Let me show how you might prove something like your first question and hopefully that will give you an idea of what this kind of proof is supposed to look like. You're trying to show that $-(-a) = a$. This seems really evident, so how do we formally prove it? I'm going to use a lot more words than is really necessary to try and provide clarity for you here, but the important thing is that you understand what is going on.

The way I would approach this is to go back to the formal definition of $-x$ in an abstract vector space. Given a vector $v$ in a vector space $V$ we define the symbol $-v$ to be the unique vector such that
$$
v+(-v) = 0
$$
This is what we mean by the symbol $-a$. The existence of such a vector is guaranteed by the vector space axioms and uniqueness is simple to prove, so the symbol makes sense. Now we are trying to show that $-(-a)) = a$. Now we know from the definition that $a+(-a) = 0$. We are trying to show that a particular vector $a$ is the inverse of $-a$, i.e. $-(-a)$. Let $b = -a$. Then the fact that we know $a+(-a) = 0$ implies that $a+b = 0$. By the uniqueness of $-b$, this means that $a = -b$, but $b = -a$ so we have that $a = -b = -(-a)$ as desired.


问:
How many sides does a circle have?
My son is in 2nd grade.  His math teacher gave the class a quiz, and one question was this:

If a triangle has 3 sides, and a rectangle has 4 sides,
  how many sides does a circle have?

My first reaction was "0" or "undefined".  But my son wrote "$\infty$" which I think is a reasonable answer.  However, it was marked wrong with the comment, "the answer is 1".

Is there an accepted correct answer in geometry?

edit: I ran into this teacher recently and mentioned this quiz problem.  She said she thought my son had written "8." She didn't know that a sideways "8" means infinity.

答: Personally I use to think a circle had infinite sides as well; however, why could it not be one side with a $360^\circ$ curve?

Title: How many sides does a circle have?


问:
Quadratic residue modulo $p$ if quadratic residue module $p^k$
I'm trying to prove this statement. I've already checked here in StackExchange but I can only find the proof of the converse.

In Wikipedia, it is stated: "A number $a$ relatively prime to an odd prime $p$ is a residue modulo any power of $p$ if and only if it is a residue modulo $p$", and I want to prove the only if part.

This problem appeared to me when I wanted to solve the following problem:

Check that the only solutions, up to congruence, of $X^2 \equiv 25 $ (mod $p^k)$, for any $k$ and for any prime $p \neq 2$, $5$, are $X \equiv \pm 5 $  (mod $p^k)$.

Any help would be appreciated.

答:
This is a special case of the primary (prime power) form of Euclid's Lemma below (put $\,B=x\!+\!5,\,C=x\!-\!5,\,$ so $\,(p,B,C) = (p,B,B\!-\!C) = (p,x\!+\!5,10)=1\,$ by $\,(p,10)=1)$

Primary Euclid Lemma $\ \color{#0a0}{(p,B,C)=1},\ p^k\mid BC\Rightarrow p^k\mid B\,$ or $\,p^k\mid C,\ $ for prime $p$

which is the special case $\,a\,$ is primary in a primitive primal form of Euclid's Lemma

$\rm\color{#0a0}{Primitive}$ Primal Euclid's Lemma $\ $ If $\,a,B,C\,$ are integers then

$$\begin{align}  &a\mid BC,\ \color{#0a0}{(a,B,C)=1}\\[.2em]
\Longrightarrow \   &a = bc,\qquad\:\! \color{#0a0}{(b,c)=1}\\ &b\mid B,\,c\mid C \end{align}$$

Proof $\,\ a = bc,\ b\mid B,\ c\mid C\,$ follows by the link (by gcd laws or $\Bbb Z$ a UFD) and the coprimality is clear: $\ d\mid b,c\,\Rightarrow d\mid B,C,bc\!=\!a\,$ so $\,d\mid\color{#0a0}{(B,C,a)=1}$


问:
Find the volume composed by a cylinder and a sphere
As the title says find the volume composed by a cylinder and a sphere

$$\left\{ \begin{array}
 xx^2 + y^2 \geq 9 \\
 x^2 + y^2 + z^2 \leq 25
\end{array} \right. $$
I want some help in this. Thank you.

答:
first get a visual idea in your head of the solid you are calculating. It is outside a cylinder but inside a sphere, so it is like a bead. The inside radius of the hole is 3, and the radius of the sphere is 5. This problem is easier to solve using cylindrical coordinates. You start by describing the solid using cylindrical coordinates (hint: you will need the z-coordinate of the intersection of the cylinder with the sphere). Then set up and evaluate the triple integral: $$V = \iiint_D r \: dr \: dz \: d\theta$$
The z-coordinates are from the bottom to the top of the hole (intersection). The r-coordinates are from 3 to the outside of the sphere (depends on z). The $\theta$ coordinates go from zero to $2\pi$. Hope this helps!


答:
try and define a function as an antiderivative:
$$F(t)-F(0)=\int_0^tf(x)dx$$
so:
$$\int_a^bf(x)dx=F(b)-F(a)$$
$$\int_a^cf(x)dx+\int_c^bf(x)dx=[F(c)-F(a)]+[F(b)-F(c)]=F(b)-F(a)$$
so that defines your top property, now:
$$\int_a^cf(x)dx-\int_c^bf(x)dx=[F(c)-F(a)]-[F(b)-F(c)]=2F(c)-[F(a)+F(b)]$$
so in short, no there is not a nice way to define this in terms of your first expression

Title: Question on basic property of definite integrals

Body:
Sorry if this is a trivial question.

We of course have the basic property that if $c\in[a,b]$, $$\int_a^bf(x) \, dx = \int_a^cf(x) \, dx+\int_c^bf(x) \, dx.$$ The question that I have is, is there any way at all to simplify an expression of the form $$\int_a^cf(x) \, dx-\int_c^bf(x) \, dx.$$

Any advice would be greatly appreciated. Thanks.


问:
How to show that $c_n=\frac{2}{n\pi i}$ for odd $n$? (Fourier series)
This 3blue1brown video on Fourier series leaves the viewer with the following exercise:

$$c_n=\int_0^{0.5}e^{-2\pi int}\,\,\mathrm{d}t\,\,+\int_{0.5}^1-e^{-2\pi int}\,\,\mathrm{d}t\tag{1}$$

Show that

$$c_n=\frac{2}{n\pi i}\tag{2}$$

for odd $n$ and $c_n=0$ otherwise.

So what I first did is find the antiderivative of $e^{-2\pi int}$ which is

$$\frac{1}{-2\pi in}e^{-2\pi int}$$

Then the sum of the integrals $(1)$ can be written as

$$c_n=\left(\frac{1}{-2\pi in}e^{-\pi in}-\frac{1}{-2\pi in}e^0\right)+\left(-1\cdot\frac{1}{-2\pi in}e^{-2\pi in}+\frac{1}{-2\pi in}e^{-\pi i n}\right)$$

which can be simplified to

$$c_n=\frac{2e^{-\pi in}-1-e^{-2\pi i n}}{-2\pi i n}$$

From there on, I am clueless on how to show that this expression reduces to $(2)$ when $n$ is an odd number or to $0$ if it isn't. So what am I missing here, is my overall approach wrong?

I apologize if this question is trivial or stupid, but as a highschool student, I only have limited knowledge of things such as a Fourier series.

答:
not bad for a high school student! Here is Euler's formula:
$$e^{i \theta} = \cos \theta + i \sin \theta$$
Apply this to each of the exponential terms in your numerator. Then think about what are the values of $\sin \pi, \sin 2\pi, \ldots$ and $\cos \pi, \cos 2\pi, \ldots$


问:
Similarity to companion matrix, uniqueness
Let $A \in \mathcal{M}_{n\times n}(\mathbf{R})$ and $b \in \mathbf{R}^n$ with $n\geqslant2$ be given.

Now should $(A,b)$ satisfy the Kalman rank condition
\begin{equation}
\text{rank}[b \,\, Ab \,\, \ldots \,\, A^{n-1}b] = n,
\end{equation}
or equivalently
\begin{equation}
\text{span}\{b, Ab, \ldots, A^{n-1}b\} = \mathbf{R}^n,
\end{equation}
then it is known that there exists $P \in \mathrm{GL}_n(\mathbf{R})$ such that
\begin{equation}
A = P\mathbf{A}P^{-1} \quad \text{ and } \quad b = Pe_n,
\end{equation}
where $e_n = [0, \ldots, 1]^T$ denotes the vector of the canonical basis of $\mathbf{R}^n$, whereas $\mathbf{A}$ is the companion matrix of $A$ (as defined in https://en.wikipedia.org/wiki/Companion_matrix):
\begin{equation}
\mathbf{A} = \begin{bmatrix}
0 & 1 & 0 &\ldots & 0\\
0 & 0 & 1 &\ldots & 0 \\
\vdots & & & \ddots &0\\
-a_n &\ldots &\ldots & \ldots & -a_1
\end{bmatrix},
\end{equation}
where $a_1, \ldots, a_n$ are the coefficients of the characteristic polynomial of $A$:
\begin{equation}
\chi_A(x) = x^n + a_1 x^{n-1} + \ldots + a_{n-1} x + a_n.
\end{equation}

My question is:

Is the matrix $P$ unique?

答:
Yes, this $P$ is unique. We can see that this holds as follows. First, suppose that $P = P_1$ and $P=P_2$ satisfy the required conditions; our goal is to show that $P_1 = P_2$. We note that
$$
A = P_1 \mathbf A P_1^{-1} = P_2 \mathbf A P_2^{-1} \implies \mathbf A = (P_1^{-1}P_2) \mathbf A (P_1^{-1}P_2)^{-1}\\
b = P_1e_n = P_2 e_n \implies P_1^{-1}P_2 e_n = e_n.
$$
Let $Q = P_1P_2^{-1}$; we want to use the above to show that $Q = I$ (the identity matrix) and thereby conclude that $P_1 = P_2$. Note the following:

Claim 1: If $\mathbf A$ is a companion matrix and $Q$ is any matrix such that $Q\mathbf A = \mathbf A Q$, then there exists a polynomial $f(x)$ such that $Q = f(\mathbf A)$.

