问:
Markov chain - probability - stochastic process
I got stuck here:

The probability that it will rain today is that it did not rain in the previous two days is $0.3$, but if it rained in one of the last two days then the probability of rain today is $0.6$.

$W(n)$ is a random variable that receives the value $1$ if in the $n\ge1$ day it rained, and the value $0$, if it's not.

I need to explain why the series $(W(n))_{n\ge 0}$ is not a Markov chain, but when they define $X_n=(W_n,W_{n-1})$, so $(W(n))_{n\ge 0}$ it is a Markov chain!

答: The point is that the last two days states fix the current day's distribution. Knowing what happend 3 day or 4 days or n days before does not add any information. Hence Xn is a marcovian.</p>


问:
Expressing $\frac{x-y}{w-z}$ as a function of $\frac{x}{w}$ and $\frac{y}{z}$
I have the expression $\frac{x-y}{w-z}$ and I would like to rewrite it as a function of two fractions, $\frac{x}{w}$ and $\frac{y}{z}$. Ideally a linear combination of $\frac{x}{w}$ and $\frac{y}{z}$ and their powers. If that is not possible, a non-linear expression in $\frac{x}{w}$ and $\frac{y}{z}$ will also do.

(Apologies in advance for possible tagging this improperly ..)

答:
You have a function in terms of four variables $f: (x, y, w, z) \mapsto \frac{x-y}{w-z}$ and you want to rewrite it in terms of two variables, $u = \frac{x}{w}, v = \frac{y}{z}$ so that $g(u, v) = f(x, y, w, z)$.

This will only be possible if the function has the property that $$f(tx, sy, tw, sz) = f(x, y, w, z) \quad \forall \ t, s, x, y, w, z \in \mathbb{R}$$
since both $(x, y, w, z)$ and $(tx, sy, tw, sz)$ map to the same $u = \frac{x}{w}, v=\frac{y}{z}$ and thus will have the same corresponding $g(u, v)$.

However,
\begin{align}
&f(tx, sy, tw, sz) = \frac{tx-sy}{tw-sz} = \frac{x-y}{w-z} =  \\
\iff& (tx-sy)(w-z) = (tw-sz)(x-y) \\
\iff& t(wx-zx)+s(zy-wy) = t(wx-wy)+s(zy-zx) \\
\iff& t(wx-zx-wx+wy) = s(zy-zx-zy+wy) \\
\iff& t(wy-zx) = s(wy-zx) \\
\iff& wy-zx = 0 \lor t=s
\end{align}
Which is definitely not always true.

Therefore, there does not exist a function $g$ such that $f(x, y, w, z) = g(u, v)$.

ContentLicense: CC BY-SA 4.0


问:
What is the overall probability of success of infinite draws with exponentially decreasing odds of success per draw?
If I flip a coin until I get "heads", sooner or later, I will get one occurrence and stop. Probability of success is 1.
But what if after each "tails" (failed) draw, odds for the next draw are divided by 2 ? Instead of having successive odds like 0.5, 0,5, 0.5, etc... I have 0.5, 0.25, 0.125, etc...
Is it possible to never get "heads"?

Trying to figure out this, I ran the following rust code:

```<code>use rand;

fn draw() -> bool {
    const MIN_MAX: f64 = 1e-20;
    let mut max = 1.0;
    while max > MIN_MAX {
        max /= 2.0;
        if rand::random::<f64>() < max {
            return true
        }
    }
    false
}

fn main() {
    const TOTAL: i32 = 10_000_000;
    let mut success = 0;
    for _ in 0..TOTAL {
        if draw() {
            success += 1;
        }
    }
    println!("{} / {}", success, TOTAL);
}
</code>```

Some comments:

I found an overall probability of 0.711

The MIN_MAXis the minimum value of individual draw odds at which the program gives up and return false. Changing this value between 1e-20 and 1e-40 does not change the result much.

I would be interested to know if the overall probablity is below 1, and if so, what are the fancy maths to explain the result.

EDIT 1 2021-03-20:

Probability to have success between 1 and n draws would be:
$\frac{1}{2}+\frac{1}{2\times4}+\frac{1\times3}{2\times4\times8}+\frac{1\times3\times7}{2\times4\times8\times16}...$
(too bad at jaxmath for formula with n)
which converges to 0.7112119049133813

EDIT 3 2021-03-20:

Googling 7112119049I found this:
https://forumserver.twoplustwo.com/47/science-math-philosophy/538-riddler-question-about-coin-flipping-game-1719467/

答:
Googling 7112119049, I found a discussion (and explanations) very close to this question: https://forumserver.twoplustwo.com/47/science-math-philosophy/538-riddler-question-about-coin-flipping-game-1719467/
Probablity of success is:

$1-phi(\frac{1}{2})$

where phi is the Euler function.


问:
Algebraic Solution of Inequalities
Ok I'm doing some Inequalities revision and I know what the signs mean and stuff but this question I'm a bit stuck on. You need to find the x such that $\frac{1}{4} (2x + 1) \leq \frac{3}{5} (3x - 1)$. </p>

答: A simple way of solving this is to find when the left hand side equals the right hand side first. This will be the only place when the left hand side can (but doesn't have to) change from being less than the right hand side to being more than the right hand side or vice versa. Hence you can now just check an x value on either side of this point to see where the left hand side is less than the right hand side (and in this case, you'll also include the point where the two sides are equal).</p>

ContentLicense: CC BY-SA 3.0

Title: Algebraic Solution of Inequalities


问:
Expected Value of Binomial Distribution
What is the expected value of the absolute value of the difference between the number of incoming tails and the number of incoming heads  when a coin is tossed 5 times?

Here is what I think :  Let  $X$  is random variable of number of incoming heads then
$p=\frac{1}{2}$ and $n=5$ so $X\sim Binom(5,\frac{1}{2})$

$X$ and $Y=5-X$ ($Y$ is random variable of number of incoming tails) then $E(K)=E(\left|X-(5-X)\right|)=E(\left|2X-5\right|)=E(2X-5)=2E(X)-5=2.\frac{5}{2}-5=0$ where $K=\left|X-Y\right|=\left|2X-5\right|$ and $E(X)=n.p=5.\frac{1}{2}=\frac{5}{2}$

But the right answer is $\frac{15}{8}$. Where am I making a mistake? Any help will be appreciated.

答:
Your random variable $Z=|X-Y|$ can take only the values

$$\{1,3,5\}$$

with probabilities

$$\left\{\frac{20}{32};\frac{10}{32};\frac{2}{32}\right\}$$

This because

$Z=5$ when you get 5 consecutive tails (or heads)

$Z=3$ when you get 4 tails (or heads)

$Z=1$ when you get 2 tails (or heads)

No other situations are possible.

Thus its expectation is

$$\mathbb{E}[Z]=1\times \frac{20}{32}+3\times\frac{10}{32}+5\times\frac{2}{32}=\frac{15}{8}$$

Title: Expected Value of Binomial Distribution


问:
Integrating a specific function
I'm struggling to integrate. Can anyone help?

$$\int \frac{1}{1+\sqrt{x+1}}dx$$

答:
Try the substitution:
$$u^2 = x+1, \ x= u^2 -1$$
$$dx = 2u \ du$$
Your integral then simplifies to:

$$\int \frac{2u}{u + 1}\ du$$

Can you finish it from here?

Title: Integrating a specific function


问:
Using multiple scale method the first term expansion for large $t$, $\epsilon y'' +y' +y =0; \ \ \ t>0$
Find the first term expansion of the solutions of the following problem that is valid for large $t$, $\epsilon y'' +y' +y =0; \ \ \ t>0$

Regular expansion:
Let us assume
\begin{eqnarray}
 y(t) \sim y_0(t) + \epsilon y_1(t)+ ... . ...(1)
\end{eqnarray}
From the given equation we have
\begin{eqnarray} \label{equ: substitution outer_expansion}
 y(t) \sim  \epsilon (y_0''+ \epsilon y_1'' +...)+ (y_0'+ \epsilon y_1' +...)+(y_0+ \epsilon y_1 +...)
\end{eqnarray}
Then,
\begin{eqnarray*}
O(1) :
\begin{cases}
y_0'+y_0=0\\
y_0(0)=0
\end{cases}
\end{eqnarray*}

Solving we get, $y_0(t)=c_1 e^{-t}$, where $c_1$ is a arbitrary constant.Using the initial condition we get,
\begin{equation*}
y_0(t)=0
\end{equation*}

Also,
\begin{eqnarray*}
O(\epsilon) :
\begin{cases}
y_0''+y'_1+y_1=0 \implies y'_1+y_1=0\\
y_1(0)=0
\end{cases}
\end{eqnarray*}

Similarly we get
\begin{equation*}
y_1(t)=0
\end{equation*}
Hence from (1) we get our solution,
\begin{equation*}
y(t) \sim 0
\end{equation*}
Exact Solution:
\begin{equation*}
y(t) = \frac{\epsilon e^{\dfrac{(\sqrt{1-4 \epsilon}+1)x}{2\epsilon} } (e^{\dfrac{(\sqrt{1-4\epsilon})x}{\epsilon}}-1)   }{\sqrt{1-4 \epsilon}}
\end{equation*}
Multiple-Scale Expansion:
Consider two time scales $t_1=t$ and $t_2=\epsilon ^\alpha t$.
\begin{equation*}
\dfrac{d}{dt} \to \dfrac{dt_1}{dt} \dfrac{\partial }{\partial t_1} +\dfrac{dt_2}{dt} \dfrac{\partial }{\partial t_2}=  \dfrac{\partial }{\partial t_1} + \epsilon^\alpha  \dfrac{\partial }{\partial t_2}
\end{equation*}
Substituting this into the given equation we get,
\begin{equation}
\epsilon \bigg (\partial^2_{t_1} + 2 \epsilon^\alpha \partial_{t_1} \partial_{t_2}+ \epsilon^{2\alpha}  \partial^2_{t_2}\bigg )y+ \bigg( \partial_{t_1}+ \epsilon^\alpha \partial_{t_2} \bigg )y+y=0 ...(2)
\end{equation}
where
\begin{equation}
 y=0 \ \text{and} \ \bigg( \partial_{t_1}+ \epsilon^\alpha \partial_{t_2} \bigg )y=1, \ \text{for} \ t_1=t_2=0.
\end{equation}
Consider the power series expansion of the form

\begin{eqnarray}
y \sim y_0(t_1, t_2) + \epsilon y_1(t_1, t_2)+ ...
\end{eqnarray}

Substituting this into (2) yields the following

\begin{equation}
\epsilon \bigg (\partial^2_{t_1} + 2 \epsilon^\alpha \partial_{t_1} \partial_{t_2}+ \epsilon^{2\alpha}  \partial^2_{t_2}\bigg )(y_0+ \epsilon y_1 +...)+ \bigg( \partial_{t_1}+ \epsilon^\alpha \partial_{t_2} \bigg )(y_0+ \epsilon y_1 +...)+(y_0+ \epsilon y_1 +...)=0
\end{equation}
\begin{eqnarray*}
 O(1) :
 \begin{cases}
 ( \partial_{t_1}+1)y_0=0,\\
  y_0=0, \ \partial_{t_1}y_0=1 \ \ \text{at}  \ \ t_1=t_2=0
 \end{cases}
\end{eqnarray*}
The general solution of the problem is
\begin{eqnarray}
y_0= c(t_2) e^{t_1}
\end{eqnarray}

Using the initial condition $y_0(0,0)=0$, we get $c(0)=0$.
\
Balancing we get $\alpha=1$.
\begin{eqnarray*}
 O(\epsilon) :
 \begin{cases}
  ( \partial_{t_1}+1)y_1=-(\partial^2_{t_1}+\partial_{t_2})y_0,\\
  y_1=0, \ \partial_{t_1}y_1=- \partial_{t_2}y_0 \ \ \text{at}  \ \ t_1=t_2=0
 \end{cases}
\end{eqnarray*}
Then we get
\begin{equation}
 ( \partial_{t_1}+1)y_1=-(c(t_2) +c'(t_2))e^{t_1}
\end{equation}
whose general solution is ,

\begin{equation}
y_1= \bigg( (-c(t_2)-c'(t_2))t_1+p(t_2)  \bigg)e^{-t_1}
\end{equation}
Using $y_1(0,0)=0$ ,we get $ p(0)=0 $.
I am stuck here. Is there any secular term. If there is any secular term, how to prevent this.

答:
WKB approach

Without assuming to know the correct balancing, start with $y=\exp(S/δ)$, set for convenience $s=S'$. Then the equation (for homogeneous basis solutions of this exponential type) becomes, multiplying with $δ^2/y$,
$$
ϵ(s^2+δs')+δs+δ^2=0
$$
$s$ and $s'$ are "normal" sized, neither going to zero or infinity as $ϵ\to 0$. This makes it obvious that the balance is for $δ=ϵ$, all the other terms are smaller by composition. Then an expansion of $s$ gives the perturbation series
\begin{align}
s^2+s&=-δ(s'+1)\\
s_0^2+s_0&=0&\implies s_0&=-1&\text{ or }~~s_0&=0\\
(2s_0+1)s_1&=-(s_0'+1),&s_1&=~~~1,& s_1&=-1\\
s_1^2+(2s_0+1)s_2&=-s_1',&  s_2&=~~~1,& s_2&=-1\\
2s_1s_2+(2s_0+1)s_3&=-s_2'&  s_3&=~~~2,& s_3&=-2
\end{align}
resulting in basis solutions
$$
y(t)=A\exp[-t/ϵ+t(1+ϵ+2ϵ^2+O(ϵ^3))]+B\exp[-t(1+ϵ+2ϵ^2+O(ϵ^3))]
$$
For $y(0)=0$ and $y'(0)=1$ one gets $A+B=0$, $(-1/ϵ+1+ϵ+...)A-(1+ϵ+...) B=1$, thus about $B=ϵ(1+2ϵ)$, $A=-B$. So in the leading parts
$$
y(t)=ϵ(1+2ϵ)(e^{-t-ϵt}-e^{-t/ϵ+t}).
$$

The roots of the characteristic polynomial
$$
ϵλ^2+λ+1=0
$$
are
$$
λ_1=\frac{-1-\sqrt{1-4ϵ}}{2ϵ}=-\frac1ϵ+1+ϵ+2ϵ^2+5ϵ^3+...
$$
and
$$
λ_1=\frac{-1+\sqrt{1-4ϵ}}{2ϵ}=-\frac{2ϵ}{1+\sqrt{1-4ϵ}}=-(1+ϵ+2ϵ^2+5ϵ^3+...)
$$


问:
Prove whether a 4x4 matrix with three eigenvalues is diagonalizable (Without given matrix).
