问:
Extreme numbers of standard deviations
How would you approximate the magnitude of the probability of an event with an extremely large number of standard deviations? All numerical approximations that I have found have far too large relative errors or do not handle large σ well.

My first rough estimate would be that the value of  in 10^- increases by half the value of σ for each additional σ.

For example, for a 670,000σ event,  would approximately be half the sum of 1 to 670,000 = 224,450,335,000/2 ≈100,000,000,000.

I'd like to know if this would be roughly correct, and if there is a better approximation.

答:
It depends on the distribution, of course.
If $X$ has a normal distribution with mean $\mu$ and standard deviation $\sigma$, then as $n \to \infty$
$$ \mathbb P(X \ge \mu + n \sigma) = \frac{\exp(-n^2/2)}{\sqrt{2 \pi}} \left(\frac{1}{n} - \frac{1}{n^3} + \frac{3}{n^5} + O(1/n^7)\right)$$
Asymptotic results for other distributions are also available.

Title: Extreme numbers of standard deviations


问:
Limit of two uncorrelated random variables
Let $\{X_n\}$ and $\{Y_n\}$ be two sequences of uncorrelated random variables with finite fourth moments. Also, let
$$
X_n \xrightarrow{d} X \text{    and    } Y_n\xrightarrow{d} Y,
$$
where $X$ and $Y$ are standard normal random variables. Can we conclude $(X_n,Y_n)\xrightarrow{d} (X',Y')$ such that $X'$ and $Y'$ are independent standard normal random variables? If not, can you please provide some sufficient conditions? Thanks for the help.

答: Since you're declining to assume the pairs $(X_i,Y_i)$ are independent of each other for different values of $i,$ the answer is no, because you can take $(X_i,Y_i)=(X_1,Y_1)$ for every i, and $X_1\sim\operatorname N(0,1)$ and $Y_1=\pm X_1$ where the choice between $\pm$ is independent of the value of $X_1$ and plus and minus are equally probable.

ContentLicense: CC BY-SA 4.0


问:
Forest graphs $F_1$ and $F_2$ with same vertices but different edges. Prove there is an edge where $F_1$ with it is still a forest
I am having quite a hard-time with this question, been thinking about it for a few hours and have not got a clue on how to even start proving this, because it is trivial but proving it has been hard for me.

Given two forests $F_1 = (V,A)$ and $F_2 = (V,B)$ with same vertices group $V$. It is also given that $|B| > |A|$. Prove: there exists an edge $e \in B \backslash A$ where $F_1 \cup \{e\}$ is still a forest.

Any help will be appreciated!

答:
Hint: The fact that $|B|>|A|$ can be used to prove that $F_1$ is not connected. Why does this imply that such an edge exists?

Some more comments: This is an intermediate result; there is an important part of the argument that requires non-connectivity (or something equivalent) to make work.

Notice that in general there is no edge that makes $F_1$ connected. Even if there is, it might not be in $B$. But we are not looking for such an edge, simply one in $B$ which does not produce a cycle in $A$.

ContentLicense: CC BY-SA 3.0


问:
$X$ is independent of $\mathcal{G}$, $f(X ,Y)$ is independent of $Y$, $Y$ is $\mathcal{G}$-measurable, then $f(X,Y)$ is independent of $\mathcal{G}$?
Let $(\Omega, \mathcal{F}, P)$ be a probability space,  $\mathcal{G}\subset \mathcal{F}$ be a sigma algebra, $X, Y : \Omega \to \mathbb{R}^n$ be random vectors, and $f$ be a measurable function on $\mathbb{R}^{2n}$.
Suppose that $X$ is independent of $\mathcal{G}$, $f(X ,Y)$ is independent of $Y$, and $Y$ is $\mathcal{G}$-measurable. Is it true that $f(X,Y)$ is independent of $\mathcal{G}$?

Motivation: This question arises from a randomized algorithm. In this algorithm, each iteration receives some randomness that is independent of the previous iterations. In my question, $\mathcal{G}$ indeed represents the randomness of the algorithm up to iteration $k$, $Y$ is a vector generated by iteration $k$, $X$ is some new randomness injected into the algorithm at iteration $k+1$, and $f(X,Y)$ is a quantity computed from $X$ and $Y$. It turns out that $f(X,Y)$ is statistically independent of $Y$. I would like to prove that $f(X,Y)$ is independent of the first $k$ iterations, which will be interesting and convenient.

Any comments or criticism will be appreciated. Thank you.

A related question: Uniform distribution on the unit sphere rotated by a random orthogonal matrix.

答:
Let $(Q_\omega)_\omega$ denote the conditional distribution of $X$ given $\mathcal{G}$. Since by assumption $X$ is independent of $\mathcal{G}$, we may take $Q_\omega = X(\mathbb{P})$ for all $\omega\in\mathcal{G}$.

The conditional distribution $(P_\omega)_\omega$ of $f(X,Y)$ given $\mathcal{G}$ is then (since $Y$ is $\mathcal{G}$-measurable) by substition given by $P_\omega = f_{Y(\omega)}(Q_\omega) = f_{Y(\omega)}(X(\mathbb{P}))$, where $f_{Y(\omega)}(x) := f(x,Y(\omega))$.

Note that $P_\omega$ only depends on $\omega$ through $Y(\omega)$. This implies that the Markov kernel $(R_y)_y$ given by $R_y = f_y(X(\mathbb{P}))$ corresponds to the conditional distribution of $f(X,Y)$ given $Y$. But $f(X,Y)$ was assumed independent of $Y$, so we may actually take $R_y$ to not depend on $y$. Since $P_\omega = R_{Y(\omega)}$, we may also take $(P_\omega)_\omega$ to not depend on $\omega$, which implies the desired, namely that $f(X,Y)$ is independent of $\mathcal{G}$.

The arguments are typical in the context of conditional distributions and various classic textbook references provide the necessary foundation. The following lecture notes offer an excellent introduction to the topic: http://web.math.ku.dk/noter/filer/beting.pdf. The results used correspond to Theorem 1.4.3, Theorem 2.1.1, and Theorem 2.1.5.


问:
Tracial topological rank of hereditary
Let $A$ be a unital simple $C^*$-algebra with $TR(A) \leq k$ (tracial topological rank). I can prove that for any unital hereditary $C^*$-subalgebra $B$ of $A$, $TR(B) \leq k$.

If $TR(A) \nleq k - 1$, when we say $TR(A) = k$. So, let $A$ be as above with $TR(A) = 0$, is it true to say that $TR(B) = 0$ for every hereditary $C^*$-subalgebra $B$ of $A$?

答: Since $TR(A) = 0 $, we have $RR(A) = 0 $ (real rank of $A$). Also, for any hereditary $C^*$-subalgebra $B$ of $A$, we have $RR(B) = 0$.(Automatically, $B$ is unital). We know that for a unital simple $C^*$-algebra with $TR(A) \leq k$, we have $TR(B) \leq k$ for any unital hereditary $C^*$-subalgebra $B$ of $A$.

Title: Tracial topological rank of hereditary


问:
Covariance between $z_t$ and its first lag
I am looking for $Cov[z_t, z_{t-1}]$

where $z_t = \delta + \psi_2 z_{t-2} + \epsilon_t$ and $\epsilon_t$ is a white noise with distribution $(0, \sigma^2)$

So I have started by doing the following:

$Cov[\delta + \psi_2 z_{t-2} + \epsilon_t, \delta + \psi_2 z_{t-3} + \epsilon_{t-1}]$

since $z_{t-1} = \delta + \psi_2 z_{t-3} + \epsilon_{t-1}$ I believe.

I believe this covariance comes out to $0$ since I think $z_t$ only depends on its even lags; that is $z_{t-2}, ..., z_{t-2n}$.

Is this line of reasoning correct or am I screwing up somewhere?

答: Well assuming $\delta$ and $\psi_2$ are constants, your $z_0, z_2, z_4, \dots$ are independent of $z_1, z_3, z_5, \dots$, so you've got that going for you. And in that case $\text{cov}(z_t, z_{t-1}) = 0$.


问:
Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas
I have this identity:

$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$

If I write this like as:

$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alpha-\cos 2\alpha)=-\sin 6\alpha \sin 2\alpha$$

I can use Werner and prostaferesis formulas and I find the identity $(1)$.

But if we suppose of not to use these formulas I have done a try writing:

$$\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha \tag 1$$

$$[\cos(2(2\alpha))]^2=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 2$$

$$\cos^4 2\alpha -2\sin^2 2\alpha\cos^22\alpha+\sin^4 2\alpha=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 3$$

Then I have abandoned because I should do long computations and I think that is not the right way.

Is there any trick without to use the prostapheresis or Werner's formulas?

答:
OK, I think I have something a little more straightforward. If I am using some of the forbidden fruit, I apologize. First of all, as hamam already noted, I'm going to let $2\alpha = x$ to save my sanity. So I want to show that
$$\cos^2 2x = \cos^2 x - \sin 3x\sin x.$$
Aside from the double-angle formula for $\cos 2x$, I will need the triple angle formula
\begin{align*}
\cos 3x &= \cos 2x\cos x -\sin 2x\sin x \\ &= (2\cos^2 x-1)\cos x - 2\sin^2x\cos x \\ &= 2\cos^3 x-\cos x - 2(1-\cos^2 x)\cos x \\ &= 4\cos^3 x-3\cos x.
\end{align*}

Note that $\cos 2x=\cos (3x-x) = \cos 3x\cos x + \sin 3x\sin x$, so the original formula is equivalent to
\begin{equation*}
\cos^2x -\cos^2 2x + \cos3x\cos x = \cos 2x. \tag{$\star$}
\end{equation*}
Well, by the triple angle formula above,
\begin{align*}
\cos 2x+ \cos^2 2x - \cos^2 x &= (2\cos^2 x-1)+(2\cos^2 x -1)^2 - \cos^2  x \\ & = 4\cos^4x - 3\cos^2 x = \cos3x\cos x,
\end{align*}
and this establishes ($\star$).

Having looked carefully at A-level student's, he's using the triple angle formula for sin, so these solutions are almost surely equivalent. :)


问:
For which angle $x$, it results $0 < \cos(x) < \sin(x)<1$?
For which angle $x$, it results $0 < \cos(x) < \sin(x) < 1$?

40°

50°

60°

70°

80°

Why is the correct answer #3 60° and not #3,4,5?

Yet, if the question asks: For which angle X, it results $0.5 < \cos(x) < \sin(x) < 1$?, the answer is #2 50°

答:
Let us start with some considerations.

To begin with, we shall emphasize that $0 < \cos(x) < 1$ and $0 < \sin(x) < 1$ iff $x\in(0,\pi/2)$.

The $\cos$ function is strictly decreasing on $(0,\pi/2)$ and the $\sin$ function is strictly increasing on $(0,\pi/2)$

Once $\cos(\pi/4) = \sin(\pi/4)$, if we take $x\in(0,\pi/4)$, then $\sin(x) < \sin(\pi/4) = \cos(\pi/4) < \cos(x)$.

On the other hand, if we take $x\in(\pi/4,\pi/2)$, then $\cos(x) < \cos(\pi/4) = \sin(\pi/4) < \sin(x)$.

So any angle $x$ which belongs to $(\pi/4,\pi/2)$ satisfies the proposed relation.

Hopefully this helps!


问:
Coin flipping in a row
If we are flipping a coin and the probability of:

Heads: $\frac{1}{3}$

Tails: $\frac{2}{3}$

$W$: 2 results in a row: $T-T$ or $H-H$

If we are flipping until we have two in a row, what is $E[W]$?

答:
Lets look at the case where we want expected number of flips to get two heads.

Let $N$ be the number of flips to get two heads in a row.

Relating back to the above comments we get:

$E[N] = (1+E[N|\text{ previous toss was Heads}])P(\text{ Heads }) + (1+E[N|\text{ previous toss was Tails}])P(\text{ Tails })$

Notice here we are only looking for the expected number of tosses for $HH$. That that gives us that $E[N|\text{ previous toss was Tails}] = E[N]$ since if we got tails on the most recent toss then it is as if we are starting fresh plus the toss we 'wasted' trying to get the desired pattern.

As for $E[N|\text{ previous toss was Heads}]$ we will either get our desired second $H$ or we get $T$ and it reverts to the case where we need to get a first $H$ again. The plus one is not a wasted toss as in the case when the first toss was $T$ however we still made a toss to get $H$; hence, we add the 1 toss.

I will let you continue here and give the following hint:

$W = N + M$ where $N :=$ number of flips to get $HH$ and $M:=$ number of flips to get $TT$. So we can use linearity of expectation and break it down into two more understandable problems.

Title: Coin flipping in a row


问:
The circumference of a circle c, is increasing at a rate of 2 cm/s, find how fast the area is increasing when r=6m?
I'm having trouble solving this related rates problem. How am I supposed to find the rate at which the area changes if I am only given the rate the circumference is changing? Thanks!</p>

答:
As $c=2\pi r$, it's given that $\frac{dc}{dt}= 2\pi \frac{dr}{dt} = 2$. (given).

$A=\pi r^2$ so $\frac{dA}{dt}= 2\pi r \frac{dr}{dt}$ by the chain rule.

Now try to combine..


答:
So, writing circumference as a function of the radius, $c(r) = 2\pi r$. Taking derivatives we write $\frac{dc}{dt} = 2\pi \frac{dr}{dt}$ where we are given $\frac{dc}{dt} = 2 cm/s$. Area as a function of the radius is $a(r) = \pi r^2$, and, again, taking derivatives we write $\frac{da}{dt} = \pi (2r\frac{dr}{dt})$.

