问:
Computing $\int_0^\infty \exp\left( -\frac{x^2}{A} -\frac{B}{x} -Cx \right){\rm d}x$
Exactly as the title suggests, I want to compute or approximate the integral of the form
$$\int_0^\infty \exp\left( -\frac{x^2}{A} -\frac{B}{x} -Cx \right){\rm d}x$$
where $A,B,C>0.$ I got the question from a friend who asked me how to evaluate it.

I was told that an asymptotic formula or a good approximation for large values of $t$ (if we integrate from $0$ to $t$ instead) is good enough, as long as it's written in terms of $A,B,C.$ I feel like contour integration might do the trick but I'm a bit too rusty on that. Thank you in advance.

答:
If we consider
$$f(a,b,c)=\int_0^\infty \exp\left( -a x^2 -\frac{b}{x} -c x \right)\,dx$$
(with $a \geq 0$, $b\geq 0$, $c\geq 0$), exploring all possibilities
$$f(a,0,0)=\frac{\sqrt{\pi }}{2 \sqrt{a}}$$
$$f(0,b,0)\qquad\qquad \text{does not converge}$$
$$f(0,0,c)=\frac 1c$$
$$f(a,b,0)=\frac b{4\sqrt \pi}G_{0,3}^{3,0}\left(\frac{a b^2}{4}|
\begin{array}{c}
 -\frac{1}{2},0,0
\end{array}
\right)$$ where appears Meijer G function
$$f(a,0,c)=\frac{\sqrt{\pi }}{2 \sqrt{a}}\,e^{\frac{c^2}{4 a}}\,\text{erfc}\left(\frac{c}{2 \sqrt{a}}\right)$$
where appears the complementary error function
$$f(0,b,c)=2 \sqrt{\frac{b}{c}} K_1\left(2 \sqrt{b c}\right)$$
where appears the modified Bessel function of the second kind.

Unfortunately, for the general case, none of the CAS I used gives an answer.

$$f(a,b,c)= \text{   } \large ???$$

Edit

If $a$ or $c$ are "small", we can expand the integrand as Taylor series and obtain results for $f(a,b,c)$. For $a$, it will be a linear combination of Bessel functions; for $c$ it will be a linear combination of Meijer G functions.


问:
Bounds on roots of polynomials
What bound is there on the roots of a given polynomial, in terms of its degree and coefficients?

Consider the polynomial $p(x) = 3x^7 – 5x^3 + 42$. Would you not agree, without doing any calculation, that one million ($10^6$) cannot be a root? It just wouldn’t be in accord with the smallness of the coefficients and the well-behavedness of polynomials. And yet I don’t recall ever having encountered anything in the literature that gave a bound on the absolute value of the roots of a polynomial in terms of the degree and coefficients of the polynomial, but I’m pretty sure such must exist, and that I simply missed it, and so I’m tagging this a a reference-request.

By the way, during my post of a question, every time, it seems, there are stray graphics on the screen, as if from someone else's question or answer. Is this happening to anyone else?

答:
If $a_n x^n + \dots + a_1 x + a_0 = 0$ we get the following useful bound by Lagrange:

$$|x| \le 2\max\left\{
\left(\frac{|a_{i}|}{|a_n|}\right)^{\frac1{n-i}} : 0 \le i < n
\right\}.$$

The following proof by Chee-Keng Yap is simple:
Let $\beta=\max\left\{\left(|a_{i}|/|a_n|\right)^{1/(n-i)} : 0 \le i < n\right\}$.
The lemma is trivial if $|x| \le \beta$; so we focus on $|x| > \beta$.

Since $a_n x^n = -(a_{n-1}x^{n-1}+\dots+a_0)$ we have
$|a_n| |x|^n \le |a_{n-1}||x|^{n-1}+\dots+|a_0|$ by the triangle inequality.
Hence
$$
1
\le \frac{1}{|a_n||x|^n}\sum_{i=1}^{n}
|a_{n-i}||x^{n-i}|
= \sum_{i=1}^{n}{\frac{|a_{n-i}|}{|a_n||x|^i}}
\le \sum_{i=1}^{n}{\frac{\beta^{i}}{|x|^{i}}}\le \frac{\beta/|x|}{1-\beta/|x|},
$$
where we bounded the sum by the infinite geometric sum, which is finite since we assumed $\beta < |x|$.
Now, $1\le \frac{\beta/|x|}{1-\beta/|x|}$ implies $1\le 2\beta/|x|$, which is what we want.

In your case this says
$$
|x|
\le 2 \max\{\sqrt[4]{5/3}, \sqrt[7]{42/3}\}
\approx 2 \max\{1.13622, 1.45792\}
< 3.
$$
So not quite as strong as Qiaochu's bound in this case, but in other cases it can be much better.

You can also try some more bounds from this Wikipedia article.

ContentLicense: CC BY-SA 3.0

Title: Bounds on roots of polynomials

ContentLicense: CC BY-SA 4.0


问:
Prove that exists an inner product such that $\{1+x,x+x^2,x^2+x^3,x^3+x^4,x^4-1\}$ is orthogonal
I've been asked to prove or disprove that exists an inner product in $\mathbb{R}_5[x]$ such that

$A = \{1+x,x+x^2,x^2+x^3,x^3+x^4,x^4-1\}$

is orthogonal.

How does one approach this?

答: Hint: $v_5 = v_4 - v_3 + v_2 - v_1 \implies$ $\langle v_5,v_j \rangle = \pm 1$ for $j=1,\dots4$.


问:
When exactly is the dual of $L^1$ isomorphic to $L^\infty$ via the natural map?
The dual space to the Banach space $L^1(\mu)$ for a sigma-finite measure $\mu$ is $L^\infty(\mu)$, given by the correspondence

$\phi \in L^\infty(\mu) \mapsto I_\phi$, where $I_\phi(f) = \int f \cdot \phi \,d\mu$ for $f \in L^1(\mu)$.

Now, we consider this for more general measures $\mu$. If $\mu$ is not semifinite, then there is a measurable subset $A$ with $\mu(A) = \infty$ but $\mu(B) = 0$ for every subset $B$ of $A$ with $\mu(B) < \infty$. Any $\phi$ supported on $A$ will then map to the zero functional, so the correspondence is not one-to-one.

If $\mu$ is semifinite, the correspondence is one-to-one, but is it onto?

Using the Radon-Nikodym Thorem, any member of $L^1(B)^*$ for $\mu(B) < \infty$ can be represented by an $L^\infty(B)$ function class $[\phi_B]$. Unlike in the $\sigma$-finite measure case, however, I do not see an easy way to form a "patchwork" $\phi$ with $\phi|_B = \phi_B$ a.e.$[\mu]$. I suspect this may have an easy answer, but it eludes me. Or perhaps there is a well-known counterexample?

答:
Let $X$ be an uncountable set, equipped with the $\sigma$-algebra $\mathcal{F}$ consisting of all countable sets and their complements, or any other $\sigma$-algebra that contains all singletons but is not $2^X$. Let $\mu$ be counting measure, which is semifinite.  Note that any function in $L^1(\mu)$ is supported on a countable set.  Let $E$ be a subset of $X$ which is not in $\mathcal{F}$, and for $f \in L^1(\mu)$ set $I(f) = \sum_{x \in E} f(x)$.  Clearly this is a continuous linear functional on $L^1(\mu)$ (with norm 1).  However, if $I(f) = \int fg\,d\mu$ it's easy to see we would have to have $g = 1_E$, but this is not in $L^\infty(\mu)$ because $E$ is not measurable.

But $\sigma$-finiteness is not necessary, either: repeat this example with $\mathcal{F} = 2^X$.


问:
The minimum number of spatial dimensions needed to understand pi (and other constants)
I wasn't too sure if I should have posted this on the mathematics or physics stack exchange, however I figured perhaps since this is in reference to fundamental constants it would be better placed here.

I have recently been thinking of fundamental constants. Specifically, the minimum number of spatial dimensions you would need to understand them. If you take pi for example, I thought at first the minimum number of spatial dimensions you would need to understand (and therefore be able to discover) pi was two. My reasoning was that, in one dimension, circles don't exist and therefore there would be no reason to hold the irrational number pi in elevated regard. However, I figured that you can have oscillations in one dimension, I.e a point moving up and down from position $-y$ to $y$, and where oscillations exist you could inevitably discover the functions sin and cos and thus the number pi. So my question is this - are there any fundamental constants that require a number of spatial dimensions > 1 to understand? Or do all fundamental constants exist in all spatial dimensions?

Apologies that this is both vague and abstract. Ι also wasn't quite sure if the tags I used were relevant. I would love to hear any thoughts on the matter.

答:
Depends really on what you mean by "understand". Many fundamental constants can be (in one way) defined in a 1-dimensional sense, in terms of their infinite sums. Sure this might not be the most intuitive or the easiest way to introduce said constants; but it can be defined this way nonetheless.

For example,
$$e = \lim_{n\to \infty} \left(1+ \frac{1}{n}\right)^n = \sum_{i=1}^\infty \frac{1}{n!}$$
$$\pi = 4\sum_{i=1}^\infty \frac{(-1)^{k+1}}{2k-1}$$
and so on...

Even
$$i^2 = -1$$
or the quaternion relations can all be described (and dare I say, understood) via summations in the real line.

In this vague sense, every fundamental constant needs only one dimension to be defined, and therefore understood via the definition. But by all means, a flavor of them exist in all spatial dimensions. Again, here the word dimension is very vague; and I am sort of interpreting it as the dimensionality of Euclidian/Vector spaces.


问:
Gini index different expressions
Let $X$ be a non-negative random variable with positive finite expectation.
The Lorenzcurve $L_X$ is defined by
$$ L_X(u) = \frac{\int_0^u F_X^{-1}(y) dy}{E(X)}, \quad 0 \leq u \leq 1,$$
where
$$
F_X^{-1}(y) = \begin{cases}
\sup\{x : F_X(x) \leq y\}, \quad 0 \leq y < 1 \\
\sup\{x : F_X(x) \leq y\}, \quad y = 1
\end{cases}
$$
is the right continuous inverse distribution function of the random variable X.
The Gini-Index $G$ is defined by $G = 1 - 2\int_0^1 L_X(u) du$.

I've stepped across the following equation
$$ G = \frac{E(X_{2:2}) - E( X_{1:2} )}{E(X_{2:2}) + E(X_{1:2})}, $$
where $X_{1:2} < X_{2:2}$ are the order statistics of a random sample of size 2 drawn from X.

Is this a true statement for any random variable X? If so, why?

In addition, I have calculated the Gini-Index in the case of $X \sim \exp(\lambda)$ for both expressions above and came to the same result, but I do fail to see any justification for the general case.

答:
First, notice that
$$
\max(x,y)+\min(x,y)=x+y,
$$
and
$$
\max(x,y)-\min(x,y)=|x-y|=(x+y)-2\min(x,y).
$$
Then, if $X$ and $Y$ are i.i.d. with cdf $F$,
\begin{align}
&\frac{\mathsf{E}\max(X,Y)-\mathsf{E}\min(X,Y)}{\mathsf{E}\max(X,Y)+\mathsf{E}\min(X,Y)}=1-\frac{\mathsf{E}\min(X,Y)}{\mathsf{E}X}\\
&\qquad=1-\frac{1}{\mathsf{E}X}\int_0^\infty[1-F(x)]^2\,dx.
\end{align}
It remains to show that the RHS equals the first definition.

Title: Gini index different expressions


问:
Integrating $\sqrt\frac{x}{{x+a}}$?
How can I integrate this expression?
$$\sqrt\frac{x}{{x+a}}$$

I substituted $x=a\tan^2t$. After substituting, I reach
$$\int \sec x\tan^2x$$
Is this way correct?

Thanks in advance.

答:
One method is to just substitute
$$t=\sqrt{\frac{x}{x+a}}$$
that is $$ x = \frac{at^2}{t^2-1}$$
$$ dx = \frac{-2at}{(t^2-1)^2} dt $$
We get
$$ \int \sqrt{\frac{x}{x+a}} dx = \int \frac{-2at^2}{(t^2-1)^2} dt$$
This is an integral of a rational function and can be solved using standard methods.

Another method is to use substitution given by the condition
$$ \sqrt{x^2+ax} = u-x$$
which can be solved to get
$$ x = \frac{u^2}{2u+a}$$
$$ \sqrt{x^2+ax} = \frac{u(u+a)}{2u+a}$$
$$ dx = \frac{2u(u+a)}{(2u+a)^2}dt $$
We get then
$$ \int \sqrt{\frac{x}{x+a}} dx  = \int \frac{x}{\sqrt{x^2+ax}} dx = \int \frac{2u^2}{(2u+a)^2}du $$
which also can be solved using standard methods.

These substitutions are special cases of so-called Euler subsitutions.

Title: Integrating $\sqrt\frac{x}{{x+a}}$?


问:
Quotient group of normal subgroups is cyclic if quotient group of intersection is cyclic
Let $M$, $N$ be normal subgroups of a group $G$.

Prove that if $G/M\cap N$ is cyclic, then $G/M$ and $G/N$ are cyclic. Give a counter example to show that the converse is not always true.

I proved until now that if $G/M\cap N$ is abelian, then $G/M$ and $G/N$ are abelian, I don't know if that helps a lot but it's the only thing I could come up with.