This claim is proved, for instance, in Matrix Analysis by Horn and Johnson. We note that $$
\mathbf A = Q \mathbf A Q^{-1} \implies \mathbf A Q = Q \mathbf A.
$$
By Claim 1, there exists a polynomial $f$ for which $Q = f(\mathbf A)$. Moreover, we have
$$
Qe_n = e_n \implies (Q - I)e_n = 0 \implies (f(\mathbf A) - I)e_n = 0.
$$

Claim 2: $(\mathbf A, e_n)$ satisfies the Kalman rank condition.

By the Cayley-Hamilton theorem, we can write $f(\mathbf A) - I$ as $g(\mathbf A)$ for some polynomial $g$ with degree at most $n-1$. Let $g(x) = c_0 + c_1 x + \cdots + c_{n-1}x^{n-1}$. We note that
$$
g(\mathbf A) e_n = 0 \implies c_0 e_1 + c_1 \mathbf A e_n + \cdots + c_{n-1} \mathbf A^{n-1}e_n = 0 \implies\\
\pmatrix{e_n & \mathbf A e_n & \cdots & \mathbf A^{n-1} e_n} \pmatrix{c_0\\ c_1 \\ \vdots \\ c_{n-1}} = 0.
$$
By Claim 2, $\pmatrix{e_n & \mathbf A e_n & \cdots & \mathbf A^{n-1}e_n}$ has full rank. Thus, we can conclude from the above that $c_0 = c_1 = \cdots = c_{n-1} = 0$. That is, we have
$$
g(\mathbf A) = f(\mathbf A) - I = 0.
$$
That is, $Q = f(\mathbf A) = I$, which is what we wanted to show.


答: We have $\int_a^cf(x) \, dx-\int_c^bf(x) \, dx$. We know that $-\int_c^bf(x) \, dx = \int_b^cf(x) \, dx$ because when you reverse the order of the bounds, you also reverse the order you subtract values to find the area, which makes the area negative. So, we get $\int_a^cf(x) \, dx-\int_c^bf(x) \, dx = \int_a^cf(x) \, dx + \int_b^cf(x) \, dx $.


问:
Which is the axiom: well ordering principle, principle of induction, both, or none?
From analysis 1 by Terence Tao, I learn that the principle of induction is a peano axiom. In many other analysis books, like analysis by Bartle and Sherbert, the well ordering principle is used to derive the principle of induction. I am thinking now: either both are equivalent statements or one of them is the axiom.</p>

答:
Strictly speaking it's an axiom schema, i.e. an infinite family of axioms, one for each unary predicate. For example, there's one for the predicate $\sum_{k=1}^n=\tfrac12n(n+1)$. There's one for $\sum_{k=1}^n=\tfrac12n(n+1)+1$, but only the former has a true base step, so, unlike the latter, it allows a proof by induction.

Anyway, the basic answer to your question is that which is the axiom (schema) is a matter of convention. Just about every text will take induction to the be the schema presented in a definition of the Peano axioms.

Something similar happens in set theory, where the axiom of choice is equivalent in ZF to all sets having a well-ordering. Combining either with ZF gives ZFC. The C is short for choice, so you can guess which version of the axiom (it's not a schema this time) is added - again, conventionally.


问:
$f$ and $g$ are holomorphic function , $A=\{z:{1\over 2}<\lvert z\rvert<1\}$, $D=\{z: \lvert z-2\rvert<1\}$
$f$ and $g$ are holomorphic function defined on $A\cup D$

$A=\{z:{1\over 2}<\lvert z\rvert<1\}$, $D=\{z:\lvert z-2\rvert <1\}$

If $f(z)g(z)=0\forall z\in A\cup D $ then either $f(z)=0\forall z\in A$ or $g(z)=0\forall z\in A$

If $f(z)g(z)=0\forall z\in D $ then either $f(z)=0\forall z\in D$ or
$g(z)=0\forall z\in D$

If $f(z)g(z)=0\forall z\in A\cup D $ then either $f(z)=0\forall z\in A\cup D$ or $g(z)=0\forall z\in A\cup D$

If $f(z)g(z)=0\forall z\in A $ then either $f(z)=0\forall z\in A$ or
$g(z)=0\forall z\in A$

Except The $1,3$, I can say $2,4$ are true beacuse of Identity Theorem right? zeroes of holomorphic function are isolated. for $3$ I can define $f(z)=z$ on $A$ and $0$ on $D$ and $g(z)=0$ on $A$ and $z^2$ on $D$

答:
You can use the identity theorem, but since that theorem doesn't directly apply to products, you should explain how you use the identity theorem.

You can, for example, argue that if $f\cdot g = 0$ on some connected open set $S$, then there are $S_1,S_2$ with $f = 0$ on $S_1$, $g = 0$ on $S_2$, $S_1 \cap S_2 = \emptyset$ and $S_1 \cup S_2 = S$. Now if $f$ isn't identically zero on $S$, then due to the identity theorem $S_1$ contains only isolated points, hence for every $x \in S_1$ you have that $B_\epsilon(x) \setminus {x} \subset S_2$ for some $\epsilon > 0$. Using the identity theorem again, you get $g = 0$ on $S$. The same applies with $f$ and $g$ reversed, thus at least one of them is identically zero on $S$.

This works, as you correctly stated, for (2) and (4). As Robert Isreal commented, you get (1) from (4), and for (3) your counter-example indeed refutes the assertion.


问:
Cantor set intersected with $\mathbf{R}\setminus\mathbf{Q}$
I am trying to construct an example of a perfect set that contains no rational.

I see in other places a construction that involves an enueration of rationals, but I thought that construction was a bit more involved than what I have (even though they bascially amount to the same idea).

My idea is to intersect the Cantor set with $\mathbf{R}\setminus\mathbf{Q}$. We know that there is a bijection from the Cantor set to the sequences of 0 and 1 by dividing every digit in the ternary expansion by 2 (or by following the path of the point in the Cantor set as being in the left or the right). Every time 2 points share the first $n$-digits in the corresponding sequences they are within $3^{-n}$ distance close, so given a point in the Cantor set and $\epsilon<3^{-n}$, by picking an irrational point in the Cantor set whose corresponding sequences share the first $n$ entries, we are done.

The only tricky part is how to actually choose such an irrational point, but that can be easily arranged by choosing any non-periodic sequence.

Is it accurate (minus the general lack of detail, of course)?

Thank you in advance!

答:
There is a nice non-constructive proof that such Cantor sets exist in large numbers:

The Cantor set is up to homeomorphism the unique compact zero-dimensional metric space without isolated points.

From this it quite quickly follows that $C \times C \simeq C$, which can also easily be seen from the fact that $C \simeq \{0,1\}^{\Bbb N}$, e.g.

So a Cantor set $C$ is a disjoint union of uncountably many homeomorphic copies of $C$ (namely the $\{c\} \times C, c \in C$ in the square) and so many of them do not contain a rational, because the rationals are only countable.


问:
Confused about partial derivatives
I am having some issues understanding what should I keep constant and what not in certain cases when I take partial derivatives. Specifically in this kind of situation: say we have a function

$$f(x,y) = x^3+7y^2$$

and we also know that $y=2x+1$ and we need to find the partial derivative with respect to $x$ at a given point, say $\frac{\partial f(2,3)}{\partial x}$. From what I understood, when taking partial derivatives with respect to a variable, you need to keep the other constant, in which case, if I do that above (keeping $y$ constant) I would get, $\frac{\partial f(x,y)}{\partial x}=3x^2$, so I get $12$. However, if I plug $x$ in $y$ explicitly I would get $$f(x,y) = x^3+7(2x+1)^2$$ so,

$$\frac{\partial f(x,y)}{\partial x}=3x^2+28(2x+1)$$

So, I get $152$. What should I do? Thank you!

答:
We split the task into two parts. At first we look at obtaining a partial derivative with respect to $x$. Then we consider the evaluation at the specific point $(2,3)$.

Chain rule revisited: In order to better see what's going on, we consider a somewhat more general situation and then we will see how OPs special case fits into this setting.

We start with OPs real-valued function $f$, namely
\begin{align*}
&f:\mathbb{R}^2\to\mathbb{R}\\
&f(x,y)=x^3+7y^2\tag{1}
\end{align*}

We put the relation $y=2x+1$ in a slighty more general context, which might help to make the application of the chain-rule more plausible. We consider $x$ and $y$ as functions of a variable $t$:
\begin{align*}
x&=x(t)=t\\
y&=y(t)=2t+1\tag{2}
\end{align*}
and define a new real-valued function $\varphi$ as
\begin{align*}
&\varphi:\mathbb{R}\to\mathbb{R}\\
&\varphi(t)=f(x(t),y(t))\tag{3}
\end{align*}

We can now write
\begin{align*}
\varphi(t)=f(x(t),y(t))=f(t,2t+1)\tag{4}
\end{align*} and have the same situation as OP when he is using the relation $y=2x+1$ and taking $f(x,y(x))=f(x,2x+1)$. The benefit is that we can now apply the chain rule using a familiar derivation:

On the one hand we obtain from (1) - (3):
\begin{align*}
\color{blue}{\frac{d\varphi}{dt}}
&=\frac{\partial \varphi}{\partial x}\,\frac{dx}{dt}+\frac{\partial \varphi}{\partial y}\,\frac{dy}{dt}\\
&=3\left(x(t)\right)^2\cdot 1+14y(t)\cdot 2\\
&=3t^2+14(2t+1)\cdot 2\\
&\,\,\color{blue}{=3t^2+56t+28}\tag{5}
\end{align*}

on the other hand we obtain from (4):
\begin{align*}
\color{blue}{\frac{d\varphi}{dt}}
&=\frac{d}{dt}f(t,2t+1)\\
&=\frac{d}{dt}\left(t^3+7(2t+1)^2\right)\\
&=3t^2+28(2t+1)\\
&\,\,\color{blue}{=3t^2+56t+28}\tag{6}
\end{align*}
in accordance with (5).

Conclusion: We conclude from (5) and (6) the following derivation of OP is valid:
\begin{align*}
\color{blue}{\frac{d}{d x}f(x,y(x))=3x^2+28(2x+1)}
\end{align*}

Evaluation at $(2,3)$:

We consider $f(x,y)=x^3+7y^2$ as above and evaluate the partial derivative wrt $x$ at $(x,y)=(2,3)$. We obtain this way
\begin{align*}
\frac{\partial}{\partial x}f(x,y)\Bigg|_{(x,y)=(2,3)}&=\frac{\partial}{\partial x}\left(x^3+7y^2\right)\Bigg|_{(x,y)=(2,3)}\\
&=\left(3x^2\right)\Bigg|_{(x,y)=(2,3)}\\
&=12
\end{align*}
but this is completely independent from the rule $y=2x+1$. In fact it cannot be conveniently connected with evaluation at a point $(x,y)=(2,3)$, since this point is not an element of the line $y=2x+1$.