You can finish this!

Title: The circumference of a circle c, is increasing at a rate of 2 cm/s, find how fast the area is increasing when r=6m?

Body: I'm having trouble solving this related rates problem. How am I supposed to find the rate at which the area changes if I am only given the rate the circumference is changing? Thanks!</p>


问:
A question of the book Elements of the Representation Theory of Associative Algebras: Volume 1
I am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1 .

I have a question on page 9, line -3 (see Page 9 here). It is said that $$h_1f_X = f_Y h_2.$$
I am trying to prove this. We have maps $h_1f_X: Xe_2 \to Xe_1 \to Ye_1$ and $f_Yh_2: Xe_2 \to Ye_2 \to Ye_1$. Let $xe_2 \in X_2$, where $x\in X$. Then $f_X(xe_2) = xe_2e_{21}e_1$. But what is the value of $h_1(xe_2e_{21}e_1)$? How to compute $h_1f_X(xe_2)$ and $f_Yh_2(xe_2)$ and show that they are equal? Thank you very much.

答: Use the fact that $h$ is an A-module homomorphism we have $h_1f_X(xe_2)=h_1(xe_2e_{21}e_1)=h(x)e_2e_{21}e_1$ and $f_Yh_2(xe_2)=f_Y(h(x)e_2)=h(x)e_2e_{21}e_1$.</p>


问:
Finding the Distance Between Two Diametrically Opposite Points on a Cone base circle
The question is as follows:

Both the slant height and the base diameter of a cone are 12 inches.  What is distace between two opposite points  on base circle of cone, if it is required that the path must lie on the lateral surface of the cone?

I am not sure of where to start with this problem, therefore, any help with how to start this will be very much appreciated. Thank you in advance.

答:
I came upon this math question while looking for something else. The question is a complicated way (perhaps meant to test one’s understanding of the terms) of addressing the very fundamental geometry of the Pythagorean theory: A squared plus B squared equals C squared-the sides A and B being the horizontal and vertical sides of a right triangle, and side C being the diagonal of the triangle. Never mind the cone, what we’re looking at is a simple right triangle. And since the question tells us the height and width are both 12”-actually we don’t even need to know the height in this case- the answer is double the base width, or 24”.
It’s a very simple question.</p>


问:
On the generalized Leibniz rule
problem definition

I have to evaluate in $z=0$ the $n$-th derivative with respect $z$ of the product $f(z)\cdot z^k$, where $f(\cdot)$ is a generic smooth function and $k$ is a given integer. I will use the short notation $F^{(n)}(z)$ to denote the $n$-th derivative of the generic function $F(z)$, so what I want is to compute for any $n\in\mathbb{N}$
\begin{equation}
[(f(z)\cdot z^k)^{(n)}]_{z=0}\triangleq \frac{\text{d}^n\left[f(z)\cdot z^k\right]}{\text{d}z^n}\Bigg|_{z=0}
\tag{1}
\end{equation}

my attempt

For the generalized Leibniz rule, holds for any $n$
\begin{equation}(f(z)\cdot z^k)^{(n)}=\sum_{i=0}^n \binom{n}{i}\cdot f^{(n-i)}(z) \cdot \left(z^k\right)^{(i)}\end{equation}
so the problem consist into compute the $i$-th derivative of the power $z^k$. If I'm not wrong,
\begin{equation}
\left(z^k\right)^{(i)}=\begin{cases}
\frac{k!}{(k-i)!}z^{k-i} & \text{if } i\leq k\\
0 & \text{otherwise}
\end{cases}
\end{equation}
This formula says that the index $i$ of the previous summation cannot exceed the value $k$. Anyway, $k$ is an external parameter and can be greater than $n$: in this case the summation stops at the value $n$,
otherwise the summation stops at value $k$.
So I would write
\begin{equation}\begin{aligned}(f(z)\cdot z^k)^{(n)}&=\sum_{i=0}^{\text{min}(n,k)} \binom{n}{i}\cdot f^{(n-i)}(z) \cdot  \frac{k!}{(k-i)!} z^{k-i}\\
&=k!\cdot\sum_{i=0}^{\text{min}(n,k)} \binom{n}{i}\cdot \frac{1}{(k-i)!}\cdot f^{(n-i)}(z) \cdot z^{k-i}\\
\end{aligned}\end{equation}
Now comes the problems. By setting $z=0$ turns out
\begin{equation}\begin{aligned}
%
[(f(z)\cdot z^k)^{(n)}]_{z=0}
&=k!\cdot\sum_{i=0}^{\text{min}(n,k)} \binom{n}{i}\cdot \frac{1}{(k-i)!}\cdot f^{(n-i)}(0) \cdot 0^{k-i}\\
\end{aligned}\end{equation}
from my prospective this expression is quite tricky because of the term $0^{k-i}$. I'm tempted to write
\begin{equation}
0^{k-i}=\begin{cases}1 & \text{if } i=k\\
0 & \text{otherwise}
\end{cases}
\end{equation}
and consequently simplify the last summation as
\begin{equation}\begin{aligned}
%
[(f(z)\cdot z^k)^{(n)}]_{z=0}
&=k!\cdot\sum_{i=0}^{\text{min}(n,k)} \binom{n}{i}\cdot \frac{1}{(k-i)!}\cdot f^{(n-i)}(0) \cdot 0^{k-i}\\
&=\begin{cases}
k!\cdot\binom{n}{k}\cdot \frac{1}{(k-k)!}\cdot f^{(n-k)}(0) \cdot 0^{k-k} & \text{if } n\geq k \\
0 & \text{otherwise}
\end{cases}\\
&=\begin{cases}
\frac{n!}{(n-k)!}\cdot f^{(n-k)}(0) & \text{if } n\geq k \\
0 & \text{otherwise}
\end{cases}\\
\end{aligned}\end{equation}

question

I don't have a precise question about my problem. Essentially I'm doubtful about my derivation because of the undefined power $0^0$.

答:
We consider  $n,k$ non-negative integers. It might be more convenient to consider a slightly simpler representation, namely
\begin{align*}
\color{blue}{\left.\left(f(z)\cdot z^k\right)^{(n)}\right|_{z=0}}
&=\left.\sum_{j=0}^n\binom{n}{j}f^{(n-j)}(z)(z^k)^{(j)}\right|_{z=0}\\
&=\left.\sum_{j=0}^n\binom{n}{j}f^{(n-j)}(z)k(k-1)\cdots(k-j+1)z^{k-j}\right|_{z=0}\tag{1}\\
&=
\begin{cases}
\binom{n}{k}f^{(n-k)}(0)k!&\ \qquad 0\leq k\leq n\\
0&\ \qquad k>n
\end{cases}\\
&\,\,\color{blue}{=
\begin{cases}
\frac{n!}{(n-k)!}f^{(n-k)}(0)&\qquad 0\leq k\leq n\\
0&\qquad k>n
\end{cases}
}
\end{align*}

In the representation (1) we see at a glance that evaluation at $z=0$ is equal to zero if $k\ne j$. This way there is no need to additionally define $0^0:=1$.

Title: On the generalized Leibniz rule


问:
If an integral is divergant with a non-negative intigrand, than the limit of the antiderivative is infinite
Somebody asked me this question in calculus 2, which I've already done, but I finished it in like a year so I don't really remember that much.

The question is : "if $f(x)$ is a non-negative and continuous function in $(a,b] $, and  $ \int_a^bf(x)dx $ is divergent, then $ \lim_{ x\to a} \int_x^bf(t)dt = \infty$".

This is assuming that it is not bounded on $(a,b]$. The definition of a divergent integral is that the limit which I described does not exist.

Intuitively I was thinking that because $f(x)$ is non-negative, than it means that the anti-derivative is non-decreasing, but I forgot some things in calculus, so I could use your help.

答:
Since $f$ is non-negative, the function $F:(a,b]\to [0,\infty)$ defined as $F(x)=\int_x^bf$ will be (weakly) decreasing. So, $\lim\limits_{x\to a^-}F(x)$ always exists in $[0,\infty]$ (in fact, it equals $\sup\limits_{x\in (a,b]}F(x)$), and the question is whether the limit is finite or infinite. But, you're told that "$\int_a^bf$ is divergent", which means either:

$\lim_{x\to a^-}\int_x^bf$ is $\pm \infty$ or

$\lim_{x\to a^-}\int_x^bf$ does not even exist in $[-\infty,\infty]$

Well, like I mentioned above, since $F$ is a (weakly) decreasing function, the second option is not possible. So, only the first option is possible. Hence
\begin{align}
\lim_{x\to a^-}F(x)=\lim_{x\to a^-}\int_x^bf = \infty.
\end{align}


问:
Splitting of marginal PDF
If we have the function:

$$f_{X,Y}\left(x,y\right)=\frac{3}{7}x,\:\text{ when }\:1\le x\le 2\:\wedge \:0\le y\le x$$

If I want to find the marginal function $f_Y\left(y\right)$.

It should be: $$f_Y\left(y\right)=\int _1^2\:\frac{3x}{7}\,dx=\frac{9}{14}.$$

Is it right or I need to split something?

答:
You can always check by integrating  (it should give you $1$, but in your case it doesn't give $1$ after integrating)

Your density is given by $$ f_{X,Y}(x,y) = \begin{cases} \frac{3}{7}x & x \in [1,2], y \in [0,x] \\ 0 & otherwise \end{cases} $$
To find marginal distribution of $Y$, you need to fix $y$ in support of your density, that is $y \in [0,2]$ and then integrate (with that fixed $y$) with respect to $x$ over whole space, that is you need to integrate:

$$ f_{Y}(y) = \int_{\mathbb R}f_{X,Y}(x,y)dx $$
Note that when $y \in [0,1)$ then $x$ can take any value between $[1,2]$ hence for $y \in [0,1)$ you indeed have $$f_{Y}(y) = \int_{1}^2 \frac{3}{7}x dx = \frac{3}{14}(2^2-1^2) = \frac{9}{14} $$
But when $y \in [1,2]$, then note that you have requirement $y \in [0,x]$, hence $x$ is not supported by whole $[1,2]$ but only $[y,2]$ (and that is okay that support of $x$ depends on $y$, our density can depend on $y$). So in case $y \in [1,2]$ you'd rather have $$ f_{Y}(y) = \int_{y}^2 \frac{3}{7}x dx = \frac{3}{14}(2^2 - y^2) = \frac{6}{7} - \frac{3y^2}{14} $$

So that marginal density is given by $$ f_{Y}(y) = \begin{cases} \frac{9}{14} & y \in [0,1) \\ \frac{6}{7} - \frac{3y^2}{14} & y \in [1,2] \\ 0 & otherwise \end{cases} $$

Title: Splitting of marginal PDF


问:
when is the region bounded by a Jordan curve "skinny"?
How can I formalize and prove the following intuition?:

Picture a very skinny rectangle, one with base length 1 and sides length $\epsilon$.  Or imagine a very flattened ellipse.  The interiors of these objects are "skinny" (or we might say $\epsilon$-skinny)  in the sense that each point in the interior is very close (or within $\epsilon$) to a point on the boundary. This notion seems easy to formalize.

Now think of the sides of the rectangle, or the boundary of the ellipse, as the image of a Jordan curve $f : S^1 \rightarrow \mathbb{R}^2$ . It's intuitively "obvious" (though not at all obvious) that what makes the bounded region skinny (in the sense of the above paragraph) is the fact that: For every $\theta \in S^1$ there exists $\theta ' \in S^1$ such that $f (\theta )$ is close to $f (\theta ' )$ but $\theta$ is not very close to $\theta '$.

I have been thinking about the right way to formalize these ideas and I'm a bit stuck.  I'd like to formulate them in the $C^0$ setting, that is to say without assuming tangent vectors or any calculus-related things.   Do you have any thoughts?  Thanks!

Also: This is motivated by the completely well-defined question here:why is an annulus close to it's boundary when it's boundary curves are close?

答:
Part I Non-parametric discussion of thinness.

each point in the interior is very close to a point on the boundary.

This says precisely that the inradius is small. The inradius (for inscribed radius) is the maximal radius of an open disk enclosed by the curve. For example, the rectangle of dimensions $a,b$ has inradius $\frac12 \min(a,b)$.

It remains to answer: small compared to what? There are two very reasonable options:

outradius: the minimal radius of a closed disk containing the curve.

diameter: the maximal distance between any pair of points on the curve.

The former is appealing because of its symmetry with inradius. The latter is easier to find in practice. It does not really matter which one you choose, because they are comparable for any planar set:
$$ \sqrt{3}  \operatorname{outradius}  \le  \operatorname{diameter} \le  2\operatorname{outradius}  \tag1$$

Part II Relation with bi-Lipschitz parameterization

Now think of the sides of the rectangle, or the boundary of the ellipse, as the image of a Jordan curve $f:S^1→\mathbb R^2$.

This changes the game a bit, because a very nice curve (e.g., a circle) has  parametrizations which horribly distort pairwise distances. Apparently, you want a description of skinniness in terms of parametrization $f$. Consider the bi-Lipschitz condition:
$$
L^{-1}\le \frac{|f(\theta)-f(\theta')|}{|\theta-\theta'|} \le L \tag2
$$
I prefer to use the chordal metric for $|\theta-\theta'|$: that is, the Euclidean distance between the two points $\theta,\theta'$. With this choice, (2) implies that the diameter of $f(S^1)$ is at most $2 L$ (obviously) and its inradius is at least $L^{-1}$ (not as obviously). Therefore, $L$ controls the diameter/inradius ratio: if $L$ is not very large, then the image is not too skinny.