A is a 4x4 matrix with three eigenvalues. One eigenspace is one-dimensional, and one of the other eigenspaces is two-dimensional. Is it possible that A is not diagonalizable?</p>

答:
A matrix is diagonalizable iff there is a basis of eigenvectors for the vector space in which we are - or equivalently: if there is an invertible matrix $S$ such that $SAS^{-1}$ is a diagonal matrix.

Now we know that eigenvectors of distinct eigenvalues are linearly independet and therefore the union of all eigenvectors is a linearly independent system. If we have one eigenspace being two-dimensional and the other two eigenspaces one-dimensional we get a system of four linearly independent eigenvectors - these form a basis because obviously $K^4$ is four-dimensional (where $K$ is the field in which the entries of the matrix lie). So the matrix is diagonalizable. (Or we could say that setting $S$ as the matrix that consists of the four eigenvectors as columns would give us that $SAS^{-1}$ is a diagonal matrix).


问:
Is my solution to this homework problem in geometry correct? Can we avoid trigonometry in solving this?
I'm trying to solve this nice homework problem in (Euclidean) geometry - https://snipboard.io/oVp3Ic.jpg, and I wrote down a detailed solution with step-by-step explanation at: https://snipboard.io/ZzdsA0.jpg (click on the + sign on the image to make it bigger). Note that in the picture of the solution, DE=7cm. Also, note that, in the picture of the problem, the drawings aren't correct, because some sides AB an CE supposed to be parallel, but they're not drawn to be so, which could be misleading.

Some steps in the solution I gave:

i) I first solved for the angle ABC

ii) I expressed BC and CD as a trigonometric functions of the angle ABC (called $\theta.)$

iii) Observed $x=BC + CD.$

I'd greatly appreciate if you could please the take a bit of time to see if my solution is correct, and if we can avoid the use of trigonometry in solving this problem? I think in the picture of the solution, if you consider the similar right-angles triangles $CDE$ and $CGB,$ then we can write:

$$\frac{DE}{CE}=\frac{CG}{BG}$$

which is, algbraically ( where $y=AG$ in the picture of the solution.):

$$\frac{7}{6+y}=\frac{3}{\sqrt{(9-y)^2 + 3^2}}.$$

答: The key is just to find the common hypotenuse of two right triangles:


问:
what do these vectors span
What do the following vectors span?

$$\left(\begin{array}{c}1\\2\\3\end{array}\right),\left(\begin{array}{c}4\\5\\1\end{array}\right),\left(\begin{array}{c}5\\7\\4\end{array}\right) $$

I have tried the gaussian elimination method and end up with the matrix :

$$A=\begin{pmatrix}1& 4& 5\\ 0& -3& -3\\ 0& 0& 0\end{pmatrix}$$

The rank of this matrix is 2 and first two vectors are independent .How do i tell what do these vectors span? In my notes its given they span a 2D subspace of R^3 but i don't understand how we conclude this

答:
You are correct in your findings. Note how you can arrive at these more easily by seeing that if your above vectors are $x,y,z$, you get $x+y=z \land \forall c \in \mathbb{R}\setminus\{{0\}}: x \neq c \cdot y$

So clearly, $x,y$ are independent, but $x,y,z$ are not. 2 independent vectors span a 2-dimensional subspace (you can imagine them forming a plane, $x,y$ are the directions that define that plane and you find particular points in that plane by evaluating linear combinations of $x,y$).

More formally you would denote this as $span(x,y)=\{\lambda_{1} x+\lambda_{2} y \mid \lambda_{1} ,\lambda_{2} \in \mathbb{R}\}$.

Title: what do these vectors span


问:
What is a suitable choice of Lyapunov function for this system?
I have the following nonlinear system:

\begin{align}\dot x&=x(2-y^4) \\ \dot y &= -3x(1+y^3) \end{align}

I first verified this system using linearization; however, the eigenvalues are $\lambda =\pm i \sqrt{2} $ which is inconclusive for the stability of $(0,0)$. I tried to find a Lyapunov function to study its stability. I started with classical one $V(x,y) = a x^2 + b y^2$ by adjusting the coefficients $a,b$. However, it takes me hours but still no result. Any hints would be appreciated!! Thanks in advance.

答:
It's obviously unstable; trajectories close to the eigendirection (2,-3) move away from the origin. Just look at the the phase portrait. And if that isn't enough to make you believe it, you can first consider the system
$$
\dot x = 2-y^4
,\qquad
\dot y = -3(1+y^3)
,
$$
whose behaviour near the origin is very obvious (since now the origin is not an equilibrium), and then ponder what the effect of that extra factor of $\color{red}{x}$ in your system
$$
\dot x = \color{red}{x}(2-y^4)
,\qquad
\dot y = -3 \color{red}{x} (1+y^3)
$$
is.


问:
Yamabe Positive iff Admits Metric of Positive Scalar Curvature
I'm trying to prove that the Yamabe invariant of a compact manifold, that is the sigma constant, is positive iff $M$ admits a metric of positive scalar curvature. I will use $$Y_{[g]} = \underset{g \in [g]}{\inf} \mathcal{E}(g)$$ to be the Yamabe constant with

$$\mathcal{E}(g) = {\frac{{\int_{M}S_g \Omega}}{\left({\int_{M} \Omega}\right)^{{\frac{n-2}{n}}}}}$$

One direction is straightforward. That is, assume $\sigma(M) > 0$, then there exists some conformal class [g] such that $Y_{[g]} > 0$. Then by the Yamabe problem there exists a metric $\tilde{g} \in [g]$ such that $(M,\tilde{g})$ has constant positive scalar curvature.

I'm struggling with the other direction. Suppose $M$ admits a metric of positive scalar curvature $g_0$, say. We want to show that $$\sigma(M) = \underset{[g]\in \mathcal{C}_M }{\sup} Y_{[g]} > 0$$ with $\mathcal{C}_{M}$ being the set of conformal classes on $M$.

So we just need to show there exists a conformal class giving a positive Yamabe constant. Given the information we have I would imagine that this supremum is achieved by $Y_{[g_0]}$. But I'm having trouble showing that this is positive. What's stopping there being some other metric, say $\tilde{g_0} \in [g_0]$ such that $\mathcal{E}(\tilde{g_0}) \leq 0$ thus giving a non-positive value for $Y_{[g_0]}$?

答:
Let $g$ be a metric of positive scalar curvature.  We will show that $Y_{[g]}>0$.  Note that if $c>0$ is a constant, then $S_{cg}=c^{-1}S_g$, so there is no loss in generality in assuming that $0<\inf S \leq \frac{4(n-1)}{n-2}$.

Let $\hat g\in[g]$.  Write $\hat g = u^{\frac{4}{n-2}}g$ for some positive function $u$.  Then
$$ \tag{$\dagger$} \int S_{\hat g}\Omega_{\hat g} = \int \left( \frac{4(n-1)}{n-2}\lvert\nabla u\rvert_g^2 + S_gu^2 \right)\Omega_g \geq (\inf S_g) \lVert u\rVert_{1,2}^2 , $$
where $\lVert u\rVert_{1,2}^2$ is the $W^{1,2}(M^n)$-norm of $u$.  (The first equality follows from the formula for $S_{\hat g}$ in terms of $g$ and $u$.)  Note that $\inf S_g>0$ by assumption.  Now, the continuity of the embedding $W^{1,2}(M^n)\subset L^{\frac{2n}{n-2}}(M^n)$ implies that there is a constant $C>0$ such that
$$ \tag{$\ddagger$} \lVert w\rVert_{\frac{2n}{n-2}} \leq C\lVert w\rVert_{1,2} $$
for all $w\in W^{1,2}(M^n)$.  Combining ($\dagger$) and ($\ddagger$) implies that
$$ \int S_{\hat g}\Omega_{\hat g} \geq C^\prime \lVert u\rVert_{\frac{2n}{n-2}}^2 = C^\prime\left( \int \Omega_{\hat g} \right)^{\frac{n-2}{n}} , $$
where $C^\prime>0$ is determined by $\inf S$ and $C$; and the last equality is just the formula for $\Omega_{\hat g}$ in terms of $\Omega_g$.  This last display proves that $Y_{[g]}>0$.


问:
Knowing $\cos\theta$, can $\cos(n\theta)=\cos(\pi k)$?
$\cos(\pi k)=1$ or $-1$. After expressing $\cos(n\theta)$ in terms of $\cos\theta$, I have found that $\cos(n\theta)=\sum^{\lfloor{\frac{n}{2}}{\rfloor}}_{l=0}{n\choose2l}(-\frac{8}{9})^{l}(-\frac{1}{3})^{n-2l}$ but I cannot proceed. Is this the right approach? Thanks very much.

答:
Without lost of generality the question can be formulated

If $\cos\theta=-\frac{1}{3}$, is there some positive integer $n$ such that $\cos(n\theta)=1$?

This in turn implies that there exists integer number $k$ such that $\theta=\frac{2k}n\pi$. But this is impossible because by the Niven's theorem the only rational values of the sine or cosine of an angle which is a rational fraction of $\pi$ are $0, \pm\frac12$, and $\pm1$. The value $-\frac13$ is not in the list.


问:
Algebra word problem : magician
A magician is playing mind tricks with the audience. He asks one member of the crowd to think of two numbers, write them on pieces of paper and throw them into his hat. Meanwhile, the magician also thinks of a number and drops it into the hat. He then waves his wand and a shiny paper comes out of the hat. The numbers the member of the audience picked are both less than the number the magician dropped into the hat. The first number is 40 percent less than the magician's number, while the second number is 50 percent less than the same. The shiny paper has the percentage comparison of the two numbers from the audience member. What is the percentage on the shiny paper?

I took the numbers as  $x$, $0.6x$ and $0.5x$, but by that logic we get the answer as 120.

Answer given is 20

答:
You seem to have done everything correctly other than the fact that you need to find the percentage difference of both numbers given by the audience and not there percentage ratio. So rather than doing, $\frac{0.6}{0.5}*100\%=120\%$ do $\frac{0.6-0.5}{0.5}*100\%=20\%$.

Hope this helps!!!

Title: Algebra word problem : magician


问:
Is this ideal maximal?
I have a problem set in which a question in its premise assumes that $<7,\sqrt{10}>$ the principal ideal generated by 7 and $\sqrt{10}$ is a maximal ideal of $\mathbb{Z}[\sqrt{10}]=\{a+b\sqrt{10}:a,b$ belong to $\mathbb{Z}\} $.

But I showed that $<7,\sqrt{10}>$ is equal to $\mathbb{Z}[\sqrt{10}]$. So it can't be maximal, since in the definition a maximal ideal must be a proper ideal.

My work: I showed that $<7,\sqrt{10}> = \{7a+10b+c\sqrt{10} : a,b, c\in \mathbb{Z} \} = \{s+t\sqrt{10} : t, s  \in \mathbb{Z}\}=\mathbb{Z}[\sqrt{10}].$

I used the fact that 7 and 10 are relatively prime, so we can write any integer $n$ as a combination of them.
Is my work false? And if it is in fact a maximal ideal, how I can show that?

Edit: in the question it asks me to show that $<7,\sqrt{10}>$ is maximal by considering the quotient ring.

答:
Presumably a typo in the question: the ideal $\langle 7,\sqrt{10}\rangle$ is actually principal, but for the simple reason that it equals $\mathbb{Z}[\sqrt{10}]$, so it's definitely not maximal.

The question is probably about the principal ideal $\langle 7+\sqrt{10}\rangle$, which is a different beast.

Consider the homomorphism $\mathbb{Z}[X]\to\mathbb{Z}[\sqrt{10}]$ that maps $X$ to $\sqrt{10}$; its kernel is generated by the polynomial $X^2-14X-39$ and is so contained in the ideal generated by $X$ and $3$. So the idea might be to show that $3+\langle7+\sqrt{10}\rangle$ is not invertible in the quotient ring.

From $3(x+y\sqrt{10})=1+(u+v\sqrt{10})(7+\sqrt{10})$ we obtain
$$
3x=1+7u+10v,\qquad 3y=u+7v
$$
Work modulo $3$ and find a contradiction.

Title: Is this ideal maximal?


问:
Evaluating the series $\sum\limits_{n=1}^\infty \frac1{4n^2+2n}$
How do we evaluate the following series:

$$\sum_{n=1}^\infty \frac1{4n^2+2n}$$

I know that it converges by the comparison test. Wolfram Alpha gives the answer $1 - \ln(2)$, but I cannot see how to get it. The Taylor series of logarithm is nowhere near this series.

答:
Rewrite the series as follows:

$$\sum_{n=1}^\infty\frac{1}{4n^2+2n}=\sum_{n=1}^\infty\frac{1}{2n(2n+1)}=\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=\sum_{m=2}^\infty\frac{(-1)^m}{m}.$$

You may now evaluate it using the Taylor series of logarithm.

Added: The third equality is justified as follows:

Write $a_N=\sum_{n=1}^N\left(\dfrac{1}{2n}-\dfrac{1}{2n+1}\right)$ and $b_M=\sum_{m=2}^M\dfrac{(-1)^m}{m}$. The sequence of partial sums $b_M$ is convergent by the Leibniz criterion. Furthermore, $a_N = b_{2N+1}$ holds for all $N\in\mathbb N$, i.e. $(a_N)_{N=1}^\infty$ is a subsequence of $(b_M)_{M=1}^\infty$. Therefore these sequences converge to the same limit, which justifies the equality.


问:
Proving the Orthogonality of linear transformation
I am having trouble with the second part of the  following problem.

This is what I have tried.

How can I show that the map is norm preserving.
This last vector is now hard to deal with. What could be the possible mistake?

Any suggestions will be really helpful.

Thanks & regards

答: If $z=\bigl(a,(b,c,d)\bigr)$, with $a^2+b^2+c^2+d^2=1$, and $v=\bigl(x,(y,z,t)\bigr)$, then$$zv=\bigl(a x-b y-c z-d t,(a y+b x+c t-d z,a z-b t+c x+d y,a t+b z-c y+dx)\bigr)$$and therefore\begin{align}\|zv\|^2&=(a^2+b^2+c^2+d^2)(x^2+y^2+z^2+t^2)\\&=x^2+y^2+z^2+t^2\\&=\|v\|^2.\end{align}So,$$\|zv\overline z\|^2=\left\|z\left(v\overline z\right)\right\|=\left\|v\overline z\right\|^2=\|v\|^2.$$


问:
What is special with the word "limit" that it describes "approaches" in calculus?