答: For a counterexample, I think the following works: $$G=\langle a,b\mid a^2=b^2=[a,b]=1\rangle$$ Take $M=\langle a\rangle$ and $N=\langle b\rangle$.</p>


问:
Understanding Zhang's result of bounded prime gaps
Here is a link on the internet:
https://www.dropbox.com/s/su3uak2a057yrqv/YitangZhang.pdf

Can someone teach me how to use trivial estimation to reach (6.1) on page 24?
Namely, how to impose $(d,P_0)<D_1$ and replace $\theta$ by $\Lambda$ with acceptable O-term?

Thanks a lot!

答:
I have asked this question in Mathoverflow and get an answer:

https://mathoverflow.net/questions/131825/a-technical-question-related-to-zhangs-result-of-bounded-prime-gaps


问:
Question about a statement involving system of linear equations and rank of matrix.
the statement is:

let $A\in \mathbb{R}^{m\times n}$,   if for every $c\in \mathbb{R}^m$ there exists a solution for $Ax=c$, then $\operatorname{rank}(A)=m$.

Now I can understand that this is a true statement, since if $\operatorname{rank}(A)<m$ we can get no solution for the equation.

What I wanted to ask if the given information was for every $c\in \mathbb{R}^m$ there exists a unique solution, could I infer that the matrix is square? ($m = n$).

This is coming from the idea that if $Ax=c\:$ has a unique solution, then $\operatorname{rank}(A) = n$, so if it has a solution for every $c$ that means $\operatorname{rank}(A) = m $ and so $m=n$.

I would appreciate any feedback and would love to know if there's more interesting stuff to infer from the given information.

答:
If the rank m>n, the system of equations is overdetermined, there are too many variables relative to the number of equations, and there are no solutions.

If the rank m<n, the system of equations is underdetermined (there are too few non-zero variables compared to the number of equations), and there are an infinite number of solutions.

There is a unique solution iff the rank m=n.


问:
Base and height of a triangle is constant, what affects how many same squares I can fit inside it?
I've been wondering this for a while now.

If the base and height of a triangle is constant, what variables affect how many same squares I can fit inside a triangle without them overlapping each other?

I think the highest amount of squares can be fit if the triangle is a right triangle and the least if the triangle is obtuse.

But if the triangle is acute, does the amount of same squares I can fit inside it get affected by the angles of the triangle or subsequently its shape?

I have tried illustrating this but I couldn't draw precise lines, and I don't really need a formal proof if there even is any, so an illustration alone would be more than adequate and appreciated.

答: You can put just as many squares in an acute triangle as in a right triangle.  The width of the triangle a given distance above the base is a constant once you have the base and altitude.  In the triangles below, the base is $3$ and the width one unit above the base is $\frac {12}5$.  In the bottom row you can fit the width of the triangle one unit above the base rounded down, so you can fit $\lfloor \frac {12}5 \rfloor=2$ triangles.  In the next row you can fit the width of the triangle two units above the base rounded down.  Here that is $\frac 95$, so you can fit $1$.  In the third row you can fit $\frac 65$ rounded down, which is again $1$.  This works for all acute and right triangles.  The reason it fails for obtuse triangles is that the width one unit above is not over part of the base.  You have to take the overlap and round it down, which may give a smaller result.


问:
bound of the function $f(x)$
I am trying to get an good estimate of the function $f(x) = \frac{\cosh(ax)}{\sinh^2(bx)}$ when $x>0$ and $a, b$ are positive constant with $a\geq b.$ I formulated $f(x)\leq \frac{a}{(bx)^2}$. I have tried graphically and have seen that not for every $a,b$ satisfying $a\geq b$, the bound hold. Supposing that $a> > b$, then the bound is false. Any suggestion how to get the estimate or any improved estimate of the function $f(x)$. Any help will be appreciated.

Update: I have tried forming the function $g(x)= bx - \sinh(bx)(\frac{cosh(ax)}{a})^{-1/2}$ and tried to show the function is decreasing and therefore $g(x)<g(0)$ and hence the result. But showing the function is decreasing with some condition on $a$ and $b$ is getting tuffer.

答:
If you use  series expansion followed by long division, you should get close to $x=0$

$$f(x) = \frac{\cosh(ax)}{\sinh^2(bx)}$$ $$f(x)=\frac{1}{b^2 x^2}+\left(\frac{a^2}{2 b^2}-\frac{1}{3}\right)+x^2
   \left(\frac{a^4}{24
   b^2}-\frac{a^2}{6}+\frac{b^2}{15}\right)+O\left(x^4\right)$$ but, stricto sensu, this is not a Taylor series.

Title: bound of the function $f(x)$


答:
If these vectors are linearly independent, then they can be extended to a basis*, which then determines a unique inner product for which the basis is orthonormal (the standard inner product in terms of coordinates with respect to this basis), and we will be done. (Since orthogonal families of nonzero vectors in an inner product space are always linearly independent, that condition is clearly necessary, so it is if and only if.)

So it remains to see whether these vectors are linearly independent. Since all $5$ are contained in the $5$ dimensional (sub-)space of polynomials of degree less than $5$, and moreover they all have zero evaluation at $x=-1$ (which is a nontrivial linear condition on that space), they lie in a $4$-dimensional subspace, so they must be linearly dependent. Therefore they cannot be orthogonal for any inner product.

Since finally it is the implication "linear dependence forbids being mutually orthogonal" for sets of nonzero vectors that solves this question, let me recall the argument proving that. One argument is by contradiction: if some vector $v$ equals a linear combination of the other vectors supposed to be orthogonal to $v$, then $v$ is orthogonal to itself, which is impossible for a nonzero vector for a (positive definite) inner product. Another argument is by contrapositive: if a family $(v_i)_{i\in I}$ of nonzero vectors is orthogonal for an inner product, then in any linear combination $v=\sum_{i\in I}c_iv_i$ one can recover the coefficients from $c_i=(v_i\mid v)/(v_i\mid v_i)$, and in particular $v=0$ implies $c_i=0$ for all $i$, proving linear independence of the family.

*It depends on what $\Bbb R_5[x]$ means; normally the notation means polynomials over $\Bbb R_5$, but since that does not designate any ring as far as I can tell, I'll assume it means polynomials in $\Bbb R[x]$ whose degree bears some relation to the number $5$. If the relation is "less than" then these vectors are already a the right number for a basis; if it is "less than or equal", then an extension would have been needed if they were linearly independent. These two relations seem to be the most plausible candidates for this unspecified notational convention.

Title: Prove that exists an inner product such that $\{1+x,x+x^2,x^2+x^3,x^3+x^4,x^4-1\}$ is orthogonal

Body:
I've been asked to prove or disprove that exists an inner product in $\mathbb{R}_5[x]$ such that

$A = \{1+x,x+x^2,x^2+x^3,x^3+x^4,x^4-1\}$

is orthogonal.

How does one approach this?


问:
Prove that $\mathbb{C}$ is a finite $\mathbb{R}$-algebra and that $\mathbb{Q}, \mathbb{R}, \mathbb{C}$ are not finite $\mathbb{Z}$-algebras
The third statement to prove would be "Prove that $\mathbb{R}, \mathbb{C}$ are not finite $\mathbb{Q}$-algebras"

I thought about proving those statements in the following way:

$\mathbb{C}$ is a finite $\mathbb{R}$-algebra

The homomorphism we use is h : $\mathbb{R}$ -> $\mathbb{R}$ and h = $id_{\mathbb{R}}$, and once we have h($\mathbb{R}$) (which is $\mathbb{R}$ itself...) we can say that $\mathbb{C}$ = h($\mathbb{R}$)[i] (because we only have to add the element i to our set h($\mathbb{R}$) and the smallest ring containing h($\mathbb{R}$) and i is $\mathbb{C}$)

$\mathbb{Q}, \mathbb{R}, \mathbb{C}$ are not finite $\mathbb{Z}$-algebras

Because $\mathbb{Q}$ ⊂ $\mathbb{R}$ ⊂ $\mathbb{C}$ we only do it for $\mathbb{Q}$

There does not exist a one argument homomorphism which gives you any rational number you want, we can only work with some kind of homomorphism which gives you an integer. So to make $\mathbb{Q}$ out of it, we'd have to add an infinite amount of elements/numbers (all the "fully" rational ones like $\frac{1}{2}$, ...) to our set h($\mathbb{Z}$) to get $\mathbb{Q}$, but that's not what we're looking for. thus it's not a finite $\mathbb{Z}$-Algebra

$\mathbb{R}, \mathbb{C}$ are not finite $\mathbb{Q}$-algebras

Because $\mathbb{R}$ ⊂ $\mathbb{C}$ we only do it for $\mathbb{R}$

Same understanding as above, there does not exist a one argument homomorphism which gives you any irrational number you'd like (and there's an infinite amount of them), and working with a homomorphism which gives you a rational number as a result, the only way to create $\mathbb{R}$ out of h($\mathbb{Q}$) is to add all the irrational numbers to the set h($\mathbb{Q}$), but that's not what we're looking for

Is this enough? Because I kind of feel that it's too little, and that some sort of mathematical explanation is lacking (some equations for example)

答:
$\mathbb{Q}, \mathbb{R}, \mathbb{C}$ are not finite $\mathbb{Z}$-algebras

Hint for $\mathbb R$ and $\mathbb C$:

A finitely generated $\mathbb Z$-algebra $A$ is the image of a ring homomorphism $\mathbb Z[x_1,\dots,x_n] \to A$ and therefore is countable.

Hint for $\mathbb Q$:

The numbers in $\mathbb Z[q_1,\dots,q_n]$ have denominators whose prime divisors come from a finite set.


问:
Debugging an analysis text
I'm having some problems understanding the following paragraph, which I read in a analysis script (hopefully I haven't made any translation errors):

"A map $f:U \rightarrow Y$, where $U$ is open and $X,Y$ are Banach spaces, is continuous at $x' \in U$ if $$f(x')=\lim_{x\rightarrow x'} f(x)=\lim_{h\rightarrow 0} f(x'+h),$$
  where $h=x-x'$. We can decompose $h$ in a "polar" fashion in $h=ts$, where $\left\Vert h \right\Vert \geq 0$ and $s=\frac{1}{\left\Vert h \right\Vert } h$. Then
  $f(x')=\lim\limits_{h\rightarrow 0} f(x'+h)$ iff $f(x'+ts)\rightarrow f(x')$ for $t \rightarrow 0^+$ uniformly with respect to $\left\Vert s \right \Vert = 1$.
  No matter from which direction $s$ with $\left\Vert s \right\Vert=1$ we approach $x'$, the value of the function has to converge to $f(x')$ with a to all $s$ common "minimal speed"  ".

What I don't understand is this:

1) What does in means to decompose anything in a "polar" fashion ?

2) I thought only sequences of function can converge uniformly...and what does it mean, if something converges uniformly with respect to another thing ?

3) What does the author mean with "common "minimal speed""

Thanks in advance.

答:
I'll only deal with (1). Like mac, I'm not sure what (3) means here.

(1) When working in polar coordinates, every point in the plane is written in the form $(r,\theta)$, where $r$ denotes the magnitude (distance to the origin) and $\theta$ denotes the argument (direction). Similarly, when we work with complex numbers, it is often very useful to write a complex number as if it were given by "polar coordinates" rather than "rectangular coordinates". The usual expression $z = a+bi$ with $a,b\in\mathbb{R}$ corresponds to the rectangular coordinates, with $a$ giving the $x$ coordinate and $b$ giving the $y$ coordinate (this is how Hamilton reified the complex numbers); then we can express $z$ instead in "polar coordinates", by writing $z = re^{i\theta}$, where $r$ is a nonnegative real, $\theta$ is a real number, and
$$e^{i\theta} = \cos\theta + i\sin\theta$$
gives the "direction." This is often called a "polar decomposition" of the complex number. The number $\theta$ is the argument, and the number $r$ is the magnitude (note that $\lVert z\rVert = r$).

By analogy, in many other circumstances where we express a quantity/object/function/transformation in terms of a "size" and a "direction", we call such an expression a "polar decomposition". This can be done for real numbers very easily: for any real number $h\neq 0$, we can write $h$ as $ts$, where $s$ is the "sign" ($s=1$ if $h\gt 0$, $s=-1$ if $h\lt 0$) and $t$ is the "magnitude" ($t=|h|$). This is the decomposition that is done here.

Title: Debugging an analysis text


问:
Show $\bmod m\!:\ ar\equiv b,\ br\equiv a\Rightarrow a^2\equiv b^2$
Suppose that $ar\equiv b$ (mod $m$) and $br\equiv a$ (mod $m$) for some $r$. Show that $a^2\equiv b^2$ (mod $m$).

I've tried quite a few different things and I'm unable to get rid of terms with $r$ in it (if that is even necessary) to show that $a^2=b^2=mz$ for some integer $z$.

I know that $ar-b=mx$ and $br-a=my$ for integers $x,y$. I tried using different mod properties like sum, difference, and product to see if it helped, but didn't have any luck. I ended up with $(a^2-b^2)(r^2-1)=m^2xz=m(mxy)$, but this is was no help and just made things more complicated. Any hints are greatly appreciated. I know there is something obvious that I haven't considered and my mind is just stuck in a certain mode of thinking and I can't escape it at the moment.

