问:
Question using Series Convergence or Divergence Tests
$$\sum_{n=2}^\infty  \frac{((\ln(n))^n}{n!}$$

So this is the series that I have to solve for whether it is convergent or divergent in my HW, but I have been pretty stumped by it because I cannot figure out how I can solve it. If I try and use the ratio test then I can't calculate the limit. Integral and alternating series tests are also out of the question. So any idea how I should solve this?

答:
We use the ratio test :
$$ \frac{(\log(n+1))^{n+1}}{(n+1)!}\frac{n!}{(\log n)^n}=\frac{\log(n+1)}{n+1}\left(\frac{\log(n+1)}{\log n}\right)^n $$
But $\lim\limits_{n\rightarrow +\infty}\frac{\log(n+1)}{n+1}=0$ and
$$ \begin{aligned} \left(\frac{\log(n+1)}{\log n}\right)^n&=\left(1+\frac{\log\left(1+\frac{1}{n}\right)}{\log n}\right)^n \\
&=\exp\left(n\log\left(1+\frac{\log\left(1+\frac{1}{n}\right)}{\log n}\right)\right) \\
&=\exp\left(n\log\left(1+\frac{1}{n\log n}+o\left(\frac{1}{n\log n}\right)\right)\right) \\
&=\exp\left(\frac{1}{\log n}+o\left(\frac{1}{\log n}\right)\right) \underset{n\rightarrow +\infty}{\longrightarrow}1
\end{aligned} $$
Therefore $\displaystyle\lim\limits_{n\rightarrow +\infty}\frac{\log(n+1)}{n+1}\left(\frac{\log(n+1)}{\log n}\right)^n=0$ and the series converges.


问:
Proof of the discrete Doob inequality
$\mathbf{Theorem:}$ Let $n \in \mathbb{N}, a,b, \in \mathbb{R}, a < b$ and $S_i$ a submartingale for $i=1,2,\dots,n$. Define $$H_n(a,b;S_1,\dots,S_n) = \text{card} \Bigg( \{ (i,j) \in \mathbb{N}^2: 1 \le i <j \le n, S_i \le a, S_j \ge b, a<S_k<b, \forall k = i+1, i+2,\dots, j-1 \} \Bigg)$$ a function that is counting the number of "upward crossings" of the interval $(a,b)$. (i.e. $H_n$ = How many $S_i$ fell below $a$ and then rose above $b$). Then

$$\mathbb{E}[H_n(a,b;S_1,\dots,S_n)] \le \frac{\mathbb{E}(S_n-a)^{+}-\mathbb{E}(S_1-a)^{+}}{b-a} \le \frac{\mathbb{E}(S_n-a)^{+}}{b-a} $$

$\mathbf{Proof:}$ Denote $Z_i = (S_i-a)^{+}$, this way our $H_n(a,b; S_1,\dots,S_n)=H_n(0,b-a; Z_1,\dots,Z_n)$ and define the stopping times $$\Lambda_k=\min\{\min\{i\in \mathbb{N}: i\ge \Psi_{k-1}, Z_i=0\},n\}$$

$$\Psi_k = \min\{ \min \{i\in \mathbb{N}: i \ge \Lambda_{k},  Z_i \ge b-a \},n\}\\ \Psi_0 \equiv 1$$

Now we can proceed proving the Doob inequality:
\begin{align}
Z_n-Z_1 &= Z_{\Lambda_{ \lfloor \frac{n}{2} \rfloor+1} } - Z_{\Psi_0}\\
&= \sum_{i=1}^{\lfloor \frac{n}{2} \rfloor +1} (Z_{\Lambda_k}-Z_{\Psi_{k-1}} )+ \sum_{i=1}^{\lfloor \frac{n}{2} \rfloor} (Z_{\Psi_k}-Z_{\Lambda_{k-1}} )  \\
& \ge \sum_{i=1}^{\lfloor \frac{n}{2} \rfloor} (Z_{\Lambda_{k+1}}-Z_{\Psi_{k}} ) + (b-a)H_n(0,b-a; Z_1,\dots,Z_n)
\end{align}
Taking the expectation of $Z_n - Z_1$ proves the inequality. $\square$

$\mathbf{Q}$: What I don't understand is the magic around the stopping times. Sure, $\Lambda_k$ denotes the first time after $\Psi_{k-1}$ (i.e. $Z_{k-1}$ rose above $b-a$ ) that $Z_k$ returns back to the origin. Conversely, $\Psi_k$ denotes the first time after $Z_{k-1}=0$ that $Z_k$ exceeded $b-a$. So this is how we play around with the notion of "interval crossing."

What I don't get, though, is why $Z_n = Z_{\Lambda_{ \lfloor \frac{n}{2} \rfloor+1}}$?? The subscript. Why the integer of $\frac{n}{2}+1$. Is this because we've got one $k$ that's taking care of the "upward" crossing and another $k$ for the "downward" crossing (so in the end for $k$ crossings, a half would be "bottom $\to$ top" and the other "top $\to$ bottom")?

答:
The equality
$Z_n  = Z_{\Lambda_{\lfloor\frac n2\rfloor + 1}}$
is simply a consequence of the fact that
$\Lambda_{\lfloor\frac n2\rfloor + 1} = n$ .  </p>

<p>Because there has to be a downcrossing between every two upcrossings we must have $\Lambda_{k+1}\geq\Lambda_k+2$  therefore $\Lambda_{\lfloor\frac n2\rfloor + 1} \geq n$ and equality must hold because $\Lambda_k$ is defined as the minimum of $n$ and the time of the $k$th upcrossing. </p>

ContentLicense: CC BY-SA 3.0

Title: Proof of the discrete Doob inequality

Body:
The equality
$Z_n  = Z_{\Lambda_{\lfloor\frac n2\rfloor + 1}}$
is simply a consequence of the fact that
$\Lambda_{\lfloor\frac n2\rfloor + 1} = n$ .

Because there has to be a downcrossing between every two upcrossings we must have $\Lambda_{k+1}\geq\Lambda_k+2$  therefore $\Lambda_{\lfloor\frac n2\rfloor + 1} \geq n$ and equality must hold because $\Lambda_k$ is defined as the minimum of $n$ and the time of the $k$th upcrossing.


问:
A confuse about Cyclic subspace
I have asked such a question before:

problem:Let $V$ be a linear space,$\mathscr A $ is a linear translation on $V$.choose $0≠α∈V$.let $S=\{\mathscr A^k(α):k≥0\}$ and $U=span(S)$, prove that:
(i) $U$ is stable under $\mathscr A$.
(ii) assume $dim(U)=r$,then ${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$ is a basis of $U$ and find out the metric of  $\mathscr A|_U$ under basis ${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$

And I gave the answer myself:

lemma: if ${α_1,α_2,\cdots,α_n}$ is linear independent,${α_1,α_2,\cdots,α_n,β}$ is linear dependent,then $β$ can be linear expression by ${α_1,α_2,\cdots,α_n}$,but $α_i$ can't be linear expression by the other.

From the comment:we know${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$ is linear dependent and ${α,\mathscr A(α),\cdots,\mathscr A^{r-1}(α),\mathscr A^r(α)}$ is linear dependent.

We write $\mathscr A^r(α)=x_0α+\cdots+x_{r-1}\mathscr A^{r-1}(α)$.

We have$(\mathscr A(α),\cdots,\mathscr A^r(α))⊆ (α,\cdots,\mathscr A^{r-1}(α))$. If ${\mathscr A(α),\cdots,\mathscr A^r(α)}$ is independent,then $α$ can be linear expression by ${\mathscr A(α),\cdots,\mathscr A^r(α)}$,by lemma,it is impossible, so ${\mathscr A(α),\cdots,\mathscr A^r(α)}$ is linear dependent,thus $\mathscr A^r(α)$ can be expression by ${\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$,so $x_0=0$.

we know${\mathscr A(α),\cdots,\mathscr A^{r-1}(α)}$ is linear dependent and ${\mathscr A(α),\cdots,\mathscr A^{r-1}(α),\mathscr A^r(α)}$ is linear dependent. Use similarly way with above,we can get $x_1=0$,this  process can continues. So $x_0=\cdots=x_{r-1}=0$.
So $\mathscr A^r(α)=0$.

Remark:$(\mathscr A(α),\cdots,\mathscr A^r(α))$ is subspace generated by${\mathscr A(α),\cdots,\mathscr A^r(α)}$.

we say set $S$ is linear independent if for any finite subset $\{α_1,\cdots,α_n\} $,if $x_1α_1+\cdots+x_nα_n=0$,then we have $x_1=\cdots=x_n=0$.if exist some finite subset $\{α_1,\cdots,α_n\} $ and Numbers that are not all zeros $x_1,\cdots,x_n$ such that $x_1α_1+\cdots+x_nα_n=0$,we say $S$ is linear dependent.

We say $α$ can be linear expression by $α_1,\cdots,α_n$ if extist $x_1,\cdots,x_n$ such that $α=x_1α_1+\cdots+x_nα_n $

Let $V$ linear space with dimension $n$

Definition: (i)let $α∈V$ the cylic subspace $C_α$ of is the smallest stable subpace  under $\mathscr A$(is: the intersection of stable subpace contain $α$)
(ii) Minimum polynomial $d_α(λ)$ of $α$ relative to to $\mathscr A$ is Polynomial with minimum degree and first term coefficient of 1 such that $d_α(\mathscr A)(α)=0$.

Here is some relatively theorem:
$C_α$ is generated by $\{α,\cdots,\mathscr A^{r}(α),\cdots \}$

My confuse comes from the following theorem :
theorem:assume $d_α(λ)=λ^r+a_{r-1}λ^{r-1}+\cdots+a_o$.then $dim C_α=r$ and
$\mathscr A^r(α)=-a_0α+\cdots-a_{r-1}\mathscr A^{r-1}(α)$.
but from my "proof" above:we have $a_0=\cdots=a_{r-1}=0$. It is seem to impossible,so Where did I go wrong?

答:
Let $r'$ be the smallest number such that $\alpha, A(\alpha),\dots,A^{r'}(\alpha)$ is linearly dependent.

Then $\alpha,\dots,A^{r'-1}(\alpha)$ are independent, so we can apply the lemma to express
$$A^{r'}(\alpha)=\sum_{i<r'}a_iA^i(\alpha)\,.$$
It doesn't follow that $a_i$ must be $0$.

But, by induction we can clearly see that $A^k(\alpha)\in{\rm span}(\alpha,\dots,A^{r'-1}(\alpha))$ for every $k\in\Bbb N$, hence $U={\rm span}(\alpha,\dots,A^{r'-1}(\alpha))$, so $r=\dim U=r'$.

Also observe that
$$A^r(\alpha)-a_{r-1}A^{r-1}(\alpha)-\dots -a_1A(\alpha)-a_0\alpha=0\,,$$
and that this polynomial $d_\alpha$ is a factor of the minimal polynomial of $A$.

ContentLicense: CC BY-SA 4.0

Title: A confuse about Cyclic subspace


问:
What do weak-star limits of normal states on von Neumann algebras look like?
Let $N$ be a von Neumann algebra and let $(\phi_\lambda)$ be a net of normal states on $N$ so that $\phi_\lambda\to\phi$ in the weak-* topology, i.e. $\phi_\lambda(x)\to\phi(x)$ for all $x\in N$. What can we infer about $\phi$ in this case? Of course, $\phi$ is going to be a state on $N$, since $\phi(x^*x)=\lim_\lambda\phi_\lambda(x^*x)\geq0$ and $\phi(1_N)=\lim_\lambda\phi_\lambda(1_N)=1$.

But can we say anything more about $\phi$? If not, can we prove that the normal states on $N$ are dense in the weak-* topology in the state space $S(N)$?

I was thinking that, if this density is true, it suffices to prove it for the case of $\mathcal{B(H)}$. Indeed, if this is true for those von Neumann algebras, then we take an arbitrary von Neumann algebra $N$ and concretely represent it on some Hilbert space, i.e. we have an inclusion of von Neumann algebras $N\subset\mathcal{B(H)}$. Now normal states of von Neumann subalgebras extend to normal states and, in general, states extend to states (in the $C^*$-algebraic setting). So we take a state $\phi$ on $N$, we extend it on a state $\Phi$ on $\mathcal{B(H)}$ and then we approximate $\Phi$ in the weak-* sense by normal states on $\mathcal{B(H)}$. We restrict those to $N$ and these are normal states on $N$ approximating $\phi$ weak-* on $N$.

I have no idea on how to prove this though, any help is appreciated.

答:
Every state on $N$ is the weak$^*$-limit of normal states.

In order to prove it, let $C$ be the weak$^*$-closure of the set of
normal states and suppose by contradiction that the (non-normal) state  $\varphi _0$ is not in $C$.  Since $C$ is clearly
weak$^*$-closed and convex,  the Hahn-Banach separation Theorem produces a weak$^*$-continuous linear functional $\Lambda $ on
$N^*$, and a real number $\alpha $,  such that
$$
  \Re(\Lambda (\varphi ))\leq  \alpha  <   \Re(\Lambda (\varphi _0)), \quad\forall \varphi \in  C.
  $$

It is well known that weak$^*$-continuous functionals must be given by evaluation at some element of $N$,  so there is
$x$ in $N$ such that $\Lambda (\varphi ) = \varphi (x)$,  for every $\varphi $ in $N^*$,  and hence
$$
  \Re(\varphi (x))\leq  \alpha  <   \Re(\varphi _0(x)), \quad\forall \varphi \in  C.
  \tag 1
  $$

Letting $a=(x+x^*)/2$,  notice that   for every self-adjoint  linear functional $\psi $ on $N$ one has
$$
  \Re(\psi (x)) =
  (\psi (x)+\overline{\psi (x)})/2 =
  (\psi (x)+\psi (x^*))/2 =
  \psi (a),
  $$
so (1) gives
$$
  \varphi (a)\leq  \alpha  <   \varphi _0(a), \quad\forall \varphi \in  C.
  $$

From the first inequality one deduces that $a\leq \alpha \cdot 1$,  but this clearly contradicts the second inequality.


问:
easy homework: Equivalence classes, how do they look?
Let's say that I got a

```<code> set = { Arnold, Harrison }
</code>```

and I want to display the equivalence class of [ Harrison ]

The actual condition for the relation doesn't matter in this case so let's just say that {Arnold} is the only relation to Harrison.

This would be displayed as:

```<code>[ Harrison ] = { Arnold }
</code>```

However, if we have this set instead:

```<code>set = { Arnold, Harrison, {Arnold, Harrison}, { Arnold, Arnold, Harrison} }
</code>```

And we say that this time, Harrison has a relation to Arnold but also {Arnold, Harrison} and even {Arnold, Arnold, Harrison}.

How is this expressed as an equivalence class?

