问:
PDF of $Y \sim U [0, X]$ conditional on $X$ having exponential distribution with $\lambda= 1$
I need to find the probability density function to calculate $P(Y>1)$.

$Y\sim U [0, X]$

and we have that:

$X \sim\text{Exp}(1)$

So I know that the PDF of a uniform distribution equals:

\begin{equation}
PDF =
\begin{cases}
\frac{1}{b-a}\text{ if }x\in [a,b] \\
0\text{ otherwise}
\end{cases}
\end{equation}

But I need to condition on $X$ being exponential, how should I continue?

答:
I have a different result.

As shown, Y-density is not necessary to calculate $P(Y>1)$

Being

$$f_{XY}(x,y)=\frac{e^{-x}}{x}\cdot\mathbb{1}_{(0;+\infty)}(x)\cdot\mathbb{1}_{(0;x]}(y) $$

We have

$$P(Y>1)=\int_1^{+\infty}\frac{e^{-x}}{x}\Bigg[\int_1^xdy\Bigg]dx=e^{-1}-\int_1^{+\infty}\frac{e^{-x}}{x}dx=e^{-1}-\Gamma(0;1)\approx0.3679-0.2194=0.1485$$


答:
You have the conditional probability density function of $Y$ given $X$, that is $f_{\small Y\mid X}(y\mid x)=\tfrac 1x\mathbf 1_{0\leqslant y\leqslant x}$ , since you were assured that $Y$ is uniformly distributed when conditioned over a given $X$.

You also have the marginal probability density function $X$, that is $f_{\small X}(x)=\mathrm e^{-x}\mathbf 1_{0\leqslant x}$ , since you were assured that $X$ is exponentially distributed with rate parameter of $1$.

This gives you the joint probability density function.$$f_{\small X,Y}(x,y)=\frac{\mathrm e^{-x}}{x}\mathbf 1_{0\leqslant y\leqslant x}$$

And the Law of Total Probability allows you to evaluate the marginal probability density function of $Y$ as:$$f_{\small Y}(y) = \mathbf 1_{0\leqslant y}\,\int_y^\infty \frac{\mathrm e^{-x}}{x}\,\mathrm d x$$

However, this does not have an elegant solution.   It is the definition for a generalised function, $\Gamma(0,y)$, known as the Incomplete Gamma Function .$$f_{\small Y}(y)=\Gamma(0,y)\,\mathbf 1_{0\leqslant y}$$

Then $$\begin{align}\mathsf P(Y>1)&=\int_1^\infty \Gamma(0,y)\,\mathrm d y\\&= -\operatorname{E}_i(-1) + \mathrm e^{-1}\\&\approx 0.148496\end{align}$$

Where $\operatorname{E}_i(y)$ is another generalised function, the Exponential Intergal function

ContentLicense: CC BY-SA 4.0

Title: PDF of $Y \sim U [0, X]$ conditional on $X$ having exponential distribution with $\lambda= 1$

Body:
I need to find the probability density function to calculate $P(Y>1)$.

$Y\sim U [0, X]$

and we have that:

$X \sim\text{Exp}(1)$

So I know that the PDF of a uniform distribution equals:

\begin{equation}
PDF =
\begin{cases}
\frac{1}{b-a}\text{ if }x\in [a,b] \\
0\text{ otherwise}
\end{cases}
\end{equation}

But I need to condition on $X$ being exponential, how should I continue?


问:
The hessian of max
I was trying to understand the Hessian matrix. I asked this question
what is the the hessian of the max and the min of two given functions. My question is the following :
what is the exact expression of this $\operatorname{Hess}(\max(u,v))$?

答: $\max$ does not have partial derivatives everywhere. Where a Hessian may be defined (specifically, on the set $\{(x,y)\in\Bbb R^2\,:\, x\ne y\}$), it's $\begin{pmatrix}0&0\\ 0&0\end{pmatrix}$.

Title: The hessian of max


问:
$\lim_{x\to \pi/2} \;\frac 1{\sec x+ \tan x}$
how to solve it answer is $0$, but $\frac 1{\infty + \infty}$ is indeterminate form

$$\lim_{x \to \pi/2} \frac 1{\sec x + \tan x}$$

答:
Clarification:

$$\lim_{x \to \left(\frac{\pi}{2}\right)^+} \frac{1}{\sec x + \tan x} \to \frac{1}{\infty} = 0$$

$$\lim_{x \to \left(\frac{\pi}{2}\right)^-} \frac{1}{\sec x + \tan x} \to -\frac{1}{\infty} = 0$$

$$\lim_{x \to \frac{\pi}{2}} \frac 1{\sec x + \tan x} \to \frac{1}{\infty} = 0$$

In other words, the limit, $\,\to \frac 1{\infty}\,$ is not of indeterminate form: the limit in both cases above is equal to zero.

I do believe that recognizing your function $$f(x) = \frac 1{\tan x + \sec x} = \frac{\cos x}{1 + \sin x}$$ makes the limit as $x \to \pi/2$ perhaps more evident.

ContentLicense: CC BY-SA 3.0


问:
Inequality of sum of error functions
I am stuck with proving the following. Let $A\geq0
, B>0$, and $\alpha \in (0,1)$. I would like to show that
$$
\text{erf}\left(\frac{\alpha A+x}{\sqrt{2} \alpha  B}\right)+\text{erf}\left(\frac{A-x}{\sqrt{2} B}\right)-\text{erf}\left(\frac{\alpha A-x}{\sqrt{2} \alpha  B}\right)-\text{erf}\left(\frac{A+x}{\sqrt{2} B}\right)\geq 0,\quad \mbox{for all } x\geq0,$$ where erf($\cdot$) denotes the error function.

In my numerical experiments I have not found a single counterexample for this inequality, but I have not been able yet to prove it (e.g., via known bounds for the error function). Anyone a clue?

答:
Let $C = \frac{A}{B\sqrt{2}}$, $y = \frac{x}{B \sqrt{2}}$ and $u = \frac{1}{\alpha} > 1$.
We need to prove that
$$\mathrm{erf}(C + uy) + \mathrm{erf}(C - y) - \mathrm{erf}(C - uy) - \mathrm{erf}(C + y)
\ge 0, \ \forall y \ge 0.$$

Let
$$f(u) = \mathrm{erf}(C + uy) + \mathrm{erf}(C - y) - \mathrm{erf}(C - uy) - \mathrm{erf}(C + y).$$
We have
$$f'(u) = \frac{2y}{\sqrt{\pi}} \mathrm{e}^{-(C+uy)^2} + \frac{2y}{\sqrt{\pi}} \mathrm{e}^{-(C - uy)^2} \ge 0.$$
Also, $f(1) = 0$. Thus, $f(u)\ge 0$ for all $u \ge 1$.

We are done.

Title: Inequality of sum of error functions


问:
Schröder–Bernstein Theorem proof help
In Understanding Analysis, there's is a guided Schröder–Bernstein Theorem proof as an exercise.

I'm trying to prove it on my own but I believe I'm confused on several points and would appreciate some aid and corrections where necessary. Here's what I have so far:

a) By partitioning the domains, if we prove that $f$ from $A$ to $B$ and $g$ from $B'$ to $A'$ are bijective, then we can write a function:
$$h(x)=\begin{cases} f(x)\text{ if}\ x\in A \\ g^{-1}(x)\text { if}\ x\in A'\\ \end{cases}$$
that is bijective and concludes the proof.

b) We know that $A_1=X\backslash g(Y)$ and all subsequent $A_n$s are $A_n\subseteq g(Y)$. Thus $A_1\ne A_2$ and inductively, as $f$ is 1-1 and so $f(Z)\ne f(Z')$ when $Z\ne Z'$ and the same goes for $g$, we have $g(f(Z))\ne g(f(Z'))$ when $Z\ne Z'$ and so $g(f(A_n))\ne g(f(A_{n-1}))$.
Thus the $A$s are disjoint pairwise and with the same logic the $B$s are also disjoint.

c) I don't get that, it seems true by definition to me...

d)For every $x\in B'$, $x$ can't be in any $f(A_n)$ for all $n$ and for all $g(x)$, as one generation of $A$ feeds the next and $g(x)$ not in $A_1$, then $g(x)$ not in any $A_n$.
(I'm not very happy about d. to be honest).

答:
As to $(a)$, note that $f$ is 1-1, so $f\restriction_A$ is too, so $f: A \to B$ is indeed bijective (as assume $f: A\to B$ is onto), and so is $g: B' \to A'$ and its inverse $g^{-1}: A' \to B'$ (which then exists!) for the same reason. As $A$ and $A'$ partition $X$, $h$ is well-defined and so $$h[X]= h[A \cup A']=h[A] \cup h[A']= f[A] \cup g^{-1}[A'] = B \cup B' = Y$$ and $h$ is onto as well. That $h$ is 1-1 follows from the facts that $f$ is 1-1 on $A$, $g^{-1}$ on $A'$ and that they are mapped into disjoint sets so 1-1 ness is preserved (if $x \neq x'$, there are three cases: both in $A$, and then $f(x)\neq f(x')$ by $f$ being 1-1 so $h(x)\neq h(x')$, both in $B'$ and then $h(x)=g^{-1}(x) \neq g^{-1}(x')=h(x')$ as $g^{-1}$ is 1-1, or (say) $x \in A$ and $x' \in A'$ and then $h(x) \in B$ and $h(x') \in B'$ so distinct too..) So $(a)$ needs actual arguments to show.

$(b)$: (parenthetical question) if $A_1 =\emptyset$, $g[Y]=X$ and $g$ is already a bijection between $Y$ and $X$ and we'd be done.

$(c)$ is true because unions and images under $f$ commute:

$$f[A]=f[\bigcup_{n=1}^\infty A_n] = \bigcup_{n=1}^\infty f[A_n] =: B$$

$(d)$ and $(b)$ need better arguments.

Title: Schröder–Bernstein Theorem proof help


问:
show convergence for x tending to zero for $(2 - a^{1/x})^x$
I have the following problem:

I need to show whether a limit exists for $F(x) = (2 - a^{1/x})^x$ for $x$ tending towards zero (and $x$ tending towards infinity) where $0<a<1$. If one of these limits exists I am supposed to calculate it.

Could anyone help me to solve this problem?

I already tried expressing $F(x)$ as a Summation according to the binomial theorem and calculating the limit of the series but without any luck :(

答:
Note that if the limit exists, then
$$
\lim_{x \to 0} F(x)
 = \exp\left(\ln \left( \lim_{x \to 0} F(x) \right)\right)
 = \exp\left(\lim_{x \to 0} \ln ( F(x) )\right).
$$


问:
Show there are elements $x,y\in K$, s.t. $x^2+y^2=-1$
I am trying to solve this problem :

Let $K$ be a field not containing $\sqrt{-1}$, and $L$ be a Galois extension of $K$ containing $\sqrt{-1}$ and having Galois group $\mathbb Z/4\mathbb Z$. Show that there exists $x,y \in K$ such that $x^2+y^2=-1$.

The way to approach the solution is as $\sqrt{-1}\in L$ we can consider $K(\sqrt{-1}):K$, as the elements over there are of the form $x+y\sqrt{-1}$, we can try to look for an element whose norm is $-1$ in $K(\sqrt{-1}):K$. But what could be that element?

答:
Let $F = K(i)$ be the quadratic subextension, and let $\sigma$ denote an extension of the nontrivial automorphism (complex conjugation) to $L/K$. Write $L = F(\sqrt{\alpha})$ for some $\alpha \in F$. Since $L/K$ is normal we must have
$\sqrt{\alpha}^\sigma = \beta \sqrt{\alpha}$ for some $\beta \in F$. Applying $\sigma$ to this equation we get $\sqrt{\alpha}^{\sigma^2} = \beta^{1+\sigma} \sqrt{\alpha}$.
Since $L/K$ is cyclic of order $4$, we must have $\sqrt{\alpha}^{\sigma^2} = - \sqrt{\alpha}$ because $\sigma^2$ must be a nontrivial automorphism, hence $\beta^{1+\sigma} = -1$. Write $\beta = x + yi$; then
$-1 = \beta^{1+\sigma} = x^2 + y^2$.


问:
Polynomial Orthogonality
I study Polynomial Orthogonality and in connection to Linear Algebra I visualize the concept this way :

$f(x)$ and $g(x)$ are polynomials

Any polynomial $p(x)$ in the interval $[a,b]$ is a vector with infinite coordinates as $[a,b]$ contains an infinite amount of numbers $[x_0,x_1,\dots,x_n]$ and it's coordinates are $(p(x_0),p(x_1),\dots,p(x_n))$

The inner product of $f(x)$ and $g(x)$ is equal to : $f(x_0)*g(x_0)+f(x_1)*g(x_1)+\dots+f(x_n)*g(x_n)$

The process above can also be expressed as : $\int_{x_0}^{x_n} f(x)*g(x) \,dx$

I got 3 questions though :

Is there a better/more accurate way to visualize the concept described above?

In most cases/examples I've seen writers limit themselves to the interval $[-1,1]$.Why is that? I tried some plotting and integration in Desmos an indeed in most cases only the interval $[-1,1]$ takes the inner product to $0$.

Besides orthogonality and in connection to Linear Algebra is there a way to show that two polynomials over an interval $[a,b]$ are as vectors linearly indpendent?

答:
It's not how it works.

First, the set of polynomials with coefficients in a field $\Bbb K$ (for instance $\Bbb K=\Bbb R$) is a vector space over the field $\Bbb K$. A natural basis of this vector space is $(1,X,X^2,\dots)$. It's also possible to consider only polynomials of degree $\le n$, which is a finite dimensional vector space, of dimension $n+1$.

Given an inner product of this vector space, it's then possible, using Gram-Schmidt orthogonalization, to build an orthogonal basis.

So, it remains to find an inner product. A common way (but not the only one), is to define

$$\left
=\int_a^bP(x)Q(x)w(x)dx$$

Where $a,b$ are fixed but arbitrary, and possibly infinite, and $w$ is a weight function. Of course, the integral must be convergent for any polynomial $P,Q$.

For instance, with $w=1$ and $[a,b]=[-1,1]$, you get Legendre polynomials. Examples on an unbounded interval include Laguerre polynomials and Hermite polynomials.

More on this on Wikipedia:

Orthogonal polynomials

Classical orthogonal polynomials

To answer your last question: two polynomials are linearly independent if their quotient is not a constant. More generally, it's possible to check the linear independance of a set of $n$ polynomials, using for instance a Gram determinant.

