问:
Moment generating function of Laplace distribution step by step
There are two similar posts but none of them helped me to get through the full derivation of the very simple MGF that should be according to Wikipedia:

$$
\frac {\exp(\mu t)}{1-b^{2}t^{2}}
$$

Here's my attempt:

Definition of MGF
$$
M_X(t) = \mathbb{E}\left[\exp(t X)\right]
$$

by LOTUS

$$
= \int \exp(t x) \cdot p(x) dx
$$

Plugging-in PDF

$$
= \int \exp(t x) \cdot \frac{1}{2b}\exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$

$$
= \frac{1}{2b} \int \exp(t x) \cdot \exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$

$$
= \frac{1}{2b} \int \exp \left(t x - \frac{\mid x - \mu \mid }{b} \right) dx
$$

getting rid of abs. value

$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(t x - \frac{ x - \mu  }{b} \right) dx
+ \frac{1}{2b} \int_{\mu}^{\infty} \exp \left(t x - \frac{- x + \mu  }{b} \right) dx
$$

the left integral first

$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x - \mu  }{b} \right) dx
$$

$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x  }{b} \right) \exp \left(\frac{- \mu  }{b} \right) dx
$$

$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu  }{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x  }{b} \right)  dx
$$

$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu  }{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{ (bt  - 1)  }{b}  x \right)  dx
$$

Now since $\int e^{ax}\,dx={\frac {1}{a}}e^{ax}+C$

$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu  }{b} \right) \cdot \left|  \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  x \right)  \right|_{-\infty}^{\mu}
$$

Let's evaluate the integral first

$$
\left|  \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  x \right) \right|_{-\infty}^{\mu}
$$

$$
= \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)
-
\frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (-\infty)\right)
$$

Since $\lim_{x \rightarrow 0} e^{-x} = 0$, the second term is 0

$$
= \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)
- 0
$$

Double check from WorlframAlpha:

$$
= \frac{b}{bt-1} \exp \left( \mu \left( t - \frac{1}{b} \right) \right)
$$

for $\frac{1}{b} < t$

Now the right-hand side integral:

$$
\frac{1}{2b} \int_{\mu}^{\infty} \exp \left(t x - \frac{- x + \mu  }{b} \right) dx
$$

$$
\frac{1}{2b} \int_{\mu}^{\infty} \exp \left(\frac{b t x + x - \mu  }{b} \right) dx
$$

$$
= \frac{1}{2b} \int_{\mu}^{\infty} \exp \left(\frac{b t x + x   }{b} \right) \exp \left(\frac{-\mu  }{b} \right)dx
$$

$$
= \frac{1}{2b} \cdot \exp \left(\frac{ -\mu  }{b} \right) \int_{\mu}^{\infty} \exp \left(\frac{b t x + x   }{b} \right) dx
$$

$$
= \frac{1}{2b} \cdot \exp \left(\frac{ -\mu  }{b} \right) \int_{\mu}^{\infty} \exp \left( \frac{b t + 1 }{b} x\right) dx
$$

Now evaluate the integral

$$
\int_{\mu}^{\infty} \exp \left( \frac{b t + 1 }{b} x\right) dx
$$

According to wolfram alpha

$$
= - \frac{b}{bt + 1}\exp\left(\mu \frac{1}{b + t}\right)
$$

for $\frac{1}{b} + t < 0$

Now sum-up these two integrals:

$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu  }{b} \right) \cdot \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)  + \\
\frac{1}{2b} \cdot \exp \left(\frac{ -\mu  }{b} \right) \cdot \left( - \frac{b}{bt + 1}\exp\left(\mu \frac{1}{b + t}\right) \right)
$$

And simplify

$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu  }{b} \right) \cdot \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)  - \\
\frac{1}{2b} \cdot \exp \left(\frac{ -\mu  }{b} \right) \cdot  \frac{b}{bt + 1} \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$

$$
= \frac{1}{2b} \cdot  \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{- \mu  }{b} \right) \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)  - \\
\frac{1}{2b} \cdot   \frac{b}{bt + 1} \cdot \exp \left(\frac{ -\mu  }{b} \right) \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$

$$
= \frac{1}{b} \cdot  \frac{1}{ (bt  - 1)  } \cdot \exp \left(\frac{- \mu  }{b} \right) \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)  - \\
\frac{1}{b} \cdot   \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu  }{b} \right) \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$

$$
= \frac{1}{b} \cdot  \frac{1}{ (bt  - 1)  } \cdot \exp \left(\frac{- \mu  }{b} +\frac{ (bt  - 1)  }{b}  (\mu) \right)  - \\
\frac{1}{b} \cdot   \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu  }{b} + \mu \frac{1}{b + t}\right)
$$

$$
= \frac{1}{b} \cdot  \frac{1}{ (bt  - 1)  } \cdot \exp \left( \mu \left(t - \frac{2}{b} \right) \right)  - \\
\frac{1}{b} \cdot   \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu  }{b} + \mu \frac{1}{b + t}\right)
$$

But I cannot move away from this ugly formula to the simple beauty. Is there just a stupid mistake above or am I doing something substantially wrong?

答:
There are a number of efficiencies that can be realized if we first consider the shifted exponential family $$f_Y(y) = \frac{1}{b} e^{-(y - \mu)/b}, \quad y \ge \mu$$ with location $\mu$ and scale $b > 0$.  The MGF is easy to calculate:  $$M_Y(t) = \int_{y=\mu}^\infty \frac{1}{b} e^{ty} e^{-y/b} e^{\mu/b} \, dy = \frac{e^{\mu/b}}{b} \int_{y=\mu}^\infty e^{-(1/b-t)y} \, dy = \frac{e^{\mu/b} e^{-(1/b - t)\mu}}{b(1/b-t)} = \frac{e^{\mu t}}{1 - bt} , \quad t < 1/b.$$  The MGF of $-Y$ is then $$M_{-Y}(t) = M_Y(-t) = \frac{e^{-\mu t}}{1 + bt}, \quad t > -1/b.$$

Now define the mixture density $$f_X(x) = \frac{1}{2} (f_{Y_1}(x) + f_{-Y_2}(x))$$ where $Y_1$ is shifted with location $\mu$, and $Y_2$ is shifted with location $-\mu$, and both share scale $b$.  Then it is easy to see $X$ is Laplace with location $\mu$ and scale $b$.  The MGF of $X$ is then $$M_X(t) = \frac{1}{2} \left( M_{Y_1}(t) + M_{-Y_2}(t) \right) = \frac{1}{2} \left( \frac{e^{\mu t}}{1 - bt} + \frac{e^{-(-\mu)t}}{1 + bt} \right) = \frac{e^{\mu t}}{1 - (bt)^2}, \quad |t| < 1/b.$$

In fact, this computation easily generalizes to the case where $Y_1$ and $Y_2$ do not have the same scale parameter, in which the distribution has nonzero skew.


问:
Using u-substitution on an integral
I have the integral of $$\int_{1}^{3} \dfrac{x^4}{\ln(x)+2}{dx}$$

I don't want to evaluate it, but I'm supposed to simply use u-substitution and let $u = \ln(x)+2$. The problem is that the correct answer is supposed to give you $\int_{2}^{ln(3)+2} \dfrac{e^{5u-10}}{u}{du}$

Now, I do understand everything up to the part of $e^{5u-10}$. When I take $u - 2 = ln(x)$ and take the equivalent of $x^4$ which is $e^{4\ln(x)}$, I get $e^{4(u-2)}$, which is then obviously $e^{4u-8}$ and not $e^{5u-10}$. Is the exercise faulty or am I making some mistake I'm not aware of?

答:
If $u=\ln(x)+2$, then $x=e^{u-2}$ and $\mathrm dx=e^{u-2}\,\mathrm du$. The integral then becomes

$$\int_1^3 \frac{x^4}{\ln(x)+2}\,\mathrm dx=\int_2^{\ln(3)+2}\frac{e^{4(u-2)}}u(e^{u-2}\,\mathrm du)=\int_2^{\ln(3)+2}\frac{e^{5u-10}}u\,\mathrm du$$

The missing link between this and your solution is accounting for how $\mathrm dx$ is transformed.

ContentLicense: CC BY-SA 4.0

Title: Using u-substitution on an integral


答:
If you take $u = \ln(x) +2$, then notice that $du = 1/x \cdot dx$. But, from the first, it follows that $x = e^{u-2}$. Thus, solving for $dx = x \cdot du$ and using that, we have

$$\int \frac{x^4}{\ln(x) + 2} dx = \int \frac{(e^{u-2})^4}{u} \cdot e^{u-2} du$$

Title: Using u-substitution on an integral

Body:
I have the integral of $$\int_{1}^{3} \dfrac{x^4}{\ln(x)+2}{dx}$$

I don't want to evaluate it, but I'm supposed to simply use u-substitution and let $u = \ln(x)+2$. The problem is that the correct answer is supposed to give you $\int_{2}^{ln(3)+2} \dfrac{e^{5u-10}}{u}{du}$

Now, I do understand everything up to the part of $e^{5u-10}$. When I take $u - 2 = ln(x)$ and take the equivalent of $x^4$ which is $e^{4\ln(x)}$, I get $e^{4(u-2)}$, which is then obviously $e^{4u-8}$ and not $e^{5u-10}$. Is the exercise faulty or am I making some mistake I'm not aware of?


问:
A particular weak subadditivity
Given a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, consider the following property.

For all $(x^1, ..., x^n) \in \left(\mathbb{R}^n \right)^n$ such that $f(x^i) \geq 0$ $\forall i \in [1,n]$, and $(\alpha_1, ..., \alpha_n) \in \left( \mathbb{R}_{\geq 0} \right)^n$, there exists $(\beta_1, ..., \beta_n) \in \left( \mathbb{R}_{\geq 0} \right)^n$ such that

$$ f\left( \sum_{i=1}^n \alpha_i x^i \right) \ \leq \ \sum_{i=1}^n \beta_i \ f( x^i ). $$

Let $\Phi$ be the set of functions $\phi$ defined as $\phi(x) := \varphi(x) + a^\top x + b$, with $\varphi$ convex and homogeneous (of some degree), $a \in \mathbb{R}^n$, $b \in \mathbb{R}$. Then any $\phi \in \Phi$ has the above property.

I am looking for a strict superset $\Phi$, i.e. $\Psi \supset \Phi$, such that any $\psi \in \Psi $ has the above "weak subadditivity" property.

答:
Note that your condition is satisfied, if the following implication holds for all $x^i \in \mathbb R^n$, $\alpha_i \in \mathbb R_{\ge0}$:
$$f(x_i) = 0 \text{ for all $i$}\quad\text{then}\quad f\big(\sum \alpha_i \,x^i\big) = 0.$$

Now, we can easily find some functions, that are not in $\Phi$, just take some function, which is everywhere positive.

ContentLicense: CC BY-SA 3.0

Title: A particular weak subadditivity


问:
f is differentiable with conditions. Show that f has a subsequence converging to a continuous function.
Let $f: \mathbb R \to \mathbb R$ be differentiable. Assume that $|f(x)| \leq 1, |f'(x)| \leq 1 \ \forall x \in \mathbb R$ and that $f(0) = 0$. Let $\{a_n\}$ be a sequence of non-zero real numbers. Prove that the sequence of functions $g_n(x) = \frac{1}{a_n} f(a_n x)$ has a subsequence converging to a continuous function.

I know that the subsequence exists because we have a bounded function. However, we are given a specific subsequence here. How do I show this is a subsequence that converges? Is there a good way to approach this? Or maybe showing that $g_n(x)$ is equicontinuous and then using Arzela-Ascoli? Any tips to do this one would be great!

答: $g_n(x)=\frac 1  {a_n}\int_0^{a_nx} f'(t)dt=\int_0^{x}f'(a_ns) ds$. This gives $|g_n(x)-g_n(y)| \leq |x-y|$ and Arzela -Ascoli Theorem gives the answer.


问:
Help to show the next proposition
Let $f:[a,b] \rightarrow R$ be a bounded function in $[a,b]$ such that $f$ is equal to zero except for a finite number of points in $[a,b]$. Show that $f$ is integrable and that its integral is zero.

I attempted the solution by induction over the number of points in which $f \neq 0$. Starting with only one point, and getting that it works for n number of points.

Proof: By induction.

Base case: Suppose that $f(x) = 0\; \forall x \neq c$ with c being a point in $[a,b]$.

But I don't know how to continue with the proof.

答:
Let $a < x_1 < x_2 < \dots < x_n < b$ be the $n$ distinct points where $f \ne 0$. (Although $a = x_1$ and $x_n = b$ are possible, just take $\hat{a} < a$ and $\hat{b} > b$ and consider the same problem on $[\hat{a}, \hat{b}]$.) Without loss of generality, assume $f \geq 0$. Let $C = \sum_{i=1}^{n} f(x_i)$.

We need to show that $L(f) = U(f)$, where $L(f) = \sup_P L(f,P)$ and $U(f) = \inf_P U(f,P)$.

First, let's show that $L(f) = 0$. To this end, let $P = \{a = y_0 < y_1 < \dots < y_m = b \}$ be any partition of $[a,b]$. Then $\inf\{f(x) : y_i \leq x \leq y_{i+1}\} = 0$ for each $i$, so $L(f,P) = 0$; hence, $L(f) =0$.

Conversely, let $\varepsilon > 0$ be sufficiently small. Consider the partition
$$
P_\varepsilon = \{ x_i \pm \varepsilon : i = 1, \dots, n \}.
$$
Then
\begin{align}
  U(f, P_\varepsilon) &= \sum_{i=1}^{n} \sup\{f(x) : |x-x_i| \leq \varepsilon \} \cdot 2\varepsilon \\
  &= \sum_{i=1}^{n} 2\varepsilon f(x_i) \\
  &= 2 C \varepsilon,
\end{align}
so $U(f) = \inf_P U(f,P) = \lim_{\varepsilon \to 0} U(f,P_\varepsilon) = 0$.

Since $U(f) = L(f)$, it follows that $f$ is Riemann integrable. Moreover, $\int_a^b f = 0$.