```<code>[ Harrison ] = {Arnold, Harrison}
</code>```

or...

```<code>[ Harrison ] = { Arnold, {Arnold, Harrison}, {Arnold, Arnold, Harrison}  }
</code>```

or... what?!

答:
Your example (or "the homework text") doesn't seem too enlightening because all the elements in the set seem to be equivalent to each other.

Remember that the equivalence class of an element $s$ in a set $S$ with an equivalence relation $\sim$ is simply the subset of $S$ whose element are the elements of $S$ equivalent to $s$:
$$
[s]=\{x\in S\,\mid\,x\sim s\}.
$$
So, for instance, if $S=\{a,b,c,d,e\}$ and

$a$ is equivalent only to $c$ and itself,

$b$ is equivalent only to $e$ and itself,

$d$ is equivalent only to itself,

then the equivalence classes are $[a]=[c]=\{a,c\}$, $[b]=[e]=\{b,e\}$ and $[d]=\{d\}$.

Furthermore, the quotient set is $S/\sim=\{[a],[b],[d]\}$.


问:
What does $f \in C^{2,1}$ mean?
As in the title, what does the symbol $C^{2,1}$ stand for in the expression $f \in C^{2,1}$?

I found such notation in this paper by Ghomi and Spruck. I bet it is a stronger condition than $f \in C^{2}$, but I could not find a definition.

EDIT: In the paper, the function $f$ is defined to be an isometric immersion of a Riemannian surface into $\mathbb{R}^{3}$. In coordinates, this is a map $U \to \mathbb{R}^{3}$, where $U$ is an open subset of $\mathbb{R}^{2}$.

答:
The answer depends on context. If you found this notation in reference to a partial differential equation or similar real analysis setting, it most likely means that $f \in C^2$ and $f''$ is Lipschitz. This is a special case of the notation $f \in C^{2,\alpha}$ to mean $f \in C^2$ and $f''$ is $\alpha$-Holder continuous. Note that $\alpha=1$ in the definition of Holder continuity just gives the definition of Lipschitz continuity.

If you found this notation in a stochastic analysis setting, it most likely means that $f = f(x, t), x \in \mathbb{R}^n, t \in \mathbb{R}_{\geq 0}$ and $\partial_{x_i}\partial_{x_j} f$ is continuous for any $i,j$ and $\partial_{t} f$ is continuous. That is, "$f$ is $C^2$ in space and $C^1$ in time."

Title: What does $f \in C^{2,1}$ mean?


问:
Prove $a^{\frac{\phi(mn)}2} \equiv 1\pmod {mn}$
I want to show that given $\gcd(m,n)=1$ with $m,n \gt2$ and $\gcd(a, mn)=1$, then $a^{\frac{\phi(mn)}2} \equiv 1\pmod {mn}$, where $\phi(mn)$ is the Euler Totient function.

My approach started using Euler's theorem, given $\gcd(a, n)= 1$ then $ a^{\phi(n)} \equiv 1\pmod {n}$. And since $(m,n)=1$ then: $$a^{\phi(mn)} \equiv 1\pmod {mn}$$

I just cannot seem to get the $\frac12$ factor to appear. What I tried doing was just raising both sides to the $\frac 12$

$$a^{\phi(mn)} \equiv 1\pmod {mn} \Rightarrow a^{\frac{\phi(mn)}2} \equiv 1^{\frac12}\pmod {mn} \equiv 1\pmod{mn}$$
But I somehow feel this is wrong. So another way I tried was again with Euler's theorem:

$$a^{\phi(n)} \equiv 1\pmod {n} \Rightarrow (a^{\phi(m)})^{\phi(n)} \equiv 1\pmod {n} \Rightarrow a^{\phi(m)\phi(n)} \equiv 1\pmod {n} $$

$$a^{\phi(m)} \equiv 1\pmod {m} \Rightarrow (a^{\phi(n)})^{\phi(m)} \equiv 1\pmod {m} \Rightarrow a^{\phi(n)\phi(m)} \equiv 1\pmod {m} $$

Multiplying one by the other yields:

$$a^{2\phi(m)\phi(n)} \equiv 1\pmod {mn} \Rightarrow a^{2\phi(mn)} \equiv 1\pmod {mn}$$

Or using the CRT:

$$a^{\phi(m)\phi(n)} \equiv 1\pmod {mn} $$

From here I have no idea on where to go or what to try, I was thinking perhaps something related to Wilson's Theorem to make the $\frac12$ factor appear, but since m and n may not be prime, they are just coprime, I don't know if I can use that. Am I going in the right direction?

答: Hint: Prove that $\phi(N)$ is even for all $N\geq 3$.  So both $\phi(m),\phi(n)$ are even, hence you have the spare factor of 2: $\operatorname{lcm}(\phi(m),\phi(n))\mid \frac12\phi(m)\phi(n)$.


问:
Doubt about the norm of the gradient vector
Is it possible to say that the gradient vector norm at a point $ (x, y, z) $ is the smallest rate of change? Considering a function $ f (x, y, z) $

答:
At a point $p = (x,y,z)$, look at all the unit vectors $u$ with their tails at $p$. We can compute the instantaneous rate of change of $f$ in the direction $u$ by taking the following derivative:
$$
\frac{d}{dt}\bigg|_{t=0}f(p + tu) = \nabla f(p) \cdot u.
$$
The equality is an exercise in the chain rule. But using this formula, we can see that the rate of change is a scalar, as expected, and it is largest (that is, most positive), when the vectors $\nabla f(p)$ and $u$ point in the same direction. That tells us that the direction of the gradient vector $\nabla f(p)$ would be the direction of most positive change, and the direction of $-\nabla f(p)$ would be the direction of most negative change.

If we pick a direction $u$ perpendicular to $\nabla f$, then the instantaneous rate of change of $f$ would be $0$ in the direction of $u$ (another consequence of the formula above, and the geometric interpretation of the dot product). Depending on how you use the language, a direction perpendicular to the gradient could fairly be said to be a direction of "smallest change" as you said in your post, because this is a direction where the change is smallest in magnitude.

The ambiguity in the way we use the words greatest and smallest makes us be more precise in order to clarify the picture we have in mind, or the idea we want to convey.


问:
limit square-root proof
If $\lim_{n\to\infty} a_n = a$, show that $\lim_{n\to\infty} \sqrt{a_n} = \sqrt{\lim_{n\to\infty} a_n} = \sqrt{a}$.

My attempt:

Assume $\lim_{n\to\infty} a_n = a$. Let $\epsilon > 0$ then there is a $N \in \mathbb{N}$ so that for $n \geq N$

$$|a_n - a| < \epsilon$$

However for $n\geq N$

$$|\sqrt{a_n} - \sqrt a| \leq (|\sqrt{a_n} - \sqrt a|)^2 \leq |\sqrt{a_n} - \sqrt a||\sqrt{a_n} + \sqrt a| = |a_n - a| < \epsilon$$

$\lim_{n\to\infty} \sqrt{a_n} = \sqrt{a}$.

QED.

Is this proof correct? Did I miss any steps?

In addition, a general question: Why do we assume $\epsilon <$ some $M$ in some of these proofs? Although I am familiar with these proofs, I have never got an explanation as to why we cannot always let $\epsilon > 0$ without further restriction. Any assistance is appreciated.

答:
Correction:

Assume $\lim_{n\to\infty} a_n = a$. Let $\epsilon > 0$ then there is a $N \in \mathbb{N}$ so that for $n \geq N$

$$|a_n - a| < \epsilon^2$$

However for $n\geq N$

$$(|\sqrt{a_n} - \sqrt a|)^2 \leq |\sqrt{a_n} - \sqrt a||\sqrt{a_n} + \sqrt a| = |a_n - a|$$

$$\implies |\sqrt{a_n} - \sqrt a| \leq  \sqrt{|a_n - a|} < \epsilon$$

$\lim_{n\to\infty} \sqrt{a_n} = \sqrt{a}$.

QED.

Title: limit square-root proof


问:
Riemann or Lebesgue integrable
Lets consider the following two functions:

$$f(x)=\begin{cases} x &,x\in[0,1]\setminus\Bbb Q \\ 0 &,x\in[0,1]\cap\Bbb Q\end{cases}$$
$$g(x)=\frac{(-1)^{[x]}}{[x]}$$

where $[x]$ is the integer part of $x$. I am trying to determine which of these integrals are Riemann or/and Lebesgue integrable; $f(x)$ over $[0,1]$ and $g(x)$ over $[1,\infty)$.

Lebesgue for $f(x)$:
$\int_{[0,1]} f dx=\int_{[0,1]\setminus\Bbb Q}x\ dx + \underbrace{\int_{[0,1]\cap\Bbb Q}0\ dx}_{=0}=\int_{[0,1]\setminus\Bbb Q}x\ dx=\int_{[0,1]}x\ dx-\underbrace{\int_{\Bbb Q}x\ dx}_{=0}=\frac12$

Riemann for $f(x)$: I assume this function is not Riemann integrable, because every interval in a partition of $[0,1]$ will contain both rationals and irrationals, but I am not sure how to conclude.

Riemann for $g(x)$: The function takes the values $-1$ on $[1,2)$, $\frac12$ on $[2,3)$, $-\frac13$ on $[3,4)$, etc. So the integral is $-1+\frac12-\frac13+\frac14-\ldots=\sum_{k=1}^\infty(-1)^k\frac1k=-\log(2)$.

But I have no idea for the Lebesgue integral of $g(x)$.

答:
$g$ is not lebesgue integrable, because the lebesgue integral of a function $g$ is defined as $$
  \int_\Omega g \,d\mu := \int_\Omega g_+ \,d\mu - \int_\Omega g_- \,d\mu
$$
where $g^+$ is the positive part of $g$, i.e. $g_+(x) = |g(x)|$ if $g(x) \geq 0$ and $g_+(x)=0$ otherwise, and $g_-$ simiarly is the negative part of $g$. $g$ is only integrable if at most one of the integrals on the RHS diverge. In your case, I believe that both $g^+$ and $g^-$ are $+\infty$.

$f$ is not riemann integrable because it's discontinuous everywhere.

Title: Riemann or Lebesgue integrable


问:
Recursively given sequence question about changing the sequence with exponent function
We have a sequence given by:

$$a_1 = 1 \\ a_2 = b \\
a_{n+2}=a_{n+1}+a_n$$

We need to find the limit of the sequence. The procedure of getting the solution is that we $a_n = p^n$ and then somehow solve the equation $p^{n+2}=\frac{1}{3}p^n+\frac{2}{3}p^{n+1}\implies p^2 - \frac{2}{3}p-\frac{1}{3} = 0$

For this quadratic equation there are two solutions, that we input:

$$a_1 = A - \frac{1}{3}B = 1, a_2 = A +\frac{1}{9}B = b \implies A = \frac{1+3b}{4}, B = \frac{9(b-1)}{4}$$

However I do not understand this process. Why can we say that $a_n = p^n$ and what follows from it. Where can we use this technique?

答:
I guess what you are asking is how the "guess" $a_n = p^n$ can be justified, i.e. why is this the way to proceed and why is it complete, i.e. it produces all solutions.

Without having to develop a theory of difference equations from scratch, we can use mighty results from linear algebra. Transform your difference equation $a_{n+2}=a_{n+1}+a_n$ into a vector equation as follows:
$$
\left ( \begin{matrix}
    a_{n+2}\\a_{n+1}
   \end{matrix} \right ) = \left ( \begin{matrix}
    1&1\\1&0
   \end{matrix} \right )
\left ( \begin{matrix}
    a_{n+1}\\a_{n}
   \end{matrix} \right )
$$
Now writing for short the vector $
x_n =
\left ( \begin{matrix}
    a_{n+1}\\a_{n}
   \end{matrix} \right )
$ and the matrix $A  = \left ( \begin{matrix}
    1&1\\1&0
   \end{matrix} \right )
$ we have $x_{n+1} = A \cdot x_n$ and hence $x_{n+1} = A^{n} \cdot x_1$. Your  $
x_1 =
\left ( \begin{matrix}
    a_{2}\\a_{1}
   \end{matrix} \right )
=\left ( \begin{matrix}
    b\\1
   \end{matrix} \right )
$ is obviously a known start vector.

So the problem has been transferred to obtaining the power of a matrix $A$. Now the linear algebra results come in. Let $u_1$ and $u_2$ be the two eigenvectors of $A$ with corresponding eigenvalues $\lambda_1$ and $\lambda_2$, then the eigenvectors form a basis and you can write any initial condition as $x_1 = c_1 u_1  + c_2 u_2$ with constants $c_1$ and $c_2$. Then you have  $x_{n+1} = A^{n} \cdot x_1 = c_1 \lambda_1^n u_1  + c_2 \lambda_2^n u_2$.

So here you have the answer to the above question. Since taking the first component of $x_{n+1}$, which is $a_{n+2}$, gives    $a_{n+2} = d_1 \lambda_1^n   + d_2 \lambda_2^n$ with constant coefficients $d_1$ and $d_2$ depending on the initial conditions and on $A$. So we can deduce:

a) the ansatz $a_{n} = b_1 \lambda_1^n   + b_2 \lambda_2^n$ is correct and complete.

b) the $\lambda_1$ and  $\lambda_2$ in this ansatz are the eigenvalues of $A$.

c) the constants  $b_1$ and  $b_2$ in this ansatz have to be matched to the initial conditions.

So it was not by chance or by some phenomenal genius that a power law was "guessed" in the beginning. Also, the characteristic equation does not  appear from out of the blue, but is exactly what is needed to find the eigenvalues of $A$.

Obviously, this derivation generalizes to homogenous linear difference equations of any finite degree.


问:
Does a limit exist at the end point of a function according to the epsilon delta definition?
According to the epsilon delta definition, the limit L of a function f(x) at a point c in the domain exists if for every $\epsilon > 0$, I can find a $\delta > 0$ such that if $0 < |x - c| < \delta$ then $|f(x) - L| < \epsilon$.

If I have a continuous function defined in $[A, B]$, and I want to calculate $\lim_{x \to A} f(x)$, the I can find the corresponding $\delta$ for for every $\epsilon$. The limit should exist and be equal to $f(A)$.

But the left hand limit of the function doesn't exist, since A is the endpoint, and a limit only exists when the left hand limit is equal to the right hand limit. Therefore the limit should not exist.

Why is it that the limit exists using one definition but not using another? Is there something wrong with how I have interpreted the $\epsilon$ $\delta$ definition?

答:
I refer to the answer provided by Adam Latosinski and reiterate the fact that we have the following result:

Proposition
If both limits

$$\lim_{x \to c^+} f(x) \quad \text{and} \quad \lim_{x \to c^-} f(x)$$

exist and coincide, the limit $\lim_{x \to c} f(x)$ exists and equals that common value.