答:
Below is a conceptual derivation, showing how to view it as arising via CM = cross multiplication criterion for fraction equivalence. If $a$ and $b$ are invertible then we can use familiar properties of (modular) fractions, where we define $\,a/b := ab^{-1}.\,$ Solving for $r$ in both we obtain

$$\begin{align} \color{#c00}b\,r\equiv \color{#c00}a\\ \color{#0a0}a\,r\equiv \color{#0a0}b\end{align}\ \Longrightarrow\ \color{#c00}{\dfrac{a}b}\equiv r \equiv \color{#0a0}{\dfrac{b}a}\,\overset{\rm CM}\Longrightarrow\, \color{#c00}a\color{#0a0}a\equiv \color{#c00}b\color{#0a0}b\qquad\qquad $$

so $\,a^2\equiv b^2\,$ by CM. We can eliminate fractions to handle the case when $a$ or $b$ is not invertible. The proof of CM amounts to scaling the fractions to put them over a  common denominator $ab$. We can do the same thing with the equations defining the fractions, where the denominator is just the coef of $\,r,\,$ e.g. $\, b\,r\equiv a \!\iff\! r\equiv a/b\,$ has denominator $\,b.\,$ So we want to scale both equations so they have the same coef of $\,r\,$ - which will necessarily be a common multiple of $a$ and $b$. We choose the simple multiple $\,ab,\,$ yielding

$$\begin{align}
&a\times [b\,r\equiv a]\to \color{#c00}{ab\,r}\equiv a^2\\
&\:\!b\times [a\,r\equiv b]\to \color{#c00}{ba\,r}\equiv b^2
\end{align}\ \Rightarrow\ a^2\equiv b^2$$

This proof amounts to scaling the proof of the CM criterion by $\,ab,\,$ so we work only with (modular) integers (vs. fractions). Of course we can do the same with the general CM rule for $\frac{a}b\equiv \frac{c}d$ to obtain

$$ \begin{align} &b\,r\equiv a\\ &d\,r\equiv c\end{align}\ \Longrightarrow\ ad\equiv bc\qquad\qquad $$

Remark $ $ More generally, from an equational point of view, we can view the above derivation as cross-multiplying the equations to eliminate $\,r.\,$ From this standpoint we can view the above as a special case of general elimination algorithms such as the Grobner basis algorithm (which is a (multivariate) generalization of both the division (with remainder) algorithm and  Gaussian elimination). Even more generally we can view it as a special case of various overlapping (unification) methods of generating consequences of equations used in term rewriting systems, e.g. the Knuth Bendix algorithm. For example, see here where I show how to derive a proof of uniqueness of inverses this way.


问:
Can there be a third type of unit other than 1 and i?
I know the question sounds vague, so let me try to explain what is on my mind. When we imagine the complex plane, one axis represents the real part and the other axis represents the imaginary part. You cannot represent the real part with the imaginary part and vice versa. (As far as I know, this represents basis vectors.) What can be an example to the third axis to these to axes, such that a point can only be represented as a linear combination of the basis vectors in such way:

$$
P(a,b,c)=a1+bi+cj \ \ \ (a,b,c \in \mathbb{R})
$$

Can it be a new set of numbers?

答:
Yes. In fact these generalized 'fields' of complex numbers have a pretty active fields too.

The first one is the Quarternions which have 1 'real' vector parts and 3 others. But these are not commutative in multiplication.

There are further levels too, like Octonions and Sedenions. But these aren't even fields (you can't divide on them properly). Thus the more you try to generalize these concepts into an 'algebraically closed' form the less algebraic they will become. At which point you might as well just use vectors and classical linear algebra.

Their applications are amongst Quantum Field Theory, generalized description of Electromagnetic fields, the specific mathematical field that tries to discover the noncommutativity of Quarternions(thus making chirality a measurable difference) is called Geometric Algebra or its older name, Clifford Algebra.

Notice how to be able to perform a vectorial product you need at least three dimensions.

$$\vec{u}\times\vec{v}=\vec{x}$$

The only other number of dimensions where the cross product is well-defined in simple vectorial notation is 7 dimensions.

To be able to calculate the cross product in any n-dimensions Grassmann introduced the wedge product, which differentiates between a product of vectors and a product of a vector and a scalar.

In classical linear algebra the product of two vectors leaves you with a vector in the $3$ dimension which is perpendicular to the plane which was created by the two vectors.

Meanwhile the product of a vector and a scalar is just our original vector which's length was changed by the scalar.

If we have a vector how do we know if it was created from two vectors or just one?

The answer: let's define a product

$$\vec{u}\wedge\vec{v}=\frac{1}{2}(\vec{u}\vec{v}-\vec{v}\vec{u})$$

where $\vec{u}\vec{v}$ is the scalar product of the two vectors.

If we switch them and take their product:
$$\vec{v}\wedge\vec{u}=\frac{1}{2}(\vec{v}\vec{u}-\vec{u}\vec{v})=-\frac{1}{2}(\vec{u}\vec{v}-\vec{v}\vec{u})=-\vec{v}\wedge\vec{u}$$

it's clear that they satisfy the anticommutativity of the original cross product.
The quantity that we get when we take the wedge product is called a bivector which is the set of two vectors and a rotational component between them.

Why is this so good? Well, it turns out this concept generalizes really well to $n$-dimensions and we can do algebra on them while not having to deal with visual downsides of the cross product.


问:
$T$ injective equivalent to $T'$ surjective
I'm currently going through Sheldon Axler's "Linear Algebra Done Right" (3rd Ed.) and struggle with the result of 3.110, where he shows that $T\in\mathcal{L}(V, W)$ is injective if and only if the dual map $T'$ is surjective. He states in the proof that null $T = {0}$ happens if the annihilator of null T (i.e. (null $T)^0)$ is equal to the dual space $V'$. However, I don't see the connection here.

答: Suppose that $U$ is a subspace of $V$ such that $U^0 = V'$. Then $\dim U^0 = \dim V'$ and cancelling these in the equation $\dim V' = \dim V = \dim U + \dim U^0$ we have $\dim U = 0$, that is, $U = \{0\}$.


问:
How to determine the integral interval? Can they be exchanged?
$\iint_D\frac{\sin y}{y}\mathrm{d}x\mathrm{d}y$, $D$ with $y=\sqrt{x}$ and $y=x$

What I'm most puzzled about is $\iint_D\frac{\sin y}{y}\mathrm{d}x\mathrm{d}y$=$\int_0^1\mathrm{d}y\int_{y^2}^y\frac{\sin y}{y}\mathrm{d}x$

Can $y$ and $y^2$ exchange their position? Can their upper and lower limits be exchanged?

When I saw the picture, the two equations intersect at one point

答: The equations $y=\sqrt x$ and $y=x$ intersect in two points, $(0,0)$ and $(1,1)$.  Over the range $0-1$ in $x$ we have $y^2 \lt y$ so the limits are correct to give a positive result.  You can interchange the limits of an integral, but it will negate the result.  You seem to be assuming that $y^2 \gt y$, but that is not the case here.


问:
AP Calculus multiple choice question
The position of a particle along a line is given by $z(t) = 2t^3 -24t^2 + 90t +7$ for $ t \geq 0$. For what values of $t$ is the speed of the particle increasing?

a) $3 < t < 4$ only

b) $t > 4$ only

c) $t > 5$ only

d) $0 < t < 3$ and $t > 5$

e) $3 < t < 4$ and $t > 5$

For this problem I said d) based on my calculations:

$z'(t) = 6t^2 - 48t + 90 = 6(t-5)(t-3)$. I then did a chart and found out that $z'(0) > 0$ ,$z'(4) < 0$ and $z'(6) > 0$ leading me to conclude d) however the textbook says e). I thought that I did everything right, what is wrong with my conclusion?

答:
First of all, realize that the speed is $|z^{'}(t)|$ and not $z^{'}(t)$. This is your mistake.

Now, graph $|z^{'}(t)|$ and you will see that your textbook is spot on.

Title: AP Calculus multiple choice question


问:
Functional derivative and first variation
Refering to the two sources given as hyperlinks in this question, there seems to me to be a discrepancy (or contradiction) in the definition of the functional derivative and its relation to the first variation which I cannot resolve.

Let $F[y]:M \rightarrow \mathbb R$ be a sufficiently regular functional with some set M of sufficiently regular functions y(x) defined on the intervall $[a,b]$. Let furthermore $\eta(x)$ be a test function and $\epsilon \in \mathbb R$ a variable. Let's assume that both y and $\eta$ are fixed.

The variation of y can be represented as $\delta y(x) = \epsilon \, \eta(x)$. According to the first source, which is an appendix to the book by Engel/Deizler, the increment $\triangle F:= F[y+\delta y]-F[y]= F[y+\epsilon \, \eta]-F[y]$ of the functional can be represented as the Taylor expansion

$ F[y+\epsilon \, \eta]-F[y]= \left[ \frac{dF[y+\epsilon \, \eta]}{d\epsilon} \right]_{\epsilon=0} \epsilon + \frac{1}{2} \left[ \frac{d^2F[y+\epsilon \, \eta]}{d\epsilon^2} \right]_{\epsilon=0} \epsilon^2 + \cdots. \qquad \qquad (1)$

Here my understanding is that the first term on the rhs represents the first variation $\delta F[y]$, etc., so that we have

$ F[y+\epsilon \, \eta]-F[y]= \delta F + \delta^2 F + \cdots. \qquad \qquad (2) $

Now in the second source, which is Wikipedia, the functional derivative $\frac{\delta F}{\delta y(x)}$ is defined by setting (in my notation)

$ \left[ \frac{dF[y+\epsilon \, \eta]}{d\epsilon} \right]_{\epsilon=0} = \int \frac{\delta F}{\delta y(x)} \eta \, dx = \delta F[y]. \qquad \qquad (3)$

The expression in the middle is explicitly identified with the differential or first variation $\delta F[y]$. It is explained that the functional derivative can be understood as something like a partial derivative in the case of infinitely many variables, and the integral, i.e. the first variation, as something like the total differential in the case of infinitely many variables.

The discrepancy is that in the Taylor expansion the first variation is the expression on the rhs of

$ \delta F[y] = \left[ \frac{dF[y+\epsilon \, \eta]}{d\epsilon} \right]_{\epsilon=0} \epsilon, \qquad \qquad (4) $

note the additional $\epsilon$.

Related to this contradiction I have a general question of understanding. If the functional derivative is the analogue to partial derivatives, and the first variation the analogue to a total differential, what, in calculus of variations, is the analogue to a total derivative? We speak of F[y] as being differentiable, but there is no total derivative.

答:
It might be, that there is a mistake in the Engl. Wiki in that the right equation of (3) is not correct. In the Italian Wiki the formula for the first variation in terms of the functional derivative is instead

$\delta F[y] = \int \frac{\delta F}{\delta y(x)} \delta y(x) \, dx $.

This would resolve the contradiction, because with $\delta y(x) = \epsilon \eta(x)$ and $\epsilon$ being constant over [a,b], the right equation of (3) is in line with (4).


答:
Using your bounds you get

$$\int_0^1 \frac{\sin y}{y}\left[ \int_{y^2}^{y}dx\right]dy=\int_0^1\sin y dy-\int_0^1y \sin y dy=1-\sin (1)$$

Title: How to determine the integral interval? Can they be exchanged?

Body:
$\iint_D\frac{\sin y}{y}\mathrm{d}x\mathrm{d}y$, $D$ with $y=\sqrt{x}$ and $y=x$

What I'm most puzzled about is $\iint_D\frac{\sin y}{y}\mathrm{d}x\mathrm{d}y$=$\int_0^1\mathrm{d}y\int_{y^2}^y\frac{\sin y}{y}\mathrm{d}x$

Can $y$ and $y^2$ exchange their position? Can their upper and lower limits be exchanged?

When I saw the picture, the two equations intersect at one point


问:
Conditional expectation given sum
There are two random variables $X$ and $Y$, both of them distributed standard normal. Based on $X$ and $Y$, define $Z = X + a  Y$ for $a$ a known parameter.

I'm interested in $E(X | Z)$.

It seems to me that a starting point could be the approach taken here. So I start with:

$ Z = E( X | Z) + a E(Y | Z) $

and then I write out the expectation terms explicitly, hoping to find an easy substitution:

$ E(X | Z) = \frac{\int_{-\infty}^{\infty}x f(x)f(\frac{Z-x}{a})dx}{\int_{-\infty}^{\infty} f(x)f(\frac{Z-x}{a})dx} $

and

$ E(Y | Z) = \frac{\int_{-\infty}^{\infty}y f(y)f(Z-ay)dy}{\int_{-\infty}^{\infty} f(y)f(Z-ay)dy} $

But here I'm stuck, I can't find an substition that would allow me to express $E(Y|Z)$ in terms of $E(X|Z)$, in order to plug in above. So how to continue from here?

Edit:

For $a =1$, one can simply use $E(X|Z) = E(Y | Z)$, plug in above, and obtain $E(X|Z) = Z/2$.

I conjecture that in general, it holds that $$E(X|Z) = \frac{Z}{1 + a^2}.$$ This is supported by several numerical simulations I made.

This would require showing that $E(Y|Z) = a E(X | Z)$. In case substituting in the integral does not help, how else could I show that?

答:
We first need to find the joint distribution of $X$ and $Z$. This can easily be found since the $cov(X,Z)=1$ and hence

$$ \left(\begin{array}{cc} X \\ Z \end{array} \right) \sim \mathcal{N}\left(\left(\begin{array}{cc} 0 \\ 0 \end{array} \right) , \left(\begin{array}{cc} 1 & 1 \\ 1 & 1+a^2 \end{array} \right)  \right)$$

Now we can simply apply the formula for the conditional expectation from a bivariate normal distribution and obtain

$$E(X | Z) = \frac{Z}{1+a^2}.$$

Some background notes:

Lecture notes on multivariate normal distribution

Marginal and conditional distributions of multivariate normal distribution

and Chapter 3 (Theorem 3.3.4) in Tong, Y.L., 2012. The multivariate normal distribution. Springer Science & Business Media.