The converse does not work: a non-skinny curve does not in general allow a bi-Lipschitz parameterization. For example, a tiniest cusp added to a disk of radius $1$ precludes the possibility of a bi-Lipschitz parameterization. To get around this, you may want to relax the bi-Lipschitz requirement by allowing an additive constant:
$$
L^{-1}|\theta-\theta'|-M\le {|f(\theta)-f(\theta')|} \le L|\theta-\theta'|+M \tag3
$$
but this is just a vague idea on my part; I don't know if it leads to a characterization of non-skinniness in terms of parametrization.

Part III Tentative parametric definition of thinness: a simple closed curve $\Gamma$ is $\epsilon$-thin if there exist two homeomorphisms $f:S^1\to \Gamma$ and $g:S^1\to\Gamma$ such that

$f$ and $g$ traverse $\Gamma$ in opposite directions (clockwise and counterclockwise)

$\max |f-g| \le \epsilon$

Indeed, suppose that 1 and 2 hold, and $a$ is a point in the interior of $\Gamma$. Then the argument I gave elsewhere applies: the curves $f$ and $g$ are not homotopic in $\mathbb R^2\setminus \{0\}$, since their winding numbers are different. In particular, the straight-line homotopy between $f$ and $g$ must pass through $a$. It follows that there exist $\theta\in S^1$ such that $f(\theta)-a$ and $g(\theta)-a$ are vectors pointing in opposite directions. Since
$$|f(\theta)-a|+|g(\theta)-a| = |f(\theta)-g(\theta)|\le \epsilon$$
it follows that the inradius of $\Gamma$ is at most $\epsilon/2$.

An equivalent, and possibly more attractive form of the above definition: for any homeomorphism $f:S^1\to \Gamma$ there exists a sense-reversing homeomorphism $\psi:S^1\to S^1$ such that $\max|f-f\circ \psi|\le \epsilon$.

It is easy to see that both ellipses and rectangles satisfy the above definition of thinness. But unfortunately, there are curves of small inradius that do not satisfy it, such as this one:


问:
Existence of a complete norm on $\ell^2$ which extends $\|۰\|_1$ (on $\ell^1$)
We know that norm-1 is a complete norm on $\ell^1.$ Also $\ell^1$ is a subspace of $\ell^2.$ Now, is there a complete norm
on $\ell^2$ which extends norm-1 (on $\ell^1$)?

答:
Let $V$ be an algebraic complement of  $\ell ^1$ within $\ell ^2$,  so that $\ell ^2 = \ell ^1 \oplus  V$.  It is easy to see that the algebraic
dimension  of $V$ is $2^{\aleph_0}$,  so $V$ is algebraically isomorphic to any separable infinite dimensional Banach space $X$.  Choosing such
an $X$ let
$$
  \varphi :V\to X
  $$
be an algebraic  isomorphism and define a norm on
$\ell ^2$
by
$$
  \|x\| = \|x_1\|_1 +  \|\varphi (x_2)\|,
  $$
where
$$
  x = x_1+x_2
  $$
is the decomposition of any given $x$  in $\ell ^2$ according to
$\ell ^2 = \ell ^1 \oplus  V$.  This norm satisfies the requirements.

EDIT.  I guess my statement above that "the algebraic
dimension  of $V$ is $2^{\aleph_0}$" is not so immediate,  so let me make up for this.

That statement is clearly equivalent to
$$
  \text{dim}(\ell ^2/\ell ^1) = 2^{\aleph_0},
  $$
so let us  prove this instead.

For each $\alpha >0$, consider the vector
$$
  x_\alpha =(1/n^\alpha )_{n\in {\mathbb N}},
  $$
and observe that $x_\alpha \in \ell ^p\Leftrightarrow p\alpha >1$.  In particular
$$
  x_\alpha \in \ell ^2,  \quad \text{and}\quad x_\alpha  \notin  \ell ^1,\quad \forall \alpha \in (1/2,1].
  $$
We next claim that the set
$$
  \{x_\alpha : \alpha \in (1/2,1]\}
  $$
is a linearly independent set in $\ell ^2/\ell ^1$.  To prove  this we must verify  that, whenever
$$
  \lambda _1x_{\alpha _1}+  \lambda _2x_{\alpha _2}+\cdots  + \lambda _nx_{\alpha _n} \in  \ell ^1,
  \tag 1
  $$
one necessarily has $\lambda _1=\lambda _2=\cdots =\lambda _n=0$.  Assuming by contradiction that this is not so, we may discard the zero terms in
the above sum and hence suppose  that the $\lambda _i$ are all nonzero.  We may also assume that  the $\alpha _i$ are organized such that
$$
  \frac 1{\alpha _1}<  \frac 1{\alpha _2}<\cdots <  \frac 1{\alpha _{n-1}}<\frac 1{\alpha _n}.
  $$
Chosing any real number $p$ in between the last to terms above, that is,
$$
  \frac 1{\alpha _{n-1}}<p<\frac 1{\alpha _n},
  $$
we then have that $p\alpha _n<1$, so $x_{\alpha _n} \notin \ell ^p$,  while $p\alpha _k>1$, for all $k=1,2,\ldots ,n-1$,  so $x_{\alpha _k} \in \ell ^p$.
However,   (1) implies  that
$$
  \lambda _nx_{\alpha _n}\in    \ell ^1-  \lambda _1x_{\alpha _1}-  \lambda _2x_{\alpha _2}-\cdots  - \lambda _{n-1}x_{\alpha _{n-1}} \subseteq  \ell ^p,
  $$
a contradiction.

This shows that
$$
  \text{dim}(\ell ^2/\ell ^1) \geq  2^{\aleph_0},
  $$
but since the cardinality of the whole of $\ell ^2$ is
$2^{\aleph_0}$,  it is evident that the opposite inequality holds as well.


问:
Image of a discrete group in a vector space is discrete and closed
Let $\Gamma \subset \operatorname{GL}_n(\mathbb R)$ be a discrete subgroup, and let $0 \neq v \in \mathbb R^n$.  Is it true that $\Gamma.v = \{ \gamma.v : \gamma \in \Gamma\}$ is discrete and closed in $\mathbb R^n$?

Writing out $v = (v_1, ... , v_n)$, and $\gamma = (\gamma_{ij})$, it suffices to show that for each $j$, the elements $\sum\limits_{i=1}^n \gamma_{ij}v_i$ run through a discrete and closed set in $\mathbb R$.

While each of the sets $\{ \gamma_{ij}v_i\} \subset \mathbb R$ is discrete, as $\gamma$ runs through the elements of $\Gamma$, it is generally not true that $E+J$ is discrete if $E$ and $J$ are discrete subsets of $\mathbb R$.

答:
No, not at all, even for cyclic groups. As a simple example, take a linear transformation $A\in GL(3, {\mathbb R})$ which has one eigenvalue $>1$ and two complex conjugate eigenvalues with absolute value $1$, and which are not roots of unity. Let $\Gamma$ be the subgroup of  $GL(3, {\mathbb R})$ generated by $A$. I will leave it to you to prove the discreteness of $\Gamma$ and to find a (nonzero) vector whose $\Gamma$-orbit is dense in an ellipse.

As a more interesting example, take $\Gamma= SL(2, {\mathbb Z})$. One can prove that for almost every vector $v\in {\mathbb R}^2$, its $\Gamma$-orbit is dense in  ${\mathbb R}^2$ (because the $\Gamma$-action is ergodic).


问:
Writing a tangent bundle in terms of the rotation group of the embedding space and connection (parallel transport)
Ok, so I'm a physics student, but this is primarily a math question, so I'm posting it here. I apologize in advance for any misuse of terminology and formality. Looking at the Tangent bundle for an m-dimensional (possibly pseudo-) Riemannian manifold $M$, we write it as the disjoint union of the tangent space at each point in the manifold.

$$TM=\dot{\bigsqcup}_{x\in M}T_{x}M$$

I can't help but wonder: is there another way to write this? Let us consider the manifold as embedded in $\mathbb{R}^{n}$. Then the tangent space at a point x $\left(T_{x}M\right)$ is related to the tangent space at another point q by a rotation in $\mathbb{R}^{n}$. Denoting such a point dependent transformation $A_{pq}$ it is obvious that $A_{pq}\in O(n)$ (or $O(n,k)$ for a pseudo-Riemannian manifold). It is clear then that our tangent bundle can be described simply by:

$$TM=\left(\dot{\bigsqcup}_{x\in M}A_{xo}\right)T_{o}M$$

Where the point o is an arbitrary fixed point on M. This is simply: $$TM=\left(\dot{\bigsqcup}_{x\in M}A_{xo}\right)\mathbb{R}^{m}$$

Which is to say we can write it in terms of the union of the tangent space at one point with all of the position dependent (a function of the embedded manifold in question) transformations which are elements of the aformentioned rotation group.

Is this correct so far? Now when we write the projection $\pi$ to $M$, how exactly do we go about doing this? Do we get a connection in the covariant derivative with values in projective representations of the rotation group on M?

For example, a Lorentzian manifold, locally has $O(3,1)$ as the rotation group, can we then say these Lorentz valued connections are projective representations from $O(n,k)$? I believe these are then spinor representations? Thank you very much in advance!

For exemplify what I'm saying, If M is flat $\mathbb{R}^{3,1}$ then the embedding and projections are trivial, but If I have say an $S^{3}xS^{1}$ space then I would have non-vanishing elements of the $O(4,2)$ appearing in a connection? Thank you very much in advance! I asked a similar question here but on the physics stackexchange they seem to be missing the point of my question.

答:
"Then the tangent space at a point $x$ ($_ $) is related to the tangent space at another point $q$ by a rotation in $\Bbb R^n$."

Consider the case of a circle in the $xyzw$-plane of $\Bbb R^4$. The tangent space at $(1, 0, 0,0)$ consists of multiples of $(0,1, 0,0)$; the tangent space at $(0, 1,0,0)$ consists of mutliples of $(-1, 0, 0,0)$. And rotation by 90 degrees in the $xy$-plane takes one of these to the other. But the rotation defined by left-multiplication by the matrix
$$
A_u = \pmatrix{
0 & -1 & 0 & 0 \\
1 & 0 & 0  & 0 \\
0 & 0 & c & -s \\
0 & 0 & s & c}
$$
where $s = \sin u, c = \cos u$ ... that rotation also takes one tangent space to the other. In particular, the $A$ in your description is not unique, and it's not even clear that there's a continuously-varying function $A : M \times M \to SO(n)$ such that $A(p,q)$ sends $T_p M $ to $T_q M$. (There surely is in this example of mine, but when you think about a Mobius strip....things get a bit dicier.


答:
How about something like this:

Definition: A Jordan curve $f\colon S^1\to\mathbb{R}^2$ is called $\epsilon$-boundary-skinny if there exists a homeomorphism $k\colon S^1 \to S^1$ such that $\|(f\circ k)(\theta)-(f\circ k)(-\theta)\|<2\epsilon$ for all $\theta\in S^1$.

Note: Here $-\theta$ is to be understood in the sense of $S^1\cong\mathbb{R}/2\pi\mathbb{Z}$, not $S^1\cong \{z\in\mathbb{C}:|z|=1\}$.

The result you are looking for would then be something like: 'Any $\epsilon$-boundary-skinny Jordan curve is $\epsilon$-skinny.'

Title: when is the region bounded by a Jordan curve "skinny"?

Body:
How can I formalize and prove the following intuition?:

Picture a very skinny rectangle, one with base length 1 and sides length $\epsilon$.  Or imagine a very flattened ellipse.  The interiors of these objects are "skinny" (or we might say $\epsilon$-skinny)  in the sense that each point in the interior is very close (or within $\epsilon$) to a point on the boundary. This notion seems easy to formalize.

Now think of the sides of the rectangle, or the boundary of the ellipse, as the image of a Jordan curve $f : S^1 \rightarrow \mathbb{R}^2$ . It's intuitively "obvious" (though not at all obvious) that what makes the bounded region skinny (in the sense of the above paragraph) is the fact that: For every $\theta \in S^1$ there exists $\theta ' \in S^1$ such that $f (\theta )$ is close to $f (\theta ' )$ but $\theta$ is not very close to $\theta '$.

I have been thinking about the right way to formalize these ideas and I'm a bit stuck.  I'd like to formulate them in the $C^0$ setting, that is to say without assuming tangent vectors or any calculus-related things.   Do you have any thoughts?  Thanks!

Also: This is motivated by the completely well-defined question here:why is an annulus close to it's boundary when it's boundary curves are close?


问:
How to find if the function is injective, surjective or both! Linear Algebra
I have an exam in Linear Algebra in a few days and there was this one question on the practice quizzes we have in our university portals! I've been trying to figure this out for a while now but I don't know how to approach this problem can you help me?

Let, $\vec{i},\vec{j},\vec{k}$ be the orthogonal vectors of the axes of $\mathbb{R}^3$. Let $\vec{v}=-\vec{j}+\vec{k}$.
Let us denote by $\times$ the vector product. Let $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ be the endomorphism defined as:  $f(U)=V\times U+2U$, for each $U\in \mathbb{R}^3$.
Which of the following is true?