Notice that the existance of both limits is an assumtion in this result. If one of them does not exist, the conclusion may not be true.

Given the quote from your book, I can see how this is confusing. This boils down to definitions. If one is very pedantic, one could take that stance that when a function is not defined on both sides of a point it cannot have a limit at that point, in which case these questions become rather cumbersome. In fact, it is much more convenient to adopt the convention that when a function is not defined to the left of a point and it has a right-sided limit, then that is its limit at that point, and vice versa. So if $f: [a, b] \to \mathbb{R}$ is a function and the limit

$$\lim_{x \to a^+} f(x)$$

exists, we will also write $\lim_{x \to a} f(x)$, where it is understood that the limit is one-sided since the left-handed limit does not make sense.

EDIT: I should also note that some author avoid this problem very elegantly by providing the following definition of a limit: Let $f: D_f \to \mathbb{R}$ a function and $a \in D_f$. We say the $f$ approaches the limit $L$ as $x$ goes to $a$, and write

$$\lim_{x \to a} f(x) = L,$$

if for every $\varepsilon > 0$ there exists $\delta > 0$ such that

$$|L-f(x)| < \varepsilon$$

whenever $x \in D_f$ and $|x-a| < \delta$.

Notice how we require $x$ to be in $D_f$, so the $x$-es outside will not matter.


问:
Algorithm that matches decimal number to expression in terms of mathematical constants
I remember finding at some point a type of algorithm which provided a decimal value with enough precision, would try to match it to expressions in terms of mathematical constants (for example, plugging in $5.85987448204$ would return $\pi+e$.)

But I cant't seem to find anything like this on the internet, so does anybody have any idea about how these algorithms are called, or where I could find something like this?

答: See Simon Plouffe's Inverse Symbolic Calculator and Robert Munafo's RIES - Find Algebraic Equations, Given Their Solution.


答: You might explore Ries, specifically about the Ries algorithm.</p>

Title: Algorithm that matches decimal number to expression in terms of mathematical constants

Body:
I remember finding at some point a type of algorithm which provided a decimal value with enough precision, would try to match it to expressions in terms of mathematical constants (for example, plugging in $5.85987448204$ would return $\pi+e$.)

But I cant't seem to find anything like this on the internet, so does anybody have any idea about how these algorithms are called, or where I could find something like this?


问:
Linearizing a constraint with a root square
I have a Mixed Integer Model. In order to solve it using CPLEX, I need to linearize the non-linear constraint stated in the following.

Let $X_{i,j,t}$ is binary variable. How to linearized the following constraint:

$m\geq \sum_{i,j \in A}\mu _{i,j}X_{i,j,t}+z_\alpha \sqrt{\sum_{i,j \in A}\sigma_{i,j}^2X_{i,j,t}}$

Where $\mu _{i,j},z_\alpha,m$ and  $\sigma_{i,j}$ are positive known parameters of the problem

答:
One approach is to rewrite as
$$m - \sum_{i,j \in A}\mu _{i,j}X_{i,j,t} \ge z_\alpha \sqrt{\sum_{i,j \in A}\sigma_{i,j}^2X_{i,j,t}},$$
square both sides,
and linearize the resulting products of binary variables in the usual way.

You might need to additionally impose
$$m \ge \sum_{i,j \in A}\mu _{i,j}X_{i,j,t}$$
to avoid extraneous solutions.


问:
Trigonometry - distance and bearing

A yacht and a catamaran leave the harbour at $09:00$ hours. The yacht sails at $24$ knots on course $037$ and the catamaran at $15$ knots on $139$. Find the bearing and the distance in nautical miles of the yacht from the catamaran at $1400$ hours ($1$ knot is a speed of $1$ nautical mile per hour).

The final answer I am getting is $135$m using the cosine rule. However this doesn’t match with the solutions. Please could someone clarify both the length and the bearing?

答:
Firstly, from 0900 to 1400 that is 5 hours so because 1 knot is 1 nautical mile we can calculate the distance both of the vessels travelled in the time period:

Yacht: 24 * 5 = 120m

Catamaran: 15 * 5 = 75m

From the two bearings at which the vessels set off from the harbour at we can calculate the angle between their paths as:

139 - 37 = 102

Now we can make a triangle with the two distances and the angle between them and the distance between the vessels as the third line.

The cosine rule states: a^2 = b^2 + c^2 - 2bc*cos(A)

So to solve the distance between the vessels we substitute in our values to get:

Distance^2 = 120^2 + 75^2 - 2 * 120 * 75 * cos(102) = 23767.41043...

Distance = sqrt(23767.41043...) = 154.1668266...m

You could alternatively do it all in a single step.

Now we have the distance you could use the sine rule to calculate the angle at the catamaran between the yacht and the harbour.

The sine rule for finding angles states: sin(A)/a = sin(B)/b = sin(C)/c

So substituting our values:

sin(angle we want)/120 = sin(102)/154.1668266...

sin(angle we want) = 120 * (sin(102)/154.166266...)

angle we want = sin^-1(120 * (sin(102)/154.166266...)) = 49.58496016...

Now to find the bearing of the yacht from the catamaran we need the angle below the triangle to 270°.

We can do this by finding the remaining angle from the path of the catamaran from the harbour to 180° as 180 - 139 = 41°

Now if you create a right angle triangle below the paths of the boats you can calculate the remaining angle below the angle at the catamaran as 180 - 90 - 41 = 49°

Finally we can calculate the bearing of the yacht from the catamaran as 270 + 49 + 49.58496016... = 368.5849602...°

:/

It seems strange but that is because it goes past north so we can fix this by subtracting 360 leaving the resulting bearing at 008.5849602...°

(don't forget the 000 as you are working in bearings)

In conclusion:

Distance = 154.1668266m

Bearing = 008.5849602°

I cannot guarantee that I am correct but I hope this helps
It is hard to explain without a diagram so it might be useful to sketch one as you follow along. :D

Edit: wow I made a stupid mistake.
I carried forward 41 instead of 49 for the angle below the catamaran!
That puzzled me for a bit.
So carrying on from that you get the correct answer. My bad.

Title: Trigonometry - distance and bearing


问:
Creating a quadratic given roots using the principle of zero products

Using the Principle of Zero Products, create a quadratic in standard form that has roots of $-1$ and $\frac{3}{2}$. Values of $a, b$, and $c$ in your final form should all be elements of the set of integers.

I really don't understand how to do this. I tried to use $(x + \frac{1}{4})2 = \frac{5}{16}$, to create a quadratic, but I can't quite seem to figure it out.

答:
You want to find $ax^2 + bx + c$ so that

$\frac 12$ and $1$ are solutions and

that $a,b,c$ are all integers.

$(x - \frac 12)(x -(-1))$  will be a quadratic with those roots[*], but it might not have integer coefficients.  To get integer coefficients we can multiply it by some constant.  ... so what $ax^2 + bx + c$ has $\frac 12$ and $-1$ as roots?

.....

[*]  If $x=\frac 12$ then $x-\frac 12= \frac 12 -\frac 12 = 0$ and so  $(x-\frac 12)(x-(-1)) =0(x-(-1))= 0$.  ANd if $x =-1$ then $(x-(-1)) = 0$ and $(x-\frac 12)(x-(-1)) = (x-\frac 12)\cdot 0 = 0$.

====

If a quadratic has roots $r_1$ and $r_2$ then $r_1, r_2$ are solutions to $(x-r_1)(x-r_2)$ and any quadratic that has those to solutions can be factored to $k(x-r_1)(x-r_1)$ where $k$ is some non-zero constant.

So if $ax^2 + bx + c$ has roots $r_1, r_2$ then $ax^2 + bx + c = k(x-r_1)(x-r_2)$.

SO we need to figure out if $k(x-\frac 12)(x - (-1))=ax^2 + bx + c$ and $a,b,c$ are all integers... what are $a,b$ and $c$.

And ... just do it.

$k(x-\frac 12)(x+1) = k(x^2 -\frac 12x + x - \frac 12)= k(x^2 +\frac 12x -\frac 12)=kx^2 + \frac 12kx -\frac 12k$.

So we need $k, \frac 12k, -\frac 12k$ to be integers.

We can do that if $k=2$.

So $2x^2 +x -1$ will have as solutions $\frac 12$ and $-1$.


问:
How can partial derivatives feature in the definition of a function?
I have a map $f(t,g,h)$ where $f:[0,1]\times C^1 \times C^1 \to \mathbb{R}.$

I want to define $$F(t,g,h) = \frac{d}{dt}f(t,g,h)$$
where $g$ and $h$ have no $t$-dependence in them. So $g(x) = t^2x$ would not be admissible if you want to calculate what $F$ is. How do I write this properly? Is it correct to write instead

Define $F$ by $F(t,\cdot,\cdot) = \frac{d}{dt}f(t,\cdot,\cdot)$.

But there is some ambiguity in the arguments. What is the best way to write it?

答:
This is called a partial derivative:

$$
F(t,g,h) = \frac{\partial}{\partial t} f(t,g,h)
$$

The definition of a partial derivative is that you differentiate with respect to one variable as usual while keeping all others fixed. The different kind of $d$ is there to remind you of that, because in this case writing $\frac{d}{dt}f(t,g,h)$ would imply that $g$ and $h$ have some kind of $t$-dependence.


问:
Building an "eventually dense" and divergent sequence in $\Bbb Q_{>0}$.
Call $R=\{R_n\}_n$ a sequence such that:

$(i)\;\;\;R_n\in\Bbb Q_{>0}$

$(ii)\;\;$strictly increasing

$(iii)\;\;R_n\to+\infty$.

Now consider a sequence of such sequences $R^{(j)}=\{R_n^{(j)}\}_n$ such that

$\bullet\;\;\;R^{(j)}\subset R^{(j+1)}\;\;\forall j\in\Bbb N$

$\bullet\;\;\;\bigcup_jR^{(j)}=\Bbb Q_{>0}\;$.

I want to construct (another!) sequence $T=\{T_n\}_n$ verifying $(i),(ii)$ and $(iii)$ such that $T_j$ is the smallest element of $R^{(j)}$ greater than $T_{j-1}$.

Clearly we can take $T_1:=R_1^{(1)}$ and inductively define $T_n$ to be the smallest element in $R^{(n)}$ strictly bigger than $T_{n-1}$.

The problem is to achieve $(iii)$.

Does exists a choice of $R^{(j)}$ and $T$ which allows this? Or does it always happen that for any choice of them $T$ turns out to be bounded?

Is the following a solution?

Define $R^{(1)}=\Bbb N$ and

$$
R^{(j+1)}=\left\{R_n^{(j)},\frac{R_n^{(j)}+R_{n+1}^{(j)}}2,R_n^{(j)}+\frac1{j+1},\;n\in\Bbb N\right\}
$$
relabelled as $\{R_n^{(j+1)}\}_n$ such that they are increasing.

Define then
$$
T_1=R_1^{(1)}
$$
$$
T_2=R_1^{(1)}+\frac12=R^{(2)}_{k_2},\;\; \exists k_2
$$
then
$$
T_3=R_{k_2}^{(2)}+\frac13=R_{k_3}^{(3)}\;\;\exists k_3
$$
and so on. In this way
$$
T_n=1+\frac12+\cdots+\frac1n\to+\infty.
$$
but I can't really see if $T_j$ is actually the smallest element of $R^{(j)}$ greater than $T_{j-1}$

答: What you have chosen may not tend to $\infty$. Instead, define $T_n$ to be the smallest element in $R^{(n)}$ strictly bigger than $\max(T_{n-1}, n)$.


问:
Is there a known closed form solution to $\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx$?
$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

Related question

Is there a known closed form solution to

\begin{align}
I_n&=
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
=?
\tag{0}\label{0}
\end{align}

It checks out numerically, for $n=1,\dots,7$ that

\begin{align}
I_1=
\int_0^1\frac{\ln(1+x^2)}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln2-\Catalan
\tag{1}\label{1}
,\\
I_2=
\int_0^1\frac{\ln(1+x^{2\cdot2})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(2+\sqrt2)-2\Catalan
\tag{2}\label{2}
,\\
I_3=
\int_0^1\frac{\ln(1+x^{2\cdot3})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln6-3\Catalan
\tag{3}\label{3}
,\\
I_4=
\int_0^1\frac{\ln(1+x^{2\cdot4})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(4+\sqrt2+2\sqrt{4+2\sqrt2})-4\Catalan
\tag{4}\label{4}
,\\
I_5=
\int_0^1\frac{\ln(1+x^{2\cdot5})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(10+4\sqrt5)-5\Catalan
=
\tfrac\pi2\,\ln(10\cot^2\tfrac\pi5)-5\Catalan
\tag{5}\label{5}
,\\
I_6=
\int_0^1\frac{\ln(1+x^{2\cdot6})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln((5\sqrt2+2\sqrt{12})(1+\sqrt2))-6\Catalan
\tag{6}\label{6}
,\\
I_7=
\int_0^1\frac{\ln(1+x^{2\cdot7})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(14\cot^2\tfrac\pi7)-7\Catalan
\tag{7}\label{7}
,
\end{align}

so \eqref{0} seems to follow the pattern

\begin{align}
I_n&=
\tfrac\pi2\,\ln(f(n))-n\Catalan
\tag{8}\label{8}
\end{align}

for some function $f$.

Items \eqref{5} and \eqref{7} look promising
as they agree to $f(n)=2n\cot^2(\tfrac\pi{n})$,
but the other fail on that.

Edit:

Also, it looks like
\begin{align}
\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\tfrac\pi2\,\ln(f(n))+n\Catalan
\tag{9}\label{9}
\end{align}

and

\begin{align}
\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\pi\,\ln(f(n))