Title: Confused about partial derivatives


问:
Is there a formula for calculating Wi and Xi for any given N in gaussian quadracture.
i've been working on a formula for numerical integration envolving gaussian quadracture but i need to calculate the weights and abcisses in a given formula for any given N.

I read that the weights follow this formula but i dont really understand what P'n(xi)² is.

答:
There are a variety of "Gaussian quadrature" methods varying in the interval of integration and an optional weighting function.

The weight you cite is for Gauss-Legendre quadrature.  Here, $P_n'(x_i)$ is the first derivative of the $n^\text{th}$ Legendre polynomial.  There are various formulae for expressing the Legendre polynomials directly.  Differentiating a polynomial is straightforward.

The $x_i$ are the zeroes of the $P_n(x)$ polynomials.  The $P_n(x)$ polynomial has $n$ zeroes and they exhibit reflection symmetry through the point $ x= 0$.  (Consequently, $x = 0$ is always a root of $P_n$ when $n$ is odd.)  The interlacing property says that the $n$ roots of $P_n(x)$ partition $(-1,1)$ into intervals, each of which contains exactly one zero of $P_{n+1}(x)$.  This can be used to numerically find the zeroes of successive $P_n(x)$s.  I mention numerically finding the zeroes because algebraically solving for the zeroes of polynomials of high degree is infeasible.

So say we are working with $n = 5$ (that is, our integration formula will evaluate the integrand $5$ times).  The $x_i$ are the zeroes of $P_5(x)$, which are the zeroes of $\frac{1}{8}(63x^5 - 70 x^3 + 15x)$.  Let's number the $x_i$ in order from least to greatest, so $x_1 < x_2 < \cdots < x_5$.  We compute $P'_5(x) = \frac{1}{8}(315x^4 - 210x^2 + 15)$.  Those zeroes are $x_1 = -\frac{1}{3}\sqrt{5+ 2\sqrt{\frac{10}{7}}}$, $x_2 = -\frac{1}{3}\sqrt{5- 2\sqrt{\frac{10}{7}}}$, $x_3 = 0$, $x_4 = \frac{1}{3}\sqrt{5- 2\sqrt{\frac{10}{7}}}$, and $x_5 = \frac{1}{3}\sqrt{5+ 2\sqrt{\frac{10}{7}}}$.  (That is, $-0.90618{\dots}$, $-0.538469{\dots}$, $0$, $0.538469{\dots}$, and $0.90618{\dots}$.)

Then \begin{align*}
w_1 = w_5 &= \frac{2}{(1-x_1^2)[P'_5(x_1)]^2}  \\
    &= \frac{2}{(1-(-0.90618{\dots})^2)[\frac{1}{8}(315(-0.90618{\dots}))^4 - 210(-0.90618{\dots}))^2 + 15)]^2}  \\
    &= \frac{322-13\sqrt{70}}{900}  \text{,}
\end{align*}
assuming no typographical or computational errors.  The weight $w_2 = w_4 = \frac{322+13\sqrt{70}}{900}$.  The weight $w_3$ (for $x_3 = 0$) is $128/225$.

Generally, the $x_i$ are sparse in the middle of the interval and get denser near the endpoints.  The weights tend to be higher in the middle and lower near the ends.  (This pattern of weights makes sense: samples near the middle represent wider intervals and samples near the ends represent narrower intervals.  If the integrand were constant, the weights would have to be greater in the middle and less near the ends to capture the widths of the area rectangles represented by the samples.)

(In response to a comment by OP.)

One can get the Legendre polynomials using Rodriguez's formula,
$$  P_n(x) = \frac{1}{2^n n!} \frac{\mathrm{d}^n}{\mathrm{d}x^n}\left[ (x^2-1)^n \right]  \text{,}  $$
but unless you have a symbolic differentiator, this is not a great plan.   A much better plan is to use any of the other formulas at the link above.  For instance, from
$$  P_n(x) = \frac{1}{2^n}\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{k} \binom{2n-2k}{n} x^{n-2k}  $$
we obtain
$$  P_n'(x) = \frac{1}{2^n}\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} (-1)^k (n-2k)\binom{n}{k} \binom{2n-2k}{n} x^{n-2k-1}  \text{.}  $$
Where we have used standard notation for the floor function.


问:
Intuition behind Pontryagin duality
We have that for a locally compact abelian group,

$$A \mapsto \operatorname{Hom}(A, \mathbb{R}/\mathbb{Z})$$

is a biduality map when endowing $\operatorname{Hom}$ with the compact-open topology.

My question is why $\mathbb{R}/\mathbb{Z}$? This sounds like a rather counter-intuitive result. I can see one why one would give $\operatorname{Hom}$ the compact-open topology (there are many categorical and topological reasons to do this) but it looks rather ad-hoc that we use the circle group.

How did this come up? Are there other bidualities using different groups?

答:
The fundamental notion behind the theory of groups is the concept of symmetries and the best place to visualize symmetries is Euclidean space and, by extension, Hilbert's space.

By a symmetry on a Hilbert space $H$ one means any map
$$
  U:H\to H
  $$
that preserves distance, and sends the origin to itself.  In fact any such map is necessarily a unitary operator.

The group $\mathscr U(H)$, formed by all unitary operators on Hilbert's space  is, according to this, the archetype of symmetry!

Given any group $G$, it therefore makes a lot of sense  to try to  model $G$ via  $\mathscr U(H)$,  and this is usually done by considering
group homomorphisms
$$
  u:G\to \mathscr U(H),
  $$
often called  group representations.

If $u_1$ and $u_2$ are representations of $G$,  one may define a direct sum representation
$$
  u=u_1\oplus u_2
  \tag 1
  $$
on $H\oplus H$,  but
everything there is to know about $u$ is already present in either $u_1$ or $u_2$.

On the other hand, should we be
interested in studying a given representation $u$, we might ask whether or not it has an expression such as (1), in
which case   it
would be better to study $u_1$ and $u_2$ separately.

This means that one should concentrate in the study of irreducible group representations, namely those  which cannot
be split as in (1).

For abelian groups, Schur's Lemma says that all irreducible unitary representations are necessarily one-dimensional,
meaning that the underlying  Hilbert space is $\mathbb C$, and since
$$
  \mathscr U(\mathbb C) = \mathbb T,
  $$
we are inexorably led to  studying homomorphisms from $G$ to $\mathbb T$!

Title: Intuition behind Pontryagin duality


问:
Solving congruences using Euclidean's algorithm or otherwise

"Solve the following congruence. Make sure that the number you enter is in the range $[0,M−1]$ where $M$ is the modulus of the congruence. If there is more than one solution, enter the answer as a list separated by commas. If there is no answer, enter $N.$"

$102x = 220 \pmod{266}$

$x = ?$

So I've tried using Euclidean's algorithm on this nearly 10 times already(no exaggeration) but every time I try getting past the point below the whole thing falls apart and I just end up getting another incorrect answer for x. Would anyone mind showing how to continue on from this point and if there are any methods other than Euclidean's that one can use to calculate such equations, thanks :)

$102x = 220\pmod{266}$

$\gcd(102,266) = 2$

$266 = (102 \times 2) + 62$

$102 = (62 \times 1) + 40$

$62 = (40 \times 1) + 22$

$40 = (22 \times 1) + 18$

$22 = (18 \times 1) + 4$

$18 = (4 \times 4) + 2$

$4 = (2 \times 2) + 0$

答:
You can solve this in the moduli of prime power factors of $266$ and recombine with the Chinese remainder theorem.

$266 = 2\cdot 7 \cdot 19$.

$102x \equiv 220 \implies 0x \equiv 0 \bmod 2$ (so $x$ unconstrained)

$102x \equiv 220 \implies 4x \equiv 3 \bmod 7$

$102x \equiv 220 \implies 7x \equiv 11 \bmod 19$

Solve the two remaining congruences which will give you a value for $x \bmod 133$ and find the two answers for $x \bmod 266$ from $x\in\{0,1\}\bmod 2$


问:
Prove that $C=\bigcup_{n=1}^{\infty}I_n$ is a compact subset of a space with a rail metric

Let $I_n$ - line segment connecting the points $(0,0)$ and $\Big( \frac{\cos (\frac{2\pi}{n})}{n}, \frac{\sin (\frac{2\pi}{n})}{n} \Big)$ for $n\in \mathbb N \setminus \{0\}$.
Prove that $C=\bigcup_{n=1}^{\infty}I_n$ is a compact subset of a space with a rail metric.

I think $C$ is a compact in rail metric because we can cover each of the line segments with a railway ball starting at $(0,0)$ and radius $1$. There will be countabless infinite numbers of these balls, because $|C|=|\mathbb N|=\aleph_0$. That's why $C$ is compact.

However I think that my proof is not very formal and results more from graphic intuition. Anyone would like to say how to prove it "elegantly"?

答:
The rail metric on $\Bbb{R}^2$ (defined as discussed in the comments above and not as in the Wikipedia article on metric spaces) agrees with the Euclidean metric on any line through the origin, so (as compactness is a topological property), each $I_n$ is compact. Also, it is easy to check from the definitions, that, if $B_R(\mathbf{v}, r)$ is the open ball around $\mathbf{v}$ of radius $r$ under the rail metric and $\mathbf{0} \in B_R(\mathbf{v}, r)$, then $r > \| v\|$ and $B_r(\mathbf{v}, r) \supseteq B_E(\mathbf{0}, r - \|v\|)$ (the open ball of radius $r - \|v\|$ around the origin $\mathbf{0}$ under the Euclidean metric). So, if $\bigcup_{n=1}^\infty I_n$ is covered by a set $\mathcal{C}$ of open balls in the rail metric, then some $B_0 \in \mathcal{C}$ has $\mathbf{0} \in B_0$ and hence contains $B_E(\mathbf{0}, \delta)$ for some $\delta > 0$. But $B_E(\mathbf{0}, \delta)$ contains all but finitely many of the $I_n$. As each of the finitely many $I_n$ that are not contained in $B_E(\mathbf{0}, \delta)$ is compact, their union is compact and hence is covered by some finite subset $\{B_1, \ldots, B_k\}$ of $\mathcal{C}$. So $\{B_0, \ldots B_n\}$ is a finite subset of $\mathcal{C}$ that covers $\bigcup_{n=1}^\infty, I_n$. This gives us that $\bigcup_{n=1}^\infty I_n$ is compact in the rail metric.