$(a)$ $f$ is surjective but not injective

$(b)$ $f$ is injective but nor surjective

$(c)$ $t(t^2 + t - 3)$ is the characteristic polynomial of $f$

$(d)$ $f$ is injective

The correct one is $(d)$ but I don't know how to arrive at that point! Can you help me?

答:
Hint:   You can establish $d)$ by proving that the kernel is trivial.

That will also disprove $a)$ and $b)$ since if an endomorphism of a finite dimensional vector space is injective or surjective, then it is an isomorphism.

$c)$ is false, because injectivity means that zero is not an eigenvalue.

To prove injectivity, note that $|a\times b|=|a|\cdot|b|\sin\theta$, where the $\theta$ is the angle between $a$ and $b$.  So we get that $f(u)=0\implies|v|\sin\theta=2\implies |v|\ge2$ a contradiction, unless $u=0$.


问:
Converge of this series
I was shown this example, but do not understand the procedure for doing this and why it's correct.

Show $\sum x^{n!}$ converges for $x \in (0,1)$.

Way it was shown: Write this as $x+x^2+0x^3+0x^4+\dots$.i.e. $\sum c_kx^k$ where $c_k$ is $1$ if $k=n!$ for some $n$, and $0$ otherwise. Now use the comparison:$|c_kx^k| \leq x^k$ so compare with geometric series.Can someone explain what is going on here with this method? Why can you write the series like this?Thanks

答:
This response is somewhat off-topic because the OP specifically asked for a proof that involves the comparison test.  However, comments that preceded this answer have provided such a proof.  Therefore, I regard it as open season to provide an alternative approach.

If you have a strictly increasing sequence of numbers, that are bounded above, then the sequence must be convergent.

You are given that $\sum_{k=0}^{\infty} x^k$ is a convergent series.  Denote its limit as $L$.

Define the sequence $a_0, a_1, a_2, \cdots,$ by $a_k = \sum_{i=0}^k x^{(i!)}.$

Then the problem reduces to showing that $\langle a_k\rangle$ is a convergent sequence.

You know that this sequence is strictly increasing.

As other responses have indicated, the sequence is bounded above by $L$.

Title: Converge of this series


问:
Is it true that the edge weight of a bridge equals to its least-squares potential difference within a graph?
Consider a (not necessarily connected) directed multigraph $G=(V,E,t,h,w,u)$, where V is the vertex set, E is the edge set,
$t: E \rightarrow V$ is the function of edge tails, $h: E \rightarrow V$ is the function of edge heads, while $w: E \rightarrow \mathbb{R}^+$ and $u: E \rightarrow \mathbb{R}$ are two (fixed) weight functions.

Suppose that $\pi: V \rightarrow \mathbb{R}$ is a potential function minimizing the following (least-squares) objective function:

$\mathit{\Omega}( \mathit{\Pi})=\sum\limits_{e \in E}{w_e{[u_e-\mathit{\Pi}(h_e)+\mathit{\Pi}(t_e)]}^2}$.

Is it true that if $e^* \in E$ is a bridge (i.e., an edge whose deletion increases the number of components within G - multiple edges are not bridges here), then $\pi(h_{e^*})-\pi(t_{e^*})=u_{e^*}$? If so, how to prove this conjecture?

答:
Suppose that $C$ is the connected component containing $h_{e^*}$ in $G - e^*$. Then adding the same constant to $\pi(v)$ for all vertices $v$ in $C$ changes $w_{e^*}[u_{e^*}  - \pi(h_{e^*}) + \pi(t_{e^*})]^2$ and no other term of $\Omega(\pi)$.

In particular, by adding the right constant, we can make $u_{e^*}  - \pi(h_{e^*}) + \pi(t_{e^*}) = 0$ and set this term to $0$, and since $\pi$ is supposed to be a minimizer, it must the the case that $u_{e^*}  - \pi(h_{e^*}) + \pi(t_{e^*})$ is already $0$.


答:
Hint:

Endomorphisms from $\mathbb{R}^3$ to $\mathbb{R}^3$ are 3x3 matrices. You can figure out this matrix by finding $f\begin{pmatrix}x\\y\\z\end{pmatrix}$ explicitly.

If you find, say $f\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2x+3y\\-x+2y-3z\\x+y+z\end{pmatrix}$, then the matrix that represents $f$ is $\begin{pmatrix}2&3&0\\-1&2&-3\\1&1&1\end{pmatrix}$ since $$\begin{pmatrix}2&3&0\\-1&2&-3\\1&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=f\begin{pmatrix}x\\y\\z\end{pmatrix}$$

Then you can find its characteristic polynomial and determine if it is surjective or injective.

Title: How to find if the function is injective, surjective or both! Linear Algebra

Body:
I have an exam in Linear Algebra in a few days and there was this one question on the practice quizzes we have in our university portals! I've been trying to figure this out for a while now but I don't know how to approach this problem can you help me?

Let, $\vec{i},\vec{j},\vec{k}$ be the orthogonal vectors of the axes of $\mathbb{R}^3$. Let $\vec{v}=-\vec{j}+\vec{k}$.
Let us denote by $\times$ the vector product. Let $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ be the endomorphism defined as:  $f(U)=V\times U+2U$, for each $U\in \mathbb{R}^3$.
Which of the following is true?

$(a)$ $f$ is surjective but not injective

$(b)$ $f$ is injective but nor surjective

$(c)$ $t(t^2 + t - 3)$ is the characteristic polynomial of $f$

$(d)$ $f$ is injective

The correct one is $(d)$ but I don't know how to arrive at that point! Can you help me?


答:
For another simple example, let $n=1$, and $\Gamma$ be the subgroup of $\text{GL}_1(\mathbb R) = \mathbb R\setminus \{0\}$, generated by any $g>1$.  Then $\Gamma$ is discrete but, for any nonzero $x$, the orbit
$$
\{g^nx:n\in \mathbb Z\}
$$
has zero in its closure.

Title: Image of a discrete group in a vector space is discrete and closed

Body:
Let $\Gamma \subset \operatorname{GL}_n(\mathbb R)$ be a discrete subgroup, and let $0 \neq v \in \mathbb R^n$.  Is it true that $\Gamma.v = \{ \gamma.v : \gamma \in \Gamma\}$ is discrete and closed in $\mathbb R^n$?

Writing out $v = (v_1, ... , v_n)$, and $\gamma = (\gamma_{ij})$, it suffices to show that for each $j$, the elements $\sum\limits_{i=1}^n \gamma_{ij}v_i$ run through a discrete and closed set in $\mathbb R$.

While each of the sets $\{ \gamma_{ij}v_i\} \subset \mathbb R$ is discrete, as $\gamma$ runs through the elements of $\Gamma$, it is generally not true that $E+J$ is discrete if $E$ and $J$ are discrete subsets of $\mathbb R$.


问:
Quantum uncertainty can explain the Riemann Hypothesis?
In the recent paper "Riemann Hypothesis as an Uncertainty Relation" (http://arxiv.org/abs/1304.2435) the author claims that the presence of zeros out of the critical line may lead to the violation of a Heisenberg-type uncertainty relation. Is that work a proof of Riemann's hypothesis?</p>

答: I developed an algorithm that is based on uncertainty principle, the result is consistent with RH. The basic idea of the algorithm is representation of objects movement in space and time using two operators. I still can not answer the question whether this could prove RH.</p>


问:
Showing that Pell's equation has at least one integer solution
I'm studying Pell's equations of the form $x^2-dy^2=1$ for $d$ a square-free natural number and $x,y$ integers. In particular, I want to show that such an equation always has a solution.

I see that this is equivalent to finding units with positive norm in $\mathbb{Z}[\sqrt{d}]$. I know that there is always a fundamental unit $\epsilon$ which generates the multiplicative group of units, and $\epsilon$ is unique if we take it as the smallest unit with $\epsilon>1$. Clearly any even power of $\epsilon$ will have positive norm since $\mathbb{Q}(\sqrt{d})$ is real. Hence any solution of Pell's equation would be some power of $\epsilon$.

But why does there always exist an $n\in\mathbb{N}$ such that $\epsilon^n\in\mathbb{Z}[\sqrt{d}]$? Clearly this is the case when $\epsilon\in\mathbb{Z}[\sqrt{d}]$, but of course if $d\equiv 1$ (mod $4$) then it may be the case that $\epsilon\in\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$.

Computation of specific examples seems to indicate that the cube of any fundamental unit will be in $\mathbb{Z}[\sqrt{d}]$. I tried proving this directly, but the amount of calculation makes me think that there is a more elegant method. I see an argument in the final question of this document, but by the heuristic law of "conservation of difficulty", I think that once one uses Dirichlet's Diophantine approximation theorem to classify the units of real quadratic number rings, there must be a method that follows easily.

Looking forward to hearing your thoughts!

答:
Question $3$ of the document you link to is not the last question of the document, so don't describe it as the "final question".

When $d \equiv 1 \bmod 4$, the cube of each unit in $\mathbf Z[(1+\sqrt{d})/2]$ is in $\mathbf Z[\sqrt{d}]$. To check the coefficients of $1$ and $\sqrt{d}$ in the cube of a unit are both integers, it is somewhat important to use the fact that you're using a unit (the norm is $\pm 1$).  The cube of a nonunit in $\mathbf Z[(1+\sqrt{d})/2]$ might not be in $\mathbf Z[\sqrt{d}]$, depending on the value of $d$.  For example, $(1+\sqrt{17})/2$ is an integer in $\mathbf Q(\sqrt{17})$ and $((1+\sqrt{17})/2)^3 = (13 + 5\sqrt{17})/2 \not\in \mathbf Z[\sqrt{17}]$. Note that it is convenient to realize that the integers of $\mathbf Q(\sqrt{d})$ for squarefree $d \equiv 1 \bmod 4$ can be described as $\{(a+b\sqrt{d})/2 : a, b \in \mathbf Z, a \equiv b \bmod 2\}$.

Another suggestion: rather than bootstrap knowledge of the unit group of the full ring of integers to tell you the structure of the unit group of $\mathbf Z[\sqrt{d}]$ when $d \equiv 1 \bmod 4$, you could run through the proof of the unit theorem to check it works for every order (subring of finite index in the full ring of integers) of a number field. The proof of the unit theorem does not use prime ideal factorization, for example. There is nothing about the proof of the unit theorem that does not carry over verbatim to orders.  In fact, accounts of the solvability of Pell's equation in elementary number theory books (not books on algebraic number theory) don't even impose conditions like "$d$ is squarefree".  The standard description of the solvability of Pell's equation $x^2 - dy^2 = 1$ is that it has an integral solution $(x,y)$ with $y \not= 0$ as long as $d$ is not a perfect square. It is completely irrelevant that $d$ may or may not be squarefree in proofs of solvability of Pell's equation.

If you want to see abstract existence of an $n \geq 1$ such that $\varepsilon^n \in \mathcal O$ when $\mathcal O$ has finite index in the full ring of integers $\mathcal O_K$ of a number field $K$ and $\varepsilon \in \mathcal O_K^\times$, think about the integral powers of $\varepsilon$ lying in the finite set $\mathcal O_K/\mathcal O$.


问:
$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} $
I have to find for which values of $a \in \Bbb N, a \ne 0$ the following limit exists and it is finite:

$$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2}  $$

Applying L'Hôpital's rule:
$$\frac{1-\cos (ax)}{x^2}\sim \frac{\sin (ax) \cdot a}{2x}\sim \frac{a^2}{2}$$
Then $$ \frac{\cos (\pi \cdot \frac{1-\cos (ax)}{x^2})}{x^2}  \sim   \frac{\cos (\pi \cdot \frac{a^2}{2})}{x^2}.$$
$$\cos (\pi \cdot \frac{a^2}{2})=0 \implies a^2=1+2k, \quad k \in \Bbb N$$
and in this case the limit is $0$.

In the book the suggested solution is $a^2=1+2k$ for which the limit is $$ \frac{(-1)^k \cdot (2k+1)^2\cdot\pi}{24}$$

but I don't understand this solution.

Trying to solve the limit with $a^2=1+2k$:

$$ \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} \sim \frac{-\sin (\pi \cdot \frac{1-\cos ax}{x^2}) \cdot \pi \cdot  \frac{\sin (ax)  a  x^2 - (1-\cos (ax))  2x}{x^4}}{2x} $$$$ =  \sin \left( \frac{\pi}{2}+k \pi\right) \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4}$$$$= (-1)^k \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4} $$

答:
Using l'Hôpital in this case is not the best strategy. You have
$$
\cos(ax)=1-\frac{a^2x^2}{2}+\frac{a^4x^4}{24}+o(x^4)
$$
and therefore
$$
\pi\frac{1-\cos(ax)}{x^2}=\pi\frac{a^2}{2}-\pi\frac{a^4x^2}{24}+o(x^2)
$$
In order for the limit to be finite, we need that
$$
\frac{a^2\pi}{2}=\frac{\pi}{2}+k\pi
$$
for some integer $k$, so $a^2=1+2k$. Now the numerator is
$$
\cos\Bigl(\frac{\pi}{2}+k\pi-\frac{a^4\pi x^2}{24}+o(x^2)\Bigr)=
(-1)^k\sin\Bigl(\frac{a^4\pi x^2}{24}+o(x^2)\Bigr)=
(-1)^k\frac{a^4\pi x^2}{24}+o(x^2)
$$
and therefore the limit is
$$
(-1)^k\frac{(2k+1)^2\pi}{24}
$$
Obviously $k$ should be a nonnegative integer.