Title: Help to show the next proposition


问:
The attractiveness of the fixed points in system $z \mapsto z + \xi(\frac{1}{2} + i z)$
In essence, I need to determine when a number is within this set: $$\Big\{ t : t \in \mathbb{R} \;\wedge\; \Big|1+ i\xi'\big(\frac12 + i t\big)\Big| < 1 \Big\},$$ where $\xi'$ is the first derivative of the Riemann xi function. This is assuming RH so it's just the critical strip.

Is there a way of simplifying this equation into something more manageable? Or is there a better way of determining this altogether?

答:
On $|\Re(s)-1/2|\le 1/4, |\Im(s)|> 1$

$|\zeta(s)|\le 2|s|$ this follows from $\zeta(s)=\frac{s}{s-1}+s\int_1^\infty(\lfloor x\rfloor-x)x^{-s-1}dx$.

$\Gamma(s/2)$ is rapidly decreasing, this is because $|\Gamma(s/2)\prod_{k=0}^{K-1}(s/2+k)|\le \Gamma(1/2+K)$

From there you get some $C=16$ such that $|\xi(s)| \le C /|s|$. This implies by the Cauchy integral formula that $|\xi'(s)|\le 32 C/|s|$.

$\xi(1/2+it)$ is real: this follows from the functional equation that $\Gamma(s/2)\pi^{-s/2}\zeta(s)$ is invariant under $s\to 1-s$ and is complex conjugated under $s\to \overline{s}$, thus $i\xi'(1/2+it)$ is real.

$\xi(1/2+it)$ changes of sign infinitely often (this is a bit less obvious) thus $i\xi'(1/2+it)$ changes of sign infinitely often. Assuming that there are  no double zeros, exactly half of non-trivial zeros $1/2+it, t> 32C$ are attractive fixed points of $s\to s+\xi(s)$. Assuming the RH there are no others in the critical strip.


问:
How to check if two data series change together?
I have a project which involves two time series of pings. One is the ping of google.com the other is the ping of my server. I'd like to compare these two time series and send an alert when the ping of my server reaches a certain threshold for a certain duration. I don't want to send an alert when it is just a short spike. I don't want to send alert either when both pings increases together, because that means it is a general network error of the monitoring tool and not that the problem is with my server. The first problem is more or less trivial, but I don't know how I should approach the second one. My first thought was correlation, but I just checked it with a few time series and it appears to be too sensitive to short spikes and jitter. Is there an algorithm that would solve this problem better?</p>

答: I checked meanwhile and I was wrong. A 5 point moving correlation appears to work in the beginning of the anomalies. Later it falls apart when we reach a plateau, but that's not a big deal. It works for short anomalies too except when there is a delay in time. So I'll us that to decide if the two series change together and I'll use a 5 point moving median to decide if we have a short or a long anomaly.</p>


问:
Notions of continuity for stochastic processes
I would like to receive some clarification regarding the difference between continuous in probability and continuous almost surely. Using the definition of the wikipedia page (that match the one I have seen in other references), we have

Continuous in probability
for all $\varepsilon>0$
\begin{equation}
\lim_{s\rightarrow t} \mathbb{P}\left(\left\{\omega \in \Omega : |X_s(\omega) - X_t(\omega) | \geq \varepsilon\right\} \right)= 0
\end{equation}

and

Continuous with probability 1 (almost surely)
\begin{equation}
\mathbb{P}\left(\left\{\omega \in \Omega : \lim_{s\rightarrow t} |X_s(\omega) - X_t(\omega) | = 0 \right\} \right)= 1
\end{equation}

Now, I can see that the second condition is stronger than the first one, but to me they seem analogous, since the first should be valid for arbitrarily close $\varepsilon$. Obviously, my intuition is wrong, but I cannot understand how the difference can be significant. Could you please explain how the two definitions give raise to different processes? Thank you!

答:
1. We first examine a much easier variant, i.e., discrete-time process. Let $Y = (Y_n)_{n\in\mathbb{N}_1}$ be a sequence of independent random variables such that

$$ \mathbb{P}(Y_n = 1) = \frac{1}{n} \qquad\text{and} \qquad \mathbb{P}(Y_n = 0) = 1 - \frac{1}{n}. $$

Then it is clear that, for each $\epsilon > 0$,

$$ \lim_{n\to\infty} \mathbb{P}( \left| Y_n - 0 \right| > \epsilon)
\leq \lim_{n\to\infty} \mathbb{P}( Y_n \neq 0)
= \lim_{n\to\infty} \frac{1}{n}
= 0. $$

So $Y_n \to 0$ in probability. However, $\sum_{n=1}^{\infty} \mathbb{P}(Y_n = 1) = \infty$, and so,  by the second Borel-Cantelli's lemma,

$$ \mathbb{P}(Y_n = 1 \text{ infinitely often}) = 1. $$

So it follows that

$$ \mathbb{P}\Bigl(\limsup_{n\to\infty} Y_n = 1\Bigr) = 1
\qquad\text{and}\qquad
\mathbb{P}\Bigl(\liminf_{n\to\infty} Y_n = 0\Bigr) = 1. $$

So $Y_n$ does not converge as $n\to\infty$ with probability one.

Here, the difference between two notions is that the probability $\mathbb{P}(\left|Y_n - 0\right| > \epsilon)$ describes what happens to a 'typical observer $\omega$', and telling that the exceptional event occurs only to a vanishing fraction of observers for each large $n$. However, for a 'fixed observer $\omega$', this exceptional event can happen infinitely often.

2. Now coming back to continuous-time case, we can borrow the above example as follows:

Let $Y=(Y_n)_{n\in\mathbb{N}_1}$ be as above, and define $X = (X_t)_{t\geq 0}$ by

$$ X_t = \begin{cases}
Y_{\lceil 1/t \rceil}, & \text{if } t > 0, \\
0, & \text{if } t = 0.
\end{cases} $$

Then $X_t \to 0$ as $t \to 0$ in probability, whereas $X_t$ does not converge as $t \to 0$ almost surely.


问:
How could I write the following expression without expanding it?
in a induction proof I stumbled over the following situation:

$$\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\frac{1}{n^{2n}}\left(-m\right)^k\left(n^2+nm+n+m\right)^{n-k}$$

The degree of the numerator is always smaller as the degree of the denominator.. so this limit goes to zero for every piece of the sum..

(I expanded it for some terms to check if this statement is indeed true or atleast should be)

How could I rewrite the sum, so that its clean and visually pleasing to see that all the pieces of the sum go to zero?

Thanks alot :)

Edit:

I came from here:

$\lim\limits_{n\longrightarrow\infty}\left(1+\frac{m+1}{n}\right)^n=\lim\limits_{n\longrightarrow\infty}\left(\frac{(n+1)(n+m)}{n^2}-\frac{m}{n^2}\right)^n=\lim\limits_{n\longrightarrow\infty}\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)-\frac{m}{n^2}\right)^n$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=0}\binom{n}{k}\left(-\frac{m}{n^2}\right)^k\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^{n-k}$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\left(-\frac{m}{n^2}\right)^k\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^{n-k}+\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^n$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\left(-\frac{m}{n^2}\right)^k\left(\frac{n^2+nm+n+m}{n^2}\right)^{n-k}+\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^n$

$=\lim\limits_{n\longrightarrow\infty}\sum\limits^n_{k=1}\binom{n}{k}\frac{1}{n^{2n}}\left(-m\right)^k\left(n^2+nm+n+m\right)^{n-k}+\left(\left(1+\frac{1}{n}\right)\left(1+\frac{m}{n}\right)\right)^n$

答:
$n^2+nm+n+m
=n(n+m)+n+m
=(n+m)(n+1)
$
so the sum is

$\sum\limits^n_{k=1}\binom{n}{k}\frac{1}{n^{2n}}\left(-m\right)^k\left((n+m)(n+1)\right)^{n-k}\\
=\sum\limits^n_{k=1}\binom{n}{k}(-1)^k\left(\frac{m}{n^2}\right)^k\left(\frac{(n+m)(n+1)}{n^2}\right)^{n-k}\\
=\sum\limits^n_{k=1}\binom{n}{k}(-1)^k\left(\frac{m}{n^2}\right)^k\left((1+m/n)(1+1/n)\right)^{n-k}
$

Are these any better?


问:
Curve lifting property
I'm reading forster's riemann surfaces book and I read this theorem: if we have a covering map between two topological spaces X,Y then it has curve lifting property, now I want to make a counter example for the inverse,but I couldn't find. Why canonical injection on open unit disk doesn't work ?could anyone help me with this problem?</p>

答: Let $p : S^1 \times \mathbb R \to S^1$ denote the projection. Then each curve $u : [0,1]  \to S^1$ has a lift, for example $\tilde u(t) = (u(t),0)$. But $p$ is not a covering map because the fibers $p^{-1}(z) = \{z\} \times   \mathbb R$ are not discrete.

Title: Curve lifting property


问:
$E(M_{n\wedge \tau}) = E(M_0)$ implies $M$ is a martingale
$(M_n)_{n\geq 0}$ is an adapted family such that $E|M_n|<\infty$, then, I need to show that $E(M_{n\wedge \tau}) = E(M_0)$ implies $M$ is a martingale, for every stopping $\tau$.

Honestly, I do not know how to start. Any hint? Thanks

答:
Consider $A \in \mathcal{F}_k$, we need to show that $\mathbb{E}[\chi_A \cdot M_k] = \mathbb{E}[\chi_A \cdot M_{k+1}]$.

Consider $\tau = k + \chi_{A}$, this is clearly a stopping time. Therefore $\mathbb{E}[M_\tau] = \mathbb{E}[M_0] = \mathbb{E}[M_k]$, i.e. $$ \mathbb{E}[M_k] = \mathbb{E}[M_k \chi_{A^c} + M_{k+1}\chi_A]$$ which implies our thesis.


问:
Earth Quake Question Logarithmic Type Problem
An earthquake off the coast of Vancouver Island was measured at 8.9 on the Richter Scale
and an earthquake off the coast of Alaska was measured at 6.5. How many times more intense, to the nearest whole number, was the earthquake off the coast of Vancouver Island than the one off the coast of Alaska?</p>

答:
Hint: $M=log(I/J)$, where M is the magnitude of the quake, I is the earthquake's intensity and J is the intensity of a standard earthquake.

Hint #2: Try converting it to exponential form: $10^{M}$=(I/J), because $\log_a x=b\implies a^b=x$.

Since you know know M, evaluate $10^{M}$. That value is equal to (I/J), so then you find (I/J) for each of the respective quakes. Hopefully that should help!


问:
Doubts on some combinatory problems.

You have 3 literature books, 2 informatics books and 1 math book. How many ways can you order them if you want the literature books close to the math ones?
The solution is: $2\cdot 3!\cdot 3!$. Why is that? For me it is $2\cdot 2$ if I want that the literature books remain compact, whereas $2\cdot 4!$ if that doesn't matter.

10 copies of a book are distributed among 5 schools. How many ways can we do that if each school receives at least one book? Solution: combination with repetition if the books are equal, $5!\cdot 5^5$ if they are different.

How many ways can we assign 8 professors to 4 different schools if each school receives at least one professor? Solution: $4^8-4\cdot 3^8-6\cdot 2^8-4$.

Which is the difference between the problems 2 and 3?

Some suggestions on the first problem?

答:
I can reverse engineer a solution to problem number 1.

First, the 3 literature books can be permuted in $3!$ ways.

Then, there are exactly $2$ slots inside the group of 3 literature books for the math book to be placed.

Once the math book is placed inside of the three literature books, you can construe the 4 books as if they were fused together into 1 unit.

Then, you have the 1 unit-fused, and 2 informatics books that each represent a separate unit.

Thus, at the end, there are $3!$ ways of permuting these 3 units.

This explains the computation of $3! \times 2 \times 3!.$

Title: Doubts on some combinatory problems.


问:
Dominated probability distributions on the real line example
I am familiar with the dominance and density on measure spaces but I am getting confused when putting it in probability distribution on the real line. I went through material in Meaning of a probability distribution being dominated by a measure but still do not understand fully on what this represents. Is there any examples for dominated probability distribution on the real line that could clarify this matter to me?

Thank you in advance!

答:
Any probability distribution with a probability density function is dominated by Lebesgue measure.  That is, if $P$ is specified by a formula $P(A)=\int_Af(x)\,dx$, for some non-negative function $f$ for which $\int_{\mathbb R}f(x)\,dx=1$, then $P$ is dominated by Lebesgue measure.  Thus, the many famous named probability distributions (the Gaussian, the Cauchy, the chi-squared, etc) with density functions are all dominated by Lebesgue measure.

One way to understand the Radon-Nikodym theorem is that it provides the converse to this assertion: all probability measures dominated by Lebesgue measure have densities.

(This is slightly different (as discussed in comments) from the condition that $P$ be continuous, that is, that the function $x\mapsto P((-\infty,x])$ is continuous.  If $P$ is dominated by Lebesgue measure it is continuous, but there exist examples of continuous distributions without density functions.)


问:
What is meant by a flat morphism of relative dimension n?
I am trying to learn about etale morphisms, and one definition that is given often is a smooth morphism of relative dimension $0$. I know that relative dimension can be defined generally in the case of flat morphisms, but there seems to be some confusion in the literature about the definition. Can anyone tell me definitively which of the following is actually meant by "relative dimension $n$":

Every non-empty fiber has dimension $n$, every non-empty fiber has pure dimension $n$, every fiber has dimension $n$, or every fiber has pure dimension $n$.

I have noticed some sources do say literally every fiber, but then a corollary of this is that any flat morphism is surjective, since otherwise some fibers would be empty. So it seems more likey that it is only referring to non-empty fibers.

答: At risk of inviting Cunningham's law down on myself, the definition I'm used to, have seen in the literature, and think is the "usual one" is the statement that a morphism of schemes $f:X\to Y$ is of relative dimension $d$ iff for all $y\in Y$, the fiber $X_y$ is equidimensional of dimension $d$. That is, all irreducible components of $X_y$ are of dimension $d$ for any $y$. In particular, we allow empty fibers since they have no irreducible components (an irreducible topological space is nonempty).


问:
How to solve a probability problem with 3 random variables?
I've never done a question with 3 random variables before. Does anyone have an idea of how to start this problem, or what equation to use, or the topic I should search up to learn it? Thanks.