I have been going through calculus class but could not figure out what does the word "limit" means here. I mean why do we use this Word not others. I found that the limit word means in Calculus, "a point or value which a sequence, function, or sum of a series can be made to approach progressively, until they are as close to it as desired". But can anybody simplify it?</p>

答:
I'll quote the entirety of the Earliest Known Uses of Some of the Words of Mathematics entry for Limit below, but I'll note here that Newton's discussion of "ultimate" velocities and ratios succinctly justifies the term thusly:

There is a limit which the velocity at the end of the motion may attain, but not exceed.

This suggests that the mathematical term is fundamentally intended to invoke the common meaning of "boundary" ... a separator between attainable and unattainable quantities.

This sense matches my head-canon (mentioned in a comment, expanded here) involving the classic scenario of taking a step forward, then a half-step, then a quarter-step, etc. In this process, there's a limit (in the English sense) to how far one can ultimately travel: namely, twice the length of that first step. You can attain any shorter distance, but none longer. And, while one can debate the (un)attainability of "twice the length of the first step" after infinitely many steps (whatever that's supposed to mean), the fact remains: that's the only distance subject to such debate, and we assign the term "limit" to that distinguished value.

(Importantly, the notion attainables and unattainables, and the separation thereof, is a bit nuanced. For instance, in considering the journey that is $\lim_{x\to 3} x$, there's nothing to keep $x$ from drifting higher or lower than ---or even onto--- the value of $3$; every value is "attainable" by the journeying $x$. However, formal definitions of limit convey that the process systematically and inevitably revokes "attainable" status from every value except $3$, so that $3$ "separates" the otherwise-empty set of attainables from the otherwise-universal set of unattainables. Again, one can debate the (un)attainability of $3$ itself at the end of the process (whatever that means), but only $3$'s status is debatable.)

From the Earliest Known Uses (emphasizing the particular sentence from Newton) ...

Gregory of St. Vincent (1584-1667) used terminus to mean the limit of a progression, according to Carl B. Boyer in The History of the Calculus and its Conceptual Development.

Isaac Newton wrote justifying limits in the Scholium to Section I of Book I of the Principia (Philosophiae Naturalis Principia Mathematica or The Mathematical Principles of Natural Philosophy) (first edition 1687)

Perhaps it may be objected, that there is no ultimate proportion, of evanescent qualities; because the proportion, before the quantities have vanished, is not the ultimate, and when they are vanished, is none. But by the same argument, it may be alledged, that a body arriving at a certain place, and there stopping, has no ultimate velocity: because the velocity, before the body comes to the place, is not its ultimate, velocity; when it has arrived, is none. But the answer is easy; for by the ultimate velocity is meant that with which the body is moved, neither before it arrives at its last place and the motion ceases, nor after, but at the very instant it arrives; that is, that velocity with which the body arrives at its last place, and with which the motion ceases. And in like manner, by the ultimate ratio of evanescent quantities is to be understood the ratio of the quantities not before they vanish, nor afterwards, but with which they vanish. In like manner the first ratio of quantities is that with which they begin to be. And the first or last sum is that with which they begin and cease to be (or to be augmented or diminished). There is a limit which the velocity at the end of the motion may attain, but not exceed. This is the ultimate velocity. And there is the like limit in all quantities and proportions that begin and cease to be. And since such limits are certain and definite, to determine the same is a problem strictly geometrical. But whatever is geometrical we may be allowed to use in determining and demonstrating any other thing that is likewise geometrical. (Translated by Andrew Motte 1729)

Katz (p. 471) comments, "A translation of Newton’s words into an algebraic statement would give a definition of limit close to, but not identical with, the modern one."

In 1821 Augustin-Louis Cauchy defined limit as follows: "If the successive values attributed to the same variables approach indefinitely a fixed value, such that they finally differ from it by as little as one wishes, this latter is called the limit of all the others." Cours d'analyse (Oeuvres II.3), p. 19. (Translation from Katz page 641) Cauchy introduced the modern ε, δ way of arguing.

This entry was contributed by John Aldrich. See also Limit and Delta and epsilon on the Earliest Use of Symbols of Calculus page.


问:
Equivalence of W^{m,p} and H^{s,p} norms
I am reading Cazenave's Semilinear Schroedinger Euqation.
On page xi the two norms are defined:

$\| u \|_{W^{m,p}} = \sum\limits_{|\alpha|\leq m} \| D^{\alpha} u \|_{L^p}$,

$\| u \|_{H^{s,p}} = \left\| \mathcal{F}^{-1} \left[ \left( 1+|\xi|^2 \right)^{\frac{s}{2}} \mathcal{F}u \right] \right\|_{L^p}$,

where $\mathcal{F}$ and $\mathcal{F}^{-1}$ are the Fourier and inverse Fourier transforms, respectively.

On page 13 (Remark 1.4.1, (iv)) it is stated that from Mihlin's multiplier theorem and for non-negative integer $m$, the equivalence between the norms of $W^{m,p}(\mathbb R^N)$ and $H^{m,p}(\mathbb R^N)$ follows.

Mihlin's multiplier theorem claims that if $\sigma$ is a function defined in $\mathbb R^N\setminus\{0\}$ which has at least $[N/2]+1$ continuous derivatives satisfying

$|\partial^{\alpha}\sigma(\xi)|\leq C_{\alpha}|\xi|^{-|\alpha|}$, for all $|\alpha|\leq [N/2]+1$ ($[l]$ is the integer part of $l$), then

$\| T_{\sigma}(u)(x) \|_{L^p}\leq C \| u \|_{L^p}$ for functions $u\in \mathcal{S}(\mathbb R^N)$, the constant $C$ depends on $\sigma$, $N$ and $p$, where

$T_{\sigma}(u)(x)=\int\limits_{\mathbb R^n}\hat{u}(\xi)\sigma(\xi)e^{2\pi i x\cdot\xi}d\xi=\mathcal{F}^{-1}(\sigma\mathcal{F}(u))(x)$.

Unfortunately I don't understand how Mihlin's theorem assures the equivalence $\| u \|_{W^{m,p}}=\| u \|_{H^{m,p}}$, i.e. $\sum\limits_{|\alpha|\leq m} \| D^{\alpha} u \|_{L^p} = \left\| \mathcal{F}^{-1} \left[ \left( 1+|\xi|^2 \right)^{\frac{s}{2}} \mathcal{F}u \right] \right\|_{L^p}$.

Can someone please share how the proof is structured? Thank you!

答:
The two norms are not equal, but they are equivalent which suffices in showing that $H^{m,p} = W^{m,p}.$ There are many sources for this, but one reference I think is quite clear is Theorem 6.2.3 of the following text:

Bergh, Jöran; Löfström, Jörgen, Interpolation spaces. An introduction, Grundlehren der mathematischen Wissenschaften. 223. Berlin-Heidelberg-New York: Springer-Verlag. (1976). ZBL0344.46071.

I will reproduce the argument here. To simplify notation we denote the associated Bessel potential operators
$$ J^{s}u = \mathcal F^{-1} \left( (1+|\xi|^2)^{\frac{s}2} \mathcal Fu(\xi)\right).$$
Note we have $\lVert u \rVert_{H^{m,p}(\Bbb R^N)} = \lVert J^mu\rVert_{L^p(\Bbb R^N)}.$

Given $|\alpha| \leq m,$ we can check that
$$ \sigma_{\alpha}(\xi) = \xi^{\alpha} (1+|\xi|^2)^{-\frac{m}2} $$
is a Mihlin multiplier; indeed it is smooth and $|\partial^{\beta}\sigma_{\alpha}| \leq C\min\{1,|\xi|^{|\alpha|-m-|\beta|}\} \leq C |\xi|^{-\beta}.$ Hence by the Mihlin multiplier theorem we have
$$ \lVert \partial^{\alpha}u \rVert_{L^p(\Bbb R^N)} = \lVert \mathcal F^{-1}\left(\sigma_{\alpha}(\xi) \mathcal F(J^mu)(\xi)\right)\rVert_{L^p(\Bbb R^N)} \leq C \lVert J^mu\rVert_{L^p(\Bbb R^N)}, $$
and so summing over all $|\alpha| \leq k$ gives
$$ \lVert u \rVert_{W^{m,p}(\Bbb R^N)} \leq C \lVert J^ku\rVert_{L^p(\Bbb R^N)} = C \lVert u \rVert_{H^{m,p}(\Bbb R^N)},$$
which proves one direction. For the converse let $\chi \in C^{\infty}(\Bbb R)$ be a cutoff such that $1_{(-1,1)} \leq 1-\chi \leq 1_{(-2,2)}$ and consider
$$\sigma_0(\xi) = (1+|\xi|^2)^{\frac{m}2} \left(1+ \sum_{j=1}^N \chi(\xi_j) |\xi_j|^m \right)^{-1}, \text{ and } \sigma_j(\xi) = \chi(\xi_j) |\xi_j|^m \xi_j^{-m}$$
for $1 \leq j \leq N.$ One can verify these are Mihlin multipliers, and that
$$ \left(1+\sum_{j=1}^N \sigma_j(\xi)\xi_j^m\right) \sigma_0(\xi) = (1+|\xi|^2)^{\frac m2}, $$
so we can compute
\begin{align*}
  \lVert J^mu \rVert_L^p(\Bbb R^N) &\leq C \left\lVert \mathcal F^{-1}\left(\left(1+\sum_{j=1}^N \sigma_j(\xi)\xi_j^m\right)\mathcal Fu\right)\right\rVert_{L^p(\Bbb R^N)} \tag{with $\sigma_0$} \\
&\leq C \left( \lVert \mathcal F^{-1}(\mathcal Fu)\Vert_{L^p(\Bbb R^N)} + \sum_{j=1}^N \lVert \mathcal F^{-1}(\sigma_j(\xi)\xi_j^m\mathcal Fu)\rVert_{L^p(\Bbb R^N)} \right) \tag{$\Delta$-inequality}\\
&= C \left( \lVert u\Vert_{L^p(\Bbb R^N)} + \sum_{j=1}^N \lVert \mathcal F^{-1}(\sigma_j(\xi)\mathcal F(\partial_j^mu))\rVert_{L^p(\Bbb R^N)} \right) \tag{Properties of $\mathcal F$}\\
&= C \left( \lVert u\Vert_{L^p(\Bbb R^N)} + \sum_{j=1}^N \lVert \partial_j^mu\rVert_{L^p(\Bbb R^N)} \right) \tag{with $\sigma_j$}\\
&\leq C \lVert u \rVert_{W^{m,p}(\Bbb R^N)},
\end{align*}
as required.

Added later: To prove the symbol estimates, the idea is not to explicitly compute the derivatives, but to separate the terms using the Leibniz rule and estimate the individual terms by an inductive argument. I will sketch the details to illustrate this for $\sigma_{\alpha};$ by the Leibniz rule we have
$$ \partial^{\beta}\sigma_{\alpha}(\xi) = \sum_{\gamma \leq \beta} {\beta \choose \gamma} \partial^{\beta-\gamma}\xi^{\alpha} \partial^{\gamma}\left( (1+|\xi|^2)^{-\frac m2} \right), $$
and we claim all the terms in the sum decay as $O(|\xi|^{-m+|\alpha|-|\beta|}).$ It is easy to see that $|\partial^{\beta-\gamma}\xi^{\alpha}| \leq C |\xi|^{|\alpha|-|\beta|+|\gamma|},$ noting that each partial derivative will decrease of the order of the polynomial if it is not already zero. Moreover noting the term vanishes if $|\alpha|-|\beta|+|\gamma|<0,$ we can also bound the term as
$$ |\partial^{\beta-\gamma}\xi^{\alpha}| \leq C(1+|\xi|^2)^{\frac{|\alpha|-|\beta|+|\gamma|}2}.$$
For the second term we claim by induction on $|\gamma|$ that for all $k$ we have
$$ \partial^{\gamma}\left( (1+|\xi|^2)^{-\frac k2} \right) \leq C(1+|\xi|^2)^{-\frac{k+|\gamma|}2}. $$
Here we write $\gamma = \gamma' + e_i,$ applying the inductive hypothesis for $\gamma'$ and $k+1.$

Putting everything together we arrive at the estimate
$$ \partial^{\beta}\sigma_{\alpha} \leq C (1+|\xi|^2)^{\frac{-m+|\alpha|-|\beta|}2} \leq C (1+|\xi|^2)^{-\frac{|\beta|}2}, $$
where we have used the fact that $|\alpha| \leq m$ in the last inequality.

For the other two terms we need to distinguish when each $|\xi_j| \leq 1$ or not; if $|\xi_j|<1$ then $\sigma_j(\xi) = 0$ and we can also since $\chi_j(\xi)=0$ the expression for $\sigma_0$ simplifies. If $|\xi_j|>1$ then $\chi_j(\xi)$ is just a smooth function whose derivatives can be bounded. One needs a similar induction argument to estimate the derivatives of $(1+\sum_{j=1}^N \chi_j(\xi)|\xi|^m)^{-1}$ (considering cases at each step), but I hope you can see how to fill in the details.


问:
Why is there this recurring pattern in the derivative of $x\cdot e^{-k\cdot x}$?
I was doing an exercise for which I needed to take the derivative(s) of the function

$$f_k(x)=x\cdot e^{-k\cdot x}$$

Doing this a few times, I noticed some sort of recurring pattern. I concluded that one can write the $n$-th  derivative of the function as

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}((-1)^nkx-(-1)^n\cdot n)$$

or perhaps, to improve readability

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}(-kx+n)$$
for odd $n$ and

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}(kx-n)$$
for even $n$.

While I know how to compute each derivative, I do not understand why the derivative follows this pattern-like structure. Is there any intuitive idea behind this?

Clarification: By "intuitive", I mean anything that does not purely rely on blindly computing each integral. A very beautiful way to represent derivatives visually is shown in this 3blue1brown video; I don't know if anything like this is possible to explain the above described pattern. Basically, I am looking for something which actually explains why there is a pattern instead of just noticing it.

Note: This is not the actual homework assignment, it just came up when doing it.

答:
The reason behind the pattern relies merely on the taylor expansion of the exponential:
$$e^{x}=\sum_{n=0}^{+\infty}\frac{x^{n}}{n!}$$
and if you take it with $-k$ in the exponent then:
$$e^{-kx}=\sum_{n=0}^{+\infty}(-1)^{n}k^{n}\frac{x^{n}}{n!}\implies xe^{-kx}=\sum_{n=0}^{+\infty}(-1)^{n}k^{n}\frac{x^{n+1}}{n!}$$
Now, if you consider the $n-th$ derivative of $xe^{-kx}$ you will see it is exactly the $n-th$ derivative of the series here above. As for intution, it is actually a weird question. Keep in mind that, for instance, in nature the concept of "continous function" doesn't really make sense: all the measures you will take in your life will be discrete quantities. The idea of continous quantity it's somthing that helps us when doing calculus. The interpretetion given by 3Blue1Brown is indeed appealing but it is actually unprecise: $dx$ is not an "infinitesimal" quantity, not since we moved from the interpretetion Leibniz gave of calculus when he basically invented it. You should see in $dx$ a way of $\textit{measure}$ something. In mathematics we measure sets, and if you will deepen your knowledge about measure theory you will see what it means exactly. Moreover you will see in which sense $dx$ is just a measure.