问:
prove that $\arcsin (3/5) /\pi $ is irrational
I was asked to prove that on the unit circle, the rational points are dense. My idea is to first show that the point (3/5,4/5) corresponds to an irrational angle (more precisely, its ratio with respect to $\pi$), then by rotation, its image will be dense on the circle. This is Kronecker's theorem?

So I have to show that the angle is irrational.

答: Hint: $\alpha/\pi$ is rational iff $e^{i \alpha}$ is an $n$th root of unity for some $n \in \Bbb{Z}$. Maybe you can show that, if $$z = \frac{3 + 4i}{5},$$ then $z^n \neq 1$ for any $n \geq 2$.


问:
How many different terms are there in the expansion of $(x_1 + x_2 + \cdots + x_m)^n$ after all terms with identical sets of exponents are added?
I just can't understand how to go about it!</p>

答:
As user545963 has commented, each pattern involves $m$ non-negative integer exponents adding up to $n$.  The number of compositions of this form can be found with stars and bars $(n$ stars and $m-1$ bars separating them into $m$ values$)$ so $${n+m-1 \choose n} \text{ possible patterns of the exponents}$$

You can also express the coefficient of each $x_1^{d_1}x_2^{d_2}\cdots x_m^{d_m}$ pattern: it is $\dfrac{n!}{d_1!\,d_2!\,\cdots d_m!}$ and if you add these all up you get the unsurprising answer of $m^n$


问:
Best Maths Books for Non-Mathematicians
I'm not a real Mathematician, just an enthusiast. I'm often in the situation where I want to learn some interesting Maths through a good book, but not through an actual Maths textbook. I'm also often trying to give people good Maths books to get them "hooked".

So the question: What are good books, for laymen, which teach interesting Mathematics, but actually does it in a "real" way. For example, "Fermat's Last Enigma" doesn't count, since it doesn't actually feature any Maths, just a story, and most textbook don't count, since they don't feature a story.

My favorite example of this is "Journey Through Genius", which is a brilliant combination of interesting storytelling and large amounts of actual Mathematics. It took my love of Maths to a whole other level.

Edit:

A few more details on what I'm looking for.

The audience of "laymen" should be anyone who has the ability (and desire) to understand actual mathematics, but does not want to learn from a textbook. Obviously I'm thinking about myself here, as a programmer who loves mathematics, I love being exposed to real maths, but I'm not going to get into it seriously. That's why books that show actual maths, but give a lot more exposition (and much clearer explanations, especially of what the intuition should be) are great.

When I say "real maths", I'm talking about actual proofs, formulas, or other mathematical theories. Specifically, I'm not talking about philosophy, nor am I talking about books which only talk about the history of maths (Simon Singh style), since they only talk about maths, they don't actually show anything. William Dunham's books and Paul J. Nahin's books are good examples.

答: I will warmly recommend Mathematics: Its Content, Methods and Meaning, edited by Alexandrov, Kolmogorov and Lavrentiev, with contributions from many others. It's excellent and very affordable.</p>

Title: Best Maths Books for Non-Mathematicians


问:
After thinking for quite a bit, I haven't been able to completely solve this induction question.
Can L tilings be created for L shaped structure (such that the L tilings are in fact trominoes). It seems to not be such a simple question to answer! Though their may be multiple ways of answering the same.</p>

答:
For the case when $n$ is odd

It is easy to divide any $3\times 2k$ rectangle into tiles.

So, divide your large L-shape into the following shapes. A and C form a border of width $3$ along the longest sides of the large L-shape.

A. Two $3\times (2n-6)$ rectangle3

B. Two $3\times (n-3)$ rectangles

C. A new L-shape with $n$ replaced by $3$

D. A new large L-shape with $n$ replaced by $n-3$.

You can now apply induction.


问:
Orthogonal Projection of $y$ onto $\operatorname{Col}(A)$
Let $A$ be an $n\times m$ matrix with linearly independent columns. Let $y$ be a vector
in $\mathbb{R}^n$. Prove that $\operatorname{Proj}_{\operatorname{Col}(A)}y=A(A^TA)^{-1}A^Ty$.

I think I am supposed to prove that $A(A^TA)^{-1}A^Ty$ is the orthogonal projection of $y$ onto $\operatorname{Col}(A)$.

I am not sure how to go about this though.

答:
Let $V:= \mathrm{Col}(A)$. I assume that the $\mathrm{Proj}_V y$ is defined as the unique vector such that
$$
(y - \mathrm{Proj}_V y)^\top v = 0
$$
for all $v \in V$. Since $V$ is the column space, we can rewrite this as finding $z \in \mathbb{R}^m$ such that
$$
(y - Az)^\top Ax = 0 \quad \forall x \in \mathbb{R}^m.
$$
Let us manipulate this equation:
$$
y^\top Ax - z^\top A^\top A x = 0 \iff (y^\top A - z^\top A^\top A)x = 0
$$
So solving $y^\top A - z^\top A^\top A = 0$ would suffice. This yields $z = (A^\top A)^{-1}A^\top y$. Plugging this in, leaves
$$
\mathrm{Proj}_V y = A(A^\top A)^{-1}A^\top y.
$$

What is left to prove is that $A^\top A$ is in fact invertible: Since the columns of $A$ are l.i., we have $\mathrm{rank}(A) = m$. Now note that trivially $\mathrm{ker}(A) \subseteq \mathrm{ker} (A^\top A)$. Furthermore if $x \in \mathrm{ker}(A^\top A)$, we have
$$
0 = x^\top A^\top Ax = \lVert Ax \rVert_2^2.
$$
But by norm properties, we get $Ax = 0$, so $x \in \mathrm{ker}(A)$. So $\mathrm{ker}(A) = \mathrm{ker} (A^\top A)$. By the rank nullity-theorem, we have:
$$
m = \dim\mathrm{ker}(A) + \mathrm{rank}(A) = \dim \mathrm{ker}(A^\top A) + m \iff \dim\mathrm{ker}(A^\top A) = 0
$$
But this means that $A^\top A$ is invertible.

There is an alternative to the first part of the proof. The projection onto $V$ could also be defined as $Ax^*$ where $x^*$ is the minimizer of $g:\mathbb{R}^m \rightarrow \mathbb{R}$
$$
g(x) := \frac{1}{2} \lVert y - Ax \rVert^2_2.
$$
One could do some calculus and find a solution to
$$
A^\top Ax - A^\top y = \nabla g(x) \overset{!}{=} 0.
$$
This is our vector $z$. Regularity of $A$ then gives positive definiteness of thw hessian $A^\top A = D^2 g(x)$ on the entire domain, i.e. that $z$ is a unique minimizer.


问:
For which angle $x$, it results $0 < \cos(x) < \sin(x)<1$?
For which angle $x$, it results $0 < \cos(x) < \sin(x) < 1$?

40°

50°

60°

70°

80°

Why is the correct answer #3 60° and not #3,4,5?

Yet, if the question asks: For which angle X, it results $0.5 < \cos(x) < \sin(x) < 1$?, the answer is #2 50°

答:
Imagine constructing a right triangle as follows:

One point is the point $A$ at $(0,0)$.

One point is the point $B$ at $(x,0)$.

One point is the point $C$ at $(x,y)$ subject to the following two constraints:

The point $(x,y)$ is a point on the unit circle, so you have that $x^2 + y^2 = 1.$

$0^{\circ} < \angle BAC < 90^\circ.$

Note that the $\cos(\angle BAC) = x$ and the $\sin(\angle BAC) = y$.  This is because the right triangle was intentionally constructed so that its hypotenuse has a length of $1$.

Therefore, the question of the comparison of the cosine of $\angle BAC$ versus the sine of $\angle BAC$ reduces to comparison of the values for the corresponding numbers $x$ and $y$.

When $\angle BAC = 45^{\circ}$, you have that the slope of the line segment $(x,y) -- (0,0)$ is equal to $1$, so when $\angle BAC = 45^{\circ}$, $y$ exactly equals $x$.

When $0^{\circ} < \angle BAC < 45^\circ,$ the slope of the line segment $(x,y) -- (0,0)$ is less than $1$, which implies that $y < x.$

When $45^{\circ} < \angle BAC < 90^\circ,$ the slope of the line segment $(x,y) -- (0,0)$ is greater than $1$, which implies that $y > x.$


问:
Let $X$ be a probability measure. If $f$ is not constant a.e, then there is $\lambda$ s.t $0<\mu(f^{-1}(\lambda,\infty))<1$
Let $X$ be a probability measure. If $f$ is not constant a.e, then there is $\lambda$ s.t $0<\mu(f^{-1}(\lambda,\infty))<1$

I found a proof but it is long and I feel like this should be a one liner.

In my proof I assume all such sets have measure either $0$ or $1$. let $A=\{\lambda|\mu(f^{-1}(\lambda,\infty))=1\}$ I prove that $A$ must bounded. Thus $\sup(A)$ exists. If $\sup(A) \in A$ $f=\sup(A)$ a.e which is a contradiction. If it does not there is a monotonic sequence $\lambda_n \to \sup(A)$. Then using continuity of measure we see that a sequence of $1$'s converge to $0$ which is a contradiction.

With all the details it is fairly lengthy. I smell IVT but not sure how to apply it here.

答:
(Assume $f$ is finite valued.)

The point is that "$\mu \circ f^{-1}$" is a Borel probability measure on $\mathbb{R}$ so it boils down to an exercise about such measures.  There is a certain amount of "boilerplate" or "legalese" making the proof necessarily a few sentences.

Define a Borel measure $\nu$ by $\nu(A) = \mu(f^{-1}(A))$.  This is a probability measure since $\nu(\mathbb{R}) = \mu(f^{-1}(\mathbb{R})) = 1$.  I claim that if $\nu((\lambda,\infty)) \in \{0,1\}$ for each $\lambda \in \mathbb{R}$, then $\nu = \delta_{\bar{\lambda}}$ for some $\bar{\lambda} \in \mathbb{R}$.

Indeed, under this assumption, we can let $\bar{\lambda} = \sup \left\{ \lambda \in \mathbb{R} \, \mid \, \nu((\lambda,\infty)) = 1 \right\}$.  $\bar{\lambda} \in \mathbb{R}$ by continuity of measure.  To see that $\nu = \delta_{\bar{\lambda}}$, observe that, for each $\epsilon > 0$,
\begin{equation*}
\nu([\bar{\lambda} - \epsilon, \bar{\lambda} + \epsilon]) = \nu((\bar{\lambda} - \epsilon,\infty)) - \nu((\bar{\lambda} + \epsilon,\infty)) = 1.
\end{equation*}
Thus, by continuity of measure, $\nu(\{\bar{\lambda}\}) = 1$.

We conclude that $\mu(f^{-1}(\{\bar{\lambda}\})) = \nu(\{\bar{\lambda}\}) = 1$, hence $f \equiv \bar{\lambda}$ ($\mu$-a.e.).


问:
355/113 and small odd cubes
An important approximation to $\pi$ is given by the convergent $\frac{355}{113}$.

The numerator and the denominator of this fraction are at the same distance of small consecutive odd cubes.

$$\frac{355}{113} = \frac{7^3+12}{5^3-12}$$

Is this a consequence of some general formula, such as a series or continued
fraction?

Attempts, drafts

$$\frac{113}{355} = \frac{1}{10}\left(2+\frac{5^3}{3+\frac{7^3}{4}}\right)-\frac{8}{355} = \frac{1}{5}\left(3-\frac{5^3}{3+\frac{7^3}{4}}\right)$$

Similarly, for $\frac{22}{7}$

$$\frac{7}{22} =\frac{1}{2}\left(\frac{3^3}{1+\frac{4^3}{2}}\right) -\frac{1}{11}$$

For the first convergent $3$,
$$\frac{3^3-6}{1^3+6}=\frac{21}{7}=3$$

The tentative expression
$$a(n)=\frac{(4n+3)^3+6(3n-1)}{(4n+1)^3-6(3n-1)}$$

has $a(0)=3$ and $a(1)=\frac{355}{113}$ but $\lim_{n \to \infty} a(n) = 1 \neq \pi$

A similar fraction for the approximation $\frac{223}{71}$ (Archimedes' lower bound) is
$$\frac{98+5^3}{98-3^3}=\frac{2·7^2+5^3}{2·7^2-3^3}=\frac{223}{71}=3+\frac{1}{7+\frac{1}{10}}$$

答:
For a coprime pair of integers $1 \le p,q$, consider the following diophantine equation:
$$\frac{x^3+z}{y^3-z}= \frac{p}{q}$$
with $x,y,z$ integer unknowns. This is equivalent to
$$z= \frac{y^3p-x^3q}{p+q}$$
Thus a solution exists if and only if there exist $x,y$ such that
$$y^3p-x^3q \equiv 0 \pmod{(p+q)}$$
Now, note that $-q \equiv p$, and that it has an inverse $\mod (p+q)$. This means that we need
$$x^3+y^3 \equiv 0 \pmod{(p+q)}$$

Now, if we pick $p=355$ and $q=113$, we get
$$p+q=468 = 7^3+5^3$$
In other words, the fraction $355/113$ has such a special property not because it is a convergent of $\pi$, but simply because the sum of numerator and denominator is a sum of two small cubes. We can do the same job with the fraction
$$\frac{203}{265} = \frac{5^3+78}{7^3-78}$$
nothing special with $\pi$.
As for $22/7$ we can find $$\frac{22}{7} = \frac{17^3+5^3}{12^3-5^3}$$
which is even more surprising. In my opinion, having three variables gives you a lot of freedom, so that yo can find such nice forms for any rational number.