Aside: under the post office metric, any singleton other than $\{\mathbf{0}\}$ is open and the $I_n$ and $\bigcup_{n=1}^\infty I_n$ are not compact. However, the union of the endpoints of the $I_n$ is compact under that metric (by a similar argument to the above).


问:
Cardinality of the set of divergent sequences
How to find the cardinality of the set of divergent sequences? Let's name this set $A$.

I know that cardinality of the set of sequences equals $2^{\lvert \mathbb{N}\rvert}$, so $\lvert A\rvert\le 2^{\lvert \mathbb{N}\rvert}$.
How to strictly prove to the other side?

Thank you for your time.

答:
Pick one divergent sequence, $a_n$, now for every $r\in\Bbb R$ let $r_n=\begin{cases} r & n=1\\ a_n & n>1\end{cases}$.

Show that this is an injection from $\Bbb R$ into $A$.

Alternatively, pick one sequence whose elements are pairwise different (e.g. $a_n=n$), and for every $K\subseteq\Bbb N$ define the sequence: $k_n=\begin{cases} a_n & n\in K\\ -a_n & n\notin K\end{cases}$

Show that this is an injection from $\mathcal P(\Bbb N)$ into $A$.


问:
Generalization of $\sqrt{n}+\sqrt{n+2005}=m$
I was solving the math Olympiad of Belgium the year $2005$, and the last problem was :

If $n$ is an integer then find all values for $n$ for which $\sqrt{n}+\sqrt{n+2005}$ is an integer as well.

My question is just to generalize this problem, so instead of $2005$ , we can solve this over all $x\in \mathbb{N}$.

My Attempt:

If you have solved the problem you will realized that this equation :
$$\sqrt{n}+\sqrt{n+x}=m$$
Has solution if and only if $m\mid x$. But I am not sure if there are some other restrictions in this equation to has a solution.

答:
In general, for non-negative integers $a, b$, the sum $\sqrt a + \sqrt b$ is an integer if and only if both $\sqrt a$ and $\sqrt b$ are integers (exercise).

Solution:

 Squaring, we see that $ab$ is the square of a rational number, and hence we may write $a = m^2 k$, $b = n^2 k$, with $k$ square-free. Then the sum $\sqrt a + \sqrt b$ becomes $(m + n)\sqrt k$, which is an integer if and only if $k = 1$.

Therefore, if $\sqrt n + \sqrt{n + x}$ is an integer, then we have $n = u^2$ and $n + x = v^2$ for some integers $u, v$.

This leads to $v^2 - u^2 = x$, which has a solution if and only if $x\not\equiv 2\mod 4$ (exercise).

Solution:

 In one direction, we have $x = (v + u)(v - u)$ and $v + u \equiv v - u \mod 2$. Hence either both factors are even and $x$ is a multiple of $4$, or both are odd and $x$ is odd. In the other direction, if $x = 2k + 1$ is odd, then $x = (k + 1)^2 - k^2$; if $x = 4k$ is multiple of $4$, then $x = (k + 1)^2 - (k - 1)^2$.

Explicit solutions:

If $x = 2k + 1$ is odd, then taking $n = k^2$ gives $\sqrt n + \sqrt {n + x} = 2k + 1$.

If $x = 4k$, then taking $n = k^2 - 2k + 1$ gives $\sqrt n + \sqrt{n + x} = 2k$.


问:
Decreasing functions $f:\mathbb{N} \rightarrow \mathbb{N}$
Let $P=\{f:\mathbb{N} \rightarrow \mathbb{N}; f(i+1)\leq f(i) \ \forall \ i \in \mathbb{N}\}.$

I need to show 2 elements of $P$ and prove that $P$ is countable (or not).

I think that $P$ only have constant functions, and if this is right, i can prove that $P$ is countable. There is some counter-example for this?

答:
I'll leave the rigorous formulations to you, here are the main ideas: (Note also that I use $0\not\in\mathbb N$; all my arguments have to be slightly adapted if you want to use $0\in\mathbb N$, but the ideas are exactly the same).

First we show that every $f\in P$ is eventually constant by contradiction. The intuition is that any non-eventually constant $f\in P$ has to go down by at least $1$ infinitely often, but $f$ can't go down by $1$ more than $f(1)$ times.

Expressed rigorously: If there exists a $f\in P$ that is not eventually constant, then there exist infinitely many $n_1<n_2<n_3<\dots$ in $\mathbb N$ such that $f(n_i+1)\le f(n_i)-1$ for every $i\in\mathbb N$.

So
\begin{split}
&f(n_{f(1)}+1)&\le &f(n_{f(1)})-1 \\
\le &f(n_{f(1)-1}+1)-1
&\le &f(n_{f(1)-1})-2 \\
\le &f(n_{f(1)-2}+1)-2 &\le &f(n_{f(1)-2})-3 \\
\le &\dots &\le &\dots \\
\le &f(n_{f(1)-(f(1)-1)}+1)-(f(1)-1)&\le & f(n_{f(1)-(f(1)-1)})-f(1), \\
\end{split}
but $f(n_{f(1)-(f(1)-1)})-f(1)=f(n_1)-f(1)\le 0$, which means that $f(n_{f(1)}+1)\le0$ which is a contradiction to $f(n_{f(1)}+1)\in\mathbb N$.

Now to show that $P$ is countable, you might want to use that the set of all eventually constant functions $f:\mathbb N\to\mathbb N$ is bijective to $\bigcup_{k\in\mathbb N} \mathbb N^k$.


答:
If you know that the irrationals are precisely the real numbers represented by infinite continued fractions, and the map

$$h:\Bbb Z\times{(\Bbb Z^+)}^{\Bbb Z^+}\to\Bbb R\setminus\Bbb Q:\langle a_n:n\ge 0\rangle\mapsto[a_0;a_1,a_2,\ldots]$$

is a homeomorphism, and that the Cantor set is homeomorphic to any product of infinitely many finite spaces each having at least two points, you can exhibit a great variety of explicit Cantor sets in the irrationals. For each $n\in\Bbb Z^+$ let $F_n\subseteq\Bbb Z^+$ be finite with at least two elements; then

$$\{[0;a_1,a_2,\ldots]:\forall n\in\Bbb Z^+(a_n\in F_n)\}$$

is a Cantor set in the irrationals.

The first answer to this question introduces the basics of continued fractions and proves in detail that $h$ is a homeomorphism; the second answer gives an easier proof that the spaces are homeomorphic.

Title: Cantor set intersected with $\mathbf{R}\setminus\mathbf{Q}$

Body:
I am trying to construct an example of a perfect set that contains no rational.

I see in other places a construction that involves an enueration of rationals, but I thought that construction was a bit more involved than what I have (even though they bascially amount to the same idea).

My idea is to intersect the Cantor set with $\mathbf{R}\setminus\mathbf{Q}$. We know that there is a bijection from the Cantor set to the sequences of 0 and 1 by dividing every digit in the ternary expansion by 2 (or by following the path of the point in the Cantor set as being in the left or the right). Every time 2 points share the first $n$-digits in the corresponding sequences they are within $3^{-n}$ distance close, so given a point in the Cantor set and $\epsilon<3^{-n}$, by picking an irrational point in the Cantor set whose corresponding sequences share the first $n$ entries, we are done.

The only tricky part is how to actually choose such an irrational point, but that can be easily arranged by choosing any non-periodic sequence.

Is it accurate (minus the general lack of detail, of course)?

Thank you in advance!


问:
Relationship between martingale and conditional expectation
I am studying a theorem about $L^p$ convergence and some equivalent conditions. What I understood is that given an $\mathcal{F}_n$-martingale $X_n$, if $X_n$ converges to $X_{\infty}$ in $L^p$, then $E[X_{\infty}\mid \mathcal{F}_n]=X_n$ for all $n\geq 0$. On the other hand, if we know that $X_n=E[X\mid \mathcal{F}_n]$ for some $X \in L^p$, then $X_n$ converges to some $X_{\infty}$ where $X_{\infty}=E[X\mid \mathcal{F}_{\infty}]$. Is my understanding correct? Am I overlooking something obvious?

答: This result appears in Lemma 4.6.6 and Theorem 4.6.8 here (pages 246-247). Specifically, if a martingale $X_n\to X$ in $L^1$, then $X_n=\mathsf{E}[X\mid \mathcal{F}_n]$ a.s. On the other hand, if $X_n=\mathsf{E}[X\mid \mathcal{F}_n]$ a.s. for some integrable r.v. $X$, then $X_n\to\mathsf{E}[X\mid \mathcal{F}_{\infty}]$ a.s. and in $L^1$, where $\mathcal{F}_{\infty}=\sigma(\cup_{n\ge 1}\mathcal{F}_n)$.


问:
Growth in direct product
I started reading with growth of groups. The problem is the following:

Problem : Let $G$ be an infinite finitely generated group.
a. If $G$ is polynomial growth then so is $G^m$ (a direct product of $G$). Moreover, the growth function $\gamma_G$ is not equivalent of $\gamma_{G^m}$ unless $m=1$.
b. If $G$ is exponential growth then so is $G^m$, and their growth functions are equivalent.

Could any one give me a hint. Thanks in advance.

答:
Let $S= \{e_1, \dots ,e_r \}$ be a generator set of $G$. For $G^m$, take the generator set $S_m= \{(e_1,1),\dots,(e_r,1),(1,e_1) ,\dots ,(1,e_r)\}$.

Claim: For $m=2$, $\ell g(x,y)=\ell g(x)+ \ell g(y)$.

If $x=w_1$ and $y=w_2$ where $w_1$ and $w_2$ are words over $S$, then $(x,y)=(w_1,1) \cdot (1,w_2)$, so $\ell g(x,y) \leq \ell g(x)+ \ell g(y)$.