问:
Showing for each zero a of f, there is an open disk D(a; r) ⊂ Ω such D(a; r) contains no other zeros of f.
*This seems a very well-known result in complex analysis, still I found it very hard to interpret.

Also f is analytic.

I try to attack this problem in the following way:

We can assume that a is a zero of f of order n, then $f(z) = (z-a)^n g(z)$.
Then g : D(a; R) → C is also an analytic function and $g(a) \not= 0.$

But here I am lost, it seems that g may have other zeros in D(a; R), and I don't see how analyticity helps us.

答: $g$ is continuous. Hence,  there exists $\delta >0$ such that $|g(z)-g(a)| <\frac {|g(a)|} 2$ for $|z-a|<\delta$. This implies such that $|g(z)| \geq \frac {|g(a)|} 2$ for $|z-a| <\delta$. There is no zero of $g$ in $B(a,\delta)$.


问:
Property of abelianization
This is related to this old MO question which wasn't answered properly, though I don't feel I phrased the question in the best way (or posted it on the right site)

Define the abelianization of a group $G$ to be the quotient group $G_{ab} := G/[G,G]$, where $[G,G]$ is the commutator subgroup. I want to know how this definition implies the following property of abelianization.

Let $\phi: G \to G_{ab}$ be the canonical surjection. For any abelian group $H$ and homomorphism $f:G\to H$, there exists a unique homomorphism $F: G_{ab} \to H$ such that $f = F\circ \phi$.

This is the ability to 'descend to a homomorphism' that I explained so badly in my initial post, though at the time it was the only terminology I heard being used.

I am aware that these two definitions are equivalent but I have not yet seen a proof, nor managed to prove it myself. Please do point me to an online proof if you know of one.

答: If $f : G \to H$ is a homomorphism to an abelian group, then $f(ab) =  f(a) f(b) = f(b) f(a) = f(ba)$, hence $[a, b] \in \ker f$, hence $[G, G] \subseteq \ker f$. Is the rest clear from here? </p>

Title: Property of abelianization


问:
Difference of limits of sequences
Let $(x_n)_{n=1}^\infty$ and $(y_n)_{n=1}^\infty$ be sequences where $x_n\to a$ $(a\in\mathbb{R})$ and $y_n\to \infty$. How can I show that $x_n-y_n\to -\infty$?

I'm guessing I have to use an $\varepsilon$-N-definition of convergent sequences but I don't know how to (or if I should) combine the definitions of convergence and divergence.

答:
Need to show that for every $M<0$, there exists $N\in\mathbb{N}$, such that
$$\forall n>N\quad x_n-y_n<M$$

For that, you can pick an $N_1\in\mathbb{N}$ such that $|x_n-a|<\epsilon$, for every $n>N_1$ and some $\epsilon>0$. Then pick an $N_2$ that is high enough to make $x_n-y_n<M$. What do you need $y_n$ to be? Notice that it can be as big as you want for high enough $N_2$, so use that. Since $M$ is arbitrary, you will have proven your claim.

Title: Difference of limits of sequences


Title: Evaluating the series $\sum\limits_{n=1}^\infty \frac1{4n^2+2n}$

Body:
How do we evaluate the following series:

$$\sum_{n=1}^\infty \frac1{4n^2+2n}$$

I know that it converges by the comparison test. Wolfram Alpha gives the answer $1 - \ln(2)$, but I cannot see how to get it. The Taylor series of logarithm is nowhere near this series.

Body: Hint:  $\frac 1{4n^2+2n} = \frac 1{2n}-\frac 1{2n+1}$


问:
Difficulty on a problem on conditional probability (Problem 3.16 of "A modern introduction to Probability and Statistics")
I am studying from the book A modern introduction to Probability and Statistics, by F.M. Dekking et al.
I have a problem with this exercise on conditional probability (Ex. 3.16 from Chapter 3):

You are diagnosed with an uncommon disease. You know that there
only is a $1$% chance of getting it. Use the letter $D$ for the event “you have the
disease” and $T$ for “the test says so.” It is known that the test is imperfect:
$P(T\mid D) = 0.98$ and $P(T^c \mid D^c) = 0.95$.

a. Given that you test positive, what is the probability that you really have
the disease?

b. You obtain a second opinion: an independent repetition of the test. You
test positive again. Given this, what is the probability that you really have
the disease?

My attempt:

a. I know from the text that $P(D)=0.01$, furthermore, because $P(\cdot \mid D^c)$ is a probability, I have that
$$P(T \mid D^c) = 1 - P(T^c \mid D^c) = 0.05$$
so I can compute $P(D\mid T)$ by Bayes' formula:
$$P(D\mid T) = \frac{P(T\mid D)P(D)}{P(T)}$$
Next I use the law of total probability to compute $P(T)$:
$$P(T) = P(T\mid D)P(D) + P(T\mid D^c)P(D^c)=0.98\cdot 0.01 + 0.05\cdot 0.99 = 0.0593$$
so that
$$P(D\mid T) = \frac{P(T\mid D)P(D)}{P(T)}=\frac{0.98\cdot 0.01}{0.0593}\simeq 0.1653$$

b. Here I don't know how to formalize the solution. I'd put sloppily
$$T_1 = \text{the first repetition of the test is positive}$$
and
$$T_2 = \text{the second repetition of the test is positive}$$
so I think I have to compute the conditional probability $$P(D\mid T_1\cap T_2)=\frac{P(D\cap T_1\cap T_2)}{P(T_1\cap T_2)}.$$
Now, I think that by the independence of the two repetitions of the test I can safely assume that
$$P(T_1\cap T_2) = P(T)^2.$$
But how to compute $P(D\cap T_1\cap T_2)$? And there is a way to better formalize the solution?

Thanks in advance for your help!

答:
It's a very standard exercise

The probabilty to have a positive test is

$$P(T)=0.01\times0.98+0.05\times0.99=0.0593$$

Thus

$$P(D|T)=\frac{0.01\times0.98}{0.01\times0.98+0.05\times0.99}\approx 16.53\%$$

If you have two consecutive independent positive tests you probability increases

$$P(D|TT)=\frac{0.01\times0.98^2}{0.01\times0.98^2+0.05^2\times0.99}\approx 79.51\%$$


答:
Here's a basic approach where the overwhelming majority of the work is squaring three 9-digit numbers. Make of that what you will!

We find that $75$ is $135$ in base 7. I'll write $\ldots$ to mean some sequence of zeros. Successively square using arithmetic in base 7 to get

\begin{align*}
135^1 &= 135 \\
135^2 &= 22254 \\
135^4 &= 22254^2 = 532640502 \\
135^8 &= 532640502^2 = 420605123423162604 < 420605124\ldots \\
135^{16} &< 420605124\ldots^2 < 243433061\ldots \\
135^{32} &< 243433061\ldots^2 < 66666663\ldots
\end{align*}

If you count the number of $0$'s in the final $\ldots$, you'll see that $135^{32}$ (in base 7) is at most 71 digits, which is less than $7^{71}$ which is 72 digits.

The $22254^2 = 532640502$ step would be tedious by hand, but really not bad. The obnoxious bit would be $532640502^2 = 420605123423162604$ and the two virtually identical calculations after it. Strictly speaking you only have to do enough calculations to assure yourself the claimed upper bounds hold, though that probably doesn't help much.

You could use Karatsuba's algorithm to make the big squaring operations less painful. For instance,

\begin{align*}
532640502^2 &= (5326\ldots + 40502)^2 \\
&= 5326\ldots^2 + 2 \cdot 5326\ldots \cdot 40502 + 40502^2 \\
&= 42051211\ldots + 630111343\ldots + 2255562604 \\
&= 4206051233(\cdots) \\
&< 420605124\ldots.
\end{align*}

In all you'd need to do 4 5-digit squares, 3 4-digit squares, 3 4-digit and 5-digit products, and some easy additions and other little things. That's actually getting pretty reasonable.

Title: Prove $7^{71}>75^{32}$

Body:
My math teacher left two questions last week, prove (1) $6^9>10^7$ and (2) $7^{71}>75^{32}.$

I did the first question: \begin{align}\frac{6^9}{10^7}&=\frac{4}{5}\times\frac{27^3}{25^3}\\&=0.8\times1.08^3\\&>0.8\times(1+3\times0.08+3\times0.08^2)\\&>0.8\times(1+3\times0.086)\\&>0.8\times1.25=1.\end{align}

But I can't work out the second,  I calculated it out on my computer, $\frac{7^{71}}{75^{32}}=1.000000949\cdots$


问:
Showing that every nonstandard model of arithmetic contains an infinite element
I am not quite sure how to prove this. Well, if $\mathfrak{N}$ is the standard model of arithmetic then by definition, a nonstandard model of arithmetic $\mathfrak{N}'$ must be elementarily equivalent to $\mathfrak{N}$ but not isomorphic to $\mathfrak{N}$.

The sentence $(\forall x)( \exists y)(x<y)$ is true in $\mathfrak{N}$ and so it must be true in $\mathfrak{N}'$ as well. But then how will I find a element $b$ in the universe of $\mathfrak{N}'$ such that $a<^{\mathfrak{N}'}b$ for each $a$ in the universe of $\mathfrak{N}'$?

I'm not sure how to proceed. Hints will be appreciated.

Here's my attempt after receiving hints from Asaf Karagila and Rob Arthan
:

Let $N$ be the set of Peano Axioms (see Example 2.8.3). Let $\mathfrak{N}$ be the standard model of arithmetic and $\mathfrak{N}'$ be a nonstandard model of arithmetic. Let $\overline{n}$ be the $\mathcal{L}_{NT}$ term given by $\overline{n}=\underbrace{SS\ldots S}_{n \text{ many}}0$. Let $\Sigma=\{ x=0 \vee x=\overline{1} \vee\ldots\vee x=\overline{n} \vee  \overline{n}<x : n\in \mathbb{Z}^{+}\}$. Let $A$ be the universe of $\mathfrak{N}'$

Claim 1. . For each $n\in\mathbb{Z}^{+}$, $N\vdash x<\overline{n} \leftrightarrow x=0 \vee x=\overline{1} \vee \ldots \vee x=\overline{n-1} $.

Proof of Claim 1. We proceed by induction on $n$. Consider the case when $n=1$. We have to prove that $N\vdash x<S0\leftrightarrow x=0$. To prove this, we use Quantifier Axioms and 10th Axiom of Peano Axioms (see Example 2.8.3). We have $N\vdash x<S0 \leftrightarrow x<0 \vee x=0$. But then by the 9th Axiom, $N\vdash \neg x<0$. So by Propositional Consequence, $N\vdash x<S0 \leftrightarrow x=0$ which is what we wanted.

Assume that we already have $N\vdash x<\overline{n} \leftrightarrow x=0 \vee x=\overline{1} \vee \ldots \vee \overline{n-1} $ for some $n\in \mathbb{Z}^{+}$. Let's prove that $N\vdash x<\overline{n+1} \leftrightarrow x=0 \vee x=\overline{1} \vee \ldots \vee x=\overline{n} $.

Now, $N\vdash x<\overline{n+1}\leftrightarrow x=\overline{n} \vee x<\overline{n}$ by the 10th Axiom and Quantifier Axiom. By Propositional Consequence and Induction hypothesis, we have what we wanted to prove.

Claim 2. For each $\sigma \in \Sigma$, $N \vdash \sigma$.

Proof of Claim 2. Let $n\in \mathbb{Z}^{+}$. We show that $N\vdash x=0 \vee x=\overline{1} \vee\ldots\vee x=\overline{n} \vee  \overline{n}<x$. By Axiom 11 and Quantifier Axioms, $N \vdash x<\overline{n} \vee x =\overline{n} \vee \overline{n}<x$. It is easy to see that Claim 1 and Propositional Consequence complete the proof of Claim 2.

Claim 3. $A\setminus \{ \underbrace{S^{\mathfrak{N}'}(S^{\mathfrak{N}'}(\ldots S^{\mathfrak{N}'}}_{n \text{ times}}(0^{\mathfrak{N}'})\ldots)) : n \ge 0 \}$ is nonempty for otherwise $\mathfrak{N}\simeq \mathfrak{N}'$.

(Proof of Claim 3 is pretty long but straightforward, so, I will not prove it here).

Now, $\mathfrak{N}\models \Sigma$ and since $\mathfrak{N}'$ is elementarily equivalent to $\mathfrak{N}$, we have $\mathfrak{N}'\models \Sigma$. Let $a$ be any nonstandard element of $\mathfrak{N}'$ (Claim 3 proves the existence of such an element). Let $s$ be any variable assignment function into $\mathfrak{N}'$ which sends $x$ to $a$.
Then for each $n\in \mathbb{N}$, $\mathfrak{N}' \models x=0 \vee x=\overline{1} \vee\ldots\vee x=\overline{n} \vee  \overline{n}<x$. So it must be that $\underbrace{S^{\mathfrak{N}'}(S^{\mathfrak{N}'}(\ldots S^{\mathfrak{N}'}}_{n \text{ times}}(0^{\mathfrak{N}'})\ldots)) <^{\mathfrak{N}'} a$ for each $n\in \mathbb{N}$.

答:
First of all, normally when we say "model of arithmetic" we mean a model of a theory which describes arithmetic, e.g. Peano, rather than true arithmetic.

Secondly, note that an "infinite element" is not $0$, or $s0$, or $ss0$, or $sss0$, or $ssss0$, or $sssss0$, and so on and so on (here $s$ is the successor operation).

Suppose that a model does not contain an infinite element. Well, it must contain $0$ and $s0$ and $ss0$ and $sss0$ and $ssss0$ and so on and so on. But we just said, that it doesn't contain any other elements. So using this, you can now show that it is isomorphic to the standard model.

(Note that your approach is inherently flawed: the statement "$x$ is a non-standard integer" is not first-order definable! It is a property of the model, not an internal property of the element.)


问:
Prove that the acceleration is either zero or parallel to the velocity given the curve lies on a plane
I'm stuck on this problem (section 14.9 exercise 13 of Apostol Calculus) :

A particle moves along a curve in such a way that the velocity vector makes a constant angle with a
given unit vector $\mathbf{c}$ . If the curve lies in a plane containing $\mathbf{c}$, prove that the acceleration vector  is either zero or parallel to the velocity

My attempt :

Define a unit vector $\displaystyle  \mathbf{T} = \frac{\mathbf{v}}{\|\mathbf{v}\|}$,  so $\displaystyle \mathbf{T} \ ' =  \frac{\mathbf{a}\|\mathbf{v}\|-\mathbf{v}\|\mathbf{v}\|'}{\|\mathbf{v}\|^2}$.