Edit:
I got something like this, but dont'w know to to advance any further. any suggeestion?

or am i doing this wrong?

答:
Comment. You have not shown any of your own work on this problem. I will try to get you started with parts (a) and (b), hoping you can then do (c) and (d) on your own.

To keep your question from being downvoted and closed, please edit it to show some of your preliminary results.

Because $A$ and $X$ both take only two values, there are (at most) four possible values for $AX.$ How about staring by listing the four possibilities and using independence to evaluate the probabilities of each. Then the normal 'fade' will cause variability around each possible value. Please edit any success you have into your question.

The simulation in R below with, $p=.6,$ of many realizations of $Y$ will give you close estimates
of $E(Y), Var(Y),$ and the distribution of $Y.$ You can compare simulation results
with your analytical results. [You can expect about two-place accuracy from the simulation.]

```<code>set.seed(2010)
m = 10^6
x = rbinom(m, 1, .6)
a = sample(c(-1,1), m, rep=T)
n = rnorm(m)
y = a*x + n
summary(y); sd(y)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
-5.759375 -0.877927  0.000663  0.001367  0.878451  5.484803
[1] 1.265227   # aprx SD(Y) = 1.2649
sqrt(1.6)
[1] 1.264911

var(a*x)
[1] 0.6004036  # aprx Var(AX)
var(n)
[1] 1.001982   # aprx Var(N)
table(x*a)/m

      -1        0        1
0.300690 0.399596 0.299714
</code>```

It seems that $AX$ takes values $-1, 0, 1$ with probabilities $0.3, 0.6, 0.3,$
respectively.

```<code>hist(y, prob=T, col="skyblue2")
 curve(dnorm(x, mean(y), sd(y)), add=T, col="orange", lwd=2)
</code>```