Title: Conditional expectation given sum


问:
What Does the Answer to a SAT Problem Look Like, is it YES or NO or is it an arrangement of Satisfying Values?
I am having difficulty understanding what a SAT problem is asking. The answer given by com here says the answer to a SAT problem is YES or NO and we don't care about the satisfying values for the literals. If this is the case can someone link me to either the definitions (preferably for 3 SAT) stating this or work showing why this is the most useful way to think about these problems? </p>

答:
The answer is TRUE / FALSE according to wikipedia: https://en.wikipedia.org/wiki/Boolean_satisfiability_problem

But you can easily use an algorithm that provides a yes/no answer to find a satisfying solution, if one exists. Simply fix one of the variables to be TRUE and use your algorithm to see if there is still a satisfying solution. If not, it means that variable must be FALSE. You can then repeat with the next variable until all your variables are fixed, which takes $O(n)$ calls to your algorithm.


问:
Suppose that y1 = 8.3, y2 = 4.9, y3 = 2.6, y4 = 6.5 is a random sample of size 4 from the two parameter uniform pdf,
Suppose that $y_1 = 8.3, y_2 = 4.9, y_3 = 2.6, y_4 = 6.5$ is a random sample of size 4 from the two-parameter uniform pdf,

$f_y(y;\theta_1,\theta_2) = \frac{1}{2\theta_2}, \theta_1 - \theta_2 \le y \le \theta_1 + \theta_2$

Use the method of moments to calculate $\theta_{1e}$ and $\theta_{2e}$

So I understand how to solve the question when it comes to one parameter. Where we essentially solve for $E(y)$. But I'm not sure how to go about solving for two parameters.

答:
This is not the usual parameterization of a uniform distribution, so looking at standard references may not be immediately helpful,

Begin by showing That $E(X) = \theta_1$ and $Var(X) = \theta_2^2/3.$

Then, for the given sample, find $\bar X = 0.375$
and $S^2 = 6.3825.$ Computations in R:

```<code>x = c(8.3, 4.1, 2.6, 6.5)
mean(x); var(x)
[1] 5.375
[1] 6.3825
</code>```

Finally, set
$E(X) = \mu = \theta_1 = \bar X$ and
$Var(X) = \sigma^2 = \theta_2^2/3 = S^2.$
Solve for $\theta_1, \theta_2$ in terms of data
to get numerical MOM estimates $\hat\theta_1, \hat\theta_2,$ respectively. Obviously, $\hat\theta_1 - \bar X =5.375.$ What is $\hat\theta_2 ?$

Note: Later you may study maximum likelihood estimates. Sometimes MOM (method-of-moments) estimators are the same as MLEs (maximum likelihood estimators), and sometimes not. When sampling from the uniform distribution of this problem, there are differences between the two kinds of estimators.

Body:
This is not the usual parameterization of a uniform distribution, so looking at standard references may not be immediately helpful,

Begin by showing That $E(X) = \theta_1$ and $Var(X) = \theta_2^2/3.$

Then, for the given sample, find $\bar X = 0.375$
and $S^2 = 6.3825.$ Computations in R:

x = c(8.3, 4.1, 2.6, 6.5)
mean(x); var(x)
[1] 5.375
[1] 6.3825

Finally, set
$E(X) = \mu = \theta_1 = \bar X$ and
$Var(X) = \sigma^2 = \theta_2^2/3 = S^2.$
Solve for $\theta_1, \theta_2$ in terms of data
to get numerical MOM estimates $\hat\theta_1, \hat\theta_2,$ respectively. Obviously, $\hat\theta_1 - \bar X =5.375.$ What is $\hat\theta_2 ?$

Note: Later you may study maximum likelihood estimates. Sometimes MOM (method-of-moments) estimators are the same as MLEs (maximum likelihood estimators), and sometimes not. When sampling from the uniform distribution of this problem, there are differences between the two kinds of estimators.


问:
Finding limit of sequence $a_n=\frac{\alpha(n)}{n}$
Given sequence $a_n=\frac{\alpha(n)}{n}$, where $\alpha(n)$=total no. of primes by which $n$ is divisible, find $\lim_{n\to \infty}a_n$.

Example: $\alpha (8)=1$, as $2$ is the only prime by which $8$ is divisible.

Observing that $\alpha(n)\lt n$ for all $n\in \mathbb N$, the given sequence is bounded. Then I tried using the fact that every $n$ has prime factorisation.

But I don't know how to proceed further. Any hints?

答:
I am posting an answer here. I would like to thank @Teresa Lisbon who's helped me a lot to solve this exercise.

Since every natural no. $n$ has a prime factorization, suppose that $n=n_1^{p_1}n_2^{p_2}\cdots n_k^{p_k}$ for prime nos. $n_i$'s where $i\in\{1,2,3,\cdots,k\}$. It follows that $\phi(n)=k$ as for every $i\in\{1,2,3,\cdots,k\}$, we have prime no. $n_i$ such that $n_i|n$.

$\begin{align}
\log_2n &=p_1\log_2n_1+p_2\log_2n_2+\cdots+p_k\log_2n_k\\
        &\ge p_1\log_22+p_2\log_22+\cdots+p_k\log_22\\
        &=p_1+p_2+\cdots+p_k \ge 1+1+\cdots+1 \text{  ($k$ times)}=k\\
\implies &\log_2n\ge k= \phi(n)\\
\implies &0\le\frac{\phi(n)}{n}\le\frac{\log_2n}{n}\end{align}$ $\tag 1$

Since $n^{\frac 1{n}}\to 1$, by continuity of $\log_2n$ we have $\frac{\log_2n}{n}=\log_2(n^{\frac 1{n}})\to \log_2(1)=0$

By squeeze theorem in $(1)$, it follows that $\frac{\phi(n)}{n}\to 0$


问:
Annihilator method, $y''-4y'-5y=x^2 e^{-x} +6e^{-x}$
The equation $y''-4y'-5y=x^2 e^{-x} +6e^{-x}$, has a characteristic polynomial  $$(r^2 -4r -5)$$

with roots $-1$, $5$. Our operator $M(y)$, which is the equation to which the RHS is a solution of, has a characteristic polynomial $$p(r)=(r+1)^3  (r+1)=(r+1)^4$$

which has roots $-1$ with multiplicity of four. The relevant polynomial for this problem is then $$(r+1)^4 (r^2 -4r-5)$$

which has roots $-1,-1,-1,-1,-1,5$. And a particular solution of the differential equation is then $$\psi_p=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}+c_4x^3e^{-x}$$

, where $c_i$ are constants to be determined. Notice that we can remove two terms because they are solutions of $L(y)=0$. Is this correct? I'm not sure how to deal with repeated roots.

答:
Make your life easier from the starting point using $y=z\, e^x$. This gives
$$z''-6 z'=x^2+6$$ Reduction of order makes the problem quite simple.


答:
$$y''-4y'-5y=x^2 e^{-x} +6e^{-x}$$
The homogeneous solution of the DE is:
$$\implies y_h=c_1e^{-x}+c_2e^{5x}$$
And the guess for the particular solution should be:
$$y_p=(ax^2+bx+c)e^{-x}\color {red}{x}$$
$$y_p=(ax^3+bx^2+cx)e^{-x}$$

Annihilator method:
$$y''-4y'-5y=x^2 e^{-x} +6e^{-x}$$
$$ (D^2 -4D-5)y=x^2 e^{-x} +6e^{-x}$$
$$ (D^2 -4D-5)y=(x^2  +6)e^{-x}$$
$$\color{red}{(D+1)^3} (D^2 -4D-5)y=0$$
$$\color{red}{(D+1)^3} (D+1)(D-5)y=0$$
With the annihilator method you should get for the particular solution:
$$y_p=c_1xe^{-x}+c_2x^2e^{-x}+c_3x^3e^{-x}$$
$$y_p=e^{-x}(c_1x+c_2x^2+c_3x^3)$$
And for the homogeneous solution:
$$y_h=Ae^{-x}+Be^{5x}$$

So your solution  looks good :
$$\psi_p=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}+c_4x^3e^{-x}$$
But the first term is absorbed by the homogeneous solution so that you only have three constant.
$$\psi_p=c_2xe^{-x}+c_3x^2e^{-x}+c_4x^3e^{-x}$$

Title: Annihilator method, $y''-4y'-5y=x^2 e^{-x} +6e^{-x}$

Body:
The equation $y''-4y'-5y=x^2 e^{-x} +6e^{-x}$, has a characteristic polynomial  $$(r^2 -4r -5)$$

with roots $-1$, $5$. Our operator $M(y)$, which is the equation to which the RHS is a solution of, has a characteristic polynomial $$p(r)=(r+1)^3  (r+1)=(r+1)^4$$

which has roots $-1$ with multiplicity of four. The relevant polynomial for this problem is then $$(r+1)^4 (r^2 -4r-5)$$

which has roots $-1,-1,-1,-1,-1,5$. And a particular solution of the differential equation is then $$\psi_p=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}+c_4x^3e^{-x}$$

, where $c_i$ are constants to be determined. Notice that we can remove two terms because they are solutions of $L(y)=0$. Is this correct? I'm not sure how to deal with repeated roots.


答: Since $ar^2 \equiv br \equiv a$, we have $b^2\equiv a^2r^2\equiv a^2 \bmod m$

Title: Show $\bmod m\!:\ ar\equiv b,\ br\equiv a\Rightarrow a^2\equiv b^2$

Body:
Suppose that $ar\equiv b$ (mod $m$) and $br\equiv a$ (mod $m$) for some $r$. Show that $a^2\equiv b^2$ (mod $m$).

I've tried quite a few different things and I'm unable to get rid of terms with $r$ in it (if that is even necessary) to show that $a^2=b^2=mz$ for some integer $z$.

I know that $ar-b=mx$ and $br-a=my$ for integers $x,y$. I tried using different mod properties like sum, difference, and product to see if it helped, but didn't have any luck. I ended up with $(a^2-b^2)(r^2-1)=m^2xz=m(mxy)$, but this is was no help and just made things more complicated. Any hints are greatly appreciated. I know there is something obvious that I haven't considered and my mind is just stuck in a certain mode of thinking and I can't escape it at the moment.


答:
In order for one new unit to exist, we have to answer this question:

what's $i\times j$?

the answer to this question has to have a particular property: the result can't be the same as any other question with either $i$ on the left or $j$ on the right and any of $1, -1, i, -i, j, -j$ filling in the other side.  so it can't be: $i\times 1 = i$, or $i\times -1 = -i$, or $i \times i = -1$ or $i \times -i = 1$, and it can't be $1 \times j = j$ or $-1 \times j = -j$, so ... we're kinda stuck.

Fortunately Hamilton went "wait what if we added yet another unit, $k$?" and then we ended up with the quaternions, which get a lot of use because unit quaternions are a double cover of 3d rotation.  There are others too, but quaternions are the ones that get the most press.

Title: Can there be a third type of unit other than 1 and i?

Body:
I know the question sounds vague, so let me try to explain what is on my mind. When we imagine the complex plane, one axis represents the real part and the other axis represents the imaginary part. You cannot represent the real part with the imaginary part and vice versa. (As far as I know, this represents basis vectors.) What can be an example to the third axis to these to axes, such that a point can only be represented as a linear combination of the basis vectors in such way:

$$
P(a,b,c)=a1+bi+cj \ \ \ (a,b,c \in \mathbb{R})
$$

Can it be a new set of numbers?


问:
Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.

Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.

I found a solution, but it involves a lot of case work. Can someone help me find a solution which doesn't involve a lot of case work?