EDIT: Just for fun, I looked up convergents of a completely different constant, namely $\gamma$: Euler-Mascheroni constant. One of its convergents is
$$\frac{71}{123}= \frac{3^3+470}{11^3-470}$$ Again, there is nothing special about this rational, simply you can find such a form for any number.

Title: 355/113 and small odd cubes


答:
You can prove the result you actually want without any extra tools like Kronecker's theorem.

Given a point $(x_0,y_0)$ on the unit circle, it will lie on the line $y = mx - 1$, where $m = \frac{y_0+1}{x_0}$. If you replace $m$ by a rational approximation to $m$, you get a line that passes through $(0,-1)$ and a rational point on the unit circle close to $(x_0,y_0)$. The rational point can be made arbitrarily close to $(x_0,y_0)$ by making the rational approximation to $m$ sufficiently good.

Title: prove that $\arcsin (3/5) /\pi $ is irrational

Body:
I was asked to prove that on the unit circle, the rational points are dense. My idea is to first show that the point (3/5,4/5) corresponds to an irrational angle (more precisely, its ratio with respect to $\pi$), then by rotation, its image will be dense on the circle. This is Kronecker's theorem?

So I have to show that the angle is irrational.


问:
Probability that a random binary matrix has no pair of same columns?
Assuming that there is a $M \times N$ matrix whose entries are ${0,1}$ binary.

For each row, there are $K$ $1$s and $N-K$ $0$s, and the positions are randomly picked.

In this case, I would like to know the probability that the matrix has no pair of two same columns. I think the probability that there is at least 1 pair of columns which are exactly the same can also give the answer to this.

答:
There are ${N \choose K}$ potential columns with $K$ ones and $N-K$ zeros.

There are therefore ${N \choose K}^M$ equally likely ways of choosing $M$ of them,
of which  $\frac{{N \choose K}!}{\left({N \choose K}-M\right)!}$ have all distinct columns, making the probability of all distinct columns $\frac{{N \choose K}}{{N \choose K}}\frac{{N \choose K}-1}{{N \choose K}}\frac{{N \choose K}-2}{{N \choose K}}\cdots\frac{{N \choose K}-(M-1)}{{N \choose K}}$     $$=\dfrac{{N \choose K}!}{{N \choose K}^M\left({N \choose K}-M\right)!}.$$

As a numerical example, if $N=14$ and $K=3$ and $M=23$ then this gives a probability of about $0.492$ (not far away from the birthday problem).


答:
Although a solution is provided (and was accepted) I wanted provide a solution using Maclaurin series of cos:

$$
cos(x)=1-x^2/2!-x^4/4!+x^6/6!-...
\\ cos(ax) = 1-a^2x^2/2!-a^4x^4/4!+a^6x^6/6!-...
\\ 1-cos(ax) = a^2x^2/2!+a^4x^4/4!-a^6x^6/6!+...
\\ \frac{(1-cos(ax)}{x^2} = a^2/2!+a^4x^2/4!-a^6x^4/6!+...
\\ \lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} = \lim_{x\to 0} \frac{\cos \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)}{x^2}
$$

Since the denominator in $\lim_{x\to 0}$ goes to zero, in order for the limit exist, the numerator also should go to zero for $\lim_{x\to 0}$:

$$
\lim_{x\to 0} {\cos \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)} = \lim_{x\to 0} {\cos \left(\pi \cdot a^2/2 \right)}
$$

for ${\cos \left(\pi \cdot a^2/2 \right)}$ to be zero $a^2$ should be $1+2k$

In this case we can apply the L'Hospital's Rule:

$$
\lim_{x\to 0} \frac{\cos \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)}{x^2} = \lim_{x\to 0} \frac{-\sin \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)(\pi (a^4(2x)/4!-4a^6x^3/6!+... ))}{2x}
= {-\sin \left(\pi \cdot a^2/2 \right)\pi a^4/4!} = \pm \pi a^4/24
$$

for which it is positive if k is even, and it is negative if k is odd.

Body:
I have to find for which values of $a \in \Bbb N, a \ne 0$ the following limit exists and it is finite:

$$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2}  $$

Applying L'Hôpital's rule:
$$\frac{1-\cos (ax)}{x^2}\sim \frac{\sin (ax) \cdot a}{2x}\sim \frac{a^2}{2}$$
Then $$ \frac{\cos (\pi \cdot \frac{1-\cos (ax)}{x^2})}{x^2}  \sim   \frac{\cos (\pi \cdot \frac{a^2}{2})}{x^2}.$$
$$\cos (\pi \cdot \frac{a^2}{2})=0 \implies a^2=1+2k, \quad k \in \Bbb N$$
and in this case the limit is $0$.

In the book the suggested solution is $a^2=1+2k$ for which the limit is $$ \frac{(-1)^k \cdot (2k+1)^2\cdot\pi}{24}$$

but I don't understand this solution.

Trying to solve the limit with $a^2=1+2k$:

$$ \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} \sim \frac{-\sin (\pi \cdot \frac{1-\cos ax}{x^2}) \cdot \pi \cdot  \frac{\sin (ax)  a  x^2 - (1-\cos (ax))  2x}{x^4}}{2x} $$$$ =  \sin \left( \frac{\pi}{2}+k \pi\right) \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4}$$$$= (-1)^k \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4} $$


问:
Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

Let $k \ge 2$, $k \in \Bbb N$. Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

My intuition tells me that we can pick $\delta=1/2$. I tried to study $u_n = \sin(xk^n)=U_{k^n}(\sin(x))=U_k(u_{n-1})$. I thought of introducing a variant of Chebyshev polynomials $U_n(\sin(x))=\sin(nx)$. But it didn't help much. Does someone have a hint? Thanks. (here $\Bbb N = \{0,1,2,\ldots\}$.)

I needed this result for solving this exercice (which is a oral exam that was posed during the competitive exams of Polytechnique, #1 engineering school in France).

Show that for $f(z)=z^k$ with $k \ge 2$
$$
\exists \delta >0, \forall x,y \in \Bbb U, x \neq y \implies \exists n \in \Bbb N, |f^{(n)}(x)-f^{(n)}(y)| \ge \delta.
$$

Suppose $x \neq 0$ and $y/x=e^{i \theta}$,
\begin{align}
| f^{(n)}(x) - f^{(n)}(y) | &= |x^{k^n} - y^{k^n}|
=
 | 1- e^{i \theta k^n}|\\
&= | e^{-i \theta k^n/2} - e^{i \theta k^n/2}|
= |2i \sin(\theta k^n/2)| = 2|\sin (\theta k^n/2)|
\end{align}
So
$$
|f^{(n)}(x)-f^{(n)}(y)| \ge \delta \iff |\sin (\theta k^n/2)| \ge \delta/2
$$

答:
First solution of the problem

We prove that for each $x \in(0,\pi)$, there exists $n,y\in \mathbb{N}$ such that
$$\pi y + \frac{\pi}{k^2} \le x k^n\le \pi y +\pi - \frac{\pi}{k^2} \iff y +\frac{1}{k^2}\le  k^n\frac{x}{\pi} \le y+1-\frac{1}{k^2} \tag{1}$$
because $$(1) \iff |\sin(xk^n)| \ge \sin \left(\frac{\pi}{k^2} \right)$$

We write $\frac{x}{\pi}$ in base-k numeral system (aka: radix system)
as follows
$$\frac{x}{\pi} =\overline{0,s_1s_2s_3...s_n...}^{(\mathbf{k})} = \sum_{i=1}^{+\infty}\frac{s_i}{k^i}=\frac{s_1}{k}+\frac{s_2}{k^2}+\frac{s_3}{k^3}+...+\frac{s_n}{k^n}+...$$
and then
$$k^n\frac{x}{\pi} =\overline{s_1s_2s_3...s_n,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$$
The integer part $\left[k^n\frac{x}{\pi}\right]$ and the fractional part $\{k^n\frac{x}{\pi}\}$ of $k^n\frac{x}{\pi}$ are defined as
$$a_n =\left[k^n\frac{x}{\pi}\right] = \overline{s_1s_2s_3...s_n}^{(\mathbf{k})}$$
$$\epsilon_n=\left\{k^n\frac{x}{\pi}\right\} = \overline{0,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$$

We notice that $0 \le \epsilon_n < 1$ and $k^n\frac{x}{\pi} = a_n + \epsilon_n $.

Case 1: $\frac{x}{\pi}$ is a terminating decimal in the base-k. Suppose $\frac{x}{\pi}$ has $N$ numbers after the $0$. Then, $\frac{x}{\pi}$ can be written as

$$\frac{x}{\pi} =\overline{0,s_1s_2s_3...s_N}^{(\mathbf{k})} = \frac{s_1}{k}+\frac{s_2}{k^2}+\frac{s_3}{k^3}+...+\frac{s_N}{k^N}$$
with $ 1 \le s_N \le (k-1)$.

Choose $y = a_{N-1}$ then $\epsilon_{N-1} = \overline{0,s_N}^{(\mathbf{k})} = \frac{s_N}{k}$. We can verify easily that $(y,n) = (a_{N-1},N-1)$ satisfy (1). Indeed, we have

$$a_{N-1}+\frac{1}{k^2} \le a_{N-1}+\frac{1}{k} \le  a_{N-1}+ \frac{s_{N-1}}{k} = a_{N-1}+ \epsilon_{N-1} = k^{N-1} \frac{x}{\pi}$$
and
$$k^{N-1} \frac{x}{\pi} = a_{N-1}+ \epsilon_{N-1} = a_{N-1}+ \frac{s_{N-1}}{k} \le a_{N-1}+ \frac{k-1}{k} \le a_{N-1}+ 1- \frac{1}{k^2}$$

Case 2-a: $\frac{x}{\pi}$ is a non-terminating decimal (aka repeating decimal) in the base-k and there exists an $n$ such that $1 \le s_{n+1} \le (k-2)$.

Then, $\epsilon_n = \overline{0,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$ satisfies $\frac{1}{k} \le \epsilon_n \le \frac{k-1}{k}$. Choose $y=a_n$, we have

$$a_n +\frac{1}{k} \le  k^n\frac{x}{k\pi} \le a_n +\frac{k-1}{k} $$
$$\iff a_n +\frac{1}{k} \le  a_n+ \epsilon_n \le a_n +\frac{k-1}{k} \tag{2}$$

From (2), we can deduce that (1) holds true. Indeed, we have

$$a_n+\frac{1}{k^2 } \le a_n+\frac{1}{k} $$
and
$$a_N+ \frac{k-1}{k} \le a_N+ 1- \frac{1}{k^2}$$

Case 2-b:  $\frac{x}{\pi}$ is a non-terminating decimal (aka repeating decimal) in the base-k and there doesn't exist an $n$ such that $1 \le s_{n+1} \le (k-2)$.

So for all $i=1,...,+\infty$ we have $s_i = 0$ or $s_i=(k-1)$

Let $u$ is the first value such that $s_u = 0$ (if $u$ doesn't existe, then $s_u = (k-1)$ for all $u$. As we suppose $\frac{x}{\pi}$ is a non-terminating decimal, we have $\frac{x}{\pi} =\sum_{i=1}^{+\infty}\frac{(k-1)}{k^i} \rightarrow 1$ which is a contradiction.).

Let $v$ is the first value such that $v>u$ and $s_v = (k-1)$. (If $v$ doesn't existe, then $s_v = 0$ for all $v>u$. Then  $\frac{x}{\pi}$ is not a non-terminating decimal $\implies$ contradiction).
Hence, $\frac{x}{\pi}$ can be written as

$$\frac{x}{\pi} = \overline{0,s_{1}s_{2}...s_{u-1}00...00(k-1)s_{v+1}....}^{(\mathbf{k})}$$
In particular,
$$k^{v-1}\frac{x}{\pi} = \overline{..s_{v-2},0(k-1)s_{v+1}..}^{(\mathbf{k})}$$

Let's $y = a_{v-1}$ and $n = v-1$, we have
$$k^n\frac{x}{\pi} = a_{v-1} +  \overline{0,0(k-1)s_{v+1}..}^{(\mathbf{k})}$$
and
$$\overline{0,0(k-1)s_{v+1}..}^{(\mathbf{k})} = \frac{k-1}{k^2} + ... \in \left(
 \frac{1}{k^2}, 1-\frac{1}{k^2} \right)$$
$$\implies a_n + \frac{1}{k^2} \le k^n\frac{x}{\pi} \le a_n + 1-\frac{1}{k^2}$$
So, (1) holds true.

Conclusion
Hence, we have for all $x\in (0,\pi), \exists n\in \mathbb{N}$ such that $|\sin(xk^n)|\ge \sin(\frac{\pi}{k^2})$.

Note: We can prove $ \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\cos(xk^n)|\ge \cos \left( \frac{\pi}{k} \right)$ by using the same method. In fact, for the $|\cos(xk^n)|$, the proof is less difficult, it suffices to prove there exists $y,n\in\mathbb{N}$ such that $y \le  k^n\frac{x}{\pi} \le y+1-\frac{1}{k}$.

Hence, we have that $\forall x \in ]0,\pi[, \exists n,m \in \Bbb N$ such that $$|\sin(xk^n)|\ge \sin\left( \frac{\pi}{k^2} \right)$$
$$|\sin(xk^m)|\le \sin\left( \frac{\pi}{k} \right)$$


问:
Using Euler's Totient Function, how do I find all values n such that, $\varphi()=14$
I just recently started working with Euler's Totient Function, and I came across the problem of solving for all possible integers $n$ such that $\varphi(n)=14$. I know there are similar questions with different numbers, I just wanted to verify if my approach was correct or I messed up since I could not find any $n$ satisfying that property.