Looking at the histogram, along with the density function of a normal
distribution with the appropriate mean and variance, one might guess
that $Y$ is normally distributed. According to a Shapiro-Wilk test,
the first 5000 simulated values of $Y$ are consistent with a random
sample from a normal distribution. [The S-W test in R will handle at
most 5000 observations.]

```<code>shapiro.test(y[1:5000])

        Shapiro-Wilk normality test

data:  y[1:5000]
W = 0.99953, p-value = 0.2663
</code>```

Body:
Comment. You have not shown any of your own work on this problem. I will try to get you started with parts (a) and (b), hoping you can then do (c) and (d) on your own.

To keep your question from being downvoted and closed, please edit it to show some of your preliminary results.

Because $A$ and $X$ both take only two values, there are (at most) four possible values for $AX.$ How about staring by listing the four possibilities and using independence to evaluate the probabilities of each. Then the normal 'fade' will cause variability around each possible value. Please edit any success you have into your question.

The simulation in R below with, $p=.6,$ of many realizations of $Y$ will give you close estimates
of $E(Y), Var(Y),$ and the distribution of $Y.$ You can compare simulation results
with your analytical results. [You can expect about two-place accuracy from the simulation.]

set.seed(2010)
m = 10^6
x = rbinom(m, 1, .6)
a = sample(c(-1,1), m, rep=T)
n = rnorm(m)
y = a*x + n
summary(y); sd(y)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
-5.759375 -0.877927  0.000663  0.001367  0.878451  5.484803
[1] 1.265227   # aprx SD(Y) = 1.2649
sqrt(1.6)
[1] 1.264911

var(a*x)
[1] 0.6004036  # aprx Var(AX)
var(n)
[1] 1.001982   # aprx Var(N)
table(x*a)/m

      -1        0        1
0.300690 0.399596 0.299714

It seems that $AX$ takes values $-1, 0, 1$ with probabilities $0.3, 0.6, 0.3,$
respectively.

hist(y, prob=T, col="skyblue2")
 curve(dnorm(x, mean(y), sd(y)), add=T, col="orange", lwd=2)

Looking at the histogram, along with the density function of a normal
distribution with the appropriate mean and variance, one might guess
that $Y$ is normally distributed. According to a Shapiro-Wilk test,
the first 5000 simulated values of $Y$ are consistent with a random
sample from a normal distribution. [The S-W test in R will handle at
most 5000 observations.]

shapiro.test(y[1:5000])

        Shapiro-Wilk normality test

data:  y[1:5000]
W = 0.99953, p-value = 0.2663


答: The relevant result is that if $p: M\to N$ is a local homeomorphism between manifolds and $p$ has the curve-lifting (usually called path-lifting) property, then $p$ is a covering map. You can find a proof for instance in do Carmo's book "Geometry of curves and surfaces." In the context of Riemann surfaces $M, N$, you probably will be using a holomorphic map $p: M\to N$ that has nowhere vanishing derivative.

Title: Curve lifting property

Body: I'm reading forster's riemann surfaces book and I read this theorem: if we have a covering map between two topological spaces X,Y then it has curve lifting property, now I want to make a counter example for the inverse,but I couldn't find. Why canonical injection on open unit disk doesn't work ?could anyone help me with this problem?</p>


问:
About the substitution in the integration
$$\int_0^1 2t \cdot \sqrt{1 + 4t^2}dt$$

I see the answer is $\frac{5\sqrt{5} -1}{6}$ but I'm not sure how we got here. Can someone remind me how to do this? My book gets right to this answer but I'm not sure what steps there were done and I haven't done integration in quite some time. Can someone show me the steps?

答:
Let
$$I = \int 2t \cdot \sqrt{1 + 4t^2}dt = 2\int t \cdot \sqrt{1 + 4t^2}dt$$
Substitute $ u=4 t^{2}+1$  and we get $d u=8 t d t$, thus
$$
I =\frac{1}{4} \int \sqrt{u} d u  =\frac{u^{3 / 2}}{6}+C=\frac{1}{6}\left(4 t^{2}+1\right)^{3 / 2} + C
$$
From Newton-Leibniz Formula, we get
$$
\int_0^1 2t \cdot \sqrt{1 + 4t^2}dt  = \frac{1}{6}\left(4 t^{2}+1\right)^{3 / 2}\mid^{1}_0 =\frac{5\sqrt{5} -1}{6}
$$
As desired.


问:
Are the elements of a Lie algebra separated by its finite-dimensional representations?
Let $\mathbb{k}$ be some field and let $\mathfrak{g}$ be a $\mathbb{k}$-Lie algebra.

Does there exist for every nonzero element $x$ of $\mathfrak{g}$ a finite-dimensional representation of $\mathfrak{g}$ on which $x$ acts nonzero?

In other words, do the finite-dimensional representations of $\mathfrak{g}$ separate the elements of $\mathfrak{g}$?

If $\mathfrak{g}$ is finite-dimensional, then this is true by Ado’s theorem.

As pointed out in an answer to a similar question, the finite-dimensional representations of $\mathfrak{g}$ even separate the points in the universal enveloping algebra $\operatorname{U}(\mathfrak{g})$ if $\mathfrak{g}$ is finite-dimensional and $\mathbb{k}$ is of characteristic zero.
(This seems to be Theorem 2.5.7 in Dixmier’s Enveloping Algebras.)

Both of the above arguments show even stronger assertions, but also need some additional assumptions and quite a bit of work.

答:
The Wikipedia says there are no finite-dimensional representations of (non-trivial) affine Lie algebras, which implies the desired result is false:
https://en.wikipedia.org/wiki/Affine_Lie_algebra</p>


问:
How can I prove that some multiple of 1234 has (in its base 10 expansion) at most 1234 digits, each of which is either 5 or 0?
I really don't know how to approach this problem. It honestly seems like I could just say "Hey, look, I have a 0. If I multiply 1234 by 0, I get 0!" This answer proves that there is SOME multiple of 1234, it has at most 1234 digits and each digit is either 5 or 0. Boom, proven. I know that this isn't an appropriate answer though, so what would be?</p>

答: By Fermat's little theorem, $1234=2\times617$ divides $\dfrac{10^{616}-1}9\times50$.


问:
Proving that $Z_n=\frac{p_b(X_1,\dots,X_n)}{p_a(X_1,\dots,X_n)}$ is a martingale iff $a=c$
Problem: For a parameter $a\in(0,1)$ let $p_a(x_1,\dots,x_n)$ denote the probability that the first $n$ terms in the i.i.d. sequence $X_1,X_2,\dots$ of $\operatorname{Bernoulli}(a)$ random variables
are exactly $x_1,\dots,x_n$. Let $a\ne b$ and
$$Z_n=\frac{p_b(X_1,\dots,X_n)}{p_a(X_1,\dots,X_n)}$$
where $X_1,X_2,\dots$ are i.i.d. $\operatorname{Bernoulli}(c)$ random variables. Prove that $Z_n$ is a martingale if and only if $c=a.$

My Attempt: We first need to compute the functions $p_a$ and $p_b$, which are the joint pmf's of $n$ Bernoulli$(a)$ and $n$ Bernoulli$(b)$ random variables, respectively. But this is not hard, thanks to independence. In particular, we have
$$p_a(x_1,\dots,x_n)=a^{x_1+\cdots+x_n}(1-a)^{n-(x_1+\cdots+x_n)},$$
$$p_b(x_1,\dots,x_n)=b^{x_1+\cdots+x_n}(1-b)^{n-(x_1+\cdots+x_n)}.$$
Therefore, we have
$$Z_n=\frac{b^{X_1+\cdots+X_n}(1-b)^{n-(X_1+\cdots+X_n)}}{a^{X_1+\cdots+X_n}(1-a)^{n-(X_1+\cdots+X_n)}},$$
where $X_1,\dots,X_n$ are i.i.d. Bernoulli$(c)$ random variables. Now we need to show that the martingale property $E[Z_{n+1}\mid\mathcal F_n]=Z_n$ is satisfied if and only if $c=a$. Using the properties of conditional expectation and the independence of the random variables,
\begin{align*}
E[Z_{n+1}\mid\mathcal F_n]
&=E\left[\frac{b^{X_1+\cdots+X_{n+1}}(1-b)^{n+1-(X_1+\cdots+X_{n+1})}}{a^{X_1+\cdots+X_{n+1}}(1-a)^{n+1-(X_1+\cdots+X_{n+1})}}\Bigg\vert\mathcal F_n\right]\\
&=\frac{b^{X_1+\cdots+X_n}(1-b)^{n-(X_1+\cdots+X_n)}}{a^{X_1+\cdots+X_n}(1-a)^{n-(X_1+\cdots+X_n)}}E\left[\left(\frac{b}{a}\right)^{X_{n+1}}\left(\frac{1-b}{1-a}\right)^{1-X_{n+1}}\right].\\
\end{align*}
Computing the expectation above, it follows that $Z_n$ is a martingale if and only if we have
$$\frac{(1-b)(1-c)}{1-a}+\frac{bc}a=1.$$
It follows that we must have $a=c.$

Do you agree with my work above?

Any help is much appreciated. Thank you for your time.

答:
You are correct.  Also, you can simplify the argument a little, and should be a little more explicit at the conclusion.

First, since $X_1,X_2,\dots$ are independent, we have that

$$Z_{n+1} = Z_n \frac{p_b (X_{n+1})}{p_a (X_{n+1})}.$$

Therefore $$E[Z_{n+1} | {\cal F}_n ] = Z_n E [ \frac{p_b(X_{n+1})}{p_a(X_{n+1})}].$$

The expectation on the righthand side is equal

$$(*) \quad  \frac{b}{a} c + \frac{1-b}{1-a} (1-c),$$

and therefore this is a martingale if and only if $(*)=1$.

Fixing $a$ and $b$, $(*)$ is a convex combination of $\frac{b}{a}$ and $\frac{1-b}{1-a}$. As one of the two numbers is necessarily $> 1$ and the other is necessarily $<1$, there exists a unique convex combination of the two giving $1$. Since the choice of $c=a$ gives $1$, this is the only solution.


问:
Does $1^0$, as a limiting form of the $n$ $n$-th roots of unity $1^{1/n}$, represent every complex value on the unit circle?
In the context of algebraic numbers, the $n$-th root has $n$ solutions in the complex plane. The equations look like: $1=x^n$, which gives $1^{1/n}=x$ (which can be rotated around the complex plane to generate all the solutions).  And then, if we let $n$ become large, we get many solutions. But then, when we let $n$ become arbitrarily large, we get $1^0 = x$, which only has one solution, namely $1$, because anything to the zeroth power gives $1$. This seems discontinuous to me, and it confuses me.

Does anyone know what I'm missing?

I have a feeling that perhaps $1^0$ actually is every value around the complex unit circle, which would be the natural extension of the limiting process. But, if that's true, there are tons of implications that don't seem right to me.

Maybe the following two facts are useful? I'm just spitballling on these two.

This reminds me of the fact that $a*0=0$ has infinitely many solutions.

And perhaps the following fact is useful: $2^0 = 2 ^ {100 * 0} =  (2 ^ {100})  ^ 0$

答:
The $n$th roots of $1$ are
$$
\exp\left(\frac{2\pi i k}{n}\right), \quad k \in \mathbb Z
$$
If $e^{i\theta}$ is any point on the unit circle, then as $n \to \infty$ we can choose $k_n \in \mathbb Z$ so that $\frac{2\pi k_n}{n} \to \theta$.  That is: if, for each $n$, we choose an appropriate $n$th root of $1$, then those roots converge to $e^{i\theta}$.


问:
Double integration problem. Am I setting this up wrong?
$$\iint_D xy dA$$ where $D$ is the region bounded by the line $y = x - 1$ and the parabola $y^2 = 2x + 6$. EValuate from left to right is the hint.

I drew this picture:

I set up my double integral this way:

$$\int_{-1}^5 \int_{-2}^{4} (y+1)(y) dydx$$

I think that's wrong but I'm not sure why.

I am told the right way to set this up is:

$$\int_{-2}^{4} \int_{(y^2-6)/2}^{1+y}xydxdy$$ is the right way to set this up but I'm not sure why. CAn someone visually explain?

答: Yoy need to set up the limits using the equations of the functions.  Using your picture the $x$ integral is $\frac{y^2-6}{2}$ to $y+1$, then the $y$ integral $-2$ to $4$.  Your limits are for a rectangle.


问:
A generalization of SLLN: Convergence of combination of iid variables
Problem: Let $(\varepsilon_i)_{i\geq 1}$ be a sequence of iid random variables. Let $(x_i)_{i\geq 1}$ be a sequence of real numbers.
Assume that:

$\mathbb E[|\varepsilon_1|]<\infty$ and $\mathbb E[\varepsilon_1]=0$.

$\frac{1}{n} \sum_{i=1}^n (x_i)^2$ is bouned for all $n$.

Prove that: $\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^n x_i \varepsilon_i=0$ with probability $1$.

Consequence: If we choose $x_i=1$ for all $i$. Then (loosely speaking) above problem recovers the strong law of large number.

My question: My teacher suggests that I can prove the problem using Kronecker's lemma. But, I have no idea how to apply it. Could anyone give me a hint. Thank you in advance!

Edit: Based on the useful comments below, the first condition is redundant. I mean: If the expectation is defined as Lebesgue Integral, then we have: $\mathbb E[\varepsilon_1]=0$ implies $\mathbb E[|\varepsilon|]<\infty$.

Therefore, the first condition should be rewritten: "$\mathbb E[\varepsilon_1]=0$ in the sense of Lebesgue Integral".

It is important to note that, the problem does not hold for Cauchy distribution. Indeed, if $\varepsilon_1$ has Cauchy distribution, then $\mathbb E[\varepsilon]$ is undefined in the sense of Lebesgue Integral (see mean of Cauchy).

答:
The assumption involves the sum of squares of $x_i$, which would naturally appear if we look at the variance of partials sums of $x_i\varepsilon_i$; however, this is not allowed because $\varepsilon_i$ do not necessarily have a finite moment of order two. This issue will be overcame in the following way.

Step 1: it suffices to prove that
$$
2^{-N}\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{i=1}^nx_i\varepsilon_i\right\rvert\to 0 \mbox{ a.s.}.
$$
Step 2: in order to do that, we use truncation: for a fixed $N$, let
$$
\varepsilon_{i,\leqslant}^N:=\varepsilon_i\mathbf{1}_{\{\lvert\varepsilon_i\rvert\leqslant 2^N \}}-\mathbb E\left[\varepsilon_i\mathbf{1}_{\{\lvert\varepsilon_i\rvert\leqslant 2^N \}}\right],
$$
$$
\varepsilon_{i,\gt}^N:=\varepsilon_i\mathbf{1}_{\{\lvert\varepsilon_i\rvert> 2^N \}}-\mathbb E\left[\varepsilon_i\mathbf{1}_{\{\lvert\varepsilon_i\rvert>2^N \}}\right].
$$
Then $\varepsilon_i=\varepsilon_{i,\leqslant}^N+\varepsilon_{i,\gt}^N$ and it thus suffices to prove that
$$\tag{1}
Y_{N}:=2^{-N}\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{i=1}^nx_i\varepsilon_{i,\leqslant}^{N}\right\rvert\to 0 \mbox{ a.s. and }
$$
$$\tag{2}
Z_N:=2^{-N}\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{i=1}^nx_i\varepsilon_{i,\gt}^{N}\right\rvert\to 0 \mbox{ a.s.}.
$$
Step 3: For (1): we show that $\sum_{N\geqslant 1}\mathbb E\left[Y_N^2\right]$ is finite. This follows from Kolomorov's maximal inequality, inequality
$$
\mathbb E\left[\left(\varepsilon_{i,\leqslant}^N\right)^2\right]\leqslant 2
\mathbb E\left[\left(\varepsilon_i\mathbf{1}_{\{\lvert\varepsilon_i\rvert\leqslant 2^N \}}\right)^2\right],
$$
the fact that $\varepsilon_i$  have the same distribution, the assumption on $x_i$ and that $\sum_{N\geqslant 1}2^{-N}\mathbf{1}_{\{1\leqslant \lvert \varepsilon_1\rvert\leqslant 2^N\}}\leqslant \frac 1X\mathbf{1}_{1\leqslant \lvert \varepsilon_1\rvert}$.