Then, because of the angle between the velocity $v$ and $c$ is constant we have

$$\mathbf{T} \cdot \mathbf{c} = \cos \alpha \implies \mathbf{T}\ ' \cdot \mathbf{c} = 0 \implies \mathbf{T} \ '=\mathbf{0} \  \lor \  \mathbf{T}\ ' \bot \ \mathbf{c}  $$

Also,

$$\mathbf{T} \cdot \mathbf{T}\ = 1 \implies \mathbf{T} \ ' \cdot v = 0 \implies    \mathbf{T} \ '=\mathbf{0} \  \lor \  \mathbf{T}\ ' \bot \ \mathbf{v}$$

If it is true that $ \mathbf{T}\ '=\mathbf{0} $ then $\displaystyle \mathbf{a}=\frac{\|\mathbf{v}\|'}{\|\mathbf{v}\|}\mathbf{v}  $ ,  so we have done.

Edit:

But as @user pointed out, $\mathbf{T} '$ cannot be perpendicular to both $\mathbf{v}$ and $\mathbf{c}$ because $\mathbf{v}$ and $\mathbf{a}$ have to lie on the same plane as $\mathbf{c}$ so does $\mathbf{T} '$, but this means it must be $\mathbf{0}$ because there isn't a non zero vector on a plane perpendicular to $\mathbf{v}$ and $\mathbf{c}$.

I think all of my problem was that I was not aware of the definition of vector lying on a plane, i thought just the spike of it was included in the plane.

答:
Consider another hypothetical particle, which starts at the origin and its position vector (which is therefore also the displacement vector) is same as the velocity vector of the given particle, at any given time.

Now since the position vector of this particle is always at constant angle to $\mathbf c$, so its trajectory is a cone, with origin as apex and $\mathbf c$ to be the axis.

Also, since the direction of movement of any particle is actually the direction of its velocity, so the velocity of the given particle (and hence the displacement of this particle) is in a plane containing $\mathbf c$.

Hence by the geometry of a cone, this particle moves in a straight line (since changing direction would either make its trajectory non-coplanar with $\mathbf c$, or will change the angle with $\mathbf c$). Hence the velocity of this particle (and thus the acceleration of the given particle) is always in the same direction as the displacement of this particle (velocity of given particle), or zero. This is what we wanted to show.


答:
$$\dfrac1{4n^2+2n} = \dfrac1{2n(2n+1)} = \dfrac1{2n} - \dfrac1{2n+1}$$
Hence,
\begin{align}
\lim_{m \to \infty} \sum_{n=1}^{m} \dfrac1{4n^2+2n} & = \lim_{m \to \infty} \sum_{n=1}^{m} \left(\dfrac1{2n} - \dfrac1{2n+1} \right)\\
& = \lim_{m \to \infty}\left(\dfrac12 - \dfrac13 + \dfrac14 - \dfrac15 \pm \cdots +\dfrac1{2m} - \dfrac1{2m+1}\right)\\
& = 1 - \lim_{m \to \infty}\left(1-\dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 \mp \cdots -\dfrac1{2m} + \dfrac1{2m+1}\right)\\
& = 1 - \lim_{m \to \infty}\left(1-\dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 \mp \cdots -\dfrac1{2m} \right) - \lim_{m \to \infty}\dfrac1{2m+1}\\
& = 1 - \lim_{m \to \infty}\dfrac1{2m+1} - \lim_{m \to \infty} \sum_{n=1}^{2m} \dfrac{(-1)^{n-1}}{n}\\
& = 1 - 0 - \log(1+1)\,\,\, \left\{\text{Since $\log(1+x) = -\displaystyle \sum_{n=1}^{\infty}\dfrac{(-x)^{n}}n$} \right\}\\
& = 1 - \log(2)
\end{align}</p>


问:
Find all entire holomorphic on $ \mathbb{C} $ which satisfies the following condition
I want to find all entire holomorphic on $ \mathbb{C} $ functions $ f=u+iv $ such that $u(z) =|z|^2 $.

Here's what I have tried;

Assume $ f\left(z\right)=|z|^{2}+iv\left(z\right)=\left(x^{2}+y^{2}\right)+iv\left(x,y\right) $. I'll try to identify $ f $.

Since $ f $ is holomorpgic, Cauchy-Riemman equation should hold. So:

$$ \begin{cases}
u_{x}=v_{y}\\
u_{y}=-v_{x}
\end{cases}\Rightarrow\begin{cases}
2x=v_{y}\\
-2y=v_{x}
\end{cases} $$

Thus integrate both sides of the equation we get $$ \begin{cases}
2xy+C_{1}=v\left(x,y\right)\\
-2xy+C_{2}=v\left(x,y\right)
\end{cases} $$

And thus if we'll sum the equations and divide by $ 2 $ we get $ v\left(x,y\right)=\frac{C_{1}+C_{2}}{2} $. That is, $v $ is a constant function, so

$$ \left\{ f:f=|z|^{2}+iv,\thinspace\thinspace f\thinspace\thinspace is\thinspace\thinspace\thinspace entire\right\} =\left\{ f:f=|z|^{2}+ic,\thinspace\thinspace\thinspace where\thinspace\thinspace\thinspace c\in\mathbb{R}\right\}  $$

I kind of treated the functions as one variable functions when I integrated in order to find $ v(x,y) $ so im not sure with my result.

My questions are:

Does my way correct?

Is there a simpler way?

Thank in advance, it is highly appreciated.

答:
Actually, there is no such function. The function $(x,y)\mapsto x^2+y^2$ is not harmonic, but the real and the imaginary parts of a holomorphic functions are always harmonic functions.

Concerning your approach, from $v_y=2x$ and $v_x=-2y$, you should deduce that $v=2xy+C_1(x)$ and $v=-2xy+C_2(y)$, for some functions $C_1$ and $C_2$. But$$2xy+C_1(x)=-2xy+C_2(y)\iff4xy=C_2(y)-C_1(x).$$And there are no two such functions $C_1$ and $C_2$.


问:
Characterising Hilbert spaces
I found this question:

Characterization of Hilbert spaces using orthogonal decomposition

It deals with the question of when pre-Hilbert spaces are Hilbert. It proves that (assuming separability)

A pre-Hilbert space $H$ is Hilbert if for all subspaces $K$
$$K^\perp = \{0\} \implies K\; \text{dense in}\; H $$

But I'm afraid I do not really understand the answer. In particular I would have thought that neither $\eta$ nor $\xi$ lie in $K$.

Additionally I would very much like a proof in the non-separable case. There is a comment saying it should be possible, but again I cannot fill in the details.

Thanks in advance!

答:
Let $H$ be a pre-Hilbert space, denote with $\mathcal H$ its completion which is Hilbert. For a subset $K$ of $H$ denote with $K'$ the orthogonal complement in $H$ and $K^\perp$ the orthogonal complement in $\mathcal H$. Its clear that $K'\subseteq K^\perp$ and even that $K'=K^\perp \cap H$.

Suppose there is a vector $v\in \mathcal H - H$ (ie that $H$ is not Hilbert), then the span $V:=\Bbb C\cdot v$ is a closed sub-space of $\mathcal H$ and define:
$$K := V^\perp \cap H.$$
Note that since $H$ is dense in $\mathcal H$ you have that $K$ is dense in $V^\perp$ in $\mathcal H$. In particular you have that
$$K^\perp = (\overline K)^\perp = (V^\perp)^\perp = V$$
since $V$ is closed. But $K'= K^\perp\cap H = V\cap H = \{0\}$. However $K$ cannot be dense in all of $H$, because then it must be dense in $\mathcal H$ and then $V^\perp$ must be dense in $\mathcal H$. But $v$ itself cannot be approximated by elements of $V^\perp$.

This shows
$$[\text{$H$ is not Hilbert}]\implies [\text{there exists a not dense subspace $K$ with $K\neq H$ and $K^\perp = \{0\}$}]$$
independently of any countability assumptions.

The other direction is elementary as $(K^\perp)^\perp = \overline K$ in a Hilbert space, and $\{0\}^\perp$ is always the entire space.

Title: Characterising Hilbert spaces


问:
Function for which $f(n) = 1$ is true only if n is prime.
I was thinking a bit about the primes so I created a function which I checked in Wolfram.

It seems that this function is doing what i expected. It returns $1$ if value is prime otherwise it returns $0$.

Let:
$$f(n) = \lceil \frac{\Gamma(n)^2}{n} \rceil- \lfloor\frac{\Gamma(n)^2}{n} \rfloor$$
$$n \in \Bbb{N}$$
Then:

If $n$ is prime then $f(n) = 1$ else $f(n) = 0$.

Function in wolfram: my function.

I know that to truly say that this is true, there is needed a proof.

Can such function be useful in math or it is just a clever useless property ?

Can we say anythink more about this function ?

Are there any other interesting functions that are strongly connected with prime numbers ?

I know that there are :

prmie counting function

rieman zetta function

formula for primes

答:
Well, $$\lceil x \rceil - \lfloor x \rfloor = \left\{
\begin{array}{ll}
      0 \ \ x \in \mathbb{Z} \\
      1 \ \ x \notin \mathbb{Z}
\end{array}
\right. $$

Note that $\frac{\Gamma(n)}{n}^2 = \frac{((n-1)!)^2}{n}$ is an integer $\iff$ $n$ is not a prime number.


问:
Derive cdf of Cauchy distribution from standard normal distribution
This post is about the process to calculate the cdf of Cauchy distribution from $X, Y \overset{i.i.d} \sim N(0, 1)$

\begin{aligned}
&P(\lbrace X/Y\leqslant t, Y>0 \rbrace\cup\lbrace X/Y\leqslant t, Y<0 \rbrace) \\
=\; &P(X/|Y|\leqslant t, Y>0) + P(-X/|Y|\leqslant t, Y<0) \quad \\
=\; &P(X/|Y|\leqslant t, Y>0) + P(X/|Y|\leqslant t, Y<0)\\
=\; &P(X/|Y|\leqslant t)
\end{aligned}

But I still have a problem to understand why
$P(\lbrace X/Y\leqslant t, Y>0 \rbrace\cup\lbrace X/Y\leqslant t, Y<0 \rbrace)$ is equivalent to $P(X/|Y|\leqslant t, Y>0) + P(-X/|Y|\leqslant t, Y<0)$

First, why is it valid to covert $Y$ to $\mid Y \mid$? It seems that it is related to the symmetry of the normal distribution, but I failed to make a mathematical proof.

Second, the reason $X$ is converted to $-X$ in second line is both $X$ and $-X$ are same distribution with $N(0, 1)$?

Please give me some detailed and answer. Thanks for your help.

Body:
$$
Z:=\frac{X}{Y}1\{Y\ne 0\}=\frac{X}{|Y|}1\{Y>0\}-\frac{X}{|Y|}1\{Y<0\}+0\cdot\{Y=0\}.
$$
Since the sets $\{Y>0\}$, $\{Y<0\}$, and $\{Y=0\}$ are disjoint,
\begin{align}
\mathsf{P}(Z\le t)&=\mathsf{P}(\{Z\le t\}\cap \{Y>0\}) \\
&\quad+\mathsf{P}(\{Z\le t\}\cap \{Y<0\}) \\
&\quad+\mathsf{P}(\{Z\le t\}\cap \{Y=0\}) \\
&=\ldots
\end{align}
Note that $Z=X/|Y|$ on $\{Y<0\}$, $Z=-X/|Y|$ on $\{Y<0\}$, and $\{Y=0\}$ is a null set.


问:
Algebraic Solution of Inequalities
Ok I'm doing some Inequalities revision and I know what the signs mean and stuff but this question I'm a bit stuck on. You need to find the x such that $\frac{1}{4} (2x + 1) \leq \frac{3}{5} (3x - 1)$. </p>

答:
$\frac{1}{4} (2x + 1) \leq \frac{3}{5} (3x - 1)$

First Steps:

1) Distribute the 1/4 and 3/5, respectively into their parentheses

2) Isolate x on one of the sides of the inequality, just as you would for an equation.

$\frac{x}{2}-\frac{9x}{5}\leq\frac{-3}{5}-\frac{1}{4}$

Can you go on from here? Note that multiplication or division by negative numbers will switch the sign of the inequality.

Title: Algebraic Solution of Inequalities


问:
Find solution to complex equation $iz^2+(3-i)z-(1+2i)=0$
Find all the complex solutions to the equation $$iz^2+(3-i)z-(1+2i)=0$$

I've tried to solve this equation with two different approaches but in both cases I couldn't arrive to anything.

1) If $P(z)=iz^2+(3-i)z-(1+2i)$, then the original problem is equivalent to finding the roots of $P$. If I consider the polynomial $P(z)\overline{P}(z)$, then this polynomial has integer coefficients, which I thought it could make things a little bit easier.

So $P(z)\overline{P}(z)=x^4-3x^3+6x^2-10x+5$, I didn't know what to do after this.

2) I've tried to solve the equation directly, $$iz^2+(3-i)z-(1+2i)=0$$ if and only if $$z(iz+3-i)=1+2i$$

The number of the left equals the number of the right if and only if they have the same module and argument. At this point I got stuck trying to actually calculate the argument and modulus of the product on the left side of the equation.

I would appreciate any help, thanks in advance.

答:
Use the AC method, just with complex numbers!

Multiply by $i$ to get $(iz)^2 + (3-i)(iz) - (i - 2) = 0$. Now let $u = iz$. Then we have $u^2 + (3-i)u - i+2 = 0$, which can be factored as $(u + 1)(u + (2-i)) = 0$. Thus $u = -1, -2+i$ and since $u=iz$, $z= \frac{-1}{i}, \frac{-2+i}{i} = i, 2i+1$.


问:
Converse to a result of convergence of $\lim\frac{1}{x}\sum_{n\leq x}a_n=k$
In this post it is proved that, if $(a_n)$ is a sequence in $\mathbb C$:

$$
\lim_{x\to\infty}\frac{1}{x}\sum_{n\leq x}a_n=k\implies\lim_{x\to\infty}\frac{1}{\log x}\sum_{n\leq x}\frac{a_n}{n}=k.
$$

Is the converse to this statement true? In other words, is there a sequence $(a_n)$ with

$$
\lim_{x\to\infty}\frac{1}{\log x}\sum_{n\leq x}\frac{a_n}{n}=k
$$

but $\lim_{x\to\infty}\frac1x\sum_{n\leq x}a_n$ converges to some other limit, or diverges? Thanks.