For this, you need an inner product: you may use one of those described above. It's also possible to use the fact that the vector space $\Bbb R_n[X]$polynomials of degree $\le n$ is isomorphic to $\Bbb R^{n+1}$ (using the values of these polynomials at $n+1$ points, thanks to the linearity of Lagrange interpolation), then use an inner product on $\Bbb R^{n+1}$.

Title: Polynomial Orthogonality

Body:
It's not how it works.

First, the set of polynomials with coefficients in a field $\Bbb K$ (for instance $\Bbb K=\Bbb R$) is a vector space over the field $\Bbb K$. A natural basis of this vector space is $(1,X,X^2,\dots)$. It's also possible to consider only polynomials of degree $\le n$, which is a finite dimensional vector space, of dimension $n+1$.

Given an inner product of this vector space, it's then possible, using Gram-Schmidt orthogonalization, to build an orthogonal basis.

So, it remains to find an inner product. A common way (but not the only one), is to define

$$\left<P,Q\right>=\int_a^bP(x)Q(x)w(x)dx$$

Where $a,b$ are fixed but arbitrary, and possibly infinite, and $w$ is a weight function. Of course, the integral must be convergent for any polynomial $P,Q$.

For instance, with $w=1$ and $[a,b]=[-1,1]$, you get Legendre polynomials. Examples on an unbounded interval include Laguerre polynomials and Hermite polynomials.

More on this on Wikipedia:

Orthogonal polynomials

Classical orthogonal polynomials

To answer your last question: two polynomials are linearly independent if their quotient is not a constant. More generally, it's possible to check the linear independance of a set of $n$ polynomials, using for instance a Gram determinant.

For this, you need an inner product: you may use one of those described above. It's also possible to use the fact that the vector space $\Bbb R_n[X]$polynomials of degree $\le n$ is isomorphic to $\Bbb R^{n+1}$ (using the values of these polynomials at $n+1$ points, thanks to the linearity of Lagrange interpolation), then use an inner product on $\Bbb R^{n+1}$.


问:
rank(A) $≤ 2 \iff A = U.V^T$
Let $A \in M_{n×n}(\mathbb R)$, with $n ≥ 2$ a positive integer. Show that the following statements are equivalent :

i) rank(A) $≤ 2$.

ii) $\exists U,V ∈ M_{n×2}(\mathbb R)$ such that $A = U.V^T$

I honestly struggled to show both ways of the equivalence but I have a few guesses (I had the idea for ii $\implies$ i while writing this):

i $\implies$ ii : If rank(A) $≤ 2$ then there are $0,1$ or $2$ independent columns in A. We want to write A as a product of two nx2 matrixes. (I tried to take 4x4 matrixes of rank 2 and tried to write them in the form we are asking. I took U to be the independent columns, and V formed by the image of the canonic basis vectors and by pure luck it worked but this is not a proof).

ii $\implies$ i : We know that $rank(A)=rank(U.V^T) ≤ min(rank(U), rank(V^T))$, U and $V^T$'s ranks are clearly $\leq 2$ so rank(A) $\leq 2$

Can someone help me to write a solid proof for i $\implies$ ii instead of a lucky one ?

答:
Hint: Via Gauss's elimination, there exists $P,Q\in M_{n\times n}(\mathbb R)$ such that

$$P\begin{bmatrix}I_r&O\\O&O\end{bmatrix}Q=A$$

where $r=\operatorname{rank}(A)$.

When $r=2$, $A=\left(P\begin{bmatrix}I_r\\O\end{bmatrix}\right)\cdot\left(\begin{bmatrix}I_r&O\end{bmatrix}Q\right)$, implying ii) as desired.

Title: rank(A) $≤ 2 \iff A = U.V^T$


问:
There exists M such that $M||\textbf{x}||_{\infty} \le ||\textbf{x}||$
Show there exists M such that $M||\textbf{x}||_{\infty} \le ||\textbf{x}||$.

M positive constant, $||\circ||_{\infty}$ uniform norm, and $||\circ||$ a norm on $\mathbb{R}^m$.

答:
$\newcommand{\norm}[1]{\Vert #1 \Vert}$
If $\Vert \cdot \Vert$ is an Euclidean norm $\Vert \cdot \Vert_2$,
$$\norm{x} _\infty = \max_{1\leq i \leq m} |x_i|
\leq
\sqrt{x_1^2 + \cdots + x_m^2}
=
\norm{x}_2
$$
Therefore, $M = 1$ will suffice in this case.


问:
Question regarding solving multivariable function continuity problems
$$f(x,y)=\begin{Bmatrix}
\frac{x+y}{x^2+y^2},(x,y)\neq (0,0)) \\0
 , (x,y)=(0,0))
\end{Bmatrix}$$

Let's say I have this problem. I use $y = mx$, solving this I get $\frac{1+m}{x(1+m^2)}$. Calculating the limit I get $\frac{1+m}{0}$. Could this prove that the limit doesn't exist?

Some follow-up questions:

When can I use th $y=mx$ method, if I can?

What is the rule of paths? I guess I can't just plug in $(1,2)$ or $(2,3)$, but can I plug in let's say $(\frac{1}{n}, \frac{1}{n})$? If yes, what should I look for when plugging in values? The 'n' has to go, but how?

If I need to prove the continuity of a function, what is the simplest way to prove it? I heard about the Squeeze Theorem, but does that work outside of sin, cos functions? At seminars we used $a^2 + b^2 >= 2ab$, but I don't find that at all clear.

EDIT:

$$f(x,y)=\begin{Bmatrix}
\frac{x^2y3}{x^2+y^2},(x,y)\neq (0,0)) \\0
 , (x,y)=(0,0))
\end{Bmatrix}$$

I have this function. Using $a^2 + b^2 >= 2ab$:

$$x^2 + y^2 >= |x^2| + |y^2| >= 2|x||y|$$

$|f(x,y)| = \frac{x^2 |y^3|}{x^2 + y^2} = \frac{x^2 |y^3|}{2|x||y|} = \frac{x y^2}{2}$ ---> this tending to 0 would prove continuity? if yes, does this method work all the time?

EDIT 2:

$$f(x,y)=\begin{Bmatrix}
\frac{3x^2y}{2x^2+5y^2},(x,y)\neq (0,0)) \\0
 , (x,y)=(0,0))
\end{Bmatrix}$$

$$|f(x,y)| = \frac{3x^2|y|}{2x^2+5y^2} <= \frac{3x^2|y|}{2\sqrt{10}|x||y|} = \frac{3}{2\sqrt{10}}*x$$

$$(a^2 + b^2 >= 2ab)  ==> 2x^2 + 5y^2 >= (|x|\sqrt{2})^2 + (|y|\sqrt{5})^2 >= 2\sqrt{10}*x$$

$f(x,y) <= \frac{3}{2\sqrt{10}}*x$(1)

$x --> 0, y --> 0$ (2)

(1)(2) --> f(x,y) --> 0 = f(0, 0).

答:
Given a function $f$ on $\mathbb{R}^2$, it does not make sense to say "the limit" of $f$: you should indicate the limit at what point.

I don't know how you get $\frac{1+m}{0}$. But restricting on $y=x$, you have $f(x,y)=\frac{1}{x}$, which tells you that the limit $\displaystyle \lim_{(x,y)\to(0,0)}f(x,y)$ does not exist.

In real analysis, there is usually no one-fit-all "method". What you often have is a certain set of "strategies". You usually try the path $y=mx$ for some fixed $m$ when studying functions of the form $f(x,y)=\frac{P}{Q}$ where $P$ and $Q$ are polynomials in $x$ and $y$.

Again, there is no "rule". You need to work on an ad hoc basis.

"If I need to prove the continuity of a function, what is the simplest way to prove it?"  There is no such thing. Sometimes, you would need to work from the first principle.

In your edited question, the estimate
$$
|\frac{x^2y^3}{x^2+y^2}|\le \frac{|x^2|y^3|}{2|x||y|}
$$
shows that $f$ is continuous at $(0,0)$.


问:
How could I interpret $\delta$ in $\mathbf{R} \in \mathbb{R}^{|S|}/\delta$?
I'm trying to implement a Paper and when looking at the pseudocode I encountered the following:

Pick a random reward vector $\mathbf{R} \in \mathbb{R}^{|S|}/\delta$

...

Repeat:

Pick a reward vector $\tilde{\mathbf{R}}$ uniformly at random from the neighbours of $\mathbf{R}$ in $\mathbb{R}^{|S|}/\delta$

...

Suppose e.g. $\delta = 0.05$ and $|S| = 5$.
What's the meaning of "$/\delta$" and how could I implement such sampling in code? I understood it has something to do with the Equivalence class but couldn't figure out how to apply this concept in practice.