Step 4: in order to prove (2), we first notice that $2^{-N}\sum_{i=1}^{2^N}\lvert x_i\rvert \mathbb E\left[\varepsilon_i\mathbf{1}_{\{\lvert\varepsilon_i\rvert>2^N \}}\right]\to 0$. By the Borel-Cantelli lemma, we are reduced to prove that for each $\delta$, $$\sum_N\mathbb P\left(2^{-N}\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{i=1}^nx_i\varepsilon_i\mathbf{1}_{\{\lvert\varepsilon_i\rvert> 2^N \}}\right\rvert>\delta\right)<\infty.$$
To this aim, we use the inclusion
$$
\left\{2^{-N}\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{i=1}^nx_i\varepsilon_i\mathbf{1}_{\{\lvert\varepsilon_i\rvert> 2^N \}}\right\rvert>\delta
\right\}\subset\bigcup_{i=1}^{2^N}\{\lvert\varepsilon_i\rvert> 2^N \}.$$ Consequently, the previously mentioned probabilities can be controlled by a union bound and convergence of the aforementioned series follows from integrability of $\varepsilon_1$.


答:
The goal is to look at what part of the integrand, once taken the derivative of, will give another part of the integrand. Remember that each time we take an integral of something, we need to reverse the process of differentiation. Doing a $u$-substitution allows us to first see where the derivative is, deal with it first, and then proceed quite normally with integration.

There is only really a few things we could consider but I'll save you the trouble of trying things out by saying that the correct substitution is $u = 1+4t^2$. Why? Well because if
$$ u = 1+4t^2 $$
then
$$ \mathrm{d}u= 8t\,\mathrm{d}t = 4 (2t\,\mathrm{d}t). $$
So doing this substitution not only simplifies the argument in the square root to something that can be dealt with, but also gets rid of the pesky $2t$ term which is multiplying it. In addition, when we do a $u$-substitution the limits of the integral change. So, it's pretty easy to see that if $t=1$ then $u=5$ and when $t=0$ then $u=1$. As such, our integral becomes,
\begin{align*} \dfrac{1}{4}\int_1^5 u^{1/2}\,\mathrm{d}u &= \dfrac{1}{4} \left[\dfrac{2u^{3/2}}{3}\right]_1^5 \\  &= \dfrac{5\sqrt{5}-1}{6}. \end{align*}

Title: About the substitution in the integration

Body:
$$\int_0^1 2t \cdot \sqrt{1 + 4t^2}dt$$

I see the answer is $\frac{5\sqrt{5} -1}{6}$ but I'm not sure how we got here. Can someone remind me how to do this? My book gets right to this answer but I'm not sure what steps there were done and I haven't done integration in quite some time. Can someone show me the steps?


问:
How to divide students into different groups?
Suppose you are a teacher who want to assign your students to different clubs. Now a student can join one or more than one club. Now what known to you is there are "n" students and there are "r" clubs.Now how many ways can you do that? If you can solve it using PIE theorem ,then I would be obliged if you also included detailed explanation behind your reasoning and also terribly sorry for my bad English .And I am new to stack so I'm pretty weak at MathJax so feel free to edit.
EDIT**:It is possible to get one or more clubs empty**

答: I'm assuming that every student must belong to a club and that a club can have more than $1$ student(otherwise why would it be called a club). Now, every student has $2$ choices pertaining to a club; join or don't join. Therefore, the number of choices on $r$ clubs is $2^r$. This includes the possibility that the student doesn't join any club. Hence, the number of choices where at least one club must be joined is $2^r-1$. Since each student can choose their clubs independently, the number of assignments of $n$ students to $r$ different clubs is $(2^r-1)^n$.


答: We can argue for this result geometrically to build intuition. For $n>1$ there are always two primitive $n$-th roots of unity that are closest to $1$ in the complex plane. Under multiplication they act on the plane by rotating each complex number $1/n$-th of a turn around the origin in either direction. As $n$ gets large, these rotations get very small. When $n$ gets infinitely large, the rotations get infinitely small, and so don't rotate the plane at all. This is the same action that $1$ has on the plane, in accordance with our intuition.

Title: Does $1^0$, as a limiting form of the $n$ $n$-th roots of unity $1^{1/n}$, represent every complex value on the unit circle?

Body:
In the context of algebraic numbers, the $n$-th root has $n$ solutions in the complex plane. The equations look like: $1=x^n$, which gives $1^{1/n}=x$ (which can be rotated around the complex plane to generate all the solutions).  And then, if we let $n$ become large, we get many solutions. But then, when we let $n$ become arbitrarily large, we get $1^0 = x$, which only has one solution, namely $1$, because anything to the zeroth power gives $1$. This seems discontinuous to me, and it confuses me.

Does anyone know what I'm missing?

I have a feeling that perhaps $1^0$ actually is every value around the complex unit circle, which would be the natural extension of the limiting process. But, if that's true, there are tons of implications that don't seem right to me.

Maybe the following two facts are useful? I'm just spitballling on these two.

This reminds me of the fact that $a*0=0$ has infinitely many solutions.

And perhaps the following fact is useful: $2^0 = 2 ^ {100 * 0} =  (2 ^ {100})  ^ 0$


问:
Prove that $\sum a_n$ converges iff $\sum 2^n a_{2^n}$ converges, but with another function instead of $2^{n}$.
Let $\{a_n\}$ be a positive, decreasing sequence of real numbers.
I know how to prove that $\sum_{n=1}^\infty a_n$ converges iff $\sum_{n=1}^\infty 2^n a_{2^n}$ converges, but is it possible that this is true for another function from naturals to naturals instead of $2^n$? In case there is no function why is it?

答:
Yes, that test generalizes.  First of all, observe that passage to the subsequence $a_{2^n}$ together with a weighting by $2^n$ is a discrete analogoue of integration by substitution on $\int f(x)\,dx$ using the substitution $x = e^u$, from which we get $\int f(x)\,dx = \int f(e^u)e^u\,du$.

The fact that $e^u$ is its own derivative (w.r.t. $u$) is analogous to the sequence $2^n$ being its own discrete difference ($2^{n+1} - 2^n = 2^n$), which suggests replacing the weighting factor $2^n$ in the convergence test with a discrete difference, and there is the following result: under suitable conditions on a sequence $u(n)$, $\sum_{n \geq 1} a_n$ with nonnegative non-increasing terms $a_n$ converges if and only if  $\sum_{n \geq 1} a_{u(n)}(\Delta u)(n)$ converges, where $(\Delta u)(n) = u(n+1) - u(n))$ for all $n \geq 1$. There is no claim that the two series are equal when they both converge, just like in the special case $u(n) = 2^n$ that you describe, so it is not quite like integration by substitution. But the analogy is still striking.

The test you write about is called Cauchy's condensation test, and the bottom of the Wikipedia page on it here states the generalization I describe above and includes some precise conditions on $u(n)$ under which you can carry out this "summation by substitution".


答: Hint: Normally, you would just take the ratios and be done but since earthquake intensities are reported on the logarithm scale- you need to take the ratio of the respective exponential values. </p>

Title: Earth Quake Question Logarithmic Type Problem

Body:
An earthquake off the coast of Vancouver Island was measured at 8.9 on the Richter Scale
and an earthquake off the coast of Alaska was measured at 6.5. How many times more intense, to the nearest whole number, was the earthquake off the coast of Vancouver Island than the one off the coast of Alaska?</p>


答:
I provide here another way to approach it for the sake of curiosity.

Let $2t = \sinh(x)$. Based on the identity $\cosh^{2}(x) - \sinh^{2}(x) = 1$ as well as on the fact that $\sinh'(x) = \cosh(x)$,
\begin{align*}
\int_{0}^{1} 2t\sqrt{1 + 4t^{2}}\mathrm{d}t & = \frac{1}{2}\int_{0}^{\ln(2+\sqrt{5})}\sinh(x)\sqrt{1+\sinh^{2}(x)}\cosh(x)\mathrm{d}x\\\\
& = \frac{1}{2}\int_{0}^{\ln(2+\sqrt{5})}\cosh^{2}(x)\sinh(x)\mathrm{d}x\\\\
& = \frac{\cosh^{3}(x)}{6}\Biggr|_{0}^{\ln(2+\sqrt{5})} = \frac{5\sqrt{5} - 1}{6}
\end{align*}

Here I have made use of the definition of $\sinh(x)$ and $\cosh(x)$. Precisely speaking,
\begin{align*}
\begin{cases}
\displaystyle\cosh(x) = \frac{e^{x} + e^{-x}}{2}\\\\
\displaystyle\sinh(x) = \frac{e^{x} - e^{-x}}{2}
\end{cases}
\end{align*}
whence we get that
\begin{align*}
\sinh^{-1}(x) = \ln(x + \sqrt{x^{2}+1})
\end{align*}

This is not the quickest way to solve the proposed integral, but it is indeed an alternative way to approach suchlike problems. Hopefully it contributes!


问:
About the conformal map $f$ of a slit disc onto a unit disc with the condition $f(i/2)=0$
The slit disc is $D$ \ $(-1,a]$ where a belongs to $(-1,1)$. The question asks me to find a conformal map from this slit disc to the unit disc and such map satisfy $f(i/2) = 0$. I currently have no certain clue how to construct such a map, but I guess I need to start with certain simpler maps and compose them together to make this $f$. So is there any general method or thought to construct such a map?

答:
With $\frac{z-a}{1-az}$ send $D\setminus(-1,a]$ to $D\setminus(-1,0]$,

with $i z^{1/2}$ send $D\setminus(-1,0]$ to $|z|<1,\Im(z)>0$,

with the inverse of $z\to \frac{z-1}{z+1}$ send $|z|<1,\Im(z)>0$, to $\Re(z)>0,\Im(z)>0$,

with $z^2$ send $\Re(z)>0,\Im(z)>0$ to $\Im(z)>0$,

which is biholomorphic to the unit disk with $\frac{z-i}{z+i}$.

Composing with an automorphism of the unit disk you get $f(i/2) = 0$


问:
How to prove a matrix $[a_{ij}]$ with $a_{ji} = 1/a_{ij}$ has no complex eigenvalues
I've been given a matrix of the form

$$\begin{pmatrix}
-1 & 1/a_{21} & \cdots & 1/a_{n1}
\\
a_{21} & -1 &\cdots & 1/a_{n2}
\\
\vdots & \vdots & \ddots & \vdots
\\
a_{n1} & a_{n2} & \cdots & -1
\end{pmatrix}$$
which satisfies
$$
\sum_{j=1}^n a_{ij} = 0,\space\space\forall\space i
$$
and I've been asked to prove whether it can have complex eigenvalues when all its coefficients are real. I've been trying to throw properties of matrices and eigevalues to it for a week but I haven't gotten anything.

For $n = 2$, I know its eigenvalues are $0$ and $-2$, its determinant is $0$ and its trace is $-2$. For $n = 3$ it also has a $0$ determinant, but this time only after applying the extra property that the sum of all the coefficients of any column must be $0$. I haven't been able to prove whether the determinant should be $0$ for any natural $n$ or not.

I don't know if there's an easy way to prove this, maybe it's possible to prove they don't even without that extra constraint. Any hint or idea I could try would be very welcome!

答:
At the very least, the second condition is necessary. Randomly generated counterexample:

$$
A = \pmatrix{
 -1.    &      0.41745274 &-1.26280132\\
  2.39548077  & -1.       &  -0.62360058\\
 -0.79189021 & -1.60359056  &-1.
}
$$
the eigenvalues of this matrix are
$$
1.28354084, \quad -2.14177042\pm 0.95442081i.
$$
It is notable that for the case of $n=3$, the matrix necessarily has rank $3$ so that the only non-zero eigenvalue is equal to the trace (i.e. $3$).


问:
Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$
Number Theory : Solve the system of congruence

(1)    $28x+17y\equiv 18 \pmod{41}$

(2)   $31x+11y\equiv 35\pmod{41}$

Attempt :

we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular arithmetic

lets multiply the first equation (1) by $31$ and the second equation(2) by $28$.

(1)    $31\cdot(28x+17y)\equiv 31\cdot18 \pmod{41}$

(1) $868x+527y\equiv 558\pmod{41}$

(2)   $28\cdot(31x+11y)\equiv 28\cdot35\pmod{41}$

(2) $868x+308y\equiv 980\pmod{41}$

So finally we can subtract equation (1) from (2) we get:

$219y\equiv -422\pmod{41}$

lets check the $\gcd(219,41)$ by Euclidian algorithm :

$219 = 41\cdot 5 + 14$

$41= 14\cdot 2 + 13$

$14= 13\cdot 1 + 1$

$\gcd(219,41)=1$

Hence, because the $\gcd$ is equal to $1$ we can find the Inverse and multiply the equation by the Inverse to find $y$.

$219y\equiv 1\pmod{41}$

$219a = 1+41k$ , the $41k$ must end with the digit of $8$ for $1+$digit $8$ will be $9$ so $k$ must be multiply of number with end digit of $8$.

I don't know how to continue from here .

答:
Let $p \in \Bbb N$ be a prime number.

Recall the following two facts from elementary number theory:

Let $p$ be a prime integer.

For all $u,v,w \in \Bbb Z$ satisfying $u\not\equiv 0 \pmod{p}$, $v\not\equiv 0 \pmod{p}$ and $w\not\equiv 0 \pmod{p}$,

$\tag 1 u \equiv v \pmod{p} \; \text{ iff } \; uw \equiv vw \pmod{p}$

If $a \in \Bbb N$ and $a\not\equiv 0 \pmod{p}$ and $a\not\equiv 1 \pmod{p}$ then there exist $q \in \Bbb Z$ and $r \in \Bbb N$ such that

$\tag 2 aq \equiv r \pmod{p} \; \text{ and } \; 1 \le r \lt p \; \text{ and } \; r \lt a$

The above theory is the counterpart to algorithms that compute the inverse

of non-zero elements in $(\mathbb{Z}/p\mathbb{Z})^\times$.

We are now ready to continue where the OP left off.

Since

$\quad 219y\equiv -422\pmod{41} \; \text{ iff } \; 14y\equiv 29\pmod{41}$

we are motivated to 'divide' (with a small negative residue) $41$ by $14$ and write ,

$\quad 41  = 14 \cdot 3 - 1$

and set $q = 3$.

Next,

$\quad 14y\equiv 29\pmod{41} \; \text{ iff } \; (3 \cdot 14) y\equiv 3 \cdot 29\pmod{41}  \; \text{ iff } \;  1y\equiv 5 \pmod{41}$

and the OP's problem is solved in one step.

By being flexible with the sign of the residue, this is a one step solution. If you can only perform Euclidean division, this is what happens:

Presented with

$\quad 14y\equiv 29\pmod{41}$

we divide $41$ by $14$ and write ,

$\quad 41  = 14 \cdot 2  + 13$

and set $q = -2$.

Next,

$\quad 14y\equiv 29\pmod{41} \; \text{ iff } \; (-2 \cdot 14) y\equiv (-2 \cdot 29)\pmod{41}  \; \text{ iff } \; 13 y\equiv 24 \pmod{41}$

and write

$\quad 41  = 13 \cdot 3  + 2$

and set $q = -3$

Next,

$\quad 13y\equiv 24\pmod{41} \; \text{ iff } \; (-3 \cdot 13) y\equiv (-3 \cdot 24)\pmod{41}  \; \text{ iff } \; 2 y\equiv 10 \pmod{41}$

and write

$\quad 41  = 2 \cdot 20  + 1$

and set $q = -20$

Next,

$\quad 2 y\equiv 10 \pmod{41} \; \text{ iff } \; (-20 \cdot 2) y\equiv (-20 \cdot 10)\pmod{41}  \; \text{ iff } \; 1y\equiv 5 \pmod{41}$

and the OP's problem is solved in $3$ steps.


问:
Prove that $\lim_{n \to \infty} \int_a^b f(x) \sin^n (x) dx = 0$.