答:
Let $x$, $y$ and $z$ be natural numbers satisfying
$$x+2y+2z=xyz.$$
If $xyz=0$ then $x=y=z=0$, so suppose $x,y,z>0$. A bit of algebra shows that
$$(xy-2)(xz-2)=x^2+4.\tag{1}$$
Of course without loss of generality $y\leq z$. If $x=1$ then the above simplifies to
$$(y-2)(z-2)=5,$$
and so $y=3$ and $z=7$. Otherwise $x\geq2$ and then from $y\leq z$ it follows that
$$(x+1)^2>x^2+4=(xy-2)(xz-2)\geq(xy-2)^2,$$
and hence that $x\geq xy-2$, or equivalently $2\geq x(y-1)$. This means that either $x=y=2$, in which case $z=3$, or $y=1$, in which case $(1)$ simplifies to
$$(x-2)(xz-2)=x^2+4.\tag{2}$$
Clearly $x\neq2$, and so from the identity
$$xz-2=\frac{x^2+4}{x-2}=x+2+\frac{8}{x-2},$$
we see that $x-2$ divides $8$, so $x-2\in\{1,2,4,8\}$. Plugging these values into $(2)$ yields
\begin{eqnarray*}
1\cdot(3z-2)&=&13,\\
2\cdot(4z-2)&=&20,\\
4\cdot(6z-2)&=&40,\\
8\cdot(10z-2)&=&104.
\end{eqnarray*}
The first three have the solutions $z=5$, $z=3$ and $z=2$, respectively, and the last has no integral solution.

In summary, up to permutation of $y$ and $z$, the solutions $(x,y,z)$ are
$$(0,0,0),\quad(1,3,7),\quad(2,2,3),\quad(3,1,5),\quad(4,1,3),\quad(6,1,2).$$


问:
What does the Gradient "with respect to a position vector" mean?
I was studying John R. Taylor's book on Classical Mechanics and he introduced a confusing concept in page 140 (2005 edition):

$$\nabla_1 = (\frac{\partial}{\partial x_1}+\frac{\partial}{\partial y_1}+\frac{\partial}{\partial z_1})$$

Where

$$\mathbf r_1 = (x_1+y_1+z_1)$$

He calls this "the gradient with respect to the coordinates of $\mathbf r_1$". In general, what is the gradient with respect to a position vector? Isn't the gradient (in physics) just dependent on where we sit our x-axis and our y-axis? In other words, I've always known the gradient as:

$$\nabla = (\frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z})$$

(If context gives a clue, he was talking about the gradient of a potential $U$ in an isolated, two body system where the positions of the two particles are $\mathbf r_1$ and $\mathbf r_2$, and the claim is that the force on particle 1 due to particle 2 $\mathbf F_1 = -\nabla_1U(\mathbf r_1-\mathbf r_2)$. The same would apply for particle 2 interchanging 1 with 2 in the last formula.)

答:
This might be outdated. But I just happened to see it so I try to show my opinion and welcome to correct me if something is not going right.
I will try to use most common notation, for instance position column vector $\mathbf r=[x,y,z]^T$,
so the gradient of the position vector
$$\begin{bmatrix}\frac{\partial {x}}{\partial x}&\frac{\partial {y}}{\partial x}&\frac{\partial {z}}{\partial x} \\ \frac{\partial {x}}{\partial y}&\frac{\partial {y}}{\partial y}&\frac{\partial {z}}{\partial y} \\ \frac{\partial {x}}{\partial z}&\frac{\partial {y}}{\partial z}& \frac{\partial {z}}{\partial z} \end{bmatrix}=[\mathbb I]=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}.$$

This can be used to prove the identity which is seemingly impossible:

$$(\mathbf v\cdot\nabla)(\mathbf r\times\mathbf c)=\mathbf v\times \mathbf c.$$ where $\mathbf c$ is a constant vector.

The proof can be $(\mathbf v\cdot\nabla)(\mathbf r\times\mathbf c)=v_\ell\partial_\ell\varepsilon_{ijk}r_jc_k=\varepsilon_{ijk}v_\ell c_k\partial_\ell r_j=\varepsilon_{ijk}v_jc_k=\mathbf v\times \mathbf c$.


问:
A question about linear application
For example, we have two vector space V, W, and X is a K-base of V.

So, therefore if we have $f,g: V \to W$ two K-linear applications. Suppose f(x)=g(x) for all $x \in X$, how to prove $f=g$?

And for all $x \in X$, if we have $w \in W$, can we show there exist a unique linear application $h: V \to W$ such that $h(x) = w$.

Sorry, it's truly a basic question, but I'm a beginner so I don't know how to solve that. Thanks in advance

答:
Look that for being $X=\{x_1,...,x_n\}$ a $K$-base of $V$, then every element $v \in V$ can be written in an unique way as
$$
v=a_1x_1+...+a_nx_n
$$
for $a_1,...,a_n \in K$.
For being $f$ and $g$ linear, then
$$
f(v)=f(a_1x_1+...+a_nx_n)=a_1f(x_1)+...+a_nf(x_n)=a_1g(x_1)+...+a_ng(x_n)=g(a_1x_1+...+a_nx_n)=g(v).
$$
Since $v$ is arbitrary, we got the result. The second question is of direct proof with this.

Title: A question about linear application


问:
Geometric Algebras
As far as I can tell there are several extensions of linear algebra which can be used to do geometry on $\mathbb{R}^n.$ There are: the Clifford Algebra, the Grassman Algebra, the Exterior Algebra, Geometric Algebra, Hamilton Algebra, "Tensor Algebra" (not really sure if this is an algebra - but you can use tensor to do a lot of the preceding stuff I believe), differential geometry/forms, and perhaps others. Edit: It seems like there is also pre-Lie algebras (and maybe just Lie algebras in general?) as well.

It's not hard for me to go to Wikipedia and look at the definition of each, but this isn't very helpful. So how are each of these defined, what are the related concepts, and how do they all fit in together? Are there any that I have missed?

Thanks.

答:
Jean-Louis Loday has written up a huge list of algebras arising in algebra, geometry, physics and many other areas, in his paper
Encyclopedia of types of algebras 2010. Note that the name of the author there, "G. W. Zinbiel" is a reference to Leibniz algebras, written from right to left, refering to the "dual" operad, see here.</p>

Title: Geometric Algebras


问:
Writing a given process as a diffusion
When is a stochastic process a diffusion process?

Wikipedia says "A diffusion process is a Markov process with continuous sample paths for which the Kolmogorov forward equation is the Fokker–Planck equation." I don't know how to interpret this; doesn't this just mean it has the typical diffusion form
\begin{equation}dX_t = \mu(X_t) dt + \sigma(X_t) dW_t, \;\; X_0 = x_0\end{equation}
and how would one check this for an arbitrary process?

I understand this means there's an average drift and a variance around that drift. But I don't know if this form is valid for an arbitrary process $X_t$, since we can always define the "drift" as $ \mu(X_t) = \mathbb{E}[X_t]$ and variance $\sigma(X_t) = \text{Var}(X_t)$.

For example, define a process $X_t$ that increases by $dX$ with probability $p \in [0,1]$ in each time interval $dt$. This resembles a Poisson process with rate $p$ and infinitesimal jump size. Since the mean and variance of Poi$(p)$ are both $p$, then I guess $\mu(X_t)dt = p dt$ and $\sigma^2 = \sqrt{p}$. But I still don't know if the Kolmogorov forward equation is the Fokker-Planck.

答:
I did some reading and here is a partial answer:

Definition: Diffusion process (see here)

A Markov process $X_t$ with transition probability $P(\Gamma, t|x,s)$ is a diffusion process if

for every $x$ and every $\epsilon>0$, $$\int_{|x-y|>\epsilon} P(dy,t|x,s) = o(t-s).$$ I.e. the probability of moving more than $\epsilon$ in a small time frame is small.

the drift is given by $a(x,s)$ such that for every $x$, and every $\epsilon >0$,
$$ \int_{|y-x|\leq \epsilon} (y-x) P(dy,t|x,s) = a(x,s)(s-t) + o(s-t)$$

Diffusion coefficient is given by $b(x,s)$ such that $$\int_{|y-x|\leq \epsilon} (y-x)^2 P(dy,t|x,s) = b(x,s)(s-t) + p(s-t).$$

I.e. the drift and diffusion coefficients specify the distribution of the movement, giving a necessary concentration of measure.

Characterization by Fokker-Planck equation
In my question I said that a diffusion process is a Markov process whose forward equation is the Fokker-Planck equation. How can this be reconciled with the above definition?

This says that if we define a pdf for the transition density, $$ P(\Gamma,t|x,s) = \int_\Gamma p(s,x,t,y)dy$$
and that assume $a, b$ are differentiable, and assume 1., 2., 3. above, then the transition probability density satisfies the forward Kolmogorov equation.

So in addition note that the equation $dX_t = a(X,t) dt + b(X_t, t) dB_t, \;\; X_0 = x_0$, does not specify a diffusion process unless the coefficients are continuous.

Something I am still confused about: I saw somewhere else that the drift is given by $$\mathbb{E}[X_{t+dt}-X_t| X_t = x] = a(x,t)dt + O(dt^2)$$ I'm not sure how this is equivalent to the integral in 2) above, and why there is a $O(dt^2)$ term (since the expectation should be fixed).

Title: Writing a given process as a diffusion


问:
Integrating $\sqrt\frac{x}{{x+a}}$?
How can I integrate this expression?
$$\sqrt\frac{x}{{x+a}}$$

I substituted $x=a\tan^2t$. After substituting, I reach
$$\int \sec x\tan^2x$$
Is this way correct?

Thanks in advance.

答: I will answer to your exact question. YES!, it is correct. The substitution transform your integral to $$2a\displaystyle\int \sec x\tan^2x\, dx.$$ Then you can simplify it further using the fact that $\sec x\tan^2x=\sec^3x-\sec x.$ Powers of secant is quit different from other integrals if that format, but can integrate using a reduction formula. First let $$I_n=\displaystyle\int \sec^nx\, dx,$$ then $I_0=x,\quad I_1=\ln|\sec x+\tan x|$ and $I_2=\tan x$ are the simplest known cases up to constants. For $n\ge 2$ use integration by parts with $u=\sec^{n-2}x$ and $dv=\sec^2x dx$ to get $$I_n=\sec^{n-2}x\tan x-(n-2)\displaystyle\int \sec^{n-2}x\tan^2 x\, dx.$$ Since $\tan^2 x=\sec^2x-1,$ we get a simplified recursive formula $$I_n=\dfrac{\sec^{n-2}x\tan x}{n-1}-\left(\dfrac{n-2}{n-1}\right)I_{n-2}.$$ Hope you can get your solution from here :).

Title: Integrating $\sqrt\frac{x}{{x+a}}$?


问:
Can you recommend some books on elliptic function?
I plan to study elliptic function.  Can you recommend  some  books? What is the relationship between elliptic function and elliptic curve？Many thanks in advance!</p>

答: There is a classical 3 volume series by C.L. Siegel. It is well-written, though the perspective is a little bit outdated. I guess (no book at hand) you can find treatments by Serge Lang in the GTM series as well. I am not sure if Stein's book on complex analysis studied this topic. </p>


问:
Semi-direct product $C_p : C_3$
I want to construct the group $C_p:C_3$ in GAP, where $p$ is prime and $p\equiv 1 \pmod 3$.

Originally, I come up with this