\tag{10}\label{10}
\end{align}

with the same $f$.

Edit

Thanks to the great answer by @Quanto,
the function $f$ can be defined as

\begin{align}
f(n)&=
2^n\!\!\!\!\!\!\!\!\!\!
\prod_{k = 1}^{\tfrac{2n-1+(-1)^n}4}
\!\!\!\!\!\!\!\!\!
\cos^2\frac{(n+1-2k)\pi}{4n}
\tag{11}\label{11}
.
\end{align}

$\endgroup$

答:
The close-form result can be expressed as

$$\color{blue}{ \int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx = -nG+\frac\pi2 n \ln 2
+ \pi \sum_{k=1}^{[\frac n2]}\ln \cos\frac{(n+1-2k)\pi}{4n} }
$$

as shown below. Note that

\begin{align}
I_n =
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
\overset{x\to\frac1x} == \frac12 J_n - nG
\end{align}

where $ \int_1^\infty\frac{\ln x}{1+x^2} \,dx=G$ and

$$J_n =\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx $$
Substitute
$$1+x^{2n} = \prod_{k=1}^{n}(1+e^{i\pi\frac{n+1-2k}n }x^2)
$$
and use the known result
$\int_0^\infty \frac{\ln(1+ax^2)}{1+x^2}dx= \pi\ln(1+a^{\frac12})
$ to integrate

\begin{align}
J_{n}& =\int_0^\infty\frac{dx}{1+x^2} \sum_{k=1}^{n}
\ln (1+e^{i\pi\frac{n+1-2k}n }x^2)
= \pi\sum_{k=1}^{n} \ln (1+e^{i\pi\frac{n+1-2k}{2n} })\\
 &=n \pi \ln 2 + 2\pi\sum_{k=1}^{[\frac n2]} \ln \cos\frac{(n+1-2k)\pi}{4n}
\end{align}
where the symmetry of the sequence is recognized in the last step.


问:
Number of ways to split balls of different colours
Given an even number of balls of different colours, find the number of ways of splitting these balls evenly between two people. The total number of balls is known, it is always even (and therefore could always be split evenly between two people), but may be large. Each ball has a colour. The colours and the quantities of each colour are known in advance, but there could be many colours. The order of the picking doesn't matter - all balls of the same colour are the same. Here are a few examples:

Example 1:
Given 2 red balls, how many ways there are to split them?
Answer: Well, just 1 - each person gets one red ball.

Example 2:
Given 20 red balls, how many ways there are to split them?
Answer: Still just 1 - each person gets 10 red balls.

Example 3:
Given 19 red balls and 1 blue ball, how many ways there are to split them?
Answer: Two ways (in fact, it doesn't matter how many red balls there are in this example, it will always be two ways) -

Person one gets 9 red a 1 blue ball and person two gets 10 red balls.

Person two gets 9 red a 1 blue ball and person one gets 10 red balls.

Example 4:
Given 4 red balls (R) and 1 blue ball (B) and 1 yellow ball (Y), how many ways there are to split them?
Answer: Four ways -

RRR / RBY

RBY / RRR

RRB / RRY

RRY / RRB

You get the idea. Now, in the general case, given W blue balls, X green balls, Y red balls, Z brown balls, ... $k$ balls of colour C, how many ways are there to split them evenly between two people? Is there a generic formula that can be applied?

I have been at it for days, but short of counting all possible permutations by hand, I cannot really see a pattern that can be generalised.

答:
Let the number of colors be $k$, the number of balls $2n$, and the number of balls of the $i$-th color be $r_i$, so that $\sum_{i=1}^k r_i=2n$

In terms of generating functions we are looking for the coefficient at $x^n$ in the product:
$$\begin{align}
\prod_{i=1}^k\sum_{j=0}^{r_i}x^j
&=\prod_{i=1}^k\frac{1-x^{r_i+1}}{1-x}\\
&=\frac{\prod_{i=1}^k(1-x^{r_i+1})}{(1-x)^{k}}\\
&=\prod_{i=1}^k(1-x^{r_i+1})\sum_{j=0}^\infty\binom{-k}{j}(-x)^j\\
&=\prod_{i=1}^k(1-x^{r_i+1})\sum_{j=0}^\infty\binom{j+k-1}{j}x^{j}\\
&=\sum_{\boldsymbol{\mu}}(-1)^{|\boldsymbol{\mu}|}x^{\boldsymbol{\mu}\cdot\boldsymbol{r}}
\sum_{j=0}^\infty\binom{j+k-1}{k-1}x^{j},
\end{align}$$
which is
$$
\sum_{\boldsymbol{\mu}}(-1)^{|\boldsymbol{\mu}|}\binom{n-\boldsymbol{\mu}\cdot\boldsymbol{r}+k-1}{k-1},\tag1
$$
where $\boldsymbol{\mu}$ are all $2^k$ binary vectors of the length $k$, $|\boldsymbol{\mu}|=\sum_{i=1}^k\mu_i$ and $\boldsymbol{r}=(r_1+1,r_2+1,\dots,r_k+1)$.

In (1) the binomial coefficient $\binom mn$ is assumed to be $0$ for $m<n$.

Solution for Example 4.

$n=3,k=3,\boldsymbol{r}=(5,2,2):$

$$\binom 52-\binom 32-\binom 32=10-3-3=4.
$$


问:
Why does the (topology given by) Hausdorff metric depend only on the topology?
If I have a compact metric space $(X,d)$, I can define the Hausdorff metric on the set $K(X)$ of all non-empty compact (equivalently, closed) subsets of $X$ as $$d_H(A,B) = \max ( \sup_{x \in A} \inf_{y \in B} d(x,y),  \sup_{y \in A} \inf_{x \in B} d(x,y))$$

Now, I'm told, "the topology on $K(X)$ depends only on the topology of $X$, as any two metrics on $X$ are equivalent". Should I interpret the any two metrics are equivalent as "they generate the same topology"? Then, the implication of the Hausdorff metrics being equivalent is not clear at all.

On the other hand, if I interpret two metrics being equivalent as $c d_1<d_2 < Cd_1$, then the Hausdorff metrics producing the same topology seems clear, but the fact that any two metrics on $X$ are equivalent is suspicious. Does the compactness of $X$ force this?

Even if I know one of the interpretation to be correct, I can then try and prove the statement. In fact, I would prefer that, over a complete proof of the statement.

As an aside, is there a standard notation for my $K(X)$? The book I'm reading uses $2^X$, although I suspect due to typographical limitations.

答:
Your first interpretation is the correct one.

The topology on $K(X)$ induced by the Hausdorff metric has a description as a hit-and-miss topology. That is, for a nonempty open set $U \subseteq X$ put
$$
U^+ = \{C \in K(X) \mid C \cap U \neq \emptyset\}
$$
and
$$
U^- = \{C \in K(X) \mid C \subseteq U\}.
$$
The set $U^+$ contains the compact sets that meet $U$ and the set $U^-$ contains the compact sets that miss $X \setminus U$.

The sets $U^+$ and $U^-$ where $U$ runs through the non-empty open sets of $X$ form a subbasis for the so-called Vietoris topology on $K(X)$. The sets $[U; V_1,\dots,V_n] := U^- \cap V_{1}^+ \cap \cdots \cap V_{n}^+$ form a convenient basis for the Vietoris topology.

The point is that one can prove that the Hausdorff metric induces the Vietoris topology on $K(X)$.

It may be helpful to prove at some point that for a countable dense subset $D$ of $X$ the collection of non-empty finite subsets of $D$ forms a countable dense subset of $K(X)$.

A nice and detailed exposition of these ideas can be found e.g. in Srivastava, A course on Borel sets, section 2.6, see Spaces of Compact sets, pages 66ff.

Added: The second interpretation is too strong. It is not very hard to show that a fat Cantor set and the usual ternary Cantor set are homeomorphic, but not bi-Lipschitz homeomorphic.


问:
example of an unstable fixed point for which the linearized dynamics are stable
What would be an example of an unstable fixed point for which the linearized dynamics are stable?

Thanks in advance

答:
It depends on what notion of stability you are using. Without loss of generality (by translating the coordinate system) we can assume our fixed point is the origin. Since you didn't indicate otherwise I'll also assume that the coefficients of your dynamical system is time independent.

Asymptotic stability

If we take the notion of stability to mean that for each initial data $f$, there exists a number $\epsilon > 0$ such that the dynamics with initial data $\epsilon f$ leads to solution $\lim_{t\to+\infty} x(t) = 0$. Then we have that

$$ \text{Linear stability } \implies \text{ nonlinear stability} $$

and the example you seek does not exist. This is due to the well-known fact that a linear system $\dot{x} = Ax$ is asymptotically stable if and only if all the eigenvalues of $A$ have strictly negative real parts. Thus for small data, the linear dynamics strictly dominates the nonlinear one, which leads to the Hartman-Grobman theorem (which can also be applied to more general hyperbolic fixed points).

Orbital stability

If, however, you are thinking in terms of orbital stability, then a simple example would be the dynamical system on $\mathbb{R}$ given by

$$ \dot{x} = x^3 $$

We have that $x(t) = 0$ is a fixed point. Its linearised dynamics is $\dot{x} = 0$, hence is trivially orbitally stable. (Recall that a fixed point $\bar{x}$ of a dynamical system is orbitally stable if for every $\epsilon > 0$ we can find a $\delta > 0$ such that if the initial data $|x_0 - \bar{x}| < \delta$ we have that the solution $|x(t) - \bar{x}| < \epsilon$ for all positive times $t$.) But solving the nonlinear ODE we get

$$ x(t) = \frac{x_0}{\sqrt{ 1 - 2 x_0^2 t}} $$

which blows-up in finite time for any non-zero initial data, and hence the nonlinear system is not orbitally stable.


问:
Outer measure of a null set
Outer measure of a null set is zero.

proof 1: By definition of outer measure, $m^*(\varnothing)\geq 0.$
Clearly $\varnothing\subseteq (-\frac{1}{n},\frac{1}{n}),\forall n\in \Bbb{N}.$
Therefore, $\varnothing\subseteq\cup_{n=1}^{\infty}(-\frac{1}{n},\frac{1}{n}).$
$m^*(\varnothing)\leq m^*(\cup_{n=1}^{\infty}(-\frac{1}{n},\frac{1}{n}))\leq \sum_{n=1}^\infty m^*(-\frac{1}{n},\frac{1}{n})=\sum_{n=1}^\infty l(-\frac{1}{n},\frac{1}{n})=\sum_{n=1}^\infty \frac{2}{n}, $

which diverges.

Proof 2:  By definition of outer measure, $m^*(\varnothing)\geq 0.$
Clearly $\varnothing\subseteq (-\frac{\epsilon}{2^{n+1}},\frac{\epsilon}{2^{n+1}}),\forall n\in \Bbb{N}$,
Therefore  $\varnothing\subseteq\cup_{n=1}^\infty(-\frac{\epsilon}{2^{n+1}},\frac{\epsilon}{2^{n+1}}).$
$m^*(\varnothing)= \text{inf}\big\{\sum_{n=1}^\infty l(-\frac{\epsilon}{2^{n+1}},\frac{\epsilon}{2^{n+1}})\big\}\leq\sum_{n=1}^\infty l(-\frac{\epsilon}{2^{n+1}},\frac{\epsilon}{2^{n+1}})=\sum_{n=1}^\infty \frac{\epsilon}{2^n}=\epsilon.$

Since $\epsilon >0$ is arbitrary. Hence $m^*(\varnothing)\leq 0$.
Thus $m^*(\varnothing)=0$.

Am I done wrong in proof1? For me proof2 sounds nicer.

If proof2 is correct, then fine. I am interested to know the proof1(only). Can choice of interval $(-\frac{1}{n},\frac{1}{n})$ is fine?

Help me with proof 1!

答: Contradiction works best with the empty set. Suppose $\mu(\emptyset)>0$. Then there should be an open interval $(a,b)\subseteq \emptyset$, and that is not true.

Title: Outer measure of a null set


问:
Show $A_{1}, \cdots, A_{n}$ is independent if and only if $P(A_{i_{1}} \cdots A_{i_{k}})=\prod_{j=1}^{k} P (A_{i_{j}})$
This is the second part of exercise question from book Probability for statistician.

b) Show that $A_{1}, \cdots, A_{n}$ are independent if and only if
$$
P\left(A_{i_{1}} \cdots A_{i_{k}}\right)=\prod_{j=1}^{k} P\left(A_{i_{j}}\right)
$$
whenever $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n$.

In part a) we are asked to prove:

Show that $P(A B)=P(A) P(B)$ if and only if $\left\{\emptyset, A, A^{c}, \Omega\right\}$ and $\left\{\emptyset, B, B^{c}, \Omega\right\}$ are independent $\sigma$ -fields.

Before the exercise, the book states the definition of independent $\sigma$-field and independent event as follow:

Consider various sub $\sigma$ -fields of $\mathcal{A}$. Call such $\sigma$ -fields $\mathcal{A}_{1}, \cdots, \mathcal{A}_{n}$ independent $\sigma$ -fields if they satisfy
$$
P\left(A_{1} \cap \cdots \cap A_{n}\right)=\prod_{1}^{n} P\left(A_{i}\right)
$$
whenever $A_{i} \in \mathcal{A}_{i}$ for $1 \leq i \leq n$.

Events $A_{1}, \cdots, A_{n}$ are independent events if $\sigma\left[A_{1}\right], \cdots$, $\sigma\left[A_{n}\right]$ are independent $\sigma$ -fields; here we let
$$
\sigma\left[A_{i}\right] \equiv\left\{\emptyset, A_{i}, A_{i}^{c}, \Omega\right\}
$$

My idea is that $P\left(A_{i_{1}} \cdots A_{i_{k}}\right)=\prod_{j=1}^{k} P\left(A_{i_{j}}\right)$ implies the $\sigma$-field generated by ay $A_{i_{1}} \cdots A_{i_{k}}$ are independent from a) part of the question, hence $A_{1}, \cdots, A_{n}$ are independent but I am not sure how to write the proof rigorously. i.e.

$P\left(A_{i_{1}} \cdots A_{i_{k}}\right)=\prod_{j=1}^{k} P\left(A_{i_{j}}\right)$
whenever $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n $ implies $\sigma[A_{i_j}]$ are independent for any $1\leq i_j\leq n$ $\Rightarrow$ $\sigma[A_i]$ are independent for each $n\geq 2$. It seems the $\Rightarrow$ has lots of gaps within it.

Could someone please address this question for me or at least gives me some clues to solve the problem? Also, the first part only involves two events while in second part there are $n$ events. Usually statement that satifies in $n=2$ case can be extended to $n=N$ where $N$ is a finite number but I don't know how the extension can be proved.