Let $f$ be a continuous function on the interval $[a,b]$. Prove that $$\displaystyle\lim_{n \to \infty} \int_a^b f(x) \sin^n (x) dx = 0.$$

Attempt:

Want to show that for a given $\epsilon > 0$, there exists $N \in \mathbb N$ s.t. if $n \geq N$, then $$\left|\int_a^b f(x) \sin^n (x) dx\right| < \epsilon.$$

Since $f$ is continuous on a closed interval, it is bounded, i.e., $|f(x)| \leq M, \forall x \in [a,b]$.

For a given $\epsilon > 0$, choose $N$ s.t. $|\sin^N(x)| < \dfrac{\epsilon}{M(b-a)}$. [Is this allowed? I am thinking of the case where $\sin(x) = 1$, then this doesn't hold.]

Hence, $$\left|\int_a^b f(x) \sin^n (x) dx\right| \leq \int_a^b |f(x)|| \sin^n(x) | dx < \epsilon$$ for $n \geq N$.

答: The bounded convergence theorem will quickly yield the desired result.  Note that since $f$ is continuous, it is bounded on any interval $[a, b]$.


问:
$\sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{\operatorname{Tr}[\rho M]}{\operatorname{Tr}[\sigma M]}$ satisfies Data Processing Inequality
$\newcommand{\Tr}{\operatorname{Tr}}$Let $\mathcal{H}_1, \mathcal{H}_2$ be a Hilbert spaces and $\rho, \sigma$ be density matrices on $\mathcal{H}_1$. Define

$$D(\rho\parallel\sigma) := \sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{\Tr[\rho M]}{\Tr[\sigma M]}.$$

(1) Show that $D$ satisfies the data processing inequality, i.e. for any density matrices $\rho, \sigma$ (positive semi-definite, hermitian, trace = 1) and for any $\mathcal{E} \in \text{CPTP}(\mathcal{H}_1 \rightarrow \mathcal{H}_2)$ (completely positive, trace preserving map) holds $D(\mathcal{E}(\rho) \parallel \mathcal{E}(\sigma)) \leq D(\rho \parallel \sigma)$

(2) Show that $D(\rho\parallel\sigma) = \inf\{c \in\mathbb{R}: ~c \sigma - \rho \geq 0\}$

Sidenote: I posted this question first on physics stack exchange, but was told to repost the question here as it seems to fit better on this SE platform.

My attempts:

For (1) we have to show that for any density matrices $\rho, \sigma$ and for any $\mathcal{E} \in \text{CPTP}(\mathcal{H}_1 \rightarrow \mathcal{H}_2)$ holds $D(\mathcal{E}(\rho) \parallel \mathcal{E}(\sigma)) \leq D(\rho \parallel \sigma)$. If the denominator is zero, the DPI is satisfied trivially. So we can assume that the denominator is not zero.

For this case, I tried to use the Kraus-decomposition of $\mathcal{E}$ and I tried to use some matrix inequalities to show the statement. My problem was that I couldn't find a suitable inequality for the denominator (in the direction we need).

For (2) I tried to prove two inequalities and I think that I found a proof for one direction. Therefore, I considered some arbitrary $c$ such that $c \sigma \geq \rho$ holds. Then we obtain

$$\sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{\Tr[\rho M]}{\Tr[\sigma M]} \leq \sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{c\Tr[\sigma M]}{\Tr[\sigma M]} = \sup_{0 \leq M \leq I_{\mathcal{H}_1}} c = c.$$

This holds for all $c$ as defined above. Therefore, we can take the infimum (with respect to c) on both sides and obtain

$$\sup_{0 \leq M \leq I_{\mathcal{H}_1}} \frac{\Tr[\rho M]}{\Tr[\sigma M]} \leq \inf\{c\in \mathbb{R}:~ c \sigma \geq \rho \}.$$

On the first glance, the other direction looked easier, as we can choose $M=I$. But then we end up with $\Tr[s \sigma - \rho] \geq 0$, where we denote the supremum by $s$. And I cannot argue why the same without the trace holds true.

I hope that I am now on the correct platform for asking this question and I am grateful for any hint or help to solve this problem!

答:
For a Hilbert space $\mathcal{H}$, let $\operatorname{Herm}(\mathcal{H})$ denote the space of Hermitian operators on $\mathcal{H}$ (i.e., space of Hermitian matrices). Given a linear mapping $\mathcal{E}:\operatorname{Herm}(\mathcal{H}_1)\to\operatorname{Herm}(\mathcal{H}_2)$, its dual map is the unique linear mapping $\mathcal{E}^*:\operatorname{Herm}(\mathcal{H}_2)\to\operatorname{Herm}(\mathcal{H}_1)$ for which
$$
\operatorname{Tr}(\mathcal{E}(A)B) = \operatorname{Tr}(A\mathcal{E}^*(B))
$$
holds for all operators $A\in\operatorname{Herm}(\mathcal{H}_1)$ and $B\in\operatorname{Herm}(\mathcal{H}_2)$. The mapping $\mathcal{E}$ is trace preserving if and only if $\mathcal{E}^*$ is unital (that is, if $\mathcal{E}^*(I_{\mathcal{H_2}})) = I_{\mathcal{H}_1}$.

To see that $D$ satisfies the data-processing inequality, let $\mathcal{E}:\operatorname{Herm}(\mathcal{H}_1)\to\operatorname{Herm}(\mathcal{H}_2)$ be a positive and trace-preserving map. For every positive operator $M\in\operatorname{Herm}(\mathcal{H}_2)$ such that $0\leq M\leq I_{\mathcal{H}_2}$, one has that $0\leq\mathcal{E}^*(M)\leq I_{\mathcal{H}_1}$ as $\mathcal{E}$ is positive and trace preserving, and thus
$$
 \frac{\operatorname{Tr}(\mathcal{E}(\rho) M)}{\operatorname{Tr}(\mathcal{E}(\sigma) M)} = \frac{\operatorname{Tr}(\rho\mathcal{E}^*(M))}{\operatorname{Tr}(\sigma\mathcal{E}^*(M))} \leq D(\rho||\sigma).
$$
It follows that $D(\mathcal{E}(\rho)||\mathcal{E}(\sigma))\leq D(\rho||\sigma)$.

Are you familiar with semidefinite programming?

Consider the following pair of optimization problems:
\begin{align*}
\underline{\text{Primal problem} }  &&&&\underline{\text{Dual problem} }\\
\text{minimize}&\quad c &&& \text{maximize} & \quad \operatorname{Tr}(\rho M)\\
\text{subject to}&\quad c\sigma\geq \rho&&& \text{subject to} & \quad \operatorname{Tr}(\sigma M)=1\\
&&&&  & \quad M\geq0
\end{align*}
It is straightforward to show that these semidefinite programs are dual to each other. One can check that Slater's conditions hold for this pair of problems, and thus strong duality holds. That is, the optimal values of the primal and dual problem are equal to each other.

I'll leave it up to you to show that the optimal value of the dual problem above is equal to $D(\rho||\sigma)$.


问:
Are there subsets of $\mathbb R$ homeomorphic to $\mathbb Z_p$ or $\mathbb Q_p$?
The title says it all. I know that the middle-third Cantor set is homeomorphic to $\mathbb Z_2$, and there is a pretty explicit homeomorphism in this case, but what about for other primes $p$?

Is there a subset of $\mathbb R$ homeomorphic to $\mathbb Q_p$?

答: Let $$f:\Bbb{Q}_p\to \Bbb{R}, f(\sum_{n=-N}^\infty a_n p^n)=\sum_{n=-N}^\infty \frac{a_n}{p} (2p)^{-n}$$ where I mean $a_n\in 0\ldots p-1$


问:
Find $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ if $a+b+c=0$
I'm stuck at this algebra problem, it seems to me that's what's provided doesn't even at all.

Provided:  $$a+b+c=0$$

Find the value of:  $$\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$

Like I'm not sure where to start, and the provided clue doesn't even make sense.  There's no way I can think of to factor this big polynomial into a form like $a\times p+b\times q+c\times r=s$ where $p,q,r,s\in\mathbb{Z}$.

Thanks!

Body:
Note that:
$$\frac{a^2}{2a^2+bc}=\frac{a^2}{a^2+a(-b-c)+bc}=\frac{a^2}{(a-b)(a-c)}$$
This applies in the same way for:
$$\frac{b^2}{2b^2+ac}=\frac{b^2}{(b-c)(b-a)}\ \text{and}\ \frac{c^2}{2c^2+ab}=\frac{c^2}{(c-a)(c-b)}$$Therefore, the original equation is equal to:
\begin{align}
&\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-c)(b-a)}+\frac{c^2}{(c-a)(c-b)}
\\
\\
&=\frac{-a^2(b-c)-b^2(c-a)-c^2(a-b)}{(a-b)(b-c)(c-a)}
\\
\\
&=\frac{-a^2b-b^2c-c^2a+a^2c+b^2a+c^2b}{(a-b)(b-c)(c-a)}
\\
\\
&=\frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}
\\
\\
&=\boxed{1}
\end{align}


问:
Proof that $S^1$ is a manifold using projections
I'm trying to understand what seems like it should be a simple proof in Spivak's text on differential geometry. In order to show that the circle, $S^1$, is indeed a manifold, he projects $S^1 \setminus \{(0,1)\}$ onto $\Bbb{R} \times \{-1\}$, as described in this image:

I can intuitively see how this gives a homeomorphism between $S^1 \setminus \{(0,1)\}$ and $\Bbb{R} \times \{-1\}$.

Next, he says that the point $(0,1)$ can be handled similarly by projecting it on  $\Bbb{R} \times \{1\}$. How does this projection work? Is the point $(0,1)$ just sent to itself?

My bigger question is, how are these two arguments combined to show that $S^1$ is indeed a 1-manifold? The first part shows "most" of $S^1$ being homeomorphic to $\Bbb{R}^1,$ and the second part shows the "rest" of $S^1$ being homeomorphic to... something? Then what?

For reference, here's the definition of manifold in the text:
A manifold is a metric space $M$ with the following property:
If $x \in M,$ then there is some neighborhood $U$ of $x$ and some integer $n \ge 0$  such that $U$ is homeomorphic to $\mathbb{R}^n.$

答:
This is sometimes called stereographic projection "from the north pole" or something like that, thinking of $(0,1)$ as the north pole of the circle. We can analogously defined stereographic projection "from the south pole" by the same type of construction, except the rays emanate from $(0,-1)$ and land on the line $\Bbb{R}\times \{1\}\subseteq \Bbb{R}^2.$ Using this projection map, $(0,1)$ is indeed sent to itself, and we get a diffeomorphism $S^1\setminus \{(0,-1)\}\to \Bbb{R}$. Combine this with the other diffeomorphism $S^1\setminus \{(0,1)\}\to \Bbb{R}$ and we get an atlas of charts for $S^1$.

What we have then shown is that there exists an open cover $\{U_1,U_2\}$ of $S^1$ with diffeomorphisms $\phi_i:U_i\to \Bbb{R}$. This proves that $S^1\subseteq \Bbb{R}^2$ is indeed a submanifold of $\Bbb{R}^2$.


问:
About the substitution in the integration
$$\int_0^1 2t \cdot \sqrt{1 + 4t^2}dt$$

I see the answer is $\frac{5\sqrt{5} -1}{6}$ but I'm not sure how we got here. Can someone remind me how to do this? My book gets right to this answer but I'm not sure what steps there were done and I haven't done integration in quite some time. Can someone show me the steps?

答:
A common method to try when seeing square root like $\sqrt{a^2+t^2}$: try substituting $t = a\tan\theta$.

Let $t = \frac 12 \tan \theta$, then $dt = \frac12 \sec^2\theta \ d\theta$.

$$\begin{align*}
\int_0^12t\sqrt{1+4t^2}dt
&= \int_0^{\arctan 2} \tan\theta\sqrt{1+\tan^2\theta}\cdot \frac12\sec^2\theta\ d\theta\\
&= \frac12 \int_0^{\arctan 2}\sec^2\theta\cdot (\tan\theta \sec\theta)\ d\theta\\
&= \frac12 \left[\frac{\sec^3\theta}{3}\right]_0^{\arctan 2}\\
&= \frac12\left[\frac{(\sqrt5)^3 - 1^3}3\right]\\
&= \frac{5\sqrt 5 - 1}6
\end{align*}$$

To calculate $\sec^3\left(\arctan 2\right)$, consider a right-angled triangle with opposite side $2$ and adjacent side $1$.


问:
Proof that $\int_0^\infty\frac{\ln x}{x^3 - 1} \, dx = \frac{4 \pi^2}{27}$
I realise this question was asked here, but I'm not able to work with any of the answers. The hint given by my professor is

Integrate around the boundary of an indented sector of aperture $\frac{2 \pi}{3}$

but when I try that I can't figure out how to deal with the (divergent) integral along the radial line at angle $2 \pi / 3$. My issue with the accepted answer is that it uses the residue theorem where it doesn't apply, at least as we've learned it, since
$$z \mapsto \frac{\log^2z}{z^3 - 1}$$
has non-isolated singularities on the closed region bounded by the proposed contour (due to branch cuts), and I am not sure how to relate the integral along the real axis to one over a contour modified to avoid the branch cut.

For a fixed $\varepsilon > 0$, and for any $\delta \in (0, 1 - \varepsilon)$, we could let $\log_{-\delta / 2}$ be the branch of the logarithmic function with a cut along the ray $\operatorname{arg}z = -\delta / 2$ and define a contour which goes along the positive real axis from $\varepsilon$ to $1 - \delta$, a semicircle in the upper half plane of radius $\delta$ around $1$, the positive real axis from $1 + \delta$ to $2$, an arc of radius $2$ around $0$ with central angle $2 \pi - \delta$, the ray $\operatorname{arg}z = 2 \pi - \delta$ from $r = 2$ to $r = \varepsilon$, and finally an arc of radius $\varepsilon$ around $0$ back to $\varepsilon$. But then, for example, I don't know how to calculate the limit of integral along the arc of radius $\varepsilon$
$$\lim_{\delta \to 0}\int_0^{2 \pi - \delta}\frac{\log_{-\delta / 2}^2(\varepsilon e^{i \theta})}{\varepsilon^3 e^{3 i \theta} - 1} \varepsilon i e^{i \theta} \, d\theta.$$

If I instead try to first use the substitution $x = e^u$ on the real integral and then compute a contour integral, I still get a divergent integral that I don't know how to handle, this time along the top of an indented rectangle.

答:
Here we discuss two ways of tackling the integral using contour integral. I added Solution 1 to help you implement the hint suggested by the professor. However, I personally recommend you to jump directly to Solution 2.

Solution 1. Consider

$$ f(z) = \frac{\log z}{z^3 - 1}, $$

where $\log(\cdot)$ is the principal complex logarithm. Now we integrate $f$ along the boundary of the indented sector of opening $2\pi/3$:

Here, the radius of the larger arc $\Gamma_2$ (resp. smaller arc $\Gamma_1$) is $R$ (resp. $\epsilon$) and $0 < \epsilon < 1 < R$. Then it is easy to check that

$$ \left| \int_{\Gamma_1} f(z) \, \mathrm{d}z \right| \leq C \epsilon ( C + \log (1/\epsilon) )
\qquad\text{and}\qquad
\left| \int_{\Gamma_2} f(z) \, \mathrm{d}z \right| \leq \frac{C(C + \log R)}{R^2} $$

for $C = 2\pi/3$, and so, the integrals along these curves vanish as $\epsilon \to 0^+$ and $R \to \infty$. So by the residue theorem,

\begin{align*}
\int_{L_1} f(z) \, \mathrm{d}z + \int_{L_2} f(z) \, \mathrm{d}z + \int_{L_3} f(z) \, \mathrm{d}z + \int_{\gamma} f(z) \, \mathrm{d}z = o(1)
\end{align*}

as $\epsilon \to 0^+$ and $R\to\infty$. However, using the fact that $\omega = e^{2\pi i/3}$ is a simple pole of $f(z)$, the function $(z - \omega)f(z)$ is analytic at $z = \omega$. So

\begin{align*}
\lim_{\epsilon \to 0^+} \int_{\gamma} f(z) \, \mathrm{d}z
&= \lim_{\epsilon \to 0^+} i\int_{2\pi/3}^{-\pi/3} \epsilon e^{i\theta} f(\omega + \epsilon e^{i\theta}) \, \mathrm{d}\theta \tag{$z=\omega+\epsilon e^{i\theta}$} \\
&= -i \pi \lim_{z \to \omega} (z - \omega)f(z)
= -i \pi \mathop{\mathrm{Res}}_{z = \omega} f(z) \\
&= \frac{2\pi^2}{9} \omega.
\end{align*}

Moreover,

\begin{align*}
\int_{L_2} f(z) \, \mathrm{d}z
&= -\omega \int_{1+\epsilon}^{R} f(\omega x) \, \mathrm{d}x