We know $$\varphi(n)= \prod_{i=0}^{k} p_{i}^{a-1}(p_i-1)$$ From there we see $p_i \leq 15$ hence the primes are $p_i=\{2, 3, 5, 7, 11, 13 \}$. Ergo, we can write $n= 2^a\cdot3^b\cdot5^c\cdot7^d\cdot11^e\cdot13^f$. Using the fact that the Function is multiplicative, we can separate $$\varphi(2^a\cdot\ldots\cdot13^f)= \varphi(2^a)\cdot\ldots\cdot \varphi(13^f) = 2^{a-1}(2-1)\cdot3^{b-1}(3-1)\cdot\ldots\cdot13^{f-1}(13-1) $$

So I ended up with $\varphi(n)= 2^{a+6}\cdot3^{b+1}\cdot5^{c}\cdot7^{d-1}\cdot11^{e-1}\cdot13^{f-1}$. Then, since 14 has no factors $3, 5, 11, 13$ their exponents are at most $1$. Furthermore since $2$ and $7$ divide both divide $14$, I tried finding any combination with different multiplicities, but none yielded $14$ as a Totative.

Was my approach correct or did I miss something?

答:
Regarding your method, you limit the primes of interest nicely, but you also need to eliminate from consideration any of those primes which do not pass the basic divisibility test of $\phi(p_i) \mid 14$. This would leave you with only $2$ and $3$ which clearly cannot produce the factor of $7$ needed to make $\phi(n)=14$.

Thinking about this problem separately from your approach, I would start with the key aspects:

a) $14+1 = 15$ is not a prime, and

b) $14=2\cdot 7$

So from (a) we do not arrive at $\phi(n)=14$ directly, and from (b) we need a factor of $7$. However that can only be supplied by a higher power of $7$ - which cannot be present as this would also include a factor of $6$ - or a larger prime (e.g. $29$), which would also include a larger factor not present.

So there is no number $n$ with $\phi(n)=14$.

The OEIS A005277 sequence gives even non-totients; $14$ is the smallest.


问:
on the application of FLT
i saw a problem which is a generalization of Beals conjecture,here
sinha conjecture,it holds a open prize of US$150,000, for the proof or disproof of a mathematical proposition . The award money was tripled on January 11, 2010.

and coming to my doubt,this is a starting step,
i saw that the there is some deep relation between the FLT,and this thing,i think we can proceed proving ,as the FLT is for the same power ,i mean its of the form
$x^a+y^a=z^a$ for some $a$,all the exponents being same, and this conjecture is of the form $x^a+y^b=z^c$,i think these two form the two sides of the argument,i mean that all the things that hold good for fermats equation may not hold good for $x^a+y^b=z^c$,so by contradiction or by inverse argument can we prove further,

what more is the relationship between FLT and sinha conjecture,and is there an intervention of taniyama shimura conjecture,relating to sinha conjecture,

thank you

答: At the given link, it's claimed Sinha has a proof of "Beal's conjecture," but won't ever publish it. Right. Anyway, I'd recommend having a look at https://mathoverflow.net/questions/28764/status-of-beal-tijdeman-zagier-conjecture to see what's been done about this circle of conjectures. </p>

Title: on the application of FLT


问:
Understanding example of a function
I am having a lot of trouble understanding the solutions given to this problem. Could someone interpret it for me.

Give an example of a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the taylor series of $f$ centered at $0$ converges for all $x$ and the taylor series converges to $f(x)$ for $x<1 $but not for $x>1$. Here is the answer word from word if someone could explain this.

In class $g(x)=e^{\frac{-1}{x^2}}, x \neq 0$, $g(x)=0$ if $x=0$. I know that this function is equal to its taylor series at $0$ only(my comment).

Here is what is confusing me

Answer: $g(x)=f(x-1)$ if $x \geq 1$ or $g(x)=0$ if $x<1$. So the taylor series is $\sum 0  x^k=0$ but $g(x) \neq 0$ for $x>1$. Could someone explain this solution. What is $f$?

答: I think there are several typos. The correct example is $g(x)=f(x-1)$ wher $f(x)=e^{-1/x^{2}}$ for $x \geq 0$ and $0$ for $x <0$. The Taylor series for $f$ converges  to $f(x)$ at $x$ iff $x \leq 0$. Hence, the Taylor series for $g$ converges  to $g(x)$ at $x$ iff $x \leq 1$.

Title: Understanding example of a function


问:
Conditioning of an event related to Brownian Motion
I'm currently reading the book Brownian Motion by Mörters and Peres and on p. 39 the following situation comes up:

We have a Brownian Motion $B(t)$ and two intervals $[a_1,b_1]$ and $[a_2,b_2]$ where $b_1<a_2$. We define $m_1 = \max_{a_1\leq t \leq b_1}B(t)$ and $m_2 = \max_{a_2\leq t \leq b_2}B(t)$. It then follows that $B(a_2)-B(b_1)$, $m_1-B(b_1)$ and $m_2-B(a_2)$ are independent.

We wan't to show that $\mathbb{P}(m_1 = m_2) = 0$ and therefore they consider the event
$$\{\omega:B(a_2)-B(b_1)= m_1-B(b_1)-(m_2-B(a_2))\}$$

They say that they condition on the values of the random variables $m_1-B(b_1)$ and $m_2-B(a_2)$ and that the output from the LHS is a continuous random variable while the RHS outputs a constant.

I don't understand what they mean by conditioning in this case. What I've seen before is conditional expectation and I believe that
$$\mathbb{E}[B(a_2)-B(b_1)\vert m_1-B(b_1)] = \mathbb{E}[B(a_2)-B(b_1)]$$
$$\mathbb{E}[m_1-B(b_1)-(m_2-B(a_2))\vert m_1-B(b_1)] =m_1-B(b_1)$$
holds by the independence however I don't see how this implies the above where they are dealing with a specific event.

答:
Define $X=B(a_2)-B(b_1)$, $Y=m_1-B(b_1)$,  $Z=m_2-B(a_2)$, and  $m=m_2-m_1=X-Y+Z$. As noted, $X,Y,Z$ are mutually independent. Let $\mu$ be the distribution of $Y$ and $\nu$ the distribution of $Z$. Then, using the independence, because $m=X-Y+Z$,
$$
P[m=0]=\int\int P[X-y+z=0]\mu(dy)\nu(dz).\qquad\qquad (1)
$$
But the random variable $X$ has a continuous distribution, so
$$
P[X=y-z]=0,
$$
for each pair $(y,z)$. It follows that the integral on the right in (1) vanishes, hence so does the left; that is, $P[m=0]=0$.


问:
Expected value of $z_t$ given all of its history including itself
So I have the following time series process AR(2):

$z_t = \delta + \psi_2 z_{t-2} + \epsilon_t$

where $\epsilon_t$ is a white noise with distribution $(0, \sigma^2)$

I am looking for the $E[z_t|z_1, ..., z_{t-1}, z_t]$.

Hence, I am looking for the conditional expectation of $z_t$ given its past history and itself.

So does this just turn into $E[z_t]$; that is, an unconditional expectation and this would just be $0$ assuming this is a mean reverting process with that mean being $0$?

Or does this just equal $z_t$ itself; that is $\delta + \psi_2 z_{t-2} + \epsilon_t$

I think it is the former but not entirely sure so any help is greatly appreciated!

答:
Whenever the value of $z_t$ is given, then that is what it is expected to be.

$$\mathsf E(z_t\mid z_t,\textit{yadda yadda yadda}) = z_t$$

(Assuming $z_t$ is measurable over the sigma algebra generated by $z_t$ and the rest.)

Note, this is not $\mathsf E(z_t)$.   The conditional expectation is the random variable itself, and that is not the expectation of the random variable (which is a constant).


问:
What is lamda? in simple terms please.
What is lamda in simple terms please i have not taken calculus, yet i need to understand it for a project i am working on.
the project i want to replicate is https://ibmathsresources.com/2019/06/19/simulating-a-football-season/
the specific part i dont understand is how to calculate lamda, i dont understand what lamda is.</p>

答:
In the project you linked "lamda" (which is a misspelling of lambda, i.e. the Greek letter $\lambda$) is  the parameter in a Poisson distribution, in
this case the distribution of the number of goals scored in a soccer game.
They get various $\lambda$'s by making some very dubious assumptions and using some
averages, since the sample mean is an estimator of the mean of the distribution, and
the mean of the Poisson distribution with parameter $\lambda$ is $\lambda$.

Title: What is lamda? in simple terms please.


问:
Population Growth Question
I am not sure how I am supposed to do this, but here is the problem and my attempt.

A farmer wants to produce chickens for the community. Let $P(t)$ represent the population of chickens at time $t$ in months. Assume the population of chickens reproduce at a rate proportional to its size. The farmer cultivates a small number of chickens for a trial run. In three months, the population of chickens increased from 75 to 300. The farmer wants to begin with an initial population of $P(0)$ chickens, and after a year, have $8000 + P(0)$ chickens, so that 8000 can be harvested, and $P(0)$ saved to start breeding again for the following year. What initial population $P(0)$ of chickens does the farmer need to start with?

Attempt:

First, since $P(t)$ represents the population at time $t$ in months, then the population growth equation can be written as: $$P(t) = P_0e^{kt}$$

Then I tried to find the relative growth rate. Since in three months, the population went from 75 to 300, then that means $$300 = 75e^{3k}$$

Or in other words, $k = \frac{\ln{4}}{3}$.

I am not exactly sure how to proceed from here, or if I am on the right track. I would like some assistance, or advice.

答: While taking the logarithm is a good general strategy, in this case it is a needless complication. You have the population quadrupling in three months. So in a year, the population will quadruple four times ($12$ months divided by $3$ months is $4$). In other words, $P_{final}=4^4P_{initial}$. You also need $P_{final}=8000+P_{initial}$. Since both $4^4P_{initial}$ and $8000+P_{initial}$ are equal to $P_{final}$, they are equal to each other: $4^4P_{initial}=8000+P_{initial}$.

Title: Population Growth Question


问:
Find the number of positive integral solutions of the equation $x_1+x_2+x_3+x_4+x_5=x_1\cdot x_2\cdot x_3\cdot x_4\cdot x_5$
Find the number of positive integral solutions of the equation $x_1+x_2+x_3+x_4+x_5=x_1\cdot x_2\cdot x_3\cdot x_4\cdot x_5$

This is one of the questions from the Indian Institution of Technology Joint Entrance Examination question. I have been struck on this problem for hours.

I was solving some intriguing maths problems on combinatorics. Then I came across this wonderful questions and I was surprised looking at it. Firstly what I thought to give it a try was to Give it a BIJECTION of Distinct Balls distribution in Identical Boxes. For example $x_1+x_2+x_3=20$ can have $(20+3-1)C(3-1)$ solutions (including zero) but how does it work here?

答:
The more general question: given size $n$, which sets of positive integers share sum and product, i.e.

$$
\prod\limits_{i=1}^n x_i=\sum\limits_{i=1}^n x_i
$$

In addition to the trivial case ($n=1$ yields the set of all integers) we also ignore non-positive $0$ solutions. The $n=2$ case may be written $x_1=\frac{x_2}{x_2-1}$ which yields the only integer with shared sum, product and exponent, namely, $2+2=2\cdot2=2^2=4$. Let's find minimum and maximum values for a brute-force search.

Since $1^{n-1} \cdot x \, =x < n - 1 + x$, start with set $\{1,1, \ldots 2,k\}$. What is $k$?

$$
\begin{align}
2 \cdot k \cdot \prod\limits_{i=1}^{n-2} 1 &= 2 + k + \sum\limits_{i=1}^{n-2} 1\\
k &= n
\end{align}
$$

Any positive integer satisfies $k=n$, so every set has a solution $\{1,1,\ldots,2,n\}$. Our first test in a brute-force analysis, then, is $\{1,1,...1,3,i\}$ starting with $i=3$. When the product becomes greater than the sum, increment the set, meaning $\{1,1,...1,j,i\}$ becomes $\{1,1,...,2,2,2\}$. If the product is greater than the sum after incrementing the set, all of the sets for a given size $n$ have been found.

Using the same logic as for the "universal" set, it's possible to construct generators for other sets. Let's try $\{1,1,\ldots,3,k\}$:

$$
\begin{align}
3 \cdot k \cdot \prod\limits_{i=1}^{n-2} 1 &= 3 + k + \sum\limits_{i=1}^{n-2} 1\\
k &= \frac{n+1}{2}
\end{align}
$$

Which means that every set with odd $n$ has a solution $\{1,1, \ldots 3,\frac{n+1}{2}\}$. More generally, if there are two non-$1$ values in the set, they relate by $k=\frac{n-2+r}{r-1}$. A clever algorithm could exploit this relationship by incrementing $r$ to find integer values of $k$, ending when $r>k$.

The process of creating generators can continue ad nauseam with the next example being $\{1,1,...,k,r, \frac{r+k+n-3}{rk-1} \}$, but these generators have diminishing value unless $n$ is quite large because we only have 1 equation with an ever increasing number of unknowns.

If we treat the sets as ordered, the drudgery of permuting each set must also be included, an exercise left to the bored.