答:
Let us use the definitions from your book and let us solve:

b) Show that $A_{1}, \cdots, A_{n}$ are independent if and only if
$$
P\left(A_{i_{1}} \cap \cdots \cap A_{i_{k}}\right)=\prod_{j=1}^{k} P\left(A_{i_{j}}\right)
$$
whenever $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n$.

(We will explicitly note the intersections $\cap$, so, for instance, $P(AB)$ will be written $P(A \cap B)$).

Solution of b: $(\Rightarrow)$ Suppose $A_{1}, \cdots, A_{n}$ are independent. Then by definition  $\sigma[A_{1}], \cdots, \sigma[A_{n}]$ are independent $\sigma$-fields. So,
$$
P\left(E_{1} \cap \cdots \cap E_{n}\right)=\prod_{1}^{n} P\left(E_{i}\right)
$$
whenever $E_{i} \in \sigma[A_{i}]$ for $1 \leq i \leq n$.

Given any set of indexes $I =\{i_1, \cdots , i_k\}$ such that $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n$, define $E_i = A_i$ if $i\in I$  and $E_i = \Omega$ if $i \notin I$. Then $E_{1} \cap \cdots \cap E_{n} = A_{i_{1}} \cap \cdots \cap A_{i_{k}}$ and $P(E_i) = 1$ if $i \notin I$, so we have

$$
P\left(A_{i_{1}} \cap \cdots \cap A_{i_{k}}\right)=P\left(E_{1} \cap \cdots \cap E_{n}\right)=\prod_{1}^{n} P\left(E_{i}\right)= \prod_{j=1}^{k} P\left(A_{i_{j}}\right)
$$

$(\Leftarrow)$
Suppose $A_{1}, \cdots, A_{n}$ are not independent. Then   $\sigma[A_{1}], \cdots, \sigma[A_{n}]$ are not independent $\sigma$-fields.

So, there are $E_{i} \in \sigma[A_{i}]$ for $1 \leq i \leq n$, such that
$$
P\left(E_{1} \cap \cdots \cap E_{n}\right)\neq\prod_{1}^{n} P\left(E_{i}\right)
$$

Then, clearly $E_{i} \neq \emptyset $ for $1 \leq i \leq n$.

So, for each $i \in   \{1, \cdots n\}$, $E_i$ has only three possibilities: $E_i=A_i$ , $E_i=A_i^c$ and $E_i=\Omega$.

Let $I= \{i : 1 \leq i \leq n \textrm{ and } E_i= A_i\} $, $J= \{i : 1 \leq i \leq n \textrm{ and } E_i= A_i^c\}$ and $K = \{i : 1 \leq i \leq n \textrm{ and } E_i= \Omega\}$.
\begin{align*}
P\left(\bigcap_{i \in I \cup J} E_{i}  \right) &=
P\left(\bigcap_{i \in I \cup J \cup K} E_{i} \right) =
P\left(E_{1} \cap \cdots \cap E_{n}\right)\neq \\
& \neq \prod_{1}^{n} P\left(E_{i}\right) =
 \prod_{i \in I \cup J \cup K} P\left(E_{i}\right)=
 \prod_{i \in I \cup J} P\left(E_{i}\right)
\end{align*}

So,
$$P\left(\bigcap_{i \in I \cup J} E_{i}  \right) \neq  \prod_{i \in I \cup J} P\left(E_{i}\right)$$

Now, suppose $j\in J$, then $E_j= A_j^c$, so we have

\begin{align*}
P\left(\bigcap_{i \in I \cup (J\setminus \{j\})} E_{i}  \right) &- P\left(\left (\bigcap_{i \in I \cup (J\setminus \{j\})} E_{i} \right ) \cap A_j \right) =
P\left(\left (\bigcap_{i \in I \cup (J\setminus \{j\})} E_{i} \right ) \cap A_j^c \right) = \\
& = P\left(\bigcap_{i \in I \cup J} E_{i}  \right) \neq \\
& \neq  \prod_{i \in I \cup J} P\left(E_{i}\right)
= \left ( \prod_{i \in I \cup (J\setminus \{j\} )} P\left(E_{i}\right)\right ) (1 -P(A_j)) == \\
& = \left ( \prod_{i \in I \cup (J\setminus \{j\} )} P\left(E_{i}\right)\right )  - \left ( \prod_{i \in I \cup (J\setminus \{j\} )} P\left(E_{i}\right)\right )P(A_j)
\end{align*}

So we have that
$$ P\left(\bigcap_{i \in I \cup (J\setminus \{j\})} E_{i}  \right) \neq \prod_{i \in I \cup (J\setminus \{j\} )} P\left(E_{i}\right)
$$
or
$$ P\left(\left (\bigcap_{i \in I \cup (J\setminus \{j\})} E_{i} \right ) \cap A_j \right) \neq \left ( \prod_{i \in I \cup (J\setminus \{j\} )} P\left(E_{i}\right)\right )P(A_j)
$$
So, in both cases we can remove $j$ from $J$ (in the first case, $j$ will be simply removed from $J$ and in second case $j$ will be removed from $J$ and added to $I$).

Since $J$ is finite, we will get  a set of index $I'$ such that

$$P\left(\bigcap_{i \in I'} E_{i}  \right) \neq  \prod_{i \in I'} P\left(E_{i}\right)$$

and, for all $i \in I'$, $E_i=A_i$. So there are $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n$ such that
$$
P\left(A_{i_{1}} \cap \cdots \cap A_{i_{k}}\right)\neq \prod_{j=1}^{k} P\left(A_{i_{j}}\right)
$$

Remark: Here is another way to prove the $(\Leftarrow)$ part of b).

Suppose, for all $1 \leq i_{1}<\cdots<i_{k} \leq n$ with $1 \leq k \leq n$,
$$
P\left(A_{i_{1}} \cap \cdots \cap A_{i_{k}}\right)=\prod_{j=1}^{k} P\left(A_{i_{j}}\right) \tag{1}
$$
We want to prove that $A_{1}, \cdots, A_{n}$ are independent. It means,  by definition,  $\sigma[A_{1}], \cdots, \sigma[A_{n}]$ are independent $\sigma$-fields, which means, for all $E_{i} \in \sigma[A_{i}]$ where $1 \leq i \leq n$,
$$
P\left(E_{1} \cap \cdots \cap E_{n}\right)=\prod_{1}^{n} P\left(E_{i}\right) \tag{2}
$$
So, given any $\{E_{i} \}_{i\in \{1, \cdots n\}}$ such that $E_{i} \in \sigma[A_{i}]$ for $1 \leq i \leq n$, if there is $i \in   \{1, \cdots n\}$ such that $E_i=\emptyset$,  then $(2)$ is trivially true and we are done. So, suppose that, for all  $i \in   \{1, \cdots n\}$,  $E_i\neq\emptyset$.

So, for each $i \in   \{1, \cdots n\}$, $E_i$ has only three possibilities: $E_i=A_i$ , $E_i=A_i^c$ and $E_i=\Omega$.

Let $I= \{i : 1 \leq i \leq n \textrm{ and } E_i= A_i\} $, $J= \{i : 1 \leq i \leq n \textrm{ and } E_i= A_i^c\}$ and $K = \{i : 1 \leq i \leq n \textrm{ and } E_i= \Omega\}$. From $(1)$,

$$P\left(\bigcap_{i \in I } E_{i}  \right) = \prod_{i \in I } P\left(E_{i}\right)$$

And then it easy to see that

$$P\left(\bigcap_{i \in I \cup K} E_{i}  \right) =  \prod_{i \in I \cup K} P\left(E_{i}\right)$$

Now, take one $j\in J$, then $E_j= A_j^c$. Note that, from $(1)$, we also have
$$
P\left(\left (\bigcap_{i \in I \cup K} E_{i} \right ) \cap A_j \right) =
\left ( \prod_{i \in I \cup K} P\left(E_{i}\right)\right )P(A_j)
$$

So, we have
\begin{align*}
P\left(\bigcap_{i \in I  \cup K \cup \{j\}} E_{i}  \right) & = P\left(\left (\bigcap_{i \in I \cup K} E_{i} \right ) \cap A_j^c \right) = \\
&=P\left(\bigcap_{i \in I \cup K} E_{i}  \right) - P\left(\left (\bigcap_{i \in I \cup K} E_{i} \right ) \cap A_j \right) = \\
& = \left ( \prod_{i \in I \cup K} P\left(E_{i}\right)\right )  - \left ( \prod_{i \in I \cup K} P\left(E_{i}\right)\right )P(A_j) =\\
& = \left ( \prod_{i \in I \cup K} P\left(E_{i}\right)\right ) (1 -P(A_j)) =  \prod_{i \in I  \cup K \cup \{j\}} P\left(E_{i}\right)
\end{align*}

Since $J$ is finite, we have by finite induction,
$$
P\left(\bigcap_{i \in I  \cup K \cup J } E_{i}  \right) =
\prod_{i \in I  \cup K \cup J } P\left(E_{i}\right) \tag{3}
$$
but $(3)$ is exactly the same as $(2)$, that we  had to prove. $\square$


答:
Below is the working using trigonometry.

$A$ represents position of yacht after $5$ hours. $B$ represents position of Catamaran after $5$ hours.  So,

$OA = 120 $ nautical miles. $OB = 75$ nautical miles.

Angle between $OA$ and $OB = 139^0 - 37^0 = 102^0$.

$AB^2 = 120^2+75^2 + 2 \times 120 \times 75 \cos 102^0 \implies AB \approx 154.167$

$\cos \angle OBA = \frac{75^2 + 154.167^2 - 120^2}{2 \times 75 \times 154.167} \implies \angle OBA \approx 49.58^0$.

So bearing of yatch from catamaran is $49.58^0 - 41^0 \approx 8.58^0$ (which is $8.58^0$ clockwise from true north).

Title: Trigonometry - distance and bearing

Body:
A yacht and a catamaran leave the harbour at $09:00$ hours. The yacht sails at $24$ knots on course $037$ and the catamaran at $15$ knots on $139$. Find the bearing and the distance in nautical miles of the yacht from the catamaran at $1400$ hours ($1$ knot is a speed of $1$ nautical mile per hour).

The final answer I am getting is $135$m using the cosine rule. However this doesn’t match with the solutions. Please could someone clarify both the length and the bearing?


问:
First Course in Linear Algebra book suggestions?
I'm starting maths degree this September and hoping to do some reading on Linear Algebra before I start. What are some good introductory books?

Is there something that starts from the very beginning but also covers stuff in depth? (Is there a book that will be useful for both 1st year and 2nd year linear algebra modules?)

It would also be great if the book has lots of exercises (proof questions!) and solutions given as well.

Thanks in advance.

答: Just an empiric, maybe a little bit personal, paranoic, exagerrated (and not so serious) piece of advice. Take any  Linear Algebra book and look at the definition the author gives of a matrix. If it is something that sounds like "A $m\times n$ matrix (over a field) is an array of scalars (numbers) with $m$ rows and $n$ columns" and so on, try to look at another book. If after a while you still can't find anything which gives a different definition for a matrix (and I doubt you will so easily), then pick one of the textbook suggested so far (actually I haven't checked if some of them provides another definition for a matrix). But if you find one which seems to disagree with the fact that a matrix has to be defined as an array of numbers, it might be worth to try and read it.</p>


问:
Is there a closed form for $\int_0^\infty \frac{\sin(x)}{x^2+a^2} dx$?
As the title says, is there a closed form for
$$
\int_0^\infty \frac{\sin(x)}{x^2+a^2} dx \,?
$$

The one with $\cos(x)$ instead of $\sin(x)$ can be calculated via a simple application of the Residue theorem, but the computation uses the fact that the function is even, so the same trick does not work for this one.

Clarification In case this was unclear, I do not care about the method used to calculate this integral.

答: Contour integration of $(e^{iz}-e^{-a})/(a^2+z^2)$ along the boundary of $|z|<R,0<\arg z<\pi/2$, with $z=ia$ a removable singularity (we assume $a>0$), yields $$\int_0^\infty\frac{\sin x}{a^2+x^2}\,dx=\int_0^\infty\frac{e^{-x}-e^{-a}}{a^2-x^2}\,dx,$$ expressible in terms of the exponential integral function (after doing partial fractions).


问:
Substitution that gives a division by $0$
When solving the following integral

$$ \int_{0}^{c}\frac{\ln(1+x^{2})}{x^{2}}dx $$

(where c is a real number; it's a constant that really doesn't matter for the problem anyway), after doing integration by parts I got

$$ \left [-\frac{ln(1+x^{2})}{x} \right ]_{0}^{c}+\left [2\arctan(x)\right ]_{0}^{c}$$

which is

$$-\frac{ln(1+c^{2})}{c} + \frac{ln(1+0^{2})}{0} +2\arctan(c) \ .$$

But there's a problem. The expression

$$\frac{ln(1)}{0} = \frac{0}{0}$$

is undefined. So what am I supposed to do in this cases?

Any help is welcome.

答:
Because this is an improper integral (the integrand isn't defined at $0$), you need to take a limit.

Specifically, we want to know

$$\lim_{x \to 0^+} \frac{\ln(1 + x^2)}{x}.$$

This is a simple application of l'Hôpital's rule.

$$\lim_{x \to 0^+} \frac{\ln(1 + x^2)}{x} = \lim_{x \to 0^+} \frac{\frac{2x}{1+x^2}}{1} = \lim_{x \to 0^+} \frac{2x}{1+x^2} = 0.$$

In general, if $f(x)$ is undefined at $x = 0$ (maybe it approaches $\infty$, maybe the limit is finite but it's still of the form $\frac{0}{0}$) then

$$\int_0^c f(x) \;dx = \lim_{a \to 0^+} \int_a^c f(x) \;dx = \lim_{a \to 0^+} [F(c) - F(a)] = F(c) - \lim_{a \to 0^+} F(a).$$

Here $F$ is an antiderivative of $f$.

So when you write

$$ \left [-\frac{\ln(1+x^{2})}{x} \right ]_{0}^{c} $$

that needs to be interpreted as

$$ \lim_{a \to 0^+} \left [-\frac{\ln(1+x^{2})}{x} \right ]_{a}^{c} = -\frac{\ln(1+c^{2})}{c} + \lim_{a \to 0^+} \frac{\ln(1+x^{2})}{x}. $$


答:
Just to put in more closed form what @Varu Vejalla has evaluated

$I=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx$

$\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=I+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx$
$\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx [t=\frac{1}{x}]=\int_0^1\frac{\ln(1+t^{2n})}{1+t^2} dt-\int_0^1\frac{\ln(t^{2n})}{1+t^2} dx=I+2nG$

$$I=\frac{1}{4}\int_{-\infty}^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx-nG$$

Branch points of $\log$ (roots of $1+x^{2n}$) are $e^{\frac{i\pi}{2n}},e^{-\frac{i\pi}{2n}}, e^{\frac{3i\pi}{2n}}, e^{\frac{-3i\pi}{2n}},... e^{\frac{(2n-1)i\pi}{2n}}, e^{-\frac{(2n-1)i\pi}{2n}} $

Next we integrate in the complex plane, closing the contour in the upper half-plane for the roots $e^{-\frac{(2k-1)i\pi}{2n}}$ and in the lower half-plane for the roots $e^{+\frac{(2k-1)i\pi}{2n}}$ - to integrate the single-valued function.

Finally we get
$$I=\frac{2\pi{i}}{4*2i}\log\Bigl((-i-e^{\frac{i\pi}{2n}})(i-e^{-\frac{i\pi}{2n}})...(-i-e^{\frac{(2n-1)i\pi}{2n}})((-i-e^{-\frac{(2n-1)i\pi}{2n}})\Bigr)-nG=$$
$$=\frac{\pi}{4}\log\Bigl((2+2\sin\frac{\pi}{2n})...(2+2\sin\frac{(2n-1)\pi}{2n})\Bigr)-nG$$

$$I=\frac{\pi}{4}\log\Bigl((1+\sin\frac{\pi}{2n})...(1+\sin\frac{(2n-1)\pi}{2n})\Bigr)+\frac{\pi{n}}{4}\log2-nG$$

Title: Is there a known closed form solution to $\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx$?