If $(x,y)=w_3$ where $w_3$ is a word over $S_2$, you can write $w_3=(w_1,1) \cdot (1,w_2)$ where $w_1$ and $w_2$ are words over $S$ (since $G \times \{1\}$ and $\{1\} \times G$ commute and have trivial intersection); in particular, $\ell g(w_3)=\ell g(w_1)+ \ell g(w_2)$. Therefore, $\ell g(x,y) \geq \ell g(x)+ \ell g(y)$.

Now by induction, you can show that:

Lemma: $\displaystyle \ell g(x_1, \dots ,x_m)= \sum\limits_{i=1}^m \ell g(x_i)$.

You deduce that $\left\{ \begin{array}{ccc} B_{G^m}(k) & \to & B_{G}(k)^m \\ (x_1,\dots,x_m) & \mapsto & (x_1,\dots,x_m) \end{array} \right.$ and $\left\{ \begin{array}{ccc} B_{G}(k)^m & \to & B_{G^m}(km) \\ (x_1,\dots,x_m) & \mapsto & (x_1,\dots,x_m) \end{array} \right.$ are well-defined and injective, hence: $$\gamma_{G^m}(k/m) \leq \gamma_{G^m}(k) \leq \gamma_G(k)^m \leq \gamma_{G^m}(km).$$

Consequently, $\gamma_{G}^m \sim \gamma_{G^m}$. Then the given assertions follow.

Title: Growth in direct product


答:
Let $C$ be the Cantor set. Consider the collection of translates of $C$, i.e. the collection $\{C + x : x \in \mathbb R\}$, where $C+x = \{c + x : c \in C\}$. Clearly every $C+x$ is a perfect set (the topological properties aren't affected by translation), so it suffices to find one that contains no rationals.

Fix $q \in \mathbb Q$. Let $X_q = \{x \in \mathbb R: q \in C + x\}$, i.e. $X_q$ is the set of offsets $x$ such that the translate $C + x$ contains $q$.

Observe that a translate $C + x$ contains $q$ if and only if $x \in q - C$, so $X_q = q - C$. In particular, the Lebesgue measure of $X_q$ equals the measure of $q - C$, which is the same as the measure of $C$, which is zero.

If we define $X = \bigcup_{q \in \mathbb Q}X_q$, then $X$ also has measure zero, as it is a countable union of sets of measure zero. But $X$ is the set of offsets $x$ such that $C+x$ contains at least one rational. Since $X$ has measure zero, it follows that almost every choice of $x \in \mathbb R$ yields a translate $C+x$ that contains no rationals.


问:
Show that a point inside an acute triangle is the orthocenter
Point $M$ lies inside an acute triangle $\triangle ABC$ such that the circumscribed circles of $\triangle ABM,\triangle BCM$ and $\triangle ABC$ have equal radii. Show that $M$ is the orthocenter.

I have been trying to solve the problem for an hour using the law of sines because we have just studied it, but I can't come up with anything. What about a solution using vectors?

答:
Here is a detailed solution using inversion. In the conditions from the OP let $D$, $E$, $F$ be the feet of the heights from $A$, $B$, $C$ respectively.

The points $A,C,D,F$ are on a circle, so let us consider the inversion denoted by $\mathcal  I=*$, so $X\to \mathcal I(X)=X^*$,

with center in $B$ and

with power $k^2:=BF\cdot BA=BD\cdot BC=ac\cos B$.

So $A^*=F$, $F=A^*$; $C^*=D$, $D^*=C$. In a picture:

Let $O$ be the circumcenter of $\Delta ABC$.
Let $B_1\in BO$ be opposed to $B$ on the circumcircle $\odot(O)=\odot(ABC)$, so $BB_1=2BO=2R$. We transform it into the point $B_1^*\in BO$. Now we have the following simple observations that will restrict the position of the point $M$ from the OP, not mentioned on the picture, but mentioned below.

The circle $\odot(ABC)=\odot(O)$ is transformed into the line $A^*C^*=DF$. It contains the point $B_1^*$, which realizes the minimal distance from $B$ to a point on $A^*C^*=DF$, since $B_1$ realizes the maximal distance from $B$ to a point of the circle $\odot O$. So $BB_1^*\perp DF$.

The circle $\odot(BMA)$ maps via $\mathcal I$ into the line $M^*A^*=M^*F$, and since this circle is obtained by a rotation around $B$ of $\odot(O)$, (same radius,) its $\mathcal I$-image is also a corresponding rotation of $DF=\odot(O)^*$, in particular also tangent to the circle
$$(\beta)$$
centered in $B$ with radius $BB_1^*$. So $M^*$ is on the tangent from $F$ to $(\beta)$.

Similarly, $M^*$ is on the tangent from $D$ to $(\beta)$.

It remains to show that these two tangents intersect in the point $E$.
(From here, $M^*$ is determined to be $E$, thus $M=M^{**}=E^*=H$, the orthocenter of the given triangle. We use $EHDC$ cyclic, thus $BE\cdot BH= BD\cdot BC=k^2$, thus $E$ transforms to $H$.)

(The reader may want to stop the lecture here, and complete the proof on her or his own.)

Let us show that $EF$ is tangent $(\beta)$. By the same inversion (which conserves orthogonality), we show equivalently that the circle $\odot(BHA)=\odot(BE^*F^*)=(EF)^*$ is tangent to $(\beta)^*$, the circle centered in $B$ with radius $BB_1=2BO=2R$.

For this, let $B_1'\in(\beta)^*$ be the reflection of $B_1$ in $AB$.
Since

$\Delta B_1'AB=\Delta B_1AB$ have each a right angle in $A$, so $BB_1$ and $BB_1'$ are respectively diameters for their circumcircles,

and $\odot(ABB_1')$ (with diameter $BB_1'$) is tangent in $B_1'$ to $(\beta)^*$ (with center $B$ and radius $BB_1'=BB_1=2R$)

it is enough to show that $H$ is also on $\odot(AB_1'B)$. This is immediate:
$$
\widehat{AB_1'B} =
\widehat{AB_1B} =
\widehat{ACB} =
\widehat{ECB} =
\widehat{DHB} =
180^\circ -\widehat{AHB} \ .
$$
$\square$


问:
What does Cohen independence theorem say?
In his celebrated paper, the independence of the continuum hypothesis, P. Cohen proved that there is a model of Z-F in which the continuum hypothesis fails. As a corollary, continuum hypothesis is independent to Z-F. This is indeed a famous story in math -- perhaps most graduate students know about it and can roughly recite the heuristic.

However, thinking a little more led me wondering what this statement really means. In particular, I'm not comfortable with the meaning of the term "model". Whatever a model is, suppose there are two models of Z-F, $M_1$ and $M_2$, in which CH holds and fails respectively. Since the $M_i$'s are models of the set theory, what is a set depends on $i$. That is to say, a set in the sense of $M_1$ is not necessarily a set in the sense of $M_2$. However, all definitions I've seen for the term "model" depend on the notion of a set.

Thus my first question: What is a model in this context? If you'd like to use the term "set" in your definition, then in which sense does that mean?

The answer, as I guess, could be that either the term "model" or "set" isn't formally defined here. That's fine, as in the end there must be some undefined terms, and this could be one of them. However, if that's the case, then in what sense can we say that such model exists? That is, we might not know what a model formally is, but at least we should know what qualifies a mathematical work to be a legit construction of a model. This is my second question.

EDIT I was benefited much from the comments, answers, and discussions below, and am grateful for it. What follows is my understanding.

By a "model" we mean a set in ZF theory. By a "set" in ZF theory we mean a class $A$ that exists in the von Neumann universe $V$, or simply denoted by $A \in V$. By "$V$" we mean the syntactic expression $\{x | x = x\}$, and by "exists in $V$" (or $A \in V$) we mean the expression given here.

Therefore, a "model" is really just a syntactic expression derivable from the ZF axioms that satisfy a bunch of syntactic requirements.. but we rather think of it as a real set that we call "set". It is the "set" that we talk about in usual mathematical context (e.g. in a finite group theory course). This answers my first question, and deprecates the second.

答: Can I try an alternative angle on this that may help. In a theory like ZF, you cannot construct a model of ZF, but you can give a perfectly rigorous definition of what is to be a model of a first-order theory like ZF. To coin a pun, ZF doesn't know much about the art of constructing models, but it knows a good model when it sees one. From this point of view, statements like "if ZF has a model, then it has a countable model" and "if ZF has a countable model, then so so does ZF+$\lnot$CH" are perfectly meaningful statements of ZF (and ZF can prove them). The "ontological" outcome of this is benign and informative. It is benign because the definitions are rigorous and all the ontological assumptions are explicit. It is informative, because if our metalanguage includes ZF and we were to make the ontological assumption that ZF has a model in our metalanguage (e.g., by allowing Grothendieck universes), then we know that our assumptions imply that ZF+$\lnot$CH would have a model too.


问:
Plotting complex numbers when the angle is complex
A few months ago I asked about having a sine equal to 10; the answer was a complex number (link).
I hoped to plot this on the usual 2-d complex plane, but couldn't figure out how.  Is there a way to graphically show complex trigonometry when the angle is complex?</p>

答:
I'm assuming you want to plot the sine function. Complex functions have 2-dimensional input and 2-dimensional output. You'd need 4 dimensions to accurately draw that graph. It's possible with an animation, but I don't think it's commonly done.

The standard approach, as far as I know, is to use color. You use the plane as input, and then you color each point in the plane according to the value of the output in polar coordinates. Black is 0, white is infinity, while hue is determined by the argument, with red along the positive real axis.

Here is a plot of the identity function $z\mapsto z$:

And here is the complex sine:

Both taken from wikimedia commons.

You can see along the real axis, that it's only red and cyan, between black dots. That's the regular real sine function. What you want are the vertical red lines, as those are the places where the sine function has positive real output greater than $1$. At some point those lines (which grow exponentially away from the real axis) are white enough to correspond to an absolute value of $10$, and that's where you have your solutions.


问:
How to prove that frame functions on Hilbert spaces are additive?
Let $\newcommand{\calH}{\mathcal{H}}\newcommand{\eff}{\operatorname{Eff}(\calH)}\calH$ be some separable Hilbert space, and denote with $\eff$ the set of effects on $\calH$, that is, the set of positive operators $E$ such that $0\le E\le I$. Define on this space the frame functions as the maps $f:\eff\to[0,1]$ such that
$$\sum_kf\left( E_k\right) = 1$$
for every set $\{E_k\}_k\subset\eff$ such that $\sum_k E_k=I$ (I'm taking this definition from Caves et al. (2005)).