答:
Without loss of generality, we set $k=1$ and define

$$
A(x)=\sum_{1\le n\le x}{a_n\over n}=\log x+R(x)
$$

wherein $R(x)=o(\log x)$. Then, it follows from partial summation that

$$
\begin{aligned}
\sum_{n\le x}a_n
&=\int_{1^-}^xt\mathrm dA(t)=xA(x)-\int_1^xA(t)\mathrm dt \\
&=x\log x+xR(x)-\int_1^x\log t\mathrm dt-\int_1^xR(t)\mathrm dt \\
&=x\log x-xR(x)-[t\log t-t]_1^x-\int_1^xR(t)\mathrm dt \\
&=x-1-xR(x)-\int_1^xR(t)\mathrm dt
\end{aligned}
$$

Certainly, more constraints are needed onto $R(x)$ in order for us to reduce the remaining terms (or otherwise we can deduce PNT directly from Mertens' first theorem).


问:
Finding the correct statements about $(5+2\sqrt{6})^{2n+1} = S + t$ with $S$ integer and $0 \leq t < 1$
Problem:

If $n$ is a positive integer and $(5+2\sqrt{6})^{2n+1} = S + t$, where $S$ is an integer and $0 \leq t < 1$, then

(a) $S$ is an odd integer

(b) $S + 1$ is not divisible by $9$

(c) The integer next above $(5+2\sqrt{6})^{2n+1}$ is divisible by $10$

(d) $S-1 = \frac{t}{t-1}$

Please guide me how to proceed with this question...

答: Hint:  note that $(5+2\sqrt 6)^{2n+1}+(5-2\sqrt 6)^{2n+1}$ is an integer, and $(5-2 \sqrt 6) \lt 1$, so a power of it will be small and positive.</p>


问:
Prove the measurability of a set related to sequence of bounded functions
Let $(X, \mathcal A, \mu)$ be measurable space, $(f_n)_{n\in \mathbb N} $ a sequence of measurable functions $f_n: X \longmapsto \mathbb C$.

I would like to show that the set $\{x\in X: f_1(x),f_2(x), \cdots \quad \text{is a bounded sequence} \}$ is measurable with respect to $\mathcal A$.

Let $A= \{x\in X: f_1(x),f_2(x), \cdots \quad \text{is a bounded sequence} \}$

and let the measurable functions
\begin{align*}
\varphi_n: X \to \mathbb C, \\
 x \longmapsto \vert f_n(x) \rvert.
\end{align*}

I hesitate to choose between this two methods:

$$
x\in A \iff \forall n \mathbb N, \, \exists R_n \geq 0, \,  \, \text{s.t,} \,\, \varphi_n (x) \leq R_n.
$$
Then
\begin{align}
A=& \bigcap_{n\in\mathbb{N}}\bigcup_{R_n \geq 0}\left\{ x\in E:\varphi_n(x) \leq R_n\right\}\\
=&  \bigcap_{n\in\mathbb{N}}\bigcup_{R_n \geq 0} \varphi_n^{-1}(]0,R_n])
\end{align}

$A = \Pi_{n=0}^\infty f^{-1}(\{x\in X: \varphi_n (x) \leq R_n\})$, where $R_n \in \mathbb R_{\geq 0}$.

Which method  enough to establish that $A$ is measurable with respect to $(X, \mathcal A, \mu)$.

答:
$$
A=\bigcup_{m\ge 1}\left\{x\in X:\sup_{n\ge 1}|f_n(x)|\le m\right\}.
$$


问:
Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$
I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$

The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.

Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?

答:
Integrate by parts

\begin{align}
\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx
&= \int_0^\infty \frac{\ln x}{2x}\>d(\frac{x^2}{x^2+1})\\
&\overset{ibp}= \frac12\int_0^\infty \underset{=\ 0}{\frac{\ln x}{x^2+1}dx }- \frac12\int_0^\infty \frac{1}{x^2+1}dx
= -\frac\pi4
\end{align}


问:
Estimating the number of divisors of a given number
I developed an algorithm which runs in $O(n\cdot d)$ where $n$ is the number and $d$ is the number of divisors (not necessarily prime divisors) of that same number. I was wondering if there is any way to estimate the number $d$ based on $n$? I'm interested in worst case scenarios, and my hunch was that $d$ should be $o(\log(n))$ (notice that this is little $o$ notation because if $n$ is prime, then $d = 1$). Is there any way to prove or disprove this?

答:
Look at these two wiki pages. For the worst case, we have for all $\varepsilon > 0$:
$$
d(n) = o(n^\varepsilon)
$$
and more presisely
$$
\limsup_{n\to\infty}\frac{\log d(n)}{{\log n}\ /\ {\log\log n}} = \log 2
\implies d(n) = e^{O\left(\frac{\log n}{\log\log n}\right)}
$$
However, the average order of $d(n)$ is $\log n$, meaning that
$$
\sum_{n\le x}d(n) \sim \sum_{n\le x}\log n
$$

For completion, for the "best" case we have $\liminf_{n\to\infty} d(n) = 2 = O(1)$ because of primes. So only in the best case is your guess $o(\log n)$ true.

Here is a simple proof that $d(n)=o(\log n)$  is false. Let $n_k = p_1\ldots p_k$ be the product of the first $k$ prime numbers. Then $d(n_k) = 2^k$ and by the prime number theorem
$$
n_k \le p_k^k \sim  (k\log k)^k
$$
so in fact for for any powers $\alpha$ and $\beta>1$ we have
$$
\log^\alpha n_k = O(k\log k)^\alpha = o(2^{\beta k}) = o(d(n_k)^\beta)
$$


问:
Cyclic property of the root of a cubic polynomial
Let $f(x) = x^3 - 3x + 1$. Let $\alpha, \beta, \gamma \in \mathbb{R}$ be the roots of $f$ with $\alpha > \beta > \gamma$. Let $g(x) = x^2 - 2$. We see that $\beta = g(\alpha)$, $\gamma = g(\beta)$ and $\alpha = g(\gamma)$ by calculation using $f$.

Why does this cyclic property hold? What is the relation between $f$ and $g$? Could you tell me its background or generalization?

答:
For any cubic polynomial $f(x)$ with three distinct roots $\alpha$, $\beta$, $\gamma$ you can find a quadratic polynomial $g(x)$ such that $g(\alpha)=\beta$, $g(\beta)=\gamma$, $g(\gamma)=\alpha$; it's just the matter of assuming $g(x)=rx^2+sx+t$ and solving three linear equations on coefficients $r$, $s$, $t$. What may be interesting is a method to find such a polynomial when you don't know the roots.

Let $\alpha$ be one of three distinct roots of $f(x) = x^3+bx^2+cx+d$. Then we can factorize $(x-\alpha)$ from $f(x)$ to get
$$ f(x) = (x-\alpha)\big(x^2 + (b + \alpha) x +(\alpha^2 + b\alpha + c)\big) $$
which means that the other two roots of $f(x)$ are the roots of the polynomial
$$ h(x) = x^2 + (b + \alpha) x +(\alpha^2 + b\alpha + c)$$
If we are looking for a binomial $g(x) = rx^2+sx+t$ such that $g(\alpha)$ is another root of $f(x)$, i.e. $g(\alpha)$ is a root of $h(x)$ , we know that it must satisfy
$$ 0 = h(g(\alpha)) = g(\alpha)^2 + (b + \alpha) g(\alpha) +(\alpha^2 + b\alpha + c)$$
\begin{align} 0 &= r^2 \alpha^4 + (2rs+r)\alpha^3 +  (s^2+2rt+br+s+1)\alpha^2 + \\
&\qquad +(2st+bs+t+b) \alpha + (t^2+bt+c) \end{align}
Remembering that $f(\alpha) =0$, that is
$$ \alpha^3 = -b\alpha^2-c\alpha-d$$
from which it follows that
$$ \alpha^4 = -b\alpha^3-c\alpha^2-d\alpha = (b^2-c)\alpha^2+(bc-d)\alpha + bd$$
that condition can be written as
\begin{align} 0 &= (r^2b^2-r^2c-2rsb+s^2+2rt+s+1)\alpha^2 +\\
&\qquad + (r^2bc-r^2d-2rsc-rc+2st+bs+t+b) \alpha + \\
&\qquad + (r^2bd-2rsd-rd + t^2+bt+c) \end{align}
If we want this to be true no matter which root $\alpha$ of $f(x)$ we start with, we need
\begin{align} 0 &= r^2b^2-r^2c-2rsb+s^2+2rt+s+1\\
0&= r^2bc-r^2d-2rsc-rc+2st+bs+t+b \\
0&= r^2bd-2rsd-rd + t^2+bt+c \end{align}
which is a set of second-degree polynomial equations on $r$,$s$, $t$ with $b$, $c$, $d$ being known parameters. It is complicated, but it has some solution (sometimes in complex numbers). For $f(x)=x^3-3x+1$, that is for $b=0$, $c=-3$, $d=1$, it can be checked that $r=1$, $s=0$, $t=-2$ is one of the solutions, and this solution gives you $g(x)=x^2-2$ from your question. Either way, if we find a solution to these equations we find a polynomial $g(x)$ which transforms each root of $f(x)$ into another.

Since this set of equations can have many solutions, we may get various polynomials $g(x)$, some of which may produce roots in a cyclic way, while others don't. For example, for $f(x)=x^3-3x^2+2x$ (with roots $0$, $1$ and $2$)  we can find that $g(x)=x^2-3x+2$ satisfies these equations, but doesn't transform the roots in a cyclic way (we have $g(0)=2$, $g(1)=g(2)=0$). There are five other solutions that transform the roots in non-cyclic way, but there are also two solutions that do that in a cyclic way: $g(x)=-\frac32x^2+\frac52x+1$ and $g(x)=\frac32x^2-\frac72 x+2$. My conjecture is that for any cubic polynomial with three distinct roots there will always be 8 solutions of the equations I derived, two of which will transform the roots in a cyclic way (in two directions). For example the polynomial that rotates the roots of $f(x)=x^3-3x+1$ in the oposite way to $x^2-2$ is $g(x)=-x^2-x+2$.

This method should work for polynomials $f$ of any degree $n$, though it quickly grows in complexity and in general it requires to solve a set of $n$ polynomial equations of degree $n-1$.


问:
Function bijective if component functions are bijective
I’ve tried searching online to find that if I have a mapping $f(x,y) = (f_1(x), f_2(y))$ and both component functions are bijective is $f$ neccessarily bijective also? I couldn’t find results for this...

答:
Yes, it is. Try to figure if $f$ is injective, I mean, supose $f(x,y)=f(x',y')$, but then
$$
(f_1(x),f_2(x))=f(x,y)=f(x',y')=(f_1(x'),f_2(x'))
$$
and by injectivity of $f_1$ and $f_2$, then $x=x'$ and $y=y'$, so $f$ is injective.

For surjectivity, in the case that the codomain of $f$ is the cartesian product of the codomains of $f_1$ and $f_2$, then you can use the same argument; as $f_1$ and $f_2$ are surjective, then it is $f$.


问:
Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers

Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers

Clearly $k$ is a rational number, so I let $k=\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \neq 0$, this gives $\frac{12b}{a+b}$ and $\frac{6b}{a-b}$ are positive integers. I tried different things here but couldn't make progress. I want to verify that $2, 3$ and $\frac{7}{5}$ are the only possible values for $k$. How to proceed from here? Thanks in advance.

答: Without loss of generality $b>0$. Since $a>b$, $12/(k+1)<6$. Setting it to each of $1,\,\cdots,\,5$ obtains $k$, hence $6/(k-1)$.


问:
Measuring Irrationality à la 3Blue1Brown vs Liouville
So, I was watching this 3B1B video on measure theory and music (as you do) and was struck by how Grant uses epsilon to measure "how irrational" a real number is.

He takes all the rationals $[0, 1]$ and assigns them progressively smaller ranges to cover as follows:

Where each range is centered on its corresponding rational and has a width corresponding to the nth number in the sequence shown above (where $\epsilon \in [0, 1]$ obviously). He then claims that these ranges won't cover $\frac {\sqrt 2}{2}$ when $\epsilon = 0.3$

So, my main question is: how did he determine that? Does $\frac {\sqrt2}{2}$ have a continuing fraction representation which shows it "hiding" in the gaps between these intervals? How does he know that some large denominator fraction doesn't happen to be an extraordinarily good approximation?

Beyond that, I was wondering if we could use this idea to measure "how irrational" a real number is. To formalize this slightly, I'll define $g(x)$ as the smallest $\epsilon$ you must have in order to cover a given real number $x$ with the intervals shown above. So, $q \in \Bbb Q \implies g(q)=0$ and the video asserts that $g(\frac {\sqrt 2}{2}) > 0.3$ (and also that $g(2^{\frac {7}{12}}-1) < 0.01$ which can be shown easily)

Has this irrationality measure (or something like it) been named or studied? What $x$s give values close to 0? Values of 1? What is this function's relationship to the Liouville-Roth Irrationality Measure? etc

Any musings welcome really. Just having some fun here

答:
I have recently got interested in continued fractions, so I'll give it a try!

CONTINUED FRACTIONS.

It is a basic property of continued fractions that if $h_n/k_n$ is the sequence of convergent to $x\in\mathbb R\setminus\mathbb Q$, then
$$
\frac 1{k_n(k_n+k_{n+1})}<\left|x-\frac{h_n}{k_n}\right|<\frac 1{k_nk_{n+1}}.
$$
It follows that for all $h,k$ coprime we have
$$
\left|x-\frac hk\right|\geq\left|x-\frac{h_n}{k_n}\right|
$$
for all $k\leq k_n$.

Now take $x=\sqrt{2}/2$, which is known to have the continued fraction $[0,1,2,2,2,...]$
$$
\frac{\sqrt{2}}2=\frac 1{1+\frac 1{2+\frac 1{2+\cdots}}}.
$$
It is possible to show by the recursion $k_{n+1}=2k_n+k_{n-1}$ that
$$
k_n=\frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}2.
$$
(Yes, it is always an integer!) Incidentally, $k_n$ is well approximated as
$$
k_n\sim\frac {(1+\sqrt 2)^n}2.
$$

THE FUNCTION $g(x)$.

Now, what is the meaning of $g(x)$ from the video. It strongly depends on the enumeration of the rationals (in the open unit interval) you choose. Let us call this enumeration $N(h/k)\in\mathbb N:=\{1,2,3,\dots\}$.

Clearly the inequality $g(x)>\epsilon$ is ensured by the inequality
$$
\left|x-\frac hk\right|>\epsilon 2^{-N(h/k)},\ \ \forall h,k\mbox{ coprime integers,}
$$
and more precisely, $g(x)$ could be defined as the sup of $\epsilon$'s making this inequality true for all $h,k$ coprime integers.