= -\omega \int_{1+\epsilon}^{R} \left( f(x) + \frac{2\pi i}{3} \cdot \frac{1}{x^3 - 1} \right) \, \mathrm{d}x
\end{align*}

and likewise

\begin{align*}
\int_{L_3} f(z) \, \mathrm{d}z
= -\omega \int_{\epsilon}^{1-\epsilon} \left( f(x) + \frac{2\pi i}{3} \cdot \frac{1}{x^3 - 1} \right) \, \mathrm{d}x.
\end{align*}

Combining altogether and using that $f(z)$ is analytic at $z = 1$,

\begin{align*}
(1 - \omega) \int_{\epsilon}^{R} f(x) \, \mathrm{d}x
- \frac{2\pi i \omega}{3} \left( \int_{\epsilon}^{1-\epsilon} \frac{1}{x^3 - 1} \, \mathrm{d}x + \int_{1+\epsilon}^{R} \frac{1}{x^3 - 1} \, \mathrm{d}x \right)
+ \frac{2\pi^2}{9} \omega = o(1).
\end{align*}

Letting $\epsilon \to 0^+$ and $R \to \infty$,

\begin{align*}
(1 - \omega) \int_{0}^{\infty} f(x) \, \mathrm{d}x
- \frac{2\pi i \omega}{3} \left( \mathop{\mathrm{PV}}\! \int_{0}^{\infty} \frac{1}{x^3 - 1} \, \mathrm{d}x \right)
+ \frac{2\pi^2}{9} \omega = 0.
\end{align*}

By noting that

\begin{align*}
\mathop{\mathrm{PV}}\! \int_{0}^{\infty} \frac{1}{x^3 - 1} \, \mathrm{d}x
&=-\frac{\pi}{3\sqrt{3}},
\end{align*}

we end up with

\begin{align*}
\int_{0}^{\infty} f(x) \, \mathrm{d}x
= \frac{\omega}{1 - \omega} \left( -\frac{\pi}{3\sqrt{3}} \cdot \frac{2\pi i}{3}
- \frac{2\pi^2}{9} \right)
= \frac{4\pi^2}{27}.
\end{align*}

Solution 2. Here is a more elegant solution. Let $\operatorname{Log}$ denote the complex logarithm chosen so that its argument lies between $0$ and $2\pi$. (Note: Using the principal complex logarithm, this can be realized by $\operatorname{Log}(z) = i\pi + \log(-z)$.) Then consider

$$ g(z) = \frac{(\operatorname{Log}(z) - 2\pi i)\operatorname{Log}(z)}{z^3 - 1}. $$

Then it is not hard to see that, for $x > 0$,

\begin{align*}
g(x + i0^+) := \lim_{\epsilon \to 0^+} g(x + i\epsilon)
&= \frac{(\log x - 2\pi i)\log x}{x^3 - 1} \\
g(x - i0^+) := \lim_{\epsilon \to 0^+} g(x - i\epsilon)
&= \frac{(\log x + 2\pi i)\log x}{x^3 - 1}.
\end{align*}

So by using the keyhole contour,

$$ \int_{0}^{\infty} \bigl( g(x + i0^+) - g(x - i0^+) \bigr) \, \mathrm{d}x
= 2\pi i \biggl( \mathop{\mathrm{Res}}_{z=e^{2\pi i/3}} g(z) + \mathop{\mathrm{Res}}_{z=e^{4\pi i/3}} g(z) \biggr) $$

Now the left-hand side is

$$ (-4\pi i) \int_{0}^{\infty} \frac{\log x}{x^3 - 1} \, \mathrm{d}x $$

and the right-hand side is

$$ 2\pi i \biggl( \frac{\bigl(\frac{2\pi i}{3} - 2\pi i \bigr)\bigl( \frac{2\pi i}{3} \bigr)}{3 e^{4\pi i}} + \frac{\bigl(\frac{4\pi i}{3} - 2\pi i \bigr)\bigl( \frac{4\pi i}{3} \bigr)}{3 e^{8\pi i}} \biggr) = 2\pi i \left( -\frac{8\pi^2}{27} \right).$$

Therefore the answer is again $\frac{4\pi^2}{27}$.


问:
Yet Another Trigonometric Proof of the Pythagorean Theorem
Let $\triangle{ABC}$ be a right-triangle with $BC=a$, $AC=b$, $AB=c$ and $\angle{ACB}=\gamma=90^\circ$. Also, let $s=\frac{a+b+c}{2}$. In the following link https://geometriadominicana.blogspot.com/2020/12/the-pythagorean-theorem-from.html I argue that the expression

$$\cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}=\frac{(a+b)^2-c^2}{4ab}=\frac{1}{2}\tag{1}$$

can be stated without risk of petitio principii, from which we have
$$\begin{align*}(a+b)^2-c^2&=2ab\\a^2+2ab+b^2-c^2&=2ab\\a^2+b^2-c^2&=0\end{align*}$$

Is this proof correct? If so, is it new?

答:
The proof is valid, but, as I suggest in comments, it's a bit unsatisfying.

OP's blog proves the relation
$$\cos^2\tfrac12C=\frac{s(s-c)}{ab} \quad\left(=\frac{(a+b+c)(a+b-c)}{4ab} \right)\tag{1}$$
for a general $\triangle ABC$ using the fact that
$$sr=|\triangle ABC| = \tfrac12ab\sin C\tag{2}$$
where $r$ is the inradius. The issue here is that we "know" that the tangent segments from $C$ to the incircle have length $s-c$; for $C$ a right angle, these segments necessarily match the triangle's inradius, so that $(2)$ immediately reduces to
$$s(s-c)=\frac12ab \qquad\to\qquad a^2+b^2=c^2 \tag{2'}$$
without ever having to consider $(1)$ or the additional fact that $\cos 45^\circ=1/\sqrt{2}$.

We can attempt to salvage the situation by providing a more-direct proof of $(1)$.

Consider the following, which invokes the Intersecting Secants Theorem for $C$ relative to the circle (ie, calculating the power of $C$ in two ways):

$$
|CB'||CA'| = |CP||CQ| \qquad\to\qquad 2a\cos\tfrac12C \cdot 2b\cos\tfrac12C = (a+b+c)(a+b-c) \tag{3}
$$

Dividing-through by $4ab$ gives us $(1)$. From here, we can consider the case with $C$ a right angle, introduce $\cos 45^\circ=1/\sqrt{2}$ by suitably Pythagorean-free means (see OP's blog), and deduce $a^2+b^2=c^2$ as OP has shown. Done. $\square$

Now, one could (and should!) ask whether this approach to proving Pythagoras is unsatisfying in its own right. Indeed, it is.

Consider a version of the above figure, with a right angle at $C$ and with $A''$ the reflection of $A'$ across diameter $\overline{PQ}$:

Here, we see that we can use the Intersecting Chords Theorem (ie, calculating the power of $A$ in two ways) to get, quite immediately,
$$|AA'||AA''| = |AP||AQ| \quad\to\quad b\cdot b = (c+a)(c-a) \quad\to\quad a^2+b^2=c^2 \tag{4}$$
So, there's no need to appeal to the value of $\cos45^\circ$ in order to reduce $2ab\cos^2\frac12C$ to $2ab$ in $(1)$ and prove Pythagoras.

Incidentally, it should perhaps be explicitly noted that, as Pythagorean proofs go, the figure above is itself overly-complicated. The elements of the proof are those contained in the circle alone. Here's a cleaner figure:

Perhaps there is some other approach to $(1)$ that doesn't just as easily, independent of $(1)$, lead to the Pythagorean Theorem, but I'll leave the search for it to the reader.


问:
Writing $x\in U\otimes V$ as $x=\sum u_i \otimes v_i$
Let $U, V$ be vector spaces.
Let $u_i,\dots,u_n \in U $ be linearly independent. For $x \in U\otimes V$ show that there exist uniquely determined $v_1,\dots,v_n \in V$ such that $x = \sum u_i \otimes v_i$.

I am curious to know if we should use the Universal Property to prove this?
And if not, what steps should we do to prove it?

答: Suppose that $U$ is a $k$-vector space, and for $(u,a) \in U \times k$ consider $u \otimes a := au$. Note that if $b : U \times k \to Z$ is bilinear, then $b(\_,1) : U \to Z$ satisfies $b = b(\_,1) \circ \otimes$, which means that $U \otimes k \cong U$. Now, it is not necessarily true that for every $x \in U$ there are unique scalars $a_1,\dots,a_n \in k$ such that $x = a_1u_1 + \cdots + a_nu_n$.


问:
looking for reference for integral inequality
Math people:

I would like a reference for the following fact (?), which I proved myself (I am 99% sure the proof is valid) but which has probably been done before.  My proof was a little messy.  If no one can supply a reference, I would appreciate an elegant proof.  Here is what I proved:

Let $\Omega$ be an open subset of $\mathbf{R}^n$, $\delta >0$, and $K >0$.  Then there exists $\kappa = \kappa(\delta, K, \Omega)>0$ with the following property: if $f , g \in C^\infty(\Omega)$ with $\inf_\Omega f \geq 0$, $\inf_\Omega g \geq 0$, $\int_\Omega f\,dx \leq K$, and $\int_\Omega g\,dx \geq \delta$, then

$$  \int_\Omega \sqrt{f^2+g^2} - f \,dx   > \kappa.$$

Please note that there are no assumptions on $\sup_\Omega f$: if there were, the proof would be easier.  My proof did not use the smoothness of $f$ or $g$, and my $\kappa$ did not involve $\Omega$ at all.  I would be OK with a $\kappa$ that did involve $\Omega$.  I do not care how sharp the inequality is.  I only need the result for $\Omega$ being a two-dimensional rectangle.

I need this as part of a larger proof and it would be better to cite this fact, if it has been done before, than to include it in the paper.

答:
One way of proving it is to write
$$\int\frac{g^2}{\sqrt{f^2+g^2}+f}$$
and use CS to end up with
$$\int\frac{g^2}{\sqrt{f^2+g^2}+f}\int\sqrt{f^2+g^2}+f\ge\left(\int g\right)^2.$$
Now $\int\sqrt{f^2+g^2}+f\le \int f+g+f\le 2K+\int g.$ So our integral bounded below by $$\frac{\left(\int g\right)^2}{K+\int g}$$ which can be easily estimated via $\delta.$</p>


问:
Does every scheme admit an affine open subset whose sections are the ring global sections?
The title basically says it all. Given an arbitrary scheme $X$ with global sections $\mathcal{O}_X(X)$, is there an affine open subset $U$ of $X$ such that $\mathcal{O}_X(U) = \mathcal{O}_X(X)$?

Obviously the scheme $Y=\operatorname{Spec}\mathcal{O}_X(X)$ has the same global sections as $X$, but it's not clear to me that you can embed $Y$ in $X$. I have been unable to produce a counter example though.

答: Consider $X = \Bbb{P}^1_k,$ where $k$ is your favorite field. We have $\mathcal{O}_X(X) = k,$ and note that $X$ is integral, so that the restriction maps $\mathcal{O}_X(U)\to\mathcal{O}_X(V)$ are injective. Now, notice that any proper open subscheme is contained in one of the two copies of $\Bbb{A}^1_k$ inside $\Bbb{P}^1_k.$ This means that if $U$ is any affine open subscheme of $X,$ we have injections $k\to k[x]\to\mathcal{O}_X(U).$ In particular, this shows that the map $\mathcal{O}_X(X)\to\mathcal{O}_X(U)$ is never an isomorphism for an affine open subscheme $U\subseteq X.$


问:
Finding the stopping time in casino game
I am attempting the following question from Grimmet & Stirzaker which I’ve attached for context:

Q) A large casino contains infinitely many gamblers Gl, G2, . . . , each with an initial fortune of \$1. A croupier tosses a coin repeatedly. For each n, gambler Gn bets as follows. Just before the nth toss he stakes his \$1 on the event that the nth toss shows heads. The game is assumed fair, so that he receives a total of \$(1/p) if he wins, where p is the probability of heads. If he wins this gamble, then he repeatedly stakes his entire current fortune on heads, at the same odds as his first gamble. At the first subsequent tail he loses his fortune and leaves the casino, penniless. Let Sn be the casino's profit (losses count negative) after the nth toss. Show that Sn is a martingale. Let N be the number of tosses before the first appearance of HHH; show that N is a stopping time and hence find E(N).

The solution provides the following expression for the casino profit after the Nth toss: $S_{N} = N - (p^{-1}+ p^{-2} + p^{-3})$ but I am having real trouble understanding this. Intuitively I understand that at N-3, we have observed a tail, thus the casino will have taken a total \$(N-3) at this point, then we have a new gambler appearing at each time step betting on heads, so at t=N-2, the casino will have profit: $(N-3)-(p^{-1})$ and at t=N-1 the casino has profit: $(N-3)-(p^{-1}) - (p^{-2} + p^{-1})$. Continuing in this way I don’t see how I can arrive at the solution in the textbook.

Further to this I cant quite figure out why $\mathbb{E}[N]= (p^{-1}+ p^{-2} + p^{-3}) $? Any help on the above would be greatly appreciated.

答:
If the game is to be fair, and we have a payoff of $-1$ with probability $p$ and a payoff of $x$ with probability $1-p$, then $-p+x(1-p)=0$, so $x=(1-p)/p=p^{-1}-1$. So if we start with $\$1$, we end up with $\$0$ with probability $p$ and $\$p^{-1}$ with probability $1-p$, but the casino has only paid us $p^{-1}-1$.  Similarly, if we win twice in a row, we now have $p^{-2}$, but the casino has only paid us $p^{-2}-1$.  If we win 3 times, we have $p^{-3}$, but the casino has only paid us $p^{-3}-1$.

Suppose that $N$ is the stopping time of the game.  Everybody except the last 3 players have bet and lost, paying the casino $1$ each.  The last player has won 3 times in a row, second to last has won twice in a row, and the third to last player has won one.  Therefore, we have switched from $N$ players all having $\$1$ to $N-3$ players having $\$0$, and the others having $p^{-1}, p^{-2}, p^{-3}$.  Thus, the casino has paid out $p^{-1}+p^{-2}+p^{-3}-N$.  Alternatively, they have paid out $-(N-3)+(p^{-1}-1)+(p^{-2}-1)+(p^{-3}-1)$.

However, all of the games are martingales, and we satisfy the conditions of the optional stopping theorem, so the stopped game is also a martingale, which means it has expecctation $0$.  Thus, $\mathbb E(p^{-1}+p^{-2}+p^{-3}-N)=0$.  Expanding and rewriting, this give $\mathbb E(N)=p^{-1}+p^{-2}+p^{-3}$.


问:
Lectures on Non-Standard Analysis Book
I'm undergrad Applied Math Student and I'm writing my thesis about Non-Standard Analysis. I find a Spriger Book named Lectures On Non-Standard Analysis which are based on short courses of lectures delivered by Machover (first part) and the Master of Science thesis of J. Hisrchfeld.

So, the construction starts by considering some set which has certain desirable closure properties. This set needs the smallest transitive set such that contains any set $X$, sometimes called the transitive closure of $X$. Here is a fragment of the construction.

I may be wrong, but I think the right set $U_0$ has to be defined as the UNION the set $\{V_i\}_{i=0}^\infty$, i.e $U_0=\bigcup\{V_i\mid i=0,1,2,\ldots\}$.

Can someone tell me if I'm right or I'm missing something?

答:
Not having read the aforementioned source, this looks like a simple but unfortunate typo to me. Of course $U_0$, as defined above, would not be the transitive closure.

For example, define $\downarrow$ as the function satisfying $\downarrow\! n = \{ y \in \mathbb{N} \:|\: y < n \}$ for each natural number $n$.

Then the set $V = \{ \downarrow\! x \:|\: x \in \mathbb{N}\}$ is clearly transitive. Thus, $V_{i+1} = V$ for each $i$, and $U_0 = \{ \emptyset, V \}$. But $U_0$ is not transitive, since e.g. $\downarrow\!2 \in V$, and $V \in U_0$ both hold, while $\downarrow\!2 \in U_0$ does not.

As you noted, one can easily fix this by writing $U_0 = \bigcup \{V_i \:|\: i \in \mathbb{N} \}$ instead.

Title: Lectures on Non-Standard Analysis Book


答: You're right. You can compare with the construction of the transitive closure of a set on English Wikipedia to check.</p>

Title: Lectures on Non-Standard Analysis Book

Body:
I'm undergrad Applied Math Student and I'm writing my thesis about Non-Standard Analysis. I find a Spriger Book named Lectures On Non-Standard Analysis which are based on short courses of lectures delivered by Machover (first part) and the Master of Science thesis of J. Hisrchfeld.

So, the construction starts by considering some set which has certain desirable closure properties. This set needs the smallest transitive set such that contains any set $X$, sometimes called the transitive closure of $X$. Here is a fragment of the construction.