A033178


问:
prove that $\arcsin (3/5) /\pi $ is irrational
I was asked to prove that on the unit circle, the rational points are dense. My idea is to first show that the point (3/5,4/5) corresponds to an irrational angle (more precisely, its ratio with respect to $\pi$), then by rotation, its image will be dense on the circle. This is Kronecker's theorem?

So I have to show that the angle is irrational.

答: By Niven's theorem the only rational values taken by $\cos$ or $\sin$ on rational multiples of $\pi$ are $\pm 1$, $\pm 1/2$ and $0$.  In particular, $3/5$ and $4/5$ are not the sine or cosine of a rational multiple of $\pi$.


问:
How many integer-sided right triangles are there whose sides are combinations?

How many integer-sided right triangles exist whose sides are combinations of the form $\displaystyle \binom{x}{2},\displaystyle \binom{y}{2},\displaystyle \binom{z}{2}$?

Attempt:

This seems like a hard question, since I can't even think of one example to this. Mathematically we have,

$$\left(\dfrac{x(x-1)}{2} \right)^2+\left (\dfrac{y(y-1)}{2} \right)^2 = \left(\dfrac{z(z-1)}{2} \right)^2\tag1$$

where we have to find all positive integer solutions $(x,y,z)$.

I find this hard to do. But here was my idea. Since we have $x^2(x-1)^2+y^2(y-1)^2 = z^2(z-1)^2$, we can try doing $x = y+1$. If we can prove there are infinitely many solutions to,

$$(y+1)^2y^2+y^2(y-1)^2 = z^2(z-1)^2\tag2$$

then we are done.

答:
For your equation, solutions include
$$(1,1,1)\quad (1,2,2)\quad (1,3,3)\quad (1,3,3)\quad
 (1,4,4)\quad (1,5,5)\quad \cdots $$
Any triple with zero or one for one of $\space x,y\space $ will work. All others would probably require a brute force search.


问:
Forest graphs $F_1$ and $F_2$ with same vertices but different edges. Prove there is an edge where $F_1$ with it is still a forest
I am having quite a hard-time with this question, been thinking about it for a few hours and have not got a clue on how to even start proving this, because it is trivial but proving it has been hard for me.

Given two forests $F_1 = (V,A)$ and $F_2 = (V,B)$ with same vertices group $V$. It is also given that $|B| > |A|$. Prove: there exists an edge $e \in B \backslash A$ where $F_1 \cup \{e\}$ is still a forest.

Any help will be appreciated!

答:
Add one edges from $B\backslash A$ to $A$ and look if there is a cycle in $A$. If there isn't the problem is solved, otherwise there is exactly one cycle in $A$. In that cycle exists en edge $e$ which is not from $B$, otherwise a cycle will be also in $B$ which is impossible, since $B$ is forest. Remove $e$ from $A$ and continue from beginning. So we can replace all edges in $A$ by edges in $B$.

If there will not be an edge in $B\backslash A$ that will contradict that $|B| > |A|$, since the size of $A$ is not changed.

That means we cant continue infinitely, so there will be an edge in $B\backslash A$ when adding it to $A$ will not form a cycle.


问:
How to force algebraic growth due to resonance - stronger than linear
I am interested in which conditions need to be fullfilled in order for a solution of an ODE system to grow algebraicly due to resonance, in best case stronger than just linearly.

Let's assume following two ODEs:

$\frac{\partial^2 u}{\partial x^2}+m^2 u=sin(n x)$

and

$\frac{\partial^2 u}{\partial x^2}+m^2 u=sin(m x)$

The solution of the first ODE reads as:

$u=C_1 sin(mx)+C2 cos(mx)+\frac{sin(nx)}{m^2-n^2}$, while the solution of the second ODE reads as:

$u=C_1 sin(mx)+C2 cos(mx)+\frac{sin(mx)-cos(mx) mx }{2 m^2}$

The second solution if obviously of algebraic growth due to the frequency of the source term (the sin-function) being the resonance frequency of the ODE.

Which conditions need to be fulfilled in order to have an algebraic grwoth which is stronger than linear, hence $x^n$ with $n>1$?

I know that for example the ODE $\frac{\partial^2 u}{\partial x^2}-2 \frac{\partial u}{\partial x}+u=e^x$, leads to $u=C_1 e^x+C_2 x e^x+x^2 \frac{e^x}{2}$, is of algebraic growth (x^2) due to the frequency 1 being a double eigenvalue of the homogeneous ODE $\frac{\partial^2 u}{\partial x^2}-2 \frac{\partial u}{\partial x}+u=0$. However for a trigonometric source terme or a source term of $e^{i x}$, I do not know which conditions an ODE or the source term neead to fulfill in order to have such a solution (x^2).

I would appreciate any help.

答:
In general, for a constant-coefficient linear equation $P(D) u = f(x)$ with
forcing term  $f(x) = c e^{i \omega x}$
where $P$ is a polynomial, in order for the solution to be a polynomial of degree $d$ times $e^{i\omega x}$ you need $i \omega$ to be a root of $P$ of multiplicity $d$.  Since for a real polynomial the complex conjugate of a root is again a root of the same multiplicity, this means if you want a quadratic you'll need at least a polynomial of degree $4$.  Thus $i$ is a root of multiplicity $2$ of $(z-i)^2 (z+i)^2 = z^4 + 2 z^2 + 1$, and the differential equation

$$P(D) u = u'''' + 2 u'' + u =  \cos(x) $$
has a solution
$$ u = \frac{2-x^2}{8} \cos(x) + \frac{3x}{8} \sin(x) $$


问:
Upper bound on the number of ways to completely fill boxes.
I have $x$ indistinguishable items and $n$ boxes with sizes $y_1,\dots,y_n$. I am interested in the number of ways of allocating those $x$ items into the boxes such that all boxes must be either fully filled (i.e., the number of items in each box equals its size) or empty. Also, I have the condition $x=o(n)$ (where $x\leq n/2$ is included in the condition), which indicate that boxes of smaller sizes might possibly lead to more number of ways.

For example, if I have 3 items and 100 boxes all with size 2, then I have 0 ways to achieve my goal. However, if one of the boxes have size 3, then I would have one way to achieve my goal. If I have 3 items and 100 boxes all with size 1, then I have ${100\choose 3}$ ways to achieve my goal.

I suspect that the upper bound for the number of ways of allocating those $x$ items into $n$ boxes such that all boxes must be fully filled is ${n\choose x}$, which is the case where all boxes have size 1. I'm having trouble arguing this rigorously. Is there a way to show this rigorously or is the claim wrong?

答:
A generating function argument is that the number of ways of doing this is the coefficient of $z^x$ in the expansion of $\left(1+z^{y_1}\right)\left(1+z^{y_2}\right)\left(1+z^{y_3}\right)\cdots \left(1+z^{y_n}\right)$

That is certainly less than or equal to the coefficient of $z^x$ in the expansion of $\left(1+z\right)^{\sum y_i}$, which is ${\sum y_i \choose x}$, but you have in effect asked if it is always less than or equal to the coefficient of $z^x$ in the expansion of $\left(1+z\right)^{n}$, which is ${n \choose  x}$.

The answer to that could clearly be no in some cases even when $x<n$.  For example if $n=10$ and $x=8$ and each $y_i=2$ then there are ${10 \choose 4}= 210$ possibilities in the box size $2$ case and ${10 \choose 8}= 45$ possibilities in the box size $1$ case.

So it may come down to precisely what you mean by $x=o(n)$.


问:
invariance property of maximum likelihood estimator when we have monotonically decreasing function
If the mapping $\theta$ to $g(\theta)$ is one-to-one, then $g(\widehat\theta)$ is the MLE of  $g(\theta)$.

I didn't understand here. If $g(\theta)$ is monotonically decreasing function, then we are minimizing $\theta$.

答:
The new parameter space is $\Gamma$, the image of $\Omega$ under the transformation g. We are interested in finding the MLE of $\psi=g(\theta)$. The likelihood function can be written

$$\begin{split}f(x|\theta)&=f(x|g^{-1}(g(\theta)))\\
&=f(x|g^{-1}(\psi))\end{split}$$

We know that the likelihood is maximized at the MLE of theta, $\hat \theta$. So we have

$$\begin{split}f(x|g^{-1}(\hat\psi))&=f(x|\hat\theta)\\
g^{-1}(\hat\psi)&=\hat\theta\\
\hat\psi&=g(\hat\theta)\end{split}$$

Therefore the MLE of $\psi=g(\theta)$ is $g(\hat\theta)$. Note that g being monotonically decreasing has nothing to do with the likelihood function; the fact that g is one-to-one allows us to take its inverse.


问:
Writing proofs in mathematics: prove that $xy+5$ is an even integer if $x$ and $y$ are odd integers
here is what I have so far:

by definition an integer is called odd if there exists an integer $k$ such that $y=2k+1$ so if $y$ and $x$ are any two odd integers the product of those two integers added to $5$ is an even integer. $(xy + 5)$ is an even integer.

I need help figuring out how to prove it correctly

答:
It seems all you've done is quote the "definition" of an odd number and just restate the conclusion, not taken steps to prove anything.

Here's a hint to get you started: let $x = 2m+1$ and $y =2n+1$. $m$ and $n$ are not necessarily equal, of course, but both $x$ and $y$ are clearly odd. Now try multiplying those expressions, add $5$ and see what you get.


答:
Good start! Let $x = 2x_1+1$ and $y = 2y_1 + 1$. Then we need to prove $(2x_1+1)(2y_1+1)+5$ is even.

Expanding, we get $$(2x_1+1)(2y_1+1)+5$$$$4x_1y_1+2x_1+2y_1+1+5$$$$4x_1y_1+2x_1+2y_1+6$$

For a number to be even it must be divisible by $2$. Luckily we can factor a two from this expression! We have $$2(2x_1y_1+x_1+y_1+3).$$

Since $x_1$ and $y_1$ are integers, the above expression must be even, and we are done.

General useful tips for proofs:

Focus on what stands out. Here, it was the odd integer values. Expressing these odd integer values differently was the key here.

Understand what to prove. Fortunately, this proof was pretty straightforward. However, later in your math career you will probably run into proofs with harder notation and concepts that may throw you off. Always make sure you understand what to prove before you actually try to prove it.

Experiment. If proving something in geometry, try drawing different diagrams. If proving number theory like this problem, express numbers in different ways. Experiment to get creative!

Title: Writing proofs in mathematics: prove that $xy+5$ is an even integer if $x$ and $y$ are odd integers

Body:
here is what I have so far:

by definition an integer is called odd if there exists an integer $k$ such that $y=2k+1$ so if $y$ and $x$ are any two odd integers the product of those two integers added to $5$ is an even integer. $(xy + 5)$ is an even integer.

I need help figuring out how to prove it correctly


问:
Given that $\cot(m)=0.75$ and $\cos(m)<0$, what is the value of $\sin(m)$?
One of my homework problems is "Given that $\cot(m)=0.75$ and $\cos(m)<0$, what is the value of $\sin(m)$?" I keep getting $\sin m=\frac{-4}{5}=-.8$ which isn't an option

My options are:

A- $\ -.5625$

B-$\ -1.25$

C-$\ -.25$

D-$\ -.8$

E- None of the above

Edit: Because cotangent is positive and cosine is negative I know the angle is in the 3rd quadrant, I then used cotangent to get two of the dimensions, $3$ as the adjacent, and $4$ as the opposite. This gave me $5$ as the hypotenuse. Since sine is opposite divided by hypotenuse, I got $\sin(m)=\sin(4/5)=-.8$

答: The cotangent is in the first or third quadrant but the cosine is less than zero so that limits it to the third quadrant where both the horizontal and vertical components are negative. It also means that $\cos\theta= -0.6$ because it's horizontal component is $-3$. Note, the hypotenuse is considered to be positive in all calculations. This means that the vertical component of the sine is $-4$ and that $\sin\theta=-0.8\space$ and you were correct in your calculation.


问:
Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas
I have this identity:

$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$

If I write this like as:

$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alpha-\cos 2\alpha)=-\sin 6\alpha \sin 2\alpha$$

I can use Werner and prostaferesis formulas and I find the identity $(1)$.

But if we suppose of not to use these formulas I have done a try writing:

$$\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha \tag 1$$

$$[\cos(2(2\alpha))]^2=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 2$$

$$\cos^4 2\alpha -2\sin^2 2\alpha\cos^22\alpha+\sin^4 2\alpha=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 3$$

Then I have abandoned because I should do long computations and I think that is not the right way.

Is there any trick without to use the prostapheresis or Werner's formulas?

答:
$$\sin^22\theta=4\sin^2\theta\ (1-\sin^2\theta)$$
$$=\sin\theta\ (\sin\theta+(3\sin\theta-4\sin^3\theta))$$
$$=\sin\theta\ (\sin\theta+\sin 3\theta)=\sin^2\theta+\sin 3\theta\sin \theta.$$
Therefore, $\cos^22\theta=\cos^2\theta-\sin 3\theta\sin\theta.$ Finally,
replace $\theta$ by $2\alpha.$


问:
Haar Measure on Semidirect Product of Unimodular Groups
Let $M, N$ be closed subgroups of a given locally compact topological group such that the product $P = MN$ is semidirect, with $M$ normalizing $N$.  Assume second countable, sigma compact, whatever so that Radon products coincide with product measures.

To make life easy, assume $M, N$ are unimodular.  How do we construct a Haar measure on $P$?

答: For future reference, Prop.s 28 & 29 on pp. 99-100 of Nachbin's The Haar Integral address exactly this problem (without the unimodularity assumption).