It turns out that such frame functions are linear. In particular, they are additive on the set of effects. I'm trying to get a better understanding of why this is the case.

Given any pair of effects $E_1,E_2\in\eff$ such that $E_1+E_2\le I$, I can easily see the additivity: defining $E_3=I-E_1 - E_2$ we see that $E_3\in\eff$ and $E_1+E_2+E_3=I$, and thus
$$1 = f(E_1 + E_2 + E_3) = f(E_1) + f(E_2) + f(E_3) = f(E_1 + E_2) + f(E_3),$$
implying $f(E_1 + E_2) = f(E_1) + f(E_2)$. This also seems the argument made in the paper above (see Eq. (4)).

However, the claim seems to be that frame functions are more generally additive on any pair of effects, and thus in particular also when $E_1+E_2\le I$ does not hold.
How is the statement proven in this more general case?

答:
Let $\newcommand{\calH}{\mathcal{H}}\newcommand{\eff}{\operatorname{Eff}(\calH)} f$ be a frame function.

Here's a breakdown of the necessary steps to prove linearity of any frame function $f$ on its domain $\eff$:

For any $E_1,E_2\in\eff$ such that $E_1+E_2\in\eff$, $f(E_1+E_2)=f(E_1)+f(E_2)$.

Proof: Define $E_3\equiv I-E_1-E_2$. Then $E_3\in\eff$. Moreover $f(E_1+E_2)+f(E_3)=f(E_1)+f(E_2)+f(E_3)=1$, thus $f(E_1+E_2)=f(E_1)+f(E_2)$.

For any $E\in\eff$ and positive integer $k\ge0$, $f(E/k)=f(E)/k$.

Proof: $E/k\in\eff$ and $k(E/k)\in \eff$, thus
$$f(E) = f(\underbrace{E/k+\cdots +E/k}_k) = k f(E/k).$$

For any $E_1,E_2\in\eff$, $f(E_1+E_2)=f(E_1)+f(E_2)$.

Proof: $E_1+E_2\le 2I$, thus $(E_1+E_2)/2\in\eff$, and the conclusion follows from the two previous points:
$$f(E_1+E_2) = f(2(E_1+E_2)/2)=2(f(E_1)/2+f(E_2)/2)= f(E_1)+f(E_2).$$

For any $q\in\mathbb{Q}\cap[0,\infty]$ and $E\in\eff$, $f(qE)=qf(E)$.

Proof: Point (3) shows that $f(kE)=k f(E)$ for any positive integer $k$, and point (2) that $f(E/k)=f(E)/k$. Together, these amount to $f(qE)=q f(E)$ for any positive rational $q$.

For any $E_1,E_2\in\eff$ such that $E_1\le E_2$, $f(E_1)\le f(E_2)$.

Proof: From the assumptions follows that $E_2-E_1\in\eff$. Thus
$$f(E_2) = f((E_2-E_1)+E_1) = f(E_2-E_1)+f(E_1) \ge f(E_1).$$

For any $\alpha\ge0$ and $E\in\eff$, $f(\alpha E)=\alpha f(E)$.

Proof: Let $(p_k)$ be an increasing sequence converging to $\alpha$ from below, and $(q_k)$ a decreasing sequence converging to $\alpha$ from above. Then $q_k E\le \alpha E\le p_k E$, and from our previous points we have
$$p_k f(E) \le f(\alpha E) \le q_k f(E).$$
Going to the limit we conclude that $\alpha f(E)=f(\alpha E)$.

One can then go a step further to extend $f$ as a linear function defined over the full set of Hermitian operators $\mathrm{Herm}(\calH)$. For the purpose, we observe that any $H\in\mathrm{Herm}(\calH)$ can be written as $H=\alpha_+ H_+-\alpha_- H_-$ for some $\alpha_\pm\ge0$ and $H_\pm\in\eff$. We then define $\tilde f:\mathrm{Herm}(\calH)\to\mathbb R$ as
$$\tilde f(H) = \alpha_+ f(H_+) - \alpha_- f(H_-).$$
Such $\tilde f$ is thus a linear functional on $\mathrm{Herm}(\calH)$.
As such, it can be represented as an inner product, which is essentially the point of (Busch's version of) Gleason's theorem.

I then also found that essentially the same argument is presented in (Busch 2003).


问:
Show that X and Y are conditionally independent given Z
Let X,Y denote the number of heads in the first and last two flips separately, and Z be the indicator random variable of the second flip. Given a fair coin and independently flip it three times.

\begin{equation}
    X =
      \begin{cases}
         0 & \text{if there are no heads in the first two flips}\\
         1 & \text{if there is exactly one head in the first two flips}\\
         2 & \text{if there is exactly two heads in the first two flips}
      \end{cases}
\end{equation}
\begin{equation}
    Y =
      \begin{cases}
         0 & \text{if there are no heads in the last two flips}\\
         1 & \text{if there is exactly one head in the last two flips}\\
         2 & \text{if there is exactly two heads in the last two flips}
      \end{cases}
\end{equation}
\begin{equation}
    Z =
      \begin{cases}
         1 & \text{if the second flip is H}\\
         0 & \text{if the second flip is T}\\
      \end{cases}
\end{equation}

Show that, X is not independent to Y, but X and Y are conditionally independent given Z.

My attempt:

To show that X is not independent to Y is easy. We can take the case where $\{x,y\}=\{0,0\}$. Then we have that
$$\mathbb{P}(X=0)\cdot\mathbb{P}(Y=0)=\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}\neq\frac{1}{8}=\mathbb{P}(X\cap Y)$$

To prove that X and Y are conditionally independent given Z, I was thinking that we need to show that $\mathbb{P}(X=x|Z=z)\mathbb{P}(Y=y|Z=z)=\mathbb{P}(X=x,Y=y|Z=z)$ is true for all combinations of x,y,z. However, this would result in showing the equality is true $3\cdot3\cdot2=18$ times, which seems like a lot of work. Is there a more efficient way of showing this independence? Thank you.

答:
Hint: When given the count for heads in the middle coin, the count for heads in the first and last two coins are each determined by the count for heads in the first and last coin, respectively, -- which are independent.

Let $U$ and $V$ be these counts, and so we have: $X=U+Z$ and $Y=V+Z$.

Thus $U,V,Z\overset{iid}\sim\mathcal U[0,1]$ and taking any $i,j,k$ from $\{0,1\}$, you may show: $$\mathsf P(X{=}i{+}k, Y{=}j{+}k\mid Z{=}k)=\mathsf P(X{=}i{+}k\mid Z{=}k)\mathsf P(Y{=}j{+}k\mid Z{=}k)$$