The natural choice made in the video is
$$
N(1/2)=1,\quad N(1/3)=2,\quad N(2/3)=3,\quad N(1/4)=4,\quad N(3/4)=5,\dots
$$
hence $N(h/k)\geq k$ for all $k\geq 4$ (absolutely not sharp estimate, but enough for what we need later on).

(Such $N$ is explicit essentially in terms of Euler totient function, but it does not matter here.)

Thus $g(x)>\epsilon$ is ensured by the inequality
$$
\left|x-\frac hk\right|>\epsilon 2^{-k},\ \ \forall h,k\mbox{ coprime},\ k\geq 4
$$
and by taking care separately of the cases with $k\leq 3$.

THE INEQUALITY $g(\sqrt 2/2)>0.3$.

Suppose that we can prove the following inequality
$$
k_n(k_n+k_{n+1})<\frac {2^{k_{n-1}}}\epsilon\qquad(\star)
$$
for all $n$ sufficiently large, say $n\geq n_0$. Then by the properties of continued fractions above we get that for $k>k_{n_0}$ we have
$$
\left|x-\frac hk\right|>\left|x-\frac{h_M}{k_M}\right|>\frac\epsilon{2^{k_{M-1}}}>\frac\epsilon{2^k}.
$$
where here $M\geq 1$ is defined by $k_{M-1}<k\leq k_M$.

In this case, the inequality $(\star)$ is clearly true at least for $n$ sufficiently large (use that $k_n\sim (1+\sqrt 2)^n/2$). Thus we only have a finite number of cases to examine.
More specifically, we look at the first numerical values
$$
\begin{array}{c|cccccccc}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \cdots
\\ \hline
k_n(k_n+k_{n+1}) &4&30&168&986&5740&33462&195024&\cdots
\end{array}
$$
and
$$
\begin{array}{c|cccccccc}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \cdots
\\ \hline
\frac{2^{k_{n-1}}}\epsilon & 6.67 & 6.67 & 26.67 & 426.67 & 436907. & 7.33\times 10^{12} & 2.11\times 10^{30}&\cdots
\end{array}
$$
So one needs to consider separately that also the neighborhoods around the "first" rationals (in the enumeration $N$) do not contain $x=\sqrt 2/2$, but I will omit this check.


问:
Is my Evaluation of $\mathcal{F}\{e^{-\alpha x^{2}}\}$, $\alpha>0$ correct?
I wish to find the Fourier transform of $f(x):=e^{-\alpha x^{2}}$ without evaluating the contour integral in the complex plane. I worked on a different approach. First, observe that :
$$
f'(x)=2\alpha xf(x)
$$
I shall take the FT of both sides of the equality :
$$
2i\pi\xi\widehat{f}(\xi)=\frac{\alpha}{i\pi}\widehat{f}'(\xi)
$$
This is regroupled to give :
$$
\widehat{f}'(\xi)+\frac{2\pi^{2}}{\alpha}\xi\widehat{f}(\xi)=0
$$
Hence general solution is expected to be of the form :
$$\widehat{f}(\xi)=\phi(\xi)\exp\left(-\frac{\pi^{2}}{\alpha}\xi^{2}\right)$$
Is the evaluation so far correct? How should we reach that $\phi(\xi)=\sqrt{\frac{\pi}{\alpha}}$?

Body: $$\widehat{f}(0)=\phi(0)\quad\implies\quad\widehat{f}(0)=\int_{\mathbb{R}}e^{-\alpha x^{2}}\;\text{d}x=\sqrt{\frac{\pi}{\alpha}}$$


问:
A general rule for L'Hôpital's Rule
In my text there are as many as $11$ theorems dealings with the different cases of L'Hôpital's Rule and due to the quantitative heaviness I often forget rules during problem session. Isn't there a general rule for L'Hôpital's Rule which is easy to remind so that I won't need to spend useless time recalling result?</p>

答:
A more general approach would be where
$$
\lim_{n\to \infty} \frac{f(x)}{g(x)} = \lim_{n\to \infty} \frac{\frac{d}{dx} f(x)}{\frac{d}{dx} g(x)}
$$
Or to 0 of course
$$
\lim_{n\to 0} \frac{f(x)}{g(x)} = \lim_{n\to 0} \frac{\frac{d}{dx} f(x)}{\frac{d}{dx} g(x)}
$$

Only when $\lim_{n\to \infty/0} f(x) = \lim_{n\to \infty/0} g(x)$ tend towards the same thing ($\infty_{+/-}, 0$)

A good example being

$$
\lim_{x->0} \frac{sin(x)}{x} = \lim_{x\to 0} \frac{cos(x)}{1} = 1
$$

Title: A general rule for L'Hôpital's Rule


问:
Cyclic subgroup of $A_5$
Consider $A_{5}$ group and let $a,b \in A_5 $ .

Find probability of the subgroup generated by a and b.
$\langle a, b\rangle$ , be
cyclic group .

I think if $a=1$ then we have 60 cyclic subgroup because $A_5$ has 60 elements, also 59 cyclic subgroup for $b=1$ so we have 60+59 cyclic subgroups and probability is $\frac{119}{60×60}$ is this true?

答:
The orders of elements of $A_5$ are

$Order \quad   Num\, Elements
\\
\\  1    \quad\quad\quad\quad       1
\\  2   \quad\quad\quad\quad            15
\\  3     \quad\quad\quad\quad          20
\\  5     \quad\quad\quad\quad          24$

Since none of these numbers are composite, $$ will  be cyclic precisely when one of these elements is a power of the other.  This is because if two elements generate a cyclic group and one is not a power of the other, then the cyclic group has composite order (in a group of prime order all elements are powers of any non-identity element, and the identity alone doesn't generate a group of prime order).  But $A_5$ has no cyclic subgroups of composite order as it has no elements of composite order.

Since the problem does not state whether $a$ and $b$ are drawn with or without replacement, it's somewhat ambiguous, and it actually makes a rather large difference: $$ is substantially more likely to be cyclic if drawn with replacement.  However, in usual mathematical usage, one does not assume $a \ne b$ unless explicitly stated, so technically we should answer the problem as though these are drawn with replacement.

Therefore, out of $3600$  (i.e., $60\cdot 60$) cases, we have

$60$ cases of $a=1$ and $b$ any.

$30 = 2\cdot 15$ cases of $o(a)=2$ and $b\in $

$60 = 3 \cdot 20$ cases of $o(a) = 3$ and $b\in $

$120 = 5 \cdot 24$ cases of $o(a) = 5$ and $b\in $

So the probability you are looking for is $270/3600$, or $7.5\%$

Also note that if you want the "draw without replacement answer" you can simply subtract the $60$ cases of $a=b$ from both the numerator and the denominator, to get $210/3540=7/118$, or just a hair over $5.9\%$.

Title: Cyclic subgroup of $A_5$

Body:
The orders of elements of $A_5$ are

$Order \quad   Num\, Elements
\\
\\  1    \quad\quad\quad\quad       1
\\  2   \quad\quad\quad\quad            15
\\  3     \quad\quad\quad\quad          20
\\  5     \quad\quad\quad\quad          24$

Since none of these numbers are composite, $<a,b>$ will  be cyclic precisely when one of these elements is a power of the other.  This is because if two elements generate a cyclic group and one is not a power of the other, then the cyclic group has composite order (in a group of prime order all elements are powers of any non-identity element, and the identity alone doesn't generate a group of prime order).  But $A_5$ has no cyclic subgroups of composite order as it has no elements of composite order.

Since the problem does not state whether $a$ and $b$ are drawn with or without replacement, it's somewhat ambiguous, and it actually makes a rather large difference: $<a,b>$ is substantially more likely to be cyclic if drawn with replacement.  However, in usual mathematical usage, one does not assume $a \ne b$ unless explicitly stated, so technically we should answer the problem as though these are drawn with replacement.

Therefore, out of $3600$  (i.e., $60\cdot 60$) cases, we have

$60$ cases of $a=1$ and $b$ any.

$30 = 2\cdot 15$ cases of $o(a)=2$ and $b\in <a>$

$60 = 3 \cdot 20$ cases of $o(a) = 3$ and $b\in <a>$

$120 = 5 \cdot 24$ cases of $o(a) = 5$ and $b\in <a>$

So the probability you are looking for is $270/3600$, or $7.5\%$

Also note that if you want the "draw without replacement answer" you can simply subtract the $60$ cases of $a=b$ from both the numerator and the denominator, to get $210/3540=7/118$, or just a hair over $5.9\%$.


问:
How to find if a number has any odd divisor (greater than 1)?
I want to build an algorithm which tells if a number has any odd divisor or not.

for example, 6 has an odd divisor 3 (greater than 1),
4 doesn't have any odd divisors, and so on.

I know that I can find all the pairs of divisors by only traversing till square root of the number.

for example, if n%i == 0 then i and n/i both are divisors.

for example, if the number given is 100 using this method I can
find all the divisors and check if any divisor is odd or not i.e. 1, 100, 2, 50, 4, 25, 5, 20, 10.

Therefore this algorithms solves the problem in O(sqrt(n)) complexity.
But I'm looking to optimize this.

Please comment if any more clarity is required from my side as I'm new to this forum.

Thanks !

答: The only possibility for any number $n>1$ not to have an odd divisor is for $n$ to be a power of two. In binary representation, this means that the number has only one $1$ bit and all other bits are $0$ (i.e. $4_{10}=100_2, 128_{10}=1000000_2$. Another property: the number $n-1$ will have all ones (i.e. $3_{10}=11_2, 127_{10}=111111_2$ but it will have one less digit than $n$. Thus, if you apply logical AND operation to $n \land (n-1)$, the result will be zero only if $n$ is a power of two. This is just one line of code regardless of $n$.


答:
Alternatively,

$$\begin{cases} \frac {12}{k+1}=m \\ \frac{6}{k-1}=n \end{cases}$$

$$ \frac{12}{m}-1=\frac 6n +1$$

$$m=6-\frac{18}{n+3}$$

$$n=3,6,15$$ which follows, $$m=3,4,5.$$

Then, you can find all values of $k$, from the formula

$$k=\frac 6n+1.$$

Body:
Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers

Clearly $k$ is a rational number, so I let $k=\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \neq 0$, this gives $\frac{12b}{a+b}$ and $\frac{6b}{a-b}$ are positive integers. I tried different things here but couldn't make progress. I want to verify that $2, 3$ and $\frac{7}{5}$ are the only possible values for $k$. How to proceed from here? Thanks in advance.


问:
Evaluating the series $\sum\limits_{n=1}^\infty \frac1{4n^2+2n}$
How do we evaluate the following series:

$$\sum_{n=1}^\infty \frac1{4n^2+2n}$$

I know that it converges by the comparison test. Wolfram Alpha gives the answer $1 - \ln(2)$, but I cannot see how to get it. The Taylor series of logarithm is nowhere near this series.

答:
Knowing the answer is helpful. Recall that
$$\ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots.$$
Thus
$$\ln 2=1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\cdots.$$
The "typical" term $\dfrac{1}{2k}-\dfrac{1}{2k+1}$ simplifies to $\dfrac{1}{2k(2k+1)}$. </p>


问:
Why can the Taylor formula be used to evaluate the limit,Why does its remainder converge to zero？
A simple example $\lim\limits_{x\to0}\frac{\cos(x)-1}{x^2}=\frac{1-\frac{1}{2}x^2+o(x^2)-1}{x^2}=-\frac{1}{2}$

If taken out alone, its limit is zero.How can I judge that the sum of infinite limits converges to zero.

Such as:$\lim\limits_{x\to0}\frac{o(x^2)}{x^2}=\lim\limits_{n\to\infty}[\sum\limits_{i=3}^{n} (\lim\limits_{x\to0}\frac{x^i}{x^2})]=?$

Clarification of the problem

For example $\lim\limits_{n\to\infty}(\frac{1}{n}+\frac{1}{n}+.....+\frac{1}{n})=\lim\limits_{n\to\infty} n*\frac{1}{n}=1$

In the case of an infinite number of items,I cannot simply think of adding an infinite number of zero

答:
Apply the rules of asymptotic analysis:
$$\frac{\cos x-1}{x^2}=\frac{1-\frac{1}{2}x^2+o(x^2)-1}{x^2}=-\frac 12+o(1).$$

Added: an elementary way to determine the limit:
$$\frac{\cos x-1}{x^2}=\frac{\cos^2 x-1}{x^2(\cos x +1)}=-\underbrace{\frac{\sin^2x}{x^2}}_{\substack{\downarrow\\1^2=1}}\,\underbrace{\frac1{\cos x+1}}_{\substack{\downarrow\\1+1=2}}$$


问:
Variable substitution & Universal Quantifier: How to arrive from $\forall n > N.|a_n-L|< \epsilon$ to $\forall n > N.| a_{n-1}-L | < \epsilon$?
I have trouble understanding variable substitution in universal statement regarding the decrementation of an index:

We have the definition: $ \forall \epsilon>0 \, \exists N\in \mathbb{N} \, \forall n > N . | a_n - L | < \epsilon. $

I am given $ \epsilon > 0 $ , so it means there exists $ N \in \mathbb{N} $ s.t. $ \forall n > N . | a_n - L | < \epsilon. $

Question: I know it now follows that $ \forall n > N . | a_{n+1} - L | < \epsilon. $  because of the following reasoning: Let $ n>N $ be arbitrary, $ n+1>N $ so ( by universal instantiation ) $ | a_{n+1} - L | < \epsilon. $ Since $ n $ was arbitrary, therefore $ \forall n > N . | a_{n+1} - L | < \epsilon. $

But I've also encountered places where it is written that $ \forall n > N . | a_{n-1} - L | < \epsilon $ and I don't know how to arrive at this universal statement using logical steps like I showed when I arrived to  $ \forall n > N . | a_{n+1} - L | < \epsilon.  $

Can you please show how to arrive to $ \forall n > N . | a_{n-1} - L | < \epsilon $ ?

For example, when learning about sums In class we had the following: ( $ S_n $ is the partial sum of sequence $ (a_n) $, we're given $ \epsilon > 0$ and that $ N \in \mathbb{N} $ exists)  $ \forall n > N . | S_n - L | < \epsilon/2. $ Therefore for all $n>N $ we have$ |S_n - S_{n-1} | = | S_n - L + L - S_{n-1} | \leq  | S_n - L | + | S_{n-1} - L | = \epsilon/2 + \epsilon/2 = \epsilon $.