```<code>for p in [1..1000] do
    if IsPrime(p) and (p mod 3)=1 then
        N:=CyclicGroup(IsPermGroup,p);
        G:=CyclicGroup(IsPermGroup,3);
        AutN:=AutomorphismGroup(N);
        for i in [1..Size(AutN)] do
            f:=GroupHomomorphismByImages(G,AutN,
                 GeneratorsOfGroup(G),[Elements(AutN)[i]]);
            if f<>fail then
                NG:=SemidirectProduct(G,f,N);
                Print(IdGroup(NG),"\n");
            fi;
        od;
    fi;#IsPrime and p mod 3 =1
od;#p
</code>```

which works fine, but then I had a thought.

At present, this will also find the direct product, which is the case f: G -> Aut(N) is the trivial homomorphism. I could get rid of this by explicitly excluding it, or alternatively, if the identity element of AutN is always Elements(AutN)[1] I could just run the i loop from [2..p-1].

We will also find each non-trivial semidirect product twice, because the map sending the generator of Cp to the automorphism x, and the map sending the generator of Cp to x^-1, will give isomorphic semidirect products.

So I revised the code a little but am still unsure of how to get rid of the non-trivial semidirect product twice and now it does not work in GAP.

```<code>for p in [1..1000] do
    if IsPrime(p) and (p mod 3)=1 then
        N:=CyclicGroup(IsPermGroup,p);
      G:=CyclicGroup(IsPermGroup,3);
        AutN:=AutomorphismGroup(N);
        for i in [2..p-1] do
            f:=GroupHomomorphismByImages(G,AutN,
                 GeneratorsOfGroup(G),[Elements(AutN)[i]]);
            if f<>fail then
                NG:=SemidirectProduct(G,f,N);
                Print(IdGroup(NG),"\n");
            fi;
        od;
    fi;#IsPrime and p mod 3=1
od;#p
</code>```

Any ideas?

答:
To make your code output every non-trivial semi-direct product exactly once, insert after the line Print(IdGroup(NG),"\n"); the command of GAP break;
But better yet, instead of the loop, use the command RootsMod( 1, 3, p);

Title: Semi-direct product $C_p : C_3$


问:
Integer solutions for this equation
What are the integers (or at least natural numbers)$x_1$ and $x_2$ which satisfy this equation:

$3x_1 + 7x_2 = 1000$.

And is there a name for such a problem?

Title: Integer solutions for this equation

Body:
This is a linear Diophantine equation.

From the Bezout relation $3(-2)+7(1)=1$, we have $3(-2000)+7(1000)=1000$.

Generally, $3(-2000+7k)+7(1000-3k)=3(-2000)+7(1000)=1000$.


问:
Is ${\mathbb Z}_p$ a free ${\mathbb Z}_{(p)}$-module?

Let $p$ be a prime. Is $\mathbb{Z}_p$ a free $\mathbb{Z}_{(p)}$ module?

Here are some more details for those who are interested:

Here ${\mathbb Z}_p$ are the ${\mathbb p}$-adic integers (so the completion of ${\mathbb Z}$ with respect to the $p$-adic topology).

The ring ${\mathbb Z}_{(p)}$ is the localization of ${\mathbb Z}$ with respect to the prime ideal $(p)$. This is a discrete valuation ring.

Note that any integer $m$ different from $p$ is invertible in ${\mathbb Z}_p$, so we have a morphism ${\mathbb Z}_{(p)} \to {\mathbb Z}_p$. In particular the latter is a ${\mathbb Z}_{(p)}$-module.

For modules of finite type over ${\mathbb Z}_{(p)}$ one can check very quickly that they are free, as soon as they are torsion free. For modules that are not of finite type, one cannot check this as fast, because it is false in general (e.g. ${\mathbb Q}$ is not of finite type and not free, yet it is torsion free over ${\mathbb Z}_{(p)}$).

Hence my question: Is ${\mathbb Z}_p$ free as a ${\mathbb Z}_{(p)}$ module?

(Note that am I not asking if it is free of finite rank, as that would be clearly false).

答:
No idea why some people voted to delete my answer, without even leaving a comment.

If $\Bbb{Z}_p$ is a free $\Bbb{Z}_{(p)}=(\Bbb{Z}-(p))^{-1}\Bbb{Z}$ module of rank $n$ (it can be $n=\infty$) then the quotient group $$\Bbb{Z}_p/p\Bbb{Z}_p\cong \Bbb{Z}_{(p)}{}^n/p\Bbb{Z}_{(p)}{}^n\cong \Bbb{F}_p{}^n$$ so that $n=1$ which is absurd.

Whence $\Bbb{Z}_p$ is a non-free $\Bbb{Z}_{(p)}$-module.

I don't think there is any easier argument. If I am missing something then there is the comment box..


答:
So how are each of these defined, what are the related concepts, and how do they all fit in together?

Of the structures you mentioned, there are really not that many in play, and they are not that complicated.

"Grassman algebra" is just another word for exterior algebra.

"Geometric algebra" is an alias or a Clifford algebra using a nondegenerate form.

Exterior algebras are a special case of a Clifford algebra (the form is the zero form.)

I don't know what you could mean by "Hamilton algebra" if it's not the quaternions, which can be looked upon as a Clifford algebra.

The tensor algebra is of a different nature than the ones above.  It is "the biggest" algebra generated by a vector space, one from which we can extract special algebras as quotients. The Clifford algebras are one of the quotients that we can construct this way.

So you are talking mainly about these things: tensor algebra, Clifford algebra, exterior algebra (the rest being redundant, and I mention exterior algebra on its own even though it is a special case of a Clifford algebra.)

I'm not going to copy the definitions here for you, I just trust that now that you basically only need to study two definitions, your task will be much easier.

Are there any that I have missed?

Yeah, there are a bunch that will probably pour in.

The counterpart of Clifford algebras that use alternating forms are called Weyl algebras.

I don't know if you count Lie algebras as being relevant, but they are certainly a main character in geometry of differentiable manifolds.

There is also a more special type of Lie algebra called a Poisson algebra that is also very important for geometry of symplectic manifolds.

There are Jordan algebras, which are of a completely different flavor.

One should also mention $C^\ast$-algebras because of their application in physics.

Title: Geometric Algebras

Body:
As far as I can tell there are several extensions of linear algebra which can be used to do geometry on $\mathbb{R}^n.$ There are: the Clifford Algebra, the Grassman Algebra, the Exterior Algebra, Geometric Algebra, Hamilton Algebra, "Tensor Algebra" (not really sure if this is an algebra - but you can use tensor to do a lot of the preceding stuff I believe), differential geometry/forms, and perhaps others. Edit: It seems like there is also pre-Lie algebras (and maybe just Lie algebras in general?) as well.

It's not hard for me to go to Wikipedia and look at the definition of each, but this isn't very helpful. So how are each of these defined, what are the related concepts, and how do they all fit in together? Are there any that I have missed?

Thanks.


答:
Note that the substitution $x=a\sinh^2t$ is more convenient:

$$\int\sqrt\frac{x}{{x+a}}dx
=a\int 2\sinh^2tdt
= \frac a2\sinh2t-at+C
$$

Title: Integrating $\sqrt\frac{x}{{x+a}}$?

Body:
How can I integrate this expression?
$$\sqrt\frac{x}{{x+a}}$$

I substituted $x=a\tan^2t$. After substituting, I reach
$$\int \sec x\tan^2x$$
Is this way correct?

Thanks in advance.


问:
Continuous function need not be Riemann-Stieltjes integrable if $\alpha$ is not monotone.
For a given function $\alpha: [0, 1]\to \mathbb{R}$, let's denote $\mathcal{R}(\alpha)$ to be the set of all Riemann-Stieltjes integral functions with respect  to $\alpha$. It is a well-known fact that (for example, see Theorem 6.8 in Rudin's Principles of Mathematical Analysis) if $f: [0,1]\to\mathbb{R}$ is a continuous function, and $\alpha: [0,1]\to\mathbb{R}$ is a monotonically increasing, then $f\in\mathcal{R}(\alpha)$.

I am interested in the following:

Question 1: What is an example of function $\alpha:[0,1]\to\mathbb{R}$ and a continuous function $f:[0,1]\to\mathbb{R}$ such that $f\not\in\mathcal{R}(\alpha)$?

Question 2: Does there exist an example of continuous function $\alpha:[0,1]\to\mathbb{R}$ and a continuous function $f:[0,1]\to\mathbb{R}$ such that $f\not\in\mathcal{R}(\alpha)$?

Thanks!

答:
The Riemann-Stieltjes integral works well as long as $\alpha$ has bounded variation (equivalently, can be written as the difference of two bounded increasing functions). The continuous function
$$\alpha(x)=\begin{cases} \sqrt{x}\cos(\pi/x),\quad &x\in (0,1] \\ 0 & x=0\end{cases} \tag1$$
has infinite variation on $[0,1]$. Indeed, $\alpha(1/n)=(-1)^n/\sqrt{n}$, which implies $\sum_{n=1}^\infty |\alpha(1/n)-\alpha(1/(n+1))|=\infty$. (Aside: all real analysis book I know use $\sin$ instead of $\cos$ in this example, and mess with extra $\pi/2$ in calculations.)

I claim that $\int_0^1 \alpha(x)\,d\alpha(x)$ does not exist. First of all, the definition that is based on decomposing $\alpha$ as  the difference of two bounded increasing functions breaks down. On the more basic level, each interval $[1/(n+1),1/n]$ contributes
$$\approx\frac{1}{\sqrt{n}} \frac{2}{\sqrt{n}}$$
to the upper sum, and
$$\approx -\frac{1}{\sqrt{n}} \frac{2}{\sqrt{n}}$$
to the lower sum.


Title: $R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$

Body:
So the question is:

$R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$.

I don't really know what to do from here. If I substitute

$$R_n = 3(2^n)-4(5^n)$$

into

$$Rn = 7R_{n-1}-10R_{n-2}$$

I end up getting

$$R_n = 7\Big(3(2^{n-1})-4(5^{n-1})\Big)-10\Big(3(2^{n-2})-4(5^{n-2})\Big)$$

Dont know what to do...

EDIT: Thanks to Zev, what I did was:

$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
3(2^n)-4(5^n)&=21(2^{n})(2^{-1})-28(5^{n})(5^{-1})-30(2^{n})(2^{-2})+40(5^{n})(5^{-2})\\\\
3(2^n)-4(5^n)&=21/2(2^{n})-28/5(5^{n})-30/2(2^{n})+40/5(5^{n})\\\\
3(2^n)-4(5^n)&=(2^{n})[21/2-30/4]+(5^{n})[40/25-28/25]\\\\
3(2^n)-4(5^n)&=(2^{n})[3]+(5^{n})[-4]\\\\
3(2^n)-4(5^n)&=3(2^{n})-4(5^{n})
\end{align*}$$

Body:
You're on the right track so far; you've used the definition of $R_n$ to express the right side of the equation. Now just do this for the left side as well. You want to show that for any $n\geq 0$,
$$3(2^n)-4(5^n)=7\bigg[3(2^{n-1})-4(5^{n-1})\bigg]-10\bigg[3(2^{n-2})-4(5^{n-2})\bigg].$$
This can be done directly:
$$\begin{align*}
3(2^n)-4(5^n)&=21(2^{n-1})-28(5^{n-1})-30(2^{n-2})+40(5^{n-2})\\\\
12(2^{n-2})-100(5^{n-2})&=42(2^{n-2})-140(5^{n-2})-30(2^{n-2})+40(5^{n-2})
\end{align*}$$


问:
Compactness - Norm on $\Bbb R^n$
I am trying to prove the question below by Heine-Borel, $K_1$ and $K_2$ are subsets of $\Bbb R^n$.

Let $K_1 $and $K_2$ be two compact sets and A = $\{\lvert x_1 - x_2\lvert: x_1 \in K_1,x_2 \in K_2\}.$ Show that $A$ is compact.

Proof:

I am proving by Heine-Borel, therefore two steps.

First Step: show that A is bounded

For any $x_1 \in K$ and $x_2 \in K$, we know that $\lvert x_1 - x_2\lvert \leq  \lvert x_1\lvert + \lvert x_2\lvert$, as $\lvert x_1\lvert $ and $\lvert x_2\lvert $ are bounded the sum is bounded, so we are done proving that $\lvert x_1 - x_2\lvert$ is bounded.

Second Step: show that $A$ is closed

I am not sure how to do the second step.

答: Take a sequence $(a_n)_{n\in\Bbb N}$ of points of $A$ which converges to some $a\in\Bbb R$; you want to prove that $a\in A$. Each $a_n$ can be writen as $|x_n-y_n|$ with $x_n\in K_1$ and $y_n\in K_2$. The sequence $(x_n)_{n\in\Bbb N}$ has a subsequence $(x_{n_k})_{k\in\Bbb N}$ which converges to some $x\in K_1$, since $K_1$ is compact. And the sequence $(y_{n_k})_{k\in\Bbb N}$ has some subsequence $(y_{m_l})_{l\in\Bbb N}$ which converges to some $y\in K_2$. But then\begin{align}a&=\lim_{l\to\infty}a_{n_l}\\&=\lim_{l\to\infty}\left|x_{n_l}-y_{n_l}\right|\\&=\left|\lim_{l\to\infty}(x_{n_l}-y_{n_l})\right|\\&=|x-y|\\&\in A.\end{align}

Title: Compactness - Norm on $\Bbb R^n$


问:
Proving that Selection Operator on Cartesian Product of Two Relations is Equivalent to Theta-Join on Those Same Realtions
As the title suggests, I am trying to prove that $\sigma_\theta(R1\times R2) =R1 \bowtie_\theta R2$.

However, as far as I know, the very definition of $\theta$-join is the LHS equation from above, so I am quite confused as to

how I should generate a (semi) formal prove here. Any suggestions will be greatly appreciated.

答: Prove that any tuple that appear in the LHS will appear in the RHS and viceversa.</p>


问:
$\int_{-\infty}^{\infty} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt$ where $\zeta$ is the Riemann zeta function
$$I=\int_{-\infty}^{\infty} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt$$
where $\zeta$ is the Riemann zeta function. It is known by Balazard Saias and Yor Paper that I is integrable and $0\leq I<\infty$.

Since by Schwarz Reflection principle, integrand is an even function, so,
$$I= 2\int_{0}^{\infty} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt  $$
Let, $0<t_1<t_2<t_3<...<t_n<t_{n+1}<...$ be infinitely many zeros of $\zeta(s)$ on the critical line $\Re(s)=1/2$.
Define ,$ \ t_0=0$.
$$I= 2 \ \sum_{n=0}^{\infty}\int_{t_n}^{t_{n+1}} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt  $$
Let, $\epsilon>0$ be arbitrarily small.
Define,
$$I_{\epsilon}=2\sum_{n=0}^{\infty}\int_{t_n+\frac{\epsilon}{4}}^{t_{n+1}-\frac{\epsilon}{4}} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt  $$
Note that $|\zeta(\frac{1}{2}+it)|$ is non zero on $[t_n+\frac{\epsilon}{4},t_{n+1}-\frac{\epsilon}{4}]$ $\forall n\geq 0$, so $\log|\zeta(\frac{1}{2}+it)|$ is well defined on $[t_n+\frac{\epsilon}{4},t_{n+1}-\frac{\epsilon}{4}]$ $\forall n\geq 0$

Question 1 Prove that
$$2\sum_{n=0}^{\infty}\lim_{\epsilon\to 0^{+}}\int_{t_n+\frac{\epsilon}{4}}^{t_{n+1}-\frac{\epsilon}{4}} \frac{\log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt =I $$
Question 2 Prove that
$$\lim_{\epsilon \to 0^{+}} I_{\epsilon}=I$$

Do I need Dominated Convergence Theorem?..

答:
As there were many comments and the situation seems to still be confused for you, I will address some issues in one post.

(edited later per comments below); the crux of the matter here is that the statement (existing in the question version as of March 8, screenshot available upon request):

"Note that $|\zeta(\frac{1}{2}+it)|$ is non zero on $[t_n+\frac{\epsilon}{4},t_{n+1}-\frac{\epsilon}{4}]$ $\forall n\geq 0$ so $\log|\zeta(\frac{1}{2}+it)|$ is well defined etc" is incorrect.

Both questions $1,2$ (as stated now) are incorrect as formulated for the simple fact that given $\epsilon >0$ fixed, (at least and if current conjectures are true, actually all for some $n(\epsilon)$ on) some of the intervals $[t_n+\frac{\epsilon}{4},t_{n+1}-\frac{\epsilon}{4}]$ are ill-defined in the sense that the left end is higher than the right end, while even if we reverse them, infinitely many such intervals will eventually still contain other RZ roots, so there is no way to define a union of intervals as you want to avoid roots of RZ.

(for once $\liminf (t_{n+1}-t_n) =0$ showing first assertion above, and for another if we could have a union of intervals whether $[t_n+\frac{\epsilon}{4},t_{n+1}-\frac{\epsilon}{4}]$ or $[t_{n+1}-\frac{\epsilon}{4}, t_n+\frac{\epsilon}{4}]$ depending which end is higher, avoiding roots it would follow that roots are spaced at least $\epsilon/4$ apart and that trivially gives a contradiction with Selberg's theorem that there at least $cT\log T$ roots on the critical line with height up to $T$).

This being said I truly have no clue about the purpose of these questions since if one wants to prove that $I$ exists is finite and non-negative, there are various rigorous ways of doing it - the Balazard et others paper gives one based on the theory of Hardy spaces, another way can be given by constructing a domain where $\log \zeta$ is analytic by excluding horizontal segments ending at zeroes and using Cauchy to move the path of integration to infinity etc. Those methods give the full expression of $I$ in terms of a sum of non-trivial zeroes with real part greater than $1/2$ which can be easily estimated above by known results to a fairly small number, so $0 \le I < \delta$ with RH true iff $I=0$.

If one just wants to show that $I$ exists and is finite, that has nothing really to do with the theory of RZ beyond trivial estimates in the strip of the type $|\zeta(s)| \le C|s|, \Re s \ge \sigma >0$, the local form of holomorphic functions at zeroes and Lebesgue integration theory as one can trivially show that $\frac{log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}$ is integrable on $[M,N]$ for $M>N \ge 0$ and that $\lim_{M >N \to {\infty}}\int_{N}^{M} \frac{log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt=0$ which gives that $\int_{0}^{\infty} \frac{log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}dt$ exists and is finite; actually pretty much same proof and estimates show that $ \frac{log|\zeta(\frac{1}{2}+it)|}{\frac{1}{4}+t^2}$ is absolutely integrable on $[0, \infty)$ though of course one doesn't have a neat form of the integral of the absolute value anymore.


答:
The product of two compact subspaces is a compact. Therefore $K_1 \times K_2$ is a compact subspace of $\mathbb R^n \times \mathbb R^n$.

Then the application $(x,y) \mapsto \Vert x-y \Vert$ is continuous from $\mathbb R^n \times \mathbb R^n$ to $\mathbb R$.

As the image under a continuous map of a compact is compact, $A$ is compact.

Title: Compactness - Norm on $\Bbb R^n$

Body:
I am trying to prove the question below by Heine-Borel, $K_1$ and $K_2$ are subsets of $\Bbb R^n$.

Let $K_1 $and $K_2$ be two compact sets and A = $\{\lvert x_1 - x_2\lvert: x_1 \in K_1,x_2 \in K_2\}.$ Show that $A$ is compact.

Proof:

I am proving by Heine-Borel, therefore two steps.

First Step: show that A is bounded

For any $x_1 \in K$ and $x_2 \in K$, we know that $\lvert x_1 - x_2\lvert \leq  \lvert x_1\lvert + \lvert x_2\lvert$, as $\lvert x_1\lvert $ and $\lvert x_2\lvert $ are bounded the sum is bounded, so we are done proving that $\lvert x_1 - x_2\lvert$ is bounded.

Second Step: show that $A$ is closed

I am not sure how to do the second step.


问:
Changing the summations involving binomial coefficient
Given the following:
$$(1)\quad\quad \sum_{k=1}^n(k-1){n \choose k-1} + \sum_{k=1}^n {n\choose k -1} = \sum_{k=0}^{n-1}k{n\choose k} + \sum_{k=0}^{n-1}{n\choose k}$$
I am trying to understand how it goes from the LHS to the RHS.

My approach so far, given the first sum summand of the LHS, we have, and taking from the concept here
$$(2)\quad\quad \sum_{k=1}^n(k-1){n \choose k-1}$$

Let $k = m + 1$, then eq (2) becomes
$$(3) \quad\quad \sum_{m=0}^nm{n\choose m}$$
Now we let $l = n +1$, the eq (3) becomes
$$(4)\quad\quad\sum_{m=0}^{l-1}m{l-1\choose m}$$
Change the symbols in (4) to resemble the first first summand of the RHS of (1), namely, let $m = k$ and $l = n$ and substitute we have:
$$(5)\quad\quad\sum_{k=0}^{n-1}k{n-1\choose k}$$
However this does need seem to be the same. Any help to what I am missing would be great!

NOTE: I have done the above process for the 2nd summand of the LHS of (1) to match the 2nd summand of the RHS of (1) and I still get $n-1$ for the top value in the binomial coefficient.

答: In (3) you made an error: the upper summation index should be set to $n-1$ ... and you are done. Exactly the same works for the second term as well.


问:
Probability of Breakdowns
Hi Math Stack Exchange Community,

I am stuck with this question and not sure how can I solve it. Any help would be much appreciated.

You work as part of the maintenance team in a manufacturing company and
keep a record of the machine breakdowns. From your past experience, it
seems that there are on average 2.5 serious breakdowns every year in your
company. Then,

a) The probability that there will be 5 breakdowns in a given year is?

b) The probability that there will be fewer than 4 breakdowns in a given year is?

c) The probability that there will be at least 2 breakdowns in a given year is?

答:
Collecting comments:

A random variable which follows a Poisson distribution takes non-negative integer values and satisfies:

$$\Pr(X=k) = \dfrac{e^{-\lambda}\lambda^k}{k!}$$

where $\lambda$ is the expected value of $X$ and $k$ is the actual number of occurrences we see in the timeframe.

The problem statement here tells us that the average number of breakdowns per year is $2.5$... in other words it tells us that $\lambda = 2.5$ here.

The first problem asks us what the probability is that we see $5$ breakdowns.  That is, it asks us for $\Pr(X=5)$ given that $\lambda = 2.5$.  That is, as you correctly found in the comments after coaxing:

$$\Pr(X=5)=\dfrac{e^{-2.5}(2.5)^5}{5!}$$

For the second problem, it asks us the probability is we see fewer than $4$ breakdowns.  That is to say, that we saw $0,1,2$ or $3$ breakdowns.  We note that $\Pr(X< 4) = \Pr(X=0)+\Pr(X=1)+\Pr(X=2)+\Pr(X=3)$, use the values of $0,1,2,3$ in place of $k$ for each of the respective calculations, all still using the lambda value of $\lambda = 2.5$, and conclude.

For the final problem, it asks us the probability that we see at least $2$ breakdowns.  That is to say, that we saw $2,3,4,5,6,7,\dots$ number of breakdowns.  In other words, that we did not see $0,$ or $1$ breakdowns.

That would be $\Pr(X\geq 2) = \Pr(X=2)+\Pr(X=3)+\Pr(X=4)+\dots$, or more easily calculatable it would be $\Pr(X\geq 2) = 1 - \Pr(X<2) = 1 - \Pr(X=0)-\Pr(X=1)$.  That is to say, some times it is easier to calculate the probability of the opposite event and subtract away from $1$.

Note: Here, we were given $\lambda$ in terms of average number of breakdowns per year and we were asked questions about the number of breakdowns we actually see in a year... we used the same timeframe for both the given information of $\lambda$ as well as for the questions we were asking.  This is important to note.  You might in future problems need to rephrase things to get them to match.  If for instance, we were asked the probability that we get $5$ breakdowns in a decade (a span of ten years) for instance, then $2.5$ being the number of breakdowns in year wouldn't work for our lambda, but it would instead become $\lambda = 25$ as that would be the average number of breakdowns in a decade.  We needed to make sure the timeframes we were referring to matched.

Title: Probability of Breakdowns


问:
Does $\pi:\mathbb{S}^2 \to \mathbb{R}P^2$ have a smooth right inverse?
Let $\pi: \mathbb S^2 \to \mathbb {RP}^2$ be the projection map. I understand $\pi$ is smooth. Ideally I should use the tangent space, but this only tells me for $x \in \mathbb{S}^2$ that
$$\pi_{*,x}:T_x \mathbb{S}^2 \to T_{\pi(x)} \mathbb{R}P^2 $$ is an isomorphism, but I dont know how to use this to construct an inverse for $\pi$.

Edit: Im sorry,The original question was too long for the title, here is a more precise statement

Let $\pi$ be the projection map $\pi:\mathbb{S}^2 \to \mathbb{R}P^2$, does there exist $s: \mathbb{R}P^2 \to \mathbb{S}^2$ such that $\pi \circ s = \text{id}_{\mathbb{R}P^2}$?

答: Hint: Try to argue by contradiction. If a smooth map $\sigma$ with the required properties would exist, then it cannot be surjective. Hence $\sigma(\mathbb RP^2)\subset S^2$ would be a non-empty, proper subset. But from the construction, it is easy to deduce that this subset is both open and closed in $S^2$ which contradicts the fact that $S^2$ is connected.


问:
Find area of the largest possible equilateral triangle inscribed in an isosceles triangle
Assume that $ΔABC$ is an isosceles triangle with base of length 3, and sides of length 4.

The main question is:

What is the area of the largest equilateral triangle that can be inscribed in $ΔABC$?

And, if we find a value, how can we prove that it's the biggest possible value and there exists no other equilateral triangle inscribed in $ΔABC$ with a larger area?

(I know that for an equilateral triangle, having the largest area is the same as having the largest side length, but the main question is to find the numerical value of the largest area)

What I have done so far:

I tried to play around with shapes in GeoGebra, in hope of finding an idea to solve the problem (something like using rotations, etc.). but it wasn't that helpful.

A problem is that the values on the image are not accurate, so I need a mathematical proof rather than only drawings, to show that a value is the largest one.

I tried to draw the isosceles triangle on a xy-plane:

(The height of the triangle is $\frac{\sqrt{55}}{2} \approx 3.7$ using the Pythagorean theorem)

First, I calculated line equation for each edge, then I chose 3 arbitrary points $P_1$ and $P_2$ on the sides (shown with $E$ and $D$ on the image) and $P_3$ on the base (point $F$ in the figure) with coordinates $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ respectively. Then I tried to rewrite $y_1$ and $y_2$ using $x$'s and the line equations. Then I used the formula for Euclidean distance and inserted $x$'s and $y$'s inside the formula and put each pair of distances in equation (since they are side lengths of an equilateral triangle). Eventually, I started to solve the equations and continue that way to establish a relation between the coordinates, however, it wasn't successful, or at least I wasn't able to derive any result.

Thanks in advance!

答:
If we set (see your first figure) $BG=x$, $CG=3-x$, $\angle ABC=\angle ACB=\alpha$, $\angle CGI=\theta$, $GI=IH=HG=l$, then by the sine rule applied to triangles $GCI$ and $BGH$ we obtain:
$$
{l\over\sin\alpha}={3-x\over\sin(\alpha+\theta)}={x\over\sin(60°+\theta-\alpha)}.
$$
We can solve for $x$ to find:
$$
x={3\over2}\left(1-{\tan(\alpha-30°)\over\tan(30°+\theta)} \right).
$$
Substituting that into the expression for $l$ and differentiating one gets
$$
{dl\over d\theta}=-{3\over2}{\sin\alpha\over\sin(60°+\theta-\alpha)}
\big(1-\tan(\alpha-30°)\big)\cot(30°+\theta).
$$
This vanishes only for $\theta=60°$, but that is a minimum (as one can deduce from the sign of $dl/d\theta$). Hence maxima are attained at the boundary values of $\theta$, which are the values for which $x=0$ or $x=3$, that is $\theta=\alpha-60°$ and $\theta=180°-\alpha$.

This gives as a maximum for $l$:
$$
l_\max={3\sin\alpha\over\sin(2\alpha-60°)}.
$$


问:
If $Q_n \to Q$ strongly in $B(L^2)$, then is it true that $|Q_n| \to |Q|$ in some appropriate sense?
Let $Q, Q_n \in B(L^2(\mathbb{R}^d))$ be self-adjoint, and $Q_n \to Q$ strongly in $B(L^2)$, that is, $Q_n f \to Q f$ in $L^2$ for all $f \in L^2$.
Then, $|Q_n| \to |Q|$ in some appropriate sense?

Observation.
$\||Q_n| f\|_2 = \|Q_n f\|_2 \to \|Q f\|_2$ for all $f \in L^2$. Therefore, for any $f\in L^2$, $(|Q_n| f)_n \subset L^2$ is bounded, and it has a subsequence $(|Q_{n(k)}| f)_k$ which weakly converges to some $g \in L^2$. I think that it is natural to expect $g=|Q|f$.

答:
More generally, if $Q,Q_n\in B(H)$ are self-adjoint and $Q_n\to Q$ strongly, then $f(Q_n)\to f(Q)$ strongly for any bounded continuous function $f$.

First note that multiplication is strongly continuous on bounded subsets of $B(H)$: If $S_\alpha\to S$ and $T_\alpha\to T$ strongly, then
$$
\|(S_\alpha T_\alpha-ST)\xi\|\leq \sup_\beta\|S_\beta\|\|(T_\alpha-T)\xi\|+\|(S_\alpha-S)T\xi\|\to 0.
$$
By induction, the continuity property from the first paragraph holds for monomials, and by linearity for arbitrary polynomials (note that $(Q_n)$ is bounded in $B(H)$ by the uniform boundedness principle).

Finally, if $f$ is an arbitrary continuous function on a compact set containing $\sigma(Q_n)$, $\sigma(Q)$, then there exists a sequence of polynomials $(p_k)$ such that $p_k\to f$ uniformly by the Stone-Weierstraß theorem. It follows that
\begin{align*}