答:
```<code>        X     Y     Z
H H H   [2    2]    1
H H T   [2    1]    1
H T H   1     1     0
H T T   1     0     0
T H H   [1    2]    1
T H T   [1    1]    1
T T H   0     1     0
T T T   0     0     0
</code>```

You don't need to show it 18 times. You can just show it 8 times. Start with Z=1.

$$P(X=2|Y=1)=P(X=2|Y=2)=P(X=2)=\frac 1 2\\
P(X=1|Y=2)=P(X=1|Y=1)=P(X=1)=\frac 1 2$$

Now do it once more for the case Z=0 and you're done. (This uses an equivalent definition of independence that $P(A|B)=P(A)$ means that A and B are independent.)

Title: Show that X and Y are conditionally independent given Z

Body:
Let X,Y denote the number of heads in the first and last two flips separately, and Z be the indicator random variable of the second flip. Given a fair coin and independently flip it three times.

\begin{equation}
    X =
      \begin{cases}
         0 & \text{if there are no heads in the first two flips}\\
         1 & \text{if there is exactly one head in the first two flips}\\
         2 & \text{if there is exactly two heads in the first two flips}
      \end{cases}
\end{equation}
\begin{equation}
    Y =
      \begin{cases}
         0 & \text{if there are no heads in the last two flips}\\
         1 & \text{if there is exactly one head in the last two flips}\\
         2 & \text{if there is exactly two heads in the last two flips}
      \end{cases}
\end{equation}
\begin{equation}
    Z =
      \begin{cases}
         1 & \text{if the second flip is H}\\
         0 & \text{if the second flip is T}\\
      \end{cases}
\end{equation}

Show that, X is not independent to Y, but X and Y are conditionally independent given Z.

My attempt:

To show that X is not independent to Y is easy. We can take the case where $\{x,y\}=\{0,0\}$. Then we have that
$$\mathbb{P}(X=0)\cdot\mathbb{P}(Y=0)=\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}\neq\frac{1}{8}=\mathbb{P}(X\cap Y)$$

To prove that X and Y are conditionally independent given Z, I was thinking that we need to show that $\mathbb{P}(X=x|Z=z)\mathbb{P}(Y=y|Z=z)=\mathbb{P}(X=x,Y=y|Z=z)$ is true for all combinations of x,y,z. However, this would result in showing the equality is true $3\cdot3\cdot2=18$ times, which seems like a lot of work. Is there a more efficient way of showing this independence? Thank you.

Body:
        X     Y     Z
H H H   [2    2]    1
H H T   [2    1]    1
H T H   1     1     0
H T T   1     0     0
T H H   [1    2]    1
T H T   [1    1]    1
T T H   0     1     0
T T T   0     0     0

You don't need to show it 18 times. You can just show it 8 times. Start with Z=1.

$$P(X=2|Y=1)=P(X=2|Y=2)=P(X=2)=\frac 1 2\\
P(X=1|Y=2)=P(X=1|Y=1)=P(X=1)=\frac 1 2$$

Now do it once more for the case Z=0 and you're done. (This uses an equivalent definition of independence that $P(A|B)=P(A)$ means that A and B are independent.)


问:
Efficient way to maximise stock profits
I got a message from my future self. Rather than a sports almanac, it contained minute-by-minute data for tomorrow's prices on a few select stocks. I want to use this to earn a bit of money. Is there an efficient way to decide when to sell and buy which stocks in order to maximize profits?

An example: Say I got the following price data:
$$
\begin{array}{|c|rr|rr|rr|}\hline
\text{Time}& \text{Company 1}&&\text{Company 2}&&\text{Company 3}\\
&\text{Sell}&\text{Buy}&\text{Sell}&\text{Buy}&\text{Sell}&\text{Buy}\\\hline
00:00:00&1.23&1.25&12.48&12.57&7.93&8.01\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\\
23:59:59&1.03&1.12&15.93&15.97&10.30&10.39\\\hline
\end{array}
$$
Now, assuming we can actually buy and sell for these exact prices at the given times (which is an approximation), how can one most efficiently navigate these data to end the day with as much money as possible? Are there any known algorithms for these? What computational complexity do you get?

I think for the case of a single company I can do linear in the number of minutes: search forward for first local minimum of buy at $b_0$. Between there and the next time the buy price is lower ($b_1$), if there is ever a sell price higher than $b_0$, then buy at $b_0$, sell as high as you can before $b_1$, and if no sell price in that time interval is higher than $b_0$, then don't buy. Then continue searching for local buy minimums from $b_1$ onward.

But for more than one company I'm not even sure how I would systematically look for the optimal solution, other than full brute force. And that quickly becomes unwieldy and exponential. Is there a more efficient algorithm?

Specifics:

I don't have to buy whole stocks. I can buy parts of one. Let's say any positive real number is valid, as long as I can afford it.

No shorting. I can't sell stocks I haven't bought earlier in the day.

I have to end up with money, not stocks. Sell before the day is over, at the latest using the $23:59:59$ prices.

I can sell for a given sell price, and then buy for a buy price within the same minute.

答:
This can be done in $O(nT)$ arithmetic operations (division, multiplication, and comparison) where $n$ is the number of companies and $T$ is the number of discrete times using an approach one could describe as dynamic programming. Index the companies by integers $0\leq i < n$ and the time steps as $0\leq t < T$. This answer assumes that we proceed in timesteps, at each of which we first have an opportunity to sell whatever stocks we like, then to buy whatever stocks we like at the given price, allowing all non-negative real numbers as quantities of stocks.

Let's define:

$b(t,i)$ is the buy price of stock $i$ at time $t$.

$s(t,i)$ is the sell price of stock $i$ at time $t$.

$f(t,i)$ is the maximum amount of money we could end with if after time step $t$ we held one stock of company $i$ (and nothing else). Loosely speaking, this is the true value of stock $i$ at time $t$.

$g(t)$ is the maximum amount of money we could end with if before time step $t$ we had one dollar (and nothing else). Loosely speaking, this is the true value of a dollar at time $t$.

We can then write equations relating these: first, for a stock, either we want to hold on to it for another step or sell it immediately:
$$f(t,i)=\max\{f(t+1,i),\,g(t+1)\cdot s(t+1,i)\}.$$
Note that a mixed strategy of "sell some" never makes sense, as it would just average the two extremes in terms of pay-off - so no mixed strategy could be the maximum. Similarly, for $g(t)$, the non-mixed strategies are "buy all of one stock" or hold on to the dollar - so
$$g(t)=\max\{g(t+1)\}\cup \left\{\frac{f(t,i)}{b(t,i)}:0\leq i < n\right\}$$
As edge cases, we also note $g(T)=1$ (as a dollar at the end is a dollar) and $f(T - 1,i)=0$ (as a stock at the end is useless).

Note that computing $f(t,i)$ only requires knowing evaluations of $g$ and $f$ at times of $t+1$ and later, and computing $g(t)$ only requires knowing $f(t,i)$ and the values of $g$ at a later time - so we can start from $t=T$ and work backwards until we compute $g(0)$, getting our answer of maximum amount we could earn. This requires $O(nT)$ operations to do - which is as fast as we possibly could get, given that there are $nT$ entries in the table to process.

We can extract the optimal strategy by seeing which branch of the maximum was used in each step - e.g. if the value of $g(0)$ is $\frac{f(0,17)}{b(0,17)}$, we know to buy stock $17$ and then if the value of $f(0,17)$ is $f(1,17)$, we know to hold on to the stock in time step $1$, but if $f(1,17)$ equals $g(2)\cdot s(2,17)$, we know to sell at time step $2$ - and then to look at what $g(2)$ equals for our next move.

Title: Efficient way to maximise stock profits


问:
I need help with a stupid webassign problem. It's about partial fraction
The question is following:

Write out the form of the partial fraction decomposition of the function (as in this example). Do not determine the numerical values of the coefficients.
$$\frac{1}{x^2+x^4}$$

I know how to solve this. but the problem is I have to do it like the example and I can't figure out what the "proper way" the Webassign is looking for.

$$\frac{x^3+x^2+1}{x(x-1)(x^2+x+1)(x^2+1)^3}=\frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^2+x+1}+\frac{Ex+F}{x^2+1}+\frac{Gx+H}{(x^2+1)^2}+\frac{Ix+J}{(x^2+1)^3}$$

答:
Hint.

Your fraction is equal to
$$
\frac{1}{x^2(x^2+1)}
$$
and $x^2+1$ is an irreducible quadratic factor.


问:
solution to a root inequality
I have the inequality
$$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc} \geq \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab}.$$I tried to do $u=a^2+b^2+c^2$ and $v=ab+ac+bc$ and $x=a^2+2bc$,  $y=b^2+2ac$,  $z=c^2+2ab$ ...but I did not find any solution. Any help is appreciated.</p>

Title: solution to a root inequality

Body:
$x=a-b,y=b-c,z=c-a,u^2=a^2+b^2+c^2,p^2=ab+bc+ac \to x+y+z=0,x^2+y^2+z^2=2u^2-2p^2,xy+yz+xz=p^2-u^2 \to x^2y^2+y^2z^2+x^2z^2=(p^2-u^2)^2,(xy)^4+(yz)^4+(xz)^4=(u^2-p^2)^2+(4p^2-4u^2)x^2y^2z^2,u^2-x^2-p^2=-yz,u^2-y^2-p^2=-xz,u^2-z^2-p^2=-xy$

$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc} \geq \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab} \iff $

$u+2p \ge \sqrt{u^2-x^2}+\sqrt{u^2-y^2}+\sqrt{u^2-z^2} \iff$

$ 2up \ge \sqrt{u^2-x^2}\sqrt{u^2-y^2}+\sqrt{u^2-x^2}\sqrt{u^2-z^2}+\sqrt{u^2-z^2}\sqrt{u^2-y^2}-p^2 \iff $

$ 4u^2p^2 \ge \sum_{cyc}(u^2-x^2)(u^2-y^2)+p^4+2\sum_{cyc}(u^2-x^2)\sqrt{u^2-z^2}\sqrt{u^2-y^2}-2p^2\sum_{cyc}\sqrt{u^2-x^2}\sqrt{u^2-y^2} \iff $

$ 4u^2p^2 \ge 3u^4+p^4+\sum_{cyc}x^2y^2-2(x^2+y^2+z^2)u^2+2\sum_{cyc}(u^2-x^2-p^2)\sqrt{u^2-z^2}\sqrt{u^2-y^2} \iff p^2(u^2-p^2) \ge -\sum_{cyc}yz\sqrt{u^2-z^2}\sqrt{u^2-y^2} \iff p^2(u^2-p^2)+xy\sqrt{u^2-x^2}\sqrt{u^2-y^2} \ge -yz\sqrt{u^2-z^2}\sqrt{u^2-y^2}-xz\sqrt{u^2-x^2}\sqrt{u^2-z^2} \iff [p^2(u^2-p^2)]^2+2x^2y^2(u^2-x^2)(u^2-y^2)-\sum_{cyc}x^2y^2(u^2-x^2)(u^2-y^2) \ge 2xy\sqrt{u^2-x^2}\sqrt{u^2-y^2}[z^2(u^2-z^2)-p^2(u^2-p^2)] \iff x^2y^2(2p^4+2z^4-u^2z^2)\ge 2x^2y^2\sqrt{u^2-x^2}\sqrt{u^2-y^2}(p^2-z^2) \iff x^4y^4[((2p^4+2z^4-u^2z^2)^2-4(p^2-z^2)^2(p^4+z^4-u^2z^2+2p^2z^2)]\ge 0 \iff (xyz)^4(4p^2-u^2)^2\ge 0 $when and only when $x=0 $ or $ y=0 $ or $z=0 $  it takes $"="$ QED.

EDIT: I add some notes for the process:

$xy>0 \to -yz\sqrt{u^2-z^2}\sqrt{u^2-y^2}-xz\sqrt{u^2-x^2}\sqrt{u^2-z^2}>0$

if $p^2 < z^2$, then the inequality already OK as $2p^4+2z^4-u^2z^2 \ge  p^4+z^4-u^2z^2+2p^2z^2=(u^2-x^2)(u^2-y^2) \ge 0$


问:
Solve the non-homogenous recurrence $h_n = 3h_{n-1} + 3(2)^n, h_0 = 1$?
I am trying to solve the non-homogenous recurrence $h_n = 3h_{n-1} + 3(2)^n, h_0 = 1$.

I can see that the solution to the homogenous $h_n = 3h_{n-1}$ is clearly $h_n = 3^n$. That being said, I have no idea how to solve $h_n - 3h_{n-1} = 3(2)^n$: I assume it's some solution of the sort $(c_1n + c_2)(2)^n$, but I don't know where we'd go from here! Can anyone give advice?

答:
Welcome to MSE!

You could always use generating functions, but that machinery isn't necessary here. Instead, there's a standard trick. Write

$$t_n = \frac{h_n}{2^n}$$

Then after dividing both sides by $2^n$ your recurrence becomes

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

Do you see how to solve this recurrence for $t$? Do you see how that will solve your original problem too?

I hope this helps ^_^


问:
Graph with $n$ vertices and $n + 1$ edges
I have the following problem: $n$ teams participate in a tournament, where everyone can play against everyone (teams obviously can't play themselves). $n + 1$ matches have already been played. Show that there is at least one team, that has played at least $3$ matches.

I've been stuck on this problem for a time now, although it feels intuitive and there should be a rather easy proof for this. I think I have found one, but I am not quite sure.

I thought about the handshake lemma, which for this case would give me the following: $\sum_{i=1}^n deg(v_i) = 2(n + 1)$. If I assume the opposite, where there are no vertices with a degree of at least three, then on the left-hand side of the handshake lemma I would basically have $n$ terms, all of which less than 3. The biggest sum this could possibly result in would be $2n$, when all vertices have a degree of 2. However, this would still leave me two short. Would this be a valid argument/proof?

答:
Your argument seems to work OK. Another way:

Each match is a "team event" for the two participating teams. So there have been a total of $2(n+1)$ "team events" for the $n$ teams. Now apply the strong pigeonhole principle.


问:
Does empty set entails everything?
I am new to propositional logic, and I am trying to figure out whether the below statement is correct:

```<code>[a |= c] or [ b |= c] → [a^b |= c]
</code>```

First of all, it is obvious that

```<code>a^b ⊆ a and a^b ⊆ b.
</code>```

Moreover, because either a |= c or b|= c are correct(or both), it seems that a^b must also entail c.
In other words:

Ma ^ b ⊆  Ma  . Also, when we say that  a |= c,

we have M a ⊆ Mc.

But what if the intersection of a and b is empty? Does this argument still hold?

答:
$a\land b\vDash c$ means that there are no interpretations(aka models) which value $a\land b$ as true but $c$ as false.

If there are no interpretations which value $a\land b$ as true, then the above holds.  (See also vacuous truths)

The intersection of the models $\mathbf M_a$ and $\mathbf M_b$ being empty means there are no models which simultaneously value $a$ and $b$ as true.   Thus $\mathbf M_{a\land b}$ will be empty.   Therefore...

Title: Does empty set entails everything?


问:
Why differentiate between continuous on $[a,b]$ and differentiable on $(a,b)$?
Before starting I want to say that my main language is not english so you may find some grammar mistakes. I will try to avoid making them as much as I can, so please forgive me.

A few months ago I decided that I wanted to learn Calculus so I got some books and started to study, during that time I faced a statement that is pretty common among math definitions related to it, and that statement is:

$f(x)$ is continuous at $[a,b]$ and differentiable at $(a,b)$

What I was thinking was that a limit needs a closed interval to compute its value, but after making some research I found that the function $f(x) = \sqrt{x}$ is continuous at $0$, so I started to make myself this question.

If we consider, for example $f(x) = x^2$. and the interval $[1,3]$. We know that the function is differentiable and continuous at its whole domain but we do we say that it has to be continuous at $[1,3]$ and differentiable at $(1,3)$, this of course, considering the function over the interval mentioned before, when in fact it is differentiable at $1$ and $3$.

Also the function $\sqrt{x}$ is considered continuous at its whole domain, even also at $x = 0$ but it is not differentiable at 0. Why?

Why do we do this difference between $[a,b]$ and $(a,b)?$

Sorry if I did not express myself the best way possible, again, English is not my main language.

答:
The classic counterexample to differentiability is $f(x) = \vert x \vert$ and its translates.

Consider e.g. $g(x) = \vert x\vert -2$. You can apply the Intermediate Value Theorem, Rolle's Theorem, Mean Value Theorem, etc. to $g$ (and indeed $f$, although that may be less interesting) on the interval from $0$ to $k$, for any positive integer $k$, despite the lack of differentiability at $(0,-2)$.

It's worth considering whether you can drop the continuity assumption at the endpoint. You cannot. Hint: Try a removable singularity, such as a step function.


答:
In general, if you have a function $f:[a,b]\to\mathbb{R}$, the differentiability of $f$ at the left endpoint is defined as
$$
f'(a):=\lim_{h\color{blue}{\downarrow} 0}\frac{f(a+h)-f(a)}{h}
$$
whenever the limit exists. If the limit does not exist, then $f$ is not differentiable at $x=a$.

For your example $f(x)=\sqrt{x}$, you can check that the following limit does not exist:
$$
\lim_{h\color{blue}{\downarrow} 0}\frac{\sqrt{h}}{h}
$$

Differentiability at a point is a local property of a function. Usually, what matters most is the interior points of the domain when talking about derivatives; note that every point in $(a,b)$ is an interior point.

Title: Why differentiate between continuous on $[a,b]$ and differentiable on $(a,b)$?

Body:
Before starting I want to say that my main language is not english so you may find some grammar mistakes. I will try to avoid making them as much as I can, so please forgive me.

A few months ago I decided that I wanted to learn Calculus so I got some books and started to study, during that time I faced a statement that is pretty common among math definitions related to it, and that statement is:

$f(x)$ is continuous at $[a,b]$ and differentiable at $(a,b)$

What I was thinking was that a limit needs a closed interval to compute its value, but after making some research I found that the function $f(x) = \sqrt{x}$ is continuous at $0$, so I started to make myself this question.

If we consider, for example $f(x) = x^2$. and the interval $[1,3]$. We know that the function is differentiable and continuous at its whole domain but we do we say that it has to be continuous at $[1,3]$ and differentiable at $(1,3)$, this of course, considering the function over the interval mentioned before, when in fact it is differentiable at $1$ and $3$.

Also the function $\sqrt{x}$ is considered continuous at its whole domain, even also at $x = 0$ but it is not differentiable at 0. Why?

Why do we do this difference between $[a,b]$ and $(a,b)?$

Sorry if I did not express myself the best way possible, again, English is not my main language.


问:
Fourier Transformation-exercice
I have this exercice:

Let $$W = \{f \in L^1(\mathbb{R}) : \hat{f} \in L^1(\mathbb{R})\}$$

1- Prove that $f \in W$ iff $\hat{f} \in W.$

2- Let $f \in W.$ $f$ is continuous? $\lim_{x\rightarrow \infty}f(x) = 0$?

Here are my answers. Can you confirm if they are correct please?

1- $f \in W \Rightarrow f \in L^1$ and $\hat{f} \in L^1.$ We have $F(\hat{f}) = \overline{F} (\check{\hat{f}}) = (2 \pi) \check{f}$

wa have $f \in L^1 \Leftrightarrow \check{f} \in L^1,$ so $\hat{\hat{f}} \in L^1(\mathbb{R})$ , so $\hat{f} \in W.$

$f = (2 \pi)^{-1} \overline{F}(\hat{f}) = (2 \pi)^{-1} F (\check{\hat{f}}) \in L^1 = (2 \pi)^{-1} \hat{\hat{f}} \in L^1,$ so $\hat{f} \in L^1$ and $f \in L^1,$ so $f \in W.$

Then the equivalence is prouved.

2- If $f \in L^1$ then $\hat{f}$ is continuous and his limit is 0 in $\infty$.

$f \in W \Leftrightarrow \hat{f} \in W \Rightarrow \hat{f} \in L^1 \Rightarrow \hat{\hat{f}}$ is continuous and $\lim_{x \rightarrow + \infty} \hat{\hat{f}} = 0$ because $f = (2 \pi)^{-1} \check{\hat{\hat{f}}}$

and $\lim f(x) = (2 \pi)^{-1} \lim \breve{\hat{f}} = \lim \hat{\hat{f}} (x) = 0.$

答: Yes, the logic of your solution is correct. The details such as multiples of $2\pi$ cannot be completely verified because you did not specify which version of the Fourier transform you are using (there are several commonly used formulas, which differ by the placing of factors $2$ and $\pi$). Regardless of the convention used, the main point is that both direct and inverse Fourier transforms map $L^1$ into $C_0(\mathbb R)$. Therefore, $W\subset L^1(\mathbb R)\cap C_0(\mathbb R)$. </p>

Title: Fourier Transformation-exercice


问:
conic section, Circle
As far as I know and have read, the following equation has no answer
x^2 + y^2 = -a^2

From the point of view of conic sections, the circle is obtained when the cutting plane is parallel to the plane of the circle producing the cone.

My question is, how is this cut that has no answer?

Because I think there is answer no matter how it is cut.

is it related to complex numbers?

答:
For conic sections, $x$ and $y$ are real numbers.  For real $x$ and $y$, $x^2 \geq 0$, $y^2 \geq 0$ and
$$  x^2 + y^2 \geq 0  \text{.}  $$

But then $-a^2 \leq 0$.  The only possibility for a solution is obtained when $a = 0$ and the conic section is a point, a circle with radius $0$.

Title: conic section, Circle


问:
Is the following proof of the continuity of a function correct?
I just started learning analysis and encountered my first task on continuity today, namely proving that $f: \mathbb{R}\to \mathbb{R}, \space f(x) = \inf\{|x-k\rvert \space \mid k\in \mathbb{Z} \}$ is continuous in every point. I attempted to prove this as follows:

Let $x_{0} \in \mathbb{R}$ be arbitrary. Let $x \in \mathbb{R}$. By the triangle inequality we have $\left|x_{0}-k\right| \leq\left|x_{0}-x\right|+|x-k|$ for all $k \in \mathbb{Z}$. Thus, one obtains
$$
\inf \left\{\left|x_{0}-k\right| \mid k \in \mathbb{Z}\right\} \leq\left|x_{0}-x\right|+\inf \{|x-k| \mid k \in \mathbb{Z}\}
$$
so by the definition of $f$
$$
f\left(x_{0}\right)-f(x) \leq\left|x_{0}-x\right|
$$
so we obtain  $f(x)-f\left(x_{0}\right) \leq$
$\left|x_{0}-x\right|$ (applying the same argumentation). We get
$$
\left|f\left(x_{0}\right)-f(x)\right| \leq\left|x_{0}-x\right|
$$
Now, let $\epsilon>0$ be arbitrary. To show that $f$ is continuous in $x_{0}$ we must find some  $\delta>0$, s.t. for all $x \in \mathbb{R}$ with $\left|x-x_{0}\right|<\delta$ the inequality $\left|f\left(x_{0}\right)-f(x)\right|<\epsilon$ holds. From the above we can clearly see that $\delta=\epsilon$ is a sufficient choice is. Since $x_{0}$ was arbitrary, we are done.

Does this attempt seem valid? Also, what are other ways one could prove this (assuming that the above works). I assume that one often can just plug in the definition of the respective function and then rearrange the inequality, but this didn‘t quite work for me in this case.

答: Your estimate appears to be correct.  If you draw a picture, this function is a "sawtooth" function.   For $n\in \mathbb{Z}$, you have $f(n) = 0$ and $f(n + 1/2) = 1/2$. The rest of the graph can be obtained by "connecting the dots."  The slopes of the connecting lines are $\pm 1$.  So, you do have $|f(x) - f(y)| \le |x - y|$ for all $x, y\in \mathbb{R}$.