Body:
$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

Related question

Is there a known closed form solution to

\begin{align}
I_n&=
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
=?
\tag{0}\label{0}
\end{align}

It checks out numerically, for $n=1,\dots,7$ that

\begin{align}
I_1=
\int_0^1\frac{\ln(1+x^2)}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln2-\Catalan
\tag{1}\label{1}
,\\
I_2=
\int_0^1\frac{\ln(1+x^{2\cdot2})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(2+\sqrt2)-2\Catalan
\tag{2}\label{2}
,\\
I_3=
\int_0^1\frac{\ln(1+x^{2\cdot3})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln6-3\Catalan
\tag{3}\label{3}
,\\
I_4=
\int_0^1\frac{\ln(1+x^{2\cdot4})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(4+\sqrt2+2\sqrt{4+2\sqrt2})-4\Catalan
\tag{4}\label{4}
,\\
I_5=
\int_0^1\frac{\ln(1+x^{2\cdot5})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(10+4\sqrt5)-5\Catalan
=
\tfrac\pi2\,\ln(10\cot^2\tfrac\pi5)-5\Catalan
\tag{5}\label{5}
,\\
I_6=
\int_0^1\frac{\ln(1+x^{2\cdot6})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln((5\sqrt2+2\sqrt{12})(1+\sqrt2))-6\Catalan
\tag{6}\label{6}
,\\
I_7=
\int_0^1\frac{\ln(1+x^{2\cdot7})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(14\cot^2\tfrac\pi7)-7\Catalan
\tag{7}\label{7}
,
\end{align}

so \eqref{0} seems to follow the pattern

\begin{align}
I_n&=
\tfrac\pi2\,\ln(f(n))-n\Catalan
\tag{8}\label{8}
\end{align}

for some function $f$.

Items \eqref{5} and \eqref{7} look promising
as they agree to $f(n)=2n\cot^2(\tfrac\pi{n})$,
but the other fail on that.

Edit:

Also, it looks like
\begin{align}
\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\tfrac\pi2\,\ln(f(n))+n\Catalan
\tag{9}\label{9}
\end{align}

and

\begin{align}
\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\pi\,\ln(f(n))
\tag{10}\label{10}
\end{align}

with the same $f$.

Edit

Thanks to the great answer by @Quanto,
the function $f$ can be defined as

\begin{align}
f(n)&=
2^n\!\!\!\!\!\!\!\!\!\!
\prod_{k = 1}^{\tfrac{2n-1+(-1)^n}4}
\!\!\!\!\!\!\!\!\!
\cos^2\frac{(n+1-2k)\pi}{4n}
\tag{11}\label{11}
.
\end{align}

$\endgroup$


问:
Show $\lim\limits_{x\to0}x\ln(x)=0$ by using Taylor series
Is it possible to show the limit $\lim\limits_{x\to0}x\ln(x)=0$ by expanding the corresponding Taylor series?

My approach:

I know that the Taylor series of $\ln(x)$ converges uniformly on a closed interval $(1-\delta,1+\delta)$ where $\delta>0$ is sufficiently small.

So let be $$S_n(x):=\sum\limits_{k=1}^n(-1)^{k+1}(x-1)^k\frac{1}{k}\\
\lim\limits_{n\to\infty}S_n(x)=\sum\limits_{k=1}^{\infty}(-1)^{k+1}(x-1)^k\frac{1}{k}=\ln(x)\\
T_n(x):=(x-1)+\sum\limits_{k=2}^n(-1)^{k}(x-1)^k\frac{1}{(k-1)k}\\
\lim\limits_{n\to\infty}T_n(x)=(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}(x-1)^k\frac{1}{(k-1)k}\overset{?}{=}x\ln(x),
$$
where $\lim\limits_{n\to\infty}S_n(x)$ denotes the Taylor series of $\ln(x)$ at $x=1$ and $\lim\limits_{n\to\infty}T_n(x)$ the Taylor series of $x\ln(x)$ at $x=1$.

What I can observe so far is that $T_n(x)$ converges uniformly on $(0,1+\delta)$ (the point $x=0$ doesn't cause problems anymore) and this allows us to swap limit taking:
$$
\lim\limits_{x\to0}\lim\limits_{n\to\infty}T_n(x)=\lim\limits_{n\to\infty}\lim\limits_{x\to0}T_n(x)=0.
$$

However, I don't know what the limit function is or in other words how to show that $\lim\limits_{n\to\infty}T_n(x)=x\ln(x)$.

Any suggestions how to complete the proof?

答:
It helps to keep in mind the geometric series $$\sum_{k=0}^\infty (1-x)^k = \frac 1{1-(1-x)} = \frac 1x$$
which is valid for $x \in (0,2)$.  You have that the Taylor expansion $$\ln x = \sum_{k=1}^\infty \frac{(-1)^{k+1}}k(x-1)^k$$ is valid on the interval $(0,2)$ but is not uniform on any interval of the form $(0,\epsilon)$. However you can conclude that $x \ln x$ is bounded on $(0,1)$ because if $0 < x < 1$ then
$$|x \ln x| = \left| \sum_{k=1}^\infty \frac{(-1)^{k+1}}kx(x-1)^k \right| \le \sum_{k=0}^\infty x(1-x)^k = 1.$$

You can finish up rather quickly if you are willing to use a law of logarithms. If $0 < x < 1$ then $0 < x^{1/2} < 1$ too and $$|x^{1/2} \ln x| = 2 |x^{1/2} \ln x^{1/2}| \le 2$$ so that $$|x \ln x| \le 2|x|^{1/2}.$$


问:
Let $f_n$ sequence of continuous functions on $[a,b]$, $\lim_{n\to \infty}f_n(x)=f(x)$ uniformly on $(a,b)$. Show $f_n\mapsto f$ uniformly on $[a,b]$
Let $f_n$ sequence of continuous functions on $[a,b]$ and $\lim_{n\to \infty}f_n(x)=f(x)$ uniformly on $(a,b)$. Show that $f_n\mapsto f$ uniformly on $[a,b]$.

I would like to know how to continue my proof, please.

First of all, as $f_n$ are continuous on closed interval, this sequence is uniformly continuous on $[a,b]$. By definition we have: $\forall \epsilon>0 \ \exists \delta_1>0 \ \forall x,y \in [a,b]$: $|x-y|<\delta_1 \implies|f_n(x)-f_n(y)|<\epsilon$

As $f_n$ converges uniformly to $f$ on $(a,b)$, $f$ is uniformly continuous on $(a,b)$. Moreover, the uniform continuity of $f$ on $(a,b)$ implies that $\lim_{x\to a^+}f(x)$ and $\lim_{x\to b^-}f(x)$ exist. So by extension by continuity we have that $f$ is uniofmrly continuous on $[a,b]$: $\forall \epsilon>0 \ \exists \delta_2>0 \ \forall x,y \in [a,b]$: $|x-y|<\delta_2 \implies |f(x)-f(y)|<\epsilon$

We would like to show the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x \in [a,b]$: $|f_n(x)-f(x)|<\epsilon$.

How can i conclude with all the hypothesises that I've done until now?

Edit: Probably I could have tried to prove by absurd and work with sequence $x_n$ which is bounded so by Bolzano-Weierstrass there is a converging sub-sequence(So cauchy) then use the uniform continuity to get a contradiction

答:
By the uniform Cauchy criterion, there exists $N(\epsilon)$ such that for all $m > n \geqslant  N(\epsilon)$ and all $x \in (a,b)$, we have $|f_m(x) - f_n(x) | < \epsilon$.

By continuity, this implies that $|f_m(a) - f_n(a)| = \lim_{x \to a+}|f_m(x) - f_n(x)| \leqslant \epsilon$, and similarly as $x \to b-$.


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答:
Morley’s Miracle: The three points of intersection of the adjacent trisectors of the angles of any triangle always form an equilateral triangle.

This is a stunning gem that slipped through the fingers of the ancients.

LastEditorDisplayName: anon


问:
Notation $\overline{K}(C)^*$ in Silverman's "The Arithmetic of Elliptic Curves"
In Joe Silverman's "The Arithmetic of Ecliptic Curves" he talks a lot about the integral domain $\overline{K}(C)$. On page $27$ he suddenly decides to chose an element $f$ from some set $\overline{K}(C)^*$. What does this notation mean? After page $27$ he begins using $\overline{K}(C)^*$ a lot, and I haven't been able to pick up what it is from context clues.

答: There is already a correct answer, but I will just add here that the notation is defined in Chapter I, in a Definition right after Example 1.3.3. where Silverman defines the affine coordinate ring and the function field of a variety.</p>


问:
Why does involution correspond to complex conjugation?
Let $K$ be a number field of degree $n$ over $\mathbb{Q}$. In Minkowski theory, we construct a map $j:K\rightarrow K_{\mathbb{C}}:=\prod_\tau\mathbb{C}$, $a\mapsto ja = (\tau a)$, where $\tau$ ranges over the $n$ embeddings of $K$ in $\mathbb{C}$. We can also identify $K\otimes_{\mathbb{Q}}\mathbb{C}\cong K_{\mathbb{C}}$ via the map $x\otimes y\mapsto (\tau(x)y)_\tau$.

Now, in Neukirch's book Algebraic Number Theory, he says that the involution $F(x\otimes y)=x\otimes\overline{y}$ corresponds to complex conjugation $F((x_\tau)_\tau)=(\overline{x}_{\overline{\tau}})_\tau$ in $K_\mathbb{C}$.

My question is, why is this true? Doesn't $x\otimes y$ map to $(\tau(x)\overline{y})_\tau$? How to get from here to the desired form?

答:
Let's take a simple example: $K = \mathbb Q(i)$.  We have $K = \mathbb Q[t]/(t^2+1)$, and

$$K \otimes_{\mathbb Q} \mathbb C = \mathbb Q[t]/(t^2+1) \otimes \mathbb C = \mathbb C[t]/(t^2+1) = \mathbb C[t]/(t-i) \times \mathbb C[t]/(t+i) = \mathbb C \times \mathbb C.$$
This last isomorphism is given by sending a pair of polynomials $(g_1(t), g_2(t))$ to $(g_1(i), g_2(-i))$.

The complex conjugation involution on $K \otimes_{\mathbb Q} \mathbb C$ sends $x \otimes \lambda$ to $x \otimes \overline{\lambda}$.  Let's see what this does to an element $(A,B)$ of $\mathbb C \times \mathbb C$. It's preimage in $\mathbb Q[t]/(t^2+1)\otimes \mathbb C$  is a sum of elementary tensors

$$\xi = h_1(t) \otimes \lambda_1 + \cdots + h_s(t) \otimes \lambda_s$$

where $$\lambda_1 h_1(i) + \cdots + \lambda_s h_s(i) = A$$$$\lambda_1 h_1(-i) + \cdots + h_s(-i) = B$$

Applying the involution to $\xi$ gives you

$$\xi^c = h_1(t) \otimes \overline{\lambda_1} + \cdots + h_s(t) \otimes \overline{\lambda_s}$$

where now

$$\overline{\lambda_1}h_1(i) + \cdots + \overline{\lambda_s} h_s(i)  = \overline{\lambda_1 h_1(-i) + \cdots + h_s(-i)} = \overline{B}$$

$$\overline{\lambda_1}h_1(-i) + \cdots + \overline{\lambda_s} h_s(-i) = \overline{\lambda_1 h_1(i) + \cdots + \lambda_s h_s(i)} = \overline{A}$$

This shows that the involution on $\mathbb C \times \mathbb C$ sends $(A,B)$ to $(\overline{B},\overline{A})$.

The computation here generalizes easily to an arbitrary number field.


问:
Series ( Should I use AST or CT?)
$$\sum_{n=1}^\infty  \frac{(-1)^{n+1}+n^2}{n^3+1}$$

So this is another series qustion that I have encountered during my HW and I have two ways of solving it in my mind. I could split the series into $\sum_{n=1}^\infty  \frac{n^2}{n^3+1}$ and $\sum_{n=1}^\infty  -\frac{(-1)^n}{n^3+1}$ but as you can probably see finding the limit for the second series is extremely difficult.

So a better way to solve this according to my notes would be to use the alternating series but I cannot get this series in the form $(-1)^{n+1} {b_n}$

Any help?

答: $$\frac{n^2+(-1)^{n+1}}{n^3+1}\geq \frac{n^2-1}{n^3+1}> \frac{1}{2 n}$$ As the last series diverges, it diverges the given one, too.

Title: Series ( Should I use AST or CT?)


问:
"In some numbers, digits are distinct" What does it mean?

Form 4 digit numbers using $\{1,2,3,4,5\}$

  a) In some of the numbers, digits are distinct.

  b) The number is even.

  Solve for just $a$, just $b$, and for both $a$ and $b$.

I could not understand what is meant in part $a$. Is not it same with the set of all $4$ digit numbers written with $\{1,2,3,4,5\} = 5 \cdot 5 \cdot 5 \cdot 5$?

答:
Solution for (just a): 5432 = 5! = 120
Solution for (just b): 5552 = 250 (Numbers may not be distinct)
Solution for (both a and b): 432*2 = 48 (Numbers are distinct digit and even)


答:
A similar example to the one given by Willie Wong  (but maybe less boring): perturbation of a center
$$\frac{d\vec x}{dt} =\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}\vec x \tag1$$
which is stable but not asymptotically stable. To (1) we should add a nonlinear term that pushes solutions away from the origin. A linear term of this kind would be $f(\vec x)=\vec x$. To make it vanish to higher order at the origin, use something like $f(\vec x)=|\vec x|^2\vec x$  (squaring is optional, used only to avoid the square root). So,
$$\frac{d\vec x}{dt} =\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}\vec x+ |\vec x|^2\vec x \tag2$$
is the desired example.

To prove that it is indeed unstable, observe that
$$\frac{d}{dt}|\vec x|^2=2\left\langle \frac{d}{dt}\vec x,\vec x\right\rangle = 2|\vec x|^4$$

Body:
What would be an example of an unstable fixed point for which the linearized dynamics are stable?

Thanks in advance


问:
Show that the following is only continuous at x=0

What is this question asking?
How does rational and irrational affect it? What are the absolute values used for in this solution?
Why is the series irrational when $c\ne0$ for any nonzero rational number but then the series is made to be all irrational values?

What is this question?

答:
hint

$ f $ is continuous at $ 0 $ because

$$(\forall x\in \Bbb R)\;\;|f(x)|=|x|$$
and
$$\lim_{x\to 0}f(x)=0=f(0)$$

Let $ a\ne 0$ rational.
Take, for $ n $ great enough,

$$a_n=a+\frac{\pi}{n}\notin \Bbb Q$$
then
$$\lim_{n\to+\infty}f(a_n)=\lim_{n\to+\infty}(-a_n)=-a\ne f(a)$$

Let $ b $ be an irrational number.
Take
$$b_n=\frac{\lfloor 10^nb\rfloor}{10^n}\in \Bbb Q$$
then

$$\lim_{n\to+\infty}f(b_n)=\lim_{n\to+\infty}b_n=b\ne f(b)$$


问:
Calculate the volume of the region in the first octant bound by $y=0$, $y=x$, $x^2+y^2+z^2=4$
I am trying to calculate the volume of the region in the first octant bound by the surfaces

$$y=0,\hspace{1em}y=x,\hspace{1em}x^2+y^2+z^2=4$$