The Paper I am referring to is Bayesian Inverse Reinforcement Learning, the pseudocode appears in Figure 3 specifically. Here we can find a (possibly incorrect) implementation:

```<code>def sample_random_rewards(n_states, step_size, r_max):
    """
    sample random rewards form gridpoint(R^{n_states}/step_size).
    :param n_states:
    :param step_size:
    :param r_max:
    :return: sampled rewards
    """
    rewards = np.random.uniform(low=-r_max, high=r_max, size=n_states)
    # move these random rewards toward a gridpoint
    # add r_max to make mod to be always positive
    # add step_size for easier clipping
    rewards = rewards + r_max + step_size
    for i, reward in enumerate(rewards):
        mod = reward % step_size
        rewards[i] = reward - mod
    # subtracts added values from rewards
    rewards = rewards - (r_max + step_size)
</code>```

Thanks

答:
As @johnny10 has mentioned in the comments, there is a definition at the beginning of section 5:

The sampling technique we use is an MCMC algorithm GridWalk (see [Vempala, 2005]) that generates a Markov chain on the intersection points of a grid of length $\delta$ in the region $\mathbb R^{|S|}$ (denoted $\mathbb R^{|S|}/\delta$).

Hence, the paper is referring to the set of all $|S|$-tuples of real numbers, where each entry is an integer multiple of $\delta$. For $\delta = 0.05$ and $|S|=3$ one example would be $(23.45, -8.10, 5.55)$, since each of the three numbers is a multiple of $0.05$. Numbers like $1.23$ would not be allowed.

The chosen notation is non-standard and a better notation would be $(\delta\mathbb Z)^{|S|}$.


问:
How is Nakagami function derived from Gamma function?
So for example if I have ∼Gamma(,)

$$\ f(y) = \frac{1}{^Γ()}y^{k-1}e^{-y/}$$

Now according to wiki and other sources, to find Nakagami distribution we just have to =√ and we get this Nakagami distributed variable like this.

$$\ f(x) = \frac{2}{\Gamma(k)\theta^k}x^{2k-1} e^{-x^2/\theta}$$

So it looks like if we bring $f(y)^{1/2}$ we should get $f(x)$. But is there some special way of how to bring gamma functions to a power. I just don't see how this transformation happens. Can anybody share the proof or derivation process please.

答:
It is very simple.

Let's start from

$$f_Y(y)=\frac{1}{\theta^k\Gamma(k)}y^{k-1}e^{-y/\theta}$$

Let's set

$$x=\sqrt{y}$$

That is

$$y=x^2$$

and its derivative is $y'=2x$

Immediately you get

$$f_X(x)=\frac{1}{\theta^k\Gamma(k)}x^{2k-2}e^{-x^2/\theta}2x=\frac{2}{\theta^k\Gamma(k)}x^{2k-1}e^{-x^2/\theta}$$

Fundamenta Tranformation Theorem

Remember that, if the transformation $y=g(x)$ is monotone, the density of Y can be derived by simply transforming the density of X with the following formula

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$


问:
Find expected value of given random Variable $X$, where $X$ equals payoff to the better.
In a state lottery a five-digit integer is selected at random. If a player
bets 1 dollar on a particular number, the payoff (if that number is selected)
is $500$ dollar minus the $1$ dollar paid for the ticket. Let X equal the payoff to the better.
Find the expected value of X.

I don't know where to begin to solve this problem? Any guidance or help will be appreciated.

答:
Hint

What is the probability a particular ticket wins?

If it does win, what is the net gain?

What is the probability a particular ticket loses?

If it does lose, what is the net gain? (it will be negative)

Combine these to find the overall expected net gain


问:
Showing that the limit of a function is $e^3$
I want to find the convergence radius of

$$ P(z) = \sum_{j=1}^{\infty}\left(\frac{j-3}{j}\right)^{j^2}z^j$$
with $z$ being a complex number, with the use of

$$e^x = \lim_{j\to\infty} (1+x/j)^j$$

Using the root test, the convergence radius is

$$R =  \lim_{j\to\infty} \frac{1}{\sqrt[j]{\left\lvert \left(\frac{j-3}{j}\right)^{j^2} \right\rvert}}$$
I tried to rewrite the expression into a form simliar to $e^x$:

$$R = \lim_{j\to\infty} \sqrt[j]{\left\lvert \left(1+\frac{3}{j-3}\right)^{j^2} \right\rvert} $$

Now it looks like the convergence radius to be $e^3$ but I am not sure how to show this in a non-intuitive way.

答: Note that$$\sqrt[j]{\left(\frac{j-3}j\right)^{j^2}}=\left(\frac{j-3}j\right)^j=\left(1-\frac3j\right)^j.$$Can you take it from here?


问:
Mary L Boas - 8.3.23

Problem states: Find the orthogonal trajectories of the family of curves
$
y=-e^{x^2}\, erf \,\,x\,+Ce^{x^2}.\,\\
$
Solve for $ x $ in terms of $ y $.

My attempt to solve this, by rearranging terms first:

$$
\begin{align}
y=-e^{x^2}\,\left(\frac{2}{\sqrt{\pi}}\int_0^x{e^{-t^2}dt}\right)+Ce^{x^2}\tag 1
\\y\,e^{-x^2}\,+\,\left(\frac{2}{\sqrt{\pi}}\int_0^x{e^{-t^2}dt}\right)=C\tag 2
\\y'-2xy=\frac{4x\,e^{-x^2}}{\sqrt{\pi}}\tag 3
\\I=-2\int x\, dx \, =\, -x^2\, \implies e^{-x^2}\tag 4
\\y\,e^{-x^2}=\int e^{-x^2}\,\frac{4x\,e^{-x^2}}{\sqrt{\pi}} + C\tag 5
\\y\,e^{-x^2}=-\frac{1}{\sqrt{\pi}}e^{-2x^2}+C\tag 6
\end{align}
$$

Should i then  differentiate $(6)$ again with respect of $x \,$, find negative reciprocal and finally get $x$ in terms of $y$?

答:
There is a mistake in equation (3), it should've been
$$
\begin{align}
e^{-x^2}\Big(y' - 2xy + \frac{2}{\sqrt{\pi}}\Big) = 0 \tag 1 \\
\text{as} \hspace{1em} \frac{\mathrm{d}}{\mathrm{d}x}\int_0^x f(t)\mathrm{d}t = f(x)
\end{align}
$$

Therefore the differential equation for the orthogonal trajectory is:
$$
\begin{align}
\boxed{y' = \frac{a}{b - xy}}, \hspace{2em} \text{where} \hspace{2em} a = \frac{1}{2}, \ b = \frac{1}{\sqrt{\pi}} \tag 2 \\
\text{which is obtained from (1), as from (1) we get:}\\
y' = 2xy - \frac{2}{\sqrt{\pi}}
\end{align}
$$

Solving equation (2) will provide the equation for the orthogonal curves.

Title: Mary L Boas - 8.3.23


问:
Closed linear span of a subset in LCS is equivalent to intersection of all closed hyperplanes containing the subset
I'm currently working through Conway's Functional Analysis, and have stumbled across a result that I can't seem to find the justification for myself:

Let $\mathcal{X}$ be a Real Locally Convex Space and $A$ a subet of $\mathcal{X}$ . The closed linear span of $A$ is the intersection of all closed hyperplanes containing $A$.

It seems to be a corollary to the result of the closed convex hull of such $A$ being the intersection of all of all the closed half-spaces containing $A$, but I can't seem to work it out myself. Any help would be appreciated.

答:
Solved myself - First note that all closed hyperplanes in $\mathcal{X}$ are kernels of elements of $\mathcal{X}^*$, and vice versa. The closed linear span of $A$ clearly lies in this intersection. To show opposite containment, we note that
since the closed linear span of $A$ is a closed convex subset of $\mathcal{X}$, it is strictly separate from any point $x_0\notin A$. ie, there exists $f\in\mathcal{X}^*$ and $\alpha\in\mathbb{R}$ such that $f(x_0)<\alpha$ and $f(a)>\alpha$ for all $a\in A$.
Then the fact that $x_0$ is not in the intersection easily follows.


问:
What does the Cauchy Criterion entail in a Hilbert space?
I'm taking an introductory course in Machine Learning and we are looking at all the requirements a Vector Space must fufill to be a Hilbert Space. I'm not a very math versed person, i had to recently learn the inner product, equivalence relations, and what it even meant to use functions as vectors. I think those concepts are fine for me now but the Cauchy Criterion is where I feel completly lost. I have watched some Youtube videos on the matter, they usued a number line from 0 to 1 and points in between from which the distance in between to consecuitive points had to be less then a value epsilon. While the Cauchy Criterion made sense in such context I fail to see how it applies to a vector space of functions,discrete functions or random valued functions.

I though that the vectors that I would encounter in a Hilbert space would be functions like sinusoids or some signal's that decay and are convergent. Are all vectors on a Hilbert Space then part of a Cauchy sequence? Can, for example, sin(x) not be part of a Hilbert space because it's not part of a Cauchy sequence? Can there be elements of a cauchy sequence and other elements that are not part of a cauchy sequence? What if there are no other elements of a cauchy sequence in my vector space and just normal linear algebra vectors, if I use a Hilbert Space inner product will it still fufill the requirements for it to be a Hilbert Space?

答:
A Hilbert space is a set with elements and some operations. In your intended application these elements could be functions. All we care about is that we can add and substract them, we have a $0$ and scalar multiplication (they form a vector space) and we have an inner product $\langle x,y\rangle$ for any $x,y$ in the set, fulfilling some axioms (inspired by those on $\Bbb R^n$, the "mother of all Hilbert spaces"). This inner product allows us to define a norm (a notion of length) by defining $\|x\|=\sqrt{\langle x,x\rangle}$ and a norm allows us to define a distance (metric) on the set by setting $d(x,y)=\|x-y\|$ for this norm from the inner product. And whenever we have a set with a distance we can define a Cauchy sequence. Also when the elements are functions and the inner product is defined by some series or integral, e.g., those are just "implemntation details", as long as the defining axioms are fulfilled (for vector space and inner product) we can talk about Cauchy sequences of those elements. Constant sequences are always Cauchy (regardless of the space), and so always is any convergent sequence (the completeness condition for a Hilbert space says that the reverse must also be true: any Cauchy sequence is convergent).

The elements themselves don't matter, the distance function on them is the crucial element.


问:
what limit is being verified and what is its value?
Given $ \epsilon > 0 $, find an interval $ \textbf{I} = (5,5 + \delta ), \delta > 0 $, such that if $x$ lies in $ \textbf{I} $ then $ \sqrt{x-5} < \epsilon $

can you help me?

I thought if $ \textbf{I} = (5,5 + \delta )$ and $x$ lies in $ \textbf{I} $  we can say that $x  \in (5,5 ; 5,5 + \delta ) $

after that if the limit exists for every number $ \epsilon >0 $ there is $ \delta >0$ such that

$ 0 < |x-a| < \delta $  then $ |f(x) - L | < \epsilon $

after that I thought  $ a = 5,5 $   because $ x \in ( 5,5 - \delta ; 5,5 + \delta $ )

so $ 0 < |x-5,5| < \delta $

and now I don't know what to do how to find the limit  $ f(x) - L = \sqrt{x-5} $

答:
What you want to show is actually that the function $f(x)=\sqrt{x-5}$ is right continuous at $x=5$. (Notice that $\sqrt{x-5}=|f(x)-f(5)|$. )

Let us assume that you have found such $\delta>0$. Then you have

$$
x-5<\epsilon^2
$$
for all $5<x<5+\delta$, i.e., $0<x-5<\delta$.

How can this be possible? Well, take $\delta=\epsilon^2$.

This is how you do an estimate in calculus.


答:
Let $\varepsilon>0$. Note that if $x\in(5,5+\delta)$ then $\sqrt{x-5}\in(0,\sqrt{\delta})$. So
$$
|\sqrt{x-5}|<\sqrt{\delta},\quad\forall x\in I
$$
Hence choosing $\delta=\epsilon^2$ will suffice and so $I=(5, 5+\varepsilon^2)$

Title: what limit is being verified and what is its value?

Body:
Given $ \epsilon > 0 $, find an interval $ \textbf{I} = (5,5 + \delta ), \delta > 0 $, such that if $x$ lies in $ \textbf{I} $ then $ \sqrt{x-5} < \epsilon $

can you help me?

I thought if $ \textbf{I} = (5,5 + \delta )$ and $x$ lies in $ \textbf{I} $  we can say that $x  \in (5,5 ; 5,5 + \delta ) $

after that if the limit exists for every number $ \epsilon >0 $ there is $ \delta >0$ such that

$ 0 < |x-a| < \delta $  then $ |f(x) - L | < \epsilon $

after that I thought  $ a = 5,5 $   because $ x \in ( 5,5 - \delta ; 5,5 + \delta $ )

so $ 0 < |x-5,5| < \delta $

and now I don't know what to do how to find the limit  $ f(x) - L = \sqrt{x-5} $


问:
Rationals : Proof contradicts problem.
I was trying to prove the following:

Prove that if x satisfies
$$
x^n+a_{n-1}x^{n-1}+...+a_0 = 0
\tag{1}
$$ for a set $(a_{n-1},...,a_0)$  then, x is irrational unless it is an integer.

I am sorry if this is stupid, but, what I did was, firstly, from $(1)$
$$
x^n+a_{n-1}x^{n-1}+...+a_0 = -a_{1} x
\tag{2}
$$
then
$$
x= \frac{-(x^n+a_{n-1}x^{n-1}+...+a_0)}{a_1}
\tag{3}
$$
so if $x \in \mathbb{Q}$, then $x = \frac{a}{b}$, where clearly, here we have a fraction representing $x$, so how can $x \notin \mathbb{Q}$? Am I just stupid?

Please feel free to change the title and tag if you want to, I don't know what to make of it. Thank you.

答: If $x\in\Bbb Q$, you can write it as $\frac ab$, with $a\in\Bbb Z$, $b\in\Bbb N$ and $a$ and $b$ coprime. And now\begin{align}x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0&\iff\left(\frac ab\right)^n+a_{n-1}\left(\frac ab\right)^{n-1}+\cdots+a_1\frac ab+a_0=0\\&\iff a^n+a_{n-1}a^{n-1}b+\cdots+a_1ab^{n-1}+a_0b^n=0\\&\iff a^n=b\bigl(-a_{n-1}a^{n-1}-\cdots-a_1ab^{n-2}-a_0b^{n-1}\bigr).\end{align}So, $b\mid a^n$. But $a$ and $b$ are coprime, and therefore $b=1$. So, $x=a\in\Bbb Z$.

Title: Rationals : Proof contradicts problem.


问:
Combinatorics existence problem. IMO shortlist
I can’t solve the problem below, I tried assigning a value to each student based on the difference between the amount of girls in the k students in front of him (him included) and the amount of girls in the k students before him but I couldn’t figure it out. I also tried Bottom down induction but that also failed. Any ideas?

答:
Let $a_i=1$ if $i$ is a girl, $0$ otherwise. Let $f(n,k)=\sum_{i=n}^{n+k-1}a_i$, so we want to show that $f(n,k)=f(n+k,k)$ for some $1\leq n\leq1000$ and $100\leq k\leq 300$.

Define $g(n,k)=f(n+k,k)-f(n,k)$, and assume for contradiction that $g(n,k)\neq0$. Then as $\lvert f(n+1,k)-f(n,k)\rvert\leq1$, it follows that $\lvert g(n+1,k)-g(n,k)\rvert\leq 2$.

Now $\sum_{n=1}^{1000}g(n,k)=0$, so somewhere $g(n,100)$ goes from $-1$ to $+1$, WLOG $g(0,100)=-1$ and $g(1,100)=1$. So this implies $a_0=0$, $a_{100}=1$, $a_{200}=0$. If $a_{201}=1$, then $g(0,101)=0$, so we must have $a_{201}=0$. Similarly, $a_{-1}=0$. Continuing like this, we see that $$a_{201}=a_{202}=\dots=a_{402}=a_{-1}=a_{-2}=\dots=a_{-200}=0,$$
but then $g(201,101)=0$, a contradiction.


问:
Quadratic sequence that converges to the closest eigenvalue
I am trying to generate a sequence $[q(k)]^TAq(k)$ where $q(k) \in \mathbb{R}^n$ and $A \in \mathbb{R}^{nxn}$ is symmetric such that this sequence converges to the eigenvalue of A closest to 2. So, I want to ask, under which conditions I can generate such a sequence and what would it be as an iterative formula?

答:
You can do the following:

Let $v(k):=(A-2I)^{-1}q(k)$, starting from a random $q(0)$.

$q(k+1):=v(k)/|v(k)|$.

Then $q(k)$ generically converges to that eigenvector $w$ whose eigenvalue $\lambda$ is closest to $2$; alternatively it can oscillate between $\pm w$, but this won't matter. (There are exceptional cases, e.g. if $A-2I$ is not invertible, or if there happen to be more than one eigenvalue equally distant from $2$.)

So $q(k)^TAq(k)\to w^TAw=\lambda$.


问:
An extension of Helly's theorem
Let $A ⊂ \Bbb R^2$ be a fixed convex set and let $X_1, \cdots , X_n ⊂ \Bbb R^2$ be any convex sets such that every three of them intersect a translation of $A$. Then there exists a translation of $A$ that intersects all sets $X_i$.

The proof of the same is given as follows:

So what I have understood is this as given below:

But how did they assume that $Y_i$ is a convex set? Don't we have to prove that it is a convex set? I want to know how to prove $Y_i$ as convex set.

答:
i) If $a$ is any interior point in the triangle
$A$, then $A'$ is a translations and $a'$ is a
corresponding point (We also use the notation $A''$ for translations)

Fix a line segment $[pq]$ Then define $D_{[pq]}^A = \bigcup_{a'\in
[pq]}\
 A'$

That is, we can view $D_{pq}^A$ as a trajectory when we push $A$
along $[pq]$. Hence the set is convex.

If $a'=p,\ a''=q$ and $A'\bigcap A''$ is nonempty, then

the
intersection is in any translation in $D_{[pq]}^A$.

ii) Fix a point $r$ so that we define
$$ R = \bigg\{ a' \bigg| A'\bigcap \{r\} \neq \emptyset \bigg\} $$

If $a',\ a''\in R$, then $[a'a'']$ is in $R$ by step i) so that $R$
is a convex set.

iii) When $X$ is a convex set, we define $Y = \bigcup_{p\in X}\ P$.

If $p,\ q\in X$, then $D_{[pq]}^R$ is contained in $Y$ so that $Y$
is convex

Here note that $Y$ coincides to the set defined in OP.

Title: An extension of Helly's theorem


问:
Find a matrix of linear transformation A
Find a matrix of linear transformation $A\in M_{n\star n}(\mathbb{R})$ and a vector $x\in \mathbb{R}^n $, such that  $x\in Ker(A)$ and $x\in Im(A)$.

If $x\in Ker(A)$  then $A\cdot x=0$. And if $x\in Im(A)$ then $\exists y\in \mathbb{R}^n $ such that $A\cdot y=x$.

$\left\{\begin{matrix}
A\cdot x=0\\
A\cdot y=x
\end{matrix}\right.$

It's only possible if $A$ is singular. I selected several singular matrices at random, but failed to find proper $x$ and $y$. Is it possible at all?

答:
Let $A=\begin{pmatrix}0&1 \\0&0
\\\end{pmatrix}$
and $y=\begin{pmatrix}0 \\1
\\\end{pmatrix}$.

Then $x=Ay=\begin{pmatrix}1 \\0
\\\end{pmatrix}$.


问:
The number of solutions of $ax^2+by^2\equiv 1\pmod{p}$ is $ p-\left(\frac{-ab}{p}\right)$
What I need to show is that

For $\gcd(ab,p)=1$ and p is a prime, the number of solutions of the equation $ax^2+by^2\equiv 1\pmod{p}$ is exactly $$p-\left(\frac{-ab}{p}\right)\,.$$

I got a hint that I have to use Legendre symbol from the answer.

I think that I may count a solution one by one.

What I did :

$$(ax)^2 \equiv a-aby^2 \pmod{p}$$
It suffices to count $y$ such that $(\frac{a-aby^2}{p})=1$.

I tried to use the complete residue system or a primitive root but it didn't work.

The factorization also didn't work.

I think that the pigeonhole principle may not work because it just says the existence.

Thanks in advance.

答:
Using Legendre symbol, the number of solutions can be written as
$$
\sum_{y=0}^{p-1} \left(1+\left(\frac{a-aby^2}{p}\right)\right),$$
since if $a-aby^2$ is nonzero square, you have to count two solutions in $x$,
and if $a-aby^2$ is zero, then you have to count one solution in $x$ (namely 0),
and if $a-aby^2$ is non-square, then no solutions.

Then rewrite the summation of second term as
$$
\sum_{y=0}^{p-1} \left(\frac{a-aby^2}{p}\right)=\left(\frac{-ab}{p}\right)\sum_{y=0}^{p-1}\left(\frac{y^2+d}{p}\right),$$
where $d=-b^*$ ($b^*$ is the inverse of $b$ modulo $p$)

The summation on the right, can be rewritten as
$$\sum_{y=0}^{p-1} \left(1+\left(\frac{y}{p}\right)\right)\left(\frac{y+d}{p}\right)$$

Then the sum over single Legendre symbol is zero, but it remains to compute
$$\sum_{y=0}^{p-1}\left(\frac{y}{p}\right)\left(\frac{y+d}{p}\right)$$

This one involves a trick: using $\left(\frac{y^*}{p}\right)$ instead of $\left(\frac{y}{p}\right)$

Then you will find that
$$\sum_{y=0}^{p-1}\left(\frac{y}{p}\right)\left(\frac{y+d}{p}\right)=-1$$

Now, putting all together, you get the desired answer.


问:
Let $A\in \mathbb{R}^{2\times 2}$, $A^2=0$. Prove that $\det(tI-A)=t^2$ for any scalar $t$ (Without the use of Cayley Hamilton theorem).
I tried alot of stuff and went far as taking a general $A$ matrix and trying to apply it and I feel that I'm missing something and I'm not in the right direction.

 Note that I would be happy to get all types of answers but this question is intended to be solved using determinant properties.
 Any help is appreciated, thanks in advance.

EDIT: I have not learnt Cayley-Hamliton theorem yet, and looking for a solution without it.

答:
You can take advantage of the small size of the matrix $A$.

All you need is to work out the following cases by hand:
$$
\begin{pmatrix}
0&0\\0&0
\end{pmatrix},\quad
\begin{pmatrix}
0&1\\0&0
\end{pmatrix}.
$$

Once you work out the simple cases, find a relation between $\det(tI-J)$ and $\det(tI-BJB^{-1})$ and show that you can write $A$ as $A=BJB^{-1}$ for some matrix $B$.


问:
Find all local extrema
I'm trying to solve an exercise that says:

Find all local extrema of the function:
$$g: \mathbb{R}^{2} \rightarrow \mathbb{R}, \quad g(x, y)=3 x^{4}-6 x^{2}-4 y^{3}+2 y^{2}+x y^{2}$$
I know I must solve the equation $\nabla g =(0,0)$. So I write:
$$12x^3-12x+y^2 = 0$$
$$-12y^2+4y+2xy=0$$
and try to solve them simultaneously. But I can't. Actually, I know a solution is (1,0). How can I find out the other solutions?

答:
If $y=0$ the second equation is satisfied, and the first becomes $x^3=x$, so we have the solutions $$(-1,0),(0,0),(1,0)$$

If $y\neq0$ we can divide the second equation through by $y$ to get $$y=\frac{x+2}6\tag1$$  Substituting in the first equation gives $$432x^3+x^2-428x+4=0$$  This is a bit messy.  Let $p(x)=432x^3+x^2-428x+4$.  $p(0)=4>0$ and clearly $\lim_{x\to-\infty}p(x)=-\infty$ so $p$ has a negative root.  We find that $$p(.01)=.000432+.0001-4.28+4<0$$ so $p$ has a real root between $0$ and $.01$.  Then the third root must also be real.

Numerical solution gives the roots as $$-1.00575068640708\\0.00926025184743472\\0.994175619744835$$  Then we can use equation $(1)$ to find the $y$ corresponding to each $x$.

This gives $6$ critical points.  They still must be tested to determine whether or not they are local extrema.

The exact values of the roots, computed with sympy, are $$
- \frac{1}{1296} + \frac{\sqrt{554689} \sin{\left(- \frac{\operatorname{arctan}{\left(\frac{46656 \sqrt{78348498}}{10909729} \right)}}{3} + \frac{\pi}{6} \right)}}{648},\\
- \frac{1}{1296} + \frac{\sqrt{554689} \sin{\left(\frac{\operatorname{arctan}{\left(\frac{46656 \sqrt{78348498}}{10909729} \right)}}{3} + \frac{\pi}{6} \right)}}{648},\\
- \frac{\sqrt{554689} \cos{\left(\frac{\operatorname{arctan}{\left(\frac{46656 \sqrt{78348498}}{10909729} \right)}}{3} \right)}}{648} - \frac{1}{1296}
$$
Personally, I find the numerical values far more informative.

Just for the record, here's my python script:

```<code>from sympy import Symbol, solve, simplify, latex

x = Symbol('x')
p = 432*x**3+x**2-428*x+4
z = solve(p,x)
for r in z:
    r = r.as_real_imag()
    print(latex(simplify(r[0])))
    print(latex(simplify(r[1])))
</code>```

Title: Find all local extrema

Body:
If $y=0$ the second equation is satisfied, and the first becomes $x^3=x$, so we have the solutions $$(-1,0),(0,0),(1,0)$$

If $y\neq0$ we can divide the second equation through by $y$ to get $$y=\frac{x+2}6\tag1$$  Substituting in the first equation gives $$432x^3+x^2-428x+4=0$$  This is a bit messy.  Let $p(x)=432x^3+x^2-428x+4$.  $p(0)=4>0$ and clearly $\lim_{x\to-\infty}p(x)=-\infty$ so $p$ has a negative root.  We find that $$p(.01)=.000432+.0001-4.28+4<0$$ so $p$ has a real root between $0$ and $.01$.  Then the third root must also be real.

Numerical solution gives the roots as $$-1.00575068640708\\0.00926025184743472\\0.994175619744835$$  Then we can use equation $(1)$ to find the $y$ corresponding to each $x$.

This gives $6$ critical points.  They still must be tested to determine whether or not they are local extrema.

The exact values of the roots, computed with sympy, are $$
- \frac{1}{1296} + \frac{\sqrt{554689} \sin{\left(- \frac{\operatorname{arctan}{\left(\frac{46656 \sqrt{78348498}}{10909729} \right)}}{3} + \frac{\pi}{6} \right)}}{648},\\
- \frac{1}{1296} + \frac{\sqrt{554689} \sin{\left(\frac{\operatorname{arctan}{\left(\frac{46656 \sqrt{78348498}}{10909729} \right)}}{3} + \frac{\pi}{6} \right)}}{648},\\
- \frac{\sqrt{554689} \cos{\left(\frac{\operatorname{arctan}{\left(\frac{46656 \sqrt{78348498}}{10909729} \right)}}{3} \right)}}{648} - \frac{1}{1296}
$$
Personally, I find the numerical values far more informative.

Just for the record, here's my python script:

from sympy import Symbol, solve, simplify, latex

x = Symbol('x')
p = 432*x**3+x**2-428*x+4
z = solve(p,x)
for r in z:
    r = r.as_real_imag()
    print(latex(simplify(r[0])))
    print(latex(simplify(r[1])))


问:
If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.
If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.

I found this in a Facebook group.

I start by doing the math in the LHS:
$8(1-x-y+xy-z+xz+yz-xyz) = 8(1-2+xy+xz+yz-xyz) = 8(xy+xz+yz-1-xyz)$.

Then we set $z=2-x-y$ and replace:

I get LHS = $x^2y+xy^2+2(x+y)-x^2-y^2-3xy-1$ but don't know how to proceed.

Thank you for your help.

答:
As $0 \leq x, y, z \leq 1, (1-x), (1-y), (1-z) \geq 0$

Using A.M - G.M,

$(1-x) + (1-y) = z \geq 2 \sqrt{(1-x)(1-y)} \ $ (as $ \ x + y + z = 2$)

Similarly for $x, y$.

Multiplying them, we have

$xyz \geq 8 (1-x)(1-y)(1-z)$


问:
$\lim_{x\to \pi/2} \;\frac 1{\sec x+ \tan x}$
how to solve it answer is $0$, but $\frac 1{\infty + \infty}$ is indeterminate form

$$\lim_{x \to \pi/2} \frac 1{\sec x + \tan x}$$

答:
In order to solve this, it would be best to convert the $\sec x$ into $\frac{1}{\cos x}$. and convert $\tan x$ to $\frac{\sin x}{\cos x}$ So we will have

$$\lim_{x\to\frac{\pi}{2}} \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}$$

Now we have to get rid of the fraction in the denominator. So multiply by $\cos x$ so we will have:

$$\lim_{x\to\frac{\pi}{2}} \frac{1}{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}$$

When then is turned into

$$\lim_{x\to\frac{\pi}{2}} \frac{\cos x}{1 + \sin x}$$

Then sub $\pi/2$ and you'll be gold.

$$\lim_{x\to\frac{\pi}{2}} \frac{\cos \frac{\pi}{2}}{1 + \sin \frac{\pi}{2}}$$

And the answer will be $0$. Therefore the limit is $0$.


答:
We let $a=1-x,b=1-y,c=1-z$ so that we have $a+b+c=1$ and $0 <  a,b,c < 1$.

Our new inequality now becomes $$8abc \leq (1-a)(1-b)(1-c)\\ \iff 8 \leq \left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right) \\ \iff 8 \leq \left(\dfrac{b+c}{a}\right)\left(\dfrac{c+a}{b}\right)\left(\dfrac{a+b}{c}\right)  $$ which means $(a+b)(b+c)(c+a)\geq 8abc$ which is true by A.M-G.M inequality.

Equality occurs at $x=y=z=\dfrac{2}{3}$.

As for your approach, it is also correct. If you are substituting the value of $z$, making it a two variable inequality, you can form a quadratic equation taking everything to one side and set it's discriminant less than or equal to $0$.

Body:
If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.

I found this in a Facebook group.

I start by doing the math in the LHS:
$8(1-x-y+xy-z+xz+yz-xyz) = 8(1-2+xy+xz+yz-xyz) = 8(xy+xz+yz-1-xyz)$.

Then we set $z=2-x-y$ and replace:

I get LHS = $x^2y+xy^2+2(x+y)-x^2-y^2-3xy-1$ but don't know how to proceed.

Thank you for your help.


答:
$$\det \begin{pmatrix} t - a & -b \\ -c & t - d \end{pmatrix} = t^2 - (a + d)t + (ad - bc)$$

Since $A^2 = 0$, we have $ad - bc = 0$.

If $a + d \neq 0$ then setting $t = a + d$ we have that $\det(tI - A) = 0$. On the other hand,

$$ (tI - A)(\alpha I + \beta A) = I$$

has a solution (try to find it). So $tI - A$ is invertible if $t \neq 0$. But then $a + d$ cannot be non-zero.

Title: Let $A\in \mathbb{R}^{2\times 2}$, $A^2=0$. Prove that $\det(tI-A)=t^2$ for any scalar $t$ (Without the use of Cayley Hamilton theorem).

Body:
I tried alot of stuff and went far as taking a general $A$ matrix and trying to apply it and I feel that I'm missing something and I'm not in the right direction.

 Note that I would be happy to get all types of answers but this question is intended to be solved using determinant properties.
 Any help is appreciated, thanks in advance.

EDIT: I have not learnt Cayley-Hamliton theorem yet, and looking for a solution without it.


问:
Questions about Trigonometry
I just learned the very basics of Trigonometry yesterday. I can answer all the questions on the homework but feel uneasy because I still have some question about it. Here are the questions:

What does it mean to take the sin, cos, or tan of an angle? I know it has do do with the ratios but how does that apply to triangles? I just don't fully understand that concept.

Also how does the calculator know for example what the sin(53) is. When finding missing side of non special triangles, at some point I type a random number into the calculator and press either sin/tan/cos. How does it know that though?

Thanks! I'm very new to this subject and so excited to learn more.

答:
Draw a right triangle abc with a b the hypotenuse of length one and angle b a c your desired angle, say 53°.

Then the sine of angle bac is defined as the length of bc, the cosine as the length of ac, and the tangent as sine/cosine.

These are just the definitions of these functions of an angle.

As the computing these functions, many ways have been derived for doing it and these have been implemented and calculators and programming languages.

Title: Questions about Trigonometry


问:
If $A=\bigl(\begin{smallmatrix} 1&1\\1&0\end{smallmatrix} \bigr)$ then $A^n=~?$

Let $A=\bigl(\begin{smallmatrix} 1&1\\1&0 \end{smallmatrix}\bigr)$.
Then $A^n~?$

I saw $A^2=\left(\begin{smallmatrix} 2&1\\1&1 \end{smallmatrix}\right)$. Is there any similarity or any recurrence relation ?

答:
Find the eigenvalues of $A$ and write $A$ as $SJS^{-1}$ where $J$ is some nice matrix.

Then find $(SJS^{-1})^n$.

The point is that it is very easy to find a "recursive" relation in $J^n$.


问:
Finite-dimensional irreducible representations
How do we show that a finite-dimensional $*$-representation of a $C^{*}$-algebra is unitary equivalent to a direct sum of irreducible representations?</p>

答:
If $M$ is a finite dimensional representation of a $C^*$ algebra $A$, then it is in fact a representation of a quotient $B=A/I$ by a closed ideal of finite codimension.

So you may just as well assume that $A$ is finite-dimensional. Then it is a direct product of matrix algebras, and it is therefore semisimple.

What you want follows at once from this.


问:
A consistency question
Let $X=P_f(\mathbb{N})$ be the set of all finite subsets of the set of natural numbers. Let $S^*$ be a binary relation over $X$, defined by: for each $A,B\in X$, $\langle A,B \rangle \in S^*$ if and only if $A\cap B=\emptyset$.