\|(f(Q_n)-f(Q))\xi\|&\leq \|(f-p_k)(Q_n)\xi\|+\|(p_k(Q_n)-p_k(Q))\xi\|+\|(p_k-f)(Q)\xi\|\\
&\leq 2\|f-p_k\|_\infty+\|(p_k(Q_n)-p_k(Q))\xi\|.
\end{align*}
As discussed above, the second summand goes to $0$ as $n\to\infty$, while the first summand converges to zero as $k\to\infty$ as a consequence of the uniform convergence of $(p_k)$ to $f$.


问:
The Board Football Problem (Part I)
The original question is here (The Board Football Problem (Probability)) and part II is here(The Board Football Problem (Part II)). I was told to segment the question in order to increase the chances of it being answered.

As kids we used to play the game "Board Football". I'm not particularly sure of the popularity of the game but the rules were pretty simple to follow. The basic objective of the game was to score as many goals as possible and you played it with nothing more than dice and a notebook.

It's a two player game, which goes on as long as the players don't bored. ie there isn't any way to win/finish the game.Both you (say A) and the opponent (say B) have a dice. Before starting the game you decide the minimum number of "passes" that need to be completed before a goal is scored(say p).

Let's assume that A starts first.A rolls the dice and gets a certain number, say 5. This means that A completes 5 passes in that round and has to complete p-5 more passes in order to progress to the "Final Stage". However, before the round is completed the opponent(B) has the opportunity to "intercept" you and obtain possession of the ball.

The way to do this is pretty simple. B rolls his dice and if he gets the same number as A, then B obtains possession of the ball and the round is completed. In the next round, B will start off from 0 passes.

We weren't particularly finicky back then so even if we rolled a number to exceed the number of passes made, the player in possession was still allowed to progress to the "Final Stage". It didn't matter in what order the passes in each round were made as long as the required passes was obtained.

Once the required passes were made and the player reached the "Final Stage", they had to flip a coin. If it turns up as heads,the player in possession scores and if it turns up as tails, he doesn't score. In both the cases, the possession is reverted back to B for the next round. We considered the coin-flipping part to take place in the same round as the round in which the player in possession completed the required passes.

For example, let a 6 sided dice be used and the required number of
passes be 9.
(That is n=6,p=9) Let us say that A has the ball in the starting.

In the first round, A rolls 4 and B rolls 5. A completes 4 passes within that round.A now has to complete at least 5 more passes in order to get an opportunity to shoot.

In the second round, A rolls 3 and B rolls 3. Possession is overturned and now B controls the ball.B now has 0 passes completed.

In the third round, B rolls 5 and A rolls 2.B has completed 5 passes in this round and needs to complete at least 4 more passes in order to get an opportunity to shoot.

In the fourth round, B rolls 6 and A rolls 4. Although B has overshot the number of passes,  B is permitted to flip the coin. B tosses the coin and gets heads. B scores a goal.B leads by one goal to nil. Now A starts
off with possession of the ball.

So my question is as follows:If we use an n-sided dice and the number of required passes is p, the what is the probability that the player in possession of the ball(starting off from 0 passes completed) scores a goal without losing possession of the ball. Is it possible to calculate the value for this in a non-computational /formulaic manner?

(BONUS QUESTION) Form a graph with the Y-Axis being the probability that the player in possession scores without losing possession of the ball and the number of required passes be the X-Axis. (Let a 10 sided dice be used for this case.)

If possible, determine how different the graph would look [in a visual sense, no need for deeper calculations] if the number of sides on the dice is made to vary. Is the graph just shifted along the x-axis or do the peaks become more gradual?).

PS This isn't a homework question/problem so please don't close it because an answer is being requested. It is an original question. Consider this to be a challenge of sorts.

答:
For the sake of completeness, I'll add a formula for the probability.

Let us construct a transition matrix of the process. We can imagine that at every step, the player is in some state and they arrive at a different state after each turn. Let there be states $0, 1, \dots, p$ and $L$, $S$ - loss and success. We start from the state $0$ and want to calculate the probability, that after a sufficient number of turns we will land at state $S$. From the state $p$ we can only go to $L$ or $S$ with the probabilities $\frac{1}{2}$, and from the states $k < p$ we can only go to $k+1, k+2, \dots, p$ or $L$, if the second player intercepts the ball. The below example illustrates the transition matrix for $n=3$  and $p=6$
\begin{pmatrix} 0 & \frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 0 & \frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 0 & 0 & \frac{2}{9} & \frac{2}{9} & \frac{1}{3} & \frac{2}{9} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{2}{9} & \frac{2}{3} & \frac{1}{9} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\\end{pmatrix}

We have probability $\frac{1}{n}\cdot\frac{n-1}{n} = \frac{n-1}{n^2}$ of going from $k_1 < p-1$ to $k_2 < p$, because there is $\frac{n-1}{n}$ probability that the ball will not intercepted. If it is intercepted, we go to the state $L$. We also assume that once we are already in $L$ or $S$, we stay in that state till the end of the game. We can also see that the ball cannot be intercepted once we reach the state $p$. From the properties of the transition matrix $P$, the probability of reaching the state $j$ from state $i$ after exactly $k$ steps is equal to $P^k_{i, j}$. The maximal number of steps that it may take to arrive at $S$ from $0$ is $p+1$, as in that case we move by one pass in each turn. Therefore, the probability may be found using the above matrix as
$$P^{(p+1)}_{0, S}$$

However, constructing the above matrix is troublesome and we haven't yet achieved anything reasonable. We can notice however, that we may divide the above matrix into four blocks:
$$ A_{(p+1) \times (p+1)} = \begin{pmatrix}
0 & \frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 & 0 & 0 \\
0 & 0 & \frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 & 0 \\
0 & 0 & 0 & \frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 \\
0 & 0 & 0 & 0 & \frac{2}{9} & \frac{2}{9} & \frac{1}{3} \\
0 & 0 & 0 & 0 & 0 & \frac{2}{9} & \frac{2}{3} \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix}  $$

$$ B_{(p+1)\times 2} = \begin{pmatrix}
\frac{1}{3} & 0 \\
\frac{1}{3} & 0 \\
\frac{1}{3} & 0 \\
\frac{2}{9} & 0 \\
\frac{1}{9} & 0 \\
0 & 0 \\
\frac{1}{2} & \frac{1}{2}
\end{pmatrix}  $$

$$C_{2\times (p+1)} = \mathbf{0}_{2\times (p+1)} $$
$$ D_{2\times 2} = I_{2 \times 2}$$

We may notice some regularities, when we calculate the powers of the transition matrix.
$$M^2 =  \begin{pmatrix}
A & B \\
C & D
\end{pmatrix}^2  = \begin{pmatrix}
A^2 & B+AB \\
C & D
\end{pmatrix}$$
$$M^k =  \begin{pmatrix}
A & B \\
C & D
\end{pmatrix}^k  = \begin{pmatrix}
A^k & B+AB+A^2B+\dots A^{(k-1)}B \\
C & D
\end{pmatrix}$$
This is in fact a known property of such stochastic processes. Using a matrix multiplication formula:
$$N^k - I = (N-I)(I+N+N^2+\dots+N^{(k-1)})$$
we achieve that
$$M^k = \begin{pmatrix}
A^k & (A-I)^{-1}(A^k-I)B \\
C & D
\end{pmatrix}$$
Let us also notice that $A$ is strongly upper triangular and therefore nilpotent. It is easy to verify that its $p+1$-power vanishes. We therefore achieve that
$$M^{(p+1)} = \begin{pmatrix}
\mathbb{0}_{(p+1)\times (p+1)} & -(A-I)^{-1}B \\
\mathbb{0}_{2\times (p+1)} & I_{2\times 2}
\end{pmatrix}$$
Let us remember that we are only interested in an element on position $0, S$ of the matrix $M^{(p+1)}$. As the second column of the matrix $B$ has only one non-zero element equal always to $\frac{1}{2}$ and indexing the rows and columns of $(A-I)^{-1}$ with $0, \dots, p$, we achieve that
$$M^{(p+1)}_{0, S} = -\frac{1}{2}(A-I)^{-1}_{0, p}$$

To further simplify that, notice that $\det{(A-I)} = (-1)^{(p+1)}$. Using the formula
$$N^{-1}=\frac{1}{\det (N)}\text{adj}{(N)}$$
we obtain that
$$M^{(p+1)}_{0, S} = -\frac{1}{2}(-1)^{(p+1)}\det (A-I)_{(1:p, 2:(p+1))}$$
where $(A-I)_{(1:p, 2:(p+1))}$ denotes the matrix consisting of the first $p$ rows and columns from $2$ to $p+1$ of $(A-I)$. Returning to the previous example, we wish to calculate the determinant of
$$ (A-I)_{(1:p, 2:(p+1))}=\begin{pmatrix}
\frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 & 0 & 0 \\
-1 & \frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 & 0 \\
0 & -1 & \frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 \\
0 & 0 & -1 & \frac{2}{9} & \frac{2}{9} & \frac{1}{3} \\
0 & 0 & 0 & -1 & \frac{2}{9} & \frac{2}{3} \\
0 & 0 & 0 & 0 & -1 & 1
\end{pmatrix}  $$

We may also use a formula for the determinant of block matrices to simplify it a bit.
For
$$N = \begin{pmatrix}
A & B \\
C & D
\end{pmatrix}$$
we have
$$\det(N) = \det(A-BD^{-1}C)\det(D)$$
Taking $D$ to be the $1\times 1$ matrix containing a single $1$, we obtain the following result.

Let $p-1, \dots, 0, \dots, -(p-1)$ be the indices of diagonals $d_p, \dots, d_{-p}$ of a matrix, where negative indices correspond to the diagonals above the main diagonal. Let us define the matrix $M_{(p-1) \times (p-1)}$:
$$M_{i, j}=\begin{cases} -1 & \text{if } M_{i, j} \text{ lies on } d_1 \\
\frac{n-1}{n^2} & \text{if } M_{i, j} \text{ lies on } d_0, \dots, d_{-(n-1)} \\
0 & \text{else.}\end{cases}$$
Let $u_{(p-1)\times 1}$ be a column vector such that:
$$u_i = \begin{cases}0 & \text{if } i < p-1 \\
-1 & \text{if } i = p-1\end{cases}$$
Finally let $v_{(p-1)\times 1}$ be a column vector such that
$$v_i = \max\{0, \frac{n-p+i}{n}\}$$
Then the seeken probability is equal to
$$(-1)^p \frac{1}{2}\det(M-uv^T)$$

Title: The Board Football Problem (Part I)


问:
Taylor series remainder equaling zero
I asked this question a while ago and I figured out my initial problem. I basically learned that convergence of series doesn't mean that it converges to that function since there could be error

My question now is, if there is no error between a function and Taylor series, then does that mean that Taylor series is equal to that function (at least on that interval?)

And, for example, for $e^x$ or $\sin(x)$. if I prove the error is zero, then do I need to prove the radius of convergence of that series (perhaps by ratio test?)?

答:
Look at the case of $e^x$. The $n$th Taylor polynomial is
$$
T_n (x) = \sum\limits_{k = 0}^n {\frac{{x^k }}{{k!}}} .
$$
It is easy to show, using the Lagrange form of the remainder, that
$$
\left| {e^x  - T_n (x)} \right| \le \frac{{x^{n + 1} }}{{(n + 1)!}}\max (1,e^x )
$$
for any real $x$. The right-hand side converges point-wise to zero as $n\to +\infty$. Thus, the function sequence $T_n(x)$ converges to $e^x$ for any real $x$, i.e., $\lim_{n\to +\infty}T_n(x)=e^x$ for any real $x$. By the definition of an infinite series
$$
\mathop {\lim }\limits_{n \to  + \infty } T_n (x) =: \sum\limits_{k = 0}^\infty  {\frac{{x^k }}{{k!}}} .
$$
Consequently,
$$
e^x  = \sum\limits_{k = 0}^\infty  {\frac{{x^k }}{{k!}}}
$$
for any real $x$. You do not need to look at the radius of convergence, because we have just shown that this equality holds for all real $x$, i.e., the series on the right-hand side converges for all real $x$ (and its sum is $e^x$). Does this answer your question?

Title: Taylor series remainder equaling zero


答:
In $(3)$, you are changing
$$
\sum_{k=1}^n(k-1)\binom{n}{k-1}
$$
via the substitution $k\mapsto m+1$. This leads to
$$
\sum_{m=0}^{\color{#C00}{n-1}}m\,\binom{n}{m}
$$
The summation is to be taken over $k\in[1,n]$, thus $m+1\in[1,n]$, which is the same as $m\in[0,n-1]$. This is the reason for the change in the limits of summation.

Note that $k$ is bound to the scope of this summation, so it is okay to substitute as is done here. However, $n$ is not bound to the scope of this formula; it could very well be referenced elsewhere. Changing $n$ probably requires a more global substitution affecting more than just this summation. Thus, substituting $l=n+1$ is not a safe thing to do.

As this procedure becomes more familiar, I find it possible to do this without changing the variable name; i.e. substituting $k\mapsto k+1$, thereby reducing the number variable names floating about.

Title: Changing the summations involving binomial coefficient

Body:
Given the following:
$$(1)\quad\quad \sum_{k=1}^n(k-1){n \choose k-1} + \sum_{k=1}^n {n\choose k -1} = \sum_{k=0}^{n-1}k{n\choose k} + \sum_{k=0}^{n-1}{n\choose k}$$
I am trying to understand how it goes from the LHS to the RHS.

My approach so far, given the first sum summand of the LHS, we have, and taking from the concept here
$$(2)\quad\quad \sum_{k=1}^n(k-1){n \choose k-1}$$

Let $k = m + 1$, then eq (2) becomes
$$(3) \quad\quad \sum_{m=0}^nm{n\choose m}$$
Now we let $l = n +1$, the eq (3) becomes
$$(4)\quad\quad\sum_{m=0}^{l-1}m{l-1\choose m}$$
Change the symbols in (4) to resemble the first first summand of the RHS of (1), namely, let $m = k$ and $l = n$ and substitute we have:
$$(5)\quad\quad\sum_{k=0}^{n-1}k{n-1\choose k}$$
However this does need seem to be the same. Any help to what I am missing would be great!

NOTE: I have done the above process for the 2nd summand of the LHS of (1) to match the 2nd summand of the RHS of (1) and I still get $n-1$ for the top value in the binomial coefficient.