I have found that $x,\ y,$ and $z$ can have a lower bound of $0$.

$z$ seems to have an upper bound of $\sqrt{4 - x^2 - y^2}$

$y$ seems to be from $0$ to $x$

$x$ seems to be form $0$ to $2$ (the radius of the sphere)

Is it correct to use the following triple integral to solve the problem?

$$\int_0^2 \int_0^x \int_0^{\sqrt{4-x^2-y^2}} dz\ dy\ dx$$

Sketch:

It seems to me that the region to find is the area shown below (the left half of the section of the sphere in the first octant). It is clear to me that the volume should be that of the sphere divided by 16, but I need to learn how to use triple integrals to solve this problem.

How would I go about solving this?

答:
Because the base of the solid is a sector of a circle, it is most natural to use either cylindrical or spherical coordinates. I don't know if you're in the American coordinate system or non-American, but notice that limiting $y$ to $0\le y\le x$ means that $0\le\theta\le\pi/4$ (with $\theta$ the polar coordinate in the plane). So you will have $0\le\theta\le\pi/4$, $0\le\phi\le\pi/2$ (because you're constrained to the first octant), and, easiest, $0\le\rho\le 2$. So the volume integral becomes
$$\int_0^{\pi/4}\int_0^{\pi/2}\int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$$
[Interchange $\theta$ and $\phi$ as required.]


问:
Lines simultaneously tangent to the curves $y = x^2$ and $y = -x^2+2x-2$
Is (are) there any straight line(s) that is (are) simultaneously tangent to both the curves given by the equations $y = x^2$ and $y = -x^2+2x-2$?

My Attempt:

Let $y = mx+b$ be any such (non-vertical) line. Then we must have
$$
mx+b = x^2 \qquad \mbox{ and } \qquad mx+b = -x^2+2x-2,
$$
or in other words,
$$
x^2 - mx-b = 0 \qquad \mbox{ and } \qquad x^2 + (m-2)x + (b+2) = 0.
$$
Moreover, each one of these two quadratics must have coincident solutions, which implies that each one of the two discriminants must be zero, that is, we must have
$$
m^2 + 4b = 0 \qquad \mbox{ and } \qquad (m-2)^2 -4 (b+2) = 0,
$$
and from the first of the preceding two relations we get
$$
b = - \frac{m^2}{4},
$$
which when substituted into the second one gives
$$
(m-2)^2 + m^2 - 8 = 0,
$$
which simplifies to
$$
2m^2 -4m -4 = 0,
$$
and hence
$$
m^2 - 2m - 2 = 0,
$$
from which we get
$$
m = \frac{ 2 \pm \sqrt{4 + 8} }{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}.
$$
For $m = 1 + \sqrt{3}$, we get
$$ b = - \frac{ (1 + \sqrt{3})^2 }{4} = - \frac{ 4 + 2 \sqrt{3} }{4} = -1 - \frac{ \sqrt{3} }{2}. $$

And, for $m = 1- \sqrt{3}$, we get
$$
b = - \frac{ (1 - \sqrt{3})^2 }{4} = - \frac{ 4 - 2 \sqrt{3} }{4} = -1 + \frac{ \sqrt{3} }{2}.
$$

Thus there are two straight lines that are simultaneously tangent to both the given curves, and these lines are given by the equations
$$
y = (1 + \sqrt{3} ) x  -1 - \frac{ \sqrt{3} }{2}
$$
and
$$
y = (1 - \sqrt{3} ) x  -1 + \frac{ \sqrt{3} }{2}.
$$

Is my solution / answer correct? If so, then is my approach to finding this solution / answer also correct? If not, then where have I erred?

Any alternative solutions, especially those using the methods of differential calculus?

答:
To determine the tangent lines is sufficient to solve for $x_1,x_2,y_1,y_2$ the system

$$
\cases{
y_1 = x_1^2\\
y_2 = -x_2^2+2x_2-2\\
y_1 = m x_1 + c\\
y_2 = m x_2 + c
}
$$

so a typical result is

$$
\left\{
\begin{array}{rcl}
x_1& = &  \frac{1}{2} \left(m-\sqrt{4 c+m^2}\right) \\
y_1 & = &  \frac{1}{2} m \left(m-\sqrt{4 c+m^2}\right)+c \\
x_2 & = & \frac{1}{2} \left(2-\sqrt{(m-4) m-4c-4}-m\right) \\
y_2 & = & c-\frac{1}{2} m \left(\sqrt{(m-4) m-4c-4}+m-2\right) \\
\end{array}
\right.
$$

from here, we solve the tangency conditions

$$
\cases{
4 c+m^2 = 0\\
(m-4) m-4c-4=0
}
$$

obtaining

$$
\cases{
m = 1\pm \sqrt{3}\\
c=-\frac 14\left(1\pm\sqrt{3}\right)^2
}
$$


问:
There are 9 blocks whats the length and width of one block. the pic has the rest of the info for the question
The rest of the questions information is here
I'm really confused. Please help me with this question.</p>

答:
Let, length=x and width=y.

Thus, from the figure
$$9xy=1620$$
$$xy=180$$

Also, from the 2nd figure,
$$4x=5y$$
$$x=\frac{5y}{4}$$
Put it in the 1st equation:
$$\frac{5y^2}{4}=180$$
$$y^2=144$$
$$y=12$$
Thus,
$$x=\frac{5y}{4}=15$$

Thus dimensions of a single block are $15*12$


问:
What is the element of unity for $\mathbb{Z}[\sqrt5]$?
What is the element of unity for $\mathbb{Z}[\sqrt5]$?

Note that $\mathbb{Z}\left[\sqrt5\right]=\{a+b\sqrt5\mid a,b\in \mathbb{Z}\}$.

I believe that I need to find an element $x$ such that $(a+b\sqrt5)x=a+b\sqrt5$, but I'm struggling to find one. Am I on the right track?

答: Since $1=1+0\times\sqrt{5}$, we have $1\in\Bbb Z[\sqrt{5}]$.


答:
Your limits are invalid. The problem is that the region defined by $(0\le x\le 2$ and $0\le y\le x)$ contains points for which $\sqrt{4-x^2-z^2}$, the upper limit for $z$, is not a real number. For instance, $(x,y)=(\frac32,\frac32)$.

It really is worth it to transform the problem into spherical coordinates!

Title: Calculate the volume of the region in the first octant bound by $y=0$, $y=x$, $x^2+y^2+z^2=4$

Body:
I am trying to calculate the volume of the region in the first octant bound by the surfaces

$$y=0,\hspace{1em}y=x,\hspace{1em}x^2+y^2+z^2=4$$

I have found that $x,\ y,$ and $z$ can have a lower bound of $0$.

$z$ seems to have an upper bound of $\sqrt{4 - x^2 - y^2}$

$y$ seems to be from $0$ to $x$

$x$ seems to be form $0$ to $2$ (the radius of the sphere)

Is it correct to use the following triple integral to solve the problem?

$$\int_0^2 \int_0^x \int_0^{\sqrt{4-x^2-y^2}} dz\ dy\ dx$$

Sketch:

It seems to me that the region to find is the area shown below (the left half of the section of the sphere in the first octant). It is clear to me that the volume should be that of the sphere divided by 16, but I need to learn how to use triple integrals to solve this problem.

How would I go about solving this?


问:
Circles and lines geometry problem

Here is my task: An arbitrary ray $OE$ crosses the circle at point $D$:
$$x ^ 2 + (y - a / 2) ^ 2 = (a / 2) ^ 2$$
and a tangent to it(at point $E$), passing through point $C$, diametrically opposite to $O$.

Through points $D$ and $E$, straight lines are drawn parallel to the axes $Ox$ and $Oy$, respectively, up to the intersection at point M.

Equate the line formed by the $M$ points and draw it (Agnesi's curl).

My attempt

I think that this problem need using Polar coordinates. But I have no idea how to start. Maybe if we assume that $M$ has coordinates $(x,y)$ we can find this point(using polar coordinates like $x=r\sin(q)$ or something like that) but how to equate the line? And of course I have already drawn it. After an hour i have idea if we find point $M$ can we use $M$-coordinates to get parametric equation of a line? If yes, how to make first step? Ok, after a few time i found $x$ like a coordinate of a point $M$ (i use as a parameter angle between $OE$ and $OC$ (q) so $x=tgqa/cosq$) how can find $OB$?

答:
Agnesi curve

Parametric equation

$\theta = angle(Ox, OD) \qquad OC = a$

triangle (OCE) : $CE = \dfrac{a}{\tan \theta} = x_{M}$

triangle (ODC) : $OD = OC\cos\Big( \dfrac{\pi}{2}-\theta \Big)= a\cos\Big( \dfrac{\pi}{2}-\theta \Big) = a\sin \theta$

triangle (OHD) : $HD = OD\cos\Big( \dfrac{\pi}{2}-\theta \Big)= OD\sin \theta  = a\sin^{2}\theta = y_{M}$

equation : $x = \dfrac{a}{\tan \theta} \qquad y = a\sin^{2}\theta \qquad \theta \in [0, \pi]$

Title: Circles and lines geometry problem


答:
Please note that your integral is incorrectly set up. If you are integrating in the order $dz$ first, then $dy$ and then $dx$, you will have to split your integral into two. So I would recommend going $dx$ before $dy$.

$\displaystyle \int_0^{\sqrt{2}} \int_y^{\sqrt{4-y^2}} \int_0^{\sqrt{4-x^2-y^2}} dz \ dx \ dy$.

The upper limit of $y$ comes from intersection of plane $y = x$ and $x^2+y^2+z^2 = 4$ at $z = 0$. Then $2y^2 = 4 \implies y = \sqrt2$.


问:
Properties of determinants.
Is this property of a determinant true?

$$|A^3| = |A|^3.$$

I haven't studied about this but while working out on a sum, wondered if this could be true, I'll check out on other sums too if this works.

Title: Properties of determinants.

Body: Hint: use $\det(AB)=\det(A)\det(B)$.


问:
Recursively given sequence question about changing the sequence with exponent function
We have a sequence given by:

$$a_1 = 1 \\ a_2 = b \\
a_{n+2}=a_{n+1}+a_n$$

We need to find the limit of the sequence. The procedure of getting the solution is that we $a_n = p^n$ and then somehow solve the equation $p^{n+2}=\frac{1}{3}p^n+\frac{2}{3}p^{n+1}\implies p^2 - \frac{2}{3}p-\frac{1}{3} = 0$

For this quadratic equation there are two solutions, that we input:

$$a_1 = A - \frac{1}{3}B = 1, a_2 = A +\frac{1}{9}B = b \implies A = \frac{1+3b}{4}, B = \frac{9(b-1)}{4}$$

However I do not understand this process. Why can we say that $a_n = p^n$ and what follows from it. Where can we use this technique?

答:
Here is some presentation without matrices, it is easy to understand for a recurrence with only two terms, but keep in mind that the generalization with many terms is easier to handle with linear algebra theory as in Andreas's answer.

Let start with the simple case $x_{n+1}=ax_n$

It can be solved by induction to $x_n=a^nx_0$

Now with $x_{n+1}=ax_n+bx_{n-1}$

Recall that the quadratic equation $$r^2-sr+p=0$$ has roots whose sum is $s$ and product $p$.

So let's call $r_1,r_2$ the roots of $r^2-ar-b=0$

We can rewrite our equation $x_{n+1}=(r_1+r_2)x_n-r_1r_2x_{n-1}\iff (\overbrace{x_{n+1}-r_1x_n}^{y_{n+1}})=r_2(\overbrace{x_n-r_1x_{n-1}}^{y_n})$

By first case then $$y_{n+1}=r_2y_n\iff y_n=C_2{r_2}^n$$

And since equation is linear, a solution is the sum of a general solution of homogeneous equation and a particular solution with RHS.

In this case $x_{n+1}-r_1x_n=\overbrace{C_2{r_2}^n}^{\text{RHS}}$

Solves to $x_n=C_1{r_1}^n+y_n=C_1{r_1}^n+C_2{r_2}^n$

Rem: here I assumed $r_1\neq r_2$, when they are equal the particular solution is different, but not let's compilate too much for the moment

For the general case $x_{n+1}=\sum\limits_{i=1}^k a_ix_{n+1-i}$

We can proceed similarly using Newton's identities to express the $a_i$ in terms of the roots $r_i$ and rearrange the terms to make an expression $y_{n+1}=r_ky_n$ and solve it recursively like in the case with $k=2$.

This is why we generally solve the characteristic equation $r^k=\sum\limits_{i=1}^{k}a_ir^{i-1}$ whose roots are $r_i$.

And that we have a solution $x_n=C_1{r_1}^n+\cdots+C_k{r_k}^n$

This at least is in the ideal case where all roots are distinct, the complications arise when some roots have higher multiplicity, in which case the general solution for a root of multiplicity $m$ is $$(\alpha_0+\alpha_1n+\cdots+\alpha_{m-1}n^{m-1})r^n\quad\text{instead of}\quad (\alpha_0\,r^n)$$


问:
Use Bayes method to solve ODE system with random noise
For ODE system $\frac{du}{dt} = \beta u$, $t>0$, $u(0)=1$, where $\beta$ is unknown. But the solution to the system at t=1 up to some noise is known: $h :=u(1) + \zeta$, where $\zeta$ is a random variable in N(0,1).

Assume we know the a prior that $\beta$ follows a Gaussian distribution N(2,1).

So estimate $\beta$ from observed h, I follow the Bayes method.

But here comes to the trouble. To find the posterior, I need the likelihood, which should be fund with the ODE and distribution of $\zeta$. But I just cannot figure out their relationship myself here.

So far, I understand that $f(\beta) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(\beta - 2)^2}{2}}$

And I got a hint which I do not understand $f(h|\beta) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(h - e^\beta)^2}{2\pi}}$

Could anyone please tell me what does this hint represent? And how can I move next step to find the posterior?

Thank you so so so much.

答:
What you're trying to do is called Maximum A Posteriori Estimation.

Namely, given a prior distribution $f$ over the parameter of interest $\beta$ and a likelihood function $f(h|\beta)$ for the quantity $h$ given $\beta$, we want to compute the most likely value of $\beta$ given our observation $h$. In other terms, we're looking for
$$\hat \beta = \underset{\beta}{\arg \max} \ \ \ f(\beta|h) $$

So, to be clear (using your notations) :

$f(\beta) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(\beta - 2)^2}{2}}$ is the prior distribution over $\beta$, i.e. we make the arbitrary assumption that it is normally distributed with mean $2$ and variance $1$.