Let $L$ be a vocabulary which contains binary relations $S$ and $R$, and contains a constant symbol $c_A$ for each $A\in X$.

Now, let $M$ be a structure to interpret $L$, defined by:

the domain is $X$, $S^M=S^*$,

$R^M$ is defined so that for each $A,B\in X$, $\langle A,B \rangle \in R^M$ if and only if $A\subseteq B$, and

$c_A^{M}=A$ for each $A\in X$.

In addition, for each $n\in \mathbb{N}$, let $A_n=\{0,1,2,...,n-1\}$ (and  $A_0=\emptyset$).

Let $T$ be the complete theory of $M$. We define $T_1$, $T_2$, and $T_3$ in $L^+=L \cup \{d\}$, where $d$ is a new constant symbol:
\begin{align*}
T_1 &= T \cup \lbrace S(c_{A_n},d)\land R(c_{\lbrace n,n+1,...,2n\rbrace},d)\mid n\in \mathbb{N}\rbrace\\
T_2&=T \cup \lbrace (S(c_{\lbrace n+1 \rbrace},d)\rightarrow R(c_{\lbrace n\rbrace},d)\mid n\in \mathbb{N}\rbrace\\
T_3&=T \cup \lbrace (S(c_{\lbrace 2n \rbrace},d)\land R(c_{\lbrace 2n+1\rbrace},d)\mid n\in \mathbb{N}\rbrace
\end{align*}

Prove that $T_1$ is inconsistent.

Prove that $T_2$ is consistent, and it has a model which is an expansion of $M$ to $L^+$.

Prove that $T_3$ is consistent, but it does not have a model which is an expansion of $M$ to $L^+$.

答:
The formulas $R(c_A,x)\to\lnot S(c_A,x)$ are valid in $M$ for every nonempty $A$, and so are $R(c_B,x)\to R(c_A,x)$ and $S(c_B,x)\to S(c_A,x)$ whenever $A\subseteq B$.

So, $T_1$ implies that both $S(c_{\{1\}},d)$ and $R(c_{\{1\}},d)$ holds (taking the formula for $n=1,2$), which is a contradiction.

and 3. You can verify that for every finite subset of the given formulas, one can find interpretation of $d$ that satisfies those. So, you can use compactness to conclude. However, a $d\in X$ in neither cases exists which satisfies all the formulas at once, as it would have infinite many elements.

Title: A consistency question


答: as $1-x=\frac{1}{2}(y+z-x)$ and similarly for others we have to prove  $$(x+y-z)(y+z-x)(x+z-y)\le xyz$$ which is schur's inequality

Title: If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.


答:
$A^n=\bigl(\begin{smallmatrix} F_{n+1}& F_n\\F_n & F_{n-1}\end{smallmatrix}\bigr)$ where $F_n=n-$th fibonacci  number.

Notice this is true for $n=1$. Now use Induction to prove my statement. You should read abot the matrix representation of fibonacci number [here]
(htttps://en.wikipedia.org/wiki/Fibonacci_number)

Title: If $A=\bigl(\begin{smallmatrix} 1&1\\1&0\end{smallmatrix} \bigr)$ then $A^n=~?$

Body:
Let $A=\bigl(\begin{smallmatrix} 1&1\\1&0 \end{smallmatrix}\bigr)$.
Then $A^n~?$

I saw $A^2=\left(\begin{smallmatrix} 2&1\\1&1 \end{smallmatrix}\right)$. Is there any similarity or any recurrence relation ?


答:
Here is yet another way to show it:

In some extension field (in this case we can just take $\Bbb C$ of course) the characteristic polynomial $\det(xI-A)$ will split into linear factors. Let $\lambda\in\Bbb C$ be any zero of it, i.e. $\det(\lambda I-A)=0$. Then $\lambda I-A$ is not invertible, hence there is a $v\in\Bbb C^2$ (an 'eigenvector') with $v\ne0$ and $(\lambda I-A)v=0$. So we get $Av=\lambda v$. Applying $A$ again gives us $$0=A^2v=A\lambda v=\lambda^2 v$$ Therefore $\lambda^2=0\Rightarrow \lambda=0$. Hence $\det(xI-A)$ has only one zero in $\Bbb C$, namely $0$ (not counting multiplicity). Therefore $\det(xI-A)=x^2$.

Of course this proof generalizes immediately to higher dimensions.


问:
Block matrix $A$
Let
\begin{align}
A = \begin{pmatrix}
 0 & a I_n \\
b I_n & 0
\end{pmatrix}
\end{align}
for $a,b\in \mathbb{R}$.  I want to prove whether this $A$ can be diagonalizable over $\mathbb{R}$ by computing characteristic polynomials and minimal polynomials.

$|A - \lambda I| = \left|\begin{pmatrix}
 -\lambda I_n & a I_n \\
b I_n & -\lambda I_n
\end{pmatrix} \right| = |\lambda I_n | | -\lambda I_n + a b \lambda I_n| =  |- \lambda I_n| | \left( - \lambda + \frac{ab}{\lambda}\right) I_n | = (\lambda^2 - ab)^n $

Here I used, for invertible matrix $A,B,C,D$,

\begin{align}
  \begin{vmatrix}
    A & B \\
    C & D
  \end{vmatrix} = |A| | D - CA^{-1} B|
\end{align}

So characteristic polynomial is $(\lambda^2 - ab)^n$. Apparently, $A^2 = ab \begin{pmatrix}
I_n & 0 \\
0 & I_n
\end{pmatrix}$ so minimal polynomial is $\lambda^2 - ab$.

Now I want to determine whether this is diagonalizable or not.  I am stuck with this. What is a good approach for showing this be diagonalizable or not?

答:
Your matrix is not always diagonalizable. Take $n=1, a=0, b=1$ to obtain $A=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ which is not diagonalizable.

Therefore, you can not prove the general diagonalizability of your matrix.

Title: Block matrix $A$


答:
Hint: "rationalize".

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

Title: $\lim_{x\to \pi/2} \;\frac 1{\sec x+ \tan x}$

Body:
how to solve it answer is $0$, but $\frac 1{\infty + \infty}$ is indeterminate form

$$\lim_{x \to \pi/2} \frac 1{\sec x + \tan x}$$


答:
Draw a right triangle with hypothenuses of length $1$ and the given angle. The sine and cosine will be the lengths of the two other sides. This is the geometric definition. (The tangent is the ratio of these.)

There is also an analytical definition, based on the complex exponential $e^{ix}$. There are tools such as the Taylor development to evaluate the functions for a given $x$ using the four arithmetic operations. This is a broad topic. You may have a look at the very clever CORDIC method. https://en.wikipedia.org/wiki/CORDIC

Title: Questions about Trigonometry

Body:
I just learned the very basics of Trigonometry yesterday. I can answer all the questions on the homework but feel uneasy because I still have some question about it. Here are the questions:

What does it mean to take the sin, cos, or tan of an angle? I know it has do do with the ratios but how does that apply to triangles? I just don't fully understand that concept.

Also how does the calculator know for example what the sin(53) is. When finding missing side of non special triangles, at some point I type a random number into the calculator and press either sin/tan/cos. How does it know that though?

Thanks! I'm very new to this subject and so excited to learn more.


问:
Volume of half torus. What's wrong with my solution?
A doughnut has been partially eaten by a meticulous person so that the portion remaining is given by rotating the half-circular region shown above about the y axis. What fraction of the original doughnut remains, assuming that the original doughnut was the volume of revolution of the full circle? Give your answer in exact form in terms of r and R.

My solution: dropbox.com/s/a816w2rgmkq223r/2013-02-12%2015.19.40.jpg

So clearly, volume of a full torus over that of above is $1\over r$

Bu it's wrong? Where did I go wrong?

EDIT: Corrected some mistakes, still getting the wrong answer:

https://www.dropbox.com/s/1sf72pxcm6dk8pp/2013-02-12%2018.45.46.jpg
https://www.dropbox.com/s/wlko0w958plo1pa/2013-02-12%2018.46.22.jpg

EDIT 2: I made a small mistake in the above solution as well. It should be $4 \pi $ and not $2 \pi$. Correcting that, I got the answer:

$$\left(\frac12\right) - \left(8r\over3 \pi R\right)$$

But this is also wrong! What should I do now?

答:
Guldin's rule, also called Pappus' centroid theorem, says the following: When you rotate a shape $S$ around an exterior axis the volume $V$ of the solid generated in this way is equal to the area of the shape times the circumference of the circle described by the centroid of $S$.

It follows that the volume of the full torus is given by $V_{\rm tot}=\pi r^2\cdot 2\pi R$.

In order to compute the remaining volume we need the centroid $(0,\rho)$ of the half circle $\{(x,y)\>|\> x^2+y^2\leq r^2,\ y\geq0\}$. It is determined by the moment equation
$${1\over2}\>\pi r^2\ \rho=\int_{-r}^r{\sqrt{r^2-x^2}\over2}\>\sqrt{r^2-x^2}\ dx={2\over3}r^3\ ,$$
from which we get ${\displaystyle \rho={4r\over 3\pi}}$.

Since by Guldin's rule the remaining volume is given by $V_{\rm rem}={1\over2}\pi r^2\cdot 2\pi(R-\rho)$ we now obtain
$${V_{\rm rem}\over V_{\rm tot}}={1\over2}-{2\over3\pi}\>{r\over R}\ .$$


问:
How to calculate expected value for a bid contract?
A company plans to open a bid of \$3000 per tonne for a contract to supply 1000 tonnes of metal. Assume that 2 competitors A and B will place bids. The probability that A places a bid less than 3000$ per tonne is 0.3 and probability that B places a bid less than \$3000 per tonne is 0.7. If the lowest bidder wins the contract, what is the Expected Value of the contract when it is assumed that the each competitor will bid independently.

I have no idea to do this problem. All I know is that if there's a specific \$ value for a known probability then I can get the expected value of it using the product of it. But in this question no such value is given but only a range. So your help is much needed. Thank you very much!

Also if anyone can provide with related tags you are most welcome.

答:
I would guess the $3000$ number means that if both $A$ and $B$ bid more than that no contract will be let and the value is $0$.  Otherwise, the value of the contract is the difference between $3000$ and the low bid.  You are asked for the expected value of the contract which you cannot find without knowing the probability distribution of $A$'s and $B$'s bids.

I can make up a probability distribution consistent with the data.  Say $A$ bids $2850$ with probability $0.3$ and $B$ bids $2900$ with probability $0.7$.  Otherwise, each bids some number greater than $3000$.  If $A$ bids $2850$ it will win the job and the contract has a value of $150$.  If $B$ wins the job (which only happens if $A$ bids high and $B$ bids low) the value of the contract is $100$.  The expected value of the contract is then
$$0.3 \cdot 150 + (1-0.3)\cdot 0.7 \cdot 100=94$$
If you have overlapping probability distributions you will need to integrate over the regions that each company wins the bid.


答:
A matrix is diagonalizable over some field $K$ if and only if its minimal polynomial is split with simple roots over $K$.

As you have correctly computed, the minimal polynomial of $A$ is $X^2 - ab$. This polynomial has degree $2$, so it is easy to discuss its behaviour over $\mathbb R$.

It splits over $\mathbb R$ if and only if $ab\geq 0$. It splits over $\mathbb R$ with simple roots if and only if $ab>0$.

With this, you know precisely when the matrix $A$ is diagonalizable over $\mathbb R$, depending on the choices of the coefficients $a,b\in \mathbb R$.

Title: Block matrix $A$

Body:
Let
\begin{align}
A = \begin{pmatrix}
 0 & a I_n \\
b I_n & 0
\end{pmatrix}
\end{align}
for $a,b\in \mathbb{R}$.  I want to prove whether this $A$ can be diagonalizable over $\mathbb{R}$ by computing characteristic polynomials and minimal polynomials.

$|A - \lambda I| = \left|\begin{pmatrix}
 -\lambda I_n & a I_n \\
b I_n & -\lambda I_n
\end{pmatrix} \right| = |\lambda I_n | | -\lambda I_n + a b \lambda I_n| =  |- \lambda I_n| | \left( - \lambda + \frac{ab}{\lambda}\right) I_n | = (\lambda^2 - ab)^n $

Here I used, for invertible matrix $A,B,C,D$,

\begin{align}
  \begin{vmatrix}
    A & B \\
    C & D
  \end{vmatrix} = |A| | D - CA^{-1} B|
\end{align}

So characteristic polynomial is $(\lambda^2 - ab)^n$. Apparently, $A^2 = ab \begin{pmatrix}
I_n & 0 \\
0 & I_n
\end{pmatrix}$ so minimal polynomial is $\lambda^2 - ab$.

Now I want to determine whether this is diagonalizable or not.  I am stuck with this. What is a good approach for showing this be diagonalizable or not?


问:
Example of endomorphism
Endomorphism $T$ is linear mapping from one linear space to itself.

Let us consider $V=span\{e^x,\sin x, \cos x, \sin(2x), \cos (2x)\}$. It is easy to verify, that vectors $e^x,\sin x, \cos x, \sin(2x), \cos (2x)$ form a basis in V. We have the mapping $T(f)=f''+f$. Is it an endomorphism?

My solution is as follow:

This mapping is linear(it is easy to verify).

To check second part of definition let $g\in V$, then $$\exists a_1,a_2,a_3,a_4,a_5\in\mathbb R: g=a_1e^x+a_2\sin x +a_3\cos x+a_4\sin(2x)+a_5\cos (2x).$$ Then
$$Tg=2a_1e^x-3a_4\sin (2x)-3a_5\cos(2x).$$ So $Im T=span\{e^x,\sin(2x),\cos(2x)\}\subset V$, and that mean that $T$ is not an endomorphism. (because $ImT\neq V$) Am I right in my consideration? I would be grateful for your hints and advices.

答:
Any linear map $V\to V$ is called an endomorphism, it needs not be surjective.

Surjective endomorphisms of a finite dimensional vector space are exactly the invertible ones, the automorphisms.