And I just didn't know how to show $ \forall n > N . | S_{n-1} - L | < \epsilon/2. $

( Although I could show that $ \forall n > N . | S_{n+1} - L | < \epsilon/2 $ and arrive to the same result just by deriving the steps from  $ |S_{n+1} - S_{n} |,  $ I still want to know how to arrive to $ \forall n > N . | S_{n-1} - L | < \epsilon/2. )$

答:
$|S_n-L|<\epsilon/2$ for $n>N$ does $\textbf{not}$ imply that $|S_{n-1}-L|<\epsilon/2$ for $n>N$.

However, if you can pick $N$ such that $|S_n-L|<\epsilon/2$ for $n>N$, then you can also pick $M$ such that $|S_{n-1}-L|<\epsilon/2$ for $n>M$. In fact, you can just take $M=N+1$. Sometimes you abuse notation and just call everything $N$ but it seems to me that this is what's being done.


问:
Probability depending on variance-to-mean ratio for a sum of normal distributions
let us consider three normally distributed random variables $ X \sim N(\mu_X,\sigma_X^2) $, $ Y \sim N(\mu_Y,\sigma_Y^2) $, $ Z \sim N(\mu_Z,\sigma_Z^2) $, all of which being (mutually) stochastically independent. Moreover, all of them have strictly positive mean values, and $ \sigma_Y^2/\mu_Y \ge \sigma_Z^2/\mu_Z $.

Let $ \alpha \in (0,0.5) $ be arbitrary but fixed, and let $ \mathbb{P}[X>1] \le \alpha $, $ \mathbb{P}[Y>1] \le \alpha $, and $ \mathbb{P}[Z>1] \le \alpha $.

My question is: If $ \mathbb{P}[X+Y>1]> \alpha $, can we have $ \mathbb{P}[X+Z>1] \le \alpha $?

(In a more abstract way: Can the considered probability decrease when $ Y $ is replaced by a random variable $ Z $ with lower or equal variance-to-mean ratio?)

答:
Yes, we can have that situation.

Since the probabilities are continuous in the parameters, you can see this by taking a situation on the boundary and then by moving the parameters off the boundary by small amounts.

So we start by taking

$\mu_x=\mu_z=0.25 \\ \mu_y=1.0 \\ \sigma_x=\sigma_z=0 \\  \sigma_y=1$

Then $P(X>1)=P(Z>1)=P(X+Z>1)=0$, $P(Y>1)=0.5$, and $P(X+Y>1)>0.5$.

By continuity we can make sufficiently small upward adjustments to $\sigma_x$ and $\sigma_z$ and a sufficiently small downward adjustment to $\mu_y$, and have
$P(X>1)$,$ P(Z>1)$, and$ P(X+Z>1)$ all less than $0.1$, $P(X+Y>1)$ still greater than $0.5$ and $0.4
1)<0.5$.

Also, you can make these changes sufficiently small that you still have $\frac{\sigma_z^2}{\mu_z} < \frac{\sigma_y^2}{\mu_y}$

Now take $\alpha$ between $P(Y>1)$ and $0.5$, and you are done.

Body:
Yes, we can have that situation.

Since the probabilities are continuous in the parameters, you can see this by taking a situation on the boundary and then by moving the parameters off the boundary by small amounts.

So we start by taking

$\mu_x=\mu_z=0.25 \\ \mu_y=1.0 \\ \sigma_x=\sigma_z=0 \\  \sigma_y=1$

Then $P(X>1)=P(Z>1)=P(X+Z>1)=0$, $P(Y>1)=0.5$, and $P(X+Y>1)>0.5$.

By continuity we can make sufficiently small upward adjustments to $\sigma_x$ and $\sigma_z$ and a sufficiently small downward adjustment to $\mu_y$, and have
$P(X>1)$,$ P(Z>1)$, and$ P(X+Z>1)$ all less than $0.1$, $P(X+Y>1)$ still greater than $0.5$ and $0.4<P(Y>1)<0.5$.

Also, you can make these changes sufficiently small that you still have $\frac{\sigma_z^2}{\mu_z} < \frac{\sigma_y^2}{\mu_y}$

Now take $\alpha$ between $P(Y>1)$ and $0.5$, and you are done.


问:
How to calculate $ \int \limits_{0}^{\infty} \frac{ t^2 e^{-t}}{(1 + e^{-t})^2} dt $?
I am trying to calculate

$$ \int \limits_{0}^{\infty} \frac{ t^2 e^{-t}}{(1 + e^{-t})^2} dt  $$

I used variable replacement $u = e^{-t} $ and got the integral

$$ \int \limits_{0}^{1} \frac{ \ln^2u}{(1 + u)^2} du  $$

Wolfram says that it is $\frac{\pi}{6}$. But I have not idea. How can I get this result?

Thank you in advance for any help!

Body:
Note

\begin{align}
\int_0^1\frac{\ln^2 u}{(u+1)^2}du
&= \int_0^1{\ln^2 u}\>d(\frac{u}{u+1})\\
&=- 2\int_0^1\frac{\ln u}{u+1}du = -2 (-\frac{\pi^2}{12})=\frac{\pi^2}6
\end{align}

where $\int_0^1 \frac{\ln u}{u+1}dt =-\frac{\pi^2}{12}$


答:
It really depends on how exact you need it. You could use the remainder term of the Taylor expansion of degree $n$ of some $f$ around some $x_0$ in its more detailed form, $$r_n(x)=\frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n+1)}(\xi),~~~\xi\in[x_0,x].$$

For the cosine with $x_0=0$ this would give for example
$$
r_3(x)=\frac{x^4}{4!}\cos(\xi)
$$

Or you could refer back to more elementary means and notice that the cosine series is alternating with falling terms if $x$ is small enough, so that the series is bounded by its partial sums,
$$
1-\frac{x^2}2\le\cos(x)\le 1-\frac{x^2}2+\frac{x^4}{4!}
$$
All this is showing that the defect in the limit formula is smaller than $\frac{x^2}{24}$ which does indeed converge to zero.

Title: Why can the Taylor formula be used to evaluate the limit,Why does its remainder converge to zero？

Body:
A simple example $\lim\limits_{x\to0}\frac{\cos(x)-1}{x^2}=\frac{1-\frac{1}{2}x^2+o(x^2)-1}{x^2}=-\frac{1}{2}$

If taken out alone, its limit is zero.How can I judge that the sum of infinite limits converges to zero.

Such as:$\lim\limits_{x\to0}\frac{o(x^2)}{x^2}=\lim\limits_{n\to\infty}[\sum\limits_{i=3}^{n} (\lim\limits_{x\to0}\frac{x^i}{x^2})]=?$

Clarification of the problem

For example $\lim\limits_{n\to\infty}(\frac{1}{n}+\frac{1}{n}+.....+\frac{1}{n})=\lim\limits_{n\to\infty} n*\frac{1}{n}=1$

In the case of an infinite number of items,I cannot simply think of adding an infinite number of zero


问:
A multiple integral problem in $6$D: $I=\iint_{B_r\times B_r}\frac{1}{|x-y|}dxdy$
Let $B_r$ to be a ball centered at origin with radius $r$ in dimension $3$. My problem is the following multiple integral:
$$I=\iint_{B_r\times B_r}\frac{1}{|x-y|}dxdy$$
The result says that $I$ is exactly equivalent to $r^5$. The lower bound for $I$ is trivial but I have trouble in the upper bound. Moreover, can this result be extended to dimension $n$?

答:
Partial solution

We can also try to find a general solution for $\textbf x$ and $\textbf y$ in $\textbf R^n$

The element of volume in spherical coordinates in $\textbf R^n$ is
$dV=r^{n-1}dr \,d\Omega$, where $d\Omega=\sin^{n-2}\theta_1\sin^{n-3}\theta_2...\sin\theta_{n-2}\,d\theta_1...d\theta_{n-2}\,d\phi$, where integration over $\theta_1...\theta_{n-2}$ goes from $0$ to $\pi$, and over $\phi$ - from $0$ to $2\pi$

If we have a full spherical symmetry in $\textbf R^n$ $\int d\Omega$ can be expressed in closed form:

$$\int_{-\infty}^{\infty}..\int_{-\infty}^{\infty}e^{-x_1^2-x_2^2-...-x_n^2}dx_1...dx_n=\pi^{\frac{n}{2}}=\int d\Omega\int_{0}^{\infty}e^{-r^2}r^{n-1}dr=\frac{1}{2}\Gamma(\frac{n}{2})\int d\Omega$$
$$ \int d\Omega=\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}$$

On the other hand

$$\int d\Omega=\int_0^\pi..\int_0^\pi\int_0^{2\pi}\sin^{n-2}\theta_1\sin^{n-3}\theta_2...\sin\theta_{n-2}\,d\theta_1...d\theta_{n-2}\,d\phi=\int_0^\pi\sin^{n-2}\theta_1\,d\theta_1\int d\Omega\,'$$

Where $\theta_1$ is the polar angle measured from axis $Z$ and $\int d\Omega\,'$ means integration over all other angles in $\textbf R^n$

$$\int_0^\pi\sin^{n-2}\theta_1\,d\theta_1=\int_0^1(1-t)^{\frac{n-3}{2}}t^{-\frac{1}{2}}dt =  B(\frac{n-1}{2},\frac{1}{2})=\frac{\sqrt{\pi}\Gamma(\frac{n-1}{2})}{\Gamma(\frac{n}{2})}$$

Thus $\int d\Omega\,'=\frac{2\pi^{\frac{n-1}{2}}}{\Gamma(\frac{n-1}{2})}$. We had to separate the integration over $\theta_1$, because we will integrate over this angle further.
$$I=\int\limits_{V_{x}}\int\limits_{V_y} d^n\textbf x\,d^n\textbf y \frac{1}{|\textbf x-\textbf y|}=\int d\Omega_x\int_0^Ar^{n-1}dr\int d\Omega_y\,'\int_0^BR^{n-1}dR\int_0^\pi\frac{\sin^{n-2}\theta \,d\theta}{\sqrt{R^2+r^2-2Rr\cos\theta}}$$

This expression can be simplified if $A=B$
$$I(A,n)=A^{2n-1}\int d\Omega_x\int_0^1x^{n-1}dx\int d\Omega_y\,'\int_0^1y^{n-1}dy\int_0^\pi\frac{\sin^{n-2}\theta \,d\theta}{\sqrt{x^2+y^2-2xy\cos\theta}}$$
$$I(A,n)=\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{2\pi^{\frac{n-1}{2}}}{\Gamma(\frac{n-1}{2})}A^{2n-1}J(n)$$
where $$J(n)=\int_0^1\int_0^1x^{n-1}y^{n-1}dxdy\int_0^\pi\frac{\sin^{n-2}\theta \,d\theta}{\sqrt{x^2+y^2-2xy\cos\theta}}$$

Using the Gamma function property $\,\Gamma(z)\Gamma(z+\frac{1}{2}) =2^{1-2z}\sqrt{\pi}\,\Gamma(2z)$
$$I(A,n)=\frac{2^n\pi^{n-1}}{(n-2)!}A^{2n-1}J(n)$$

To evaluate the integral $J(n)$, we will use the decomposition of the square root in a series of Legendre polynomials
$$J(n)=\int_0^\pi\sin^{n-2}\theta \,d\theta\int_0^1x^{n-1}dx\biggl(\int_0^x\frac{y^{n-1}dy}{x\sqrt{1+(\frac{y}{x})^2-2\frac{y}{x}\cos\theta}}+\int_x^1\frac{y^{n-1}dy}{y\sqrt{1+(\frac{x}{y})^2-2\frac{x}{y}\cos\theta}}\biggr)$$
$$=2\int_0^\pi\sin^{n-2}\theta \,d\theta\int_0^1x^{n-2}dx\int_0^x\frac{y^{n-1}dy}{\sqrt{1+(\frac{y}{x})^2-2\frac{y}{x}\cos\theta}}$$

But $$\frac{1}{\sqrt{1+t^2-2t\cos\theta}}=\sum_{k=0}^{\infty} \,t^k \,P_k(\cos\theta)$$
where $t<1$, and $P_k\,(z)=\frac{1}{k!2^k}\frac{d^k}{dz^k}(z^2-1)^k$ - Legendre polynomial of degree $k$

$$J(n)=2\sum_{k=0}^{\infty}\int_0^\pi\sin^{n-2}\theta \,P_k\,(\cos\theta)\,d\theta\int_0^1x^{n-2-k}dx\int_0^xy^{n-1+k}dy$$ $$=\frac{2}{2n-1}\sum_{k=0}^{\infty}\frac{1}{n+k}\int_{-1}^1(1-z^2)^{\frac{n-3}{2}}P_k(z)dz$$

I'm afraid there is no a closed form for this integral, though it can be evaluated for every given $n$. For instance, for odd $n\,$ there is the most simple case, because $\,k\leqslant n-3$ (integral with higher $k$ vanishes).

For $n=3$ the only surviving term is with $k=0$; $\,J(n=3)=\frac{2}{5}\frac{2}{3}=\frac{4}{15}$ and $I(A, n=3)=\frac{32\pi^2}{15}A^5$ - as was evaluated above for $R^3$

For $n=5$ there are two surviving terms ($k=2$ and $k=0$); $\,J(n=5)=\frac{2}{9}(\frac{4}{15}-\frac{4}{15}\frac{1}{7})=\frac{2}{9}\frac{8}{35}$ and $I(a,n=5)=\frac{2^8}{35*3^3}\pi^4A^9$

For arbitrary $n$:

$$I(A,n)=\frac{\pi^{n-1}2^{n+1}}{(n-2)!(2n-1)}A^{2n-1}\sum_{k=0}^{\infty}\frac{1}{n+k}\int_{-1}^1(1-z^2)^{\frac{n-3}{2}}P_k(z)dz$$
As the last point we can evaluate asymptotics for big $n>>1$
$$\int_{-1}^1(1-z^2)^{\frac{n-3}{2}}P_k(z)dz=\int_{-1}^1e^{\frac{n-3}{2}\log(1-z^2)}P_k(z)dz\sim\int_{-\infty}^\infty e^{-\frac{n-3}{2}z^2}P_k(z)dz\sim\sqrt{\frac{2\pi}{n-3}}P_k(0)$$
$$J(n)\sim\frac{2}{2n-1}\sqrt{\frac{2\pi}{n-3}}\sum_{k=0}^{\infty}\frac{1}{n+k}P_k(0)\sim\frac{\sqrt{2\pi}}{n^{\frac{5}{2}}}\sum_{k=0}^{\infty}P_k(0)=\frac{\sqrt{\pi}}{n^{\frac{5}{2}}}$$
$$I(A,n)\sim\frac{\pi^{n-\frac{1}{2}}2^n}{(n-2)!\,n^{\frac{5}{2}}}A^{2n-1}\,, \,n>>1$$