$f(h|\beta) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(h - e^\beta)^2}{2\pi}}$ is the likelihood function. It represents (loosely speaking) the probability to observe that value of $h$ when you know the value of $\beta$. Here for instance, $u(1) = e^\beta$ so $h = u(1) + \zeta$ is a standard normal random variable "shifted" by $e^\beta$. In other words, given the value of $\beta$, $h$ is a normal random variable with mean $e^\beta$ and variance $1$, which explains the formula you're given for its distribution.

Finally, $f(\beta |h)$ is the posterior distribution over $\beta$ given $h$. It represents (loosely speaking once again) the probability of the value of $\beta$ given the value of $h$. We can compute it using Bayes's theorem :
$$f(\beta |h) = \frac{f(h|\beta)f(\beta)}{f(h)} \propto f(h|\beta)f(\beta) $$
The $f(h)$ term in the denominator represents the probability distribution of $h$, and we don't know how to compute it. However, it doesn't matter too much since it is independent of $\beta$, so it being here or not won't influence the $\arg\max$.

So we're looking for the value $\hat\beta$ that maximizes the posterior distribution written above, and that is the value that we'll take as our estimate of the true $\beta$. You now have all the elements you need to compute it using standard analysis tools :
$$ \begin{align}\hat \beta &= \underset{\beta}{\arg \max} \ \ \ f(\beta|h) \\
&= \underset{\beta}{\arg \max}  \ \ \ f(h|\beta)f(\beta) \\
&= \underset{\beta}{\arg \max}  \ \ \ \frac{1}{\sqrt{2\pi}}e^{-\frac{(h - e^\beta)^2}{2\pi}} \times \frac{1}{\sqrt{2\pi}}e^{-\frac{(\beta - 2)^2}{2}} \end{align}$$


问:
Floor function identity for $[\sqrt{n^2-n}]$
From my textbook, where [n] is the greatest integer less than or equal to n.

Since $n^2-n=(n-\frac{1}{2})^2-\frac{1}{4}$, it follows, $[\sqrt{n^2-n}]=n-1$

This one feels like a great leap, I really don't see it at all and it doesn't feel like the kind of question where I can say "I tried doing ..."

答:
It's a bit of a joke in the mathematical community that the phrase "it follows" doesn't necessarily give any indication of how elementary the implication is.  Ideally, an author has an appropriate sense of their audience when they say that, but sometimes it takes a wall of algebra to justify a conclusion that an established mathematician would regard as "obvious".  As you grow as a student, you will find your sense of obviousness will expand.

In this case, one presumes that $n$ is positive integer.  Then

$$(n-1)^2=n^2-2n+1\le n^2-n<n^2$$

and since every expression there is positive we can take the square root of everything to get

$$n-1\le\sqrt{n^2-n}<n$$

And the stated implication follows immediately from the definition of the greatest integer function.

What was the author going for with the statement they cited?  I suppose they were saying that $\sqrt{n^2-n}\approx n-\frac12$ as $n$ becomes large.  Frankly, I don't think that's a particularly strong argument since you don't how a priori how the square root behaves before $n$ becomes large, but one assumes that it is "obvious" to the author.


问:
irreducible over $\mathbb R[x,y]$
Let $A = \mathbb R[x,y]$ where $x^2 + y^2 = 1$

Show that $A$ is an integral domain.

My thoughts are to show that $x^2 + y^2 = 1$ is irreducible over  $\mathbb R[x,y] =  \mathbb R[x][y],$ I am guessing that I can use Einstein criterion, is that correct? if so how?

Could someone push me in the right direction please?

EDIT:

I also know that:

1-Irreducibles are primes in a UFD.

2-$\mathbb R[x], \mathbb R[y]$ are UFD because $R$ is a field. And $<x^2 + y^2 -1>$ is a prime ideal in $B[y]$(where B = R[x] and where $B[y]$ is a UFD) iff $<x^2 + y^2 -1>$ is irreducible in $B[y]$

3- $I$ is prime iff $R/I$ is an integral domain.

答:
I may as well make this an answer. The ring $A$ is simply $\mathbb R[X,Y]/(X^2+Y^2-1)$ which is an integral domain because the ideal $(X^2+Y^2-1)$ is prime. It's prime because $\mathbb R[X,Y]$ is an integral domain, so an ideal $(f(X,Y))$ is prime iff $f(X,Y)$ is irreducible. Indeed, $X^2+Y^2-1 = X^2+(Y+1)(Y-1)$ which is Eisenstein at the primes $Y\pm 1$.

A very similar argument shows that the ring $\mathbb R[X,Y]/(X^n+Y^n-1)$ is an integral domain. If you've seen any algebraic geometry, this is the same as the fact that the Fermat curve $V(X^n+Y^n-1)$ is irreducible, i.e. an algebraic variety. This time, write $X^n+(Y-1)(1+Y+\cdots+Y^{n-1})$ which is still Eisenstein at $Y-1$.

Title: irreducible over $\mathbb R[x,y]$


问:
Spivak, Ch 2 Prologue, Problem 13: how to prove $\sqrt{3}$ is irrational using fact that $k^2$ divisible by 3 means k is also divisible by 3?
Problem 13 in Spivak's chapter 2 Prologue is to prove $\sqrt{3}$, $\sqrt{5}$, and $\sqrt{6}$ are irrational with the hint that to treat $\sqrt{3}$ you can use the fact that every integer can be written in the form $3n$, $3n+1$, or $3n+2$.

Let's just consider the $\sqrt{3}$ case, because the other two are analogous. The question here is simply about one step in the proof, namely the part with divisibility by three explained below.

The solution at the back of Spivak for $\sqrt{3}$ is as follows:

$(3n+1)^2 = 9n^2+6n+1 = 3(3n^2+2n) + 1$

$(3n+2)^2 = 9n^2+12n+4 = 3(3n^2+4n + 1) + 1$

From the above, it follows that if $k^2$ is divisible by 3, then $k$ is divisible by 3. Main question: why does this follow?

Assume $\sqrt{3}$ is rational. $\sqrt{3} = p/q$ where $p$ and $q$ have no common factor.
$$p^2=3q^2$$
So $p^2$ is divisible by 3, and therefore $p$ is too. $$p=3p'$$ for some natural number $p'$ (shouldn't it be some integer $p'$?)

Consequently,

$$(3p')^2=3q^2$$
$$q^2=3p'^2$$

Therefore, $q$ is divisible by 3, a contradiction since this means $p$ and $q$ have a common factor.

答:
If a number $k$ is not a multiple of $3$, then it is of the form $3n+1$ or $3n+2$ for some integer $n$, but then those computations show that $k^2$ is a multiple of $3$ plus $1$; therefore, $k^2$ is not a multiple of $3$.

Concerning your other question, if $\sqrt3$ was rational, it could be written as $\frac pq$, with $p,q\in\Bbb N$. And, if $3\mid p$, then $\frac p3(=p')$ would also be a natural number.


答:
It follows because we have shown that $(3n+1)^2$ is of the form $3k+1$, meaning that it cannot be of the form $3k$ (i.e. it cannot be divisible by $3$). The same reasoning applies for $(3n+2)^2$.

Notice that we are assuming that $p,q$ are positive, and since $\frac{p}{3}$ is positive, so is $p'$. It follows because  the natural numbers are the set of positive integers.

Title: Spivak, Ch 2 Prologue, Problem 13: how to prove $\sqrt{3}$ is irrational using fact that $k^2$ divisible by 3 means k is also divisible by 3?

Body:
Problem 13 in Spivak's chapter 2 Prologue is to prove $\sqrt{3}$, $\sqrt{5}$, and $\sqrt{6}$ are irrational with the hint that to treat $\sqrt{3}$ you can use the fact that every integer can be written in the form $3n$, $3n+1$, or $3n+2$.

Let's just consider the $\sqrt{3}$ case, because the other two are analogous. The question here is simply about one step in the proof, namely the part with divisibility by three explained below.

The solution at the back of Spivak for $\sqrt{3}$ is as follows:

$(3n+1)^2 = 9n^2+6n+1 = 3(3n^2+2n) + 1$

$(3n+2)^2 = 9n^2+12n+4 = 3(3n^2+4n + 1) + 1$

From the above, it follows that if $k^2$ is divisible by 3, then $k$ is divisible by 3. Main question: why does this follow?

Assume $\sqrt{3}$ is rational. $\sqrt{3} = p/q$ where $p$ and $q$ have no common factor.
$$p^2=3q^2$$
So $p^2$ is divisible by 3, and therefore $p$ is too. $$p=3p'$$ for some natural number $p'$ (shouldn't it be some integer $p'$?)

Consequently,

$$(3p')^2=3q^2$$
$$q^2=3p'^2$$

Therefore, $q$ is divisible by 3, a contradiction since this means $p$ and $q$ have a common factor.


问:
How can partial derivatives feature in the definition of a function?
I have a map $f(t,g,h)$ where $f:[0,1]\times C^1 \times C^1 \to \mathbb{R}.$

I want to define $$F(t,g,h) = \frac{d}{dt}f(t,g,h)$$
where $g$ and $h$ have no $t$-dependence in them. So $g(x) = t^2x$ would not be admissible if you want to calculate what $F$ is. How do I write this properly? Is it correct to write instead

Define $F$ by $F(t,\cdot,\cdot) = \frac{d}{dt}f(t,\cdot,\cdot)$.

But there is some ambiguity in the arguments. What is the best way to write it?

答:
Since $g$ and $h$ do not depend on $t$ one has
$$F(t,g,h) := \frac{d}{dt}f(t,g,h)=f_{.1}(t,g,h)\ ,$$
where $f_{.1}$ denotes differentiation with respect to the first variable, whatever its name.</p>


问:
Estimating size of minimum vertex cover from randomly sampled subgraph
I would like to estimate the size of the minimum vertex cover (MVC) for a very large graph $G=(V,E)$. To be exact, I would like to estimate the following proportion:

$$p(G)=\frac{|MVC|_G}{|V|}$$

As computing a MVC is quite expensive, I want to sample a subgraph $G_s=(V',E')$ with $n$ vertices from $G$ and compute $p(G')$. Finally, my goal is to estimate $p(G)$ from $p(G')$ with some error and/or confidence depending on $n$.

After a lot of research, I wasn't able to get a convincing result and almost no literature. I linked this problem to the estimation of the independence number but, once again, its estimation from a randomly sampled subgraph is absent.

I'm open on any parameter of the problem and if you have even a small idea of a potentially interesting paper to read, I am all ears. Thank you.

答:
If there exists a vertex cover $U$ of order $|U| = p(G) |V|$, then any subgraph of size $n$ is going to see $p(G) n$ of $U$'s vertices in expectation, and this is tightly concentrated, central limit theorem and all. This means $p(G_s) \lessapprox p(G)$; we can work out what the bounds look like, but they're not the big problem here.

The big problem is that $G_s$ could have a really good vertex cover that does not come from $U$. To give one bad example, suppose $G$ is the cycle $C_N$, with $p(G) = \frac12$. Then a random sample of $n$ vertices contains a fixed edge with probability $\binom{N-2}{n-2}/\binom Nn = \frac{n(n-1)}{N(N-1)}$, so the expected number of edges is $\frac{n(n-1)}{N-1}$.

For $n \ll \sqrt{N}$, we'll see none of the edges of $G$, and get $p(G_s) = 0$, with high probability. By a similar calculation, for $n \ll N^{2/3}$, we'll just see a bunch of isolated edges and isolated vertices; in this range, the expected size of a vertex cover of $G_s$ is just the expected number of edges, and the expected value of $p(G_s)$ is about $\frac{n-1}{N-1}$, which is tiny. We see similar behavior for all $n \ll N$.

Even for $n = \frac12 N$, which is ridiculous as a sampling strategy, most components of $G_s$ are paths of constant length, and for every path of even length or isolated vertex, we reduce the size of the minimum vertex cover. So $p(G_s)$ is off by a constant from $p(G)$; I compute that more precisely, the expected value of $p(G_s)$ is about $\frac13$, but the actual computation is annoying and doesn't matter.


问:
Apollonius circle
I'm given two points, $A$ and $B$, and two lengths, $b$ and $c$.  I need to find the locus of point $C$ such that $BC:AC=b:c$.

This link describes Apollonius circle of first type, but I can't seem to find anywhere how to construct the center of the Apollonius circle.

答:
If you are given the point $D$ that fulfils the ratio on the line between $A$ and $B$, you can find the locus of similar points (the Apollonius circle of first type) through the following construction:

Construct the circle $\color{#0B2}{\Gamma}$ centred on $D$ through the nearer of $A$ and $B$, which I will assume is point $B$. Also construct a line perpendicular to $AB$ at $B$. Then form the tangent to circle $\Gamma$ from $A$. This intersects the perpendicular at $E$ which is on the Apollonius circle. Since $AE$ is in fact tangent to this circle, you can construct the perpendicular at $E$ and the intersect with $AB$ will be the centre of the Apollonius circle $O$ (going though both $D$ and $E$).

Note that $E$ is on the Apollonius circle because $\triangle AEB$ is similar to the right triangle $\triangle ADT$ (with $T$ the tangent point of $A$ on $\Gamma$), so $|AE|:|EB| = |AD|:|DT| = |AD|:|BD|$ as required.

Since $|AE|:|EB| = |AD|:|BD|$, $ED$ bisects $\angle BEA$ giving $\angle BED = \angle DEA$. Also $\triangle DOE$ is isosceles so $\angle OED = \angle ODE$. Thus $\angle OEA = \angle OED + \angle DEA = \angle ODE + \angle BED = \color{red}{\perp}$ due to $\triangle BED$.

Title: Apollonius circle