Title: Example of endomorphism


问:
Does $\lim_{x\to 0}(\frac{1}{x}-\frac{\sin x}{x^2})$ exist?
$$\lim_{x\to0}\left(\frac{1}{x}-\frac{\sin x}{x^2}\right)=\lim_{x\to0}\left(\frac{x-\sin x}{x^2}\right)$$
By L'Hopital's Rule, we have
$$\lim_{x\to0}\left(\frac{x-\sin x}{x^2}\right)=\lim_{x\to0}\frac{1-\cos x}{2x}=\lim_{x\to0}\frac{\sin x}{2}=0$$
This is what I did, but the answer in my textbook says that the limit does not exist. Can someone explain to me where have I done it wrong? Thx!

答:
As others have pointed out, the OP's answer is correct; the limit is amenable to L'Hopital, and equals $0$. It's worth noting, though, that one can avoid L'Hopital, using instead the inequality $\sin x\ge x-{1\over3}x^3$ for $x\ge0$. This inequality and the symmetry $\sin(-x)=-\sin x$ give us

$$|x-\sin x|\lt{1\over3}|x|^3$$

for all $x$, hence

$$\left|{1\over x}-{\sin x\over x^2}\right|\lt{1\over3}|x|\to0$$

as $x\to0$.


问:
$G\otimes \mathbb Q=0$ implies torsion group
let $G$ be an abelian group. suppose that $G\otimes \mathbb Q=0$. Does this imply that
$G$ is necessarily a torsion group?</p>

答: Yes. Note that $G\otimes_\mathbb{Z}\mathbb{Q}\cong (\mathbb{Z}\setminus 0 )^{-1}G$ as $\mathbb{Z}$-modules. Now $\frac{g}{n}=\frac{0}{1}$ by definition means $zg=0$ in $G$ for some $z\in\mathbb{Z}\setminus 0$.</p>


问:
Calculate the limit of an integration domain
Consider the set $D(\varepsilon)=\{(x,y)\in\mathbb{R}^2|0<x^2+y^2\leq\varepsilon^2\}$, which is the filled disk of radius $\varepsilon$ with a hole at the origin. Calculate the limit

$$\lim_{\varepsilon\rightarrow0}\dfrac{1}{\varepsilon}\int_{D(\varepsilon)}\dfrac{1+\sin x+\sin y}{\sqrt{x^2+y^2}}d(x,y).$$

My first instinct was to do a change of variables into polar coordinates which would give $D(\varepsilon)=\{(x,y)\in\mathbb{R}^2|0<r^2\leq\varepsilon^2; 0\leq\theta\leq2\pi\}$ and

$$\lim_{\varepsilon\rightarrow0}\dfrac{1}{\varepsilon}\int_{D(\varepsilon)}[1+\sin(r\cos\theta)+\sin(r\sin\theta)]d(r,\theta),$$

But this does need seem any simpler than before. Any hints are welcome.

答:
Both $\frac{\sin x}{\sqrt{x^2+y^2}}$ and $\frac{\sin y}{\sqrt{x^2+y^2}}$ are odd functions and given the shape of the domain integrate to $0$ by symmetry. This means the limit equals

$$\lim_{\epsilon\to0}\frac{1}{\epsilon}\int_0^{2\pi}\int_0^{\epsilon} drd\theta = 2\pi$$


问:
Finding Pythagorean triple where the squares of sides are weighted
Suppose $a, b, c \in \mathbb{Z}_{>0}$. The following is a variation of a Pythagorean triple, but with weighted squares:

$$a^2 + 3 b^2 = 4 c^2$$

Can I find something like Euclid's formula that can generate all solutions for this formula? I'm not sure how to approach this problem.

答:
As I said in the comment, one needs to verify that no other solutions exist but the found ones. I do not have the time now to verify this (or possibly I might not able to do that anyway), but I want to share my modified pair of Euclid's formula for your case:

$a = 2(m^2 - 3n^2)$

$b = 4mn$

$c = m^2 + 3n^2$

or

$a = 2(3m^2 - n^2)$

$b = 4mn$

$c = 3m^2 + n^2$

Update:

Addressing the comment below: a trivial solution is when

$a = b = c$

Update 2:

Further addressing the comment about $(a, b, c) = (1, 1, 1)$:

if $(a, b, c)$ is a solution of $a^2 + 3 b^2 = 4 c^2$ and all elements in the triplet are even, then dividing all of them by 2 is also a solution, i.e. with

$a = 2a'$

$b = 2b'$

$c = 2c'$

we have

$(2a')^2 + 3(2b')^2 = 4(2c')^2$

$4a'^2 + 3*4b'^2 = 4*4c'^2$

$a'^2 + 3b'^2 = 4c'^2$

so we get the original statement, just with $(a', b', c')$ which therefore is a solution triplet, and this reduction can continue until one element becomes odd.

(And this can be done with any common divisors of all elements, until arriving to a $gcd$ of 1. And with the same logic, if a triplet is a solution, multiplying all elements with the same number is also a solution - so only coprime triplets we should be looking for, but I leave this condition out.)

Now if we set $m = n$ in my second triplet definition above we get

$a = 2(3m^2 - n^2) = 4m^2$

$b = 4mn = 4m^2$

$c = 3m^2 + n^2 = 4m^2$

i.e. we have $a = b = c$. Now setting $m = 1$ we get a solution triplet of $(4, 4, 4)$, and based on the above $(2, 2, 2)$ and $(1, 1, 1)$ are also solutions.

I still can not verify what solution triplets this system is not covering.

Update 3:

In case in any doubt that the solutions so far do not cover all solutions, I pedantically collected and compared them.

First of all, in individ's answer the two systems are equivalent, by substituting $(p + s)$ for $p$ in the second system. The second system as it is now is equivalent with the OP's result with the method of finding rational points on the ellipse (though now it is incorrectly shown, as $a$ and $b$ in the question correspond to a factors of $1$ and $3$, respectively, so $a = \frac{k}{d} \cdot (3 s^2 - 6 s t - t^2)$ and $b = \frac{k}{d} \cdot (3 s^2 + 2 s t - t^2)$ and $c = \frac{k}{d} \cdot (3 s^2 + t^2)$) with the reduction/extension factor of $\frac{k}{d}$ added.

My second system's definiton of $c$ is similar, while $a$ and $b$ are not:

$a = 2(3m^2 - n^2)$

$b = 4mn$

$c = 3m^2 + n^2$

With this system one solution triplet is $(94, 16, 49)$, whose elements are relative primes.

Using $k = d = 1$ with the other answers' system, we see that if we wanted to get the triplet $(94, 16, 49)$, we can use the same values $(m, n)$ that generated the last element as the definition is the same, and these are those values: $(s, t) \in \{(4, 1), (-4, 1), (-4, -1), (4, -1)\}$. Calculating $a$'s and $b$'s for that system we get the following values for each $(s, t)$:

\begin{array}{c|c|c|c|c}
{} & {(4, 1)} & {(-4, 1)}
& {(-4, -1)} & {(4, -1)}\\\hline
a & 23 & 71 & 23 & 71 \\\hline
b & 55 & 39 & 55 & 39
\end{array}

As we see, the $(94, 16, 49)$ triplet can not be generated by any $(s, t)$ that gives $49$ as the last element. I wanted to prove that even if we do not set $k = d = 1$ that system can not generate this triplet when I recognised that the answers are lacking the absoulte value in the definitions, however that is valid as all are squared and that might add otherwise not producible triplets to the soution set. I stop the investigation at this point. The problem with those answers is at the moment, that with a lack of proof/theorem which shows how to find all solutions, all identities leaves one with "good luck finding that or a counterexample".

Finally, the paper linked by the OP's answer can generate both $(94, 16, 49)$ and $(71, 39, 49)$, but also lacks the previously mentioned proof/theorem of full coverage, as I commented so.


问:
Using Mean Value Theorem, Prove that ${\tan x\over x}>{x\over\sin x}$
Prove using Mean Value Theorem,

$${\tan x\over x}>{x\over\sin x} \space\forall \space x \space\in(0, \pi/2) $$

Attempt::

$f(x) = x-\sin x$

$f^\prime(x) = 1-\cos x > 0 $

Hence $x-\sin x>0, {x\over\sin x}>1$

Similarly, I got ${\tan x\over x}>1$

But how do I compare them and get the required inequality?

答:
Suppose we have a function $f(t)=\sqrt{\sin{t}\tan{t}}$. Now, applying the mean value theorem in $(0, \frac{\pi}{2})$, we have $$\frac{f(x)-f(0)}{x-0}=f^\prime(c)$$ Or $$\frac{f(x)}{x}=f^\prime(c)\quad (where\quad 0<c<x<\frac{\pi}{2})$$ Or $$\frac{\sqrt{\sin{x}\tan{x}}}{x}=f^\prime(c)\tag{1}$$ Now we proceed to derivation of $f(t)$: $$f^\prime(t) = \frac{\sin{t}\sec^2{t}+\tan{t}\cos{t}}{2\sqrt{\sin{t}\tan{t}}}$$ $$=\frac{\tan{t}(\sec{t}+\cos{t})}{2\sqrt{\sin{t}\tan{t}}}$$ $$=\frac{\tan{t}(\sec{t}+\cos{t})\sqrt{\cos{t}}}{2{\sin{t}}}$$ $$=\frac{\sec{t}+\cos{t}}{2\sqrt{\cos{t}}}$$ $$=\frac{\sec{t}\sqrt{\sec{t}}+\sqrt{\cos{t}}}{2}$$ Now, $$f^{\prime\prime}(t)=\frac{3\sec{t}\tan{t}-\sin{t}}{2{\sqrt{\cos{t}}}}$$ Since $f^{\prime\prime}(t) = 0$ at $ t = 0$ we get that $f^\prime(t)$ is minimum at $t = 0$. As $f^\prime(0) = 1$, we conclude that for $c\in(0, \frac{\pi}{2})$, $f^\prime(c)>1$. Therefore, from (1) we have, $$\frac{\sqrt{\sin{x}\tan{x}}}{x}>1\quad x\in(0, \frac{\pi}{2})$$ or $$\frac{\sin{x}\tan{x}}{x^2}>1$$ or $$\frac{\tan{x}}{x}>\frac{x}{\sin{x}}\quad
 if \quad 0<x<\frac{\pi}{2}$$
$$***$$


问:
Find x for which for every "a" the equation has solution
$$a^{31x} \equiv a \mod 271$$

I need to find x variable, for which the equation has solution with any a. How can I do this?

Generaly, modular equations have solutions when $GCD(a^{31x}, 271) = 1$, or $GCD(a^{31x}, 271) = d > 1; d|a$

It also looks like I could use Little Fermat's Theorem...

But I can't come up with anything...

答:
Since 271 is prime, we can indeed apply Fermat's little theorem.
Using this theorem we know that $a^{n}\equiv a \pmod{271}$ for any $n\equiv 1 \pmod{270}$.
Hence, we need to find an $x$ such that $31x\equiv 1\pmod{270}$.

To find an appropriate $x$ we may use the extended Euclidean algorithm:

$ 0\cdot 31+1\cdot 270 = 270$

$1\cdot 31+0\cdot 270 = 31$

$ -8\cdot 31+1\cdot 270 = 22$

$ 9\cdot 31-1\cdot 270 = 9$

$-26\cdot 31+3\cdot 270 = 4$

$61\cdot 31-7\cdot 270 = 1$

Thus $61\cdot 31\equiv 1 \pmod{270}$ and hence any $x\equiv 61\pmod {270}$ will do.


问:
Condition for polynomial factoring
I encountered a sum , prove that $x²+px+p²$ would be a factor of $(x+p)^n–x^n–p^n$ if $n$ be odd and not a multiple of $3$. I tried breaking $(x+p)^n$ by binomial but it became more complicated and also tried factoring the dividend polynomial into $(x+p)[(x+p)^{n-1}–(x^{n-1}–......p^{n-1})]$ but to no avail . Can anyone please help by telling which way of viewing this problem am i overlooking or not thinking. Can anyone please enlighten me? I don't want a full detailed solution i can work that on my own i just want the path to do it.

答:
Hint:
$$(x^2+px+p^2)(x-p)=x^3-p^3$$

Now, can you do it?

Title: Condition for polynomial factoring


问:
Does this ring isomorphism hold for rings that aren't necessarily fields?
The problem I'm working on is: Show that $R[x,y]/(x^2-y^3)\cong A$ where $A=\{\sum a_ix^i\in R[x]:a_1=0\}$. A similar question has been posted on the website multiple times, but the explanation is either confusing (to me at least) or requires $R$  to be a field (which is not assumed to necessarily be true in the problem).

I have constructed a homomorphism $\varphi:R[x,y]\to A; \varphi(x):=x^3;\varphi(y):=x^2;\varphi(r):=r,r\in R$. There is a unique homomorphism for which all of the previous equalities hold and it is surjective. To prove the claim it is sufficient to show that $ker\varphi\subset(x^2-y^3)$. The other inclusion is trivial. Once proven, the hypothesis easily follows from the  first isomorphism theorem. The inclusion above could be proven using division with rest if $R$ were a field, however that isn't necessarily the case.