I may be wrong, but I think the right set $U_0$ has to be defined as the UNION the set $\{V_i\}_{i=0}^\infty$, i.e $U_0=\bigcup\{V_i\mid i=0,1,2,\ldots\}$.

Can someone tell me if I'm right or I'm missing something?


问:
Want to show $\sum_{n>N}n^{-1/2} \ll N^{-1/2}$
I feel like I can use the result $$\sum_{n=x}^N \frac{a_n}{n^s} = A(N)N^{-s} + s \int_x^N A(t)t^{-s-1}dt$$ where $s=1/2$ to verify the $\ll$ approximation.

If I pick $A(n)=\sum_{x\leq t\leq n}\chi(t)$, then I know $|A(n)|\ll 1$. I think that will make the first term become $O(N^{-1/2})$, but am not sure how I should go about handling the integral term. Thanks!

Thanks for the help in advance!

答:
$$\sum_{n=x}^N \frac{a_n}{n^s} = A(N)N^{-s} + s \int_x^N A(t)t^{-s-1}dt$$
gives for $\Re(s) >0$ $$|\sum_{n=x}^\infty \frac{\chi(n)}{n^s}| =| s \int_x^\infty (\sum_{x\leq n\leq t}\chi(n))t^{-s-1}dt|\le  |s| \int_x^\infty q t^{-\Re(s)-1}dt= \frac{q|s| x^{-\Re(s)}}{\Re(s)}$$


问:
Calculating permutations if the sequences have to be in ascending order?
How would you go about calculating the number of permutations in ascending order.

Obviously if you had (a set of) 3 numbers you have $ 3! $ permutations:

(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1)

But only one of these is in ascending order (1,2,3).

Consider the lottery - picking 6 numbers from 49. Normally the order for this doesn't matter you just calculate '49 choose 6' but if the numbers have to be in ascending order how would you calculate that?

答: In the same way: The number of (strictly) ascending sequences of $6$ numbers chosen from $49$ is $\binom{49}{6}$. For as you have pointed out, once the $6$ numbers have been chosen, there is precisely one way to arrange them in ascending order. </p>


问:
Conjugacy Classes, Normal Subgroups and Guaranteed Existence
Suppose we have a group $G$ of order $24$, and we know apriori that this group has $5$ conjugacy classes, or sizes $1$, $3$, $6$, $6$, and $8$.

The question I'm presented with is whether or not one can conclude that $G$ has exactly two normal subgroups.

I know that normal subgroups are the union of conjugacy classes. Moreover, this union must include the singleton conjugacy class for the identity, and the order of the union should divide the order of the group. It seems plausible in this case then that we could take the conjugacy classes and obtain two normal subgroups of order $1 + 3 = 4$ and $1 + 3 + 8 = 12$. But is this guaranteed?

答:
If you take the classification of groups of order 24, you find that there is only one group with 5 conjugacy classes (namely $S_4$) and it has exactly two proper normal subgroups. So (somewhat casuistic) the naive answer is "Yes".

But the existence of normal subgroups does not follow just from the class sizes. It is not hard to come up with examples of groups and class sizes, where a union of classes satisfies the conditions, but does not form a normal subgroup. So as soon as you ask the same question in the generic case, the answer is "No".


问:
Every simple planar graph with $\delta\geq 3$ has an adjacent pair with $\deg(u)+\deg(v)\leq 13$

Claim: Every simple planar graph with minimum degree at least three has an edge $uv$ such that $\deg(u) + \deg(v)\leq 13$.  Furthermore, there exists an example showing that $13$ cannot be replaced by $12$.

This seems related to Hard planar graph problem.

By Euler's formula, we know that $|V|-|E|+|F|=2$.  Rearranging, we have $|F| = 2+|E|-|V|$

As $\delta\geq3$ we know $|V|\geq 4$ and $|E|\geq 6$ and since each face has at least 3 edges bounding it and each edge is on the boundary of at most 2 faces we have then that $3|F|\leq 2|E|$.  Substituting into Euler's formula, we have $$2|E|\geq 3|F| = 3(2+|E|-|V|)\Rightarrow 3|V|-6\geq |E|$$

From this we have from the handshaking lemma that avgdegree = $\sum\frac{\deg(v)}{|V|} = 2\frac{|E|}{|V|} \leq 6 - \frac{12}{|V|}\lneq 6$.  And so, the average degree of a simple planar graph is strictly less than 6.

Suppose there was a smallest counterexample, I.e., some graph such that every edge $uv$ has $\deg(u)+\deg(v)>13\dots$

It is at this point that I feel as though I am stuck.  It seems that you should be able to reach a contradiction about either the average degree, the planarity, or $\delta\geq 3$.  In the hint in the linked similar problem, they suggest to show that you can find a stricter bound for the minimum degree of the smallest counterexample from below, but it doesn't seem to apply here.  They then suggest to show that a substantial proportion of the vertices are of low degree (in this problem would translate to degree less than 6), and then for the punchline to show that the number of edges is enough that there are more than #of vertices of high degree choose 2, meaning there must necessarily be an edge between two vertices of low degree thus proving the claim.

答:
This may be better as a simple comment, but I lack the reputation.

It can be assumed that we are working with a triangulation, but we need to be careful about which edges we add when triangulating. Suppose that the graph has minimum degree three and there is no edge $uv$ with $\deg(u) + \deg(v) \leq 13$. Now suppose there were a face of degree at least four. Let $x$ be the vertex of minimum degree on that face. In any case, both neighbours of $x$ on the face will have degree at least $7$. Add an edge between those two vertices and continue this process until you are left with only triangular faces.

I stole this trick from https://faculty.math.illinois.edu/~west/pubs/dischnew.pdf where it is used at the start of Lemma 3.2. Incidentally, that lemma proves something stronger than what is required here: Every plane graph with minimum degree $3$ has an edge with weight at most $11$ or a $4$-cycle through two degree $3$ vertices and a common neighbour of degree at most $10$.


问:
Local and global extremes
I Wrote problems and solutions, I need just few explanations.

1.Let

$$J(x)=\int_{0}^{1}x'^{2}dt,\quad x(0)=0, x(1)=1. $$
   Find the extrema value for $J$.

I'm doing this using Euler equation $\frac{\partial f}{\partial x}-\frac{d}{dt} \frac{\partial f}{\partial x'}=0$, where $f=x'^2$.

Applying Euler equation, I got:

$$-2x''=0  \Rightarrow x''=0  \Rightarrow x'=C_1 \Rightarrow x=C_1 t+C_2.  $$

Using initial conditions, I got that $x=t$. This is candidate for extreme.

Let $h\in\ C^1[0,1]$ such that $h(0)=h(1)=0$. Then,

$$J(x(t)+h(t))-J(x(t))= J(t+h(t))-J(t)=  \int_{0}^{1}(1+h')^{2}dt-\int_{0}^{1}dt$$
$$=  \int_{0}^{1}h'^{2}dt\ge 0.  $$

My question is: Why is $x=t$ weak global minimum of $J$?

2.Let

$$J(x)=\int_{0}^{1}x'^{3}dt ,\quad x(0)=0, x(1)=1.  $$

Doing the same procedure, I got that $x=t$ is a candidate for extreme.  Then I got :
$$J(x(t)+h(t))-J(x(t))=  J(t+h(t))-J(t)=  \int_{0}^{1}(1+h')^{3}dt-\int_{0}^{1}dt=  \int_{0}^{1}(3h'^2+h'^3)dt=  3\int_{0}^{1}h'^2(1+\frac{1}{3} h')^{2}dt.$$

Question for this part is when $J(x(t)+h(t))-J(x(t)) \geq 0$  ?
(I have solution that it $J(x(t)+h(t))-J(x(t)) \geq0$  when $||h||_1\leq3$ and that for such $h$ I have weak local minimum). I don't know when it is local/global and weak/strong.

答:
If OK I give a more proper example. Let
$$L(y)=\int_0^2dx\,y'^2(1-y')^2$$
with b.c. $y(0)=0$ and $y(2)=1$.

For weak variation we can use Euler-Lagrange equation such as
$$\frac{d\,F}{d\,y'}=2y'(1-y')^2-y'^2(1-y')=k$$
where $k$ is any constant and $F=y'^2(1-y')^2$. Solving the nonlinear ordinary differential (you can use WolframAlpha) and using boundary conditions you can find the solution
$$y=\frac x2\Rightarrow L=\frac 18$$
For strong variation we don't need a smooth function and we can break the the integral into integral with $0\le c\le 2$
$$L(y)=\int_0^cdx\,y'^2(1-y')^2+\int_c^2dx\,y'^2(1-y')^2$$
By solving Euler Lagrange equations we can find two curves
$$y_1=m_1\,x+d_1\qquad 0\le x\le c$$
$$y_2=m_2\,x+d_2\qquad c\le x\le 2$$
Applying boundary conditions $y(0)=0$ and $y(2)=1$ we get
$$y_1=m_1\,x\qquad 0\le x\le c$$
$$y_2=m_2\,(x-2)+1\qquad c\le x\le 2$$
We now need corner point conditions and continuity equation
$$\lim_{x\to c_-} F-y'F_{y'}=\lim_{x\to c_+} F-y'F_{y'}\Rightarrow m_1^2(1-m_1)(1-3m_1)=m_2^2(1-m_2)(1-3m_2)$$
$$\lim_{x\to c_-} F_{y'}=\lim_{x\to c_+} F_{y'}\Rightarrow m_1(1-m_1)(1-2m_1)=m_2(1-m_2)(1-2m_2)$$
$$m_1\,c=m_2(c-2)+1$$
By solving three equations we find two solution sets such as
$$m_1=1\quad m_2=0\quad c=1$$
$$m_1=0\quad m_2=1\quad c=1$$
Let's replace the first set into $L(y)$ to see the result
$$L=\int_0^1dx\,1^2(1-1)^2+\int_1^2dx\,0^2(1-0)^2=0$$
which is a global minimum.

Title: Local and global extremes


问:
Rank of adjoint operator
$
\newcommand{\span}{\operatorname{span}}
\newcommand{\rank}{\operatorname{rank}}
\newcommand{\ip}[1]{ \left \langle #1 \right \rangle }
$
Suppose $T: V \rightarrow W$ is a linear transformation, where $v$ and $W$ are finite-dimensional vector spaces. If $T^*$ is the adjoint of $T$, show that:

$$\rank(T) = \rank(T^*)$$

My attempt:

Suppose $B = \{v_1, ..., v_m\}$ and $C = \{w_1, ..., w_n\}$ are bases of $V$ and $W$ respectively. Applying the Gram-Schmidt process, $B' = \{s_1, ..., s_m\}$ and $C' = \{t_1, ..., t_m\}$ are orthonormal bases of $V$ and $W$ respectively. Suppose $v \in V$ and $w \in W$, then $v \in \span(s_1, ..., s_m)$ and $w \in \span(t_1, ..., t_n)$. So

$$v = a_1s_1 + ... a_ms_m$$ for scalars $a_1, ..., a_m$

, and

$$w = b_1t_1 + ... b_nt_n$$ for scalars $b_1, ..., b_m$.

By definition, $\ip{ Tv, w }$ = $\ip{ v, T^*w }$ for $v \in V$, $w \in W$. Hence,

\begin{align*}
&\ip{ T\sum_{i=1}^{m} a_is_i, \sum_{j=1}^{n} b_jt_j } = \ip { \sum_{i=1}^{m} a_is_i, T^*\sum_{j=1}^{n} b_jt_j }\\
&\implies \ip{ \sum_{i=1}^{m} a_iTs_i, \sum_{j=1}^{n} b_jt_j }= \ip{ \sum_{i=1}^{m} a_is_i, \sum_{j=1}^{n} b_jT^*t_j }\\
&\implies \sum_{i=1}^{m} \sum_{j=1}^{n} \ip{a_iTs_i, b_jt_j} = \sum_{i=1}^{m} \sum_{j=1}^{n}\ip{a_is_i, b_jT^*t_j}
\end{align*}

From here, I'm stuck. I'm not sure how this would simplify to the point of implying that $\rank(T) = \rank(T^*)$.

Any assistance is much appreciated. I'm also open to other proofs...

答:
Here are some hints that may help you avoid getting down too much in the weeds:

$\tag 1\operatorname{dim} V = \operatorname{dim}N(T) + \operatorname{dim} N(T)^{\perp}$ and

$\tag2 \operatorname{dim} V = \operatorname{dim}N(T) + \operatorname{dim} R(T)$
so

$\tag3 \operatorname{dim} N(T)^{\perp}=\operatorname{dim} R(T)$

which means that it suffices to prove that

$\tag4 \operatorname{dim} R(T^*)=\operatorname{dim} N(T)^{\perp}.$ And for this it is enough to show that

$\tag5 R(T^*)^{\perp}=N(T).$

So, let $x\in R(T^*)^{\perp}.$ Then, $0=\langle x,T^*y\rangle$ for all $y\in W$ so...

and if $x\in N(T)$ then $0=\langle T(x)*y\rangle$ for all $y\in W$ so...

Title: Rank of adjoint operator


问:
Trouble proving the trigonometric identity $\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$
I have become stuck while solving a trig identity. It is:

$$\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$$

I have simplified the left side as far as I can:

\begin{align}
\frac{1-2\sin(x)}{\sec(x)}
&=\frac{1-2\sin(x)}{1/\cos(x)}=(1-2\sin(x))\cos(x)\\
&=\cos(x)-2\sin(x)\cos(x)=\cos(x)-\sin(2x)
\end{align}

However, I'm not sure what to do on the right side. I know I can use a compound angle formula to break $\cos(3x)$ into $\cos(2x)\cos(x)-\sin(2x)\sin(x)$; however, I do not know where to go after that. My main problem is with the denominator of the right side, I can't figure out how to get rid of it, either by multiplying, or by using a trig identity. Any help in solving this identity would be greatly appreciated!

答:
First, in case you don't know how, I'll expand $\cos{3x}$.
$$\require{cancel}\begin{aligned}\cos{3x}&=\cos\left(x+\left(x+x\right)\right)\\&=\cos x\cos\left(x+x\right)-\sin x\sin\left(x+x\right)\\&=\cos{x}\left(\cos^{2}x-\sin^{2}x\right)-\sin{x}\left(2\sin{x}\cos{x}\right)\\&=\cos^3 x-\sin^{2}{x}\cos{x}-2\sin^{2}{x}\cos{x}\\&=\cos^3 x-3\sin^{2}{x}\cos{x}\\&=\cos^2 x\cos{x}-3\sin^{2}{x}\cos{x}\\&=\cos{x}\left(1-\sin^2{x}\right)-3\sin^{2}{x}\cos{x}\\&=\cos{x}-\sin^2{x}\cos{x}-3\sin^{2}{x}\cos{x}\\&=\cos{x}-4\sin^2{x}\cos{x}\end{aligned}$$
Let's rock!
$$\begin{aligned}\frac{1-2\sin{x}}{\sec{x}}&=\frac{1-2\sin{x}}{\frac{1}{\cos x}}\\&=\cos{x}\left(1-2\sin x\right)\\&=\cos x-2\sin{x}\cos{x}\\&=\cos x-2\sin{x}\cos{x}\cdot\frac{1+2\sin x}{1+2\sin x}\\&=\frac{\cos x\cancel{+2\sin{x}\cos{x}}\cancel{-2\sin{x}\cos{x}}-4\sin^{2}{x}\cos{x}}{1+2\sin x}\\&=\frac{\cos x-4\sin^{2}{x}\cos{x}}{1+2\sin x}\\&=\frac{\cos{3x}}{1+2\sin{x}}\end{aligned}$$


答:
You could argue that $G$ has at most $2$ proper non-trivial normal subgroups since the combinations you found for writing the size of a normal subgroup as a sum of class sizes are exhaustive.

But a group of size $24$ is not simple, so one of the two possibilities must occur. If $G$ has a normal subgroup of order $4$, say $N$, look at the quotient $G/N$ which has order $6$. By Sylow (or Cauchy), $G/N$ has a normal subgroup $H/N$ of order $3$ and thus $H$ is normal in $G$ and of order $12$. A similar argument will allow you to deduce the existence of a normal subgroup of order $4$ given the existence of a normal subgroup of order $12$.

In general, however, as pointed out in the other answer it does not follow that if the sum of some class sizes is a divisor of the order of the group then a normal subgroup of that order must exist.

Title: Conjugacy Classes, Normal Subgroups and Guaranteed Existence

Body:
Suppose we have a group $G$ of order $24$, and we know apriori that this group has $5$ conjugacy classes, or sizes $1$, $3$, $6$, $6$, and $8$.

The question I'm presented with is whether or not one can conclude that $G$ has exactly two normal subgroups.

I know that normal subgroups are the union of conjugacy classes. Moreover, this union must include the singleton conjugacy class for the identity, and the order of the union should divide the order of the group. It seems plausible in this case then that we could take the conjugacy classes and obtain two normal subgroups of order $1 + 3 = 4$ and $1 + 3 + 8 = 12$. But is this guaranteed?