I've gotten this far on my own:

We can split the terms of any $f\in R[x,y]$ in the following way: $f=g_0+g_1+...+g_n$ where $\varphi(g_k)=c_kx^k$ for some $c_k\in R$ and all $k$. This partition of terms is clearly unique. Let $f\in ker\varphi$. In this case $c_k=0$ for all $k$. The goal is to prove that $g_k(x,y)=(x^2-y^3)h_k(x,y)$ for some polynomial $h_k.$ It is easy to see that $g_0$ consists of only constant terms, therefor $g_0=0$. To find the others let us take a look at the following equation: $\varphi(x^iy^j)=x^k$. This equation clearly holds if and only if $3i+2j=k$. It's easy to find all of the solutions of this equation. If $k$ is even then $i=0,j=\frac{k}{2}$ is a solution. If, on the other hand $k$ is odd, then $i=1,j=\frac{k-3}{2}$ is a solution (we're obviously looking for solutions where $k\neq1$). All the other solutions can be found by subtracting $3$ from $j$ and adding $2$ to $i$. Using this method we can find that:

\begin{equation}
\begin{split}
&g_0=a_{0,0}\\
&g_1=0\\
&g_2=a_{2,0}y\\
&g_3=a_{3,0}x\\
&g_4=a_{4,0}y^2\\
&g_5=a_{5,0}xy\\
&g_6=a_{6,0}x^2+a_{6,1}y^3...
\end{split}
\end{equation}

Where $a_{i,j}\in R$. From here we can see why $g_0=g_1=...=g_5=0$. For example $0=\varphi(g_2)=a_{2,0}x^2\Rightarrow a_{2,0}=0\Rightarrow g_2=0$. Similarily, we can conclude that $g_k=0$ whenever the equation $3i+2j=k$ has a unique solution in the natural numbers (holds also for $k=7$ and $k=8$). If the equation has two solutions (like in case $k=6$) we can do the following:

$$
0=\varphi(g_6)=(a_{6,0}+a_{6,1})x^6
$$
This means that $g_6(x,y)=a_{6,0}x^2-a_{6,0}y^3=(x^2-y^3)a_{6,0}$. So the claim holds for $k=6$. Similar can be shown for all other $k$ for which the equation above has only two solutions (like for $k=9,10,11$) using the equality $cx^iy^j-cx^{i-2}y^{j+3}=(x^2-y^3)\cdot cx^{i-2}y^j$. Furthermore it can also be easily factored in the case where the equation has three solutions ($k=12$ for example). That's because if

$$
g_k(x,y)=c_0x^iy^j+c_1x^{i-2}y^{j+3}+c_2x^{i-4}y^{j+6}
$$

then $\varphi(g_k)=0\Rightarrow c_0+c_1+c_2=0$ therefor

\begin{equation}
g_k(x,y) = c_0x^iy^j - c_2x^{i-2}y^{j+3} - c_0x^{i-2}y^{j+3} + c_2x^{i-4}y^{j+6} = \\
(x^2-y^2)(c_0x^{i-2}y^j - c_2x^{i-4}y^{j+3})
\end{equation}

This is as far as I've come. I can't find a solution for when the equation with $k$ has four solutions. All of the attempts at factoring the general case have failed.

答:
We want to show that the kernel of $\varphi$ is $(x^2-y^3)=: I$, with $I$ being a subset of said kernel.

Let $h \in R[x,y] \backslash I$ with $\varphi(h)=0$ (assume there exists one for the sake of contradiction). We can write $h \in g+I$, with $\varphi(g)=0$, and $g(x,y)=g_1(x)+yg_2(x)+y^2g_3(x)$, $g_1,g_2,g_3 \in R[x]$, and thus $g \notin I$.

But $0=\varphi(g)=g_1(x^3)+x^2g_2(x^3)+xh_3(x^3)$ where $h_3(x)=xf_3(x)$.

Each term of the RHS only contributes to disjoint classes of powers of $x$: from left to right, the $x^{3d}$, the $x^{3d+2}$, the $x^{3d+1}$. As the RHS is zero, it follows that all three terms are zero and thus $g=0 \in I$, a contradiction.


问:
A discrete random variable question

The probability distribution of the discrete random variable $X$ is
$$
\mathbb{P}[X=x] = \begin{cases}
k/x,    & x \in \{1,2,3\}\\
m/(2x), & x \in \{6,9\}\\
0,      & \text{otherwise}
\end{cases}
$$
where $k,m$ are positive constants. Given that $\mathbb{E}[X] = 3.8$, find $\mathbb{Var}(X)$.

How does one do this question? I tried but am stuck, could somebody please help?

答:
HINT

So your distribution is in terms of two parameters, $k$ and $m$, and you are given $\mathbb{E}[X]$. Let's use it.

Find the total weight of the distribution. If you sum over all possible values of $X$, the total weight should be a well-known constant. That will give you one relationship between $k$ and $m$.

Find $\mathbb{E}[X]$ from the distribution in terms of $k,m$. This will give you a second relationship between $k$ and $m$.

Solve the $2 \times 2$ system to find $k,m$.

Compute the variance.

Title: A discrete random variable question


问:
Why is the Hilbert curve called like that?
Everywhere the function $H\colon\mathcal{I}\to Q=[0,1]\times[0,1]$ is called the Hilbert curve. Is this a historical thing because normally one would call $H\colon\mathcal{I}\to Q$ maybe the Hilbert path and $H(\mathcal{I})$ the Hilbert curve.

Right now I am writing my bachelor thesis and I am unsure wether I should explain the term "Hilbert curve" as the function itself and not the image which would be just $[0,1]\times[0,1]$.

答:
In some contexts people are careful to distinguish between a function and its graph, while in others they aren't.  Moreover, even in the former circumstance, which of the two is known as the "curve" also varies according to context.  In topology a "curve" is typically a function $\gamma : I \to X$, whereas in geometry that function would be called a "parameterization" and its image would typically be called a curve.

The upshot is that you can typically count on definitions to be consistent within a particular paper or book, but not across mathematics in general.  This may not be an ideal state of affairs, but it's what we have to work with.

What this means for your talk is that you can call the two whatever you prefer but you should take care to explain the convention you're using if it's important to the talk.


问:
Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$

Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.

I tried solving this question through substitution, e.g. $a_1=\sqrt[n]{x_1},...,a_n=\sqrt[n]{x_n}$, hence $a_1...a_n=1$. Also for $k=1,2,...,n$ we have that $x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}$.

This is where I got stuck. My intuition tells me that I should be able to finish it off from here using AM-GM, however it isn't working out for me. Could you please explain to me how to finish the question off?

答:
The case $x_1=x_2=...=x_n=1$ is trivial, and for $n >2$, is the only case when equality holds. We prove this below.

Suppose the $x_i$s are not all $1$. Then, there exists $i,j \in \{1,2,..,n\}, i \neq j$, such that $x_i<1<x_j$. Replace the pair $(x_i  \ ,x_j)$ with $(x_i'\ ,x_j')$ , such that:

$$x_i'=1, \ x_j'=x_ix_j.$$
$x_i'$ and $x_j'$ have the same product as $x_i$ and $x_j$. Consider the sum

$$\dfrac{1}{n-1+x_i} + \dfrac{1}{n-1+x_j} = \dfrac{x_i+x_j+2(n-1)}{(n-1)^2 + (n-1)(x_i+x_j) + x_ix_j}.$$

It would be ideal if the above sum was strictly smaller than $\dfrac{1}{n-1+x_i'} + \dfrac{1}{n-1+x_j'} = \dfrac{1}{n} + \dfrac{1}{n-1+x_ix_j}.$ (By our replacement, we increase the number of $x_i$s which are equal to $1$, while strictly increasing the L.H.S. of the given inequality and preserving the given constraint. Eventually, all the $x_i$s will equal $1$, and the inequality becomes an equality.) Attempting to proceed along this line,

\begin{align}
& \dfrac{1}{n-1+x_i} + \dfrac{1}{n-1+x_j} < \dfrac{1}{n} + \dfrac{1}{n-1+x_ix_j} \\
& \iff \dfrac{1}{n-1+x_i} - \dfrac{1}{n} < \dfrac{1}{n-1+x_ix_j} - \dfrac{1}{n-1+x_j} \\
& \iff \dfrac{1-x_i}{n(n-1+x_i)} < \dfrac{x_j-x_ix_j}{(n-1+x_ix_j)(n-1+x_j)} \\
& \iff \dfrac{1}{n(n-1+x_i)} < \dfrac{x_j}{(n-1+x_ix_j)(n-1+x_j)} \\
& \iff (n-1+x_ix_j)(n-1+x_j) < nx_j(n-1+x_i) \\
& \iff x_ix_j^2 - x_ix_j + nx_j - x_j + n^2-2n+1 - n^2x_j + nx_j < 0 \\
& \iff n^2(1-x_j) - x_ix_j(1-x_j) - 2n(1-x_j) + (1-x_j) < 0 \\
& \iff n^2-x_ix_j -2n + 1 >0 \\
& \iff x_ix_j < (n-1)^2.
\end{align}
Unfortunately, it is possible to have $x_ix_j \geq (n-1)^2$. Fortunately, this issue can be circumvented. W.L.O.G. let $x_1 \leq x_2 ... \leq x_n , x_1 < 1 < x_n$. There are $2$ cases:

Case $1$:
$x_1x_n \geq (n-1)^2 \Rightarrow \dfrac{x_1+x_n+2(n-1)}{(n-1)^2 + (n-1)(x_1+x_n) + x_1x_n } \leq \dfrac{x_1+x_n+2(n-1)}{2(n-1)^2 + (n-1)(x_1+x_n)} = \dfrac{1}{n-1}.$ Since $\dfrac{1}{n-1+x_n}$ is positive, we conclude that $\dfrac{1}{n-1+x_1} < \dfrac{1}{n-1}$. But the largest term in the sum $ \displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i}$ is $\dfrac{1}{n-1+x_1}$. Thus,
$$\displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i} \leq \dfrac{1}{n-1} + \displaystyle \sum_{i=2}^{n-1} \dfrac{1}{n-1+x_i} < \dfrac{1}{n-1} + \dfrac{n-2}{n-1} =1. $$

Case $2$: $x_1x_n < (n-1)^2 \Rightarrow \dfrac{1}{n-1+x_1} + \dfrac{1}{n-1+x_n} < \dfrac{1}{n} + \dfrac{1}{n-1+x_1x_n}$. Letting $x_1'=1, x_n'=x_1x_n$ and $x_j'=x_j \ \forall \ j \notin \{1,n\},$ we have $\displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i} < \displaystyle  \sum_{i=1}^{n} \dfrac{1}{n-1+x_i'}.$ And we are done by a repeated application of this "smoothing" procedure! (At each stage,  W.L.O.G. re-order the  variables $x_i$s from smallest to largest, since the previous round of "smoothing" disrupted the original order.) If, at any stage, the product of the smallest term with the largest term is greater than or equal to $(n-1)^2$, we are done by Case $1$. Otherwise, we continually increase the L.H.S. and eventually end up with all the $x_is$ equal to $1$, giving the equality case L.H.S. $=1$.


问:
Let $A\in \mathbb{R}^{2\times 2}$, $A^2=0$. Prove that $\det(tI-A)=t^2$ for any scalar $t$ (Without the use of Cayley Hamilton theorem).
I tried alot of stuff and went far as taking a general $A$ matrix and trying to apply it and I feel that I'm missing something and I'm not in the right direction.

 Note that I would be happy to get all types of answers but this question is intended to be solved using determinant properties.
 Any help is appreciated, thanks in advance.

EDIT: I have not learnt Cayley-Hamliton theorem yet, and looking for a solution without it.

答:
If $A^2=\pmatrix{a&b\\c&d}\pmatrix{a&b\\c&d}=\pmatrix{a^2+bc&b(a+d)\\c(a+d)&bc+d^2}=\pmatrix{0&0\\0&0},$

then $b=0$, in which case $a=d=0$, or $b\ne0$, in which case $a=-d$ and $c=\dfrac{-a^2}b$,

so $A=\pmatrix{0&0\\c&0}$ or $A=\pmatrix{a&b\\\dfrac{-a^2}b&-a}$.

Can you take it from here and compute $\det(tI-A)$ now?


问:
Regarding the values that n and r can take in $\binom{n}{r}$
I am thinking what values can n and r take in the following notation,
$\binom{n}{r} = {n!\over r!(n-r)!}$ ?

I mean which one among the following is correct and why?

$0≤r≤n$ where $r,n\in Z_0^+---------(1)$

$0≤r≤n$ where $r,n\in Z^+---------(2)$

$0<r≤n$ where $r,n\in Z_0^+---------(3)$

$0<r≤n$ where $r,n\in Z^+---------(4)$

Furthermore, can you kindly say if $\binom{0}{0}$ exists. If it does exist, I can not find it holding a physical meaning even though we can obtain 1 by simplifying.

I would like to know which is correct for nPr from the above 4 connections I have written. 1, 2, 3 or 4 and why?

Thanks in advance.

答:
From a combinatorial definition, one has that $\binom{n}{r}$ is the answer to the question of "How many $r$-element subsets are there of an $n$-element set?"

Under this definition and interpretation, one has that the question is invalid and makes no sense when $n$ is not a non-negative integer (There do not exist any sets with $-2$ number of elements of $\pi$ number of elements, etc...) and similarly makes no sense when $r$ is not a non-negative integer (similarly there exist no subsets of size $-2$ or size $\pi$ etc...)

As such, we either treat $\binom{n}{r}$ as being equal to zero or being undefined in the case that $n$ or $r$ is not a non-negative integer.

The counting question is perfectly valid from a syntactical standpoint when the values are  both non-negative integers.  Some of these will be easy to answer immediately as "there are none" for example the question of "How many $5$ element subsets are there of a $2$-element set?"  There are no five-element subsets of a two-element set so the answer here is zero.  The notation $\binom{2}{5}$ is perfectly valid in this case as notating the answer to this question and equals zero.

As to the question of $\binom{0}{0}$, this is also a perfectly valid question!  The question being "How many zero-element subsets of a zero-element set exist?"  The answer to this question... is one!  There is in fact a zero-element subset of a zero-element set... Recall that $\emptyset\subseteq \emptyset$ is true.  We have then $\binom{0}{0}=1$

We have for this interpretation the following results:

$\binom{n}{r}$ is

$0$ or undefined (depending on preference) whenever $n$ or $r$ is not a non-negative integer

$1$ when $r=0$ or $r=n$ and $n$ is a non-negative integer (including when $n=0$)

$0$ when $r>n$ and both $n$ and $r$ are non-negative integers

a positive integer when $0\leq r\leq n$ and $r$ and $n$ are both non-negative integers

There also exists a purely algebraic interpretation.  $\binom{n}{r}$ is the coefficient of the $X^r$ term in the expansion of $(1+X)^n$.  We have the binomial theorem, that for non-negative integer $n$ we have $(1+X)^n = \sum\limits_{r=0}^n \binom{n}{r}X^r$.  This is further generalized to complex $n$.  The generalized binomial theorem is for $\alpha\in\Bbb C$ and $|X|<1$ we have

$$(1+X)^\alpha = \sum\limits_{r=0}^\infty \binom{\alpha}{r}X^r$$

where here $\binom{\alpha}{r} = \dfrac{\alpha\frac{r}{~}}{r!}$ where $\alpha\frac{r}{~}=\underbrace{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-r+1)}_{r~\text{terms in the product}}$ is the falling factorial.  Note that here we do not require anything of $\alpha$ but we do still require $r$ to be a non-negative integer.

The values now of things such as $\binom{2}{5}$ and $\binom{\pi}{2}$ which made little sense before now do make sense.  $\binom{2}{5}=\dfrac{2\cdot 1\cdot 0\cdot (-1)\cdot (-2)}{5!}=0$ and $\binom{\pi}{2}=\dfrac{\pi(\pi-1)}{2!}$

The generalized binomial coefficient gives the same results for all non-negative integers.  You still have the same identities such as Paschal's identity.  There are just now many more cases which give non-zero defined results.


问:
Best strategy to optimize probability of winning in card drawing game with opponent
I am curious about the solution to a probability question that was asked in a trading interview:

You and your opponent choose a suit (of a standard 52 cards deck). Then one card after another is drawn (without replacement) until either your suit or your opponents suit have appeared 5 times (not necessarily in a row). You win if your suit is the one which was drawn 5 times first. Now, your opponent lets you decide if you want to choose your suit before the game starts, after 1, or after 2 cards are revealed and he will choose his suit afterwards. Which strategy gives you the best chances to win?

So as an example, I decide to choose after the first card, it comes hearts, I choose hearts, he chooses any of the other suits and from that point I will only need 4 more hearts whereas he still needs 5 of his suit.

This makes it obvious that choosing before the game starts is not optimal, as you get an advantage by choosing the suit of the first card.

But I am not sure how to determine whether choosing after the first or after the second card is better. In the scenario where I choose after the second there are two possibilities: Either the first 2 cards have the same suit, then I will only need 3 more or they have a different suit. In that case I choose one of the two suits and my opponent chooses the other and our chances will be equal again.

To summarize, my chances of winning if I choose after the second card are
$$ P(\textrm{first 2 cards have same suit}) P(\textrm{3 out of 11 before 5 out of 13}) \\
+ \frac{1}{2}P(\textrm{first 2 cards have different suit}) \\
= \frac{12}{51}P(\textrm{3 out of 11 before 5 out of 13}) + \frac{1}{2}\frac{39}{51}
$$
And if I choose after the first card my chances are simply
$$
P(\textrm{4 out of 12 before 5 out of 13})
$$

But in both cases I have no idea how to come up with solutions for the missing probabilities, especially because the game is played without replacement and I can not use a binomial distribution approach as it were possible if the game was for example played with coin tosses and you can choose head or tails.

答:
Let $f(i, j)$ be probability of you winning if there already are $i$ of your cards and $j$ of your opponents ($0 \leq i, j \leq 5$, at most one of $i$ and $j$ is equal to $5$). $f(5, j) = 1$ and $f(i, 5) = 0$ by definition, also $f(i, i) = \frac{1}{2}$ by symmetry.

Probability of winning if you choose before any card draw is $f(0, 0)$, if you choose after the first - $f(1, 0)$ and after the second - $p \cdot f(1, 1) + (1 - p) \cdot f(2, 0)$, where $p = \frac{39}{51}$ is probability of first two cards having different suits.

Your question is if $f(1, 0) > \frac{39}{102} + \frac{24}{102} \cdot f(2, 0)$.

It's not too hard to calculate using $f(i, j) = \frac{13 - i}{26 - i - j} \cdot f(i + 1, j) + \frac{13 - j}{26 - i - j} \cdot f(i, j + 1)$. In theory, it can be calculated by hands if needed, but I'll cheat a bit and just print the table:

$$\begin{array}{|c|c|c|c|c|c|}
\hline j\backslash i & 4 & 3 & 2 & 1 & 0 \\
\hline 4 & \frac{1}{2} & \frac{5}{19} & \frac{11}{76} & \frac{11}{133} & \frac{13}{266} \\
\hline 3 & \frac{14}{19} & \frac{1}{2} & \frac{44}{133} & \frac{29}{133} & \frac{442}{3059} \\
\hline 2 & \frac{65}{76} & \frac{89}{133} & \frac{1}{2} & \frac{1117}{3059} & \frac{923}{3496} \\
\hline 1 & \frac{122}{133} & \frac{104}{133} & \frac{1942}{3059} & \frac{1}{2} & \frac{169}{437} \\
\hline 0 & \frac{253}{266} & \frac{2617}{3059} & \frac{2573}{3496} & \frac{268}{437} & \frac{1}{2} \\ \hline
\end{array}$$

(we actually need just left-bottom half of the table)

So $f(1, 0) = \frac{268}{437} \approx 0.61$, $f(2, 0) = \frac{2573}{3496}$ and $\frac{39}{102} + \frac{24}{102} \cdot f(2, 0) \approx 0.56$.


问:
Clarify if the topological sum on $\Bbb R^n\times\Bbb R^n$ is equal to the usual topology.

For a collection $\mathscr{X}=\{(X_j ,\mathcal{T}_j):j\in J\}$ of topological spaces let be $X:=\bigcup_{j\in J}X_j$ and for each $j\in J$ let be $\iota_j:X_j\rightarrow X$ the inclusion map defined through the condition
$$
\iota_j(x)=x
$$
for any $x\in X$ and thus let be $\mathcal{F}=\{\iota_j:j\in J\}$ the collection of such functions. So we say that $X$ is the topological sum of the collection $\mathscr X$ if it is equipped with the final topology induced by the collection $\mathscr F$ as above defined.

So now I ask to clarify if it is true that the usual product toplogy on $\Bbb R^n\times\Bbb R^n$ is equal to the topological sum induced by the collection $\mathscr{X}:=\Big\{\{x\}\times\Bbb R^n:x\in\Bbb R^n\Big\}$ where $\{x\}\times\Bbb R^n$ is obviously equipped with the subspace topology induced by the product topology of $\Bbb R^n\times\Bbb R^n$. So could someone help me, please?

答: Obviously not, as the topological sum is disconnected and the product topology on $\Bbb R^n \times \Bbb R^n$ is not (it's just $\Bbb R^{2n}$ so connected as can be).


问:
If $F$ is a field and $\phi:F[x] \to F[x]$ is a ring automorphism,with the property that $\forall \lambda \in F:\phi (\lambda)=\lambda$
If $F$ is a field and $\phi:F[x] \to F[x]$ is a ring automorphism,with the property that $\forall \lambda \in F:\phi (\lambda)=\lambda$,then prove that exist $a,b \in F:\phi(x)=ax+b$.

答:
Hint

Let $\phi(x) = p(x)$. $\phi$ is completely defined by $p$ and the condition $\phi(\lambda) = \lambda$ for all $\lambda \in F$.

Prove that if the degree of $p$ is greater than one, then $\phi$ can't be onto. And that if the degree of $p$ is equal to zero, then $\phi$ isn't one-to-one.

Therefore $\phi(x) = ax+b$ which leads to a ring automorphism.


问:
Is the Fourier transform akin to a basis change?
So the Hilbert space $L^2(R,dx)$ is separable and hence admits a countable (Schauder) basis.  However, the Fourier transform rewrites the function as an integral over an uncountable set of functions.

In QM we also thing if position eigenstates $|x\rangle$ as a basis, however, this is also uncountable.

I guess, are Fourier transforms really a new basis, or does the fact that all basis are of the same size break down in infinte vector spaces?

答:
At the intuitive level, it is indeed good to think of this as a change of basis. However, one has to be careful when implementing this intuition in a rigorous treatment (e.g., with Gelfand triples, generalized eigenvectors, the spectral theorem for unbounded operators and whatnot). Definitions like that of (Hamel) bases and theorems about them like, all having the same cardinality, are not to be taken too seriously in this context.

Note that when one chooses a basis in a finite-dimensional vector space $V$, this automatically endows tensor powers $V^{\otimes n}$. Change of bases maps define (in a functorial way) similar maps for tensor powers. This is also true of the Fourier transform and is used a lot in the quantum field theory literature where one often switches from $x$-space representation to momentum space representation for Feynman diagrams and correlation functions. One can also switch to discrete bases, e.g., the basis of Hermite functions. One area where this is particularly useful is group field theory as well as noncommutative field theory (see for example the work of Grosse and Wulkenhaar which reformulates NCQFT in the so-called matrix basis).


问:
Proof that the interior of any union of closed sets with empty interior in a compact Hausdorff space is empty
The question is pretty much in the title, I need to show that given $X$ is a compact Hausdorff space and $\left\{ A_n\right\}_{n=1}^\infty$ is a collection of closed subsets of $X$ each with empty interior, then $\bigcup_{n\in\mathbb{N}}A_n$ has empty interior as well.

Out of curiosity only I was wondering whether the fact the collection is countable is necessary for the claim or if it's true for any such collection of subsets.

Help would be appreciated :)

答:
A space with the property that the union of countably many nowhere dense closed sets has empty interior is called a Baire space.

It's easier if you dualize and show that the intersection of countably many open dense sets $D_n$ is dense. For if $A$ is closed and has empty interior, its complement $D$ is open and dense. Start with an open set $U$ and intersect it with $D_1$. Take a point $x_1$ in the intersection. There is an open neighborhood $U_1$ with compact closure $C_1\subset U\cap D_1$. Then intersect $U_1$ with $D_2$ and repeat this process for a point $x_2$ in $U_1\cap D_2$.


问:
Is there an accepted term for the opposite of mode in statistics?
In descriptive statistics, there are terms for all sorts of things.  The mean, median, and mode for a set of data are each three very frequently thrown around examples.

The mode in particular is defined as the result that has occurred the most frequently.

Is there a name for the opposite of this?  Is there a name for the result which has occurred the least frequently (but still a nonzero number of times)?

If I were to guess at a name or come up with one myself, perhaps a name like "anti-mode" might be appropriate, but I was wondering if there was an already accepted name.

答:
I prefer the term

least frequent score

which is from @Armaud's answer. That term is easily understood by anyone without needing to know any special definitions.

In the special case of a bimodal distribution, you could refer to the 'peaks' (the two modes) and 'trough' (the lowest point). Example here:

the extent of the difference between the peaks and trough of the (bimodal) distribution


问:
There is no endomorphism $\phi : \mathbb R^5 \to \mathbb R^5$ with $\ker \phi = \text{Im}\phi$
Is it true, that we haven't any linear mappings (Let it be called $\phi$) between $\mathbb{R}^n \to \mathbb{R}^n$, where $n = 5$ and Ker $\ \phi$ = Im$\ \phi$

Where:

Ker $\ \phi = \{v \  | \ \phi(v) = 0, v \in \mathbb{R}^n\}$

Im$\ \phi = \{\phi(v) \  | \ v \in \mathbb{R}^n\}$

答:
According to  rank-nullity theorem this would lead to the contradiction

$$\dim \ker \phi + \dim \text{Im} \phi = 2\dim \ker \phi = 5$$


问:
How to draw a closed curve in cartesian coordinates based on its area
I've posted a question before asking how I would derive the length of a closed curve that I found in a paper. I am returning to now ask how I could draw my shape in Cartesian coordinates. From my first post about this paper we have that the authors use a particular case o the Gielis formula where, in polar coordinates, $\rho$ is defined as such:

$$
\begin{align}
\rho = r_m(\varphi)
&= \frac{1}{2}\lim_{n\to+\infty}\left(|\cos{\frac{m\varphi}{4}|^n}+|\sin{\frac{m\varphi}{4}}|^n\right)^{-\frac{1}{n}}\\
&=\frac{1}{2}\min{\left(\left|\sec{\frac{m\varphi}{4}}\right|,\left|\csc{\frac{m\varphi}{4}}\right|\right)}\label{1}
\tag{1}\end{align}$$

where m is a positive integer that defines the number of pseudovertices of a closed curve $C_m$. They then show that from the closed curve equations we have that

$$A_m = \frac{1}{2}\int_{0}^{2\pi} r_m^2(\varphi) \,d\varphi = 1 \tag{2}$$

$$
\begin{align}
l_m &=  \int_{0}^{2\pi} \sqrt{r_m^2(\varphi)+\left[\frac{dr_m(\varphi)}{d\varphi}\right]^2} \,d\varphi \\
&= \sqrt{8+\frac{m^2}{2}}+4\left[F\left(\frac{\pi}{4}\bigg|1-\frac{m^2}{16}\right)-E\left(\frac{\pi}{4}\bigg|1-\frac{m^2}{16}\right)\right] \tag{3}
\end{align}$$

What I did then was stard from a square and define the area I want to have on my practical implementation (which is why I need to know how to draw the curves with respect to m). My square 2mm x 2mm, so my $A_m = 4~mm^2$. So now how would I proceed to create my geometry on a Cartesian framework?

I am sorry if this sounds a bit stupid, I've never attempted anything remotely close to this. Thank you in advance

答:
Since $x = \rho \cos (\varphi)$ and $y = \rho \sin (\varphi)$ we can just plug your expression

$$\rho(\varphi) = \frac{1}{2} \min\left(\left|\sec \frac{m \varphi}{4}\right|, \left|\csc \frac{m \varphi}{4}\right|\right)$$

in, giving the Cartesian parametrization

$$x(t) = \frac{1}{2} \min\left(\left|\sec \frac{m t}{4}\right|, \left|\csc \frac{m t}{4}\right|\right) \cos(t)$$
$$y(t) = \frac{1}{2} \min\left(\left|\sec \frac{m t}{4}\right|, \left|\csc \frac{m t}{4}\right|\right) \sin(t)$$

Update: in MATLAB (actually, GNU Octave),

```<code>m = 5;
t = 0:0.01:2*pi;
x = min(abs(sec(m * t / 4)), abs(csc(m * t / 4))) .* cos(t) / 2;
y = min(abs(sec(m * t / 4)), abs(csc(m * t / 4))) .* sin(t) / 2;
plot(x, y)
</code>```

Body:
Since $x = \rho \cos (\varphi)$ and $y = \rho \sin (\varphi)$ we can just plug your expression

$$\rho(\varphi) = \frac{1}{2} \min\left(\left|\sec \frac{m \varphi}{4}\right|, \left|\csc \frac{m \varphi}{4}\right|\right)$$

in, giving the Cartesian parametrization

$$x(t) = \frac{1}{2} \min\left(\left|\sec \frac{m t}{4}\right|, \left|\csc \frac{m t}{4}\right|\right) \cos(t)$$
$$y(t) = \frac{1}{2} \min\left(\left|\sec \frac{m t}{4}\right|, \left|\csc \frac{m t}{4}\right|\right) \sin(t)$$

Update: in MATLAB (actually, GNU Octave),

m = 5;
t = 0:0.01:2*pi;
x = min(abs(sec(m * t / 4)), abs(csc(m * t / 4))) .* cos(t) / 2;
y = min(abs(sec(m * t / 4)), abs(csc(m * t / 4))) .* sin(t) / 2;
plot(x, y)